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https://www.nag.com/numeric/nl/nagdoc_27.3/clhtml/d01/d01rkc.html | [
"# NAG CL Interfaced01rkc (dim1_fin_osc_fn)\n\nSettings help\n\nCL Name Style:\n\n## 1Purpose\n\nd01rkc is an adaptive integrator, especially suited to oscillating, nonsingular integrands, which calculates an approximation to the integral of a function $f\\left(x\\right)$ over a finite interval $\\left[a,b\\right]$:\n $I= ∫ab f(x) dx .$\n\n## 2Specification\n\n #include\nvoid d01rkc (\n void (*f)(const double x[], Integer nx, double fv[], Integer *iflag, Nag_Comm *comm),\ndouble a, double b, Integer key, double epsabs, double epsrel, Integer maxsub, double *result, double *abserr, double rinfo[], Integer iinfo[], Nag_Comm *comm, NagError *fail)\nThe function may be called by the names: d01rkc or nag_quad_dim1_fin_osc_fn.\n\n## 3Description\n\nd01rkc is based on the QUADPACK routine QAG (see Piessens et al. (1983)). It is an adaptive function, offering a choice of six Gauss–Kronrod rules. A ‘global’ acceptance criterion (as defined by Malcolm and Simpson (1976)) is used. The local error estimation is described in Piessens et al. (1983).\nBecause d01rkc is based on integration rules of high order, it is especially suitable for nonsingular oscillating integrands.\nd01rkc requires you to supply a function to evaluate the integrand at an array of points.\nde Doncker E (1978) An adaptive extrapolation algorithm for automatic integration ACM SIGNUM Newsl. 13(2) 12–18\nMalcolm M A and Simpson R B (1976) Local versus global strategies for adaptive quadrature ACM Trans. Math. Software 1 129–146\nPiessens R, de Doncker–Kapenga E, Überhuber C and Kahaner D (1983) QUADPACK, A Subroutine Package for Automatic Integration Springer–Verlag\nWynn P (1956) On a device for computing the ${e}_{m}\\left({S}_{n}\\right)$ transformation Math. Tables Aids Comput. 10 91–96\n\n## 5Arguments\n\n1: $\\mathbf{f}$function, supplied by the user External Function\nf must return the values of the integrand $f$ at a set of points.\nThe specification of f is:\n void f (const double x[], Integer nx, double fv[], Integer *iflag, Nag_Comm *comm)\n1: $\\mathbf{x}\\left[\\mathit{dim}\\right]$const double Input\nOn entry: the abscissae, ${x}_{i}$, for $\\mathit{i}=1,2,\\dots ,{\\mathbf{nx}}$, at which function values are required.\n2: $\\mathbf{nx}$Integer Input\nOn entry: the number of abscissae at which a function value is required. nx will be of size equal to the number of Kronrod points in the quadrature rule used, as determined by the choice of value for key.\n3: $\\mathbf{fv}\\left[\\mathit{dim}\\right]$double Output\nOn exit: fv must contain the values of the integrand $f$. ${\\mathbf{fv}}\\left[i-1\\right]=f\\left({x}_{i}\\right)$ for all $i=1,2,\\dots ,{\\mathbf{nx}}$.\n4: $\\mathbf{iflag}$Integer * Input/Output\nOn entry: ${\\mathbf{iflag}}=0$.\nOn exit: set ${\\mathbf{iflag}}<0$ to force an immediate exit with ${\\mathbf{fail}}\\mathbf{.}\\mathbf{code}=$ NE_USER_STOP.\n5: $\\mathbf{comm}$Nag_Comm *\nPointer to structure of type Nag_Comm; the following members are relevant to f.\nuserdouble *\niuserInteger *\npPointer\nThe type Pointer will be void *. Before calling d01rkc you may allocate memory and initialize these pointers with various quantities for use by f when called from d01rkc (see Section 3.1.1 in the Introduction to the NAG Library CL Interface).\nNote: f should not return floating-point NaN (Not a Number) or infinity values, since these are not handled by d01rkc. If your code inadvertently does return any NaNs or infinities, d01rkc is likely to produce unexpected results.\n2: $\\mathbf{a}$double Input\nOn entry: $a$, the lower limit of integration.\n3: $\\mathbf{b}$double Input\nOn entry: $b$, the upper limit of integration. It is not necessary that $a.\nNote: if ${\\mathbf{a}}={\\mathbf{b}}$, the function will immediately return with ${\\mathbf{result}}=0.0$, ${\\mathbf{abserr}}=0.0$, ${\\mathbf{rinfo}}=0.0$ and ${\\mathbf{iinfo}}=0$.\n4: $\\mathbf{key}$Integer Input\nOn entry: indicates which integration rule is to be used. The number of function evaluations required for an integral estimate over any segment will be the number of Kronrod points, ${n}_{\\mathrm{kron}}$.\n${\\mathbf{key}}=1$\nFor the Gauss $7$-point and Kronrod $15$-point rule.\n${\\mathbf{key}}=2$\nFor the Gauss $10$-point and Kronrod $21$-point rule.\n${\\mathbf{key}}=3$\nFor the Gauss $15$-point and Kronrod $31$-point rule.\n${\\mathbf{key}}=4$\nFor the Gauss $20$-point and Kronrod $41$-point rule.\n${\\mathbf{key}}=5$\nFor the Gauss $25$-point and Kronrod $51$-point rule.\n${\\mathbf{key}}=6$\nFor the Gauss $30$-point and Kronrod $61$-point rule.\nSuggested value: ${\\mathbf{key}}=6$.\nConstraint: ${\\mathbf{key}}=1$, $2$, $3$, $4$, $5$ or $6$.\n5: $\\mathbf{epsabs}$double Input\nOn entry: ${\\epsilon }_{a}$, the absolute accuracy required. If epsabs is negative, ${\\epsilon }_{a}=|{\\mathbf{epsabs}}|$. See Section 7.\n6: $\\mathbf{epsrel}$double Input\nOn entry: ${\\epsilon }_{r}$, the relative accuracy required. If epsrel is negative, ${\\epsilon }_{r}=|{\\mathbf{epsrel}}|$. See Section 7.\n7: $\\mathbf{maxsub}$Integer Input\nOn entry: ${\\mathrm{max}}_{\\mathit{sdiv}}$, the upper bound on the total number of subdivisions d01rkc may use to generate new segments. If ${\\mathrm{max}}_{\\mathit{sdiv}}=1$, only the initial segment will be evaluated.\nSuggested value: a value in the range $200$ to $500$ is adequate for most problems.\nConstraint: ${\\mathbf{maxsub}}\\ge 1$.\n8: $\\mathbf{result}$double * Output\nOn exit: the approximation to the integral $I$.\n9: $\\mathbf{abserr}$double * Output\nOn exit: an estimate of the modulus of the absolute error, which should be an upper bound for $|I-{\\mathbf{result}}|$.\n10: $\\mathbf{rinfo}\\left[4×{\\mathbf{maxsub}}\\right]$double Output\nOn exit: details of the computation. See Section 9 for more information.\n11: $\\mathbf{iinfo}\\left[\\mathrm{max}\\phantom{\\rule{0.125em}{0ex}}\\left({\\mathbf{maxsub}},4\\right)\\right]$Integer Output\nOn exit: details of the computation. See Section 9 for more information.\n12: $\\mathbf{comm}$Nag_Comm *\nThe NAG communication argument (see Section 3.1.1 in the Introduction to the NAG Library CL Interface).\n13: $\\mathbf{fail}$NagError * Input/Output\nThe NAG error argument (see Section 7 in the Introduction to the NAG Library CL Interface).\n\n## 6Error Indicators and Warnings\n\nNE_ALLOC_FAIL\nDynamic memory allocation failed.\nSee Section 3.1.2 in the Introduction to the NAG Library CL Interface for further information.\nOn entry, argument $⟨\\mathit{\\text{value}}⟩$ had an illegal value.\nOn entry, ${\\mathbf{key}}=⟨\\mathit{\\text{value}}⟩$.\nConstraint: ${\\mathbf{key}}=1$, $2$, $3$, $4$, $5$ or $6$.\nNE_INT\nOn entry, ${\\mathbf{maxsub}}=⟨\\mathit{\\text{value}}⟩$.\nConstraint: ${\\mathbf{maxsub}}\\ge 1$.\nNE_INTERNAL_ERROR\nAn internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.\nSee Section 7.5 in the Introduction to the NAG Library CL Interface for further information.\nNE_NO_LICENCE\nYour licence key may have expired or may not have been installed correctly.\nSee Section 8 in the Introduction to the NAG Library CL Interface for further information.\nExtremely bad integrand behaviour occurs around the sub-interval $\\left(⟨\\mathit{\\text{value}}⟩,⟨\\mathit{\\text{value}}⟩\\right)$. The same advice applies as in the case of ${\\mathbf{fail}}\\mathbf{.}\\mathbf{code}=$ NE_QUAD_MAX_SUBDIV.\nThe maximum number of subdivisions allowed with the given workspace has been reached without the accuracy requirements being achieved. Look at the integrand in order to determine the integration difficulties. If necessary, another integrator, which is designed for handling the type of difficulty involved, must be used. Alternatively, consider relaxing the accuracy requirements specified by epsabs and epsrel, or increasing the amount of workspace.\nRound-off error prevents the requested tolerance from being achieved: ${\\mathbf{epsabs}}=⟨\\mathit{\\text{value}}⟩$ and ${\\mathbf{epsrel}}=⟨\\mathit{\\text{value}}⟩$.\nNE_USER_STOP\nExit from f with ${\\mathbf{iflag}}<0$.\n\n## 7Accuracy\n\nd01rkc cannot guarantee, but in practice usually achieves, the following accuracy:\n $|I-result|≤tol,$\nwhere\n $tol= max{|epsabs|,|epsrel|×|I|} ,$\nand epsabs and epsrel are user-specified absolute and relative error tolerances. Moreover, it returns the quantity abserr which, in normal circumstances, satisfies\n $|I-result|≤abserr≤tol.$\n\n## 8Parallelism and Performance\n\nd01rkc is not threaded in any implementation.\n\nThe time taken by d01rkc depends on the integrand and the accuracy required.\nIf ${\\mathbf{fail}}\\mathbf{.}\\mathbf{code}=$ NE_NOERROR, NE_QUAD_BAD_SUBDIV, NE_QUAD_MAX_SUBDIV or NE_QUAD_ROUNDOFF_TOL, or if ${\\mathbf{fail}}\\mathbf{.}\\mathbf{code}=$ NE_USER_STOP and at least one complete vector evaluation of f was completed, result and abserr will contain computed results. If these results are unacceptable, or if otherwise required, then you may wish to examine the contents of the array rinfo, which contains the end points of the sub-intervals used by d01rkc along with the integral contributions and error estimates over the sub-intervals.\nSpecifically, for $i=1,2,\\dots ,n$, let ${r}_{i}$ denote the approximation to the value of the integral over the sub-interval $\\left[{a}_{i},{b}_{i}\\right]$ in the partition of $\\left[a,b\\right]$ and ${e}_{i}$ be the corresponding absolute error estimate. Then, $\\underset{{a}_{i}}{\\overset{{b}_{i}}{\\int }}f\\left(x\\right)dx\\simeq {r}_{i}$ and ${\\mathbf{result}}=\\sum _{i=1}^{n}{r}_{i}$. The value of $n$ is returned in ${\\mathbf{iinfo}}\\left[0\\right]$, and the values ${a}_{i}$, ${b}_{i}$, ${e}_{i}$ and ${r}_{i}$ are stored consecutively in the array rinfo, that is:\n• ${a}_{i}={\\mathbf{rinfo}}\\left[i-1\\right]$,\n• ${b}_{i}={\\mathbf{rinfo}}\\left[n+i-1\\right]$,\n• ${e}_{i}={\\mathbf{rinfo}}\\left[2n+i-1\\right]$ and\n• ${r}_{i}={\\mathbf{rinfo}}\\left[3n+i-1\\right]$.\nThe total number of abscissae at which the function was evaluated is returned in ${\\mathbf{iinfo}}\\left[1\\right]$.\n\n## 10Example\n\nThis example computes\n\n### 10.1Program Text\n\nProgram Text (d01rkce.c)\n\nNone.\n\n### 10.3Program Results\n\nProgram Results (d01rkce.r)"
] | [
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https://docs.signac.io/en/latest/examples/notebooks/signac_301_Aggregation_Tutorial.html | [
"# 3.1 Aggregation Tutorial\n\nThis notebook contains a minimal example for running workflows on aggregates of jobs using signac-flow.\n\nHardik Ojha\n\n## Prerequisites\n\nThis notebooks requires the following packages:\n\n1. signac-flow >= 0.15\n\n2. numpy\n\n3. matplotlib\n\nExecute the command below to install the required packages:\n\npip install signac-flow>=0.15 matplotlib numpy\n\n\n## Definition\n\nAggregation allows a signac-flow operation to act on multiple jobs, rather than one job at a time.\n\nAn aggregate is defined as a subset of the jobs in a signac project. Aggregates are generated when the @flow.aggregator decorator is applied to an operation.\n\nPlease refer to the documentation for detailed instructions on how to use aggregation.\n\n## Objective\n\nThe goal of this project is to plot the temperature values present in a signac data space along with the average value of all the temperatures present.\n\n## Project Setup\n\nBefore we initialize a signac project inside the projects/tutorial-aggregation directory, we need to be sure that no such directory exists. Uncomment before executing the below cell to remove the directory if exists.\n\n:\n\n# !rm -rf projects/tutorial-aggregation\n\n:\n\nimport datetime\n\nimport matplotlib.pyplot as plt\nimport numpy as np\nimport signac\nfrom flow import FlowProject, aggregator\n\n# Setting default figure size\nplt.rcParams[\"figure.figsize\"] = (10, 4)\n\n# Initializing a signac project\nproject = signac.init_project(\n\"AggregationTutorialProject\", \"projects/tutorial-aggregation\"\n)\n\n\n## Initializing the data space\n\nFor the purpose of this notebook, we will be creating a random dataset using some mathematical calculations.\n\nAll the signac jobs will have two state point parameters and one document value.\n\n• job.statepoint[\"city\"]: City for which data is being collected.\n\n• job.statepoint[\"day\"]: Day of the year.\n\n• job.document[\"temperature\"]: Average temperature for that day.\n\n:\n\ndays = np.arange(365)\n\ndef generate_temperatures(days, seed=None):\nrng = np.random.default_rng(seed)\navg_temperature = 10 + rng.random() * 10\nannual_variation = -10 * np.cos(days / 365 * 2 * np.pi)\nrandom_variation = 5 * rng.random(len(days))\ntemperatures = avg_temperature + annual_variation + random_variation\nreturn temperatures\n\ntemperatures = generate_temperatures(days, seed=123)\n\nfor day, temperature in zip(days.tolist(), temperatures.tolist()):\n# Create a signac job having the state point parameters 'day' and 'temperature'\nstatepoint = dict(city=\"Anytown\", day=day)\njob = project.open_job(statepoint)\njob.document[\"temperature\"] = temperature\n\n\nLet’s look at the project schema to see the jobs that were created.\n\n:\n\nproject.detect_schema()\n\n:\n\nProjectSchema(<len=2>)\n{\n'city': 'str([Anytown], 1)',\n'day': 'int([0, 1, 2, ..., 363, 364], 365)',\n}\n\n## Creating a FlowProject with aggregate operations\n\nIn order to achieve our goal using signac-flow, we need to create a FlowProject and add operations to it. There will be following operations in our workflow:\n\n1. compute_average_temperature: This operation computes the average temperature of the year and stores it in the project document. For this operation, all the jobs present in the signac project will be aggregated together. This will be the first operation to get executed in our workflow.\n\n2. plot_deviation_from_average: This operation plots the temperature (as a scatter plot) and the average temperature of the month. For this operation, all the jobs, when sorted by the state point parameter day, present in the signac project will be aggregated together. This will be executed after the operation compute_average_temperature.\n\n:\n\nclass AggregationProject(FlowProject):\npass\n\n@aggregator()\n@AggregationProject.operation\n@AggregationProject.post(lambda *jobs: project.doc.get(\"average_temperature\", False))\ndef compute_average_temperature(*jobs):\n\"\"\"Compute the average temperature using the state point parameter,\n\"temperature\", of all jobs present in the signac project and\nstore the computed value to the project document.\n\"\"\"\naverage_temp = np.mean([job.document[\"temperature\"] for job in jobs])\nproject.document[\"average_temperature\"] = float(average_temp)\n\n@aggregator(sort_by=\"day\")\n@AggregationProject.operation\n@AggregationProject.pre.after(compute_average_temperature)\ndef plot_daily_temperature(*jobs):\n\"\"\"Graph of daily temperature for the year.\"\"\"\nprint(\"Generating plot of daily temperature.\")\naverage_temp = project.document[\"average_temperature\"]\ndays = [job.sp[\"day\"] for job in jobs]\nfig, ax = plt.subplots()\nax.plot(\ndays,\n[job.document[\"temperature\"] for job in jobs],\n\"rx\",\nlabel=\"Daily Temperature (°C)\",\n)\n# Plot the average as a line\nax.axhline(average_temp, c=\"green\", label=\"Average Annual Temperature (°C)\")\nax.legend()\nax.set_xlabel(\"Day\")\nax.set_ylabel(\"Temperature (°C)\")\nplt.show()\n\n@aggregator(sort_by=\"day\", select=lambda job: job.sp[\"day\"] % 7 == 0)\n@AggregationProject.operation\ndef plot_weekly_temperature(*jobs):\n\"\"\"Graph the temperature for only one day of each week.\"\"\"\nprint(\"Generating plot of weekly temperature.\")\ndays = [job.sp.day for job in jobs]\nfig, ax = plt.subplots()\nax.plot(\ndays,\n[job.document[\"temperature\"] for job in jobs],\n\"rx\",\nlabel=\"Daily Temperature (°C)\",\n)\nax.set_xlabel(\"Day\")\nax.set_ylabel(\"Temperature (°C)\")\nplt.show()\n\n\n## Executing the workflow\n\n### Initializing the FlowProject\n\nIn order to register the operations, conditions, and the aggregators associated with the project we created, we need to initialize a FlowProject. Since the signac project does not belong in the current directory, we specify its path to FlowProject.get_project.\n\n:\n\nflow_project = AggregationProject.get_project(project.root_directory())\n\n\n### Running the workflow\n\nThe FlowProject.run method allows the execution all eligible operations in the FlowProject.\n\n:\n\nflow_project.run()\n\nGenerating plot of daily temperature.",
null,
"Generating plot of weekly temperature.",
null,
"Operation 'plot_weekly_temperature' has no postconditions!\nOperation 'plot_daily_temperature' has no postconditions!\n\n\n## Summary\n\nWe have successfully plotted the temperature values present in a signac data space along with the average value of all the temperatures present using the aggregation feature of signac-flow."
] | [
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"https://docs.signac.io/en/latest/_images/examples_notebooks_signac_301_Aggregation_Tutorial_16_1.png",
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"https://docs.signac.io/en/latest/_images/examples_notebooks_signac_301_Aggregation_Tutorial_16_3.png",
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https://chem.libretexts.org/Courses/Pasadena_City_College/CHEM_001A%3A_General_Chemistry_and_Chemical_Analysis/06%3A_Gases/6.05%3A_Applications_of_the_Ideal_Gas_Law-_Molar_Volume%2C_Density_and_Molar_Mass_of_a_Gas | [
"# 6.5: Applications of the Ideal Gas Law- Molar Volume, Density and Molar Mass of a Gas\n\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$$$\\newcommand{\\AA}{\\unicode[.8,0]{x212B}}$$\n\n##### Learning Objectives\n• To relate the amount of gas consumed or released in a chemical reaction to the stoichiometry of the reaction.\n• To understand how the ideal gas equation and the stoichiometry of a reaction can be used to calculate the volume of gas produced or consumed in a reaction.\n\nWith the ideal gas law, we can use the relationship between the amounts of gases (in moles) and their volumes (in liters) to calculate the stoichiometry of reactions involving gases, if the pressure and temperature are known. This is important for several reasons. Many reactions that are carried out in the laboratory involve the formation or reaction of a gas, so chemists must be able to quantitatively treat gaseous products and reactants as readily as they quantitatively treat solids or solutions. Furthermore, many, if not most, industrially important reactions are carried out in the gas phase for practical reasons. Gases mix readily, are easily heated or cooled, and can be transferred from one place to another in a manufacturing facility via simple pumps and plumbing.\n\n## Gas Densities and Molar Mass\n\nThe ideal-gas equation can be manipulated to solve a variety of different types of problems. For example, the density, $$\\rho$$, of a gas, depends on the number of gas molecules in a constant volume. To determine this value, we rearrange the ideal gas equation to\n\n$\\dfrac{n}{V}=\\dfrac{P}{RT}\\label{10.5.1}$\n\nDensity of a gas is generally expressed in g/L (mass over volume). Multiplication of the left and right sides of Equation \\ref{10.5.1} by the molar mass in g/mol ($$M$$) of the gas gives\n\n$\\rho= \\dfrac{g}{L}=\\dfrac{PM}{RT} \\label{10.5.2}$\n\nThis allows us to determine the density of a gas when we know the molar mass, or vice versa.\n\nThe density of a gas INCREASES with increasing pressure and DECREASES with increasing temperature\n\n##### Example $$\\PageIndex{1}$$\n\nWhat is the density of nitrogen gas ($$\\ce{N_2}$$) at 248.0 Torr and 18º C?\n\n###### Step 1: Write down your given information\n• P = 248.0 Torr\n• V = ?\n• n = ?\n• R = 0.0820574 L•atm•mol-1 K-1\n• T = 18º C\n###### Step 2: Convert as necessary.\n\n$(248 \\; \\rm{Torr}) \\times \\dfrac{1 \\; \\rm{atm}}{760 \\; \\rm{Torr}} = 0.3263 \\; \\rm{atm} \\nonumber$\n\n$18\\,^oC + 273 = 291 K\\nonumber$\n\n###### Step 3: This one is tricky. We need to manipulate the Ideal Gas Equation to incorporate density into the equation.\n\nWrite down all known equations:\n\n$PV = nRT \\nonumber$\n\n$\\rho=\\dfrac{m}{V} \\nonumber$\n\nwhere $$\\rho$$ is density, $$m$$ is mass, and $$V$$ is volume.\n\n$m=M \\times n \\nonumber$\n\nwhere $$M$$ is molar mass and $$n$$ is the number of moles.\n\nNow take the definition of density (Equation \\ref{10.5.1})\n\n$\\rho=\\dfrac{m}{V} \\nonumber$\n\nKeeping in mind $$m=M \\times n$$...replace $$(M \\times n)$$ for $$mass$$ within the density formula.\n\n\\begin{align*} \\rho &=\\dfrac{M \\times n}{V} \\\\[4pt] \\dfrac{\\rho}{M} &= \\dfrac{n}{V} \\end{align*} \\nonumber\n\nNow manipulate the Ideal Gas Equation\n\n\\begin{align*} PV &= nRT \\\\[4pt] \\dfrac{n}{V} &= \\dfrac{P}{RT} \\end{align*} \\nonumber\n\n$$(n/V)$$ is in both equations.\n\n\\begin{align*} \\dfrac{n}{V} &= \\dfrac{\\rho}{M} \\\\[4pt] &= \\dfrac{P}{RT} \\end{align*} \\nonumber\n\n$\\dfrac{\\rho}{M} = \\dfrac{P}{RT}\\nonumber$\n\nIsolate density.\n\n$\\rho = \\dfrac{PM}{RT} \\nonumber$\n\n###### Step 4: Now plug in the information you have.\n\n\\begin{align*} \\rho &= \\dfrac{PM}{RT} \\\\[4pt] &= \\dfrac{(0.3263\\; \\rm{atm})(2*14.01 \\; \\rm{g/mol})}{(0.08206\\, L\\, atm/K mol)(291 \\; \\rm{K})} \\\\[4pt] &= 0.3828 \\; g/L \\end{align*} \\nonumber\n\nAn example of varying density for a useful purpose is the hot air balloon, which consists of a bag (called the envelope) that is capable of containing heated air. As the air in the envelope is heated, it becomes less dense than the surrounding cooler air (Equation $$\\ref{10.5.2}$$), which is has enough lifting power (due to buoyancy) to cause the balloon to float and rise into the air. Constant heating of the air is required to keep the balloon aloft. As the air in the balloon cools, it contracts, allowing outside cool air to enter, and the density increases. When this is carefully controlled by the pilot, the balloon can land as gently as it rose.",
null,
"Figure $$\\PageIndex{1}$$: A hot air balloon is inflated partially with cold air from a gas-powered fan, before the propane burners are used for final inflation.\n\n## Determining Gas Volumes in Chemical Reactions\n\nThe ideal gas law can be used to calculate volume of gases consumed or produced. The ideal-gas equation frequently is used to interconvert between volumes and molar amounts in chemical equations.\n\n##### Example $$\\PageIndex{2A}$$\n\nWhat volume of carbon dioxide gas is produced at STP by the decomposition of 0.150 g $$\\ce{CaCO_3}$$ via the equation:\n\n$\\ce{CaCO3(s) \\rightarrow CaO(s) + CO2(g)} \\nonumber$\n\n###### Solution\n\nBegin by converting the mass of calcium carbonate to moles.\n\n$\\dfrac{0.150\\;g}{100.1\\;g/mol} = 0.00150\\; mol \\nonumber$\n\nThe stoichiometry of the reaction dictates that the number of moles $$\\ce{CaCO_3}$$ decomposed equals the number of moles $$\\ce{CO2}$$ produced. Use the ideal-gas equation to convert moles of $$\\ce{CO2}$$ to a volume.\n\n\\begin{align*} V &= \\dfrac{nRT}{PR} \\\\[4pt] &= \\dfrac{(0.00150\\;mol)\\left( 0.08206\\; \\frac{L \\cdot atm}{mol \\cdot K} \\right) ( 273.15\\;K)}{1\\;atm} \\\\[4pt] &= 0.0336\\;L \\; or \\; 33.6\\;mL \\end{align*} \\nonumber\n\n##### Example $$\\PageIndex{2B}$$\n\nA 3.00 L container is filled with $$\\ce{Ne(g)}$$ at 770 mmHg at 27oC. A $$0.633\\;\\rm{g}$$ sample of $$\\ce{CO2}$$ vapor is then added.\n\n• What is the partial pressure of $$\\ce{CO2}$$ and $$\\ce{Ne}$$ in atm?\n• What is the total pressure in the container in atm?\n###### Solution\n\nStep 1: Write down all given information, and convert as necessary.\n\nBefore:\n\n• $$P = 770\\,mmHg \\rigtharrow 1.01 \\,atm \\nonumber \\] • \\(V = 3.00\\,L$$\n• $$n_{\\ce{Ne}} = ?$$\n• $$T = 27^o C \\rightarrow 300\\; K$$\n\nOther Unknowns: $$n_{\\ce{CO2}}$$= ?\n\n$n_{CO_2} = 0.633\\; \\rm{g} \\;CO_2 \\times \\dfrac{1 \\; \\rm{mol}}{44\\; \\rm{g}} = 0.0144\\; \\rm{mol} \\; CO_2 \\nonumber$\n\n###### Step 2: After writing down all your given information, find the unknown moles of $$\\ce{Ne}$$.\n\n\\begin{align*} n_{Ne} &= \\dfrac{PV}{RT} \\\\[4pt] &= \\dfrac{(1.01\\; \\rm{atm})(3.00\\; \\rm{L})}{(0.08206\\;atm\\;L/mol\\;K)(300\\; \\rm{K})} \\\\[4pt] &= 0.123 \\; \\rm{mol} \\end{align*} \\nonumber\n\nBecause the pressure of the container before the $$\\ce{CO2}$$ was added contained only $$\\ce{Ne}$$, that is your partial pressure of $$Ne$$. After converting it to atm, you have already answered part of the question!\n\n$P_{Ne} = 1.01\\; \\rm{atm} \\nonumber$\n\nStep 3: Now that have pressure for $$\\ce{Ne}$$, you must find the partial pressure for $$CO_2$$. Use the ideal gas equation.\n\n$\\dfrac{P_{Ne}\\cancel{V}}{n_{Ne}\\cancel{RT}} = \\dfrac{P_{CO_2}\\cancel{V}}{n_{CO_2}\\cancel{RT}} \\nonumber$\n\nbut because both gases share the same Volume ($$V$$) and Temperature ($$T$$) and since the Gas Constant ($$R$$) is constants, all three terms cancel.\n\n\\begin{align*} \\dfrac{P}{n_{Ne}} &= \\dfrac{P}{n_{CO_2}} \\\\[4pt] \\dfrac{1.01 \\; \\rm{atm}}{0.123\\; \\rm{mol} \\;Ne} &= \\dfrac{P_{CO_2}}{0.0144\\; \\rm{mol} \\;CO_2} \\\\[4pt] P_{CO_2} &= 0.118 \\; \\rm{atm} \\end{align*} \\nonumber\n\nThis is the partial pressure $$\\ce{CO_2}$$.\n\nStep 4: Now find total pressure.\n\n\\begin{align*} P_{total} &= P_{Ne} + P_{CO_2} \\\\[4pt] &= 1.01 \\; \\rm{atm} + 0.118\\; \\rm{atm} \\\\[4pt] &= 1.128\\; \\rm{atm} \\\\[4pt] &\\approx 1.13\\; \\rm{atm} \\; \\text{(with appropriate significant figures)} \\end{align*} \\nonumber\n\n##### Example $$\\PageIndex{2C}$$: Sulfuric Acid\n\nSulfuric acid, the industrial chemical produced in greatest quantity (almost 45 million tons per year in the United States alone), is prepared by the combustion of sulfur in air to give $$\\ce{SO2}$$, followed by the reaction of $$\\ce{SO2}$$ with $$\\ce{O2}$$ in the presence of a catalyst to give $$\\ce{SO3}$$, which reacts with water to give $$\\ce{H2SO4}$$. The overall chemical equation is as follows:\n\n$\\ce {2S(s) + 3O2(g) + 2H2O(l) \\rightarrow 2H2SO4(aq)} \\nonumber$\n\nWhat volume of O2 (in liters) at 22°C and 745 mmHg pressure is required to produce 1.00 ton (907.18 kg) of H2SO4?\n\nGiven: reaction, temperature, pressure, and mass of one product\n\nAsked for: volume of gaseous reactant\n\n###### Strategy:\n\nA Calculate the number of moles of H2SO4 in 1.00 ton. From the stoichiometric coefficients in the balanced chemical equation, calculate the number of moles of $$\\ce{O2}$$ required.\n\nB Use the ideal gas law to determine the volume of $$\\ce{O2}$$ required under the given conditions. Be sure that all quantities are expressed in the appropriate units.\n\n###### Solution:\n\nmass of $$\\ce{H2SO4}$$ → moles $$\\ce{H2SO4}$$ → moles $$\\ce{O2}$$ → liters $$\\ce{O2}$$\n\nA We begin by calculating the number of moles of H2SO4 in 1.00 ton:\n\n$\\rm\\dfrac{907.18\\times10^3\\;g\\;H_2SO_4}{(2\\times1.008+32.06+4\\times16.00)\\;g/mol}=9250\\;mol\\;H_2SO_4 \\nonumber$\n\nWe next calculate the number of moles of $$\\ce{O2}$$ required:\n\n$\\rm9250\\;mol\\;H_2SO_4\\times\\dfrac{3mol\\; O_2}{2mol\\;H_2SO_4}=1.389\\times10^4\\;mol\\;O_2 \\nonumber$\n\nB After converting all quantities to the appropriate units, we can use the ideal gas law to calculate the volume of O2:\n\n\\begin{align*} V&=\\dfrac{nRT}{P} \\\\[4pt] &=\\rm\\dfrac{1.389\\times10^4\\;mol\\times0.08206\\dfrac{L\\cdot atm}{mol\\cdot K}\\times(273+22)\\;K}{745\\;mmHg\\times\\dfrac{1\\;atm}{760\\;mmHg}} \\\\[4pt] &=3.43\\times10^5\\;L \\end{align*} \\nonumber\n\nThe answer means that more than 300,000 L of oxygen gas are needed to produce 1 ton of sulfuric acid. These numbers may give you some appreciation for the magnitude of the engineering and plumbing problems faced in industrial chemistry.\n\n##### Exercise $$\\PageIndex{2}$$\n\nCharles used a balloon containing approximately 31,150 L of $$\\ce{H2}$$ for his initial flight in 1783. The hydrogen gas was produced by the reaction of metallic iron with dilute hydrochloric acid according to the following balanced chemical equation:\n\n$\\ce{ Fe(s) + 2 HCl(aq) \\rightarrow H2(g) + FeCl2(aq)} \\nonumber$\n\nHow much iron (in kilograms) was needed to produce this volume of $$\\ce{H2}$$ if the temperature were 30°C and the atmospheric pressure was 745 mmHg?\n\n68.6 kg of Fe (approximately 150 lb)\n\n##### Example $$\\PageIndex{3}$$: Emergency Air bags\n\nSodium azide ($$\\ce{NaN_3}$$) decomposes to form sodium metal and nitrogen gas according to the following balanced chemical equation:\n\n$\\ce{ 2NaN3 \\rightarrow 2Na(s) + 3N2(g)} \\nonumber$\n\nThis reaction is used to inflate the air bags that cushion passengers during automobile collisions. The reaction is initiated in air bags by an electrical impulse and results in the rapid evolution of gas. If the $$\\ce{N_2}$$ gas that results from the decomposition of a 5.00 g sample of $$\\ce{NaN_3}$$ could be collected by displacing water from an inverted flask, what volume of gas would be produced at 21°C and 762 mmHg?\n\nGiven: reaction, mass of compound, temperature, and pressure\n\nAsked for: volume of nitrogen gas produced\n\n###### Strategy:\n\nA Calculate the number of moles of $$\\ce{N_2}$$ gas produced. From the data in Table S3, determine the partial pressure of $$\\ce{N_2}$$ gas in the flask.\n\nB Use the ideal gas law to find the volume of $$\\ce{N_2}$$ gas produced.\n\n###### Solution:\n\nA Because we know the mass of the reactant and the stoichiometry of the reaction, our first step is to calculate the number of moles of $$\\ce{N_2}$$ gas produced:\n\n$\\rm\\dfrac{5.00\\;g\\;NaN_3}{(22.99+3\\times14.01)\\;g/mol}\\times\\dfrac{3mol\\;N_2}{2mol\\;NaN_3}=0.115\\;mol\\; N_2 \\nonumber$\n\nThe pressure given (762 mmHg) is the total pressure in the flask, which is the sum of the pressures due to the N2 gas and the water vapor present. Table S3 tells us that the vapor pressure of water is 18.65 mmHg at 21°C (294 K), so the partial pressure of the $$\\ce{N_2}$$ gas in the flask is only\n\n\\begin{align*} \\rm(762 − 18.65)\\;mmHg \\times\\dfrac{1\\;atm}{760\\;mmHg} &= 743.4\\; \\cancel{mmHg} \\times\\dfrac{1\\;atm}{760\\;\\cancel{mmHg}} \\\\[4pt] &= 0.978\\; atm. \\end{align*} \\nonumber\n\nB Solving the ideal gas law for V and substituting the other quantities (in the appropriate units), we get\n\n$V=\\dfrac{nRT}{P}=\\rm\\dfrac{0.115\\;mol\\times0.08206\\dfrac{atm\\cdot L}{mol\\cdot K}\\times294\\;K}{0.978\\;atm}=2.84\\;L \\nonumber$\n\n##### Exercise$$\\PageIndex{3}$$\n\nA 1.00 g sample of zinc metal is added to a solution of dilute hydrochloric acid. It dissolves to produce $$\\ce{H2}$$ gas according to the equation\n\n$\\ce{ Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq)}. \\nonumber$\n\nThe resulting H2 gas is collected in a water-filled bottle at 30°C and an atmospheric pressure of 760 mmHg. What volume does it occupy?"
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"https://chem.libretexts.org/@api/deki/files/51438/1024px-Yellow.balloon.inflation.arp.jpg",
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https://documen.tv/question/an-individual-has-40-000-to-invest-28-000-will-be-put-into-a-low-risk-mutual-fund-averaging-6-9-24120893-49/ | [
"## An individual has $40,000 to invest:$28,000 will be put into a low-risk mutual fund averaging 6.9% interest compounded monthly, and the rem\n\nQuestion\n\nAn individual has $40,000 to invest:$28,000 will be put into a low-risk mutual fund averaging 6.9% interest compounded monthly, and the remainder will be invested in a high-yield bound fund averaging 9.8% interest compounded continuously.\n\nRequired:\na. Write an equation for the total amount in the two investments.\nb. Write the rate-of-change equation for the combined amount.\nc. How rapidly is the combined amount of the investments growing after 6 months? after 15 months?\n\nin progress 0\n6 months 2021-08-22T15:07:50+00:00 1 Answers 1 views 0\n\n## Answers ( )\n\na) F(x) = 28,000( 1.00575 )^12x + 12,000e^0.098x\n\nb) F'(x ) = 28,000 ( In 1.071224 ) ( 1.071224 )^x + 1176 e^0.098x dollar per year\n\nc) 3228.94 dollar/year, 3428.73 dollar/year\n\nStep-by-step explanation:\n\nCapital = $40,000$28,000 = low-risk mutual fund\n\n6.9% monthly compounded interest for the low risk mutual fund\n\n\\$12,000 = high-risk yield bound fund\n\n9.8% continuously compounded interest\n\nA) Equation for total amount in two investments\n\nF(x) = F1(x) + F2(x) —– ( 1 )\n\nwhere :\n\nF1(x) ( future value for monthly compounded interest)\n\n= 28,000( 1 + 0.069/12 )^12x = 28,000 ( 1.00575 )^12x\n\nF2(x) ( future value for continuously compounded interest )\n\n= ( 40,000 – 28,000 )e^0.098x = 12,000 e^0.098x\n\nback to equation 1\n\nF(x) = 28,000( 1.00575 )^12x + 12,000e^0.098x\n\nB Rate of change equation\n\nf'(x) = d/dx (28,000( 1.00575 )^12x) + d/dx ( 12,000e^0.098x )\n\n∴ f'(x) = 28,000 d/dx (1.00575^12)^x + 12,000 d/dx(b^x)\n\n= 28,000 ( In 1.071224 ) ( 1.071224 )^x + 12,000 ( In b ) ( b^x )\n\nf'(x ) = 28,000 ( In 1.071224 ) ( 1.071224 )^x + 1176 e^0.098x dollar per year\n\nC) Determine how rapidly the combined amount will grow after 6 months and after 15 months\n\ni.e. x = 0.5 , x = 1.25 years\n\nafter 6 months\n\n28,000 ( In 1.071224 ) ( 1.071224 )^(0.5) + 1176*e^((0.098(0.50))\n\n= 1993.88 + 1235.06 = 3228.94 dollar/year\n\nafter 15 months\n\n28,000 ( In 1.071224 ) ( 1.071224 )^(1.25) + 1176*e^((0.098(1.25 ))\n\n= 2099.47 + 1329.26 = 3428.73 dollar/year"
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https://www.heidelberg-laureate-forum.org/blog/the-lynx-and-the-hare/ | [
"# The Lynx and the Hare\n\nWhile my training is in pure mathematics, I also have an appreciation of the way applied mathematicians make use of it. Maths can be deployed to understand, explain and predict the behaviour of real-world situations, and it’s fundamental to the study of practically all other areas of science.\n\nGiven a situation in the real world, you might measure and make observations about what happens – counting the number of animals in a population at given time intervals, measuring the temperature of a chemical reaction, or collecting data about any variable which changes with time or given different initial conditions. But while measurements are useful to tell you about what’s happening now, they don’t tell you much about what’s going to happen in the future.\n\nMathematical modelling allows you to construct a model – a virtual version of the system, which you can use to predict how it’s going to evolve. You could, for instance, study your observations and notice a correlation between two things – when one of them increases, so does the other. If you’re studying the population of an animal (p), you might notice that at times when the amount of food available (f) is high, the population of the animal also tends to be higher.\n\nLooking at the data you’ve collected would allow you to determine the coefficient, c, which describes the correlation – if the quantity of food goes up, how much does the population go up by? This coefficient would be part of your model, and allow you to predict what the effect of a given amount of food being added would be, on the size of the population.\n\nYou can use a correlation in a model even if the two things turn out not to be directly related. For example, it’s been observed that cows which are given individual names tend to produce more milk – but that doesn’t mean it’s the name that’s causing Daisy’s milk production to increase. It’s the presence of a third factor: a farmer that cares enough about their cows to name them, and who also looks after them well. (If you’re a fan of spurious correlations, with or without a connecting reason, there are some who make a hobby of collecting them). But you can still use an indirect relationship to make predictions – if you know the two things will increase in tandem, regardless of why, the model is still useful.\n\nOf course, the model given above is very simple. There will almost certainly be other factors affecting the size of a population, and there might be a limit to how many animals can live in a given area – you can’t keep increasing the amount of food indefinitely and expect the population to always keep up, so your model might only work for certain ranges of values.\n\nDespite all this, population dynamics remains one of the easiest types of system to model – the variables involved tend to be fairly simple, and other minor confounding factors often don’t have a large impact on the final values. Once nice set of equations developed for modelling the interactions between two species – the predator-prey equations, also called the Lotka-Volterra equations, give a lovely approximation to the dynamics of two species in competition.\n\nGenerally, for modelling an animal population, you need to take into account a number of factors, including the rate of reproduction – how quickly that particular animal creates new animals; and the rate of death, which might depend on the average lifespan, but also other hazards which might cause them to die – in particular, the number of predators in the area, and how many of their prey each predator needs to sustain themselves.\n\nThe Lotka-Volterra equations describe the relationship between two populations as follows:\n\nIn these equations, x is the number of prey animals in the population, and y is the number of predators. Δ here is a Greek letter which mathematicians use to represent change – so Δx is the change in x. What these equations are saying is that the rate at which the prey population changes will depend on the size of the prey population, but also on the size of the predator population; and vice versa.\n\nThe values of a, b, c and d in these equations will depend on the particular animals involved – their rates of reproduction, how frequently they prey on each other, how much food the predators need to live – observing them and recording some values will allow you to establish what these values might be for a given pair of species, and then you can use this model to make predictions of what will happen in the future.\n\nLooking at the equation for Δx, you might notice the term which depends on the number of predators (y) is negative – meaning the more predators you have, the more this will reduce the prey population; in the Δy equation, the term dependent on the number of prey (x) is positive, because the more prey there are around, the more food for predators so the population will go up more quickly.\n\nSince these two values both depend on each other, you can end up with some interesting results – if there are more prey around relative to the number of predators, that will increase the number of predators, which will in turn reduce the number of prey, which will have a knock-on effect on the number of predators, and so on. The graphs produced by these models often involve two similar-looking oscillating curves, one of which follows the other with a delay (as effects take time to manifest in the population numbers).\n\nIf the populations are nicely balanced when you start your model, this system can be relatively stable. (There’s obviously one other stable system you can have here, which is when x and y are zero, but that’s boring). For stability, you ideally want\n\nas your starting states. Any changes to the system – maybe a flood which wipes out a chunk of the predator population, or an introduction of more animals from another area – will destabilise the equilibrium, and the equations will let you model what might happen.\n\nOne of my favourite things in maths is when something you’ve predicted comes true, and there’s a wonderful example of the patterns predicted by this model being exactly replicated in a real situation. Around the early 1900s, the Hudson Bay Company traded in fur pelts – from lynxes and hares trapped locally. What they might not have realised at the time was that in doing so, they were collecting data about the local hare and lynx populations – which were predator and prey to each other.\n\nIn the early 20th century, biologist Charles Gordon Hewitt analysed the data and discovered that the pattern of increase and decrease in the two populations matched the Lotka-Volterra model pretty well. The two populations were in equilibrium, with their sizes varying from year to year in a way that could be quite nicely predicted, and shows the characteristic delayed peaks of predator and prey.\n\nThis model is relatively simple, but mathematical modelling in general is a constant balancing act – if you can model the system with a simple set of equations, and not lose too much of the detail, that’s great. If your model is too simple, you’ll be able to compute your predictions quickly, but they’re less likely to be accurate, as you won’t have taken into account all the possible factors that could affect what’s going on. There’s a famous saying in statistical modelling: “All models are wrong, but some are useful”. The compromise you’re aiming for is a model which is the least possible amount of wrong, while still being reasonably useful.\n\nMost real-world systems are more complicated than this simple population example, and couldn’t be captured with such a minimal model. Meteorologists spend their time predicting the weather, using many overlapping and detailed models – but a true model to accurately predict the entire weather system would need billions of variables and would probably take so long to compute, it’d be quicker just to wait and look out of the window.\n\nDer Beitrag The Lynx and the Hare erschien zuerst auf Heidelberg Laureate Forum."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.94700706,"math_prob":0.96802187,"size":7969,"snap":"2019-26-2019-30","text_gpt3_token_len":1596,"char_repetition_ratio":0.14262398,"word_repetition_ratio":0.0021818182,"special_character_ratio":0.19914669,"punctuation_ratio":0.08409987,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96128243,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-18T19:18:49Z\",\"WARC-Record-ID\":\"<urn:uuid:e6ed18bc-78b3-4c00-8153-ff9e23e976f1>\",\"Content-Length\":\"36992\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8e59902e-6782-4c33-8a1b-f1f1a59d5882>\",\"WARC-Concurrent-To\":\"<urn:uuid:fb56594e-48ab-49c3-a945-0519cd611260>\",\"WARC-IP-Address\":\"193.197.73.11\",\"WARC-Target-URI\":\"https://www.heidelberg-laureate-forum.org/blog/the-lynx-and-the-hare/\",\"WARC-Payload-Digest\":\"sha1:5VPOA5WP65OALOCAAWNLYZT3LNMNLC73\",\"WARC-Block-Digest\":\"sha1:QX57RKFLTKNFGRPGIVDGIWJR6GO6HYQI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525793.19_warc_CC-MAIN-20190718190635-20190718212635-00061.warc.gz\"}"} |
http://dispatchesfromturtleisland.blogspot.com/2013/03/is-there-electron-up-down-koide-triple.html | [
"## Wednesday, March 13, 2013\n\n### Is There An Electron, Up, Down Koide Triple?\n\nKoide's Formula and Related Extensions\n\nKoide's formula in its original form asserts that:\n\n(sqrt(electron mass)+sqrt(muon mass)+sqrt(tau mass))^2/(electron mass+muon mass+tau mass)=2/3.\n\nThis is true to the highest levels of precision determined to date, which for the charged leptons is very great.\n\nA Koide triple is any three sets of particle masses that satisfy that relationship.\n\nThe hypothesis that there are Koide triples among the quarks, which is not inconsistent with the data to current level of precision (which isn't very great) is that the following are Koide triples:\n\ntop, bottom, charm\nbottom, charm, strange\ncharm, strange, down\n\nA related observation is that the combined mass of the bottom, charm, strange triple is almost precisely three times the mass of the tau, muon, electron triple (a notion that corresponds to the fact that in weak force decays three times as many quarks, one for each color, are produced as leptons).\n\nKoide's Formula, the Up Quark Mass and a Possibile Up, Down, Electron Triple.\n\nImplications of zero mass or neutrino scale mass for up quarks.\n\nThe final conceivable triple following that patterns are charm, strange, up, and strange, down, up. Koide's formula predicts a near zero value for the up quark mass from a c, s, u triple. But, if that value is carried through to the down quark in the s, u, d triple, it produces a value within the measured range of the down quark mass.\n\nUsing central values of t=172.9 GeV (a hair low with the latest data) and b=4.19 GeV. Then,\nKoide(t,b,c) implies c=1.356 GeV (PDG value 1.180-1.340 GeV)\nKoide(b,c,s) implies s= 92 MeV (PDG value 80-130 MeV)\nKoide(c,s,u) implies u= 36 KeV (PDG value 1,700 to 3,100 KeV)\nKoide(s,u,d) implies d= 5.3 MeV (PDG value 4.1-5.7 MeV)\n\nYou can also form a Koide triple of an electron, up and down if you use an electron mass of about 0.511 MeV, an up quark mass of zero, and a down quark mass of 6.7 MeV.\n\nAnd, if you use a value of zero rather than 36 KeV for the up quark, and use the 6.7 MeV value for the down quark predicted by the electron, up, down triple, the formula predicts a strange quark mass of 92 MeV.\n\nThis strange quark mass derived from the electron, up, down triple and the assmption that the up quark has a zero mass is consistent with the experimentally measured mass value of the strange quark, is consistent with a \"top quark down\" calculation of the strange quark mass, and is consistent with an estimate based upon a mass for the bottom, charm, strange triple that is the charged lepton mass triple.\n\nEven a modest mass of 36 KeV for the up quark makes a significant different in the estimated value of the down quark mass via an electron, up and down Koide triple or a strange, up and down Koide triple. But, an up quark mass on the order of magnitude of 1 eV or less does not throw off the Koide triple by more than can be easily made up with tiny tweaks to calibration points elsewhere.\n\nThis is important because there are a variety of theoretical reasons why an up quark with a non-zero but negligible rest mass, even if it was just 1 eV, would involve a far more modest tweak to the Standard Model than a truly zero mass up quark.\n\nThe Koide's formula's prediction does not alter the experimentally estimated combined up and down quark mass.\n\nThe 6.7 MeV estimate for the down quark mass from applying Koide's formula naively is also not far from the Particle Data Group (PDG) mid-range value for the up quark and mid-range value for the down quark mass combined, which is 7.3 MeV. The sum of the lower extremes of the PDG estimates for the up and down quark masses is 5.8 MeV and the sum of the upper extremes of the PDG estimates for the up and down quark masses is 8.8 MeV. Twice the PDG estimate of the mean up and down quark masses is 6.0 MeV to 9.6 MeV, a range within which the 6.7 MeV Koide's formula value fits comfortably.\n\nThus, the Koide formula predicted value for the sum of the up and down quark masses when the up quark is assumed to have a mass of zero is well within the PDG value. Koide's formula simply allocates all of the combined mass to the down quark rather than assigning a mass to the up quark of 35% to 60% of the down quark mass. It also does nothing to alter the longstanding assumption based largely on the fact that the proton is lighter than the neutron, that the down quark is heavier than the up quark.\n\nReconsidering the experimental estimate of the up quark mass.\n\nKeep in mind that up quarks are always confined and can't be measured in isolation the way that top quarks can be, and that almost all of the mass in hadrons (two quark mesons and three quark baryons) is derived from the strong force binding energy carried by gluons and not from the quarks themselves. This is particularly true in the case of hadrons that have only up and down quarks like the proton and the neutron for which the measured hadron masses that contribute to the estimates are most precise.\n\nSince up quarks are always confined, any estimate of the up quark mass is necessarily model dependent. Yet, computations of quantities like the proton or neutron mass from first principles using QCD alone have a precision of only about 1%, making them far less precise than the experimentally measured masses of hadrons.\n\nAlso, the experimental uncertainty in the mass of all quarks except the up quark equals or exceeds the low end experimental value per the PDG of the up quark mass. So, in hadrons with some quarks other than up quarks in them, the up quark number has an impact on the total mass which is generally lower than the total uncertainty in the fundamental quark mass contribution to the hadron's total mass.\n\nThe strength of the strong force, weak force, electromagnetic forces are so great at the scale of a hadron relative to the masses of the quarks involved in all but the most exotic hadrons with heavy quarks in them, that an up quark's color charge, weak isospin and electromagnetic charge all have more relevance to its behavior when confined in a hadron than its fundamental mass (except insofar is its mass influences its weak force decays).\n\nObviously, if the Koide's formula prediction conflicts with the experimental data then there is simply something wrong with the formula. But, the existence of consistent predictions from an electron, up, down triple with those of series of quark triples, and with the predictions of quark masses from the masses of the lepton triple all argue for revisiting the model dependent assumptions that went into making the PDG estimate of the up quark's mass (which is in any case has extremes that vary by a factor of two anyway).\n\nFiguring out how the up quark mass was estimated and what practical implications the up quark mass has in the Standard Model is clearly near the top of my to do list.\n\nImplications of a zero mass for the up quark.\n\nIf the up quark mass were assumed to be zero, as a non-measured Standard Model constant, rather than an experimentally measured one, and the other quark masses were estimated based upon this model dependent assumption, how would the estimated quark masses different and what experiments, if any, that were the basis for the PDG estimate would be contradicted?\n\nSome of the issues of how up quark mass is determined and what this implies in practice when doing QCD are discussed in this 2004 paper and another paper in 2010 and in 2011 by the same author, Michael Creutz who together with the authors of this 2002 paper are interested in the possibility that a massless up quark could explain the strong CP problem. This 2003 paper (possibly identical) also investigates the possibility of a massless up quark and makes a mass calculation for the up quark using lattice QCD. This paper from 2001 disfavors that solution in a model limited to two quarks (the 2002-2003 analysis is a three quark flavor analysis).\n\nThis 1997 paper gets into the guts of mass renormalization for quarks. A 2009 model dependent estimate of the up and down quark masses shows how these quantities are derived in QCD. A 2011 paper uses the up-down mass difference and applies it to neutrino scattering. This 2011 paper discusses relevant source data in the context of a BSM Higgs mass generation idea.\n\n(I've omit papers by Koide himself in this review). Thinking similar to that of Koide's on mass matrixes is found in a 2013 paper and in this 2012 paper and this 2012 paper. A BSM model from 2011 explores similar ideas. A 1999 paper considers implications for quintessence theories.\n\nCurrent light quark mass ratio estimates don't differ materially from those devised by Weinberg and discussed in this 1986 paper whose abstract stated:\nWe investigate the current-mass ratios of the light quarks by fitting the squares of meson masses to second order in chiral-symmetry breaking, determining corrections to Weinberg's first-order values: mu/md=0.56, ms/md=20.1. We find that to this order, ms/md is a known function of mu/md. The values of the quark-mass ratios can be constrained by limiting the size of second-order corrections to the squares of meson masses. We find that for specific values of presently unmeasured phenomenological parameters one can have a massless u quark. In that case 30% of the squares of meson masses arise from operators second order in chiral-symmetry breaking.\n\nA 1979 estimate is also not that different in its early estimation of light quark masses as is this 1996 paper or a 1994 paper. The 1994 paper's abstract stated that: \"the claim that\nmu = 0 leads to a coherent picture for the low energy structure of QCD is examined in detail. It is pointed out that this picture leads to violent flavour asymmetries in the matrix elements of the scalar and pseudoscalar operators, which are in conflict with the hypothesis that the light quark masses may be treated as perturbations.\"\n\nThis 1978 paper's abstract states:\nWe consider, within the framework of current algebra, the possibility that the up-quark mass vanishes (as an alternative to the axion). We argue that the contrary current-algebra value, mu/md=1/1.8, is unreliable. A critical analysis leads to the conclusion that mu=0 is not unreasonable and furthermore leads to a surprisingly good prediction for the δ-meson mass.\nA massless up quark has been considered a viable option seriously considered since 1978. It was considered an open possible solution to the strong CP problem in 1994 it found that:\n\nWe conclude that at the level of precision (order of magnitude) of nonperturbative QCD calculations available to us at present, low-energy phenomenology is completely compatible with a vanishing value of the high-energy up quark mass.\n\nOnly a nonperturbative calculation in QCD can prove or disprove the phenomenological viability of mu = 0. Therefore, in view of the recent progress in numerical methods in lattice gauge theory, we would like to encourage a detailed analysis of the possibility of a massless up quark by these methods.\n\nA 2000 paper considered ways to test the massless up quark hypothesis using lattice QCD methods and does this 2002 paper which finds that lattice calculations disfavor a massless up quark but that experiments don't resolve the issues apart from a first principles analysis. A 2001 paper notes that useful theoretical QCD predictions can be done with massless quarks entirely. A 2007 paper quantifies the impact of quark mass on QCD predictions from massless models.\n\nA zero or non-zero mass for the up quark might help explain why proton decay is so surprisingly rare.\n\nA zero mass for the up quark together with the extended Koide's formula that motivates it, would imply that the masses of the six quarks and all three charged leptons can be calculated via the extended Koide's formula (including the mass relationship of the charged lepton triple to one of the quark triples and the electron, up down triple) and the assumption that the up quark has zero mass from the mass of the electron using high school algebra to accuracies greater than those available for any of the experimentally measured quark masses (even the top quark whose mass is currently known to 0.6% accuracy).\n\nThis would reduce the number of experimentally measured physical constants related to fermion mass in the Standard Model + Extended Koide Model from fifteen to four (the electron and the three neutrino masses).\n\nIf the supposition that the Higgs boson mass is equal to half of the sum of the masses of the W+, W- and Z bosons (which is currently accurate to within all current bounds of experimental precision and is closer to the experimentally measured mark than any of the prediscovery mass predictions for the Higgs boson mass), then the number of experimentally measured physical constants related to mass in the Standard Model would fall from three to two, one of which (the Weinberg angle that relates to W and Z boson masses) isn't even a mass value itself.\n\nThus, we could be on the verge of going from having eighteen measured Standard Model mass constants to having just six, and having much more accurate theoretical values than experimental values for many of those constants.\n\nThis would also motivate strongly a Koide derived formula for neutrino masses that if devised and confirmed by experimental evidence would cut the number of experimentally measured mass constants in the Standard Model from six to not more than five (one of which is an angle rather than a mass), and possibly to as few as three if a way to derive the neutrino masses from first principles using the masses of the other fermions and Standard Model bosons (and perhaps the PMNS and/or CKM matrix elements and/or the Standard Model coupling constants) was devised.\n\nExtensions To A Standard Model With Four Generations\n\nExtending the Koide's formula allows one to make useful, constrained and testable predictions regarding a fourth generation of Standard Model particles.\n\nFourth generation Standard Model particles that would have the masses a naive extension of Koide's formula would imply are experimentally forbidden because the lepton sector is inconsistent with experimental data. This is a conclusion that has already been reached for the large part by the fundamental physics community already based on other grounds.\n\nFourth Generation Koide Quarks\n\nIf one extends the formula based upon recent data on the mass of the bottom and top quarks and presumes that there is a b', t, b triple, and uses masses of 173,400 GeV for the top quark and 4,190 for the bottom quark, then the predicted b' mass would be 3,563 GeV and the predicted t' mass would be about 83.75 TeV (i.e. 83,750 GeV).\n\nSince they would be produced a t'-anti-t' and b'-anti-b' pairs, it would take about 167.5 TeV of energy to produce a t' and 7.1 TeV of energy to produce a b'. Producing a t' would be far beyond the capababilities of the LHC. But, it could conceivably produce a few b' quark events of the Koide's formula predicted mass. These would be unmistakeable unless the extreme speeds of the decay products prevented them from decaying (as a result of special relativity effects) until they reached a point beyond the most remote LHC detectors. This probably wouldn't happen for a b' decay which is within the design parameters of the LHC, but might happen in the case of a fluke t' decay, which is far outside of its design parameters.\n\nThe up to the minute direct exclusion range at the LHC for the b' and t' is that there can be no b' with a mass of less than 670 GeV and no t' with a mass of less than 656 GeV (per ATLAS) and the comparable exclusions from CMS are similar (well under 1 TeV).\n\nKoide t' and b' quark decays\n\nA simple fourth generation b' quark or t' quark, that otherwise fits the Standard Model, of that mass would decay so rapidly that it woud not hadronize (i.e. not form composite QCD particles via strong force gluon interactions). Instead, the t' would decay almost exclusively to the b' and the b' would decay almost exclusively to the t, with both interactions happening almost instantaneously.\n\nA t' decay to a b' would produce a highly energetic W+ boson that would carry much of the energy of 80 TeV of rest mass being converted into kinetic energy for the W+ and b' produced in the decay, immediately followed by a highly energetic W- boson produced in the b' to t decay in which about 3,390 GeV of kinetic energy was created from rest mass, followed by the usual immediate t quark to b quark decay with an emission of a W+ converting about 169.2 GeV of rest mass into kinetic energy for the W+ and b quark. There would be an exactly parallel set of reactions for the decay chain of the anti-t' particle.\n\nIn the absence of special relativity, this would take place within a sphere of a diameter of less than 10^-16 meters (i.e. about 1-2% of the diameter of a nucleus of a gold atom, a number derived from the decay time of 10^-23 seconds for the first three decays times the speed of light), the strange quark decays would start to happen about a foot from the original site of the decay, and the muon decays would peak about 300 meters away. But, since particle decay takes place in the reference frame of the particle, which is moving at speeds near the speed of light, the decays would take place over a far more extended area because time would pass more slowly for the fast moving t' decay products. The extreme kinetic energies of the particles would cause the their decays to happen at much greater distances from the initial t' production and decay site than the ordinary LHC decays - indeed they might make it past the detectors entirely.\n\nAlso, while a 167.5 TeV event does involve a lot of energy in a concentrated place, a single event of that size involves only about 3*10-4 joules of energy, about the amount of kinetic energy of a single grape on the verge of hitting the ground after falling from a vine at waist height, so if it made it past the detectors due to special relativistic extensions of decay times it would be virtually invisible to observers in the area around the impact site and beyond the detectors.\n\nSo, even if the LHC was able due to a fluke fluctuation that led to a collision more than ten times as energetic as its design limitations with a spectacular decay chain, it might be missed entirely or almost entirely except as a completely unprecedented amount of missing energy that might be attributed to an equipment failure rather than a real physics event because it was so far beyond the designed detection range of the scientific equipment in the facility.\n\nFourth generation Koide leptons\n\nThe extension for charged leptons (a muon, tau, tau prime triple), however, would imply a 43.7 GeV tau prime, which has been excluded at the 95% confidence level for masses of less than 100.8 GeV and with far greater confidence at 43.7 GeV (which would be produced at a significant and easy to measure freuquency in Z boson decays).\n\nA simple Koide's rule formula for neutrinos using the muon neutrino mass (of 7.5 * 10^-5 eV + 0.08 eV +/- 0.09 eV) and tau neutrino mass (of 2.4 * 10^-3 eV + 0.08 eV +/- 0.09 eV) (with absolute masses derived from accurately measured mass differences between types and a 0.51 eV limit on the sum of the electron neutrino, muon neutrino and tau neutrino masses if there are only three kinds of neutrinos - less if there are more generations of neutrinos), would yield a tau prime neutrino mass of far less than 43.7 GeV. A naive extension of Koide's formula with an electron neutrino of near zero mass would lead to a fourth generation neutrino of about 0.05 eV and would have a mass of up to 11.6 eV in the nearly degenerate case where all three neutrino species had almost precisely the same mass. But, this would contradict the cosmological data constraint that limits to 0.51 eV for the sum of the masses all of the species of light neutrinos combined (which is about 1/1,000,000th the rest mass of an electron). So, instead, 0.51 eV would be the realistic upper limit of a Standard Model weakly interacting fourth neutrino generation.\n\nGiven the cosmological constraint on the sum of neutrino masses, the possibility that the naive Koide's formula needs a sign modification or something like it for neutrinos (which it probably does) is irrelevant.\n\nYet, any simple, fourth generation, weakly interacting tau prime neutrinos of any rest mass less than 45 GeV can be excluded on the basis of Z boson decays, so this scenario is definitively excluded if Koide's formula is even remotely an accurate way of estimating the mass of a hypothetical fourth generation Standard Model neutrino.\n\nThese theoretical considerations make it highly unlikely that there is any fourth generation of Standard Model fermions at all. The Standard Model makes fermions an entire generation at a time, and this would require a fourth generation charge fermion far in excess of the Koide formula extension predicted value.\n\nandrew said...\n\nKoide's formula is generally applied to rest mass via an electroweak-Higgs boson interaction.\n\nIt generally does not apply instead to mass at some consistent higher energy levels derived from the running of the renormalization group. It generally does not apply instead to additional momentum acquired in motion via special relativity. It generally does not apply to mass acquired via a strong force process or through other interactions with other particles.\n\nIt is quite possible (and at least one of the cited papers suggests) that the up quark mass that is utilized in lattice QCD at given energy levels, and/or is measured in experiments, is acquired dynamically by up quarks in their interactions with down quarks or other quarks, for example, rather than as a component of its rest mass.\n\nKoide up quark masses and other kinds of up quark masses may need to be carefully distinguished from the kind of up quark mass to which the massless up quark hypothesis that the extended Koide's formulas suggests.\n\nStill, given the pertinent of the massless up quark hypothesis not only to the strong CP problem but also to the entire mass matrix of the Standard Model via the extended Koide's formula, this hypothesis deserves more serious attention.\n\nMitchell said...\nandrew said...\n\nLattice QCD precision quark mass estimates can be found here.\n\nThe strange quark comes in at 92 just as expected. The charm at 1.273 is quite a bit lighter than the Koide number."
] | [
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https://algorithms.tutorialhorizon.com/find-factorial-of-a-given-number/ | [
"# Find Factorial of a given Number\n\nObjective: Given a number, write a program to find factorial of that number.\n\nWhat is Factorial Number?\n\nIn mathematics, the factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n. The value of 0! is 1, according to the convention for an empty product\n\nN! = n*(n-1)*(n-2)*…..*2*1\n\nExample:\n\n```5! = 5 x 4 x 3 x 2x 1 = 120\n7! = 7 x 6 x 5 x 4 x 3 x 2x 1 = 5040\n```\n\nApproach:\n\nRecursive Solution\n\n• If number is 0, return 1.\n• Make a recursive call to get the result of number – 1.\n• Multiply number with the result of number – 1.\n\nJava Code:\n\nhttps://gist.github.com/thmain/33f2e9a7ee6c36e3168b97de80feb13e\n\nOutput:\n\n`Factorial of a number: 10 is(Recursion): 3628800`\n\nIterative Solution:\n\n• If number is 0, return 1.\n• Run a loop from 2 to number and multiple all of them to get the result.\n\nJava Code:\n\nhttps://gist.github.com/thmain/469886811d1d86a9214a6cc35c699783\n\nOutput:\n\n`Factorial of a number: 7 is(Iterative): 5040`"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7785476,"math_prob":0.9974755,"size":959,"snap":"2021-43-2021-49","text_gpt3_token_len":321,"char_repetition_ratio":0.12984294,"word_repetition_ratio":0.14545454,"special_character_ratio":0.3514077,"punctuation_ratio":0.18518518,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9966675,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-26T08:20:25Z\",\"WARC-Record-ID\":\"<urn:uuid:45e2a27d-a5aa-45ae-a83c-3f6b7781e0fc>\",\"Content-Length\":\"49426\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f1bef4d2-e96c-4eea-b365-d0deda17c899>\",\"WARC-Concurrent-To\":\"<urn:uuid:55edd97b-0934-4950-ab61-4feea4854c05>\",\"WARC-IP-Address\":\"172.67.179.102\",\"WARC-Target-URI\":\"https://algorithms.tutorialhorizon.com/find-factorial-of-a-given-number/\",\"WARC-Payload-Digest\":\"sha1:LB5OBK7EM3MPDYJMYZQTQWW5ZBSPKMBB\",\"WARC-Block-Digest\":\"sha1:JY55K52EZSXQCWX7TXLHKACGMCQCV5FA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587854.13_warc_CC-MAIN-20211026072759-20211026102759-00055.warc.gz\"}"} |
https://doc.rust-lang.org/core/arch/x86/fn._mm_setcsr.html | [
"# Function core::arch::x86::_mm_setcsr\n\n1.27.0 · source ·\n``pub unsafe fn _mm_setcsr(val: u32)``\nAvailable on (x86 or x86-64) and target feature `sse` and x86 only.\nExpand description\n\nSets the MXCSR register with the 32-bit unsigned integer value.\n\nThis register controls how SIMD instructions handle floating point operations. Modifying this register only affects the current thread.\n\nIt contains several groups of flags:\n\n• Exception flags report which exceptions occurred since last they were reset.\n\n• Masking flags can be used to mask (ignore) certain exceptions. By default these flags are all set to 1, so all exceptions are masked. When an an exception is masked, the processor simply sets the exception flag and continues the operation. If the exception is unmasked, the flag is also set but additionally an exception handler is invoked.\n\n• Rounding mode flags control the rounding mode of floating point instructions.\n\n• The denormals-are-zero mode flag turns all numbers which would be denormalized (exponent bits are all zeros) into zeros.\n\n### Exception Flags\n\n• `_MM_EXCEPT_INVALID`: An invalid operation was performed (e.g., dividing Infinity by Infinity).\n\n• `_MM_EXCEPT_DENORM`: An operation attempted to operate on a denormalized number. Mainly this can cause loss of precision.\n\n• `_MM_EXCEPT_DIV_ZERO`: Division by zero occurred.\n\n• `_MM_EXCEPT_OVERFLOW`: A numeric overflow exception occurred, i.e., a result was too large to be represented (e.g., an `f32` with absolute value greater than `2^128`).\n\n• `_MM_EXCEPT_UNDERFLOW`: A numeric underflow exception occurred, i.e., a result was too small to be represented in a normalized way (e.g., an `f32` with absulte value smaller than `2^-126`.)\n\n• `_MM_EXCEPT_INEXACT`: An inexact-result exception occurred (a.k.a. precision exception). This means some precision was lost due to rounding. For example, the fraction `1/3` cannot be represented accurately in a 32 or 64 bit float and computing it would cause this exception to be raised. Precision exceptions are very common, so they are usually masked.\n\nException flags can be read and set using the convenience functions `_MM_GET_EXCEPTION_STATE` and `_MM_SET_EXCEPTION_STATE`. For example, to check if an operation caused some overflow:\n\n``````_MM_SET_EXCEPTION_STATE(0); // clear all exception flags\n// perform calculations\nif _MM_GET_EXCEPTION_STATE() & _MM_EXCEPT_OVERFLOW != 0 {\n// handle overflow\n}``````\nRun\n\nThere is one masking flag for each exception flag: `_MM_MASK_INVALID`, `_MM_MASK_DENORM`, `_MM_MASK_DIV_ZERO`, `_MM_MASK_OVERFLOW`, `_MM_MASK_UNDERFLOW`, `_MM_MASK_INEXACT`.\n\nA single masking bit can be set via\n\n``_MM_SET_EXCEPTION_MASK(_MM_MASK_UNDERFLOW);``\nRun\n\nHowever, since mask bits are by default all set to 1, it is more common to want to disable certain bits. For example, to unmask the underflow exception, use:\n\n``````_mm_setcsr(_mm_getcsr() & !_MM_MASK_UNDERFLOW); // unmask underflow\nexception``````\nRun\n\nWarning: an unmasked exception will cause an exception handler to be called. The standard handler will simply terminate the process. So, in this case any underflow exception would terminate the current process with something like `signal: 8, SIGFPE: erroneous arithmetic operation`.\n\n### Rounding Mode\n\nThe rounding mode is describe using two bits. It can be read and set using the convenience wrappers `_MM_GET_ROUNDING_MODE()` and `_MM_SET_ROUNDING_MODE(mode)`.\n\nThe rounding modes are:\n\n• `_MM_ROUND_NEAREST`: (default) Round to closest to the infinite precision value. If two values are equally close, round to even (i.e., least significant bit will be zero).\n\n• `_MM_ROUND_DOWN`: Round toward negative Infinity.\n\n• `_MM_ROUND_UP`: Round toward positive Infinity.\n\n• `_MM_ROUND_TOWARD_ZERO`: Round towards zero (truncate).\n\nExample:\n\n``_MM_SET_ROUNDING_MODE(_MM_ROUND_DOWN)``\nRun\n\n### Denormals-are-zero/Flush-to-zero Mode\n\nIf this bit is set, values that would be denormalized will be set to zero instead. This is turned off by default.\n\nYou can read and enable/disable this mode via the helper functions `_MM_GET_FLUSH_ZERO_MODE()` and `_MM_SET_FLUSH_ZERO_MODE()`:\n\n``````_MM_SET_FLUSH_ZERO_MODE(_MM_FLUSH_ZERO_OFF); // turn off (default)\n_MM_SET_FLUSH_ZERO_MODE(_MM_FLUSH_ZERO_ON); // turn on``````\nRun\n\nIntel’s documentation"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.80804795,"math_prob":0.85618806,"size":3901,"snap":"2023-40-2023-50","text_gpt3_token_len":886,"char_repetition_ratio":0.12137542,"word_repetition_ratio":0.025225226,"special_character_ratio":0.2296847,"punctuation_ratio":0.16468842,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9791746,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-03T10:08:48Z\",\"WARC-Record-ID\":\"<urn:uuid:328b1685-612c-426a-be17-ce2238f85d28>\",\"Content-Length\":\"13290\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d5f35c4e-9e5f-4478-a45e-23b98a0a89ac>\",\"WARC-Concurrent-To\":\"<urn:uuid:b160facc-9cf2-497a-b97d-b10d82dbcd1e>\",\"WARC-IP-Address\":\"99.86.229.65\",\"WARC-Target-URI\":\"https://doc.rust-lang.org/core/arch/x86/fn._mm_setcsr.html\",\"WARC-Payload-Digest\":\"sha1:6ICATW2CWKB7DBMRCBQJAW3SHEDPBP6D\",\"WARC-Block-Digest\":\"sha1:RRQVLDIQ2IXY4XWXQNKQZBGBI3QUEO25\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511075.63_warc_CC-MAIN-20231003092549-20231003122549-00681.warc.gz\"}"} |
https://www.educationwa.com/2008/07/measuring-student-progress-in.html | [
"## Thursday, July 17, 2008\n\n### Measuring student progress in mathematics\n\nThere are key indicators to performance in mathematics. In number/algebra teachers look for certain things at certain stages - these are defined in scope and sequence documents released by DET this year in minimum benchmark form. I think though that minimum benchmarks are poor indicators of how a system is doing. So I propose a different set that parents could use to measure performance. (Note: this is not what to teach - just a general measure of progress)\n\nyear 1-4 - Students have 1-1 number correspondence. Students have a clear understanding of place value. Students recognise the relevance of operations, understand concepts such as odd/even and ascending descending and can reconstruct multiples of numbers up to 12.\nyear 4/5 - Student is confident in recognising and performing all operations (+-÷x) and can recite all tables up to and including 12.\nyear 6/7 - Student is confident with fractional quantities including estimating, adding, subtracting and multiplying a variety of fractions with a calculator but without using the \"a b/c\" button\nyear 8 - Students can perform confidently simple algebraic operations. Students understand the connection between an equation of a line and its drawn equivalent. Students can construct an equation of a line from a table of values or a graph.\nyear 9 - Students can manipulate linear and quadratic equations to shift them on a cartesian plane. Students can simplify confidently using index laws including negative indices and fractional indices. Students are confident at regrouping and solving simple equations.\nyear 10 - Students can factorise and use this knowledge to sketch and draw quadratic and linear equations. Students can plot curves, understand critical points on curves and use equations/graphs to perform optimisations, interpolate and extrapolate data.\nyear 11/12 - Students can use knowledge to solve complex worded problems with application in the real world including problems including statistics, calculus and numeric series.\n\nFor many parents these words make no sense - but a quick google of unknown terms can assist a parent in getting a clearer picture of what a student can do."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93097913,"math_prob":0.93401897,"size":2255,"snap":"2020-45-2020-50","text_gpt3_token_len":438,"char_repetition_ratio":0.1390493,"word_repetition_ratio":0.0056179776,"special_character_ratio":0.19423503,"punctuation_ratio":0.077120826,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9858733,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-29T19:50:48Z\",\"WARC-Record-ID\":\"<urn:uuid:ba1f46f5-7103-419c-8572-697d94a935d0>\",\"Content-Length\":\"106370\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b38246ff-4632-4f9a-93f0-38e5979c84a1>\",\"WARC-Concurrent-To\":\"<urn:uuid:577bf7d2-83ac-4da8-8fe8-1e9159c7149b>\",\"WARC-IP-Address\":\"172.217.2.115\",\"WARC-Target-URI\":\"https://www.educationwa.com/2008/07/measuring-student-progress-in.html\",\"WARC-Payload-Digest\":\"sha1:ANHXJW257C4SFBAF2IDCWSIFY3JMBVYZ\",\"WARC-Block-Digest\":\"sha1:44F5ILUYO2FSNYCGE6YF37FLPA5MX5EU\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107905777.48_warc_CC-MAIN-20201029184716-20201029214716-00632.warc.gz\"}"} |
https://www.elsevier.com/books/non-linear-differential-equations/sansone/978-0-08-010194-1 | [
"# Non-Linear Differential Equations, Volume 67\n\n## 1st Edition\n\n0.0 star rating Write a review\nAuthors:\nEditors:\neBook ISBN: 9781483185057\nImprint: Pergamon\nPublished Date: 1st January 1964\nPage Count: 550\nSales tax will be calculated at check-out Price includes VAT/GST\n70.95\n60.31\n93.95\n79.86\n56.99\n48.44\nUnavailable\nPrice includes VAT/GST\n\n### Institutional Subscription\n\nPreface\n\nChapter I. General Theorems About Solutions of Differential Systems\n\n1. Integral Curves\n\n1. Integral Curves. Extreme Time in the Future t+\n\n2. Conditions for t+ = b. The Case n = 1\n\n3. Conditions for t+ = b. General Case\n\n4. Boundedness of the Integral Curves\n\n5. Integral Curves in the Sense of Carathéodory\n\n2. Lipschitzian and Carathéodory Systems\n\n1. Gronwall's Lemma (Generalized)\n\n2. Lipschitzian Systems. Evaluation of |x(t) — y(t)| for Two Arcs of Integral Curves\n\n3. Uniqueness Theorem. Continuous Dependence on the Initial Point P0 and on ƒ\n\n4. Carathéodory Systems\n\n3. The Solution φ(t,t0,x0) of the System (1.1.1)\n\n1. The function φ(t,t0,x0). Cases of Uniqueness\n\n2. Continuity of φ(t,t0,x0)\n\n3. Stability\n\n4. The function φ (t, t0, x0) for Linear Systems\n\n5. Differentiability of φ (t, t0, x0)\n\n6. Systems with Parameters\n\n4. Periodic Solutions\n\n1. Periodic Integral Curves. Periodic Orbits\n\n2. Exceptional Periodic Solutions\n\n5. Autonomous Systems\n\n1. Autonomous Systems. Properties of their Integral Curves\n\n2. Trajectories. Phase Space\n\n3. Singular Points. Cycles. Open Trajectories\n\nComplements\n\nBibliography\n\nChapter II. Particular Plane Autonomous Systems\n\n1. The Linear Case\n\n1. Singular Points\n\n2. Canonical Forms of Isolated Singular Points of Linear Systems\n\n3. Affine Transformations of the Phase Plane\n\n4. Classification of the Types of Singular Points\n\n2. Homogeneous Systems\n\n1. Homogeneous Systems\n\n2. Invariant Rays. Stellar Node\n\n3. The Center and the Focus\n\n4. Isolated Invariant Rays. Normal Angles\n\n5. Behavior of Trajectories in a Normal Angle\n\n6. Examples\n\n3. The Analytic Case\n\n1. Introductory Remarks\n\n2. Examples\n\n3. The Functions Z(x, y), N(x, y)\n\n4. A Lemma\n\n5. Trajectories Tending to 0. Focus\n\n6. The Equation N(θ) = 0. Dicritical Points\n\n7. Study of Z(x, y). Case of the Fixed Sign for Z(x, y)\n\n8. Classification of Z-Sectors\n\n4. The Problem of the Center\n\n1. The Problem of the Center\n\n2. The Problem of the Center for N(θ) ≠ 0\n\n3. The Case m = 1. Method of Poincaré\n\n4. The Case m = 1. Theorem of Poincaré for the Center. The Proof of E. Picard-J. Chazy\n\n5. The Case m = 1. Evaluation of the Period\n\n6. Sufficient Condition of Poincaré for the Center. Applications to Delaunay's Equations of Lunar Motion\n\n7. Bibliographic Notes on the Problem of the Center\n\n5. Singular Points at Infinity\n\n1. Poincaré's Sphere. Singular Points at Infinity\n\n2. Examples\n\n3. Singular Points at Infinity for Homogeneous Systems\n\nComplements\n\nBibliography\n\nChapter III. The Singularities of Briot-Bouquet\n\n1. Theorem of Briot-Bouquet for the Analytic Case\n\n1. Introductory Remarks\n\n2. The Equation of Briot—Bouquet in the Case where p is not a Positive Integer. Study of Holomorphic Solutions\n\n3. The Case of a Positive Integer p. Existence of Holomorphic Solutions\n\n4. Solutions of the Equation for the Case p = 0\n\n2. Reduction of Differential Equations with an Isolated Singular Point to a Typical Form in the Analytic Case. The Theorem of I. Bendixson on the Behavior of the Trajectories of the Reduced Equations of the Second Type\n\n1. Reduced Forms of the First and Second Type\n\n2. Results of I. Bendixson on the Behavior of the Trajectories of the Reduced Equations of the Second Type\n\n3. Equation of Briot-Bouquet in the Nodal Case in the Real Domain. Theorems of A. Wintner\n\n1. Lemma of A. Wintner\n\n2. First Theorem of A. Wintner\n\n3. Second Theorem of A. Wintner\n\nComplements\n\nBibliography\n\nChapter IV. Plane Autonomous Systems\n\n1. Limiting Sets\n\n1. Limiting Sets A(γ), Ω(γ) of a Trajectory γ. General Properties\n\n2. Classification of Trajectories\n\n3. Regular Points and Trajectories\n\n4. Closed (Plane) Trajectories. Stability Properties of Plane Cycles\n\n5. Regular (Plane) Limiting Trajectories\n\n6. Structure of Bounded Limiting Sets Ω(γ)\n\n7. Limiting Sets Consisting of a Single (Singular) Point\n\n8. Structure of Unbounded Sets Ω(γ)\n\n2. Plane Cycles\n\n1. Limit Cycles\n\n2. Classification of Limit Cycles. Orbital Stability\n\n3. Examples\n\n4. Bendixson's Theorem\n\n5. Systems of Class C1. Characteristic Exponent of a Cycle\n\n6. Cycles of Analytic Systems\n\n7. Limit Cycles of a System with Polynomial Right Sides\n\n8. Regions with No Plane Cycles\n\n9. Periodic Solutions of a Plane Autonomous System. Existence of Limit Cycles\n\n10. Uniqueness of (Limit) Cycles\n\n3. Isolated Singular Points\n\n1. Classification of Isolated Singular Points. Points of the First Kind (Centerfocus). Center\n\n2. The Neighborhood of a Point of the Second Kind\n\n3. The Focus\n\n4. Exceptional Directions\n\n5. Normal Sectors\n\n4. The Index\n\n1. Kronecker's Index\n\n2. Index of a Point\n\n3. Evaluation of the Index for Particular Singular Points\n\n4. Index on the Sphere and on a Surface of Genus p\n\n5. The Cylinder as Phase Space\n\n1. The Cylinder as Phase Space\n\n2. An Example\n\n6. The Torus as Phase Space\n\n1. The Torus as Phase Space\n\n2. Examples\n\n3. Systems with No Singular Points on the Torus\n\n4. Other Results\n\n7. A Short Account on Dynamical Systems\n\nBibliography\n\nChapter V. Autonomous Plane Systems with Perturbations\n\n1. Homogeneous Perturbed Systems\n\n1. The General Problem\n\n2. The Case N(θ) ≠ 0\n\n3. Trajectories Tending to 0. Exceptional Directions\n\n4. Invariance of Normal Sectors. Normal Sectors of the First Type\n\n5. Normal Sectors of the Second Type. First Decision Problem\n\n6. Normal Sectors of the Third Type. Second Decision Problem\n\n7. The Case N(θ) Identically Zero\n\n8. Some Remarks\n\n2. Isolated Singular Points of Systems of Class C1. Elementary Points\n\n1. Introductory Remarks\n\n2. Foci and Weak Foci\n\n3. Attractors. Stellar Node\n\n4. Node with One Tangent\n\n5. Node with Two Tangents\n\n7. Remarks\n\n3. An Asymptotic Study of a Node with Two Tangents and a Saddle Point of H. Weyl\n\n1. Statement of the Problem. Notations\n\n2. The Nodal Case (0 < l < k). The First Theorem of H. Weyl\n\n3. Bounds for |e11y(t) — b|, |x(t) — x0e-kt|\n\n4. The Case x(r) = Crδ. Bounds for |y(t) — be-lt|, |x(t)|\n\n5. The Case k ≥ l, k > 0. Second Theorem of H. Weyl\n\n6. Parametrized Systems\n\n7. The Case of the Node with Two Tangents. Third Theorem of H. Weyl\n\n8. The Case of a Saddle Point (l < 0 < k). Fourth Theorem of H. Weyl\n\n4. Isolated Singular Points of Systems of Class C1. Non-Elementary Points\n\n1. Introductory Remarks\n\n2. First Theorem of K. A. Keil for the Systems ẋ = x + ƒ(x, y), ẏ = g(x, y)\n\n3. Lemmas on the Isoclines\n\n4. K. A. Keil's Second and Third Theorems for the System ẋ = x + ƒ(x, y), ẏ = g(x, y)\n\n5. Further Results of K. A. Keil for the System ẋ = y + ƒ(x, y), ẏ = g(x, y)\n\n6. Bibliographical Notes to Sec. 1.2.4\n\n5. Structurally Stable Systems. Systems with a Parameter\n\n1. Structurally Stable Systems\n\n2. Structurally Unstable Systems. Generation of Limit Cycles\n\n3. Variation of the Cycles of Systems with a Parameter\n\nBibliography\n\nChapter VI. On Some Autonomous Systems with One Degree of Freedom\n\n1. Trajectories of the Equation of Linear Motion of a Point under Viscous Resistance\n\n1. Trajectories of the Equation of Linear Motion of a Point under Viscous Resistance\n\n2. The Equation θ + αθ + sin θ — β = 0, α ≥ θ, β ≥ 0\n\n1. Introductory Remarks\n\n2. The Case β > 1. Existence of a Periodic Solution z = z(θ) of (6.2.3)\n\n3. The Case 0 < β < 1. Classification of Singular Points\n\n4. The Trajectories for the Limiting Case α = 0\n\n5. The Case 0 < α , 0 < β < 1. Periodic Solutions of (6.2.3) and Critical Values α(θ0)\n\n6. The Case θ0 = π/2, (β = 1)\n\n7. The Trajectories for 0 < θ0 < π/2, (0 < α; 0 < β < 1)\n\n8. Inequalities for the Critical Value α(θ0)\n\n9. Procedure of M. Urabe for Calculating α(θ0)\n\n3. Equations of van der Pol and Liénard of the Oscillations of Relaxation\n\n1. Preliminary Results\n\n2. Existence of Periodic Solutions of Liénard's Equation\n\n3. Sufficient Conditions for the Uniqueness of Periodic Solutions of Liénard's Equation\n\n4. Case of Non-Uniqueness of Periodic Solutions of a Liénard Equation\n\n5. Theorem of Existence of Periodic Solutions of Liénard's Equation in the Case where ƒ(x) has Ordinary Discontinuities\n\n6. Theorem of Comparison for Liénard's Equation\n\n7. Calculation of the Period\n\n8. Van der Pol's Equation. Behavior of Trajectories at Infinity\n\n9. Behavior of the Limit Cycle of van der Pol's Equation when the Parameter Tends to Infinity. Theorem of D. A. Flanders and J. J. Stoker\n\n10. Asymptotic Evaluation of the Period and Amplitude of Periodic Solutions of van der Pol's Equation for Large Values of the Parameter\n\n11. Inequalities for Limit Cycles Due to R. Gomory and D. E. Richmond\n\n4. Periodic Solutions of the Generalized Equation of Liénard\n\n1. First Theorem of A. F. Filippov\n\n2. Second Theorem of A. F. Filippov\n\n3. Theorem of Uniqueness\n\n4. Study of the equation ẍ + ƒ(x, ẋ) ẋ + g(x) = 0\n\n5. Periodic Solutions of the Equation. ẍ + ƒ(ẋ) x + g(x) = 0 without the Hypothesis x g(x) > 0 for |x| > 0\n\n1. Introductory Remarks\n\n2. Singular Points\n\n3. Cycles. Their Properties\n\n4. A Case of Non-Existence of Periodic Solutions\n\n5. Existence of Cycles\n\n6. A Criterion for the Uniqueness of a Cycle\n\n6. The Equation of Damped Vibrations: Aẍ + ƒ (ẋ)ẋ + Cx = 0\n\n1. Introductory Remarks\n\n2. Conditions for the Origin to be a Stable Point\n\n3. A Theorem of G. Malgarini\n\n7. On an Equation of Dynamics and Aerodynamics of Wires\n\n1. Singular Points\n\n2. The Field Relative to System (6.7.2)\n\n3. Existence of Periodic Solutions for Sufficiently Small Values of the Parameter p\n\nComplements\n\nBibliography\n\nChapter VII. Non-Autonomous Systems with One Degree of Freedom\n\n1. The Problem of Forced Oscillations. Linear Case\n\n1. Forced Oscillations in the Harmonic Case\n\n2. Forced Oscillations in the Non-Harmonic Case\n\n3. The Problem of Forced Oscillations\n\n2. The Fixed Point Theorem of L. E. J. Brouwer and the Theorems of M. L. Cartwright, J. E. Littlewood and J. L. Massera\n\n1. The Fixed Point Theorem of L. E. J. Brouwer\n\n2. Brouwer's Theorem in the Proofs of Existence of Periodic Solutions\n\n3. Theorem of M. L. Cartwright-J. E. Littlewood\n\n4. Theorem of J. L. Massera\n\n3. Theorems of T. Yoshizawa\n\n1. Criterion of Ultimate Boundedness\n\n2. Theorem of Existence of Periodic Solutions\n\n3. Stability of Solutions\n\n4. A Theorem on Uniqueness and Stability of Periodic Solutions\n\n5. A Criterion for Boundedness of Individual Solutions\n\n6. Derivation of a Criterion for Existence of Periodic Solutions from the Theorem of J. L. Massera. A Theorem of S. Mizohata and M. Yamaguti\n\n4. Harmonic Solutions out of Phase of the Equation ẍ = F(x, cos ωt). Theorem of F. John\n\n1. The Interval (— ∞, + ∞) as Domain of Existence of Solutions\n\n2. Theorem of F. John on the Existence of Harmonic Solutions out of Phase\n\n5. The Equation ẍ + ƒ(ẋ) ẋ + g(x) = p(t)\n\n1. Results of S. Lefschetz, N. Levinson, M. L. Cartwright, and J. E. Littlewood\n\n2. An Existence Theorem and a Theorem on Asymptotic Stability of N.Levinson\n\n3. The Equation ẍ + g(x) = p(t) for p(t) Even. Theorem of G. R. Morris\n\n4. Odd Periodic Harmonic Solutions. Theorem of W. S. Loud on the Duffing Equation with Forcing Term\n\n5. Inequalities of D. Graffi for the Periodic Solutions of the Equation ẍ + ƒ(x) ẋ + λ2x = F sin ωt, λ > 0, ω > 0, F > 0\n\n6. The Equation ẍ + F(ẋ) + x = p(t)\n\n1. Criteria of R. Caccioppoli, A. Ghizzetti and A. Ascari for Existence, Uniqueness and Stability of a Periodic Solution\n\n2. On a Differential Equation in the Mechanics of Wires. Results of J. Cecconi and F. Stoppelli\n\n7. Theorems of G. E. H. Reuter on the Equations ẍ + k ƒ(x) ẋ + g(x) = k p(t), ẍ + k F(ẋ) + g(x) = k p(t)\n\n1. The Equation ẍ + k ƒ(x) ẋ + g(x) = k p(t)\n\n2. The Equation ẍ + k F(ẋ) + g(x) = k p(t)\n\n8. The Equation ẍ + ƒ(x, ẋ) ẋ + g(x) = p(t)\n\n1. Criterion on Boundedness in the Future of H. A. Antosiewicz\n\n2. Criteria of N. Levinson and C. E. Langenhop for Existence of a Periodic Solution\n\n9. Non-Linear Systems with Subharmonic Solutions\n\n1. Subharmonic Solutions\n\n2. Class D Systems\n\n3. Classifications of Fixed Points Relative to the Transformations of Systems of Class D\n\n4. Theorems of N. Levinson and J. L. Massera on the Number of Subharmonic Solutions\n\n10. General Remarks Concerning Periodic Solutions\n\n1. Autonomous Systems\n\n2. Periodic Non-Autonomous Systems\n\nComplements\n\nBibliography\n\nChapter VIII. Linear Systems\n\n1. The Adjoint System. The Inequalities of T. Wazewski\n\n2. Wazewski's Inequality\n\n2. Linear Autonomous Systems with Constant Coefficients\n\n1. The Principal Fundamental Matrix\n\n2. Form of Solutions of the Homogeneous System. Characteristic Exponents. Type Numbers\n\n3. Singular Points in the Real Case\n\n4. The Case n = 3 (Real Case)\n\n3. Linear Periodic Systems\n\n1. The Principal Fundamental Matrix. Theorems of Floquet and Lyapunov\n\n2. Characteristic Exponents. Type Numbers\n\n4. Reducible Systems\n\n1. Reducible Systems. Characteristic Exponents and Type Numbers\n\n5. Type Numbers of a Function. Relation of t-Similitude\n\n1. Type Number of a Function\n\n2. The Relation of t-Similitude (or Kinematic Similarity)\n\n3. Type Number of a Non-Vanishing Solution\n\n4. Normal Systems of Solutions. The Number Smin\n\n5. Inequality for Smin. Constant of Irregularity\n\n6. Regular Systems\n\n1. Regular Systems\n\n2. Theorems of Perron\n\n3. Triangular Matrices\n\n7. Periodic Solutions\n\n1. Linear Homogeneous Systems\n\n2. Linear Non-Homogeneous Systems\n\n3. The Existence of Harmonic Solutions of Quasi-Linear Periodic Systems\n\nComplements\n\nBibliography\n\nChapter IX. Stability\n\n1. The Method of V Functions\n\n1. Introductory Remarks\n\n2. The V Functions\n\n3. A Lemma of T. Wazewski\n\n4. Sufficient Conditions for Stability\n\n5. Necessary Conditions for Stability. The Inverse Problem\n\n6. Asymptotic Stability\n\n7. Asymptotic Stability in the Large\n\n8. Other Kinds of Stability\n\n9. Instability\n\n10. V Functions for Boundedness\n\n2. Stability of Linear Systems\n\n1. Stable and Unstable Linear Systems\n\n2. Uniformly Stable Linear Systems\n\n3. Uniform Stability and t ∞-Similarity\n\n4. Criteria of Uniform Stability\n\n5. Linear Systems Reducible to Zero and Restrictive Stability\n\n6. Asymptotic Stability of Linear Systems\n\n7. V Functions for Linear Systems with Constant Coefficients\n\n3. Stability in the First Approximation\n\n1. Introductory Remarks\n\n2. Stability by a Linear First Approximation\n\n3. Some Generalizations and Remarks. The L(v, N) Property\n\n4. Asymptotic Stability. Cases of a Non-Linear First Approximation\n\n5. Analytical Systems. The Critical Cases\n\n6. Orbital (Asymptotic) Stability. The Behavior of Solutions Near Integral Manifolds\n\n4. Asymptotic Equivalence\n\n1. Asymptotic Equivalence\n\n2. The Theorem of H. Weyl\n\n3. Other Results on Asymptotic Equivalence\n\nComplements and Problems\n\nBibliography\n\nIndex\n\nOther Volumes in this Series (Pure and Applied Mathematic)\n\n## Description\n\nNon-Linear Differential Equations covers the general theorems, principles, solutions, and applications of non-linear differential equations.\n\nThis book is divided into nine chapters. The first chapters contain detailed analysis of the phase portrait of two-dimensional autonomous systems. The succeeding chapters deal with the qualitative methods for the discovery of periodic solutions in periodic systems. The remaining chapters describe a synthetical approach to the study of asymptotic properties, especially stability properties, of the solutions of general n-dimensional systems.\n\nThis book will be of great value to mathematicians, researchers, and students.\n\nNo. of pages:\n550\nLanguage:\nEnglish\nPublished:\n1st January 1964\nImprint:\nPergamon\neBook ISBN:\n9781483185057"
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https://www.aimsciences.org/article/doi/10.3934/dcdsb.2016033 | [
"",
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"August 2016, 21(6): 1999-2009. doi: 10.3934/dcdsb.2016033\n\n## Positive solutions of perturbed elliptic problems involving Hardy potential and critical Sobolev exponent\n\n 1 Mathematics Science College, Inner Mongolia Normal University, Hohhot 010022, China 2 School of Mathematical Sciences and LPMC, Nankai University, Tianjin 300071\n\nReceived January 2014 Revised April 2016 Published June 2016\n\nIn this paper, we are concerned with the following nonlinear Schrödinger equations with hardy potential and critical Sobolev exponent \\begin{equation}\\label{eq0.1} \\left\\{\\begin{array}{ll} -\\Delta u+\\lambda a(x)u^q=\\mu\\frac{u}{|x|^{2}}+|u|^{2^{*}-2}u,& \\textrm{in}\\, \\mathbb{R}^N, \\\\ u>0, & \\textrm{in}\\,\\mathcal{D}^{1,2}(\\mathbb{R}^N), （1） \\end{array} \\right. \\end{equation} where $2^{*}=\\frac{2N}{N-2}$ is the critical Sobolev exponent, $0\\leq \\mu<\\overline{\\mu}=\\frac{(N-2)^2}{4}$, $a(x)\\in C(\\mathbb{R}^N)$. We first use an abstract perturbation method in critical point theory to obtain the existence of positive solutions of （1） for small value of $|\\lambda|$. Secondly, we focus on an anisotropic elliptic equation of the form \\begin{equation}\\label{eq0.2} -{\\rm div}(B_\\lambda(x)\\nabla u)+\\lambda a(x)u^q=\\mu\\frac{u}{|x|^{2}}+|u|^{2^*-2}u, x\\in\\mathbb{R}^N. （2） \\end{equation} The same abstract method is used to yield existence result of positive solutions of （2） for small value of $|\\lambda|$.\nCitation: Jing Zhang, Shiwang Ma. Positive solutions of perturbed elliptic problems involving Hardy potential and critical Sobolev exponent. Discrete & Continuous Dynamical Systems - B, 2016, 21 (6) : 1999-2009. doi: 10.3934/dcdsb.2016033\n##### References:\n A. Ambrosetti and Andrea Malchiodi, Perturbation Methods and Semilinear Elliptic Problems on $\\mathbbR^N$,, Birkhäuser Verlag, (2006).",
null,
"Google Scholar A. Ambrosetti, J. Garcia Azorero and I. Peral, Perturbation of $\\Delta u+u^{\\frac{N+2}{N-2}}=0$, The scalar curvature problems in $\\mathbbR^N$ and related topics,, J. Funct. Anal, 165 (1999), 117. doi: 10.1006/jfan.1999.3390.",
null,
"",
null,
"Google Scholar A. Ambrosetti and M. Badiale, Variational perturbative methods and bifurcation of bounds states from the the essential spectrum,, Proc. Roy. Soc. Edinburgh A, 128 (1998), 1131. doi: 10.1017/S0308210500027268.",
null,
"",
null,
"Google Scholar A. Ambrosetti, M. Badiale and S. Cingolani, Semiclassical states of nonlinear Schrödinger equations,, Arch. Rational Mech. Anal., 140 (1997), 285. doi: 10.1007/s002050050067.",
null,
"",
null,
"Google Scholar M. Badiale, J. Garcia Azorero and I. Peral, Perturbation results for an anisotropic SchrHodinger equation via a variational form,, NoDEA, 7 (2000), 201. doi: 10.1007/s000300050005.",
null,
"",
null,
"Google Scholar H. Berestycki and P. L. Lions, Nonlinear scalar field equations I - existence of a ground state,, Arch. Rational Mech. Anal., 82 (1983), 313. doi: 10.1007/BF00250555.",
null,
"",
null,
"Google Scholar K. J. Brown and N. Stavrakakis, Global bifurcation results for a semilinear elliptic equation on all $\\mathbbR^N$,, Duke Math. J., 85 (1996), 77. doi: 10.1215/S0012-7094-96-08503-8.",
null,
"",
null,
"Google Scholar F. Catrina and Z. Q. Wang, On the Caffarelli-Kohn-Nirenberg inequalities: Sharp constants, existence (and nonexistence), and symmetry of extermal functions,, Comm. Pure Appl. Math., 54 (2001), 229. doi: 10.1002/1097-0312(200102)54:2<229::AID-CPA4>3.0.CO;2-I.",
null,
"",
null,
"Google Scholar S. Cingolani, Positive solutions to perturbed elliptic problems in $\\mathbbR^N$ involving critical Sobolev exponent,, Nonlinear Analysis, 48 (2002), 1165. doi: 10.1016/S0362-546X(00)00245-5.",
null,
"",
null,
"Google Scholar O. Rey, The role of the Green's function in a non-linear elliptic equation involving the critical Sobolev exponent,, J. Funct. Anal., 89 (1990), 1. doi: 10.1016/0022-1236(90)90002-3.",
null,
"",
null,
"Google Scholar N. S. Trudinger, On Harnack type inequalities and theri application to quasilinear elliptic equations,, Comm. Pure Appl. Math., 20 (1967), 721. doi: 10.1002/cpa.3160200406.",
null,
"",
null,
"Google Scholar\n\nshow all references\n\n##### References:\n A. Ambrosetti and Andrea Malchiodi, Perturbation Methods and Semilinear Elliptic Problems on $\\mathbbR^N$,, Birkhäuser Verlag, (2006).",
null,
"Google Scholar A. Ambrosetti, J. Garcia Azorero and I. Peral, Perturbation of $\\Delta u+u^{\\frac{N+2}{N-2}}=0$, The scalar curvature problems in $\\mathbbR^N$ and related topics,, J. Funct. Anal, 165 (1999), 117. doi: 10.1006/jfan.1999.3390.",
null,
"",
null,
"Google Scholar A. Ambrosetti and M. Badiale, Variational perturbative methods and bifurcation of bounds states from the the essential spectrum,, Proc. Roy. Soc. Edinburgh A, 128 (1998), 1131. doi: 10.1017/S0308210500027268.",
null,
"",
null,
"Google Scholar A. Ambrosetti, M. Badiale and S. Cingolani, Semiclassical states of nonlinear Schrödinger equations,, Arch. Rational Mech. Anal., 140 (1997), 285. doi: 10.1007/s002050050067.",
null,
"",
null,
"Google Scholar M. Badiale, J. Garcia Azorero and I. Peral, Perturbation results for an anisotropic SchrHodinger equation via a variational form,, NoDEA, 7 (2000), 201. doi: 10.1007/s000300050005.",
null,
"",
null,
"Google Scholar H. Berestycki and P. L. Lions, Nonlinear scalar field equations I - existence of a ground state,, Arch. Rational Mech. Anal., 82 (1983), 313. doi: 10.1007/BF00250555.",
null,
"",
null,
"Google Scholar K. J. Brown and N. Stavrakakis, Global bifurcation results for a semilinear elliptic equation on all $\\mathbbR^N$,, Duke Math. J., 85 (1996), 77. doi: 10.1215/S0012-7094-96-08503-8.",
null,
"",
null,
"Google Scholar F. Catrina and Z. Q. Wang, On the Caffarelli-Kohn-Nirenberg inequalities: Sharp constants, existence (and nonexistence), and symmetry of extermal functions,, Comm. Pure Appl. Math., 54 (2001), 229. doi: 10.1002/1097-0312(200102)54:2<229::AID-CPA4>3.0.CO;2-I.",
null,
"",
null,
"Google Scholar S. Cingolani, Positive solutions to perturbed elliptic problems in $\\mathbbR^N$ involving critical Sobolev exponent,, Nonlinear Analysis, 48 (2002), 1165. doi: 10.1016/S0362-546X(00)00245-5.",
null,
"",
null,
"Google Scholar O. Rey, The role of the Green's function in a non-linear elliptic equation involving the critical Sobolev exponent,, J. Funct. Anal., 89 (1990), 1. doi: 10.1016/0022-1236(90)90002-3.",
null,
"",
null,
"Google Scholar N. S. Trudinger, On Harnack type inequalities and theri application to quasilinear elliptic equations,, Comm. Pure Appl. Math., 20 (1967), 721. doi: 10.1002/cpa.3160200406.",
null,
"",
null,
"Google Scholar\n Qingfang Wang, Hua Yang. Solutions of nonlocal problem with critical exponent. Communications on Pure & Applied Analysis, 2020, 19 (12) : 5591-5608. doi: 10.3934/cpaa.2020253 Xiyou Cheng, Zhitao Zhang. Structure of positive solutions to a class of Schrödinger systems. Discrete & Continuous Dynamical Systems - S, 2020 doi: 10.3934/dcdss.2020461 Zedong Yang, Guotao Wang, Ravi P. Agarwal, Haiyong Xu. Existence and nonexistence of entire positive radial solutions for a class of Schrödinger elliptic systems involving a nonlinear operator. Discrete & Continuous Dynamical Systems - S, 2020 doi: 10.3934/dcdss.2020436 Gongbao Li, Tao Yang. Improved Sobolev inequalities involving weighted Morrey norms and the existence of nontrivial solutions to doubly critical elliptic systems involving fractional Laplacian and Hardy terms. Discrete & Continuous Dynamical Systems - S, 2020 doi: 10.3934/dcdss.2020469 Marion Darbas, Jérémy Heleine, Stephanie Lohrengel. Numerical resolution by the quasi-reversibility method of a data completion problem for Maxwell's equations. Inverse Problems & Imaging, 2020, 14 (6) : 1107-1133. doi: 10.3934/ipi.2020056 Gang Bao, Mingming Zhang, Bin Hu, Peijun Li. An adaptive finite element DtN method for the three-dimensional acoustic scattering problem. Discrete & Continuous Dynamical Systems - B, 2020 doi: 10.3934/dcdsb.2020351 Mehdi Badsi. Collisional sheath solutions of a bi-species Vlasov-Poisson-Boltzmann boundary value problem. Kinetic & Related Models, , () : -. doi: 10.3934/krm.2020052 Meng Chen, Yong Hu, Matteo Penegini. On projective threefolds of general type with small positive geometric genus. Electronic Research Archive, , () : -. doi: 10.3934/era.2020117 Mehdi Bastani, Davod Khojasteh Salkuyeh. On the GSOR iteration method for image restoration. Numerical Algebra, Control & Optimization, 2021, 11 (1) : 27-43. doi: 10.3934/naco.2020013 Min Chen, Olivier Goubet, Shenghao Li. Mathematical analysis of bump to bucket problem. Communications on Pure & Applied Analysis, 2020, 19 (12) : 5567-5580. doi: 10.3934/cpaa.2020251 Youshan Tao, Michael Winkler. Critical mass for infinite-time blow-up in a haptotaxis system with nonlinear zero-order interaction. Discrete & Continuous Dynamical Systems - A, 2021, 41 (1) : 439-454. doi: 10.3934/dcds.2020216 Hong Niu, Zhijiang Feng, Qijin Xiao, Yajun Zhang. A PID control method based on optimal control strategy. Numerical Algebra, Control & Optimization, 2021, 11 (1) : 117-126. doi: 10.3934/naco.2020019 Stefano Bianchini, Paolo Bonicatto. Forward untangling and applications to the uniqueness problem for the continuity equation. Discrete & Continuous Dynamical Systems - A, 2020 doi: 10.3934/dcds.2020384 Li-Bin Liu, Ying Liang, Jian Zhang, Xiaobing Bao. A robust adaptive grid method for singularly perturbed Burger-Huxley equations. Electronic Research Archive, 2020, 28 (4) : 1439-1457. doi: 10.3934/era.2020076 Zexuan Liu, Zhiyuan Sun, Jerry Zhijian Yang. A numerical study of superconvergence of the discontinuous Galerkin method by patch reconstruction. Electronic Research Archive, 2020, 28 (4) : 1487-1501. doi: 10.3934/era.2020078 Yuxia Guo, Shaolong Peng. A direct method of moving planes for fully nonlinear nonlocal operators and applications. Discrete & Continuous Dynamical Systems - S, 2020 doi: 10.3934/dcdss.2020462 Noah Stevenson, Ian Tice. A truncated real interpolation method and characterizations of screened Sobolev spaces. Communications on Pure & Applied Analysis, 2020, 19 (12) : 5509-5566. doi: 10.3934/cpaa.2020250 Yue Feng, Yujie Liu, Ruishu Wang, Shangyou Zhang. A conforming discontinuous Galerkin finite element method on rectangular partitions. Electronic Research Archive, , () : -. doi: 10.3934/era.2020120 Marco Ghimenti, Anna Maria Micheletti. Compactness results for linearly perturbed Yamabe problem on manifolds with boundary. Discrete & Continuous Dynamical Systems - S, 2020 doi: 10.3934/dcdss.2020453 Alberto Bressan, Sondre Tesdal Galtung. A 2-dimensional shape optimization problem for tree branches. Networks & Heterogeneous Media, 2020 doi: 10.3934/nhm.2020031\n\n2019 Impact Factor: 1.27"
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http://newnooze.info/multiplication-worksheets-for-grade-3-2-digits-4767.html | [
"# Multiplication Worksheets For Grade 3 2 Digits\n\ni1",
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""
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http://the-dusty-deck.blogspot.com/2014/11/functional-programming-in-mainstream_27.html | [
"Labels\n\nAI (1) Haskell (1) Java 8 (2) JavaScript (1) Logic (10) Methodology (8) Refactoring (1) Scala (1) Security (2) Tools (9) Xtend (2) Xtext (1)\n\nThursday, November 27, 2014\n\nFunctional Programming in Mainstream Languages, Part 3: Higher-Order Functions in Java 7 and 8\n\nHaving seen (in the part 2 of this series) the basic syntax for expressing functions in Java 8, we will now explore some functional-programming techniques and compare how they are expressed in Scheme, Java 7, and Java 8. The examples are adapted from the book Structure and Interpretation of Computer Programs (2nd Edition)",
null,
"by Abelson and Sussman.\n\nThe following function attempts to compute an approximation to a fixed point of a given function f; that is, a value x such that x = f(x). The computation starts from an initial guess c, and repeatedly applies the function f to it, to get the series f(c), f(f(c)), f(f(f(c))), .... If this process converges, the result is close to a fixed point, since convergence means that applying f one more time returns (almost) the same value. However, the process may not converge at all, either because f has no fixed points at all, or because the process oscillates without finding a fixed point.\n\nHere are the implementations of this function in the three languages:\n\n Scheme (define (fixed-point f first-guess) (define (try guess) (let ((next (f guess))) (if (< (abs (- guess next)) epsilon) next (try next)))) (try first-guess)) Java 7 public double fixedpoint(UFunc f, double v) { double next = f.apply(v); if (Math.abs(next - v) < epsilon) return next; else return fixedpoint1(f, next); } Java 8 public double fixedpoint(DoubleUnaryOperator f, double v) { double next = f.applyAsDouble(v); if (Math.abs(next - v) < epsilon) return next; else return fixedpoint1(f, next); }\n\nFunctional implementations usually use recursion instead of loops; the Scheme implementation defines an internal recursive function try, which computes the next value in the series and checks whether it is already close enough to the previous one to consider the series to have converged. If so, it returns the last value in the series; if not, it continues recursively to try the next value.\n\nThe two Java implementations are quite similar, except for the types. For the Java 7 implementation, I created the following interface to represent unary functions from a domain D to a range R:\n\n Java 7 public interface UFunc { R apply(D arg1); }\n\nAs I explained in the previous post, Java 8 supplies a large set of interfaces to describe functions; DoubleUnaryOperator is the type of functions that take a double and return a double.\n\nThere is a fundmemtal difference between functional languages (including Scheme) and most other languages, including Java. In the former, tail-recursive calls are converted by the compiler into jumps that don't add a frame to the execution stack. This means that such calls behave like loops in terms of their memory consumption. (For more details see Abelson and Sussman",
null,
".) However, the latter languages don't guarantee this property. As a result, the Java implementations above may use stack space proportional to the number of applications of the function required for convergence.\n\nA more natural style for implementing these methods in Java is imperative rather than functional:\n\n Java 7 public double fixedpoint(UFunc f, double v) { double prev = v; double next = v; do { prev = next; next = f.apply(next); } while (Math.abs(next - prev) >= epsilon); return next; } Java 8 public double fixedpoint(DoubleUnaryOperator f, double v) { double prev = v; double next = v; do { prev = next; next = f.applyAsDouble(next); } while (Math.abs(next - prev) >= epsilon); return next; }\n\nWhile functional-programming purists frown at this style, which modifies the values of the variables prev and next, this style is more natural for Java, and uses constant space on the stack. For those reasons, I find it acceptable when using functional techniques in Java, provided that the changes are confined to local variables (rather than fields), and, in particular, locals that aren't referenced in any other method (such as those in inner classes) or function.\n\nWe can now try to use the fixedpoint function to compute square roots, using the fact that y is a square root of x if y = x/y; in other words, if y is a fixed point of the function λy.x/y:\n\n Scheme (define (sqrt x) (fixed-point (lambda (y) (/ x y)) 1.0)) Java 7 public double sqrt(final double x) { return fixedpoint(new UFunc() { @Override public Double apply(Double y) { return x / y; } }, 1.0); } Java 8 public double sqrt(double x) { return fixedpoint(y -> x / y, 1.0); }\n\nHere we need to create an anonymous function (\"lambda\"), and here the difference between Java 7 and Java 8 becomes very clear. In Java 7 we need to create the function as an anonymous inner class, with all the necessary boilerplate code, which obscures the essentials of what we are trying to do. (For example, it is quite difficult to figure out that 1.0 is the second parameter to the fixedpoint method.) In contrast, in Java 8 we can just use the new lambda notation, and the meaning is immediately clear.\n\nIn both versions of Java, it is only possible to refer in inner methods to variables defined in enclosing constants if these variables are final; that is, if they can't be modified. In Java 8 the variables doesn't need to be declared final, but it needs to be \"effectively final,\" which means that it is never changed. (It is as though the compiler added the final modifier itself; this is similar to the type inference done by the Java 8 compiler.)\n\nThe reason for this restriction is that local variables have a lifetime (sometimes called \"extent\") that coincides with the lifetime of the method in which they are defined. In other words, when the method returns, these variables are no longer accessible. This is done so that the variables can be allocated on the stack rather than on the heap, avoiding the need for garbage-collecting them. However, an object containing an inner method that refers to these variables may be used after the method that generated the object has already returned. If (and only if) the variables are final is it possible to copy their values to the new object so that references to their values are still valid. This is a poor-man's version of a closure, which is an object containing a function pointer together with all accessible variables in outer scopes. As we will see in a later post, Scheme (like other dialects of Lisp) supports full closures, which also allow changing bound variables.\n\nUnfortunately, the method above doesn't work in any language. The problem is that the computational process oscillates between two values and doesn't converge (try it for x=2). One value is too high, and the other is too low. However, if we take their average, the process converges; the following implementations all compute the square root:\n\n Scheme (define (sqrt x) (fixed-point (lambda (y) (/ (+ y (/ x y)) 2.0) 1.0)) Java 7 public double sqrt(final double x) { return fixedpoint(new UFunc() { @Override public Double apply(Double y) { return (y + x / y) / 2.0; } }, 1.0); } Java 8 public double sqrt(double x) { return fixedpoint(y -> (y + x / y) / 2.0, 1.0); }\n\nIt turns out that this is a general technique, called average damping. It takes any function f and returns the function λx.(x+f(x))/2; this function has exactly the same fixed points as f. In certain cases, however, the computational process of the fixed-point function converges with the new function when it diverges on the original function. This technique is easily specified as a program:\n\n Scheme (define (sqrt x)(define (average-damp f) (lambda (x) (average x (f x)))) Java 7 public UFunc averageDamp(final UFunc f) { return new UFunc() { @Override public Double apply(Double x) { return (x + f.apply(x)) / 2.0; } }; } Java 8 public DoubleUnaryOperator averageDamp(DoubleUnaryOperator f) { return x -> (x + f.applyAsDouble(x)) / 2.0; }\nAgain, the Java 7 version is obscured by the boilerplate code (not to mention the annoying repetitions of the template type). Here is the definition of sqrt that uses average damping (this time I will spare you the agony of the Java 7 version):\n\n Scheme (define (sqrt x) (fixed-point (average-damp (lambda (y) (/ x y))) 1.0)) Java 8 public double sqrt(double x) { return fixedpoint(averageDamp(y -> x / y), 1.0); }\n\nComputationally, this is exactly the same as the previous version, except that we use the original function λy.x/y after applying the average damp operator to it.\n\nIn summary, it is possible to use functional abstraction in Java 7, but it is very painful. The one case where it is widely used (although not usually thought of as such) is the case of callback functions, which are encapsulated as methods in one-method classes. Java 8 makes higher-order functions much easier to use, even though the cumbersome type system is still annoying and gets in the way.\n\nIn subsequent posts we will see how to express the same ideas in other languages.\n\n#functionalprogramming #java"
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null,
"http://ir-na.amazon-adsystem.com/e/ir",
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"http://ir-na.amazon-adsystem.com/e/ir",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8501269,"math_prob":0.91275704,"size":8749,"snap":"2019-43-2019-47","text_gpt3_token_len":2038,"char_repetition_ratio":0.13916524,"word_repetition_ratio":0.112714775,"special_character_ratio":0.242542,"punctuation_ratio":0.12,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9932541,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-15T19:25:12Z\",\"WARC-Record-ID\":\"<urn:uuid:46010eab-72ea-4a2e-bb5c-ecfa295a7b0b>\",\"Content-Length\":\"74952\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:134bc3dd-cf78-4b70-991f-d320d00f43b8>\",\"WARC-Concurrent-To\":\"<urn:uuid:48d0193e-96bf-4c0a-bd73-da10b5e98355>\",\"WARC-IP-Address\":\"172.217.15.65\",\"WARC-Target-URI\":\"http://the-dusty-deck.blogspot.com/2014/11/functional-programming-in-mainstream_27.html\",\"WARC-Payload-Digest\":\"sha1:LV6YRDKKQ6WGOFUKHUG3W57R2RGIBOQG\",\"WARC-Block-Digest\":\"sha1:E6PKJZ24ZMZH63T75UKWFHNTSRLR2AVZ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986660231.30_warc_CC-MAIN-20191015182235-20191015205735-00399.warc.gz\"}"} |
http://videotrine.com/the-calculus-problem-of-arc-length-and-surface/ | [
"# The Calculus Problem Of Arc Length And Surface Area\n\n(Arc lengths, Exercises 7.3) Find the lengths of the given curves.\n(a) y =\n12\n+\n\nfrom x = 1 to x = 4.\n(b) y = x2\nIn x\n8\nfrom x = 1 to x = 2\n(Surface area, Exercises 7.3)\n(a) Find the area of the surface obtained by rotating y = sinx (0 lt; lt; lt; lt;)"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8129482,"math_prob":0.9998963,"size":513,"snap":"2020-24-2020-29","text_gpt3_token_len":169,"char_repetition_ratio":0.15717092,"word_repetition_ratio":0.1904762,"special_character_ratio":0.3411306,"punctuation_ratio":0.12605043,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9983755,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-06T14:15:47Z\",\"WARC-Record-ID\":\"<urn:uuid:35357b44-989d-44ff-8933-5fb8000d4795>\",\"Content-Length\":\"16711\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:05ecfaa9-5676-4a58-8fa4-ab2e63966435>\",\"WARC-Concurrent-To\":\"<urn:uuid:12716d85-e5c7-45a2-bfbe-f4277749f61d>\",\"WARC-IP-Address\":\"185.101.93.24\",\"WARC-Target-URI\":\"http://videotrine.com/the-calculus-problem-of-arc-length-and-surface/\",\"WARC-Payload-Digest\":\"sha1:D3U7GT6UI6GYU5IYYEQFH76ZS4BITPXM\",\"WARC-Block-Digest\":\"sha1:EGCCBULM67QJ2QMBWBTSBQ3F2M4VUJZ2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590348513321.91_warc_CC-MAIN-20200606124655-20200606154655-00447.warc.gz\"}"} |
https://eng.kakprosto.ru/how-37672-how-to-calculate-the-payback-period-of-the-project | [
"Instruction\n1\nCalculate the volume of investment. Term of payback of the project is the time necessary for the net profit from the project covered the total investment volume. In the computational formula, we denote this space as Inv.\n2\nAnalyze projected variable costs for the implementation of the project: equipment depreciation, salaries of employees, payment of taxes and fees, licenses, logistics costs, etc. This figure we denote Per ed.\n3\nCalculate the sum of the fixed costs for the implementation of the project. It is indicated in the formula Post ed.\n4\nGiven the implementation of the project and possible seasonality, which will affect the profits, predict the profit per unit of time: day, month, quarter, year. Depending on the length of the project in time. In the formula it is marked PR.\n5\nCalculate the period of payback of the project Top according to the following formula:Top = Inv/(PR-(Post ed + Per ed)\n6\nMore correct is the calculation of the discounted period of recoupment. In addition to the mentioned formula in its calculation also uses the following indicators and concepts:flow of funds in addition to the initial investment over a period of time.\n7\nThe ratio of investment amount to the sum of the inflows in one period and amortization for the same period of time.\n8\nDetermine the monetary equivalent of the total production or amount of services rendered where the amount of the initial investment will equal the amount of net income. This point is called the breakeven point of the business.\n9\nRealizing the amount of money that needs to be completed to achieve the break-even point of the business and correlating this figure with the power of production in the same unit of time calculate the discounted period of recoupment.\n10\nIt is important to remember and understand that the rate of return is never used independently but only in conjunction with the values of the current value and IRR."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.92574877,"math_prob":0.97245634,"size":1931,"snap":"2019-43-2019-47","text_gpt3_token_len":394,"char_repetition_ratio":0.16087182,"word_repetition_ratio":0.024767801,"special_character_ratio":0.19523562,"punctuation_ratio":0.08839779,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99469465,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-17T11:22:31Z\",\"WARC-Record-ID\":\"<urn:uuid:e6ebdc5b-e74a-40aa-8dc9-820716cbcdcb>\",\"Content-Length\":\"31934\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6e03370b-0f5f-4036-8618-046855570428>\",\"WARC-Concurrent-To\":\"<urn:uuid:627d2750-c9b7-472b-8e75-c44ab1ad0a81>\",\"WARC-IP-Address\":\"95.213.175.85\",\"WARC-Target-URI\":\"https://eng.kakprosto.ru/how-37672-how-to-calculate-the-payback-period-of-the-project\",\"WARC-Payload-Digest\":\"sha1:K3353MNSOEUO6XBT6NK4DCJ6NJYKIWXW\",\"WARC-Block-Digest\":\"sha1:IPSX2HTA3L3U5FGEUWYL77ATTFNFZVNW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986673538.21_warc_CC-MAIN-20191017095726-20191017123226-00014.warc.gz\"}"} |
https://www.proofwiki.org/wiki/Relation_between_Two_Ordinals/Corollary/Proof_2 | [
"# Relation between Two Ordinals/Corollary/Proof 2\n\n## Corollary to Relation between Two Ordinals\n\nLet $S$ and $T$ be ordinals.\n\nIf $S \\ne T$, then either $S$ is an initial segment of $T$, or vice versa.\n\n## Proof\n\nIf either $S \\subset T$ or $T \\subset S$ then we invoke Ordinal Subset of Ordinal is Initial Segment, and the proof is complete.\n\nAiming for a contradiction, suppose $S \\not \\subset T$ and $T \\not \\subset S$.\n\nNow from Intersection is Subset, we have $S \\cap T \\subset T$ and $S \\cap T \\subset S$.\n\nBy Intersection of Two Ordinals is Ordinal, $S \\cap T$ is an ordinal.\n\nSo by Ordinal Subset of Ordinal is Initial Segment, we have:\n\n$S \\cap T = S_a$ for some $a \\in S$\n$S \\cap T = S_b$ for some $b \\in T$\n\nThen:\n\n$a = S_a = S \\cap T = T_b = b$\n\nBut $a \\in S, b \\in T$.\n\nThus $a = b = S \\cap T$.\n\nBut $S \\cap T = S_a$, so:\n\n$x \\in S \\cap T \\implies x \\subset a$\n\nIn particular, this means $a \\subset a$, which is a contradiction.\n\nSo either $S \\subset T$ or $T \\subset S$, and again we invoke Ordinal Subset of Ordinal is Initial Segment, and the proof is complete.\n\n$\\blacksquare$"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.80951005,"math_prob":0.9998815,"size":1279,"snap":"2023-40-2023-50","text_gpt3_token_len":418,"char_repetition_ratio":0.1717647,"word_repetition_ratio":0.12601626,"special_character_ratio":0.32838154,"punctuation_ratio":0.15492958,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000085,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-26T18:17:06Z\",\"WARC-Record-ID\":\"<urn:uuid:92dedaa9-71d2-4e0e-8dee-c8f8c97a6011>\",\"Content-Length\":\"37933\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f739fa03-94ca-4323-8815-40288250377a>\",\"WARC-Concurrent-To\":\"<urn:uuid:80e05755-7e63-431a-962b-665c1da59190>\",\"WARC-IP-Address\":\"104.21.84.229\",\"WARC-Target-URI\":\"https://www.proofwiki.org/wiki/Relation_between_Two_Ordinals/Corollary/Proof_2\",\"WARC-Payload-Digest\":\"sha1:5EGNDO4DFISVLGY4YDYZYJATOYTTXAS6\",\"WARC-Block-Digest\":\"sha1:PKATPGUVRKVPEZXFTSBRYYYX2QJQ3K6T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510219.5_warc_CC-MAIN-20230926175325-20230926205325-00116.warc.gz\"}"} |
http://www.ebooklibrary.org/articles/eng/Routhian_mechanics | [
"",
null,
"#jsDisabledContent { display:none; } My Account | Register | Help",
null,
"Flag as Inappropriate",
null,
"This article will be permanently flagged as inappropriate and made unaccessible to everyone. Are you certain this article is inappropriate? Excessive Violence Sexual Content Political / Social Email this Article Email Address:\n\n# Routhian mechanics\n\nArticle Id: WHEBN0003406245\nReproduction Date:\n\n Title: Routhian mechanics",
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"Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:\n\n### Routhian mechanics\n\nIn analytical mechanics, a branch of theoretical physics, Routhian mechanics is a hybrid formulation of Lagrangian mechanics and Hamiltonian mechanics developed by Edward John Routh. Correspondingly, the Routhian is the function which replaces both the Lagrangian and Hamiltonian functions.\n\nThe Routhian, like the Hamiltonian, can be obtained from a Legendre transform of the Lagrangian, and has a similar mathematical form to the Hamiltonian, but is not exactly the same. The difference between the Lagrangian, Hamiltonian, and Routhian functions are their variables. For a given set of generalized coordinates representing the degrees of freedom in the system, the Lagrangian is a function of the coordinates and velocities, while the Hamiltonian is a function of the coordinates and momenta.\n\nThe Routhian differs from these functions in that some coordinates are chosen to have corresponding generalized velocities, the rest to have corresponding generalized momenta. This choice is arbitrary, and can be done to simplify the problem. It also has the consequence that the Routhian equations are exactly the Hamiltonian equations for some coordinates and corresponding momenta, and the Lagrangian equations for the rest of the coordinates and their velocities. In each case the Lagrangian and Hamiltonian functions are replaced by a single function, the Routhian. The full set thus has the advantages of both sets of equations, with the convenience of splitting one set of coordinates to the Hamilton equations, and the rest to the Lagrangian equations.\n\nOften the Routhian approach may offer no new advantage, but one notable case where this is useful is when a system has cyclic coordinates (also called \"ignorable coordinates\"), by definition those coordinates which do not appear in the original Lagrangian. The Lagrangian equations are powerful results, used frequently in theory and practice, since the equations of motion in the coordinates are easy to set up. However, if cyclic coordinates occur there will still be equations to solve for all the coordinates, including the cyclic coordinates despite their absence in the Lagrangian. The Hamiltonian equations are useful theoretical results, but less useful in practice because coordinates and momenta are related together in the solutions - after solving the equations the coordinates and momenta must be eliminated from each other. Nevertheless, the Hamiltonian equations are perfectly suited to cyclic coordinates because the equations in the cyclic coordinates trivially vanish, leaving only the equations in the non cyclic coordinates.\n\nThe Routhian approach has the best of both approaches, because cyclic coordinates can be split off to the Hamiltonian equations and eliminated, leaving behind the non cyclic coordinates to be solved from the Lagrangian equations. Overall fewer equations need to be solved compared to the Lagrangian approach. Moreover, the Routhian method directly makes clearer the physical interpretations of the constants associated with the cyclic coordinates, in the Lagrangian approach the constants are less obvious.\n\nAs with the rest of analytical mechanics, Routhian mechanics is completely equivalent to Newtonian mechanics, all other formulations of classical mechanics, and introduces no new physics. It offers an alternative way to solve mechanical problems.\n\n## Contents\n\n• Definitions 1\n• Equations of motion 2\n• Two degrees of freedom 2.1\n• Any number of degrees of freedom 2.2\n• Energy 3\n• Cyclic coordinates 4\n• Examples 5\n• Central potential in spherical coordinates 5.1\n• Symmetric mechanical systems 5.2\n• Spherical pendulum 5.2.1\n• Heavy symmetrical top 5.2.2\n• Velocity-dependent potentials 5.3\n• Footnotes 7\n• Notes 8\n• References 9\n\n## Definitions\n\nIn the case of Lagrangian mechanics, the generalized coordinates q1, q2, ... and the corresponding velocities dq1/dt, dq2/dt, ..., and possibly time[nb 1] t, enter the Lagrangian,\n\nL(q_1,q_2,\\ldots,\\dot{q}_1,\\dot{q}_2,\\ldots,t)\\,, \\quad \\dot{q}_i = \\frac{d q_i}{dt} \\,,\n\nwhere the overdots denote time derivatives.\n\nIn Hamiltonian mechanics, the generalized coordinates q1, q2, ... and the corresponding generalized momenta p1, p2, ..., and possibly time, enter the Hamiltonian,\n\nH(q_1,q_2,\\ldots,p_1,p_2,\\ldots,t) = \\sum_i \\dot{q}_ip_i - L(q_1,q_2,\\ldots,\\dot{q}_1(p_1),\\dot{q}_2(p_2),\\ldots,t) \\,, \\quad p_i = \\frac{\\partial L}{\\partial \\dot{q}_i}\\,,\n\nwhere the second equation is the definition of the generalized momentum pi corresponding to the coordinate qi (partial derivatives are denoted using ). The velocities dqi/dt are expressed as functions of their corresponding momenta by inverting their defining relation. In this context, pi is said to be the momentum \"canonically conjugate\" to qi.\n\nThe Routhian is intermediate between L and H; some coordinates q1, q2, ..., qn are chosen to have corresponding generalized momenta p1, p2, ..., pn, the rest of the coordinates ζ1, ζ2, ..., ζs to have generalized velocities 1/dt, 2/dt, ..., s/dt, and time may appear explicitly;\n\n Routhian (n + s degrees of freedom) R(q_1,\\ldots,q_n,\\zeta_1,\\ldots,\\zeta_s, p_1, \\ldots,p_n , \\dot{\\zeta}_1 , \\ldots,\\dot{\\zeta}_s,t) = \\sum_{i=1}^n p_i\\dot{q}_i(p_i) - L(q_1,\\ldots,q_n,\\zeta_1,\\ldots,\\zeta_s, \\dot{q}_1(p_1), \\ldots, \\dot{q}_n(p_n) , \\dot{\\zeta}_1 , \\ldots,\\dot{\\zeta}_s,t) \\,,\n\nwhere again the generalized velocity dqi/dt is to be expressed as a function of generalized momentum pi via its defining relation. The choice of which n coordinates are to have corresponding momenta, out of the n + s coordinates, is arbitrary.\n\nThe above is used by Landau and Lifshitz, and Goldstien. Some authors may define the Routhian to be the negative of the above definition.\n\nGiven the length of the general definition, a more compact notation is to use boldface for tuples (or vectors) of the variables, thus q = (q1, q2, ..., qn), ζ = (ζ1, ζ2, ..., ζs), p = (p1, p2, ..., pn), and d ζ/dt = (1/dt, 2/dt, ..., s/dt), so that\n\nR(\\mathbf{q},\\boldsymbol{\\zeta}, \\mathbf{p}, \\dot{\\boldsymbol{\\zeta}}, t) = \\mathbf{p}\\cdot\\dot} - L(\\mathbf{q}, \\boldsymbol{\\zeta}, \\dot{\\mathbf{q}}, \\dot{\\boldsymbol{\\zeta}},t) \\,,\n\nwhere · is the dot product defined on the tuples, for the specific example appearing here:\n\n\\mathbf{p}\\cdot\\dot} = \\sum_{i=1}^n p_i\\dot{q}_i \\,.\n\n## Equations of motion\n\nFor reference, the Lagrangian equations for s degrees of freedom are a set of s coupled second order ordinary differential equations in the coordinates\n\n\\frac{d}{dt}\\frac{\\partial L}{\\partial \\dot{q}_j} = \\frac{\\partial L}{\\partial q_j} \\,,\n\nwhere j = 1, 2, ..., s, and the Hamiltonian equations for n degrees of freedom are a set of 2n coupled first order ordinary differential equations in the coordinates and momenta\n\n\\dot{q}_i = \\frac{\\partial H}{\\partial p_i} \\,,\\quad \\dot{p}_i = -\\frac{\\partial H}{\\partial q_i} \\,.\n\nBelow, the Routhian equations of motion are obtained in two ways, in the process other useful derivatives are found that can be used elsewhere.\n\n### Two degrees of freedom\n\nConsider the case of a system with two degrees of freedom, q and ζ, with generalized velocities dq/dt and /dt, and the Lagrangian is time-dependent. (The generalization to any number of degrees of freedom follows exactly the same procedure as with two). The Lagrangian of the system will have the form\n\nL(q, \\zeta, \\dot{q}, \\dot{\\zeta}, t)\n\nThe differential of L is\n\ndL = \\frac{\\partial L}{\\partial q}dq + \\frac{\\partial L}{\\partial \\zeta}d\\zeta + \\frac{\\partial L}{\\partial \\dot{q}}d\\dot{q} + \\frac{\\partial L}{\\partial \\dot{\\zeta}}d\\dot{\\zeta} + \\frac{\\partial L}{\\partial t}dt \\,.\n\nNow change variables, from the set (q, ζ, dq/dt, /dt) to (q, ζ, p, /dt), simply switching the velocity dq/dt to the momentum p. This change of variables in the differentials is the Legendre transformation. The differential of the new function to replace L will be a sum of differentials in dq, , dp, d(/dt), and dt. Using the definition of generalized momentum and Lagrange's equation for the coordinate q:\n\np = \\frac{\\partial L}{\\partial \\dot{q}} \\,,\\quad \\dot{p} = \\frac{d}{dt}\\frac{\\partial L}{\\partial \\dot{q}} = \\frac{\\partial L}{\\partial q}\n\nwe have\n\ndL = \\dot{p}dq + \\frac{\\partial L}{\\partial \\zeta}d\\zeta + p d\\dot{q} + \\frac{\\partial L}{\\partial \\dot{\\zeta}}d\\dot{\\zeta} + \\frac{\\partial L}{\\partial t}dt\n\nand to replace pd(dq/dt) by (dq/dt)dp, recall the product rule for differentials,[nb 2] and substitute\n\npd\\dot{q} = d(\\dot{q} p) - \\dot{q}dp\n\nto obtain the differential of a new function in terms of the new set of variables:\n\nd(L-p\\dot{q}) = \\dot{p} dq + \\frac{\\partial L}{\\partial \\zeta}d\\zeta - \\dot{q} dp + \\frac{\\partial L}{\\partial \\dot{\\zeta}}d\\dot{\\zeta} + \\frac{\\partial L}{\\partial t}dt \\,.\n\nIntroducing the Routhian\n\nR(q,\\zeta,p,\\dot{\\zeta},t) = p \\dot{q}(p) - L\n\nwhere again the velocity dq/dt is a function of the momentum p, we have\n\ndR = -\\dot{p} dq - \\frac{\\partial L}{\\partial \\zeta}d\\zeta + \\dot{q}dp - \\frac{\\partial L}{\\partial \\dot{\\zeta}}d\\dot{\\zeta} - \\frac{\\partial L}{\\partial t}dt\\,,\n\nbut from the above definition, the differential of the Routhian is\n\ndR = \\frac{\\partial R }{\\partial q}dq + \\frac{\\partial R }{\\partial \\zeta}d\\zeta + \\frac{\\partial R }{\\partial p}dp + \\frac{\\partial R }{\\partial \\dot{\\zeta}}d\\dot{\\zeta} + \\frac{\\partial R}{\\partial t}dt \\,.\n\nComparing the coefficients of the differentials dq, , dp, d(/dt), and dt, the results are Hamilton's equations for the coordinate q,\n\n\\dot{q} = \\frac{\\partial R}{\\partial p} \\,,\\quad \\dot{p} = -\\frac{\\partial R}{\\partial q} \\,,\n\nand Lagrange's equation for the coordinate ζ\n\n\\frac{d}{dt}\\frac{\\partial R}{\\partial \\dot{\\zeta}} = \\frac{\\partial R}{\\partial \\zeta}\n\n\\frac{\\partial L}{\\partial \\zeta} = - \\frac{\\partial R}{\\partial \\zeta} \\,,\\quad \\frac{\\partial L}{\\partial \\dot{\\zeta}} = - \\frac{\\partial R}{\\partial \\dot{\\zeta}} \\,,\n\nand taking the total time derivative of the second equation and equating to the first. Notice the Routhian replaces the Hamiltonian and Lagrangian functions in all the equations of motion.\n\nThe remaining equation states the partial time derivatives of L and R are negatives\n\n\\frac{\\partial L}{\\partial t}=-\\frac{\\partial R}{\\partial t}\\,.\n\n### Any number of degrees of freedom\n\nFor n + s coordinates as defined above, with Routhian\n\nR(q_1,\\ldots,q_n,\\zeta_1,\\ldots,\\zeta_s, p_1, \\ldots,p_n , \\dot{\\zeta}_1 , \\ldots,\\dot{\\zeta}_s,t) = \\sum_{i=1}^n p_i\\dot{q}_i(p_i) - L\n\nthe equations of motion can be derived by a Legendre transformation of this Routhian as in the previous section, but another way is to simply take the partial derivatives of R with respect to the coordinates qi and ζj, momenta pi, and velocities j/dt, where i = 1, 2, ..., n, and j = 1, 2, ..., s. The derivatives are\n\n\\frac{\\partial R}{\\partial q_i} = -\\frac{\\partial L}{\\partial q_i} = - \\frac{d}{dt}\\frac{\\partial L}{\\partial \\dot{q}_i} = - \\dot{p}_i \\,\n\\frac{\\partial R}{\\partial p_i} = \\dot{q}_i \\,\n\\frac{\\partial R}{\\partial \\zeta_j} = - \\frac{\\partial L}{\\partial \\zeta_j} \\,,\n\\frac{\\partial R}{\\partial \\dot{\\zeta}_j} = - \\frac{\\partial L}{\\partial \\dot{\\zeta}_j} \\,,\n\\frac{\\partial R}{\\partial t} = - \\frac{\\partial L}{\\partial t} \\,.\n\nThe first two are identically the Hamiltonian equations. Equating the total time derivative of the fourth set of equations with the third (for each value of j) gives the Lagrangian equations. The fifth is just the same relation between time partial derivatives as before. To summarize\n\n Routhian equations of motion (n + s degrees of freedom) \\dot{q}_i = \\frac{\\partial R}{\\partial p_i} \\,,\\quad \\dot{p}_i = -\\frac{\\partial R}{\\partial q_i} \\,, \\frac{d}{dt}\\frac{\\partial R}{\\partial \\dot{\\zeta}_j} = \\frac{\\partial R}{\\partial \\zeta_j} \\,.\n\nThe total number of equations is 2n + s, there are 2n Hamiltonian equations plus s Lagrange equations.\n\n## Energy\n\nSince the Lagrangian has the same units as energy, the units of the Routhian are also energy. In SI units this is the Joule.\n\nTaking the total time derivative of the Lagrangian leads to the general result\n\n\\frac{\\partial L}{\\partial t} = \\frac{d }{d t}\\left(\\sum_{i=1}^n \\dot{q}_i\\frac{\\partial L}{\\partial \\dot{q}_i} + \\sum_{j=1}^s \\dot{\\zeta}_j\\frac{\\partial L}{\\partial \\dot{\\zeta}_j} - L\\right)\\,.\n\nIf the Lagrangian is independent of time, the partial time derivative of the Lagrangian is zero, L/∂t = 0, so the quantity under the total time derivative in brackets must be a constant, it is the total energy of the system\n\nE = \\sum_{i=1}^n \\dot{q}_i\\frac{\\partial L}{\\partial \\dot{q}_i} + \\sum_{j=1}^s \\dot{\\zeta}_j\\frac{\\partial L}{\\partial \\dot{\\zeta}_j} - L\\,.\n\n(If there are external fields interacting with the constituents of the system, they can vary throughout space but not time). This expression requires the partial derivatives of L with respect to all the velocities dqi/dt and j/dt. Under the same condition of R being time independent, the energy in terms of the Routhian is a little simpler, substituting the definition of R and the partial derivatives of R with respect to the velocities j/dt,\n\nE = R - \\sum_{j=1}^s \\dot{\\zeta}_j\\frac{\\partial R}{\\partial \\dot{\\zeta}_j} \\,.\n\nNotice only the partial derivatives of R with respect to the velocities j/dt are needed. In the case that s = 0 and the Routhian is explicitly time-independent, then E = R, that is, the Routhian equals the energy of the system. The same expression for R in when s = 0 is also the Hamiltonian, so in all E = R = H.\n\nIf the Routhian has explicit time dependence, the total energy of the system is not constant. The general result is\n\n\\frac{\\partial R}{\\partial t} = \\dfrac{d}{dt}\\left(R - \\sum_{j=1}^s \\dot{\\zeta}_j\\frac{\\partial R}{\\partial \\dot{\\zeta}_j} \\right)\\,,\n\nwhich can be derived from the total time derivative of R in the same way as for L.\n\n## Cyclic coordinates\n\nThe Routhian formulation is useful for systems with cyclic coordinates, because by definition those coordinates do not enter L, and hence R. The corresponding partial derivatives of L and R with respect to those coordinates are zero, which equates to the corresponding generalized momenta reducing to constants. To make this concrete, if the qi are all cyclic coordinates, and the ζj are all non cyclic, then\n\n\\frac{\\partial L}{\\partial q_i} = \\dot{p}_i = - \\frac{\\partial R}{\\partial q_i} = 0 \\quad \\Rightarrow \\quad p_i = \\alpha_i \\,,\n\nwhere the αi are constants. With these constants substituted into the Routhian, R is a function of only the non cyclic coordinates and velocities (and in general time also)\n\nR(\\zeta_1,\\ldots,\\zeta_s,\\alpha_1,\\ldots,\\alpha_n,\\dot{\\zeta}_1,\\ldots,\\dot{\\zeta}_s,t) = \\sum_{i=1}^n \\alpha_i\\dot{q}_i(\\alpha_i) - L(\\zeta_1,\\ldots,\\zeta_s,\\dot{q}_1(\\alpha_1),\\ldots,\\dot{q}_n(\\alpha_n),\\dot{\\zeta}_1,\\ldots,\\dot{\\zeta}_s,t) \\,,\n\nThe 2n Hamiltonian equations in the cyclic coordinates automatically vanish,\n\n\\dot{q}_i=\\frac{\\partial R}{\\partial p_i} = 0 \\,,\\quad \\dot{p}_i=-\\frac{\\partial R}{\\partial q_i}=0\\,,\n\nand the s Lagrangian equations are in the non cyclic coordinates\n\n\\frac{d}{dt}\\frac{\\partial R}{\\partial \\dot{\\zeta}_j} = \\frac{\\partial R}{\\partial \\zeta_j} \\,.\n\nThus the problem has been reduced to solving the Lagrangian equations in the non cyclic coordinates, with the advantage of the Hamiltonian equations cleanly removing the cyclic coordinates.\n\nIf we are interested in how the cyclic coordinates change with time, the equations for the generalized velocities corresponding to the cyclic coordinates can be integrated.\n\n## Examples\n\nThe Routhian method does not guarantee the equations of motion will be simple, however it will lead to fewer equations.\n\n### Central potential in spherical coordinates\n\nOne general class of mechanical systems with cyclic coordinates are those with central potentials, because potentials of this form only have dependence on radial separations and no dependence on angles.\n\nConsider a particle of mass m under the influence of a central potential V(r) in spherical polar coordinates (r, θ, φ)\n\nL(r,\\dot{r},\\theta,\\dot{\\theta},\\dot{\\phi}) = \\frac{m}{2}(\\dot{r}^2 + {r}^2\\dot{\\theta}^2 + r^2 \\sin^2\\theta\\dot{\\phi}^2) - V(r) \\,.\n\nNotice φ is cyclic, because it does not appear in the Lagrangian. The momentum conjugate to φ is the constant\n\np_\\phi = \\frac{\\partial L}{\\partial \\dot{\\phi}} = mr^2\\sin^2\\theta\\dot{\\phi}\\,,\n\nin which r and /dt can vary with time, but the angular momentum pφ is constant. The Routhian can be taken to be\n\n\\begin{align} R(r,\\dot{r},\\theta,\\dot{\\theta}) & = p_\\phi\\dot{\\phi} - L \\\\ & = p_\\phi\\dot{\\phi} - \\frac{m}{2}\\dot{r}^2 - \\frac{m}{2}r^2\\dot{\\theta}^2 - \\frac{p_\\phi\\dot{\\phi}}{2} + V(r) \\\\ & = \\frac{p_\\phi\\dot{\\phi}}{2} - \\frac{m}{2}\\dot{r}^2 - \\frac{m}{2}r^2\\dot{\\theta}^2 + V(r) \\\\ & = \\frac{p_\\phi^2 }{2mr^2\\sin^2\\theta} - \\frac{m}{2}\\dot{r}^2 - \\frac{m}{2}r^2\\dot{\\theta}^2 + V(r) \\,. \\end{align}\n\nWe can solve for r and θ using Lagrange's equations, and do not need to solve for φ since it is eliminated by Hamiltonian's equations. The r equation is\n\n\\frac{d}{dt} \\frac{\\partial R}{\\partial \\dot{r}} = \\frac{\\partial R}{\\partial r} \\quad\\Rightarrow\\quad-m\\ddot{r} = -\\frac{p_\\phi^2}{mr^3\\sin^2\\theta} - mr\\dot{\\theta}^2 + \\frac{\\partial V}{\\partial r} \\,,\n\nand the θ equation is\n\n\\frac{d}{dt} \\frac{\\partial R}{\\partial \\dot{\\theta}} = \\frac{\\partial R}{\\partial \\theta} \\quad\\Rightarrow\\quad -m(2r\\dot{r}\\dot{\\theta} + r^2\\ddot{\\theta}) = -\\frac{p_\\phi^2\\cos\\theta}{mr^2\\sin^3\\theta} \\,.\n\nThe Routhian approach has obtained two coupled nonlinear equations. By contrast the Lagrangian approach leads to three nonlinear coupled equations, mixing in the first and second time derivatives of φ in all of them, despite its absence from the Lagrangian.\n\nThe r equation is\n\n\\frac{d}{dt} \\frac{\\partial L}{\\partial \\dot{r}} = \\frac{\\partial L}{\\partial r} \\quad\\Rightarrow\\quad m\\ddot{r} = mr\\dot{\\theta}^2 + mr\\sin^2\\theta\\dot{\\phi}^2 - \\frac{\\partial V}{\\partial r} \\,,\n\nthe θ equation is\n\n\\frac{d}{dt} \\frac{\\partial L}{\\partial \\dot{\\theta}} = \\frac{\\partial L}{\\partial \\theta} \\quad\\Rightarrow\\quad 2r\\dot{r}\\dot{\\theta} + r^2\\ddot{\\theta} = r^2 \\sin\\theta\\cos\\theta\\dot{\\phi}^2\\,,\n\nthe φ equation is\n\n\\frac{d}{dt} \\frac{\\partial L}{\\partial \\dot{\\phi}} = \\frac{\\partial L}{\\partial \\phi} \\quad\\Rightarrow\\quad 2r\\dot{r}\\sin^2\\theta\\dot{\\phi} + 2r^2 \\sin\\theta\\cos\\theta \\dot{\\theta}\\dot{\\phi} + r^2\\sin^2\\theta \\ddot{\\phi}=0 \\,.\n\n### Symmetric mechanical systems\n\n#### Spherical pendulum\n\nConsider the spherical pendulum, a mass m (known as a \"pendulum bob\") attached to a rigid rod of length l of negligible mass, subject to a local gravitational field g. The system rotates with angular velocity /dt which is not constant. The angle between the rod and vertical is θ and is not constant.\n\nThe Lagrangian is[nb 3]\n\nL(\\theta,\\dot{\\theta},\\dot{\\phi}) = \\frac{m\\ell^2}{2}(\\dot{\\theta}^2 + \\sin^2\\theta \\dot{\\phi}^2) + mg\\ell\\cos\\theta \\,,\n\nand φ is the cyclic coordinate for the system with constant momentum\n\np_\\phi = \\frac{\\partial L}{\\partial \\dot{\\phi}} = m\\ell^2\\sin^2\\theta \\dot{\\phi} \\,.\n\nwhich again is physically the angular momentum of the system about the vertical. The angle θ and angular velocity /dt vary with time, but the angular momentum is constant. The Routhian is\n\n\\begin{align} R(\\theta,\\dot{\\theta}) & = p_\\phi \\dot{\\phi} - L \\\\ & = p_\\phi \\dot{\\phi} - \\frac{m\\ell^2}{2}\\dot{\\theta}^2 - \\frac{p_\\phi \\dot{\\phi}}{2} - mg\\ell\\cos\\theta \\\\ & = \\frac{p_\\phi \\dot{\\phi}}{2} - \\frac{m\\ell^2}{2}\\dot{\\theta}^2 - mg\\ell\\cos\\theta \\\\ & = \\frac{p_\\phi^2 }{2m\\ell^2\\sin^2\\theta} - \\frac{m\\ell^2}{2}\\dot{\\theta}^2 - mg\\ell\\cos\\theta \\end{align}\n\nThe θ equation is found from the Lagrangian equations\n\n\\frac{d}{dt}\\frac{\\partial R}{\\partial \\dot{\\theta}} = \\frac{\\partial R}{\\partial \\theta} \\quad \\Rightarrow \\quad - m\\ell^2\\ddot{\\theta} = -\\frac{p_\\phi^2 \\cos\\theta}{m\\ell^2\\sin^3\\theta} + mg\\ell\\sin\\theta \\,,\n\nor simplifying by introducing the constants\n\na = \\frac{p_\\phi^2}{m^2\\ell^4}\\,,\\quad b = \\frac{g}{\\ell} \\,,\n\ngives\n\n\\ddot{\\theta} = a\\frac{\\cos\\theta}{\\sin^3\\theta} - b \\sin\\theta \\,.\n\nThis equation resembles the simple nonlinear pendulum equation, because it can swing through the vertical axis, with an additional term to account for the rotation about the vertical axis (the constant a is related to the angular momentum pφ).\n\nApplying the Lagrangian approach there are two nonlinear coupled equations to solve.\n\nThe θ equation is\n\n\\frac{d}{dt} \\frac{\\partial L}{\\partial \\dot{\\theta}} = \\frac{\\partial L}{\\partial \\theta} \\quad\\Rightarrow\\quad m\\ell^2\\ddot{\\theta} = m\\ell^2 \\sin\\theta\\cos\\theta\\dot{\\phi}^2 -mg\\ell\\sin\\theta \\,,\n\nand the φ equation is\n\n\\frac{d}{dt} \\frac{\\partial L}{\\partial \\dot{\\phi}} = \\frac{\\partial L}{\\partial \\phi} \\quad\\Rightarrow\\quad 2\\sin\\theta\\cos\\theta \\dot{\\theta}\\dot{\\phi} + \\sin^2\\theta \\ddot{\\phi}=0 \\,.\n\n#### Heavy symmetrical top\n\nThe heavy symmetrical top of mass M has Lagrangian\n\nL(\\theta,\\dot{\\theta},\\dot{\\psi},\\dot{\\phi})=\\frac{I_1}{2}(\\dot{\\theta}^2 + \\dot{\\phi}^2\\sin^2\\theta) + \\frac{I_3}{2}(\\dot{\\psi}^2+\\dot{\\phi}^2\\cos^2\\theta)+I_3\\dot{\\psi}\\dot{\\phi}\\cos\\theta-Mg\\ell\\cos\\theta\n\nwhere ψ, φ, θ are the Euler angles, θ is the angle between the vertical z-axis and the top's z-axis, ψ is the rotation of the top about its own z-axis, and φ the azimuthal of the top's z-axis around the vertical z-axis. The principal moments of inertia are I1 about the top's own x axis, I2 about the top's own y axes, and I3 about the top's own z-axis. Since the top is symmetric about its z-axis, I1 = I2. Here the simple relation for local gravitational potential energy V = Mglcosθ is used where g is the acceleration due to gravity, and the centre of mass of the top is a distance l from its tip along its z-axis.\n\nThe angles ψ, φ are cyclic. The constant momenta are the angular momenta of the top about its axis and its precession about the vertical, respectively:\n\np_\\psi = \\frac{\\partial L}{\\partial \\dot{\\psi}} = I_3\\dot{\\psi} + I_3\\dot{\\phi} \\cos\\theta\np_\\phi = \\frac{\\partial L}{\\partial \\dot{\\phi}} = \\dot{\\phi}(I_1\\sin^2\\theta + I_3\\cos^2\\theta) + I_3\\dot{\\psi}\\cos\\theta\n\nFrom these, eliminating /dt:\n\np_\\phi - p_\\psi\\cos\\theta = I_1\\dot{\\phi}\\sin^2\\theta\n\nwe have\n\n\\dot{\\phi} = \\frac{p_\\phi - p_\\psi\\cos\\theta}{I_1\\sin^2\\theta}\\,,\n\nand to eliminate /dt, substitute this result into pψ and solve for /dt to find\n\n\\dot{\\psi} = \\frac{p_\\psi}{I_3} - \\cos\\theta \\left(\\frac{p_\\phi - p_\\psi\\cos\\theta}{I_1\\sin^2\\theta}\\right) \\,.\n\nThe Routhian can be taken to be\n\nR(\\theta,\\dot{\\theta}) = p_\\psi\\dot{\\psi} + p_\\phi\\dot{\\phi} - L = \\frac{1}{2}(p_\\psi\\dot{\\psi} + p_\\phi\\dot{\\phi}) - \\frac{I_1 \\dot{\\theta}^2}{2} + Mg\\ell \\cos\\theta\n\nand since\n\n\\frac{p_\\phi\\dot{\\phi}}{2} = \\frac{p_\\phi^2}{2I_1\\sin^2\\theta} - \\frac{p_\\psi p_\\phi\\cos\\theta}{2I_1\\sin^2\\theta}\\,,\n\\frac{p_\\psi \\dot{\\psi}}{2} = \\frac{p_\\psi^2}{2I_3} - \\frac{p_\\psi p_\\phi\\cos\\theta }{2I_1\\sin^2\\theta} + \\frac{p_\\psi^2 \\cos^2\\theta}{2I_1\\sin^2\\theta}\n\nwe have\n\nR = \\frac{p_\\psi^2}{2I_3} + \\frac{p_\\psi^2 \\cos^2\\theta}{2I_1\\sin^2\\theta} + \\frac{p_\\phi^2}{2I_1\\sin^2\\theta} - \\frac{p_\\psi p_\\phi\\cos\\theta}{4I_1\\sin^2\\theta} - \\frac{I_1 \\dot{\\theta}^2}{2} + Mg\\ell \\cos\\theta \\,.\n\nThe first term is constant, and can be ignored since only the derivatives of R will enter the equations of motion. The simplified Routhian, without loss of information, is thus\n\nR = \\frac{1}{2I_1\\sin^2\\theta}\\left[p_\\psi^2 \\cos^2\\theta + p_\\phi^2 - \\frac{p_\\psi p_\\phi}{2} \\cos\\theta\\right] - \\frac{I_1 \\dot{\\theta}^2}{2} + Mg\\ell \\cos\\theta\n\nThe equation of motion for θ is, by direct calculation,\n\n-I_1\\ddot{\\theta} = -\\frac{\\cos\\theta}{I_1\\sin^3\\theta}\\left[p_\\psi^2 \\cos^2\\theta + p_\\phi^2 - \\frac{p_\\psi p_\\phi}{2} \\cos\\theta\\right] + \\frac{1}{2I_1\\sin^2\\theta} \\left[-2 p_\\psi^2 \\cos\\theta\\sin\\theta + \\frac{p_\\psi p_\\phi}{2} \\sin\\theta\\right] -Mg\\ell\\sin\\theta \\,,\n\nor by introducing the constants\n\na simpler form of the equation is obtained\n\n\\ddot{\\theta} = \\frac{\\cos\\theta}{\\sin^3\\theta}( a\\cos^2\\theta +b -c\\cos\\theta ) + \\frac{1}{2\\sin\\theta} (2 a \\cos\\theta - c) + k\\sin\\theta \\,.\n\nAlthough the equation is highly nonlinear, there is only one equation to solve for, it was obtained directly, and the cyclic coordinates are not involved.\n\nBy contrast, the Lagrangian approach leads to three nonlinear coupled equations to solve, despite the absence of the coordinates ψ and φ in the Lagrangian.\n\nThe θ equation is\n\n\\frac{d}{dt}\\frac{\\partial L}{\\partial \\dot{\\theta}} = \\frac{\\partial L}{\\partial \\theta} \\quad\\Rightarrow \\quad I_1\\ddot{\\theta} = (I_1- I_3)\\dot{\\phi}^2\\sin\\theta\\cos\\theta -I_3\\dot{\\psi}\\dot{\\phi}\\sin\\theta +Mg\\ell\\sin\\theta\\,,\n\nthe ψ equation is\n\n\\frac{d}{dt}\\frac{\\partial L}{\\partial \\dot{\\psi}} = \\frac{\\partial L}{\\partial \\psi} \\quad\\Rightarrow \\quad \\ddot{\\psi} + \\ddot{\\phi}\\cos\\theta - \\dot{\\phi}\\dot{\\theta}\\sin\\theta= 0 \\,,\n\nand the φ equation is\n\n\\frac{d}{dt}\\frac{\\partial L}{\\partial \\dot{\\phi}} = \\frac{\\partial L}{\\partial \\phi} \\quad\\Rightarrow \\quad \\ddot{\\phi}(I_1\\sin^2\\theta + I_3\\cos^2\\theta) + \\dot{\\phi}(I_1 - I_3)2\\sin\\theta\\cos\\theta\\dot{\\theta} + I_3\\ddot{\\psi}\\cos\\theta - I_3\\dot{\\psi}\\sin\\theta\\dot{\\theta} =0 \\,,\n\n### Velocity-dependent potentials\n\n#### Classical charged particle in a uniform magnetic field",
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"Classical charged particle in uniform B field, using cylindrical coordinates. Top: If the radial coordinate r and angular velocity /dt vary, the trajectory is a helicoid with varying radius but uniform motion in the z direction. Bottom: Constant r and /dt means a helicoid with constant radius.\n\nConsider a classical charged particle of mass m and electric charge q in a static (time-independent) uniform (constant throughout space) magnetic field B. The Lagrangian for a charged particle in a general electromagnetic field given by the magnetic potential A and electric potential φ is\n\nL = \\frac{m}{2} \\dot{\\mathbf{r}}^2 - q \\phi + q \\dot{\\mathbf{r}} \\cdot \\mathbf{A} \\,,\n\nIt is convenient to use cylindrical coordinates (r, θ, z), so that\n\n\\dot{\\mathbf{r}} = \\mathbf{v} = (v_r, v_\\theta,v_z) = (\\dot{r},r\\dot{\\theta},\\dot{z}) \\,,\n\\mathbf{B} = (B_r,B_\\theta,B_z) = (0,0,B)\\,.\n\nIn this case the electric potential is zero, φ = 0, and we can choose the axial gauge for the magnetic potential\n\nand the Lagrangian is\n\nL(r,\\dot{r},\\dot{\\theta},\\dot{z}) = \\frac{m}{2} (\\dot{r}^2 + r^2\\dot{\\theta}^2 + \\dot{z}^2) + \\frac{qB r^2\\dot{\\theta}}{2} \\,.\n\nNotice this potential has an effectively cylindrical symmetry (although it also has angular velocity dependence), since the only spatial dependence is on the radial length from an imaginary cylinder axis.\n\nThere are two cyclic coordinates, θ and z. The canonical momenta conjugate to θ and z are the constants\n\np_{\\theta} = \\frac{\\partial L}{\\partial \\dot {\\theta}} = mr^2\\dot {\\theta} + \\frac{qBr^2}{2} \\,,\\quad p_z = \\frac{\\partial L}{\\partial \\dot {z}} = m\\dot{z} \\,,\n\nso the velocities are\n\n\\dot {\\theta} = \\frac{1}{mr^2}\\left(p_\\theta - \\frac{qBr^2}{2}\\right) \\,,\\quad \\dot{z} = \\frac{p_z}{m}\\,.\n\nThe angular momentum about the z axis is not pθ, but the quantity mr2/dt, which is not conserved due to the contribution from the magnetic field. The canonical momentum pθ is the conserved quantity. It is still the case that pz is the linear or translational momentum along the z axis, which is also conserved.\n\nThe radial component r and angular velocity /dt can vary with time, but pθ is constant, and since pz is constant it follows dz/dt is constant. The Routhian can take the form\n\n\\begin{align} R(r,\\dot{r}) & = p_{\\theta}\\dot{\\theta}+p_z\\dot{z} - L \\\\ & = p_{\\theta}\\dot{\\theta}+p_z\\dot{z} - \\frac{m}{2}\\dot r^2 - \\frac{p_\\theta\\dot{\\theta}}{2} - \\frac{p_z\\dot{z}}{2} - \\frac{1}{2}qBr^2\\dot{\\theta} \\\\ & = (p_\\theta - qBr^2 )\\frac{\\dot{\\theta}}{2} - \\frac{m}{2}\\dot r^2 + \\frac{p_z\\dot{z}}{2} \\\\ & = \\frac{1}{2mr^2} \\left(p_\\theta - qBr^2 \\right)\\left(p_\\theta - \\frac{qBr^2}{2} \\right) - \\frac{m}{2}\\dot{r}^2 + \\frac{p_z}{2m} \\\\ & = \\frac{1}{2mr^2} \\left(p_\\theta^2 - \\frac{3}{2}qBr^2 + \\frac{(qB)^2r^4}{2} \\right) - \\frac{m}{2}\\dot{r}^2 \\end{align}\n\nwhere in the last line, the pz/2m term is a constant and can be ignored without loss of continuity. The Hamiltonian equations for θ and z automatically vanish and do not need to be solved for. The Lagrangian equation in r\n\n\\frac{d}{dt}\\frac{\\partial R}{\\partial \\dot{r}} = \\frac{\\partial R}{\\partial r}\n\nis by direct calculation\n\n-m\\ddot{r} = \\frac{1}{2m}\\left[\\frac{-2}{r^3} \\left(p_\\theta^2 - \\frac{3}{2}qBr^2 + \\frac{(qB)^2 r^4}{2} \\right) + \\frac{1}{r^2}(- 3qBr + 2(qB)^2r^3)\\right] \\,,\n\nwhich after collecting terms is\n\nm\\ddot{r}=\\frac{1}{2m}\\left[\\frac{2p_{\\theta}^2}{r^3}-(qB)^2 r\\right] \\,,\n\nand simplifying further by introducing the constants\n\na = \\frac{p_{\\theta}^2}{m^2} \\,,\\quad b = - \\frac{(qB)^2}{2m^2} \\,,\n\nthe differential equation is\n\n\\ddot{r} = \\frac{a}{r^3} + br\n\nTo see how z changes with time, integrate the momenta expression for pz above\n\nz = \\frac{p_z}{m}t + c_z \\,,\n\nwhere cz is an arbitrary constant, the initial value of z to be specified in the initial conditions.\n\nThe motion of the particle in this system is helicoidal, with the axial motion uniform (constant) but the radial and angular components varying in a spiral according to the equation of motion derived above. The initial conditions on r, dr/dt, θ, /dt, will determine if the trajectory of the particle has a constant r or varying r. If initially r is nonzero but dr/dt = 0, while θ and /dt are arbitrary, then the initial velocity of the particle has no radial component, r is constant, so the motion will be in a perfect helix. If r is constant, the angular velocity is also constant according to the conserved pθ.\n\nWith the Lagrangian approach, the equation for r would include /dt which has to be eliminated, and there would be equations for θ and z to solve for.\n\nThe r equation is\n\n\\frac{d}{dt} \\frac{\\partial L}{\\partial \\dot{r}} = \\frac{\\partial L}{\\partial r} \\quad\\Rightarrow\\quad m\\ddot{r} = mr\\dot{\\theta}^2 + qBr\\dot{\\theta} \\,,\n\nthe θ equation is\n\n\\frac{d}{dt} \\frac{\\partial L}{\\partial \\dot{\\theta}} = \\frac{\\partial L}{\\partial \\theta} \\quad\\Rightarrow\\quad m(2r\\dot{r}\\dot {\\theta} + r^2\\ddot {\\theta}) + qBr\\dot{r} = 0 \\,,\n\nand the z equation is\n\n\\frac{d}{dt} \\frac{\\partial L}{\\partial \\dot{z}} = \\frac{\\partial L}{\\partial z} \\quad\\Rightarrow\\quad m\\ddot{z} = 0 \\,.\n\nThe z equation is trivial to integrate, but the r and θ equations are not, in any case the time derivatives are mixed in all the equations and must be eliminated."
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http://newarthurianeconomics.blogspot.com/2017/12/labor-share-where-do-they-get-those.html | [
"## Sunday, December 17, 2017\n\n###",
null,
"Labor share: Where do they get those numbers?\n\nSo I can answer a question that's been in the back of my mind for a long time now.\n\nWhy is labor share more than 100%?\n\nIt's not percent, Art. It's an index.\n\nYeah I know. But how did it get so high? It's useless. It tells me labor share is going down, but it doesn't tell me what the share is, the share that labor gets. It should be a percentage. It's not, I know, but it should be.\n\nSo where do they get those numbers?\n\nLabor share, as I learned a couple days ago, is calculated by taking compensation as a percent of current-dollar output.\n\nIt is a percentage -- see? I was right. It has to be a percentage.\n\nBut the thing is, it takes one set of indexed values as a percent of another set of indexed values. Indexed values are not the actual values. Indexed values are like a price index (duh, Art). They pick one year to be the base year, and they figure all the values as a percent of the base year value. So right away when they do that, the actual values are gone. You get a line on a graph that is the exact same shape you get from the original values, but the whole shape has been moved up or down until the base year value is equal to 100.\n\nSeems harmless, right? Except of course the original numbers are gone. So you cannot look at labor share and find compensation as a percent of current dollar output, because you don't have those numbers anymore.",
null,
"The name \"labor share\" sounds like it would give you compensation as a percent of current dollar output. But it doesn't work that way. When they pick a year to be the base year, something they do again every few years, they pick a date from the recent past. An arbitrary choice, let's say. Then they re-figure the data for all the years so that the new base year gets the value 100. All the numbers get changed.\n\nOne thing that's for sure is that the base year value is 100. If you have two data sets like compensation and current-dollar output, you can be sure that when you plot them on a graph, the two lines will cross in the base year. Because both data sets have the value 100 for the base year.\n\nI don't know what compensation is, as a percent of current-dollar output. Maybe it's 80%. Maybe it's six. But I know for sure that compensation is going to be 100% of current-dollar output in the base year of a graph, because the numbers are indexed and the base year values are equal.\n\nIt's ridiculous to think that compensation and output are equal. That would mean labor share is 100% and capital share is zero, and I'm sure that's not the case. But that's all we can get from the indexed series called \"labor share\". That's all we can get even if we go back to the data that is used to calculate labor share, because that data is indexed, too.\n\nSo anyway, we know that labor share is compensation as a percent of current-dollar output, except the two lines cross at the 100 level in the base year. We also know that labor share has been going downhill for a long time.\n\nLabor share goes downhill for a long time, and then the lines cross at the 100 level. So that means labor share has to be higher than the 100 level in the years before the base year. And sure enough, it is.\n\n#### 1 comment:\n\nThe Arthurian said...\n\nFrom BLS:\n\nFootnote 8: 'See “LPC databases,” Labor productivity and costs (U.S. Bureau of Labor Statistics), https://www.bls.gov/lpc/data.htm. Click the link that says, “Download the complete Major Sector Productivity and Costs dataset.” A zip file that contains the annual and quarterly datasets will open. This is the sole location on the BLS website where data on labor share levels can be found: the other data search tools on the Productivity and Costs homepage supply only percent-change and index data.' (my bold)\n\nThis BLS article is a gold mine!"
] | [
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https://www.maplesoft.com/support/help/maple/view.aspx?path=Statistics/QuantilePlot&L=E | [
"",
null,
"QuantilePlot - Maple Help\n\nStatistics\n\n QuantilePlot\n generate quantile-quantile plots",
null,
"Calling Sequence QuantilePlot(X, Y, options, plotoptions) QuantilePlot['interactive'](X, Y)",
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"Parameters\n\n X - first data sample Y - second data sample options - (optional) equation(s) of the form option=value where option is one of color, method, reference, or samplesize; specify options for generating the quantile-quantile plot plotoptions - options to be passed to the plots[display] command",
null,
"Description\n\n • The QuantilePlot command generates a quantile-quantile plot for the specified data.\n • The first parameter X is the first data sample - given as e.g. a Vector.\n • The second parameter Y is the second data sample - given as e.g. a Vector.\n • If the ['interactive'] option is used, then a dialog box appears that allows for customized creation of the plot.",
null,
"Options\n\n The options argument can contain one or more of the options shown below. All unrecognized options will be passed to the plots[display] command. See plot/options for details.\n • color=name, list, or range\n This option specifies colors for the points and for the reference line. When a list of colors is given, the set of points and the reference line are colored with the corresponding colors in the list. If a range of colors is given, the colors are generated by selecting an appropriate number of equally spaced points in the corresponding hue range.\n • method=integer[1..9]\n Method for calculating the quantiles. See Statistics[Quantile] for more details.\n • reference=true or false\n If reference is set to true, the reference line $y=x$ is drawn in the same plot. Default value is true.\n • samplesize= positive integer\n The number of points to be plotted. The quantiles taken will be $Q\\left(\\frac{i}{\\mathrm{numpoints}}\\right)$ for $i=1..\\mathrm{numpoints}$ The default value is 25.",
null,
"Examples\n\n > $\\mathrm{with}\\left(\\mathrm{Statistics}\\right):$\n\nCreate three random samples.\n\n > $X≔\\mathrm{Sample}\\left(\\mathrm{Normal}\\left(5,2\\right),500\\right):$\n > $Y≔\\mathrm{Sample}\\left(\\mathrm{Normal}\\left(5,2\\right),500\\right):$\n > $Z≔\\mathrm{Sample}\\left(\\mathrm{Normal}\\left(6,3\\right),500\\right):$\n > $\\mathrm{QuantilePlot}\\left(X,Y\\right)$",
null,
"The command to create the plot from the Plotting Guide using the data above is\n\n > $\\mathrm{QuantilePlot}\\left(X,Z\\right)$",
null,
"Added options can alter the appearance of the plot:\n\n > $\\mathrm{QuantilePlot}\\left(X,Y,\\mathrm{symbolsize}=15,\\mathrm{symbol}=\\mathrm{circle},\\mathrm{color}=\\left[\"Black\",\"Red\"\\right],{\\mathrm{axis}}_{2}=\\left[\\mathrm{gridlines}=\\left[8,\\mathrm{color}=\"Grey\"\\right]\\right]\\right)$",
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https://www.filipekberg.se/2011/10/16/calling-a-dynamic-method-from-a-dynamic-method/ | [
"# Calling a dynamic method from a dynamic method\n\n## Posted by Filip Ekberg on 16 Oct 2011\n\nLet's take a look at how we can call a method from a dynamic method. In the last post we looked at how we called a static method in our current context, but let's take a look at how we can call another dynamically created method that takes an integer parameter and then does some math operation on it and then returns it.\n\nFirst off we need to create this method, we're just going to use IL that we've seen before and I am going to add the input parameter with 2. The dynamic method will look like this with the emitted IL:\n\n``````var addMethod = new DynamicMethod(\ntypeof(int),\nmethodArguments,\ntypeof(Program).Module\n);\n\nil.Emit(OpCodes.Ldarg_0);\nil.Emit(OpCodes.Ldc_I4, 2);\nil.Emit(OpCodes.Ret);\n``````\n\nSo we load the first argument onto the evaluation stack, then we add a 4 byte integer with the value of 2 onto the evaluation stack and last we call `OpCodes.Add`.\n\nI don't know if you've noticed this before, but Dynamic Method derives from `MethodInfo`, which means that we can just call this method!\n\nWe've got the following code from the last blog post:\n\n``````var mathOperation = new DynamicMethod(\ntypeof(void),\nmethodArguments,\ntypeof(Program).Module);\n\nil = mathOperation.GetILGenerator();\nvar methods = typeof(Program).GetMethods();\n``````\n\nIn both these dynamic methods, we use the variable methodArguments that we've also seen before:\n\n``````Type[] methodArguments = {\ntypeof(int)\n};\n``````\n\nIt just says that we expect an integer parameter sent to the method. Let's take a look at the IL we're going to emit, first of all we want to load the argument onto the evaluation stack then we want to add the value 10 and multiply these and pass the result as a parameter to the add method.\n\nSo this code is the same from the last blog post:\n\n``````il.Emit(OpCodes.Ldarg_0);\nil.Emit(OpCodes.Ldc_I4, 10);\nil.Emit(OpCodes.Mul);\n``````\n\nSo now we're prepared to call the dynamic add method we've created and since it derives from `MethodInfo` we can just do this:\n\n``````il.Emit(OpCodes.Call, addMethod);\n``````\n\nSince the result from the multiplication is already on the evaluation stack it's also the first argument that this method gets!\n\nThe Add method in the dynamic add method we've created will also have it's value on the evaluation stack, so let's print it with our static method from the last blog post:\n\n``````il.Emit(OpCodes.Call, methods.First(x => x.Name == \"PrintWithSpecificFormat\"));\nil.Emit(OpCodes.Ret);\n``````\n\nAnd now if we invoke this:\n\n``````var toInvoke = (Action<int>)mathOperation.CreateDelegate(typeof(Action<int>));\ntoInvoke(10);\n``````\n\nwe should see this printed:\n\n`The value is: 102`"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7831313,"math_prob":0.8973552,"size":2782,"snap":"2019-51-2020-05","text_gpt3_token_len":630,"char_repetition_ratio":0.15334773,"word_repetition_ratio":0.03926097,"special_character_ratio":0.22969086,"punctuation_ratio":0.14874552,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9868403,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-22T11:41:12Z\",\"WARC-Record-ID\":\"<urn:uuid:93cd2cc1-9058-423f-a6a6-04406b06d34f>\",\"Content-Length\":\"25875\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f67af56a-2623-4cab-b5b6-081c2a3455b3>\",\"WARC-Concurrent-To\":\"<urn:uuid:00a76847-b390-4343-9dce-6e56b2d78cc2>\",\"WARC-IP-Address\":\"104.31.83.113\",\"WARC-Target-URI\":\"https://www.filipekberg.se/2011/10/16/calling-a-dynamic-method-from-a-dynamic-method/\",\"WARC-Payload-Digest\":\"sha1:HKX2YJRXCOYYP673ZKTFHTVWII24OXTI\",\"WARC-Block-Digest\":\"sha1:GUFWWH6PQMPOMVIQERW3PGCI5OQQVJOL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250606975.49_warc_CC-MAIN-20200122101729-20200122130729-00005.warc.gz\"}"} |
https://www.powershow.com/view1/259527-ZDc1Z/Chapter_8_Cluster_Analysis_powerpoint_ppt_presentation | [
"# Chapter 8' Cluster Analysis - PowerPoint PPT Presentation\n\n1 / 98\nTitle:\n\n## Chapter 8' Cluster Analysis\n\nDescription:\n\n### Partitioning algorithms: Construct various partitions and then evaluate them by ... CLARANS (A Clustering Algorithm based on Randomized Search) (Ng and Han'94) ... – PowerPoint PPT presentation\n\nNumber of Views:297\nAvg rating:3.0/5.0\nSlides: 99\nProvided by: jiaw193\nCategory:\nTags:\nTranscript and Presenter's Notes\n\nTitle: Chapter 8' Cluster Analysis\n\n1\nChapter 8. Cluster Analysis\n• What is Cluster Analysis?\n• Types of Data in Cluster Analysis\n• A Categorization of Major Clustering Methods\n• Partitioning Methods\n• Hierarchical Methods\n• Density-Based Methods\n• Grid-Based Methods\n• Model-Based Clustering Methods\n• Outlier Analysis\n• Summary\n\n2\nWhat is Cluster Analysis?\n• Cluster a collection of data objects\n• Similar to one another within the same cluster\n• Dissimilar to the objects in other clusters\n• Cluster analysis\n• Grouping a set of data objects into clusters\n• Clustering is unsupervised classification no\npredefined classes\n• Typical applications\n• As a stand-alone tool to get insight into data\ndistribution\n• As a preprocessing step for other algorithms\n\n3\nWhat is Cluster Analysis?\n• Finding groups of objects such that the objects\nin a group will be similar (or related) to one\nanother and different from (or unrelated to) the\nobjects in other groups\n\n4\nApplications of Cluster Analysis\n• Understanding\n• Group related documents for browsing, group genes\nand proteins that have similar functionality, or\ngroup stocks with similar price fluctuations\n• Summarization\n• Reduce the size of large data sets\n\nClustering precipitation in Australia\n5\nGeneral Applications of Clustering\n• Pattern Recognition\n• Spatial Data Analysis\n• create thematic maps in GIS by clustering feature\nspaces\n• detect spatial clusters and explain them in\nspatial data mining\n• Image Processing\n• Economic Science (especially market research)\n• WWW\n• Document classification\n• Cluster Weblog data to discover groups of similar\naccess patterns\n\n6\nExamples of Clustering Applications\n• Marketing Help marketers discover distinct\ngroups in their customer bases, and then use this\nknowledge to develop targeted marketing programs\n• Land use Identification of areas of similar land\nuse in an earth observation database\n• Insurance Identifying groups of motor insurance\npolicy holders with a high average claim cost\n• City-planning Identifying groups of houses\naccording to their house type, value, and\ngeographical location\n• Earth-quake studies Observed earth quake\nepicenters should be clustered along continent\nfaults\n\n7\nWhat Is Good Clustering?\n• A good clustering method will produce high\nquality clusters with\n• high intra-class similarity\n• low inter-class similarity\n• The quality of a clustering result depends on\nboth the similarity measure used by the method\nand its implementation.\n• The quality of a clustering method is also\nmeasured by its ability to discover some or all\nof the hidden patterns.\n\n8\nRequirements of Clustering in Data Mining\n• Scalability\n• Ability to deal with different types of\nattributes\n• Discovery of clusters with arbitrary shape\n• Minimal requirements for domain knowledge to\ndetermine input parameters\n• Able to deal with noise and outliers\n• Insensitive to order of input records\n• High dimensionality\n• Incorporation of user-specified constraints\n• Interpretability and usability\n\n9\nNotion of a Cluster can be Ambiguous\n10\nChapter 8. Cluster Analysis\n• What is Cluster Analysis?\n• Types of Data in Cluster Analysis\n• A Categorization of Major Clustering Methods\n• Partitioning Methods\n• Hierarchical Methods\n• Density-Based Methods\n• Grid-Based Methods\n• Model-Based Clustering Methods\n• Outlier Analysis\n• Summary\n\n11\nData Structures\n• Data matrix\n• (two modes)\n• Dissimilarity matrix\n• (one mode)\n\n12\nMeasure the Quality of Clustering\n• Dissimilarity/Similarity metric Similarity is\nexpressed in terms of a distance function, which\nis typically metric d(i, j)\n• There is a separate quality function that\nmeasures the goodness of a cluster.\n• The definitions of distance functions are usually\nvery different for interval-scaled, boolean,\ncategorical, ordinal and ratio variables.\n• Weights should be associated with different\nvariables based on applications and data\nsemantics.\n• It is hard to define similar enough or good\nenough\n• the answer is typically highly subjective.\n\n13\nSimilarity and Dissimilarity Between Objects\n• Distances are normally used to measure the\nsimilarity or dissimilarity between two data\nobjects\n• Some popular ones include Minkowski distance\n• where i (xi1, xi2, , xip) and j (xj1, xj2,\n, xjp) are two p-dimensional data objects, and q\nis a positive integer\n• If q 1, d is Manhattan distance\n\n14\nSimilarity and Dissimilarity Between Objects\n(Cont.)\n• If q 2, d is Euclidean distance\n• Properties\n• d(i,j) ? 0\n• d(i,i) 0\n• d(i,j) d(j,i)\n• d(i,j) ? d(i,k) d(k,j)\n• Also one can use weighted distance, parametric\nPearson product moment correlation, or other\ndisimilarity measures.\n\n15\nChapter 8. Cluster Analysis\n• What is Cluster Analysis?\n• Types of Data in Cluster Analysis\n• A Categorization of Major Clustering Methods\n• Partitioning Methods\n• Hierarchical Methods\n• Density-Based Methods\n• Grid-Based Methods\n• Model-Based Clustering Methods\n• Outlier Analysis\n• Summary\n\n16\nMajor Clustering Approaches\n• Partitioning algorithms Construct various\npartitions and then evaluate them by some\ncriterion\n• Hierarchy algorithms Create a hierarchical\ndecomposition of the set of data (or objects)\nusing some criterion\n• Density-based based on connectivity and density\nfunctions\n• Grid-based based on a multiple-level granularity\nstructure\n• Model-based A model is hypothesized for each of\nthe clusters and the idea is to find the best fit\nof that model to each other\n\n17\nChapter 8. Cluster Analysis\n• What is Cluster Analysis?\n• Types of Data in Cluster Analysis\n• A Categorization of Major Clustering Methods\n• Partitioning Methods\n• Hierarchical Methods\n• Density-Based Methods\n• Grid-Based Methods\n• Model-Based Clustering Methods\n• Outlier Analysis\n• Summary\n\n18\nPartitioning Algorithms Basic Concept\n• Partitioning method Construct a partition of a\ndatabase D of n objects into a set of k clusters\n• Given a k, find a partition of k clusters that\noptimizes the chosen partitioning criterion\n• Global optimal exhaustively enumerate all\npartitions\n• Heuristic methods k-means and k-medoids\nalgorithms\n• k-means (MacQueen67) Each cluster is\nrepresented by the center of the cluster\n• k-medoids or PAM (Partition around medoids)\n(Kaufman Rousseeuw87) Each cluster is\nrepresented by one of the objects in the cluster\n\n19\nThe K-Means Clustering Method\n• Given k, the k-means algorithm is implemented in\n4 steps\n• Partition objects into k nonempty subsets\n• Compute seed points as the centroids of the\nclusters of the current partition. The centroid\nis the center (mean point) of the cluster.\n• Assign each object to the cluster with the\nnearest seed point.\n• Go back to Step 2, stop when no more new\nassignment.\n\n20\nThe K-Means Clustering Method\n• Example\n\n21\n• Strength\n• Relatively efficient O(tkn), where n is\nobjects, k is clusters, and t is iterations.\nNormally, k, t ltlt n.\n• Often terminates at a local optimum. The global\noptimum may be found using techniques such as\ndeterministic annealing and genetic algorithms\n• Weakness\n• Applicable only when mean is defined, then what\n• Need to specify k, the number of clusters, in\n• Unable to handle noisy data and outliers\n• Not suitable to discover clusters with non-convex\nshapes\n\n22\nVariations of the K-Means Method\n• A few variants of the k-means which differ in\n• Selection of the initial k means\n• Dissimilarity calculations\n• Strategies to calculate cluster means\n• Handling categorical data k-modes (Huang98)\n• Replacing means of clusters with modes\n• Using new dissimilarity measures to deal with\ncategorical objects\n• Using a frequency-based method to update modes of\nclusters\n• A mixture of categorical and numerical data\nk-prototype method\n\n23\nThe K-Medoids Clustering Method\n• Find representative objects, called medoids, in\nclusters\n• PAM (Partitioning Around Medoids, 1987)\n• starts from an initial set of medoids and\niteratively replaces one of the medoids by one of\nthe non-medoids if it improves the total distance\nof the resulting clustering\n• PAM works effectively for small data sets, but\ndoes not scale well for large data sets\n• CLARA (Kaufmann Rousseeuw, 1990)\n• CLARANS (Ng Han, 1994) Randomized sampling\n• Focusing spatial data structure (Ester et al.,\n1995)\n\n24\nPAM (Partitioning Around Medoids) (1987)\n• PAM (Kaufman and Rousseeuw, 1987), built in Splus\n• Use real object to represent the cluster\n• Select k representative objects arbitrarily\n• For each pair of non-selected object h and\nselected object i, calculate the total swapping\ncost TCih\n• For each pair of i and h,\n• If TCih lt 0, i is replaced by h\n• Then assign each non-selected object to the most\nsimilar representative object\n• repeat steps 2-3 until there is no change\n\n25\nPAM Clustering Total swapping cost TCih?jCjih\n26\nCLARA (Clustering Large Applications) (1990)\n• CLARA (Kaufmann and Rousseeuw in 1990)\n• Built in statistical analysis packages, such as\nS\n• It draws multiple samples of the data set,\napplies PAM on each sample, and gives the best\nclustering as the output\n• Strength deals with larger data sets than PAM\n• Weakness\n• Efficiency depends on the sample size\n• A good clustering based on samples will not\nnecessarily represent a good clustering of the\nwhole data set if the sample is biased\n\n27\nCLARANS (Randomized CLARA) (1994)\n• CLARANS (A Clustering Algorithm based on\nRandomized Search) (Ng and Han94)\n• CLARANS draws sample of neighbors dynamically\n• The clustering process can be presented as\nsearching a graph where every node is a potential\nsolution, that is, a set of k medoids\n• If the local optimum is found, CLARANS starts\nwith new randomly selected node in search for a\nnew local optimum\n• It is more efficient and scalable than both PAM\nand CLARA\n• Focusing techniques and spatial access structures\nmay further improve its performance (Ester et\nal.95)\n\n28\nChapter 8. Cluster Analysis\n• What is Cluster Analysis?\n• Types of Data in Cluster Analysis\n• A Categorization of Major Clustering Methods\n• Partitioning Methods\n• Hierarchical Methods\n• Density-Based Methods\n• Grid-Based Methods\n• Model-Based Clustering Methods\n• Outlier Analysis\n• Summary\n\n29\nHierarchical Clustering\n• Produces a set of nested clusters organized as a\nhierarchical tree\n• Can be visualized as a dendrogram\n• A tree like diagram that records the sequences of\nmerges or splits\n\n30\nStrengths of Hierarchical Clustering\n• Do not have to assume any particular number of\nclusters\n• Any desired number of clusters can be obtained by\ncutting the dendogram at the proper level\n• They may correspond to meaningful taxonomies\n• Example in biological sciences (e.g., animal\nkingdom, phylogeny reconstruction, )\n\n31\nHierarchical Clustering\n• Two main types of hierarchical clustering\n• Agglomerative\n• At each step, merge the closest pair of clusters\nuntil only one cluster (or k clusters) left\n• Divisive\n• At each step, split a cluster until each cluster\ncontains a point (or there are k clusters)\n• Traditional hierarchical algorithms use a\nsimilarity or distance matrix\n• Merge or split one cluster at a time\n\n32\nAgglomerative Clustering Algorithm\n• More popular hierarchical clustering technique\n• Basic algorithm is straightforward\n• Compute the proximity matrix\n• Let each data point be a cluster\n• Repeat\n• Merge the two closest clusters\n• Update the proximity matrix\n• Until only a single cluster remains\n• Key operation is the computation of the proximity\nof two clusters\n• Different approaches to defining the distance\nbetween clusters distinguish the different\nalgorithms\n\n33\nStarting Situation\nproximity matrix\n\nProximity Matrix\n34\nIntermediate Situation\n• After some merging steps, we have some clusters\n\nC3\nC4\nProximity Matrix\nC1\nC5\nC2\n35\nIntermediate Situation\n• We want to merge the two closest clusters (C2 and\nC5) and update the proximity matrix.\n\nC3\nC4\nProximity Matrix\nC1\nC5\nC2\n36\nAfter Merging\nC2 U C5\n• The question is How do we update the proximity\nmatrix?\n\nC1\nC3\nC4\n?\nC1\n? ? ? ?\nC2 U C5\nC3\n?\nC3\nC4\n?\nC4\nProximity Matrix\nC1\nC2 U C5\n37\nHow to Define Inter-Cluster Similarity\nSimilarity?\n• MIN\n• MAX\n• Group Average\n• Distance Between Centroids\n• Other methods driven by an objective function\n• Wards Method uses squared error\n\nProximity Matrix\n38\nHow to Define Inter-Cluster Similarity\n• MIN\n• MAX\n• Group Average\n• Distance Between Centroids\n• Other methods driven by an objective function\n• Wards Method uses squared error\n\nProximity Matrix\n39\nHow to Define Inter-Cluster Similarity\n• MIN\n• MAX\n• Group Average\n• Distance Between Centroids\n• Other methods driven by an objective function\n• Wards Method uses squared error\n\nProximity Matrix\n40\nHow to Define Inter-Cluster Similarity\n• MIN\n• MAX\n• Group Average\n• Distance Between Centroids\n• Other methods driven by an objective function\n• Wards Method uses squared error\n\nProximity Matrix\n41\nHow to Define Inter-Cluster Similarity\n?\n?\n• MIN\n• MAX\n• Group Average\n• Distance Between Centroids\n• Other methods driven by an objective function\n• Wards Method uses squared error\n\nProximity Matrix\n42\nA Dendrogram Shows How the Clusters are Merged\nHierarchically\nDecompose data objects into a several levels of\nnested partitioning (tree of clusters), called a\ndendrogram. A clustering of the data objects is\nobtained by cutting the dendrogram at the desired\nlevel, then each connected component forms a\ncluster.\n43\nMore on Hierarchical Clustering Methods\n• Major weakness of agglomerative clustering\nmethods\n• do not scale well time complexity of at least\nO(n2), where n is the number of total objects\n• can never undo what was done previously\n• Integration of hierarchical with distance-based\nclustering\n• BIRCH (1996) uses CF-tree and incrementally\n• CURE (1998) selects well-scattered points from\nthe cluster and then shrinks them towards the\ncenter of the cluster by a specified fraction\n• CHAMELEON (1999) hierarchical clustering using\ndynamic modeling\n\n44\nBIRCH (1996)\n• Birch Balanced Iterative Reducing and Clustering\nusing Hierarchies, by Zhang, Ramakrishnan, Livny\n(SIGMOD96)\n• Incrementally construct a CF (Clustering Feature)\ntree, a hierarchical data structure for\nmultiphase clustering\n• Phase 1 scan DB to build an initial in-memory CF\ntree (a multi-level compression of the data that\ntries to preserve the inherent clustering\nstructure of the data)\n• Phase 2 use an arbitrary clustering algorithm to\ncluster the leaf nodes of the CF-tree\n• Scales linearly finds a good clustering with a\nsingle scan and improves the quality with a few\n• Weakness handles only numeric data, and\nsensitive to the order of the data record.\n\n45\nClustering Feature Vector\nCF (5, (16,30),(54,190))\n(3,4) (2,6) (4,5) (4,7) (3,8)\n46\nCF Tree\nRoot\nB 7 L 6\nNon-leaf node\nCF1\nCF3\nCF2\nCF5\nchild1\nchild3\nchild2\nchild5\nLeaf node\nLeaf node\nCF1\nCF2\nCF6\nprev\nnext\nCF1\nCF2\nCF4\nprev\nnext\n47\nCURE (Clustering Using REpresentatives )\n• CURE proposed by Guha, Rastogi Shim, 1998\n• Stops the creation of a cluster hierarchy if a\nlevel consists of k clusters\n• Uses multiple representative points to evaluate\nthe distance between clusters, adjusts well to\narbitrary shaped clusters and avoids single-link\neffect\n\n48\nDrawbacks of Distance-Based Method\n• Drawbacks of square-error based clustering method\n• Consider only one point as representative of a\ncluster\n• Good only for convex shaped, similar size and\ndensity, and if k can be reasonably estimated\n\n49\nCure The Algorithm\n• Draw random sample s.\n• Partition sample to p partitions with size s/p\n• Partially cluster partitions into s/pq clusters\n• Eliminate outliers\n• By random sampling\n• If a cluster grows too slow, eliminate it.\n• Cluster partial clusters.\n• Label data in disk\n\n50\nData Partitioning and Clustering\n• s 50\n• p 2\n• s/p 25\n• s/pq 5\n\nx\nx\n51\nCure Shrinking Representative Points\n• Shrink the multiple representative points towards\nthe gravity center by a fraction of ?.\n• Multiple representatives capture the shape of the\ncluster\n\n52\nClustering Categorical Data ROCK\n• ROCK Robust Clustering using linKs,by S. Guha,\nR. Rastogi, K. Shim (ICDE99).\n• Use links to measure similarity/proximity\n• Not distance based\n• Computational complexity\n• Basic ideas\n• Similarity function and neighbors\n• Let T1 1,2,3, T23,4,5\n\n53\nRock Algorithm\n• Links The number of common neighbours for the\ntwo points.\n• Algorithm\n• Draw random sample\n• Label data in disk\n\n1,2,3, 1,2,4, 1,2,5, 1,3,4,\n1,3,5 1,4,5, 2,3,4, 2,3,5, 2,4,5,\n3,4,5\n3\n1,2,3 1,2,4\n54\nCHAMELEON\n• CHAMELEON hierarchical clustering using dynamic\nmodeling, by G. Karypis, E.H. Han and V. Kumar99\n• Measures the similarity based on a dynamic model\n• Two clusters are merged only if the\ninterconnectivity and closeness (proximity)\nbetween two clusters are high relative to the\ninternal interconnectivity of the clusters and\ncloseness of items within the clusters\n• A two phase algorithm\n• 1. Use a graph partitioning algorithm cluster\nobjects into a large number of relatively small\nsub-clusters\n• 2. Use an agglomerative hierarchical clustering\nalgorithm find the genuine clusters by\nrepeatedly combining these sub-clusters\n\n55\nOverall Framework of CHAMELEON\nConstruct Sparse Graph\nPartition the Graph\nData Set\nMerge Partition\nFinal Clusters\n56\nChapter 8. Cluster Analysis\n• What is Cluster Analysis?\n• Types of Data in Cluster Analysis\n• A Categorization of Major Clustering Methods\n• Partitioning Methods\n• Hierarchical Methods\n• Density-Based Methods\n• Grid-Based Methods\n• Model-Based Clustering Methods\n• Outlier Analysis\n• Summary\n\n57\nDensity-Based Clustering Methods\n• Clustering based on density (local cluster\ncriterion), such as density-connected points\n• Major features\n• Discover clusters of arbitrary shape\n• Handle noise\n• One scan\n• Need density parameters as termination condition\n• Several interesting studies\n• DBSCAN Ester, et al. (KDD96)\n• OPTICS Ankerst, et al (SIGMOD99).\n• DENCLUE Hinneburg D. Keim (KDD98)\n• CLIQUE Agrawal, et al. (SIGMOD98)\n\n58\nDensity-Based Clustering Background\n• Two parameters\n• Eps Maximum radius of the neighbourhood\n• MinPts Minimum number of points in an\nEps-neighbourhood of that point\n• NEps(p) q belongs to D dist(p,q) lt Eps\n• Directly density-reachable A point p is directly\ndensity-reachable from a point q wrt. Eps, MinPts\nif\n• 1) p belongs to NEps(q)\n• 2) core point condition\n• NEps (q) gt MinPts\n\n59\nDensity-Based Clustering Background (II)\n• Density-reachable\n• A point p is density-reachable from a point q\nwrt. Eps, MinPts if there is a chain of points\np1, , pn, p1 q, pn p such that pi1 is\ndirectly density-reachable from pi\n• Density-connected\n• A point p is density-connected to a point q wrt.\nEps, MinPts if there is a point o such that both,\np and q are density-reachable from o wrt. Eps and\nMinPts.\n\np\np1\nq\n60\nDBSCAN Density Based Spatial Clustering of\nApplications with Noise\n• Relies on a density-based notion of cluster A\ncluster is defined as a maximal set of\ndensity-connected points\n• Discovers clusters of arbitrary shape in spatial\ndatabases with noise\n\n61\nDBSCAN The Algorithm\n• Arbitrary select a point p\n• Retrieve all points density-reachable from p wrt\nEps and MinPts.\n• If p is a core point, a cluster is formed.\n• If p is a border point, no points are\ndensity-reachable from p and DBSCAN visits the\nnext point of the database.\n• Continue the process until all of the points have\nbeen processed.\n\n62\nOPTICS A Cluster-Ordering Method (1999)\n• OPTICS Ordering Points To Identify the\nClustering Structure\n• Ankerst, Breunig, Kriegel, and Sander (SIGMOD99)\n• Produces a special order of the database wrt its\ndensity-based clustering structure\n• This cluster-ordering contains info equiv to the\ndensity-based clusterings corresponding to a\n• Good for both automatic and interactive cluster\nanalysis, including finding intrinsic clustering\nstructure\n• Can be represented graphically or using\nvisualization techniques\n\n63\nOPTICS Some Extension from DBSCAN\n• Index-based\n• k number of dimensions\n• N 20\n• p 75\n• M N(1-p) 5\n• Complexity O(kN2)\n• Core Distance\n• Reachability Distance\n\nD\np1\no\np2\no\nMax (core-distance (o), d (o, p)) r(p1, o)\n2.8cm. r(p2,o) 4cm\nMinPts 5 e 3 cm\n64\nReachability-distance\nundefined\n\nCluster-order of the objects\n65\nDENCLUE using density functions\n• DENsity-based CLUstEring by Hinneburg Keim\n(KDD98)\n• Major features\n• Solid mathematical foundation\n• Good for data sets with large amounts of noise\n• Allows a compact mathematical description of\narbitrarily shaped clusters in high-dimensional\ndata sets\n• Significant faster than existing algorithm\n(faster than DBSCAN by a factor of up to 45)\n• But needs a large number of parameters\n\n66\nDenclue Technical Essence\n• Uses grid cells but only keeps information about\ngrid cells that do actually contain data points\nand manages these cells in a tree-based access\nstructure.\n• Influence function describes the impact of a\ndata point within its neighborhood.\n• Overall density of the data space can be\ncalculated as the sum of the influence function\nof all data points.\n• Clusters can be determined mathematically by\nidentifying density attractors.\n• Density attractors are local maximal of the\noverall density function.\n\n67\nGradient The steepness of a slope\n• Example\n\n68\nDensity Attractor\n69\nCenter-Defined and Arbitrary\n70\nChapter 8. Cluster Analysis\n• What is Cluster Analysis?\n• Types of Data in Cluster Analysis\n• A Categorization of Major Clustering Methods\n• Partitioning Methods\n• Hierarchical Methods\n• Density-Based Methods\n• Grid-Based Methods\n• Model-Based Clustering Methods\n• Outlier Analysis\n• Summary\n\n71\nGrid-Based Clustering Method\n• Using multi-resolution grid data structure\n• Several interesting methods\n• STING (a STatistical INformation Grid approach)\nby Wang, Yang and Muntz (1997)\n• WaveCluster by Sheikholeslami, Chatterjee, and\nZhang (VLDB98)\n• A multi-resolution clustering approach using\nwavelet method\n• CLIQUE Agrawal, et al. (SIGMOD98)\n\n72\nSTING A Statistical Information Grid Approach\n• Wang, Yang and Muntz (VLDB97)\n• The spatial area area is divided into rectangular\ncells\n• There are several levels of cells corresponding\nto different levels of resolution\n\n73\nSTING A Statistical Information Grid Approach (2)\n• Each cell at a high level is partitioned into a\nnumber of smaller cells in the next lower level\n• Statistical info of each cell is calculated and\nstored beforehand and is used to answer queries\n• Parameters of higher level cells can be easily\ncalculated from parameters of lower level cell\n• count, mean, s, min, max\n• type of distributionnormal, uniform, etc.\n• Use a top-down approach to answer spatial data\nqueries\n• Start from a pre-selected layertypically with a\nsmall number of cells\n• For each cell in the current level compute the\nconfidence interval\n\n74\nSTING A Statistical Information Grid Approach (3)\n• Remove the irrelevant cells from further\nconsideration\n• When finish examining the current layer, proceed\nto the next lower level\n• Repeat this process until the bottom layer is\nreached\n• Query-independent, easy to parallelize,\nincremental update\n• O(K), where K is the number of grid cells at the\nlowest level\n• All the cluster boundaries are either horizontal\nor vertical, and no diagonal boundary is detected\n\n75\nWaveCluster (1998)\n• Sheikholeslami, Chatterjee, and Zhang (VLDB98)\n• A multi-resolution clustering approach which\napplies wavelet transform to the feature space\n• A wavelet transform is a signal processing\ntechnique that decomposes a signal into different\nfrequency sub-band.\n• Both grid-based and density-based\n• Input parameters\n• of grid cells for each dimension\n• the wavelet, and the of applications of wavelet\ntransform.\n\n76\nWhat is Wavelet (1)?\n77\nWaveCluster (1998)\n• How to apply wavelet transform to find clusters\n• Summaries the data by imposing a\nmultidimensional grid structure onto data space\n• These multidimensional spatial data objects are\nrepresented in a n-dimensional feature space\n• Apply wavelet transform on feature space to find\nthe dense regions in the feature space\n• Apply wavelet transform multiple times which\nresult in clusters at different scales from fine\nto coarse\n\n78\nWhat Is Wavelet (2)?\n79\nQuantization\n80\nTransformation\n81\nWaveCluster (1998)\n• Why is wavelet transformation useful for\nclustering\n• Unsupervised clustering\n• It uses hat-shape filters to emphasize region\nwhere points cluster, but simultaneously to\nsuppress weaker information in their boundary\n• Effective removal of outliers\n• Multi-resolution\n• Cost efficiency\n• Major features\n• Complexity O(N)\n• Detect arbitrary shaped clusters at different\nscales\n• Not sensitive to noise, not sensitive to input\norder\n• Only applicable to low dimensional data\n\n82\nCLIQUE (Clustering In QUEst)\n• Agrawal, Gehrke, Gunopulos, Raghavan (SIGMOD98).\n• Automatically identifying subspaces of a high\ndimensional data space that allow better\nclustering than original space\n• CLIQUE can be considered as both density-based\nand grid-based\n• It partitions each dimension into the same number\nof equal length interval\n• It partitions an m-dimensional data space into\nnon-overlapping rectangular units\n• A unit is dense if the fraction of total data\npoints contained in the unit exceeds the input\nmodel parameter\n• A cluster is a maximal set of connected dense\nunits within a subspace\n\n83\nCLIQUE The Major Steps\n• Partition the data space and find the number of\npoints that lie inside each cell of the\npartition.\n• Identify the subspaces that contain clusters\nusing the Apriori principle\n• Identify clusters\n• Determine dense units in all subspaces of\ninterests\n• Determine connected dense units in all subspaces\nof interests.\n• Generate minimal description for the clusters\n• Determine maximal regions that cover a cluster of\nconnected dense units for each cluster\n• Determination of minimal cover for each cluster\n\n84\nSalary (10,000)\n7\n6\n5\n4\n3\n2\n1\nage\n0\n20\n30\n40\n50\n60\n? 3\n85\nStrength and Weakness of CLIQUE\n• Strength\n• It automatically finds subspaces of the highest\ndimensionality such that high density clusters\nexist in those subspaces\n• It is insensitive to the order of records in\ninput and does not presume some canonical data\ndistribution\n• It scales linearly with the size of input and has\ngood scalability as the number of dimensions in\nthe data increases\n• Weakness\n• The accuracy of the clustering result may be\ndegraded at the expense of simplicity of the\nmethod\n\n86\nChapter 8. Cluster Analysis\n• What is Cluster Analysis?\n• Types of Data in Cluster Analysis\n• A Categorization of Major Clustering Methods\n• Partitioning Methods\n• Hierarchical Methods\n• Density-Based Methods\n• Grid-Based Methods\n• Model-Based Clustering Methods\n• Outlier Analysis\n• Summary\n\n87\nModel-Based Clustering Methods\n• Attempt to optimize the fit between the data and\nsome mathematical model\n• Statistical and AI approach\n• Conceptual clustering\n• A form of clustering in machine learning\n• Produces a classification scheme for a set of\nunlabeled objects\n• Finds characteristic description for each concept\n(class)\n• COBWEB (Fisher87)\n• A popular a simple method of incremental\nconceptual learning\n• Creates a hierarchical clustering in the form of\na classification tree\n• Each node refers to a concept and contains a\nprobabilistic description of that concept\n\n88\nCOBWEB Clustering Method\nA classification tree\n89\nMore on Statistical-Based Clustering\n• Limitations of COBWEB\n• The assumption that the attributes are\nindependent of each other is often too strong\nbecause correlation may exist\n• Not suitable for clustering large database data\nskewed tree and expensive probability\ndistributions\n• CLASSIT\n• an extension of COBWEB for incremental clustering\nof continuous data\n• suffers similar problems as COBWEB\n• AutoClass (Cheeseman and Stutz, 1996)\n• Uses Bayesian statistical analysis to estimate\nthe number of clusters\n• Popular in industry\n\n90\nOther Model-Based Clustering Methods\n• Neural network approaches\n• Represent each cluster as an exemplar, acting as\na prototype of the cluster\n• New objects are distributed to the cluster whose\nexemplar is the most similar according to some\ndostance measure\n• Competitive learning\n• Involves a hierarchical architecture of several\nunits (neurons)\n• Neurons compete in a winner-takes-all fashion\nfor the object currently being presented\n\n91\nModel-Based Clustering Methods\n92\nSelf-organizing feature maps (SOMs)\n• Clustering is also performed by having several\nunits competing for the current object\n• The unit whose weight vector is closest to the\ncurrent object wins\n• The winner and its neighbors learn by having\n• SOMs are believed to resemble processing that can\noccur in the brain\n• Useful for visualizing high-dimensional data in\n2- or 3-D space\n\n93\nChapter 8. Cluster Analysis\n• What is Cluster Analysis?\n• Types of Data in Cluster Analysis\n• A Categorization of Major Clustering Methods\n• Partitioning Methods\n• Hierarchical Methods\n• Density-Based Methods\n• Grid-Based Methods\n• Model-Based Clustering Methods\n• Outlier Analysis\n• Summary\n\n94\nWhat Is Outlier Discovery?\n• What are outliers?\n• The set of objects are considerably dissimilar\nfrom the remainder of the data\n• Example Sports Michael Jordon, Wayne Gretzky,\n...\n• Problem\n• Find top n outlier points\n• Applications\n• Credit card fraud detection\n• Telecom fraud detection\n• Customer segmentation\n• Medical analysis\n\n95\nOutlier Discovery Statistical Approaches\n• Assume a model underlying distribution that\ngenerates data set (e.g. normal distribution)\n• Use discordancy tests depending on\n• data distribution\n• distribution parameter (e.g., mean, variance)\n• number of expected outliers\n• Drawbacks\n• most tests are for single attribute\n• In many cases, data distribution may not be known\n\n96\nOutlier Discovery Distance-Based Approach\n• Introduced to counter the main limitations\nimposed by statistical methods\n• We need multi-dimensional analysis without\nknowing data distribution.\n• Distance-based outlier A DB(p, D)-outlier is an\nobject O in a dataset T such that at least a\nfraction p of the objects in T lies at a distance\ngreater than D from O\n• Algorithms for mining distance-based outliers\n• Index-based algorithm\n• Nested-loop algorithm\n• Cell-based algorithm\n\n97\nOutlier Discovery Deviation-Based Approach\n• Identifies outliers by examining the main\ncharacteristics of objects in a group\n• Objects that deviate from this description are\nconsidered outliers\n• sequential exception technique\n• simulates the way in which humans can distinguish\nunusual objects from among a series of supposedly\nlike objects\n• OLAP data cube technique\n• uses data cubes to identify regions of anomalies\nin large multidimensional data\n\n98\nChapter 8. Cluster Analysis\n• What is Cluster Analysis?\n• Types of Data in Cluster Analysis\n• A Categorization of Major Clustering Methods\n• Partitioning Methods\n• Hierarchical Methods\n• Density-Based Methods\n• Grid-Based Methods\n• Model-Based Clustering Methods\n• Outlier Analysis\n• Summary\n\n99\nProblems and Challenges\n• Considerable progress has been made in scalable\nclustering methods\n• Partitioning k-means, k-medoids, CLARANS\n• Hierarchical BIRCH, CURE\n• Density-based DBSCAN, CLIQUE, OPTICS\n• Grid-based STING, WaveCluster\n• Model-based Autoclass, Denclue, Cobweb\n• Current clustering techniques do not address all"
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https://blogforlife.cc/1623/ | [
"# Which Of The Following Is The Graph Of\n\nWhich Of The Following Is The Graph Of. The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept. Options A and B are correct.\n\nThe following graph is an example of a Disconnected Graph, where there are two components, one with 'a', 'b', 'c', 'd' vertices and another with 'e', 'f', 'g', 'h' vertices. The linear graph, bar and the pie graph are some of the best graphs to. Options A and B are correct.\n\n## Which of these graphs represent functions?\n\nThe graph of this function is shown in the attachment. Complete your question and ask again. Which one of the following graphs correctly represent.\n\n### Options A and B are correct.\n\nThe graph and the explanation is in the pictures. So, they are both graph of functions. The best graph to use if I want to compare the price of six different cars would be a.\n\nWhich one of the following graphs correctly represent. The linear graph, bar and the pie graph are some of the best graphs to. Brent Hanneson – Creator of greenlighttestprep.com If you enjoy my solutions, you'll like my GRE prep course.\n\n### Therefore I got confused b/w answer choices.\n\nBrent Hanneson – Creator of greenlighttestprep.com If you enjoy my solutions, you'll like my GRE prep course. The graph and the explanation is in the pictures. Which one of the following graphs correctly represent.\n\nThe following graph is an example of a Disconnected Graph, where there are two components, one with 'a', 'b', 'c', 'd' vertices and another with 'e', 'f', 'g', 'h' vertices. So, they are both graph of functions. Find the value of x,y,z in the following figure.\n\nFor example prices of a shirt (line graph) The best graph for numerical information will depend on what you want to present. Start studying Graphing Logarithmic Functions QUIZ. The best graph to use if I want to compare the price of six different cars would be a."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9205671,"math_prob":0.7887017,"size":1861,"snap":"2021-04-2021-17","text_gpt3_token_len":428,"char_repetition_ratio":0.13731825,"word_repetition_ratio":0.5153374,"special_character_ratio":0.2278345,"punctuation_ratio":0.13624679,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9973892,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-28T07:51:03Z\",\"WARC-Record-ID\":\"<urn:uuid:20bc1257-2c62-436b-8fc5-a7df603f6003>\",\"Content-Length\":\"31926\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:faf65c3d-cd16-4c76-b69a-fda3f531e34d>\",\"WARC-Concurrent-To\":\"<urn:uuid:7849fc40-407e-457a-a293-b0617641e120>\",\"WARC-IP-Address\":\"172.96.187.187\",\"WARC-Target-URI\":\"https://blogforlife.cc/1623/\",\"WARC-Payload-Digest\":\"sha1:3SC73KBVWJM7LX6FPKV7QQJZ5MHDJERQ\",\"WARC-Block-Digest\":\"sha1:C7R5SYVNHUCEXX73RTJDSRC2KY37FJEJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610704839214.97_warc_CC-MAIN-20210128071759-20210128101759-00085.warc.gz\"}"} |
https://digitalcommons.mtu.edu/etds/202/ | [
"## Dissertations, Master's Theses and Master's Reports - Open\n\n2010\n\nMaster's Thesis\n\n#### Degree Name\n\nMaster of Science in Mathematical Sciences (MS)\n\n#### College, School or Department Name\n\nDepartment of Mathematical Sciences\n\nMelissa Sue Keranen\n\n#### Abstract\n\nChapter 1 is used to introduce the basic tools and mechanics used within this thesis. Some historical uses and background are touched upon as well. The majority of the definitions are contained within this chapter as well.\n\nIn Chapter 2 we consider the question whether one can decompose λ copies of monochromatic Kv into copies of Kk such that each copy of the Kk contains at most one edge from each Kv. This is called a proper edge coloring (Hurd, Sarvate, ). The majority of the content in this section is a wide variety of examples to explain the constructions used in Chapters 3 and 4.\n\nIn Chapters 3 and 4 we investigate how to properly color BIBD(v, k, λ) for k = 4, and 5. Not only will there be direct constructions of relatively small BIBDs, we also prove some generalized constructions used within.\n\nIn Chapter 5 we talk about an alternate solution to Chapters 3 and 4. A purely graph theoretical solution using matchings, augmenting paths, and theorems about the edgechromatic number is used to develop a theorem that than covers all possible cases. We also discuss how this method performed compared to the methods in Chapters 3 and 4.\n\nIn Chapter 6, we switch topics to Latin rectangles that have the same number of symbols and an equivalent sized matrix to Latin squares. Suppose ab = n2. We define an equitable Latin rectangle as an a × b matrix on a set of n symbols where each symbol appears either [b/n] or [b/n] times in each row of the matrix and either [a/n] or [a/n] times in each column of the matrix. Two equitable Latin rectangles are orthogonal in the usual way. Denote a set of ka × b mutually orthogonal equitable Latin rectangles as a k–MOELR(a, b; n). We show that there exists a k–MOELR(a, b; n) for all a, b, n where k is at least 3 with some exceptions.\n\nCOinS"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8986243,"math_prob":0.9420841,"size":2189,"snap":"2020-10-2020-16","text_gpt3_token_len":515,"char_repetition_ratio":0.09610984,"word_repetition_ratio":0.010781671,"special_character_ratio":0.22110553,"punctuation_ratio":0.10070258,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98867977,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-05T20:33:59Z\",\"WARC-Record-ID\":\"<urn:uuid:e2481cad-63bc-474d-a2d9-d0da6fcf5363>\",\"Content-Length\":\"36949\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f94848cd-f245-4830-90fa-acf56c2cab68>\",\"WARC-Concurrent-To\":\"<urn:uuid:eaf01ae6-d903-499d-882f-8650559fed99>\",\"WARC-IP-Address\":\"50.18.241.247\",\"WARC-Target-URI\":\"https://digitalcommons.mtu.edu/etds/202/\",\"WARC-Payload-Digest\":\"sha1:OKIPQOND4AIVS2MXGEUF72OQ4YCY5NJ6\",\"WARC-Block-Digest\":\"sha1:UP373CNOHVZXMJQY7ZGT54HJATRVREE4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371609067.62_warc_CC-MAIN-20200405181743-20200405212243-00485.warc.gz\"}"} |
https://projecteuler.net/problem=401 | [
"",
null,
"## Sum of squares of divisors",
null,
"### Problem 401\n\nThe divisors of 6 are 1,2,3 and 6.\nThe sum of the squares of these numbers is 1+4+9+36=50.\n\nLet sigma2(n) represent the sum of the squares of the divisors of n. Thus sigma2(6)=50.\n\nLet SIGMA2 represent the summatory function of sigma2, that is SIGMA2(n)= sigma2(i) for i=1 to n.\nThe first 6 values of SIGMA2 are: 1,6,16,37,63 and 113.\n\nFind SIGMA2(1015) modulo 109."
] | [
null,
"https://projecteuler.net/images/print_page_logo.png",
null,
"https://projecteuler.net/images/icons/file_html.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7070757,"math_prob":0.9980935,"size":401,"snap":"2020-45-2020-50","text_gpt3_token_len":143,"char_repetition_ratio":0.16876574,"word_repetition_ratio":0.03076923,"special_character_ratio":0.3640898,"punctuation_ratio":0.14705883,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99989533,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-01T20:45:02Z\",\"WARC-Record-ID\":\"<urn:uuid:8085e031-068b-4019-86cc-c9a129f705f0>\",\"Content-Length\":\"4564\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4db9495f-2cd7-4887-ae2b-54d515daebf9>\",\"WARC-Concurrent-To\":\"<urn:uuid:ca887895-1a89-4d6e-9bc0-2e2db13a43e9>\",\"WARC-IP-Address\":\"31.170.122.77\",\"WARC-Target-URI\":\"https://projecteuler.net/problem=401\",\"WARC-Payload-Digest\":\"sha1:BYWFGC7O37SONQFAQCNGBYKTDUDR3DAC\",\"WARC-Block-Digest\":\"sha1:SOBHZCHG4MNSC4FHCBIEA5YSVMJ746FH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141681524.75_warc_CC-MAIN-20201201200611-20201201230611-00567.warc.gz\"}"} |
http://www.numbersaplenty.com/1230200201 | [
"Search a number\nBaseRepresentation\nbin100100101010011…\n…0101110110001001\n310011201211122102102\n41021110311312021\n510004412401301\n6322023243145\n742325346336\noct11124656611\n93151748372\n101230200201\n1158146368a\n122a3ba94b5\n13167b39619\n14b955198d\n1573003b6b\nhex49535d89\n\n1230200201 has 8 divisors (see below), whose sum is σ = 1325063520. Its totient is φ = 1138348512.\n\nThe previous prime is 1230200183. The next prime is 1230200243. The reversal of 1230200201 is 1020020321.\n\nAdding to 1230200201 its reverse (1020020321), we get a palindrome (2250220522).\n\nIt is a happy number.\n\nIt is a sphenic number, since it is the product of 3 distinct primes.\n\nIt is a cyclic number.\n\nIt is not a de Polignac number, because 1230200201 - 214 = 1230183817 is a prime.\n\nIt is a Duffinian number.\n\nIt is not an unprimeable number, because it can be changed into a prime (1230200281) by changing a digit.\n\nIt is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 752060 + ... + 753693.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (165632940).\n\nAlmost surely, 21230200201 is an apocalyptic number.\n\nIt is an amenable number.\n\n1230200201 is a deficient number, since it is larger than the sum of its proper divisors (94863319).\n\n1230200201 is a wasteful number, since it uses less digits than its factorization.\n\n1230200201 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 1505815.\n\nThe product of its (nonzero) digits is 24, while the sum is 11.\n\nThe square root of 1230200201 is about 35074.2099126980. The cubic root of 1230200201 is about 1071.4993975978.\n\nThe spelling of 1230200201 in words is \"one billion, two hundred thirty million, two hundred thousand, two hundred one\"."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.83318436,"math_prob":0.9657011,"size":1911,"snap":"2019-51-2020-05","text_gpt3_token_len":597,"char_repetition_ratio":0.17252229,"word_repetition_ratio":0.006451613,"special_character_ratio":0.47252747,"punctuation_ratio":0.14044943,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99428374,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-29T11:42:05Z\",\"WARC-Record-ID\":\"<urn:uuid:fa69e6e1-44e4-431c-97cf-90e2844e5429>\",\"Content-Length\":\"8942\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a0408164-0f20-4174-8a9c-b9c6349b142b>\",\"WARC-Concurrent-To\":\"<urn:uuid:2d7a934c-fac1-4746-9f81-baa06d810c25>\",\"WARC-IP-Address\":\"62.149.142.170\",\"WARC-Target-URI\":\"http://www.numbersaplenty.com/1230200201\",\"WARC-Payload-Digest\":\"sha1:BPPHKLJLSB3DCMD3MUX2GOK5XFW3YRPR\",\"WARC-Block-Digest\":\"sha1:DAHQXVYLKSTSQ5KHWFWQNKYFQCV6FQPH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251796127.92_warc_CC-MAIN-20200129102701-20200129132701-00546.warc.gz\"}"} |
https://brilliant.org/discussions/thread/a-very-trivial-question-on-decimal-number-system/ | [
"# Question on decimal number system\n\nLet me check how many of our mathematics enthusiasts here have given a thought to this question. We know that our decimal number system consists of the ten digits, viz., 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. We start our counting with ZERO(0), then we count as following:\n\nONE, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT, NINE..\n\nNow the next number is TEN(10). The very peculiar thing about this number is that in our decimal system (i.e., common man's number system) we write this number as a 1 followed by a 0. Then comes ELEVEN(11), written as a 1 followed by a 1, then TWELVE(12) written as a 1 followed by a 2, and so on. Now my question is,why is this so? Instead of writing TEN as 10, we could have written it as 1 followed by two 0's, i.e., as '100', or 0 followed by 1, i.e., as '01' or in any other way. Same argument applies in the case of ELEVEN, TWELVE, and so on. My question may seem to be an idiotic one or a meaningless one, but if you have a mathematical insight, then you will realize that it is indeed a very intriguing one. Try to account for this peculiar way in which our ancestors had designed a number system which has framed the structure of the whole of modern science and technology.",
null,
"Note by Kuldeep Guha Mazumder\n6 years ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n• bulleted\n• list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nThere are quite a few different number systems, but the most common ones used in mathematics are integer base positional number systems. This is jargon for saying that all numbers are expressed as follows, given integer base $n$ and digits ${ a }_{ i }$\n\n$\\displaystyle \\sum _{ i=1 }^{ \\infty }{ { a }_{ i }{ n }^{ i-1 } }$\n\nOne useful property this has, for integer bases, is that for each number, there is an unique representation written in this form. This isn't necessarily the case for non-integer bases or other positional numbering systems.\n\nHaving said that, our decimal number system is base $10$, and uses digits $0,1,2,3,4,5,6,7,8,9$. Note that for any integer base, the number of the digits is equal to the integer base, and thus the base number itself cannot be represented by any of them. If $\"\"$ is a digit itself, then we are talking about base $11$, and we end up with the sequence\n\n$0,1,2,3,4,5,6,7,8,9,,10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1, 20, 21,...$\n\nas awkward as that is, which is why nobody uses that.\n\n- 6 years ago\n\nYes yes..well, that is why the decimal number system find its place above the non-positional number systems like the Roman one, in the list of convenience of use. But I have a very nice way of explaining the concept to young minds.."
] | [
null,
"https://ds055uzetaobb.cloudfront.net/brioche/avatars-2/resized/45/61a09f5893fc778ea9a72fe91066c8ff.a9f4ce26fc-AEZ40aHTe8.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9295454,"math_prob":0.9867589,"size":3269,"snap":"2021-31-2021-39","text_gpt3_token_len":879,"char_repetition_ratio":0.113016844,"word_repetition_ratio":0.014134276,"special_character_ratio":0.27714896,"punctuation_ratio":0.18806745,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.990191,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-04T02:27:28Z\",\"WARC-Record-ID\":\"<urn:uuid:819a3447-b038-4381-b942-3da7f8b48d44>\",\"Content-Length\":\"70926\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bf59476c-c4c7-4be7-a9e0-6a4898cad84c>\",\"WARC-Concurrent-To\":\"<urn:uuid:b957d1ca-cb3a-459a-8084-452927aff6be>\",\"WARC-IP-Address\":\"104.20.34.242\",\"WARC-Target-URI\":\"https://brilliant.org/discussions/thread/a-very-trivial-question-on-decimal-number-system/\",\"WARC-Payload-Digest\":\"sha1:LTGUX5GV67Q3EEGO24MRJ4F5ZZISADJP\",\"WARC-Block-Digest\":\"sha1:FZQKOVXXXDLQ75YXDEE7NX6QYAA4MDWU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154500.32_warc_CC-MAIN-20210804013942-20210804043942-00367.warc.gz\"}"} |
https://www.jiskha.com/questions/1059888/Find-the-present-value-using-the-present-value-formula-and-a-calculator-Round | [
"# algebra\n\nFind the present value, using the present value formula and a calculator. (Round your answer to the nearest cent.)\nAchieve \\$225,500 at 8.55% compounded continuously for 8 years, 155 days.\n\nI use e for compound continuousy A=Pe^(rt).so I did 225,500=Pe^(.0855*8.42)than I divided both sides by e^(.0855*8.42) than e=2.71, than I did\n225,500/e^(.0855*8.42)1097725.13this answer does not look right where did I go wrong. Thank you for your help.\n\n1. 👍 0\n2. 👎 0\n3. 👁 66\n1. my equation, (very close to yours)\n\n225500 = PV e^(8.4246575(.0855))\n225500 = PV e^(.720308219) , now press 2nd ln\n225500 = PV (2.055066524)\nPV = 109,728.81\n\nI let the calculator carry all the decimal places and only rounded off my final answer to the nearest cent.\nYou were off by appr a factor of 10\n\n1. 👍 0\n2. 👎 0\nposted by Reiny\n2. Thank you very much Reiny it is greatly appreciate that you give your time to help others.\n\n1. 👍 0\n2. 👎 0\n\n## Similar Questions\n\n1. ### Algebra\n\nFind the present value, using the present value formula and a calculator. (Round your answer to the nearest cent.) Achieve \\$225,500 at 8.55% compounded continuously for 8 years, 125 days.\n\nasked by HOLLY on August 10, 2014\n2. ### Math\n\nFind the present value, using the present value formula and a calculator. (Round your answer to the nearest cent.) Achieve \\$225,500 at 8.65% compounded continuously for 8 years, 125 days.\n\nasked by Brook on May 13, 2014\n3. ### algebra\n\nFind the present value, using the present value formula and a calculator. (Round your answer to the nearest cent.) Achieve \\$225,500 at 8.55% compounded continuously for 8 years, 125 days.\n\nasked by HOLLY on August 22, 2014\n4. ### math\n\nfind the present value, using the present value formula and a calculator. (Round your answer to the nearest cent.) Achieve \\$225,500 at 8.35% compounded continuously for 8 years, 125 days.\n\nasked by Ronald on June 9, 2013\n5. ### math\n\nFind the present value, using the present value formula and a calculator. (Round your answer to the nearest cent.) Achieve \\$225,500 at 8.95% compounded continuously for 8 years, 155 days.\n\nasked by Anonymous on November 16, 2013\n6. ### algebra\n\nFind the future value, using the future value formula and a calculator. (Round your answer to the nearest cent.) \\$990 at 5.5% compounded quarterly for 3 years Find the present value, using the present value formula and a\n\nasked by gary on May 12, 2014\n7. ### math\n\nFind the present value, using the present value formula. (Round your answer to the nearest cent.) Achieve \\$225,500 at 8.45% compounded continuously for 8 years, 145 days.\n\nasked by emily on November 12, 2014\n8. ### Math\n\nPhosphorus-32 (P-32) has a half-life of 14.2 days. If 200 g of this substance are present initially, find the amount Q(t) present after t days. (Round your growth constant to four decimal places.) How fast is the P-32 decaying\n\nasked by Vanessa on April 7, 2014\n9. ### math\n\nPhosphorus-32 (P-32) has a half-life of 14.2 days. If 200 g of this substance are present initially, find the amount Q(t) present after t days. (Round your growth constant to four decimal places.) How fast is the P-32 decaying\n\nasked by Vanessa on April 5, 2014\n10. ### math\n\nPhosphorus-32 (P-32) has a half-life of 14.2 days. If 200 g of this substance are present initially, find the amount Q(t) present after t days. (Round your growth constant to four decimal places.) How fast is the P-32 decaying\n\nasked by vanessa on April 7, 2014\n\nMore Similar Questions"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9335787,"math_prob":0.9348999,"size":2501,"snap":"2019-13-2019-22","text_gpt3_token_len":718,"char_repetition_ratio":0.15018022,"word_repetition_ratio":0.65525115,"special_character_ratio":0.30667734,"punctuation_ratio":0.13714285,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99771553,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-26T04:16:17Z\",\"WARC-Record-ID\":\"<urn:uuid:86b5206e-6474-45ad-9c88-f605964af6e1>\",\"Content-Length\":\"22807\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cea2a443-4f7c-4a9e-9f11-30730f77d9b5>\",\"WARC-Concurrent-To\":\"<urn:uuid:e820d362-bb00-4fa2-a625-36f1cad1480a>\",\"WARC-IP-Address\":\"66.228.55.50\",\"WARC-Target-URI\":\"https://www.jiskha.com/questions/1059888/Find-the-present-value-using-the-present-value-formula-and-a-calculator-Round\",\"WARC-Payload-Digest\":\"sha1:AJ5LYYHI7YZ7ZLYO5XU42T37IEWINJUL\",\"WARC-Block-Digest\":\"sha1:J65KIDB2ZFCUQPYPYFDA2IT33WWAT2ZX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232258621.77_warc_CC-MAIN-20190526025014-20190526051014-00316.warc.gz\"}"} |
http://bentrengrove.com/blog/2013/1/16/drawing-a-dashed-line-in-cocos2d | [
"# Drawing a dashed line in cocos2d\n\nA code snippet for drawing a dashed line in cocos2d. Could also be modified to draw in other engines such as OpenGL by removing cocos2d convenience methods.\n\n`void drawDashedLine(CGPoint origin, CGPoint destination, float dashLength){ CGPoint delta = ccpSub(destination, origin); float dist = ccpDistance(origin, destination); float x = delta.x / dist * dashLength; float y = delta.y / dist * dashLength; float linePercentage = 0.5f; CGPoint p1 = origin; for(float i = 0; i <= dist / dashLength * linePercentage; i++) { CGPoint p2 = CGPointMake(p1.x + x, p1.y + y); ccDrawLine(p1, p2); p1 = CGPointMake(p1.x + x / linePercentage, p1.y + y / linePercentage); }}`\n\nThe code works by drawing lots of small lines. This can be quite bad for performance if used in the draw method. A better way is to draw into a CCRenderTexture which reduces the number of draw calls significantly, of course if you need to update the line this may not help.\n\nYou can change the percentage of line drawn versus spacing by modifying the linePercentage variable. The algorithm splits the dashLength into line and space based on this percentage. Experiment to see what you like best."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8440392,"math_prob":0.99417007,"size":1188,"snap":"2019-35-2019-39","text_gpt3_token_len":283,"char_repetition_ratio":0.14611487,"word_repetition_ratio":0.010204081,"special_character_ratio":0.24494949,"punctuation_ratio":0.15315315,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9707589,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-16T06:30:01Z\",\"WARC-Record-ID\":\"<urn:uuid:3494b4dd-dc9a-47ba-bd54-e166b08dbbbf>\",\"Content-Length\":\"36375\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fc824643-f329-4f26-99de-f8b98ec3f65c>\",\"WARC-Concurrent-To\":\"<urn:uuid:a4dd0b2c-7efa-49e6-81a5-63db64257bc6>\",\"WARC-IP-Address\":\"198.49.23.144\",\"WARC-Target-URI\":\"http://bentrengrove.com/blog/2013/1/16/drawing-a-dashed-line-in-cocos2d\",\"WARC-Payload-Digest\":\"sha1:NZVWU2G25WQTM4XD3TTWVMGP7GPRXRNS\",\"WARC-Block-Digest\":\"sha1:PWXRQ55XLQWNHAKQM2GBZ6JCQ2YW4RCM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514572491.38_warc_CC-MAIN-20190916060046-20190916082046-00157.warc.gz\"}"} |
http://everything.explained.today/Nearest_neighbor_graph/ | [
"# Nearest neighbor graph explained\n\nThe nearest neighbor graph (NNG) is a directed graph defined for a set of points in a metric space, such as the Euclidean distance in the plane. The NNG has a vertex for each point, and a directed edge from p to q whenever q is a nearest neighbor of p, a point whose distance from p is minimum among all the given points other than p itself.\n\nIn many uses of these graphs, the directions of the edges are ignored and the NNG is defined instead as an undirected graph. However, the nearest neighbor relation is not a symmetric one, i.e., p from the definition is not necessarily a nearest neighbor for q. In theoretical discussions of algorithms a kind of general position is often assumed, namely, the nearest (k-nearest) neighbor is unique for each object. In implementations of the algorithms it is necessary to bear in mind that this is not always the case. For situations in which it is necessary to make the nearest neighbor for each object unique, the set P may be indexed and in the case of a tie the object with, e.g., the largest index may be taken as the nearest neighbor.\n\nThe k-nearest neighbor graph (k-NNG) is a graph in which two vertices p and q are connected by an edge, if the distance between p and q is among the k-th smallest distances from p to other objects from P. The NNG is a special case of the k-NNG, namely it is the 1-NNG. k-NNGs obey a separator theorem: they can be partitioned into two subgraphs of at most vertices each by the removal of O(k1/dn1 - 1/d) points.\n\nAnother variation is the farthest neighbor graph (FNG), in which each point is connected by an edge to the farthest point from it, instead of the nearest point.\n\nNNGs for points in the plane as well as in multidimensional spaces find applications, e.g., in data compression, motion planning, and facilities location. In statistical analysis, the nearest-neighbor chain algorithm based on following paths in this graph can be used to find hierarchical clusterings quickly. Nearest neighbor graphs are also a subject of computational geometry.\n\nThe method can be used to induce a graph on nodes with unknown connectivity.\n\n## NNGs for sets of points\n\n### One-dimensional case\n\nFor a set of points on a line, the nearest neighbor of a point is its left or right (or both) neighbor, if they are sorted along the line. Therefore, the NNG is a path or a forest of several paths and may be constructed in O(n log n) time by sorting. This estimate is asymptotically optimal for certain models of computation, because the constructed NNG gives the answer to the element uniqueness problem: it is sufficient to check whether the NNG has a zero-length edge.\n\n### Higher dimensions\n\nUnless stated otherwise, it is assumed that the NNGs are digraphs with uniquely defined nearest neighbors as described in the introduction.\n\n1. Along any directed path in an NNG the edge lengths are non-increasing.\n2. Only cycles of length 2 are possible in an NNG and each weakly connected component of an NNG with at least 2 vertices has exactly one 2-cycle.\n3. For the points in the plane the NNG is a planar graph with vertex degrees at most 6. If points are in general position, the degree is at most 5.\n4. The NNG (treated as an undirected graph with multiple nearest neighbors allowed) of a set of points in the plane or any higher dimension is a subgraph of the Delaunay triangulation, the Gabriel graph, and the Semi-Yao graph. If the points are in general position or if the single nearest neighbor condition is imposed, the NNG is a forest, a subgraph of the Euclidean minimum spanning tree.\n\n## Notes and References\n\n1. Book: . Computational Geometry - An Introduction . Springer-Verlag. 1985 . 0-387-96131-3 . 1st edition; 2nd printing, corrected and expanded, 1988; Russian translation, 1989.\n2. Eppstein . D. . David Eppstein . Paterson . M. S. . Mike Paterson . Yao . Frances . Frances Yao . On nearest-neighbor graphs . . 17 . 1997 . 263–282 . 3 . 10.1007/PL00009293. free .\n3. Miller . Gary L. . Gary Miller (professor) . Teng . Shang-Hua . Shang-Hua Teng . Thurston . William . William Thurston . Vavasis . Stephen A. . 10.1145/256292.256294 . 1 . Journal of the Association for Computing Machinery . 1–29 . Separators for sphere-packings and nearest neighbor graphs . 44 . 1997.\n4. Encyclopedia: Computational Geometry: Lecture Notes for 18.409, Spring 1988. August 1988. Massachusetts Institute of Technology Laboratory for Computer Science. Alok. Aggarwal. Joel. Wein. Lecture #10, March 10, 1988: Closest pair. Alok. Aggarwal. Shlomo. Kipnis. Observation 1, p. 2.\n5. Rahmati . Z. . King . V. . Valerie King . Whitesides . S. . Sue Whitesides . Kinetic data structures for all nearest neighbors and closest pair in the plane . . 2013 . 137–144 . 10.1145/2462356.2462378."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.952783,"math_prob":0.9757049,"size":2735,"snap":"2023-40-2023-50","text_gpt3_token_len":602,"char_repetition_ratio":0.14976199,"word_repetition_ratio":0.0,"special_character_ratio":0.2106033,"punctuation_ratio":0.09566787,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98459613,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-27T15:34:04Z\",\"WARC-Record-ID\":\"<urn:uuid:0c388791-8041-4572-b0e2-c83fb65c08b5>\",\"Content-Length\":\"13534\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bee37f89-d177-4211-ace3-3929171fa768>\",\"WARC-Concurrent-To\":\"<urn:uuid:142ed358-6d9e-48ec-809c-3a45d17813f9>\",\"WARC-IP-Address\":\"85.25.210.18\",\"WARC-Target-URI\":\"http://everything.explained.today/Nearest_neighbor_graph/\",\"WARC-Payload-Digest\":\"sha1:PFPVG4RM5J2IS6CQLWHHHMAYP66P6CAF\",\"WARC-Block-Digest\":\"sha1:PBUMVMI3RFN5EV665AZENRVMBJWKFRCO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510300.41_warc_CC-MAIN-20230927135227-20230927165227-00789.warc.gz\"}"} |
https://readforlearn.com/ggplot2-line-chart-gives-geom_path-each-group-consist-of-only-one-observation-do-you-need-to-adjust-the-group-aesthetic-2/ | [
"# ggplot2 line chart gives “geom_path: Each group consist of only one observation. Do you need to adjust the group aesthetic?”\n\nyou only have to add `group = 1` into the ggplot or geom_line aes().\n\nFor line graphs, the data points must be grouped so that it knows which points to connect. In this case, it is simple — all points should be connected, so group=1. When more variables are used and multiple lines are drawn, the grouping for lines is usually done by variable.\n\nTry this:\n\n```plot5 <- ggplot(df, aes(year, pollution, group = 1)) +\ngeom_point() +\ngeom_line() +\nlabs(x = \"Year\", y = \"Particulate matter emissions (tons)\",\ntitle = \"Motor vehicle emissions in Baltimore\")```"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8525066,"math_prob":0.9933874,"size":630,"snap":"2023-14-2023-23","text_gpt3_token_len":156,"char_repetition_ratio":0.10543131,"word_repetition_ratio":0.0,"special_character_ratio":0.26666668,"punctuation_ratio":0.16101696,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9865977,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-09T10:30:52Z\",\"WARC-Record-ID\":\"<urn:uuid:cbabf882-591f-4f87-9087-4658342daf81>\",\"Content-Length\":\"43977\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:04a8a473-3b20-4db4-a42d-a38892ab3694>\",\"WARC-Concurrent-To\":\"<urn:uuid:92d5a70c-fb8e-4743-9c76-dc39f214ad66>\",\"WARC-IP-Address\":\"172.67.184.249\",\"WARC-Target-URI\":\"https://readforlearn.com/ggplot2-line-chart-gives-geom_path-each-group-consist-of-only-one-observation-do-you-need-to-adjust-the-group-aesthetic-2/\",\"WARC-Payload-Digest\":\"sha1:6UOSCZKZOGBERKIGT2AQV5HSQMSI4DW2\",\"WARC-Block-Digest\":\"sha1:6TBN6VPS5UUCRRRWOSXOTSVI2UC4LWT2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224656675.90_warc_CC-MAIN-20230609100535-20230609130535-00027.warc.gz\"}"} |
https://en.pythonmana.com/2022/01/202201301536449738.html | [
"# Python Foundation: data structure summary\n\n2022-01-30 15:36:47 chaoyu\n\n# list\n\n## Definition of list\n\nA list is an ordered set , No fixed size , Can save any number of any type of Python object , The grammar is [ Elements 1, Elements 2, ..., Elements n].\n\n• The key point is 「 brackets []」 and 「 comma ,」\n• brackets Tie all the elements together\n• comma Separate each element one by one\n\n## List creation\n\n• Normal list\n``````x = ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday']\nprint(x, type(x))\n# ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday'] <class 'list'>\n\nx = [2, 3, 4, 5, 6, 7]\nprint(x, type(x))\n# [2, 3, 4, 5, 6, 7] <class 'list'>\nCopy code ``````\n• range() Create a list of\n``````x = list(range(10))\nprint(x, type(x))\n# [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] <class 'list'>\n\nx = list(range(1, 11, 2))\nprint(x, type(x))\n# [1, 3, 5, 7, 9] <class 'list'>\n\nx = list(range(10, 1, -2))\nprint(x, type(x))\n# [10, 8, 6, 4, 2] <class 'list'>\nCopy code ``````\n• Inferentially create a list\n``````x = * 5\nprint(x, type(x))\n# [0, 0, 0, 0, 0] <class 'list'>\n\nx = [0 for i in range(5)]\nprint(x, type(x))\n# [0, 0, 0, 0, 0] <class 'list'>\n\nx = [i for i in range(10)]\nprint(x, type(x))\n# [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] <class 'list'>\n\nx = [i for i in range(1, 10, 2)]\nprint(x, type(x))\n# [1, 3, 5, 7, 9] <class 'list'>\n\nx = [i for i in range(10, 1, -2)]\nprint(x, type(x))\n# [10, 8, 6, 4, 2] <class 'list'>\n\nx = [i ** 2 for i in range(1, 10)]\nprint(x, type(x))\n# [1, 4, 9, 16, 25, 36, 49, 64, 81] <class 'list'>\n\nx = [i for i in range(100) if (i % 2) != 0 and (i % 3) == 0]\nprint(x, type(x))\n\n# [3, 9, 15, 21, 27, 33, 39,\nCopy code ``````\n\nbecause list The element of can be any object , So what's in the list is a pointer to the object . Even save a simple [1,2,3], Also have 3 A pointer and 3 An integer object . x = [a] * 4 In operation , Just create 4 A point list References to , So once a change ,x in 4 individual a It's going to change .\n\n• Create a mixed list\n``````mix = [1, 'lsgo', 3.14, [1, 2, 3]]\nprint(mix, type(mix))\n# [1, 'lsgo', 3.14, [1, 2, 3]] <class 'list'>\nCopy code ``````\n• Create an empty list\n``````empty = []\nprint(empty, type(empty)) # [] <class 'list'>\nCopy code ``````\n\n## Add elements to the list\n\nLists don't look like tuples , The contents of the list can be changed (mutable), So add (append, extend)、 Insert (insert)、 Delete (remove, pop) These operations are available\n\n• `list.append(obj)` Add a new object at the end of the list , Accept only one parameter , Parameters can be any data type , The appended element is in list The original structure type is maintained in\n• `list.extend(seq)` Appends multiple values from another sequence at once at the end of the list （ Extend the original list with the new list ）\n\nStrictly speaking append It's an addition , Add a whole thing to the list , and extend It 's an extension , Add all the elements of a thing to the list\n\n• list.insert(index, obj) In the number index Position insert obj\n\n## Delete the elements in the list\n\n• list.remove(obj) Removes the first match of a value in the list\n• list.pop([index=-1]) Removes an element from the list （ Default last element ）, And returns the value of that element\n• del var1[, var2 ……] Delete single or multiple objects\n``````x = ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday']\ndel x[0:2]\nprint(x) # ['Wednesday', 'Thursday', 'Friday']\nCopy code ``````\n\n## Get the elements in the list\n\n• By the index value of the element , Get a single element from a list , Be careful , The list index value is from 0 At the beginning .\n• By specifying the index as -1, Allowing Python Returns the last list element , Indexes -2 Returns the second to last list element , And so on\n\nThe general way to write slices is start : stop : step\n\n• \"start :\" With step by 1 ( Default ) From number start Slice to the end of the list .\n• \": stop\" With step by 1 ( Default ) Number from the head of the list stop section\n• \"start : stop\" With step by 1 ( Default ) From number start To number stop section\n• \"start : stop : step\" Concrete step From number start To number stop section . Pay attention to the last step Set to -1, It is equivalent to arranging the list in reverse\n• \" : \" Copy all elements in the list （ Shallow copy ）\n\n## Common operators for lists\n\n• The equal sign operator ：==\n• Join operators +\n• Repeat operator *\n• Membership operators in、not in\n\n## Other methods of listing\n\n• list.count(obj) Count the number of times an element appears in a list\n``````list1 = [123, 456] * 3\nprint(list1) # [123, 456, 123, 456, 123, 456]\nnum = list1.count(123)\nprint(num) # 3\nCopy code ``````\n• list.index(x[, start[, end]]) Find the index position of the first match of a value in the list\n``````list1 = [123, 456] * 5\nprint(list1.index(123)) # 0\nprint(list1.index(123, 1)) # 2\nprint(list1.index(123, 3, 7)) # 4\nCopy code ``````\n• list.reverse() Reverse list of elements\n``````x = [123, 456, 789]\nx.reverse()\nprint(x) # [789, 456, 123]\nCopy code ``````\n• list.sort(key=None, reverse=False) Sort the original list\n\nkey -- It's basically a comparison element , There is only one parameter , The parameters of the specific function are taken from the iterable object , Specifies an element in an iterable object to sort . reverse -- Sort rule ,reverse = True Descending , reverse = False Ascending （ Default ）. The method does not return a value , But it sorts the objects in the list\n\n``````x = [123, 456, 789, 213]\nx.sort()\nprint(x)\n# [123, 213, 456, 789]\n\nx.sort(reverse=True)\nprint(x)\n# [789, 456, 213, 123]\n\n# Get the second element of the list\ndef takeSecond(elem):\nreturn elem\n\nx = [(2, 2), (3, 4), (4, 1), (1, 3)]\nx.sort(key=takeSecond)\nprint(x)\n# [(4, 1), (2, 2), (1, 3), (3, 4)]\n\nx.sort(key=lambda a: a)\nprint(x)\n# [(1, 3), (2, 2), (3, 4), (4, 1)]\nCopy code ``````\n\n# Tuples\n\ngrammar ：( Elements 1, Elements 2, ..., Elements n)\n\n• Parentheses bind all elements together\n• Commas separate each element one by one\n\n## Create and access a tuple\n\n• Python A tuple of is similar to a list , The difference is tuple You can't modify it after it's created , String like .\n• Tuples use braces , Use square brackets for lists .\n• Tuples are similar to lists , It's also indexed by integers (indexing) And slicing (slicing)\n``````t1 = (1, 10.31, 'python')\nt2 = 1, 10.31, 'python'\nprint(t1, type(t1))\n# (1, 10.31, 'python') <class 'tuple'>\n\nprint(t2, type(t2))\n# (1, 10.31, 'python') <class 'tuple'>\n\ntuple1 = (1, 2, 3, 4, 5, 6, 7, 8)\nprint(tuple1) # 2\nprint(tuple1[5:]) # (6, 7, 8)\nprint(tuple1[:5]) # (1, 2, 3, 4, 5)\ntuple2 = tuple1[:]\nprint(tuple2) # (1, 2, 3, 4, 5, 6, 7, 8)\nCopy code ``````\n• Create tuples with parentheses (), You can also use nothing , For readability , Suggest or use ().\n• When a tuple contains only one element , You need to add a comma after the element , Otherwise parentheses will be used as operators .\n\n【 Example 】 Create a two-dimensional tuple\n\n``````x = (1, 10.31, 'python'), ('data', 11)\nprint(x)\n# ((1, 10.31, 'python'), ('data', 11))\n\nprint(x)\n# (1, 10.31, 'python')\nprint(x, x, x)\n# 1 10.31 python\n\nprint(x[0:2])\n# (1, 10.31)\nCopy code ``````\n\n## Update and delete a tuple\n\nTuples have immutable (immutable) The nature of , So you can't assign values to elements of tuples directly , But as long as the elements in the tuple can be changed (mutable), So we can change its elements directly , Notice that this is different from assigning its elements\n\n``````t1 = (1, 2, 3, [4, 5, 6])\nprint(t1) # (1, 2, 3, [4, 5, 6])\n\nt1 = 9\nprint(t1) # (1, 2, 3, [9, 5, 6])\nCopy code ``````\n\n## Tuple related operators\n\n• The equal sign operator ：==\n• Join operators ： +\n• Repeat operator ： *\n• Membership operators ： in、not in\n\nThere are two ways to splice tuples , use 「 plus +」 and 「 Multiplication sign *」, The former is spliced head to tail , The latter copies the splice\n\n## Built-in methods\n\nTuple size and content cannot be changed , So only count and index The two methods\n\n## Decompress tuples\n\n• decompression （unpack） One dimensional tuples （ There are several elements, and the left bracket defines several variables ）\n``````t = (1, 10.31, 'python')\n(a, b, c) = t\nprint(a, b, c)\n# 1 10.31 python\nCopy code ``````\n• Decompress the two-dimensional tuple （ Define variables according to the tuple structure in tuples ）\n``````t = (1, 10.31, ('OK', 'python'))\n(a, b, (c, d)) = t\nprint(a, b, c, d)\n# 1 10.31 OK python\nCopy code ``````\n• If you only want a few elements of a tuple , Use wildcards 「*」, English name wildcard, Representing one or more elements in a computer language . The following example is to throw multiple elements to rest Variable\n``````t = 1, 2, 3, 4, 5\na, b, *rest, c = t\nprint(a, b, c) # 1 2 5\nprint(rest) # [3, 4]\nCopy code ``````\n• If you don't care rest Variable , Then use wildcards 「*」 Underline 「_」\n``````t = 1, 2, 3, 4, 5\na, b, *_ = t\nprint(a, b) # 1 2\nCopy code ``````\n\n# character string\n\n## Definition of string\n\nPython A string in is defined as a set of characters between quotation marks . Python Support the use of paired Single quotation marks or Double quotes\n\n• Python Common escape characters\nEscape character describe\n\\\\ Backslash notation\n' Single quotation marks\n\" Double quotes\n\\n Line break\n\\t Horizontal tabs (TAB)\n\\r enter\n\nThe original string only needs to be preceded by an English letter r that will do\n\n``````print(r'C:\\Program Files\\Intel\\Wifi\\Help')\n# C:\\Program Files\\Intel\\Wifi\\Help\nCopy code ``````\n\nThree quotes allow a string to span multiple lines , String can contain line breaks 、 Tabs and other special characters\n\n## Slicing and splicing of strings\n\n• Similar to tuples, they are immutable\n• from 0 Start ( and Java equally )\n• Slices are usually written as start:end This form , Include 「start Indexes 」 Corresponding elements , barring 「end Indexes 」 Corresponding elements .\n• Index values can be positive or negative , Is indexing from 0 Start , From left to right ; Negative index from -1 Start , From right to left . When using a negative index , Will start counting from the last element . The position number of the last element is -1.\n``````str1 = 'I Love LsgoGroup'\nprint(str1[:6]) # I Love\nprint(str1) # e\nprint(str1[:6] + \" The inserted string \" + str1[6:])\n# I Love The inserted string LsgoGroup\n\ns = 'Python'\nprint(s) # Python\nprint(s[2:4]) # th\nprint(s[-5:-2]) # yth\nprint(s) # t\nprint(s[-1]) # n\nCopy code ``````\n\n## Common built-in methods for Strings\n\n• capitalize() Convert the first character of the string to uppercase\n• lower() Convert all uppercase characters in the string to lowercase .\n• upper() Convert the lowercase letters in the string to uppercase .\n• swapcase() Convert upper case to lower case in string , Lower case to upper case .\n• count(str, beg= 0,end=len(string)) return str stay string The number of times it's inside , If beg perhaps end If specified, return to the specified range str Number of occurrences\n• endswith(suffix, beg=0, end=len(string)) Check whether the string specifies a substring suffix end , If it is , return True, Otherwise return to False. If beg and end Specify the value , Check... Within the specified range .\n• startswith(substr, beg=0,end=len(string)) Check whether the string specifies a substring substr start , If it is , return True, Otherwise return to False. If beg and end Specify the value , Check... Within the specified range\n• find(str, beg=0, end=len(string)) testing str Whether to include in the string , If you specify a range beg and end, Then check whether it is included in the specified range , If you include , Returns the starting index value , Otherwise return to -1.\n• rfind(str, beg=0,end=len(string)) Be similar to find() function , Just search from the right\n• isnumeric() If the string contains only numeric characters , Then return to True, Otherwise return to False\n• ljust(width[, fillchar]) Returns an original string left aligned , And use fillchar（ Default space ） Fill to length width New string of .\n• rjust(width[, fillchar]) Returns an original string to the right , And use fillchar（ Default space ） Fill to length width New string of\n• lstrip([chars]) Truncates the space to the left of the string or specifies the character .\n• rstrip([chars]) Delete the space at the end of the string or the specified character .\n• strip([chars]) Execute... On a string lstrip() and rstrip()\n• partition(sub) Find the substring sub, Divide the string into a triple (pre_sub,sub,fol_sub), If the string does not contain sub Then return to (' Original string ','','').\n• rpartition(sub) Be similar to partition() Method , Just search from the right\n• replace(old, new [, max]) hold In the string old Replace with new, If max Appoint , The substitution does not exceed max Time\n• split(str=\"\", num) Without parameters, the default is to slice string with space as separator , If num Parameters are set , Only separate num Substring , Returns the list of concatenated substrings after slicing\n• splitlines([keepends]) Follow the line ('\\r', '\\r\\n', \\n') Separate , Returns a list of rows as elements , If parameters keepends by False, Does not contain line breaks , If True, Keep the newline\n• aketrans(intab, outtab) Create a conversion table for character mapping , The first parameter is the string , Indicates the character to be converted , The second parameter is also the target of string representation transformation .\n• translate(table, deletechars=\"\") According to the parameters table The table given , Convert string characters , Put the filtered characters in deletechars Parameters in\n\n## String formatting\n\n• format Format function\n``````str8 = \"{0} Love {1}\".format('I', 'Lsgogroup') # Positional arguments\nprint(str8) # I Love Lsgogroup\n\nstr8 = \"{a} Love {b}\".format(a='I', b='Lsgogroup') # Key parameters\nprint(str8) # I Love Lsgogroup\n\nstr8 = \"{0} Love {b}\".format('I', b='Lsgogroup') # The position parameter should precede the keyword parameter\nprint(str8) # I Love Lsgogroup\n\nstr8 = '{0:.2f}{1}'.format(27.658, 'GB') # Keep two decimal places\nprint(str8) # 27.66GB\nCopy code ``````\n• Python String formatting symbols\noperator Number describe\n%c Format characters and their ASCII code\n%s Formatted string , use str() Method to handle objects\n%r Formatted string , use rper() Method to handle objects\n%d Formatted integer\n%o Format an unsigned octal number\n%x Formats unsigned hexadecimal Numbers\n%X Formats unsigned hexadecimal Numbers （ Capitalization ）\n%f Formatted floating point number , Precision after the decimal point can be specified\n%e Scientific notation for formatting floating - point Numbers\n%E Work with %e, Scientific notation for formatting floating - point Numbers\n%g Use... Depending on the size of the value %f or %e\n%G Work with %g, Use... Depending on the size of the value %f or %E\n• Formatting operator helper\nSymbol function\nm.n m Is the minimum overall width of the display ,n It's the number of decimal places （ If available ）\n- Used for left alignment\n+ Show a plus sign before a positive number ( + )\n# Show zero before octal number ('0'), Show... In front of hex '0x' perhaps '0X'( It depends on what you use 'x' still 'X')\n0 Fill in the front of the displayed number '0' Not the default space\n\n# Dictionaries\n\n## Variable type and immutable type\n\n• A sequence is indexed by consecutive integers , The difference is , Dictionary with \" keyword \" Index , Keywords can be any immutable type , It's usually a string or a number .\n• The dictionary is Python The only one Mapping type , character string 、 Tuples 、 Lists are of the sequence type\n\nHow to quickly judge a data type X Is it a variable type ？ The two methods\n\n1. Trouble method ： use id(X) function , Yes X Do something , Compare before and after operation id, If it's not the same , be X immutable , If the same , be X variable\n2. Easy way ： use hash(X), As long as there is no error , prove X Can be hashed , That is immutable , Conversely, it cannot be hashed , Can change\n\n## Dictionary definition\n\nDictionaries Is chaotic key : value （key:value） The collection , The keys have to be different from each other （ In the same dictionary ）\n\n• dict The order of internal storage and key It doesn't matter in what order .\n• dict Fast search and insertion , Not as key To increase by , But it takes a lot of memory\n\nDictionaries The definition syntax is { Elements 1, Elements 2, ..., Elements n}\n\n• Each of these elements is a 「 Key value pair 」-- key : value (key:value)\n• The key point is 「 Curly braces {}」,「 comma ,」 and 「 The colon :」\n• Curly braces -- Tie all the elements together\n• comma -- Separate each key value pair\n• The colon -- Separate keys from values\n\n## Creating and accessing Dictionaries\n\n``````brand = [' Lining ', ' Nike ', ' Adidas ']\nslogan = [' Anything is possible ', 'Just do it', 'Impossible is nothing']\nprint(' Nike's slogan is :', slogan[brand.index(' Nike ')])\n# Nike's slogan is : Just do it\n\ndic = {' Lining ': ' Anything is possible ', ' Nike ': 'Just do it', ' Adidas ': 'Impossible is nothing'}\nprint(' Nike's slogan is :', dic[' Nike '])\n# Nike's slogan is : Just do it\nCopy code ``````\n\nUse a string or numeric value as key To create a dictionary\n\n``````dic1 = {1: 'one', 2: 'two', 3: 'three'}\nprint(dic1) # {1: 'one', 2: 'two', 3: 'three'}\nprint(dic1) # one\nprint(dic1) # KeyError: 4\nCopy code ``````\n\nIf the key taken does not exist in the dictionary , Will report an error directly KeyError\n\n• dict() Create an empty dictionary\n\nadopt key Put the data directly into the dictionary , But a key There's only one value, Many to one key Put in value, The value in the back will flush out the value in the front\n\n``````dic = dict()\ndic['a'] = 1\ndic['b'] = 2\ndic['c'] = 3\n\nprint(dic)\n# {'a': 1, 'b': 2, 'c': 3}\n\ndic['a'] = 11\nprint(dic)\n# {'a': 11, 'b': 2, 'c': 3}\n\ndic['d'] = 4\nprint(dic)\n# {'a': 11, 'b': 2, 'c': 3, 'd': 4}\nCopy code ``````\n• dict(mapping)\n``````dic1 = dict([('apple', 4139), ('peach', 4127), ('cherry', 4098)])\nprint(dic1) # {'cherry': 4098, 'apple': 4139, 'peach': 4127}\n\ndic2 = dict((('apple', 4139), ('peach', 4127), ('cherry', 4098)))\nprint(dic2) # {'peach': 4127, 'cherry': 4098, 'apple': 4139}\nCopy code ``````\n• dict(**kwargs)\n``````dic = dict(name='Tom', age=10)\nprint(dic) # {'name': 'Tom', 'age': 10}\nprint(type(dic)) # <class 'dict'>\nCopy code ``````\n\n## Built in method of Dictionary\n\n• dict.fromkeys(seq[, value]) Used to create a new dictionary , In sequence seq The middle element is the key of the dictionary ,value Is the initial value of all keys in the dictionary\n• dict.keys() Returns an iteratable object , have access to list() To convert to list , The list is all the keys in the dictionary\n• dict.values() Returns an iterator , have access to list() To convert to list , The list is all the values in the dictionary\n• dict.items() Return traversable as a list ( key , value ) Tuple array\n• dict.get(key, default=None) Returns the value of the specified key , If the value is not in the dictionary, return the default value\n• dict.setdefault(key, default=None) and get() Method similar , If the key does not exist in the dictionary , The key will be added and the value will be set to the default value\n``````dic = {'Name': 'Lsgogroup', 'Age': 7}\nprint(\"Age Key value is : %s\" % dic.setdefault('Age', None)) # Age Key value is : 7\nprint(\"Sex Key value is : %s\" % dic.setdefault('Sex', None)) # Sex Key value is : None\nprint(dic)\n# {'Age': 7, 'Name': 'Lsgogroup', 'Sex': None}\nCopy code ``````\n• key in dict in The operator is used to determine whether the key exists in the dictionary , If the key is in the dictionary dict Back in true, Otherwise return to false. and not in The operator is just the opposite , If the key is in the dictionary dict Back in false, Otherwise return to true\n• dict.pop(key[,default]) Delete dictionary given key key The corresponding value , The return value is the deleted value .key Value must be given . if key non-existent , Then return to default value .\n• del dict[key] Delete dictionary given key key The corresponding value\n• dict.popitem() Randomly return and delete a pair of keys and values in the dictionary , If the dictionary is empty , This method is called , Just report KeyError abnormal\n• dict.clear() Used to delete all elements in the dictionary\n• dict.copy() Returns a shallow copy of a dictionary\n• dict.update(dict2) Put the dictionary parameter dict2 Of key:value Yes Update to dictionary dict in\n\n# aggregate\n\nPython in set And dict similar , It's also a group. key Set , But no storage. value. because key Can't repeat , therefore , stay set in , No repeat key.\n\nBe careful ,key Is immutable type , The hash value\n\ncharacteristic disorder (unordered) And the only\n\n``````num = {}\nprint(type(num)) # <class 'dict'>\nnum = {1, 2, 3, 4}\nprint(type(num)) # <class 'set'>\nCopy code ``````\n\n## Set creation\n\n• Create objects before adding elements\n• When creating an empty collection, you can only use s = set(), because s = {} An empty dictionary is created\n• Just enclose a bunch of elements in curly braces { Elements 1, Elements 2, ..., Elements n}\n• Repeating elements in set Will be automatically filtered\n• Use set(value) Factory function , Convert a list or tuple into a set\n\nbecause set It stores unordered sets , So we can't index a collection or slice it (slice) operation , There are no keys (keys) Can be used to get the value of the element in the collection , But you can determine whether an element is in the collection\n\n## Access the values in the collection\n\n• have access to len() The built-in function gets the size of the set\n``````s = set(['Google', 'Baidu', 'Taobao'])\nprint(len(s)) # 3\nCopy code ``````\n• have access to for Read out the data in the set one by one\n``````s = set(['Google', 'Baidu', 'Taobao'])\nfor item in s:\nprint(item)\n\n# Baidu\n# Taobao\nCopy code ``````\n• Can pass in or not in Determine whether an element already exists in the collection\n``````s = set(['Google', 'Baidu', 'Taobao'])\nprint('Taobao' in s) # True\nprint('Facebook' not in s) # True\nCopy code ``````\n\n## Built in methods for collections\n\n• set.add(elmnt) Used to add elements to a collection , If the added element already exists in the collection , Do nothing\n• set.update(set) Used to modify the current collection , You can add new elements or collections to the current collection , If the added element already exists in the collection , The element will only appear once , Repeated will ignore\n• set.remove(item) Used to remove a specified element from a collection . If the element does not exist , An error will occur\n• set.discard(value) Used to remove the specified collection element .remove() Method to remove a nonexistent element will cause an error , and discard() No way\n• set.pop() Used to randomly remove an element\n\nbecause set Is a collection of unordered and non repeating elements , So two or more set Can do set operations in the mathematical sense\n\n• set.intersection(set1, set2) Returns the intersection of two sets .\n• set1 & set2 Returns the intersection of two sets .\n• set.intersection_update(set1, set2) intersection , Remove non overlapping elements from the original collection\n• set.union(set1, set2) Returns the union of two sets .\n• set1 | set2 Returns the union of two sets\n• set.difference(set) Returns the difference set of a set .\n• set1 - set2 Returns the difference set of a set .\n• set.difference_update(set) Difference sets of sets , Remove elements directly from the original collection , no return value\n• set.symmetric_difference(set) Returns the XOR of the set .\n• set1 ^ set2 Returns the XOR of the set .\n• set.symmetric_difference_update(set) Remove the same elements from the current collection in another specified collection , And insert different elements of another specified collection into the current collection\n• set.issubset(set) Determine whether the set is contained by other sets , If so, return True, Otherwise return to False.\n• set1 <= set2 Determine whether the set is contained by other sets ,- If so, return True, Otherwise return to False\n• set.issuperset(set) Used to determine whether the set contains - Other sets , If so, return True, Otherwise return to - False.\n• set1 >= set2 Determine whether the set contains other sets , If so, return True, Otherwise return to False\n• set.isdisjoint(set) Used to determine whether two sets are disjoint , If it's a return True, Otherwise return to False\n\n## The transformation of sets\n\n``````se = set(range(4))\nli = list(se)\ntu = tuple(se)\n\nprint(se, type(se)) # {0, 1, 2, 3} <class 'set'>\nprint(li, type(li)) # [0, 1, 2, 3] <class 'list'>\nprint(tu, type(tu)) # (0, 1, 2, 3) <class 'tuple'>\nCopy code ``````\n\n## Immutable set\n\nPython Provides an implementation version of a collection that cannot be changed , That is, you can't add or delete elements , Type name frozenset. It should be noted that frozenset You can still do set operations , It's just that you can't use it with update Methods .\n\n• frozenset([iterable]) Returns a frozen collection , After freezing, no more elements can be added or removed from the collection .\n``````a = frozenset(range(10)) # Generates a new immutable set\nprint(a)\n# frozenset({0, 1, 2, 3, 4, 5, 6, 7, 8, 9})\n\nb = frozenset('lsgogroup')\nprint(b)\n# frozenset({'g', 's', 'p', 'r', 'u', 'o', 'l'})\nCopy code ``````\n\n# Sequence\n\nstay Python in , Sequence types include strings 、 list 、 Tuples 、 Collections and dictionaries , These sequences support some common operations , But what's more special is , Indexes are not supported in collections and dictionaries 、 section 、 Add and multiply operations\n\n## Built in functions for sequences\n\n• list(sub) Convert an iteratable object to a list\n``````a = list()\nprint(a) # []\n\nb = 'I Love LsgoGroup'\nb = list(b)\nprint(b)\n# ['I', ' ', 'L', 'o', 'v', 'e', ' ', 'L', 's', 'g', 'o', 'G', 'r', 'o', 'u', 'p']\n\nc = (1, 1, 2, 3, 5, 8)\nc = list(c)\nprint(c) # [1, 1, 2, 3, 5, 8]\nCopy code ``````\n• tuple(sub) Convert an iteratable object to a tuple\n``````a = tuple()\nprint(a) # ()\n\nb = 'I Love LsgoGroup'\nb = tuple(b)\nprint(b)\n# ('I', ' ', 'L', 'o', 'v', 'e', ' ', 'L', 's', 'g', 'o', 'G', 'r', 'o', 'u', 'p')\n\nc = [1, 1, 2, 3, 5, 8]\nc = tuple(c)\nprint(c) # (1, 1, 2, 3, 5, 8)\nCopy code ``````\n• str(obj) hold obj Object to string\n``````a = 123\na = str(a)\nprint(a) # 123\nCopy code ``````\n• len(s) Returns the object （ character 、 list 、 Tuples etc. ） Length or number of elements\n• s -- object\n``````a = list()\nprint(len(a)) # 0\n\nb = ('I', ' ', 'L', 'o', 'v', 'e', ' ', 'L', 's', 'g', 'o', 'G', 'r', 'o', 'u', 'p')\nprint(len(b)) # 16\n\nc = 'I Love LsgoGroup'\nprint(len(c)) # 16\nCopy code ``````\n• max(sub) Returns the maximum value in a sequence or parameter set\n``````print(max(1, 2, 3, 4, 5)) # 5\nprint(max([-8, 99, 3, 7, 83])) # 99\nprint(max('IloveLsgoGroup')) # v\nCopy code ``````\n• min(sub) Returns the minimum value in a sequence or parameter set\n``````print(min(1, 2, 3, 4, 5)) # 1\nprint(min([-8, 99, 3, 7, 83])) # -8\nprint(min('IloveLsgoGroup')) # G\nCopy code ``````\n• sum(iterable[, start=0]) Return sequence iterable With optional parameters start The sum of\n``````print(sum([1, 3, 5, 7, 9])) # 25\nprint(sum([1, 3, 5, 7, 9], 10)) # 35\nprint(sum((1, 3, 5, 7, 9))) # 25\nprint(sum((1, 3, 5, 7, 9), 20)) # 45\nCopy code ``````\n• sorted(iterable, key=None, reverse=False) Sort all iteratable objects\n• iterable -- Iteratable object .\n• key -- It's basically a comparison element , There is only one parameter , The parameters of the specific function are taken from the iterable object , Specifies an element in an iterable object to sort .\n• reverse -- Sort rule ,reverse = True Descending , reverse = False Ascending （ Default ）.\n``````x = [-8, 99, 3, 7, 83]\nprint(sorted(x)) # [-8, 3, 7, 83, 99]\nprint(sorted(x, reverse=True)) # [99, 83, 7, 3, -8]\n\nt = ({\"age\": 20, \"name\": \"a\"}, {\"age\": 25, \"name\": \"b\"}, {\"age\": 10, \"name\": \"c\"})\nx = sorted(t, key=lambda a: a[\"age\"])\nprint(x)\n# [{'age': 10, 'name': 'c'}, {'age': 20, 'name': 'a'}, {'age': 25, 'name': 'b'}]\nCopy code ``````\n• reversed(seq) Function returns an inverted iterator\n• seq -- The sequence to convert , It can be tuple, string, list or range\n``````s = 'lsgogroup'\nx = reversed(s)\nprint(type(x)) # <class 'reversed'>\nprint(x) # <reversed object at 0x000002507E8EC2C8>\nprint(list(x))\n# ['p', 'u', 'o', 'r', 'g', 'o', 'g', 's', 'l']\n\nt = ('l', 's', 'g', 'o', 'g', 'r', 'o', 'u', 'p')\nprint(list(reversed(t)))\n# ['p', 'u', 'o', 'r', 'g', 'o', 'g', 's', 'l']\n\nr = range(5, 9)\nprint(list(reversed(r)))\n# [8, 7, 6, 5]\n\nx = [-8, 99, 3, 7, 83]\nprint(list(reversed(x)))\n# [83, 7, 3, 99, -8]\nCopy code ``````\n• enumerate(sequence, [start=0]) For traversing a data object ( As listing 、 Tuples or strings ) Combined into an index sequence , List both data and data index , Generally used in for Cycle of\n``````seasons = ['Spring', 'Summer', 'Fall', 'Winter']\na = list(enumerate(seasons))\nprint(a)\n# [(0, 'Spring'), (1, 'Summer'), (2, 'Fall'), (3, 'Winter')]\n\nb = list(enumerate(seasons, 1))\nprint(b)\n# [(1, 'Spring'), (2, 'Summer'), (3, 'Fall'), (4, 'Winter')]\n\nfor i, element in a:\nprint('{0},{1}'.format(i, element))\n# 0,Spring\n# 1,Summer\n# 2,Fall\n# 3,Winter\nCopy code ``````\n• zip(iter1 [,iter2 [...]])\n• Used to take iteratable objects as parameters , Package the corresponding elements in the object into tuples , And then return the objects made up of these tuples , The advantage is that it saves a lot of memory .\n• have access to list() Convert to output list .\n• If the number of elements in each iterator is inconsistent , Returns a list of the same length as the shortest object , utilize * The operator , Tuples can be unzipped into lists .\n``````a = [1, 2, 3]\nb = [4, 5, 6]\nc = [4, 5, 6, 7, 8]\n\nzipped = zip(a, b)\nprint(zipped) # <zip object at 0x000000C5D89EDD88>\nprint(list(zipped)) # [(1, 4), (2, 5), (3, 6)]\nzipped = zip(a, c)\nprint(list(zipped)) # [(1, 4), (2, 5), (3, 6)]\n\na1, a2 = zip(*zip(a, b))\nprint(list(a1)) # [1, 2, 3]\nprint(list(a2)) # [4, 5, 6]\nCopy code ``````"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5964098,"math_prob":0.90411824,"size":33514,"snap":"2022-40-2023-06","text_gpt3_token_len":9536,"char_repetition_ratio":0.1435094,"word_repetition_ratio":0.122624055,"special_character_ratio":0.32028407,"punctuation_ratio":0.16644454,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98167795,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-05T08:42:45Z\",\"WARC-Record-ID\":\"<urn:uuid:5e7beb63-b939-48ef-8891-c1da217abda0>\",\"Content-Length\":\"235432\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:352a0fb4-ce89-4fa3-b616-b030df53aa14>\",\"WARC-Concurrent-To\":\"<urn:uuid:2c1e7839-c1f5-4219-a5c1-c28e93caf399>\",\"WARC-IP-Address\":\"172.67.133.238\",\"WARC-Target-URI\":\"https://en.pythonmana.com/2022/01/202201301536449738.html\",\"WARC-Payload-Digest\":\"sha1:BSLDFXOL4I3OFT3PU3RLXGD3OIM46PPS\",\"WARC-Block-Digest\":\"sha1:3OEWV6HD4YG5TYF2LJ3OH7GF2DLDSD3P\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337595.1_warc_CC-MAIN-20221005073953-20221005103953-00786.warc.gz\"}"} |
https://qna.talkjarvis.com/199786/spring-stretched-application-required-stretch-spring-through-stretching-spring-through | [
"# A spring 40mm ling is stretched by the application of force. If 10N force is required to stretch the spring through 1mm, then the work done in stretching the spring through 40mm is?\n\n+1 vote\nA spring 40mm ling is stretched by the application of force. If 10N force is required to stretch the spring through 1mm, then the work done in stretching the spring through 40mm is?\n\n(a) 23J\n\n(b) 68J\n\n(c) 84J\n\n(d) 8J\n\nThis question was addressed to me in class test.\n\nThis question is from Forces of Nature topic in portion Gravitation of Engineering Physics I\n\n+1 vote\nby (1.4m points)\nselected by\n\nThe correct answer is (d) 8J\n\nThe best explanation: Spring constant, k = F/x = 10N/1mm = 10N/(10^-3 m)= 10^4 N/m\n\nWork done in stretching the spring through 40m,\n\nW = 1/2 kx^2=1/2×10^4×(40×10^-3)^2 = 8J.\n\n+1 vote\n+1 vote\n+1 vote\n+1 vote"
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https://archive.cmih.maths.cam.ac.uk/events-archive/low-regularity-fourier-integrators-for-the-nonlinear-schrodinger-equation/ | [
"# Low-regularity Fourier integrators for the nonlinear Schrödinger equation\n\nA large toolbox of numerical schemes for the nonlinear Schrödinger equation has been established, based on different discretization techniques such as discretizing the variation-of-constants formula (e.g., exponential integrators) or splitting the full equation into a series of simpler subproblems (e.g., splitting methods). In many situations these classical schemes allow a precise and efficient approximation. This, however, drastically changes whenever “non-smooth’’ phenomena enter the scene such as for problems at low-regularity and high oscillations. Classical schemes fail to capture the oscillatory parts within the solution which leads to severe instabilities and loss of convergence. In this talk I present a new class of Fourier integrators for the nonlinear Schrödinger equation at low-regularity. The key idea in the construction of the new schemes is to tackle and hardwire the underlying structure of resonances into the numerical discretization. These terms are the cornerstones of theoretical analysis of the long time behaviour of differential equations and their numerical discretizations (cf. modulated Fourier Expansion; Hairer, Lubich & Wanner) and offer the new schemes strong geometric structure at low regularity."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8771756,"math_prob":0.915373,"size":1606,"snap":"2023-14-2023-23","text_gpt3_token_len":335,"char_repetition_ratio":0.12109863,"word_repetition_ratio":0.055555556,"special_character_ratio":0.18244085,"punctuation_ratio":0.11673152,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9839133,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-24T06:38:18Z\",\"WARC-Record-ID\":\"<urn:uuid:5ac830d4-de00-46f5-a386-f91c14791938>\",\"Content-Length\":\"85646\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:11b4bcb5-0e02-45a6-b1ac-b515d484b39c>\",\"WARC-Concurrent-To\":\"<urn:uuid:a36da1a3-c665-4388-8c8d-fd9d0e1eb29b>\",\"WARC-IP-Address\":\"52.30.235.45\",\"WARC-Target-URI\":\"https://archive.cmih.maths.cam.ac.uk/events-archive/low-regularity-fourier-integrators-for-the-nonlinear-schrodinger-equation/\",\"WARC-Payload-Digest\":\"sha1:NESD2YVMMXVTGL455FSSMGJAHQZ4TW4Y\",\"WARC-Block-Digest\":\"sha1:7LKDXYMYCDQU5JQMRMVQPVS4DUTTNPVO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945248.28_warc_CC-MAIN-20230324051147-20230324081147-00660.warc.gz\"}"} |
http://foodgram.info/area-and-distributive-property-worksheets/using-the-distributive-property-worksheet-answers-grade-common-core-finding-area-worksheets-3rd-and-distribut/ | [
"### Using The Distributive Property Worksheet Answers Grade Common Core Finding Area Worksheets 3rd And Distribut",
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"using the distributive property worksheet answers grade common core finding area worksheets 3rd and distribut.\n\narea with distributive property worksheets teaching using an model school finding 3rd grade,distributive property worksheets grade and fresh area collection of model worksheet pdf using,area using distributive property worksheet model the reteach for with worksheets,distributive property worksheet grade finding area using worksheets 3rd of rectangle,distributive property area model worksheet pdf using worksheets 3rd grade,using distributive property worksheet worksheets graders grade finding area 3,worksheets for distributive property algebra area model worksheet pdf using finding 3rd grade,area distributive property grade 3 worksheets 3rd pdf model worksheet multiplication using,area distributive property worksheet with worksheets 3rd grade and,area model distributive property worksheet pdf using grade worksheets."
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"http://foodgram.info/wp-content/uploads/2019/05/using-the-distributive-property-worksheet-answers-grade-common-core-finding-area-worksheets-3rd-and-distribut.jpg",
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http://cran.r-project.org/web/packages/splines2/readme/README.html | [
"# splines2\n\nThe R package splines2 provides functions to construct basis matrix of\n\n• B-splines\n• M-splines\n• I-splines\n• convex splines (C-splines)\n• generalized Bernstein polynomials\n• their integrals (except C-splines) and derivatives of given order by close-form recursive formulas\n\nIn addition to the R interface, splines2 also provides a C++ header-only library integrated with Rcpp, which allows construction of spline basis matrix directly in C++ with the help of Rcpp and RcppArmadillo. So it can also be treated as one of the Rcpp* packages. A toy example package that uses the C++ interface is available here.\n\n## Installation of CRAN Version\n\nYou can install the released version from CRAN.\n\n``install.packages(\"splines2\")``\n\n## Development\n\nThe latest version of package is under development at GitHub. If it is able to pass the building check by Travis CI, one may install it by\n\n``````if (! require(remotes)) install.packages(\"remotes\")\nremotes::install_github(\"wenjie2wang/splines2\")``````\n\n## Getting Started\n\nOnline document provides reference for all functions and contains the following vignettes:\n\n## Performance\n\nSince v0.3.0, the implementation of the main functions has been rewritten in C++ with the help of the Rcpp and RcppArmadillo package. The computational performance has thus been boosted.\n\nSome benchmarks with the splines package (version 4.0.1) are provided for reference as follows:\n\n``````library(microbenchmark)\nlibrary(splines)\nlibrary(splines2)\n\nx <- seq.int(0, 1, 0.001)\ndegree <- 3\nord <- degree + 1\nknots <- seq.int(0.1, 0.9, 0.1)\nb_knots <- range(x)\nall_knots <- sort(c(knots, rep(b_knots, ord)))\n\n## check equivalency of outputs\nmy_check <- function(values) {\nall(sapply(values[- 1], function(x) {\nall.equal(unclass(values[]), x, check.attributes = FALSE)\n}))\n}``````\n\nFor B-splines, function `splines2::bSpline()` provides equivalent results with `splines::bs()` and `splines::splineDesign()`, and is about 3x faster than `bs()` and 2x faster than `splineDesign()`.\n\n``````## B-splines\nmicrobenchmark(\n\"splines::bs\" = bs(x, knots = knots, degree = degree,\nintercept = TRUE, Boundary.knots = b_knots),\n\"splines::splineDesign\" = splineDesign(x, knots = all_knots, ord = ord),\n\"splines2::bSpline\" = bSpline(x, knots = knots, degree = degree,\nintercept = TRUE, Boundary.knots = b_knots),\ncheck = my_check,\ntimes = 1e3\n)``````\n``````Unit: microseconds\nexpr min lq mean median uq max neval cld\nsplines::bs 335.703 353.810 387.53 362.81 381.259 3015.9 1000 c\nsplines::splineDesign 204.151 213.133 244.16 216.05 226.820 2342.8 1000 b\nsplines2::bSpline 84.866 91.677 108.45 95.46 99.399 2149.9 1000 a ``````\n\nSimilarly, for derivatives of B-splines, `splines2::dbs()` provides equivalent results with `splines::splineDesign()`, and is more than 2x faster.\n\n``````## Derivatives of B-splines\nderivs <- 2\nmicrobenchmark(\n\"splines::splineDesign\" = splineDesign(x, knots = all_knots,\nord = ord, derivs = derivs),\n\"splines2::dbs\" = dbs(x, derivs = derivs, knots = knots, degree = degree,\nintercept = TRUE, Boundary.knots = b_knots),\ncheck = my_check,\ntimes = 1e3\n)``````\n``````Unit: microseconds\nexpr min lq mean median uq max neval cld\nsplines::splineDesign 274.066 285.540 330.04 295.3 327.12 4143.4 1000 b\nsplines2::dbs 88.085 94.344 127.73 99.0 107.18 2639.1 1000 a ``````\n\nThe splines package does not provide function producing integrals of B-splines. So we instead performed a comparison with package ibs (version 1.4), where the function `ibs::ibs()` was also implemented in Rcpp.\n\n``````## integrals of B-splines\nset.seed(123)\ncoef_sp <- rnorm(length(all_knots) - ord)\nmicrobenchmark(\n\"ibs::ibs\" = ibs::ibs(x, knots = all_knots, ord = ord, coef = coef_sp),\n\"splines2::ibs\" = as.numeric(\nsplines2::ibs(x, knots = knots, degree = degree,\nintercept = TRUE, Boundary.knots = b_knots) %*% coef_sp\n),\ncheck = my_check,\ntimes = 1e3\n)``````\n``````Unit: microseconds\nexpr min lq mean median uq max neval cld\nibs::ibs 2445.25 2666.93 3259.59 3213.59 3342.26 113446.1 1000 b\nsplines2::ibs 264.84 319.18 363.78 338.62 360.94 2826.4 1000 a ``````\n\nThe function `ibs::ibs()` returns the integrated B-splines instead of the integrals of spline bases. So we applied the same coefficients to the bases from `splines2::ibs()` for equivalent results, which was still much faster than `ibs::ibs()`."
] | [
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http://www.gsherpa.com/rylst/3534a3-pie-chart-worksheets-for-grade-3 | [
"All grade 6 math concepts are combined and will evaluate the students' math knowledge and skills. These worksheets are pdf files. Students are given a simple fraction and asked to color in the pie charts. Second graders need to be able to construct different types of graphs—specifically picture graphs, line plots, and bar graphs—using data sets including up to four categories. Displaying top 8 worksheets found for - 3rd Grade Pie Chart. You can & download or print using the browser document reader options. The worksheets might be less challenging than those where the angles must be measured or drawn, but still are great sixth grade graphing material. Showing top 8 worksheets in the category - Pie Chart. Parenting » Worksheets » 3rd grade bar graph worksheets . (a) How many students took part in the survey? Pie Chart Worksheet - shading in segments. Mathster; Corbett Maths Creating Pie Charts #1; Creating Pie Charts #2; Creating Pie Charts #3; More help with making pie charts. Created: Apr 5, 2012. Worksheet will open in a new window. Found worksheet you are looking for? Some of the worksheets for this concept are Rock restaurant surveyed a sample of customers on their, Pie graph, The pie graph shows the information on the number of farm, Pie charts, Grade 4 fractions work, Lesson plan 3 pie graphs pie graphs, Pie charts, Summer camp activites. Common Core Math Standards push for students in first grade to practice organizing and reasoning with data that is divided into up to three categories. (3 marks) 6. Convert the Data into either Fraction or Percent Study the pie graph and answer the questions by converting the data into either fraction or percentage accordingly. The \"Basic Pie Graphs\" require students to have a basic understanding of fractions. See the pie chart to answer this question. (Approx grade levels: 6, 7, 8) Line Graph Worksheets. Feb 4, 2018 - Read & interpret data on pie graphs (circle graphs). 2nd grade To understand a pie chart (circle graph) and to be able to deduce information about the relative size of the parts shown in it. Advertisement. Pie Chart Worksheets. The pie chart below shows the percentages of each continent. Read more. Access some of these handouts for free! A. Pie graph for Beginners Put your learners in charge as they work their way up analyzing a pie graph, with this themed free printable worksheet for 3rd grade children. Free pdf worksheets from K5 Learning's online reading and math program. Pie Chart Practice: Colored Pencils Your child might not be ready for statistics, but she can practice using pie charts with this fun, cute worksheet. Read, create, and interpret bar graphs with these worksheets. Some of the worksheets displayed are Pie charts, Pie graph, Graphing exercise create a pie graph by selecting a, Introducing pie charts maths work from urbrainy, Year 6 summer term week 6 to 7, Mathematics linear 1ma0 pie charts, Lifestyle balance pie, Name gcse 1 9 pie charts. 100: C. 60: D. 10 Worksheet will open in a new window. Found worksheet you are looking for? The pie chart shows some information about their favourite holiday. Some of the worksheets for this concept are Rock restaurant surveyed a sample of customers on their, Pie graph, The pie graph shows the information on the number of farm, Pie charts, Grade 4 fractions work, Lesson plan 3 pie graphs pie graphs, Pie charts, Summer camp activites. ..... (2) Noreen chooses one of these students at random. This lesson plan is aligned to Standard 3.NF.3 (Third Grade Numbers and Operations– Fractions, item 3 ) in the Common Core. arrow_back Back to Pie Charts Pie Charts: Worksheets with Answers. Some of the worksheets displayed are Summer camp activites, Pie charts, Grade 4 fractions work, Lesson plan 3 pie graphs pie graphs, Mathematics linear 1ma0 pie charts, Grade 4 fractions work, Pie graph, The pie graph shows the information on the number of farm. Read and create line graphs with these worksheets. 3rd grade bar graph worksheets. If you are in search of Multiplication Worksheets For Grade 3, you are on the right site.In this site, you will find multiplication worksheet for grade 3 which includes basic multiplication questions, the meaning of multiplication that multiplication is repeated addition for example:- 5 × 10 = 5+5+5+5+5, 3 × 9 = 3+3+3+3+3+3+3+3+3. (Approx. The pie graph worksheets are designed for students of grade 3 through grade 7. Mar 2, 2017 - Grade 3 math worksheets on identifying equivalent fractions using pie charts. Pie Graphing Worksheets For Printable Download. The intent is to visually reinforce the meaning of \"equivalent\" for fractions. Share on Pinterest. Tci History Alive The Ancient World Answer Key, Compound Sentence And Simple Sentence Grade 6, Addin Subtracting Multiplying Polynomials. Pie Chart Grade 3 Showing top 8 worksheets in the category - Pie Chart Grade 3 . Some of the worksheets for this concept are Rock restaurant surveyed a sample of customers on their, Pie graph, The pie graph shows the information on the number of farm, Pie charts, Grade 4 fractions work, Lesson plan 3 pie graphs pie graphs, Pie charts, Summer camp activites. In other words, the entire circle represents $100 \\%$ of a whole, while the sectors represent portions of the whole. Pie Charts Questions Main. A middle school surveyed its students to find out how they commute to school everyday. Circle Graphs Lesson. Easter Worksheets For 3Rd Grade and Easter Activities For Rd Grade - Easter Multiplication. Some of the worksheets for this concept are Summer camp activites, Pie charts, Grade 4 fractions work, Lesson plan 3 pie graphs pie graphs, Mathematics linear 1ma0 pie charts, Grade 4 fractions work, Pie graph, The pie graph shows the information on the number of farm. Worksheet to support lower ability pupils in pie charts. These math pie graph exercises come with grade 6 ratio and percent calculations. This page has several pie graph worksheets. A protractor is a device that measures angles. Easter Worksheets For 3Rd Grade and Easter Coloring Pages For Rd Grade First Grade Easter Math. These worksheets are highly recommended for students of grade 1 through grade 6. Loading... Save for later. 5 students said that walking is their favourite holiday. Below are six versions of our grade 3 math worksheet on equivalent fractions; students are asked to color in the pie charts which represent each fraction. a) What is the area of Asia? A KS3 worksheet on pie charts with differentiated questions. Homework that was awarded a grade B can be expressed as the following fraction: \\dfrac{5}{24}. Contents. Each worksheet contains a unique theme to clearly understand the usage and necessity of a bar graph in real-life. Displaying top 8 worksheets found for - Pie Chart Grade 7. 50: B. 3rd Grade Pie Chart - Displaying top 8 worksheets found for this concept. Free. Prior to dealing with Pie Chart Worksheets, be sure to realize that Education and learning is definitely all of our crucial for a greater next week, as well as learning doesn’t just halt when the education bell rings.Of which being said, all of us provide number of very simple nevertheless useful content articles plus design templates designed suitable for every academic purpose. To download/print, click on pop-out icon or print icon to worksheet to print or download. When the sections of the pie chart are not equal in size you need a protractor to read a pie chart if you want details. Bar Graph For 3Rd Grade Worksheets and Bar Graphs Rd Grade. The pie graph in these printable worksheets for grade 5, grade 6, and grade 7 require conversion of a whole number into percentage. This free printable worksheet features a pie chart with sections for the eight traditional parts of speech: nouns, pronouns, adjectives, verbs, adverbs, prepositions, conjunctions, and interjections. Bar Graph Worksheets. Pie charts and percentages and angles. 3rd grade bar graph worksheets let your child practice reading and interpreting data presented in graphs, a key math skill. These printables feature basic pie graphs with basic fractions, as well as advanced ones with percentages. Some of the worksheets for this concept are Rock restaurant surveyed a sample of customers on their, Pie graph, The pie graph shows the information on the number of farm, Pie charts, Grade 4 fractions work, Lesson plan 3 pie graphs pie graphs, Pie charts, Summer camp activites. Using a pie chart already divided every 10 degrees students can learn and practice drawing pie charts. Pie charts often label each segment with a percentage, so it is vital that children understand percentages before they can properly interpret pie charts.. For example: a pie chart showing the favourite fruit of children in one class, may have 3 segments, showing strawberries as 55%, bananas as 39% and blueberries without a percentage. In order to compare informations, pie chart uses percentages. Pie Chart Grade 3 - Displaying top 8 worksheets found for this concept.. The \"Advanced Pie Graphs\" require students to understand percentages./p> ... Use the information in the summer camp pie graph to answer the questions. Line Plot (Dot Plot) Worksheets You can & download or print using the browser document reader options. Worksheets: identifying equivalent fractions using pie charts. Noreen carries out a survey of some students. The first time using this worksheet, kids give definitions and examples of each part of sheet. c) How much bigger is Africa than Europe? The data of the survey are presented in the pie chart shown below. These worksheets are pdf files. Therefore \\frac{1}{4} of the pie will be the slice for grade A so one quarter of the whole pie chart will be a slice with a right-angle (90\\degree) since, 360\\degree \\div 4 = 90\\degree . To download/print, click on pop-out icon or print icon to worksheet to print or download. 4.9 12 customer reviews. page 2. april 26. help your second grader learn how to interpret a pie graph with this worksheet all about kids and their pets. You will find more about measuring angles in degrees here and there is more about equivalent fractions here. Easter Worksheets For 3Rd Grade and April In Kindergarten - Freebies. Students create and interpret pie graphs based on animals. About this resource. Some of the worksheets for this concept are Mathematics linear 1ma0 pie charts, Data handling grade 4 7, Grade 4 fractions work, Graphs and charts, Pie charts, Pie graph, Bar charts histograms line graphs pie charts, Summer camp activites. Box Plots (Box-and-Whisker Plots) Create box plots on a number line using the values for Q1, median, Q3, minimum, and maximum. Favorite Animal at the Zoo. Bar Graph For 3Rd Grade Worksheets and Making Bar Graph Worksheet - Free Printable Educational. A good real life pie chart … Info. Pie chart (or pie graph) is a way of displaying data in a circular graph which is divided into sectors.Furthermore, each pie sector represents a certain category. b) What is the area Europe? The worksheet is then kept as a \"cheat sheet.\" Solution to Example 4: a) 33.2% × 134 = 44.5 million square kilometers b) 7.5% × 134 = 10.5 million square kilometers c) (22.3% - 7.5%) × 134 = 19.8 million square kilometers If 200 students responded to the survey, how many are either bike or car riders? Pie Graph Worksheets 3rd Grade Math worksheets graphs and charts free graphs and chart worksheets. Below are six versions of our grade 3 math worksheet on identifying equivalent fractions using pie charts; students are shown sets of two pie charts representing equivalent fractions and are asked to write numerators and denominators for each fraction. 3rd Grade Pie Chart - Displaying top 8 worksheets found for this concept. come with answers. The three worksheets below will provide practice with calculating the angles that are used to create pie charts. Bar Graph For 3Rd Grade Worksheets and Bar Graphs Rd Grade. Preview and details Files included (1) docx, 278 KB. Whether you want a homework, some cover work, or a lovely bit of extra practise, this is the place for you. Pie Graph - Color, Tally and GraphThis is a first pie graph worksheet set - most suited to lower grades (kindergarten, grade one, grade two)There are 9 worksheets in this set.Each worksheet follows the same pattern, you color the animals according to the key, you count and tally them up. Requires knowledge of fractions. 3rd Grade Pie Chart Displaying top 8 worksheets found for - 3rd Grade Pie Chart . Pie and Circle Graph Worksheets Click the buttons to print each worksheet and answer key. And best of all they all (well, most!) To download/print, click on pop-out icon or print using the browser document reader options find... 2 ; Creating pie charts: worksheets with Answers are presented in the survey, how students... Sentence Grade 6 worksheet all about kids and their pets as well advanced... And percent calculations Grade Easter math worksheets and Making bar Graph worksheet - free Printable Educational as well advanced. Files included ( 1 ) docx, 278 KB Sentence Grade 6 ratio and percent calculations Chart below the... And April in Kindergarten - Freebies can be expressed as the following fraction: \\dfrac { 5 {. Displaying top 8 worksheets found for - 3rd Grade pie Chart Grade 7 mathster ; Corbett Maths pie Graph this. With these worksheets are designed for students of Grade 1 through Grade 7 worksheets the! And practice drawing pie charts # 3 ; more help with Making pie charts their... And will evaluate the students ' math knowledge and skills worksheet on pie graphs based animals. Grader learn how to interpret a pie Chart ( a ) how much bigger Africa! A ) how many students took part in the category - pie Chart shows some information their... Graphing worksheets for Printable download part in the Common Core each part of sheet. and best of all all! The worksheet is then kept as a cheat sheet. Chart Grade 3 math on. Worksheet is then kept as a cheat sheet. Pages for Grade. For you a ) how many students took part in the pie charts and. Category - pie Chart below shows the percentages of each continent { 24 } Grade 7 3.NF.3 Third... April in Kindergarten - Freebies key, Compound Sentence and simple Sentence Grade 6 math concepts are combined and evaluate... ) docx, 278 KB with percentages Standard 3.NF.3 ( Third Grade Numbers and Operations– fractions as! Meaning of equivalent '' for fractions with percentages provide practice with calculating the angles are. Ability pupils in pie charts pie charts pie charts: worksheets with Answers Maths pie Graph worksheets 3rd and! Practice drawing pie charts # 2 ; Creating pie charts worksheet - free Printable Educational 3. Cover work, or a lovely bit of extra practise, this is the place for you color. First time using this worksheet all about kids and their pets Read & interpret data pie! Basic pie graphs with these worksheets students at random 278 KB and simple Sentence Grade 6 and bar... Lower ability pupils in pie charts come with Grade 6, create, and interpret bar Rd. Grade 6 ratio and percent calculations of extra practise, this is the place for you 7, 8 Line! Your second grader learn how to interpret a pie Graph exercises come with Grade 6 ratio percent... - Freebies using pie charts: worksheets with Answers icon or print the... In degrees here and there is more about measuring angles in degrees here there. ( a ) how many students took part in the Common Core K5 Learning online! Pdf worksheets from K5 Learning 's online reading and math program Alive Ancient... Based on animals degrees here and there is more about measuring angles in here...: 6, 7, 8 ) Line Graph worksheets click the buttons to print worksheet. Bigger is Africa than Europe homework, some cover work, or a lovely bit of extra practise this... Either bike or car riders fractions using pie charts, 8 ) Graph... Divided every 10 degrees students can learn and practice drawing pie charts differentiated... Pupils in pie charts the percentages of each part of sheet. Creating pie charts browser document options... Second grader learn how to interpret a pie Graph worksheets are designed for students of Grade 1 Grade. Pie and Circle Graph worksheets lovely bit of extra practise, this is the place for you Third Grade and... Degrees here and there is more about measuring angles in degrees here there... Simple Sentence Grade 6 ratio and percent calculations this is the place for you math pie Graph this. Worksheet and answer key the survey are presented in the pie Graph click... Can & download or print using the browser document reader options cover,. Making pie charts Approx Grade levels: 6, 7, 8 ) Line Graph worksheets click the buttons print. To worksheet to print or download Easter math can be expressed as the following:... Based on animals to compare informations, pie Chart students of Grade 1 through Grade 6 math concepts are and... Color in the category - pie Chart uses percentages online reading and math program presented in the pie -. April 26. help your second grader learn how to interpret a pie Graph exercises come with 6! Highly recommended for students of Grade 1 through Grade 7 worksheet, kids give definitions and examples of part. These printables feature basic pie graphs based on animals and interpret bar graphs these... ) Line Graph worksheets are highly recommended for students of Grade 1 through Grade 6 Operations– fractions, as as! Page 2. April 26. help your second grader learn how to interpret a pie Graph this... Three worksheets below will provide practice with calculating the angles that are used to pie. Worksheet - free Printable Educational and Operations– fractions, item 3 ) in the Core. Understanding of fractions some information about their favourite holiday 3 - Displaying top 8 worksheets found for this concept as... Sheet. c ) how much bigger is Africa than Europe or download as! To create pie charts your second grader learn how to interpret a pie Graph worksheets designed! Definitions and examples of each continent math worksheets graphs and Chart worksheets \\dfrac. Used to create pie charts the survey are presented in the pie Chart … pie Graphing worksheets for Grade! Can & download or print icon to worksheet to support lower ability pupils in pie charts worksheets on equivalent. And there is more about equivalent fractions using pie charts Displaying top 8 worksheets for. For this concept Graphing worksheets for Printable download bit of extra practise, this is the for... - 3rd Grade bar Graph for 3rd Grade worksheets and bar graphs Grade. Differentiated questions students responded to the survey are presented in the category - pie Chart Multiplying... To support lower ability pupils in pie charts and asked to color the... Real life pie Chart - Displaying top 8 worksheets found for - 3rd Grade and Easter Coloring Pages Rd! Pie Chart shows some information about their favourite holiday worksheet to print or.! Grade Numbers and Operations– fractions, as well as advanced ones with percentages and Operations– fractions item. Found for - 3rd Grade bar Graph worksheets are designed for students of Grade through. And Circle Graph worksheets each worksheet and answer key, Compound Sentence and simple Sentence Grade.! These printables feature basic pie graphs with these worksheets good real life pie Chart Displaying top worksheets... At random 1 ) docx, 278 KB included ( 1 ) docx 278. Charts # 2 ; Creating pie charts lower ability pupils in pie charts Rd. 3.Nf.3 ( Third Grade Numbers and Operations– fractions, as well as advanced with! The first time using this worksheet, kids give definitions and examples of each part of sheet. &!\n\nSudbury Radio Contest, How To Apply Calcium Nitrate To Tomato Plants, Korean Peel Off Mask, 1965 Chevy Van For Sale Craigslist, 1965 Chevy Van For Sale Craigslist, Isolved User Guide, The Daily Object Show Lava Bucket,"
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https://www.interviewkickstart.com/problems/validate-binary-search-tree | [
"",
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"",
null,
"# Binary Search Tree Problem\n\nGiven a binary tree, check if it is a binary search tree (BST). A valid BST does not have to be complete or balanced.\n\nConsider the below definition of a BST:\n\n1. All nodes values of left subtree are less than or equal to parent node value\n2. All nodes values of right subtree are greater than or equal to parent node value\n3. Both left subtree and right subtree must be a BST\n4. By definition, NULL tree is a BST\n5. By definition, trees having a single node or leaf nodes are BST.\n\nExample One\n\nInput:\n\nOutput: false\n\nLeft child value 200 is greater than the parent node value 100; violates the definition of BST.\n\nExample Two\n\nInput:\n\nOutput: true\n\nNotes\n\nInput Parameters: There is only one argument named root denoting the root of the input tree.\n\nOutput: Return true if the input tree is a BST or false otherwise\n\nConstraints:\n\n• 0 <= number of nodes <= 100000\n• -10^9 <= values stored in the nodes <= 10^9\n• Return value must be boolean\n\n### Solutions\n\nWe provided two solutions, both are optimal in terms of time and space complexity, though algorithms are different.\n\n### 1) other_solution.cpp:\n\nIn the following three examples, tree A and B is a BST, where tree C is not:\n\nIn other_solution.cpp we use the definition of BST quite literally. For every node, recursively, we check that all values in the left subtree are <= current node value and all values in the right subtree are >= current node value. We also propagate all the \"<=\" and \">=\" conditions down the recursion tree (they get more specific, more narrow as we go down the tree), effectively forming a valid range for every node we check. For example, in the following tree,\n\nthe requirements for X are both X>=1 and X<=4. Further, valid range for Y is Y>=X and Y<=4. Notice that because X>=1 holds true (X is checked before Y, higher up in the recursion tree), the effective valid range for Y is narrower than one for X; that way the range gets narrower as we go deeper down the tree. For better understanding please look at the solution.\n\nTime Complexity:\n\nO(n) where n denotes the number of nodes.\n\nAs we are traversing all the nodes of the tree to check if the node value lies within a range or not, hence we have to iterate through all the nodes and edges of the tree. A tree with n nodes have (n-1) edges. So, we have to iterate n + (n-1) → 2n-1 times can be represented as O(n) complexity.\n\nAuxiliary Space:\n\nO(n) where n denotes the number of nodes.\n\nAs we are calling functions in recursion, so in the worst case functional stack can have n number of function calls which is equal to the number of nodes of the given tree. Hence auxiliary space for that is O(n)\n\nSpace complexity:\n\nO(n).\n\n``````\n// -------- START --------\n\nbool isBSTHelper(TreeNode *root, int min, int max){\n// NULL node check\nif(root == NULL){\nreturn true;\n}\n// current node value is not within valid range\nif(root->val>max || root->valleft_ptr, min, root->val) && isBSTHelper(root->right_ptr, root->val, max);\n}\n\nbool isBST(TreeNode* root){\n// empty or null tree check\nif(root==NULL){\nreturn true;\n}\nint min = INT_MIN;\nint max = INT_MAX;\nreturn isBSTHelper(root, min, max);\n}\n\n// -------- END --------\n``````\n\n### 2) optimal_solution.cpp:\n\nIn optimal_solution.cpp solution we have used in-order traversal property of a BST to validate whether it is a BST or not. If we store the values in an array during in-order traversal, it will be a sorted array if the given tree is BST. To check if an array is sorted or not, we just need to compare an element with the previous element of the array. We can do this while traversing the tree instead of storing the values in an array which will reduce space complexity. To achieve this, we have used a variable which stores the last visited nodes value at any time. So, when we are in a node, we can compare if the current node value is greater or equal to the previous node value. If this condition fails for any node then the given tree is not a BST. Otherwise it is.\n\nTime Complexity:\n\nO(n) where n denotes the number of nodes.\n\nAs we are traversing all the nodes of the tree to check if the current node value is greater or equal to the previous node value, hence we have to iterate through all the nodes and edges of the tree. A tree with n nodes have (n-1) edges. So, we have to iterate n + (n-1) → 2n-1 times can be represented as O(n) complexity.\n\nAuxiliary Space:\n\nO(n) where n denotes the number of nodes.\n\nAs we are calling functions in recursion, so in the worst case functional stack can have n number of function calls which is equal to the number of nodes of the given tree. Hence auxiliary space for that is O(n)\n\nSpace Complexity:\n\nO(n).\n\n``````\n// -------- START --------\n\nbool isBSTHelper(TreeNode *root, int &prev){\n// NULL node check\nif(root == NULL){\nreturn true;\n}\n\n// check if left subtree is bst or not\nbool isLeftSubtreeBST = isBSTHelper(root->left_ptr, prev);\n\n// Check if current node value is greater or equal to the max value of left subtree nodes\nif(root->val < prev){\nreturn false;\n}\n\n// update the prev variable by current node value as each value of right subtree must be greater or equal\n// to the current root value\nprev = root->val;\n// true when both left and right subtrees are valid BST\nbool isRightSubtreeBST = isBSTHelper(root->right_ptr, prev);\n\n// Bitwise AND operation to return true only when both subtree is BST, otherwise false\nreturn (isLeftSubtreeBST && isRightSubtreeBST);\n}\n\nbool isBST(TreeNode* root){\n// empty or null tree check\nif(root==NULL){\nreturn true;\n}\nint min = INT_MIN;\nreturn isBSTHelper(root, min);\n}\n\n// -------- END --------\n``````\n\n### Try yourself in the Editor\n\nNote: Input and Output will already be taken care of.\n\n# Binary Search Tree Problem\n\nGiven a binary tree, check if it is a binary search tree (BST). A valid BST does not have to be complete or balanced.\n\nConsider the below definition of a BST:\n\n1. All nodes values of left subtree are less than or equal to parent node value\n2. All nodes values of right subtree are greater than or equal to parent node value\n3. Both left subtree and right subtree must be a BST\n4. By definition, NULL tree is a BST\n5. By definition, trees having a single node or leaf nodes are BST.\n\nExample One\n\nInput:\n\nOutput: false\n\nLeft child value 200 is greater than the parent node value 100; violates the definition of BST.\n\nExample Two\n\nInput:\n\nOutput: true\n\nNotes\n\nInput Parameters: There is only one argument named root denoting the root of the input tree.\n\nOutput: Return true if the input tree is a BST or false otherwise\n\nConstraints:\n\n• 0 <= number of nodes <= 100000\n• -10^9 <= values stored in the nodes <= 10^9\n• Return value must be boolean\n\n### Solutions\n\nWe provided two solutions, both are optimal in terms of time and space complexity, though algorithms are different.\n\n### 1) other_solution.cpp:\n\nIn the following three examples, tree A and B is a BST, where tree C is not:\n\nIn other_solution.cpp we use the definition of BST quite literally. For every node, recursively, we check that all values in the left subtree are <= current node value and all values in the right subtree are >= current node value. We also propagate all the \"<=\" and \">=\" conditions down the recursion tree (they get more specific, more narrow as we go down the tree), effectively forming a valid range for every node we check. For example, in the following tree,\n\nthe requirements for X are both X>=1 and X<=4. Further, valid range for Y is Y>=X and Y<=4. Notice that because X>=1 holds true (X is checked before Y, higher up in the recursion tree), the effective valid range for Y is narrower than one for X; that way the range gets narrower as we go deeper down the tree. For better understanding please look at the solution.\n\nTime Complexity:\n\nO(n) where n denotes the number of nodes.\n\nAs we are traversing all the nodes of the tree to check if the node value lies within a range or not, hence we have to iterate through all the nodes and edges of the tree. A tree with n nodes have (n-1) edges. So, we have to iterate n + (n-1) → 2n-1 times can be represented as O(n) complexity.\n\nAuxiliary Space:\n\nO(n) where n denotes the number of nodes.\n\nAs we are calling functions in recursion, so in the worst case functional stack can have n number of function calls which is equal to the number of nodes of the given tree. Hence auxiliary space for that is O(n)\n\nSpace complexity:\n\nO(n).\n\n``````\n// -------- START --------\n\nbool isBSTHelper(TreeNode *root, int min, int max){\n// NULL node check\nif(root == NULL){\nreturn true;\n}\n// current node value is not within valid range\nif(root->val>max || root->valleft_ptr, min, root->val) && isBSTHelper(root->right_ptr, root->val, max);\n}\n\nbool isBST(TreeNode* root){\n// empty or null tree check\nif(root==NULL){\nreturn true;\n}\nint min = INT_MIN;\nint max = INT_MAX;\nreturn isBSTHelper(root, min, max);\n}\n\n// -------- END --------\n``````\n\n### 2) optimal_solution.cpp:\n\nIn optimal_solution.cpp solution we have used in-order traversal property of a BST to validate whether it is a BST or not. If we store the values in an array during in-order traversal, it will be a sorted array if the given tree is BST. To check if an array is sorted or not, we just need to compare an element with the previous element of the array. We can do this while traversing the tree instead of storing the values in an array which will reduce space complexity. To achieve this, we have used a variable which stores the last visited nodes value at any time. So, when we are in a node, we can compare if the current node value is greater or equal to the previous node value. If this condition fails for any node then the given tree is not a BST. Otherwise it is.\n\nTime Complexity:\n\nO(n) where n denotes the number of nodes.\n\nAs we are traversing all the nodes of the tree to check if the current node value is greater or equal to the previous node value, hence we have to iterate through all the nodes and edges of the tree. A tree with n nodes have (n-1) edges. So, we have to iterate n + (n-1) → 2n-1 times can be represented as O(n) complexity.\n\nAuxiliary Space:\n\nO(n) where n denotes the number of nodes.\n\nAs we are calling functions in recursion, so in the worst case functional stack can have n number of function calls which is equal to the number of nodes of the given tree. Hence auxiliary space for that is O(n)\n\nSpace Complexity:\n\nO(n).\n\n``````\n// -------- START --------\n\nbool isBSTHelper(TreeNode *root, int &prev){\n// NULL node check\nif(root == NULL){\nreturn true;\n}\n\n// check if left subtree is bst or not\nbool isLeftSubtreeBST = isBSTHelper(root->left_ptr, prev);\n\n// Check if current node value is greater or equal to the max value of left subtree nodes\nif(root->val < prev){\nreturn false;\n}\n\n// update the prev variable by current node value as each value of right subtree must be greater or equal\n// to the current root value\nprev = root->val;\n// true when both left and right subtrees are valid BST\nbool isRightSubtreeBST = isBSTHelper(root->right_ptr, prev);\n\n// Bitwise AND operation to return true only when both subtree is BST, otherwise false\nreturn (isLeftSubtreeBST && isRightSubtreeBST);\n}\n\nbool isBST(TreeNode* root){\n// empty or null tree check\nif(root==NULL){\nreturn true;\n}\nint min = INT_MIN;\nreturn isBSTHelper(root, min);\n}\n\n// -------- END --------\n``````\n\n## Worried About Failing Tech Interviews?\n\nAttend our free webinar to amp up your career and get the salary you deserve.",
null,
"Hosted By\nRyan Valles\nFounder, Interview Kickstart",
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https://learn.careers360.com/ncert/question-find-the-sum-of-all-natural-numbers-lying-between-100-and-1000-which-are-multiples-of-5/ | [
"Q\n\n# Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.\n\n2. Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.\n\nViews\n\nNumbers divisible by 5 from 100 to 1000 are\n\nThis sequence is an A.P.\n\nHere , first term =a =105\n\ncommon difference = 5.\n\nWe know ,\n\nThe sum of numbers divisible by 5 from 100 to 1000 is 98450.\n\nExams\nArticles\nQuestions"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93399864,"math_prob":0.9971133,"size":281,"snap":"2020-10-2020-16","text_gpt3_token_len":84,"char_repetition_ratio":0.12635379,"word_repetition_ratio":0.11111111,"special_character_ratio":0.36654803,"punctuation_ratio":0.13846155,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9982184,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-27T21:31:57Z\",\"WARC-Record-ID\":\"<urn:uuid:386e461f-e6f0-4574-ba84-8338f55cadd0>\",\"Content-Length\":\"764113\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8edf121b-7c34-4c7e-a04f-640c08d42004>\",\"WARC-Concurrent-To\":\"<urn:uuid:4a60360d-dc0b-48ca-9888-4fe69b3c627b>\",\"WARC-IP-Address\":\"13.126.158.197\",\"WARC-Target-URI\":\"https://learn.careers360.com/ncert/question-find-the-sum-of-all-natural-numbers-lying-between-100-and-1000-which-are-multiples-of-5/\",\"WARC-Payload-Digest\":\"sha1:RP5YKQ6TLG2KINOPIZOPOZO5KXZL6AVW\",\"WARC-Block-Digest\":\"sha1:Y5C7HOZTGPMUH22HKYE4LNKGLAEWPGXL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146809.98_warc_CC-MAIN-20200227191150-20200227221150-00357.warc.gz\"}"} |
https://discourse.julialang.org/t/int-numerical-calculation-speed-slower-than-float/34705 | [
"# Int numerical calculation speed slower than Float?\n\nWhy the calculation speed of “Int” matrix is much slower than “Float” matrix?\n\nIDE: Jupyter\nJulia version: 1.1.0\n\nCode for Float:\ns=1000\nd1=rand(s,s)\nd2=rand(s,s)\n@time (d1*d2);\nResult: 0.032192 seconds (6 allocations: 7.630 MiB)\n\nCode for Int:\ns=1000\nd3=rand(Int,s,s)\nd4=rand(Int, s,s)\n@time (d3*d4);\nResult: 1.131875 seconds (12 allocations: 7.630 MiB)\n\nFloat matrices use blas. Int uses a generic fallback. Making the fallback method multithreaded for large matrices would fix much of the problem.\n\n2 Likes\n\nI think that for floats, BLAS is used, while for integers it is native Julia code. The latter could probably be made faster, but it is not a common use case so it is waiting for someone to do it.\n\nDo you think it would be worth it for mixed type matmul to convert arguments before multiplying? I think that should speed things up a lot (with some memory downsides). The other big thing the fallback needs is better cache aware looping.\n\nNo, I would not convert. First, integers have specific overflow semantics in Julia different from float, so I am not sure what is intended and what isn’t.\n\nSecond (and more importantly), you really have to go out of your way to get a matrix with a non-concrete element type when writing idiomatic code, so I am not sure it is a common use case. I would leave it up to the user to promote if that is needed.\n\nWe should have a specialized very effective Int multiplication kernel though.\n\n4 Likes\n\nOne way to do this is LoopVectorization’s example, which is faster but still not as fast as floats:\n\n``````julia> C1 = Matrix{Int}(undef, M, N); A = rand(1:100, M, K); B = rand(1:100, K, N);\n\njulia> C2 = similar(C1); C3 = similar(C1);\n\njulia> @btime mygemmavx!(\\$C1, \\$A, \\$B)\n77.412 μs (0 allocations: 0 bytes)\n\njulia> @btime mygemm!(\\$C2, \\$A, \\$B)\n245.869 μs (0 allocations: 0 bytes)\n\njulia> @btime mul!(\\$C3, \\$A, \\$B); # julia's generic_matmul\n164.278 μs (6 allocations: 336 bytes)\n``````\n\ncompared to Float64:\n\n``````julia> @btime mygemmavx!(\\$C1, \\$A, \\$B)\n14.599 μs (0 allocations: 0 bytes)\n\njulia> @btime mygemm!(\\$C2, \\$A, \\$B)\n290.296 μs (0 allocations: 0 bytes)\n\njulia> @btime mul!(\\$C3, \\$A, \\$B); # openblas, not MKL\n22.635 μs (0 allocations: 0 bytes)\n``````\n\nNote BTW that in your example, `rand(Int, ...)` produces lots of large numbers, which will overflow:\n\n``````julia> (float(d3) * float(d4)) .- (d3 * d4) |> extrema\n(-4.2418818662328814e39, 4.4115065145744565e39)\n\njulia> extrema(A)\n(1, 100)\n\njulia> (float(A) * float(B)) .- (A * B) |> extrema\n(0.0, 0.0)\n``````\n4 Likes\n\nUsing\n\n``````julia> M, K, N = 72, 75, 71;\n``````\n\nMy results with `Float64` are:\n\n``````julia> BLAS.set_num_threads(1)\n\njulia> @btime mygemmavx!(\\$C1, \\$A, \\$B)\n7.380 μs (0 allocations: 0 bytes)\n\njulia> @btime mygemm!(\\$C2, \\$A, \\$B)\n231.900 μs (0 allocations: 0 bytes)\n\njulia> @btime mul!(\\$C3, \\$A, \\$B); # julia's generic_matmul\n6.780 μs (0 allocations: 0 bytes)\n``````\n\nAnd with `Int`:\n\n``````julia> @btime mygemmavx!(\\$C1, \\$A, \\$B)\n26.158 μs (0 allocations: 0 bytes)\n\njulia> @btime mygemm!(\\$C2, \\$A, \\$B)\n190.645 μs (0 allocations: 0 bytes)\n\njulia> @btime mul!(\\$C3, \\$A, \\$B); # julia's generic_matmul\n101.748 μs (6 allocations: 336 bytes)\n``````\n\nSo `Int` is about 3.5x slower than Float for me, while it is 5.3x slower for you.\nWith integers, it uses the `vpmullq` instruction for integer multiplication. But this instruction appears to be slow, with a reciprocal throughput of around 1.5-3, while the vpaddq instruction is around 0.33 or 0.5.\nThe floating point versions use fused multiply-add instructions to combine both the multiplication and addition, and have a reciprical throughput of about 0.5.\nYou can think of “reciprical throughput” as how many clock cycles it takes per completed instruction if a core is executing many simultaneously. It generally takes many more clock cycles to complete any given instruction (e.g., 4 for the fma instructions), but a core can work on many simultaneously, thus the rate at which they’re completed can be much faster.\n\n3 Likes\n\nRealize that implementing a highly optimized matrix–matrix multiplication is nontrivial. Optimized BLAS libraries typically involve tens of thousands of lines of code and painstaking performance tuning. While there is no theoretical reason why this cannot be replicated in Julia, it is a huge undertaking.\n\n1 Like\n\nThanks, interesting. And the reason for this difference in the first place, perhaps that there’s just more demand for vectorised floating point stuff, which justifies spending a lot of silicon on it?\n\nI have no doubt. “We should” in open source can perhaps only mean “it would be appreciated”.\n\nNot having BLAS implementation for a type really hurts.\n\nThis is why Arraymancer (a BLAS written in Nim that supports a bunch of types/is generic?),\ntrashes everyone at integer matmul\n\nOut of curiosity, how comes that JuliaBLAS is closing in on BLAS for floats but is not fundamentally faster than generic matmul for ints?"
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https://hoopercharles.wordpress.com/2009/12/05/ | [
"## SQL – Retain Specific Sort Order\n\n5 12 2009\n\nDecember 5, 2009\n\nIs there a simple way to use a previous result in an ORDER BY clause?\n\nI have a rather complicate query that filters, sorts and returns a series of IDs:\n\n```FOO_ID\n======\n1\n98\n12\n33```\n\nThen, I use these IDs to fetch further information about the items they represent:\n\n```SELECT .......\nFROM FOO\nLEFT JOIN BAR .......\nWHERE FOO_ID IN (1, 98, 12, 33)```\n\nI keep the two queries separate to avoid excessive complexity. I compose the SQL code using PHP. Right now, the second query comes unsorted from Oracle: I use PHP to sort it at a later stage in my application (my PHP skills are better than my SQL ones). Would it be possible to use the ID list to sort the second query inside Oracle?\n\nServer runs Oracle9i Enterprise Edition Release 9.2.0.1.0\n\nNo need to make this too difficult. If you were on 10g, you could so some fancy things with regexp_substr. A simple example which should work on 9i and below:\nCreate a testing table for this demonstration named T1, think of this as your FOO table:\n\n```CREATE TABLE T1 AS\nSELECT\nROWNUM C1\nFROM\nDUAL\nCONNECT BY\nLEVEL<=100;```\n\nNow the first step, just retrieve the rows you want:\n\n```SELECT\nC1\nFROM\nT1\nWHERE\nC1 IN (1,98,12,33);\n\nC1\n----------\n1\n12\n33\n98```\n\nNow, sort the rows:\n\n```SELECT\nC1\nFROM\nT1\nWHERE\nC1 IN (1,98,12,33)\nORDER BY\nINSTR('1,98,12,33,' , TO_CHAR(C1)||',');\n\nC1\n----------\n1\n98\n12\n33```\n\nNote in the INSTR, the sequence of the numbers must end in a comma, and we tell INSTR to locate the number in the list with a comma appended to the end of the value of C1.\nYour SQL statement would look like this:\n\n```SELECT .......\nFROM FOO\nLEFT JOIN BAR .......\nWHERE FOO_ID IN (1, 98, 12, 33)\nORDER BY\nINSTR(',1,98,12,33,' , ','||TO_CHAR(FOO_ID)||',');```\n\nOr:\n\n```SELECT .......\nFROM\nFOO,\nBAR\nWHERE\nFOO.FOO_ID=BAR.FOO_ID(+)\nAND FOO_ID IN (1, 98, 12, 33)\nORDER BY\nINSTR(',1,98,12,33,' , ','||TO_CHAR(FOO_ID)||',');```\n\nBefore deciding to use the above technique, determine if there is a better way to do *everything* in a single SQL statement. You might be able to do this by wrapping your complicated SQL statement into an inline view, and joining to that just as if it were a regular table:\n\n```SELECT .......\nFROM\nFOO,\nBAR,\n( complicated SQL here ) V\nWHERE\nV.FOO_ID=FOO.ID\nAND FOO.FOO_ID=BAR.FOO_ID(+)\nORDER BY\nV.RN;```\n\nThe RN column would be generated inside the inline view V, possibly like this, if there is an ORDER BY clause in the inline view:\n\n` ROWNUM RN`\n\n## SQL – Recursive Summing of Related Entities\n\n5 12 2009\n\nDecember 5, 2009\n\nTable dir_size stores the mbytes of storage used in a given directory. Table directories stores various directory names which may or may not exist in table dir_size.\nFor every directory in table directories, report the cumulative storage in that directory and all its subdirectories. This solution uses a cartesian join. I imagine it will not scale well.\n\n```create table dir_size (\ndir_name varchar2(40),\nmbytes number\n);\n\ncreate table directories (\ndir_name varchar2(40)\n);\n\ninsert into dir_size values ('c:\\aaa\\bbb\\ccc\\ddd', 100);\ninsert into dir_size values ('c:\\aaa\\bbb\\ccc', 100);\ninsert into dir_size values ('c:\\aaa\\bbb', 100);\ninsert into dir_size values ('c:\\aaa', 100);\ninsert into dir_size values ('c:\\', 100);\ninsert into directories values ('c:\\aaa\\bbb\\ccc\\ddd');\ninsert into directories values ('c:\\aaa\\bbb\\ccc');\ninsert into directories values ('c:\\aaa\\bbb');\ninsert into directories values ('c:\\aaa');\ninsert into directories values ('c:\\');\ninsert into directories values ('c:\\xxx\\yyy\\zzz');\ncommit;\n\nselect dir_name, sum(mbytes) from (\nselect directories.dir_name,\ninstr(dir_size.dir_name, directories.dir_name) INSTR,\nmbytes\nfrom directories, dir_size\n)\nwhere INSTR = 1\ngroup by dir_name\norder by 1;\n\nDIR_NAME SUM(MBYTES)\n---------------------------------------- -----------\nc:\\ 500\nc:\\aaa 400\nc:\\aaa\\bbb 300\nc:\\aaa\\bbb\\ccc 200\nc:\\aaa\\bbb\\ccc\\ddd 100```\n\nThis appears to be a hard problem. To avoid headaches, make certain that each of the DIR_NAMES ends with “\\”\n\nLet’s start here:\n\n```SELECT\n'c:\\aaa\\bbb\\ccc\\ddd\\' DIR_NAME,\n100 MBYTES\nFROM\nDUAL;\n\nDIR_NAME MBYTES\n-------------------- ----------\nc:\\aaa\\bbb\\ccc\\ddd\\ 100```\n\nIn your example, you would like to put 100MB into the following directories based on the above:\n\n```c:\\\nc:\\aaa\\\nc:\\aaa\\bbb\\\nc:\\aaa\\bbb\\ccc\\\nc:\\aaa\\bbb\\ccc\\ddd\\```\n\nYou somehow need to be able to break that one row into 5 rows. The following might help\n\n```SELECT\nLEVEL L\nFROM\nDUAL\nCONNECT BY\nLEVEL<=20;\n\nL\n---\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n16\n17\n18\n19\n20```\n\nIf we join those two row sources together we might be able to create 5 rows from the one row:\n\n```SELECT\nSUBSTR(DIR_NAME,1,INSTR(DIR_NAME,'\\',1,L)) DIR_NAME2,\nMBYTES\nFROM\n(SELECT\n'c:\\aaa\\bbb\\ccc\\ddd\\' DIR_NAME,\n100 MBYTES\nFROM\nDUAL) DIR_SIZE,\n(SELECT\nLEVEL L\nFROM\nDUAL\nCONNECT BY\nLEVEL<=20) C\nWHERE\nSUBSTR(DIR_NAME,1,INSTR(DIR_NAME,'\\',1,L)) IS NOT NULL;\n\nDIR_NAME2 MBYTES\n-------------------- ----------\nc:\\ 100\nc:\\aaa\\ 100\nc:\\aaa\\bbb\\ 100\nc:\\aaa\\bbb\\ccc\\ 100\nc:\\aaa\\bbb\\ccc\\ddd\\ 100```\n\nNow, if we performed the same process for all of the rows in the DIR_SIZE table, grouping on DIR_NAME2, we might be able to find the SUM of the MBYTES column.\n\n(Note that I did not provide an exact/final answer to the original poster – my post was intended to push the OP in the right direction of a solution.)\n\nThe OP followed up with this comment:\n\nThanks for the suggestion. I suspect the best way will involve some kind of recursive processing. The tricky bit is the matching of the rows in the directories table to the rows in the dir_size table. We need to do a “like” (which we can’t, of course) which is why I thought of the instr.\n\nThe LIKE keyword is not necessary.\n\nNotice how closely the output of the following SQL statement:\n\n```SELECT\n'c:\\aaa\\bbb\\ccc\\ddd\\' DIR_NAME,\n100 MBYTES\nFROM\nDUAL;```\n\nMatches the row created by one of your insert statements:\n\n`insert into dir_size values ('c:\\aaa\\bbb\\ccc\\ddd', 100);`\n\nYou might try replacing in the above examples:\n\n```SELECT\n'c:\\aaa\\bbb\\ccc\\ddd\\' DIR_NAME,\n100 MBYTES\nFROM\nDUAL;```\n\nWith a SQL statement that selects all of the rows from your DIR_SIZE table – the results might surprise you IF each of the DIR_NAME values end with a “\\”.\nYou really need more variety in the insert statements to see what is happening, for example:\n\n```insert into dir_size values ('c:\\ddd\\', 800);\ninsert into dir_size values ('c:\\ddd\\kkk\\', 300);```\n\nThe first of the above SQL statements will increase the calculated SUM in the c:\\ directory by 800, and the second insert statement will increase the SUM in both of the c:\\ and c:\\ddd\\ directories by 300 if you modify my original example to use the DIR_SIZE table rather than the DUAL table.\nThe final part that I did not provide to the OP is below:\n\n```TRUNCATE TABLE DIR_SIZE;\n\ninsert into dir_size values ('c:\\aaa\\bbb\\ccc\\ddd\\', 100);\ninsert into dir_size values ('c:\\aaa\\bbb\\ccc\\', 100);\ninsert into dir_size values ('c:\\aaa\\bbb\\', 100);\ninsert into dir_size values ('c:\\aaa\\', 100);\ninsert into dir_size values ('c:\\', 100);\ninsert into dir_size values ('c:\\ddd\\', 800);\ninsert into dir_size values ('c:\\ddd\\kkk\\', 300);```\n\nWorking with the hints provided and the final SQL statement in my post, we start with the following:\n\n```SELECT\nSUBSTR(DIR_NAME,1,INSTR(DIR_NAME,'\\',1,L)) DIR_NAME2,\nMBYTES\nFROM\n(SELECT\n*\nFROM\nDIR_SIZE) DIR_SIZE,\n(SELECT\nLEVEL L\nFROM\nDUAL\nCONNECT BY\nLEVEL<=20) C\nWHERE\nSUBSTR(DIR_NAME,1,INSTR(DIR_NAME,'\\',1,L)) IS NOT NULL;\n\nDIR_NAME2 MBYTES\n-------------------- ----------\nc:\\ 100\nc:\\ 100\nc:\\ 100\nc:\\ 100\nc:\\ 100\nc:\\ 800\nc:\\ 300\nc:\\aaa\\ 100\nc:\\aaa\\ 100\nc:\\aaa\\ 100\nc:\\aaa\\ 100\nc:\\ddd\\ 800\nc:\\ddd\\ 300\nc:\\aaa\\bbb\\ 100\nc:\\aaa\\bbb\\ 100\nc:\\aaa\\bbb\\ 100\nc:\\ddd\\kkk\\ 300\nc:\\aaa\\bbb\\ccc\\ 100\nc:\\aaa\\bbb\\ccc\\ 100\nc:\\aaa\\bbb\\ccc\\ddd\\ 100```\n\n```SELECT\nSUBSTR(DIR_NAME,1,INSTR(DIR_NAME,'\\',1,L)) DIR_NAME2,\nSUM(MBYTES) MBYTES\nFROM\nDIR_SIZE,\n(SELECT\nLEVEL L\nFROM\nDUAL\nCONNECT BY\nLEVEL<=20) C\nWHERE\nSUBSTR(DIR_NAME,1,INSTR(DIR_NAME,'\\',1,L)) IS NOT NULL\nGROUP BY\nSUBSTR(DIR_NAME,1,INSTR(DIR_NAME,'\\',1,L))\nORDER BY\nSUBSTR(DIR_NAME,1,INSTR(DIR_NAME,'\\',1,L));\n\nDIR_NAME2 MBYTES\n-------------------- ----------\nc:\\ 1600\nc:\\aaa\\ 400\nc:\\aaa\\bbb\\ 300\nc:\\aaa\\bbb\\ccc\\ 200\nc:\\aaa\\bbb\\ccc\\ddd\\ 100\nc:\\ddd\\ 1100\nc:\\ddd\\kkk\\ 300```\n\n## SQL – Methods of Reformatting into Equivalent Forms 6\n\n5 12 2009\n\nDecember 5, 2009\n\nA recent post in the comp.database.oracle.server Usenet group asked the following question:\n\nIs there way to force the randomize within a set of number in Oracle?\nSay I have a set of integers ( 2,8,6,5) and I want to force randomize function to randomly pick within that set only. I know about the DBMS_RANDOM package, sample and seed clause but none help what I need to do here.\n\nany thoughts?\n\nIt was not clear if the integers will be present in table rows, supplied in a comma separated list, or through another method. Additionally, it was not clear if the solution should be provided in SQL, PL/SQL, or another programming language. A couple people offered potential solutions.\n\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n\nStraightforward way: just generate random numbers between 0 and 1, divide the 0..1 range into 4 subranges and then pick a number from\nyour list depending on which subrange your generated random value belongs, like this:\n\n```with rnd as (select dbms_random.value val from dual)\nselect\ncase\nwhen val < 0.25 then 2\nwhen val >= 0.25 and val < 0.5 then 8\nwhen val >= 0.5 and val < 0.75 then 6\nwhen val >= 0.75 then 5\nend x\nfrom rnd```\n\n(note that this query does not work correctly in 9.2.0.8 for some reason – returns wrong results. In 10.2.0.4 it returns expected results.) Obviously, you can partition 0..1 range into as many subranges as there are numbers in your list and apply the same technique. A function that will do this automatically given an array of possible return values is not too hard to write.\n\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n\nTiago offered the following solution:\n\n```SET serveroutput ON\nDECLARE\nType tabIntegers IS TABLE OF NUMBER ;\nIntegers tabIntegers ;\nminInteger NUMBER ;\nmaxInteger NUMBER ;\nrndInteger NUMBER ;\nrndIntOk BOOLEAN := False ;\nBEGIN\nintegers := tabIntegers( 2, 5, 10, 7, 3, 8, 11, 25, 0 ) ;\nFOR i IN integers.first..integers.last\nLOOP\nminInteger := Least(NVL(minInteger,integers(i)), integers(i));\nmaxInteger := Greatest(NVL(maxInteger,integers(i)), integers(i));\nEND LOOP ;\nLOOP\nrndInteger := TRUNC( dbms_random.value(minInteger,maxInteger) ) ;\nFOR i IN integers.first..integers.last\nLOOP\nIF rndInteger = integers(i) THEN\nrndIntOk := true ;\nEXIT ;\nEND IF ;\nEND LOOP ;\nEXIT WHEN rndIntOk ;\nEND LOOP ;\ndbms_output.put_line(rndInteger);\nEND ;\n```\n\nTiago offered a follow-up solution:\n\nsimplified, don’t know what I was thinking when did version 1.0.\n\n```SET serveroutput ON\nDECLARE\nType tabIntegers\nIS\nTABLE OF NUMBER ;\nIntegers tabIntegers ;\nrndInteger NUMBER ;\nBEGIN\nintegers := tabIntegers( 2, 5, 10, 7, 3, 8, 11, 25, 0 ) ;\nrndInteger := integers(TRUNC( dbms_random.value(1,integers.last) ) ) ;\ndbms_output.put_line(rndInteger);\nEND ;```\n\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n\nMark Powell offered the following advice:\n\nCK, instead of using a fixed range as Vladimir had in his example I was thinking I might use a MOD division of the random number returned from dbms_random to generate the index key into my table of values. This would make the code flexible for variable length lists. If you have a fixed number of entries then Vlad’s solution is simple and easy to understand. If your number of variables varies then I think this would fit the bill.\n\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n\nI offered the following:\n\nHere is another way to do it, if you do not know how many elements will be in the list.\n\nFirst, we return a row from DUAL with the rows of interest with a comma appended at the start and end:\n\n```SELECT\n',2,8,6,5,54,100,67,7778,6,' ITEMS\nFROM\nDUAL;\n\nITEMS\n--------------------------\n,2,8,6,5,54,100,67,7778,6,```\n\nNext, we need to determine the number of elements and pick an element position at random:\n\n```SELECT\nITEMS,\nSUM(SIGN(INSTR(ITEMS, ',',1,ROWNUM)))-1 NUM_ITEMS,\n(TRUNC(DBMS_RANDOM.VALUE(0,SUM(SIGN(INSTR(ITEMS ,',',1,ROWNUM)))-1)) +1) SEL_ITEM\nFROM\n(SELECT\n',2,8,6,5,54,100,67,7778,6,' ITEMS\nFROM\nDUAL)\nCONNECT BY\nLEVEL<20;\n\nITEMS NUM_ITEMS SEL_ITEM\n-------------------------- ---------- ----------\n,2,8,6,5,54,100,67,7778,6, 9 6```\n\nFinally, we push the above SQL statement into an inline view, search for the specified number of commas according to SEL_ITEM column to determine the starting position of the element, and then search for the next comma to determine the ending position of the element:\n\n```SELECT\nITEMS,\nSEL_ITEM,\nSUBSTR(ITEMS,INSTR(ITEMS,',',1,SEL_ITEM)+1,(INSTR(ITEMS,',', 1,SEL_ITEM+1)) - (INSTR(ITEMS,',',1,SEL_ITEM)) -1) ITEM\nFROM\n(SELECT\nITEMS,\nSUM(SIGN(INSTR(ITEMS, ',',1,ROWNUM)))-1 NUM_ITEMS,\n(TRUNC(DBMS_RANDOM.VALUE(0,SUM(SIGN(INSTR(ITEMS ,',',1,ROWNUM)))-1)) +1) SEL_ITEM\nFROM\n(SELECT\n',2,8,6,5,54,100,67,7778,6,' ITEMS\nFROM\nDUAL)\nCONNECT BY\nLEVEL<20);\n\nITEMS SEL_ITEM ITEM\n-------------------------- ---------- ----\n,2,8,6,5,54,100,67,7778,6, 6 100```\n\nYou might also be able to do something like this:\n\n```SELECT\n'2,8,6,5,54,100,67,7778,6' ITEMS,\nDBMS_RANDOM.VALUE(0,1) PERCENT\nFROM\nDUAL;\n\nITEMS PERCENT\n------------------------ ----------\n2,8,6,5,54,100,67,7778,6 .582165524```\n\n```SELECT\nROWNUM ITEM_NUMBER,\nREGEXP_SUBSTR(ITEMS,'\\w+',1,LEVEL) ITEM,\nPERCENT\nFROM\n(SELECT\n'2,8,6,5,54,100,67,7778,6' ITEMS,\nDBMS_RANDOM.VALUE(0,1) PERCENT\nFROM\nDUAL)\nCONNECT BY\nREGEXP_SUBSTR(ITEMS,'\\w+',1,LEVEL) IS NOT NULL;\n\nITEM_NUMBER ITEM PERCENT\n----------- ----- ----------\n1 2 .104480002\n2 8 .81670697\n3 6 .826051929\n4 5 .477132421\n5 54 .89077554\n6 100 .640842927\n7 67 .145088893\n8 7778 .252241096\n9 6 .490905924```\n\nAs you can see from the above, we have a problem in that the random percent changes for each row, which will cause a problem for us if we try to use it in a WHERE clause.\n\n```SELECT\nMAX(ITEM_NUMBER) OVER () NUM_ITEMS,\nPERCENT_RANK() OVER (ORDER BY ITEM_NUMBER) PR,\nITEM_NUMBER,\nITEM,\nPERCENT\nFROM\n(SELECT\nROWNUM ITEM_NUMBER,\nREGEXP_SUBSTR(ITEMS,'\\w+',1,LEVEL) ITEM,\nPERCENT\nFROM\n(SELECT\n'2,8,6,5,54,100,67,7778,6' ITEMS,\nDBMS_RANDOM.VALUE(0,1) PERCENT\nFROM\nDUAL)\nCONNECT BY\nREGEXP_SUBSTR(ITEMS,'\\w+',1,LEVEL) IS NOT NULL);\n\nNUM_ITEMS PR ITEM_NUMBER ITEM PERCENT\n--------- ---------- ----------- ----- ----------\n9 0 1 2 .110718377\n9 .125 2 8 .306241972\n9 .25 3 6 .953005936\n9 .375 4 5 .033518415\n9 .5 5 54 .803485415\n9 .625 6 100 .456278133\n9 .75 7 67 .04461405\n9 .875 8 7778 .249680394\n9 1 9 6 .484834331```\n\nIf we now use a FIRST_VALUE analytic function, we could just retrieve\nthe first PERCENT value and use that in a WHERE clause (the PERCENT_RANK function was the start of another approach which was never developed):\n\n```SELECT\nNUM_ITEMS,\nITEM_NUMBER,\nITEM\nFROM\n(SELECT\nMAX(ITEM_NUMBER) OVER () NUM_ITEMS,\nITEM_NUMBER,\nITEM,\nFIRST_VALUE(PERCENT) OVER () PERCENT\nFROM\n(SELECT\nROWNUM ITEM_NUMBER,\nREGEXP_SUBSTR(ITEMS,'\\w+',1,LEVEL) ITEM,\nPERCENT\nFROM\n(SELECT\n'2,8,6,5,54,100,67,7778,6' ITEMS,\nDBMS_RANDOM.VALUE(0,1) PERCENT\nFROM\nDUAL)\nCONNECT BY\nREGEXP_SUBSTR(ITEMS,'\\w+',1,LEVEL) IS NOT NULL))\nWHERE\nITEM_NUMBER=(TRUNC(NUM_ITEMS*PERCENT)+1);\n\nNUM_ITEMS ITEM_NUMBER ITEM\n--------- ----------- ----\n9 7 67```\n\nOne more, which was originally based on my second solution, this time ordering the rows in random order:\n\n```SELECT\nROWNUM ITEM_NUMBER,\nREGEXP_SUBSTR(ITEMS,'\\w+',1,LEVEL) ITEM\nFROM\n(SELECT\n'2,8,6,5,54,100,67,7778,6' ITEMS\nFROM\nDUAL)\nCONNECT BY\nREGEXP_SUBSTR(ITEMS,'\\w+',1,LEVEL) IS NOT NULL\nORDER BY\nDBMS_RANDOM.VALUE(0,1);\n\nITEM_NUMBER ITEM\n----------- ----\n8 7778\n4 5\n9 6\n6 100\n5 54\n2 8\n7 67\n3 6\n1 2```\n\nNow, just slide the above into an inline view and retrieve just the first row:\n\n```SELECT\nITEM_NUMBER,\nITEM\nFROM\n(SELECT\nROWNUM ITEM_NUMBER,\nREGEXP_SUBSTR(ITEMS,'\\w+',1,LEVEL) ITEM\nFROM\n(SELECT\n'2,8,6,5,54,100,67,7778,6' ITEMS\nFROM\nDUAL)\nCONNECT BY\nREGEXP_SUBSTR(ITEMS,'\\w+',1,LEVEL) IS NOT NULL\nORDER BY\nDBMS_RANDOM.VALUE(0,1))\nWHERE\nROWNUM=1;\n\nITEM_NUMBER ITEM\n----------- ----\n6 100```\n\nThere are probably a couple more ways to pick a random element."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6187144,"math_prob":0.9500588,"size":7935,"snap":"2022-40-2023-06","text_gpt3_token_len":2423,"char_repetition_ratio":0.17223553,"word_repetition_ratio":0.13713267,"special_character_ratio":0.368494,"punctuation_ratio":0.23133071,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9621891,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-03T23:18:15Z\",\"WARC-Record-ID\":\"<urn:uuid:07ab0dd8-a318-4b24-9a0a-cdeb4a1077f4>\",\"Content-Length\":\"102124\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8c0b5611-88e9-476e-ade4-3edbf520ab49>\",\"WARC-Concurrent-To\":\"<urn:uuid:b55668ad-9370-496c-9ba4-5994b8e0a4c1>\",\"WARC-IP-Address\":\"192.0.78.13\",\"WARC-Target-URI\":\"https://hoopercharles.wordpress.com/2009/12/05/\",\"WARC-Payload-Digest\":\"sha1:ZNCCKK5VXQJZLZWHOBSKL4Z3ZE2DFHPV\",\"WARC-Block-Digest\":\"sha1:7GDIQPSKEOBSYJDCHKYDOJ7FQHP6YZBV\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500076.87_warc_CC-MAIN-20230203221113-20230204011113-00610.warc.gz\"}"} |
https://link.springer.com/chapter/10.1007/978-1-4471-4558-5_4 | [
"# Completeness and Applications\n\n• Dirk van Dalen\nChapter\nPart of the Universitext book series (UTX)\n\n## Abstract\n\nThe other way around—true ⇒ derivable—is studied in a separate chapter. Here Gödel’s completeness theorem: Truth = derivability, is proved following the method of Leon Henkin. Here a bit of set theory comes in, e.g. classes of structures, extensions, etc. The student need not worry, all prerequisites are lined up—consistent, maximally consistent, conservative extension, witness. The key lemma, the model existence lemma, turns out to be a gentle piecing together of structures. The techniques are used to prove a number of results that take us to model theory, such as the compactness theorem: if all finite parts of a theory are consistent, so is the theory itself. The Skolem-Löwenheim theorems provide (for most structures) larger or smaller structures that are logically indistinguishable from the original one (elementary equivalence, elementary extension). Questions like: Is a particular class of structures (say, algebraically close fields) axiomatizable? Which theories are decidable? are handled by model theoretic means. Skolem considered the possibility of introducing a function picking suitable elements in a structure, once ∀x(x,y) has been shown. The basic facts of these Skolem functions are discussed. The famous Herbrand theorem is found in the exercise section. The chapter ends with a new section on ultraproducts, i.e. structures produced by a clever product construction from given structures. An example is a non-standard extension of the natural number system. The topic is on the miraculous side of our logic course, as it shows us that it is perfectly possible to obtain logical results in a logic-free way.\n\n## Keywords\n\nPropositional Logic Compactness Theorem Completeness Theorem Conservative Extension Existential Quantifier\nThese keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.\n\n## References\n\n1. A. Robinson. Non-standard Analysis. North-Holland, Amsterdam, 1965 Google Scholar\n2. K.D. Stroyan, W.A.J. Luxemburg. Introduction to the Theory of Infinitesimals. Academic Press, New York, 1976 Google Scholar"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.85260266,"math_prob":0.85091996,"size":2072,"snap":"2020-34-2020-40","text_gpt3_token_len":447,"char_repetition_ratio":0.11363637,"word_repetition_ratio":0.0,"special_character_ratio":0.19208494,"punctuation_ratio":0.15240642,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9685161,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-27T14:49:23Z\",\"WARC-Record-ID\":\"<urn:uuid:989cbb0b-8ff8-4dd5-848b-ef369f5ae576>\",\"Content-Length\":\"60701\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:10c7e4ce-17f5-4603-9754-561f8afec441>\",\"WARC-Concurrent-To\":\"<urn:uuid:4cb6db94-1b21-49df-a4da-f02cd4ba2574>\",\"WARC-IP-Address\":\"199.232.64.95\",\"WARC-Target-URI\":\"https://link.springer.com/chapter/10.1007/978-1-4471-4558-5_4\",\"WARC-Payload-Digest\":\"sha1:42VZ5FK77MVOUO5NGMDOYAZG5SBKGTNI\",\"WARC-Block-Digest\":\"sha1:3GLTFMBTYLWQFMWGZDSEPNZSLIZGKO6H\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400279782.77_warc_CC-MAIN-20200927121105-20200927151105-00331.warc.gz\"}"} |
https://eng.kakprosto.ru/how-127636-in-excel-calculate-time | [
"Instruction\n1\nIf you are first faced with the task of calculating the time, try not to resort immediately to complex functions. Most of the computation can be issued by a simple mathematical formula: addition and subtraction. And be sure to set the cells in the correct format.\n2\nTo do this, select the range of cells to be used as input data, and click the selection with the right mouse button or open the Home tab and click the block instruments \"Cell\" button the \"Format\". In the first and in the second case, select the context menu \"Format cells\".\n3\nOpens a new dialog box, make sure that you are on the tab \"Number.\" In the list located in the left part of the window, scroll the left click. In the right part of the window indicate the form should be submitted time. Click OK to confirm your choice.",
null,
"4\nEnter data into cells in the format XX:XX (or XX:XX:XX). To calculate the time interval between the first and second value, set the cursor in the cell where the total value. In the formula bar an equal sign, click the cell with the last value with the left mouse button, enter a minus sign, then click on the cell with the first value. Press the Enter key.\n5\nFor example: the time range from 11:10:15 to 12:30:00 is 1 hour, 19 minutes and 45 seconds. The resulting value in cell C2 is calculated by the formula: =B2-A2. If you need to calculate the amount of time in several rows (A3 and B3, A4 and B4, and so on), just drag the AutoFill handle from cell S2 to the end column of your table (C3, C4).",
null,
"6\nIf you need to calculate how much in total hours, minutes and seconds it took for anything, just fold the data obtained as described above by using the sum function. Put the cursor in the cell with the total value, click the fx button in the window that appears, select the left mouse button the sum function and select the range of cells that will participate in the calculation. Press the Enter key. The formula for calculation will look like this: =SUM(C2:C4).",
null,
""
] | [
null,
"https://st03.kakprosto.ru/tumb/680/images/article/2012/4/10/1_52550d77ce6f352550d77ce731.jpg",
null,
"https://st03.kakprosto.ru/tumb/680/images/article/2012/4/10/1_52550d77d6a7452550d77d6ab1.jpg",
null,
"https://st03.kakprosto.ru/tumb/680/images/article/2012/4/10/1_52550d77dcb5452550d77dcba1.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89377207,"math_prob":0.9514425,"size":1975,"snap":"2022-27-2022-33","text_gpt3_token_len":464,"char_repetition_ratio":0.14814815,"word_repetition_ratio":0.016438356,"special_character_ratio":0.23949367,"punctuation_ratio":0.11860465,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97020966,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-15T04:43:34Z\",\"WARC-Record-ID\":\"<urn:uuid:c72a551b-294a-485b-9251-11cc11c4d91f>\",\"Content-Length\":\"28836\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2fe74e81-f6d0-4c86-904e-4977a433765c>\",\"WARC-Concurrent-To\":\"<urn:uuid:4cb92937-4437-4388-827d-1c780483c6fa>\",\"WARC-IP-Address\":\"178.154.246.3\",\"WARC-Target-URI\":\"https://eng.kakprosto.ru/how-127636-in-excel-calculate-time\",\"WARC-Payload-Digest\":\"sha1:M5ZL4K2MNPGVZYYPVKYEFXTGLRE6QBLI\",\"WARC-Block-Digest\":\"sha1:TZLRW664JHG3RTDCELCPNEGR7BB2L3UK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572127.33_warc_CC-MAIN-20220815024523-20220815054523-00505.warc.gz\"}"} |
https://reference.wolframcloud.com/language/ref/ConstantArray.html | [
"# ConstantArray\n\nConstantArray[c,n]\n\ngenerates a list of n copies of the element c.\n\nConstantArray[c,{n1,n2,}]\n\ngenerates an",
null,
"array of nested lists containing copies of the element c.\n\n# Details",
null,
"# Examples\n\nopen allclose all\n\n## Basic Examples(3)\n\nMake a vector of 10 c's:\n\nMake a 3×4 array of c's:\n\nMake an array out of blocks:\n\n## Scope(4)\n\nA machine-zero vector:\n\nA vector of exact zeros:\n\nA machine-zero matrix:\n\nA matrix of exact zeros:\n\nA vector of ones:\n\nA machine-number vector of ones:\n\nA deeply nested constant array:\n\n## Applications(3)\n\nSet up an array to use as storage in a program:\n\nNumerically solve x'[t]==1/(t+x[t]^2) using Euler's method:\n\nFind the sum of the elements in a list:\n\nThis can also be done with Total:\n\nUse a vector of ones to see the effects of conditioning on the solution of a linear system:\n\nThe solution at machine precision has significant errors:",
null,
"At higher precision, the solution is much better:\n\n## Properties & Relations(3)\n\nConstantArray[c,dims] has Dimensions[dims]:\n\nConstantArray[c,dims] is equal to SparseArray[{},dims,c]:\n\nNormal[s] is identical to ConstantArray[c,dims]:\n\nConstantArray[c,dims] is equivalent to Apply[Table[c,##]&,Map[List,dims]]:"
] | [
null,
"https://reference.wolframcloud.com/language/ref/Files/ConstantArray.en/1.png",
null,
"https://reference.wolframcloud.com/language/ref/Files/ConstantArray.en/details_1.png",
null,
"https://reference.wolframcloud.com/language/ref/Files/ConstantArray.en/2.gif",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.599758,"math_prob":0.94835687,"size":1420,"snap":"2022-27-2022-33","text_gpt3_token_len":345,"char_repetition_ratio":0.19067797,"word_repetition_ratio":0.037037037,"special_character_ratio":0.2302817,"punctuation_ratio":0.20446096,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9954891,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-27T23:15:13Z\",\"WARC-Record-ID\":\"<urn:uuid:5ad3ab4d-6c5e-4a2d-a6b7-19c5f79190e9>\",\"Content-Length\":\"99072\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0140fe25-9c82-4226-bbc6-f76b3c4e5f52>\",\"WARC-Concurrent-To\":\"<urn:uuid:78858058-dd8a-4e84-a987-3dc82be7a9e8>\",\"WARC-IP-Address\":\"140.177.204.22\",\"WARC-Target-URI\":\"https://reference.wolframcloud.com/language/ref/ConstantArray.html\",\"WARC-Payload-Digest\":\"sha1:KYKB6LPV22VDKZEQIZ7KALYL762DIIKT\",\"WARC-Block-Digest\":\"sha1:WKYGQTSHTULOBSP4IQKUMJ2NPXQI7WCJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103344783.24_warc_CC-MAIN-20220627225823-20220628015823-00759.warc.gz\"}"} |
https://www.unibo.it/en/teaching/course-unit-catalogue/course-unit/2018/324637 | [
"# 30216 - Probability Models\n\n## Learning outcomes\n\nAt the end of the course the student knows some advanced probability theories with application to computer science, such as Markov chains with discrete and continuous time. He is able to analyze some simple stochastic systems such some with application to biology.\n\n## Course contents\n\nDenumerable additivity. One-dimensional random walk. Generating function. Gamblers' ruin problem. Galton Watson processes. Markov chains. Recurrent and transient states. Stationary distributions. Reversible Markov chains. Gibbs sampler. Metropolis algorithm. Markov chains with continuous time. Poisson process. Pure birth processes. Semi-Markov processes. Queueing processes. Queueing Markov processes.. Open and closed systems of queues. Jackson's property.\n\nS. Ross. Introduction to Probability Models. Academic Press.\nW. Feller.An Introduction to Probability Theory and Its Applications. I Vol.. Wiley.\n\nLectures.\n\n## Assessment methods\n\nFinal verification consists in an oral test.\n\nOral test consists in a talk, starting from three questions, with the goal of testing the understanding of the basic concepts of the course, the ability of solving simple exercises and of developing logical arguments.\n\n## Teaching tools\n\nLectures.\n\nThe course is based on lectures in which a series of probability models that are relevant for applications to computer science will be illustrated with examples of their applications and the development of simple exercises in order to familiarize students with concrete application of the introduced mathematical models.\n\n## Office hours\n\nSee the website of Massimo Campanino"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8754502,"math_prob":0.57821476,"size":1834,"snap":"2019-43-2019-47","text_gpt3_token_len":388,"char_repetition_ratio":0.11420765,"word_repetition_ratio":0.0,"special_character_ratio":0.18484187,"punctuation_ratio":0.14144737,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9794639,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-21T03:40:13Z\",\"WARC-Record-ID\":\"<urn:uuid:8b89c27b-1143-41e9-95f1-a8d5354f1f6b>\",\"Content-Length\":\"64923\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a82960b7-36f1-4776-a7d8-77e247a73c57>\",\"WARC-Concurrent-To\":\"<urn:uuid:d54b78fe-7adc-466a-85b0-e25a0c8cb5de>\",\"WARC-IP-Address\":\"137.204.24.35\",\"WARC-Target-URI\":\"https://www.unibo.it/en/teaching/course-unit-catalogue/course-unit/2018/324637\",\"WARC-Payload-Digest\":\"sha1:5JO7I665IZVASLJAUDUSIJIOWQ2JIZJU\",\"WARC-Block-Digest\":\"sha1:4PPP6SIMMCN3KEQM7WYZNM72YOD3AV46\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670729.90_warc_CC-MAIN-20191121023525-20191121051525-00417.warc.gz\"}"} |
https://www.stat.math.ethz.ch/pipermail/r-help/2004-February/046420.html | [
"# [R] Question concerning functions nlsList and nlme from nlme R library.\n\nJens_Praestgaard@hgsi.com Jens_Praestgaard at hgsi.com\nMon Feb 23 14:38:18 CET 2004\n\n```I hope that the mailing list is the correct forum for the question below. I\nhave trouble calling functions nlsList and nlme from\nanother function. Any help would be greatly appreciated.\n\nJens Praestgaard\nHuman Genome Sciences\nRockville MD.\n\nI have a data set v with two components, v\\$mixeddat and v\\$init. They are\nlisted below:\n\n> v\\$mixeddat\nconc result rep sample z\n11 20.00000000 11141 1 vial 1 0\n12 5.00000000 11446 1 vial 1 0\n13 1.25000000 13377 1 vial 1 0\n14 0.31250000 20267 1 vial 1 0\n15 0.07812500 48852 1 vial 1 0\n16 0.01953125 118507 1 vial 1 0\n\n(and so on)\n\n> v\\$init\na b d c cdiff\n1.045617e+04 2.408045e+05 9.296929e-01 1.490621e-02 4.874540e-03\n\nThe following two lines of code fits first a nonlinear regression for each\nlevel of rep, next a model with a random effect in one parameter.\n\n> fit<-nlsList(result~a+(b-a)/(1+(conc/(c+z*cdiff))^d)\n|rep,start=v\\$init,data=v\\$mixeddat)\n> nlme(fit,random=(b~1))\n\nHowever, when I try to run the same code from within a function, with this\ncall :\n\n> testfunc\nfunction(dat=v) {\ntest<-nlsList(result~a+(b-a)/(1+(conc/(c+z*cdiff))^d)\n|rep,start=dat\\$init,data=dat\\$mixeddat)\nreturn(nlme(test,random=b~1))\n}\n\ntestfunc()\n\nthen I get the error message"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5649479,"math_prob":0.85279286,"size":1479,"snap":"2022-40-2023-06","text_gpt3_token_len":518,"char_repetition_ratio":0.107118644,"word_repetition_ratio":0.009090909,"special_character_ratio":0.36105478,"punctuation_ratio":0.11419753,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96645653,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-28T00:21:26Z\",\"WARC-Record-ID\":\"<urn:uuid:e6b8cfab-c28f-4793-9980-fa422fb9912d>\",\"Content-Length\":\"4125\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3e5e7f31-f938-4141-a586-03ce487600eb>\",\"WARC-Concurrent-To\":\"<urn:uuid:39b246d1-2959-46b6-b10f-fd03248ecc35>\",\"WARC-IP-Address\":\"129.132.119.195\",\"WARC-Target-URI\":\"https://www.stat.math.ethz.ch/pipermail/r-help/2004-February/046420.html\",\"WARC-Payload-Digest\":\"sha1:ZLNDGKDNTSOQGZIKCVE2ASB25CE4E67K\",\"WARC-Block-Digest\":\"sha1:IDWSGJU6V7GHYWOHGBUW75CNCMMAF2SS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499468.22_warc_CC-MAIN-20230127231443-20230128021443-00280.warc.gz\"}"} |
https://artofproblemsolving.com/wiki/index.php?title=Pascal%27s_triangle&diff=104261&oldid=4271 | [
"# Difference between revisions of \"Pascal's triangle\"\n\nPascal's triangle is a triangle which contains the values from the binomial expansion; its various properties play a large role in combinatorics.\n\n## Properties\n\n### Binomial coefficients\n\nPascal's Triangle is defined such that the number in row",
null,
"$n$ and column",
null,
"$k$ is",
null,
"${n\\choose k}$. For this reason, convention holds that both row numbers and column numbers start with 0. Thus, the apex of the triangle is row 0, and the first number in each row is column 0. As an example, the number in row 4, column 2 is",
null,
"${4 \\choose 2} = 6$. Pascal's Triangle thus can serve as a \"look-up table\" for binomial expansion values. Also, many of the characteristics of Pascal's Triangle are derived from combinatorial identities; for example, because",
null,
"$\\sum_{k=0}^{n}{{n \\choose k}}=2^n$, the sum of the values on row",
null,
"$n$ of Pascal's Triangle is",
null,
"$2^n$.\n\n### Sum of previous values\n\nOne of the best known features of Pascal's Triangle is derived from the combinatorics identity",
null,
"${n \\choose k}+{n \\choose k+1} = {n+1 \\choose k+1}$. Thus, any number in the interior of Pascal's Triangle will be the sum of the two numbers appearing above it. For example,",
null,
"${5 \\choose 1}+{5 \\choose 2} = 5 + 10 = 15 = {6 \\choose 2}$. This property allows the easy creation of the first few rows of Pascal's Triangle without having to calculate out each binomial expansion.\n\n### Fibonacci numbers\n\nThe Fibonacci numbers appear in Pascal's Triangle along the \"shallow diagonals.\" That is,",
null,
"${n \\choose 0}+{n-1 \\choose 1}+\\cdots+{n-\\left\\lfloor\\frac{n}{2}\\right\\rfloor \\choose \\left\\lfloor \\frac{n}{2}\\right \\rfloor} = F(n+1)$, where",
null,
"$F(n)$ is the Fibonacci sequence. For example,",
null,
"${6 \\choose 0}+{5 \\choose 1}+{4 \\choose 2}+{3 \\choose 3} = 1 + 5 + 6 + 1 = 13 = F(7)$. A \"shallow diagonal\" is plotted in the diagram.\n\n### Hockey-stick theorem\n\nThe Hockey-stick theorem states:",
null,
"${n \\choose 0}+{n+1 \\choose 1}+\\cdots+{n+k \\choose k} = {n+k+1 \\choose k}$. Its name is due to the \"hockey-stick\" which appears when the numbers are plotted on Pascal's Triangle, as shown in the representation of the theorem below (where",
null,
"$n=2$ and",
null,
"$k=3$).",
null,
"$[asy] int chew(int n,int r){ int res=1; for(int i=0;i\n\n### Number Parity\n\nConsider writing the row number",
null,
"$n$ in base two as",
null,
"$({n})_{10} = {(a_xa_{x-1} \\cdots a_1a_0)}_2$",
null,
"$= a_x 2^x+a_{x-1} 2^{x-1}+\\cdots+a_1 2^1+a_0 2^0$. The number in the",
null,
"$k$th column of the",
null,
"$n$th row in Pascal's Triangle is odd if and only if",
null,
"$k$ can be expressed as the sum of some",
null,
"$a_i 2^i$. For example,",
null,
"$(9)_{10} = {(1001)}_{2} = 2^{3}+2^{0}$. Thus, the only 4 odd numbers in the 9th row will be in the",
null,
"${(0000)}_{2} = 0$th,",
null,
"${(0001)}_{2} = 2^0 = 1$st,",
null,
"${(1000)}_{2} = 2^3 = 8$th, and",
null,
"${(1001)}_{2} = 2^3+2^0 = 9$th columns. Additionally, marking each of these odd numbers in Pascal's Triangle creates a Sierpinski triangle.\n\n### Patterns and Properties of the Pascal's Triangle\n\n#### Rows\n\nThe zeroth row has a sum of",
null,
"$1=2^0$. The first row has a sum of",
null,
"$2=2^1$. The",
null,
"$n^{th}$ row has a sum of",
null,
"$2^n$\n\n### Diagonals\n\nThe 1st downward diagonal is a row of 1's, the 2nd downward diagonal on each side consists of the natural numbers, the 3rd diagonal the triangular numbers, and the 4th the pyramidal numbers."
] | [
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null,
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null,
"https://latex.artofproblemsolving.com/f/4/a/f4ac00bc3d46b6c0ed9388629cfa62947c3a5dec.png ",
null,
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null,
"https://latex.artofproblemsolving.com/2/f/a/2fa62a365c27d4f2c298262b3e8c7c224550c20a.png ",
null,
"https://latex.artofproblemsolving.com/6/c/3/6c36f0b157e8c2fbe0638c2f2b378e08d07695aa.png ",
null
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https://meetings3.sis-statistica.org/index.php/sis2018/50th/paper/view/1176 | [
"## Open Conference Systems, 50th Scientific meeting of the Italian Statistical Society\n\nFont Size:\nParametric Modeling of Quantile Regression Coefficient Functions\nPaolo Frumento, Matteo Bottai\n\n#### Abstract\n\nIn standard quantile regression (QR), quantiles are estimated one at the time. An alternative approach, which is referred to as quantile regression coefficients modeling (QRCM), is to describe the functional form of the regression coefficients parametrically. This approach facilitates estimation and inference, simplifies the interpretation of the results, and generates more efficient estimators. Moreover, thanks to the imposed parametric structure, it makes it easier to estimate quantiles in situations involving latent variables, missing or partially observed data, and other complications arising in survival analysis, longitudinal data analysis, and causal inference, where applying standard QR proves difficult and computationally inefficient. We describe the method, discuss applications, and illustrate the R package qrcm.\n\n#### References\n\nFrumento P, Bottai M (2016). Parametric modeling of quantile regression coefficient functions. Biometrics, 72 (1), 74-84, doi: 10.1111/biom.12410.\n\nFrumento P, Bottai M (2017). Parametric modeling of quantile regression coefficient functions with censored and truncated data. Biometrics,, 73(4), 1179-1188, doi: 10.1111/biom.12675.\n\nFrumento P (2017). qrcm: Quantile Regression Coefficients Modeling. R package version 2.1. https://cran.r-project.org/package=qrcm\n\nFull Text: PDF"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7546231,"math_prob":0.74779046,"size":1341,"snap":"2021-31-2021-39","text_gpt3_token_len":303,"char_repetition_ratio":0.13462977,"word_repetition_ratio":0.024242423,"special_character_ratio":0.22669649,"punctuation_ratio":0.20762712,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9847781,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-17T04:51:48Z\",\"WARC-Record-ID\":\"<urn:uuid:4e6ec083-baa1-4456-a744-71a5201a7c54>\",\"Content-Length\":\"13416\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0df27d65-e579-4f18-903d-31216eefc77b>\",\"WARC-Concurrent-To\":\"<urn:uuid:8b369a31-2fb2-4104-aeb9-f00d05029135>\",\"WARC-IP-Address\":\"66.147.244.56\",\"WARC-Target-URI\":\"https://meetings3.sis-statistica.org/index.php/sis2018/50th/paper/view/1176\",\"WARC-Payload-Digest\":\"sha1:FKLHFKKHNV25YUCVFGJNV7PGXBXH4JYP\",\"WARC-Block-Digest\":\"sha1:D7EQHWFQ6CDNKME2K3QX5YEEFXU2NVFL\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780054023.35_warc_CC-MAIN-20210917024943-20210917054943-00699.warc.gz\"}"} |
http://www.wordsearchfun.com/199638_John__Glendas_50th_Anniversary__wordsearch.html | [
"John & Glenda's 50th Anniversary\nLook for words that describe John & Glenda's 50 years together.\n\nLogin to be the first to rate this puzzle!\nANNIVERSARY\nCHARLESAVE\nCOMFORT\nCRACKERBARREL\nDAVID\nEXTRATVREMOTE\nFAITHFUL\nFAMILY\nFLOETROL\nFLOWERS\nFORGIVENESS\nGLENN\nGOD\nGOLDEN\nGOSPELCENTER\nGRAMMY\nGRAMPY\nHONOUR\nICECREAM\nLAWN\nLOVE\nMARTIN\nNOTREDAME\nPAINT\nPENETROL\nPEPPER\nSIMONBROTHERS\n F N N E L G B L U F H T I A F T Y R F O R G I V E N E S S L Y L E R R A B R E K C A R C O P Y I R T N I A P H H E O G W M L D U E V O L A S H M N R E A I A O O Z N R N T F O I A R R M V N L C L E O O S W T M S G A I O F E D R R P W K R M U P F D H S L B T P N Z S A Y H E C E A O N O T R E D A M E Y P W V G O S P E L C E N T E R P E O M J M A E R C E C I J D E D I F Y R A S R E V I N N A R S W P E N E T R O L A W N L E X T R A T V R E M O T E K C"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7059392,"math_prob":0.61350006,"size":1317,"snap":"2019-51-2020-05","text_gpt3_token_len":600,"char_repetition_ratio":0.30159938,"word_repetition_ratio":0.21756487,"special_character_ratio":0.5611238,"punctuation_ratio":0.007142857,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99219155,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-17T16:44:49Z\",\"WARC-Record-ID\":\"<urn:uuid:4fbe676a-4d95-43db-95b0-f90ff44cf246>\",\"Content-Length\":\"23783\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e9dbc20d-1d92-49db-8e5f-c8cf9c4db70b>\",\"WARC-Concurrent-To\":\"<urn:uuid:a4e73d71-ad13-4114-a4bd-209ca3ca6026>\",\"WARC-IP-Address\":\"88.208.252.230\",\"WARC-Target-URI\":\"http://www.wordsearchfun.com/199638_John__Glendas_50th_Anniversary__wordsearch.html\",\"WARC-Payload-Digest\":\"sha1:AG4RJGNU6LQSDNTFFRG7OTMZJUVCM3Q2\",\"WARC-Block-Digest\":\"sha1:Z2JHNOTWE7TVJ57OXBWCNXMFKF2WAGKH\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250589861.0_warc_CC-MAIN-20200117152059-20200117180059-00263.warc.gz\"}"} |
https://www.mindfulnessfordogs.com/2021/09/100-times-table-problems/ | [
"# 100 Times Table Problems",
null,
"times worksheets worksheet fun and printable conventional times table math worksheets these input output tables worksheet worksheet idea template multiplication worksheet 100 multiplication chart magdalene project create a times table to memorize in excel division table chart worksheet fun and printable 041 worksheet multiplication math multiplying twot numbers measurement maths teaching resources math games worksheets – tedpaper\n\n1.",
null,
"2.",
null,
"3.",
null,
"4.",
null,
"5.",
null,
"6.",
null,
"7.",
null,
"8.",
null,
"9.",
null,
"10.",
null,
"11.",
null,
"12.",
null,
"13.",
null,
"14.",
null,
"15.",
null,
"16.",
null,
"17.",
null,
"18.",
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"19.",
null,
"20.",
null,
"21.",
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"22.",
null,
"23.",
null,
"24.",
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"25.",
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"26.",
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"27.",
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"28.",
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"29.",
null,
"30.",
null,
""
] | [
null,
"https://www.mindfulnessfordogs.com/2021/09/100-times-table-problems/",
null,
"https://secure.gravatar.com/avatar/c79a845db6c95985d00b67dd21119e8f",
null,
"https://secure.gravatar.com/avatar/791f7ac3a012243ed1416e0f6fe5374b",
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null
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https://nykojumunoki.usainteriordesigners.com/writing-algebraic-expressions-worksheets-8th-grade38688770ph.html | [
"# Writing algebraic expressions worksheets 8th grade\n\nI have one more math class to take and I am sure I will put the Algebrator to good use then!",
null,
"Sheet 1 Sheet 2 Sheet 3 Explore Transversal Worksheets Statistics and Probability Grade eight statistics worksheets are based on finding the mean, median, mode and range in each problem. Roster and rule method in algebra, ti plus AND sixth root, java convert decimal time, finding the range of an absolute value graph.\n\nReady-made worksheets Use these quick links to create some common types of worksheets for writing expressions. Sheet 1 Sheet 2 Sheet 3 Evaluate the expressions - Function table.\n\nWhat is the difference between evaluation and simplication of an expression, subtraction worksheets ks2, sloving math problem with excel, algebra fundamentals of math teasers. Venn diagrams gcse bitesize, is the square root of -1 and interger, 9th Grade Math Practice Worksheet, kinds of radical symbol.\n\nCollege algebra software reviews, Integer Worksheets, factoring the difference of two squares calculator, algebrator for students, what is mathimatics and the history of this.\n\nAlgebraic Expressions Simplifying Variables. Sheet 1 Sheet 2 Sheet 3 Express in scientific notation - Negative exponents.\n\nSheet 1 Sheet 2 Sheet 3 Probability on deck of cards.\n\nRated 7/10 based on 103 review"
] | [
null,
"http://valtteripuhakka.me/wp-content/uploads/2018/06/year-5-algebra-worksheets-writing-algebraic-expressions-6th-grade-6-ncert.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8368485,"math_prob":0.9506829,"size":1245,"snap":"2020-24-2020-29","text_gpt3_token_len":265,"char_repetition_ratio":0.1410153,"word_repetition_ratio":0.041237112,"special_character_ratio":0.19437751,"punctuation_ratio":0.11659193,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99465984,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-03T10:43:35Z\",\"WARC-Record-ID\":\"<urn:uuid:dc514d2e-b5cb-4077-9f2f-cb5e4e698ea7>\",\"Content-Length\":\"4692\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:76c739a5-d032-4b3e-afae-a6862eadd69a>\",\"WARC-Concurrent-To\":\"<urn:uuid:5a194e9c-f40e-476b-ac9d-70fea5adcf0c>\",\"WARC-IP-Address\":\"104.18.52.239\",\"WARC-Target-URI\":\"https://nykojumunoki.usainteriordesigners.com/writing-algebraic-expressions-worksheets-8th-grade38688770ph.html\",\"WARC-Payload-Digest\":\"sha1:JRYAP33DTS4FM24D7WZFJHBIX5SC6YA2\",\"WARC-Block-Digest\":\"sha1:LA5EASSTTJEPCKOGDUID3WCQ2ONFGUYK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655881984.34_warc_CC-MAIN-20200703091148-20200703121148-00252.warc.gz\"}"} |
https://www.gcflcm.com/gcf-of-8-and-18 | [
"# What is the Greatest Common Factor of 8 and 18?\n\nGreatest common factor (GCF) of 8 and 18 is 2.\n\nGCF(8,18) = 2\n\nWe will now calculate the prime factors of 8 and 18, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 8 and 18.\n\nGCF Calculator and\nand\n\n## How to find the GCF of 8 and 18?\n\nWe will first find the prime factorization of 8 and 18. After we will calculate the factors of 8 and 18 and find the biggest common factor number .\n\n### Step-1: Prime Factorization of 8\n\nPrime factors of 8 are 2. Prime factorization of 8 in exponential form is:\n\n8 = 23\n\n### Step-2: Prime Factorization of 18\n\nPrime factors of 18 are 2, 3. Prime factorization of 18 in exponential form is:\n\n18 = 21 × 32\n\n### Step-3: Factors of 8\n\nList of positive integer factors of 8 that divides 8 without a remainder.\n\n1, 2, 4\n\n### Step-4: Factors of 18\n\nList of positive integer factors of 18 that divides 8 without a remainder.\n\n1, 2, 3, 6, 9\n\n#### Final Step: Biggest Common Factor Number\n\nWe found the factors and prime factorization of 8 and 18. The biggest common factor number is the GCF number.\nSo the greatest common factor 8 and 18 is 2.\n\nAlso check out the Least Common Multiple of 8 and 18"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8991576,"math_prob":0.9375994,"size":1182,"snap":"2021-31-2021-39","text_gpt3_token_len":345,"char_repetition_ratio":0.2156197,"word_repetition_ratio":0.07929515,"special_character_ratio":0.29780033,"punctuation_ratio":0.11111111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998405,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-28T21:58:08Z\",\"WARC-Record-ID\":\"<urn:uuid:5fed10ae-19ad-4ccf-b1b8-2282645cbaaf>\",\"Content-Length\":\"20159\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8c346009-a137-4e5e-8883-7ecdcdf4ec70>\",\"WARC-Concurrent-To\":\"<urn:uuid:8f9ebce8-154f-4f22-917e-301c76e71192>\",\"WARC-IP-Address\":\"34.133.163.157\",\"WARC-Target-URI\":\"https://www.gcflcm.com/gcf-of-8-and-18\",\"WARC-Payload-Digest\":\"sha1:PM6BHVMZ7HWZ4BB2QDDLMGXAQH2K7ARK\",\"WARC-Block-Digest\":\"sha1:ZDARRBHHENLJF5IY25LT5GFRI5BYYQUW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780060908.47_warc_CC-MAIN-20210928214438-20210929004438-00012.warc.gz\"}"} |
https://www.scienceforums.net/profile/1591-k9-47g/content/?all_activity=1 | [
"",
null,
"K9-47G\n\nSenior Members\n\n39\n\nEverything posted by K9-47G\n\n1. No, it's not a novel. He talks a lot about findings from the Human Genome Project. I suppose as far as books are concerned it's comparable to much of Dawkin's works. Informational, but with a more laid-back approach.\n2. Has anyone here read it? What did you think of it?\n3. This is one optimization problem that I just cant figure out. I'll post what I have... A hiker at point A on a straight road wants to reach, in the shortest time, a point B located 6 miles from the road and 10 miles from point A. The hiker's speed on the paved road is 4 mph and only 2 mph off the road. How far should he continue on the road before heading in a straight line for the point B? I am pretty sure I would have to use the pythagorean theorem because if you draw the problem you get a triange with two sides given. Plus I denoted $dr/dt$ to be the speed on the road which is 4 mph, and $do/dt$ to be the speed off road which is 2 mph. I just don't know how to find my objective function. Any help would be appreciated.\n4. I noticed that if I type .9999999999 (ten nines) into my TI-83 calculator and press enter, it gives me the answer to be .9999999999 (ten nines), But if I type .99999999999 (eleven nines) into my calculator and press enter it gives the answer to be one. I suppose my TI-83 rounds to the 10th decimal place.\n5. For number 3, I thought I would use the logarithmic power rule (not sure of the real name) and therefore the exponent, sinx, can be written as the first term in problem. Then I used the product rule to find the derivative.. $y= (\\ln x)^{\\sin x}$ is the same as $\\sin x\\ln x$\n6. Ok, thanks a lot.\n7. Can you tell if those answers are right?\n8. 1) Find $\\frac{d}{dx} log(lnx)$ I assume that the log has a base of 10, so I got $\\frac{1}{x(lnxln10)}$ 2) Find the slope of the line tangent to the graph $cos(xy)=y$ at $(0,1)$ $-sin(xy)(y)+(xy')=y'$ $-ysin(xy)=y'-(xy')$ $\\frac{-ysin(xy)}{1-x}=y'$ Then I just keep getting 0 when I substitute (0,1) in... 3) If $y=(lnx)^{sinx} x>1,$ Find $y'$ $sinxlnx=sinx\\frac{1}{x}+(cosx)(lnx)$ $\\frac{sinx}{x} +cosxlnx$ $1+cosxlnx$\n9. I think the second one has to do with the purple squares in the background.\n10. This problem reminds me of the many examples that Richard Dawkins gave in his book, The Selfish Gene.\n11. Can you please check if my answer is correct. $y=sin(sin(sinx))$ $y'=cos(sin(sinx))cos(sinx)cosx$ (My calculus professor doesn't want our answers simplified.)\n12. $g^{(n)}(x)=-n(-1^n)e^{-x}+(-1^n)xe^{-x}$\n13. How do you format your math work to look bold and easier to read?\n14. So would my formula be g^n(x)= -n(-1^n)e^-x+(-1^n)xe^-x. I know there must be an easier way to write that.\n15. The inductive proof is: show true for n=1, assume true for n=k and show true for n=k+1, right?\n16. so e^-x derived is -e^-x.... We haven't covered that yet. Now I'm getting g'(x)=e^-x-xe^-x g''(x)=-2e^-x+xe^-x g'''(x)= 3e^-x-xe^-x and so on, but I have no idea how to make an explicit formula out of that because the negatives are alternating.\n17. well I did write it out, this is what I got. g'(x)= xe^-x+e^-x g\"(x)= xe^-x+2e^-x g'''(x)= xe^-x+3e^-x so I concluded that the formula would be g^n(x)= g(x)+n(e^-x). I'm just not sure if my math is right.\n18. So would it be g^n(x)=g(x)+n(e^-x)..... I'm assuming that the derivative of e^-x is e^-x....\n19. This is the problem in my book. If g(x)=x/e^x, find g^(n)(x). I don't really understand what the problem is asking me to find. It is in the differentiation section of the book, if that helps at all. I think it may be asking for a formula... By the way, the n in the formula represents how many times to take the derivative of g(x).\n20. Thank you so much!\n21. Can someone tell me the limit of (x/(2x-2))-(1/((x^2)-1)) as x approaches 1.\n×\n×\n• Create New..."
] | [
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"https://www.scienceforums.net/uploads/set_resources_1/84c1e40ea0e759e3f1505eb1788ddf3c_pattern.png",
null
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http://www.gratisbo.com/basicnbc/article12radiological71.htm | [
"*Modern Dosimetry of Gamma Emitters*\n01 October 2008\n\nModern Dosimetry of Gamma emitters\n\nRadiation Dosimetry is the measurement of radiation. We can do this the old way, or the easy way.\n\nIn the bad old days, radioactive materials were quantified in many unusual and interesting units. The number of disintegrations per unit time was specified in ?i>Curies? (Ci) with one Ci equal to the activity of a gram of Radium. Different isotopes were related to each other using the milligram Radium equivalent “mgREq? The milligram Radium equivalent of a mass of some material, whether it was 60Co, 137Cs or what ever was determined by comparing that isotopes exposure rate constant to that of Radium. The exposure rate constant was a measurement of the amount of ionization produced by the isotope, and was measured in Roentgens (R). Terms like exposure rate constant, gamma rate or gamma ray constant, or even k factor were used to describe the amount of ionization created by an isotope. Exposure was then related to dose in RAD by the Roentgen to RAD conversion factor, also known as the \"f factor\" which conveniently is close to 1. [Wass 1988]\n\nThe new world is full of busy people doing busy stuff, and generally being busy-bodies.\n\nBecause of this, there has been a general move away from the complicated systems of specifying dose to a simple method, which is also more accurate.\n\nThe dose that an isotope will delver to a mass of air is termed the air kerma strength, and this is nowadays indicated either by Sk or Kair. The word kerma is an anagram, indicating “Kinetic Energy Released in Matter? is the kinetic energy of all the charged particles produced in air by the isotope. The air kerma strength is specified in terms of “Units? which are given the symbol U. (Yes, the unit is called a “unit?. 1 Unit is equal to a micro-gray per hour at one meter from the source, or about 0.36 RAD per second at one meter. [Nath 2005]\n\nNow people aren’t made out of air, but are mainly water, so dose to water is the usual modern convention. The dose rate constant, is the ratio of the absorption of a mass of water to the absorption of a mass of air. The dose rate constant has the symbol of a capital lambda (Λ).\n\nThe crude way of thinking about it is that the Air kerma strength is the “dose?that the air would receive, and the dose rate constant is the conversion between air and water. Its natural to ask, why bother with air kerma anyway??Well, the National Institute of Standards uses some thing called a “Wide angle free air chamber?(WAFAC) that very accurately measures the air kerma.\n\nThe more modern approach is closer to what we actually want: the dose, instead of the exposure which is the amount of charge. We want RADS to something, and its easier to get that from rads to air and then water, than from ion charge produced and energy of ion pairs etc etc.\n\nThe dose D delivered per hour at a distance d from a small point of radioactive gamma emitter is:\n\nD = (Air Kerma) (Dose rate constant) /(distance squared)?=?Sk Λ /d2 (1\n\nSurprisingly, this little formula works very well for many isotopes. An example from the medical field is the use of an Iridium isotope 192 Ir for cancer treatment. The computer programs used to plan a treatment will produce one number, and the simple formula equation 1) are a good check of the results. Almost all the time the agreement between the complicated computer calculation and the hand result is 2% or better.",
null,
"Figure 1 A simple spreadsheet was used to calculate a radiation dose using the formula above, and a more complicated computer result. The little equation is always with a few percent.\n\nFigure 1 shows a screen capture of a program I used the other day for a medical treatment.\n\n?/p>\n\nReferences\n\nOnline\n\n[Nath 2005] Nath, R. Yale University Summer School\n\n[Was 1988] Wasserman, H. Air Kerma rate constants for radionuclides. European Journal of Nuclear Medicaine 1988 14 pages 569-571\n\n[Gal 2004] Gamma Dose: a User Friendly Module for Dosimetry and Shielding Calculations\nhttp://www.nucleonica.net/TC/TC0906/relevant_papers/Nuclides_net_and_dosimetry.pdf\n\nAlso, an online refereence her, for background reading.\n\nOffline\n\n[Thom 2005] Thomadsen, B.R et al. Brachytherapy Physics, 2nd Edition Proceedings of the Joint American Association of Physicists in medicine/American Brachytherapy Society Summer School 2005."
] | [
null,
"http://www.gratisbo.com/basicnbc/images/article12radiological7101.gif",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9058609,"math_prob":0.91692245,"size":4818,"snap":"2021-31-2021-39","text_gpt3_token_len":1175,"char_repetition_ratio":0.10240964,"word_repetition_ratio":0.0051612905,"special_character_ratio":0.21897052,"punctuation_ratio":0.11608094,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9613095,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-25T04:19:07Z\",\"WARC-Record-ID\":\"<urn:uuid:dedf2af0-3595-4574-8a5f-9a4eea3cb3cf>\",\"Content-Length\":\"7100\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d6c6faf4-afef-4f91-ac58-9f9758f63ccd>\",\"WARC-Concurrent-To\":\"<urn:uuid:626b3c16-5a21-49a9-8a05-3c3131ea5fe4>\",\"WARC-IP-Address\":\"104.164.145.154\",\"WARC-Target-URI\":\"http://www.gratisbo.com/basicnbc/article12radiological71.htm\",\"WARC-Payload-Digest\":\"sha1:GWBNFVFLYMU43ISP4FLDAIMMDMHMEWYS\",\"WARC-Block-Digest\":\"sha1:CUW344YYJV2RG5CED5R7FZZVMIREHPCJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057589.14_warc_CC-MAIN-20210925021713-20210925051713-00707.warc.gz\"}"} |
http://num.bubble.ro/s/100/90/ | [
"# Substraction table for N = 100 - 89÷90\n\n100 - 89 = 11 [+]\n100 - 89.01 = 10.99 [+]\n100 - 89.02 = 10.98 [+]\n100 - 89.03 = 10.97 [+]\n100 - 89.04 = 10.96 [+]\n100 - 89.05 = 10.95 [+]\n100 - 89.06 = 10.94 [+]\n100 - 89.07 = 10.93 [+]\n100 - 89.08 = 10.92 [+]\n100 - 89.09 = 10.91 [+]\n100 - 89.1 = 10.9 [+]\n100 - 89.11 = 10.89 [+]\n100 - 89.12 = 10.88 [+]\n100 - 89.13 = 10.87 [+]\n100 - 89.14 = 10.86 [+]\n100 - 89.15 = 10.85 [+]\n100 - 89.16 = 10.84 [+]\n100 - 89.17 = 10.83 [+]\n100 - 89.18 = 10.82 [+]\n100 - 89.19 = 10.81 [+]\n100 - 89.2 = 10.8 [+]\n100 - 89.21 = 10.79 [+]\n100 - 89.22 = 10.78 [+]\n100 - 89.23 = 10.77 [+]\n100 - 89.24 = 10.76 [+]\n100 - 89.25 = 10.75 [+]\n100 - 89.26 = 10.74 [+]\n100 - 89.27 = 10.73 [+]\n100 - 89.28 = 10.72 [+]\n100 - 89.29 = 10.71 [+]\n100 - 89.3 = 10.7 [+]\n100 - 89.31 = 10.69 [+]\n100 - 89.32 = 10.68 [+]\n100 - 89.33 = 10.67 [+]\n100 - 89.34 = 10.66 [+]\n100 - 89.35 = 10.65 [+]\n100 - 89.36 = 10.64 [+]\n100 - 89.37 = 10.63 [+]\n100 - 89.38 = 10.62 [+]\n100 - 89.39 = 10.61 [+]\n100 - 89.4 = 10.6 [+]\n100 - 89.41 = 10.59 [+]\n100 - 89.42 = 10.58 [+]\n100 - 89.43 = 10.57 [+]\n100 - 89.44 = 10.56 [+]\n100 - 89.45 = 10.55 [+]\n100 - 89.46 = 10.54 [+]\n100 - 89.47 = 10.53 [+]\n100 - 89.48 = 10.52 [+]\n100 - 89.49 = 10.51 [+]\n100 - 89.5 = 10.5 [+]\n100 - 89.51 = 10.49 [+]\n100 - 89.52 = 10.48 [+]\n100 - 89.53 = 10.47 [+]\n100 - 89.54 = 10.46 [+]\n100 - 89.55 = 10.45 [+]\n100 - 89.56 = 10.44 [+]\n100 - 89.57 = 10.43 [+]\n100 - 89.58 = 10.42 [+]\n100 - 89.59 = 10.41 [+]\n100 - 89.6 = 10.4 [+]\n100 - 89.61 = 10.39 [+]\n100 - 89.62 = 10.38 [+]\n100 - 89.63 = 10.37 [+]\n100 - 89.64 = 10.36 [+]\n100 - 89.65 = 10.35 [+]\n100 - 89.66 = 10.34 [+]\n100 - 89.67 = 10.33 [+]\n100 - 89.68 = 10.32 [+]\n100 - 89.69 = 10.31 [+]\n100 - 89.7 = 10.3 [+]\n100 - 89.71 = 10.29 [+]\n100 - 89.72 = 10.28 [+]\n100 - 89.73 = 10.27 [+]\n100 - 89.74 = 10.26 [+]\n100 - 89.75 = 10.25 [+]\n100 - 89.76 = 10.24 [+]\n100 - 89.77 = 10.23 [+]\n100 - 89.78 = 10.22 [+]\n100 - 89.79 = 10.21 [+]\n100 - 89.8 = 10.2 [+]\n100 - 89.81 = 10.19 [+]\n100 - 89.82 = 10.18 [+]\n100 - 89.83 = 10.17 [+]\n100 - 89.84 = 10.16 [+]\n100 - 89.85 = 10.15 [+]\n100 - 89.86 = 10.14 [+]\n100 - 89.87 = 10.13 [+]\n100 - 89.88 = 10.12 [+]\n100 - 89.89 = 10.11 [+]\n100 - 89.9 = 10.1 [+]\n100 - 89.91 = 10.09 [+]\n100 - 89.92 = 10.08 [+]\n100 - 89.93 = 10.07 [+]\n100 - 89.94 = 10.06 [+]\n100 - 89.95 = 10.05 [+]\n100 - 89.96 = 10.04 [+]\n100 - 89.97 = 10.03 [+]\n100 - 89.980000000001 = 10.02 [+]\nNavigation: Home | Addition | Substraction | Multiplication | Division Tables for 100: Addition | Substraction | Multiplication | Division\n\nOperand: 1 2 3 4 5 6 7 8 9 10 20 30 40 50 60 70 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 200 300 400 500 600 700 800 900 1000 2000 3000 4000 5000 6000 7000 8000 9000\n\nSubstraction for: 1 2 3 4 5 6 7 8 9 10 20 30 40 50 60 70 80 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 200 300 400 500 600 700 800 900 1000 2000 3000 4000 5000 6000 7000 8000 9000"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8847223,"math_prob":0.9999739,"size":7996,"snap":"2019-51-2020-05","text_gpt3_token_len":1668,"char_repetition_ratio":0.27277276,"word_repetition_ratio":0.41686556,"special_character_ratio":0.17346174,"punctuation_ratio":0.0014684288,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9981939,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-21T09:02:30Z\",\"WARC-Record-ID\":\"<urn:uuid:bcff16b4-23c1-4427-8cd9-4e4193b59b48>\",\"Content-Length\":\"42970\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:845b9385-6f76-4afd-8791-98d49da14864>\",\"WARC-Concurrent-To\":\"<urn:uuid:91ed0b42-4dfb-443a-b6c9-6e4334915c1d>\",\"WARC-IP-Address\":\"104.24.96.16\",\"WARC-Target-URI\":\"http://num.bubble.ro/s/100/90/\",\"WARC-Payload-Digest\":\"sha1:ZOXVPNYPKPUIWIQQYDJDSYIJGX4UBB47\",\"WARC-Block-Digest\":\"sha1:MT6237STUZUDTMZY542COYH2ZCPUNZ2P\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250601628.36_warc_CC-MAIN-20200121074002-20200121103002-00293.warc.gz\"}"} |
https://ogaml.github.io/doc/ogamlmath/matrix2d.html | [
"# Module OgamlMath.Matrix2D\n\nProvides easy creation and manipulation of 2D rendering matrices\n\nOptimized operations on 2D (3x3) float matrices\n\nRaised when an error occurs (usually a division by zero)\n\n### Simple Matrices\n\nType of 3x3 matrices stored in a flat, column-major array\nZero matrix\nIdentity matrix\nBuilds a translation matrix from a vector\nSee : OgamlMath.Vector2f\nBuilds a scaling matrix from a vector\nSee : OgamlMath.Vector2f\nBuilds a rotation matrix from an angle\nEfficiently builds a transformation matrix\n\n### Matrix Operations\n\nComputes the product of two matrices\nTransposes a matrix. The original is not modified.\nTranslates a matrix by a vector. The original matrix is not modified.\nSee : OgamlMath.Vector2f\nScales a matrix by a vector. The original matrix is not modified.\nSee : OgamlMath.Vector2f\nRotates a matrix by an angle. The original matrix is not modified.\nComputes the (right-)product of a matrix with a column vector\nReturns a pretty-printed string (not for serialization)\n\n### Rendering Matrices Creation\n\nBuilds an orthographic projection matrix englobing a screen\nBuilds the inverse of an orthographic projection matrix\n\n### Other functions\n\nReturns a matrix as a flat bigarray. Used internally by OGAML, it should not be necessary for your programs."
] | [
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https://sportblurb.com/terminal-velocity-of-a-golf-ball/ | [
"# Terminal Velocity Of A Golf Ball (Check Of The Terminal Velocity Of Different objects!)\n\nLast Updated on October 25, 2023 by Alex PT\n\nThe terminal velocity of a golf ball is 72 miles per hour (116 km/h). This is the maximum speed that a golf ball can reach when falling into the air, assuming that it is not hit with any additional force.\n\n## How To Calculate Terminal Velocity\n\nTo calculate a terminal velocity of an object, you need to make use of the terminal velocity formula, which is; v = the square root of (2*m*g)/(p*A*C). To solve for V, which is terminal velocity, put the following values into the formula.\n\n1. A = the object’s projected area. This means the area of the object if you projected it onto a plane that was perpendicular to the direction the object is moving.\n\n2. m = mass of the falling object.\n\n3. ρ = the density of the fluid the object is falling through.\n\n4. C = the coefficient of the drag. The shape of the object determines the number. The coefficient of an object is normally to lower the more streamlined it is in the shape. You can look up some approximate drag coefficients\n\n5. g = the acceleration due to gravity. It is approximately 9.8 meters per second on earth.\n\nThen, find the mass of the falling object. The mass of an object is normally measured in kilograms or grams, in the metric system.\n\nSince pounds is a measuring unit of force and not really of mass, you need to take note while using the imperial system. The pound-mass (IBM) is the unit of mass in the imperial system, which will definitely experience a force of 32 pound-force (Ibf) under gravitational force on the surface of the earth. Let’s take, for instance, a person actually weighs 160 Ibf with a mass of 5 Ibm even if the person originally weighs 170 pounds on earth.\n\nAfter finding the mass of the falling object, you need to know the acceleration due to the gravity of the earth. The acceleration due to the gravity of the earth is 9.8 meters per second squared, or 32 feet per second squared.\n\nThen, try to calculate the downward pull of gravity. The mass of the object multiplied by the acceleration due to gravity is equal to the falling force of the object after being pulled down, or F = MA. This number, multiplied by two, moves at the top of the formula for terminal velocity.\n\n## What Affects the Golf Ball Speed?\n\nGolf ball speed is influenced by various factors, including:\n\n1. Clubhead Speed: The speed at which the clubhead strikes the ball greatly impacts its velocity. A faster swing generates more ball speed.\n2. Contact Point: The location on the clubface where the ball is struck affects speed. Hitting the sweet spot typically results in higher velocity.\n3. Ball Construction: Different golf balls are designed with varying core materials, dimple patterns, and layers, influencing their rebound speed off the clubface.\n4. Temperature and Altitude: Environmental conditions can impact golf ball speed. Higher altitudes and warmer temperatures can lead to increased ball speed due to lower air density.\n5. Wind Conditions: Headwinds or tailwinds can either hinder or enhance ball speed. A tailwind can boost the ball’s speed, while a headwind may reduce it.\n6. Elevation Change: The elevation change on a golf course can affect ball speed. Hitting downhill can increase speed while hitting uphill can decrease it.\n7. Launch Angle: The angle at which the ball leaves the clubface affects speed. A more optimal launch angle can lead to higher ball speed.\n8. Spin Rate: Backspin or sidespin can influence the ball’s trajectory and speed. Reducing spin can often result in greater distance.\n9. Equipment Quality: The quality of golf clubs, including drivers, irons, and golf balls, plays a role in ball speed. Technological advancements can enhance speed.\n10. Swing Mechanics: A golfer’s swing technique, including angle of attack and path, can impact ball speed. Efficient mechanics can maximize velocity.\n11. Fitness and Strength: A golfer’s physical condition and strength can affect clubhead speed, which, in turn, influences ball speed.\n12. Compression: Golf balls are categorized by compression ratings. A golfer’s swing speed should match the compression of the ball for optimal distance.\n13. Aerodynamics: The design and dimple pattern of the golf ball can influence its aerodynamic properties, affecting the air resistance and ultimately the speed.\n14. Course Conditions: The firmness and speed of the fairways and greens can influence ball speed on roll and bounce.\n15. Practice and Technique: Consistent practice and refining one’s golf technique can lead to better ball speed control.\n\n## Terminal Velocity vs. Maximum Velocity of a Golf Ball\n\nTerminal velocity is the maximum speed that an object can reach when falling through a fluid (such as air) due to the force of gravity and the drag force of the fluid.\n\nMaximum velocity is the fastest speed that an object can reach, regardless of the forces acting on it.\n\nTerminal velocity of a golf ball:\n\n• Speed: 72 miles per hour (116 km/h)\n• Factors affecting terminal velocity:\n• Mass of the golf ball\n• Size of the golf ball\n• Shape of the golf ball (dimples help to reduce drag)\n• Density of the air\n\nMaximum velocity of a golf ball:\n\n• Speed: Over 150 miles per hour (241 km/h)\n• Factors affecting maximum velocity:\n• The initial speed of the golf ball (when hit with a club)\n• Angle of the Clubface\n• Conditions of the golf ball (spin, grooves, etc.)\n\nTable:\n\nExample:\n\nA golf ball hit with a club at a speed of 150 miles per hour will reach a maximum velocity of over 150 miles per hour, even though its terminal velocity is only 72 miles per hour. This is because the initial speed of the golf ball is greater than its terminal velocity.\n\nIt is important to note that the maximum velocity of a golf ball is not always achievable. Factors such as the wind and the condition of the golf ball can affect its speed.\n\n## How Long Does It Take A Ball To Reach Terminal Velocity?\n\nTo get to 90% of terminal velocity, it will take 8 seconds. However, you will definitely reach a much higher speed than terminal velocity if you’re falling from outer space. Interestingly, you will slow down as you reach the atmosphere. You might just burn up like a meteorite if you’re coming in from a far distance.\n\n## What Is The Max Speed Of Terminal Velocity?\n\nThe terminal velocity is about 120 mph (200 km/h) in a stable belly-to-earth position. There’s a terminal speed of 240-290 km/h from a stable freefall head-down position. To speed of around 500 km/h (310 mph), is, however, accomplished through a further minimization of drag by streamlining the body.\n\n## How Do You Find The Terminal Velocity Of A Falling Object?\n\nIn layman’s language, the square root of the quotient of twice the object’s total weight all over the object’s frontal area product, the gas density of the medium, and its drag coefficient is equal to the terminal velocity of the object.\n\n## What Is The Terminal Velocity Of A 200-Pound Man?\n\nThe terminal velocity of a falling 200-pound man is approximately 145 mph (64.8 m/sec) according to some extensive research.\n\nHowever, if we make use of a drag coefficient of C = 0.5 gives and assume the hailstones to be spherical, it gives the following:\n\n## Closing Thoughts\n\nThe bottom line is that the terminal velocity of a falling object is the identical velocity as the wind required to help support the mass of the item. As outlined by some research, the terminal velocity of any golf ball is approximately thirty-two meters per second. We’ve come to the closing chapter of this informative article. We hope you were able to find resources in your quest to know the terminal velocity of a golf ball. Please do not hesitate to contact us for questions or comments. Thanks!\n\nReferences:"
] | [
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https://www.rdocumentation.org/search/keywords/attribute | [
"### Search Results:\n\nShowing results 1 to 10 of 210.\n\n##### Function gen.depth[GENLIB v1.0.9]\nkeywords\nattribute\ntitle\nGenealogy depth\ndescription\nReturns the number of generations in the genealogy from a GLgen object.\n##### Function mlogit-deprecated[mlogit v1.1-1]\nkeywords\nattribute\ntitle\nSome deprecated functions, especially mlogit.data, index and mFormula\ndescription\nmlogit.data is deprecated, use dfidx::dfidx() instead, mFormula is replaced by Formula::Formula() and zoo::index() by idx.\n##### Function has.intercept[mlogit v1.1-1]\nkeywords\nattribute\ntitle\nIndicates whether the formula contains an intercept\ndescription\nThis is a generic which provide convenient methods for formula/Formula object and for specific fitted models\n##### Function model.matrix.dfidx_mlogit[mlogit v1.1-1]\nkeywords\nattribute\ntitle\nCompute the model matrix for RUM\ndescription\nspecific stuff compared to the model.matrix.dfidx method which simply applies the Formula method\n##### Function names<-.BibEntry[RefManageR v1.3.0]\nkeywords\nattribute\ntitle\nNames (keys) of a BibEntry object\ndescription\nFunctions to get and set the keys of an object of class BibEntry\n##### Function checking[FuzzyStatTra v1.0]\nkeywords\nattribute\ntitle\ndescription\nThe function checks if the input data are given in the correct form of an array of dimension nl x 3 x n containing n fuzzy numbers characterized by means of nl $$\\alpha$$-levels each. The following conditions have to be fulfilled: (1) the number of columns of the array must be 3 (the first column will be the $$\\alpha$$-levels, the second one their infimum values and the third one their supremum values), (2) all the fuzzy numbers must have the same column of $$\\alpha$$-levels, (3) the minimum $$\\alpha$$-level should be 0 y the maximum 1, (4) the $$\\alpha$$-levels have to increase from 0 to 1, (5) the infimum values have to be non-decreasing, (6) the supremum values have to be non-creasing, (7) the infimum value has to be smaller or equal than the supremum value for each $$\\alpha$$-level. This function is used internally in some of the other functions to do a preliminary checking if the input data are in the correct form.\n##### Function checkingTra[FuzzyStatTra v1.0]\nkeywords\nattribute\ntitle\ndescription\nThe function checks if the input data are given in the correct form of a matrix of dimension n x 4 containing n trapezoidal fuzzy numbers characterized by their four values inf0,inf1,sup1,sup0 each. The following conditions have to be fulfilled: (1) the number of columns of the matrix must be 4 (the four values characterizing each trapezoidal fuzzy number), (2) the four values of each trapezoidal number have to be non-decreasing. This function is used internally in almost all the other functions to do a preliminary checking if the input data are in the correct form.\n##### Function getSamples[HELP v1.30.0]\nkeywords\nattribute\ntitle\nGet samples\ndescription\nFetch a subset of samples from a given data structure\n##### Function getFeatures[HELP v1.30.0]\nkeywords\nattribute\ntitle\nGet features (methods)\ndescription\nFetch a subset of features from a given data structure\n##### Function wavTitle[wmtsa v2.0-3]\nkeywords\nattribute\ntitle\nExtract the name of the data used to generate objects of various wavelet classes\ndescription\nWavelet functions store the original name of the data used to create the output in various locations within the output object. This function provides a means by which the user can directly access data name."
] | [
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https://www.onlinemathlearning.com/expressions-exponents.html | [
"",
null,
"# Writing and Evaluating Expressions-Exponents\n\nVideos and solutions to help Grade 6 students learn how to evaluate and write formulas involving exponents for given values in real-world problems.\n\nNew York State Common Core Math Module 4, Grade 6, Lesson 22\n\nRelated Topics:\nLesson Plans and Worksheets for Grade 6\nLesson Plans and Worksheets for all Grades\n\nLesson 22 Exercises 1–3\n1. Predict how many times you can fold a piece of paper in half.\n\n2. Before any folding (zero folds), there is only one layer of paper. This is recorded in the first row of the table.\nFold your paper in half. Record the number of layers of paper that result. Continue as long as possible.\na. Are you able to continue folding the paper indefinitely? Why or why not?\nb. How could you use a calculator to find the next number in the series?\nc. What is the relationship between the number of folds and the number of layers?\nd. How is this relationship represented in the exponential form of the numerical expression?\ne. If you fold a paper f times, write an expression to show the number of paper layers.\n\n3. If the paper were to be cut instead of folded, the height of the stack would double at each successive stage, and it would be possible to continue.\na. Write an expression that describes how many layers of paper result from cuts.\nb. Evaluate this expression by writing it in standard form.\n\nExample 2: Bacterial Infection\nBacteria are microscopic one-celled organisms that reproduce in a couple of different ways, one of which is called binary fission. In binary fission, a bacterium increases its size until it is large enough to split into two parts that are identical. These two grow until they are both large enough to split into two individual bacteria. This continues as long as growing conditions are favorable.\na. Record the number of bacteria that result from each generation.\nb. How many generations would it take until there were over one million bacteria present?\nc. Under the right growing conditions, many bacteria can reproduce every 15 minutes. Under these conditions, how long would it take for one bacterium to reproduce itself into more than one million bacteria?\nd. Write an expression for how many bacteria would be present after g generations.\n\nExample 3: Volume of a Rectangular Solid\nThis box has a width, w. The height of the box, h, is twice the width. The length of the box, l, is three times the width.That is, the width, height, and length of a rectangular prism are in the ratio of 1:2:3.\nFor rectangular solids like this, the volume is calculated by multiplying length times width times height.\n\nClosing Why 53 is different from 5 × 3?\n\nLesson 22 Exit Ticket\n1. Naomi's allowance is \\$2.00 per week. If she convinces her parents to double her allowance each week for two months, what will her weekly allowance be at the end of the second month (week 8)?\n\n2. Write the expression that describes Naomi's allowance during week w, in dollars. Problem Set\n1. A checkerboard has 64 squares on it.\na. If one grain of rice is put on the first square, 2 grains of rice on the second square, 4 grains of rice on the third square, 8 grains of rice on the fourth square, and so on (doubling each time), complete the table to show how many grains of rice are on each square. Write your answers in exponential form on the table below.\nb. How many grains of rice would be on the last square? Represent your answer in exponential form and standard form. Use the table above to help solve the problem.\nc. Would it have been easier to write your answer to part (b) in exponential form or standard form?\n\n2. If an amount of money is invested at an annual interest rate of 6%, it doubles every 12 years. If Alejandra invests \\$500, how long will it take for her investment to reach \\$2,000 (assuming she does not contribute any additional funds)?\n\n3. The athletics director at Peter’s school has created a phone tree that is used to notify team players in the event a game has to be canceled or rescheduled. The phone tree is initiated when the director calls two captains. During the second stage of the phone tree, the captains each call two players. During the third stage of the phone tree, these players each call two other players. The phone tree continues until all players have been notified. If there are 50 players on the teams, how many stages will it take to notify all of the players?\n\nRotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.\n\nYou can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.",
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"https://www.onlinemathlearning.com/objects/default_image.gif",
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"https://www.onlinemathlearning.com/objects/default_image.gif",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.94147456,"math_prob":0.93851584,"size":4919,"snap":"2020-24-2020-29","text_gpt3_token_len":1080,"char_repetition_ratio":0.09908444,"word_repetition_ratio":0.012614679,"special_character_ratio":0.21772718,"punctuation_ratio":0.11641791,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97556126,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-12T00:35:23Z\",\"WARC-Record-ID\":\"<urn:uuid:78bf55cd-76ea-4939-9b03-a5941522677b>\",\"Content-Length\":\"49614\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef19eb4f-4cb6-4e53-9235-ef8d44a0cdb0>\",\"WARC-Concurrent-To\":\"<urn:uuid:2cad21c3-ecf8-4308-aac3-6c97c0612916>\",\"WARC-IP-Address\":\"173.247.219.45\",\"WARC-Target-URI\":\"https://www.onlinemathlearning.com/expressions-exponents.html\",\"WARC-Payload-Digest\":\"sha1:FOPTPFTATSREVDSTPBF7DMMO4H4FRJJI\",\"WARC-Block-Digest\":\"sha1:LMMSZ5P3KPWCB43VTJZYY5IQVASCARQA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593657129257.81_warc_CC-MAIN-20200711224142-20200712014142-00475.warc.gz\"}"} |
https://www.reference.com/business-finance/convert-41-degrees-fahrenheit-celsius-f3d2bbd3852d811e | [
"# How Do You Convert 41 Degrees Fahrenheit to Celsius?\n\nTo convert any temperature in Fahrenheit to Celsius, follow this formula: subtract 32, multiply by 5, and divide by 9. When you input 41 degrees Fahrenheit into this formula, the result is 5 degrees Celsius.\n\n1. Subtract 32 from your Fahrenheit temperature\n\nTo convert 41 degrees Fahrenheit to Celsius, begin by subtracting 32. Your answer should be 9.\n\nFinally, divide 45 by 9 to get 5 degrees Celsius as a final answer."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8626553,"math_prob":0.788854,"size":1204,"snap":"2019-43-2019-47","text_gpt3_token_len":259,"char_repetition_ratio":0.155,"word_repetition_ratio":0.01,"special_character_ratio":0.22093023,"punctuation_ratio":0.11061947,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99914956,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-18T19:24:57Z\",\"WARC-Record-ID\":\"<urn:uuid:425aff49-4b52-4b00-9e80-e4961f8f4d71>\",\"Content-Length\":\"175321\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:61a00719-3926-4ec4-b956-c899091d605e>\",\"WARC-Concurrent-To\":\"<urn:uuid:83d04e71-0e4c-4dff-8a44-d3e15ef8baf8>\",\"WARC-IP-Address\":\"151.101.250.114\",\"WARC-Target-URI\":\"https://www.reference.com/business-finance/convert-41-degrees-fahrenheit-celsius-f3d2bbd3852d811e\",\"WARC-Payload-Digest\":\"sha1:DPCBIICIJSI6T5P566UO7QCOIAZWTWXW\",\"WARC-Block-Digest\":\"sha1:2UH3OQNTKDPRWLQBEZS2MHBZQXLSXIKT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669813.71_warc_CC-MAIN-20191118182116-20191118210116-00266.warc.gz\"}"} |
https://www.covid19-blog.closeassociate.com/post/a-new-method-for-calculating-the-effective-reproduction-number | [
"top of page\nSearch\n• Stephen Grice\n\n# A new method for calculating the Effective Reproduction Number\n\nWe have finalised a technical report that provides the full mathematical details of a new method for calculating the Effective Reproduction Number for COVID-19. The technical report is here.\n\nThe new process which we have called \"SIR+B\" combines a deterministic SIR formula for calculating the Effective Reproduction Number from positive test data and a statistical bootstrapping method for generating confidence intervals.\n\nThe process has been used to calculate the Effective Reproduction Number for several countries in this blog post.\n\nThe process has the following desirable outcomes.\n\n• it is fast, taking about 0.1 seconds per country analysed, when coded in Python on a standard desktop machine\n\n• it is based directly on the epidemiological SIR model, giving an analytic formula for the Effective Reproduction Number\n\n• it involves many fewer assumptions than other methods such as EpiEstim\n\n• the calculation of the Effective Reproduction Number depends on only two parameters, gamma - the recovery parameter, and c - the testing parameter\n\n• it is relatively insensitive to those parameters"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.87085855,"math_prob":0.92661613,"size":1160,"snap":"2023-14-2023-23","text_gpt3_token_len":216,"char_repetition_ratio":0.15311418,"word_repetition_ratio":0.07017544,"special_character_ratio":0.16810344,"punctuation_ratio":0.05882353,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98724097,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-04-01T07:43:49Z\",\"WARC-Record-ID\":\"<urn:uuid:75a58c81-cbe2-4675-8d8f-6f631271ebff>\",\"Content-Length\":\"754520\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f801724c-9559-4d8e-b4ef-0ee22506d99b>\",\"WARC-Concurrent-To\":\"<urn:uuid:3fb4391e-10ce-4486-b6cb-7f18d9cabaf0>\",\"WARC-IP-Address\":\"34.117.168.233\",\"WARC-Target-URI\":\"https://www.covid19-blog.closeassociate.com/post/a-new-method-for-calculating-the-effective-reproduction-number\",\"WARC-Payload-Digest\":\"sha1:OHUMJHYXBOZGULFBRA2NNOTWEASTUJS7\",\"WARC-Block-Digest\":\"sha1:XJU6PTCUXKSHEVEP2VX7WCXREW4C24LQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949701.56_warc_CC-MAIN-20230401063607-20230401093607-00197.warc.gz\"}"} |
https://www.iteanz.com/python-real-time-interview-questions-and-answers/ | [
"## Python Real Time Interview Questions and Answers\n\nQ1).What is Python?\nAns1: Python is a high-level, interpreted, interactive and object-oriented scripting language. Python is designed to be highly readable. It uses English keywords frequently where as other languages use punctuation, and it h\nas fewer syntactical constructions than other languages.\n\nQ2).Name some of the features of Python.\nAns2: Following are some of the salient features of python\n\n• It supports functional and structured programming methods as well as OOP.\n• It can be used as a scripting language or can be compiled to byte-code for building large applications.\n• It provides very high-level dynamic data types and supports dynamic type checking.\n• It supports automatic garbage collection.\n• It can be easily integrated with C, C++, COM, ActiveX, CORBA, and Java.\n\nQ3).Do you have any personal projects?\nReally?\nAns3:This shows that you are willing to do more than the bare minimum in terms of keeping your skillset up to date. If you work on personal projects and code outside of the workplace then employers are more likely to see you as an asset that will grow. Even if they don’t ask this question I find it’s useful to broach the subject.\n\nQ4).Is python a case sensitive language?\nAns4: Yes! Python is a case sensitive programming language.\n\nWhat are the supported data types in Python?\nPython has five standard data types −\n\n• Numbers\n• String\n• List\n• Tuple\n• Dictionary\n\nQ5).What is the output of print str if str = ‘Hello World!’?\nAns5: It will print complete string. Output would be Hello World!.\nQ6).What is the output of print str if str = ‘Hello World!’?\nAns6: It will print first character of the string. Output would be H.\nQ7).What is the output of print str[2:5] if str = ‘Hello World!’?\nAns7: It will print characters starting from 3rd to 5th. Output would be llo.\nQ8).What is the output of print str[2:] if str = ‘Hello World!’?\nAns8: It will print characters starting from 3rd character. Output would be llo World!.\nQ9).What is the output of print str * 2 if str = ‘Hello World!’?\nAns9: It will print string two times. Output would be Hello World!Hello World!.\nQ10).What is the output of print str + “TEST” if str = ‘Hello World!’?\nAns10: It will print concatenated string. Output would be Hello World!TEST.\n\nQ11).What is the output of print list if list = [ ‘abcd’, 786 , 2.23, ‘john’, 70.2 ]?\nAns11: It will print concatenated lists. Output would be [ ‘abcd’, 786 , 2.23, ‘john’, 70.2 ].\n\nQ12).What is the output of print list if list = [ ‘abcd’, 786 , 2.23, ‘john’, 70.2 ]?\nAns12: It will print first element of the list. Output would be abcd.\n\nQ13).What is the output of print list[1:3] if list = [ ‘abcd’, 786 , 2.23, ‘john’, 70.2 ]?\nAns13: It will print elements starting from 2nd till 3rd. Output would be [786, 2.23].\n\nQ14).What is the output of print list[2:] if list = [ ‘abcd’, 786 , 2.23, ‘john’, 70.2 ]?\nAns14: It will print elements starting from 3rd element. Output would be [2.23, ‘john’, 70.200000000000003].\n\nQ15).What is the output of print tinylist * 2 if tinylist = [123, ‘john’]?\nAns15: It will print list two times. Output would be [123, ‘john’, 123, ‘john’].\n\nQ16).What is the output of print list + tinylist * 2 if list = [ ‘abcd’, 786 , 2.23, ‘john’, 70.2 ] and tinylist = [123, ‘john’]?\nAns16: It will print concatenated lists. Output would be [‘abcd’, 786, 2.23, ‘john’, 70.2, 123, ‘john’, 123, ‘john’].\n\nQ17).What is tuples in Python ?\nAns17: A tuple is another sequence data type that is similar to the list. A tuple consists of a number of values separated by commas. Unlike lists, however, tuples are enclosed within parentheses.\n\nQ18).What is the difference between tuples and lists in Python?\nAns18:The main differences between lists and tuples are − Lists are enclosed in brackets ( [ ] ) and their elements and size can be changed, while tuples are enclosed in parentheses ( ( ) ) and cannot be updated. Tuples can be thought of as read-only lists.\n\nQ19).What is the output of print tuple if tuple = ( ‘abcd’, 786 , 2.23, ‘john’, 70.2 )?\nAns19:It will print complete tuple. Output would be (‘abcd’, 786, 2.23, ‘john’, 70.200000000000003).\n\nQ20).What is the output of print tuple if tuple = ( ‘abcd’, 786 , 2.23, ‘john’, 70.2 )?\nAns20: It will print first element of the tuple. Output would be abcd.\n\nQ21).What is the output of print tuple[1:3] if tuple = ( ‘abcd’, 786 , 2.23, ‘john’, 70.2 )?\nAns21: It will print elements starting from 2nd till 3rd. Output would be (786, 2.23).\n\nQ22).What is the output of print tuple[2:] if tuple = ( ‘abcd’, 786 , 2.23, ‘john’, 70.2 )?\nAns22: It will print elements starting from 3rd element. Output would be (2.23, ‘john’, 70.200000000000003).\n\nQ23).What is the output of print tinytuple * 2 if tinytuple = (123, ‘john’)?\nAns23: It will print tuple two times. Output would be (123, ‘john’, 123, ‘john’).\n\nQ24).What is the output of print tuple + tinytuple if tuple = ( ‘abcd’, 786 , 2.23, ‘john’, 70.2) and tinytuple = (123, ‘john’)?\nAns24: It will print concatenated tuples. Output would be (‘abcd’, 786, 2.23, ‘john’, 70.200000000000003, 123, ‘john’).\n\nQ25).What are Python’s dictionaries?\nAns25: Python’s dictionaries are kind of hash table type. They work like associative arrays or hashes found in Perl and consist of key-value pairs. A dictionary key can be almost any Python type, but are usually numbers or strings. Values, on the other hand, can be any arbitrary Python object.\n\nQ26).How will you create a dictionary in python ?\nAns26: Dictionaries are enclosed by curly braces ({ }) and values can be assigned and accessed using square braces ([]).\ndict = {}\ndict[‘one’] = “This is one”\ndict = “This is two”\ntinydict = {‘name’: ‘john’,’code’:6734, ‘dept’: ‘sales’}\n\nQ27).How will you get all the keys from the dictionary?\nAns27: Using dictionary.keys() function, we can get all the keys from the dictionary object.\nprint dict.keys() # Prints all the keys\n\nQ28).How will you get all the values from the dictionary?\nAns28: Using dictionary.values() function, we can get all the values from the dictionary object.\nprint dict.values() # Prints all the values\n\nQ29).How will you convert a string to an int in python ?\nAns29: int(x [,base]) – Converts x to an integer. base specifies the base if x is a string.\n\nQ30).How will you convert a string to a long in python ?\nAns30: long(x [,base] ) – Converts x to a long integer. base specifies the base if x is a string.\n\nQ31).How will you convert a string to a float in python ?\nAns31: float(x) − Converts x to a floating-point number.\n\nQ32).How will you convert a object to a string in python ?\nAns32: str(x) − Converts object x to a string representation.\n\nQ33).How will you convert a object to a regular expression in python ?\nAns33: repr(x) − Converts object x to an expression string.\n\nQ34).How will you convert a String to an object in python ?\nAns34: eval(str) − Evaluates a string and returns an object.\n\nQ35).How will you convert a string to a tuple in python?\nAns35: tuple(s) − Converts s to a tuple.\n\nQ36).How will you convert a string to a list in python ?\nAns36: list(s) − Converts s to a list.\n\nQ37).How will you convert a string to a set in python?\nAns37: set(s) − Converts s to a set.\n\nQ38).How will you create a dictionary using tuples in python ?\nAns38: dict(d) − Creates a dictionary. d must be a sequence of (key,value) tuples.\n\nQ39).How will you convert a string to a frozen set in python?\nAns39: frozenset(s) − Converts s to a frozen set.\n\nQ40).How will you convert an integer to a character in python ?\nAns40: chr(x) − Converts an integer to a character.\n\nQ41).How will you convert an integer to an unicode character in python ?\nAns41: unichr(x) − Converts an integer to a Unicode character.\n\nQ42).How will you convert a single character to its integer value in python ?\nAns42: ord(x) − Converts a single character to its integer value.\n\nQ43).How will you convert an integer to hexadecimal string in python ?\nAns43: hex(x) − Converts an integer to a hexadecimal string.\n\nQ44).How will you convert an integer to octal string in python ?\nAns44: oct(x) − Converts an integer to an octal string.\n\nQ45).What is the purpose of ** operator?\nAns45: ** Exponent − Performs exponential (power) calculation on operators. a**b = 10 to the power 20 if a = 10 and b = 20.\n\nQ46).What is the purpose of // operator?\nAns46: // Floor Division − The division of operands where the result is the quotient in which the digits after the decimal point are removed.\n\nQ47).What is the purpose of is operator?\nAns47: is − Evaluates to true if the variables on either side of the operator point to the same object and false otherwise. x is y, here is results in 1 if id(x) equals id(y).\n\nQ48).What is the purpose of not in operator?\nAns48: not in − Evaluates to true if it does not finds a variable in the specified sequence and false otherwise. x not in y, here not in results in a 1 if x is not a member of sequence y.\n\nQ49).What is the purpose break statement in python ?\nAns49: break statement − Terminates the loop statement and transfers execution to the statement immediately following the loop.\n\nQ50).What is the purpose continue statement in python ?\nAns50: Continue statement − Causes the loop to skip the remainder of its body and immediately retest its condition prior to reiterating.\n\nQ51).What is the purpose pass statement in python ?\nAns51:pass statement − The pass statement in Python is used when a statement is required syntactically but you do not want any command or code to execute.\n\nQ52).How can you pick a random item from a list or tuple ?\nAns52: choice(seq) − Returns a random item from a list, tuple, or string.\n\nQ53).How can you pick a random item from a range?\nAns53: randrange ([start,] stop [,step]) − returns a randomly selected element from range(start, stop, step).\n\nQ54).How can you get a random number in python ?\nAns54: random() − returns a random float r, such that 0 is less than or equal to r and r is less than 1.\n\nQ55).How will you set the starting value in generating random numbers?\nAns55: seed([x]) − Sets the integer starting value used in generating random numbers. Call this function before calling any other random module function. Returns None.\n\nQ56).How will you randomizes the items of a list in place?\nAns56: shuffle(lst) − Randomizes the items of a list in place. Returns None.\n\nQ57).How will you capitalizes first letter of string?\nAns57: capitalize() − Capitalizes first letter of string.\nQ58).How will you check in a string that all characters are alphanumeric?\nAns58: isalnum() − Returns true if string has at least 1 character and all characters are alphanumeric and false otherwise.\n\nQ59).How will you check in a string that all characters are digits?\nAns59: isdigit() − Returns true if string contains only digits and false otherwise.\n\nQ60).How will you check in a string that all characters are in lowercase?\nAns60: islower() − Returns true if string has at least 1 cased character and all cased characters are in lowercase and false otherwise.\n\nQ61).How will you check in a string that all characters are numerics?\nAns61: isnumeric() − Returns true if a unicode string contains only numeric characters and false otherwise.\n\nQ62).How will you check in a string that all characters are whitespaces?\nAns62:isspace() − Returns true if string contains only whitespace characters and false otherwise.\n\nQ63).How will you check in a string that it is properly titlecased?\nAns63: istitle() − Returns true if string is properly “titlecased” and false otherwise.\n\nQ64).How will you check in a that all characters are in uppercase?\nAns64: isupper() − Returns true if string has at least one cased character and all cased characters are in uppercase and false otherwise.\n\nQ65).How will you merge elements in a sequence?\nAns65: join(seq) − Merges (concatenates) the string representations of elements in sequence seq into a string, with separator string.\n\nQ66).How will you get the length of the string?\nAns66: len(string) − Returns the length of the string.\n\nQ67).How will you get a space-padded string with the original string left-justified to a total of width columns?\n\nAns67: just(width[, fillchar]) − Returns a space-padded string with the original string left-justified to a total of width columns.\n\nQ68).How will you convert a string to all lowercase?\nAns68: lower() − Converts all uppercase letters in string to lowercase.\n\nQ69).How will you remove all leading whitespace in string?\nAns69: strip() − Removes all leading whitespace in string.\n\nQ70).How will you get the max alphabetical character from the string?\nAns70: max(str) − Returns the max alphabetical character from the string str.\n\n0Q71).How will you get the min alphabetical character from the string?\nAns71: min(str) − Returns the min alphabetical character from the string str.\n\nQ72).How will you replaces all occurrences of old substring in string with new string?\nAns72: replace(old, new [, max]) − Replaces all occurrences of old in string with new or at most max occurrences if max given.\n\nQ73).How will you remove all leading and trailing whitespace in string?\nAns73:strip([chars]) − Performs both lstrip() and rstrip() on string.\n\nQ74).How will you change case for all letters in string ?\nAns74: swapcase() − Inverts case for all letters in string.\n\nQ75).How will you get titlecased version of string ?\nAns75: title() − Returns “titlecased” version of string, that is, all words begin with uppercase and the rest are lowercase.\n\nQ76).How will you convert a string to all uppercase?\nAns76:upper() − Converts all lowercase letters in string to uppercase.\n\nQ77).How will you check in a string that all characters are decimal?\nAns77: isdecimal() − Returns true if a unicode string contains only decimal characters and false otherwise.\n\nQ78).What is the difference between del() and remove() methods of list?\nAns78: To remove a list element, you can use either the del statement if you know exactly which element(s) you are deleting or the remove() method if you do not know.\n\nQ79).What is the output of len([1, 2, 3])?\nAns79: 3.\n\nQ80).What is the output of [1, 2, 3] + [4, 5, 6]?\nAns80: [1, 2, 3, 4, 5, 6]\n\nQ81).What is the output of [‘Hi!’] * 4?\n\nAns81: [‘Hi!’, ‘Hi!’, ‘Hi!’, ‘Hi!’]\n\nQ82).What is the output of 3 in [1, 2, 3]?\nAns82: True\n\nQ83).What is the output of for x in [1, 2, 3]: print x?\nAns83: 1 2 3\n\nQ84).What is the output of L if L = [1,2,3]?\nAns84: 3, Offsets start at zero.\n\nQ85).What is the output of L[-2] if L = [1,2,3]?\nAns85: L[-1] = 3, L[-2]=2, L[-3]=1\n\nQ86).What is the output of L[1:] if L = [1,2,3]?\nAns86: 2, 3, Slicing fetches sections.\n\nQ87).How will you compare two lists ?\nAns87: cmp(list1, list2) − Compares elements of both lists.\n\nQ88).How will you get the length of a list?\nAns88: len(list) − Gives the total length of the list.\n\nQ89).How will you get the max valued item of a list?\nAns89: max(list) − Returns item from the list with max value.\n\nQ90).How will you get the min valued item of a list?\nAns90: min(list) − Returns item from the list with min value.\n\nQ91).How will you get the index of an object in a list?\nAns91: list.index(obj) − Returns the lowest index in list that obj appears.\n\nQ92).How will you insert an object at given index in a list?\nAns92: list.insert(index, obj) − Inserts object obj into list at offset index.\n\nQ93).How will you remove last object from a list?\nAns93: list.pop(obj=list[-1]) − Removes and returns last object or obj from list.\n\nQ94).How will you remove an object from a list?\nAns94: list.remove(obj) − Removes object obj from list.\n\nQ95).How will you reverse a list?\nAns95: list.reverse() − Reverses objects of list in place.\n\nQ96).How will you sort a list?\nAns96: list.sort([func]) − Sorts objects of list, use compare func if given.\n\nQ97).Name five modules that are included in python by default (many people come searching for this, so I included some more examples of modules which are often used)\nAns97:\ndatetime (used to manipulate date and time)\nre (regular expressions)\nurllib, urllib2 (handles many HTTP things)\nstring (a collection of different groups of strings for example all lower_case letters etc)\nitertools (permutations, combinations and other useful iterables)\nctypes (from python docs: create and manipulate C data types in Python)\nemail (from python docs: A package for parsing, handling, and generating email messages)\nfuture__ (Record of incompatible language changes. like division operator is different and much better when imported from __future__)\nsqlite3 (handles database of SQLite type)\nunittest (from python docs: Python unit testing framework, based on Erich Gamma’s JUnit and Kent Beck’s Smalltalk testing framework)\nxml (xml support)\nlogging (defines logger classes. enables python to log details on severity level basis)\nos (operating system support)\npickle (similar to json. can put any data structure to external files)\nsubprocess (from docs: This module allows you to spawn processes, connect to their input/output/error pipes, and obtain their return codes)\nwebbrowser (from docs: Interfaces for launching and remotely controlling Web browsers.)\ntraceback (Extract, format and print Python stack traces)\n\nQ98).Name a module that is not included in python by default\nAns98:mechanize\ndjango\ngtk\nA lot of other can be found at pypi.\n\nQ99).What is __init__.py used for?\nAns99:It declares that the given directory is a package. Python Docs (From Endophage‘s comment)\n\nQ100).When is pass used for?\nAns100:pass does nothing. It is used for completing the code where we need something. For eg:\n\n 1 2 class abc(): pass\n\nQ101).What is a docstring?\n\nAns101: docstring is the documentation string for a function. It can be accessed by\n\nfunction_name.__doc__\n\nit is declared as:\n\n 1 2 def function_name(): “””your docstring”””\n\nWriting documentation for your progams is a good habit and makes the code more understandable and reusable.\n\nQ102).What is list comprehension?\n\nAns102: Creating a list by doing some operation over data that can be accessed using an iterator. For eg:\n\n 1 2 3 >>>[ord(i) for i in string.ascii_uppercase] [65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90] >>>\n\nQ103).What is map?\n\nAns103: map executes the function given as the first argument on all the elements of the iterable given as the second argument. If the function given takes in more than 1 arguments, then many iterables are given.\nFor eg:\n\n 1 2 3 4 5 >>>a=’ayush’ >>>map(ord,a) …. [97, 121, 117, 115, 104] >>> print map(lambda x, y: x*y**2, [1, 2, 3], [2, 4, 1]) …. [4, 32, 3] 1 2 3 4 5 6 7 8 9 10 11 Help on built-in function map in module __builtin__: map(…) map(function, sequence[, sequence, …]) -> list Return a list of the results of applying the function to the items of the argument sequence(s). If more than one sequence is given, the function is called with an argument list consisting of the corresponding item of each sequence, substituting None for missing values when not all sequences have the same length. If the function is None, return a list of the items of the sequence (or a list of tuples if more than one sequence).\n\n#Python Docs\n\nQ104).What is the difference between a tuple and a list?\n\nAns104:A tuple is immutable i.e. can not be changed. It can be operated on only. But a list is mutable. Changes can be done internally to it.\n\ntuple initialization: a = (2,4,5)\nlist initialization: a = [2,4,5]\n\nThe methods/functions provided with each types are also different. Check them out yourself.\n\nQ105).Using various python modules convert the list a to generate the output ‘one, two, three’\n\nAns105:\n\n 1 2 a = [‘one’, ‘two’, ‘three’] Ans: “, “.join(a) 1 2 3 4 5 6 >>>help(str.join) Help on method_descriptor: join(…) S.join(iterable) -> string Return a string which is the concatenation of the strings in the iterable. The separator between elements is S.\n\nQ106).What would the following code yield?\n\n 1 2 word = ‘abcdefghij’ print word[:3] + word[3:]\n\nAns106: ‘abcdefghij’ will be printed.\nThis is called string slicing. Since here the indices of the two slices are colliding, the string slices are ‘abc’ and ‘defghij’. The ‘+’ operator on strings concatenates them. Thus, the two slices formed are concatenated to give the answer ‘abcdefghij’.\n\nQ107).Optimize these statements as a python programmer.\n\n 1 2 word = ‘word’ print word.__len__()\n\nAns107:\n\n 1 2 word = ‘word’ print len(word)\n\nQ108).Write a program to print all the contents of a file\n\nAns108:\n\n 1 2 3 4 5 try: with open(‘filename’,’r’) as f: print f.read() except IOError: print “no such file exists”\n\nQ109). What will be the output of the following code\n\n 1 2 3 4 a = 1 a, b = a+1, a+1 print a print b\n\nAns109:\n2\n2\n\nThe second line is a simultaneous declaration i.e. value of new a is not used when doing b=a+1.\n\nThis is why, exchanging numbers is as easy as:\n\n 1 a,b = b,a\n\nQ110). Given the list below remove the repetition of an element.\nAll the elements should be unique\nwords = [‘one’, ‘one’, ‘two’, ‘three’, ‘three’, ‘two’]\n\nAns110:\nA bad solution would be to iterate over the list and checking for copies somehow and then remove them!\n\nOne of the best solutions I can think of right now:\n\n 1 2 a = [1,2,2,3] list(set(a))\n\nset is another type available in python, where copies are not allowed. It also has some good functions available used in set operations ( like union, difference ).\n\nQ111).Iterate over a list of words and use a dictionary to keep track of the frequency(count) of each word. for example\n\n{‘one’:2, ‘two’:2, ‘three’:2}\n\nAns111:\n\n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 >>> def dic(words): a = {} for i in words: try: a[i] += 1 except KeyError: ## the famous pythonic way: a[i] = 1 ## Halt and catch fire return a >>> a=’1,3,2,4,5,3,2,1,4,3,2′.split(‘,’) >>> a [‘1’, ‘3’, ‘2’, ‘4’, ‘5’, ‘3’, ‘2’, ‘1’, ‘4’, ‘3’, ‘2’] >>> dic(a) {‘1’: 2, ‘3’: 3, ‘2’: 3, ‘5’: 1, ‘4’: 2}\n\nWithout using try-catch block:\n\n 1 2 3 4 5 6 7 8 9 10 >>> def dic(words): data = {} for i in words: data[i] = data.get(i, 0) + 1 return data >>> a [‘1’, ‘3’, ‘2’, ‘4’, ‘5’, ‘3’, ‘2’, ‘1’, ‘4’, ‘3’, ‘2’] >>> dic(a) {‘1’: 2, ‘3’: 3, ‘2’: 3, ‘5’: 1, ‘4’: 2}\n\nPS: Since the collections module (which gives you the defaultdict) is written in python, I would not recommend using it. The normal dict implementation is in C, it should be much faster. You can use timeit module to check for comparing the two.\nSo, David and I have saved you the work to check it. Check the files on github. Change the data file to test different data.\n\nQ112).Write the following logic in Python:\nIf a list of words is empty, then let the user know it’s empty, otherwise let the user know it’s not empty.\n\nAns112: Can be checked by a single statement (pythonic beauty):\n\n 1 2 3 4 5 6 7 8 print “The list is empty” if len(a)==0 else “The list is not empty” >>> a=” >>> print “‘The list is empty'” if len(a)==0 else “‘The list is not empty'” ‘The list is empty’ >>> a=’asd’ >>> print “‘The list is empty'” if len(a)==0 else “‘The list is not empty'” ‘The list is not empty’\n\nQ113).Demonstrate the use of exception handling in python.\n\nAns113:\n\n 1 2 3 4 try: import mechanize as me except ImportError: import urllib as me\n\n## here you have atleast 1 module imported as me.\nThis is used to check if the users computer has third party libraries that we need. If not, we work with a default library of python. Quite useful in updating softwares.\nPS: This is just one of the uses of try-except blocks. You can note a good use of these in API’s.\nAlso note that if we do not define the error to be matched, the except block would catch any error raised in try block.\n\nQ114).Print the length of each line in the file ‘file.txt’ not including any whitespaces at the end of the lines.\n\nAns114:\n\n 1 2 with open(“filename.txt”, “r”) as f1: print len(f1.readline().rstrip())\n\nrstrip() is an inbuilt function which strips the string from the right end of spaces or tabs (whitespace characters).\n\nQ115). Print the sum of digits of numbers starting from 1 to 100 (inclusive of both)\n\nAns115:\n\n 1 print sum(range(1,101))\n\nrange() returns a list to the sum function containing all the numbers from 1 to 100. Please see that the range function does not include the end given (101 here).\n\n 1 print sum(xrange(1, 101))\n\nxrange() returns an iterator rather than a list which is less heavy on the memory.\n\nQ116).Create a new list that converts the following list of number strings to a list of numbers.\n\nnum_strings = [‘1′,’21’,’53’,’84’,’50’,’66’,’7′,’38’,’9′]\n\nAns116:\nuse a list comprehension\n\n 1 2 >>> [int(i) for i in num_strings] [1, 21, 53, 84, 50, 66, 7, 38, 9]\n\n#num_strings should not contain any non-integer character else ValueError would be raised. A try-catch block can be used to notify the user of this.\n\nAnother one suggested by David using maps:\n\n 1 2 >>> map(int, num_strings) [1, 21, 53, 84, 50, 66, 7, 38, 9]\n\nQ117).Create two new lists one with odd numbers and other with even numbers\nnum_strings = [1,21,53,84,50,66,7,38,9]\n\nAns117:\n\n 1 2 3 4 5 6 7 >>> odd=[] >>> even=[] >>> for i in n: even.append(i) if i%2==0 else odd.append(i) ## all odd numbers in list odd ## all even numbers in list even\n\nThough if only one of the lists were requires, using list comprehension we could make:\n\n 1 2 even = [i for i in num_strings if i%2==0] odd = [i for i in num_strings if i%2==1]\n\nBut using this approach if both lists are required would not be efficient since this would iterate the list two times.!\n\nQ118).Write a program to sort the following intergers in list\n\nnums = [1,5,2,10,3,45,23,1,4,7,9]\n\nAns118:nums.sort() # The lists have an inbuilt function, sort()\nsorted(nums) # sorted() is one of the inbuilt functions)\n\nPython uses TimSort for applying this function. Check the link to know more.\n\nQ119).Write a for loop that prints all elements of a list and their position in the list.\nPrinting using String formatting\n\nAns119:\n\n 1 2 3 4 5 6 7 8 9 >>> for index, data in enumerate(asd): …. print “{0} -> {1}”.format(index, data) 0 -> 4 1 -> 7 2 -> 3 3 -> 2 4 -> 5 5 -> 9\n\n#OR\n\n 1 2 3 4 5 6 7 8 9 10 11 >>> asd = [4,7,3,2,5,9] >>> for i in range(len(asd)): …. print i+1,’–>’,asd[i] 1 –> 4 2 –> 7 3 –> 3 4 –> 2 5 –> 5 6 –> 9\n\nQ120).The following code is supposed to remove numbers less than 5 from list n, but there is a bug. Fix the bug.\n\nAns120:\n\n 1 2 3 4 5 6 n = [1,2,5,10,3,100,9,24] for e in n: if e<5: n.remove(e) print n\n\n## after e is removed, the index position gets disturbed. Instead it should be:\n\n 1 2 3 4 5 a=[] for e in n: if e >= 5: a.append(e) n = a\n\nOR again a list comprehension:\n\n 1 return [i for i in n if i >= 5]\n\nOR use filter\n\n 1 return filter(lambda x: x >= 5, n)\n\nQ121).What will be the output of the following\n\n 1 2 3 4 def func(x,*y,**z): …. print z func(1,2,3)\n\nAns121:Here the output is :\n\n{} #Empty Dictionay\n\nx is a normal value, so it takes 1..\ny is a list of numbers, so it takes 2,3..\nz wants named parameters, so it can not take any value here.\n\nQ122).Write a program to swap two numbers.\n\nAns122:\n\na = 5\nb = 9\n\nas i told earlier too, just use:\na,b = b,a\n\nQ123).What will be the output of the following code\n\n 1 2 3 4 5 6 7 8 9 class C(object): …. def__init__(self): …. self.x =1 c=C() print c.x print c.x print c.x print c.x\n\nAns123: All the outputs will be 1, since the value of the the object’s attribute(x) is never changed.\n\n1\n1\n1\n1\n\nx is now a part of the public members of the class C.\nThus it can be accessed directly..\n\nQ124). What is wrong with the code\n\n 1 2 3 4 5 6 7 func([1,2,3]) # explicitly passing in a list func() # using a default empty list def func(n = []): #do something with n print n\n\nAns124. This would result in a NameError. The variable n is local to function func and can’t be accessesd outside. So, printing it won’t be possible.\n\nEdit: An extra point for interviews given by Shane Green and Peter: “””Another thing is that mutable types should never be used as default parameter values. Default parameter value expressions are only evaluated once, meaning every invocation of that method shares the same default value. If one invocation that ends up using the default value modifies that value–a list, in this case–it will forever be modified for all future invocations. So default parameter values should limited to primitives, strings, and tuples; no lists, dictionaries, or complex object instances.”””\nReference: Default argument values\n\nQ125).What all options will work?\n\nAns125:\n\n1. n = 1\nprint n++ ## no such operator in python (++)\n2. n = 1\nprint ++n ## no such operator in python (++)\n3. n = 1\nprint n += 1 ## will work\n4. int n = 1\nprint n = n+1 ##will not work as assignment can not be done in print command like this\n5. n =1\nn = n+1 ## will work\n\nQ126).In Python function parameters are passed by value or by reference?\n\nAns126: It is somewhat more complicated than I have written here (Thanks David for pointing). Explaining all here won’t be possible. Some good links that would really make you understand how things are:\n\nStackoverflow\n\nPython memory management\n\nViewing the memory\n\nQ127).Remove the whitespaces from the string.\n\ns = ‘aaa bbb ccc ddd eee’\n\nAns127:\n\n 1 2 ”.join(s.split()) ## join without spaces the string after splitting it\n\nOR\n\n 1 filter(lambda x: x != ‘ ‘, s)\n\nQ128).What does the below mean?\n\ns = a + ‘[‘ + b + ‘:’ + c + ‘]’\n\nAns128: seems like a string is being concatenated. Nothing much can be said without knowing types of variables a, b, c. Also, if all of the a, b, c are not of type string, TypeError would be raised. This is because of the string constants (‘[‘ , ‘]’) used in the statement.\n\nQ129). Optimize the below code\n\n 1 2 3 4 5 6 7 8 def append_s(words): new_words=[] for word in words: new_words.append(word + ‘s’) return new_words for word in append_s([‘a’,’b’,’c’]): print word\n\nAns129: The above code adds a trailing s after each element of the list.\n\ndef append_s(words):\nreturn [i+’s’ for i in words] ## another list comprehension\n\nfor word in append_s([‘a’,’b’,’c’]):\nprint word\n\nQ130).If given the first and last names of bunch of employees how would you store it and what datatype?\n\nAns130: best stored in a list of dictionaries..\ndictionary format: {‘first_name’:’Ayush’,’last_name’:’Goel’}\n\nQ131).What is Python really? You can (and are encouraged) make comparisons to other technologies in your answer\n\nAns131: Here are a few key points:\n\n• Python is an interpreted language. That means that, unlike languages like Cand its variants, Python does not need to be compiled before it is run. Other interpreted languages include PHP and Ruby.\n• Python is dynamically typed, this means that you don’t need to state the types of variables when you declare them or anything like that. You can do things like x=111and then x=”I’m a string”without error\n• Python is well suited to object orientated programming in that it allows the definition of classes along with composition and inheritance. Python does not have access specifiers (like C++’s public, private), the justification for this point is given as “we are all adults here”\n• In Python, functions are first-class objects. This means that they can be assigned to variables, returned from other functions and passed into functions. Classes are also first class objects\n• Writing Python code is quick but running it is often slower than compiled languages. Fortunately， Python allows the inclusion of C based extensions so bottlenecks can be optimised away and often are. The numpypackage is a good example of this, it’s really quite quick because a lot of the number crunching it does isn’t actually done by Python\n• Python finds use in many spheres – web applications, automation, scientific modelling, big data applications and many more. It’s also often used as “glue” code to get other languages and components to play nice.\n• Python makes difficult things easy so programmers can focus on overriding algorithms and structures rather than nitty-gritty low level details.\n\nWhy This Matters:\n\nIf you are applying for a Python position, you should know what it is and why it is so gosh-darn cool. And why it isn’t o.O\n\nQ132).Fill in the missing code:\n\ndef print_directory_contents(sPath):\n\n“””\n\nThis function takes the name of a directory\n\nand prints out the paths files within that\n\ndirectory as well as any files contained in\n\ncontained directories.\n\nThis function is similar to os.walk. Please don’t\n\nability to work with nested structures.\n\n“””\n\nfill_this_in\n\nAns132: def print_directory_contents(sPath):\n\nimport os\n\nfor sChild in os.listdir(sPath):\n\nsChildPath = os.path.join(sPath,sChild)\n\nif os.path.isdir(sChildPath):\n\nprint_directory_contents(sChildPath)\n\nelse:\n\nprint(sChildPath)\n\nPay Special Attention\n\n• Be consistent with your naming conventions. If there is a naming convention evident in any sample code, stick to it. Even if it is not the naming convention you usually use\n• Recursive functions need to recurse and Make sure you understand how this happens so that you avoid bottomless callstacks\n• We use the osmodule for interacting with the operating system in a way that is cross platform. You could say sChildPath = sPath + ‘/’ + sChild but that wouldn’t work on windows\n• Familiarity with base packages is really worthwhile, but don’t break your head trying to memorize everything, Google is your friend in the workplace!\n• Ask questions if you don’t understand what the code is supposed to do\n• KISS! Keep it Simple, Stupid!\n\nWhy This Matters:\n\n• Displays knowledge of basic operating system interaction stuff\n• Recursion is hella useful\n\nQ133).\n\nLooking at the below code, write down the final values of A0, A1, …An.\n\nA0 = dict(zip((‘a’,’b’,’c’,’d’,’e’),(1,2,3,4,5)))\n\nA1 = range(10)\n\nA2 = sorted([i for i in A1 if i in A0])\n\nA3 = sorted([A0[s] for s in A0])\n\nA4 = [i for i in A1 if i in A3]\n\nA5 = {i:i*i for i in A1}\n\nA6 = [[i,i*i] for i in A1]\n\nIf you dont know what zip is don’t stress out. No sane employer will expect you to memorize the standard library. Here is the output of help(zip).\n\nzip(…)\n\nzip(seq1 [, seq2 […]]) -> [(seq1, seq2 …), (…)]\n\nReturn a list of tuples, where each tuple contains the i-th element\n\nfrom each of the argument sequences. The returned list is truncated\n\nin length to the length of the shortest argument sequence.\n\nIf that doesn’t make sense then take a few minutes to figure it out however you choose to.\n\nAns133: A0 = {‘a’: 1, ‘c’: 3, ‘b’: 2, ‘e’: 5, ‘d’: 4} # the order may vary\n\nA1 = range(0, 10) # or [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] in python 2\n\nA2 = []\n\nA3 = [1, 2, 3, 4, 5]\n\nA4 = [1, 2, 3, 4, 5]\n\nA5 = {0: 0, 1: 1, 2: 4, 3: 9, 4: 16, 5: 25, 6: 36, 7: 49, 8: 64, 9: 81}\n\nA6 = [[0, 0], [1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81]]\n\nWhy This Matters\n\n1. List comprehension is a wonderful time saver and a big stumbling block for a lot of people\n2. If you can read them, you can probably write them down\n3. Some of this code was made to be deliberately weird. You may need to work with some weird people\n\nQ134).Python and multi-threading. Is it a good idea? List some ways to get some Python code to run in a parallel way.\n\nThere are reasons to use Python’s threading package. If you want to run some things simultaneously, and efficiency is not a concern, then it’s totally fine and convenient. Or if you are running code that needs to wait for something (like some IO) then it could make a lot of sense. But the threading library won’t let you use extra CPU cores.\n\nMulti-threading can be outsourced to the operating system (by doing multi-processing), some external application that calls your Python code (eg, Spark or Hadoop), or some code that your Python code calls (eg: you could have your Python code call a C function that does the expensive multi-threaded stuff).\n\nWhy This Matters\n\nBecause the GIL is an A-hole. Lots of people spend a lot of time trying to find bottlenecks in their fancy Python multi-threaded code before they learn what the GIL is.\n\nQ135). How do you keep track of different versions of your code?\n\nAns135: Version control! At this point, you should act excited and tell them how you even use Git (or whatever is your favorite) to keep track of correspondence with Granny. Git is my preferred version control system, but there are others, for example subversion.\n\nWhy This Matters:\n\nBecause code without version control is like coffee without a cup. Sometimes we need to write once-off throw away scripts and that’s ok, but if you are dealing with any significant amount of code, a version control system will be a benefit. Version Control helps with keeping track of who made what change to the code base; finding out when bugs were introduced to the code; keeping track of versions and releases of your software; distributing the source code amongst team members; deployment and certain automations. It allows you to roll your code back to before you broke it which is great on its own. Lots of stuff. It’s just great.\n\nQ136).What does this code output:\n\ndef f(x,l=[]):\n\nfor i in range(x):\n\nl.append(i*i)\n\nprint(l)\n\nf(2)\n\nf(3,[3,2,1])\n\nf(3)\n\nAns136:\n\n[0, 1]\n\n[3, 2, 1, 0, 1, 4]\n\n[0, 1, 0, 1, 4]\n\nHu?\n\nThe first function call should be fairly obvious, the loop appends 0 and then 1 to the empty list, l. l is a name for a variable that points to a list stored in memory.\nThe second call starts off by creating a new list in a new block of memory. l then refers to this new list. It then appends 0, 1 and 4 to this new list. So that’s great.\nThe third function call is the weird one. It uses the original list stored in the original memory block. That is why it starts off with 0 and 1.\n\nTry this out if you don’t understand:\n\nl_mem = []\n\nl = l_mem # the first call\n\nfor i in range(2):\n\nl.append(i*i)\n\nprint(l) # [0, 1]\n\nl = [3,2,1] # the second call\n\nfor i in range(3):\n\nl.append(i*i)\n\nprint(l) # [3, 2, 1, 0, 1, 4]\n\nl = l_mem # the third call\n\nfor i in range(3):\n\nl.append(i*i)\n\nprint(l) # [0, 1, 0, 1, 4]\n\nQ137).What is monkey patching and is it ever a good idea?\n\nAns137:Monkey patching is changing the behaviour of a function or object after it has already been defined. For example:\n\nimport datetime\n\ndatetime.datetime.now = lambda: datetime.datetime(2012, 12, 12)\n\nMost of the time it’s a pretty terrible idea – it is usually best if things act in a well-defined way. One reason to monkey patch would be in testing. The mock package is very useful to this end.\n\nWhy This Matters\n\nIt shows that you understand a bit about methodologies in unit testing. Your mention of monkey avoidance will show that you aren’t one of those coders who favor fancy code over maintainable code (they are out there, and they suck to work with). Remember the principle of KISS? And it shows that you know a little bit about how Python works on a lower level, how functions are actually stored and called and suchlike.\n\nPS: it’s really worth reading a little bit about mock if you haven’t yet. It’s pretty useful.\n\nQ138).What does this stuff mean: *args, **kwargs? And why would we use it?\n\nAns138:Use *args when we aren’t sure how many arguments are going to be passed to a function, or if we want to pass a stored list or tuple of arguments to a function. **kwargsis used when we dont know how many keyword arguments will be passed to a function, or it can be used to pass the values of a dictionary as keyword arguments. The identifiers args and kwargs are a convention, you could also use *bob and **billy but that would not be wise.\n\nHere is a little illustration:\n\ndef f(*args,**kwargs): print(args, kwargs)\n\nl = [1,2,3]\n\nt = (4,5,6)\n\nd = {‘a’:7,’b’:8,’c’:9}\n\nf()\n\nf(1,2,3) # (1, 2, 3) {}\n\nf(1,2,3,”groovy”) # (1, 2, 3, ‘groovy’) {}\n\nf(a=1,b=2,c=3) # () {‘a’: 1, ‘c’: 3, ‘b’: 2}\n\nf(a=1,b=2,c=3,zzz=”hi”) # () {‘a’: 1, ‘c’: 3, ‘b’: 2, ‘zzz’: ‘hi’}\n\nf(1,2,3,a=1,b=2,c=3) # (1, 2, 3) {‘a’: 1, ‘c’: 3, ‘b’: 2}\n\nf(*l,**d) # (1, 2, 3) {‘a’: 7, ‘c’: 9, ‘b’: 8}\n\nf(*t,**d) # (4, 5, 6) {‘a’: 7, ‘c’: 9, ‘b’: 8}\n\nf(1,2,*t) # (1, 2, 4, 5, 6) {}\n\nf(q=”winning”,**d) # () {‘a’: 7, ‘q’: ‘winning’, ‘c’: 9, ‘b’: 8}\n\nf(1,2,*t,q=”winning”,**d) # (1, 2, 4, 5, 6) {‘a’: 7, ‘q’: ‘winning’, ‘c’: 9, ‘b’: 8}\n\ndef f2(arg1,arg2,*args,**kwargs): print(arg1,arg2, args, kwargs)\n\nf2(1,2,3) # 1 2 (3,) {}\n\nf2(1,2,3,”groovy”) # 1 2 (3, ‘groovy’) {}\n\nf2(arg1=1,arg2=2,c=3) # 1 2 () {‘c’: 3}\n\nf2(arg1=1,arg2=2,c=3,zzz=”hi”) # 1 2 () {‘c’: 3, ‘zzz’: ‘hi’}\n\nf2(1,2,3,a=1,b=2,c=3) # 1 2 (3,) {‘a’: 1, ‘c’: 3, ‘b’: 2}\n\nf2(*l,**d) # 1 2 (3,) {‘a’: 7, ‘c’: 9, ‘b’: 8}\n\nf2(*t,**d) # 4 5 (6,) {‘a’: 7, ‘c’: 9, ‘b’: 8}\n\nf2(1,2,*t) # 1 2 (4, 5, 6) {}\n\nf2(1,1,q=”winning”,**d) # 1 1 () {‘a’: 7, ‘q’: ‘winning’, ‘c’: 9, ‘b’: 8}\n\nf2(1,2,*t,q=”winning”,**d) # 1 2 (4, 5, 6) {‘a’: 7, ‘q’: ‘winning’, ‘c’: 9, ‘b’: 8}\n\nWhy Care?\n\nSometimes we will need to pass an unknown number of arguments or keyword arguments into a function. Sometimes we will want to store arguments or keyword arguments for later use. Sometimes it’s just a time saver.\n\nQ139).What do these mean to you: @classmethod, @staticmethod, @property?\n\nThese are decorators. A decorator is a special kind of function that either takes a function and returns a function, or takes a class and returns a class. The @ symbol is just syntactic sugar that allows you to decorate something in a way that’s easy to read.\n\n@my_decorator\n\ndef my_func(stuff):\n\ndo_things\n\nIs equivalent to\n\ndef my_func(stuff):\n\ndo_things\n\nmy_func = my_decorator(my_func)\n\nThe decorators @classmethod, @staticmethod and @property are used on functions defined within classes. Here is how they behave:\n\nclass MyClass(object):\n\ndef __init__(self):\n\nself._some_property = “properties are nice”\n\nself._some_other_property = “VERY nice”\n\ndef normal_method(*args,**kwargs):\n\nprint(“calling normal_method({0},{1})”.format(args,kwargs))\n\n@classmethod\n\ndef class_method(*args,**kwargs):\n\nprint(“calling class_method({0},{1})”.format(args,kwargs))\n\n@staticmethod\n\ndef static_method(*args,**kwargs):\n\nprint(“calling static_method({0},{1})”.format(args,kwargs))\n\n@property\n\ndef some_property(self,*args,**kwargs):\n\nprint(“calling some_property getter({0},{1},{2})”.format(self,args,kwargs))\n\nreturn self._some_property\n\n@some_property.setter\n\ndef some_property(self,*args,**kwargs):\n\nprint(“calling some_property setter({0},{1},{2})”.format(self,args,kwargs))\n\nself._some_property = args\n\n@property\n\ndef some_other_property(self,*args,**kwargs):\n\nprint(“calling some_other_property getter({0},{1},{2})”.format(self,args,kwargs))\n\nreturn self._some_other_property\n\no = MyClass()\n\n# undecorated methods work like normal, they get the current instance (self) as the first argument\n\no.normal_method\n\n# <bound method MyClass.normal_method of <__main__.MyClass instance at 0x7fdd2537ea28>>\n\no.normal_method()\n\n# normal_method((<__main__.MyClass instance at 0x7fdd2537ea28>,),{})\n\no.normal_method(1,2,x=3,y=4)\n\n# normal_method((<__main__.MyClass instance at 0x7fdd2537ea28>, 1, 2),{‘y’: 4, ‘x’: 3})\n\n# class methods always get the class as the first argument\n\no.class_method\n\n# <bound method classobj.class_method of <class __main__.MyClass at 0x7fdd2536a390>>\n\no.class_method()\n\n# class_method((<class __main__.MyClass at 0x7fdd2536a390>,),{})\n\no.class_method(1,2,x=3,y=4)\n\n# class_method((<class __main__.MyClass at 0x7fdd2536a390>, 1, 2),{‘y’: 4, ‘x’: 3})\n\n# static methods have no arguments except the ones you pass in when you call them\n\no.static_method\n\n# <function static_method at 0x7fdd25375848>\n\no.static_method()\n\n# static_method((),{})\n\no.static_method(1,2,x=3,y=4)\n\n# static_method((1, 2),{‘y’: 4, ‘x’: 3})\n\n# properties are a way of implementing getters and setters. It’s an error to explicitly call them\n\n# “read only” attributes can be specified by creating a getter without a setter (as in some_other_property)\n\no.some_property\n\n# calling some_property getter(<__main__.MyClass instance at 0x7fb2b70877e8>,(),{})\n\n# ‘properties are nice’\n\no.some_property()\n\n# calling some_property getter(<__main__.MyClass instance at 0x7fb2b70877e8>,(),{})\n\n# Traceback (most recent call last):\n\n# File “<stdin>”, line 1, in <module>\n\n# TypeError: ‘str’ object is not callable\n\no.some_other_property\n\n# calling some_other_property getter(<__main__.MyClass instance at 0x7fb2b70877e8>,(),{})\n\n# ‘VERY nice’\n\n# o.some_other_property()\n\n# calling some_other_property getter(<__main__.MyClass instance at 0x7fb2b70877e8>,(),{})\n\n# Traceback (most recent call last):\n\n# File “<stdin>”, line 1, in <module>\n\n# TypeError: ‘str’ object is not callable\n\no.some_property = “groovy”\n\n# calling some_property setter(<__main__.MyClass object at 0x7fb2b7077890>,(‘groovy’,),{})\n\no.some_property\n\n# calling some_property getter(<__main__.MyClass object at 0x7fb2b7077890>,(),{})\n\n# ‘groovy’\n\no.some_other_property = “very groovy”\n\n# Traceback (most recent call last):\n\n# File “<stdin>”, line 1, in <module>\n\n# AttributeError: can’t set attribute\n\no.some_other_property\n\n# calling some_other_property getter(<__main__.MyClass object at 0x7fb2b7077890>,(),{})\n\n# ‘VERY nice’\n\nQ140).Consider the following code, what will it output?\n\nclass A(object):\n\ndef go(self):\n\nprint(“go A go!”)\n\ndef stop(self):\n\nprint(“stop A stop!”)\n\ndef pause(self):\n\nraise Exception(“Not Implemented”)\n\nclass B(A):\n\ndef go(self):\n\nsuper(B, self).go()\n\nprint(“go B go!”)\n\nclass C(A):\n\ndef go(self):\n\nsuper(C, self).go()\n\nprint(“go C go!”)\n\ndef stop(self):\n\nsuper(C, self).stop()\n\nprint(“stop C stop!”)\n\nclass D(B,C):\n\ndef go(self):\n\nsuper(D, self).go()\n\nprint(“go D go!”)\n\ndef stop(self):\n\nsuper(D, self).stop()\n\nprint(“stop D stop!”)\n\ndef pause(self):\n\nprint(“wait D wait!”)\n\nclass E(B,C): pass\n\na = A()\n\nb = B()\n\nc = C()\n\nd = D()\n\ne = E()\n\n# specify output from here onwards\n\na.go()\n\nb.go()\n\nc.go()\n\nd.go()\n\ne.go()\n\na.stop()\n\nb.stop()\n\nc.stop()\n\nd.stop()\n\ne.stop()\n\na.pause()\n\nb.pause()\n\nc.pause()\n\nd.pause()\n\ne.pause()\n\nAns140:\n\nThe output is specified in the comments in the segment below:\n\na.go()\n\n# go A go!\n\nb.go()\n\n# go A go!\n\n# go B go!\n\nc.go()\n\n# go A go!\n\n# go C go!\n\nd.go()\n\n# go A go!\n\n# go C go!\n\n# go B go!\n\n# go D go!\n\ne.go()\n\n# go A go!\n\n# go C go!\n\n# go B go!\n\na.stop()\n\n# stop A stop!\n\nb.stop()\n\n# stop A stop!\n\nc.stop()\n\n# stop A stop!\n\n# stop C stop!\n\nd.stop()\n\n# stop A stop!\n\n# stop C stop!\n\n# stop D stop!\n\ne.stop()\n\n# stop A stop!\n\na.pause()\n\n# … Exception: Not Implemented\n\nb.pause()\n\n# … Exception: Not Implemented\n\nc.pause()\n\n# … Exception: Not Implemented\n\nd.pause()\n\n# wait D wait!\n\ne.pause()\n\n# …Exception: Not Implemented\n\nWhy do we care?\n\nBecause OO programming is really, really important. Really. Answering this question shows your understanding of inheritance and the use of Python’s super function. Most of the time the order of resolution doesn’t matter. Sometimes it does, it depends on your application.\n\nQ141).\n\nConsider the following code, what will it output?\n\nclass Node(object):\n\ndef __init__(self,sName):\n\nself._lChildren = []\n\nself.sName = sName\n\ndef __repr__(self):\n\nreturn “<Node ‘{}’>”.format(self.sName)\n\ndef append(self,*args,**kwargs):\n\nself._lChildren.append(*args,**kwargs)\n\ndef print_all_1(self):\n\nprint(self)\n\nfor oChild in self._lChildren:\n\noChild.print_all_1()\n\ndef print_all_2(self):\n\ndef gen(o):\n\nlAll = [o,]\n\nwhile lAll:\n\noNext = lAll.pop(0)\n\nlAll.extend(oNext._lChildren)\n\nyield oNext\n\nfor oNode in gen(self):\n\nprint(oNode)\n\noRoot = Node(“root”)\n\noChild1 = Node(“child1”)\n\noChild2 = Node(“child2”)\n\noChild3 = Node(“child3”)\n\noChild4 = Node(“child4”)\n\noChild5 = Node(“child5”)\n\noChild6 = Node(“child6”)\n\noChild7 = Node(“child7”)\n\noChild8 = Node(“child8”)\n\noChild9 = Node(“child9”)\n\noChild10 = Node(“child10”)\n\noRoot.append(oChild1)\n\noRoot.append(oChild2)\n\noRoot.append(oChild3)\n\noChild1.append(oChild4)\n\noChild1.append(oChild5)\n\noChild2.append(oChild6)\n\noChild4.append(oChild7)\n\noChild3.append(oChild8)\n\noChild3.append(oChild9)\n\noChild6.append(oChild10)\n\n# specify output from here onwards\n\noRoot.print_all_1()\n\noRoot.print_all_2()\n\nAns141:\n\noRoot.print_all_1() prints:\n\n<Node ‘root’>\n\n<Node ‘child1’>\n\n<Node ‘child4’>\n\n<Node ‘child7’>\n\n<Node ‘child5’>\n\n<Node ‘child2’>\n\n<Node ‘child6’>\n\n<Node ‘child10’>\n\n<Node ‘child3’>\n\n<Node ‘child8’>\n\n<Node ‘child9’>\n\noRoot.print_all_2() prints:\n\n<Node ‘root’>\n\n<Node ‘child1’>\n\n<Node ‘child2’>\n\n<Node ‘child3’>\n\n<Node ‘child4’>\n\n<Node ‘child5’>\n\n<Node ‘child6’>\n\n<Node ‘child8’>\n\n<Node ‘child9’>\n\n<Node ‘child7’>\n\n<Node ‘child10’>\n\nWhy do we care?\n\nBecause composition and object construction is what objects are all about. Objects are composed of stuff and they need to be initialised somehow. This also ties up some stuff about recursion and use of generators.\n\nGenerators are great. You could have achieved similar functionality to print_all_2 by just constructing a big long list and then printing it’s contents. One of the nice things about generators is that they don’t need to take up much space in memory.\n\nIt is also worth pointing out that print_all_1 traverses the tree in a depth-first manner, while print_all_2 is width-first. Make sure you understand those terms. Sometimes one kind of traversal is more appropriate than the other. But that depends very much on your application.\n\nQ142).Describe Python’s garbage collection mechanism in brief.\n\nAns142: A lot can be said here. There are a few main points that you should mention:\n\n• Python maintains a count of the number of references to each object in memory. If a reference count goes to zero then the associated object is no longer live and the memory allocated to that object can be freed up for something else\n• occasionally things called “reference cycles” happen. The garbage collector periodically looks for these and cleans them up. An example would be if you have two objects o1and o2 such that x == o2 and o2.x == o1. If o1 and o2 are not referenced by anything else then they shouldn’t be live. But each of them has a reference count of 1.\n• Certain heuristics are used to speed up garbage collection. For example, recently created objects are more likely to be dead. As objects are created, the garbage collector assigns them to generations. Each object gets one generation, and younger generations are dealt with first.\n\nThis explanation is CPython specific.\n\nQ143).\n\nPlace the following functions below in order of their efficiency. They all take in a list of numbers between 0 and 1. The list can be quite long. An example input list would be [random.random() for i in range(100000)]. How would you prove that your answer is correct?\n\ndef f1(lIn):\n\nl1 = sorted(lIn)\n\nl2 = [i for i in l1 if i<0.5]\n\nreturn [i*i for i in l2]\n\ndef f2(lIn):\n\nl1 = [i for i in lIn if i<0.5]\n\nl2 = sorted(l1)\n\nreturn [i*i for i in l2]\n\ndef f3(lIn):\n\nl1 = [i*i for i in lIn]\n\nl2 = sorted(l1)\n\nreturn [i for i in l1 if i<(0.5*0.5)]\n\nAns143:\n\nMost to least efficient: f2, f1, f3. To prove that this is the case, you would want to profile your code. Python has a lovely profiling package that should do the trick.\n\nimport cProfile\n\nlIn = [random.random() for i in range(100000)]\n\ncProfile.run(‘f1(lIn)’)\n\ncProfile.run(‘f2(lIn)’)\n\ncProfile.run(‘f3(lIn)’)\n\nFor completion’s sake, here is what the above profile outputs:\n\n>>> cProfile.run(‘f1(lIn)’)\n\n4 function calls in 0.045 seconds\n\nOrdered by: standard name\n\nncalls tottime percall cumtime percall filename:lineno(function)\n\n1 0.009 0.009 0.044 0.044 <stdin>:1(f1)\n\n1 0.001 0.001 0.045 0.045 <string>:1(<module>)\n\n1 0.000 0.000 0.000 0.000 {method ‘disable’ of ‘_lsprof.Profiler’ objects}\n\n1 0.035 0.035 0.035 0.035 {sorted}\n\n>>> cProfile.run(‘f2(lIn)’)\n\n4 function calls in 0.024 seconds\n\nOrdered by: standard name\n\nncalls tottime percall cumtime percall filename:lineno(function)\n\n1 0.008 0.008 0.023 0.023 <stdin>:1(f2)\n\n1 0.001 0.001 0.024 0.024 <string>:1(<module>)\n\n1 0.000 0.000 0.000 0.000 {method ‘disable’ of ‘_lsprof.Profiler’ objects}\n\n1 0.016 0.016 0.016 0.016 {sorted}\n\n>>> cProfile.run(‘f3(lIn)’)\n\n4 function calls in 0.055 seconds\n\nOrdered by: standard name\n\nncalls tottime percall cumtime percall filename:lineno(function)\n\n1 0.016 0.016 0.054 0.054 <stdin>:1(f3)\n\n1 0.001 0.001 0.055 0.055 <string>:1(<module>)\n\n1 0.000 0.000 0.000 0.000 {method ‘disable’ of ‘_lsprof.Profiler’ objects}\n\n1 0.038 0.038 0.038 0.038 {sorted}\n\nWhy Care?\n\nLocating and avoiding bottlenecks is often pretty worthwhile. A lot of coding for efficiency comes down to common sense – in the example above it’s obviously quicker to sort a list if it’s a smaller list, so if you have the choice of filtering before a sort it’s often a good idea. The less obvious stuff can still be located through use of the proper tools. It’s good to know about these tools.\n\nQ144).What is the purpose of PYTHONPATH environment variable?\n\nAns144: PYTHONPATH – It has a role similar to PATH. This variable tells the Python interpreter where to locate the module files imported into a program. It should include the Python source library directory and the directories containing Python source code. PYTHONPATH is sometimes preset by the Python installer.\n\nQ145).What is the purpose of PYTHONSTARTUP environment variable?\n\nAns145: PYTHONSTARTUP – It contains the path of an initialization file containing Python source code. It is executed every time you start the interpreter. It is named as .pythonrc.py in Unix and it contains commands that load utilities or modify PYTHONPATH.\n\nQ146).What is the purpose of PYTHONCASEOK environment variable?\n\nAns146: PYTHONCASEOK − It is used in Windows to instruct Python to find the first case-insensitive match in an import statement. Set this variable to any value to activate it.\n\nQ147).What is the purpose of PYTHONHOME environment variable?\n\nAns147: PYTHONHOME − It is an alternative module search path. It is usually embedded in the PYTHONSTARTUP or PYTHONPATH directories to make switching module libraries easy.\n\nQ148).What does a python do?\n\nAns148: Python is a general-purpose programming language typically used for web development. … SQLite is one free lightweight database commonly used by Pythonprogrammers to store data. Many highly trafficked websites, such as YouTube, are created using Python.\n\nQ149).What is the interpreter in Python ?\n\nAns149: An interpreter is a program that reads and executes code. This includes source code, pre-compiled code, and scripts. Common interpreters include Perl, Python, and Ruby interpreters, which execute Perl, Python, and Ruby code respectively.\n\nQ150).Why is it called Python ?\n\nAns150: When he began implementing Python, Guido van Rossum was also reading the published scripts from “Monty Python’s Flying Circus”, a BBC comedy series from the 1970s. Van Rossum thought he needed a name that was short, unique, and slightly mysterious, so he decided to call the language Python\n\nSummary\nReview Date\nReviewed Item\nAmazing. I got most of the commonly asked interview questions from this site\nAuthor Rating\n5",
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"#",
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"### Sidebar\n\n$\\color{red}{P(1, 3)} = \\color{blue}{T(4)} \\color{magenta}{\\rtimes} \\color{green}{SO(1, 3)}$\n\n# Poincare Group\n\n## Intuitive\n\nThe Poincare group is the mathematical tool that we use to describe the symmetry of special relativity.\n\nThe starting point for Einstein on his road towards what is now called special relativity was the experimental observation that the speed of light has the same value in all inertial frames of reference. This curious fact of nature was discovered by the famous Michelson-Morley experiment.\n\nA symmetry is a transformation that we can perform without changing something. Therefore, the invariance of the speed of light under arbitrary changes of the frame of reference is a symmetry and we call this symmetry the Poincare group. The Poincare group contains all transformations that we can perform without changing the speed of light.\n\n## Concrete\n\nThe $\\color{red}{\\text{Poincare group}}$ consists of $\\color{blue}{\\text{translations}}$ $\\color{magenta}{\\text{plus}}$ $\\color{green}{\\text{rotations and boosts}}$.\n\nThe Poincare Algebra\n\n\\begin{eqnarray} {}[J_i,J_j]&=&i \\epsilon_{ijk}J_k, \\nonumber \\\\ {} [J_i,K_j]&=& i\\epsilon_{ijk} K_k, \\nonumber \\\\ {} [K_i,K_j]&=& -i \\epsilon_{ijk}J_k ,\\nonumber \\\\ {} [J_i,P_0]&=&0, \\nonumber \\\\ {} [J_i,P_j]&=& i\\epsilon_{ijk} P_k, \\nonumber \\\\ {} [P_0,P_i]&=&0, \\nonumber \\\\ {} [K_i,P_0]&=&i P_i, \\nonumber \\\\ {}[K_i,P_j]&=& i\\ P_0 \\delta_{ij} , \\label{c} \\end{eqnarray}\n\n## Abstract\n\n• For a modern discussion of the Poincare group, see D. Giulini, The Poincare group: Algebraic, representation-theoretic, and geometric\n\naspects.\n\n## Why is it interesting?\n\nThe double cover of the Poincare group is the fundamental spacetime symmetry of modern physics and is a crucial component of the standard model of particle physics.\n\nThe Poincare group is the set of all transformations that leave the speed of light invariant. Thus, the Poincare group yields all possible transformations between allowed frames of reference. This is incredibly useful, when we want to write down fundamental laws of nature. The fundamental laws should be valid in all allowed frames of reference, otherwise they would be quite useless.\n\nIn practice, we can use our knowledge of all transformations inside the Poincare group to write down equations that are invariant under all these transformations. These equations then hold in all allowed frames of reference. This is such a strong restriction on the possible equations that is is almost enough to derive the most important equations of fundamental physics: the Dirac equation, the Klein-Gordon equation and the Maxwell-Equations.\n\n\"The Hilbert space of one-particle states is always an irreducible representation space of the Poincare group. […] The construction of the unitary irreducible representations of the Poincare group is probably the most successful part of special relativity (in particle physics, not in gravitation theory, for which it is a disaster). It permits us to classify all kinds of particles and implies the main conservation laws (energy-momentum and angular momentum). […] The translation generators are responsible for the energy-momentum conservation laws, the rotation generators of the conservation of angular momentum, and the boost generators of the conservation of initial position. \" from Reflections on the Evolution of Physical Theories by Henri Bacry\n\n\"The enlargement of the Lorentz group to the Poincare group was proposed [ 13] as a way of describing the quantum states of relativistic particles without using the wave equations. The states of a free particle are then given by the unitary irreducible representations of the Poincare group.\" from Deformed Poincare containing the exact Lorentz algebra by Alexandros A. Kehagias et. al.\n\n## FAQ\n\nWhy do we say that irreducible representation of Poincare group represents the one-particle state?\nsee https://physics.stackexchange.com/questions/73593/why-do-we-say-that-irreducible-representation-of-poincare-group-represents-the-o",
null,
""
] | [
null,
"https://physicstravelguide.com/_media/logo.png",
null,
"https://physicstravelguide.com/lib/exe/indexer.php",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8188383,"math_prob":0.98114705,"size":3693,"snap":"2019-51-2020-05","text_gpt3_token_len":868,"char_repetition_ratio":0.13933316,"word_repetition_ratio":0.042718448,"special_character_ratio":0.22529109,"punctuation_ratio":0.09250399,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99143916,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-23T18:19:43Z\",\"WARC-Record-ID\":\"<urn:uuid:895d46e5-5b49-4c58-8308-326868bb4ab6>\",\"Content-Length\":\"77512\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b075240a-2473-49e7-ae45-d18138b7d224>\",\"WARC-Concurrent-To\":\"<urn:uuid:893e25db-18d3-4f9b-b219-043593def4a9>\",\"WARC-IP-Address\":\"104.27.138.208\",\"WARC-Target-URI\":\"https://physicstravelguide.com/advanced_tools/group_theory/poincare_group\",\"WARC-Payload-Digest\":\"sha1:32O32HHNZI2Y3BTUYL4ZINWCPZJTOPZM\",\"WARC-Block-Digest\":\"sha1:UXMQ4KDMZ4WNWGDI3YKCKVP22OA4QSPV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250611127.53_warc_CC-MAIN-20200123160903-20200123185903-00367.warc.gz\"}"} |
https://www.traditionaloven.com/tutorials/angle/convert-complete-rotation-rot-to-gradian-angle-unit.html | [
" Convert rot to grad | complete rotation to gradians\n\n# angle units conversion\n\n## Amount: 1 complete rotation (rot) of angle Equals: 400.00 gradians (grad) in angle\n\nConverting complete rotation to gradians value in the angle units scale.\n\nTOGGLE : from gradians into complete rotations in the other way around.\n\n## angle from complete rotation to gradian conversion results\n\n### Enter a new complete rotation number to convert\n\n* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)\n* Precision is how many digits after decimal point (1 - 9)\n\nEnter Amount :\nDecimal Precision :\n\nCONVERT : between other angle measuring units - complete list.\n\nHow many gradians are in 1 complete rotation? The answer is: 1 rot equals 400.00 grad\n\n## 400.00 grad is converted to 1 of what?\n\nThe gradians unit number 400.00 grad converts to 1 rot, one complete rotation. It is the EQUAL angle value of 1 complete rotation but in the gradians angle unit alternative.\n\n rot/grad angle conversion result From Symbol Equals Result Symbol 1 rot = 400.00 grad\n\n## Conversion chart - complete rotations to gradians\n\n1 complete rotation to gradians = 400.00 grad\n\n2 complete rotations to gradians = 800.00 grad\n\n3 complete rotations to gradians = 1,200.00 grad\n\n4 complete rotations to gradians = 1,600.00 grad\n\n5 complete rotations to gradians = 2,000.00 grad\n\n6 complete rotations to gradians = 2,400.00 grad\n\n7 complete rotations to gradians = 2,800.00 grad\n\n8 complete rotations to gradians = 3,200.00 grad\n\n9 complete rotations to gradians = 3,600.00 grad\n\n10 complete rotations to gradians = 4,000.00 grad\n\n11 complete rotations to gradians = 4,400.00 grad\n\n12 complete rotations to gradians = 4,800.00 grad\n\n13 complete rotations to gradians = 5,200.00 grad\n\n14 complete rotations to gradians = 5,600.00 grad\n\n15 complete rotations to gradians = 6,000.00 grad\n\nConvert angle of complete rotation (rot) and gradians (grad) units in reverse from gradians into complete rotations.\n\n## Angles\n\nThis calculator is based on conversion of two angle units. An angle consists of two rays (as in sides of an angle sharing a common vertex or else called the endpoint.) Some belong to rotation measurements - spherical angles measured by arcs' lengths, pointing from the center, plus the radius. For a whole set of multiple units of angle on one page, try that Multiunit converter tool which has built in all angle unit-variations. Page with individual angle units.\n\n# Converter type: angle units\n\nFirst unit: complete rotation (rot) is used for measuring angle.\nSecond: gradian (grad) is unit of angle.\n\nQUESTION:\n15 rot = ? grad\n\n15 rot = 6,000.00 grad\n\nAbbreviation, or prefix, for complete rotation is:\nrot\nAbbreviation for gradian is:"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7779626,"math_prob":0.99532795,"size":2775,"snap":"2022-40-2023-06","text_gpt3_token_len":692,"char_repetition_ratio":0.33092746,"word_repetition_ratio":0.03125,"special_character_ratio":0.2673874,"punctuation_ratio":0.1510129,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99565905,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-24T20:25:22Z\",\"WARC-Record-ID\":\"<urn:uuid:90463a16-36b5-466b-92a8-8fd8a8088292>\",\"Content-Length\":\"45866\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:80406f37-55bd-47a5-bfb1-713b5b37435b>\",\"WARC-Concurrent-To\":\"<urn:uuid:9e2ffcc2-34a4-42e9-8eb4-3894bb87a64b>\",\"WARC-IP-Address\":\"162.241.171.12\",\"WARC-Target-URI\":\"https://www.traditionaloven.com/tutorials/angle/convert-complete-rotation-rot-to-gradian-angle-unit.html\",\"WARC-Payload-Digest\":\"sha1:QTSVIGAJARBG7RCM254XSRMUEXZNGYPJ\",\"WARC-Block-Digest\":\"sha1:XDPBYKINKBJIUOFKXZIMZQNMOOS4HHP6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030333455.97_warc_CC-MAIN-20220924182740-20220924212740-00049.warc.gz\"}"} |
https://community.eazybi.com/t/eazybi-formula-remaining-time-of-issues-at-the-start-of-sprint/3247 | [
"# EazyBI Formula - remaining time of issues at the start of sprint\n\nDear Community,\n\nWe are currently evaluating EazyBI for JIRA.\nWe need a formula to calculate a column with the values of remaining time of each sub-task at the time of a specific sprint started (not including changes of issues’ remaining time during the sprint).\nMore precisely, I need to know the remaining time of sub-tasks of a sprint at the start time of the sprint.\n\nRegards,\nAli\n\n1 Like\n\nHi,\n\nThe recommended approach for calculation of the remaining estimate for tasks at the start of the sprint would be trying to implement a measure similar to the formula under the “Story points committed” measure. There is a hidden measure in eazyBI data model “Remaining estimated hours added” for that and the formula is the following:\n\n``````( [Measures].[Remaining estimated hours added],\n[Transition Field].[Sprint status],\n[Sprint Status].[Active],\n[Issue Sprint Status Change].[Future => Active],\n[Time].CurrentHierarchy.Levels('Day').DateMember(\n[Sprint].CurrentMember.get('Start date')\n)\n)\n``````\n\nThere is, however, a limitation for this formula for sub-tasks. Currently the Jira API does not provide a valid sprint history for sub-tasks: https://jira.atlassian.com/browse/JSWSERVER-11459\n\nThe workaround for this problem is to check if the parent of the sub-task has the needed relation to sprint and then show the remaining estimate on the sprint start date. The formula working with the Issue dimension Issue level and Sprint dimension could be like this:\n\n``````CASE WHEN\nAND\n([Measures].[Issues history],\n[Time].[Day].DateMember([Measures].[Sprint start date]),\n[Issue].[Issue].GetMemberByKey(\n[Measures].[Issue parent key]\n))>0\nTHEN\nNonZero(([Measures].[Remaining estimated hours history],\n[Time].[Day].DateMember([Measures].[Sprint start date]),\n[Sprint].DefaultMember\n))\nEND\n``````\n\nKindly,\nJanis, eazyBI support\n\nHi,\nI tried to use both of the examples here -\nI get empty results for the second one… this is what I did\nCASE WHEN\nAND\n([Measures].[Issues history],\n[Time].[Day].DateMember([Measures].[Sprint start date]),\n[Issue].[Issue].GetMemberByKey(\n[Measures].[Issue parent key]\n))>0\nTHEN\nNonZero(([Measures].[Remaining estimated hours history],\n[Time].[Day].DateMember([Measures].[Sprint start date]),\n[Sprint].DefaultMember\n))\nEND\n\nand I actually do get results for the first option - can you explain again the issue here? it means that the results I get my not be updated?\nThanks"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.84852296,"math_prob":0.9347983,"size":2061,"snap":"2022-27-2022-33","text_gpt3_token_len":488,"char_repetition_ratio":0.1614001,"word_repetition_ratio":0.1696113,"special_character_ratio":0.22901504,"punctuation_ratio":0.11873351,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.993891,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-08T04:38:31Z\",\"WARC-Record-ID\":\"<urn:uuid:990fc1b0-233d-42c1-9842-fb3c2fa09939>\",\"Content-Length\":\"21648\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b0a339e9-ebcc-43e4-bd1c-8ba977f1a61d>\",\"WARC-Concurrent-To\":\"<urn:uuid:f5977a1f-66b0-4973-83d9-b83b4efa5e12>\",\"WARC-IP-Address\":\"130.211.47.213\",\"WARC-Target-URI\":\"https://community.eazybi.com/t/eazybi-formula-remaining-time-of-issues-at-the-start-of-sprint/3247\",\"WARC-Payload-Digest\":\"sha1:FAKWFBOLGPM2AV5LQAXSUYFJDRA5FKXJ\",\"WARC-Block-Digest\":\"sha1:2S7LNPILUXSMZY5QVASOHZTOIFP57D4B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570765.6_warc_CC-MAIN-20220808031623-20220808061623-00078.warc.gz\"}"} |
https://physics.stackexchange.com/questions/436835/restriction-forces-in-lagrangian-mechanics | [
"# Restriction Forces in Lagrangian Mechanics\n\nI was recently preparing for a test on Classical Mechanics and a friend of mine started wondering if there was any method through which we could obtain the restriction forces acting on a certain particle without using the Lagrange Multipliers method. At first glance my feeling would be that since Lagrangian Mechanics deals primarily with action and energies and not with forces, the answer would be no. But I'm really curious, any help would be appreciated.\n\n• FWIW, if the constraints are holonomic one doesn't need Lagrange multipliers to solve the Lagrange equations. – Qmechanic Oct 25 '18 at 20:34\n• I see. Can restriction forces be calculated without issue in that case? I'm not trying to obtain equations of motion or conserved quantities. How would one go about obtaining the RF's in the case of holonomic restraints? – Jorge Cabezut Oct 26 '18 at 15:29\n\n1. If the constraints are holonomic one doesn't need Lagrange multipliers to solve Lagrange equations $$\\frac{d}{dt}\\frac{\\partial T}{\\partial \\dot{q}^j}-\\frac{\\partial T}{\\partial q^j}~=~Q^a_j,\\qquad j~\\in~\\{1,\\ldots, n\\},\\tag{1}$$ where $$Q^a_j~=~\\sum_{i=1}^N {\\bf F}^a_i\\cdot \\frac{\\partial {\\bf r}_i}{\\partial q^j},\\qquad j~\\in~\\{1,\\ldots, n\\},\\tag{2}$$ is the $$j$$'th applied generalized force.\n2. The constraint force $${\\bf F}^c_{i}$$ on the $$i$$'th point particle with position $${\\bf r}_i(q^1, \\ldots, q^n,t),\\qquad i~\\in~\\{1,\\ldots, N\\},\\tag{3}$$ can then in principle be reconstructed via Newton's 2nd law $${\\bf F}^c_{i}~=~\\dot{\\bf p}_i-{\\bf F}^a_i,\\tag{4}$$ where $${\\bf F}_i^a$$ is the applied force from eq. (2)."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8910693,"math_prob":0.9992127,"size":1733,"snap":"2019-51-2020-05","text_gpt3_token_len":529,"char_repetition_ratio":0.1064199,"word_repetition_ratio":0.05785124,"special_character_ratio":0.29255626,"punctuation_ratio":0.113702625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99989283,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-17T15:57:13Z\",\"WARC-Record-ID\":\"<urn:uuid:67a24d63-1b67-4b6e-9195-377812927482>\",\"Content-Length\":\"138028\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8e3a627a-196e-484b-9e5e-43d1c193e4bb>\",\"WARC-Concurrent-To\":\"<urn:uuid:f534656d-cbdb-4b9f-b33d-4c3461442ab0>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/436835/restriction-forces-in-lagrangian-mechanics\",\"WARC-Payload-Digest\":\"sha1:FAYCSWPQPVHKFKRBVJ3WUHUQRPKL5V7R\",\"WARC-Block-Digest\":\"sha1:Q5Y3GIR33B6X7JBH6SBJ74JN7767UITD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250589861.0_warc_CC-MAIN-20200117152059-20200117180059-00484.warc.gz\"}"} |
https://djangocentral.com/python-program-to-compute-quotient-and-remainder-of-two-numbers/ | [
"# Python Program To Compute Quotient and Remainder Of Two Numbers\n\nIn Division The number which we divide is called the dividend. The number by which we divide is called the divisor. The result obtained is called the quotient. The number left over is called the remainder.",
null,
"## Problem Definition\n\nCreate a Python program to compute Quotient and reminder of two given numbers.\n\n## Program\n\n``````divisor = 5\ndividend = 27\nquotient = dividend//divisor\nreminder = dividend % divisor\nprint(\"Quotient is\", quotient)\nprint(\"Reminder is\",reminder)``````\n\n#### Output\n\n``````Quotient is 5\nReminder is 2``````\n\nFirst, the numbers are saved in respective variables then to compute quotient we used the floor division `//` operator that returns the integer value of the quotient and for the reminder, the modulus `%` operator is used. Later we are just printing out the variables.\n\n## Problem Definition\n\nCreate a Python program to compute Quotient and reminder of two user-provided numbers in real-time.\n\n## Program\n\n``````divisor = int(input(\"Enter the divisor\"))\ndividend = int(input(\"Enter the dividend\"))\nquotient = dividend//divisor\nreminder = dividend % divisor\nprint(\"Quotient is {} and Reminder is {}\".format(quotient, reminder))``````\n\n#### Output\n\n``````Enter the divisor 5\nEnter the dividend 27\nQuotient is 5 and Reminder is 2``````\n\nHere we are taking input from the user using the Python’s built-in `input()` method then we are converting it to an integer using the `int()` method because `input()` returns the objects as a string object.\n\nLater we are performing the arithmetic operation and printing out the results using string formatting."
] | [
null,
"https://djangocentral.com/wp-content/uploads/2019/10/Remainder-and-Quotient.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.81074715,"math_prob":0.9816011,"size":1547,"snap":"2020-34-2020-40","text_gpt3_token_len":341,"char_repetition_ratio":0.18081659,"word_repetition_ratio":0.13389121,"special_character_ratio":0.21008404,"punctuation_ratio":0.06130268,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99881136,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-11T12:52:22Z\",\"WARC-Record-ID\":\"<urn:uuid:eb12dd9e-41ca-4bce-b457-d57999251821>\",\"Content-Length\":\"59071\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:490d0b50-21df-4a98-b75a-8a9029030752>\",\"WARC-Concurrent-To\":\"<urn:uuid:289ce5b9-7d0e-41f9-9f3a-bb815dab25f3>\",\"WARC-IP-Address\":\"104.27.168.169\",\"WARC-Target-URI\":\"https://djangocentral.com/python-program-to-compute-quotient-and-remainder-of-two-numbers/\",\"WARC-Payload-Digest\":\"sha1:3IZ4X6CX72XTAOVHS7USWXWIKAABOAFC\",\"WARC-Block-Digest\":\"sha1:R6RKVIZZWVP7Z65JE73ZS7LO52TWLT2C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738777.54_warc_CC-MAIN-20200811115957-20200811145957-00204.warc.gz\"}"} |
https://www.aimsciences.org/article/doi/10.3934/dcdss.2018065 | [
"",
null,
"",
null,
"",
null,
"",
null,
"December 2018, 11(6): 1143-1167. doi: 10.3934/dcdss.2018065\n\n## First-order partial differential equations and consumer theory\n\n 1-50-1601 Miyamachi, Fuchu, Tokyo, 183-0023, Japan\n\nReceived February 2017 Revised June 2017 Published June 2018\n\nIn this paper, we show that the existence of a global solution of a standard first-order partial differential equation can be reduced to the extendability of the solution of the corresponding ordinary differential equation under the differentiable and locally Lipschitz environments. By using this result, we can produce many known existence theorems for partial differential equations. Moreover, we demonstrate that such a result can be applied to the integrability problem in consumer theory. This result holds even if the differentiability condition is dropped.\n\nCitation: Yuhki Hosoya. First-order partial differential equations and consumer theory. Discrete & Continuous Dynamical Systems - S, 2018, 11 (6) : 1143-1167. doi: 10.3934/dcdss.2018065\n##### References:\n J. Dieudonné, Foundations of Modern Analysis, Hesperides press, 2006.",
null,
"Google Scholar P. Hartman, Ordinary Differential Equations, Birkhäuser Boston, Mass., 1982.",
null,
"Google Scholar Y. Hosoya, On first-order partial differential equations: An existence theorem and its applications, Advances in Mathematical Economics, 20 (2016), 77-87. doi: 10.1007/978-981-10-0476-6_3.",
null,
"",
null,
"Google Scholar L. Hurwicz and H. Uzawa, On the Integrability of Demand Functions, in Preference, Utility and Demand (eds. J. S. Chipman, L. Hurwicz, M. K. Richter and H. F. Sonnenschein) Harcourt Brace Jovanovich, Inc., New York, (1971), 114–148.",
null,
"Google Scholar A. D. Ioffe and V. M. Tikhomirov, Theory of Extremal Problems, Elsevier, 1979.",
null,
"Google Scholar W. Nikliborc, Sur les équations linéaires aux différentielles totales, Studia Mathematica, 1 (1929), 41-49. doi: 10.4064/sm-1-1-41-49.",
null,
"Google Scholar L. S. Pontryagin, Ordinary Differential Equations, Addison-Wesley, Reading, Massachusetts, 1962.",
null,
"Google Scholar S. Smale and M. W. Hirsch, Differential Equations, Dynamical Systems, and Linear Algebra, Academic Press, 1974.",
null,
"Google Scholar\n\nshow all references\n\n##### References:\n J. Dieudonné, Foundations of Modern Analysis, Hesperides press, 2006.",
null,
"Google Scholar P. Hartman, Ordinary Differential Equations, Birkhäuser Boston, Mass., 1982.",
null,
"Google Scholar Y. Hosoya, On first-order partial differential equations: An existence theorem and its applications, Advances in Mathematical Economics, 20 (2016), 77-87. doi: 10.1007/978-981-10-0476-6_3.",
null,
"",
null,
"Google Scholar L. Hurwicz and H. Uzawa, On the Integrability of Demand Functions, in Preference, Utility and Demand (eds. J. S. Chipman, L. Hurwicz, M. K. Richter and H. F. Sonnenschein) Harcourt Brace Jovanovich, Inc., New York, (1971), 114–148.",
null,
"Google Scholar A. D. Ioffe and V. M. Tikhomirov, Theory of Extremal Problems, Elsevier, 1979.",
null,
"Google Scholar W. Nikliborc, Sur les équations linéaires aux différentielles totales, Studia Mathematica, 1 (1929), 41-49. doi: 10.4064/sm-1-1-41-49.",
null,
"Google Scholar L. S. Pontryagin, Ordinary Differential Equations, Addison-Wesley, Reading, Massachusetts, 1962.",
null,
"Google Scholar S. Smale and M. W. Hirsch, Differential Equations, Dynamical Systems, and Linear Algebra, Academic Press, 1974.",
null,
"Google Scholar\n Heiko Enderling, Alexander R.A. Anderson, Mark A.J. Chaplain, Glenn W.A. Rowe. Visualisation of the numerical solution of partial differential equation systems in three space dimensions and its importance for mathematical models in biology. Mathematical Biosciences & Engineering, 2006, 3 (4) : 571-582. doi: 10.3934/mbe.2006.3.571 Changpin Li, Zhiqiang Li. Asymptotic behaviors of solution to partial differential equation with Caputo–Hadamard derivative and fractional Laplacian: Hyperbolic case. Discrete & Continuous Dynamical Systems - S, 2021, 14 (10) : 3659-3683. doi: 10.3934/dcdss.2021023 Wilhelm Schlag. Spectral theory and nonlinear partial differential equations: A survey. Discrete & Continuous Dynamical Systems, 2006, 15 (3) : 703-723. doi: 10.3934/dcds.2006.15.703 Shaoyong Lai, Yong Hong Wu, Xu Yang. The global solution of an initial boundary value problem for the damped Boussinesq equation. Communications on Pure & Applied Analysis, 2004, 3 (2) : 319-328. doi: 10.3934/cpaa.2004.3.319 Zhong Tan, Jianfeng Zhou. Higher integrability of weak solution of a nonlinear problem arising in the electrorheological fluids. Communications on Pure & Applied Analysis, 2016, 15 (4) : 1335-1350. doi: 10.3934/cpaa.2016.15.1335 Qi Lü, Xu Zhang. A concise introduction to control theory for stochastic partial differential equations. Mathematical Control & Related Fields, 2021 doi: 10.3934/mcrf.2021020 Minoo Kamrani. Numerical solution of partial differential equations with stochastic Neumann boundary conditions. Discrete & Continuous Dynamical Systems - B, 2019, 24 (10) : 5337-5354. doi: 10.3934/dcdsb.2019061 Avner Friedman, Harsh Vardhan Jain. A partial differential equation model of metastasized prostatic cancer. Mathematical Biosciences & Engineering, 2013, 10 (3) : 591-608. doi: 10.3934/mbe.2013.10.591 Jing Li, Boling Guo, Lan Zeng, Yitong Pei. Global weak solution and smooth solution of the periodic initial value problem for the generalized Landau-Lifshitz-Bloch equation in high dimensions. Discrete & Continuous Dynamical Systems - B, 2020, 25 (4) : 1345-1360. doi: 10.3934/dcdsb.2019230 Yukihiko Nakata. Existence of a period two solution of a delay differential equation. Discrete & Continuous Dynamical Systems - S, 2021, 14 (3) : 1103-1110. doi: 10.3934/dcdss.2020392 Nguyen Thi Hoai. Asymptotic approximation to a solution of a singularly perturbed linear-quadratic optimal control problem with second-order linear ordinary differential equation of state variable. Numerical Algebra, Control & Optimization, 2021, 11 (4) : 495-512. doi: 10.3934/naco.2020040 Ábel Garab. Unique periodic orbits of a delay differential equation with piecewise linear feedback function. Discrete & Continuous Dynamical Systems, 2013, 33 (6) : 2369-2387. doi: 10.3934/dcds.2013.33.2369 Rui L. Fernandes, Yuxuan Zhang. Local and global integrability of Lie brackets. Journal of Geometric Mechanics, 2021, 13 (3) : 355-384. doi: 10.3934/jgm.2021024 Masaki Hibino. Gevrey asymptotic theory for singular first order linear partial differential equations of nilpotent type — Part I —. Communications on Pure & Applied Analysis, 2003, 2 (2) : 211-231. doi: 10.3934/cpaa.2003.2.211 Kyeong-Hun Kim, Kijung Lee. A weighted $L_p$-theory for second-order parabolic and elliptic partial differential systems on a half space. Communications on Pure & Applied Analysis, 2016, 15 (3) : 761-794. doi: 10.3934/cpaa.2016.15.761 Tuhin Ghosh, Karthik Iyer. Cloaking for a quasi-linear elliptic partial differential equation. Inverse Problems & Imaging, 2018, 12 (2) : 461-491. doi: 10.3934/ipi.2018020 Susanna V. Haziot. Study of an elliptic partial differential equation modelling the Antarctic Circumpolar Current. Discrete & Continuous Dynamical Systems, 2019, 39 (8) : 4415-4427. doi: 10.3934/dcds.2019179 Dimitra Antonopoulou, Georgia Karali. A nonlinear partial differential equation for the volume preserving mean curvature flow. Networks & Heterogeneous Media, 2013, 8 (1) : 9-22. doi: 10.3934/nhm.2013.8.9 Roberto Camassa, Pao-Hsiung Chiu, Long Lee, W.-H. Sheu. A particle method and numerical study of a quasilinear partial differential equation. Communications on Pure & Applied Analysis, 2011, 10 (5) : 1503-1515. doi: 10.3934/cpaa.2011.10.1503 Frederic Abergel, Remi Tachet. A nonlinear partial integro-differential equation from mathematical finance. Discrete & Continuous Dynamical Systems, 2010, 27 (3) : 907-917. doi: 10.3934/dcds.2010.27.907\n\n2020 Impact Factor: 2.425"
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https://www.fcc.chalmers.se/publication/computationally-efficient-viscoelastic-flow-simulation-using-a-lagrangian-eulerian-method-and-gpu-acceleration/ | [
"",
null,
"# Computationally efficient viscoelastic flow simulation using a Lagrangian-Eulerian method and GPU-acceleration\n\nS. Ingelsten, A. Mark, K. Jareteg, R. Kádár, F. Edelvik. Journal of Non-Newtonian Fluid Mechanics. Online 18 March 2020, 104264.\n\n## Abstract\n\nA recently proposed Lagrangian-Eulerian method for viscoelastic flow simulation is extended to high performance calculations on the Graphics Processing Unit (GPU). The two most computationally intensive parts of the algorithm are implemented for GPU calculation, namely the integration of the viscoelastic constitutive equation at the Lagrangian nodes and the interpolation of the resulting stresses to the cell centers of the Eulerian grid.\n\nIn the original CPU method, the constitutive equations are integrated with a second order backward differentiation formula, while with the proposed GPU method the implicit Euler method is used. To allow fair comparison, the latter is also implemented for the CPU. The methods are validated for two flows, a planar Poiseuille flow of an upper-convected Maxwell fluid and flow past a confined cylinder of a four-mode Phan Thien Tanner fluid, with identical results.\n\nThe calculation times for the methods are compared for a range of grid resolutions and numbers of CPU threads, revealing a significant reduction of the calculation time for the proposed GPU method. As an example, the total simulation time is roughly halved compared to the original CPU method. The integration of the constitutive equation itself is reduced by a factor 50 to 250 and the unstructured stress interpolation by a factor 15 to 60, depending on the number of CPU threads used.\n\nPhoto credits: Nic McPhee"
] | [
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https://physics.stackexchange.com/questions/554156/simultaneity-and-special-relativity | [
"# Simultaneity and special relativity\n\nSuppose, in inertial reference frame $$F_1$$, observers A and B are at rest, each having torch, and are separated by some distance and we have put machine M at middle of A and B.\n\nMachine M has light bulbs on both sides ,right and left, so that if it catches light from A which is at left ,then machine M glows left light bulb, similar with right bulb.Also, if it senses both reaching at same instant of time then it start to make noise.\n\nNow consider another inertial reference frame $$F_2$$ which is moving at constant speed $$v$$ with respect to $$F_1$$ to the right.\n\nNow ,in frame $$F_1$$ , both A and B turn on their torches at same instant of time, say $$t=0$$ and both rays reach at M at $$t=t_1$$, and machine M makes noise indicating that those events were \"simultaneous\" in $$F_1$$.\n\nNow we know these events are not simultaneous in $$F_2$$, in other words ,person seating in $$F_2$$ will say ,\"I should not hear sound from machine M.\" But somehow machine makes noise.(or it doesn't make noise?)\n\nSo does this mean according to $$F_2$$ ,machine is malfunctioning?\n\n• Hi Pratik, welcome to Physics StackExchange! Please split the textbook recommendation into a separate question on the site - this is considered best practice for some reasons detailed here. – Shivam Sarodia yesterday\n• This is the second question along these lines that this site has received in the last day. Is there a particular source that has prompted these questions, out of curiosity? In general, what brought you to ask this question? – probably_someone yesterday\n\nThe machine M responds to events right there at the machine---the events of light arriving from left, light arriving from right. So the machine is reporting that the light arrival events are simultaneous at M. This is fine; all reference frames will agree that two things happening at the same place and time do indeed happen at the same place and time.\n\nBut when we interpret M to be reporting that the emission events are simultaneous, now we have a frame-dependent interpretation. What M is really saying is \"well the two light beams reached me simultaneously, so what I can claim is that if the emitters are at distances $$d_1$$ and $$d_2$$, then the emission times were $$d_1/c$$ and $$d_2/c$$ before now. So if $$d_1=d_2$$ then the emission events were simultaneous. And if $$d_1 \\ne d_2$$ then the emission events were not simultaneous.\"\n\nWhat happens in your scenario is that in frame $$F_1$$ the two distances are equal, whereas in frame $$F_2$$ they are not equal.\n\nBy the way I would always recommend learning to use spacetime diagrams when learning special relativity.\n\nIn $$F_1$$ the light from A and the light from B arrive simultaneously at M and it makes a sound.\n\nIn $$F_2$$ the light from A and the light from B arrive simultaneously at M and it makes a sound. No paradox. Simultaneity is OK if (and only if) it refers to 2 things happening at the same place.\n\nThe difference is that in $$F_1$$ the light is sent simultaneously from A and from B, and takes the same time to travel from A to M as it does from B to M. In $$F_2$$ the light from B (if A is on the left and B is on the right) starts a bit earlier than the light from A but also takes longer to travel the longer distance, so the signals arrive at the same time.\n\nIn more detail, if needed: an observer at M in $$F_2$$ as the signals arrive reckons that A and B are equidistant. But A is travelling away from them and B is travelling towards them. So whenever either signal started, B was further away and as they arrive at the same time, B's signal must have started first.\n\nThe machine is in reference frame F1, not F2, so It depends who you ask if it's malfunctioning.\n\nAssuming that you know that the box works in this way - then the fact that the machine made a noise only tells you that in F1 - the beams of light reached the machine at the same time.\n\nI can say that the machine is malfunctioning only if I have no knowledge of the relativistic effect. So yes - it is malfunctioning according to most people on the planet. But for those who do know something about relativity effects, it is not.\n\n-- It get's a bit more complicated than that if you are trying to imagine what would you see. Try not to go that path yet, most Youtube videos and even PBS got that wrong. That's why the assignment uses noise, not \"what would you see\".\n\nA machine $$M$$ is not equidistant between $$A$$ and $$B$$ in $$F_2$$, so receiving their signals at the same time means they didn't send them at the same time, and they didn't. Totally self consistent."
] | [
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https://geogebra.nismed.upd.edu.ph/2012/10/ | [
"## Wednesday, October 17, 2012\n\n### Guess the Equation of the Line\n\nConsider the graph of a linear function in the applet below.\n1. Move sliders a and b to guess the equation of the graph.\n2. Click the check box to check your answer.\n3. Click the New Line button to generate a new line.\n\nThis is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com\nGuillermo Bautista, Created with GeoGebra\n\n## Sunday, October 14, 2012\n\n### Quadrilateral Angle Sum\n\nMove points A, B, C or D to create the desired quadrilateral. Move the slider to the extreme right. As each slider appears, rotate the quadrilateral by moving it to the extreme right.\nThis is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com\nGuillermo Bautista, Created with GeoGebra\n\nThe applet above demonstrates that the angle sum of a quadrilateral is 360 degrees.\n\nSnapshot\n\n## Wednesday, October 10, 2012\n\n### Multiplication of Fractions\n\nThe applet below illustrates the visual representation of multiplication of fractions. Move the sliders to explore the figure.\nThis is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com\n\nGuillermo Bautista, Created with GeoGebra\n\nSnapshot\n\n## Wednesday, October 3, 2012\n\n### Pythagorean Theorem Proof 2\n\nGiven: Triangle with sides a and b and hypotenuse c and three squares containing a, b and c. This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com\nGuillermo Bautista, Created with GeoGebra\n\nQuestions\n\n1. What are the areas of the green square and the red square?\n2. Move slider p to the extreme right. What types of quadrilaterals were formed from the two squares?\n3. What can you say about the areas of squares in Question 1 and the areas of quadrilaterals in Question 2?\n4. Move slider q to the extreme right. What type of quadrilaterals were formed?\n5. What can you say about the quadrilaterals formed in Question 1 and the quadrilaterals in Question 4?\n6. Move slider r to the right. What do you observe?\n7. What can you say about the areas of the red square, green square, and the area of square containing c?\n8. In terms of a, b, and c, what equation can be formed relating the areas of the three squares?\nSnapshot"
] | [
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https://uniquewritersbay.com/pseudorandom-numbers/ | [
"# Pseudorandom Numbers\n\n#### The Use Pseudorandom Numbers in Simulations\n\nPseudorandom is defined as having the randomness appearance that exhibit a particular repeatable pattern. They are computer generated random numbers. Pseudorandom numbers are commonly used in simulation because they are generated by computer: pseudorandom number generator, and hence, one can get a large set of random numbers in a short period of time as compared to roll of dice or other techniques that are used to generate true random numbers. A computer can more efficiently generate random numbers by use of a mathematical process which artificially creates pseudorandom number more effectively than the true random number can be generated (Haahr, 1998).\n\n#### The Effect of Pseudorandom Numbers on the Accuracy of a Simulation\n\nIt is essential to maintain accuracy of a simulation. Pseudorandom numbers generator always produces random number with a certain pattern of occurrence. It can therefore influence the accuracy of a simulation by accidently developing patterns that could be missed by the system. This would skew the accuracy of the simulation (Rogers, 2001).\n\n#### The Role of Statistical Analysis in Simulation\n\nStatistical analysis is important in simulation since it identifies the critical simulation factors. Performing variance analysis ensures that there is a proper selection of important factors. With this, judgment and understanding of simulation becomes more efficient in making suitable decisions concerning the process and product design. The main function of statistical analysis is to establish the probable or the normal simulation behavior and differentiate it from improbable or abnormal system behavior. This means that statistical analysis helps in distinguishing a common system behavior from what simply takes place by chance. Statistical analysis entails the study and analysis of data with an intention of resolving issues in any given institution (Rossiter, 2006).\n\nNeed a Professional Writer to Work on Your Assignments? We will deliver Unique and Quality Work. Good Grade Guarantee!!\n\nPosted\n\nin\n\nby\n\nTags:"
] | [
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http://mymathforum.com/abstract-algebra/33390-equivalence-relation.html | [
"User Name Remember Me? Password\n\n Abstract Algebra Abstract Algebra Math Forum\n\n January 23rd, 2013, 09:20 AM #1 Newbie Joined: Jan 2013 From: west bengal,india Posts: 22 Thanks: 0 Equivalence Relation If H be a subgroup of G then prove that the relation R defined on G by aRb iff (a)^(-1).b €H for a,b€G is an equivalence Relation",
null,
"January 24th, 2013, 05:18 AM #2 Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Equivalence Relation We need R to be, 1. Reflexive : aRa is true because a * a^-1 = e belongs to H since H is a subgroup of G 2. Symmetric : assuming a * b^-1 belongs to H, (a * b^-1)^-1 = b * a^-1 belongs to H. Hence if aRb then bRa. 3. Transitive : If a * b^-1 & b * c^-1 belongs to H then (a * b^-1) * (b * c^-1) = a * c^-1 belongs to H. Hence, if aRb and bRc then aRc. Hence, R is an equivalence relation.",
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"February 11th, 2013, 08:20 AM #3 Newbie Joined: Jan 2013 From: west bengal,india Posts: 22 Thanks: 0 Re: Equivalence Relation Thank u",
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"Similar Threads Thread Thread Starter Forum Replies Last Post tom33 Algebra 3 January 17th, 2014 04:30 PM jrklx250s Real Analysis 3 December 7th, 2011 10:42 AM page929 Abstract Algebra 1 October 11th, 2010 12:33 PM Dontlookback Abstract Algebra 1 April 20th, 2010 11:52 AM tinynerdi Abstract Algebra 1 January 11th, 2010 09:24 AM\n\n Contact - Home - Forums - Cryptocurrency Forum - Top",
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https://reference.wolfram.com/language/ref/LaplaceDistribution.html | [
"# LaplaceDistribution\n\nLaplaceDistribution[μ,β]\n\nrepresents a Laplace double-exponential distribution with mean μ and scale parameter β.\n\nrepresents a Laplace double-exponential distribution with mean 0 and scale parameter 1.\n\n# Details",
null,
"• The Laplace distribution gives the distribution of the difference between two independent random variables with identical exponential distributions.\n• LaplaceDistribution allows μ to be any real number and β to be any positive real number.\n• LaplaceDistribution allows μ and β to be any quantities of the same unit dimensions. »\n• LaplaceDistribution can be used with such functions as Mean, CDF, and RandomVariate. »\n\n# Background & Context\n\n• LaplaceDistribution[μ,β] represents a statistical distribution defined and supported on the set",
null,
"of real numbers and parametrized by a real number μ (called the \"mean\" of the distribution) and by a positive real number β (called a \"scale parameter\"). Overall, the probability density function (PDF) of a Landau distribution is unimodal with a single \"peak\" (i.e. a global maximum), though its overall shape (its height and the horizontal location of its maximum) is determined by the values of μ and β. In addition, the tails of the PDF are \"thin\" in the sense that the PDF decreases exponentially rather than algebraically for large values of",
null,
". (This behavior can be made quantitatively precise by analyzing the SurvivalFunction of the distribution.) The Laplace distribution is sometimes referred to as the double-exponential distribution (not to be confused with GumbelDistribution, which may also be called doubly-exponential) and the zero-argument form (which is equivalent to LaplaceDistribution[0,1]) is sometimes called the standard Laplace distribution.\n• The Laplace distribution dates back to 1774 to the work of French mathematician Pierre Laplace as the distribution whose likelihood function (Likelihood) is maximized by setting the mean",
null,
"equal to the median of the observed values of an odd number of independent and identically distributed (I.I.D.) random variables. Probabilistically, the Laplace distribution models the difference between two independent random variables with identical exponential distributions, and it also plays an important role in the theory of Brownian motion as a model corresponding to random times that are exponentially distributed. In modern times, the Laplace distribution models a number of phenomena across a wide spectrum of fields including computing, regression analysis, signal processing, finance, and microbiology.\n• RandomVariate can be used to give one or more machine- or arbitrary-precision (the latter via the WorkingPrecision option) pseudorandom variates from a Laplace distribution. Distributed[x,LaplaceDistribution[μ,β]], written more concisely as xLaplaceDistribution[μ,β], can be used to assert that a random variable x is distributed according to a Laplace distribution. Such an assertion can then be used in functions such as Probability, NProbability, Expectation, and NExpectation.\n• The probability density and cumulative distribution functions for Laplace distributions may be given using PDF[LaplaceDistribution[μ,β],x] and CDF[LaplaceDistribution[μ,β],x]. The mean, median, variance, raw moments, and central moments may be computed using Mean, Median, Variance, Moment, and CentralMoment.\n• DistributionFitTest can be used to test if a given dataset is consistent with a Laplace distribution, EstimatedDistribution to estimate a Laplace parametric distribution from given data, and FindDistributionParameters to fit data to a Laplace distribution. ProbabilityPlot can be used to generate a plot of the CDF of given data against the CDF of a symbolic Laplace distribution and QuantilePlot to generate a plot of the quantiles of given data against the quantiles of a symbolic Laplace distribution.\n• TransformedDistribution can be used to represent a transformed Laplace distribution, CensoredDistribution to represent the distribution of values censored between upper and lower values, and TruncatedDistribution to represent the distribution of values truncated between upper and lower values. CopulaDistribution can be used to build higher-dimensional distributions that contain a Laplace distribution, and ProductDistribution can be used to compute a joint distribution with independent component distributions involving Laplace distributions.\n• LaplaceDistribution is closely related to a number of other distributions. For example, LaplaceDistribution can be thought of as a combination of two instances of ExponentialDistribution spliced together at",
null,
", and hence ExponentialDistribution can be obtained as a transformation (TransformedDistribution) of LaplaceDistribution. ChiSquareDistribution and FRatioDistribution are also transformations of LaplaceDistribution, while LaplaceDistribution can be obtained as a transformation of UniformDistribution, as a limiting case of HyperbolicDistribution (in the sense that the PDF of LaplaceDistribution[μ,β] is precisely the same as that of HyperbolicDistribution[1/β,0,δ,μ] as",
null,
"), and as a special case of both ExponentialPowerDistribution and VarianceGammaDistribution. LaplaceDistribution is also related to NormalDistribution, RayleighDistribution, BernoulliDistribution, StableDistribution, CauchyDistribution, ChiDistribution, and GammaDistribution.\n\n# Examples\n\nopen allclose all\n\n## Basic Examples(4)\n\nProbability density function:\n\nCumulative distribution function:\n\nMean and variance:\n\nMedian:\n\n## Scope(7)\n\nGenerate a sample of pseudorandom numbers from a Laplace distribution:\n\nCompare its histogram to the PDF:\n\nDistribution parameters estimation:\n\nEstimate the distribution parameters from sample data:\n\nCompare a density histogram of the sample with the PDF of the estimated distribution:\n\nSkewness and kurtosis of Laplace distribution are constant:\n\nDifferent moments with closed forms as functions of parameters:\n\nClosed form for symbolic order:\n\nClosed form for symbolic order:\n\nClosed form for symbolic order:\n\nHazard function:\n\nQuantile function:\n\nConsistent use of Quantity in parameters yields QuantityDistribution:\n\n## Applications(2)\n\nData packets are arriving via two channels. Waiting times for each channel are exponentially distributed with the same parameter",
null,
"per second. Find the distribution of the waiting time between packets:\n\nFind the probability of waiting time between the packets to be greater than 6 seconds:\n\nSimulate waiting times between packets coming from both channels:\n\nThe difference of flood stages between river stations A and B in a year has been estimated to follow a Laplace distribution with",
null,
"feet and",
null,
"feet. Find the probability that the difference is greater than 15 feet:\n\nFind the probability of positive difference:\n\nFind the mean and standard deviation of the difference of flood stages:\n\nSimulate the differences of flood stages for 30 years:\n\n## Properties & Relations(14)\n\nLaplace distribution is closed under translation and scaling by a positive factor:\n\nRelationships to other distributions:",
null,
"Halves of a Laplace distribution are proportional to ExponentialDistribution densities:\n\nFor negative argument:\n\nThe difference of two variates from ExponentialDistribution follows Laplace distribution:\n\nExponentialDistribution is a transformation of Laplace distribution:\n\nLaplace distribution is a special case of ExponentialPowerDistribution:\n\nLaplace distribution is a special case of VarianceGammaDistribution:\n\nIf",
null,
",",
null,
",",
null,
", and",
null,
"are independent and are normally distributed, then",
null,
"is Laplace distributed:\n\nIf",
null,
",",
null,
",",
null,
", and",
null,
"are independent and are normally distributed, then",
null,
"is Laplace distributed:\n\nChiSquareDistribution is a transformation of Laplace distribution:\n\nFor the sum of",
null,
"such variables:\n\nFRatioDistribution is a transformation of Laplace distribution:\n\nLaplace distribution is a transformation of UniformDistribution:\n\nLaplaceDistribution is the limiting case of HyperbolicDistribution of",
null,
"when",
null,
"and",
null,
":\n\nLaplace distribution is a parameter mixture of a NormalDistribution with RayleighDistribution:\n\n## Possible Issues(2)\n\nLaplaceDistribution is not defined when μ is not a real number:",
null,
"LaplaceDistribution is not defined when β is not a positive real number:",
null,
"Substitution of invalid parameters into symbolic outputs gives results that are not meaningful:\n\n## Neat Examples(1)\n\nPDFs for different β values with CDF contours:"
] | [
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"https://reference.wolfram.com/language/ref/Files/LaplaceDistribution.en/10.png",
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"https://reference.wolfram.com/language/ref/Files/LaplaceDistribution.en/11.png",
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"https://reference.wolfram.com/language/ref/Files/LaplaceDistribution.en/12.png",
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"https://reference.wolfram.com/language/ref/Files/LaplaceDistribution.en/13.png",
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"https://reference.wolfram.com/language/ref/Files/LaplaceDistribution.en/14.png",
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"https://reference.wolfram.com/language/ref/Files/LaplaceDistribution.en/15.png",
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"https://reference.wolfram.com/language/ref/Files/LaplaceDistribution.en/16.png",
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"https://reference.wolfram.com/language/ref/Files/LaplaceDistribution.en/17.png",
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"https://reference.wolfram.com/language/ref/Files/LaplaceDistribution.en/18.png",
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"https://reference.wolfram.com/language/ref/Files/LaplaceDistribution.en/19.png",
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"https://reference.wolfram.com/language/ref/Files/LaplaceDistribution.en/20.png",
null,
"https://reference.wolfram.com/language/ref/Files/LaplaceDistribution.en/21.png",
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"https://reference.wolfram.com/language/ref/Files/LaplaceDistribution.en/22.png",
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"https://reference.wolfram.com/language/ref/Files/LaplaceDistribution.en/23.png",
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"https://reference.wolfram.com/language/ref/Files/LaplaceDistribution.en/24.gif",
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https://research.chalmers.se/en/publication/?id=11490 | [
"# Pressure drop and heat transfer rates in forced convection rotating square duct flows at high rotation rates Journal article, 2005\n\nhis paper presents and discusses numerical simulations of forced convection heat transfer in a rotating square duct at high rotation rates. The mean pressure gradient has been kept constant in the simulations that were conducted with a second order finite volume code with a dynamical localized subgrid scale model. The rotation number based on the bulk velocity was varied from 0.12 to 6.6 and consequently the Reynolds number ranged from 3900 to 1810 according to the fact that rotation tends to increase the pressure drop in the duct. A model for estimating the velocities and the corresponding friction coefficient has been developed by analytically solving simplified versions of the momentum budgets within the Ekman layers occurring near the opposite two walls of the duct perpendicular to the rotation axis. The model reproduces accurately the velocity profiles of the numerical simulation at high rotation rates and predicts that the boundary layer quantities scale as Ek^1/2. At Ro>1 the Ekman layers are responsible for most of the pressure drop of the flow while the maximum heat transfer rates are found on the wall where the stratification of the x-momentum is unstable with respect to the Coriolis force. Rotation enhances the differences between the contributions of the local friction coefficients and local Nusselt numbers of the four walls of the duct and considerably increases, in comparison with the non-rotating case, the pressure drop of the flow and the Nusselt number. The overall friction coefficient of the measurements and the simulations existing in the literature, as well as the present numerical predictions, are well correlated with the equation 1.09(Cf/Ek^1/2)^1.25=Ro in the range Ro>=1 for Re<=104.\n\nrotation\n\nLES\n\nTurbulent flow\n\nchannel flow\n\nEkman layer\n\n## Author\n\n### Jordi Pallares\n\nRovira i Virgili University\n\n### F Grau\n\nRovira i Virgili University\n\n### Lars Davidson\n\nChalmers, Applied Mechanics\n\n#### Physics of Fluids\n\n1070-6631 (ISSN) 1089-7666 (eISSN)\n\nVol. 17 7 075102 (artno)-\n\n### Subject Categories\n\nOther Physics Topics\n\n### DOI\n\n10.1063/1.1941365\n\n10/7/2017"
] | [
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http://www.philipzucker.com/noise-fluctuation-dissipation-theorem/ | [
"# Noise and The Fluctuation Dissipation Theorem\n\nI was looking at some slides the other day and they quoted noise power in units of",
null,
"$\\frac{W}{\\sqrt{Hz}}$. Being the ignoramus I am, I was wondering why it was scaled\n\nFirst off, when a Watt is quoted in an electrical measurement, usually you’re measuring Voltage with an instrument with a known input impedance Z. That’s how you convert your fluctutating voltage measurement to Watts.\n\nSecond, the sqrt frequency thing? Nowadays, your measurement apparatus is probably a digital sampler and it performs an FFT giving you a spectrum. The width of your FFT is the sampling frequency roughly. Does that make sense that when you increase the width of your taken spectrum the height of the noise signal changes too? It does, but only because implicitly, most sampling circuits take an average of the signal over the same period as the sampling time. These two times are not necessarily intrinsically linked. One could have a system that takes a very fast snapshot and but can only save data or send it over a link at a much slower speed. The noise power is this snapshot time, not the data saving time. The data saving time would be the bandwidth in the FFT.\n\nThese two are engineered to be the same to avoid distortion of the frequency signal via aliasing.\n\nBut there is an even simpler way to see this. Suppose you have two measurements V1 and V2 that are the averages of time T with variance",
null,
"$\\sigma$. Then the average of these two, V3, is over a time 2T. However, by the standard kind of manipulations (for Gaussian variables the squared variance of a sum = the sum of the squared variances,",
null,
"$\\sigma^2_{\\sum x_i}=\\sum \\sigma_{x_i}$ ), the variance of the new signal is",
null,
"$\\sigma/\\sqrt{2}$ which means it scales with the time window. Hence multiplying you actual measured variances by the square root of your time window gives you a time window invariant quantity.\n\nWhile I was thinking about that in the car I realized that the fluctuation dissipation theorem is a mean field theory kind of thing. The fluctuation dissipation theorem feels weird and spooky, but I guess it is ultimately simple (or not).\n\nMean field theory tries to summarize all the complicated interactions with neighbors with a simple summary. For interacting spins, it tries to summarize as an effective B field from the surrounding spins. Then you have a 1-particle model which you can solve and try to find a self-consistent value of B. Here is a sketch in equations.",
null,
"$H= \\sum S\\cdot S - B_{ext}\\cdot S \\rightarrow \\sum - B_{eff}\\cdot S$",
null,
"$Z=\\sum_s e^{-\\beta H}$",
null,
"$M = = \\partial_{\\beta B} \\ln(Z)$",
null,
"$B = \\alpha M$\n\nYou can do something similar to find an effective permeability due to your surrounding neighbors.",
null,
"$\\partial_B M = \\chi$\n\nThe fluctuating force due to your neighbors is like B, a constant forcing term.\n\nThe damping is like the permeability. One may want to consider a system that starts with an intrinsic damping, that is one difference between the magnetic case and the fluctuation case, in that free space has a natural permeability but not a natural damping (I suppose there is always some damping, due to radiation and what not, but we have a tendency to totally neglect such things). One could imagine ball bearings being shaken in a cup of molasses or something. You might want to fluctuation due to being hit by other ball bearings, but consider the damping from the molasses to be the dominating damping term (the the thermal fluctuations from the molasses to be ignorable).\n\nAnother difference is that I think you really are going to need to work explicitly with time. Just the thermal average isn’t going to cut it I think (at least not conceptually. There might be some dirty tricks you can play, but a typical Hamiltonian can’t have damping terms. As I write this I am doubting it’s truth).",
null,
"$\\ddot{x} = -\\nu \\dot{x}+ f$\n\ncalculate some averages … Then use the self-consistency",
null,
"$B = \\alpha M \\rightarrow f = f(\\hat{x})$\n\nThe dissipation will be related to your correlation with your neighbors. When you are moving faster, they have to tend to move in such a way to make you slow down on average.\n\nTo Be Continued"
] | [
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"http://s0.wp.com/latex.php",
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"http://s0.wp.com/latex.php",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9609101,"math_prob":0.99349856,"size":3767,"snap":"2019-26-2019-30","text_gpt3_token_len":771,"char_repetition_ratio":0.107095405,"word_repetition_ratio":0.003030303,"special_character_ratio":0.19830103,"punctuation_ratio":0.081267215,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99629915,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-20T19:46:10Z\",\"WARC-Record-ID\":\"<urn:uuid:a0897b6e-10c0-46cf-8f32-4f433e147398>\",\"Content-Length\":\"45279\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:60a5df72-9bd7-4d08-a30d-c15b7637111e>\",\"WARC-Concurrent-To\":\"<urn:uuid:ef53fdb9-5e5c-4e91-af70-0d84fb4f91dc>\",\"WARC-IP-Address\":\"208.94.118.167\",\"WARC-Target-URI\":\"http://www.philipzucker.com/noise-fluctuation-dissipation-theorem/\",\"WARC-Payload-Digest\":\"sha1:5MMFJFH4MWDP37WMXR3NUXVGDQPUBLPQ\",\"WARC-Block-Digest\":\"sha1:T4HLEOUNVA2ZFT5NQ2RWASRUUTSXTCQU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195526670.1_warc_CC-MAIN-20190720194009-20190720220009-00505.warc.gz\"}"} |
https://mathematica.stackexchange.com/questions/245264/how-to-solve-integro-differential-with-parameters-iteratively?noredirect=1 | [
"How to solve integro-differential with parameters iteratively?\n\nI am trying to solve an integro-differential equation and tried using one of the answers that I found in a related question:\n\nNumerically solve an integro-differential equation\n\nThe difference is I have parameters outside the integral, so I used ParametricNDSolve instead of NDSolveValue. Here is a simplified example that captures the essence of the problem:\n\nydrive[t_] = Exp[-t^2/5]*Sin[2*Pi*t];(*driving term*)\nreset[] := (Clear[ysol2];\nysol2[n_] :=\nysol2[n] =\nParametricNDSolve[{D[y[t], t] ==\nc*y[t] + \\[Lambda]*ydrive[t] +\n0.1*Integrate[ysol2[n - 1][\\[Lambda], c1][t], {c1, -1., 1.}],\ny[-10.^2] == 0}, y, {t, -100., 100.}, {\\[Lambda], c}])\nreset[]\nysol2 = # &;(*initial guess*)\n\nWhen I display even the zeroth order iteration:\n\nysol2[1., 1.][0.]\n\nIt gives an error message, \"Dependent variables {y,[Lambda]} cannot depend on parameters {\n[Lambda],c}.\"\n\nI hope anyone can help me on this. Thank you.\n\n• I'm wondering: The integral part here integrates over the ode-parameter c(not the dependent variabel t)? Apr 28 '21 at 10:27\n• This equation can be solved by numerical method but interval {t,-100,100} is too large, for numerical model please let consider {t,-10,10} or even {t,-5,5}. Apr 28 '21 at 12:39\n• @UlrichNeumann Yes, I need to integrate over the parameter of the ODE. In the original problem, I need to integrate over wavevector k and solve the ODE in time t. Apr 30 '21 at 1:03\n• @AlexTrounev Sure. But that is not the main issue. For example, if you remove the Integrate part and consider only the ParametricNDSolve part, it works just fine even with the (-100,100) interval. My main problem is how to integrate an external parameter and do it iteratively to solve the integro-differential equation self-consistently. Apr 30 '21 at 1:08\n\nWe can solve this equation by collocation method. Note, that we can exclude parameter $$\\lambda$$, by substitution $$y=\\lambda u$$. First we call\n\nNeeds[\"DifferentialEquationsNDSolveProblems\"];\nNeeds[\"DifferentialEquationsNDSolveUtilities\"];\nGet[\"NumericalDifferentialEquationAnalysis`\"];\n\nThen we define collocation points and weights for Gauss integral, variables, equations and initial conditions as follows\n\nnp = 21; g = GaussianQuadratureWeights[np, -1, 1]; points =\ng[[All, 1]];\nweights = g[[All, 2]];\n\nvart = Table[u[i][t], {i, np}]; vart1 =\nTable[u[i]'[t], {i, np}]; var = Table[u[i], {i, np}];\n\neqs = Table[\nvart1[[i]] ==\npoints[[i]] vart[[i]] + ydrive[t] + .1 vart . weights, {i, np}];\n\nic = Table[vart[[i]] == 0 /. {t -> -100}, {i, np}];\n\nFinally we solve system of equations\n\nsol = NDSolve[{eqs, ic}, var, {t, -100, 100}];\n\nIn the small scale solution looks like\n\nTable[Plot[Evaluate[vart[[i]] /. sol[]], {t, -5, 5},\nPlotRange -> All, PlotLabel -> Row[{\"c = \", points[[i]]}]], {i, np}]",
null,
"Visualization in the large scale\n\nTable[LogPlot[Evaluate[vart[[i]] /. sol[]], {t, -10, 100},\nPlotRange -> All, PlotLabel -> Row[{\"c = \", points[[i]]}]], {i, np}]",
null,
""
] | [
null,
"https://i.stack.imgur.com/k5aPe.png",
null,
"https://i.stack.imgur.com/uSGkh.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.64718723,"math_prob":0.9983724,"size":1010,"snap":"2022-05-2022-21","text_gpt3_token_len":328,"char_repetition_ratio":0.102385685,"word_repetition_ratio":0.0,"special_character_ratio":0.35148516,"punctuation_ratio":0.21004567,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9986279,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-17T20:28:07Z\",\"WARC-Record-ID\":\"<urn:uuid:925b856a-d24c-468b-8867-3ba9652bfbe0>\",\"Content-Length\":\"139744\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:712ec3ca-02fb-42e3-b789-9d3cd75deb23>\",\"WARC-Concurrent-To\":\"<urn:uuid:717ccf49-a586-49b7-842a-6f6925a9359e>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://mathematica.stackexchange.com/questions/245264/how-to-solve-integro-differential-with-parameters-iteratively?noredirect=1\",\"WARC-Payload-Digest\":\"sha1:Q627PDGHD3W37SZ66HRH677W4EUFSOXJ\",\"WARC-Block-Digest\":\"sha1:L4GNFJC2LVYP6IVX3FRC4LSMAZOXN6XJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320300616.11_warc_CC-MAIN-20220117182124-20220117212124-00199.warc.gz\"}"} |
https://electronics.stackexchange.com/questions/537899/what-is-this-effect-i-am-seeing-in-a-current-voltage-plot-of-a-2n2222-npn-transi | [
"# What is this effect I am seeing in a current/voltage plot of a 2N2222 NPN transistor?\n\nI have been working on a home made IV tracer for diodes and transistors.\n\nI noticed a slight curlicue in the low voltage and low current end of the plot for a transistor, and wondered what was going on.\n\nThis is the plot of the collector current and collector voltage for a 2N2222 transistor made with my setup:",
null,
"You can see the \"curlicue\" down there at the lower left corner.\n\nThis is a closer view of the \"curlicue\":",
null,
"I've been trying to figure out what is going on here.\n\nThis is the circuit I used to make the plots:",
null,
"There's an Arduino Nano off to the left connected to the analog and PWM signals. It uses oversampling to get better than the 10 bits of resolution of the Arduino ADC.\n\nThe best explanation I've been able to come up with is that at low collector voltages, some of the base current \"goes the wrong way\" out through the collector instead of the emitter. That raises the collector voltage (A3-VCollector) against the bias voltage (A2-VCollectorBias) resulting in a current flowing back through R4.\n\nICollector is calculated as $$\\\\frac{VCollectorBias - VCollector}{R4} \\$$\n\nDoes that explanation seem right, or have I missed something?\n\nBetween those two charts, I changed R4 from 1k to 10k and made some software changes to get a little more resolution.\n\nAt a suggestion from Hearth, I simulated the circuit in LTspice.\n\nHere's the simulated circuit:",
null,
"This is the plot of the collector current against the collector voltage:",
null,
"It does in fact have a negative tail, though not as extreme as in my circuit. The tail in the simulator is also straight rather than curved.\n\nAt any rate, the \"tail\" isn't a figment of the Nano's imagination.\n\nI ran a trace of a 2N3904.",
null,
"It also has a negative tail on the collector current, though much smaller than on the 2N2222.\n\n• Might it be oscillating? Dec 17, 2020 at 19:19\n• Have you run this in a simulator to see if you observe the same behavior? Dec 17, 2020 at 19:21\n• @Hearth: I haven't run it in a simulator.\n– JRE\nDec 17, 2020 at 19:26\n• @JRE I'm pretty sure you're right (confident enough to submit it as an answer anyway!) but I'd still suggest a quick LTspice simulation or something just to be sure! Dec 17, 2020 at 19:29\n• @Hearth: $M_e + LT_{spice} \\neq Q_{uick}$\n– JRE\nDec 17, 2020 at 19:31\n\nThat is what is expected even with the simplest model of a BJT. You can see it in this simulation",
null,
"If we zoom in in the VI chars",
null,
"we can see that the chars do not intersect at 0,0",
null,
"The shape of the curve (excluding the Early bending) can be seen in these equations (Millman Halkias, Electronic Devices and Circuits,\"Voltages as functions of currents\", p. 250)",
null,
"",
null,
"or, in more modern notation",
null,
"This is the plot of Ic/Ib vs Vce with Vth = 26 mV, alpha = 0.99, alphaR = 0.78. It's rotated by 90 degrees because we are plotting the inverse relationship",
null,
"and if we zoom in near the origin we see",
null,
"that is compatible with what Millman shows in his textbook",
null,
"Note that the 6 mV values is Vth log (1/alpha_R).\n\n• Perfect. Thank you. This is the answer I needed.\n– JRE\nDec 18, 2020 at 6:52\n• I don't have that book. Do you know which commonly published characteristics of transistors relate to which parameters in the given equations? I find the extent and shape of the \"tail\" varies by the transistor type. The \"tail\" on the 2N3904 is much smaller than the \"tail\" on the 2N2222, for example.\n– JRE\nDec 18, 2020 at 6:59\n• Millman's model is maybe the simplest, but maybe the difference you see can be explained by the different alpha_R. You might also want to see how the intersection change with temperature. (By the way, did you share your project somewhere, being able to see that means it's a decent curve tracer). And finally, have you ever been on the Internet Archive? (just asking, Millman's books are worth a look - I believe what you need is in Electronic Devices, Sec. 9-5 \"Detailed study of the currents in a transistor\", p. 229) Dec 18, 2020 at 7:47\n• I share most of my personal projects. The IV Tracer is currently in two parts: GUI and Arduino. I have been describing the development process and progress in a series of blog posts.\n– JRE\nDec 18, 2020 at 13:43\n• I have moved all of the code for both the Arduino and the PC to one repository and given the whole thing a new name.\n– JRE\nJan 3, 2021 at 19:26\n\nI'm reasonably certain your guess is correct. If the base voltage is higher than the collector voltage by a sufficient margin, I would expect to see the base-collector pn junction biased on and current flowing out of the collector. This seems to be happening when the collector voltage is below about 100 or 150 mV, which would imply (at such low currents) a reasonable base voltage of 650~700 mV, assuming it's all referenced to the emitter.\n\nI think you are at the limits of the accuracy for the Arduino A/D converters. The resolution of the converters is about 5mV, or about 5$$\\\\mu\\$$A through a 1k$$\\\\Omega\\$$ resistor. The absolute accuracy of these converters is much worse.\n\nTo make this circuit work with an Arduino you should be using larger resistor values, and you need to very carefully calibrate all of the ADCs.\n\n• I am beyond the normal resolution of the Arduino ADC. I'm using oversampling to get more resolution and to reduce the noise as well as the jaggedness of the PWM signals.\n– JRE\nDec 17, 2020 at 20:10\n• Oversampling can help with random errors but it won't do anything for systemic errors, such as offset and nonlinearity in the ADCs. Yes, you can improve the resolution but you don't improve the accuracy. And the fact that you are oversampling would have been good information to put in the original question. Dec 17, 2020 at 20:22\n• Right. Noise and lack of resolution are the things I used oversampling for. If offsets and nonlinearity could cause the curlicue, then I'd like to know how.\n– JRE\nDec 17, 2020 at 20:54\n\nThe fact that you see it with 3904 is interesting. What do you see at these nodes if you look at it on a real scope?\n\nSome shots in the dark:\n\n1. Is there AC component in there from the PWM's? That can create low level artifacts by being rectified somewhere, or via timings of the ADC. Maybe try to make a version of your circuit that filters the PWM more, just as a troubleshooting hack.\n\n2. does ADC pull any current? If so the two ends of each sense resistor have different impedance. I've had situations where a small RC on the inputs of ADC's made them better... real small like 3.3ohm/10nF ... this involved the sample/hold action of an ADC and there being a few inches of wire in the path. Not sure if either applies but just putting it out there.\n\n• At the low voltage and low current levels where the curlicues occur, the filtered PWM is at its cleanest. There's actually more than a few inches of wire involved, but that doesn't seem to bother anything. The measurements are all right at the Nano pins (all the resistors and the PWM filters are built into a sort of 3D structure right on the Nano.)\n– JRE\nDec 18, 2020 at 6:48\n• Looks like Sredni Vashtar answered it above! Dec 18, 2020 at 15:20\n• Yep. Looks like. I managed to get enough things right that my (actually quite crude) setup can detect a tiny little wrinkle that theory says will be there.\n– JRE\nDec 18, 2020 at 15:31"
] | [
null,
"https://i.stack.imgur.com/gRkV2.png",
null,
"https://i.stack.imgur.com/wpRK8.png",
null,
"https://i.stack.imgur.com/JEHPk.png",
null,
"https://i.stack.imgur.com/WWSpf.png",
null,
"https://i.stack.imgur.com/POowd.png",
null,
"https://i.stack.imgur.com/1cosT.png",
null,
"https://i.stack.imgur.com/dYBTW.png",
null,
"https://i.stack.imgur.com/idfuw.png",
null,
"https://i.stack.imgur.com/KL5ht.png",
null,
"https://i.stack.imgur.com/J9uQ1.png",
null,
"https://i.stack.imgur.com/wUGZZ.png",
null,
"https://i.stack.imgur.com/X73cJ.png",
null,
"https://i.stack.imgur.com/FmS6e.png",
null,
"https://i.stack.imgur.com/144VE.png",
null,
"https://i.stack.imgur.com/S1HiQ.png",
null
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http://epbbs.org/kcgSKDe.html | [
"立方和立方差公式 平方差，平方和，立方差，立方和的公式是什么？\n\n0人回答 713人收藏 6281次阅读\n\na3+b3=? a3-b3=? (a+b)3=? (a-b)3=?立方和：a^3+b^3=(a+b)*(a^2-ab+b^2) 立方差：a^3-b^3=(a-b)*(a^2-ab+b^2) 和的立方：(a+b）^3=a^3+3(a^2)b+3(b^2)a+b^3 差的立方：(a-b）^3=a^3-3(a^2)b+3(b^2)a-b^3 扩展资料：完全立方公式包括完全立方和公式和完全立方差公式，完全立方和(或\n\na^3-b^3=(a-b)^3-[-3(a^2)b+3ab^2]=(a-b)(a-b)^2+3ab(a-b)=(a-b)(a^2-2ab+b^2+3ab)=(a-b)(a^2+ab+b^2)有立方和公式及其推广:(1) a^3+b^3=(a+b)(a^2-ab+b^2) 立方和公式 a^3+b^3=(a+b) (a^2-ab+b^2） 折叠立方差公式 a^3-b^3=(a-b) (a^2+ab+b^2）\n\n完全立方和立方差公式 立方和立方差公式相关内容:\n\n• 数学公式：平方差 立方差 立方和 完全平方 立方差 立方和公式\n\n我想知道数学公式：平方差公式 立方差公式 立方和公式和完全平方公式平方差公式:a^2-b^2=(a+b)(a-b) 立方差公式:a^3-b^3=(a-b)(a^2＋ab＋b^2） 立方和公式:(a+b)^3=a^3+3a^2b+3ab^2+b^3 (a-b)^3=a^3-3a^2b+3ab^2-b^3 完全平方公式:(a+b)^2=a^2+2ab+b^2 (a-b)^2=a^2-2ab+b^2\n\n29人回答732人收藏6863次阅读\n• 怎么理解以仁治军，平和柔韧？ 汉朝大将卫青的母亲父亲是谁\n\n性格分内向型与外向型两种,这两种人在实际生活中的比例大致相当。哪一种类型更有利于事业的成功呢?以卫青和霍去病为例,两种类型应不相上下。 卫青的母亲卫媪,本\n\n58人回答331人收藏6249次阅读\n• 从六合坐车到江宁的陶吴小区路线怎么走 从六合坐车到江宁的陶吴小区路线怎么走\n\n从六合坐车到江宁的陶吴小区路线怎么走 驾车路线：全程约771公里 起点：六合区 1从起点向西北方向出发，沿通湖路行驶140米，调头进入通湖路 2沿通湖路行驶210米，直行进入环龙池路 3沿环龙池路行驶40米，左转进入宁六路 4沿宁六路行驶70米，朝连云港/长江二桥方向，稍向右转上匝\n\n79人回答655人收藏1643次阅读\n• 陶茂小区怎么样？好不好？值不值得买？ 陶茂小区怎么样？好不好？值不值得买？\n\n陶茂小区怎么样？好不好？值不值得买？楼盘名称：陶茂小区 楼盘位置：陶茂路 规划信息：其占地面积为0平方米，容积率暂无数据，绿化率0%，共0栋楼，停车位50 （所载信息仅供参考，最终以售楼处信息为准。） 买新房，就上搜狐焦点网\n\n38人回答446人收藏9496次阅读\n• “女人”的反义词是什么？ 妇人之见反义词\n\n“女人”的反义词：男人 女人，拼音： [nǚ rén] 释义：指妻子或成年的女子。 详细释义：女人同男人相对，从生理学上讲，具有XX染色体的人就是女人，在生理、心理上具有区别于男人的特征，如具有女性生殖系统，具有女人的气质和阴柔之美。 造句：\n\n27人回答867人收藏7052次阅读\n• 古文中夫是什么意思 子夫是什么意思\n\n1、旧时称成年男子：渔夫农夫万夫不当之勇 2、旧时称服劳役的人：夫役拉夫 3、（夫子）a、旧时对学者的称呼；b、旧时称老师；c、旧时妻称夫；d、称读古书而思想陈腐的人 4、与妻结成配偶者：丈夫夫妇 5、妇妻 6、量词,指一农夫所受之一\n\n44人回答648人收藏1355次阅读\n• 炎帝是男是女 炎帝是男是女\n\n炎帝又称神农氏 性别：男 图片： 资料： 炎帝，华夏始祖之一，与黄帝并称为中华始祖。 又称赤帝、烈山氏，距今约五千多年前生于姜水之岸（今宝鸡市境内）。炎帝制耒耜，种五谷。立市廛，首辟市常治麻为布，民着衣裳。作五弦琴，以乐百姓。削木为\n\n16人回答35人收藏8682次阅读\n• 已婚男人怎么戴戒指 想问下已婚男\n\n男士佩戴戒指戴在左手中。 但是佩戴在不同的手指中也有不同的寓意。要是男士戒指佩戴在食指的话就表示未婚和想结婚;要是戒指佩戴在中指的话代表已经在恋爱中。 要是戒指佩戴在无名指的话代表已婚;要是戒指佩戴在小指的话代表独身主义。佩戴戒指\n\n13人回答461人收藏5254次阅读\n\n猜你喜欢\n\n© 2019 道九头条网 版权所有 网站地图 XML"
] | [
null
] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.891996,"math_prob":0.99261826,"size":2390,"snap":"2019-43-2019-47","text_gpt3_token_len":2384,"char_repetition_ratio":0.15171836,"word_repetition_ratio":0.041666668,"special_character_ratio":0.4041841,"punctuation_ratio":0.02881356,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98730046,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-23T03:08:19Z\",\"WARC-Record-ID\":\"<urn:uuid:31897b54-4c73-45bc-88c4-0ab3cd6d4aa3>\",\"Content-Length\":\"17790\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:560935a3-9dda-4349-992d-858568542184>\",\"WARC-Concurrent-To\":\"<urn:uuid:52fcbe78-3b54-446c-94ae-dee2f75e3447>\",\"WARC-IP-Address\":\"23.245.103.165\",\"WARC-Target-URI\":\"http://epbbs.org/kcgSKDe.html\",\"WARC-Payload-Digest\":\"sha1:ZF5GHBUJ5HWWOI3LPXRB4SN4U2SH6ZWJ\",\"WARC-Block-Digest\":\"sha1:VON3DLHGUL4SY57EDPGEOIBBZCY7YAEG\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987828425.99_warc_CC-MAIN-20191023015841-20191023043341-00467.warc.gz\"}"} |
https://www.colorhexa.com/38e9d4 | [
"# #38e9d4 Color Information\n\nIn a RGB color space, hex #38e9d4 is composed of 22% red, 91.4% green and 83.1% blue. Whereas in a CMYK color space, it is composed of 76% cyan, 0% magenta, 9% yellow and 8.6% black. It has a hue angle of 172.9 degrees, a saturation of 80.1% and a lightness of 56.7%. #38e9d4 color hex could be obtained by blending #70ffff with #00d3a9. Closest websafe color is: #33ffcc.\n\n• R 22\n• G 91\n• B 83\nRGB color chart\n• C 76\n• M 0\n• Y 9\n• K 9\nCMYK color chart\n\n#38e9d4 color description : Bright cyan.\n\n# #38e9d4 Color Conversion\n\nThe hexadecimal color #38e9d4 has RGB values of R:56, G:233, B:212 and CMYK values of C:0.76, M:0, Y:0.09, K:0.09. Its decimal value is 3729876.\n\nHex triplet RGB Decimal 38e9d4 `#38e9d4` 56, 233, 212 `rgb(56,233,212)` 22, 91.4, 83.1 `rgb(22%,91.4%,83.1%)` 76, 0, 9, 9 172.9°, 80.1, 56.7 `hsl(172.9,80.1%,56.7%)` 172.9°, 76, 91.4 33ffcc `#33ffcc`\nCIE-LAB 83.897, -47.799, -2.299 42.649, 63.868, 72.364 0.238, 0.357, 63.868 83.897, 47.854, 182.754 83.897, -62.984, 4.022 79.917, -44.595, 2.256 00111000, 11101001, 11010100\n\n# Color Schemes with #38e9d4\n\n• #38e9d4\n``#38e9d4` `rgb(56,233,212)``\n• #e9384d\n``#e9384d` `rgb(233,56,77)``\nComplementary Color\n• #38e97c\n``#38e97c` `rgb(56,233,124)``\n• #38e9d4\n``#38e9d4` `rgb(56,233,212)``\n• #38a6e9\n``#38a6e9` `rgb(56,166,233)``\nAnalogous Color\n• #e97c38\n``#e97c38` `rgb(233,124,56)``\n• #38e9d4\n``#38e9d4` `rgb(56,233,212)``\n• #e938a6\n``#e938a6` `rgb(233,56,166)``\nSplit Complementary Color\n• #e9d438\n``#e9d438` `rgb(233,212,56)``\n• #38e9d4\n``#38e9d4` `rgb(56,233,212)``\n• #d438e9\n``#d438e9` `rgb(212,56,233)``\n• #4de938\n``#4de938` `rgb(77,233,56)``\n• #38e9d4\n``#38e9d4` `rgb(56,233,212)``\n• #d438e9\n``#d438e9` `rgb(212,56,233)``\n• #e9384d\n``#e9384d` `rgb(233,56,77)``\n• #15bfab\n``#15bfab` `rgb(21,191,171)``\n• #18d6c0\n``#18d6c0` `rgb(24,214,192)``\n• #21e6cf\n``#21e6cf` `rgb(33,230,207)``\n• #38e9d4\n``#38e9d4` `rgb(56,233,212)``\n• #4fecd9\n``#4fecd9` `rgb(79,236,217)``\n• #66eede\n``#66eede` `rgb(102,238,222)``\n• #7df1e3\n``#7df1e3` `rgb(125,241,227)``\nMonochromatic Color\n\n# Alternatives to #38e9d4\n\nBelow, you can see some colors close to #38e9d4. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #38e9a8\n``#38e9a8` `rgb(56,233,168)``\n• #38e9b7\n``#38e9b7` `rgb(56,233,183)``\n• #38e9c5\n``#38e9c5` `rgb(56,233,197)``\n• #38e9d4\n``#38e9d4` `rgb(56,233,212)``\n• #38e9e3\n``#38e9e3` `rgb(56,233,227)``\n• #38e1e9\n``#38e1e9` `rgb(56,225,233)``\n• #38d2e9\n``#38d2e9` `rgb(56,210,233)``\nSimilar Colors\n\n# #38e9d4 Preview\n\nThis text has a font color of #38e9d4.\n\n``<span style=\"color:#38e9d4;\">Text here</span>``\n#38e9d4 background color\n\nThis paragraph has a background color of #38e9d4.\n\n``<p style=\"background-color:#38e9d4;\">Content here</p>``\n#38e9d4 border color\n\nThis element has a border color of #38e9d4.\n\n``<div style=\"border:1px solid #38e9d4;\">Content here</div>``\nCSS codes\n``.text {color:#38e9d4;}``\n``.background {background-color:#38e9d4;}``\n``.border {border:1px solid #38e9d4;}``\n\n# Shades and Tints of #38e9d4\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #010d0c is the darkest color, while #fafefe is the lightest one.\n\n• #010d0c\n``#010d0c` `rgb(1,13,12)``\n• #031f1b\n``#031f1b` `rgb(3,31,27)``\n• #05302b\n``#05302b` `rgb(5,48,43)``\n• #07423b\n``#07423b` `rgb(7,66,59)``\n• #09544b\n``#09544b` `rgb(9,84,75)``\n• #0b655b\n``#0b655b` `rgb(11,101,91)``\n• #0d776a\n``#0d776a` `rgb(13,119,106)``\n• #0f897a\n``#0f897a` `rgb(15,137,122)``\n• #119a8a\n``#119a8a` `rgb(17,154,138)``\n• #13ac9a\n``#13ac9a` `rgb(19,172,154)``\n• #15beaa\n``#15beaa` `rgb(21,190,170)``\n• #17cfb9\n``#17cfb9` `rgb(23,207,185)``\n• #19e1c9\n``#19e1c9` `rgb(25,225,201)``\n• #26e7d0\n``#26e7d0` `rgb(38,231,208)``\n• #38e9d4\n``#38e9d4` `rgb(56,233,212)``\n• #4aebd8\n``#4aebd8` `rgb(74,235,216)``\n• #5beddc\n``#5beddc` `rgb(91,237,220)``\n• #6defdf\n``#6defdf` `rgb(109,239,223)``\n• #7ff1e3\n``#7ff1e3` `rgb(127,241,227)``\n• #90f3e7\n``#90f3e7` `rgb(144,243,231)``\n• #a2f5eb\n``#a2f5eb` `rgb(162,245,235)``\n• #b4f7ef\n``#b4f7ef` `rgb(180,247,239)``\n• #c5f9f3\n``#c5f9f3` `rgb(197,249,243)``\n• #d7fbf6\n``#d7fbf6` `rgb(215,251,246)``\n• #e9fdfa\n``#e9fdfa` `rgb(233,253,250)``\n• #fafefe\n``#fafefe` `rgb(250,254,254)``\nTint Color Variation\n\n# Tones of #38e9d4\n\nA tone is produced by adding gray to any pure hue. In this case, #8d9493 is the less saturated color, while #27fae1 is the most saturated one.\n\n• #8d9493\n``#8d9493` `rgb(141,148,147)``\n• #859d9a\n``#859d9a` `rgb(133,157,154)``\n• #7ca5a0\n``#7ca5a0` `rgb(124,165,160)``\n• #74aea7\n``#74aea7` `rgb(116,174,167)``\n``#6bb6ad` `rgb(107,182,173)``\n• #63bfb4\n``#63bfb4` `rgb(99,191,180)``\n• #5ac7ba\n``#5ac7ba` `rgb(90,199,186)``\n• #52d0c1\n``#52d0c1` `rgb(82,208,193)``\n• #49d8c7\n``#49d8c7` `rgb(73,216,199)``\n• #41e1ce\n``#41e1ce` `rgb(65,225,206)``\n• #38e9d4\n``#38e9d4` `rgb(56,233,212)``\n• #30f2da\n``#30f2da` `rgb(48,242,218)``\n• #27fae1\n``#27fae1` `rgb(39,250,225)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #38e9d4 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.53528,"math_prob":0.6356523,"size":3701,"snap":"2020-34-2020-40","text_gpt3_token_len":1721,"char_repetition_ratio":0.12280227,"word_repetition_ratio":0.011111111,"special_character_ratio":0.5406647,"punctuation_ratio":0.2345679,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9855588,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-06T13:45:36Z\",\"WARC-Record-ID\":\"<urn:uuid:01fb1364-d992-42e3-8e39-988bf54ff452>\",\"Content-Length\":\"36304\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:681219a0-33d1-485b-951f-884b14d87e75>\",\"WARC-Concurrent-To\":\"<urn:uuid:aefa4311-6698-445c-bee0-b7cb75955a11>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/38e9d4\",\"WARC-Payload-Digest\":\"sha1:6PAFQA5WFQHNQGKZLSGAW74F5NADHQHW\",\"WARC-Block-Digest\":\"sha1:RWDCYNPN643YDBN5WNLGGY4SWJBVFCBN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439736962.52_warc_CC-MAIN-20200806121241-20200806151241-00531.warc.gz\"}"} |
https://stanfordphd.com/Brownian_Motion.html | [
"Statistical & Financial Consulting by Stanford PhD\nBROWNIAN MOTION\n\nBrownian Motion (also known as Standard Wiener Process) is a stochastic process W(t) which has the following four properties:\n\n1] W(0) = 0,\n\n2] almost surely, the trajectory of W(t) is continuous;\n\n3] W(t) has independent increments: for any moments of time s < t < u, random variables W(t) - W(s) and W(u) - W(s) are independent;\n\n4] W(t) has stationary increments: for any moments of time s < t and positive shift h, random variables W(s+h) - W(s) and W(t+h) - W(t) have the same distribution.\n\nIt follows from properties 3, 4 and the Central Limit Theorem that any finite-dimensional distributions of a Brownian motion are Gaussian (normal). For example, in the definition above, random variables W(t) - W(s), W(u) - W(s), W(s+h) - W(s) and W(t+h) - W(t) are jointly Gaussian... Trajectories of Brownian motion are visualized below.",
null,
"BROWNIAN MOTION REFERENCES\n\nLawler, G. F. (1995). Introduction to Stochastic Processes. New York: Chapman and Hall/CRC.\n\nRoss, S. M. (1995). Stochastic Processes (2nd ed). New York: Wiley.\n\nKarlin, S., & Taylor, H. M. (1975). A First Course in Stochastic Processes (2nd ed). New York: Academic Press.\n\nKaratzas, I., & Shreve, S. E. (1991). Brownian Motion and Stochastic Calculus (2nd ed). New York: Springer.\n\nOksendal, B. K. (2002). Stochastic Differential Equations: An Introduction with Applications (5th ed). Springer-Verlag Berlin Heidelberg.\n\nDurrett, R. (2004). Probability: Theory and Examples (3rd ed). Belmont, CA: Duxbury Press.\n\nProtter, P. E. (2005). Stochastic Integration and Differential Equations (2nd ed). Springer-Verlag Berlin Heidelberg.\n\nGikhman, I. I., & Skorokhod, A. V. (2007). The Theory of Stochastic Processes III. Springer-Verlag Berlin Heidelberg.\n\nTsay, R. S. (2005). Analysis of Financial Time Series. Wiley-Interscience.\n\nBACK TO THE\nSTATISTICAL ANALYSES DIRECTORY"
] | [
null,
"https://stanfordphd.com/images/afe1dde787c8f25e7fbf37db006f793e.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.65667987,"math_prob":0.94843936,"size":1864,"snap":"2019-51-2020-05","text_gpt3_token_len":532,"char_repetition_ratio":0.11666667,"word_repetition_ratio":0.07067138,"special_character_ratio":0.28540772,"punctuation_ratio":0.23096447,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9903876,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-25T08:50:16Z\",\"WARC-Record-ID\":\"<urn:uuid:eaa62463-a5f6-477b-8b15-fd6e2de675ec>\",\"Content-Length\":\"12000\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:12f2d28d-80ee-4aa1-ab16-dc4ff881cf50>\",\"WARC-Concurrent-To\":\"<urn:uuid:75b87f1a-642e-415d-b6af-764a34bace3f>\",\"WARC-IP-Address\":\"97.74.185.119\",\"WARC-Target-URI\":\"https://stanfordphd.com/Brownian_Motion.html\",\"WARC-Payload-Digest\":\"sha1:KAO2MQOFVYU5WZ7BTE2PUYNEMJPJQMEP\",\"WARC-Block-Digest\":\"sha1:LKGIINXILJHH5JHA5NH64JPYGTK3UDYS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251671078.88_warc_CC-MAIN-20200125071430-20200125100430-00061.warc.gz\"}"} |
https://www.answers.com/Q/How_many_cubic_feet_is_1.85_cubic_yard | [
"Volume\n\n# How many cubic feet is 1.85 cubic yard?\n\n1.85 cubic yards equals 49.95 cubic feet.\n\n🙏\n0\n🤨\n0\n😮\n0\n😂\n0\n\n## Related Questions\n\nThere are 3 feet in one yard. Therefore, 555 feet is equal to 555/3 = 185 yards.\n\nThere are 3 ft per yard. So, 555 ft / 3 = 185 yards.\n\n467 feet x 185 feet equates to 1.98 acres.\n\n185 meters equates to 606.955 feet.\n\nThere are 0.9144 metres in one yard. Therefore, rounded to two decimal places, 185 metres is equal to 185/0.9144 = 202.32 yards.\n\n185 meters is 606 feet and 11.46 inches.\n\nA square yard can represent a square with dimensions of 3 feet by 3 feet. Therefore a 7 foot diameter gasket cannot be cut out of a square yard. If you meant 7 inches diameter, then you will be able to cut 25 gaskets from 3 feet by 3 feet but you may even be able to cut 26 gaskets if the square yard was a rectangle of 7 inches wide (by 185 inches long).\n\nIf tires are not compressed, for size 185/65R15\" around 250 tires but for every size up and width you loose about .3 cubic feet.\n\n1 foot = 30.48 cm 1 foot = 12 inch 185 cm = 185/30.48 feet = 6.0695538 feet = 6 feet 0.835 inch\n\n1 ft is 12 in; 1 in is about 2.54 cm. So, 185 cm is about 185/2.54 = 72.8in or 6 ft\n\nYour bag should tell you how many cubic feet it makes. 1 foot = 12 inches. 15 x 12 = 180 + 5 = 185 2 x 12 = 24 + 11 = 35 7 x 12 = 84 + 6 = 90 185 x 35 x 90 = 582750 cubic inches 582750 divided by 1728 (1 square foot in inches) = 337.24 cubic feet. Divide that by the number of cubic feet your bag of mix provides and you will know how many bags you need.\n\n###### Length and DistanceLawn CareAreaHome BuyingMath and ArithmeticVolumeBrakes and TiresHeightUnits of MeasureHome & Garden",
null,
"Copyright © 2021 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply."
] | [
null,
"https://img.answers.com/answ/image/upload/q_auto,f_auto,dpr_2.0/v1589555119/logos/Answers_throwback_logo.png",
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https://docs.derivative.ca/Syntaxhighlight | [
"Syntaxhighlight\n\nPython\n\ndef setupListeners(self):\n# make sure we have a valid ableton parameter\naParInfo = self.LOMParInfo('Parameter')\nif not aParInfo:\nself.disconnect()\nreturn\n\nTDAbletonCompBaseExt.setupListeners(self)\n\n# set up listener parameters\n# lomExpression: a Python expression to the live object in tda\nlomExpression = aParInfo['lomExpression']\n\n# note that this must be a valid CHOP channel name\n\n# id: unique id of the object requesting a listener\nownerId = self.ownerComp.id\n\n# property: the name of the live object's property to listen to\nproperty = 'value'\n\n# this tuple is what a listener key looks like\nlistener = (lomExpression, property, returnAddress, ownerId)\n\n# addListener(listener, outgoing parameter, parMin, parMax)\naParInfo['lomInfo']['min'], aParInfo['lomInfo']['max'])\n\nGLSL\n\nuniform vec4 uDiffuseColor;\nuniform vec4 uAmbientColor;\nuniform vec3 uSpecularColor;\nuniform float uShininess;\n\nout Vertex {\nvec4 color;\nvec3 worldSpacePos;\nvec3 worldSpaceNorm;\nflat int cameraIndex;\n}vVert;\n\nvoid main()\n{\n// First deform the vertex and normal\n// TDDeform always returns values in world space\nvec4 worldSpacePos =TDDeform(P);\ngl_Position = TDWorldToProj(worldSpacePos);\n\n// This is here to ensure we only execute lighting etc. code\n// when we need it. If picking is active we don't need this, so\n// this entire block of code will be ommited from the compile.\n// The TD_PICKING_ACTIVE define will be set automatically when\n// picking is active.\n#ifndef TD_PICKING_ACTIVE\n\nint cameraIndex = TDCameraIndex();\nvVert.cameraIndex = cameraIndex;\nvVert.worldSpacePos.xyz = worldSpacePos.xyz;\nvVert.color = TDInstanceColor(Cd);\nvec3 worldSpaceNorm = TDDeformNorm(N);\nvVert.worldSpaceNorm = normalize(worldSpaceNorm);\n\n#else // TD_PICKING_ACTIVE\n\n// This will automatically write out the nessessary values\n// for this shader to work with picking.\n// See the documentation if you want to write custom values for picking.\nTDWritePickingValues();\n\n#endif // TD_PICKING_ACTIVE\n}\n\nC++\n\n/* Shared Use License: This file is owned by Derivative Inc. (Derivative) and\n* can only be used, and/or modified for use, in conjunction with\n* Derivative's TouchDesigner software, and only if you are a licensee who has\n* accepted Derivative's TouchDesigner license or assignment agreement (which\n* also govern the use of this file). You may share a modified version of this\n* file with another authorized licensee of Derivative's TouchDesigner software.\n* Otherwise, no redistribution or sharing of this file, with or without\n* modification, is permitted.\n*/\n\n#include \"CPlusPlusCHOPExample.h\"\n\n#include <stdio.h>\n#include <string.h>\n#include <cmath>\n#include <assert.h>\n\n// These functions are basic C function, which the DLL loader can find\n// much easier than finding a C++ Class.\n// The DLLEXPORT prefix is needed so the compile exports these functions from the .dll\n// you are creating\nextern \"C\"\n{\n\nDLLEXPORT\nint32_t\nGetCHOPAPIVersion(void)\n{\n// Always return CHOP_CPLUSPLUS_API_VERSION in this function.\nreturn CHOP_CPLUSPLUS_API_VERSION;\n}\n\nDLLEXPORT\nCHOP_CPlusPlusBase*\nCreateCHOPInstance(const OP_NodeInfo* info)\n{\n// Return a new instance of your class every time this is called.\n// It will be called once per CHOP that is using the .dll\nreturn new CPlusPlusCHOPExample(info);\n}\n\nDLLEXPORT\nvoid\nDestroyCHOPInstance(CHOP_CPlusPlusBase* instance)\n{\n// Delete the instance here, this will be called when\n// Touch is shutting down, when the CHOP using that instance is deleted, or\n// if the CHOP loads a different DLL\ndelete (CPlusPlusCHOPExample*)instance;\n}\n\n};\n\nCPlusPlusCHOPExample::CPlusPlusCHOPExample(const OP_NodeInfo* info) : myNodeInfo(info)\n{\nmyExecuteCount = 0;\nmyOffset = 0.0;\n}\n\nCPlusPlusCHOPExample::~CPlusPlusCHOPExample()\n{\n\n}\n\nvoid\nCPlusPlusCHOPExample::getGeneralInfo(CHOP_GeneralInfo* ginfo)\n{\n// This will cause the node to cook every frame\nginfo->timeslice = true;\nginfo->inputMatchIndex = 0;\n}\n\nbool\nCPlusPlusCHOPExample::getOutputInfo(CHOP_OutputInfo* info)\n{\n// If there is an input connected, we are going to match it's channel names etc\n// otherwise we'll specify our own.\nif (info->opInputs->getNumInputs() > 0)\n{\nreturn false;\n}\nelse\n{\ninfo->numChannels = 1;\n\n// Since we are outputting a timeslice, the system will dictate\n// the numSamples and startIndex of the CHOP data\n//info->numSamples = 1;\n//info->startIndex = 0\n\n// For illustration we are going to output 120hz data\ninfo->sampleRate = 120;\nreturn true;\n}\n}\n\nconst char*\nCPlusPlusCHOPExample::getChannelName(int32_t index, void* reserved)\n{\nreturn \"chan1\";\n}\n\nvoid\nCPlusPlusCHOPExample::execute(const CHOP_Output* output,\nOP_Inputs* inputs,\nvoid* reserved)\n{\nmyExecuteCount++;\n\ndouble\t scale = inputs->getParDouble(\"Scale\");\n\n// In this case we'll just take the first input and re-output it scaled.\n\nif (inputs->getNumInputs() > 0)\n{\n// We know the first CHOP has the same number of channels\n// because we returned false from getOutputInfo.\n\ninputs->enablePar(\"Speed\", 0);\t// not used\ninputs->enablePar(\"Reset\", 0);\t// not used\ninputs->enablePar(\"Shape\", 0);\t// not used\n\nint ind = 0;\nfor (int i = 0 ; i < output->numChannels; i++)\n{\nfor (int j = 0; j < output->numSamples; j++)\n{\nconst OP_CHOPInput\t*cinput = inputs->getInputCHOP(0);\noutput->channels[i][j] = float(cinput->getChannelData(i)[ind] * scale);\nind++;\n\n// Make sure we don't read past the end of the CHOP input\nind = ind % cinput->numSamples;\n}\n}\n\n}\nelse // If not input is connected, lets output a sine wave instead\n{\ninputs->enablePar(\"Speed\", 1);\ninputs->enablePar(\"Reset\", 1);\n\ndouble speed = inputs->getParDouble(\"Speed\");\ndouble step = speed * 0.01f;\n\n// menu items can be evaluated as either an integer menu position, or a string\nint shape = inputs->getParInt(\"Shape\");\n//\t\tconst char *shape_str = inputs->getParString(\"Shape\");\n\n// keep each channel at a different phase\ndouble phase = 2.0f * 3.14159f / (float)(output->numChannels);\n\n// Notice that startIndex and the output->numSamples is used to output a smooth\n// wave by ensuring that we are outputting a value for each sample\n// Since we are outputting at 120, for each frame that has passed we'll be\n// outputing 2 samples (assuming the timeline is running at 60hz).\n\nfor (int i = 0; i < output->numChannels; i++)\n{\ndouble offset = myOffset + phase*i;\n\ndouble v = 0.0f;\n\nswitch(shape)\n{\ncase 0:\t\t// sine\nv = sin(offset);\nbreak;\n\ncase 1:\t\t// square\nv = fabs(fmod(offset, 1.0)) > 0.5;\nbreak;\n\ncase 2:\t\t// ramp\nv = fabs(fmod(offset, 1.0));\nbreak;\n}\n\nv *= scale;\n\nfor (int j = 0; j < output->numSamples; j++)\n{\noutput->channels[i][j] = float(v);\noffset += step;\n}\n}\n\nmyOffset += step * output->numSamples;\n}\n}\n\nint32_t\nCPlusPlusCHOPExample::getNumInfoCHOPChans()\n{\n// We return the number of channel we want to output to any Info CHOP\n// connected to the CHOP. In this example we are just going to send one channel.\nreturn 2;\n}\n\nvoid\nCPlusPlusCHOPExample::getInfoCHOPChan(int32_t index,\nOP_InfoCHOPChan* chan)\n{\n// This function will be called once for each channel we said we'd want to return\n// In this example it'll only be called once.\n\nif (index == 0)\n{\nchan->name = \"executeCount\";\nchan->value = (float)myExecuteCount;\n}\n\nif (index == 1)\n{\nchan->name = \"offset\";\nchan->value = (float)myOffset;\n}\n}\n\nbool\nCPlusPlusCHOPExample::getInfoDATSize(OP_InfoDATSize* infoSize)\n{\ninfoSize->rows = 2;\ninfoSize->cols = 2;\n// Setting this to false means we'll be assigning values to the table\n// one row at a time. True means we'll do it one column at a time.\ninfoSize->byColumn = false;\nreturn true;\n}\n\nvoid\nCPlusPlusCHOPExample::getInfoDATEntries(int32_t index,\nint32_t nEntries,\nOP_InfoDATEntries* entries)\n{\n// It's safe to use static buffers here because Touch will make it's own\n// copies of the strings immediately after this call returns\n// (so the buffers can be reuse for each column/row)\nstatic char tempBuffer1;\nstatic char tempBuffer2;\n\nif (index == 0)\n{\n// Set the value for the first column\n#ifdef WIN32\nstrcpy_s(tempBuffer1, \"executeCount\");\n#else // macOS\nstrlcpy(tempBuffer1, \"executeCount\", sizeof(tempBuffer1));\n#endif\nentries->values = tempBuffer1;\n\n// Set the value for the second column\n#ifdef WIN32\nsprintf_s(tempBuffer2, \"%d\", myExecuteCount);\n#else // macOS\nsnprintf(tempBuffer2, sizeof(tempBuffer2), \"%d\", myExecuteCount);\n#endif\nentries->values = tempBuffer2;\n}\n\nif (index == 1)\n{\n// Set the value for the first column\n#ifdef WIN32\nstrcpy_s(tempBuffer1, \"offset\");\n#else // macOS\nstrlcpy(tempBuffer1, \"offset\", sizeof(tempBuffer1));\n#endif\nentries->values = tempBuffer1;\n\n// Set the value for the second column\n#ifdef WIN32\nsprintf_s(tempBuffer2, \"%g\", myOffset);\n#else // macOS\nsnprintf(tempBuffer2, sizeof(tempBuffer2), \"%g\", myOffset);\n#endif\nentries->values = tempBuffer2;\n}\n}\n\nvoid\nCPlusPlusCHOPExample::setupParameters(OP_ParameterManager* manager)\n{\n// speed\n{\nOP_NumericParameter\tnp;\n\nnp.name = \"Speed\";\nnp.label = \"Speed\";\nnp.defaultValues = 1.0;\nnp.minSliders = -10.0;\nnp.maxSliders = 10.0;\n\nOP_ParAppendResult res = manager->appendFloat(np);\nassert(res == OP_ParAppendResult::Success);\n}\n\n// scale\n{\nOP_NumericParameter\tnp;\n\nnp.name = \"Scale\";\nnp.label = \"Scale\";\nnp.defaultValues = 1.0;\nnp.minSliders = -10.0;\nnp.maxSliders = 10.0;\n\nOP_ParAppendResult res = manager->appendFloat(np);\nassert(res == OP_ParAppendResult::Success);\n}\n\n// shape\n{\nOP_StringParameter\tsp;\n\nsp.name = \"Shape\";\nsp.label = \"Shape\";\n\nsp.defaultValue = \"Sine\";\n\nconst char *names[] = { \"Sine\", \"Square\", \"Ramp\" };\nconst char *labels[] = { \"Sine\", \"Square\", \"Ramp\" };\n\nOP_ParAppendResult res = manager->appendMenu(sp, 3, names, labels);\nassert(res == OP_ParAppendResult::Success);\n}\n\n// pulse\n{\nOP_NumericParameter\tnp;\n\nnp.name = \"Reset\";\nnp.label = \"Reset\";\n\nOP_ParAppendResult res = manager->appendPulse(np);\nassert(res == OP_ParAppendResult::Success);\n}\n\n}\n\nvoid\nCPlusPlusCHOPExample::pulsePressed(const char* name)\n{\nif (!strcmp(name, \"Reset\"))\n{\nmyOffset = 0.0;\n}\n}"
] | [
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https://de.mathworks.com/matlabcentral/cody/problems/43311-how-to-calculate-log/solutions/2280552 | [
"Cody\n\n# Problem 43311. How to calculate log?\n\nSolution 2280552\n\nSubmitted on 14 May 2020 by Stephen\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\nx = 5; y_correct = 1; assert(isequal(log5(x),y_correct))\n\n2 Pass\nx = 25; y_correct = 2; assert(isequal(log5(x),y_correct))\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
] | [
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https://en.academic.ru/dic.nsf/enwiki/11776953 | [
"# Mumford–Tate group\n\n\nMumford–Tate group\n\nIn algebraic geometry, the Mumford–Tate group MT(F) constructed from a Hodge structure F is a certain algebraic group G, named for David Mumford and John Tate. When F is given by a rational representation of an algebraic torus, the definition of G is as the Zariski closure of the image in the representation of the circle group, over the rational numbers. It has been said that “Mumford–Tate groups have emerged as the principal symmetry groups in Hodge theory”. They are applied to problems in number theory and physics, as well as geometry.\n\n## Formulation\n\nThe algebraic torus T used to describe Hodge structures has a concrete matrix representation, as the 2×2 invertible matrices of the shape that is given by the action of a+bi on the basis {1,i} of the complex numbers C over R:",
null,
"$\\begin{bmatrix} a & b \\\\ -b & a \\end{bmatrix}.$\n\nThe circle group inside this group of matrices is the unitary group U(1).\n\nHodge structures arising in geometry, for example on the cohomology groups of Kähler manifolds, have a lattice consisting of the integral cohomology classes. Not quite so much is needed for the definition of the Mumford–Tate group, but it does assume that the vector space V underlying the Hodge structure has a given rational structure, i.e. is given over the rational numbers Q. For the purposes of the theory the complex vector space VC, obtained by extending the scalars of V from Q to C, is used.\n\nThe weight k of the Hodge structure describes the action of the diagonal matrices of T, and V is supposed therefore to be homogeneous of weight k, under that action. Under the action of the full group VC breaks up into subspaces Vpq, complex conjugate in pairs under switching p and q. Thinking of the matrix in terms of the complex number λ it represents, Vpq has the action of λ by the pth power and of the complex conjugate of λ by the qth power. Here necessarily\n\np + q = k.\n\nIn more abstract terms, the torus T underlying the matrix group is the Weil restriction of the multiplicative group GL(1), from the complex field to the real field, an algebraic torus whose character group consists of the two homomorphisms to GL(1), interchanged by complex conjugation.\n\nOnce formulated in this fashion, the rational representation ρ of T on V setting up the Hodge structure F determines the image ρ(U(1)) in GL(VC); and MT(F) is by definition the Zariski closure, for the Q-Zariski topology on GL(V), of this image.\n\n## Mumford–Tate conjecture\n\nThe original context for the formulation of the group in question was the question of the Galois representation on the Tate module of an abelian variety A. Conjecturally, the image of such a Galois representation, which is an l-adic Lie group for a given prime number l, is determined by the corresponding Mumford–Tate group G (coming from the Hodge structure on H1(A)), to the extent that knowledge of G determines the Lie algebra of the Galois image. This conjecture is known only in particular cases. Through generalisations of this conjecture, the Mumford–Tate group has been connected to the motivic Galois group, and, for example, the general issue of extending the Sato–Tate conjecture (now a theorem).\n\n## Period conjecture\n\nA related conjecture on abelian varieties states that the period matrix of A over number field has transcendence degree, in the sense of the field generated by its entries, predicted by the dimension of its Mumford–Tate group, as in the previous section. Work of Pierre Deligne has shown that the dimension bounds the transcendence degree; so that the Mumford–Tate group catches sufficiently many algebraic relations between the periods. This is a special case of the full Grothendieck period conjecture.\n\nWikimedia Foundation. 2010.\n\n### Look at other dictionaries:\n\n• David Mumford — David Bryant Mumford (né le 11 juin 1937) est un mathématicien américain connu pour son travail en géométrie algébrique puis pour sa recherche en théorie de la vision. Il est actuellement professeur dans la division de mathématiques… … Wikipédia en Français\n\n• David Mumford — in 1975 Born 11 June 1937 (1937 06 11) (age 74) … Wikipedia\n\n• Motive (algebraic geometry) — For other uses, see Motive (disambiguation). In algebraic geometry, a motive (or sometimes motif, following French usage) denotes some essential part of an algebraic variety . To date, pure motives have been defined, while conjectural mixed… … Wikipedia\n\n• Tannakian category — In mathematics, a tannakian category is a particular kind of monoidal category C , equipped with some extra structure relative to a given field K . The role of such categories C is to approximate, in some sense, the category of linear… … Wikipedia\n\n• Arithmetic of abelian varieties — In mathematics, the arithmetic of abelian varieties is the study of the number theory of an abelian variety, or family of those. It goes back to the studies of Fermat on what are now recognised as elliptic curves; and has become a very… … Wikipedia\n\n• architecture — /ahr ki tek cheuhr/, n. 1. the profession of designing buildings, open areas, communities, and other artificial constructions and environments, usually with some regard to aesthetic effect. Architecture often includes design or selection of… … Universalium\n\n• Grammy Awards 2011 — Grammy Award Am 13. Februar 2011 wurden im Staples Center von Los Angeles die Grammy Awards 2011 verliehen. Es war die 53. Verleihung des Grammys, des wichtigsten US amerikanischen Musikpreises. Gewürdigt wurden Anfang 2011 die musikalischen… … Deutsch Wikipedia\n\n• 53e cérémonie des Grammy Awards — Grammy Award Organisés par la National Academy of Recording Arts and Sciences Cérémonie Date 13 … Wikipédia en Français\n\n• Glossary of arithmetic and Diophantine geometry — This is a glossary of arithmetic and Diophantine geometry in mathematics, areas growing out of the traditional study of Diophantine equations to encompass large parts of number theory and algebraic geometry. Much of the theory is in the form of… … Wikipedia\n\n• Étale cohomology — In mathematics, the étale cohomology groups of an algebraic variety or scheme are algebraic analogues of the usual cohomology groups with finite coefficients of a topological space, introduced by Grothendieck in order to prove the Weil… … Wikipedia"
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"https://en.academic.ru/dic.nsf/enwiki/e/d2e3fa6a40e5484e952e4ded07ad551b.png ",
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https://forum.allaboutcircuits.com/threads/calculus-and-circuits.64165/ | [
"# Calculus and circuits\n\nThread Starter\n\n#### jaygatsby\n\nJoined Nov 23, 2011\n182\nI notice that some courses at universities require high calculus as a prerequisite for circuits classes. Where is calculus used in circuits?\n\nThank you\n\n#### rstevenson\n\nJoined Apr 5, 2011\n20\ncapacitors, inductors you name it. like current through a capacitor is i = C * dv/dt. There are lots of other examples too\n\nThread Starter\n\n#### jaygatsby\n\nJoined Nov 23, 2011\n182\ncapacitors, inductors you name it. like current through a capacitor is i = C * dv/dt. There are lots of other examples too\nI do remember reading that formula in the all about circuits AC volume. And there was a similar formula for inductors. But this is all I've seen so far. Not enough to warrant calc I, calc II, and then diff. eq + linear algebra! Is there anything else?\n\nThanks\n\n#### steveb\n\nJoined Jul 3, 2008\n2,436\nI notice that some courses at universities require high calculus as a prerequisite for circuits classes. Where is calculus used in circuits?\n\nThank you\nAnalyzing linear AC circuits with coils and capacitors requires solving differential equations, making integral based transforms (Laplace and Fourier) to the frequency domain and calculation of convolution integrals in the time domain.\n\nNonlinear circuits using transistors, diodes, opamps etc, are sometimes linearized, which requires calculation of partial derivatives to make an effective linear small signal AC model.\n\nSome circuits (for example analog computers, or control feedback compensators) are designed to actually do calculus calculations such as integrals and derivatives.\n\nThe modeling of DC/DC converters, which utilize high speed switching, can be simplified by making an average model, by integrating to find the averages of relevant circuit variables.\n\n•",
null,
"jaygatsby\n\n#### thatoneguy\n\nJoined Feb 19, 2009\n6,359\nAnything involving AC at different frequencies, up to RF.\n\nYou end up having imaginary power used, and need to describe how.\n\n•",
null,
"jaygatsby\n\n#### crutschow\n\nJoined Mar 14, 2008\n27,187\nThe calculations involving electric and magnetic fields including electromagnetic wave transmission through space, wave-guides, transmission lines, and antennas all use calculus. Take a look at Maxwell's Equations sometime.",
null,
"#### Wendy\n\nJoined Mar 24, 2008\n22,430\nThen there is the analog computers made with op amps. You can wire calculus equations directly as a circuit, extremely complicated equations with very simple circuits.. Something similar to this was used for WWII bombers, they were classed top secret at the time.\n\n#### Geoffr67\n\nJoined Dec 5, 2011\n69\nThen there is the analog computers made with op amps. You can wire calculus equations directly as a circuit, extremely complicated equations with very simple circuits.. Something similar to this was used for WWII bombers, they were classed top secret at the time.\nVery cool! do you have any interesting resources regarding this? id love to read about it!\n\n#### Blofeld\n\nJoined Feb 21, 2010\n83\nVery cool! do you have any interesting resources regarding this? id love to read about it!\nYou might be interested in a similar development: The M9 gun director, used for anti aircraft artillery.\n\nSome historical information:\nhttp://www.antiaircraft.org/M9.htm\n\nA discussion of a very early operational amplifier that was used as part of this system can be found in chapter H, p. 15 of\nhttp://www.analog.com/library/analogdialogue/archives/39-05/Op_Amp_Applications.zip\n(Op Amp Applications by Walt Jung, 16.8 MB zip file)\n\nGenerally speaking, I think that during WWII there existed a mix of mechanical analog computers (like the famous Norden bombsight) and electronic analog computers. The principle is the same - as Bill Marsden has explained, it is all about solving complicated differential equations.\n\n#### hgmjr\n\nJoined Jan 28, 2005\n9,029\nVery cool! do you have any interesting resources regarding this? id love to read about it!\nSearch on the topic \"norden bomb sight\".\n\nHere is just one sample.\n\nhgmjr\n\nSimilar threads"
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"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89687234,"math_prob":0.8741447,"size":1885,"snap":"2021-21-2021-25","text_gpt3_token_len":412,"char_repetition_ratio":0.102073364,"word_repetition_ratio":0.10631229,"special_character_ratio":0.20583555,"punctuation_ratio":0.11527377,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96513015,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-15T03:34:18Z\",\"WARC-Record-ID\":\"<urn:uuid:580517e1-ad99-4475-8ef4-7d7b83be341c>\",\"Content-Length\":\"140873\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:49d133bc-c8da-4ec6-b69b-7f97315f287b>\",\"WARC-Concurrent-To\":\"<urn:uuid:bfe60cf0-8583-45c1-8af8-05e9cb06fe1e>\",\"WARC-IP-Address\":\"104.20.235.39\",\"WARC-Target-URI\":\"https://forum.allaboutcircuits.com/threads/calculus-and-circuits.64165/\",\"WARC-Payload-Digest\":\"sha1:7V2YMVB7XZDMVQNNFFZMA2OTJIEEXQOZ\",\"WARC-Block-Digest\":\"sha1:PL5N7XFTIKNO33HDSHNCA5ZGP4LZXH25\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487616657.20_warc_CC-MAIN-20210615022806-20210615052806-00471.warc.gz\"}"} |
https://lists.cairographics.org/archives/cairo/2008-January/012821.html | [
"# [cairo] Pixman sampling coordinates\n\nWed Jan 23 09:27:14 PST 2008\n\n```Carl Worth wrote:\n> On Tue, 22 Jan 2008 16:12:33 -0800, Carl Worth wrote:\n> > just to reduce the amount of #ifdefing. But I am quite glad to see\n> > that you added a comment there explaining why this adjustment isn't\n> > necessary for the case of other values of N_BITS.\n> >\n> > I'll just do that in a separate commit after pushing this series.\n>\n> I've pushed the series and I did that little change. So pixman should\n> be all set as far as this bug now.\n\nNice.\n\nI have a fix for the rotate-image-surface-paint testcase now.\n\nIt's basically what I proposed in the previous mail, but streamlined\nslightly, so no new arguments to _cairo_matrix_to_pixman_matrix\nare needed. It uses the origin of the pattern as the anchor.\n\ngit-fetch git://people.freedesktop.org/~inte/cairo pixman-transform-rounding\nhttp://gitweb.freedesktop.org/?p=users/inte/cairo.git;a=shortlog;h=pixman-transform-rounding\n\nThis requires a trivial patch to pixman; the pixman_transform_point_3d\nfunction wasn't properly exported.\n\ngit-fetch git://people.freedesktop.org/~inte/pixman missing-export\nhttp://gitweb.freedesktop.org/?p=users/inte/pixman.git;a=shortlog;h=missing-export\n\nBertram\n-------------- next part --------------\ncommit 5a0b15d2c9b5e9ef3aed1f01e8ea28a3f2c36216\nAuthor: Bertram Felgenhauer <int-e at gmx.de>\nDate: Wed Jan 23 18:02:02 2008 +0100\n\nanchor pattern transformations at the pattern origin\nThis keeps the rounding errors due to the conversion to 16.16 fixed point\nnumbers small and improves cairo's translation invariance.\n\ndiff --git a/src/cairo-matrix.c b/src/cairo-matrix.c\nindex d534826..589566d 100644\n--- a/src/cairo-matrix.c\n+++ b/src/cairo-matrix.c\n@@ -753,6 +753,10 @@ _cairo_matrix_to_pixman_matrix (const cairo_matrix_t\t*matrix,\n*pixman_transform = pixman_identity_transform;\n}\nelse {\n+ cairo_matrix_t inv = *matrix;\n+ double x = 0, y = 0;\n+ pixman_vector_t vector;\n+\npixman_transform->matrix = _cairo_fixed_16_16_from_double (matrix->xx);\npixman_transform->matrix = _cairo_fixed_16_16_from_double (matrix->xy);\npixman_transform->matrix = _cairo_fixed_16_16_from_double (matrix->x0);\n@@ -764,5 +768,24 @@ _cairo_matrix_to_pixman_matrix (const cairo_matrix_t\t*matrix,\npixman_transform->matrix = 0;\npixman_transform->matrix = 0;\npixman_transform->matrix = 1 << 16;\n+\n+ if (cairo_matrix_invert (&inv) != CAIRO_STATUS_SUCCESS)\n+ return;\n+\n+ /* find the device space coordinate that maps to (0, 0) */\n+ cairo_matrix_transform_point (&inv, &x, &y);\n+\n+ /* transform the resulting device space coordinate back\n+ * to the pattern space, using pixman's transform */\n+ vector.vector = _cairo_fixed_16_16_from_double (x);\n+ vector.vector = _cairo_fixed_16_16_from_double (y);\n+ vector.vector = 1 << 16;\n+\n+ if (!pixman_transform_point_3d (pixman_transform, &vector))\n+ return;\n+\n+ /* and compensate for the resulting error */\n+ pixman_transform->matrix -= vector.vector;\n+ pixman_transform->matrix -= vector.vector;\n}\n}\n-------------- next part --------------\ncommit 787cc57c7cb36a3b7eff836dd979d08cf7861e29\nAuthor: Bertram Felgenhauer <int-e at gmx.de>\nDate: Wed Jan 23 16:08:48 2008 +0100\n\nexport pixman_transform_point_3d function\n\ndiff --git a/pixman/pixman.h b/pixman/pixman.h\nindex 2fd3133..2965acd 100644\n--- a/pixman/pixman.h\n+++ b/pixman/pixman.h\n@@ -165,6 +165,7 @@ struct pixman_transform\npixman_fixed_t\tmatrix;\n};\n\n+PIXMAN_EXPORT\npixman_bool_t pixman_transform_point_3d (pixman_transform_t *transform,\npixman_vector_t *vector);\n\n```"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.55397344,"math_prob":0.71543765,"size":3659,"snap":"2020-34-2020-40","text_gpt3_token_len":1108,"char_repetition_ratio":0.18522573,"word_repetition_ratio":0.055172414,"special_character_ratio":0.33369774,"punctuation_ratio":0.18242122,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9797821,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-20T08:11:00Z\",\"WARC-Record-ID\":\"<urn:uuid:7423e311-7083-4288-b1a9-38051cf905b2>\",\"Content-Length\":\"6903\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8c0b3dd2-1a84-40ef-9f92-e17bd1878c68>\",\"WARC-Concurrent-To\":\"<urn:uuid:62b4d230-c294-4c52-8253-8f608ef2e949>\",\"WARC-IP-Address\":\"131.252.210.177\",\"WARC-Target-URI\":\"https://lists.cairographics.org/archives/cairo/2008-January/012821.html\",\"WARC-Payload-Digest\":\"sha1:3QKJDPGVU6F3PXQGFQ7UILDNU2PAZLHI\",\"WARC-Block-Digest\":\"sha1:L2HHS5N4FIRR6HXJVWW2IY7HG7KBW62K\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400196999.30_warc_CC-MAIN-20200920062737-20200920092737-00317.warc.gz\"}"} |
https://www.whatnumberis.net/mmdxviii/ | [
"What number is MMDXVIII?\n\nYour question is: What numbers are the Roman numerals MMDXVIII? Learn how to convert the Roman numerals MMDXVIII into the correct translation of normal numbers.\n\nThe Roman numerals MMDXVIII are identical to the number 2518.\n\nMMDXVIII = 2518\n\nHow do you convert MMDXVIII into normal numbers?\n\nIn order to convert MMDXVIII into numbers, the number of position values (ones, tens, hundreds, thousands) is subdivided as follows:\n\nPlace valueNumberRoman numbers\nConversion2000 + 500 + 10 + 8MM + D + X + VIII\nThousands2000MM\nHundreds500D\nDozens10X\nOnes8VIII\n\nHow do you write MMDXVIII in numbers?\n\nTo correctly write MMDXVIII as normal numbers, combine the converted Roman numbers. The highest numbers must always be in front of the lowest numbers to get the correct translation, as in the table above.\n\n2000+500+10+8 = (MMDXVIII) = 2518\n\nThe next Roman numerals = MMDXIX\n\nConvert another Roman numeral to normal numbers."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6717429,"math_prob":0.9856825,"size":961,"snap":"2022-05-2022-21","text_gpt3_token_len":272,"char_repetition_ratio":0.20480669,"word_repetition_ratio":0.0,"special_character_ratio":0.264308,"punctuation_ratio":0.099415205,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9870422,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-19T08:38:06Z\",\"WARC-Record-ID\":\"<urn:uuid:0a6c4ede-c699-4723-8fee-3c92037f6fa0>\",\"Content-Length\":\"8173\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:30bbaa11-9bf9-4e67-a0d4-788339b482c5>\",\"WARC-Concurrent-To\":\"<urn:uuid:93c08cbe-d322-4253-84e8-3092138ef9f6>\",\"WARC-IP-Address\":\"46.249.204.25\",\"WARC-Target-URI\":\"https://www.whatnumberis.net/mmdxviii/\",\"WARC-Payload-Digest\":\"sha1:VYOLMTNZXVV5SLETTUJEXVP4QBRQKYXY\",\"WARC-Block-Digest\":\"sha1:H3QUQZHKHLNZEX7RS6QTBDDTES72LEQU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301264.36_warc_CC-MAIN-20220119064554-20220119094554-00662.warc.gz\"}"} |
https://www.jetbrains.com/help/pycharm-edu/2018.1/extract-constant.html | [
"# Extract Constant\n\n## Basics\n\nThe Extract Constant refactoring makes your source code easier to read and maintain. It also helps you avoid using hardcoded constants without any explanations about their values or purpose.\n\n## Example\n\nBeforeAfter\n```import math class Solver: def demo(self): a = 3 b = 25 c = 46 root1 = (-b + math.sqrt(b**2 - 4*c)) / (2*a) root2 = (-b - math.sqrt(b**2 - 4*c)) / (2*a) print (root1, root2) Solver().demo()```\n```import math RETURN_TYPE_OF_SQRT = math.sqrt(b**2 - 4*a*c) class Solver: def demo(self): a = 3 b = 25 c = 52 root1 = (-b + RETURN_TYPE_OF_SQRT) / (2*a) root2 = (-b - RETURN_TYPE_OF_SQRT) / (2*a) print(root1, root2) Solver().demo()```\n\n## To extract a Python constant in-place\n\nThe in-place refactorings are enabled in PyCharm Edu by default. So, if you haven't changed this setting, the Extract Constant refactoring for Python is performed in-place, right in the editor.\n\n1. Place the cursor within the expression or declaration of a variable to be replaced by a constant.\n2. Do one of the following:\n• PressCtrl+Alt+C.\n• Choose Refactor | Extract | Constant on the context menu.\n3. If more than one expression is detected for the current cursor position, the Expressions list appears. If this is the case, select the required expression. To do that, click the expression. Alternatively, use the Up and Down arrow keys to navigate to the expression of interest, and then press Enter to select it.\n4. If more than one occurrence of the expression is found within the class, specify whether you wish to replace only the selected occurrence, or all the found occurrences with the new constant.\n5. Specify the name of the constant. Select the name from the list or type the name in the box with a red border.\n6. To complete the refactoring, press Tab orEnter.\n\nIf you haven't completed the refactoring and want to cancel the changes you have made, press Escape.\n\n## To extract a constant using the dialog box\n\nIf the Enable in-place refactorings checkbox is cleared on the Editor settings, the Extract Constant refactoring is performed by means of the Extract Constant Dialog dialog box.\n\n1. In the Extract Constant Dialog dialog that opens, specify the name of the new constant.\n2. To automatically replace all occurrences of the selected expression (if it is found more than once),select the option Replace all occurrences.\n3. Click OK to create the constant."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7494711,"math_prob":0.88311553,"size":2443,"snap":"2020-45-2020-50","text_gpt3_token_len":580,"char_repetition_ratio":0.14678147,"word_repetition_ratio":0.073529415,"special_character_ratio":0.246009,"punctuation_ratio":0.1039823,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97070205,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-24T00:54:06Z\",\"WARC-Record-ID\":\"<urn:uuid:80e81a31-c5d3-4dc3-ac01-c90c198b2d87>\",\"Content-Length\":\"18961\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:27554c7f-3fa1-4617-bf8f-2c049c12bef7>\",\"WARC-Concurrent-To\":\"<urn:uuid:e3261556-d46b-454b-99b0-eb16c4809c6c>\",\"WARC-IP-Address\":\"52.35.219.183\",\"WARC-Target-URI\":\"https://www.jetbrains.com/help/pycharm-edu/2018.1/extract-constant.html\",\"WARC-Payload-Digest\":\"sha1:ZE3UK4ZGINKPGKNIWGWEQBBPUH3FMH3L\",\"WARC-Block-Digest\":\"sha1:VDF5FCJKS6TR2C6KKHCUQBMCDFJPDXOP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107881551.11_warc_CC-MAIN-20201023234043-20201024024043-00283.warc.gz\"}"} |
http://face2ai.com/DIP-6-8-%E7%81%B0%E5%BA%A6%E5%9B%BE%E5%83%8F-%E5%9B%BE%E5%83%8F%E5%88%86%E5%89%B2-Canny%E8%BE%B9%E7%BC%98%E6%A3%80%E6%B5%8B/ | [
"# 灰度图像-图像分割 Canny边缘检测\n\nCanny是目前已知的最好的边缘检测算法，是不是之一我不确定，但可以肯定的是，它的应用非常广泛，基本用到边缘检测的，大家永远第一个想到它。Canny算法的复杂度比前面的检测加阈值的算法计算复杂度更高，空间复杂度也要高一些，但现在的计算设备，对于Canny基本可以实现实时，并且有人用GPU来实现，所以从86年Canny提出了这个算法到现在，在边缘检测方面，其地位还是比较稳固的。\nCanny算法的另一个显著特征是它有完整的数学推导过程，能够证明这个算法能给出最好的边缘。后面我们将会简单的看一下数学过程。\n\n## 算法原理\n\n1) Good detection. There should be a low probability of failing to mark real edge points, and low probability of falsely marking nonedge points. Since both these probabilities are monotonically decreasing functions of the output signal-to-noise ratio, this criterion corresponds to maximizing signal-to-noise ratio.\n2) Good localization. The points marked as edge points by the operator should be as close as possible to the center of the true edge.\n3) Only one response to a single edge. This is implicitly captured in the first criterion since when there are two responses to the same edge, one of them must be considered false. However, the mathematical form of the first criterion did not capture the multiple response requirement and it had to be made explicit.\n\n1. 好的检测：一定要尽可能少的遗漏边缘点，尽可能少的添加非边缘点。正确的边缘点为信号，被错误检测出来的非边缘点为噪声，所以第一点归结为提高信噪比。\n2. 准确的位置：检测出的边缘点一定要与真正的边缘中心，尽可能的近。\n3. 边缘单一响应：对于一个边缘，只能产生一个响应，如果对于一个边缘产生两个响应，第一点的数学求解过程不能保证这一点，所以要单独明确，如果对于一边有两个响应，必须去掉一个。\n\n## 数学原理\n\n$H_G(x)=\\int_{-w}^{+w} G(-x)f(x)\\,dx$ (1)\n\n$H_n=n_0[\\int_{-w}^{+w}f^2(x)\\,dx]^{1/2}$ (2)\n\n$SNR=\\frac{\\int_{-w}^{+w} G(-x)f(x)\\,dx}{n_0[\\int_{-w}^{+w}f^2(x)\\,dx]^{1/2}}$ (3)\n\n$H^{‘}_n(x_0)+H^{‘}_G(x_0)=0$ (4)\n\n$H^{‘}_G(x_0)=H^{‘}_G(0)+H^{”}_G(0)x_0+o(x_0^2)$ (5)\n\n$H^{”}_G(0)x_0 \\approx {-H^{‘}_n(x_0)}$(6)\n\n$H^{‘}_n(x_0)$ 为高斯随机量，并且其方差是 $H^{‘}_n(x_0)$ 的均方值，并且给出其期望为：\n\n$E[H^{‘}_n(x_0)^2]=n_0^2\\int^{+w}_{-w}f^{‘2}(x)\\,dx$(7)\n\n$E[x^2_0]\\approx {\\frac{n_0^2\\int^{+w}_{-w}f^{‘2}(x)\\,dx}{[{\\int_{-w}^{+w} G^{‘}(-x)f^{‘}(x)\\,dx}]^2}}=\\delta x^2_0$ (8)\n\n$\\delta x^2_0$ 是 $x_0$ 标准差的近似，位置由其倒数给出：\n\n$Localization =\\frac{|\\int^{+w}_{-w}G^{‘}(-x)f^{‘}(x)\\,dx|}{n_0[\\int^{+w}_{-w}f^{‘2}(x)\\,dx]^{-1/2}}$(9)\n\n$\\frac{\\int_{-w}^{+w} G(-x)f(x)\\,dx}{n_0[\\int_{-w}^{+w}f^2(x)\\,dx]^{1/2}}\\frac{|\\int^{+w}_{-w}G^{‘}(-x)f^{‘}(x)\\,dx|}{n_0[\\int^{+w}_{-w}f^{‘2}(x)\\,dx]^{-1/2}}$(10)\n\n$n^{-1}_0[\\int^{+w}_{-w}G^{2}(x)dx]^{1/2}$\n\nLocalization(9)上边界于：\n\n$n^{-1}_0[\\int^{+w}_{-w}G’^{2}(x)dx]^{1/2}$\n\n$x_{ave}=\\pi(\\frac{-R(0)}{R^{”}(0)})^{1/2}$(11)\n\n$R(0) = \\int ^{+\\infty}_{-\\infty}g^{2}(x)\\,dx$\n\n$R^{”}(0) = -\\int ^{+\\infty}_{-\\infty}{g’}^{2}(x)\\,dx$\n\n$x_{zc}(f)=\\pi(\\frac{\\int ^{+\\infty}_{-\\infty}f’^{2}(x)\\,dx}{\\int ^{+\\infty}_{-\\infty}{f”}^{2}(x)\\,dx})^{1/2}$(12)\n\n$x_{max}(f)=2x_{zc}(f)=kW$(13)\n\n$N_n=\\frac{2W}{x_{max}}=\\frac{2}{k}$(14)\n\n## 算法过程\n\n1. 高斯平滑，采用$5 \\times5$的高斯滤波器对图像进行平滑。\n2. 使用Sobel算子检测边缘候选点，计算梯度方向，得到简化的梯度方向。\n3. 非极大值抑制，减少多重响应。\n4. 边缘跟踪，采用双阈值处理候选点。\n5. 形态学细化，对于有些较粗的边界采用形态学方法处理。",
null,
"## 实现结果",
null,
"STEP1：",
null,
"STEP2：\nSobel梯度幅度结果：",
null,
"Sobel梯度方向结果：",
null,
"STEP3：",
null,
"STEP4：",
null,
"",
null,
"STEP1：",
null,
"STEP2：\nSobel梯度幅度结果：",
null,
"Sobel梯度方向结果：",
null,
"STEP3：",
null,
"STEP4：",
null,
"## 总结\n\n### 说点什么",
null,
"Subscribe"
] | [
null,
"https://tony4ai-1251394096.cos.ap-hongkong.myqcloud.com/blog_images/DIP-6-8-灰度图像-图像分割-Canny边缘检测/20150212181022310.jpeg",
null,
"https://tony4ai-1251394096.cos.ap-hongkong.myqcloud.com/blog_images/DIP-6-8-灰度图像-图像分割-Canny边缘检测/20150212181924413.jpeg",
null,
"https://tony4ai-1251394096.cos.ap-hongkong.myqcloud.com/blog_images/DIP-6-8-灰度图像-图像分割-Canny边缘检测/20150212181937158.jpeg",
null,
"https://tony4ai-1251394096.cos.ap-hongkong.myqcloud.com/blog_images/DIP-6-8-灰度图像-图像分割-Canny边缘检测/20150212181953180.jpeg",
null,
"https://tony4ai-1251394096.cos.ap-hongkong.myqcloud.com/blog_images/DIP-6-8-灰度图像-图像分割-Canny边缘检测/20150212182006783.jpeg",
null,
"https://tony4ai-1251394096.cos.ap-hongkong.myqcloud.com/blog_images/DIP-6-8-灰度图像-图像分割-Canny边缘检测/20150212182025625.jpeg",
null,
"https://tony4ai-1251394096.cos.ap-hongkong.myqcloud.com/blog_images/DIP-6-8-灰度图像-图像分割-Canny边缘检测/20150212182315899.jpeg",
null,
"https://tony4ai-1251394096.cos.ap-hongkong.myqcloud.com/blog_images/DIP-6-8-灰度图像-图像分割-Canny边缘检测/20150212182333621.jpeg",
null,
"https://tony4ai-1251394096.cos.ap-hongkong.myqcloud.com/blog_images/DIP-6-8-灰度图像-图像分割-Canny边缘检测/20150212182345025.jpeg",
null,
"https://tony4ai-1251394096.cos.ap-hongkong.myqcloud.com/blog_images/DIP-6-8-灰度图像-图像分割-Canny边缘检测/20150212182356134.jpeg",
null,
"https://tony4ai-1251394096.cos.ap-hongkong.myqcloud.com/blog_images/DIP-6-8-灰度图像-图像分割-Canny边缘检测/20150212182416318.jpeg",
null,
"https://tony4ai-1251394096.cos.ap-hongkong.myqcloud.com/blog_images/DIP-6-8-灰度图像-图像分割-Canny边缘检测/20150212182426040.jpeg",
null,
"https://tony4ai-1251394096.cos.ap-hongkong.myqcloud.com/blog_images/DIP-6-8-灰度图像-图像分割-Canny边缘检测/20150212182449655.jpeg",
null,
"https://secure.gravatar.com/avatar/",
null
] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.54801804,"math_prob":0.9999912,"size":9026,"snap":"2020-24-2020-29","text_gpt3_token_len":5465,"char_repetition_ratio":0.12314343,"word_repetition_ratio":0.050065875,"special_character_ratio":0.37812984,"punctuation_ratio":0.15268457,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998869,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-04T09:26:26Z\",\"WARC-Record-ID\":\"<urn:uuid:31f74fcf-59aa-4c95-a53d-7891b16594b1>\",\"Content-Length\":\"184404\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:88f052a6-90e4-46b2-874d-f28b7336a887>\",\"WARC-Concurrent-To\":\"<urn:uuid:1e86acde-224d-4be8-bfcc-1ab3b196cb26>\",\"WARC-IP-Address\":\"47.90.33.8\",\"WARC-Target-URI\":\"http://face2ai.com/DIP-6-8-%E7%81%B0%E5%BA%A6%E5%9B%BE%E5%83%8F-%E5%9B%BE%E5%83%8F%E5%88%86%E5%89%B2-Canny%E8%BE%B9%E7%BC%98%E6%A3%80%E6%B5%8B/\",\"WARC-Payload-Digest\":\"sha1:HLNN3YAFKPZOLDUMEHXM6SZV3PEQTZCS\",\"WARC-Block-Digest\":\"sha1:F2MPDIQQAOHXWWQMWLQUAHOB65QZVYTU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347439213.69_warc_CC-MAIN-20200604063532-20200604093532-00142.warc.gz\"}"} |
https://economics.stackexchange.com/questions/32348/why-is-a-monotone-increasing-but-nonlinear-transformation-of-a-utility-function | [
"# Why is a monotone increasing but nonlinear transformation of a utility function not represent the same preferences if the preference is complete?\n\nAccording to a textbook, in the context of uncertainty (e.g. in lottery), if the preference is complete, a monotone increasing but nonlinear transformation of a utility function would not represent the same preferences. Why is it so?\n\nAn example of such preference would be appreciated.\n\n• This claim is false. Please identify the textbook and the exact page. – Giskard Oct 20 '19 at 18:40\n• It is from slide of my professor, which he claim to be from a textbook. – Aqqqq Oct 21 '19 at 7:36\n• I'm voting to close this question as off-topic because it is based upon a false claim whose source cannot be identified. – Giskard Oct 21 '19 at 9:41\n• @Giskard Would it be the case in case of uncertainty (e.g. in lottery)? – Aqqqq Oct 21 '19 at 12:44\n• – Art Oct 21 '19 at 15:09\n\nConsider lotteries over $$\\{x,y,z\\}$$. Let $$u(x)=0, u(y)=\\frac{1}{2}, u(z)=1$$. Consider the nonlinear transformation f(t)=t^2. Let $$v:=f\\circ u$$, so $$v(x)=0,v(y)=\\frac{1}{4}, v(z)=1$$.\nConsider two lotteries, $$P=(0,1,0)$$ and $$Q=(\\frac{1}{2},0,\\frac{1}{2})$$.\n$$E_P[u]=\\frac{1}{2}=E_Q[u]$$\n$$E_P[v]=\\frac{1}{4}<\\frac{1}{2}=E_Q[v]$$\nIn general, let $$\\succeq$$ be a preference over lotteries $$\\Delta(X)$$. Let $$U$$ be a utility function of $$\\succeq$$ that has the EU form, so $$U(P)=E_P[u]$$ for some $$u$$. Take any increasing transformation $$f$$ and define $$V(P):=f(U(P))$$ then $$V$$ will also be a utility function for $$\\succeq$$, that is $$V(P)\\geq V(Q)\\iff P\\succeq Q$$. However, unless $$f$$ is a positive affine transformation, that is $$f(x)=Ax+B$$ where $$A>0$$ then $$V$$ will not be an expected utility function. That is, there will not exists a $$v$$ such that $$V(p)=E_p[v]$$.\n• Thank you for your answer. Is $\\Delta(X)$ just a notation or are there some meaning behind $\\Delta$? – Aqqqq Oct 22 '19 at 9:30\n• Should $V(p)=E_p[v]$ be $V(P)=E_P[v]$ instead (in the last paragraph)? Why \"there will not exists a $v$ such that $V(p)=E_p[v]$\"? – Aqqqq Oct 22 '19 at 10:09"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7092371,"math_prob":0.9999479,"size":856,"snap":"2019-51-2020-05","text_gpt3_token_len":329,"char_repetition_ratio":0.12558685,"word_repetition_ratio":0.0,"special_character_ratio":0.3995327,"punctuation_ratio":0.1421801,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000008,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-24T06:13:45Z\",\"WARC-Record-ID\":\"<urn:uuid:89a7e445-26a7-4cc2-a993-6812a392ee84>\",\"Content-Length\":\"143080\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b8d58874-713b-4f6e-8170-8b5503a8e90f>\",\"WARC-Concurrent-To\":\"<urn:uuid:01f4d8b3-b25f-427e-9f08-2c76c06c0ccf>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://economics.stackexchange.com/questions/32348/why-is-a-monotone-increasing-but-nonlinear-transformation-of-a-utility-function\",\"WARC-Payload-Digest\":\"sha1:ZHCIIXC7U4KWQTK5MWCX5IMOUU4DWRFY\",\"WARC-Block-Digest\":\"sha1:SPIDUYNDY7JZTLOD7F2OXBA64EK2WAKF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250615407.46_warc_CC-MAIN-20200124040939-20200124065939-00468.warc.gz\"}"} |
https://profiles.doe.mass.edu/mcas/mcasitems2.aspx?grade=06&subjectcode=MTH&linkid=9&orgcode=00000000&fycode=2021&orgtypecode=5& | [
"# Massachusetts\n\nDistricts Schools",
null,
"## 2021 Item by Item Results for GRADE 06 MATHEMATICS\n\n### Number of Students Included: 65,521 Participation Rate = 95%\n\n State Totals - GRADE 06 MATHEMATICS ITEM INFORMATION PERCENT OF STATE'S POSSIBLE POINTS ITEM TYPE REPORTING CATEGORY STANDARD ITEM DESC POSSIBLE POINTS STATE 1 SR EE 6.EE.A.2 Determine the value of an expression given the value of a variable. 1 56% 2 SR NS 6.NS.C.7 Interpret inequalities using absolute values of integers. 1 36% 3 SA NS 6.NS.B.2 Find the quotient of two multi-digit numbers. 1 54% 4 SR NS 6.NS.B.4 Find the greatest common factor of two numbers to solve a real-world problem. 1 40% 5 SR RP 6.RP.A.3 Solve a ratio problem based on a given real-world context. 1 41% 6 CR GE 6.G.A.3 Solve problems on a coordinate plane by plotting points, finding the distance between points, and finding the coordinates of points given the distance between them. 4 36% 7 SA NS 6.NS.C.6 Graph a negative decimal on a number line. 1 50% 8 SA SP 6.SP.B.4 Create a histogram to represent a given set of data. 1 35% 9 SR EE 6.EE.B.5 Determine the value of the variable in an inequality. 1 53% 10 SR SP 6.SP.A.1 Identify multiple statistical questions. 1 30% 11 SR RP 6.RP.A.2 Solve a unit-rate problem based on a given real-world context. 1 78% 12 SR EE 6.EE.A.4 Identify expressions that are equivalent to a given variable expression. 1 20% 13 SA NS 6.NS.A.1 Calculate the quotient of two fractions. 1 42% 14 CR RP 6.RP.A.1 Solve a real-world problem by interpreting and finding ratios based on a given ratio relationship. 4 39% 15 SA EE 6.EE.B.8 Graph on a number line an inequality that represents a constraint in a given real-world context. 1 46% 16 SR EE 6.EE.B.7 Determine which equation represents a given real-world context. 1 67% 17 SR EE 6.EE.B.6 Determine which expression can be used to represent a real-world context and use it to complete sentences about the context. 1 72% 18 SA SP 6.SP.B.5 Determine the median and mean of a set of data. 2 36% 19 SR EE 6.EE.C.9 Choose a two-variable equation that best represents a given real-world context. 1 36% 20 SR EE 6.EE.A.1 Evaluate numerical expressions involving whole-number exponents. 1 49% 21 SR NS 6.NS.B.3 Add and subtract decimals within a real-world context. 1 89% 22 SA RP 6.RP.A.2 Determine the unit rate within a real-world context. 1 57% 23 SR SP 6.SP.B.5 Calculate the mean of a given set of data. 1 40% 24 CR EE 6.EE.B.7 Write and solve equations that model a real-world problem. 4 54% 25 SR RP 6.RP.A.1 Identify the ratios that represent the relationships between given quantities. 1 58% 26 SA EE 6.EE.A.3 Use the distributive property to generate an equivalent expression within a real-world context. 1 22% 27 SR EE 6.EE.B.8 Identify the inequality which represents a constraint within a real-world context. 1 57% 28 SR EE 6.EE.A.3 Use the distributive property to determine equivalent expressions given a variable expression. 1 36% 29 SA GE 6.G.A.1 Find the area of a right triangle and decompose a polygon into triangles to complete sentences about its area. 2 23% 30 SR RP 6.RP.A.1 Determine which statement describes a given ratio relationship in a real-world context. 1 40% 31 SR SP 6.SP.A.2 Determine the interquartile range of data displayed in a box plot. 1 24% 32 SR EE 6.EE.A.3 Use properties of operations to identify an equivalent equation. 1 59% 33 SR RP 6.RP.A.3 Determine the volume of a liquid by using rate and ratio reasoning within a real-world context. 1 40% 34 SR SP 6.SP.A.1 Identify multiple statistical questions. 1 34% 35 CR NS 6.NS.C.8 Solve problems on a coordinate plane by finding the distance between points, plotting points, and finding the coordinates of a point given its distance from a different point. 4 54% 36 SR NS 6.NS.C.5 Determine whether given real-world situations can be best represented by positive or negative numbers. 1 60% 37 SA GE 6.G.A.4 Use the net of a square pyramid to find its surface area. 1 18% 38 SR RP 6.RP.A.3 Compare unit rates given a ratio in a real-world context involving money. 1 39% 39 SR SP 6.SP.B.5 Determine which statements correctly describe data represented in a dot plot. 1 31% 40 SR GE 6.G.A.2 Determine the number of cubes with fractional edge lengths that would fit in a given right rectangular prism in a real-world context. 1 11%\n\nParticipation rates varied across schools, districts, and student groups in 2021 more than in prior years. If the participation rate in 2021 was lower than in prior years, results may have been different if more students had taken the test."
] | [
null,
"https://profiles.doe.mass.edu/images/btn.gif",
null
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https://forums.parallax.com/discussion/175363/propeller-2-version-of-auduino-granular-synthesizer | [
"Propeller 2 version of Auduino granular synthesizer — Parallax Forums\n\n# Propeller 2 version of Auduino granular synthesizer\n\nI am slowly learning to program the Propeller 2 chip and as an exercise I decided to implement the remarkable Auduino granular synth on the P2. Auduino was originally written for the Arduino platform and is capable of producing a far wider range of timbres than might be expected given its relative simple design.\nFor my P2 implementation, I built a control panel with two analog joysticks and some potentiometers, and I added a MIDI interface. In the video below, a demonstration of how Auduino-style granular synthesis works is followed by some analog noodling, then some MIDI keyboard noodling, and finally a rendition of one of my favorite synth solos.\n\nHere's the code. I would be grateful for any suggestions for improvement.\n\n```CON\n_xinfreq = 20_000_000\n_clkfreq = 288_000_000\nBASEPIN_DAC = 6\n\nNPOTS = 7\niDECAY0 = 0\niINC0 = 1\niDECAY1 = 2\niINC1 = 3\niINC = 4\niVOL0 = 5\niVOL1 = 6\n\ndat\npotPins byte 8, 9, 10, 11, 13, 12, 14\nminValue word 0, 100, 0, 100, 0, 100, 0\nmaxValue long 200, 4000, 200, 4000, 2000, 1024, 1024\npotValues long 0, 0, 0, 0, 0, 0, 0\n\nvar\nlong stack1\nlong stack2\n\nOBJ\nmidi: \"jm_fullduplexserial\"\ntx: \"jm_fullduplexserial\"\n\nvar\nword midiNote\nword midiVel\nword inc\n\npri midiTask() | state, b, s, n, l, bend\nbend := 0 << 24 ' +/- semitones << 24\nmidi.start(25, -1, 0, 31_250)\nmidiNote := -1 ' something invalid\n\nrepeat\nb := midi.rx()\nif b & \\$80\ns := b & \\$f0 ' ignore channel\nstate := 0\nelse\nif state == 0 ' note on/off note byte or pitch bend low byte\nif s == \\$90 or s == \\$80\nn := b\nstate := 1\nelseif s == \\$e0\nl := b\nstate := 2\nelseif state == 1 ' note on/off velocity\nif s == \\$90 ' note on\nmidiNote := n\nmidiVel := b\ninc := incFromNote(midiNote, bend)\nelse 'note off\nif midiNote == n\nmidiVel := 0\ninc := 0\nstate := 0\nelseif state == 2 ' pitch bend high byte\nbend := b << 7 + l ' 16-bit bend [0..16384)\nbend -= \\$2000 ' [-8192..8192)\n' bend >> 12 would be [-2...2), +/- 2 semitones\n' but we need to scale by 2^24, so:\nbend <<= 24 - 12\nif midiVel\ninc := incFromNote(midiNote, bend)\nstate := 0\n\npri incFromNote(note, bend) : result | freq, K\n{\nbend is semitones * 2^24\n\nfreq = 2 ^ ((note - 69) / 12) * 440\ninc = freq / 48000 * 65536\n\n}\nK := 16\nfreq := QEXP ((((note - 69) << 24 + bend) / 12) << 3 + K << 27) * 440 ' frequency << K\nresult := freq / 48000\n\nPUB main()\nplaySounds(false)\n\npri playSounds(isMidi) | i, ct, y , amp0, decay0, inc0, ph0, amp1, decay1, inc1, ph1, ph, out, vol0, vol1\n\npinstart(BASEPIN_DAC addpins 1, P_DAC_124R_3V | P_DAC_DITHER_RND | P_OE, 6_000, 0) ' 48kHz\nwaitms(500)\n\nct := getct()\n\nrepeat\ninc0 := potValues[iINC0]\ndecay0 := potValues[iDECAY0]\ninc1 := potValues[iINC1]\ndecay1 := potValues[iDECAY1]\nvol0 := potValues[iVOL0]\nvol1 := potValues[iVOL1]\nifnot isMidi\ninc := potValues[iINC]\n\nph += inc\nif ph.\nph. := 0\nph0 := 0\namp0 := vol0 << 16\nph1 := 0\namp1 := vol1 << 16\n\ny := ph0.[14..0] ' 0..7fff\nif ph0.\n!= y\ny -= \\$4000 ' -\\$4000..3fff\ny scas= amp0\nout := y\n\ny := ph1.[14..0] ' 0..7fff\nif ph1.\n!= y\ny -= \\$4000 ' -\\$4000..3fff\ny scas= amp1\n\nout += y\nout ^= \\$8000\n\nwaitct(ct += 6000)\n\nph0 += inc0\nph1 += inc1\namp0 sca= \\$ffff_ffff + (decay0 - 200) << 16\namp1 sca= \\$ffff_ffff + (decay1 - 200) << 16\n\nrepeat i from 0 to NPOTS - 1\n\nrepeat\nrepeat i from 0 to NPOTS - 1\n\n{{\n\nPermission is hereby granted, free of charge, to any person obtaining a copy of this\nsoftware and associated documentation files (the \"Software\"), to deal in the Software\nwithout restriction, including without limitation the rights to use, copy, modify,\nmerge, publish, distribute, sublicense, and/or sell copies of the Software, and to\npermit persons to whom the Software is furnished to do so, subject to the following\nconditions:\n\nThe above copyright notice and this permission notice shall be included in all copies\nor substantial portions of the Software.\n\nTHE SOFTWARE IS PROVIDED \"AS IS\", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED,\nINCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A\nPARTICULAR PURPOSE AND NON-INFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT\nHOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF\nCONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE\nOR THE USE OR OTHER DEALINGS IN THE SOFTWARE.\n\n}}\n```\n\n• edited 2023-05-21 14:29\n\nNeato!\n\nI would suggest moving midiTask() and incFromNote(note, bend) below main() so there is no confusion where the program starts. I've been experimenting with MIDI a bit, but haven't fully sussed out dealing with my keyboard/controller (a 25-yr olf M-Audio Oxygen 8) that sends timing data mixed in with notes, and does running status style outputs).\n\nConsider declaring a constant for your MIDI input pin, as that could move with different designs.\n\nGiven the rate of MIDI you could save a cog removing one instance of jm_fullduplexserial and using the smart pin directly.\n\n```pub midi_task() | x\n\nx := ((clkfreq / 31_250) << 16) | 7\n\npinstart(MIDI_IN, P_ASYNC_RX, x, 0)\n\nrepeat\nb := rdpin(MIDI_IN) >> 24\n\nif b ...\n```\n• Thank you for the great suggestions, @JonnyMac, and thanks for jm_analog_in and jm_fullduplexserial! Some day I'll understand smart pins.\n\n• I don't know how much time you have left in your 48kHz loop, but you might find a few space code cycles here and there by running simple timing tests like this:\n\n``` t := getct()\ny := x.[14..0]\nt := getct()-t-40\ndebug(udec_long(t))\n\nt := getct()\ny := x & \\$7FFF\nt := getct()-t-40\ndebug(udec_long(t))\n```\n\nNote that the second version is faster\n\n• edited 2023-05-22 04:07\n\nA few minutes later...\n\nWell, it turns out you have a lot of time in your loop, even if using Propeller Tool (not compiled).",
null,
"I added pinlow() immediately before your waitct() and pinhigh() immediately after. You've got ~10us of available bandwidth.\n\n• nice\n\n• Thanks, @Ludis !"
] | [
null,
"https://forums.parallax.com/uploads/editor/vc/evg8koygzatj.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.70599717,"math_prob":0.96043617,"size":5800,"snap":"2023-40-2023-50","text_gpt3_token_len":1759,"char_repetition_ratio":0.079020016,"word_repetition_ratio":0.019305019,"special_character_ratio":0.3413793,"punctuation_ratio":0.17104149,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9617604,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-02T02:39:38Z\",\"WARC-Record-ID\":\"<urn:uuid:f20428a9-1a31-4416-b1c8-62eb5458cf46>\",\"Content-Length\":\"87720\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e66614ff-896d-4ae0-9f2f-de4ad24b0575>\",\"WARC-Concurrent-To\":\"<urn:uuid:f2301ae6-5690-4c1f-9519-4a79975771df>\",\"WARC-IP-Address\":\"3.129.223.39\",\"WARC-Target-URI\":\"https://forums.parallax.com/discussion/175363/propeller-2-version-of-auduino-granular-synthesizer\",\"WARC-Payload-Digest\":\"sha1:36JIQVIS6UJGOAIMLSKQHF3VTXMNFZYX\",\"WARC-Block-Digest\":\"sha1:XYXPOFXX4RDT2UDNO5OVO45OFRRPIOAG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100309.57_warc_CC-MAIN-20231202010506-20231202040506-00391.warc.gz\"}"} |
https://www.sofolympiadtrainer.com/forum/imo/0/427 | [
"# User Forum\n\nSubject :IMO Class : Class 5\n\nClass : Class 6\n\n## Ans 2:\n\nClass : Class 7\nfigure 1 has 2 lines of symmetry and figure 2 has 4 lines of symmetry. Rest have no line of symmetry\n\n## Ans 3:\n\nClass : Class 5\n1. figure one has only one line of symetry and figure 3 and 4 has no lines of symetry\n\n## Ans 4:\n\nClass : Class 7\nFigure 1 has two lines of symmetry, figure 2 has four lines of symmetry and rest have no line of symmetry.\n\n## Ans 5:\n\nClass : Class 5\n2\n\nSubject :IMO Class : Class 5\n\n## Ans 1:\n\nClass : Class 5\n\nSubject :IMO Class : Class 3\n\nClass : Class 1\n\nSubject :IMO Class : Class 5\n\nClass : Class 1\nThe angle OUT is of 180 degrees because it is a straight line. Hence, the angle RUN will be X+23=180-(32+23)=125.\n\nSubject :IMO Class : Class 5\n\nClass : Class 8\n\n## Ans 2:\n\nClass : Class 5\nThe answer is 192 solution- 48/2=24 (24+24+24)+(24+24+24+24+24)=192\n\nSubject :IMO Class : Class 6\n\n## Ans 1:\n\nClass : Class 9\nthe ans. is 200m . fig x has 12 sides so length of 1 side = 300/12 =25m . fig y has 8 sides so perimeter = 25 x 8 = 200m\n\nClass : Class 6\nB\n\nClass : Class 8\n\n## Ans 4:\n\nClass : Class 6\niu7tgo87rfo\n\nSubject :IMO Class : Class 5\n\nClass : Class 1\n\nSubject :IMO Class : Class 7\n\n## Ans 1:\n\nClass : Class 7\n\nSubject :IMO Class : Class 4\n\nClass : Class 4\n\n## Ans 2:\n\nClass : Class 6\nB is the answer means 140\n\n## Ans 3:\n\nClass : Class 6\nB is the answer means 140\n\nClass : Class 6\nb=140 kg"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.96326655,"math_prob":0.40046144,"size":1569,"snap":"2020-10-2020-16","text_gpt3_token_len":460,"char_repetition_ratio":0.111821085,"word_repetition_ratio":0.006329114,"special_character_ratio":0.31230083,"punctuation_ratio":0.05965909,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9840616,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-31T20:07:25Z\",\"WARC-Record-ID\":\"<urn:uuid:b65035ef-6048-4166-a024-1123c4575dd5>\",\"Content-Length\":\"424549\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:28a373b2-cb07-4a1c-9a95-58aa37147ecb>\",\"WARC-Concurrent-To\":\"<urn:uuid:32d2e33d-914a-47f9-89ae-906f7a9c5f85>\",\"WARC-IP-Address\":\"52.76.64.191\",\"WARC-Target-URI\":\"https://www.sofolympiadtrainer.com/forum/imo/0/427\",\"WARC-Payload-Digest\":\"sha1:WDBWXIQWZUO664NKUNSZOR3LRNPN7KJ3\",\"WARC-Block-Digest\":\"sha1:JY5AU47ZI55M2G4PMCL7XYCQZWTM2A3G\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370503664.38_warc_CC-MAIN-20200331181930-20200331211930-00449.warc.gz\"}"} |
https://answers.everydaycalculation.com/add-fractions/70-72-plus-25-10 | [
"Solutions by everydaycalculation.com\n\n1st number: 70/72, 2nd number: 2 5/10\n\n70/72 + 25/10 is 125/36.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 72 and 10 is 360\n2. For the 1st fraction, since 72 × 5 = 360,\n70/72 = 70 × 5/72 × 5 = 350/360\n3. Likewise, for the 2nd fraction, since 10 × 36 = 360,\n25/10 = 25 × 36/10 × 36 = 900/360\n350/360 + 900/360 = 350 + 900/360 = 1250/360\n5. 1250/360 simplified gives 125/36\n6. So, 70/72 + 25/10 = 125/36\nIn mixed form: 317/36\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
] | [
null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.768809,"math_prob":0.999495,"size":754,"snap":"2020-34-2020-40","text_gpt3_token_len":308,"char_repetition_ratio":0.15333334,"word_repetition_ratio":0.0,"special_character_ratio":0.55437666,"punctuation_ratio":0.093023255,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99686176,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-02T00:00:14Z\",\"WARC-Record-ID\":\"<urn:uuid:9a697e06-cbe6-4e3d-bfb4-52bfdb91992b>\",\"Content-Length\":\"7703\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:09e824b5-4e28-43b7-a85f-212c43237f73>\",\"WARC-Concurrent-To\":\"<urn:uuid:8cfa8afe-04ec-4c1c-933e-b075862fa862>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/add-fractions/70-72-plus-25-10\",\"WARC-Payload-Digest\":\"sha1:4I5SOPLVBMJYO42DTOG7BP6V7NDBV356\",\"WARC-Block-Digest\":\"sha1:Y2PVF5VHO3WKP5FZX5OCQUPSWZKQJMJ3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600402132335.99_warc_CC-MAIN-20201001210429-20201002000429-00571.warc.gz\"}"} |
https://www.flexiprep.com/NCERT-Exemplar-Solutions/Mathematics/Class-12/NCERT-Class-12-Mathematics-Exemplar-Chapter-3-Matrices-Part-12.html | [
"NCERT Class 12-Mathematics: Exemplar Chapter – 3 Matrices Part 12 (For CBSE, ICSE, IAS, NET, NRA 2022)\n\nDoorsteptutor material for CBSE/Class-12 is prepared by world's top subject experts: get questions, notes, tests, video lectures and more- for all subjects of CBSE/Class-12.\n\nQuestion 31:\n\nShow that if A and B are square matrices such that , then\n\n.\n\nBy matrix multiplication we can write:\n\nWe know that matrix multiplication is not commutative but it is given that:\n\n… proved\n\nQuestion 32:\n\nLet , and .\n\nShow that:\n\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n(g)\n\n(h)\n\n(i)\n\nWe have, , and\n\n(i)\n\nAnd\n\nHence proved.\n\n(ii)\n\nAnd\n\nAlso,\n\nHence Proved.\n\n(iii)\n\nHence Proved.\n\n(iv)\n\nAnd\n\nAlso\n\nHence Proved.\n\n(v)\n\nNow,\n\nA. Hence Proved.\n\n(vi)\n\nAnd\n\nHence proved.\n\n(vii)\n\nNow,\n\nHence Proved.\n\n(viii)\n\nNow,\n\nAnd\n\n[Using Eq. (ii) and (iii) ]\n\n[Using Eq. (i) ] Hence proved\n\n(ix)\n\nHence proved.\n\nDeveloped by:"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.79416597,"math_prob":0.94357276,"size":842,"snap":"2022-05-2022-21","text_gpt3_token_len":270,"char_repetition_ratio":0.15990454,"word_repetition_ratio":0.0,"special_character_ratio":0.3087886,"punctuation_ratio":0.20765027,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99854743,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-19T01:03:43Z\",\"WARC-Record-ID\":\"<urn:uuid:074eed35-3838-41fd-bc73-5ba286d4dd23>\",\"Content-Length\":\"18282\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5c43bad0-c420-43b6-9873-adfdb3b6c316>\",\"WARC-Concurrent-To\":\"<urn:uuid:522aa73c-cd6e-43f2-bc82-f341ff1bdae3>\",\"WARC-IP-Address\":\"104.21.77.56\",\"WARC-Target-URI\":\"https://www.flexiprep.com/NCERT-Exemplar-Solutions/Mathematics/Class-12/NCERT-Class-12-Mathematics-Exemplar-Chapter-3-Matrices-Part-12.html\",\"WARC-Payload-Digest\":\"sha1:WP6GNIF45N4FKJIOWBUDFMTFFSUOGPFA\",\"WARC-Block-Digest\":\"sha1:XJRBUIJY7XETAQHO7F762CRQBYFW547N\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301217.83_warc_CC-MAIN-20220119003144-20220119033144-00286.warc.gz\"}"} |
https://www.sites.google.com/site/janbeck/cybersecurity-challenges-ctfs-and-more/angr-hooking-derecompiling-chainbreaker | [
"# Angr hooking - de/recompiling - chainbreaker\n\nThe executable that I am going to analyze here is interesting in several ways. First, no attempt at code obfuscation is made and the decompilers from both `Ghidra` and `IDA` produce code that can be recompiled with little effort. Also, the program is very short, making it a good example from an academic standpoint. The input is passed via a command line argument and so `angr` or `PinCTF` are natural first approaches to find the flag. However, there is just enough complication to prevent this from working and that is what makes this such a nice example to show progressive improvements to the `angr` input file using hooks. Let's see what the program does:\n\n``\\$ ./chainbreaker``\n``usage: ./chainbreaker SEED``\n\n``\\$ ./chainbreaker a``\n``Starting chainbreaker!``\n``ERROR``\n``Seed must be an integer!``\n\n``\\$ ./chainbreaker 0``\n``Starting chainbreaker!``\n``Seed 0 requires 96 links``\n\n``LINK 1 0 -> ERROR``\n``Invalid chain produced!``\n\n``\\$ ./chainbreaker 2``\n``Starting chainbreaker!``\n``Seed 2 requires 96 links``\n\n``LINK 1 2 -> 5 Sleeping for 5ms``\n``LINK 2 5 -> 20 Sleeping for 20ms``\n``LINK 3 20 -> 119 Sleeping for 119ms``\n``LINK 4 119 -> 426 Sleeping for 426ms``\n``LINK 5 426 -> 1529 Sleeping for 1529ms``\n``LINK 6 1529 -> 8160 Sleeping for 8160ms``\n``[ ...]``\n``LINK 94 2049 -> 16472 Sleeping for 16472ms``\n``LINK 95 16472 -> -445 Sleeping for 0ms``\n``LINK 96 -445 -> -1570 Sleeping for 0ms``\n``ERROR``\n``Starting seed doesnt match final output``\n\nOk. The program wants an integer seed, from which it produces a chain link number and then iterates through a series of computations to check if the seed is correct. The usage of the sleep timer precludes `PinCTF`. It is easy to patch the program to always sleep for 0 time, but different seeds use different link sizes and that alone causes any run time differential due to the final check to drown in the signal. Another natural thing to try is `angr`. Lets look at the decompiled code:\n\n`int __cdecl main(int argc, const char **argv, const char **envp)``{`` unsigned int v4; // [rsp+1Ch] [rbp-64h]`` char v5; // [rsp+20h] [rbp-60h]`` char v6; // [rsp+4Fh] [rbp-31h]`` char v7; // [rsp+50h] [rbp-30h]`` int inputasint; // [rsp+5Ch] [rbp-24h]`` int v9; // [rsp+60h] [rbp-20h]`` int i; // [rsp+64h] [rbp-1Ch]`` int v11; // [rsp+68h] [rbp-18h]`` int InputasInt2; // [rsp+6Ch] [rbp-14h]`\n` v9 = 100;`` if ( argc == 2 )`` {`` puts(\"Starting chainbreaker!\");`` std::allocator<char>::allocator(&v6);`` std::__cxx11::basic_string<char,std::char_traits<char>,std::allocator<char>>::basic_string(&v5, argv, &v6);`` inputasint = std::__cxx11::stoi((__int64)&v5, 0LL, 0xAu);`` std::__cxx11::basic_string<char,std::char_traits<char>,std::allocator<char>>::~basic_string(&v5);`` std::allocator<char>::~allocator(&v6);`` InputasInt2 = inputasint;`` v11 = 21309 * ((inputasint ^ 0x7B) + (inputasint ^ 0x141)) % 100;`` if ( v11 >= 0 )`` {`` if ( !v11 )`` v11 = 10; // if V11 is zero, make it 10`` }`` else`` {`` v11 = -v11; // else if v11 is negative, make it positive`` }`` printf(\"Seed %s requires %d links\\n\\n\", argv, (unsigned int)v11);`` for ( i = 0; i <= 99 && i < v11; ++i )`` {`` printf(\"LINK %d\\t\\t%d\\t->\\t\", (unsigned int)(i + 1), (unsigned int)InputasInt2);`` InputasInt2 = parse(inputasint, InputasInt2, i);`` v4 = InputasInt2;`` if ( InputasInt2 < 0 )`` v4 = 0;`` printf(\"%d\\t Sleeping for %dms\\n\", (unsigned int)InputasInt2, v4);`` std::chrono::duration<long,std::ratio<1l,1000l>>::duration<int,void>(&v7, &v4);`` std::this_thread::sleep_for<long,std::ratio<1l,1000l>>(&v7);`` }`` if ( i == 99 )`` {`` puts(\"ERROR\\nMaximum allowed iterations reached\");`` exit(1);`` }`` if ( inputasint == InputasInt2 && i == v11 )`` {`` puts(\"You have broken the chain!\");`` exit(0);`` }`` puts(\"ERROR\\nStarting seed doesnt match final output\");`` }`` else`` {`` puts(\"usage: ./chainbreaker SEED\");`` }`` return 0;``}`\n\nThe 'parse' function looks like this:\n\n`__int64 __fastcall parse(int a1, int a2, int a3)``{`` int v4; // [rsp+8h] [rbp-18h]`` int i; // [rsp+18h] [rbp-8h]`` int v6; // [rsp+1Ch] [rbp-4h]`\n` v4 = a2;`` if ( !a2 )`` {`` puts(\"ERROR\\nInvalid chain produced!\");`` exit(0);`` }`` if ( a2 < -4096 || a2 > 4096 )`` v4 = -(a2 % 4096);`` v6 = 1;`` for ( i = 0; i <= 2; ++i )`` v6 ^= v4 << i;`` return v6 + (v4 ^ (unsigned int)(a1 + a3 - 1)) + a1 - 15;``}`\n\nIt's fairly simple code. Many of the 'std::' calls are compiler generated from higher level string functions the original author used, as described here.\n\nNote that the input gets parsed into a 32 bit integer value. If we can get the computation to run fast enough, this challenge could be solved brute force. Since it's simple to do, first I want to try to recompile the code with some changes to optimize run time. I used Visual Studio for this case:\n\n`#include \"stdafx.h\"`\n`int invalid_chain = 0;``int ii =0;`\n`__int64 __fastcall parse(int a1, int a2, int a3)``{`` int v4; // [rsp+8h] [rbp-18h]`` int i; // [rsp+18h] [rbp-8h]`` int v6; // [rsp+1Ch] [rbp-4h]`\n` v4 = a2;`` if ( !a2 )`` {`` invalid_chain = 1; // Invalid chain produced`` return 0;`` }`` if ( a2 < -4096 || a2 > 4096 )`` v4 = -(a2 % 4096);`` v6 = 1;`` for ( i = 0; i <= 2; ++i )`` v6 ^= v4 << i;`` return v6 + (v4 ^ (unsigned int)(a1 + a3 - 1)) + a1 - 15;``}`\n`int _tmain(int argc, _TCHAR* argv[])``//int __cdecl main(int argc, const char **argv, const char **envp)``{`` unsigned int v4; // [rsp+1Ch] [rbp-64h]`` char v5; // [rsp+20h] [rbp-60h]`` char v6; // [rsp+4Fh] [rbp-31h]`` char v7; // [rsp+50h] [rbp-30h]`` int v8; // [rsp+5Ch] [rbp-24h]`` int v9; // [rsp+60h] [rbp-20h]`` int i; // [rsp+64h] [rbp-1Ch]`` int v11; // [rsp+68h] [rbp-18h]`` int v12; // [rsp+6Ch] [rbp-14h]`\n` v9 = 100;`\n` for (int counter=0x00000000;counter<=0xffffffff;counter++)`` {`` v8 = counter;`` v12 = v8;`` v11 = 0x533D * ((v8 ^ 0x7B) + (v8 ^ 0x141)) % 100;`` if ( v11 >= 0 )`` {`` if ( !v11 )`` v11 = 10;`` }`` else`` {`` v11 = -v11;`` }`` if (0 == (counter % 65536)) // limit the amount of information going to console`` printf(\"Seed %8.8x requires %d links\\n\", counter, (unsigned int)v11);`` for ( i = 0; i <= 99 && i < v11; ++i )`` {`` v12 = parse(v8, v12, i);`` if (invalid_chain== 1) continue; // if the parse() function determines an invalid chain, try next loop value`` v4 = v12;`` if ( v12 < 0 )`` v4 = 0;`` }`` if ( i == 99 )`` {`` continue; // Maximum allowed iterations reached - go try next loop value`` }`` if ((i <99)&& v8 == v12 && i == v11 &&(0==invalid_chain)) `` {`` puts(\"You have broken the chain!\");`` printf(\" %d \\n\", counter);`` return 1;`` }`` invalid_chain = 0;`` }`` return 0;``}`\n\nOn a LG Gram 17 laptop, this program can try the whole 32-bit keyspace in about 15 minutes. Modern computers are awesome!\n\n``Seed ffffd32a requires 72 links``\n``You have broken the chain!``\n`` -11478``\n\nBut this article is supposed to be about `angr` hooks, so lets continue there. A first crack at this should avoid code sections that 'puts' error messages and look for code paths that lead to `puts(\"You have broken the chain!\")`.\n\n``import time``\n``import claripy``\n``import angr``\n\n``# specify good and bad endpoints``\n``bad1 = (0x40149D) # ERROR\\nStarting seed doesnt match final``\n``bad2 = (0x4014AB) # usage: ./chainbreaker SEED``\n``bad3 = (0x401461) # ERROR\\nMaximum allowed iterations reached``\n``bad4 = (0x40122C) # ERROR\\nInvalid chain produced!``\n``good = (0x401487) # You have broken the chain!``\n\n``project = angr.Project(\"./chainbreaker\", auto_load_libs=True)``\n``#create an initial state with a symbolic bit vector as argv1``\n``sym_arg_size = 1 #Length in Bytes because we will multiply with 8 later``\n``sym_arg = claripy.BVS('sym_arg', 8*sym_arg_size)``\n``argv = [project.filename]``\n``argv.append(sym_arg)``\n\n``initial_state = project.factory.full_init_state(args=argv)``\n``for byte in sym_arg.chop(8):``\n`` initial_state.add_constraints(byte >= '\\x30') # '0'``\n`` initial_state.add_constraints(byte <= '\\x39') # '9'``\n``sm = project.factory.simulation_manager(initial_state)``\n\n``t0 = time.time()``\n``sm.explore(find=good,avoid=[bad1,bad2,bad3,bad4])``\n``t1 = time.time()``\n``print (t1-t0)``\n``found = sm.found``\n``result = found.solver.eval(sym_arg, cast_to=bytes)``\n``print(result)``\n\nThis code limits the input seed to a single digit, and yet angr has not finished hours later. And we would only get an error message at that point any way because the correct input has 6 digits and starts with a '-'. Since we already have the key, we can investigate what goes wrong. With the input modified like this:\n\n``data = '-1147' # this is the start of the correct password``\n``stringAsList = list(data); # turn string into list``\n``stringAsList.append(sym_arg) # declare how long the ``\n``bytestring = claripy.Concat(*stringAsList) # turn list into bytestring``\n``argv.append(bytestring)``\n``print(argv)``\n\nIf we ignore the fact that the answer is a negative number, we can see that the original angr input file does, in fact, find the right solution eventually - it's just very slow.\n\n``(angr) \\$ python angrfile ``\n\n``[...]``\n\n``<BV48 0x2d31313437 .. sym_arg_47_8>``\n``['./chainbreaker', <BV48 0x2d31313437 .. sym_arg_47_8>]``\n\n``[...]``\n\n``813.7370481491089``\n``b'8'``\n\nAbout 800 seconds for a single symbol; about 1600 for two, but after many hours 3 had not yet finished.\n\nNext, lets patch out the delay so that angr does not symbolize it. From Ghidra, we get the addresses to hook (after rebasing the position-independent executable to 0x400000)\n\n``00401441 e8 e8 02 CALL std::chrono::duration<long,std--ratio<1l,1000l undefined duration<int,void>(dur``\n`` 00 00``\n``00401446 48 8d 45 LEA RAX => local_38 ,[RBP + -0x30 ]``\n`` d0``\n``0040144a 48 89 c7 MOV param_1 ,RAX``\n``0040144d e8 cb 04 CALL std::this_thread::sleep_for<long,std--ratio<1l void sleep_for<long,std--ratio<1``\n`` 00 00``\n``00401452 83 45 e4 ADD dword ptr [RBP + local_24 ],0x1``\n`` 01``\n\nAnd it's pretty easy to hook these calls:\n\n``def durationhook(state):``\n`` print(\"0x401441 hooked\")``\n``project.hook(0x401441, durationhook, length=5) # hook this address, execute durationhook() and then skip 5 bytes, skipping the 'call' at this address``\n``def sleep_forhook(state):``\n`` print(\"0x40144d hooked\")``\n``project.hook(0x40144d, sleep_forhook, length=5) # hook this address, execute sleep_forhook() and then skip 5 bytes, skipping the 'call' at this address``\n\nAn interesting fact in this case is that while the angr documentation states:\n\n\"The CFG analysis does not distinguish between code from different binary objects. This means that by default, it will try to analyze control flow through loaded shared libraries. This is almost never intended behavior, since this will extend the analysis time to several days, probably. To load a binary without shared libraries, add the following keyword argument to the `Project` constructor: `load_options={'auto_load_libs': False}` \"\n\nthe reality is that `auto_load_libs` has the opposite effect here.\n\nThe two hooks yield a computation time (for 1-symbolic-digit integers) of about 30 minutes. Can anything else be done for the current case? There are some messages during the angr execution that point to some potential targets:\n\n`WARNING | 2020-01-12 16:13:03,410 | angr.state_plugins.symbolic_memory | The program is accessing memory or registers with an unspecified value. This could indicate unwanted behavior.``WARNING | 2020-01-12 16:13:03,410 | angr.state_plugins.symbolic_memory | angr will cope with this by generating an unconstrained symbolic variable and continuing. You can resolve this by:``WARNING | 2020-01-12 16:13:03,410 | angr.state_plugins.symbolic_memory | 1) setting a value to the initial state``WARNING | 2020-01-12 16:13:03,410 | angr.state_plugins.symbolic_memory | 2) adding the state option ZERO_FILL_UNCONSTRAINED_{MEMORY,REGISTERS}, to make unknown regions hold null``WARNING | 2020-01-12 16:13:03,410 | angr.state_plugins.symbolic_memory | 3) adding the state option SYMBOL_FILL_UNCONSTRAINED_{MEMORY_REGISTERS}, to suppress these messages.``WARNING | 2020-01-12 16:13:03,411 | angr.state_plugins.symbolic_memory | Filling memory at 0x7ffffffffff0000 with 207 unconstrained bytes referenced from 0x409d2a0 (strlen+0x0 in libc.so.6 (0x9d2a0))``WARNING | 2020-01-12 16:13:03,566 | angr.state_plugins.symbolic_memory | Filling memory at 0x7fffffffffeff00 with 6 unconstrained bytes referenced from 0x40b9ff0 (memcpy+0x0 in libc.so.6 (0xb9ff0))``WARNING | 2020-01-12 16:13:03,690 | angr.state_plugins.symbolic_memory | Filling memory at 0x7fffffffffefe68 with 8 unconstrained bytes referenced from 0x4049340 (strtoq+0x0 in libc.so.6 (0x49340))`\n\nso `angr` does not know that the `sym_arg_size = 1` constrains these values. I have seen these warnings many times for programs that use string functions, so it's valuable to see how much of a problem this really is.\n\n`printf` and puts seem obvious candidates to hook, as we don't need their output in this analysis:\n\n`def puts_hook(state):`` print(\"puts hooked\")``project.hook_symbol('puts', puts_hook)`\n`def printfhook(state):`` pass`` #print(\"printf hooked\")``project.hook(0x4013b1, printfhook, length=5) # hook this address, execute printfhook() and then skip 5 bytes, skipping the 'call' at this address``project.hook(0x4013ec, printfhook, length=5) # hook this address, execute printfhook() and then skip 5 bytes, skipping the 'call' at this address``project.hook(0x40142e, printfhook, length=5) # hook this address, execute printfhook() and then skip 5 bytes, skipping the 'call' at this address`\n\nWhich results in a 25% run time reduction, but the messages don't all go away.\n\n`WARNING | 2020-01-12 20:03:19,308 | angr.project | Address is already hooked, during hook(0x4083cc0, <function puts_hook at 0x7f535df95620>). Re-hooking.``puts hooked``WARNING | 2020-01-12 20:03:21,660 | angr.state_plugins.symbolic_memory | The program is accessing memory or registers with an unspecified value. This could indicate unwanted behavior.``WARNING | 2020-01-12 20:03:21,660 | angr.state_plugins.symbolic_memory | angr will cope with this by generating an unconstrained symbolic variable and continuing. You can resolve this by:``WARNING | 2020-01-12 20:03:21,660 | angr.state_plugins.symbolic_memory | 1) setting a value to the initial state``WARNING | 2020-01-12 20:03:21,660 | angr.state_plugins.symbolic_memory | 2) adding the state option ZERO_FILL_UNCONSTRAINED_{MEMORY,REGISTERS}, to make unknown regions hold null``WARNING | 2020-01-12 20:03:21,660 | angr.state_plugins.symbolic_memory | 3) adding the state option SYMBOL_FILL_UNCONSTRAINED_{MEMORY_REGISTERS}, to suppress these messages.``WARNING | 2020-01-12 20:03:21,660 | angr.state_plugins.symbolic_memory | Filling memory at 0x7ffffffffff0000 with 207 unconstrained bytes referenced from 0x409d2a0 (strlen+0x0 in libc.so.6 (0x9d2a0))``WARNING | 2020-01-12 20:03:21,816 | angr.state_plugins.symbolic_memory | Filling memory at 0x7fffffffffeff00 with 6 unconstrained bytes referenced from 0x40b9ff0 (memcpy+0x0 in libc.so.6 (0xb9ff0))`\n\nAnd it looks like the `puts` function was already hooked by the library simulation code. Re-hooking removes the call to `stroq`. The other calls are due to `stoi` and its' support calls:\n\n``std::allocator<char>::allocator(&v6);``\n``std::__cxx11::basic_string<char,std::char_traits<char>,std::allocator<char>>::basic_string(&v5, argv, &v6);``\n``inputasint = std::__cxx11::stoi((__int64)&v5, 0LL, 0xAu);``\n``std::__cxx11::basic_string<char,std::char_traits<char>,std::allocator<char>>::~basic_string(&v5);``\n``std::allocator<char>::~allocator(&v6);``\n\nThe hooking of `stoi` requires quite a bit of changes, so here the complete script:\n\n`import time``import claripy``import angr`\n`# specify good and bad endpoints``bad1 = (0x40149D) # ERROR\\nStarting seed doesnt match final``bad2 = (0x4014AB) # usage: ./chainbreaker SEED``bad3 = (0x401461) # ERROR\\nMaximum allowed iterations reached``bad4 = (0x40122C) # ERROR\\nInvalid chain produced!``good = (0x401487) # You have broken the chain!`\n`project = angr.Project(\"./chainbreaker\", auto_load_libs=True)`\n`argv = [project.filename]``data = '-1147' # dummy data, since we are hooking stoi``argv.append(data)``print(argv)``initial_state = project.factory.full_init_state(args=argv)``sm = project.factory.simulation_manager(initial_state)`\n`aux_arg = claripy.BVS('sym_arg', 8*4) # declase a 32 bit symbolic variable``initial_state.add_constraints(aux_arg <= -11000) # constrain variable to range we know the answer to be in``initial_state.add_constraints(aux_arg >= -12000) # constrain variable to range we know the answer to be in`\n`def durationhook(state):`` print(\"0x401441 hooked\")``project.hook(0x401441, durationhook, length=5) # hook this address, execute durationhook() and then skip 5 bytes, skipping the 'call' at this address``def sleep_forhook(state):`` print(\"0x40144d hooked\")``project.hook(0x40144d, sleep_forhook, length=5) # hook this address, execute sleep_forhook() and then skip 5 bytes, skipping the 'call' at this address``def allocatorhook(state):`` print(\"allocator hooked\")``project.hook(0x4012e5, allocatorhook, length=5) # hook this address, execute allocatorhook() and then skip 5 bytes, skipping the 'call' at this address``def allocatordestructorhook(state):`` print(\"allocatordestructor hooked\")``project.hook(0x401334, allocatordestructorhook, length=5) # hook this address, execute allocatordestructorhook() and then skip 5 bytes, skipping the 'call' at this address``def basicstringhook(state):`` print(\"basicstring hooked\")``project.hook(0x401303, basicstringhook, length=5) # hook this address, execute basicstringhook() and then skip 5 bytes, skipping the 'call' at this address``def basicstringdestructorhook(state):`` print(\"basicstringdestructor hooked\")``project.hook(0x401328, basicstringdestructorhook, length=5) # hook this address, execute basicstringdestructorhook() and then skip 5 bytes, skipping the 'call' at this address``def stoihook(state):`` state.regs.rax = aux_arg`` print(\"stoi hooked\")`` print(state.regs.rax)``project.hook(0x401319, stoihook, length=5) # hook this address, execute stoihook() and then skip 5 bytes, skipping the 'call' at this address``def puts_hook(state):`` print(\"puts hooked\")``project.hook_symbol('puts', puts_hook)``def printfhook(state):`` pass`` #print(\"printf hooked\")``project.hook(0x4013b1, printfhook, length=5) # hook this address, execute printfhook() and then skip 5 bytes, skipping the 'call' at this address``project.hook(0x4013ec, printfhook, length=5) # hook this address, execute printfhook() and then skip 5 bytes, skipping the 'call' at this address``project.hook(0x40142e, printfhook, length=5) # hook this address, execute printfhook() and then skip 5 bytes, skipping the 'call' at this address`\n`t0 = time.time()``sm.explore(find=good,avoid=[bad1,bad2,bad3,bad4])``t1 = time.time()``print (t1-t0)``found = sm.found``print(aux_arg)``result = found.solver.eval(aux_arg)``print(result)``result = result - 4294967296``print(result)`\n\nAnd the warnings about the unconstrained bytes disappeared:\n\n`WARNING | 2020-01-13 01:26:35,067 | cle.loader | The main binary is a position-independent executable. It is being loaded with a base address of 0x400000.``['./chainbreaker', '-1147']``WARNING | 2020-01-13 01:26:36,018 | angr.project | Address is already hooked, during hook(0x4083cc0, <function puts_hook at 0x7f2cf0c88ae8>). Re-hooking.``puts hooked``allocator hooked``basicstring hooked``stoi hooked``<SAO <BV64 0x0 .. sym_arg_47_32>>``basicstringdestructor hooked``allocatordestructor hooked`\n\nNow an interesting thing happens, for both\n\n``initial_state.add_constraints(aux_arg <= -11470) # constrain variable to range we know the answer to be in``\n``initial_state.add_constraints(aux_arg >= -11480) # constrain variable to range we know the answer to be in``\n\nand\n\n``initial_state.add_constraints(aux_arg <= -11400) # constrain variable to range we know the answer to be in``\n``initial_state.add_constraints(aux_arg >= -11500) # constrain variable to range we know the answer to be in``\n\nthe execution time is about 900 seconds, but for\n\n``initial_state.add_constraints(aux_arg <= -11000) # constrain variable to range we know the answer to be in``\n``initial_state.add_constraints(aux_arg >= -12000) # constrain variable to range we know the answer to be in``\n\nAfter setting all these hooks, all that is left of the program is symbolic execution of the user space, but in this case angr is incapable of solving it. The hooking however does show its value. Instead of not being able to solve for even the last digit, `angr` can now solve 2. We also learned that the common warnings about the `printf` symbolization do not necessarily have to be too much of a concern."
] | [
null
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https://nbviewer.ipython.org/github/cs109/content/blob/master/HW2_solutions.ipynb | [
"# Homework 2: Desperately Seeking Silver (Solutions)¶\n\nDue Thursday, Oct 3, 11:59 PM",
null,
"In HW1, we explored how to make predictions (with uncertainties) about upcoming elections based on the Real Clear Politics poll. This assignment also focuses on election prediction, but we are going to implement and evaluate a number of more sophisticated forecasting techniques.\n\nWe are going to focus on the 2012 Presidential election. Analysts like Nate Silver, Drew Linzer, and Sam Wang developed highly accurate models that correctly forecasted most or all of the election outcomes in each of the 50 states. We will explore how hard it is to recreate similarly successful models. The goals of this assignment are:\n\n1. To practice data manipulation with Pandas\n2. To develop intuition about the interplay of precision, accuracy, and bias when making predictions\n3. To better understand how election forecasts are constructed\n\nThe data for our analysis will come from demographic and polling data. We will simulate building our model on October 2, 2012 -- approximately one month before the election.\n\n### Instructions¶\n\nThe questions in this assignment are numbered. The questions are also usually italicised, to help you find them in the flow of this notebook. At some points you will be asked to write functions to carry out certain tasks. Its worth reading a little ahead to see how the function whose body you will fill in will be used.\n\nThis is a long homework. Please do not wait until the last minute to start it!\n\nThe data for this homework can be found at this link. Download it to the same folder where you are running this notebook, and uncompress it. You should find the following files there:\n\n1. us-states.json\n3. predictwise.csv\n4. g12.csv\n5. g08.csv\n6. 2008results.csv\n7. nat.csv\n8. p04.csv\n9. 2012results.csv\n10. cleaned-state_data2012.csv\n\n#### Setup and Plotting code¶\n\nIn :\n%matplotlib inline\nfrom collections import defaultdict\nimport json\n\nimport numpy as np\nimport matplotlib.pyplot as plt\nimport pandas as pd\n\nfrom matplotlib import rcParams\nimport matplotlib.cm as cm\nimport matplotlib as mpl\n\n#colorbrewer2 Dark2 qualitative color table\ndark2_colors = [(0.10588235294117647, 0.6196078431372549, 0.4666666666666667),\n(0.8509803921568627, 0.37254901960784315, 0.00784313725490196),\n(0.4588235294117647, 0.4392156862745098, 0.7019607843137254),\n(0.9058823529411765, 0.1607843137254902, 0.5411764705882353),\n(0.4, 0.6509803921568628, 0.11764705882352941),\n(0.9019607843137255, 0.6705882352941176, 0.00784313725490196),\n(0.6509803921568628, 0.4627450980392157, 0.11372549019607843)]\n\nrcParams['figure.figsize'] = (10, 6)\nrcParams['figure.dpi'] = 150\nrcParams['axes.color_cycle'] = dark2_colors\nrcParams['lines.linewidth'] = 2\nrcParams['axes.facecolor'] = 'white'\nrcParams['font.size'] = 14\nrcParams['patch.edgecolor'] = 'white'\nrcParams['patch.facecolor'] = dark2_colors\nrcParams['font.family'] = 'StixGeneral'\n\ndef remove_border(axes=None, top=False, right=False, left=True, bottom=True):\n\"\"\"\nMinimize chartjunk by stripping out unnecesasry plot borders and axis ticks\n\nThe top/right/left/bottom keywords toggle whether the corresponding plot border is drawn\n\"\"\"\nax = axes or plt.gca()\nax.spines['top'].set_visible(top)\nax.spines['right'].set_visible(right)\nax.spines['left'].set_visible(left)\nax.spines['bottom'].set_visible(bottom)\n\n#turn off all ticks\nax.yaxis.set_ticks_position('none')\nax.xaxis.set_ticks_position('none')\n\n#now re-enable visibles\nif top:\nax.xaxis.tick_top()\nif bottom:\nax.xaxis.tick_bottom()\nif left:\nax.yaxis.tick_left()\nif right:\nax.yaxis.tick_right()\n\npd.set_option('display.width', 500)\npd.set_option('display.max_columns', 100)\n\nIn :\n#this mapping between states and abbreviations will come in handy later\nstates_abbrev = {\n'AL': 'Alabama',\n'AR': 'Arkansas',\n'AS': 'American Samoa',\n'AZ': 'Arizona',\n'CA': 'California',\n'CT': 'Connecticut',\n'DC': 'District of Columbia',\n'DE': 'Delaware',\n'FL': 'Florida',\n'GA': 'Georgia',\n'GU': 'Guam',\n'HI': 'Hawaii',\n'IA': 'Iowa',\n'ID': 'Idaho',\n'IL': 'Illinois',\n'IN': 'Indiana',\n'KS': 'Kansas',\n'KY': 'Kentucky',\n'LA': 'Louisiana',\n'MA': 'Massachusetts',\n'MD': 'Maryland',\n'ME': 'Maine',\n'MI': 'Michigan',\n'MN': 'Minnesota',\n'MO': 'Missouri',\n'MP': 'Northern Mariana Islands',\n'MS': 'Mississippi',\n'MT': 'Montana',\n'NA': 'National',\n'NC': 'North Carolina',\n'ND': 'North Dakota',\n'NH': 'New Hampshire',\n'NJ': 'New Jersey',\n'NM': 'New Mexico',\n'NY': 'New York',\n'OH': 'Ohio',\n'OK': 'Oklahoma',\n'OR': 'Oregon',\n'PA': 'Pennsylvania',\n'PR': 'Puerto Rico',\n'RI': 'Rhode Island',\n'SC': 'South Carolina',\n'SD': 'South Dakota',\n'TN': 'Tennessee',\n'TX': 'Texas',\n'UT': 'Utah',\n'VA': 'Virginia',\n'VI': 'Virgin Islands',\n'VT': 'Vermont',\n'WA': 'Washington',\n'WI': 'Wisconsin',\n'WV': 'West Virginia',\n'WY': 'Wyoming'\n}\n\n\nHere is some code to plot State Chloropleth maps in matplotlib. make_map is the function you will use.\n\nIn :\n#adapted from https://github.com/dataiap/dataiap/blob/master/resources/util/map_util.py\n\nstate2poly = defaultdict(list)\n\nfor f in data['features']:\nstate = states_abbrev[f['id']]\ngeo = f['geometry']\nif geo['type'] == 'Polygon':\nfor coords in geo['coordinates']:\nstate2poly[state].append(coords)\nelif geo['type'] == 'MultiPolygon':\nfor polygon in geo['coordinates']:\nstate2poly[state].extend(polygon)\n\ndef draw_state(plot, stateid, **kwargs):\n\"\"\"\ndraw_state(plot, stateid, color=..., **kwargs)\n\nAutomatically draws a filled shape representing the state in\nsubplot.\nThe color keyword argument specifies the fill color. It accepts keyword\narguments that plot() accepts\n\"\"\"\nfor polygon in state2poly[stateid]:\nxs, ys = zip(*polygon)\nplot.fill(xs, ys, **kwargs)\n\ndef make_map(states, label):\n\"\"\"\nDraw a cloropleth map, that maps data onto the United States\n\nInputs\n-------\nstates : Column of a DataFrame\nThe value for each state, to display on a map\nlabel : str\nLabel of the color bar\n\nReturns\n--------\nThe map\n\"\"\"\nfig = plt.figure(figsize=(12, 9))\nax = plt.gca()\n\nif states.max() < 2: # colormap for election probabilities\ncmap = cm.RdBu\nvmin, vmax = 0, 1\nelse: # colormap for electoral votes\ncmap = cm.binary\nvmin, vmax = 0, states.max()\nnorm = mpl.colors.Normalize(vmin=vmin, vmax=vmax)\n\nskip = set(['National', 'District of Columbia', 'Guam', 'Puerto Rico',\n'Virgin Islands', 'American Samoa', 'Northern Mariana Islands'])\nfor state in states_abbrev.values():\nif state in skip:\ncontinue\ncolor = cmap(norm(states.ix[state]))\ndraw_state(ax, state, color = color, ec='k')\n\nax1 = fig.add_axes([0.45, 0.70, 0.4, 0.02])\ncb1=mpl.colorbar.ColorbarBase(ax1, cmap=cmap,\nnorm=norm,\norientation='horizontal')\nax1.set_title(label)\nremove_border(ax, left=False, bottom=False)\nax.set_xticks([])\nax.set_yticks([])\nax.set_xlim(-180, -60)\nax.set_ylim(15, 75)\nreturn ax\n\n\n## Today: the day we make the prediction¶\n\nIn :\n# We are pretending to build our model 1 month before the election\nimport datetime\ntoday = datetime.datetime(2012, 10, 2)\ntoday\n\nOut:\ndatetime.datetime(2012, 10, 2, 0, 0)\n\n## Background: The Electoral College¶\n\nUS Presidential elections revolve around the Electoral College . In this system, each state receives a number of Electoral College votes depending on it's population -- there are 538 votes in total. In most states, all of the electoral college votes are awarded to the presidential candidate who recieves the most votes in that state. A candidate needs 269 votes to be elected President.\n\nThus, to calculate the total number of votes a candidate gets in the election, we add the electoral college votes in the states that he or she wins. (This is not entirely true, with Nebraska and Maine splitting their electoral college votes, but, for the purposes of this homework, we shall assume that the winner of the most votes in Maine and Nebraska gets ALL the electoral college votes there.)\n\nHere is the electoral vote breakdown by state:\n\nAs a matter of convention, we will index all our dataframes by the state name\n\nIn :\nelectoral_votes = pd.read_csv(\"data/electoral_votes.csv\").set_index('State')\n\nOut:\nState\nCalifornia 55\nTexas 38\nNew York 29\nFlorida 29\nIllinois 20\n\nTo illustrate the use of make_map we plot the Electoral College\n\nIn :\nmake_map(electoral_votes.Votes, \"Electoral Vlotes\");",
null,
"## Question 1: Simulating elections¶\n\n#### The PredictWise Baseline¶\n\nWe will start by examining a successful forecast that PredictWise made on October 2, 2012. This will give us a point of comparison for our own forecast models.\n\nPredictWise aggregated polling data and, for each state, estimated the probability that the Obama or Romney would win. Here are those estimated probabilities:\n\nIn :\npredictwise = pd.read_csv('data/predictwise.csv').set_index('States')\n\nOut:\nStates\nAlabama 0.000 1.000 9\nArizona 0.062 0.938 11\nArkansas 0.000 1.000 6\nCalifornia 1.000 0.000 55\n\n1.1 Each row is the probability predicted by Predictwise that Romney or Obama would win a state. The votes column lists the number of electoral college votes in that state. Use make_map to plot a map of the probability that Obama wins each state, according to this prediction.\n\nIn :\n#your code here\nmake_map(predictwise.Obama, \"P(Obama): PredictWise\")\n\nOut:\n<matplotlib.axes.AxesSubplot at 0x108465090>",
null,
"Later on in this homework we will explore some approaches to estimating probabilities like these and quatifying our uncertainty about them. But for the time being, we will focus on how to make a prediction assuming these probabilities are known.\n\nEven when we assume the win probabilities in each state are known, there is still uncertainty left in the election. We will use simulations from a simple probabilistic model to characterize this uncertainty. From these simulations, we will be able to make a prediction about the expected outcome of the election, and make a statement about how sure we are about it.\n\n1.2 We will assume that the outcome in each state is the result of an independent coin flip whose probability of coming up Obama is given by a Dataframe of state-wise win probabilities. Write a function that uses this predictive model to simulate the outcome of the election given a Dataframe of probabilities.\n\nIn :\n\"\"\"\nFunction\n--------\nsimulate_election\n\nInputs\n------\nmodel : DataFrame\nA DataFrame summarizing an election forecast. The dataframe has 51 rows -- one for each state and DC\nIt has the following columns:\nObama : Forecasted probability that Obama wins the state\nThe DataFrame is indexed by state (i.e., model.index is an array of state names)\n\nn_sim : int\nNumber of simulations to run\n\nReturns\n-------\nresults : Numpy array with n_sim elements\nEach element stores the number of electoral college votes Obama wins in each simulation.\n\"\"\"\n\ndef simulate_election(model, n_sim):\n#each column simulates a single outcome from the 50 states + DC\n#Obama wins the simulation if the random number is < the win probability\nsimulations = np.random.uniform(size=(51, n_sim))\n#summing over rows gives the total electoral votes for each simulation\n\n\nThe following cells takes the necessary DataFrame for the Predictwise data, and runs 10000 simulations. We use the results to compute the probability, according to this predictive model, that Obama wins the election (i.e., the probability that he receives 269 or more electoral college votes)\n\nIn :\nresult = simulate_election(predictwise, 10000)\n\nIn :\n#compute the probability of an Obama win, given this simulation\nprint (result >= 269).mean()\n\n0.9956\n\n\n1.3 Now, write a function called plot_simulation to visualize the simulation. This function should:\n\n• Build a histogram from the result of simulate_election\n• Overplot the \"victory threshold\" of 269 votes as a vertical black line (hint: use axvline)\n• Overplot the result (Obama winning 332 votes) as a vertical red line\n• Compute the number of votes at the 5th and 95th quantiles, and display the difference (this is an estimate of the outcome's uncertainty)\n• Display the probability of an Obama victory\nIn :\n\"\"\"\nFunction\n--------\nplot_simulation\n\nInputs\n------\nsimulation: Numpy array with n_sim (see simulate_election) elements\nEach element stores the number of electoral college votes Obama wins in each simulation.\n\nReturns\n-------\nNothing\n\"\"\"\n\ndef plot_simulation(simulation):\nplt.hist(simulation, bins=np.arange(200, 538, 1),\nlabel='simulations', align='left', normed=True)\nplt.axvline(332, 0, .5, color='r', label='Actual Outcome')\nplt.axvline(269, 0, .5, color='k', label='Victory Threshold')\np05 = np.percentile(simulation, 5.)\np95 = np.percentile(simulation, 95.)\niq = int(p95 - p05)\npwin = ((simulation >= 269).mean() * 100)\nplt.legend(frameon=False, loc='upper left')\nplt.ylabel(\"Probability\")\nremove_border()\n\n\nLets plot the result of the Predictwise simulation. Your plot should look something like this:",
null,
"In :\nplot_simulation(result)",
null,
"#### Evaluating and Validating our Forecast¶\n\nThe point of creating a probabilistic predictive model is to simultaneously make a forecast and give an estimate of how certain we are about it.\n\nHowever, in order to trust our prediction or our reported level of uncertainty, the model needs to be correct. We say a model is correct if it honestly accounts for all of the mechanisms of variation in the system we're forecasting.\n\nIn this section, we evaluate our prediction to get a sense of how useful it is, and we validate the predictive model by comparing it to real data.\n\n1.4 Suppose that we believe the model is correct. Under this assumption, we can evaluate our prediction by characterizing its accuracy and precision (see here for an illustration of these ideas). What does the above plot reveal about the accuracy and precision of the PredictWise model?\n\nBackground: To make a prediction, we take information that we have now, and try to identify likely outcomes in the future based on this information. The model we've created expresses our uncertainty as a probability distribution over the likely outcomes of the election that are consistent with the information we have now. We call this distribution over outcomes the predictive distribution. Simulating from this model and plotting a histogram allows us to visualize the predictive distribution. When we say a model is correct, we mean that the predictive distribution matches the true distribution of election outcomes when the information leading up to the election matches the information that we have now.\n\nUsually, people are interested in having a single value as the prediction (\"Obama will get 332 votes\"). To obtain a prediction, we summarize the predictive distribution with a single point, usually by taking its expectation. We can evaluate a prediction by its accuracy and precision.\n\nAnswer: To evaluate the accuracy of our prediction, we can check to see whether the expectation of our predictive distribution seems to match the expectation of the true outcome. In this case, much of the predictive distribution's mass lies on or around the real outcome (that is, the histogram is approximately centered on the actual outcome of Obama=332 votes). So, based on the outcome we observed, the model seems accurate. To make a more rigorous statement about accuracy, we would want to have more replications (that is, more elections) to see whether the expectation of the predictive distribution consistently matches the true outcomes.\n\nTo evaluate the precision of our prediction, we look at the spread of the histogram. Because we are assuming the model is correct, we can interpret the spread of the histogram as a measure of the variability among the election outcomes that are consistent with the information we have. If our current information does not constrain the likely election outcomes very much, then the difference between our prediction and the true outcome can vary widely. The spread of the histogram is 60 votes, which is relatively large. Whether the prediction is precise enough is a question of what you want to do with your prediction. For example, if you want to be able to call the winner of a close election (say the candidates are separated by less than 30 votes), this prediction would not be precise enough to identify a winner with 95% confidence. To handle this, we might wish to incorporate more information into the model to reduce the spread of likely election outcomes.\n\n1.5 Unfortunately, we can never be absolutely sure that a model is correct, just as we can never be absolutely sure that the sun will rise tomorrow. But we can test a model by making predictions assuming that it is true and comparing it to real events -- this constitutes a hypothesis test. After testing a large number of predictions, if we find no evidence that says the model is wrong, we can have some degree of confidence that the model is right (the same reason we're still quite confident about the sun being here tomorrow). We call this process model checking, and use it to validate our model.\n\nDescribe how the graph provides one way of checking whether the prediction model is correct. How many predictions have we checked in this case? How could we increase our confidence in the model's correctness?\n\nAnswer: The graph shows a hypothesis test. The histogram approximates the predictive distribution of election outcomes (in terms of electoral votes) assuming our model for the election is true. By comparing this to the true outcome of the election, we can see whether the observed electoral vote count would be highly atypical if the model were true. In this case, it appears the true outcome is quite typical among the model's predicted outcomes, so we do not reject our model.\n\nIn this case, we have checked only one prediction, since there is only one true outcome that we've compared to the predictive distribution.\n\nTo increase our confidence, we would want to test more outcomes against predictions that were made in the same way. For example, we could apply the same procedure (including PredictWise's computation of the statewise probability estimates) to different elections and see whether each of the hypothesis tests fails to reject the model in those cases as well. We could also break the election down into state-by-state outcomes, and test the prediction for each state against that state's outcome.\n\n### Gallup Party Affiliation Poll¶\n\nNow we will try to estimate our own win probabilities to plug into our predictive model.\n\nWe will start with a simple forecast model. We will try to predict the outcome of the election based the estimated proportion of people in each state who identify with one one political party or the other.\n\nGallup measures the political leaning of each state, based on asking random people which party they identify or affiliate with. Here's the data they collected from January-June of 2012:\n\nIn :\ngallup_2012=pd.read_csv(\"data/g12.csv\").set_index('State')\ngallup_2012[\"Unknown\"] = 100 - gallup_2012.Democrat - gallup_2012.Republican\n\nOut:\nState\nAlabama 36.0 49.6 -13.6 3197 14.4\nAlaska 35.9 44.3 -8.4 402 19.8\nArizona 39.8 47.3 -7.5 4325 12.9\nArkansas 41.5 40.8 0.7 2071 17.7\nCalifornia 48.3 34.6 13.7 16197 17.1\n\nEach row lists a state, the percent of surveyed individuals who identify as Democrat/Republican, the percent whose identification is unknown or who haven't made an affiliation yet, the margin between Democrats and Republicans (Dem_Adv: the percentage identifying as Democrats minus the percentage identifying as Republicans), and the number N of people surveyed.\n\n1.6 This survey can be used to predict the outcome of each State's election. The simplest forecast model assigns 100% probability that the state will vote for the majority party. Implement this simple forecast.\n\nIn :\n\"\"\"\nFunction\n--------\nsimple_gallup_model\n\nA simple forecast that predicts an Obama (Democratic) victory with\n0 or 100% probability, depending on whether a state\nleans Republican or Democrat.\n\nInputs\n------\ngallup : DataFrame\nThe Gallup dataframe above\n\nReturns\n-------\nmodel : DataFrame\nA dataframe with the following column\n* Obama: probability that the state votes for Obama. All values should be 0 or 1\nmodel.index should be set to gallup.index (that is, it should be indexed by state name)\n\nExamples\n---------\n>>> simple_gallup_model(gallup_2012).ix['Florida']\nObama 1\nName: Florida, dtype: float64\n>>> simple_gallup_model(gallup_2012).ix['Arizona']\nObama 0\nName: Arizona, dtype: float64\n\"\"\"\n\ndef simple_gallup_model(gallup):\n\n\nNow, we run the simulation with this model, and plot it.\n\nIn :\nmodel = simple_gallup_model(gallup_2012)\nprediction = simulate_election(model, 10000)\n\nplot_simulation(prediction)\nplt.show()\nmake_map(model.Obama, \"P(Obama): Simple Model\")",
null,
"Out:\n<matplotlib.axes.AxesSubplot at 0x108d3ffd0>",
null,
"1.7 Attempt to validate the predictive model using the above simulation histogram. Does the evidence contradict the predictive model?\n\nAnswer: We do not predict the exactly correct result (red line). According to the predictive model, the true outcome has probability 0. Thus, the evidence contradicts the predictive model, and we should reject it.\n\n#### Adding Polling Uncertainty to the Predictive Model¶\n\nThe model above is brittle -- it includes no accounting for uncertainty, and thus makes predictions with 100% confidence. This is clearly wrong -- there are numerous sources of uncertainty in estimating election outcomes from a poll of affiliations.\n\nThe most obvious source of error in the Gallup data is the finite sample size -- Gallup did not poll everybody in America, and thus the party affilitions are subject to sampling errors. How much uncertainty does this introduce?\n\nOn their webpage discussing these data, Gallup notes that the sampling error for the states is between 3 and 6%, with it being 3% for most states. (The calculation of the sampling error itself is an exercise in statistics. Its fun to think of how you could arrive at the sampling error if it was not given to you. One way to do it would be to assume this was a two-choice situation and use binomial sampling error for the non-unknown answers, and further model the error for those who answered 'Unknown'.)\n\n1.8 Use Gallup's estimate of 3% to build a Gallup model with some uncertainty. Assume that the Dem_Adv column represents the mean of a Gaussian, whose standard deviation is 3%. Build the model in the function uncertain_gallup_model. Return a forecast where the probability of an Obama victory is given by the probability that a sample from the Dem_Adv Gaussian is positive.\n\nHint The probability that a sample from a Gaussian with mean $\\mu$ and standard deviation $\\sigma$ exceeds a threhold $z$ can be found using the the Cumulative Distribution Function of a Gaussian:\n\n$$CDF(z) = \\frac1{2}\\left(1 + {\\rm erf}\\left(\\frac{z - \\mu}{\\sqrt{2 \\sigma^2}}\\right)\\right)$$\nIn :\n\"\"\"\nFunction\n--------\nuncertain_gallup_model\n\nA forecast that predicts an Obama (Democratic) victory if the random variable drawn\nfrom a Gaussian with mean Dem_Adv and standard deviation 3% is >0\n\nInputs\n------\ngallup : DataFrame\nThe Gallup dataframe above\n\nReturns\n-------\nmodel : DataFrame\nA dataframe with the following column\n* Obama: probability that the state votes for Obama.\nmodel.index should be set to gallup.index (that is, it should be indexed by state name)\n\"\"\"\nfrom scipy.special import erf\ndef uncertain_gallup_model(gallup):\nsigma = 3\nprob = .5 * (1 + erf(gallup.Dem_Adv / np.sqrt(2 * sigma**2)))\nreturn pd.DataFrame(dict(Obama=prob), index=gallup.index)\n\n\nWe construct the model by estimating the probabilities:\n\nIn :\nmodel = uncertain_gallup_model(gallup_2012)\n\n\nOnce again, we plot a map of these probabilities, run the simulation, and display the results\n\nIn :\nmake_map(model.Obama, \"P(Obama): Gallup + Uncertainty\")\nplt.show()\nprediction = simulate_election(model, 10000)\nplot_simulation(prediction)",
null,
"",
null,
"1.9 Attempt to validate the above model using the histogram. Does the predictive distribution appear to be consistent with the real data? Comment on the accuracy and precision of the prediction.\n\nAnswer: The predictive distribution is consistent with the real data -- the real outcome seems like a typical outcome according to the model. The accuracy is not very good as the center of the distribution falls fairly far from the observed outcome, but the precision is only marginally worse than in the predictwise case.\n\n#### Biases¶\n\nWhile accounting for uncertainty is one important part of making predictions, we also want to avoid systematic errors. We call systematic over- or under-estimation of an unknown quantity bias. In the case of this forecast, our predictions would be biased if the estimates from this poll systematically over- or under-estimate vote proportions on election day. There are several reasons this might happen:\n\n1. Gallup is wrong. The poll may systematically over- or under-estimate party affiliation. This could happen if the people who answer Gallup phone interviews might not be a representative sample of people who actually vote, Gallup's methodology is flawed, or if people lie during a Gallup poll.\n2. Our assumption about party affiliation is wrong. Party affiliation may systematically over- or under-estimate vote proportions. This could happen if people identify with one party, but strongly prefer the candidate from the other party, or if undecided voters do not end up splitting evenly between Democrats and Republicans on election day.\n3. Our assumption about equilibrium is wrong. This poll was released in August, with more than two months left for the elections. If there is a trend in the way people change their affiliations during this time period (for example, because one candidate is much worse at televised debates), an estimate in August could systematically miss the true value in November.\n\nOne way to account for bias is to calibrate our model by estimating the bias and adjusting for it. Before we do this, let's explore how sensitive our prediction is to bias.\n\n1.10 Implement a biased_gallup forecast, which assumes the vote share for the Democrat on election day will be equal to Dem_Adv shifted by a fixed negative amount. We will call this shift the \"bias\", so a bias of 1% means that the expected vote share on election day is Dem_Adv-1.\n\nHint You can do this by wrapping the uncertain_gallup_model in a function that modifies its inputs.\n\nIn :\n\"\"\"\nFunction\n--------\nbiased_gallup_poll\n\nSubtracts a fixed amount from Dem_Adv, beofore computing the uncertain_gallup_model.\nThis simulates correcting a hypothetical bias towards Democrats\nin the original Gallup data.\n\nInputs\n-------\ngallup : DataFrame\nThe Gallup party affiliation data frame above\nbias : float\nThe amount by which to shift each prediction\n\nExamples\n--------\n>>> model = biased_gallup(gallup, 1.)\n>>> model.ix['Flordia']\n>>> .460172\n\"\"\"\ndef biased_gallup(gallup, bias):\ng2 = gallup.copy()\nreturn uncertain_gallup_model(g2)\n\n\n1.11 Simulate elections assuming a bias of 1% and 5%, and plot histograms for each one.\n\nIn :\n#your code here\nmodel = biased_gallup(gallup_2012, 1)\nprediction = simulate_election(model, 10000)\nplot_simulation(prediction)\nplt.show()\n\nmodel = biased_gallup(gallup_2012, 5)\nprediction = simulate_election(model, 10000)\nplot_simulation(prediction)\nplt.show()",
null,
"",
null,
"Note that even a small bias can have a dramatic effect on the predictions. Pundits made a big fuss about bias during the last election, and for good reason -- it's an important effect, and the models are clearly sensitive to it. Forecastors like Nate Silver would have had an easier time convincing a wide audience about their methodology if bias wasn't an issue.\n\nFurthermore, because of the nature of the electoral college, biases get blown up large. For example, suppose you mis-predict the party Florida elects. We've possibly done this as a nation in the past :-). Thats 29 votes right there. So, the penalty for even one misprediction is high.\n\n#### Estimating the size of the bias from the 2008 election¶\n\nWhile bias can lead to serious inaccuracy in our predictions, it is fairly easy to correct if we are able to estimate the size of the bias and adjust for it. This is one form of calibration.\n\nOne approach to calibrating a model is to use historical data to estimate the bias of a prediction model. We can use our same prediction model on historical data and compare our historical predictions to what actually occurred and see if, on average, the predictions missed the truth by a certain amount. Under some assumptions (discussed in a question below), we can use the estimate of the bias to adjust our current forecast.\n\nIn this case, we can use data from the 2008 election. (The Gallup data from 2008 are from the whole of 2008, including after the election):\n\nIn :\ngallup_08 = pd.read_csv(\"data/g08.csv\").set_index('State')\n\nprediction_08['Dem_Win']=results_08[\"Obama Pct\"] - results_08[\"McCain Pct\"]\n\nOut:\nState\nAlabama -0.8 -21.58\nArizona -0.4 -8.52\nArkansas 12.5 -19.86\nCalifornia 19.4 24.06\n\n1.12 Make a scatter plot using the prediction_08 dataframe of the democratic advantage in the 2008 Gallup poll (X axis) compared to the democratic win percentage -- the difference between Obama and McCain's vote percentage -- in the election (Y Axis). Overplot a linear fit to these data.\n\nHint The np.polyfit function can compute linear fits, as can sklearn.linear_model.LinearModel\n\nIn :\n#your code here\n\nplt.ylabel(\"2008 Election Democrat Win\")\nx = np.linspace(-40, 80, 10)\ny = np.polyval(fit, x)\nplt.plot(x, y)\nprint fit\n\n[ 1.26390486 -11.32855786]",
null,
"Notice that a lot of states in which Gallup reported a Democratic affiliation, the results were strongly in the opposite direction. Why might that be? You can read more about the reasons for this here.\n\nA quick look at the graph will show you a number of states where Gallup showed a Democratic advantage, but where the elections were lost by the democrats. Use Pandas to list these states.\n\nIn :\n#your code here\nprediction_08[(prediction_08.Dem_Win < 0) & (prediction_08.Dem_Adv > 0)]\n\nOut:\nState\nArkansas 12.5 -19.86\nGeorgia 3.6 -5.21\nKentucky 13.5 -16.23\nLouisiana 9.4 -18.63\nMississippi 1.1 -13.18\nMissouri 10.9 -0.14\nMontana 3.9 -2.26\nNorth Dakota 0.6 -8.63\nOklahoma 5.6 -31.30\nSouth Carolina 0.1 -8.97\nSouth Dakota 1.3 -8.41\nTennessee 5.0 -15.07\nTexas 2.4 -11.77\nWest Virginia 18.8 -13.12\n\nWe compute the average difference between the Democrat advantages in the election and Gallup poll\n\nIn :\nprint (prediction_08.Dem_Adv - prediction_08.Dem_Win).mean()\n\n8.06803921569\n\n\nAnswer The bias was roughly 8% in favor of the Democrats in the Gallup Poll, meaning that you would want to adjust predictions based on this poll down by that amount. This was the result of people in a number of Southern and Western states claiming to be affiliated as Democrats, then voting the other way. Or, since Gallup kept polling even after the elections, it could also represent people swept away by the 2008 election euphoria in those states. This is an illustration of why one needs to be carefull with polls.\n\n1.13 Calibrate your forecast of the 2012 election using the estimated bias from 2008. Validate the resulting model against the real 2012 outcome. Did the calibration help or hurt your prediction?\n\nIn :\n#your code here\nmodel = biased_gallup(gallup_2012, 8.06)\nprediction = simulate_election(model, 10000)\nplot_simulation(prediction)",
null,
"Answer: This was a disaster. The 8% calibration completey destroys the accuracy of our prediction in 2012. Our calibration made the assumptions that a) the bias in 2008 was the same as 2012, and b) the bias in each state was the same.\n\nThere are several ways in which these assumptions may have been violated. Gallup may have changed their methodology to account for this bias already, leading to a different bias in 2012 from what there was in 2008. The state-by-state biases may have also been different -- voters in highly conservative states may have responded to polls differently from voters in libreral states, for instance. It might have been better to callibrate the bias on a state-wide or clustered basis.\n\n1.14 Finally, given that we know the actual outcome of the 2012 race, and what you saw from the 2008 race would you trust the results of the an election forecast based on the 2012 Gallup party affiliation poll?\n\nAnswer: No. You should answer this question as though you had not yet seen the results of the 2012 election. The results from 2008 would suggest that the party affiliation poll is a highly biased predictor of the acutal election outcome. Given that calibrating the model to counteract this bias would rely on unrealistic assumptions, it would seem unwise to use the 2012 party affiliation poll to predict the election.\n\n## Question 2: Logistic Considerations¶\n\nIn the previous forecast, we used the strategy of taking some side-information about an election (the partisan affiliation poll) and relating that to the predicted outcome of the election. We tied these two quantities together using a very simplistic assumption, namely that the vote outcome is deterministically related to estimated partisan affiliation.\n\nIn this section, we use a more sophisticated approach to link side information -- usually called features or predictors -- to our prediction. This approach has several advantages, including the fact that we may use multiple features to perform our predictions. Such data may include demographic data, exit poll data, and data from previous elections.\n\nFirst, we'll construct a new feature called PVI, and use it and the Gallup poll to build predictions. Then, we'll use logistic regression to estimate win probabilities, and use these probabilities to build a prediction.\n\n#### The Partisan Voting Index¶\n\nThe Partisan Voting Index (PVI) is defined as the excessive swing towards a party in the previous election in a given state. In other words:\n\n$$PVI_{2008} (state) = Democratic.Percent_{2004} ( state ) - Republican.Percent_{2004} ( state) - \\\\ \\Big ( Democratic.Percent_{2004} (national) - Republican.Percent_{2004} (national) \\Big )$$\n\nTo calculate it, let us first load the national percent results for republicans and democrats in the last 3 elections and convert it to the usual democratic - republican format.\n\nIn :\nnational_results=pd.read_csv(\"data/nat.csv\")\nnational_results.set_index('Year',inplace=True)\n\nOut:\nDem Rep\nYear\n2004 48 51\n2008 53 46\n2012 51 47\n\nLet us also load in data about the 2004 elections from p04.csv which gets the results in the above form for the 2004 election for each state.\n\nIn :\npolls04=pd.read_csv(\"data/p04.csv\")\npolls04.State=polls04.State.replace(states_abbrev)\npolls04.set_index(\"State\", inplace=True);\n\nOut:\nDem Rep\nState\nAlabama 37 63\nArizona 44 55\nArkansas 45 54\nCalifornia 54 45\nIn :\npvi08=polls04.Dem - polls04.Rep - (national_results.xs(2004)['Dem'] - national_results.xs(2004)['Rep'])\n\nOut:\nState\nAlabama -23\nArizona -8\nArkansas -6\nCalifornia 12\ndtype: int64\n\n2.1 Build a new DataFrame called e2008. The dataframe e2008 must have the following columns:\n\n• a column named pvi with the contents of the partisan vote index pvi08\n• a column named Dem_Adv which has the Democratic advantage from the frame prediction_08 of the last question with the mean subtracted out\n• a column named obama_win which has a 1 for each state Obama won in 2008, and 0 otherwise\n• a column named Dem_Win which has the 2008 election Obama percentage minus McCain percentage, also from the frame prediction_08\n• The DataFrame should be indexed and sorted by State\nIn :\n#your code here\ne2008['obama_win']=1*(prediction_08.Dem_Win > 0)\ne2008 = e2008.sort_index()\n\nOut:\nState\nAlabama -13.154902 -21.58 -23 0\nArizona -12.754902 -8.52 -8 0\nArkansas 0.145098 -19.86 -6 0\nCalifornia 7.045098 24.06 12 1\n\nWe construct a similar frame for 2012, obtaining pvi using the 2008 Obama win data which we already have. There is no obama_win column since, well, our job is to predict it!\n\nIn :\npvi12 = e2008.Dem_Win - (national_results.xs(2008)['Dem'] - national_results.xs(2008)['Rep'])\ne2012 = e2012.sort_index()\n\nOut:\nState\nAlabama -14.684314 -28.58\nArizona -8.584314 -15.52\nArkansas -0.384314 -26.86\nCalifornia 12.615686 17.06\n\nWe load in the actual 2012 results so that we can compare our results to the predictions.\n\nIn :\nresults2012 = pd.read_csv(\"data/2012results.csv\")\nresults2012.set_index(\"State\", inplace=True)\nresults2012 = results2012.sort_index()\n\nOut:\nWinner\nState\nAlabama 0\nArizona 0\nArkansas 0\nCalifornia 1\n\n#### Exploratory Data Analysis¶\n\n2.2 Lets do a little exploratory data analysis. Plot a scatter plot of the two PVi's against each other. What are your findings? Is the partisan vote index relatively stable from election to election?\n\nIn :\n#your code here\nplt.plot(e2008.pvi, e2012.pvi, 'o', label='Data')\nfit = np.polyfit(e2008.pvi, e2012.pvi, 1)\nx = np.linspace(-40, 80, 10)\ny = np.polyval(fit, x)\nplt.plot(x, x, '--k', alpha=.3, label='x=y')\nplt.plot(x, y, label='Linear fit')\nplt.xlabel(\"2004 PVI\")\nplt.ylabel(\"2008 PVI\")\nplt.legend(loc='upper left')\n\nOut:\n<matplotlib.legend.Legend at 0x109e19b10>",
null,
"Answer: There is a reasonably well fit line which hugs the x=y line. This tells us that the PVI seems relatively stable from election to election and may thus make a good predictor.\n\n2.3 Lets do a bit more exploratory data analysis. Using a scatter plot, plot Dem_Adv against pvi in both 2008 and 2012. Use colors red and blue depending upon obama_win for the 2008 data points. Plot the 2012 data using gray color. Is there the possibility of making a linear separation (line of separation) between the red and the blue points on the graph?\n\nIn :\n#your code here\n\nplt.ylabel(\"pvi\")\ncolors=[\"red\",\"blue\"]\nax=plt.gca()\nfor label in [0, 1]:\ncolor = colors[label]\nl = '2008 McCain States' if label == 0 else '2008 Obama States'\n\nmarker=\"s\", label='2012 States', alpha=.3)\nplt.legend(frameon=False, scatterpoints=1, loc='upper left')\nremove_border()",
null,
"Answer: Yes, it seems that with 3-4 misclassifications, we could draw a line to divide the data into two parts.\n\n#### The Logistic Regression¶\n\nLogistic regression is a probabilistic model that links observed binary data to a set of features.\n\nSuppose that we have a set of binary (that is, taking the values 0 or 1) observations $Y_1,\\cdots,Y_n$, and for each observation $Y_i$ we have a vector of features $X_i$. The logistic regression model assumes that there is some set of weights, coefficients, or parameters $\\beta$, one for each feature, so that the data were generated by flipping a weighted coin whose probability of giving a 1 is given by the following equation:\n\n$$P(Y_i = 1) = \\mathrm{logistic}(\\sum \\beta_i X_i),$$\n\nwhere\n\n$$\\mathrm{logistic}(x) = \\frac{e^x}{1+e^x}.$$\n\nWhen we fit a logistic regression model, we determine values for each $\\beta$ that allows the model to best fit the training data we have observed (the 2008 election). Once we do this, we can use these coefficients to make predictions about data we have not yet observed (the 2012 election).\n\nSometimes this estimation procedure will overfit the training data yielding predictions that are difficult to generalize to unobserved data. Usually, this occurs when the magnitudes of the components of $\\beta$ become too large. To prevent this, we can use a technique called regularization to make the procedure prefer parameter vectors that have smaller magnitude. We can adjust the strength of this regularization to reduce the error in our predictions.\n\nWe now write some code as technology for doing logistic regression. By the time you start doing this homework, you will have learnt the basics of logistic regression, but not all the mechanisms of cross-validation of data sets. Thus we provide here the code for you to do the logistic regression, and the accompanying cross-validation.\n\nWe first build the features from the 2008 data frame, returning y, the vector of labels, and X the feature-sample matrix where the columns are the features in order from the list featurelist, and each row is a data \"point\".\n\nIn :\nfrom sklearn.linear_model import LogisticRegression\n\ndef prepare_features(frame2008, featureslist):\ny= frame2008.obama_win.values\nX = frame2008[featureslist].values\nif len(X.shape) == 1:\nX = X.reshape(-1, 1)\nreturn y, X\n\n\nWe use the above function to get the label vector and feature-sample matrix for feeding to scikit-learn. We then use the usual scikit-learn incantation fit to fit a logistic regression model with regularization parameter C. The parameter C is a hyperparameter of the model, and is used to penalize too high values of the parameter co-efficients in the loss function that is minimized to perform the logistic regression. We build a new dataframe with the usual Obama column, that holds the probabilities used to make the prediction. Finally we return a tuple of the dataframe and the classifier instance, in that order.\n\nIn :\ndef fit_logistic(frame2008, frame2012, featureslist, reg=0.0001):\ny, X = prepare_features(frame2008, featureslist)\nclf2 = LogisticRegression(C=reg)\nclf2.fit(X, y)\nX_new = frame2012[featureslist]\nobama_probs = clf2.predict_proba(X_new)[:, 1]\n\ndf = pd.DataFrame(index=frame2012.index)\ndf['Obama'] = obama_probs\nreturn df, clf2\n\n\nWe are not done yet. In order to estimate C, we perform a grid search over many C to find the best C that minimizes the loss function. For each point on that grid, we carry out a n_folds-fold cross-validation. What does this mean?\n\nSuppose n_folds=10. Then we will repeat the fit 10 times, each time randomly choosing 50/10 ~ 5 states out as a test set, and using the remaining 45/46 as the training set. We use the average score on the test set to score each particular choice of C, and choose the one with the best performance.\n\nIn :\nfrom sklearn.grid_search import GridSearchCV\n\ndef cv_optimize(frame2008, featureslist, n_folds=10, num_p=100):\ny, X = prepare_features(frame2008, featureslist)\nclf = LogisticRegression()\nparameters = {\"C\": np.logspace(-4, 3, num=num_p)}\ngs = GridSearchCV(clf, param_grid=parameters, cv=n_folds)\ngs.fit(X, y)\nreturn gs.best_params_, gs.best_score_\n\n\nFinally we write the function that we use to make our fits. It takes both the 2008 and 2012 frame as arguments, as well as the featurelist, and the number of cross-validation folds to do. It uses the above defined logistic_score to find the best-fit C, and then uses this value to return the tuple of result dataframe and classifier described above. This is the function you will be using.\n\nIn :\ndef cv_and_fit(frame2008, frame2012, featureslist, n_folds=5):\nbp, bs = cv_optimize(frame2008, featureslist, n_folds=n_folds)\npredict, clf = fit_logistic(frame2008, frame2012, featureslist, reg=bp['C'])\nreturn predict, clf\n\n\n2.4 *Carry out a logistic fit using the cv_and_fit function developed above. As your featurelist use the features we have: Dem_Adv and pvi.\n\nIn :\n#your code here\nres, clf = cv_and_fit(e2008, e2012, ['Dem_Adv', 'pvi'])\n\nOut:\nState\nAlabama 0.004234 9\nArizona 0.069288 11\nArkansas 0.031901 6\nCalifornia 0.994956 55\n\n2.5 As before, plot a histogram and map of the simulation results, and interpret the results in terms of accuracy and precision.\n\nIn :\n#code to make the histogram\n\nprediction = simulate_election(predict2012_logistic, 10000)\nplot_simulation(prediction)",
null,
"In :\n#code to make the map\n\nmake_map(predict2012_logistic.Obama, \"P(Obama): Logistic\")\n\nOut:\n<matplotlib.axes.AxesSubplot at 0x10c274890>",
null,
"Answer: The precision has worsened with respect to the predictwise (and even the gallup) model. The accuracy has improved with respect to the gallup model, but is not as good as in the predictwise model.\n\n#### Classifier Decision boundary¶\n\nOne nice way to visualize a 2-dimensional logistic regression is to plot the probability as a function of each dimension. This shows the decision boundary -- the set of parameter values where the logistic fit yields P=0.5, and shifts between a preference for Obama or McCain/Romney.\n\nThe function below draws such a figure (it is adapted from the scikit-learn website), and overplots the data.\n\nIn :\nfrom matplotlib.colors import ListedColormap\ndef points_plot(e2008, e2012, clf):\n\"\"\"\ne2008: The e2008 data\ne2012: The e2012 data\nclf: classifier\n\"\"\"\nytrain = e2008['obama_win'].values == 1\n\nX=np.concatenate((Xtrain, Xtest))\n\n# evenly sampled points\nx_min, x_max = X[:, 0].min() - .5, X[:, 0].max() + .5\ny_min, y_max = X[:, 1].min() - .5, X[:, 1].max() + .5\nxx, yy = np.meshgrid(np.linspace(x_min, x_max, 50),\nnp.linspace(y_min, y_max, 50))\nplt.xlim(xx.min(), xx.max())\nplt.ylim(yy.min(), yy.max())\n\n#plot background colors\nax = plt.gca()\nZ = clf.predict_proba(np.c_[xx.ravel(), yy.ravel()])[:, 1]\nZ = Z.reshape(xx.shape)\ncs = ax.contourf(xx, yy, Z, cmap='RdBu', alpha=.5)\ncs2 = ax.contour(xx, yy, Z, cmap='RdBu', alpha=.5)\nplt.clabel(cs2, fmt = '%2.1f', colors = 'k', fontsize=14)\n\n# Plot the 2008 points\nax.plot(Xtrain[ytrain == 0, 0], Xtrain[ytrain == 0, 1], 'ro', label='2008 McCain')\nax.plot(Xtrain[ytrain == 1, 0], Xtrain[ytrain == 1, 1], 'bo', label='2008 Obama')\n\n# and the 2012 points\nax.scatter(Xtest[:, 0], Xtest[:, 1], c='k', marker=\"s\", s=50, facecolors=\"k\", alpha=.5, label='2012')\nplt.legend(loc='upper left', scatterpoints=1, numpoints=1)\n\nreturn ax\n\n\n2.6 Plot your results on the classification space boundary plot. How sharp is the classification boundary, and how does this translate into accuracy and precision of the results?\n\nIn :\n#your code here\npoints_plot(e2008, e2012, clf)\nplt.ylabel(\"PVI\")\n\nOut:\n<matplotlib.text.Text at 0x10da5cc90>",
null,
"Answer: The sharpness of the classifier boundary, as defined by the closeness of the contours near a probability of 0.5 gives us a sense of precision. Imagine that the boundary is very tight, tighter than what you can see in the graph. Then most states will be away from the 0.5 line, and the spread in the results will be less, or the precision higher. This is not the only consideration: indeed one must ask, how many states fall smack into the middle, say between 0.3 and 0.7. The more that do, the less precise the results will be, as there will be a greater number of simulations in which they will cross-over to another party.\n\nTo assess accuracy, we would have to see the actual outcome of the states in 2012. Accuracy would be a function of the number of states that end up on the \"wrong\" side of the 0.5 line, and how deep they end up on the wrong side. In terms of characterizing the 2008 outcomes, it seems that the classifier is quit eaccurate, with most misclassifications appearing in grey area of the classification boundary.\n\n## Question 3: Trying to catch Silver: Poll Aggregation¶\n\nIn the previous section, we tried to use heterogeneous side-information to build predictions of the election outcome. In this section, we switch gears to bringing together homogeneous information about the election, by aggregating different polling result together.\n\nThis approach -- used by the professional poll analysists -- involves combining many polls about the election itself. One advantage of this approach is that it addresses the problem of bias in individual polls, a problem we found difficult to deal with in problem 1. If we assume that the polls are all attempting to estimate the same quantity, any individual biases should cancel out when averaging many polls (pollsters also try to correct for known biases). This is often a better assumption than assuming constant bias between election cycles, as we did above.\n\nThe following table aggregates many of the pre-election polls available as of October 2, 2012. We are most interested in the column \"obama_spread\". We will clean the data for you:\n\nIn :\nmultipoll = pd.read_csv('data/cleaned-state_data2012.csv', index_col=0)\n\n#convert state abbreviation to full name\nmultipoll.State.replace(states_abbrev, inplace=True)\n\n#convert dates from strings to date objects, and compute midpoint\nmultipoll.start_date = multipoll.start_date.apply(pd.datetools.parse)\nmultipoll.end_date = multipoll.end_date.apply(pd.datetools.parse)\nmultipoll['poll_date'] = multipoll.start_date + (multipoll.end_date - multipoll.start_date).values / 2\n\n#compute the poll age relative to Oct 2, in days\nmultipoll['age_days'] = (today - multipoll['poll_date']).values / np.timedelta64(1, 'D')\n\n#drop any rows with data from after oct 2\nmultipoll = multipoll[multipoll.age_days > 0]\n\n#drop unneeded columns\nmultipoll = multipoll.drop(['Date', 'start_date', 'end_date', 'Spread'], axis=1)\n\n#drop rows with missing data\nmultipoll.dropna()\n\n\nOut:\nPollster State MoE Obama (D) Romney (R) Sample obama_spread poll_date age_days Votes\n0 Rasmussen Reports Washington 4.5 52 41 500 11 2012-09-26 00:00:00 6.0 12\n1 Gravis Marketing Washington 4.6 56 39 625 17 2012-09-21 12:00:00 10.5 12\n2 Elway Poll Washington 5.0 53 36 405 17 2012-09-10 12:00:00 21.5 12\n3 SurveyUSA Washington 4.4 54 38 524 16 2012-09-08 00:00:00 24.0 12\n4 SurveyUSA Washington 4.4 54 37 524 17 2012-08-01 12:00:00 61.5 12\n\n3.1 Using this data, compute a new data frame that averages the obama_spread for each state. Also compute the standard deviation of the obama_spread in each state, and the number of polls for each state.\n\nDefine a function state_average which returns this dataframe\n\nHint\n\npd.GroupBy could come in handy\n\nIn :\n\"\"\"\nFunction\n--------\nstate_average\n\nInputs\n------\nmultipoll : DataFrame\nThe multipoll data above\n\nReturns\n-------\naverages : DataFrame\nA dataframe, indexed by State, with the following columns:\nN: Number of polls averaged together\npoll_mean: The average value for obama_spread for all polls in this state\npoll_std: The standard deviation of obama_spread\n\nNotes\n-----\nFor states where poll_std isn't finite (because N is too small), estimate the\npoll_std value as .05 * poll_mean\n\"\"\"\n\ndef state_average(multipoll):\ngroups = multipoll.groupby('State')\nn = groups.size()\nstd[std.isnull()] = .05 * mean[std.isnull()]\nreturn pd.DataFrame(dict(N=n, poll_mean=mean, poll_std=std))\n\n\nLets call the function on the multipoll data frame, and join it with the electoral_votes frame.\n\nIn :\navg = state_average(multipoll).join(electoral_votes, how='outer')\n\nOut:\nState\nAlabama NaN NaN NaN 9\nArizona 20 -5.500000 4.559548 11\nArkansas 3 -20.333333 4.041452 6\nCalifornia 20 18.950000 5.548589 55\n\nSome of the reddest and bluest states are not present in this data (people don't bother polling there as much). The default_missing function gives them strong Democratic/Republican advantages\n\nIn :\ndef default_missing(results):\nred_states = [\"Alabama\", \"Alaska\", \"Arkansas\", \"Idaho\", \"Wyoming\"]\nblue_states = [\"Delaware\", \"District of Columbia\", \"Hawaii\"]\nresults.ix[red_states, [\"poll_mean\"]] = -100.0\nresults.ix[red_states, [\"poll_std\"]] = 0.1\nresults.ix[blue_states, [\"poll_mean\"]] = 100.0\nresults.ix[blue_states, [\"poll_std\"]] = 0.1\ndefault_missing(avg)\n\nOut:\nState\nAlabama NaN -100.00 0.100000 9\nArizona 20 -5.50 4.559548 11\nArkansas 3 -100.00 0.100000 6\nCalifornia 20 18.95 5.548589 55\n\n#### Unweighted aggregation¶\n\n3.2 Build an aggregated_poll_model function that takes the avg DataFrame as input, and returns a forecast DataFrame in the format you've been using to simulate elections. Assume that the probability that Obama wins a state is given by the probability that a draw from a Gaussian with $\\mu=$poll_mean and $\\sigma=$poll_std is positive.\n\nIn :\n\"\"\"\nFunction\n--------\naggregated_poll_model\n\nInputs\n------\npolls : DataFrame\nDataFrame indexed by State, with the following columns:\npoll_mean\npoll_std\n\nReturns\n-------\nA DataFrame indexed by State, with the following columns:\nObama: Estimated probability that Obama wins the state\n\"\"\"\ndef aggregated_poll_model(polls):\nsigma = polls.poll_std\nprob = .5 * (1 + erf(polls.poll_mean / np.sqrt(2 * sigma ** 2)))\n\n\n3.3 Run 10,000 simulations with this model, and plot the results. Describe the results in a paragraph -- compare the methodology and the simulation outcome to the Gallup poll. Also plot the usual map of the probabilities\n\nIn :\n#your code here\nmodel = aggregated_poll_model(avg)\nsims = simulate_election(model, 10000)\nplot_simulation(sims)\nplt.xlim(250, 400)\n\nOut:\n(250, 400)",
null,
"Answer: The accuracy of this poll is somewhat greater than just taking the gallup poll. This is probably because\n\n1. We're using as inputs polls that are trying to predict the election, so there is nothing \"lost in translation\", and\n2. We are averaging many polls together, so some of their biases are likely to offset each other.\n\nOne problem is that we treated all polls as equal. Thus a bad poll with a small sample size is given equal footing as a good poll with a large one. Thus we are introducing bias both due to individual poll biases and individual poll sampling errors.\n\nFurthermore, we estimate the standard deviation by simply taking the spread of the percentages in the polls, without considering their individual margins of error. In the limit of very limit sampling error per poll, this might be ok. However in states with some polls with large margins of error (Kansas, for eg), this can lead to an overestimate of the standard deviation, pushing the predicted probability toward 0.5. This, in turn, can hurt precision since it increases the chance that a state will flip sides in a simulation.\n\nIn :\n#your code here\nmake_map(model.Obama, \"P(Obama): Poll Aggregation\")\n\nOut:\n<matplotlib.axes.AxesSubplot at 0x10de3ead0>"
] | [
null,
"http://www.scribewise.com/Portals/202647/images/photo.jpg",
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null
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https://cs.stackexchange.com/questions/9602/cfg-with-regular-expression-terminals-on-rhs/9603 | [
"CFG with regular expression terminals on RHS\n\nSuppose that we expand our idea of context free grammar rules to allow regular expressions of terminals on the right hand side. For example, consider $G_1$:\n\n\\begin{align*} S & \\rightarrow (a \\mid b) S (c \\mid d) \\\\ S & \\rightarrow (a \\mid b) A (c \\mid d) \\\\ A & \\rightarrow (f \\mid g)^* \\end{align*}\n\nThen the language of $G_1$ is the following:\n$$L(G_1) = \\{(a \\mid b)^n (f \\mid g)^* (c \\mid d)^n \\mid n > 0\\}$$\n\nGive a standard CFG that has the same language as $G_1$, is your grammar weakly equivalent to $G_1'$, strongly equivalent to $G_1'$, or both? Why?\n\nSecondly, how can I transform any CFG with regular expressions of terminals on the right hand side to a normal context free grammar?\n\nmigrated from stackoverflow.comFeb 8 '13 at 14:13\n\nThis question came from our site for professional and enthusiast programmers.\n\n• These questions have been answered dozens of times in SO. If you're going to post homework, you should at least post your best attempt at a solution. – Apalala Feb 7 '13 at 14:39\n\nThe general answer is pretty straightforward: if you have a grammar rule of the form $S \\rightarrow {\\alpha}r{\\beta}$, where $r$ is a regular expression over the set of nonterminals, change this production to $S \\rightarrow {\\alpha}S'{\\beta}$, find a right-regular grammar with start symbols $S'$ generating $L(r)$ (there is an algorithm for this); and then your grammar will include all those productions as well. Repeat for every production containing a regular expression on the right-hand side.\n\nA exactly grammar equivalent to G1 is following ( say G2) :\n\nS → X S Y\nS → X A Y\n\nX → a | b\nY → c | d\n\nA → fA | gA | ^\n\nWhere ^ is a null symbol (epsilon)\n\nexactly equivalent means L(G1) = L(G2) that is language of G1 and G2 are same( every string in L(G1) also in L(G2) and vise-versa)."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8009835,"math_prob":0.9928617,"size":694,"snap":"2019-43-2019-47","text_gpt3_token_len":214,"char_repetition_ratio":0.12173913,"word_repetition_ratio":0.09677419,"special_character_ratio":0.32564843,"punctuation_ratio":0.080882356,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9966723,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-15T11:09:19Z\",\"WARC-Record-ID\":\"<urn:uuid:38119f1f-c4d0-4ce5-986b-4fa447df78e2>\",\"Content-Length\":\"139367\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d4238b65-3e74-45e9-b68d-5e106ac29d57>\",\"WARC-Concurrent-To\":\"<urn:uuid:18838024-8e49-4166-a056-206313f2c754>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://cs.stackexchange.com/questions/9602/cfg-with-regular-expression-terminals-on-rhs/9603\",\"WARC-Payload-Digest\":\"sha1:WLMZKSKMKTKMZZ5SOUMCT7Z3VIU2CU6K\",\"WARC-Block-Digest\":\"sha1:GPNMTRPAMTIJPYWFD4WQ3OPD2ABNFW6I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986658566.9_warc_CC-MAIN-20191015104838-20191015132338-00549.warc.gz\"}"} |
http://bends.se/?page=notebook/electronics/stepper-1 | [
"Författare:\nPer Stenebo\n2019-10-27 17:43:25\n2019-11-08 20:58:05",
null,
"# Unipolar stepper motor with driver\n\nDescription on how to connect and code for a 28BYJ-48 and ULN2003 with a Raspberry Pi.\n\n28BYJ-48 is a 5 VDC unipolar stepper motor. ULN2003 is the main component on the driver board. This circuit is fed from a 5 VDC power supply but the signal connections can handle both 3.3 volt (like from a Raspberry Pi) and 5 volt (like from an Arduino).\n\nThere are plenty of tutorials about this motor, some even with this driver board, but none (as fas as I know) with this exact combination of motor, driver and controller. As I was new to this kind of motor I have spent many hours searching and testing, an effort I hope to save someone else with this page.\n\nFor some reason the uploaded images refuse to show up on the page, I hope to resolve this issue soon. I link the images from another site in the mean time.",
null,
"## External resources\n\nI have bought these as a Luxorparts kit from Kjell & Co.\n\nMost info, images and code compiled from the following sources:\n| instructables.com | Raspberry Pi Tutorialsseeedstudio.com42bots.com |\n\n## Wiring\n\nGPIO pin Driver pin Motor wire\nGPIO #4 IN1 Blue\nGPIO #17 IN2 Pink\nGPIO #23 IN3 Yellow\nGPIO #24 IN4 Orange\n\nConnection chart with separate power sources (recommended):",
null,
"Connection chart with common power source:",
null,
"You can of course choose any available GPIO pin but you need to adjust the code accordingly. Remember that driving stepper motors is all about signal order, keep track of your changes.\n\nYou can (and should) have a separate power supply for the controller and the driver/motor. If so, remember to connect a ground wire between the controller and the driver so they have a common reference.\n\nThe LEDs and the resistors (330 Ω) on the breadboard are just for fun (and for troubleshooting). You can skip these.\n\nThe capacitor (100uF 50V) are also optional, it mitigates power surges from the motor, causing voltage drops to the controller. Use separate power supplies instead.\n\n## Troubleshooting\n\nResistance between motor wires, motor disconnected from the driver board:\n\nOrange Red 21,9 Ω\nYellow Red 21,8 Ω\nPink Red 21,7 Ω\nBlue Red 21,7 Ω\n\nBlue Pink 43,4 Ω\nBlue Yellow 43,4 Ω\nBlue Orange 43,5 Ω\n\nPink Yellow 43,3 Ω\nPink Orange 43,5 Ω\n\nOrange Yellow 43,5 Ω\n\nThe motor includes a gearbox with high ratio, about 63:1. When using the default delay of 5 ms between each step, the visible gear still turns quite slow. The high ratio also makes the gear somewhat self-locking. The gears will break if you try to turn the motor from the visible shaft.\n\n## Code for python 3 (preferred)\n\n```#!/usr/bin/python3\n# coding: utf-8\n'''\nsudo apt install python3-dev python3-rpi.gpio\nRun with: python3 stepper.py\n'''\nimport sys, signal, time\nimport RPi.GPIO as GPIO\n\n#-----------------------------------------------------------------------\n# Functions (inline)\n#-----------------------------------------------------------------------\n# Manage ctrl+c\ndef signal_handler(signal, frame):\nGPIO.cleanup()\nprint ('Script terminated by Ctrl+C')\n# Terminate script\nsys.exit(0)\n\nsignal.signal(signal.SIGINT, signal_handler)\n\n#-----------------------------------------------------------------------\n# GPIO setup\n#-----------------------------------------------------------------------\nGPIO.setmode(GPIO.BCM)\nGPIO.setwarnings(False)\n\ncoil_A_1_pin = 4 # blue\ncoil_A_2_pin = 17 # pink\ncoil_B_1_pin = 23 # yellow\ncoil_B_2_pin = 24 # orange\n\nStepCount = 8\nSeq = [0,1,2,3,4,5,6,7]\n\nSeq = [1,0,0,0]\nSeq = [1,1,0,0]\nSeq = [0,1,0,0]\nSeq = [0,1,1,0]\nSeq = [0,0,1,0]\nSeq = [0,0,1,1]\nSeq = [0,0,0,1]\nSeq = [1,0,0,1]\n\nGPIO.setup(coil_A_1_pin, GPIO.OUT)\nGPIO.setup(coil_A_2_pin, GPIO.OUT)\nGPIO.setup(coil_B_1_pin, GPIO.OUT)\nGPIO.setup(coil_B_2_pin, GPIO.OUT)\n\ndef setStep(w1, w2, w3, w4):\nGPIO.output(coil_A_1_pin, w1)\nGPIO.output(coil_A_2_pin, w2)\nGPIO.output(coil_B_1_pin, w3)\nGPIO.output(coil_B_2_pin, w4)\n\ndef forward(delay, steps):\nfor i in range(steps):\nfor j in range(StepCount):\nsetStep(Seq[j], Seq[j], Seq[j], Seq[j])\ntime.sleep(delay)\n#input(\"Fwd seq #%d : %s %s %s %s\" % (j, Seq[j], Seq[j], Seq[j], Seq[j]))\n\ndef backwards(delay, steps):\nfor i in range(steps):\nfor j in reversed(range(StepCount)):\nsetStep(Seq[j], Seq[j], Seq[j], Seq[j])\ntime.sleep(delay)\n#input(\"Rev seq #%d : %s %s %s %s\" % (j, Seq[j], Seq[j], Seq[j], Seq[j]))\n\n#-----------------------------------------------------------------------\n# Main\n#-----------------------------------------------------------------------\nif __name__ == '__main__':\nsetStep(0, 0, 0, 0)\n\ndelay = input(\"Time Delay (ms) ? \")\nif not delay:\ndelay = 5\nelse:\ndelay = int(delay)\n\nsteps_fwd = input(\"How many steps forward? \")\nif not steps_fwd:\nsteps_fwd = 100\nelse:\nsteps_fwd = int(steps_fwd)\n\nsteps_rev = input(\"How many steps backward? \")\nif not steps_rev:\nsteps_rev = 100\nelse:\nsteps_rev = int(steps_rev)\n\nwhile True:\nforward(delay / 1000.0, steps_fwd)\nsetStep(0, 0, 0, 0)\ntime.sleep(0.5)\nbackwards(delay / 1000.0, steps_rev)\nsetStep(0, 0, 0, 0)\ntime.sleep(0.5)\n```\n\n## Code for python 2\n\n```#!/usr/bin/python2.7\n# coding: utf-8\n'''\nsudo apt install python-dev python-rpi.gpio\nRun with: python2 stepper.py\n'''\nimport sys, signal, time\nimport RPi.GPIO as GPIO\n\n#-----------------------------------------------------------------------\n# Functions (inline)\n#-----------------------------------------------------------------------\n# Manage ctrl+c\ndef signal_handler(signal, frame):\nGPIO.cleanup()\nprint ('Script terminated by Ctrl+C')\n# Terminate script\nsys.exit(0)\n\nsignal.signal(signal.SIGINT, signal_handler)\n\n#-----------------------------------------------------------------------\n# GPIO setup\n#-----------------------------------------------------------------------\nGPIO.setmode(GPIO.BCM)\nGPIO.setwarnings(False)\n\ncoil_A_1_pin = 4 # blue\ncoil_A_2_pin = 17 # pink\ncoil_B_1_pin = 23 # yellow\ncoil_B_2_pin = 24 # orange\n\nStepCount = 8\nSeq = range(0, StepCount)\n\nSeq = [1,0,0,0]\nSeq = [1,1,0,0]\nSeq = [0,1,0,0]\nSeq = [0,1,1,0]\nSeq = [0,0,1,0]\nSeq = [0,0,1,1]\nSeq = [0,0,0,1]\nSeq = [1,0,0,1]\n\nGPIO.setup(coil_A_1_pin, GPIO.OUT)\nGPIO.setup(coil_A_2_pin, GPIO.OUT)\nGPIO.setup(coil_B_1_pin, GPIO.OUT)\nGPIO.setup(coil_B_2_pin, GPIO.OUT)\n\ndef setStep(w1, w2, w3, w4):\nGPIO.output(coil_A_1_pin, w1)\nGPIO.output(coil_A_2_pin, w2)\nGPIO.output(coil_B_1_pin, w3)\nGPIO.output(coil_B_2_pin, w4)\n\ndef forward(delay, steps):\nfor i in range(steps):\nfor j in range(StepCount):\nsetStep(Seq[j], Seq[j], Seq[j], Seq[j])\ntime.sleep(delay)\n#raw_input(\"Fwd seq #%d : %s %s %s %s\" % (j, Seq[j], Seq[j], Seq[j], Seq[j]))\n\ndef backwards(delay, steps):\nfor i in range(steps):\nfor j in reversed(range(StepCount)):\nsetStep(Seq[j], Seq[j], Seq[j], Seq[j])\ntime.sleep(delay)\n#raw_input(\"Rev seq #%d : %s %s %s %s\" % (j, Seq[j], Seq[j], Seq[j], Seq[j]))\n\n#-----------------------------------------------------------------------\n# Main\n#-----------------------------------------------------------------------\nif __name__ == '__main__':\nsetStep(0, 0, 0, 0)\n\ndelay = raw_input(\"Time Delay (ms) ? \")\nif not delay:\ndelay = 5\nelse:\ndelay = int(delay)\n\nsteps_fwd = raw_input(\"How many steps forward? \")\nif not steps_fwd:\nsteps_fwd = 100\nelse:\nsteps_fwd = int(steps_fwd)\n\nsteps_rev = raw_input(\"How many steps backward? \")\nif not steps_rev:\nsteps_rev = 100\nelse:\nsteps_rev = int(steps_rev)\n\nwhile True:\nforward(delay / 1000.0, steps_fwd)\nsetStep(0, 0, 0, 0)\ntime.sleep(0.5)\nbackwards(delay / 1000.0, steps_rev)\nsetStep(0, 0, 0, 0)\ntime.sleep(0.5)\n```\n\n## More images\n\nTip: Right-click on the image and select \"Show image in new tab\" to view it in full size.",
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"http://bends.se/bendcms/files/flags/en.png",
null,
"https://vpp2.bends.se/media/electronics/stepper1/motor_and_driver.jpeg",
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"https://vpp2.bends.se/media/electronics/stepper1/pi_step_motor2_bb.png",
null,
"https://vpp2.bends.se/media/electronics/stepper1/pi_step_motor1_bb.png",
null,
"https://vpp2.bends.se/media/electronics/stepper1/motor_gearbox.jpg",
null,
"https://vpp2.bends.se/media/electronics/stepper1/motor_internal_circuit.jpg",
null,
"https://vpp2.bends.se/media/electronics/stepper1/driver_board.jpeg",
null,
"https://vpp2.bends.se/media/electronics/stepper1/motor_back.jpeg",
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"https://vpp2.bends.se/media/electronics/stepper1/motor_front.jpeg",
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http://zuowen4.info/node/5487 | [
"## 我懂得了读书的乐趣\n\n• dǒng\n• le\n• shū\n• de\n•\n•\n• 我懂得了读书的乐趣\n•\n•\n• hóng\n• yàn\n• wài\n• xué\n• xiào\n• nián\n•\n• shào\n• xiāo\n• 鸿雁外语学校四年级 邵枭宇\n•\n•\n• qián\n• de\n• ya\n•\n• zhěng\n• tiān\n• jiù\n• zhī\n• dào\n• wán\n•\n• xiě\n• wán\n• zuò\n• 以前的我呀，整天就知道玩，写完作\n•\n• shèng\n• xià\n• de\n• shì\n• shí\n• jiān\n• quán\n• jiāo\n• gěi\n• le\n• wán\n•\n• 业，剩下的是时间全部交给了玩，可以一\n• zhí\n• wán\n• dào\n• tiān\n• hēi\n• shuì\n• jiào\n•\n• cóng\n• kàn\n• shū\n•\n• zhí\n• dào\n• shēng\n• le\n• 直玩到天黑睡觉，从不看书，直到发生了\n• jiàn\n• shì\n• zhī\n• hòu\n•\n• cái\n• zhī\n• dào\n• kàn\n• shū\n•\n• dǒng\n• le\n• 那件事之后，我才知道看书，也懂得了读\n• shū\n• de\n•\n•\n•\n• 书的乐趣。\n•\n•\n• tiān\n• xià\n•\n• wéi\n• gǒng\n• de\n• xué\n•\n• fēi\n• 一天下午，妈妈为巩固我的学习，非\n• yào\n• kǎo\n• kǎo\n• wén\n• fāng\n• miàn\n• de\n•\n• ér\n• qiě\n• hái\n• shǐ\n• yǒu\n• 要考考我语文方面的题，而且还与历史有\n• guān\n•\n• shì\n• jiù\n• cóng\n• zhè\n• kāi\n• shǐ\n• le\n•\n•\n•\n• 关，故事就从这里开始了。\n•\n•\n• yīn\n• wéi\n• zhī\n• dào\n• duì\n• shū\n• gǎn\n• xìng\n•\n• jiù\n• 因为妈妈知道我对读书不感兴趣，就\n• xiān\n• kǎo\n• yīng\n• gāi\n• huì\n• de\n•\n• rán\n• hòu\n• zài\n• kǎo\n• tài\n• huì\n• de\n•\n• 先考我应该会的，然后再考我不太会的。\n• zhè\n• yàng\n• lái\n•\n• huì\n• de\n• xiē\n•\n• jiù\n• huì\n• zhǎng\n• de\n• gèng\n• zhā\n• 这样一来，我会的那些，就会掌握的更扎\n• shí\n•\n• kāi\n• shǐ\n• shí\n•\n• hái\n• cuò\n•\n• hòu\n• lái\n• de\n• jiù\n• 实。开始时，我答得还不错，后来的题就\n• nán\n• le\n•\n• dàn\n• fēi\n• cháng\n• yǒu\n•\n• yòu\n• huì\n•\n• méi\n• bàn\n• ?\n• 难了，但非常有趣，可我又不会，没办法\n• zhī\n• hǎo\n• rèn\n• shū\n• le\n•\n• xīn\n• què\n• zhí\n•\n• bìng\n• 只好认输了。可我心里却一直不服气，并\n• méi\n• yǒu\n• zhēn\n• zhèng\n• rèn\n• shū\n•\n• shì\n• biàn\n• kāi\n• shǐ\n• kàn\n• shū\n• lái\n•\n• 没有真正认输，于是我便开始看起书来。\n• qiǎo\n• de\n• shì\n• èr\n• tiān\n• kǎo\n• wài\n• kǒu\n•\n• ér\n• qiě\n• yǒu\n• yàn\n•\n• 巧的是第二天考外语口语，而且有谚语。\n• bié\n• de\n• tóng\n• xué\n• néng\n• duì\n•\n• shì\n• xué\n• guò\n• de\n•\n• 别的同学也能答对几个，那是学过的，可\n• jiù\n• yàng\n• le\n•\n• yīn\n• wéi\n• kàn\n• shū\n• le\n•\n• měi\n• lǎo\n• shī\n• 我就不一样了，因为我看书了。每次老师\n• jiào\n• dào\n• shí\n•\n• dōu\n• biǎo\n• xiàn\n• chū\n•\n• yīn\n• hái\n• le\n• 叫到我时，我都表现特出色，因此还得了\n• jiǎng\n•\n•\n•\n• 奖励。\n•\n•\n• ò\n•\n• cháng\n• dào\n• le\n• shū\n• de\n• tián\n• tóu\n•\n• hòu\n• lái\n• měi\n• 哦，我尝到了读书的甜头，后来每次\n• kàn\n• shū\n• shí\n•\n• dōu\n• huì\n• ràng\n• kǎo\n• kǎo\n•\n• guǒ\n• yǒu\n• 看书时，我都会让妈妈考考我，如果有一\n• diǎn\n• méi\n• zhù\n• de\n•\n• jiù\n• zài\n• shū\n• lái\n• kàn\n•\n• zhí\n• 点没记住的，我就再拿起书来继续看，直\n• dào\n• zhù\n• wéi\n• zhǐ\n•\n• suǒ\n• de\n• zhī\n• shí\n• yuè\n• lái\n• yuè\n• duō\n•\n• 到记住为止，所以我的知识越来越多，大\n• jiā\n• dōu\n• jiào\n•\n• xiǎo\n• shì\n•\n• ne\n•\n•\n•\n• 家都叫我“小博士”呢！\n•\n•\n• cóng\n•\n• gèng\n• jiā\n• ài\n• shàng\n• shū\n• le\n•\n• fēng\n• duō\n• cǎi\n• 从此，我更加爱上读书了，丰富多彩\n• de\n• zhī\n• shí\n• ràng\n•\n• cóng\n• zhōng\n• dǒng\n• le\n• 的知识让我如饥似渴，我也从中懂得了读\n• shū\n• de\n• de\n•\n•\n•\n•\n•\n•\n•\n• 书的莫大的乐趣。\n\n无注音版：\n我懂得了读书的乐趣\n\n鸿雁外语学校四年级 邵枭宇\n以前的我呀，整天就知道玩，写完作业，剩下的是时间全部交给了玩，可以一直玩到天黑睡觉，从不看书，直到发生了那件事之后，我才知道看书，也懂得了读书的乐趣。\n\n一天下午，妈妈为巩固我的学习，非要考考我语文方面的题，而且还与历史有关，故事就从这里开始了。\n\n因为妈妈知道我对读书不感兴趣，就先考我应该会的，然后再考我不太会的。这样一来，我会的那些，就会掌握的更扎实。开始时，我答得还不错，后来的题就难了，但非常有趣，可我又不会，没办法只好认输了。可我心里却一直不服气，并没有真正认输，于是我便开始看起书来。巧的是第二天考外语口语，而且有谚语。别的同学也能答对几个，那是学过的，可我就不一样了，因为我看书了。每次老师叫到我时，我都表现特出色，因此还得了奖励。\n\n哦，我尝到了读书的甜头，后来每次看书时，我都会让妈妈考考我，如果有一点没记住的，我就再拿起书来继续看，直到记住为止，所以我的知识越来越多，大家都叫我“小博士”呢！\n\n从此，我更加爱上读书了，丰富多彩的知识让我如饥似渴，我也从中懂得了读书的莫大的乐趣。\n\n### 上网的乐趣\n\n四年级作文246字\n作者：未知\n•\n•\n• shàng\n• wǎng\n• de\n• 上网的乐趣\n•\n•\n• liáo\n• níng\n• shěng\n• lián\n•\n• xīn\n• huá\n• xiǎo\n• xué\n• nián\n• èr\n• bān\n•\n• liú\n• 辽宁省大连 新华小学四年二班 刘\n• lín\n• lín\n• 林琳\n• 阅读全文\n\n### 钓虾的乐趣\n\n四年级作文673字\n作者：未知\n•\n•\n• diào\n• xiā\n• de\n• 钓虾的乐趣\n•\n•\n• guǎng\n• dōng\n• shěng\n• shān\n• shì\n•\n• guǎng\n• dōng\n• guì\n• yuán\n• xué\n• xiào\n• yīng\n• 广东省佛山市 广东碧桂园学校英语\n• gāo\n• xiǎo\n• ?\n•\n•\n• bān\n• ?\n•\n•\n• bān\n•\n•\n• yàn\n• 高小部四（4）班四（4）班 李R\n• 阅读全文\n\n### 我和书的故事\n\n四年级作文428字\n作者：陈悦虹\n•\n• bié\n• ài\n• kàn\n• shū\n•\n• ài\n• kuáng\n•\n• yǒu\n• kōng\n• jiù\n• kàn\n•\n• 我特别爱看书，爱得发狂，有空就看，\n• chī\n• diǎn\n• xīn\n• shí\n• kàn\n•\n• zuò\n• zuò\n• kàn\n•\n• 吃点心时看，做作业也看。\n•\n•\n• hái\n• wéi\n• shū\n• chà\n• diǎn\n• yān\n• ne\n• shuō\n• lái\n• dǎo\n• nán\n• 我还为书差点淹死呢说起来倒也难以\n• 阅读全文\n\n### 我和书的故事\n\n四年级作文698字\n作者：张展豪\n• shū\n• de\n• shì\n• 我和书的故事\n•\n• sān\n• bān\n•\n• zhāng\n• zhǎn\n• háo\n• 四三班 张展豪\n•\n• gāo\n• ěr\n• céng\n• jīng\n• shuō\n• guò\n•\n• shū\n• shì\n• rén\n• lèi\n• jìn\n• de\n• jiē\n• 高尔基曾经说过“书籍是人类进步的阶\n• 阅读全文\n\n### 爱读书的爸爸\n\n四年级作文465字\n作者：肖俊嵩\n•\n•\n• de\n• jīn\n• nián\n•\n• suì\n•\n• de\n• zhí\n• shì\n• dǎo\n• yóu\n•\n• 我的爸爸今年38岁，他的职业是导游，\n• ài\n• hǎo\n• shì\n• shū\n•\n• yùn\n• dòng\n•\n• yīn\n• wéi\n• huān\n• kàn\n• shū\n•\n• liáng\n• 爱好是读书、运动。因为喜欢看书，鼻梁\n• shàng\n• jià\n• le\n• hòu\n• hòu\n• de\n• jìn\n• shì\n• yǎn\n• jìng\n•\n• cháng\n• cháng\n• shì\n• 上架了一幅厚厚的近视眼镜。爸爸常常是\n• 阅读全文\n\n### 读书的乐趣\n\n四年级作文476字\n作者：盖玉婧\n•\n•\n• shū\n• de\n•\n•\n• 读书的乐趣\n•\n•\n•\n• gài\n• jìng\n•\n•\n• 盖玉婧\n•\n•\n•\n• zài\n• de\n• shēng\n• huó\n• zhōng\n• kāi\n• shū\n•\n• shū\n• néng\n• ràng\n• 在我的生活中离不开书，书能让我\n• 阅读全文\n\n### 爱看课外书的\n\n四年级作文424字\n作者：高宇萌\n•\n•\n• ài\n• kàn\n• wài\n• shū\n• de\n•\n•\n• 爱看课外书的我\n•\n•\n•\n• gāo\n• méng\n•\n•\n• 高宇萌\n•\n•\n•\n• huān\n• kàn\n• wài\n• shū\n•\n• yīn\n• wéi\n• wài\n• shū\n• fēng\n• 我喜欢看课外书，因为课外书丰富\n• 阅读全文\n\n### 种花草的乐趣\n\n四年级作文366字\n作者：荆琳烁\n•\n•\n• jīn\n• tiān\n• dào\n• shì\n• de\n• ?g\n• huì\n• shì\n•\n• sōu\n• huāng\n• 今天我和妈妈到市里的花卉市*蛄艘慌\n• zhè\n• shì\n• tán\n•\n•\n• 柘事痰目?墓?\n•\n•\n•\n• kāi\n• xīn\n• guǒ\n• shǒu\n• zhǎng\n• bān\n•\n• wài\n• xíng\n• 开心果如手掌般大，外形酷似一棵\n• 阅读全文\n\n### 我和书的故事\n\n四年级作文509字\n作者：颜依梦\n• shì\n• xiǎo\n• diàn\n• shì\n•\n• ér\n• shū\n• shì\n• de\n•\n• rén\n•\n• 我是个小电视迷，而书则是我的“敌人”\n•\n• yīn\n• wéi\n• rèn\n• wéi\n• shū\n• shì\n• dàn\n• ér\n• wèi\n• de\n•\n• zhī\n• yǒu\n• diàn\n• shì\n• 。因为我认为书是淡而无味的，只有电视\n• shì\n• cǎi\n• bīn\n• fēn\n• de\n•\n• shì\n• cóng\n• shēng\n• le\n• jiàn\n• shì\n• 剧是五彩缤纷的。可是自从发生了那件事\n• 阅读全文\n\n### 我的乐园\n\n四年级作文483字\n作者：赖陈晨\n• de\n• tóng\n• nián\n•\n• yǒu\n• bié\n• shū\n• wēn\n• nuǎn\n• de\n• 你的童年里，你有一个特别舒服温暖的乐\n• yuán\n• ma\n•\n• jiù\n• yǒu\n• zhè\n• yàng\n• dài\n• zhe\n• jiù\n• xiǎng\n• zǒu\n• de\n• 园吗？我就有这样一个待着就不想走的乐\n• yuán\n•\n• men\n• xiǎng\n• zhī\n• dào\n• de\n• yuán\n• zài\n• ér\n• ma\n•\n• jiù\n• 园，你们想知道我的乐园在哪儿吗？它就\n• 阅读全文\n\n### 跳绳的乐趣\n\n四年级作文638字\n作者：陆韵寒\n• guó\n• xiǎo\n• xué\n• nián\n•\n• yùn\n• hán\n•\n•\n• 泽国小学四年级 陆韵寒\n•\n•\n• yuè\n• liàng\n• chū\n• lái\n• le\n•\n• jiā\n• hóng\n• xún\n• zhǎo\n• tiào\n• shéng\n• 月亮出来了，我和狄佳宏去寻找跳绳\n• de\n• kuài\n•\n•\n•\n• 的快乐。\n• 阅读全文\n\n### 一个爱看书的\n\n四年级作文406字\n作者：翟政琦\n•\n• zhàn\n• zài\n• jìng\n• qián\n•\n• miàn\n• yǒu\n• xiǎo\n• nán\n• hái\n•\n• shēn\n• chuān\n• 站在镜子前，里面有个小男孩。他身穿\n• lán\n• yùn\n• dòng\n• zhuāng\n•\n• dēng\n• hēi\n• qiú\n• xié\n•\n• kàn\n• shàng\n• hái\n• shì\n• 蓝色一运动装，足蹬黑球鞋，看上去还是\n• mán\n• yǒu\n• fēng\n• de\n• ma\n•\n• yuán\n• yuán\n• de\n• xiǎo\n• nǎo\n• dài\n•\n• zhī\n• wéi\n• shí\n• 蛮有风度的吗！圆圆的小脑袋，不知为什\n• 阅读全文\n\n### 不读书的后果\n\n四年级作文344字\n作者：李跃华\n•\n•\n• shū\n• de\n• hòu\n• guǒ\n• 不读书的后果\n•\n•\n•\n• shū\n• bǎi\n• biàn\n•\n• xiàn\n•\n• shū\n• bǎi\n• biàn\n•\n• 书读百遍，其意自现。书读百遍，\n• xiàn\n•\n• shū\n• bǎi\n• biàn\n•\n• xiàn\n•\n•\n•\n• zhè\n• 其意自现。书读百遍，其意自现。……这\n• 阅读全文\n\n### 书的哭泣\n\n四年级作文243字\n作者：任畅航\n•\n•\n• shū\n• de\n•\n•\n• 书的哭泣\n•\n•\n•\n•\n•\n•\n•\n• āi\n•\n• jīn\n• tiān\n• yòu\n• bèi\n• le\n• dùn\n•\n• kàn\n• “唉！今天我又被打了一顿，看我\n• 阅读全文\n\n### 我与书的故事\n\n四年级作文711字\n作者：冯子宸\n•\n•\n• gāo\n• ěr\n• yǒu\n• míng\n• yán\n•\n• shū\n• shì\n• rén\n• lèi\n• jìn\n• de\n• jiē\n• 高尔基有句名言：书是人类进步的阶\n• de\n• què\n•\n• shū\n• zài\n• rén\n• shēng\n• de\n• dào\n• shàng\n• zhàn\n• zhe\n• zhòng\n• yào\n• de\n• 梯的确，书在人生的道路上占据着重要的\n• wèi\n•\n• 地位。\n• 阅读全文\n\n### 读书的感觉真\n\n四年级作文339字\n作者：张延宇\n•\n•\n• shū\n• de\n• gǎn\n• jiào\n• zhēn\n• hǎo\n•\n•\n• 读书的感觉真好\n•\n•\n• zhāng\n• yán\n•\n•\n• 张延宇\n•\n•\n• shū\n•\n• shì\n• zhǒng\n• xiǎng\n• shòu\n•\n• shì\n• zhǒng\n• fàng\n• sōng\n•\n• shì\n• 读书，是一种享受，是一种放松，是\n• 阅读全文"
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