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https://www.brainbell.com/tutorials/ms-office/excel/Compare_Two_Excel_Ranges.htm	[ "# Compare Two Excel Ranges\n\nSpotting the differences between two large tables of data can be a very time-consuming task. Fortunately, there are at least two ways in which you can automate what would otherwise be a very tedious manual process.\n\nThe two methods you will use are methods we have used in the past when we received an updated copy of a spreadsheet and we needed to identify which cells in the updated copy differed from the ones in the original copy. Both methods save hours of tedious manual checking and, more importantly, eliminate the possibility of mistakes.\n\nFor the following examples, we copied the newer data onto the same sheet as the older data beforehand. Figure below shows how the data is presented as two ranges. Note that for easier viewing, we boldfaced the cells in Table 2 that are not the same as their counterparts in Table 1.\n\n##### Figure. Two ranges to be compared", null, "#### Method 1: Using True or False\n\nThe first method involves entering a simple formula into another range of the same size and shape. The best part about this method is that you can add the formula in one step without having to copy and paste.\n\nTo compare the ranges shown in the figure, select the range E1:G7, starting from cell E1. This ensures that E1 is the active cell in the selection. With this range selected, click in the Formula bar and type the following:\n\n```=A1=A9\n```\n\nEnter the preceding formula by pressing Ctrl-Enter at the same time. In doing so, you are entering the relative reference formula into each cell of the selection. This is the standard method of entering a formula into an array of cells and having their references change appropriately.\n\nThe range E1:G7 should be filled with True (the same) and False (not the same) values.\n\nIf your two sets of data reside on different worksheets, you can use a third worksheet to store the True/False values simply by array-entering the formula. For example, assuming the second table of data is on Sheet2 and starts in cell A9, and the original table of data is on Sheet1 and starts in cell A1, on a third worksheet you can array-enter this formula:\n\n```=Sheet1!A1=Sheet2!A9\n```\n\nYou might find it useful to adjust your zoom downward when working with large amounts of data.\n\nTo delete an array-entered formula, you must select and delete the whole range. You cannot delete part of it.\n\n#### Method 2: Using Conditional Formatting\n\nThe second method is often preferred, as it is easier to make any needed changes once the comparison is made. However, with this method, both sets of data must reside on the same worksheet, which should entail only a simple copy and paste.\n\nAgain, assuming we're comparing the preceding two ranges, select the range A1:C7, starting from cell A1. This ensures that A1 is the active cell in the selection.\n\nWith this range selected, select Format » Conditional Formatting.... Select Formula Is and then type the following formula:\n\n```=NOT(A1=A9)\n```\n\nClick the Format button, shown in the figure, and choose the format with which you want to highlight the differences.\n\n##### Figure. Conditional formatting dialog", null, "Click OK and all the differences will be formatted according to the format you chose.\n\nWhen or if you make any changes to your data, the cells' format will automatically revert back to normal if the cell content is the same as the cell in the other table." ]	[ null, "https://www.brainbell.com/tutorials/ms-office/excel/images/059600625X/figs/exhk_0601.gif", null, "https://www.brainbell.com/tutorials/ms-office/excel/images/059600625X/figs/exhk_0602.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9117056,"math_prob":0.8439743,"size":3066,"snap":"2019-26-2019-30","text_gpt3_token_len":655,"char_repetition_ratio":0.13520575,"word_repetition_ratio":0.030245747,"special_character_ratio":0.20939335,"punctuation_ratio":0.10491803,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95849985,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-23T11:56:31Z\",\"WARC-Record-ID\":\"<urn:uuid:a624970b-a7df-497c-adb0-f67e616a0d01>\",\"Content-Length\":\"35574\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ffbf757e-df46-47f5-9794-cad05763561d>\",\"WARC-Concurrent-To\":\"<urn:uuid:dfacf53d-92c6-4329-bf34-41ae6ec4803d>\",\"WARC-IP-Address\":\"34.237.101.58\",\"WARC-Target-URI\":\"https://www.brainbell.com/tutorials/ms-office/excel/Compare_Two_Excel_Ranges.htm\",\"WARC-Payload-Digest\":\"sha1:3TGL45BMB6QMO34BBFD2HUV7UO2EMYIC\",\"WARC-Block-Digest\":\"sha1:TOGCQC37BSQE253XYDDM5QPBUDK7VN5K\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195529276.65_warc_CC-MAIN-20190723105707-20190723131707-00166.warc.gz\"}"}
https://it.mathworks.com/help/stats/gscatter.html	[ "# gscatter\n\nScatter plot by group\n\n## Syntax\n\n``gscatter(x,y,g)``\n``gscatter(x,y,g,clr,sym,siz)``\n``gscatter(x,y,g,clr,sym,siz,doleg)``\n``gscatter(x,y,g,clr,sym,siz,doleg,xnam,ynam)``\n``gscatter(ax,___)``\n``h = gscatter(___)``\n\n## Description\n\nexample\n\n````gscatter(x,y,g)` creates a scatter plot of `x` and `y`, grouped by `g`. The inputs `x` and `y` are vectors of the same size.```\n\nexample\n\n````gscatter(x,y,g,clr,sym,siz)` specifies the marker color `clr`, symbol `sym`, and size `siz` for each group.```\n````gscatter(x,y,g,clr,sym,siz,doleg)` controls whether a legend is displayed on the graph. `gscatter` creates a legend by default.```\n\nexample\n\n````gscatter(x,y,g,clr,sym,siz,doleg,xnam,ynam)` specifies the names to use for the x-axis and y-axis labels. If you do not provide `xnam` and `ynam`, and the `x` and `y` inputs are variables with names, then `gscatter` labels the axes with the variable names.```\n\nexample\n\n````gscatter(ax,___)` uses the plot axes specified by the axes object `ax`. Specify `ax` as the first input argument followed by any of the input argument combinations in the previous syntaxes.```\n\nexample\n\n````h = gscatter(___)` returns graphics handles corresponding to the groups in `g`.You can pass in `[]` for `clr`, `sym`, `siz`, and `doleg` to use their default values.```\n\n## Examples\n\ncollapse all\n\nLoad the `carsmall` data set.\n\n`load carsmall`\n\nPlot the `Displacement` values on the x-axis and the `Horsepower` values on the y-axis. `gscatter` uses the variable names as the default labels for the axes. Group the data points by `Model_Year`.\n\n`gscatter(Displacement,Horsepower,Model_Year)`", null, "Load the `discrim` data set.\n\n`load discrim`\n\nThe data set contains ratings of cities according to nine factors such as climate, housing, education, and health. The matrix `ratings` contains the ratings information.\n\nPlot the relationship between the ratings for climate (first column) and housing (second column) grouped by city size in the matrix `group`. Choose different colors and plotting symbols for each group.\n\n```gscatter(ratings(:,1),ratings(:,2),group,'br','xo') xlabel('climate') ylabel('housing')```", null, "Load the `hospital` data set.\n\n`load hospital`\n\nPlot the ages and weights of the hospital patients. Group the patients according to their gender and smoker status. Use the `o` symbol to represent nonsmokers and the `` symbol to represent smokers.\n\n```x = hospital.Age; y = hospital.Weight; g = {hospital.Sex,hospital.Smoker}; gscatter(x,y,g,'rkgb','o',8,'on','Age','Weight') legend('Location','northeastoutside')```", null, "Load the `carsmall` data set. Create a figure with two subplots and return the `axes` objects as `ax1` and `ax2`. Create a scatter plot in each set of axes by referring to the corresponding `Axes` object. In the left subplot, group the data using the `Model_Year` variable. In the right subplot, group the data using the `Cylinders` variable. Add a title to each plot by passing the corresponding `Axes` object to the `title` function.\n\n```load carsmall color = lines(6); % Generate color values ax1 = subplot(1,2,1); % Left subplot gscatter(ax1,Acceleration,MPG,Model_Year,color(1:3,:)) title(ax1,'Left Subplot (Model Year)') ax2 = subplot(1,2,2); % Right subplot gscatter(ax2,Acceleration,MPG,Cylinders,color(4:6,:)) title(ax2,'Right Subplot (Cylinders)')```", null, "Load the `carbig` data set.\n\n`load carbig`\n\nCreate a scatter plot comparing `Acceleration` to `MPG`. Group data points based on `Origin`.\n\n`h = gscatter(Acceleration,MPG,Origin)`\n```h = 7x1 Line array: Line (USA) Line (France) Line (Japan) Line (Germany) Line (Sweden) Line (Italy) Line (England) ```\n\nDisplay the `Line` object corresponding to the group labeled `(Japan)`.\n\n`jgroup = h(3)`\n```jgroup = Line (Japan) with properties: Color: [0.2857 1 0] LineStyle: 'none' LineWidth: 0.5000 Marker: '.' MarkerSize: 15 MarkerFaceColor: 'none' XData: [1x79 double] YData: [1x79 double] ZData: [1x0 double] Show all properties ```\n\nChange the marker color for the `Japan` group to black.\n\n`jgroup.Color = 'k';`", null, "## Input Arguments\n\ncollapse all\n\nx-axis values, specified as a numeric vector. `x` must have the same size as `y`.\n\nData Types: `single` \| `double`\n\ny-axis values, specified as a numeric vector. `y` must have the same size as `x`.\n\nData Types: `single` \| `double`\n\nGrouping variable, specified as a categorical vector, logical vector, numeric vector, character array, string array, or cell array of character vectors. Alternatively, `g` can be a cell array containing several grouping variables (such as ```{g1 g2 g3}```), in which case observations are in the same group if they have common values of all grouping variables. Points in the same group appear on the scatter plot with the same marker color, symbol, and size.\n\nThe number of rows in `g` must be equal to the length of `x`.\n\nExample: `species`\n\nExample: `{Cylinders,Origin}`\n\nData Types: `categorical` \| `logical` \| `single` \| `double` \| `char` \| `string` \| `cell`\n\nMarker colors, specified as either a character vector or string scalar of colors recognized by the `plot` function or a matrix of RGB triplet values. Each RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color, respectively. Each intensity must be in the range [0,1].\n\nThis table lists the available color characters and their equivalent RGB triplet values.\n\nLong NameShort NameRGB Triplet\nYellow`'y'``[1 1 0]`\nMagenta`'m'``[1 0 1]`\nCyan`'c'``[0 1 1]`\nRed`'r'``[1 0 0]`\nGreen`'g'``[0 1 0]`\nBlue`'b'``[0 0 1]`\nWhite`'w'``[1 1 1]`\nBlack`'k'``[0 0 0]`\n\nIf you do not specify enough values for all groups, then `gscatter` cycles through the specified values as needed.\n\nExample: `'rgb'`\n\nExample: `[0 0 1; 0 0 0]`\n\nData Types: `char` \| `string` \| `single` \| `double`\n\nMarker symbols, specified as a character vector or string scalar of symbols recognized by the `plot` function. This table lists the available marker symbols.\n\nValueDescription\n`'o'`Circle\n`'+'`Plus sign\n`''`Asterisk\n`'.'`Point\n`'x'`Cross\n`'s'`Square\n`'d'`Diamond\n`'^'`Upward-pointing triangle\n`'v'`Downward-pointing triangle\n`'>'`Right-pointing triangle\n`'<'`Left-pointing triangle\n`'p'`Five-pointed star (pentagram)\n`'h'`Six-pointed star (hexagram)\n`'none'`No markers\n\nIf you do not specify enough values for all groups, then `gscatter` cycles through the specified values as needed.\n\nExample: `'o+v'`\n\nData Types: `char` \| `string`\n\nMarker sizes, specified as a positive numeric vector in points. The default value is determined by the number of observations. If you do not specify enough values for all groups, then `gscatter` cycles through the specified values as needed.\n\nExample: `[6 12]`\n\nData Types: `single` \| `double`\n\nOption to include a legend, specified as either `'on'` or `'off'`. By default, the legend is displayed on the graph.\n\nx-axis label, specified as a character vector or string scalar.\n\nData Types: `char` \| `string`\n\ny-axis label, specified as a character vector or string scalar.\n\nData Types: `char` \| `string`\n\nAxes for the plot, specified as an `Axes` or `UIAxes` object. If you do not specify `ax`, then `gscatter` creates the plot using the current axes. For more information on creating an axes object, see `axes` and `uiaxes`.\n\n## Output Arguments\n\ncollapse all\n\nGraphics handles, returned as an array of `Line` objects. Each `Line` object corresponds to one of the groups in `g`. You can use dot notation to query and set properties of the line objects. For a list of `Line` object properties, see Line Properties." ]	[ null, "https://it.mathworks.com/help/examples/stats/win64/ScatterPlotWithDefaultSettingsExample_01.png", null, "https://it.mathworks.com/help/examples/stats/win64/ScatterPlotOfClimateAndHousingRatingsExample_01.png", null, "https://it.mathworks.com/help/examples/stats/win64/ScatterPlotWithMultipleGroupingVariablesAndAllOptionsExample_01.png", null, "https://it.mathworks.com/help/examples/stats/win64/SpecifyAxesForScatterPlotExample_01.png", null, "https://it.mathworks.com/help/examples/stats/win64/ModifyScatterPlotAfterCreationExample_01.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6098117,"math_prob":0.92451906,"size":4940,"snap":"2021-31-2021-39","text_gpt3_token_len":1264,"char_repetition_ratio":0.13452189,"word_repetition_ratio":0.13801453,"special_character_ratio":0.24352227,"punctuation_ratio":0.12577319,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9751806,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-21T14:20:22Z\",\"WARC-Record-ID\":\"<urn:uuid:93fc0523-4694-4357-bae4-04362bf022b1>\",\"Content-Length\":\"124325\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cf1e46e8-d797-4ec9-9ef5-60e460d0b6d0>\",\"WARC-Concurrent-To\":\"<urn:uuid:8d540084-87b7-43ca-84cd-3b950a71b5ab>\",\"WARC-IP-Address\":\"23.197.108.134\",\"WARC-Target-URI\":\"https://it.mathworks.com/help/stats/gscatter.html\",\"WARC-Payload-Digest\":\"sha1:XO7MEROMNE42JUGYKCTK3NJT44UEH5HD\",\"WARC-Block-Digest\":\"sha1:2BGFOKNUYHUWOSZHIBRIAUXHBCWG4RBN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057225.38_warc_CC-MAIN-20210921131252-20210921161252-00073.warc.gz\"}"}
https://www.iclub.com/faq/Home/Article?id=345&category=6&parent=0	[ "", null, "## Section 1 of the SSG in Toolkit 6\n\nThis page covers several smaller questions that we have seen about Section 1 (The Front Page) of the Stock Study in Toolkit 6.\n\nQ: How does Toolkit calculate PTP numbers? I don't see where they are input into the Annual Data and the Quarterly Data numbers do not always add up to the Annual number. Is there a calculation involved using other numbers in the Data page?\n\nA: The data includes annual Net Profit and the Tax Rate. Toolkit calculates pre-tax profit from those (i.e., multiply the net profit by the “not-tax” rate).\nThe quarterly and annual figures are separate, so they may not add up precisely, because of rounding or because of normalization that is done on the annual figures.\n\nQ: The % Institutions and % Insiders figures add up to more than 100%, is there something wrong with the data?\n\nA: Not necessarily; by Morningstar's standards, any institution that owns more than 5% of a company's stock is also classified as an insider. This means that those two figures can end up higher than 100%.\n\nQ: How does Toolkit figure out Debt to Capital?\n\nA: Debt to capital in Toolkit is [Last Quarterly Total Debt] / [Total Capital\n\n• Last Quarterly Total Debt is calculated as [Last Quarterly Short Term Debt (Commercial Paper + Current Portion of Long-Term Debt)] + [Last Quarterly Long Term Debt]\n• Total Capital is [Last Quarterly Shareholder’s Equity] + [Last Quarterly Total Debt]\n\nIn cases where a Total Capital figure is not available (such as in preliminary data), Toolkit will calculate it as follows:\n\n[Last Quarterly Total Debt] / [Last Quarterly Total Debt + Last Quarterly Preferred Stock + (Last Annual Book Value per Share * Last Quarterly Diluted Shares))\n\nQ: What is Free Cash Flow?\n\nA: For an explanation of Free Cash Flow, we suggest an article by Carol Theine in the Puget Sound Chapter’s newsletter about cash flow analysis basics." ]	[ null, "https://www.iclub.com/faq/Content/images/iclub_logo.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91114676,"math_prob":0.9232725,"size":1662,"snap":"2023-14-2023-23","text_gpt3_token_len":363,"char_repetition_ratio":0.11761158,"word_repetition_ratio":0.007246377,"special_character_ratio":0.2202166,"punctuation_ratio":0.110410094,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96158004,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-05T10:59:29Z\",\"WARC-Record-ID\":\"<urn:uuid:931ffe78-597f-4660-a94d-52004a074c6a>\",\"Content-Length\":\"11036\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2d766ab9-027d-4f34-9952-96615bef2d2d>\",\"WARC-Concurrent-To\":\"<urn:uuid:79bb5512-d0a5-4995-8006-badccad66b7d>\",\"WARC-IP-Address\":\"174.136.147.4\",\"WARC-Target-URI\":\"https://www.iclub.com/faq/Home/Article?id=345&category=6&parent=0\",\"WARC-Payload-Digest\":\"sha1:2Y6ZVQMUZWT47I643AN72BHKYUHQX7CK\",\"WARC-Block-Digest\":\"sha1:VKAIFCLPBGMZCGHZM5FSFORRMO7RQ37A\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224651815.80_warc_CC-MAIN-20230605085657-20230605115657-00628.warc.gz\"}"}
http://rtcmix.org/reference/scorefile/pickwrand.php	[ "RTcmix an open-source digital signal processing and sound synthesis language about · links · contact\npickwrand - return a weighted random choice from a set of numbers\n\n### Synopsis\n\nvalue = pickwrand(number1, probability1 [, number2, probability2, ... numberN, probabilityN ])\n\nParameters inside the [brackets] are optional.\n\n### Description\n\nCall pickwrand to choose randomly among several numbers that you specify, with a probability for each number.\n\nIt's a good idea to call srand once to seed the random number generator before using pickwrand. Otherwise, a seed of 1 will be used.\n\n### Arguments\n\nnumber\nprobability\nThere can be as many number, probability pairs as you like, as long as there is at least one pair.\n\nA probability argument determines how likely it is that pickwrand will choose the corresponding number argument. The higher the probability, the more likely.\n\nThe total probability is the sum of all the probability arguments.\n\n### RETURN VALUE\n\nOne of the number arguments to pickwrand, selected randomly in accordance with the given probabilities.\n\n### Examples\n\n``` srand(0)\nwhile (outskip < ending_time) {\nstereo_loc = pickwrand(0.0, 10, 0.5, 80, 1.0, 10)\nWAVETABLE(outskip, dur, amp, frequency, stereo_loc)\noutskip = outskip + 0.2\n}\n```\n\nplays WAVETABLE notes, panning them in accordance with the following probabilities: 10% of the notes will pan to hard left, 10% of the notes will pan to hard right, and 80% of the notes will pan to the center." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.71729505,"math_prob":0.9844724,"size":1261,"snap":"2021-31-2021-39","text_gpt3_token_len":332,"char_repetition_ratio":0.14001592,"word_repetition_ratio":0.052083332,"special_character_ratio":0.24742268,"punctuation_ratio":0.18502203,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97726667,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-25T15:23:59Z\",\"WARC-Record-ID\":\"<urn:uuid:01527487-bcaf-4c69-82f1-1769a48e1e91>\",\"Content-Length\":\"3626\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1b81f227-29d7-4470-9bb5-1cfb04b117c0>\",\"WARC-Concurrent-To\":\"<urn:uuid:230c57bd-ec05-4ebd-a232-fd7a2806e03e>\",\"WARC-IP-Address\":\"74.208.236.100\",\"WARC-Target-URI\":\"http://rtcmix.org/reference/scorefile/pickwrand.php\",\"WARC-Payload-Digest\":\"sha1:SR2PNTITAEL5MP2IEOGHTVF4Q4AHFUTQ\",\"WARC-Block-Digest\":\"sha1:OSDY5RYTYKIFJ2CPMQRCA2GHD233WE7U\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057687.51_warc_CC-MAIN-20210925142524-20210925172524-00435.warc.gz\"}"}
https://www.brainkart.com/article/Vector-Control-of-AC-Induction-Machines_12662/	[ "Home \| \| Solid State Drives \| Vector Control of AC Induction Machines\n\n# Vector Control of AC Induction Machines\n\nVector control is the most popular control technique of AC induction motors. In special reference frames, the expression for the electromagnetic torque of the smooth-air-gap machine is similar to the expression for the torque of the separately excited DC machine.\n\nVector Control of AC Induction Machines\n\nVector control is the most popular control technique of AC induction motors. In special reference frames, the expression for the electromagnetic torque of the smooth-air-gap machine is similar to the expression for the torque of the separately excited DC machine. In the case of induction machines, the control is usually performed in the reference frame (d-q) attached to the rotor flux space vector. That’s why the implementation of vector control requires information on the modulus and the space angle (position) of the rotor flux space vector. The stator currents of the induction machine are separated into flux- and torque-producing components by utilizing transformation to the d-q coordinate system, whose direct axis (d) is aligned with the rotor flux space vector. That means that the q-axis component of the rotor flux space vector is always zero:", null, "The rotor flux space vector calculation and transformation to the d-q coordinate system require the high computational power of a microcontroller. The digital signal processor is suitable for this task. The following sections describe the space vector transformations and the rotor flux space vector calculation.\n\n## Block Diagram of the Vector Control\n\nShows the basic structure of the vector control of the AC induction motor. To perform vector control, it is necessary to follow these steps:\n\nMeasure the motor quantities (phase voltages and currents)\n\nTransform them to the 2-phase system (α,β) using a Clarke transformation\n\nCalculate the rotor flux space vector magnitude and position angle\n\nTransform stator currents to the d-q coordinate system using a Park transformation\n\nThe stator current torque and flux producing components are separately controlled\n\nThe output stator voltage space vector is calculated using the decoupling block\n\nThe stator voltage space vector is transformed by an inverse Park transformation back from the d-q coordinate system to the 2-phase system fixed with the stator\n\nUsing the space vector modulation, the output 3-phase voltage is generated", null, "Block Diagram of the AC Induction Motor Vector Control\n\n## Forward and Inverse Clarke Transformation (a,b,c to α,β and backwards)\n\nThe forward Clarke transformation converts a 3-phase system a,b,c to a 2-phase coordinate system α,β. Figure shows graphical construction of the space vector and projection of the space vector to the quadrature-phase components α,β.", null, "The inverse Clarke transformation goes back from a 2-phase (α,β) to a 3-phase isa, isb, isc system. For constant k=2/3, it is given by the following equations:", null, "## Forward and Inverse Park Transformation (α,β to d-q and backwards)\n\nThe components isα and isβ, calculated with a Clarke transformation, are attached to the stator reference frame α, β. In vector control, it is necessary to have all quantities expressed in the same reference frame. The stator reference frame is not suitable for the control process. The space vector isβ is rotating at a rate equal to the angular frequency of the phase currents. The components isα and isβ depend on time and speed. We can transform these components from the stator reference frame to the d-q reference frame rotating at the same speed as the angular frequency of the phase currents. Then the isd and isq components do not depend on time and speed. If we consider the d-axis aligned with the rotor flux, the transformation is illustrated in Figure where θfield is the rotor flux position.", null, "The inverse Park transformation from the d-q to α,β coordinate system is given by the following equations:", null, "## Rotor Flux Model\n\nKnowledge of the rotor flux space vector magnitude and position is key information for the AC induction motor vector control. With the rotor magnetic flux space vector, the rotational coordinate system (d-q) can be established. There are several methods for obtaining the rotor magnetic flux space vector. The implemented flux model utilizes monitored rotor speed and stator voltages and currents. It is calculated in the stationary reference frame (α,β) attached to the stator. The error in the calculated value of the rotor flux, influenced by the changes in temperature, is negligible for this rotor flux model.\n\nThe rotor flux space vector is obtained by solving the differential equations (EQ 4-2) and (EQ 4-3), which are resolved into the α and β components. The equations are derived from the equations of the AC induction motor model", null, "", null, "Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail\nSolid State Drives : Induction Motor Drives : Vector Control of AC Induction Machines \|" ]	[ null, "https://img.brainkart.com/imagebk13/3yfmL6I.jpg", null, "https://img.brainkart.com/imagebk13/oRWXbA6.jpg", null, "https://img.brainkart.com/imagebk13/kcBWeHT.jpg", null, "https://img.brainkart.com/imagebk13/vkcKLdg.jpg", null, "https://img.brainkart.com/imagebk13/dhlwx7M.jpg", null, "https://img.brainkart.com/imagebk13/5xbSMlM.jpg", null, "https://img.brainkart.com/imagebk13/hwc9LbV.jpg", null, "https://img.brainkart.com/imagebk13/yDnnO0S.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8497274,"math_prob":0.9805192,"size":4426,"snap":"2022-40-2023-06","text_gpt3_token_len":882,"char_repetition_ratio":0.1781999,"word_repetition_ratio":0.06303725,"special_character_ratio":0.19023949,"punctuation_ratio":0.07954545,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99165785,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-28T00:15:06Z\",\"WARC-Record-ID\":\"<urn:uuid:16555997-3a88-447f-82ce-a5d107af76cb>\",\"Content-Length\":\"42069\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9420c2bd-7eb6-446d-9582-1568a3a5905c>\",\"WARC-Concurrent-To\":\"<urn:uuid:8c662b75-09b1-4d69-9c7b-6bfc6cf07db5>\",\"WARC-IP-Address\":\"68.178.145.35\",\"WARC-Target-URI\":\"https://www.brainkart.com/article/Vector-Control-of-AC-Induction-Machines_12662/\",\"WARC-Payload-Digest\":\"sha1:BONTKKKCSVH4QXDNOU7ZKBTNEI76IEBT\",\"WARC-Block-Digest\":\"sha1:3J4QD74MDC4TXCRGI6EYERBTIEZ22QXG\",\"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-20230127231443-20230128021443-00820.warc.gz\"}"}
https://blog.schoolconnectonline.com/ncert-solutions-class-6-maths-chapter-1-knowing-our-numbers/	[ "# NCERT Solutions For Class 6 Maths Chapter 1: Knowing Our Numbers", null, "Class 6 Maths Chapter 1 Knowing Our Numbers NCERT Solutions:\n\nIn Chapter 1 Knowing our numbers, we discuss about Comparing Numbers Worksheet, Large Numbers In Practice, Using Brackets and Roman Numerals Chart. You can find the other\n\n## NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers\n\nNCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Exercise 1.1\n\n### Class 6 Maths Chapter 1 Ex 1.1\n\n#### Ex 1.1 Class 6 Maths Question 1.\n\n1. Fill in the blanks:\n\n(a) 1 lakh = ………….. ten thousand.\n\nSol : 1,00,000\n\n(b) 1 million = ………… hundred thousand.\n\nSol : 10,00,000\n\n(c) 1 crore = ………… ten lakh.\n\nSol : 1,00,00,000\n\n(d) 1 crore = ………… million.\n\nSol : 1,00,00,000\n\n(e) 1 million = ………… lakh.\n\nSol : 1,000,000\n\n#### Ex 1.1 Class 6 Maths Question 2.\n\n2. Place commas correctly and write the numerals:\n\n(a) Seventy three lakh seventy five thousand three hundred seven.\n\nSol : The numeral of seventy three lakh seventy five thousand three hundred seven is 73,75,307\n\n(b) Nine crore five lakh forty one.\n\nSol : The numeral of nine crore five lakh forty one is 9,05,00,041\n\n(c) Seven crore fifty two lakh twenty one thousand three hundred two.\n\nSol : The numeral of seven crore fifty two lakh twenty one thousand three hundred two is 7,52,21,302\n\n(d) Fifty eight million four hundred twenty three thousand two hundred two.\n\nSol : The numeral of fifty eight million four hundred twenty three thousand two hundred two is 5,84,23,202\n\n(e) Twenty three lakh thirty thousand ten.\n\nSol : The numeral of twenty three lakh thirty thousand ten is 23,30,010\n\n#### Ex 1.1 Class 6 Maths Question 3.\n\n3. Insert commas suitably and write the names according to Indian System of Numeration:\n\n(a) 87595762\n\nSol : Eight crore seventy five lakh ninety five thousand seven hundred sixty two\n\n(b) 8546283\n\nSol : Eighty five lakh forty six thousand two hundred eighty three\n\n(c) 99900046\n\nSol : Nine crore ninety nine lakh forty six\n\n(d) 98432701\n\nSol : Nine crore eighty four lakh thirty two thousand seven hundred one\n\n#### Ex 1.1 Class 6 Maths Question 4.\n\n4. Insert commas suitably and write the names according to International System of Numeration:\n\n(a) 78921092\n\nSol : Seventy eight million nine hundred twenty one thousand ninety two\n\n(b) 7452283\n\nSol : Seven million four hundred fifty-two thousand two hundred eighty three\n\n(c) 99985102\n\nSol : Ninety-nine million nine hundred eighty five thousand one hundred two\n\n(d) 48049831\n\nSol : Forty-eight million forty-nine thousand eight hundred thirty-one\n\n### Class 6 Maths Chapter 1 Ex 1.2\n\nNCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Exercise 1.2\n\n#### Ex 1.2 Class 6 Maths Question 1.\n\n1. A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.\n\nSol :\n\nNumber of tickets sold on 1st day = 1094\n\nNumber of tickets sold on 2nd day = 1812\n\nNumber of tickets sold on 3rd day = 2050\n\nNumber of tickets sold on 4th day = 2751\n\nHence, number of tickets sold on all the four days = 1094 + 1812 + 2050 + 2751 = 7707 tickets\n\n2. Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?\n\nSol :\n\nShekhar scored = 6980 runs\n\nHe want to complete = 10000 runs\n\nRuns need to score more = 10000 – 6980 = 3020\n\nHence, he need 3020 more runs to score\n\n3. In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?\n\nSol :\n\nNo. of votes secured by the successful candidate = 577500\n\nNo. of votes secured by his rival = 348700\n\nMargin by which he won the election = 577500 – 348700 = 228800 votes\n\n∴ Successful candidate won the election by 228800 votes\n\n4. Kirti bookstore sold books worth Rs 2,85,891 in the first week of June and books worth Rs 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?\n\nSol :\n\nPrice of books sold in June first week = Rs 285891\n\nPrice of books sold in June second week = Rs 400768\n\nNo. of books sold in both weeks together = Rs 285891 + Rs 400768 = Rs 686659\n\nThe sale of books is the highest in the second week\n\nDifference in the sale in both weeks = Rs 400768 – Rs 285891 = Rs 114877\n\n∴ Sale in the second week was greater by Rs 114877 than in the first week.\n\n5. Find the difference between the greatest and the least 5-digit number that can be written using the digits 6, 2, 7, 4, 3 each only once.\n\nSol :\n\nDigits given are 6, 2, 7, 4, 3\n\nGreatest 5-digit number = 76432\n\nLeast 5-digit number = 23467\n\nDifference between the two numbers = 76432 – 23467 = 52965\n\n∴ The difference between the two numbers is 52965\n\n6. A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?\n\nSol :\n\nNumber of screws manufactured in a day = 2825\n\nSince January month has 31 days\n\nHence, number of screws manufactured in January = 31 × 2825 = 87575\n\nHence, machine produce 87575 screws in the month of January 2006\n\n7. A merchant had Rs 78,592 with her. She placed an order for purchasing 40 radio sets at Rs 1200 each. How much money will remain with her after the purchase?\n\nSol :\n\nTotal money the merchant had = Rs 78592\n\nCost of each radio set = Rs 1200\n\nSo, cost of 40 radio sets = Rs 1200 × 40 = Rs 48000\n\nMoney left with the merchant = Rs 78592 – Rs 48000 = Rs 30592\n\nHence, money left with the merchant after purchasing radio sets is Rs 30592\n\n8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?\n\nSol :\n\nDifference between 65 and 56 i.e (65 – 56) = 9\n\nThe difference between the correct and incorrect answer = 7236 × 9 = 65124\n\n9. To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain?\n\nSol :\n\nGiven\n\nTotal length of the cloth = 40 m\n\n= 40 × 100 cm = 4000 cm\n\nCloth required to stitch one shirt = 2 m 15 cm\n\n= 2 × 100 + 15 cm = 215 cm\n\nNumber of shirts that can be stitched out of 4000 cm = 4000 / 215 = 18 shirts\n\nHence 18 shirts can be stitched out of 40 m and 1m 30 cm of cloth is left out\n\n10. Medicine is packed in boxes, each weighing 4 kg 500g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?\n\nSol :\n\nWeight of one box = 4 kg 500 g = 4 × 1000 + 500\n\n= 4500 g\n\nMaximum weight carried by the van = 800 kg = 800 × 1000\n\n= 800000 g\n\nHence, number of boxes that can be loaded in the van = 800000 / 4500 = 177 boxes\n\n11. The distance between the school and a student’s house is 1 km 875 m. Everyday she walks both ways. Find the total distance covered by her in six days.\n\nSol :\n\nDistance covered between school and house = 1 km 875 m = 1000 + 875 = 1875 m\n\nSince, the student walked both ways.\n\nHence, distance travelled by the student in one day = 2 × 1875 = 3750 m\n\nDistance travelled by the student in 6 days = 3750 m × 6 = 22500 m = 22 km 500 m\n\n∴ Total distance covered by the student in six days is 22 km and 500 m\n\n12. A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?\n\nSol :\n\nQuantity of curd in the vessel = 4 l 500 ml = 4 × 1000 + 500 = 4500 ml\n\nCapacity of 1 glass = 25 ml\n\n∴ Number of glasses that can be filled with curd = 4500 / 25 = 180 glasses\n\nHence, 180 glasses can be filled with curd.\n\n### Class 6 Maths Chapter 1 Ex 1.3\n\nNCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3\n\n#### Ex 1.3 Class 6 Maths Question 1\n\n1. Estimate each of the following using general rule:\n\n(a) 730 + 998 (b) 796 – 314 (c) 12904 + 2888 (d) 28292 – 21496\n\nMake ten more such examples of addition, subtraction and estimation of their outcome.\n\nSolutions:\n\n(a) 730 + 998\n\nRound off to hundreds\n\n730 rounds off to 700\n\n998 rounds off to 1000\n\nHence, 730 + 998 = 700 + 1000 = 1700\n\n(b) 796 – 314\n\nRound off to hundreds\n\n796 rounds off to 800\n\n314 rounds off to 300\n\nHence, 796 – 314 = 800 – 300 = 500\n\n(c) 12904 + 2888\n\nRound off to thousands\n\n12904 rounds off to 13000\n\n2888 rounds off to 3000\n\nHence, 12904 + 2888 = 13000 + 3000 = 16000\n\n(d) 28292 – 21496\n\nRound off to thousands\n\n28292 round off to 28000\n\n21496 round off to 21000\n\nHence, 28292 – 21496 = 28000 – 21000 = 7000\n\nTen more such examples are\n\n(i) 330 + 280 = 300 + 300 = 600\n\n(ii) 3937 + 5990 = 4000 + 6000 = 10000\n\n(iii) 6392 – 3772 = 6000 – 4000 = 2000\n\n(iv) 5440 – 2972 = 5000 – 3000 = 2000\n\n(v) 2175 + 1206 = 2000 + 1000 = 3000\n\n(vi) 1110 – 1292 = 1000 – 1000 = 0\n\n(vii) 910 + 575 = 900 + 600 = 1500\n\n(viii) 6400 – 4900 = 6000 – 5000 = 1000\n\n(ix) 3731 + 1300 = 4000 + 1000 = 5000\n\n(x) 6485 – 4319 = 6000 – 4000 = 2000\n\n2. Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens):\n\n(a) 439 + 334 + 4317 (b) 108734 – 47599 (c) 8325 – 491 (d) 489348 – 48365\n\nMake four more such examples.\n\nSolutions:\n\n(a) 439 + 334 + 4317\n\nRounding off to nearest hundreds\n\n439 + 334 + 4317 = 400 + 300 + 4300\n\n= 5000\n\nRounding off to nearest tens\n\n439 + 334 + 4317 = 440 + 330 + 4320\n\n= 5090\n\n(b) 108734 – 47599\n\nRounding off to nearest hundreds\n\n108734 – 47599 = 108700 – 47600\n\n= 61100\n\nRounding off to nearest tens\n\n108734 – 47599 = 108730 – 47600\n\n= 61130\n\n(c) 8325 – 491\n\nRounding off to nearest hundreds\n\n8325 – 491 = 8300 – 500\n\n= 7800\n\nRounding off to nearest tens\n\n8325 – 491 = 8330 – 490\n\n= 7840\n\n(d) 489348 – 48365\n\nRounding off to nearest hundreds\n\n489348 – 48365 = 489300 – 48400\n\n= 440900\n\nRounding off to nearest tens\n\n489348 – 48365 = 489350 – 48370\n\n= 440980\n\nFour more examples are as follows\n\n(i) 4853 + 662\n\nRounding off to nearest hundreds\n\n4853 + 662 = 4800 + 700\n\n= 5500\n\nRounding off to nearest tens\n\n4853 + 662 = 4850 + 660\n\n= 5510\n\n(ii) 775 – 390\n\nRounding off to nearest hundreds\n\n775 – 390 = 800 – 400\n\n= 400\n\nRounding off to nearest tens\n\n775 – 390 = 780 – 400\n\n= 380\n\n(iii) 6375 – 2875\n\nRounding off to nearest hundreds\n\n6375 – 2875 = 6400 – 2900\n\n= 3500\n\nRounding off to nearest tens\n\n6375 – 2875 = 6380 – 2880\n\n= 3500\n\n(iv) 8246 – 6312\n\nRounding off to nearest hundreds\n\n8246 – 6312 = 8200 – 6300\n\n= 1900\n\nRounding off to nearest tens\n\n8246 – 6312 = 8240 – 6310\n\n= 1930\n\n3. Estimate the following products using general rule:\n\n(a) 578 × 161\n\n(b) 5281 × 3491\n\n(c) 1291 × 592\n\n(d) 9250 × 29\n\nMake four more such examples.\n\nSolutions:\n\n(a) 578 × 161\n\nRounding off by general rule\n\n578 and 161 rounded off to 600 and 200 respectively\n\n600\n\n× 200\n\n____________\n\n120000\n\n_____________\n\n(b) 5281 × 3491\n\nRounding off by general rule\n\n5281 and 3491 rounded off to 5000 and 3500 respectively\n\n5000\n\n× 3500\n\n_________\n\n17500000\n\n_________\n\n(c) 1291 × 592\n\nRounding off by general rule\n\n1291 and 592 rounded off to 1300 and 600 respectively\n\n1300\n\n× 600\n\n_____________\n\n780000\n\n______________\n\n(d) 9250 × 29\n\nRounding off by general rule\n\n9250 and 29 rounded off to 9000 and 30 respectively\n\n9000\n\n× 30\n\n_____________\n\n270000\n\n______________\n\n## CBSE Notes for Class 6 Maths Free Download for All Chapters\n\nOur Blog Site – https://bestlearns.in/\n\nLearn with best notes,free videos,practice questions and mock tests with School Connect Online for free demo click here", null, "#### By School Connect Online\n\nSchool Connect Online is an Integrated Learning Program for Academic Institution,and supported and mentored by StartUp Oasis,an inititive of CIIE.CO Please visit school.schoolconnectonline.com" ]	[ null, "https://blog.schoolconnectonline.com/wp-content/uploads/2021/01/afe81-capture-32.png", null, "https://secure.gravatar.com/avatar/b55007e8bcf50b1263dc759709107879", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8361928,"math_prob":0.99393797,"size":13582,"snap":"2023-40-2023-50","text_gpt3_token_len":4316,"char_repetition_ratio":0.16541465,"word_repetition_ratio":0.09438969,"special_character_ratio":0.390885,"punctuation_ratio":0.07992352,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9951868,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-11T09:15:55Z\",\"WARC-Record-ID\":\"<urn:uuid:0e69b777-6fb4-4f65-8ee9-621c2cd0980c>\",\"Content-Length\":\"323035\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:75f58d00-b64f-44a7-aee5-4776a55ad024>\",\"WARC-Concurrent-To\":\"<urn:uuid:84d07240-a7e7-4314-9409-a5ef25b20a43>\",\"WARC-IP-Address\":\"192.0.78.12\",\"WARC-Target-URI\":\"https://blog.schoolconnectonline.com/ncert-solutions-class-6-maths-chapter-1-knowing-our-numbers/\",\"WARC-Payload-Digest\":\"sha1:A7SMWPVNTORGMGP4GB3GN6P73LXS2UKS\",\"WARC-Block-Digest\":\"sha1:GKAARKAWEKXGB5EKTSRYBONBMEDBXVRT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679103810.88_warc_CC-MAIN-20231211080606-20231211110606-00167.warc.gz\"}"}
https://www.ncertsolutionsfor.com/class-9-maths-chapter-15-probability/	[ "# NCERT Solutions for Class 9 Maths Chapter 15 – Probability\n\nHere we provide NCERT Solutions for Class 9 Maths Chapter 15 – Probability for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Class 9 Maths solution Chapter 15 pdf. Now you will get step by step solution to each question.\n\nTotal numbers of students = 135 + 65 = 200\n\n(i) Numbers of students who like statistics = 135\nRequired probability = 135/200 = 27/40\n\n(ii) Numbers of students who does not like statistics = 65\nRequired probability = 65/200 = 13/40\n\n8. Refer to Q.2, Exercise 14.2. What is the empirical probability that an engineer lives:\n(i) less than 7 km from her place of work?\n(ii) more than or equal to 7 km from her place of work?\n(iii) within 1/2 km from her place of work?\n\nThe distance (in km) of 40 engineers from their residence to their place of work were found as follows:\n5 3 10 20 25 11 13 7 12 31 19 10 12 17 18 11 3 2 17 16 2 7 9 7 8 3 5 12 15 18 3 12 14 2 9 6 15 15 7 6 12\n\nTotal numbers of engineers = 40\n(i) Numbers of engineers living less than 7 km from her place of work = 9\nRequired probability = 9/40\n\n(ii) Numbers of engineers living less than 7 km from her place of work = 40 – 9 = 31\nRequired probability = 31/40\n\n(iii) Numbers of engineers living less than 7 km from her place of work = 0\nRequired probability = 0/40 = 0\n\nPage No: 285\n\n11. Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg):\n4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00\nFind the probability that any of these bags chosen at random contains more than 5 kg of flour.\n\nTotal numbers of bags = 11\nNumbers of bags containing more than 5 kg of flour = 7\nRequired probability = 7/11\n\n12. In Q.5, Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12-0.16 on any of these days.\nThe data obtained for 30 days is as follows:\n0.03 0.08 0.08 0.09 0.04 0.17 0.16 0.05 0.02 0.06 0.18 0.20 0.11 0.08 0.12 0.13 0.22 0.07 0.08 0.01 0.10 0.06 0.09 0.18 0.11 0.07 0.05 0.07 0.01 0.04\n\nTotal numbers of days data recorded = 30 days\nNumbers of days in which sulphur dioxide in the interval 0.12-0.16 = 2\nRequired probability = 2/30 = 1/15\n\n13. In Q.1, Exercise 14.2, you were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.\nThe blood groups of 30 students of Class VIII are recorded as follows:\nA, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.\n\nTotal numbers of students = 30\nNumbers of students having blood group AB = 3\nRequired probability = 3/30 = 1/10\n\nAll Chapter NCERT Solutions For Class 9 Maths\n\n—————————————————————————–\n\nAll Subject NCERT Solutions For Class 9\n\n*************************************************\n\nI think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9032516,"math_prob":0.97467905,"size":3355,"snap":"2023-14-2023-23","text_gpt3_token_len":1034,"char_repetition_ratio":0.14413609,"word_repetition_ratio":0.122257054,"special_character_ratio":0.36542475,"punctuation_ratio":0.14987406,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.99376655,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-28T05:46:22Z\",\"WARC-Record-ID\":\"<urn:uuid:ddf363f6-b509-4b2b-a50c-0fe7462bebee>\",\"Content-Length\":\"245078\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:79c2c3b8-904e-483d-931c-f20c382724f9>\",\"WARC-Concurrent-To\":\"<urn:uuid:1ff8f1a0-c986-4ef6-94d6-2f870a06814a>\",\"WARC-IP-Address\":\"104.21.84.244\",\"WARC-Target-URI\":\"https://www.ncertsolutionsfor.com/class-9-maths-chapter-15-probability/\",\"WARC-Payload-Digest\":\"sha1:MQVKON2YWAWVP3TZS54L3V67OYOU666P\",\"WARC-Block-Digest\":\"sha1:FW6W6HAGZV3T36RB2CHU7QWGDAO66G2N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296948765.13_warc_CC-MAIN-20230328042424-20230328072424-00438.warc.gz\"}"}
https://mathbits.org/MathBits/TISection/Trig/degrees.html	[ "", null, "Degrees, Radians, and Trig Angle Entries\n\nRemember: The calculator defaults to Radian measure.\n\n Working with Degrees:\n\nThere are two ways to engage \"Degrees\" when working with the calculator:\n1. Set the MODE to Degree and all further calculations will be in degrees.\n2. If in Radian MODE, use the Degree Symbol (2nd APPS (Angle) and select #1 ° ).\nNote: The degree status will be applied to that one calculation only. All other calculations\n\nFind the radian measure of an angle whose measure is 500°.\n\n Method 1: (in Radian or Degree mode) Convert by multiplying by", null, ". Mode set to Radian", null, "Method 2: (in Radian mode) With the mode set to Radian, type 500°. Notice the degree symbol. Hit Enter.\n\nNOTE: If in Degree mode, you can force an answer to radians by using\n2nd APPS(Angle) #3 r.\n\n In Degree mode, but ......", null, "gives the answer 0.0548036651 (WRONG!!!)", null, "gives the correct answer of 0.\n\nFind the degree measure of an angle whose measure is", null, "Method 1: (in Radian or Degree mode) Convert by multiplying by", null, ". Mode set to Degree", null, "Method 2: (in Degree mode) In Degree mode, the r symbol can be used to convert radians to degrees. The r is found 2nd APPS(Angle) #3 r. With the mode set to Degree, type", null, "Convert 57° 45' 17'' to decimal degrees:\n\nIn either Radian or Degree Mode: Type 57° 45' 17'' and hit Enter.\n° is under Angle (above APPS) #1\n' is under Angle (above APPS) #2\n'' use ALPHA (green) key with the quote symbol above the + sign.\n\n Convert 57° 45' 17'' to radians:\n\nWith the mode set to Radian: Type 57° 45' 17'' ° and hit Enter. Answer: 1.008010061\n** Note the use of the additional degree symbol at the end. Without it, conversion would be to decimal degrees, even though the mode is set to radians.\n\n Convert 48.555° to degrees, minutes, seconds:\n\nType 48.555 ►DMS Answer: 48° 33' 18''\nThe ►DMS is #4 on the Angle menu (2nd APPS). This function works even if Mode is set to Radian.\n\n Given cos A = .0258. Find ∠A expressed in degree, minutes, seconds.\n\nWith the mode set to Degree: Type cos-1(.0258). Hit Enter.\nEngage ►DMS\n(Be careful here to be in the correct mode!!)\n\n Find sin 57° 45' 17'':\n\nEnter sin(57° 45' 17'' °). This problem is clearly dealing with degrees. The entry shown here works even if MODE is set to Radian. Answer: .8457717984\nIf MODE is set to Degree, the degree symbol at the end is not needed.\n\n Find cos(-250° 21'):\n\nIf MODE is set to Degree, the degree symbol at the end is not needed.\n\n Find sec(25° 40'):\n\nThere is no key for secant. But, since", null, ", enter: 1/cos(25° 40' °) when in Radian mode. If MODE is set to Degree, the degree symbol at the end is not needed.\n\n Find", null, ":\n\nEnter directly:", null, "when in Radian mode. Answer: -.7660444431\nIf in Degree mode, enter", null, ".\n\n Find", null, ":\n\nThere is no key for cosecant. But, since", null, ", enter:", null, "when in Radian mode. Answer: 1.015426612\n\nIn Degree mode, enter", null, ".", null, "" ]	[ null, "https://mathbits.org/MathBits/TISection/Trig/logoTrig.gif", null, "https://mathbits.org/MathBits/TISection/Trig/degree1.gif", null, "https://mathbits.org/MathBits/TISection/Trig/degree3.gif", null, "https://mathbits.org/MathBits/TISection/Trig/degree4.gif", null, "https://mathbits.org/MathBits/TISection/Trig/degree2.gif", null, "https://mathbits.org/MathBits/TISection/Trig/degree6.gif", null, "https://mathbits.org/MathBits/TISection/Trig/degree7.gif", null, "https://mathbits.org/MathBits/TISection/Trig/degree8.gif", null, "https://mathbits.org/MathBits/TISection/Trig/degree10.gif", null, "https://mathbits.org/MathBits/TISection/Trig/degree24.gif", null, "https://mathbits.org/MathBits/TISection/Trig/degree25.gif", null, "https://mathbits.org/MathBits/TISection/Trig/degree26.gif", null, "https://mathbits.org/MathBits/TISection/Trig/degree27.gif", null, "https://mathbits.org/MathBits/TISection/Trig/degree28.gif", null, "https://mathbits.org/MathBits/TISection/Trig/degree29.gif", null, "https://mathbits.org/MathBits/TISection/Trig/degree30.gif", null, "https://mathbits.org/MathBits/TISection/Trig/degree31.gif", null, "https://mathbits.org/MathBits/TISection/Trig/tidivider.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7500282,"math_prob":0.9762952,"size":2945,"snap":"2022-05-2022-21","text_gpt3_token_len":889,"char_repetition_ratio":0.16286977,"word_repetition_ratio":0.14417745,"special_character_ratio":0.3263158,"punctuation_ratio":0.18489985,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9785002,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36],"im_url_duplicate_count":[null,null,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,10,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-21T16:33:41Z\",\"WARC-Record-ID\":\"<urn:uuid:b402a737-9401-496f-b64a-cdcb8f00357b>\",\"Content-Length\":\"16609\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:122155e3-0cad-431f-ad00-f34e534a0308>\",\"WARC-Concurrent-To\":\"<urn:uuid:79f94841-6171-4c4d-9d5f-ffeef61aaebe>\",\"WARC-IP-Address\":\"67.43.11.130\",\"WARC-Target-URI\":\"https://mathbits.org/MathBits/TISection/Trig/degrees.html\",\"WARC-Payload-Digest\":\"sha1:5QU76V4LX4EC6DBK3LSWS2UMRF4ZAHPY\",\"WARC-Block-Digest\":\"sha1:Y7KHMDZJNTCIM5ECRYXWV2TND2IEOK6D\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320303512.46_warc_CC-MAIN-20220121162107-20220121192107-00608.warc.gz\"}"}
https://www.colorhexa.com/1ca58b	[ "#1ca58b Color Information\n\nIn a RGB color space, hex #1ca58b is composed of 11% red, 64.7% green and 54.5% blue. Whereas in a CMYK color space, it is composed of 83% cyan, 0% magenta, 15.8% yellow and 35.3% black. It has a hue angle of 168.6 degrees, a saturation of 71% and a lightness of 37.8%. #1ca58b color hex could be obtained by blending #38ffff with #004b17. Closest websafe color is: #339999.\n\n• R 11\n• G 65\n• B 55\nRGB color chart\n• C 83\n• M 0\n• Y 16\n• K 35\nCMYK color chart\n\n#1ca58b color description : Dark cyan.\n\n#1ca58b Color Conversion\n\nThe hexadecimal color #1ca58b has RGB values of R:28, G:165, B:139 and CMYK values of C:0.83, M:0, Y:0.16, K:0.35. Its decimal value is 1877387.\n\nHex triplet RGB Decimal 1ca58b `#1ca58b` 28, 165, 139 `rgb(28,165,139)` 11, 64.7, 54.5 `rgb(11%,64.7%,54.5%)` 83, 0, 16, 35 168.6°, 71, 37.8 `hsl(168.6,71%,37.8%)` 168.6°, 83, 64.7 339999 `#339999`\nCIE-LAB 60.799, -40.78, 3.664 18.593, 29.019, 29.046 0.243, 0.379, 29.019 60.799, 40.945, 174.866 60.799, -47.721, 11.387 53.87, -32.664, 5.74 00011100, 10100101, 10001011\n\nColor Schemes with #1ca58b\n\n• #1ca58b\n``#1ca58b` `rgb(28,165,139)``\n• #a51c36\n``#a51c36` `rgb(165,28,54)``\nComplementary Color\n• #1ca547\n``#1ca547` `rgb(28,165,71)``\n• #1ca58b\n``#1ca58b` `rgb(28,165,139)``\n• #1c7ba5\n``#1c7ba5` `rgb(28,123,165)``\nAnalogous Color\n• #a5471c\n``#a5471c` `rgb(165,71,28)``\n• #1ca58b\n``#1ca58b` `rgb(28,165,139)``\n• #a51c7b\n``#a51c7b` `rgb(165,28,123)``\nSplit Complementary Color\n• #a58b1c\n``#a58b1c` `rgb(165,139,28)``\n• #1ca58b\n``#1ca58b` `rgb(28,165,139)``\n• #8b1ca5\n``#8b1ca5` `rgb(139,28,165)``\n• #36a51c\n``#36a51c` `rgb(54,165,28)``\n• #1ca58b\n``#1ca58b` `rgb(28,165,139)``\n• #8b1ca5\n``#8b1ca5` `rgb(139,28,165)``\n• #a51c36\n``#a51c36` `rgb(165,28,54)``\n• #116454\n``#116454` `rgb(17,100,84)``\n• #157966\n``#157966` `rgb(21,121,102)``\n• #188f79\n``#188f79` `rgb(24,143,121)``\n• #1ca58b\n``#1ca58b` `rgb(28,165,139)``\n• #20bb9d\n``#20bb9d` `rgb(32,187,157)``\n• #23d1b0\n``#23d1b0` `rgb(35,209,176)``\n• #31dcbc\n``#31dcbc` `rgb(49,220,188)``\nMonochromatic Color\n\nAlternatives to #1ca58b\n\nBelow, you can see some colors close to #1ca58b. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #1ca569\n``#1ca569` `rgb(28,165,105)``\n• #1ca574\n``#1ca574` `rgb(28,165,116)``\n• #1ca580\n``#1ca580` `rgb(28,165,128)``\n• #1ca58b\n``#1ca58b` `rgb(28,165,139)``\n• #1ca596\n``#1ca596` `rgb(28,165,150)``\n• #1ca5a2\n``#1ca5a2` `rgb(28,165,162)``\n• #1c9da5\n``#1c9da5` `rgb(28,157,165)``\nSimilar Colors\n\n#1ca58b Preview\n\nThis text has a font color of #1ca58b.\n\n``<span style=\"color:#1ca58b;\">Text here</span>``\n#1ca58b background color\n\nThis paragraph has a background color of #1ca58b.\n\n``<p style=\"background-color:#1ca58b;\">Content here</p>``\n#1ca58b border color\n\nThis element has a border color of #1ca58b.\n\n``<div style=\"border:1px solid #1ca58b;\">Content here</div>``\nCSS codes\n``.text {color:#1ca58b;}``\n``.background {background-color:#1ca58b;}``\n``.border {border:1px solid #1ca58b;}``\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, #020e0c is the darkest color, while #fcfffe is the lightest one.\n\n• #020e0c\n``#020e0c` `rgb(2,14,12)``\n• #051f1a\n``#051f1a` `rgb(5,31,26)``\n• #083028\n``#083028` `rgb(8,48,40)``\n• #0b4036\n``#0b4036` `rgb(11,64,54)``\n• #0e5144\n``#0e5144` `rgb(14,81,68)``\n• #116252\n``#116252` `rgb(17,98,82)``\n• #137361\n``#137361` `rgb(19,115,97)``\n• #16836f\n``#16836f` `rgb(22,131,111)``\n• #19947d\n``#19947d` `rgb(25,148,125)``\n• #1ca58b\n``#1ca58b` `rgb(28,165,139)``\n• #1fb699\n``#1fb699` `rgb(31,182,153)``\n• #22c7a7\n``#22c7a7` `rgb(34,199,167)``\n• #25d7b5\n``#25d7b5` `rgb(37,215,181)``\n• #33dcbc\n``#33dcbc` `rgb(51,220,188)``\n• #44dfc2\n``#44dfc2` `rgb(68,223,194)``\n• #55e2c7\n``#55e2c7` `rgb(85,226,199)``\n• #65e5cd\n``#65e5cd` `rgb(101,229,205)``\n• #76e8d2\n``#76e8d2` `rgb(118,232,210)``\n• #87ebd8\n``#87ebd8` `rgb(135,235,216)``\n• #98eddd\n``#98eddd` `rgb(152,237,221)``\n• #a8f0e3\n``#a8f0e3` `rgb(168,240,227)``\n• #b9f3e8\n``#b9f3e8` `rgb(185,243,232)``\n• #caf6ee\n``#caf6ee` `rgb(202,246,238)``\n• #dbf9f3\n``#dbf9f3` `rgb(219,249,243)``\n• #ecfcf9\n``#ecfcf9` `rgb(236,252,249)``\n• #fcfffe\n``#fcfffe` `rgb(252,255,254)``\nTint Color Variation\n\nTones of #1ca58b\n\nA tone is produced by adding gray to any pure hue. In this case, #5f6262 is the less saturated color, while #06bb99 is the most saturated one.\n\n• #5f6262\n``#5f6262` `rgb(95,98,98)``\n• #576a66\n``#576a66` `rgb(87,106,102)``\n• #50716b\n``#50716b` `rgb(80,113,107)``\n• #49786f\n``#49786f` `rgb(73,120,111)``\n• #418074\n``#418074` `rgb(65,128,116)``\n• #3a8779\n``#3a8779` `rgb(58,135,121)``\n• #328f7d\n``#328f7d` `rgb(50,143,125)``\n• #2b9682\n``#2b9682` `rgb(43,150,130)``\n• #239e86\n``#239e86` `rgb(35,158,134)``\n• #1ca58b\n``#1ca58b` `rgb(28,165,139)``\n• #15ac90\n``#15ac90` `rgb(21,172,144)``\n• #0db494\n``#0db494` `rgb(13,180,148)``\n• #06bb99\n``#06bb99` `rgb(6,187,153)``\nTone Color Variation\n\nColor Blindness Simulator\n\nBelow, you can see how #1ca58b 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" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.57696575,"math_prob":0.74660695,"size":3691,"snap":"2019-26-2019-30","text_gpt3_token_len":1652,"char_repetition_ratio":0.121779226,"word_repetition_ratio":0.011111111,"special_character_ratio":0.55296665,"punctuation_ratio":0.23337092,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9885351,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-17T03:22:36Z\",\"WARC-Record-ID\":\"<urn:uuid:2af0ffc4-2806-4029-a28b-1a43f7842cd6>\",\"Content-Length\":\"36426\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4dfb7c54-bca6-4b27-b143-1877e94774a9>\",\"WARC-Concurrent-To\":\"<urn:uuid:ed1ffd47-1770-486b-975e-38a049e81227>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/1ca58b\",\"WARC-Payload-Digest\":\"sha1:A2K65VPOENNHTGVVO4QT5CDJGZZ6OI7Z\",\"WARC-Block-Digest\":\"sha1:74YY2MGWDR63I5ZHUS5AY4GNRDAAWWFU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998369.29_warc_CC-MAIN-20190617022938-20190617044938-00266.warc.gz\"}"}
https://www.quantum-inspire.com/kbase/full-adder/	[ "# Code example: Quantum full adder\n\n## Backends\n\nThis example is written for the emulator backend. Spin-2 has only two qubits and does not support this example. The starmon-5 backend can be used to execute this example, however, since it does not support the Toffoli gate, the code needs to be rewritten into two-qubit and single-qubit operations.\n\n## What is the quantum full adder\n\nJust like in classical electronics, where you can make different types of binary adders like half adders, full adders, ripple carry adders etcetera, you can make adders in quantum circuits as well. In this example we will show how a quantum full adder is created and how this adder acts on superposition states.\n\nThe full adder adds to input bits A and B plus a carry_input bit and produces the sum and carry_output bits as output. In classical control electronics the full adder has therefore three inputs and four outputs.\n\nSince quantum circuits are reversible, they have an equal amount of input and output qubits, therefore we define a 4-qubit function, where the input qubits are A,B, CarryIn and (zero) and the output qubits are A,B, Sum and CarryOut, see the figure below.\n\nOne possible implementation of a 2-bit full adder, using CNOT gates and Toffoli gates is the following:\n\nFor completeness we show the truth table for the full adder:\n\nInputs: q0 = A ; q1 = B ; q2= CarryIn\n\nOutputs: q0 = A ; q1 = B ; q2= SumOut ; q3 = CarryOut\n\n. q3 q2 q1 q0 . q3 q2 q1 q0\n. . Ci B A . Co S B A\n. . . . . . . . . .\nT1 0 0 0 0 - 0 0 0 0\nT2 0 1 0 0 - 0 1 0 0\nT3 0 0 1 0 - 0 1 1 0\nT4 0 1 1 0 - 1 0 1 0\nT5 0 0 0 1 - 0 1 0 1\nT6 0 1 0 1 - 1 0 0 1\nT7 0 0 1 1 - 1 0 1 1\nT8 0 1 1 1 - 1 1 1 1\n\n## How does it work\n\nThe code below shows what happens when we use the quantum full adder to add three qubit states. You can copy and paste this code in your own editor and see what happens when you change the input states A, B and CarryIn to either $\\left\\lvert 0 \\right\\rangle$ or $\\left\\lvert 1 \\right\\rangle$ using the X-gate in the initialization function. In this example we set A = 1, B = 0 and CarryIn = 1, equal to T6 in the truth table.\n\n## Examination of the code and results\n\nWhen we execute the code, setting the number of shots to 1, we get the histogram as shown below the code, which is equal to the expected output (compare to T6 in the truth table). Note: one shot is enough to determine the probability distribution\n\n \nversion 1.0\n\nqubits 4\n\n# qubit definitions\n# q --> A\n# q --> B\n# q --> CarryIn_SumOut\n# q --> CarryOut\n\n# initialize inputs to some values\n.init\n#initialize inputs A=1, B=0 and carry_in=1\n{x q \| x q}\n\ntoffoli q,q,q\ncnot q,q\ntoffoli q,q,q\ncnot q,q\ncnot q,q\n\n\n\nWe can also use superposition states and entangled states, such as in the following code where we set A and B to a Bell state and CarryIn to a superposition state. This is similar to setting the input to states T1, T2, T7 and T8 at the same time.\n\nComparing again the truth table to the histogram of probability amplitudes, we indeed see that all 4 output states that we would expect are generated.\n\n## Want to know more?\n\nIn this example we showed how to create a simple quantum full adder. In the literature you can find other examples to add qubit states, subtract qubit states and execute more complex operations on qubit states." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.78323275,"math_prob":0.9846736,"size":3440,"snap":"2020-45-2020-50","text_gpt3_token_len":1090,"char_repetition_ratio":0.14988358,"word_repetition_ratio":0.039156627,"special_character_ratio":0.32877907,"punctuation_ratio":0.10785619,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9897343,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-30T17:37:16Z\",\"WARC-Record-ID\":\"<urn:uuid:a8b02acc-8796-49ca-84d2-67e5d1b6c2b9>\",\"Content-Length\":\"141983\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dd504982-ca33-49f6-b550-1babdb788d42>\",\"WARC-Concurrent-To\":\"<urn:uuid:b3be8ddd-7234-44f7-a651-af868c817d50>\",\"WARC-IP-Address\":\"104.248.63.231\",\"WARC-Target-URI\":\"https://www.quantum-inspire.com/kbase/full-adder/\",\"WARC-Payload-Digest\":\"sha1:3QUI3SSFODHY2ZRWTUHGOX3XW62PMGFC\",\"WARC-Block-Digest\":\"sha1:74DWZQGV22LBURQ5YUFL44GGPWSFG6RQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141216897.58_warc_CC-MAIN-20201130161537-20201130191537-00225.warc.gz\"}"}
https://www.numbersaplenty.com/122368	[ "Search a number\nBaseRepresentation\nbin11101111000000000\n320012212011\n4131320000\n512403433\n62342304\n71016521\noct357000\n9205764\n10122368\n1183a34\n125a994\n134390c\n1432848\n15263cd\nhex1de00\n\n122368 has 20 divisors (see below), whose sum is σ = 245520. Its totient is φ = 60928.\n\nThe previous prime is 122363. The next prime is 122387. The reversal of 122368 is 863221.\n\nAdding to 122368 its reverse (863221), we get a palindrome (985589).\n\nTogether with 123152 it forms an amicable pair .\n\nIt is a plaindrome in base 10.\n\nIt is a congruent number.\n\nIt is not an unprimeable number, because it can be changed into a prime (122363) by changing a digit.\n\nIt is a pernicious number, because its binary representation contains a prime number (7) of ones.\n\nIt is a polite number, since it can be written as a sum of consecutive naturals, namely, 393 + ... + 631.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (12276).\n\n2122368 is an apocalyptic number.\n\nIt is an amenable number.\n\nIt is a practical number, because each smaller number is the sum of distinct divisors of 122368, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (122760).\n\n122368 is an abundant number, since it is smaller than the sum of its proper divisors (123152).\n\nIt is a pseudoperfect number, because it is the sum of a subset of its proper divisors.\n\n122368 is an frugal number, since it uses more digits than its factorization.\n\n122368 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 257 (or 241 counting only the distinct ones).\n\nThe product of its digits is 576, while the sum is 22.\n\nThe square root of 122368 is about 349.8113777452. The cubic root of 122368 is about 49.6465743592.\n\nThe spelling of 122368 in words is \"one hundred twenty-two thousand, three hundred sixty-eight\"." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91206455,"math_prob":0.9931904,"size":1774,"snap":"2020-45-2020-50","text_gpt3_token_len":468,"char_repetition_ratio":0.17740113,"word_repetition_ratio":0.006329114,"special_character_ratio":0.33370912,"punctuation_ratio":0.13896458,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9963673,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-28T02:38:29Z\",\"WARC-Record-ID\":\"<urn:uuid:973e2828-929d-46c5-b18f-51549009e39f>\",\"Content-Length\":\"9313\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c12aecfe-f3c7-4229-968e-c9c4e8f02f2c>\",\"WARC-Concurrent-To\":\"<urn:uuid:d1202f93-0bce-4e09-95d1-ffbdcda33b5e>\",\"WARC-IP-Address\":\"62.149.142.170\",\"WARC-Target-URI\":\"https://www.numbersaplenty.com/122368\",\"WARC-Payload-Digest\":\"sha1:WNTRHD26VW4BHT5YRZZW7V7DBSIRCZ2Z\",\"WARC-Block-Digest\":\"sha1:P7YDNFUANFLEG7WVO6X6H2VHCBWNGAOI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107896048.53_warc_CC-MAIN-20201028014458-20201028044458-00008.warc.gz\"}"}
https://scicomp.stackexchange.com/questions/23385/what-is-the-worst-case-complexity-of-conjugate-gradient/23390	[ "# What is the worst case complexity of Conjugate Gradient?\n\nLet $A\\in \\mathbb{R}^{n\\times n}$, symmetric and positive definite. Suppose it takes $m$ units of work to multiply a vector by $A$. It is well known that performing the CG algorithm on $A$ with condition number $\\kappa$ requires $\\mathcal{O} (m\\sqrt{\\kappa})$, units of work.\n\nNow, of course, being a $\\mathcal{O}$ statement this is an upper-bound. And the CG algorithm can always terminate in zero steps with a lucky initial guess.\n\nDo we know if there exists a RHS and an initial (unlucky) guess that will require $\\mathcal{\\Theta}(\\sqrt{\\kappa})$ steps? Put another way, is worst-case work-complexity of CG really $\\Theta( m \\sqrt{\\kappa})$?\n\nThis question arises when I tried to determine if the benefit of a preconditioner (lower $\\kappa$) outweighed its cost (higher $m$). Right now, I am working with toy problems and would like to have a better idea before I implement anything in a compiled language.\n\n• You could presumably construct a pessimal initial guess by running the CG algorithm \"backwards\" and putting suitable energy into each of the $A$-orthogonal search directions that the algorithm requires all the steps. – origimbo Mar 17 '16 at 16:11\n\nThe answer is a resounding yes. The convergence rate bound of $(\\sqrt{\\kappa}-1) / (\\sqrt{\\kappa}+1)$ is sharp over the set of symmetric positive definite matrices with condition number $\\kappa$. In other words, knowing nothing more about $A$ than its condition number, CG really can take $\\sim\\sqrt{\\kappa}$ iterations to converge. Loosely speaking, the upper-bound is attained if the eigenvalues of $A$ are uniformly distributed (i.e. \"peppered\") within an interval of condition number $\\kappa$.\n\nHere's a more rigorous statement. Deterministic versions are more involved but work using the same principles.\n\nTheorem (Worst-case choice of $A$). Pick any random orthogonal matrix $U$, let $\\lambda_1,\\ldots,\\lambda_n$ be $n$ real numbers uniformly sampled from the real interval $[1,\\kappa]$, and let $b=[b_1;\\ldots;b_n]$ be $n$ real numbers sampled i.i.d. from the standard Gaussian. Define $$A=U\\mathrm{diag}(\\lambda_1,\\ldots,\\lambda_n)U^T.$$ Then in the limit $n\\to\\infty$, conjugate gradients will convergence with probability one to an $\\epsilon$ accurate solution of $Ax=b$ in no less than $\\Omega(\\sqrt{\\kappa}\\log\\epsilon^{-1})$ iterations.\n\nProof. The standard proof is based on optimal Chebyshev polynomial approximations, using techniques found in a number of places, such as Greenbaum's book or Saad's book.\n\n• The bound is not sharp, as the answer explains later, If the eigenvalues are not uniformly distributed, cg converges faster, since it is not a stationalry iteration. Thus, we need to know more about the matrix. – Guido Kanschat Mar 20 '16 at 3:39\n• @GuidoKanschat: Good point, and I've fixed the statement to clarify that sharpness is attained over all $A$ with condition $\\kappa$. – Richard Zhang Mar 21 '16 at 15:50\n• The proof boils down to minimizing $\\\|p(A)\\\|$ in the space of order-$k$ polynomials satisfying $p(0)=1$. Equivalently this is $\\min_p \\max_{\\lambda\\in\\Lambda(A)} \|p(\\lambda)\|$. In the stated limit, $\\Lambda(A)\\to[1,\\kappa]$, and the solution for the minimax problem is then the Chebyshev polynomial, whose error converges as $\\sim\\sqrt{\\kappa}$ – Richard Zhang Mar 21 '16 at 21:27\n\nTaking this as my original question: Do we know if there exists a RHS and an initial (unlucky) guess that will require $\\Theta(\\sqrt{\\kappa})$ steps?\n\nThe answer to the question is \"no\". The idea of this answer comes from the comment from Guido Kanschat.\n\nClaim: For any given condition number $k$, there exists a matrix $A$, with that condition number for which the CG algorithm will terminate in at most two steps (for any given RHS and initial guess).\n\nConsider $A\\in \\mathbb{R}^{n\\times n}$ where $A=\\mathrm{diag}(1,\\kappa,\\kappa,\\ldots, \\kappa)$. Then the condition number of $A$ is $\\kappa$. Let $b\\in \\mathbb{R}^n$ be the RHS, and denote the eigenvalues of $A$ as $\\lambda_i$ where $$\\lambda_i = \\left\\{\\begin{array}{ll}1 & i=1\\\\ \\kappa & i\\not= 1 \\end{array} \\right. .$$\n\nWe first consider the case where $x^{(0)} \\in \\mathbb{R}^n$, the initial guess, is zero. Denote $x^{(2)}\\in \\mathbb{R}^n$ as the second estimate of $A^{-1}b$ from the CG algorithm. We show that $x^{(2)} =A^{-1}b$ by showing $\\langle x^{(2)}-A^{-1}b, A(x^{(2)}-A^{-1}b)\\rangle =0$. Indeed, we have\n\n\\begin{align} \\langle x^{(2)}-A^{-1}b, A(x^{(2)}-A^{-1}b)\\rangle &= \\left\\\| x^{(2)}-A^{-1}b \\right\\\|_A^2 \\\\ &=\\min_{p\\in \\mathrm{poly}_{1} } \\left\\\| (p(A)-A^{-1}) b \\right\\\|_A^2\\\\ &=\\min_{p\\in \\mathrm{poly}_{1} } \\sum_{i=1}^n (p(\\lambda_i) - \\lambda_i^{-1})^2 \\lambda_i b_i^2 \\\\ &\\le \\sum_{i=1}^n (\\widehat{p}(\\lambda_i) - \\lambda_i^{-1})^2 \\lambda_i b_i^2 = 0 \\end{align}\n\nWhere we use the first order polynomial $\\widehat{p}$ defined as $\\widehat{p}(x)= (1+\\kappa-x)/\\kappa$. So we proven the case for $x^{(0)}= 0$.\n\nIf $x^{(0)} \\not = 0$, then $x^{(2)}= \\overline{x^{(2)}}+ x^{(0)}$ where $\\overline{x^{(2)} }$ is the second estimate of the CG algorithm with $b$ replaced with $\\overline{b} = b-A x^{(0)}$. So we have reduced this case to the previous one.\n\n• How much of this is robust to finite precision arithmetic? – origimbo Mar 21 '16 at 16:28\n• @origimbo If your question was directed to me, the answer is, \"I don't know.\" – fred Mar 21 '16 at 16:36" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6056781,"math_prob":0.999811,"size":1834,"snap":"2019-51-2020-05","text_gpt3_token_len":690,"char_repetition_ratio":0.1273224,"word_repetition_ratio":0.0,"special_character_ratio":0.3942203,"punctuation_ratio":0.079155676,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999992,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-17T13:58:50Z\",\"WARC-Record-ID\":\"<urn:uuid:64ffa04e-fd31-4c06-b4a5-29216ffa389c>\",\"Content-Length\":\"148434\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0c4c8d18-696b-43c9-a4b5-45adce953650>\",\"WARC-Concurrent-To\":\"<urn:uuid:4a81b51a-88b1-417a-ab72-7a2891e439e4>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://scicomp.stackexchange.com/questions/23385/what-is-the-worst-case-complexity-of-conjugate-gradient/23390\",\"WARC-Payload-Digest\":\"sha1:27ZHMETMZ2OBFTU6WZFF3CNOHCRSQ4KG\",\"WARC-Block-Digest\":\"sha1:KKVJXKNXY2NK2IS2SJTE5KVPV4EENC2J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250589560.16_warc_CC-MAIN-20200117123339-20200117151339-00322.warc.gz\"}"}
https://openstax.org/books/elementary-algebra/pages/1-8-the-real-numbers	[ "Elementary Algebra\n\n# 1.8The Real Numbers\n\nElementary Algebra1.8 The Real Numbers\n\n### Learning Objectives\n\nBy the end of this section, you will be able to:\n\n• Simplify expressions with square roots\n• Identify integers, rational numbers, irrational numbers, and real numbers\n• Locate fractions on the number line\n• Locate decimals on the number line\n\n### Be Prepared 1.8\n\nA more thorough introduction to the topics covered in this section can be found in the Prealgebra chapters, Decimals and Properties of Real Numbers.\n\n### Simplify Expressions with Square Roots\n\nRemember that when a number n is multiplied by itself, we write $n2n2$ and read it “n squared.” The result is called the square of n. For example,\n\n$82read‘8squared’6464is called thesquareof8.82read‘8squared’6464is called thesquareof8.$\n\nSimilarly, 121 is the square of 11, because $112112$ is 121.\n\n### Square of a Number\n\nIf $n2=m,n2=m,$ then m is the square of n.\n\n### Manipulative Mathematics\n\nDoing the Manipulative Mathematics activity “Square Numbers” will help you develop a better understanding of perfect square numbers.\n\nComplete the following table to show the squares of the counting numbers 1 through 15.", null, "The numbers in the second row are called perfect square numbers. It will be helpful to learn to recognize the perfect square numbers.\n\nThe squares of the counting numbers are positive numbers. What about the squares of negative numbers? We know that when the signs of two numbers are the same, their product is positive. So the square of any negative number is also positive.\n\n$(−3)2=9(−8)2=64(−11)2=121(−15)2=225(−3)2=9(−8)2=64(−11)2=121(−15)2=225$\n\nDid you notice that these squares are the same as the squares of the positive numbers?\n\nSometimes we will need to look at the relationship between numbers and their squares in reverse. Because $102=100,102=100,$ we say 100 is the square of 10. We also say that 10 is a square root of 100. A number whose square is $mm$ is called a square root of m.\n\n### Square Root of a Number\n\nIf $n2=m,n2=m,$ then n is a square root of m.\n\nNotice $(−10)2=100(−10)2=100$ also, so $−10−10$ is also a square root of 100. Therefore, both 10 and $−10−10$ are square roots of 100.\n\nSo, every positive number has two square roots—one positive and one negative. What if we only wanted the positive square root of a positive number? The radical sign, $m,m,$ denotes the positive square root. The positive square root is called the principal square root. When we use the radical sign that always means we want the principal square root.\n\nWe also use the radical sign for the square root of zero. Because $02=0,02=0,$ $0=0.0=0.$ Notice that zero has only one square root.\n\n### Square Root Notation\n\n$mm$ is read “the square root of m", null, "If $m=n2,m=n2,$ then $m=n,m=n,$ for $n≥0.n≥0.$\n\nThe square root of m, $m,m,$ is the positive number whose square is m.\n\nSince 10 is the principal square root of 100, we write $100=10.100=10.$ You may want to complete the following table to help you recognize square roots.", null, "### Example 1.108\n\nSimplify: $2525$ $121.121.$\n\n### Try It 1.215\n\nSimplify: $3636$ $169.169.$\n\n### Try It 1.216\n\nSimplify: $1616$ $196.196.$\n\nWe know that every positive number has two square roots and the radical sign indicates the positive one. We write $100=10.100=10.$ If we want to find the negative square root of a number, we place a negative in front of the radical sign. For example, $−100=−10.−100=−10.$ We read $−100−100$ as “the opposite of the square root of 10.”\n\n### Example 1.109\n\nSimplify: $−9−9$ $−144.−144.$\n\n### Try It 1.217\n\nSimplify: $−4−4$ $−225.−225.$\n\n### Try It 1.218\n\nSimplify: $−81−81$ $−100.−100.$\n\n### Identify Integers, Rational Numbers, Irrational Numbers, and Real Numbers\n\nWe have already described numbers as counting numbers, whole numbers, and integers. What is the difference between these types of numbers?\n\n$Counting numbers1,2,3,4,…Whole numbers0,1,2,3,4,…Integers…−3,−2,−1,0,1,2,3,…Counting numbers1,2,3,4,…Whole numbers0,1,2,3,4,…Integers…−3,−2,−1,0,1,2,3,…$\n\nWhat type of numbers would we get if we started with all the integers and then included all the fractions? The numbers we would have form the set of rational numbers. A rational number is a number that can be written as a ratio of two integers.\n\n### Rational Number\n\nA rational number is a number of the form $pq,pq,$ where p and q are integers and $q≠0.q≠0.$\n\nA rational number can be written as the ratio of two integers.\n\nAll signed fractions, such as $45,−78,134,−20345,−78,134,−203$ are rational numbers. Each numerator and each denominator is an integer.\n\nAre integers rational numbers? To decide if an integer is a rational number, we try to write it as a ratio of two integers. Each integer can be written as a ratio of integers in many ways. For example, 3 is equivalent to $31,62,93,124,155…31,62,93,124,155…$\n\nAn easy way to write an integer as a ratio of integers is to write it as a fraction with denominator one.\n\n$3=31−8=−810=013=31−8=−810=01$\n\nSince any integer can be written as the ratio of two integers, all integers are rational numbers! Remember that the counting numbers and the whole numbers are also integers, and so they, too, are rational.\n\nWhat about decimals? Are they rational? Let’s look at a few to see if we can write each of them as the ratio of two integers.\n\nWe’ve already seen that integers are rational numbers. The integer $−8−8$ could be written as the decimal $−8.0.−8.0.$ So, clearly, some decimals are rational.\n\nThink about the decimal 7.3. Can we write it as a ratio of two integers? Because 7.3 means $7310,7310,$ we can write it as an improper fraction, $7310.7310.$ So 7.3 is the ratio of the integers 73 and 10. It is a rational number.\n\nIn general, any decimal that ends after a number of digits (such as 7.3 or $−1.2684)−1.2684)$ is a rational number. We can use the place value of the last digit as the denominator when writing the decimal as a fraction.\n\n### Example 1.110\n\nWrite as the ratio of two integers: $−27−27$ 7.31.\n\n### Try It 1.219\n\nWrite as the ratio of two integers: $−24−24$ 3.57.\n\n### Try It 1.220\n\nWrite as the ratio of two integers: $−19−19$ 8.41.\n\nLet’s look at the decimal form of the numbers we know are rational.\n\nWe have seen that every integer is a rational number, since $a=a1a=a1$ for any integer, a. We can also change any integer to a decimal by adding a decimal point and a zero.\n\n$Integer−2−10123Decimal form−2.0−1.00.01.02.03.0Integer−2−10123Decimal form−2.0−1.00.01.02.03.0$\n$These decimal numbers stop.These decimal numbers stop.$\n\nWe have also seen that every fraction is a rational number. Look at the decimal form of the fractions we considered above.\n\n$Ratio of integers45−78134−203The decimal form0.8−0.8753.25−6.666…−6.6–Ratio of integers45−78134−203The decimal form0.8−0.8753.25−6.666…−6.6–$\n$These decimals either stop or repeat.These decimals either stop or repeat.$\n\nWhat do these examples tell us?\n\nEvery rational number can be written both as a ratio of integers, $(pq,(pq,$ where p and q are integers and $q≠0),q≠0),$ and as a decimal that either stops or repeats.\n\nHere are the numbers we looked at above expressed as a ratio of integers and as a decimal:\n\nFractions Integers\nNumber $4545$ $−78−78$ $134134$ $−203−203$ $−2−2$ $−1−1$ $00$ $11$ $22$ $33$\nRatio of Integers $4545$ $−78−78$ $134134$ $−203−203$ $−21−21$ $−11−11$ $0101$ $1111$ $2121$ $3131$\nDecimal Form $0.80.8$ $−0.875−0.875$ $3.253.25$ $−6.6–−6.6–$ $−2.0−2.0$ $−1.0−1.0$ $0.00.0$ $1.01.0$ $2.02.0$ $3.03.0$\n\n### Rational Number\n\nA rational number is a number of the form $pq,pq,$ where p and q are integers and $q≠0.q≠0.$\n\nIts decimal form stops or repeats.\n\nAre there any decimals that do not stop or repeat? Yes!\n\nThe number $ππ$ (the Greek letter pi, pronounced “pie”), which is very important in describing circles, has a decimal form that does not stop or repeat.\n\n$π=3.141592654...π=3.141592654...$\n\nWe can even create a decimal pattern that does not stop or repeat, such as\n\n$2.01001000100001…2.01001000100001…$\n\nNumbers whose decimal form does not stop or repeat cannot be written as a fraction of integers. We call these numbers irrational.\n\n### Irrational Number\n\nAn irrational number is a number that cannot be written as the ratio of two integers.\n\nIts decimal form does not stop and does not repeat.\n\nLet’s summarize a method we can use to determine whether a number is rational or irrational.\n\n### Rational or Irrational?\n\nIf the decimal form of a number\n\n• repeats or stops, the number is rational.\n• does not repeat and does not stop, the number is irrational.\n\n### Example 1.111\n\nGiven the numbers $0.583–,0.47,3.605551275...0.583–,0.47,3.605551275...$ list the rational numbers irrational numbers.\n\n### Try It 1.221\n\nFor the given numbers list the rational numbers irrational numbers: $0.29,0.816–,2.515115111….0.29,0.816–,2.515115111….$\n\n### Try It 1.222\n\nFor the given numbers list the rational numbers irrational numbers: $2.63–,0.125,0.418302…2.63–,0.125,0.418302…$\n\n### Example 1.112\n\nFor each number given, identify whether it is rational or irrational: $3636$ $44.44.$\n\n### Try It 1.223\n\nFor each number given, identify whether it is rational or irrational: $8181$ $17.17.$\n\n### Try It 1.224\n\nFor each number given, identify whether it is rational or irrational: $116116$ $121.121.$\n\nWe have seen that all counting numbers are whole numbers, all whole numbers are integers, and all integers are rational numbers. The irrational numbers are numbers whose decimal form does not stop and does not repeat. When we put together the rational numbers and the irrational numbers, we get the set of real numbers.\n\n### Real Number\n\nA real number is a number that is either rational or irrational.\n\nAll the numbers we use in elementary algebra are real numbers. Figure 1.15 illustrates how the number sets we’ve discussed in this section fit together.\n\nFigure 1.15 This chart shows the number sets that make up the set of real numbers. Does the term “real numbers” seem strange to you? Are there any numbers that are not “real,” and, if so, what could they be?\n\nCan we simplify $−25?−25?$ Is there a number whose square is $−25?−25?$\n\n$()2=−25?()2=−25?$\n\nNone of the numbers that we have dealt with so far has a square that is $−25.−25.$ Why? Any positive number squared is positive. Any negative number squared is positive. So we say there is no real number equal to $−25.−25.$\n\nThe square root of a negative number is not a real number.\n\n### Example 1.113\n\nFor each number given, identify whether it is a real number or not a real number: $−169−169$ $−64.−64.$\n\n### Try It 1.225\n\nFor each number given, identify whether it is a real number or not a real number: $−196−196$ $−81.−81.$\n\n### Try It 1.226\n\nFor each number given, identify whether it is a real number or not a real number: $−49−49$ $−121.−121.$\n\n### Example 1.114\n\nGiven the numbers $−7,145,8,5,5.9,−64,−7,145,8,5,5.9,−64,$ list the whole numbers integers rational numbers irrational numbers real numbers.\n\n### Try It 1.227\n\nFor the given numbers, list the whole numbers integers rational numbers irrational numbers real numbers: $−3,−2,0.3–,95,4,49.−3,−2,0.3–,95,4,49.$\n\n### Try It 1.228\n\nFor the given numbers, list the whole numbers integers rational numbers irrational numbers real numbers: $−25,−38,−1,6,121,2.041975…−25,−38,−1,6,121,2.041975…$\n\n### Locate Fractions on the Number Line\n\nThe last time we looked at the number line, it only had positive and negative integers on it. We now want to include fractions and decimals on it.\n\n### Manipulative Mathematics\n\nDoing the Manipulative Mathematics activity “Number Line Part 3” will help you develop a better understanding of the location of fractions on the number line.\n\nLet’s start with fractions and locate $15,−45,3,74,−92,−5,and8315,−45,3,74,−92,−5,and83$ on the number line.\n\nWe’ll start with the whole numbers $33$ and $−5.−5.$ because they are the easiest to plot. See Figure 1.16.\n\nThe proper fractions listed are $15and−45.15and−45.$ We know the proper fraction $1515$ has value less than one and so would be located between $0 and 1.0 and 1.$ The denominator is 5, so we divide the unit from 0 to 1 into 5 equal parts $15,25,35,45.15,25,35,45.$ We plot $15.15.$ See Figure 1.16.\n\nSimilarly, $−45−45$ is between 0 and $−1.−1.$ After dividing the unit into 5 equal parts we plot $−45.−45.$ See Figure 1.16.\n\nFinally, look at the improper fractions $74,−92,83.74,−92,83.$ These are fractions in which the numerator is greater than the denominator. Locating these points may be easier if you change each of them to a mixed number. See Figure 1.16.\n\n$74=134−92=−41283=22374=134−92=−41283=223$\n\nFigure 1.16 shows the number line with all the points plotted.\n\nFigure 1.16\n\n### Example 1.115\n\nLocate and label the following on a number line: $4,34,−14,−3,65,−52,and73.4,34,−14,−3,65,−52,and73.$\n\n### Try It 1.229\n\nLocate and label the following on a number line: $−1,13,65,−74,92,5,−83.−1,13,65,−74,92,5,−83.$\n\n### Try It 1.230\n\nLocate and label the following on a number line: $−2,23,75,−74,72,3,−73.−2,23,75,−74,72,3,−73.$\n\nIn Example 1.116, we’ll use the inequality symbols to order fractions. In previous chapters we used the number line to order numbers.\n\n• a < ba is less than b” when a is to the left of b on the number line\n• a > ba is greater than b” when a is to the right of b on the number line\n\nAs we move from left to right on a number line, the values increase.\n\n### Example 1.116\n\nOrder each of the following pairs of numbers, using < or >. It may be helpful to refer Figure 1.17.\n\n$−23___−1−23___−1$ $−312___−3−312___−3$ $−34___−14−34___−14$ $−2___−83−2___−83$\n\nFigure 1.17\n\n### Try It 1.231\n\nOrder each of the following pairs of numbers, using < or >:\n\n$−13___−1−13___−1$ $−112___−2−112___−2$ $−23___−13−23___−13$ $−3___−73.−3___−73.$\n\n### Try It 1.232\n\nOrder each of the following pairs of numbers, using < or >:\n\n$−1___−23−1___−23$ $−214___−2−214___−2$ $−35___−45−35___−45$ $−4___−103.−4___−103.$\n\n### Locate Decimals on the Number Line\n\nSince decimals are forms of fractions, locating decimals on the number line is similar to locating fractions on the number line.\n\n### Example 1.117\n\nLocate 0.4 on the number line.\n\n### Try It 1.233\n\nLocate on the number line: 0.6.\n\n### Try It 1.234\n\nLocate on the number line: 0.9.\n\n### Example 1.118\n\nLocate $−0.74−0.74$ on the number line.\n\n### Try It 1.235\n\nLocate on the number line: $−0.6.−0.6.$\n\n### Try It 1.236\n\nLocate on the number line: $−0.7.−0.7.$\n\nWhich is larger, 0.04 or 0.40? If you think of this as money, you know that $0.40 (forty cents) is greater than$0.04 (four cents). So,\n\n$0.40>0.040.40>0.04$\n\nAgain, we can use the number line to order numbers.\n\n• a < ba is less than b” when a is to the left of b on the number line\n• a > ba is greater than b” when a is to the right of b on the number line\n\nWhere are 0.04 and 0.40 located on the number line? See Figure 1.20.\n\nFigure 1.20\n\nWe see that 0.40 is to the right of 0.04 on the number line. This is another way to demonstrate that 0.40 > 0.04.\n\nHow does 0.31 compare to 0.308? This doesn’t translate into money to make it easy to compare. But if we convert 0.31 and 0.308 into fractions, we can tell which is larger.\n\n 0.31 0.308 Convert to fractions. $3110031100$ $30810003081000$ We need a common denominator to compare them.", null, "", null, "$31010003101000$ $30810003081000$\n\nBecause 310 > 308, we know that $3101000>3081000.3101000>3081000.$ Therefore, 0.31 > 0.308.\n\nNotice what we did in converting 0.31 to a fraction—we started with the fraction $3110031100$ and ended with the equivalent fraction $3101000.3101000.$ Converting $31010003101000$ back to a decimal gives 0.310. So 0.31 is equivalent to 0.310. Writing zeros at the end of a decimal does not change its value!\n\n$31100=3101000and0.31=0.31031100=3101000and0.31=0.310$\n\nWe say 0.31 and 0.310 are equivalent decimals.\n\n### Equivalent Decimals\n\nTwo decimals are equivalent if they convert to equivalent fractions.\n\nWe use equivalent decimals when we order decimals.\n\nThe steps we take to order decimals are summarized here.\n\n### How To\n\n#### Order Decimals.\n\n1. Step 1. Write the numbers one under the other, lining up the decimal points.\n2. Step 2. Check to see if both numbers have the same number of digits. If not, write zeros at the end of the one with fewer digits to make them match.\n3. Step 3. Compare the numbers as if they were whole numbers.\n4. Step 4. Order the numbers using the appropriate inequality sign.\n\n### Example 1.119\n\nOrder $0.64___0.60.64___0.6$ using $<<$ or $>.>.$\n\n### Try It 1.237\n\nOrder each of the following pairs of numbers, using $:0.42___0.4.:0.42___0.4.$\n\n### Try It 1.238\n\nOrder each of the following pairs of numbers, using $:0.18___0.1.:0.18___0.1.$\n\n### Example 1.120\n\nOrder $0.83___0.8030.83___0.803$ using $<<$ or $>.>.$\n\n### Try It 1.239\n\nOrder the following pair of numbers, using $:0.76___0.706.:0.76___0.706.$\n\n### Try It 1.240\n\nOrder the following pair of numbers, using $:0.305___0.35.:0.305___0.35.$\n\nWhen we order negative decimals, it is important to remember how to order negative integers. Recall that larger numbers are to the right on the number line. For example, because $−2−2$ lies to the right of $−3−3$ on the number line, we know that $−2>−3.−2>−3.$ Similarly, smaller numbers lie to the left on the number line. For example, because $−9−9$ lies to the left of $−6−6$ on the number line, we know that $−9<−6.−9<−6.$ See Figure 1.21.\n\nFigure 1.21\n\nIf we zoomed in on the interval between 0 and $−1,−1,$ as shown in Example 1.121, we would see in the same way that $−0.2>−0.3and−0.9<−0.6.−0.2>−0.3and−0.9<−0.6.$\n\n### Example 1.121\n\nUse $<<$ or $>>$ to order $−0.1___−0.8.−0.1___−0.8.$\n\n### Try It 1.241\n\nOrder the following pair of numbers, using < or >: $−0.3___−0.5.−0.3___−0.5.$\n\n### Try It 1.242\n\nOrder the following pair of numbers, using < or >: $−0.6___−0.7.−0.6___−0.7.$\n\n### Section 1.8 Exercises\n\n#### Practice Makes Perfect\n\nSimplify Expressions with Square Roots\n\nIn the following exercises, simplify.\n\n659.\n\n$36 36$\n\n660.\n\n$4 4$\n\n661.\n\n$64 64$\n\n662.\n\n$169 169$\n\n663.\n\n$9 9$\n\n664.\n\n$16 16$\n\n665.\n\n$100 100$\n\n666.\n\n$144 144$\n\n667.\n\n$− 4 − 4$\n\n668.\n\n$− 100 − 100$\n\n669.\n\n$− 1 − 1$\n\n670.\n\n$− 121 − 121$\n\nIdentify Integers, Rational Numbers, Irrational Numbers, and Real Numbers\n\nIn the following exercises, write as the ratio of two integers.\n\n671.\n\n5 3.19\n\n672.\n\n8 1.61\n\n673.\n\n$−12−12$ 9.279\n\n674.\n\n$−16−16$ 4.399\n\nIn the following exercises, list the rational numbers, irrational numbers\n\n675.\n\n$0.75 , 0.22 3 – , 1.39174 0.75 , 0.22 3 – , 1.39174$\n\n676.\n\n$0.36 , 0.94729 … , 2.52 8 – 0.36 , 0.94729 … , 2.52 8 –$\n\n677.\n\n$0.4 5 – , 1.919293 … , 3.59 0.4 5 – , 1.919293 … , 3.59$\n\n678.\n\n$0.1 3 – , 0.42982 … , 1.875 0.1 3 – , 0.42982 … , 1.875$\n\nIn the following exercises, identify whether each number is rational or irrational.\n\n679.\n\n$2525$ $3030$\n\n680.\n\n$4444$ $4949$\n\n681.\n\n$164164$ $169169$\n\n682.\n\n$225225$ $216216$\n\nIn the following exercises, identify whether each number is a real number or not a real number.\n\n683.\n\n$−81−81$ $−121−121$\n\n684.\n\n$−64−64$ $−9−9$\n\n685.\n\n$−36−36$ $−144−144$\n\n686.\n\n$−49−49$ $−144−144$\n\nIn the following exercises, list the whole numbers, integers, rational numbers, irrational numbers, real numbers for each set of numbers.\n\n687.\n\n$−8 , 0 , 1.95286 … , 12 5 , 36 , 9 −8 , 0 , 1.95286 … , 12 5 , 36 , 9$\n\n688.\n\n$−9 , −3 4 9 , − 9 , 0.40 9 – , 11 6 , 7 −9 , −3 4 9 , − 9 , 0.40 9 – , 11 6 , 7$\n\n689.\n\n$− 100 , −7 , − 8 3 , −1 , 0.77 , 3 1 4 − 100 , −7 , − 8 3 , −1 , 0.77 , 3 1 4$\n\n690.\n\n$−6 , − 5 2 , 0 , 0. 714285 ——— , 2 1 5 , 14 −6 , − 5 2 , 0 , 0. 714285 ——— , 2 1 5 , 14$\n\nLocate Fractions on the Number Line\n\nIn the following exercises, locate the numbers on a number line.\n\n691.\n\n$3 4 , 8 5 , 10 3 3 4 , 8 5 , 10 3$\n\n692.\n\n$1 4 , 9 5 , 11 3 1 4 , 9 5 , 11 3$\n\n693.\n\n$3 10 , 7 2 , 11 6 , 4 3 10 , 7 2 , 11 6 , 4$\n\n694.\n\n$7 10 , 5 2 , 13 8 , 3 7 10 , 5 2 , 13 8 , 3$\n\n695.\n\n$2 5 , − 2 5 2 5 , − 2 5$\n\n696.\n\n$3 4 , − 3 4 3 4 , − 3 4$\n\n697.\n\n$3 4 , − 3 4 , 1 2 3 , −1 2 3 , 5 2 , − 5 2 3 4 , − 3 4 , 1 2 3 , −1 2 3 , 5 2 , − 5 2$\n\n698.\n\n$1 5 , − 2 5 , 1 3 4 , −1 3 4 , 8 3 , − 8 3 1 5 , − 2 5 , 1 3 4 , −1 3 4 , 8 3 , − 8 3$\n\nIn the following exercises, order each of the pairs of numbers, using < or >.\n\n699.\n\n$−1 ___ − 1 4 −1 ___ − 1 4$\n\n700.\n\n$−1 ___ − 1 3 −1 ___ − 1 3$\n\n701.\n\n$−2 1 2 ___ −3 −2 1 2 ___ −3$\n\n702.\n\n$−1 3 4 ___ −2 −1 3 4 ___ −2$\n\n703.\n\n$− 5 12 ___ − 7 12 − 5 12 ___ − 7 12$\n\n704.\n\n$− 9 10 ___ − 3 10 − 9 10 ___ − 3 10$\n\n705.\n\n$−3 ___ − 13 5 −3 ___ − 13 5$\n\n706.\n\n$−4 ___ − 23 6 −4 ___ − 23 6$\n\nLocate Decimals on the Number Line In the following exercises, locate the number on the number line.\n\n707.\n\n0.8\n\n708.\n\n$−0.9 −0.9$\n\n709.\n\n$−1.6 −1.6$\n\n710.\n\n3.1\n\nIn the following exercises, order each pair of numbers, using < or >.\n\n711.\n\n$0.37 ___ 0.63 0.37 ___ 0.63$\n\n712.\n\n$0.86 ___ 0.69 0.86 ___ 0.69$\n\n713.\n\n$0.91 ___ 0.901 0.91 ___ 0.901$\n\n714.\n\n$0.415 ___ 0.41 0.415 ___ 0.41$\n\n715.\n\n$−0.5 ___ −0.3 −0.5 ___ −0.3$\n\n716.\n\n$−0.1 ___ −0.4 −0.1 ___ −0.4$\n\n717.\n\n$−0.62 ___ −0.619 −0.62 ___ −0.619$\n\n718.\n\n$−7.31 ___ −7.3 −7.31 ___ −7.3$\n\n#### Everyday Math\n\n719.\n\nField trip All the 5th graders at Lincoln Elementary School will go on a field trip to the science museum. Counting all the children, teachers, and chaperones, there will be 147 people. Each bus holds 44 people.\n\nHow many busses will be needed?\nWhy must the answer be a whole number?\nWhy shouldn’t you round the answer the usual way, by choosing the whole number closest to the exact answer?\n\n720.\n\nChild care Serena wants to open a licensed child care center. Her state requires there be no more than 12 children for each teacher. She would like her child care center to serve 40 children.\n\nHow many teachers will be needed?\nWhy must the answer be a whole number?\nWhy shouldn’t you round the answer the usual way, by choosing the whole number closest to the exact answer?\n\n#### Writing Exercises\n\n721.\n\nIn your own words, explain the difference between a rational number and an irrational number.\n\n722.\n\nExplain how the sets of numbers (counting, whole, integer, rational, irrationals, reals) are related to each other.\n\n#### Self Check\n\nAfter completing the exercises, use this checklist to evaluate your mastery of the objective of this section.", null, "On a scale of $1−10,1−10,$ how would you rate your mastery of this section in light of your responses on the checklist? 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https://www.daniweb.com/programming/game-development/threads/471199/retro-game-qix-logic	[ "Hello. First time posting on these forums as I am really stuck on my code. The language I am using is GML (Game Maker Language) and the program Game Maker: Studio. As I could not get any help on GM forums I am seeking help elsewhere.\n\nThe game I am making is a remake of a retro game called QIX.\n\"Qix is an old game where you are drawing boxes in order to fill in the screen with a color or image while avoiding enemies. To pass a level you must fill in a certain percentage of the screen. A more modern game series with this gameplay would be the Gals Panic series.\"\nWikipedia: http://en.wikipedia.org/wiki/Qix\n\nI have made it work, but the problem is that it is extremely slow. It takes about a minute for larger area to be drawn. What I am doing:\n-After the player finishes his move I have an array of the current lines and it adds the new lines that the player just made to the array.\n-It takes two points on each side of the first line and checks a line from one point to the QIX. Depending on the lines crossed it calculates which point is in the closed area without the QIX. This can turn out wrong in some cases, not only because the check (if even number of lines crossed, QIX is in the same area) is wrong, but also the way I check if the lines crossing is wrong. Before drawing I fix this by checking if the QIX is on revealed pixel and if yes, I swap all values(explained later). So suggestions on how to check, which side of the line the QIX is on are welcomed too.\n-Then it sets the point that is supposedly not on the side of the QIX to 2 in a 2D array. I also have min and max x and y values of the 2D array, which at the start also equal the point's x and y.\n- After that it checks the 2D array from min to max values until there are no values that equal 2. If a value = 2, it checks if a line crosses it. If no, it sets the top, bottom, left and right points to 2, and changes min and max x and y values if needed. This one point that was checked is set to 1.\n\nThis way I have a 2D array with 0 for covered and 1 for uncovered points. And here is the code that does that:\n\n``````do //This is the code after I have a starting value in the 2D array - pointd\n{\nenough=1\nfor(for1=minx; for1<=maxx; for1+=1) //min and max x and y values of the points for checking\nfor(for2=miny; for2<=maxy; for2+=1) //so it doesn't check everything everytime\n{\nif(pointd[for1,for2]=2)for(aba=1;aba<=thln;aba+=1) //If it is 2, check it with every line\n{\nif(((for1>=thlposx[aba] and for1<=thlendx[aba]) or (for1<=thlposx[aba] and for1>=thlendx[aba]))\nand((for2>=thlposy[aba] and for2<=thlendy[aba]) or (for2<=thlposy[aba] and for2>=thlendy[aba])))\n{pointd[for1,for2]=1; aba=thln+1} //If it is on a line, set it to 1, to be drawn and stop checking\n}\nif(pointd[for1,for2]=2) //If it is not on a line, set adjacent point for checking\n{\nenough=0 //stop value to 0, continue with do until\npointd[for1,for2]=1 //set point to be drawn\nif(pointd[for1-1,for2]=0)pointd[for1-1,for2]=2 //set points to be checked\nif(pointd[for1+1,for2]=0)pointd[for1+1,for2]=2\nif(pointd[for1,for2-1]=0)pointd[for1,for2-1]=2\nif(pointd[for1,for2+1]=0)pointd[for1,for2+1]=2\nif(minx=for1 and for1>1)minx=for1-1 //if this is min or max x or y change it\nif(miny=for2 and for2>1)miny=for2-1\nif(maxx=for1 and for1<319)maxx=for1+1\nif(maxy=for2 and for2<374)maxy=for2+1\n}\n}\n}\nuntil enough=1\nif(pointd[objQIX.x,objQIX.y]=1) //if it got the wrong part of the screen (the wrong side of the line), change all\n{\nfor(aaaa=0; aaaa<=320; aaaa+=1)\nfor(aaaaa=0; aaaaa<=375; aaaaa+=1)\nif(pointd[aaaa,aaaaa]=1)pointd[aaaa,aaaaa]=0\nelse pointd[aaaa,aaaaa]=1\n}\n``````\n\nIt appears that the do until cycle spins less than 100 times in simple cases and 200-400 times for more complicated ones. The problem seems to be this cycle:\n\n``````if(pointd[for1,for2]=2)for(aba=1;aba<=thln;aba+=1) //If it is 2, check it with every line\n{\nif(((for1>=thlposx[aba] and for1<=thlendx[aba]) or (for1<=thlposx[aba] and for1>=thlendx[aba]))\nand((for2>=thlposy[aba] and for2<=thlendy[aba]) or (for2<=thlposy[aba] and for2>=thlendy[aba])))\n{pointd[for1,for2]=1; aba=thln+1} //If it is on a line, set it to 1, to be drawn and stop checking\n}\n``````\n\nThat for cycle does over 100,000 spins in most cases, and even over 1,000,000 for bigger areas.\n\nAny ideas or code on how to speed it up or change it to a more efficient algorithm are welcome. Thank You in advance!\n\nAre you running through every pixel at a time?\n\nHow about using a scaling algorithm rather? Let's say, each entry in your grid equals 4 real pixels. Then you already have 1/4 of the number of calculations to do. I don't think that on a game like that, you really …\n\nNot sure if the language supports it, but can't you use regions?\n\nBasically this. I know you're not using that, but that's the idea.\n\nThis is quite an interesting conundrum. What are the values of the maximum x and maximum y? I'm trying to get a grasp for the scope of the loops.\n\nIs that the area in pixels, or is the area from 2 - 320 by 2 - 374 pixels?\n\nThere's got to be a way to short-circuit the test algorithm :)\n\n## All 18 Replies\n\nAre you running through every pixel at a time?\n\nHow about using a scaling algorithm rather? Let's say, each entry in your grid equals 4 real pixels. Then you already have 1/4 of the number of calculations to do. I don't think that on a game like that, you really need to handle each pixel as a single block.\n\nThe original game was a text-mode game, so it worked on 80x25 characters! I guess things have changed a lot since then :)\n\nNot sure if the language supports it, but can't you use regions?\n\nThank You for the replies!\n\n@Ewald Horn\nI was thinking about that, but in some cases 1/4 is still a lot. I changed it to see how it will work out, but I am trying to get rid of few bugs that it brought up.\n\n@pritaeas\nWhat exactly do you mean by regions? I don't know what that is, sorry.\n\nBasically this. I know you're not using that, but that's the idea.\n\nSo I got rid of the bugs. It is faster, but it still takes 27 seconds for the whole field to be calculated. That is the time it takes on my laptop and it is meant to be a mobile game.\n\n@pritaeas\nI can use rectangles, but I don't see how I can do it with rectangles. The only thing that came to my mind is have each border, create a rectangle from itself to the opposite border. This would be really fast, but there is just one problem. I don't know how to get them all to check on the same side of the shape. If one creates rectangle in the zone with QIX and other in the zone without QIX, it will completely mess up.\n\nThis is quite an interesting conundrum. What are the values of the maximum x and maximum y? I'm trying to get a grasp for the scope of the loops.\n\nThe total area is 2,2,320,374 set by border lines.\n\nIs that the area in pixels, or is the area from 2 - 320 by 2 - 374 pixels?\n\nThere's got to be a way to short-circuit the test algorithm :)\n\nThose are the x and y coordinates of the rectangle formed by the border lines. The area starts at 2,2 and ends at 320,374. The whole room size is 320x480, but the calculations happen only for the area in the border lines.\n\nAh, that's great, so it's not such a large array at all. As you said, the problem is with the nested loops, there's a lot of expensive looping.\n\nThank You so much for the help!\n\nI thought about using rectangles before, but I could not come with a solution on how to make it exactly. Last night I came up with this and I think this should finally work.\n\n``````// s - starting point of path, e - end point\n//if one line - opposite of QIX\n//Else if on opposite walls\n{\n//if towards e(line direction) – away QIX(rectangle from line), towards s – towards QIX, don’t need the other 2\n// need to continue past collision with the parts of the lines that are not in collision.\n}\n//else\n{\n//if away s – towards e, towards s – away e , away e – away s, towards e – towards s\n}\n``````\n\nI have not tested it yet.\n\nLet us know, there's nothing like finding a solution to something that's been bugging you for a while :)\n\nGreat job on not giving up by the way, I admire that.\n\nI ran into a problem and I remembered that I saw that the first time I thought on how to use rectangles to do it.", null, "Oh, but there's more to fill, isn't there?\n\nyes\n\nHello,\n\nHave you tried the floodFill Algorythm to do the fill and area calc? I'm trying to make the same game, but mine will be on Construct2.\n\nSo far i'm still working in the main code (draw and fill the smaller area) the fastest way possible.\n\nSince Construct2 can be extended with javascript plugins, i left Construct2 for a while to mess on javascript code first.\n\nI'm still ignoring the enemies cause thats made on Construct2. For now i just want to draw.\n\nUsing floodFill algorythm i have managed to quickly check the whole array on both sides of the path (using last known path point and player movement direction to flood both sides).\n\nWorking demo here:\nhttp://luis.peralta.pt/sandbox/js/flood.html\n\nYou'll notice a slow fill, but thats because of the HUGE helper table that demo is using (the code sets a class for each cell to help me visualize what's happening in the array).\n\nMy next move will be get rid of the table and start drawing HTML canvas polys based on the vertices found on the array. But that's yet to think how to do.\n\nIn short, what that code do?\n\nAs you'll see, my 2D array uses the following values:\nU: unfilled\nF: filled\nP: path\nT1: temporary area being flooded at the same time area 2 is\nT2: temporary area being flooed at the same time area 1 is\n\nThe game starts with the perimeter set to F so the player can move safely around the array.\n\nAs soon the player gets out of the safe area (F) it starts drawing the path setting (P) as it goes. If the player inverts the direction, it starts removing P and setting U again (erasing the path).\n\nAn array with the cardeal points to where the player is moving is saved. So i can kill the player when he colides with the path without the exception mentioned above (move back).\n\nWhen the player reaches the F (safe/filled area), i check the direction the player was having and from the lastX and lastY (which will be the last path point) i run the floodFill algorythm on the pixels on each side.\n\nthe first flood will set T1 (temp area 1) on all adjacent U's of the first pixel. The second flood will set T2 (temp area 2) on all adjacent U's of the second pixel. While flooding i save the flood count. That will give me the area of both floods.\n\nThis process can have 3 possible endings:\n\nIf both starting pixels where U, the whole array will be flooded. and F + T1 + T2 = layout area. Then you just fill (set to F) the smaller Tx, or the area where Qix isn't.\n\nIf one of the pixels was F, only one flood will happen. In this case, where F + T1 + T2 != layout area, something is yet to temporary fill and calc. That means that the U's yet to calc are the area of the other Tx. More over, this temp fill is the one to pass from Tx to F.\n\nIf both pixels were already F, this is yet to code. ;) but will be just a check to get the first U, do the flood, and apply the previous process... (i think)\n\nFeel free to contact me if you have any doubt about that JS code (it's not a mess but might not be easy to read.. sorry).\n\nI didn't exactly understand everything you wrote, but I think you are doing what I have already done. I also used flood fill algorithm(see code in first post), but my area is much bigger and it takes too much time.\n\nP.S. I have new idea with rectangles, but I need to test it out to make sure there aren't any flaws, but I don't have the time right now.\n\nSo I have been struggling for 2 weeks to get my idea to work, but with no success. I am posting and hoping some good soul can see where I am making a mistake and help me fix my code.\n\nWhat my code does is when the player completes a shape, it takes the first line and creates a rectangle from that line in certain direction. The rectangle gets as big as it can in every direction until it hits a line. Then it checks each side of the rectangle and how that side fits with a line. If the line does not completely cover the rectangle side, the part of the rectangle side that is not covered, gets added as line to an array. After we are done with this rectangle, we get the next line in the array and create a rectangle for that one too. We do this until there are no more lines in the array.\n\nIt used to work, but it did not make correct rectangles. I went thought that code so many times, changed it so many times and I still can't get it to work. The last change I made even made it add lines to the array until the array gets too big and it shows error and crashes.\n\nHere is the latest code:\n\n``````if(draw_points=1)\n{\nllc=1\nllx1=thlposx[firstnew] //thl starting values are the drawn lines, all in the array\nlly1=thlposy[firstnew] //firstnew is the first drawn line for the last shape, the shape which will be checked\nllx2=thlendx[firstnew]\nlly2=thlendy[firstnew]\n//lld 1 <- 2 ^ 3 -> 4 v\n//if lld is 1 we have right line of rectangle, 2 - bottom, 3 - left, 4 - top\nif(thlposx[firstnew]=thlendx[firstnew])lld=1\nelse lld=2\nfor(ap=1; ap<=llc; ap+=1)\n{\nrectx1=0; recty1=0; rectx2=322; recty2=375 //Border lines form rectangle with coordinates 2,2,320,374\nif(lld[ap]=1){recty1=min(lly1[ap],lly2[ap]); rectx2=llx2[ap]; recty2=max(lly1[ap],lly2[ap]);}\nelse if(lld[ap]=2){rectx1=min(llx1[ap],llx2[ap]); recty2=lly2[ap]; rectx2=max(llx1[ap],llx2[ap])}\nelse if(lld[ap]=3){recty1=min(lly1[ap],lly2[ap]); rectx1=llx1[ap]; recty2=max(lly1[ap],lly2[ap]);}\nelse if(lld[ap]=4){rectx1=min(llx1[ap],llx2[ap]); recty1=lly1[ap]; rectx2=max(llx1[ap],llx2[ap])}\nvallld=lld[ap]\nif(vallld=1 or vallld=3) //Have x2 or x1\n{\nif(vallld=1)for(aba=1;aba<=thln;aba+=1) //Find x1\n{\nif(thlposx[aba]=thlendx[aba] and thlposx[aba]>rectx1 and thlposx[aba]<rectx2 and (\n(min(thlposy[aba],thlendy[aba])>=min(recty1,recty2) and min(thlposy[aba],thlendy[aba])<max(recty1,recty2))\nor (max(thlposy[aba],thlendy[aba])>min(recty1,recty2) and max(thlposy[aba],thlendy[aba])<=max(recty1,recty2)))){rectx1=thlposx[aba]; aban=aba}\n}\nif(vallld=3)for(aba=1;aba<=thln;aba+=1) //Find x2\n{\nif(thlposx[aba]=thlendx[aba] and thlposx[aba]>rectx1 and thlposx[aba]<rectx2 and (\n(min(thlposy[aba],thlendy[aba])>=min(recty1,recty2) and min(thlposy[aba],thlendy[aba])<max(recty1,recty2))\nor (max(thlposy[aba],thlendy[aba])>min(recty1,recty2) and max(thlposy[aba],thlendy[aba])<=max(recty1,recty2)))){rectx2=thlposx[aba]; aban=aba}\n}\nif(rectx1!=0 and rectx2!=322)\n{\nif!(min(thlposy[aban],thlendy[aban])<=min(recty1,recty2) and max(thlposy[aban],thlendy[aban])>=max(recty1,recty2))\n{\nif(min(thlposy[aban],thlendy[aban])<=min(recty1,recty2) and max(thlposy[aban],thlendy[aban])<max(recty1,recty2) and max(thlposy[aban],thlendy[aban])>min(recty1,recty2))\n{ llc+=1; llx1[llc]=thlposx[aban]; lly1[llc]=max(thlposy[aban],thlendy[aban]); llx2[llc]=thlposx[aban]; lly2[llc]=max(recty1,recty2); lld[llc]=vallld}\nelse if(min(thlposy[aban],thlendy[aban])>min(recty1,recty2) and min(thlposy[aban],thlendy[aban])<max(recty1,recty2) and max(thlposy[aban],thlendy[aban])>=max(recty1,recty2))\n{ llc+=1; llx1[llc]=thlposx[aban]; lly1[llc]=min(recty1,recty2); llx2[llc]=thlposx[aban]; lly2[llc]=min(thlposy[aban],thlendy[aban]); lld[llc]=vallld}\nelse if(min(thlposy[aban],thlendy[aban])>min(recty1,recty2) and max(thlposy[aban],thlendy[aban])<max(recty1,recty2))\n{llc+=1; llx1[llc]=thlposx[aban]; lly1[llc]=min(recty1,recty2); llx2[llc]=thlposx[aban]; lly2[llc]=min(thlposy[aban],thlendy[aban]); lld[llc]=vallld\nllc+=1; llx1[llc]=thlposx[aban]; lly1[llc]=max(thlposy[aban],thlendy[aban]); llx2[llc]=thlposx[aban]; lly2[llc]=max(recty1,recty2); lld[llc]=vallld}\nelse {llc+=1; llx1[llc]=thlposx[aban]; lly1[llc]=min(recty1,recty2); llx2[llc]=thlposx[aban]; lly2[llc]=max(recty1,recty2); lld[llc]=vallld}\n}\nfor(aba=1;aba<=thln;aba+=1) //This should check how rectangle line fits drawn lines\n{\nif(thlposy[aba]=thlendy[aba] and thlposy[aba]=recty1 and (min(rectx1,rectx2)<max(thlposx[aba],thlendx[aba]) and max(rectx1,rectx2)>min(thlposx[aba],thlendx[aba])) or (min(rectx1,rectx2)<max(thlposx[aba],thlendx[aba]) and max(rectx1,rectx2)>=max(thlposx[aba],thlendx[aba])))\n{\nif!(min(thlposx[aba],thlendx[aba])<=min(rectx1,rectx2) and max(thlposx[aba],thlendx[aba])>=max(rectx1,rectx2))\n{ //And create a new rectangle to be checked if it is not perfect\nif(min(thlposx[aba],thlendx[aba])<=min(rectx1,rectx2) and max(thlposx[aba],thlendx[aba])<max(rectx1,rectx2) and max(thlposx[aba],thlendx[aba])>min(rectx1,rectx2))\n{llc+=1; llx1[llc]=max(thlposx[aba],thlendx[aba]); lly1[llc]=thlposy[aba]; llx2[llc]=max(rectx1,rectx2); lly2[llc]=thlposy[aba]; lld[llc]=2}\nelse if(min(thlposx[aba],thlendx[aba])>min(rectx1,rectx2) and min(thlposx[aba],thlendx[aba])<max(rectx1,rectx2) and max(thlposx[aba],thlendx[aba])>=max(rectx1,rectx2))\n{llc+=1; llx1[llc]=min(rectx1,rectx2); lly1[llc]=thlposy[aba]; llx2[llc]=min(thlposx[aba],thlendx[aba]); lly2[llc]=thlposy[aba]; lld[llc]=2}\nelse if(min(thlposx[aba],thlendx[aba])>min(rectx1,rectx2) and max(thlposx[aba],thlendx[aba])<max(rectx1,rectx2))\n{llc+=1; llx1[llc]=min(rectx1,rectx2); lly1[llc]=thlposy[aba]; llx2[llc]=min(thlposx[aba],thlendx[aba]); lly2[llc]=thlposy[aba]; lld[llc]=2\nllc+=1; llx1[llc]=max(thlposx[aba],thlendx[aba]); lly1[llc]=thlposy[aba]; llx2[llc]=max(rectx1,rectx2); lly2[llc]=thlposy[aba]; lld[llc]=2}\nelse {llc+=1; llx1[llc]=min(rectx1,rectx2); lly1[llc]=thlposy[aba]; llx2[llc]=max(rectx1,rectx2); lly2[llc]=thlposy[aba]; lld[llc]=2}\n}\n}\n}\nfor(aba=1;aba<=thln;aba+=1) //This should check how rectangle line fits drawn lines\n{\nif(thlposy[aba]=thlendy[aba] and thlposy[aba]=recty2 and (min(rectx1,rectx2)<max(thlposx[aba],thlendx[aba]) and max(rectx1,rectx2)>min(thlposx[aba],thlendx[aba])) or (min(rectx1,rectx2)<max(thlposx[aba],thlendx[aba]) and max(rectx1,rectx2)>=max(thlposx[aba],thlendx[aba])))\n{\nif!(min(thlposx[aba],thlendx[aba])<=min(rectx1,rectx2) and max(thlposx[aba],thlendx[aba])>=max(rectx1,rectx2))\n{ //And create a new rectangle to be checked if it is not perfect\nif(min(thlposx[aba],thlendx[aba])<=min(rectx1,rectx2) and max(thlposx[aba],thlendx[aba])<max(rectx1,rectx2) and max(thlposx[aba],thlendx[aba])>min(rectx1,rectx2))\n{llc+=1; llx1[llc]=max(thlposx[aba],thlendx[aba]); lly1[llc]=thlposy[aba]; llx2[llc]=max(rectx1,rectx2); lly2[llc]=thlposy[aba]; lld[llc]=4}\nelse if(min(thlposx[aba],thlendx[aba])>min(rectx1,rectx2) and min(thlposx[aba],thlendx[aba])<max(rectx1,rectx2) and max(thlposx[aba],thlendx[aba])>=max(rectx1,rectx2))\n{llc+=1; llx1[llc]=min(rectx1,rectx2); lly1[llc]=thlposy[aba]; llx2[llc]=min(thlposx[aba],thlendx[aba]); lly2[llc]=thlposy[aba]; lld[llc]=4}\nelse if(min(thlposx[aba],thlendx[aba])>min(rectx1,rectx2) and max(thlposx[aba],thlendx[aba])<max(rectx1,rectx2))\n{llc+=1; llx1[llc]=min(rectx1,rectx2); lly1[llc]=thlposy[aba]; llx2[llc]=min(thlposx[aba],thlendx[aba]); lly2[llc]=thlposy[aba]; lld[llc]=4\nllc+=1; llx1[llc]=max(thlposx[aba],thlendx[aba]); lly1[llc]=thlposy[aba]; llx2[llc]=max(rectx1,rectx2); lly2[llc]=thlposy[aba]; lld[llc]=4}\nelse {llc+=1; llx1[llc]=min(rectx1,rectx2); lly1[llc]=thlposy[aba]; llx2[llc]=max(rectx1,rectx2); lly2[llc]=thlposy[aba]; lld[llc]=4}\n}\n}\n}\n}\n}\nelse if(vallld=2 or vallld=4) //Have y2 or y1\n{\nif(vallld=2)for(aba=1;aba<=thln;aba+=1) //Find y1\n{\nif(thlposy[aba]=thlendy[aba] and thlposy[aba]>recty1 and thlposy[aba]<recty2 and (\n(min(thlposx[aba],thlendx[aba])>=min(rectx1,rectx2) and min(thlposx[aba],thlendx[aba])<max(rectx1,rectx2))\nor (max(thlposx[aba],thlendx[aba])>min(rectx1,rectx2) and max(thlposx[aba],thlendx[aba])<=max(rectx1,rectx2)))){recty1=thlposy[aba]; aban=aba}\n}\nif(vallld=4)for(aba=1;aba<=thln;aba+=1) //Find y2\n{\nif(thlposy[aba]=thlendy[aba] and thlposy[aba]>recty1 and thlposy[aba]<recty2 and (\n(min(thlposx[aba],thlendx[aba])>=min(rectx1,rectx2) and min(thlposx[aba],thlendx[aba])<max(rectx1,rectx2))\nor (max(thlposx[aba],thlendx[aba])>min(rectx1,rectx2) and max(thlposx[aba],thlendx[aba])<=max(rectx1,rectx2)))){recty2=thlposy[aba]; aban=aba}\n}\nif(recty1!=0 and recty2!=375)\n{\nif!(min(thlposx[aban],thlendx[aban])<=min(rectx1,rectx2) and max(thlposx[aban],thlendx[aban])>=max(rectx1,rectx2))\n{\nif(min(thlposx[aban],thlendx[aban])<=min(rectx1,rectx2) and max(thlposx[aban],thlendx[aban])<max(rectx1,rectx2) and max(thlposx[aban],thlendx[aban])>min(rectx1,rectx2))\n{llc+=1; llx1[llc]=max(thlposx[aban],thlendx[aban]); lly1[llc]=thlposy[aban]; llx2[llc]=max(rectx1,rectx2); lly2[llc]=thlposy[aban]; lld[llc]=vallld}\nelse if(min(thlposx[aban],thlendx[aban])>min(rectx1,rectx2) and min(thlposx[aban],thlendx[aban])<max(rectx1,rectx2) and max(thlposx[aban],thlendx[aban])>=max(rectx1,rectx2))\n{llc+=1; llx1[llc]=min(rectx1,rectx2); lly1[llc]=thlposy[aban]; llx2[llc]=min(thlposx[aban],thlendx[aban]); lly2[llc]=thlposy[aban]; lld[llc]=vallld}\nelse if(min(thlposx[aban],thlendx[aban])>min(rectx1,rectx2) and max(thlposx[aban],thlendx[aban])<max(rectx1,rectx2))\n{llc+=1; llx1[llc]=min(rectx1,rectx2); lly1[llc]=thlposy[aban]; llx2[llc]=min(thlposx[aban],thlendx[aban]); lly2[llc]=thlposy[aban]; lld[llc]=vallld\nllc+=1; llx1[llc]=max(thlposx[aban],thlendx[aban]); lly1[llc]=thlposy[aban]; llx2[llc]=max(rectx1,rectx2); lly2[llc]=thlposy[aban]; lld[llc]=vallld}\nelse {llc+=1; llx1[llc]=min(rectx1,rectx2); lly1[llc]=thlposy[aban]; llx2[llc]=max(rectx1,rectx2); lly2[llc]=thlposy[aban]; lld[llc]=vallld}\n}\nfor(aba=1;aba<=thln;aba+=1)\n{\nif(thlposx[aba]=thlendx[aba] and thlposx[aba]=rectx1 and (min(recty1,recty2)<max(thlposy[aba],thlendy[aba]) and max(recty1,recty2)>min(thlposy[aba],thlendy[aba])) or (min(recty1,recty2)<max(thlposy[aba],thlendy[aba]) and max(recty1,recty2)>=max(thlposy[aba],thlendy[aba])))\n{\nif!(min(thlposy[aba],thlendy[aba])<=min(recty1,recty2) and max(thlposy[aba],thlendy[aba])>=max(recty1,recty2))\n{\nif(min(thlposy[aba],thlendy[aba])<=min(recty1,recty2) and max(thlposy[aba],thlendy[aba])<max(recty1,recty2) and max(thlposy[aba],thlendy[aba])>min(recty1,recty2))\n{llc+=1; llx1[llc]=thlposx[aba]; lly1[llc]=max(thlposy[aba],thlendy[aba]); llx2[llc]=thlposx[aba]; lly2[llc]=max(recty1,recty2); lld[llc]=1}\nelse if(min(thlposy[aba],thlendy[aba])>min(recty1,recty2) and min(thlposy[aba],thlendy[aba])<max(recty1,recty2) and max(thlposy[aba],thlendy[aba])>=max(recty1,recty2))\n{llc+=1; llx1[llc]=thlposx[aba]; lly1[llc]=min(recty1,recty2); llx2[llc]=thlposx[aba]; lly2[llc]=min(thlposy[aba],thlendy[aba]); lld[llc]=1}\nelse if(min(thlposy[aba],thlendy[aba])>min(recty1,recty2) and max(thlposy[aba],thlendy[aba])<max(recty1,recty2))\n{llc+=1; llx1[llc]=thlposx[aba]; lly1[llc]=min(recty1,recty2); llx2[llc]=thlposx[aba]; lly2[llc]=min(thlposy[aba],thlendy[aba]); lld[llc]=1\nllc+=1; llx1[llc]=thlposx[aba]; lly1[llc]=max(thlposy[aba],thlendy[aba]); llx2[llc]=thlposx[aba]; lly2[llc]=max(recty1,recty2); lld[llc]=1}\nelse {llc+=1; llx1[llc]=thlposx[aba]; lly1[llc]=min(recty1,recty2); llx2[llc]=thlposx[aba]; lly2[llc]=max(recty1,recty2); lld[llc]=1}\n}\n}\n}\nfor(aba=1;aba<=thln;aba+=1)\n{\nif(thlposx[aba]=thlendx[aba] and thlposx[aba]=rectx2 and (min(recty1,recty2)<max(thlposy[aba],thlendy[aba]) and max(recty1,recty2)>min(thlposy[aba],thlendy[aba])) or (min(recty1,recty2)<max(thlposy[aba],thlendy[aba]) and max(recty1,recty2)>=max(thlposy[aba],thlendy[aba])))\n{\nif!(min(thlposy[aba],thlendy[aba])<=min(recty1,recty2) and max(thlposy[aba],thlendy[aba])>=max(recty1,recty2))\n{\nif(min(thlposy[aba],thlendy[aba])<=min(recty1,recty2) and max(thlposy[aba],thlendy[aba])<max(recty1,recty2) and max(thlposy[aba],thlendy[aba])>min(recty1,recty2))\n{llc+=1; llx1[llc]=thlposx[aba]; lly1[llc]=max(thlposy[aba],thlendy[aba]); llx2[llc]=thlposx[aba]; lly2[llc]=max(recty1,recty2); lld[llc]=3}\nelse if(min(thlposy[aba],thlendy[aba])>min(recty1,recty2) and min(thlposy[aba],thlendy[aba])<max(recty1,recty2) and max(thlposy[aba],thlendy[aba])>=max(recty1,recty2))\n{llc+=1; llx1[llc]=thlposx[aba]; lly1[llc]=min(recty1,recty2); llx2[llc]=thlposx[aba]; lly2[llc]=min(thlposy[aba],thlendy[aba]); lld[llc]=3}\nelse if(min(thlposy[aba],thlendy[aba])>min(recty1,recty2) and max(thlposy[aba],thlendy[aba])<max(recty1,recty2))\n{llc+=1; llx1[llc]=thlposx[aba]; lly1[llc]=min(recty1,recty2); llx2[llc]=thlposx[aba]; lly2[llc]=min(thlposy[aba],thlendy[aba]); lld[llc]=3\nllc+=1; llx1[llc]=thlposx[aba]; lly1[llc]=max(thlposy[aba],thlendy[aba]); llx2[llc]=thlposx[aba]; lly2[llc]=max(recty1,recty2); lld[llc]=3}\nelse {llc+=1; llx1[llc]=thlposx[aba]; lly1[llc]=min(recty1,recty2); llx2[llc]=thlposx[aba]; lly2[llc]=max(recty1,recty2); lld[llc]=3}\n}\n}\n}\n}\n//End of rectangle creating\n``````\nBe a part of the DaniWeb community\n\nWe're a friendly, industry-focused community of 1.20 million developers, IT pros, digital marketers, and technology enthusiasts learning and sharing knowledge." ]	[ null, "https://static.daniweb.com/attachments/4/ee79fb47b58a1cc436fb26c8e9d2bd04.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87159204,"math_prob":0.9850967,"size":4434,"snap":"2020-45-2020-50","text_gpt3_token_len":1439,"char_repetition_ratio":0.13656884,"word_repetition_ratio":0.12760055,"special_character_ratio":0.2947677,"punctuation_ratio":0.09860558,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9863222,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-26T23:41:12Z\",\"WARC-Record-ID\":\"<urn:uuid:3a9862e1-2069-4613-9589-67c37a4c3351>\",\"Content-Length\":\"144270\",\"Content-Type\":\"application/http; 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https://www.directhardwaresupply.com/store/page/30/?filter_product-group=memory&query_type_product-group=or	[ "Showing 465–480 of 897 results\n\n'; for (i = 0; i < msgs.length; i++) { alert_box += '' + msgs[i] + '\n'; if(form != '') { form.closest('form').append(alert_box); } else { \\$('#form-5e8136f606281').append(alert_box); } } function addConditionClass(field_id, cond_class, form_fields_wrapper) { \\$(field_id).each(function(){ if (\\$(this).is(':input') \|\| \\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class); \\$(this).children().each(function(){ addConditionClass(\\$(this), cond_class, form_fields_wrapper); }) }); return false; } function compareRule(objs, cmp_operator, cmp_value, cmp_id, \\$form_part_0) { var comp_res = false; var areOperandsCb = false; // Stores true if both operands are checkboxes. switch(cmp_operator) { case 'is': if (cmp_value.startsWith('Checkbox_')) { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { break; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } \\$(objs).each(function(){ if (areOperandsCb) { comp_res = false; } \\$cmp1 = \\$(this).val(); \\$(test).each(function(){ \\$cmp2 = \\$(this).val(); if (\\$cmp1 == \\$cmp2) { comp_res = true; if (!areOperandsCb) { return; } } }); if (areOperandsCb && false == comp_res) { return; } }); break; case 'is-not': if (cmp_value.startsWith('Checkbox_')) { test = \\$form_part_0.find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { return true; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } for(let objsElement of objs) { comp_res = false; \\$cmp1 = \\$(objsElement).val(); for(let testElement of test) { \\$cmp2 = \\$(testElement).val(); if (\\$cmp1 != \\$cmp2) { comp_res = true; // return; } else if(areOperandsCb) { comp_res = false; break; } } if(areOperandsCb && true == comp_res) { break; } } break; case 'less-than': \\$(objs).each(function(){ // Return if current element is non-relevant input field inside 'Rating' field. if ('undefined' != typeof \\$(this).attr('id') && 'Rating_' != \\$(this).attr('id').match(/^Rating_/) && \\$(this).closest('div[id^=Rating_]')) { return; } // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() < cmp_value) { comp_res = true; return; } }); break; case 'greater-than': \\$(objs).each(function(){ // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() > cmp_value) { comp_res = true; return; } }); break; case 'starts-with': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) == 0) { comp_res = true; return; } }); break; case 'contains': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) != -1) { comp_res = true; return; } }); break; case 'ends-with': \\$(objs).each(function(){ indexPoint = (\\$(this).val().length - cmp_value.length); if (indexPoint >=0 && \\$(this).val().indexOf(cmp_value, indexPoint) == indexPoint) { comp_res = true; return; } }); break; default: comp_res = false; break; } return comp_res; } function applyRule(field_id) { \\$('.cond_filler_'+field_id).each(function(){ var this_conditions = \\$('#'+field_id).attr('data-cond-fields').split('\|'); var this_action = \\$('#'+field_id).attr('data-cond-action').split(':'); var cmp_res = this_action == 'all' ? true : false; for (i=0 ; i\n'; for (i = 0; i < msgs.length; i++) { alert_box += '' + msgs[i] + '\n'; if(form != '') { form.closest('form').append(alert_box); } else { \\$('#form-5e8136f607c27').append(alert_box); } } function addConditionClass(field_id, cond_class, form_fields_wrapper) { \\$(field_id).each(function(){ if (\\$(this).is(':input') \|\| \\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class); \\$(this).children().each(function(){ addConditionClass(\\$(this), cond_class, form_fields_wrapper); }) }); return false; } function compareRule(objs, cmp_operator, cmp_value, cmp_id, \\$form_part_0) { var comp_res = false; var areOperandsCb = false; // Stores true if both operands are checkboxes. switch(cmp_operator) { case 'is': if (cmp_value.startsWith('Checkbox_')) { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { break; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } \\$(objs).each(function(){ if (areOperandsCb) { comp_res = false; } \\$cmp1 = \\$(this).val(); \\$(test).each(function(){ \\$cmp2 = \\$(this).val(); if (\\$cmp1 == \\$cmp2) { comp_res = true; if (!areOperandsCb) { return; } } }); if (areOperandsCb && false == comp_res) { return; } }); break; case 'is-not': if (cmp_value.startsWith('Checkbox_')) { test = \\$form_part_0.find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { return true; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } for(let objsElement of objs) { comp_res = false; \\$cmp1 = \\$(objsElement).val(); for(let testElement of test) { \\$cmp2 = \\$(testElement).val(); if (\\$cmp1 != \\$cmp2) { comp_res = true; // return; } else if(areOperandsCb) { comp_res = false; break; } } if(areOperandsCb && true == comp_res) { break; } } break; case 'less-than': \\$(objs).each(function(){ // Return if current element is non-relevant input field inside 'Rating' field. if ('undefined' != typeof \\$(this).attr('id') && 'Rating_' != \\$(this).attr('id').match(/^Rating_/) && \\$(this).closest('div[id^=Rating_]')) { return; } // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() < cmp_value) { comp_res = true; return; } }); break; case 'greater-than': \\$(objs).each(function(){ // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() > cmp_value) { comp_res = true; return; } }); break; case 'starts-with': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) == 0) { comp_res = true; return; } }); break; case 'contains': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) != -1) { comp_res = true; return; } }); break; case 'ends-with': \\$(objs).each(function(){ indexPoint = (\\$(this).val().length - cmp_value.length); if (indexPoint >=0 && \\$(this).val().indexOf(cmp_value, indexPoint) == indexPoint) { comp_res = true; return; } }); break; default: comp_res = false; break; } return comp_res; } function applyRule(field_id) { \\$('.cond_filler_'+field_id).each(function(){ var this_conditions = \\$('#'+field_id).attr('data-cond-fields').split('\|'); var this_action = \\$('#'+field_id).attr('data-cond-action').split(':'); var cmp_res = this_action == 'all' ? true : false; for (i=0 ; i\n'; for (i = 0; i < msgs.length; i++) { alert_box += '' + msgs[i] + '\n'; if(form != '') { form.closest('form').append(alert_box); } else { \\$('#form-5e8136f609619').append(alert_box); } } function addConditionClass(field_id, cond_class, form_fields_wrapper) { \\$(field_id).each(function(){ if (\\$(this).is(':input') \|\| \\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class); \\$(this).children().each(function(){ addConditionClass(\\$(this), cond_class, form_fields_wrapper); }) }); return false; } function compareRule(objs, cmp_operator, cmp_value, cmp_id, \\$form_part_0) { var comp_res = false; var areOperandsCb = false; // Stores true if both operands are checkboxes. switch(cmp_operator) { case 'is': if (cmp_value.startsWith('Checkbox_')) { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { break; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } \\$(objs).each(function(){ if (areOperandsCb) { comp_res = false; } \\$cmp1 = \\$(this).val(); \\$(test).each(function(){ \\$cmp2 = \\$(this).val(); if (\\$cmp1 == \\$cmp2) { comp_res = true; if (!areOperandsCb) { return; } } }); if (areOperandsCb && false == comp_res) { return; } }); break; case 'is-not': if (cmp_value.startsWith('Checkbox_')) { test = \\$form_part_0.find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { return true; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } for(let objsElement of objs) { comp_res = false; \\$cmp1 = \\$(objsElement).val(); for(let testElement of test) { \\$cmp2 = \\$(testElement).val(); if (\\$cmp1 != \\$cmp2) { comp_res = true; // return; } else if(areOperandsCb) { comp_res = false; break; } } if(areOperandsCb && true == comp_res) { break; } } break; case 'less-than': \\$(objs).each(function(){ // Return if current element is non-relevant input field inside 'Rating' field. if ('undefined' != typeof \\$(this).attr('id') && 'Rating_' != \\$(this).attr('id').match(/^Rating_/) && \\$(this).closest('div[id^=Rating_]')) { return; } // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() < cmp_value) { comp_res = true; return; } }); break; case 'greater-than': \\$(objs).each(function(){ // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() > cmp_value) { comp_res = true; return; } }); break; case 'starts-with': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) == 0) { comp_res = true; return; } }); break; case 'contains': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) != -1) { comp_res = true; return; } }); break; case 'ends-with': \\$(objs).each(function(){ indexPoint = (\\$(this).val().length - cmp_value.length); if (indexPoint >=0 && \\$(this).val().indexOf(cmp_value, indexPoint) == indexPoint) { comp_res = true; return; } }); break; default: comp_res = false; break; } return comp_res; } function applyRule(field_id) { \\$('.cond_filler_'+field_id).each(function(){ var this_conditions = \\$('#'+field_id).attr('data-cond-fields').split('\|'); var this_action = \\$('#'+field_id).attr('data-cond-action').split(':'); var cmp_res = this_action == 'all' ? true : false; for (i=0 ; i\n'; for (i = 0; i < msgs.length; i++) { alert_box += '' + msgs[i] + '\n'; if(form != '') { form.closest('form').append(alert_box); } else { \\$('#form-5e8136f60ad68').append(alert_box); } } function addConditionClass(field_id, cond_class, form_fields_wrapper) { \\$(field_id).each(function(){ if (\\$(this).is(':input') \|\| \\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class); \\$(this).children().each(function(){ addConditionClass(\\$(this), cond_class, form_fields_wrapper); }) }); return false; } function compareRule(objs, cmp_operator, cmp_value, cmp_id, \\$form_part_0) { var comp_res = false; var areOperandsCb = false; // Stores true if both operands are checkboxes. switch(cmp_operator) { case 'is': if (cmp_value.startsWith('Checkbox_')) { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { break; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } \\$(objs).each(function(){ if (areOperandsCb) { comp_res = false; } \\$cmp1 = \\$(this).val(); \\$(test).each(function(){ \\$cmp2 = \\$(this).val(); if (\\$cmp1 == \\$cmp2) { comp_res = true; if (!areOperandsCb) { return; } } }); if (areOperandsCb && false == comp_res) { return; } }); break; case 'is-not': if (cmp_value.startsWith('Checkbox_')) { test = \\$form_part_0.find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { return true; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } for(let objsElement of objs) { comp_res = false; \\$cmp1 = \\$(objsElement).val(); for(let testElement of test) { \\$cmp2 = \\$(testElement).val(); if (\\$cmp1 != \\$cmp2) { comp_res = true; // return; } else if(areOperandsCb) { comp_res = false; break; } } if(areOperandsCb && true == comp_res) { break; } } break; case 'less-than': \\$(objs).each(function(){ // Return if current element is non-relevant input field inside 'Rating' field. if ('undefined' != typeof \\$(this).attr('id') && 'Rating_' != \\$(this).attr('id').match(/^Rating_/) && \\$(this).closest('div[id^=Rating_]')) { return; } // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() < cmp_value) { comp_res = true; return; } }); break; case 'greater-than': \\$(objs).each(function(){ // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() > cmp_value) { comp_res = true; return; } }); break; case 'starts-with': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) == 0) { comp_res = true; return; } }); break; case 'contains': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) != -1) { comp_res = true; return; } }); break; case 'ends-with': \\$(objs).each(function(){ indexPoint = (\\$(this).val().length - cmp_value.length); if (indexPoint >=0 && \\$(this).val().indexOf(cmp_value, indexPoint) == indexPoint) { comp_res = true; return; } }); break; default: comp_res = false; break; } return comp_res; } function applyRule(field_id) { \\$('.cond_filler_'+field_id).each(function(){ var this_conditions = \\$('#'+field_id).attr('data-cond-fields').split('\|'); var this_action = \\$('#'+field_id).attr('data-cond-action').split(':'); var cmp_res = this_action == 'all' ? true : false; for (i=0 ; i\n'; for (i = 0; i < msgs.length; i++) { alert_box += '' + msgs[i] + '\n'; if(form != '') { form.closest('form').append(alert_box); } else { \\$('#form-5e8136f60c6dc').append(alert_box); } } function addConditionClass(field_id, cond_class, form_fields_wrapper) { \\$(field_id).each(function(){ if (\\$(this).is(':input') \|\| \\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class); \\$(this).children().each(function(){ addConditionClass(\\$(this), cond_class, form_fields_wrapper); }) }); return false; } function compareRule(objs, cmp_operator, cmp_value, cmp_id, \\$form_part_0) { var comp_res = false; var areOperandsCb = false; // Stores true if both operands are checkboxes. switch(cmp_operator) { case 'is': if (cmp_value.startsWith('Checkbox_')) { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { break; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } \\$(objs).each(function(){ if (areOperandsCb) { comp_res = false; } \\$cmp1 = \\$(this).val(); \\$(test).each(function(){ \\$cmp2 = \\$(this).val(); if (\\$cmp1 == \\$cmp2) { comp_res = true; if (!areOperandsCb) { return; } } }); if (areOperandsCb && false == comp_res) { return; } }); break; case 'is-not': if (cmp_value.startsWith('Checkbox_')) { test = \\$form_part_0.find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { return true; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } for(let objsElement of objs) { comp_res = false; \\$cmp1 = \\$(objsElement).val(); for(let testElement of test) { \\$cmp2 = \\$(testElement).val(); if (\\$cmp1 != \\$cmp2) { comp_res = true; // return; } else if(areOperandsCb) { comp_res = false; break; } } if(areOperandsCb && true == comp_res) { break; } } break; case 'less-than': \\$(objs).each(function(){ // Return if current element is non-relevant input field inside 'Rating' field. if ('undefined' != typeof \\$(this).attr('id') && 'Rating_' != \\$(this).attr('id').match(/^Rating_/) && \\$(this).closest('div[id^=Rating_]')) { return; } // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() < cmp_value) { comp_res = true; return; } }); break; case 'greater-than': \\$(objs).each(function(){ // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() > cmp_value) { comp_res = true; return; } }); break; case 'starts-with': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) == 0) { comp_res = true; return; } }); break; case 'contains': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) != -1) { comp_res = true; return; } }); break; case 'ends-with': \\$(objs).each(function(){ indexPoint = (\\$(this).val().length - cmp_value.length); if (indexPoint >=0 && \\$(this).val().indexOf(cmp_value, indexPoint) == indexPoint) { comp_res = true; return; } }); break; default: comp_res = false; break; } return comp_res; } function applyRule(field_id) { \\$('.cond_filler_'+field_id).each(function(){ var this_conditions = \\$('#'+field_id).attr('data-cond-fields').split('\|'); var this_action = \\$('#'+field_id).attr('data-cond-action').split(':'); var cmp_res = this_action == 'all' ? true : false; for (i=0 ; i\n'; for (i = 0; i < msgs.length; i++) { alert_box += '' + msgs[i] + '\n'; if(form != '') { form.closest('form').append(alert_box); } else { \\$('#form-5e8136f60e066').append(alert_box); } } function addConditionClass(field_id, cond_class, form_fields_wrapper) { \\$(field_id).each(function(){ if (\\$(this).is(':input') \|\| \\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class); \\$(this).children().each(function(){ addConditionClass(\\$(this), cond_class, form_fields_wrapper); }) }); return false; } function compareRule(objs, cmp_operator, cmp_value, cmp_id, \\$form_part_0) { var comp_res = false; var areOperandsCb = false; // Stores true if both operands are checkboxes. switch(cmp_operator) { case 'is': if (cmp_value.startsWith('Checkbox_')) { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { break; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } \\$(objs).each(function(){ if (areOperandsCb) { comp_res = false; } \\$cmp1 = \\$(this).val(); \\$(test).each(function(){ \\$cmp2 = \\$(this).val(); if (\\$cmp1 == \\$cmp2) { comp_res = true; if (!areOperandsCb) { return; } } }); if (areOperandsCb && false == comp_res) { return; } }); break; case 'is-not': if (cmp_value.startsWith('Checkbox_')) { test = \\$form_part_0.find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { return true; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } for(let objsElement of objs) { comp_res = false; \\$cmp1 = \\$(objsElement).val(); for(let testElement of test) { \\$cmp2 = \\$(testElement).val(); if (\\$cmp1 != \\$cmp2) { comp_res = true; // return; } else if(areOperandsCb) { comp_res = false; break; } } if(areOperandsCb && true == comp_res) { break; } } break; case 'less-than': \\$(objs).each(function(){ // Return if current element is non-relevant input field inside 'Rating' field. if ('undefined' != typeof \\$(this).attr('id') && 'Rating_' != \\$(this).attr('id').match(/^Rating_/) && \\$(this).closest('div[id^=Rating_]')) { return; } // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() < cmp_value) { comp_res = true; return; } }); break; case 'greater-than': \\$(objs).each(function(){ // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() > cmp_value) { comp_res = true; return; } }); break; case 'starts-with': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) == 0) { comp_res = true; return; } }); break; case 'contains': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) != -1) { comp_res = true; return; } }); break; case 'ends-with': \\$(objs).each(function(){ indexPoint = (\\$(this).val().length - cmp_value.length); if (indexPoint >=0 && \\$(this).val().indexOf(cmp_value, indexPoint) == indexPoint) { comp_res = true; return; } }); break; default: comp_res = false; break; } return comp_res; } function applyRule(field_id) { \\$('.cond_filler_'+field_id).each(function(){ var this_conditions = \\$('#'+field_id).attr('data-cond-fields').split('\|'); var this_action = \\$('#'+field_id).attr('data-cond-action').split(':'); var cmp_res = this_action == 'all' ? true : false; for (i=0 ; i\n'; for (i = 0; i < msgs.length; i++) { alert_box += '' + msgs[i] + '\n'; if(form != '') { form.closest('form').append(alert_box); } else { \\$('#form-5e8136f60fd36').append(alert_box); } } function addConditionClass(field_id, cond_class, form_fields_wrapper) { \\$(field_id).each(function(){ if (\\$(this).is(':input') \|\| \\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class); \\$(this).children().each(function(){ addConditionClass(\\$(this), cond_class, form_fields_wrapper); }) }); return false; } function compareRule(objs, cmp_operator, cmp_value, cmp_id, \\$form_part_0) { var comp_res = false; var areOperandsCb = false; // Stores true if both operands are checkboxes. switch(cmp_operator) { case 'is': if (cmp_value.startsWith('Checkbox_')) { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { break; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } \\$(objs).each(function(){ if (areOperandsCb) { comp_res = false; } \\$cmp1 = \\$(this).val(); \\$(test).each(function(){ \\$cmp2 = \\$(this).val(); if (\\$cmp1 == \\$cmp2) { comp_res = true; if (!areOperandsCb) { return; } } }); if (areOperandsCb && false == comp_res) { return; } }); break; case 'is-not': if (cmp_value.startsWith('Checkbox_')) { test = \\$form_part_0.find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { return true; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } for(let objsElement of objs) { comp_res = false; \\$cmp1 = \\$(objsElement).val(); for(let testElement of test) { \\$cmp2 = \\$(testElement).val(); if (\\$cmp1 != \\$cmp2) { comp_res = true; // return; } else if(areOperandsCb) { comp_res = false; break; } } if(areOperandsCb && true == comp_res) { break; } } break; case 'less-than': \\$(objs).each(function(){ // Return if current element is non-relevant input field inside 'Rating' field. if ('undefined' != typeof \\$(this).attr('id') && 'Rating_' != \\$(this).attr('id').match(/^Rating_/) && \\$(this).closest('div[id^=Rating_]')) { return; } // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() < cmp_value) { comp_res = true; return; } }); break; case 'greater-than': \\$(objs).each(function(){ // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() > cmp_value) { comp_res = true; return; } }); break; case 'starts-with': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) == 0) { comp_res = true; return; } }); break; case 'contains': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) != -1) { comp_res = true; return; } }); break; case 'ends-with': \\$(objs).each(function(){ indexPoint = (\\$(this).val().length - cmp_value.length); if (indexPoint >=0 && \\$(this).val().indexOf(cmp_value, indexPoint) == indexPoint) { comp_res = true; return; } }); break; default: comp_res = false; break; } return comp_res; } function applyRule(field_id) { \\$('.cond_filler_'+field_id).each(function(){ var this_conditions = \\$('#'+field_id).attr('data-cond-fields').split('\|'); var this_action = \\$('#'+field_id).attr('data-cond-action').split(':'); var cmp_res = this_action == 'all' ? true : false; for (i=0 ; i\n'; for (i = 0; i < msgs.length; i++) { alert_box += '' + msgs[i] + '\n'; if(form != '') { form.closest('form').append(alert_box); } else { \\$('#form-5e8136f6114f6').append(alert_box); } } function addConditionClass(field_id, cond_class, form_fields_wrapper) { \\$(field_id).each(function(){ if (\\$(this).is(':input') \|\| \\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class); \\$(this).children().each(function(){ addConditionClass(\\$(this), cond_class, form_fields_wrapper); }) }); return false; } function compareRule(objs, cmp_operator, cmp_value, cmp_id, \\$form_part_0) { var comp_res = false; var areOperandsCb = false; // Stores true if both operands are checkboxes. switch(cmp_operator) { case 'is': if (cmp_value.startsWith('Checkbox_')) { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { break; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } \\$(objs).each(function(){ if (areOperandsCb) { comp_res = false; } \\$cmp1 = \\$(this).val(); \\$(test).each(function(){ \\$cmp2 = \\$(this).val(); if (\\$cmp1 == \\$cmp2) { comp_res = true; if (!areOperandsCb) { return; } } }); if (areOperandsCb && false == comp_res) { return; } }); break; case 'is-not': if (cmp_value.startsWith('Checkbox_')) { test = \\$form_part_0.find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { return true; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } for(let objsElement of objs) { comp_res = false; \\$cmp1 = \\$(objsElement).val(); for(let testElement of test) { \\$cmp2 = \\$(testElement).val(); if (\\$cmp1 != \\$cmp2) { comp_res = true; // return; } else if(areOperandsCb) { comp_res = false; break; } } if(areOperandsCb && true == comp_res) { break; } } break; case 'less-than': \\$(objs).each(function(){ // Return if current element is non-relevant input field inside 'Rating' field. if ('undefined' != typeof \\$(this).attr('id') && 'Rating_' != \\$(this).attr('id').match(/^Rating_/) && \\$(this).closest('div[id^=Rating_]')) { return; } // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() < cmp_value) { comp_res = true; return; } }); break; case 'greater-than': \\$(objs).each(function(){ // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() > cmp_value) { comp_res = true; return; } }); break; case 'starts-with': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) == 0) { comp_res = true; return; } }); break; case 'contains': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) != -1) { comp_res = true; return; } }); break; case 'ends-with': \\$(objs).each(function(){ indexPoint = (\\$(this).val().length - cmp_value.length); if (indexPoint >=0 && \\$(this).val().indexOf(cmp_value, indexPoint) == indexPoint) { comp_res = true; return; } }); break; default: comp_res = false; break; } return comp_res; } function applyRule(field_id) { \\$('.cond_filler_'+field_id).each(function(){ var this_conditions = \\$('#'+field_id).attr('data-cond-fields').split('\|'); var this_action = \\$('#'+field_id).attr('data-cond-action').split(':'); var cmp_res = this_action == 'all' ? true : false; for (i=0 ; i\n'; for (i = 0; i < msgs.length; i++) { alert_box += '' + msgs[i] + '\n'; if(form != '') { form.closest('form').append(alert_box); } else { \\$('#form-5e8136f612f1a').append(alert_box); } } function addConditionClass(field_id, cond_class, form_fields_wrapper) { \\$(field_id).each(function(){ if (\\$(this).is(':input') \|\| \\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class); \\$(this).children().each(function(){ addConditionClass(\\$(this), cond_class, form_fields_wrapper); }) }); return false; } function compareRule(objs, cmp_operator, cmp_value, cmp_id, \\$form_part_0) { var comp_res = false; var areOperandsCb = false; // Stores true if both operands are checkboxes. switch(cmp_operator) { case 'is': if (cmp_value.startsWith('Checkbox_')) { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { break; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } \\$(objs).each(function(){ if (areOperandsCb) { comp_res = false; } \\$cmp1 = \\$(this).val(); \\$(test).each(function(){ \\$cmp2 = \\$(this).val(); if (\\$cmp1 == \\$cmp2) { comp_res = true; if (!areOperandsCb) { return; } } }); if (areOperandsCb && false == comp_res) { return; } }); break; case 'is-not': if (cmp_value.startsWith('Checkbox_')) { test = \\$form_part_0.find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { return true; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } for(let objsElement of objs) { comp_res = false; \\$cmp1 = \\$(objsElement).val(); for(let testElement of test) { \\$cmp2 = \\$(testElement).val(); if (\\$cmp1 != \\$cmp2) { comp_res = true; // return; } else if(areOperandsCb) { comp_res = false; break; } } if(areOperandsCb && true == comp_res) { break; } } break; case 'less-than': \\$(objs).each(function(){ // Return if current element is non-relevant input field inside 'Rating' field. if ('undefined' != typeof \\$(this).attr('id') && 'Rating_' != \\$(this).attr('id').match(/^Rating_/) && \\$(this).closest('div[id^=Rating_]')) { return; } // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() < cmp_value) { comp_res = true; return; } }); break; case 'greater-than': \\$(objs).each(function(){ // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() > cmp_value) { comp_res = true; return; } }); break; case 'starts-with': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) == 0) { comp_res = true; return; } }); break; case 'contains': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) != -1) { comp_res = true; return; } }); break; case 'ends-with': \\$(objs).each(function(){ indexPoint = (\\$(this).val().length - cmp_value.length); if (indexPoint >=0 && \\$(this).val().indexOf(cmp_value, indexPoint) == indexPoint) { comp_res = true; return; } }); break; default: comp_res = false; break; } return comp_res; } function applyRule(field_id) { \\$('.cond_filler_'+field_id).each(function(){ var this_conditions = \\$('#'+field_id).attr('data-cond-fields').split('\|'); var this_action = \\$('#'+field_id).attr('data-cond-action').split(':'); var cmp_res = this_action == 'all' ? true : false; for (i=0 ; i\n'; for (i = 0; i < msgs.length; i++) { alert_box += '' + msgs[i] + '\n'; if(form != '') { form.closest('form').append(alert_box); } else { \\$('#form-5e8136f6148de').append(alert_box); } } function addConditionClass(field_id, cond_class, form_fields_wrapper) { \\$(field_id).each(function(){ if (\\$(this).is(':input') \|\| \\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class); \\$(this).children().each(function(){ addConditionClass(\\$(this), cond_class, form_fields_wrapper); }) }); return false; } function compareRule(objs, cmp_operator, cmp_value, cmp_id, \\$form_part_0) { var comp_res = false; var areOperandsCb = false; // Stores true if both operands are checkboxes. switch(cmp_operator) { case 'is': if (cmp_value.startsWith('Checkbox_')) { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { break; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } \\$(objs).each(function(){ if (areOperandsCb) { comp_res = false; } \\$cmp1 = \\$(this).val(); \\$(test).each(function(){ \\$cmp2 = \\$(this).val(); if (\\$cmp1 == \\$cmp2) { comp_res = true; if (!areOperandsCb) { return; } } }); if (areOperandsCb && false == comp_res) { return; } }); break; case 'is-not': if (cmp_value.startsWith('Checkbox_')) { test = \\$form_part_0.find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { return true; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } for(let objsElement of objs) { comp_res = false; \\$cmp1 = \\$(objsElement).val(); for(let testElement of test) { \\$cmp2 = \\$(testElement).val(); if (\\$cmp1 != \\$cmp2) { comp_res = true; // return; } else if(areOperandsCb) { comp_res = false; break; } } if(areOperandsCb && true == comp_res) { break; } } break; case 'less-than': \\$(objs).each(function(){ // Return if current element is non-relevant input field inside 'Rating' field. if ('undefined' != typeof \\$(this).attr('id') && 'Rating_' != \\$(this).attr('id').match(/^Rating_/) && \\$(this).closest('div[id^=Rating_]')) { return; } // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() < cmp_value) { comp_res = true; return; } }); break; case 'greater-than': \\$(objs).each(function(){ // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() > cmp_value) { comp_res = true; return; } }); break; case 'starts-with': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) == 0) { comp_res = true; return; } }); break; case 'contains': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) != -1) { comp_res = true; return; } }); break; case 'ends-with': \\$(objs).each(function(){ indexPoint = (\\$(this).val().length - cmp_value.length); if (indexPoint >=0 && \\$(this).val().indexOf(cmp_value, indexPoint) == indexPoint) { comp_res = true; return; } }); break; default: comp_res = false; break; } return comp_res; } function applyRule(field_id) { \\$('.cond_filler_'+field_id).each(function(){ var this_conditions = \\$('#'+field_id).attr('data-cond-fields').split('\|'); var this_action = \\$('#'+field_id).attr('data-cond-action').split(':'); var cmp_res = this_action == 'all' ? true : false; for (i=0 ; i\n'; for (i = 0; i < msgs.length; i++) { alert_box += '' + msgs[i] + '\n'; if(form != '') { form.closest('form').append(alert_box); } else { \\$('#form-5e8136f6162df').append(alert_box); } } function addConditionClass(field_id, cond_class, form_fields_wrapper) { \\$(field_id).each(function(){ if (\\$(this).is(':input') \|\| \\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class); \\$(this).children().each(function(){ addConditionClass(\\$(this), cond_class, form_fields_wrapper); }) }); return false; } function compareRule(objs, cmp_operator, cmp_value, cmp_id, \\$form_part_0) { var comp_res = false; var areOperandsCb = false; // Stores true if both operands are checkboxes. switch(cmp_operator) { case 'is': if (cmp_value.startsWith('Checkbox_')) { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { break; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } \\$(objs).each(function(){ if (areOperandsCb) { comp_res = false; } \\$cmp1 = \\$(this).val(); \\$(test).each(function(){ \\$cmp2 = \\$(this).val(); if (\\$cmp1 == \\$cmp2) { comp_res = true; if (!areOperandsCb) { return; } } }); if (areOperandsCb && false == comp_res) { return; } }); break; case 'is-not': if (cmp_value.startsWith('Checkbox_')) { test = \\$form_part_0.find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { return true; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } for(let objsElement of objs) { comp_res = false; \\$cmp1 = \\$(objsElement).val(); for(let testElement of test) { \\$cmp2 = \\$(testElement).val(); if (\\$cmp1 != \\$cmp2) { comp_res = true; // return; } else if(areOperandsCb) { comp_res = false; break; } } if(areOperandsCb && true == comp_res) { break; } } break; case 'less-than': \\$(objs).each(function(){ // Return if current element is non-relevant input field inside 'Rating' field. if ('undefined' != typeof \\$(this).attr('id') && 'Rating_' != \\$(this).attr('id').match(/^Rating_/) && \\$(this).closest('div[id^=Rating_]')) { return; } // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() < cmp_value) { comp_res = true; return; } }); break; case 'greater-than': \\$(objs).each(function(){ // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() > cmp_value) { comp_res = true; return; } }); break; case 'starts-with': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) == 0) { comp_res = true; return; } }); break; case 'contains': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) != -1) { comp_res = true; return; } }); break; case 'ends-with': \\$(objs).each(function(){ indexPoint = (\\$(this).val().length - cmp_value.length); if (indexPoint >=0 && \\$(this).val().indexOf(cmp_value, indexPoint) == indexPoint) { comp_res = true; return; } }); break; default: comp_res = false; break; } return comp_res; } function applyRule(field_id) { \\$('.cond_filler_'+field_id).each(function(){ var this_conditions = \\$('#'+field_id).attr('data-cond-fields').split('\|'); var this_action = \\$('#'+field_id).attr('data-cond-action').split(':'); var cmp_res = this_action == 'all' ? true : false; for (i=0 ; i\n'; for (i = 0; i < msgs.length; i++) { alert_box += '' + msgs[i] + '\n'; if(form != '') { form.closest('form').append(alert_box); } else { \\$('#form-5e8136f617d3c').append(alert_box); } } function addConditionClass(field_id, cond_class, form_fields_wrapper) { \\$(field_id).each(function(){ if (\\$(this).is(':input') \|\| \\$(this).is('select')) \\$(this).addClass('cond_filler_'+cond_class); \\$(this).children().each(function(){ addConditionClass(\\$(this), cond_class, form_fields_wrapper); }) }); return false; } function compareRule(objs, cmp_operator, cmp_value, cmp_id, \\$form_part_0) { var comp_res = false; var areOperandsCb = false; // Stores true if both operands are checkboxes. switch(cmp_operator) { case 'is': if (cmp_value.startsWith('Checkbox_')) { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { break; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } \\$(objs).each(function(){ if (areOperandsCb) { comp_res = false; } \\$cmp1 = \\$(this).val(); \\$(test).each(function(){ \\$cmp2 = \\$(this).val(); if (\\$cmp1 == \\$cmp2) { comp_res = true; if (!areOperandsCb) { return; } } }); if (areOperandsCb && false == comp_res) { return; } }); break; case 'is-not': if (cmp_value.startsWith('Checkbox_')) { test = \\$form_part_0.find('#'+cmp_value+' :input:checked'); areOperandsCb = cmp_id.startsWith('Checkbox_') ? true : false; if (areOperandsCb && objs.length != test.length) { return true; } } else { test = objs.closest('#form_part_0').find('#'+cmp_value+' :input'); } for(let objsElement of objs) { comp_res = false; \\$cmp1 = \\$(objsElement).val(); for(let testElement of test) { \\$cmp2 = \\$(testElement).val(); if (\\$cmp1 != \\$cmp2) { comp_res = true; // return; } else if(areOperandsCb) { comp_res = false; break; } } if(areOperandsCb && true == comp_res) { break; } } break; case 'less-than': \\$(objs).each(function(){ // Return if current element is non-relevant input field inside 'Rating' field. if ('undefined' != typeof \\$(this).attr('id') && 'Rating_' != \\$(this).attr('id').match(/^Rating_/) && \\$(this).closest('div[id^=Rating_]')) { return; } // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() < cmp_value) { comp_res = true; return; } }); break; case 'greater-than': \\$(objs).each(function(){ // if cmp_value is number, convert it into number type data. if (!isNaN(cmp_value)) { cmp_value = Number(cmp_value); } if (\\$(this).val() > cmp_value) { comp_res = true; return; } }); break; case 'starts-with': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) == 0) { comp_res = true; return; } }); break; case 'contains': \\$(objs).each(function(){ if (\\$(this).val().indexOf(cmp_value) != -1) { comp_res = true; return; } }); break; case 'ends-with': \\$(objs).each(function(){ indexPoint = (\\$(this).val().length - cmp_value.length); if (indexPoint >=0 && \\$(this).val().indexOf(cmp_value, indexPoint) == indexPoint) { comp_res = true; return; } }); break; default: comp_res = false; break; } return comp_res; } function applyRule(field_id) { \\$('.cond_filler_'+field_id).each(function(){ var this_conditions = \\$('#'+field_id).attr('data-cond-fields').split('\|'); var this_action = \\$('#'+field_id).attr('data-cond-action').split(':'); var cmp_res = this_action == 'all' ? true : false; for (i=0 ; i\n'; for (i = 0; i < msgs.length; i++) { alert_box += '' + msgs[i] + '" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.905009,"math_prob":0.9946754,"size":304,"snap":"2020-10-2020-16","text_gpt3_token_len":95,"char_repetition_ratio":0.10333333,"word_repetition_ratio":0.115384616,"special_character_ratio":0.34868422,"punctuation_ratio":0.04918033,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9967659,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-30T00:05:57Z\",\"WARC-Record-ID\":\"<urn:uuid:9dcf0cfd-3281-4647-9139-af78c0408ae5>\",\"Content-Length\":\"877569\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f625c093-0aa9-4eef-aaec-6765d1abf3e5>\",\"WARC-Concurrent-To\":\"<urn:uuid:9860bfff-ac60-4954-a471-1fe71e8dc747>\",\"WARC-IP-Address\":\"185.134.29.27\",\"WARC-Target-URI\":\"https://www.directhardwaresupply.com/store/page/30/?filter_product-group=memory&query_type_product-group=or\",\"WARC-Payload-Digest\":\"sha1:CE5HYRQNNAFVMU3PDPLPZ5OLV4S6XVLZ\",\"WARC-Block-Digest\":\"sha1:GWSE3SP35YWZQQ55GN7D2TMUY6MRLNB6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370496330.1_warc_CC-MAIN-20200329232328-20200330022328-00305.warc.gz\"}"}
https://hcl.ucd.ie/wiki/index.php/PGF/Tikz	[ "# Write a figure\n\nThe preamble of the latex file must contain:\n\\usepackage{tikz}\n\nSome optional libraries could be add like this:\n\\usetikzlibrary{calc}\n\nTo start a figure, the code must be inside the tikzpicture environment like this:\n\\begin{tikzpicture} ... TikZ code here... \\end{tikzpicture}\n\n\n# Exemple\n\n% Author: Quintin Jean-Noël\n% <http://moais.imag.fr/membres/jean-noel.quintin/>\n\\documentclass{article}\n\\usepackage{tikz}\n\\usetikzlibrary[topaths]\n% A counter, since TikZ is not clever enough (yet) to handle\n% arbitrary angle systems.\n\\newcount\\mycount\n\\begin{document}\n\\begin{tikzpicture}[transform shape]\n%the multiplication with floats is not possible. Thus I split the loop\n%in two.\n\\foreach \\number in {1,...,8}{\n% Computer angle:\n\\mycount=\\number\n\\multiply\\mycount by 45\n\\node[draw,circle,inner sep=0.125cm] (N-\\number) at (\\the\\mycount:5.4cm) {};\n}\n\\foreach \\number in {9,...,16}{\n% Computer angle:\n\\mycount=\\number\n\\multiply\\mycount by 45\n\\node[draw,circle,inner sep=0.125cm] (N-\\number) at (\\the\\mycount:5.4cm) {};\n}\n\\foreach \\number in {1,...,15}{\n\\mycount=\\number" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6164023,"math_prob":0.9165815,"size":1373,"snap":"2022-27-2022-33","text_gpt3_token_len":467,"char_repetition_ratio":0.1577794,"word_repetition_ratio":0.18064517,"special_character_ratio":0.29788783,"punctuation_ratio":0.21428572,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.997434,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-10T15:11:24Z\",\"WARC-Record-ID\":\"<urn:uuid:33da666f-c1dd-4ad5-8654-c86b59d21acd>\",\"Content-Length\":\"14959\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b3175386-36d2-42ae-a5a4-762ae2effafb>\",\"WARC-Concurrent-To\":\"<urn:uuid:79f36af8-7c02-44d4-b9ca-5a1bfa0fcafc>\",\"WARC-IP-Address\":\"137.43.92.117\",\"WARC-Target-URI\":\"https://hcl.ucd.ie/wiki/index.php/PGF/Tikz\",\"WARC-Payload-Digest\":\"sha1:LLRNDFUVYND3PVPUP3H4YYHT2GYA7BNR\",\"WARC-Block-Digest\":\"sha1:LAQY27SLL6APP4YWHMAQBGVJFKOJA7VO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571190.0_warc_CC-MAIN-20220810131127-20220810161127-00680.warc.gz\"}"}
https://ww2.mathworks.cn/matlabcentral/cody/problems/43158-solve-system-of-equation/solutions/2641150	[ "Cody\n\n# Problem 43158. Solve system of equation!\n\nSolution 2641150\n\nSubmitted on 1 Jul 2020 by Ramesh Kumar V\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 = [2 1 3; 1 1 2]; y_correct = x(:,1:end-1)\\x(:,end); assert(isequal(solvesystem(x),y_correct))\n\n2 Pass\nx = [1 1 0; 1 -1 0]; y_correct = x(:,1:end-1)\\x(:,end); assert(isequal(solvesystem(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!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.56394064,"math_prob":0.9741023,"size":487,"snap":"2021-04-2021-17","text_gpt3_token_len":171,"char_repetition_ratio":0.13043478,"word_repetition_ratio":0.028985508,"special_character_ratio":0.40246406,"punctuation_ratio":0.25396827,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98134434,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-16T03:21:52Z\",\"WARC-Record-ID\":\"<urn:uuid:ebc7ab92-2c8f-4fa6-9cfd-7ab3ea4790e8>\",\"Content-Length\":\"82596\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ce048c00-12cc-4dae-8537-80e963468f77>\",\"WARC-Concurrent-To\":\"<urn:uuid:4405dc97-a233-4aeb-8e29-931175e41a4b>\",\"WARC-IP-Address\":\"104.127.224.102\",\"WARC-Target-URI\":\"https://ww2.mathworks.cn/matlabcentral/cody/problems/43158-solve-system-of-equation/solutions/2641150\",\"WARC-Payload-Digest\":\"sha1:OL2SBJJQ6I426OPE7ETFEDKWX2TXAFQJ\",\"WARC-Block-Digest\":\"sha1:MOIHDNPDSC62U2TC7YLSNEKTZDB5DYAN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703499999.6_warc_CC-MAIN-20210116014637-20210116044637-00681.warc.gz\"}"}
https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-laws-of-motion/in-in-class11th-physics-normal-force/v/more-on-normal-force-shoe-on-floor	[ "If you're seeing this message, it means we're having trouble loading external resources on our website.\n\nIf you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked.\n\n# More on Normal force (shoe on floor)\n\nDavid explains how to determine the normal force for a variety of scenarios (extra forces, diagonal forces, acceleration) involving a shoe on the floor. Created by David SantoPietro.\n\n## Want to join the conversation?\n\n• What about the instant when a falling shoe hits the ground? What would be the normal force then?", null, "• I saw a question: a boy of mass M_ holding a box of mass _m above his head jumps from a building. What is the force exterted by the box on his head during the freefall? Does it increase during the period he balances himself after hitting the ground?\n\nIs there a normal force when two bodies in contact are in freefall? Will they stay in contact?", null, "•", null, "Great question! i don't want to give a straight up answer because you'd appreciate it more when you'll come to conclusion yourself. Let's think about what Galileo told us about object in a free fall. A stone and a feather, ignoring any air resistance, will fall from a building at the same rate! Of course, you knew that already. But it's the key to your answer. Imagine yourself holding a ball in your palm and jumping from a high diving board. Both you and ball are falling at the same rate. Which means if you removed your palm from underneath the ball, it would still keep falling the same way. Think another case: Even if you didn't have the ball on your palm and both you and the ball had been dropped side by side from the diving board, the picture would still be the same.\n\nSo is there a normal force on the ball in free fall?\n\nIf you don't get the answer, ask again. I promise to tell the answer but you have to give the problem your best effort. Read the chapter again.\n• what about the force on the horizontal direction? what happens to it?", null, "•", null, "• In the 'shoe falling through the air' case, what if we take in consideration the air resistance? Could that be viewed as a normal force? (it still is a contact force, since we're talking about the air molecules reacting to the shoe in the opposite direction when they come in contact).\nOf course, in this case the normal force (air resistance) would be a lot smaller than the force that caused it (mg), but then again that's why it's falling through the air, not levitating.", null, "• Seeing as how this was made 7 years ago, you might not care anymore but I would like to still give my answer.\n\nAir resistance like normal force is a kind of contact force, forces acting upon objects in contact. Air resistance is really a special kind of frictional force(also a contact force). The difference between frictional and normal force that you will see in diagrams is that the normal force is perpendicular to the place of contact while friction is along it or parallel.\n\nNormal forces are also described as the reactionary force of the compounds or atoms of a material to not allow another object to pass through it. This is an important distinction in comparison to frictional forces which does \"let an object pass through it\" or does not completely stop an object despite being in contact with it. For example gases(air resistance) and liquids(water resistance) and solids whose point of contact is parallel to the force acting upon it.\n\nHowever, as you've mentioned isn't this kind of the same fundamental force. And you are right! fundamentally, they are really the same forces, intermolecular ones at points of contact. However, because of the fact that macroscopically normal force, air resistance, water resistance, and the usual kind of friction are so different looking and whose emergent properties differ, it is much easier to deal with them as different kinds of forces.\n• Hi, If acceleration = force/mass doesn't that mean that the object with more mass falls slower than object with less mass? now here doesn't mass means mass of the object as f=ma so we're taking force on the object. But that doesn't happen according to experiments. Where am I wrong?", null, "• In the above video what will happen to the horizontal force F3x ?", null, "• When he gives the example of Fsub2 being applied in the upward direction,the equation will become\nFn=mg-F2\nor\nmg=Fn+F2\nDoes this means that the weight of shoe will be increased? (I'm referring this concept from the weightlessness in an elevator)\nPlease clearify this to me.I'm finding this very confusing.In case of an elevator it was obvious but here it is not making sense..", null, "• No, the weight remains constant, remember weight cannot change, as long as you are near the surface of the Earth. Because remember, weight is just the measure of the force of gravity on an object. No, the weight of the shoe will not increase, the normal force will decrease ( assuming the acceleration is 0 ), because the F2 relieved some of the force that was needed to be applied by the normal force.\nI hope this helps.\nComment if you still have some doubts.\n-ƙαɾƚιƙҽყҽ ʂԋɾιʋαʂƚαʋα™\n• Doesn't newton say \"Every reaction has an equal and opposite reaction\"?doesn't the normal force represents to the opposite reaction?So, the normal force shouldn't be equal to the force is given?please clear me that.{When the surface is tilted}\n(1 vote)", null, "• Nope. Action and reaction forces operate on different objects, not the same object. The reaction force to the normal force the surface exerts on the object (up) is the normal force the object exerts on the surface (down). The reaction force of the gravity earth exerts on the object (down) is the gravity the object exerts on earth (up). The reason earth isn’t affected much by this gravity is that its mass is much much greater, so its acceleration is negligible.\n\nSo to sum up, Action and reaction forces operate on different objects and are the same type of force (reaction force of normal force is still normal force, reaction force of gravity is still gravity, reaction force of friction is still friction). In addition to that is Newtons third law.", null, "" ]	[ null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/badges/moon/good-answer-40x40.png", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null, "https://cdn.kastatic.org/images/avatars/svg/blobby-green.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9611878,"math_prob":0.89211446,"size":18556,"snap":"2022-40-2023-06","text_gpt3_token_len":4257,"char_repetition_ratio":0.17238034,"word_repetition_ratio":0.017096493,"special_character_ratio":0.22720414,"punctuation_ratio":0.099921934,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9551171,"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\":\"2023-02-05T23:24:29Z\",\"WARC-Record-ID\":\"<urn:uuid:fbd42d7c-7d05-4468-8162-7afd72ac79a4>\",\"Content-Length\":\"1049415\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3290c38f-e3d3-44b2-b577-64ea906080d3>\",\"WARC-Concurrent-To\":\"<urn:uuid:88fefaea-e25d-40a6-a905-f4a78fa11c15>\",\"WARC-IP-Address\":\"146.75.37.42\",\"WARC-Target-URI\":\"https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-laws-of-motion/in-in-class11th-physics-normal-force/v/more-on-normal-force-shoe-on-floor\",\"WARC-Payload-Digest\":\"sha1:Z54V2JJDYTSAKQ6Q7TCFALRESTNENXHK\",\"WARC-Block-Digest\":\"sha1:6XBCN32ZT4DJ6WPG5OWX2FBVAMD6YPOP\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500294.64_warc_CC-MAIN-20230205224620-20230206014620-00836.warc.gz\"}"}
http://sciences.heptic.fr/archives/432	[ "# Gogeometry Problem 336\n\nFrom Gogeometry", null, "", null, "Using :", null, "", null, "• In ΔDMF : DM^2=DF^2+FM^2\n• (1) DM^2= (r-x/2)^2+x^2\n• r^2= r^2+ DM^2 -2xr\n• (2) DM^2=2xr\n• (1) and (2) 2xr=(r-x/2)^2+x^2\n• 2xr=r^2-xr+x^2/4+x^2\n• 3xr=r^2+5x^2/4\n• Dividing by x^2 (x not equal to 0) =>3r/x=5/4+(r/x)^2\n• Let y=r/x : y^2-3y+5/4=0\n• Delta = 4 and y=5/2 or y=1/2\n\nTherefore x=2r/5 or x=2r\n\n• Since the only definition of M and H are\n• M belongs to the square and circle A,\n• and H belongs to the square and Circle B\n• Therefore, there are 2 different solutions:\n• x=2r/5 (blue square)\n• or x=2r (green square)", null, "" ]	[ null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6299138,"math_prob":0.9993722,"size":610,"snap":"2022-40-2023-06","text_gpt3_token_len":297,"char_repetition_ratio":0.10891089,"word_repetition_ratio":0.018348623,"special_character_ratio":0.47540984,"punctuation_ratio":0.043209877,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999976,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-04T18:22:43Z\",\"WARC-Record-ID\":\"<urn:uuid:ba63a977-581c-4f38-b1b4-d5c6c097476f>\",\"Content-Length\":\"36602\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:37e1efe2-4187-4eae-97b7-c791f4c4f1d6>\",\"WARC-Concurrent-To\":\"<urn:uuid:4cd6b2ab-c7af-4ac2-a6e6-202e3202817a>\",\"WARC-IP-Address\":\"217.160.0.239\",\"WARC-Target-URI\":\"http://sciences.heptic.fr/archives/432\",\"WARC-Payload-Digest\":\"sha1:JXUVOLKN37E4WOB4BBTMAGNECSLWWUKL\",\"WARC-Block-Digest\":\"sha1:PDQE3MAR7FFQCECREIWADNBB7BMQEBFW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500151.93_warc_CC-MAIN-20230204173912-20230204203912-00851.warc.gz\"}"}
https://sttv-depannage-antenne.fr/12760-Jun-2015.html	[ "##### Impact Force Calculator\n\nUsing the impact force calculator; Impact force formula; Formula accuracy; Calculation examples Using the impact force calculator. This versatile impact force calculator is useful for estimating the impact forces involved in collisions of different kinds. For example, it can be used to calculate the impact force of a vehicle (car, truck, train), plane, football, of birds hitting a plane or ...\n\n##### A Reliability Calculations and Statistics\n\n2017-8-25 · 362 A Reliability Calculations and Statistics Table A.1. Conﬁdence levels γ and corresponding values of c γ (%) c 80 1.28 90 1.65 95 1.96 98 2.33 99 2.58 which contains the real probability p with a chosen conﬁdence level γ.If we set γ very close to 1, this interval becomes very large. It depends on\n\n##### How to Calculate CapEx\n\nExample of the CapEx calculation in Excel. Here is an example of how to calculate capital expenditures, as it applies to financial modeling in Excel: Image: CFI''s Financial Modeling Courses. In the above example, let''s look at CapEx in 2018 and the following information: 2018 Depreciation is \\$15,005 on the income statement\n\n##### How to Calculate the Impact and Probability of Business ...\n\n2018-12-2 · We can use this tool to calculate whether negative outcomes will happen, and if so how destructive the effects could be. This is done using a numbered scoring method and color-coded indicators. Scoring. Our scoring is done when we select a level of Impact (1 to 5), and a level of probability (1 to 5).\n\n##### Appendix 9C: Design Calculations for Electrical Design\n\n2021-1-11 · Chapter 9 Electrical Design Appendix 9C Design Calculations for Electrical Design 2 SPU Design Standards and Guidelines November 2020 3. CALCULATION MATRIX Project calculations serve as formal documentation of the project electrical design.\n\n##### DESIGN AND ANALYSIS OF IMPACT CRUSHERS\n\n2012-7-4 · Chapter 2: Design and Calculation 2.1 Design of V-Belt drive 7 2.2 Design of Shaft 10 2.3 Design of Hammers 14 2.3.1 Using Impact Bending 14 2.3.2 Using Strain Energy method 21 2.4 Design of conveyor belt 22 Chapter 3: A performance model for impact crusher\n\n##### Car Crash Example\n\n2017-2-3 · Example of Force on Car (This initial example is cast in U.S. common units because most U.S. readers can make comparisons to known forces more easily in those terms. The calculation provides the results in SI units as well.) Vary the parameters of …\n\n##### Inflation Formula \| Calculator (Example with Excel Template)\n\n2021-8-27 · Inflation Formula – Example #2. Let us take the example of Dylan, who is an economist and wishes to compute the inflation rate in his state. He has created a CPI basket that includes food, cloth, fuel and education and has considered 2010 as the …\n\n##### Risk Assessment Calculation Formula\n\nMatrix Model was first introduced in 1850 by James Joseph Sylvester. Matrix Model was invented in 1857/1858 by Cayley.. Risk assessment is the process of evaluating risks to worker''s safety and health, assets from workplace hazards.\n\n##### DESIGN AND ANALYSIS OF A HORIZONTAL SHAFT …\n\n2011-5-10 · An impact crusher can be further classified as Horizontal impact crusher (HSI) and vertical shaft impact crusher (VSI) based on the type of arrangement of the impact rotor and shaft. Horizontal shaft impact crusher These break rock by impacting the rock with hammers/blow bars that are fixed upon the outer edge of a spinning rotor.\n\n##### impact crusher calculation\n\nimpact crusher production calculation. Impact Crushers Design And Calculations 2015-7-14 ensp 0183 ensp Impact crusher in the production process the main parameters of the equipment has an important significance for normal operation generally includes three basic parameters the rotor speed productivity and motor power So for these impact crusher parameter selection and calculation method of ...\n\n##### impact crusher power calculation\n\n2020-9-28 · Impact Crusher Calculations Urlaub im Kroatien. Calculation of impact crusher capacity and power impactor crusher capacity formula aug 29 2016 dear all one of my jaw crushers is the bottleneck of a certain calculations on capacity of a cone crusher calculation model of a cone crusher …\n\n##### Determining Sample Size Page 2\n\n2021-8-13 · to calculate sample sizes. This formula was used to calculate the sample sizes in Tables 2 and 3 and is shown below. A 95% confidence level andP=.5are assumed for Equation 5. Where n is the sample size, N is the population size, and e is the level of precision. When this formula is applied to the above sample, we get Equation 6. Formula For ...\n\n##### Car Crash Example\n\n2016-11-9 · Force on Driver in Example Car Crash. For the car crash scenario where a car stops in 1 foot from a speed of 30 mi/hr, what is the force on the driver? Assume a 160 lb (mass = 5 slugs) driver. If firmly held in non-stretching seatbelt harness: …\n\n##### Math Calculations in Pharmacology: Formulas & Conversions ...\n\nLiter, often seen as L, for volume. Meter, usually written as m, for length. Important conversions between these units that you need to be aware of include: 1 g = 1000 mg. 1 mg = 1000 mcg. 1 L ...\n\n##### How can I calculate sample size for multiple groups and ...\n\n1992-12-1 · Before sending article I want to know about the impact factor of journals. View. Is there a formula to calculate the sample size of a study? Question. 14 answers. Asked 13th Aug, 2018;\n\n##### Sample Size Calculator\n\nThis free sample size calculator determines the sample size required to meet a given set of constraints. Learn more about population standard deviation, or explore other statistical calculators, as well as hundreds of other calculators addressing math, finance, health, …\n\n##### Analysis of Impact Force Equations\n\n2006-7-17 · Analysis of Impact Force Equations Prepared for the International Technical Rescue Symposium, November 2002 By Chuck Weber Abstract This paper compares the actual impact forces measured during controlled testing to the values calculated by a commonly accepted rope force-predicting equation. All tests studied were\n\n##### Sample Size Calculator: Understanding Sample Sizes ...\n\nIf the sample size calculator says you need more respondents, we can help. Tell us about your population, and we''ll find the right people to take your surveys. With millions of qualified respondents, SurveyMonkey Audience makes it easy to get survey …\n\n##### Population Proportion\n\nFormula. This calculator uses the following formula for the sample size n: n = NX / (X + N – 1), where, X = Z α/22 p* (1-p) / MOE 2, and Z α/2 is the critical value of the Normal distribution at α/2 (e.g. for a confidence level of 95%, α is 0.05 and the critical value is 1.96), MOE is the margin of error, p is the sample proportion ...\n\n##### Sample Size Calculator\n\n2020-6-27 · Sample Size Calculator Terms: Confidence Interval & Confidence Level. The confidence interval (also called margin of error) is the plus-or-minus figure usually reported in newspaper or television opinion poll results. For example, if you use a confidence interval of 4 and 47% percent of your sample picks an answer you can be \"sure\" that if you had asked the question of the entire relevant ...\n\n##### Sample Size Formulas for our Sample Size Calculator ...\n\n2019-9-11 · Sample Size Formulas for our Sample Size Calculator. Here are the formulas used in our Sample Size Calculator: Sample Size . ss = Z 2 * (p) * (1-p) c 2: Where: Z = Z value (e.g. 1.96 for 95% confidence level) p = percentage picking a choice, expressed as decimal (.5 used for sample …\n\n##### Adverse Impact: What is it? How do you calculate it?\n\n2008-7-8 · 3) Calculate impact ratios by dividing the selection rate of each group by that of the highest group 4) Determine if the selection rates are substantially different (i.e., impact ratio < .80) Source: Uniform Guidelines Q&A 12\n\n##### Carbon footprint calculations\n\nThe carbon footprint calculator converts each of these energy, travel, and waste sources from their original unit of measure, for example, kWh or miles, to tons CO 2 per year by applying emission conversion factors. The calculator uses the most current energy, travel, and waste data from the energy log, the travel log, and the waste log.\n\n##### Sample Size (Definition, Formula) \| Calculate Sample Size\n\n2021-8-28 · Sample size calculation is important to understand the concept of the appropriate sample size because it is used for the validity of research findings. In case it is too small, it will not yield valid results, while a sample is too large may be a waste of …\n\n##### FORMULAS FROM EPIDEMIOLOGY KEPT SIMPLE (3e) …\n\n2015-6-19 · = (independent samples only; for matched-pairs and tuples data, see text) • Rounding: Basic measures should be reported with 2 or 3 significant digit accuracy. Carry 4 or 5 significant digits to derive a final answer that is accurate to 2 or 3 significant digits, respectively. 3.3 Measures of Potential Impact\n\n##### Crusher Efficiency Calculations\n\n2014-9-15 · The screen area needed under the jaw crusher is 38/1.89 = 20.1 sq ft. For the 1-in. screen below the roll crusher the capacity has no correction factor and the area needed is 32/2.1 = 15.2 sq ft. To handle the output from a 40 x 24 roll crusher the screen will have to be at least 24 in. wide.\n\n##### How to empirically calculate impact sound insulation\n\n2017-3-8 · Adding this value to the formula, we can obtain: L''n, w = Ln, w, eq – ∆Lw + K – Kt, (dB) If you want to obtain the calculation template, tell us what you want to calculate and we can help you with some of our impact solutions. Manuel Taborga How to empirically calculate impact sound insulation …\n\n##### Statistical Significance Calculator\n\nExample of a statistical significance calculation and its steps. Let''s test the significance occurrence for two sample sizes (s 1) of 25 and (s 2) of 50 having a percentage of response (r 1) of 5%, respectively (r 2) of 7%: Step 1: Substitute the figures from the above example in the formula of comparative error:\n\n##### How to Calculate Force of Impact \| Sciencing\n\n2020-12-6 · Now suppose you want to know the impact force of a 2,200-kilogram car traveling at 20 meters per second that crashes into a wall during a safety test. The stop distance in this example is the crumple zone of the car, or the distance by which the car shortens on impact.\n\n##### Sample size calculation and development of sampling plan\n\nCalculate sample size – single survey using two-stage cluster sampling To estimate sample size, you need to know: Estimate of the prevalence of the key indicator (e.g. rate of stunting) Precision desired (for example: ± 5%) Level of confidence (always use 95%) Expected response rate Population For nutrition surveys: number of eligible individuals per\n\n##### IMPACT\n\nFormula. IBW Estimated ideal body weight in (kg) Males: IBW = 50 kg + 2.3 kg for each inch over 5 feet. Females: IBW = 45.5 kg + 2.3 kg for each inch over 5 feet. ABW Estimated adjusted body weight in (kg) If the actual body weight is greater than 30% of the calculated IBW, calculate the adjusted body weight (ABW): ABW = IBW + 0.4 (actual ..." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8777977,"math_prob":0.9760702,"size":9659,"snap":"2022-27-2022-33","text_gpt3_token_len":2334,"char_repetition_ratio":0.13940963,"word_repetition_ratio":0.008383233,"special_character_ratio":0.25913656,"punctuation_ratio":0.11715915,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99099624,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-10T20:25:11Z\",\"WARC-Record-ID\":\"<urn:uuid:c161c355-5d97-408a-bcd5-72d1a691dd02>\",\"Content-Length\":\"19901\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f23b6424-7524-4608-9b7d-edaa5382c0bd>\",\"WARC-Concurrent-To\":\"<urn:uuid:327ec013-ef25-4aeb-b8b0-8c65045c8663>\",\"WARC-IP-Address\":\"104.21.3.64\",\"WARC-Target-URI\":\"https://sttv-depannage-antenne.fr/12760-Jun-2015.html\",\"WARC-Payload-Digest\":\"sha1:4WHHVOZD6IHE52VSGF32KPYPJUMQMP5I\",\"WARC-Block-Digest\":\"sha1:SCKJFDSI4MKUUUF65M4KKUTKY445H57Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571210.98_warc_CC-MAIN-20220810191850-20220810221850-00430.warc.gz\"}"}
https://www.semanticscholar.org/paper/The-Dehn-functions-of-Out(F_n)-and-Aut(F_n)-Bridson-Vogtmann/4e4ef55a13839cd3c5dc9c055c0afcdb4cb3b63f	[ "# The Dehn functions of Out(F_n) and Aut(F_n)\n\n@article{Bridson2010TheDF,\ntitle={The Dehn functions of Out(F\\_n) and Aut(F\\_n)},\nauthor={Martin R. Bridson and Karen Vogtmann},\njournal={arXiv: Group Theory},\nyear={2010}\n}\n• Published 5 November 2010\n• Mathematics\n• arXiv: Group Theory\nFor n > 2, the Dehn functions of Aut(F_n) and Out(F_n) are exponential. Hatcher and Vogtmann proved that they are at most exponential, and the complementary lower bound in the case n=3 was established by Bridson and Vogtmann. Handel and Mosher completed the proof by reducing the lower bound for n>4 to the case n=3. In this note we give a shorter, more direct proof of this last reduction.\n13 Citations\nLipschitz retraction and distortion for subgroups of Out(Fn)\n• Mathematics\n• 2010\nGiven a free factor A of the rank n free group Fn , we characterize when the subgroup of Out.Fn/ that stabilizes the conjugacy class of A is distorted in Out.Fn/. We also prove that the image of the\nThe Geometry of the Handlebody Groups II: Dehn Functions\n• Mathematics\n• 2018\nWe show that the Dehn function of the handlebody group is exponential in any genus $g\\geq 3$. On the other hand, we show that the handlebody group of genus $2$ is cubical, biautomatic, and therefore\nThe topology and geometry of automorphism groups of free groups\nIn the 1970s Stallings showed that one could learn a great deal about free groups and their automorphisms by viewing the free groups as fundamental groups of graphs and modeling their automorphisms\nTrain Tracks on Graphs of Groups and Outer Automorphisms of Hyperbolic Groups\nStallings remarked that an outer automorphism of a free group may be thought of as a subdivision of a graph followed by a sequence of folds. In this thesis, we prove that automorphisms of fundamental\nSubspace arrangements, BNS invariants, and pure symmetric outer automorphisms of right-angled Artin groups\n• Mathematics\n• 2015\nWe introduce a homology theory for subspace arrangements, and use it to extract a new system of numerical invariants from the Bieri-Neumann-Strebel invariant of a group. We use these to characterize\nSpotted disk and sphere graphs I\nThe disk graph of a handlebody H of genus g ≥ 2 with m ≥ 0 marked points on the boundary is the graphwhose vertices are isotopy classes of disks disjoint from themarked points andwhere two vertices\nHierarchically hyperbolic spaces II: Combination theorems and the distance formula\n• Mathematics\nPacific Journal of Mathematics\n• 2019\nWe introduce a number of tools for finding and studying \\emph{hierarchically hyperbolic spaces (HHS)}, a rich class of spaces including mapping class groups of surfaces, Teichmuller space with either\nLocal topological properties of asymptotic cones of groups\n• Mathematics\n• 2014\nWe define a local analogue to Gromov’s loop division property which we use to give a sufficient condition for an asymptotic cone of a complete geodesic metric space to have uncountable fundamental\nIsoperimetric inequalities for the handlebody groups\n• Mathematics\n• 2011\nWe show that the mapping class group of a handlebody of genus at least 2 has a Dehn function of at most exponential growth type.\nGeometry of graphs of discs in a handlebody\nFor a handlebody H of genus g>1 we use surgery to identify a graph whose vertices are discs and which is quasi-isometrically embedded in the curve graph of the boundary surface." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8556628,"math_prob":0.8537519,"size":5949,"snap":"2022-27-2022-33","text_gpt3_token_len":1427,"char_repetition_ratio":0.16703112,"word_repetition_ratio":0.10341261,"special_character_ratio":0.21902841,"punctuation_ratio":0.058195926,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9787614,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-06T04:19:41Z\",\"WARC-Record-ID\":\"<urn:uuid:2a608a47-fd08-40a5-a070-326743c5a104>\",\"Content-Length\":\"346213\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:54ad1c60-f41d-479c-8971-6a4071b183e5>\",\"WARC-Concurrent-To\":\"<urn:uuid:5f19343b-c5fa-4c1b-b323-b675bf812b65>\",\"WARC-IP-Address\":\"18.67.65.61\",\"WARC-Target-URI\":\"https://www.semanticscholar.org/paper/The-Dehn-functions-of-Out(F_n)-and-Aut(F_n)-Bridson-Vogtmann/4e4ef55a13839cd3c5dc9c055c0afcdb4cb3b63f\",\"WARC-Payload-Digest\":\"sha1:W7R22KHWI2JC3ZKLPTLOH7Q4IDP3JSUY\",\"WARC-Block-Digest\":\"sha1:J6RMGODVK5DGLGQ7MRD3SBF2N6SVWLBO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104660626.98_warc_CC-MAIN-20220706030209-20220706060209-00586.warc.gz\"}"}
http://ying.ninja/?cat=3	[ "# Count Univalue Subtrees\n\nCount Univalue Subtrees\n\nGiven a binary tree, count the number of uni-value subtrees.\n\nA Uni-value subtree means all nodes of the subtree have the same value.\n\nFor example:\nGiven binary tree,\n\n``` 5\n/ \\\n1 5\n/ \\ \\\n5 5 5\n```\n\nreturn `4`.\n\nSolution:\n\n```public int countUnivalSubtrees(TreeNode root) {\nint[] count = new int;\nisUnivalSubtrees(root, count);\nreturn count;\n}\n\npublic boolean isUnivalSubtrees(TreeNode root, int[] count) {\nif (root == null) {\nreturn false;\n}\nboolean left = isUnivalSubtrees(root.left, count);\nboolean right = isUnivalSubtrees(root.right, count);\nif (!left && !right) {\nif (root.left == null && root.right == null) {\ncount++;\nreturn true;\n}\n} else if (left && right) {\nif (root.left.val == root.val && root.right.val == root.val) {\ncount++;\nreturn true;\n}\n} else if (left && !right) {\nif (root.right == null && root.left.val == root.val) {\ncount++;\nreturn true;\n}\n} else if (!left && right) {\nif (root.left == null && root.right.val == root.val) {\ncount++;\nreturn true;\n}\n}\nreturn false;\n}```\n\n# Wildcard Matching\n\nImplement wildcard pattern matching with support for `'?'` and `''`.\n\n• `'?'` Matches any single character.\n• `''` Matches any sequence of characters (including the empty sequence).\n\nThe matching should cover the entire input string (not partial).\n\nExample\n```<code>isMatch(\"aa\",\"a\") → false\nisMatch(\"aa\",\"aa\") → true\nisMatch(\"aaa\",\"aa\") → false\nisMatch(\"aa\", \"\") → true\nisMatch(\"aa\", \"a\") → true\nisMatch(\"ab\", \"?\") → true\nisMatch(\"aab\", \"cab\") → false</code>```\n\nSolution1. Dynamic Programming\n\nisMatch[i][j] = if first i chars for s can match first j chars of p\n\nj==’’: OR(isMatch[0~i][j-1])\n\nj==’?’: isMatch[i-1][j-1]\n\nelse: isMatch[i-1][j-1] && s.charAt(i-1)==p.charAt(j-1)\n\n```public boolean isMatch(String s, String p) {\n//dp version\nif (s == null \|\| p == null) {\nreturn false;\n}\nboolean[][] isMatch = new boolean[s.length() + 1][p.length() + 1];\nisMatch = true;\nfor (int i = 1; i <= s.length(); i++) {\nisMatch[i] = false;\n}\nfor (int j = 1; j <= p.length(); j++) {\nisMatch[j] = isMatch[j - 1] && p.charAt(j - 1) == '';\n}\nfor (int i = 1; i <= s.length(); i++) {\nfor (int j = 1; j <= p.length(); j++) {\nif (p.charAt(j - 1) == '') {\nfor (int k = 0; k <= i; k++) {\nif (isMatch[k][j - 1]) {\nisMatch[i][j] = true;\nbreak;\n}\n}\n} else if (p.charAt(j - 1) == '?') {\nisMatch[i][j] = isMatch[i - 1][j - 1];\n} else {\nisMatch[i][j] = isMatch[i - 1][j - 1] && s.charAt(i - 1) == p.charAt(j - 1);\n}\n}\n}\nreturn isMatch[s.length()][p.length()];\n}```\n\n# Leetcode: Set Matrix Zeroes\n\nGiven a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.\n\nFollow up:Did you use extra space?\nA straight forward solution using O(mn) space is probably a bad idea.\nA simple improvement uses O(m + n) space, but still not the best solution.\nCould you devise a constant space solution?1. O(m+n) space solutionAnalysis:\nTo achieve O(m+n) space, you need two array to store information while traversal the matrix for the first time checking if it is zero or not. You can also combine two arrays into one, but still cost O(m+n) space. So I just use two to store row and column information.\n```public class Solution {\npublic void setZeroes(int[][] matrix) {\nif (matrix.length == 0 \|\| matrix.length == 0) {\nreturn;\n}\nboolean[] zeroRows = new boolean[matrix.length];\nboolean[] zeroCols = new boolean[matrix.length];\n//traversal the entire matrix to check if it is zero or not\nfor (int i = 0; i < matrix.length; i++) {\nfor (int j = 0; j < matrix[i].length; j++) {\nif (matrix[i][j] == 0) {\nzeroRows[i] = true;\nzeroCols[j] = true;\n}\n}\n}\n//now set matrix zero based on the two boolean arrays\nfor (int i = 0; i < matrix.length; i++) {\nfor (int j = 0; j < matrix[i].length; j++) {\nif (zeroRows[i] \|\| zeroCols[j]) {\nmatrix[i][j] = 0;\n}\n}\n}\n}\n}```\n\n2. Optimal O(1) space solution\n\nAnalysis:\nTo save space, we can just make use of the space inside the matrix. Which means, we use the first row and first column to store the information.\nIn the first traversal, we set the corresponding cell on the first row or first column to 0 if there is any cell which is zero in this row or column.\nIn the second traversal, we set the matrix to zeros based on the first row and first column.\n\n``` public void setZeroes(int[][] matrix) {\nif (matrix.length == 0 \|\| matrix.length == 0) {\nreturn;\n}\nboolean firstRowHasZero = false;\nboolean firstColHasZero = false;\n//traversal the entire matrix to check if it is zero or not\nfor (int i = 0; i < matrix.length; i++) {\nfor (int j = 0; j < matrix[i].length; j++) {\nif (matrix[i][j] == 0) {\nif (i == 0) {\nfirstRowHasZero = true;\n}\nif (j == 0) {\nfirstColHasZero = true;\n}\nmatrix[j] = 0;\nmatrix[i] = 0;\n}\n}\n}\n//now set matrix zero based on first row and first column\nfor (int i = 1; i < matrix.length; i++) {\nfor (int j = 1; j < matrix[i].length; j++) {\nif (matrix[j] == 0 \|\| matrix[i] == 0) {\nmatrix[i][j] = 0;\n}\n}\n}\n//deal with the first column\nif (firstColHasZero) {\nfor (int i = 1; i < matrix.length; i++) {\nmatrix[i] = 0;\n}\n}\n//deal with the first row\nif (firstRowHasZero) {\nfor (int i = 1; i < matrix.length; i++) {\nmatrix[i] = 0;\n}\n}\n}```\n\nNote: The tricky part of this algorithm is, if without those two boolean variables to keep track of if there is any zero in the first row or first column , when matrix==0, we have no idea if the first row contains zero, or it is the first column contains zero, or both. also, we should deal with the rest part first based on the first row and first column and then deal with the first row and first column based on the two boolean values.\n\n# Leetcode: Minimum Path Sum\n\nGiven a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.\n\nNote: You can only move either down or right at any point in time.\n\n1. Recursive Solution\n\n``` public int minPathSum(int[][] grid) {\nif (grid.length == 0 \|\| grid.length == 0) {\nreturn 0;\n}\nint[][] resultMap = new int[grid.length][grid.length];\nreturn minPathSum(grid, resultMap, grid.length-1, grid.length-1);\n}\n\npublic int minPathSum(int[][] grid, int[][] resultMap, int row, int col) {\nif (row < 0 \|\| col < 0) {\nreturn Integer.MAX_VALUE;\n}\nif (row == 0 && col == 0) {\nreturn grid;\n}\nif (resultMap[row][col] == 0) {\nresultMap[row][col] = grid[row][col] + Math.min(minPathSum(grid, resultMap, row - 1, col), minPathSum(grid, resultMap, row, col - 1));\n}\nreturn resultMap[row][col];\n}```\n\nNote:\n\nThis is a very typical dp problem. where dp[i][j] = grid[i][j] + min( dp[i-1][j], dp[i][j-1] ); To prevent same cell being visited and calculated multiple times, we have a resultMap 2d array to store the result. We also need to check the edge carefully to ensure the index is not out of bounds of the given array.\n\n2. Iterative Solution (O(mn) space)\n\n```public class Solution {\npublic int minPathSum(int[][] grid) {\nif(grid.length==0\|\|grid.length==0){\nreturn 0;\n}\n//2D map -- O(mn) space\nint[][] resultMap = new int[grid.length][grid.length];\nresultMap = grid;\n//Calculate the first row\nfor(int i = 1;i<grid.length;i++){\nresultMap[i] = grid[i] + resultMap[i-1];\n}\n//Calculate the first col\nfor(int i = 1;i<grid.length;i++){\nresultMap[i] = grid[i] + resultMap[i-1];\n}\n//Calculate the rest\nfor(int i=1;i<grid.length;i++){\nfor(int j=1;j<grid.length;j++){\nresultMap[i][j] = grid[i][j]+Math.min(resultMap[i-1][j],resultMap[i][j-1]);\n}\n}\nreturn resultMap[grid.length-1][grid.length-1];\n}\n}```\n\nNote: First calculate first row and first col which is there own grid value, then calculate other cells based on it.\n\n3. Optimal Iterative Solution (O(n) space)\n\n```public class Solution {\npublic int minPathSum(int[][] grid) {\nif (grid.length == 0 \|\| grid.length == 0) {\nreturn 0;\n}\nint rows = grid.length;\nint cols = grid.length;\n\n//1D map -- O(n) space, n is # of cols\nint[] resultMap = new int[cols];\nresultMap = grid;\n\n//Calculate the first row\nfor (int i = 1; i < grid.length; i++) {\nresultMap[i] = grid[i] + resultMap[i - 1];\n}\n\nfor (int i = 1; i < rows; i++) {\nresultMap = resultMap + grid[i];\nfor (int j = 1; j < cols; j++) {\nresultMap[j] = grid[i][j] + Math.min(resultMap[j], resultMap[j - 1]);\n}\n}\nreturn resultMap[cols - 1];\n}\n}```\n\nNote: You don’t necessarily need a 2d map to store which cost O(n) space where n is the # of columns. You can just use one row to stored the result of previous row, and replace the value with the current row from left to right one by one. One thing worth mentioning is this code is more space efficient though, it has lower readability.\n\nMore information and further thoughts for minimum path sum questions can be found here.\n\nhttp://www.cnblogs.com/hiddenfox/p/3408931.html\n\nhttp://www.cnblogs.com/wuyuegb2312/p/3183214.html\n\n# Leetcode: Best Time to Buy and Sell Stock\n\nSay you have an array for which the ith element is the price of a given stock on day i.\n\nIf you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.\n\nSolution: One Pass O(n) time.\n\nFor each day i,\n\n1. think of if sell it at day i, compare the profit with maxProfit.\n\n2. check if day i is a better day than any previous day to buy a stock, meaning the price is cheaper than any previous day, then we buy the stock at day i.\n\n```public int maxProfit(int[] prices) {\nif (prices == null \|\| prices.length <= 1) {\nreturn 0;\n}\nint maxProfit = 0;\nint sum = 0;\nfor (int i = 1; i < prices.length; i++) {\nmaxProfit = Math.max(maxProfit, prices[i] - prices[buyDate]);\n}\n}\nreturn maxProfit;\n}```\n\n# Leetcode: Search a 2D Matrix\n\nWrite an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:\n\n• Integers in each row are sorted from left to right.\n• The first integer of each row is greater than the last integer of the previous row.\n\nFor example,\n\nConsider the following matrix:\n\n```[\n[1, 3, 5, 7],\n[10, 11, 16, 20],\n[23, 30, 34, 50]\n]\n```\n\nGiven target = `3`, return `true`.\n\n```public class Solution {\npublic boolean searchMatrix(int[][] matrix, int target) {\nif (matrix.length == 0 \|\| matrix.length == 0) {\nreturn false;\n}\nint m = matrix.length;\nint n = matrix.length;\nint start = 0;\nint end = m * n - 1;\nwhile (start <= end) {\nint mid = (start + end) / 2;\nint midVal = matrix[mid / n][mid % n];\nif (midVal == target) {\nreturn true;\n} else if (midVal > target) {\nend = mid - 1;\n} else {\nstart = mid + 1;\n}\n}\nreturn false;\n}\n}```\n\nNote: O(log(mn))\n\n# Leetcode: Binary Tree Level Order Traversal II\n\nGiven a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).\n\nFor example:\nGiven binary tree `{3,9,20,#,#,15,7}`,\n\n``` 3\n/ \\\n9 20\n/ \\\n15 7\n```\n\nreturn its bottom-up level order traversal as:\n\n```[\n[15,7],\n[9,20],\n\n]\n```\n\nconfused what `\"{1,#,2,3}\"` means? > read more on how binary tree is serialized on OJ.\n\n```public class Solution {\npublic List<List<Integer>> levelOrderBottom(TreeNode root) {\nif (root == null) {\nreturn result;\n}\nwhile (currLevel.size() > 0) {\nwhile (currLevel.size() > 0) {\nTreeNode runner = currLevel.remove(0);\nif (runner.left != null) {\n}\nif (runner.right != null) {\n}\n}\ncurrLevel = nextLevel;\n}\nreturn result;\n}\n}```\n\n# Leetcode: Container With Most Water\n\nGiven n non-negative integers a1, a2, …, an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.\n\nNote: You may not slant the container.\n\n``` public int maxArea(int[] height) {\nint maxArea = 0;\nint left = 0;\nint right = height.length - 1;\nwhile (left < right) {\nmaxArea = Math.max(maxArea, calculateArea(height, left, right));\nif (height[left] < height[right]) {\nleft++;\n} else {\nright--;\n}\n}\nreturn maxArea;\n}\n\npublic int calculateArea(int[] height, int left, int right) {\nreturn Math.abs(right - left) Math.min(height[left], height[right]);\n}```\n\nNote:\n\nFor any container, its volume depends on the shortest board.\n\nTwo-pointer scan. And always move with shorter board index.\n\n# Leetcode: Rotate Image\n\nYou are given an n x n 2D matrix representing an image.\n\nRotate the image by 90 degrees (clockwise).\n\nCould you do this in-place?\n\n```public class Solution {\npublic void rotate(int[][] matrix) {\nint n = matrix.length;\nif (n <= 1 \|\| matrix.length <= 1 \|\| n != matrix.length) {\nreturn;\n}\nint mid = (n - 1) / 2;\nint offset = 0;\nwhile (offset <= mid) {\nfor (int i = offset; i < n - 1 - offset; i++) {\nint tmp = matrix[offset][i];\nmatrix[offset][i] = matrix[n - 1 - i][offset];\nmatrix[n - 1 - i][offset] = matrix[n - 1 - offset][n - 1 - i];\nmatrix[n - 1 - offset][n - 1 - i] = matrix[i][n - 1 - offset];\nmatrix[i][n - 1 - offset] = tmp;\n}\noffset++;\n}\n\n}\n}```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.62865525,"math_prob":0.9971627,"size":3063,"snap":"2023-40-2023-50","text_gpt3_token_len":929,"char_repetition_ratio":0.18339327,"word_repetition_ratio":0.10229645,"special_character_ratio":0.3467189,"punctuation_ratio":0.1559633,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99917424,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T09:14:21Z\",\"WARC-Record-ID\":\"<urn:uuid:0290a7ab-b777-4c0f-82f7-69eb55f4e6b3>\",\"Content-Length\":\"82472\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:313a1bdb-9f49-46aa-a41b-e33014c7f024>\",\"WARC-Concurrent-To\":\"<urn:uuid:4fcad4ca-e0b1-4583-ba40-d8221352fcb3>\",\"WARC-IP-Address\":\"166.62.77.128\",\"WARC-Target-URI\":\"http://ying.ninja/?cat=3\",\"WARC-Payload-Digest\":\"sha1:ZLKKPGB2USNUU76EOPEXWL4TFOJUX2YP\",\"WARC-Block-Digest\":\"sha1:E6WF2YBYV2VFUZBUEBZDKY22USRSWIOG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100873.6_warc_CC-MAIN-20231209071722-20231209101722-00054.warc.gz\"}"}
https://www.nag.com/numeric/nl/nagdoc_26/nagdoc_fl26/html/c06/c06intro.html	[ "# NAG Library Chapter Introduction\n\n## 1Scope of the Chapter\n\nThis chapter is concerned with the following tasks.\n (a) Calculating the discrete Fourier transform of a sequence of real or complex data values. (b) Calculating the discrete convolution or the discrete correlation of two sequences of real or complex data values using discrete Fourier transforms. (c) Calculating the inverse Laplace transform of a user-supplied subroutine. (d) Calculating the fast Gauss transform approximation to the discrete Gauss transform. (e) Direct summation of orthogonal series. (f) Acceleration of convergence of a sequence of real values.\n\n## 2Background to the Problems\n\n### 2.1Discrete Fourier Transforms\n\n#### 2.1.1Complex transforms\n\nMost of the routines in this chapter calculate the finite discrete Fourier transform (DFT) of a sequence of $n$ complex numbers ${z}_{\\mathit{j}}$, for $\\mathit{j}=0,1,\\dots ,n-1$. The direct transform is defined by\n $z^k = 1n ∑ j=0 n-1 zj exp -i 2πjk n$ (1)\nfor $k=0,1,\\dots ,n-1$. Note that equation (1) makes sense for all integral $k$ and with this extension ${\\stackrel{^}{z}}_{k}$ is periodic with period $n$, i.e., ${\\stackrel{^}{z}}_{k}={\\stackrel{^}{z}}_{k±n}$, and in particular ${\\stackrel{^}{z}}_{-k}={\\stackrel{^}{z}}_{n-k}$. Note also that the scale-factor of $\\frac{1}{\\sqrt{n}}$ may be omitted in the definition of the DFT, and replaced by $\\frac{1}{n}$ in the definition of the inverse.\nIf we write ${z}_{j}={x}_{j}+i{y}_{j}$ and ${\\stackrel{^}{z}}_{k}={a}_{k}+i{b}_{k}$, then the definition of ${\\stackrel{^}{z}}_{k}$ may be written in terms of sines and cosines as\n $ak = 1n ∑ j=0 n-1 xj cos 2πjk n + yj sin 2πjk n$\n $bk = 1n ∑ j= 0 n- 1 yj cos 2πjk n - xj sin 2πjk n .$\nThe original data values ${z}_{j}$ may conversely be recovered from the transform ${\\stackrel{^}{z}}_{k}$ by an inverse discrete Fourier transform:\n $zj = 1n ∑ k=0 n-1 z^k exp +i 2πjk n$ (2)\nfor $j=0,1,\\dots ,n-1$. If we take the complex conjugate of (2), we find that the sequence ${\\stackrel{-}{z}}_{j}$ is the DFT of the sequence ${\\stackrel{-}{\\stackrel{^}{z}}}_{k}$. Hence the inverse DFT of the sequence ${\\stackrel{^}{z}}_{k}$ may be obtained by taking the complex conjugates of the ${\\stackrel{^}{z}}_{k}$; performing a DFT, and taking the complex conjugates of the result. (Note that the terms forward transform and backward transform are also used to mean the direct and inverse transforms respectively.)\nThe definition (1) of a one-dimensional transform can easily be extended to multidimensional transforms. For example, in two dimensions we have\n $z^ k1 k2 = 1 n1 n2 ∑ j1=0 n1-1 ∑ j2=0 n2-1 z j1 j2 exp -i 2 π j1 k1 n1 exp -i 2 π j2 k2 n2 .$ (3)\nNote: definitions of the discrete Fourier transform vary. Sometimes (2) is used as the definition of the DFT, and (1) as the definition of the inverse.\n\n#### 2.1.2Real transforms\n\nIf the original sequence is purely real valued, i.e., ${z}_{j}={x}_{j}$, then\n $z^k = ak + i bk = 1n ∑ j=0 n-1 xj exp -i 2πjk n$\nand ${\\stackrel{^}{z}}_{n-k}$ is the complex conjugate of ${\\stackrel{^}{z}}_{k}$. Thus the DFT of a real sequence is a particular type of complex sequence, called a Hermitian sequence, or half-complex or conjugate symmetric, with the properties\n $a n-k = ak b n-k = -bk b0 = 0$\nand, if $n$ is even, ${b}_{n/2}=0$.\nThus a Hermitian sequence of $n$ complex data values can be represented by only $n$, rather than $2n$, independent real values. This can obviously lead to economies in storage, with two schemes being used in this chapter. In the first (deprecated) scheme, which will be referred to as the real storage format for Hermitian sequences, the real parts ${a}_{k}$ for $0\\le k\\le n/2$ are stored in normal order in the first $n/2+1$ locations of an array x of length $n$; the corresponding nonzero imaginary parts are stored in reverse order in the remaining locations of x. To clarify, if x is declared with bounds $\\left(0:n-1\\right)$ in your calling subroutine, the following two tables illustrate the storage of the real and imaginary parts of ${\\stackrel{^}{z}}_{k}$ for the two cases: $n$ even and $n$ odd.\nIf $n$ is even then the sequence has two purely real elements and is stored as follows:\n Index of x 0 1 2 $\\dots$ $n/2$ $\\dots$ $n-2$ $n-1$ Sequence ${a}_{0}$ ${a}_{1}+{ib}_{1}$ ${a}_{2}+{ib}_{2}$ $\\dots$ ${a}_{n/2}$ $\\dots$ ${a}_{2}-{ib}_{2}$ ${a}_{1}-{ib}_{1}$ Stored values ${a}_{0}$ ${a}_{1}$ ${a}_{2}$ $\\dots$ ${a}_{n/2}$ $\\dots$ ${b}_{2}$ ${b}_{1}$\n $xk = ak , for k= 0, 1, …, n/2 , and xn-k = bk , for k= 1, 2, …, n/2-1 .$\nIf $n$ is odd then the sequence has one purely real element and, letting $n=2s+1$, is stored as follows:\n Index of x 0 1 2 $\\dots$ $s$ $s+1$ $\\dots$ $n-2$ $n-1$ Sequence ${a}_{0}$ ${a}_{1}+{ib}_{1}$ ${a}_{2}+{ib}_{2}$ $\\dots$ ${a}_{s}+i{b}_{s}$ ${a}_{s}-i{b}_{s}$ $\\dots$ ${a}_{2}-{ib}_{2}$ ${a}_{1}-{ib}_{1}$ Stored values ${a}_{0}$ ${a}_{1}$ ${a}_{2}$ $\\dots$ ${a}_{s}$ ${b}_{s}$ $\\dots$ ${b}_{2}$ ${b}_{1}$\n $xk = ak , for k= 0, 1, …, s , and xn-k = bk , for k= 1, 2, …, s .$\nThe second (recommended) storage scheme, referred to in this chapter as the complex storage format for Hermitian sequences, stores the real and imaginary parts ${a}_{k},{b}_{k}$, for $0\\le k\\le n/2$, in consecutive locations of an array x of length $n+2$. If x is declared with bounds $\\left(0:n+1\\right)$ in your calling subroutine, the following two tables illustrate the storage of the real and imaginary parts of ${\\stackrel{^}{z}}_{k}$ for the two cases: $n$ even and $n$ odd.\nIf $n$ is even then the sequence has two purely real elements and is stored as follows:\n Index of x 0 1 2 3 $\\dots$ $n-2$ $n-1$ $n$ $n+1$ Stored values ${a}_{0}$ ${b}_{0}=0$ ${a}_{1}$ ${b}_{1}$ $\\dots$ ${a}_{n/2-1}$ ${b}_{n/2-1}$ ${a}_{n/2}$ ${b}_{n/2}=0$\n $x2×k = ak , for k= 0, 1, …, n/2 , and x2×k+1 = bk , for k= 0, 1, …, n/2 .$\nIf $n$ is odd then the sequence has one purely real element and, letting $n=2s+1$, is stored as follows:\n Index of x 0 1 2 3 $\\dots$ $n-2$ $n-1$ $n$ $n+1$ Stored values ${a}_{0}$ ${b}_{0}=0$ ${a}_{1}$ ${b}_{1}$ $\\dots$ ${b}_{s-1}$ ${a}_{s}$ ${b}_{s}$ $0$\n $x2×k = ak , for k= 0, 1, …, s , and x2×k+1 = bk , for k= 0, 1, …, s .$\nAlso, given a Hermitian sequence, the inverse (or backward) discrete transform produces a real sequence. That is,\n $xj = 1n a0 + 2 ∑ k=1 n/2-1 ak cos 2πjk n - bk sin 2πjk n + a n/2$\nwhere ${a}_{n/2}=0$ if $n$ is odd.\nFor real data that is two-dimensional or higher, the symmetry in the transform persists for the leading dimension only. So, using the notation of equation (3) for the complex two-dimensional discrete transform, we have that ${\\stackrel{^}{z}}_{{k}_{1}{k}_{2}}$ is the complex conjugate of ${\\stackrel{^}{z}}_{\\left({n}_{1}-{k}_{1}\\right)\\left({n}_{2}-{k}_{2}\\right)}$. It is more convenient for transformed data of two or more dimensions to be stored as a complex sequence of length $\\left({n}_{1}/2+1\\right)×{n}_{2}×\\cdots ×{n}_{d}$ where $d$ is the number of dimensions. The inverse discrete Fourier transform operating on such a complex sequence (Hermitian in the leading dimension) returns a real array of full dimension (${n}_{1}×{n}_{2}×\\cdots ×{n}_{d}$).\n\n#### 2.1.3Real symmetric transforms\n\nIn many applications the sequence ${x}_{j}$ will not only be real, but may also possess additional symmetries which we may exploit to reduce further the computing time and storage requirements. For example, if the sequence ${x}_{j}$ is odd, $\\left({x}_{j}={-x}_{n-j}\\right)$, then the discrete Fourier transform of ${x}_{j}$ contains only sine terms. Rather than compute the transform of an odd sequence, we define the sine transform of a real sequence by\n $x^k = 2n ∑j=1 n-1 xj sin πjkn ,$\nwhich could have been computed using the Fourier transform of a real odd sequence of length $2n$. In this case the ${x}_{j}$ are arbitrary, and the symmetry only becomes apparent when the sequence is extended. Similarly we define the cosine transform of a real sequence by\n $x^k = 2n 12 x0 + ∑ j=1 n-1 xj cos πjkn + 12 -1k xn$\nwhich could have been computed using the Fourier transform of a real even sequence of length $2n$.\nIn addition to these ‘half-wave’ symmetries described above, sequences arise in practice with ‘quarter-wave’ symmetries. We define the quarter-wave sine transform by\n $x^k = 1n ∑ j=1 n-1 xj sin π j 2k-1 2n + 12 -1 k-1 xn$\nwhich could have been computed using the Fourier transform of a real sequence of length $4n$ of the form\n $0,x1,…,xn,xn-1 ,…,x1,0,-x1,…,-xn, -x n-1 ,…, -x 1 .$\nSimilarly we may define the quarter-wave cosine transform by\n $x^k = 1n 12 x0 + ∑j= 1 n- 1 xj cos π j 2k- 1 2n$\nwhich could have been computed using the Fourier transform of a real sequence of length $4n$ of the form\n $x0,x1,…, x n-1 ,0, -x n-1 ,…,-x0,-x1,…, -x n-1 ,0, x n-1 ,…,x1 .$\n\n#### 2.1.4Fourier integral transforms\n\nThe usual application of the discrete Fourier transform is that of obtaining an approximation of the Fourier integral transform\n $F s = ∫ -∞ ∞ f t exp -i 2 π s t dt$\nwhen $f\\left(t\\right)$ is negligible outside some region $\\left(0,c\\right)$. Dividing the region into $n$ equal intervals we have\n $F s ≅ cn ∑ j=0 n-1 fj exp -i 2 π s j c n$\nand so\n $Fk ≅ cn ∑ j= 0 n- 1 fj exp -i 2 π jk n$\nfor $k=0,1,\\dots ,n-1$, where ${f}_{j}=f\\left(jc/n\\right)$ and ${F}_{k}=F\\left(k/c\\right)$.\nHence the discrete Fourier transform gives an approximation to the Fourier integral transform in the region $s=0$ to $s=n/c$.\nIf the function $f\\left(t\\right)$ is defined over some more general interval $\\left(a,b\\right)$, then the integral transform can still be approximated by the discrete transform provided a shift is applied to move the point $a$ to the origin.\n\n#### 2.1.5Convolutions and correlations\n\nOne of the most important applications of the discrete Fourier transform is to the computation of the discrete convolution or correlation of two vectors $x$ and $y$ defined (as in Brigham (1974)) by\n• convolution: ${z}_{k}=\\sum _{j=0}^{n-1}{x}_{j}{y}_{k-j}$\n• correlation: ${w}_{k}=\\sum _{j=0}^{n-1}{\\stackrel{-}{x}}_{j}{y}_{k+j}$\n(Here $x$ and $y$ are assumed to be periodic with period $n$.)\nUnder certain circumstances (see Brigham (1974)) these can be used as approximations to the convolution or correlation integrals defined by\n $z s = ∫ -∞ ∞ x t y s-t dt$\nand\n $w s = ∫ -∞ ∞ x- t y s+t dt , -∞ < s < ∞ .$\nFor more general advice on the use of Fourier transforms, see Hamming (1962); more detailed information on the fast Fourier transform algorithm can be found in Gentleman and Sande (1966) and Brigham (1974).\n\n#### 2.1.6Applications to solving partial differential equations (PDEs)\n\nA further application of the fast Fourier transform, and in particular of the Fourier transforms of symmetric sequences, is in the solution of elliptic PDEs. If an equation is discretized using finite differences, then it is possible to reduce the problem of solving the resulting large system of linear equations to that of solving a number of tridiagonal systems of linear equations. This is accomplished by uncoupling the equations using Fourier transforms, where the nature of the boundary conditions determines the choice of transforms – see Section 3.3. Full details of the Fourier method for the solution of PDEs may be found in Swarztrauber (1977) and Swarztrauber (1984).\n\n### 2.2Inverse Laplace Transforms\n\nLet $f\\left(t\\right)$ be a real function of $t$, with $f\\left(t\\right)=0$ for $t<0$, and be piecewise continuous and of exponential order $\\alpha$, i.e.,\n $ft ≤ M eαt$\nfor large $t$, where $\\alpha$ is the minimal such exponent.\nThe Laplace transform of $f\\left(t\\right)$ is given by\n $F s = ∫0∞ e-st f t dt , t>0$\nwhere $F\\left(s\\right)$ is defined for $\\mathrm{Re}\\left(s\\right)>\\alpha$.\nThe inverse transform is defined by the Bromwich integral\n $f t = 12πi ∫ a-i∞ a+i∞ est F s ds , t>0 .$\nThe integration is performed along the line $s=a$ in the complex plane, where $a>\\alpha$. This is equivalent to saying that the line $s=a$ lies to the right of all singularities of $F\\left(s\\right)$. For this reason, the value of $\\alpha$ is crucial to the correct evaluation of the inverse. It is not essential to know $\\alpha$ exactly, but an upper bound must be known.\nThe problem of determining an inverse Laplace transform may be classified according to whether (a) $F\\left(s\\right)$ is known for real values only, or (b) $F\\left(s\\right)$ is known in functional form and can therefore be calculated for complex values of $s$. Problem (a) is very ill-defined and no routines are provided. Two methods are provided for problem (b).\n\n### 2.3Fast Gauss Transform\n\nGauss transforms have applications in areas including statistics, machine learning, and numerical solution of the heat equation. The discrete Gauss transform (DGT), $G\\left(y\\right)$, evaluated at a set of target points $y\\left(j\\right)$, for $j=1,2,\\dots ,m\\in {ℝ}^{d}$, is defined as:\n $G yj = ∑ i=1 n qi e - yj - xi 2 2 / h i 2 , j=1,…,m$\nwhere ${x}_{i}$, for $i=1,2,\\dots ,n\\in {ℝ}^{d}$, are the Gaussian source points, ${q}_{i}$, for $i=1,2,\\dots ,n\\in {ℝ}^{+}$, are the source weights and ${h}_{i}$, for $i=1,2,\\dots ,n\\in {ℝ}^{+}$, are the source standard deviations (alternatively source scales or source bandwidths).\nThe fast Gauss transform (FGT) algorithm presented in Raykar and Duraiswami (2005) approximates the DGT by using two Taylor series and clustering of the source points.\n\n### 2.4Direct Summation of Orthogonal Series\n\nFor any series of functions ${\\varphi }_{i}$ which satisfy a recurrence\n $ϕr+1 x + αr x ϕr x + βr x ϕr-1 x =0$\nthe sum\n $∑ r= 0 n ar ϕr x$\nis given by\n $∑ r=0 n ar ϕr x = b0 x ϕ0 x + b1 x ϕ1 x + α0 x ϕ0 x$\nwhere\n $br x + αr x br+ 1 x + βr+ 1 x br+ 2 x = ar b n+ 1 x = b n+ 2 x = 0 .$\nThis may be used to compute the sum of the series. For further reading, see Hamming (1962).\n\n### 2.5Acceleration of Convergence\n\nThis device has applications in a large number of fields, such as summation of series, calculation of integrals with oscillatory integrands (including, for example, Hankel transforms), and root-finding. The mathematical description is as follows. Given a sequence of values $\\left\\{{s}_{n}\\right\\}$, for $\\mathit{n}=m,\\dots ,m+2l$, then, except in certain singular cases, arguments, $a$, ${b}_{i}$, ${c}_{i}$ may be determined such that\n $sn = a + ∑ i=1 l bi cin .$\nIf the sequence $\\left\\{{s}_{n}\\right\\}$ converges, then $a$ may be taken as an estimate of the limit. The method will also find a pseudo-limit of certain divergent sequences – see Shanks (1955) for details.\nTo use the method to sum a series, the terms ${s}_{n}$ of the sequence should be the partial sums of the series, e.g., ${s}_{n}=\\sum _{k=1}^{n}{t}_{k}$, where ${t}_{k}$ is the $k$th term of the series. The algorithm can also be used to some advantage to evaluate integrals with oscillatory integrands; one approach is to write the integral (in this case over a semi-infinite interval) as\n $∫0∞ f x dx = ∫0a1 f x dx + ∫ a1 a2 f x dx + ∫ a2 a3 f x dx + …$\nand to consider the sequence of values\n $s1 = ∫ 0 a1 f x dx , s2 = ∫ 0 a2 f x dx = s1 + ∫ a1 a2 f x dx , etc.,$\nwhere the integrals are evaluated using standard quadrature methods. In choosing the values of the ${a}_{k}$, it is worth bearing in mind that c06baf converges much more rapidly for sequences whose values oscillate about a limit. The ${a}_{k}$ should thus be chosen to be (close to) the zeros of $f\\left(x\\right)$, so that successive contributions to the integral are of opposite sign. As an example, consider the case where $f\\left(x\\right)=M\\left(x\\right)\\mathrm{sin}x$ and $M\\left(x\\right)>0$: convergence will be much improved if ${a}_{k}=k\\pi$ rather than ${a}_{k}=2k\\pi$.\n\n## 3Recommendations on Choice and Use of Available Routines\n\nThe fast Fourier transform algorithm ceases to be ‘fast’ if applied to values of $n$ which cannot be expressed as a product of small prime factors. All the FFT routines in this chapter are particularly efficient if the only prime factors of $n$ are $2$, $3$ or $5$.\n\n### 3.1One-dimensional Fourier Transforms\n\nThe choice of routine is determined first of all by whether the data values constitute a real, Hermitian or general complex sequence. It is wasteful of time and storage to use an inappropriate routine.\n\n#### 3.1.1Real and Hermitian data\n\nc06paf transforms a single sequence of real data onto (and in-place) a representation of the transformed Hermitian sequence using the complex storage scheme described in Section 2.1.2. c06paf also performs the inverse transform using the representation of Hermitian data and transforming back to a real data sequence.\nAlternatively, the two-dimensional routine c06pvf can be used (on setting the second dimension to 1) to transform a sequence of real data onto an Hermitian sequence whose first half is stored in a separate Complex array. The second half need not be stored since these are the complex conjugate of the first half in reverse order. c06pwf performs the inverse operation, transforming the the Hermitian sequence (half-)stored in a Complex array onto a separate real array.\n\n#### 3.1.2Complex data\n\nc06pcf transforms a single complex sequence in-place; it also performs the inverse transform. c06psf transforms multiple complex sequences, each stored sequentially; it also performs the inverse transform on multiple complex sequences. This routine is designed to perform several transforms in a single call, all with the same value of $n$.\nIf extensive use is to be made of these routines and you are concerned about efficiency, you are advised to conduct your own timing tests.\n\n### 3.2Half- and Quarter-wave Transforms\n\nFour routines are provided for computing fast Fourier transforms (FFTs) of real symmetric sequences. c06ref computes multiple Fourier sine transforms, c06rff computes multiple Fourier cosine transforms, c06rgf computes multiple quarter-wave Fourier sine transforms, and c06rhf computes multiple quarter-wave Fourier cosine transforms.\n\n### 3.3Application to Elliptic Partial Differential Equations\n\nAs described in Section 2.1.6, Fourier transforms may be used in the solution of elliptic PDEs.\nc06ref may be used to solve equations where the solution is specified along the boundary.\nc06rff may be used to solve equations where the derivative of the solution is specified along the boundary.\nc06rgf may be used to solve equations where the solution is specified on the lower boundary, and the derivative of the solution is specified on the upper boundary.\nc06rhf may be used to solve equations where the derivative of the solution is specified on the lower boundary, and the solution is specified on the upper boundary.\nFor equations with periodic boundary conditions the full-range Fourier transforms computed by c06paf are appropriate.\n\n### 3.4Multidimensional Fourier Transforms\n\nThe following routines compute multidimensional discrete Fourier transforms of real, Hermitian and complex data stored in complex arrays:\n real Hermitian complex 2 dimensions c06pvf c06pwf c06puf 3 dimensions c06pyf c06pzf c06pxf any number of dimensions c06pjf\nThe Hermitian data, either transformed from or being transformed to real data, is compacted (due to symmetry) along its first dimension when stored in Complex arrays; thus approximately half the full Hermitian data is stored.\nc06puf and c06pxf should be used in preference to c06pjf for two- and three-dimensional transforms, as they are easier to use and are likely to be more efficient.\nThe transform of multidimensional real data is stored as a complex sequence that is Hermitian in its leading dimension. The inverse transform takes such a complex sequence and computes the real transformed sequence. Consequently, separate routines are provided for performing forward and inverse transforms.\nc06pvf performs the forward two-dimensionsal transform while c06pwf performs the inverse of this transform.\nc06pyf performs the forward three-dimensional transform while c06pzf performs the inverse of this transform.\nThe complex sequences computed by c06pvf and c06pyf contain roughly half of the Fourier coefficients; the remainder can be reconstructed by conjugation of those computed. For example, the Fourier coefficients of the two-dimensional transform ${\\stackrel{^}{z}}_{\\left({n}_{1}-{k}_{1}\\right){k}_{2}}$ are the complex conjugate of ${\\stackrel{^}{z}}_{{k}_{1}{k}_{2}}$ for ${k}_{1}=0,1,\\dots ,{n}_{1}/2$, and ${k}_{2}=0,1,\\dots ,{n}_{2}-1$.\n\n### 3.5Convolution and Correlation\n\nc06fkf computes either the discrete convolution or the discrete correlation of two real vectors.\nc06pkf computes either the discrete convolution or the discrete correlation of two complex vectors.\n\n### 3.6Inverse Laplace Transforms\n\nTwo methods are provided: Weeks' method (c06lbf) and Crump's method (c06laf). Both require the function $F\\left(s\\right)$ to be evaluated for complex values of $s$. If in doubt which method to use, try Weeks' method (c06lbf) first; when it is suitable, it is usually much faster.\nTypically the inversion of a Laplace transform becomes harder as $t$ increases so that all numerical methods tend to have a limit on the range of $t$ for which the inverse $f\\left(t\\right)$ can be computed. c06laf is useful for small and moderate values of $t$.\nIt is often convenient or necessary to scale a problem so that $\\alpha$ is close to $0$. For this purpose it is useful to remember that the inverse of $F\\left(s+k\\right)$ is $\\mathrm{exp}\\left(-kt\\right)f\\left(t\\right)$. The method used by c06laf is not so satisfactory when $f\\left(t\\right)$ is close to zero, in which case a term may be added to $F\\left(s\\right)$, e.g., $k/s+F\\left(s\\right)$ has the inverse $k+f\\left(t\\right)$.\nSingularities in the inverse function $f\\left(t\\right)$ generally cause numerical methods to perform less well. The positions of singularities can often be identified by examination of $F\\left(s\\right)$. If $F\\left(s\\right)$ contains a term of the form $\\mathrm{exp}\\left(-ks\\right)/s$ then a finite discontinuity may be expected in the inverse at $t=k$. c06laf, for example, is capable of estimating a discontinuous inverse but, as the approximation used is continuous, Gibbs' phenomena (overshoots around the discontinuity) result. If possible, such singularities of $F\\left(s\\right)$ should be removed before computing the inverse.\n\n### 3.7Fast Gauss Transform\n\nThe only routine available is c06saf. If the dimensionality of the data is low or the number of source and target points is small, however, it may be more efficient to evaluate the discrete Gauss transform directly.\n\n### 3.8Direct Summation of Orthogonal Series\n\nThe only routine available is c06dcf, which sums a finite Chebyshev series\n $∑ j=0 n cj Tj x , ∑ j=0 n cj T 2j x or ∑ j=0 n cj T 2j+1 x$\ndepending on the choice of argument.\n\n### 3.9Acceleration of Convergence\n\nThe only routine available is c06baf.\n\n## 4Decision Trees\n\n### Tree 1: Fourier Transform of Discrete Complex Data\n\n Is the data one-dimensional? Multiple vectors? Stored as rows? c06prf yes yes yes no no no Stored as columns? c06psf yes c06pcf Is the data two-dimensional? c06puf yes no Is the data three-dimensional? c06pxf yes no Transform on one dimension only? c06pff yes no Transform on all dimensions? c06pjf yes\n\n### Tree 2: Fourier Transform of Real Data or Data in Complex Hermitian Form Resulting from the Transform of Real Data\n\n Quarter-wave sine (inverse) transform? c06rgf yes no Quarter-wave cosine (inverse) transform? c06rhf yes no Sine (inverse) transform? c06ref yes no Cosine (inverse) transform? c06rff yes no Is the data three-dimensional? Forward transform on real data? c06pyf yes yes no no Inverse transform on Hermitian data? c06pzf yes Is the data two-dimensional? Forward transform on real data? c06pvf yes yes no no Inverse transform on Hermitian data? c06pwf yes Is the data multi one-dimensional? Sequences stored by row? c06ppf yes yes no no Sequences stored by column? c06pqf yes c06paf\n\n## 5Functionality Index\n\n Acceleration of convergence c06baf\n Convolution or Correlation,\n complex vectors c06pkf\n real vectors,\n time-saving c06fkf\n Discrete Fourier Transform,\n multidimensional,\n complex sequence,\n complex storage c06pjf\n real storage c06fjf\n multiple half- and quarter-wave transforms,\n Fourier cosine transforms,\n simple use c06rbf\n Fourier cosine transforms, simple use c06rff\n Fourier sine transforms,\n simple use c06raf\n Fourier sine transforms, simple use c06ref\n quarter-wave cosine transforms,\n simple use c06rdf\n quarter-wave cosine transforms, simple use c06rhf\n quarter-wave sine transforms,\n simple use c06rcf\n quarter-wave sine transforms, simple use c06rgf\n one-dimensional,\n multiple transforms,\n complex sequence,\n complex storage by columns c06psf\n complex storage by rows c06prf\n Hermitian/real sequence,\n complex storage by columns c06pqf\n complex storage by rows c06ppf\n multi-variable,\n complex sequence,\n complex storage c06pff\n real storage c06fff\n single transforms,\n complex sequence,\n time-saving,\n complex storage c06pcf\n real storage c06fcf\n Hermitian/real sequence,\n time-saving,\n complex storage c06paf\n Hermitian sequence,\n time-saving,\n real storage c06fbf\n real sequence,\n time-saving,\n real storage c06faf\n three-dimensional,\n complex sequence,\n complex storage c06pxf\n real storage c06fxf\n Hermitian/real sequence,\n complex-to-real c06pzf\n real-to-complex c06pyf\n two-dimensional,\n complex sequence,\n complex storage c06puf\n Hermitian/real sequence,\n complex-to-real c06pwf\n real-to-complex c06pvf\n Fast Gauss Transform c06saf\n Inverse Laplace Transform,\n Crump's method c06laf\n Weeks' method,\n compute coefficients of solution c06lbf\n evaluate solution c06lcf\n Summation of Chebyshev series c06dcf\n\nNone.\n\n## 7Routines Withdrawn or Scheduled for Withdrawal\n\nThe following lists all those routines that have been withdrawn since Mark 19 of the Library or are scheduled for withdrawal at one of the next two marks.\n WithdrawnRoutine Mark ofWithdrawal Replacement Routine(s) c06dbf 25 c06dcf c06eaf 26 c06paf c06ebf 26 c06paf c06ecf 26 c06pcf c06ekf 26 c06fkf c06fpf 28 c06pqf c06fqf 28 c06pqf c06frf 26 c06psf c06fuf 26 c06puf c06gbf 26 No replacement required c06gcf 26 No replacement required c06gqf 26 No replacement required c06gsf 26 No replacement required c06haf 26 c06ref c06hbf 26 c06rff c06hcf 26 c06rgf c06hdf 26 c06rhf\nBrigham E O (1974) The Fast Fourier Transform Prentice–Hall\nDavies S B and Martin B (1979) Numerical inversion of the Laplace transform: A survey and comparison of methods J. Comput. Phys. 33 1–32\nFox L and Parker I B (1968) Chebyshev Polynomials in Numerical Analysis Oxford University Press\nGentleman W S and Sande G (1966) Fast Fourier transforms for fun and profit Proc. Joint Computer Conference, AFIPS 29 563–578\nHamming R W (1962) Numerical Methods for Scientists and Engineers McGraw–Hill\nRaykar V C and Duraiswami R (2005) Improved Fast Gauss Transform With Variable Source Scales University of Maryland Technical Report CS-TR-4727/UMIACS-TR-2005-34\nShanks D (1955) Nonlinear transformations of divergent and slowly convergent sequences J. Math. Phys. 34 1–42\nSwarztrauber P N (1977) The methods of cyclic reduction, Fourier analysis and the FACR algorithm for the discrete solution of Poisson's equation on a rectangle SIAM Rev. 19(3) 490–501\nSwarztrauber P N (1984) Fast Poisson solvers Studies in Numerical Analysis (ed G H Golub) Mathematical Association of America\nSwarztrauber P N (1986) Symmetric FFT's Math. Comput. 47(175) 323–346\nWynn P (1956) On a device for computing the ${e}_{m}\\left({S}_{n}\\right)$ transformation Math. Tables Aids Comput. 10 91–96\n© The Numerical Algorithms Group Ltd, Oxford, UK. 2017" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84232295,"math_prob":0.99956065,"size":21237,"snap":"2021-21-2021-25","text_gpt3_token_len":5016,"char_repetition_ratio":0.18791504,"word_repetition_ratio":0.14756407,"special_character_ratio":0.22818665,"punctuation_ratio":0.098426245,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99973243,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-21T10:02:40Z\",\"WARC-Record-ID\":\"<urn:uuid:873938a1-8b74-49d6-929f-3631781ef45c>\",\"Content-Length\":\"126198\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:df6c1927-4f43-4d12-b8d5-f981dacd9a76>\",\"WARC-Concurrent-To\":\"<urn:uuid:54c73a25-a110-4547-adb0-30796c2f3006>\",\"WARC-IP-Address\":\"78.129.168.4\",\"WARC-Target-URI\":\"https://www.nag.com/numeric/nl/nagdoc_26/nagdoc_fl26/html/c06/c06intro.html\",\"WARC-Payload-Digest\":\"sha1:3EOYMN3X37BERSTDKQMH4P3PHCMMCBTY\",\"WARC-Block-Digest\":\"sha1:MDWJJZO2PRP2RJH4FLTPGWJ4I5S4VGJF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488269939.53_warc_CC-MAIN-20210621085922-20210621115922-00390.warc.gz\"}"}
https://www.tu-chemnitz.de/informatik/KI/projects/ANNarchy/index.php	[ "# Navigation\n\nSpringe zum Hauptinhalt\n\nNeurosimulator ANNarchy\nANNarchy (Artificial Neural Networks architect)\n\n# ANNarchy (Artificial Neural Networks architect)\n\nANNarchy is a parallel simulator for distributed rate-coded or spiking neural networks. The definition of the network is declared in Python, but the library generates optimized C++ code to actually run the simulation in parallel, using either openMP (on multicore CPU architectures) or CUDA (on GPUs). The current stable version is 4.6 and is released under the GNU GPL v2 or later.\n\nThe code is available at:\n\nhttp://bitbucket.org/annarchy/annarchy\n\nThe documentation is available at:\n\nhttp://annarchy.readthedocs.org\n\n## Core principles\n\nANNarchy separates the description of a neural network from its simulation. The description is declared in a Python script, offering high flexibility and readability of the code, and allowing to use the huge ecosystem of scientific libraries available with Python (Numpy, Scipy, Matplotlib, Sympy, Cython). Using Python furthermore reduces the programming effort to a minimum, letting the modeller concentrate on network design and data analysis.", null, "A neural network is defined as a collection of interconnected populations of neurons. Each population comprises a set of similar artificial neurons (rate-coded or spiking point-neurons), whose activity is ruled by one or many ordinary differential equations. The activity of a neuron depends on the activity of other neurons through synapses, whose strength can evolve with time depending on pre- or post-synaptic activities (synaptic or plasticity). Populations are interconnected with each other through projections, which contain synapses between two populations.\n\nANNarchy provides a set of classical neuron or synapse models, but also allows the definition of specific models. The ordinary differential equations (ODE) governing neural or synaptic dynamics have to be specified by the modeler. Contrary to other simulators (except Brian) which require to code these modules in a low-level language, ANNarchy provides a mathematical equation parser which can generate optimized C++ code depending on the chosen parallel framework. Bindings from C++ to Python are generated thanks to Cython (C-extensions to Python), which is a static compiler for Python. These bindings allow the Python script to access all data generated by the simulation (neuronal activity, connection weights) as if they were simple Python attributes. However, the simulation itself is independent from Python and its relatively low performance.\n\n## Example of a pulse-coupled network of Izhikevich neurons\n\nTo demonstrate the simplicity of the interface of ANNarchy, here is the Hello, World! of spiking networks: the pulse-coupled network of Izhikevich neurons (Izhikevich, 2003). It can be defined in ANNarchy as:\n\nfrom ANNarchy import \n\n# Create the excitatory and inhibitory population\npop = Population(geometry=1000, neuron=Izhikevich)\nExc = pop[:800] ; Inh = pop[800:]\n\n# Set the population parameters\nre = np.random.random(800) ; ri = np.random.random(200)\nExc.noise = 5.0 ; Inh.noise = 2.0\nExc.a = 0.02 ; Inh.a = 0.02 + 0.08 ri\nExc.b = 0.2 ; Inh.b = 0.25 - 0.05 * ri\nExc.c = -65.0 + 15.0 * re*2 ; Inh.c = -65.0\nExc.d = 8.0 - 6.0 re*2 ; Inh.d = 2.0\nExc.v = -65.0 ; Inh.v = -65.0\nExc.u = Exc.v Exc.b ; Inh.u = Inh.v * Inh.b\n\n# Create the projections\nexc_proj = Projection(pre=Exc, post=pop, target='exc')\nexc_proj.connect_all_to_all(weights=Uniform(0.0, 0.5))\n\ninh_proj = Projection(pre=Inh, post=pop, target='inh')\ninh_proj.connect_all_to_all(weights=Uniform(0.0, 1.0))\n\n# Compile\ncompile()\n\n# Start recording the spikes in the network to produce the plots\nM = Monitor(pop, ['spike', 'v'])\n\n# Simulate 1 second\nsimulate(1000.0, measure_time=True)\n\n# Retrieve the spike recordings and the membrane potential\nspikes = M.get('spike')\nv = M.get('v')\n\n# Compute the raster plot\nt, n = M.raster_plot(spikes)\n\n# Compute the population firing rate\nfr = M.histogram(spikes)\n\n# Plot the results\nimport matplotlib.pyplot as plt\nax = plt.subplot(3,1,1)\nax.plot(t, n, 'b.', markersize=1.0)\nax = plt.subplot(3,1,2)\nax.plot(v[:, 15])\nax = plt.subplot(3,1,3)\nax.plot(fr)\nplt.show()", null, "## Releases\n\n• 1.0: Initial version, purely C++.\n• 1.1: Management of exceptions.\n• 1.3: Parallelization of the computation using openMP.\n• 2.0: Optimized version with separated arrays for typed connections.\n• 2.1: Parallelization using CUDA.\n• 2.2: Optimized parallelization using openMP.\n• 3.0: Python interface to the C++ core using Boost::Python.\n• 4.0: Python-only version using Cython for the interface to the generated C++ code.\n\n## Relevant Publications\n\nVitay, J., Dinkelbach, H., Hamker, F.H. (2015). ANNarchy: a code generation approach to neural simulations on parallel hardware. Frontiers in Neuroinformatics 9, 19. doi:10.3389/fninf.2015.00019. PDF-document\n\nDinkelbach, H., Vitay, J., Beuth, F., Hamker, F.H. (2012). Comparison of GPU- and CPU-implementations of mean-firing rate neural networks on parallel hardware. Network: Computation in Neural Systems, 23(4): 212-236. PDF-document" ]	[ null, "https://www.tu-chemnitz.de/informatik/KI/projects/ANNarchy/annarchy.png", null, "https://www.tu-chemnitz.de/informatik/KI/projects/ANNarchy/izhikevich.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.73961645,"math_prob":0.91130364,"size":4957,"snap":"2019-35-2019-39","text_gpt3_token_len":1273,"char_repetition_ratio":0.10922673,"word_repetition_ratio":0.002793296,"special_character_ratio":0.26003632,"punctuation_ratio":0.20440882,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98512673,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-25T04:11:15Z\",\"WARC-Record-ID\":\"<urn:uuid:c27364e5-f972-477e-9e2e-6986f723aca9>\",\"Content-Length\":\"43921\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cd330c0e-5dd0-404c-a710-bda4e1d11885>\",\"WARC-Concurrent-To\":\"<urn:uuid:fc7ee8fb-0760-4cd1-be06-2fe5265f2fee>\",\"WARC-IP-Address\":\"134.109.226.8\",\"WARC-Target-URI\":\"https://www.tu-chemnitz.de/informatik/KI/projects/ANNarchy/index.php\",\"WARC-Payload-Digest\":\"sha1:AGTPVPGMOH6VRUZVMQAGVVIDZXOU4CVA\",\"WARC-Block-Digest\":\"sha1:2QBMUPZUO7QQWIA4QF6GDDZWETAA224Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027322170.99_warc_CC-MAIN-20190825021120-20190825043120-00193.warc.gz\"}"}
https://visualfractions.com/percentage-calculator/1-is-what-percent-of-1231/	[ "# 1 is what percent of 1231?\n\nIf you want to find out what 1 as a percentage of 1231 is then this is the article for you. In this quick and straightforward guide, we'll show you exactly how to find out exactly what one number as a percentage of another is. It's very simple and will help you to calculate future percentage problems like this yourself without the need to search for it.\n\nIn a rush and just need to know the answer? 1 is 0.08% of 1231.\n\nis what % of\n\n## 1 is what % of 1231?\n\nUsing percentages is something all of us do every single day, whether we realise it or not. It's such a common task but often we forget just how easy it is to work it out for ourselves and reach for Siri or a quick Google search. That's totally fine, and you probably found this page doing just that, but it is very helpful to know how to do it yourself.\n\nSo, for example, let's say you just ate a meal at a restaurant and the bill comes in at \\$108. You have a \\$10 bill that you want to leave as a tip. What percentage of the check are you tipping? Is it enough or should you tip more? By working out the percentage of one number to another, you can easily figure that out.\n\nLet's dive into this problem and explain exactly how to solve it. First of all, whenever we want to know what percentage 1 is of 1231 what we're really talking about is a fraction:\n\n1 / 1231\n\nSo to work out the percentage from a fraction, the first step is to divide the number above the line (the numerator) by the number below the line (the denominator):\n\n1 / 1231 = 0.0008123476848091\n\nOur fraction is now in decimal format, so how do we get to a percentage from there? If you said \"multiply 0.0008123476848091 by 100\" then you deserve a cookie, my friend!\n\n0.0008123476848091 x 100 = 0.08%\n\nAnd there you have it! You just successfully calculated the percentage of 1 to 1231. You can now go forth and use this method to work out and calculate the percentages of any numbers.\n\nIf you end up having to do this will large numbers than generate a lot of decimal places, you may still need a calculator to work out the percentage, but you definitely shouldn't need to Google or ask Siri again now that you have the knowledge to manually work it out.\n\nHead back to the percentage calculator to work out any more calculations you need to make or be brave and give it a go by hand. Hopefully this article has shown you that it's easier than you might think!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9364069,"math_prob":0.91119367,"size":3153,"snap":"2022-40-2023-06","text_gpt3_token_len":763,"char_repetition_ratio":0.1527469,"word_repetition_ratio":0.011257036,"special_character_ratio":0.2607041,"punctuation_ratio":0.11497731,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9838805,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-28T04:26:54Z\",\"WARC-Record-ID\":\"<urn:uuid:6e8d6dfc-f9bb-4d8f-9ff4-6efdd612a996>\",\"Content-Length\":\"22479\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4bf81de7-7308-4c9e-adce-ac7290f5471e>\",\"WARC-Concurrent-To\":\"<urn:uuid:3bcf6dc6-4d7c-4647-814a-e9ad269888b3>\",\"WARC-IP-Address\":\"104.21.87.217\",\"WARC-Target-URI\":\"https://visualfractions.com/percentage-calculator/1-is-what-percent-of-1231/\",\"WARC-Payload-Digest\":\"sha1:AZ5NSPTLCLMHAIVT4GZ3E2IER7JVHRBK\",\"WARC-Block-Digest\":\"sha1:IJGFY6VUGXS2QKUEPTECA737NCFT2CEB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335059.43_warc_CC-MAIN-20220928020513-20220928050513-00121.warc.gz\"}"}
https://www.mis.mpg.de/publications/preprint-repository/article/2012/issue-67	[ "", null, "", null, "# MiS Preprint Repository\n\nDelve into the future of research at MiS with our preprint repository. Our scientists are making groundbreaking discoveries and sharing their latest findings before they are published. Explore repository to stay up-to-date on the newest developments and breakthroughs.\n\nMiS Preprint\n67/2012\n\n### Fast Convolution Quadrature for Wave Equation in Three Dimensions\n\nLehel Banjai and Maryna Kachanovska\n\n#### Abstract\n\nIn this work the question of efficient solution of an external boundary value problem for the wave equation in three dimensions is addressed. The problem is reformulated in terms of time domain boundary integral equations; the corresponding convolution equations are discretized with the help of Runge-Kutta convolution quadrature. The resulting lower triangular Toeplitz system of size $N$ is solved recursively, constructing $O(N)$ discretizations of boundary single-layer operator of Helmholtz equation.\n\nSince the problem is posed in odd dimension, Huygens principle holds true and convolution weights of Runge-Kutta convolution quadrature $w_{n}^{h}(d)$ exhibit exponential decay outside of a neighborhood of the diagonal $d \\approx nh$, where $h$ is a time step. Therefore, only a constant number of discretizations of boundary integral operators has to contain the near-field and for the rest only the far-field can be constructed. We combine this property with a use of data-sparse techniques, namely $\\mathcal{H}$-matrices and high-frequency fast multipole method, to design an efficient recursive algorithm. Issues specific to the application of data-sparse techniques to the convolution quadrature are also addressed. Numerical experiments indicate the efficiency of the proposed approach.\n\nNov 9, 2012\nPublished:\nNov 12, 2012\nMSC Codes:\n65M38, 35L05\nKeywords:\ndata-sparse techniques, wave equation, time-domain boundary integral equations, Runge-Kutta convolution quadrature\n\n### Related publications\n\ninJournal\n2014 Repository Open Access\nLehel Banjai and Maryna Kachanovska\n\n### Fast convolution quadrature for wave equation in three dimensions\n\nIn: Journal of computational physics, 279 (2014), pp. 103-126" ]	[ null, "data:image/svg+xml;base64,<svg width="186" height="30" viewBox="0 0 186 30" fill="none" xmlns="http://www.w3.org/2000/svg">
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https://xmphysics.com/2023/01/02/appendix-a-standing-wave-resonance-from-first-principle-beyond-syllabus/	[ "# Appendix A: Standing Wave Resonance from First Principle (Beyond Syllabus)\n\nWhat does the H2 syllabus require to know? Not much really. Just that repeated reflections at both ends of the string causes incident and reflected waves to superpose. At resonant frequencies, a standing wave of large amplitude is formed, with nodes at both ends of the string.\n\nI have noticed that even university textbooks do not provide much detail about what’s really happening in the string. So I have had to form my own thoughts, which I share in this section.\n\nLet’s call the wave that has just departed from the vibrator the 1st generation incident wave. When it returns after the reflection, it becomes the 1st generation reflected wave. The 1st generation incident wave superposes with the 1st generation reflected wave to form, well, the 1st generation standing wave. Assuming no energy loss, the amplitudes of the incident, reflected and standing wave would be A, A and 2A respectively.\n\nOf course, the 1G reflected wave will undergo reflection and sets off as the 2G incident wave, returns as the 2G reflected wave, superposing to form the 2G standing wave. As long as the vibrator keeps on oscillating the string, the waves in the string will keep piling up. So we have the 1G, 2G, 3G, 4G … NG standing waves all in the string at the same time. \n\nAnd they all superpose of course. At most frequencies, all these generations of standing waves would have a progressive phase difference between generations. The resulting destructive interference produces a standing wave of zero (or close to zero) amplitude.\n\nHowever, at resonant frequencies, all these generations of standing waves would be in-phase with one another. The resulting constructive interference produces a standing wave of amplitude much larger than A. Assuming that there are N generations in the string, and no attenuation occurs during reflections, no energy lost to dissipative force, the amplitude of the standing wave would be 2NA.\n\nSo what’s the condition for resonance to occur? Since every incident wave must go forth and back along the length L of the string before setting off as the next generation incident wave, the path difference between generations is simply 2L.", null, "$2L=n\\lambda$ ,", null, "$n=1,2,3,...$\n\nThis means that the resonant wavelengths are", null, "$\\displaystyle \\lambda =\\frac{{2L}}{n},n=1,2,3,...$\n\nSince", null, "$v=f\\lambda$, the resonant frequencies are", null, "$\\displaystyle f=n\\frac{v}{{2L}},n=1,2,3,...$\n\nClosed Pipe\n\nActually, the reflection at a fixed end adds a phase change of π rad. But on a guitar string, an incident wave must undergo two reflections before setting off as the next generation incident wave, so a total phase change of 2π rad phase per round trip. This allows use to ignore the phase change caused by reflections. If we are talking about a closed pipe, where the total phase change caused by reflections is only π rad (because the reflection at the open end incurs no phase change), the condition for resonance would have been\n\nThis means that the resonant wavelengths are", null, "$\\displaystyle \\lambda =\\frac{{4L}}{{2n-1}},n=1,2,3,...$\n\nSince", null, "$v=f\\lambda$, the resonant frequencies are", null, "$\\displaystyle f=(2n-1)\\frac{v}{{4L}},n=1,2,3,...$\n\n I created this “wave generations” nomenclature myself. You won’t read about iterations of waves described as 1G, 2G, 3G and so on in any textbook." ]	[ null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9312076,"math_prob":0.9776816,"size":3014,"snap":"2023-14-2023-23","text_gpt3_token_len":646,"char_repetition_ratio":0.16644518,"word_repetition_ratio":0.08300395,"special_character_ratio":0.20238885,"punctuation_ratio":0.10350877,"nsfw_num_words":2,"has_unicode_error":false,"math_prob_llama3":0.9808441,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-07T18:49:11Z\",\"WARC-Record-ID\":\"<urn:uuid:07614ddc-6a8b-47a9-840d-e5913e84f109>\",\"Content-Length\":\"88406\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bca57344-6b41-4f1f-9538-544df6b9f831>\",\"WARC-Concurrent-To\":\"<urn:uuid:8645daf9-0bb6-4819-8ebe-f15d8c8181c9>\",\"WARC-IP-Address\":\"192.0.78.24\",\"WARC-Target-URI\":\"https://xmphysics.com/2023/01/02/appendix-a-standing-wave-resonance-from-first-principle-beyond-syllabus/\",\"WARC-Payload-Digest\":\"sha1:7HMWS5SAWBOOTGYXO2LCZT4Y3NBUEJJI\",\"WARC-Block-Digest\":\"sha1:DOMRI4WTDAGRQV6BDH7IBMV4FG56UDM7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224654012.67_warc_CC-MAIN-20230607175304-20230607205304-00272.warc.gz\"}"}
https://www.arxiv-vanity.com/papers/quant-ph/0406196/	[ "arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org.\n\n# Improved Simulation of Stabilizer Circuits\n\nScott Aaronson MIT Daniel Gottesman Perimeter Institute\n###### Abstract\n\nThe Gottesman-Knill theorem says that a stabilizer circuit—that is, a quantum circuit consisting solely of CNOT, Hadamard, and phase gates—can be simulated efficiently on a classical computer. This paper improves that theorem in several directions. First, by removing the need for Gaussian elimination, we make the simulation algorithm much faster at the cost of a factor- increase in the number of bits needed to represent a state. We have implemented the improved algorithm in a freely-available program called CHP (CNOT-Hadamard-Phase), which can handle thousands of qubits easily. Second, we show that the problem of simulating stabilizer circuits is complete for the classical complexity class , which means that stabilizer circuits are probably not even universal for classical computation. Third, we give efficient algorithms for computing the inner product between two stabilizer states, putting any -qubit stabilizer circuit into a “canonical form” that requires at most gates, and other useful tasks. Fourth, we extend our simulation algorithm to circuits acting on mixed states, circuits containing a limited number of non-stabilizer gates, and circuits acting on general tensor-product initial states but containing only a limited number of measurements.\n\n###### pacs:\n03.67.Lx, 03.67.Pp, 02.70.-c\n\n## I Introduction\n\nAmong the many difficulties that quantum computer architects face, one of them is almost intrinsic to the task at hand: how do you design and debug circuits that you can’t even simulate efficiently with existing tools? Obviously, if a quantum computer output the factors of a -digit number, then you wouldn’t need to simulate it to verify its correctness, since multiplying is easier than factoring. But what if the quantum computer didn’t work? Ordinarily architects might debug a computer by adding test conditions, monitoring registers, halting at intermediate steps, and so on. But for a quantum computer, all of these standard techniques would probably entail measurements that destroy coherence. Besides, it would be nice to design and debug a quantum computer using classical CAD tools, before trying to implement it!\n\nQuantum architecture is one motivation for studying classical algorithms to simulate and manipulate quantum circuits, but it is not the only motivation. Chemists and physicists have long needed to simulate quantum systems, and they have not had the patience to wait for a quantum computer to be built. Instead, they have developed limited techniques such as Quantum Monte-Carlo (QMC) suzuki for computing properties of certain ground states. More recently, several general-purpose quantum computer simulators have appeared, including Oemer’s quantum programming language QCL oemer , the QuIDD (Quantum Information Decision Diagrams) package of Viamontes et al. vmh ; vrmh , and the parallel quantum computer simulator of Obenland and Despain od . The drawback of such simulators, of course, is that their running time grows exponentially in the number of qubits. This is true not only in the worst case but in practice. For example, even though it uses a variant of binary decision diagrams to avoid storing an entire amplitude vector for some states, Viamontes et al. vmh report that the QuIDD package took more than 22 hours to simulate Grover’s algorithm on 40 qubits. With a general-purpose package, then, simulating hundreds or thousands of qubits is out of the question.\n\nA different direction of research has sought to find nontrivial classes of quantum circuits that can be simulated efficiently on a classical computer. For example, Vidal vidal showed that, so long as a quantum computer’s state at every time step has polynomially-bounded entanglement under a measure related to Schmidt rank, the computer can be simulated classically in polynomial time. Notably, in a follow-up paper vidal2 , Vidal actually implemented his algorithm and used it to simulate -dimensional quantum spin chains consisting of hundreds of spins. A second example is a result of Valiant valiant , which reduces the problem of simulating a restricted class of quantum computers to that of computing the Pfaffian of a matrix. The latter is known to be solvable in classical polynomial time. Terhal and DiVincenzo td have shown that Valiant’s class corresponds to a model of noninteracting fermions.\n\nThere is one class of quantum circuits that is known to be simulable in classical polynomial time, that does not impose any limit on entanglement, and that arises naturally in several applications. This is the class of stabilizer circuits introduced to analyze quantum error-correcting codes bdsw ; crss ; gottesman ; gottesman2 . A stabilizer circuit is simply a quantum circuit in which every gate is a controlled-NOT, Hadamard, phase, or -qubit measurement gate. We call a stabilizer circuit unitary if it does not contain measurement gates. Unitary stabilizer circuits are also known as Clifford group circuits.", null, "Figure 1: The four types of gate allowed in the stabilizer formalism\n\nStabilizer circuits can be used to perform the encoding and decoding steps for a quantum error-correcting code, and they play an important role in fault-tolerant circuits. However, the stabilizer formalism used to describe these circuits has many other applications. This formalism is rich enough to encompass most of the “paradoxes” of quantum mechanics, including the GHZ (Greenberger-Horne-Zeilinger) experiment ghz , dense quantum coding bw , and quantum teleportation bbcjpw . On the other hand, it is not so rich as to preclude efficient simulation by a classical computer. That conclusion, sometimes known as the Gottesman-Knill theorem, is the starting point for the contributions of this paper.\n\nOur results are as follows. In Section III we give a new tableau algorithm for simulating stabilizer circuits that is faster than the algorithm directly implied by the Gottesman-Knill theorem. By removing the need for Gaussian elimination, this algorithm enables measurements to be simulated in steps instead of (where is the number of qubits), at a cost of a factor- increase in the number of bits needed to represent a quantum state.\n\nSection IV describes CHP, a high-performance stabilizer circuit simulator that implements our tableau algorithm. We present the results of an experiment designed to test how CHP’s performance is affected by properties of the stabilizer circuit being simulated. CHP has already found application in simulations of quantum fault-tolerance circuits cross .\n\nSection V proves that the problem of simulating stabilizer circuits is complete for the classical complexity class . Informally, this means that any stabilizer circuit can be simulated using CNOT gates alone; the availability of Hadamard and phase gates provides at most a polynomial advantage. This result removes some of the mystery about the Gottesman-Knill theorem by showing that stabilizer circuits are unlikely to be capable even of universal classical computation.\n\nIn Section VI we prove a canonical form theorem that we expect will have many applications to the study of stabilizer circuits. The theorem says that given any stabilizer circuit, there exists an equivalent stabilizer circuit that applies a round of Hadamard gates, followed by a round of phase gates, followed by a round of CNOT gates, and so on in the sequence H-C-P-C-P-C-H-P-C-P-C (where H, C, P stand for Hadamard, CNOT, Phase respectively). One immediate corollary, building on a result by Patel, Markov, and Hayes pmh and improving one by Dehaene and De Moor dm , is that any stabilizer circuit on qubits has an equivalent circuit with only gates.\n\nFinally, Section VII extends our simulation algorithm to situations beyond the usual one considered in the Gottesman-Knill theorem. For example, we show how to handle mixed states, without keeping track of pure states from which the mixed states are obtainable by discarding qubits. We also show how to simulate circuits involving a small number of non-stabilizer gates; or involving arbitrary tensor-product initial states, but only a small number of measurements. Both of these latter two simulations take time that is polynomial in the number of qubits, but exponential in the number of non-stabilizer gates or measurements. Presumably this exponential dependence is necessary, since otherwise we could simulate arbitrary quantum computations in classical subexponential time.\n\nWe conclude in Section VIII with some directions for further research.\n\n## Ii Preliminaries\n\nWe assume familiarity with quantum computing. This section provides a crash course on the stabilizer formalism, confining attention to those aspects we will need. See Section 10.5.1 of Nielsen and Chuang nc for more details.\n\nThroughout this paper we will use the following four Pauli matrices:\n\n I=(1001)X=(0110)Y=(0−ii0)Z=(100−1)\n\nThese matrices satisfy the following identities:\n\n X2=Y2=Z2=I XY=iZYZ=iXZX=iYYX=−iZZY=−iXXZ=−iY\n\nIn particular, every two Pauli matrices either commute or anticommute. The rule for whether to include a minus sign is the same as that for quaternions, if we replace by .\n\nWe define the group of -qubit Pauli operators to consist of all tensor products of Pauli matrices, together with a multiplicative factor of or (so the total number of operators is ). We omit tensor product signs for brevity; thus should be read (we will use to represent the Pauli group operation). Given two Pauli operators and , it is immediate that commutes with if and only if the number of indices such that anticommutes with is even; otherwise anticommutes with . Also, for all , if has a phase of then , whereas if has a phase of then .\n\nGiven a pure quantum state , we say a unitary matrix stabilizes if is an eigenvector of with eigenvalue , or equivalently if where we do not ignore global phase. To illustrate, the following table lists the Pauli matrices and their opposites, together with the unique -qubit states that they stabilize:\n\nThe identity matrix stabilizes all states, whereas stabilizes no states.\n\nThe key idea of the stabilizer formalism is to represent a quantum state , not by a vector of amplitudes, but by a stabilizer group, consisting of unitary matrices that stabilize . Notice that if and both stabilize then so do and , and thus the set of stabilizers of is a group. Also, it is not hard to show that if then . But why does this strange representation buy us anything? To write down generators for (even approximately) still takes exponentially many bits in general by an information-theoretic argument. Indeed stabilizers seem worse than amplitude vectors, since they require about parameters to specify instead of about !\n\nRemarkably, though, a large and interesting class of quantum states can be specified uniquely by much smaller stabilizer groups—specifically, the intersection of with the Pauli group gottesman ; gottesman2 ; crss . This class of states, which arises in quantum error correction and many other settings, is characterized by the following theorem.\n\n###### Theorem 1\n\nGiven an -qubit state , the following are equivalent:\n\n1. can be obtained from by CNOT, Hadamard, and phase gates only.\n\n2. can be obtained from by CNOT, Hadamard, phase, and measurement gates only.\n\n3. is stabilized by exactly Pauli operators.\n\n4. is uniquely determined by , or the group of Pauli operators that stabilize .\n\nBecause of Theorem 1, we call any circuit consisting entirely of CNOT, Hadamard, phase, and measurement gates a stabilizer circuit, and any state obtainable by applying a stabilizer circuit to a stabilizer state. As a warmup to our later results, the following proposition counts the number of stabilizer states.\n\n###### Proposition 2\n\nLet be the number of pure stabilizer states on qubits. Then\n\n N=2nn−1∏k=0(2n−k+1)=2(1/2+o(1))n2.\n\nProof. We have , where is the total number of generating sets and is the number of equivalent generating sets for a given stabilizer . To find , note that there are choices for the first generator (ignoring overall sign), because it can be anything but the identity. The second generator must commute with and cannot be or , so there are choices for . Similarly, must commute with and , but cannot be in the group generated by them, so there are choices for it, and so on. Hence, including overall signs,\n\n G=2nn−1∏k=0(4n2k−2k)=2n(n+1)/2n−1∏k=0(4n−k−1).\n\nSimilarly, to find , note that given , there are choices for , choices for , choices for , and so on. Thus\n\n A=n−1∏k=0(2n−2k)=2n(n−1)/2n−1∏k=0(2n−k−1).\n\nTherefore\n\n N=GA=2nn−1∏k=0(4n−k−12n−k−1)=2nn−1∏k=0(2n−k+1).\n\n## Iii Efficient Simulation of Stabilizer Circuits\n\nTheorem 1 immediately suggests a way to simulate stabilizer circuits efficiently on a classical computer. A well-known fact from group theory says that any finite group has a generating set of size at most . So if is a stabilizer state on qubits, then the group of Pauli operators that stabilize has a generating set of size . Each generator takes bits to specify: bits for each of the Pauli matrices, and bit for the phase 111If , then can only have a phase of , not : for in the latter case would be in , but we saw that does not stabilize anything.. So the total number of bits needed to specify is . What Gottesman and Knill showed, furthermore, is that these bits can be updated in polynomial time after a CNOT, Hadamard, phase, or measurement gate is applied to . The updates corresponding to unitary gates are very efficient, requiring only time for each gate.\n\nHowever, the updates corresponding to measurements are not so efficient. We can decide in time whether a measurement of qubit will yield a deterministic or random outcome. If the outcome is random, then updating the state after the measurement takes time, but if the outcome is deterministic, then deciding whether the outcome is or seems to require inverting an matrix, which takes time in theory cw but order time in practice. What that complexity means is that simulations of, say, -qubit systems would already be prohibitive on a desktop PC, given that measurements are frequent.\n\nThis section describes a new simulation algorithm, by which both deterministic and random measurements can be performed in time. The cost is a factor- increase in the number of bits needed to specify a state. For in addition to the stabilizer generators, we now store “destabilizer” generators, which are Pauli operators that together with the stabilizer generators generate the full Pauli group . So the number of bits needed is .\n\nThe algorithm represents a state by a tableau consisting of binary variables for all , , and for all 222Dehaene and De Moor dm came up with something like this tableau representation independently, though they did not use it to simulate measurements in time.:\n\nRows to of the tableau represent the destabilizer generators , and rows to represent the stabilizer generators . If , then bits determine the Pauli matrix : means , means , means , and means . Finally, is if has negative phase and if has positive phase. As an example, the -qubit state is stabilized by the Pauli operators and , so a possible tableau for is\n\nIndeed, we will take the obvious generalization of the above “identity matrix” to be the standard initial tableau.\n\nThe algorithm uses a subroutine called , which sets generator equal to . Its purpose is to keep track, in particular, of the phase bit , including all the factors of that appear when multiplying Pauli matrices. The subroutine is implemented as follows.\n\nrowsum: Let be a function that takes bits as input, and that returns the exponent to which is raised (either , , or ) when the Pauli matrices represented by and are multiplied. More explicitly, if then ; if then ; if and then ; and if and then . Then set if\n\n 2rh+2ri+n∑j=1g(xij,zij,xhj,zhj)≡0(mod4),\n\nand set if the sum is congruent to mod (it will never be congruent to or ). Next, for all , set and set (here and throughout, denotes exclusive-OR).\n\nWe now give the algorithm. It will be convenient to add an additional row for “scratch space.” The initial state has for all , and and for all and , where is if and otherwise. The algorithm proceeds through the gates in order; for each one it does one of the following depending on the gate type.\n\nCNOT from control to target . For all , set , , and .\n\nHadamard on qubit . For all , set and swap with .\n\nPhase on qubit . For all , set and then set .\n\nMeasurement of qubit in standard basis. First check whether there exists a such that .\n\nCase I: Such a exists (if more than one exists, then let be the smallest). In this case the measurement outcome is random, so the state needs to be updated. This is done as follows. First call for all such that and . Second, set entire the row equal to the row. Third, set the row to be identically , except that is or with equal probability, and . Finally, return as the measurement outcome.\n\nCase II: Such an does not exist. In this case the outcome is determinate, so measuring the state will not change it; the only task is to determine whether or is observed. This is done as follows. First set the row to be identically . Second, call for all such that . Finally return as the measurement outcome.\n\nOnce we interpret the , , and bits for as representing generators of , and as representing the group operation in , the correctness of the CNOT, Hadamard, phase, and random measurement procedures follows immediately from previous analyses by Gottesman gottesman2 . It remains only to explain why the determinate measurement procedure is correct. Observe that commutes with if the symplectic inner product\n\n Rh⋅Ri=xh1zi1⊕⋯⊕xhnzin⊕xi1zh1⊕⋯⊕xinzhn\n\nequals , and anticommutes with if . Using that fact it is not hard to show the following.\n\n###### Proposition 3\n\nThe following are invariants of the tableau algorithm:\n\n• generate , and generate .\n\n• commute.\n\n• For all , anticommutes with .\n\n• For all such that , commutes with .\n\nNow suppose that a measurement of qubit yields a determinate outcome. Then the operator must commute with all elements of the stabilizer, so\n\n n∑h=1chRh+n=±Za\n\nfor a unique choice of . Our goal is to determine the ’s, since then by summing the appropriate ’s we can learn whether the phase representing the outcome is positive or negative. Notice that for all ,\n\n ci≡n∑h=1ch(Ri⋅Rh+n)≡Ri⋅n∑h=1chRh+n≡Ri⋅Za(mod2)\n\nby Proposition 3. Therefore by checking whether anticommutes with —which it does if and only if —we learn whether and thus whether needs to be called.\n\nWe end this section by explaining how to compute the inner product between two stabilizer states and , given their full tableaus. The inner product is if the stabilizers contain the same Pauli operator with opposite signs. Otherwise it equals , where is the minimum, over all sets of generators for and for , of the number of for which . For example, and have inner product , since . The proof is easy: it suffices to observe that neither the inner product nor is affected if we transform and to and respectively, for some unitary such that has the trivial stabilizer. This same observation yields an algorithm to compute the inner product: first transform the tableau of to that of using Theorem 8; then perform Gaussian elimination on the tableau of to obtain . Unfortunately, this algorithm takes order steps.\n\n## Iv Implementation and Experiments\n\nWe have implemented the tableau algorithm of Section III in a C program called CHP (CNOT-Hadamard-Phase), which is available for download 333At www.scottaaronson.com/chp. CHP takes as input a program in a simple “quantum assembly language,” consisting of four instructions: c (apply CNOT from control to target ), h (apply Hadamard to ), p (apply phase gate to ), and m (measure in the standard basis, output the result, and update the state accordingly). Here and are nonnegative integers indexing qubits; the maximum or that occurs in any instruction is assumed to be , where is the number of qubits. As an example, the following program demonstrates the famous quantum teleportation protocol of Bennett et al. bbcjpw :\n\nEPR pair is prepared (qubit is Alice’s half; qubit is Bob’s half)\n\nAlice interacts qubit (the state to be teleported) with her half of the EPR pair\n\nAlice sends classical bits to Bob\n\nBob uses the bits from Alice to recover the teleported state\n\nWe also have available CHP programs that demonstrate the Bennett-Wiesner dense quantum coding protocol bw , the GHZ (Greenberger-Horne-Zeilinger) experiment ghz , Simon’s algorithm simon , and the Shor -qubit quantum error-correcting code shor .\n\nOur main design goal for CHP was high performance with a large number of qubits and frequent measurements. The only reason to use CHP instead of a general-purpose quantum computer simulator such as QuIDD vmh or QCL oemer is performance, so we wanted to leverage that advantage and make thousands of qubits easily simulable rather than just hundreds. Also, the results of Section V suggest that classical postprocessing is unavoidable for stabilizer circuits, since stabilizer circuits are not even universal for classical computation. So if we want to simulate (for example) Simon’s algorithm, then one measurement is needed for each bit of the first register. CHP’s execution time will be dominated by these measurements, since as discussed in Section III, each unitary gate takes only time to simulate.\n\nOur experimental results, summarized in Figure 2, show that CHP makes practical the simulation of arbitrary stabilizer circuits on up to about qubits. Since the number of bits needed to represent qubits grows quadratically in , the main limitation is available memory. On a machine with 256MB of RAM, CHP can handle up to about qubits before virtual memory is needed, in which case thrashing makes its performance intolerable. The original version of CHP required ~ bits for memory; we were able to reduce this to ~ bits, enabling a 41% increase in the number of qubits for a fixed memory size. More trivially, we obtained an eightfold improvement in memory by storing bits to each byte instead of . Not only did that change increase the number of storable qubits by 183%, but it also made CHP about 50% faster—presumably because (1) the subroutine now needed to exclusive-OR only as many bytes, and (2) the memory penalty was reduced. Storing the bits in -bit words yielded a further 10% performance gain, presumably because of (1) rather than (2) (since even with byte-addressing, a whole memory line is loaded into the cache on a cache miss).\n\nAs expected, the experimentally measured execution time per unitary gate grows linearly in , whereas the time per measurement grows somewhere between linearly and quadratically, depending on the states being measured. Thus the time needed for measurements generally dominates execution time. So the key question is this: what properties of a circuit determine whether the time per measurement is linear, quadratic, or somewhere in between? To investigate this question we performed the following experiment.\n\nWe randomly generated stabilizer circuits on qubits, for ranging from to in increments of . For each , we used the following distribution over circuits: Fix a parameter ; then choose random unitary gates: a CNOT from control to target , a Hadamard on qubit , or a phase gate on qubit , each with probability , where and are drawn uniformly at random from subject to . Then measure qubit for each in sequence.\n\nWe simulated the resulting circuits in CHP. For each circuit, we counted the number of seconds needed for all measurement steps (ignoring the time for unitary gates), then divided by to obtain the number of seconds per measurement. We repeated the whole procedure for ranging from to in increments of .\n\nThere were several reasons for placing measurements at the end of a circuit rather than interspersing them with unitary gates. First, doing so models how many quantum algorithms actually work (apply unitary gates, then measure, then perform classical postprocessing); second, it allowed us to ignore the effect of measurements on subsequent computation; third, it ‘standardized’ the measurement stage, making comparisons between different circuits more meaningful; and fourth, it made simulation harder by increasing the propensity for the measurements to be nontrivially correlated.\n\nThe decision to make the number of unitary gates proportional to was based on the following heuristic argument. The time needed to simulate a measurement is determined by how many times the procedure is called, which in turn is determined by how many ’s there are such that (where is the qubit being measured). Initially if and only if , so a measurement takes time. For a random state, by contrast, the expected number of ’s such that is by symmetry, so a measurement takes order time. In general, the more ’s there are in the tableau, the longer measurements take. But where does the transition from linear to quadratic time occur, and how sharp is it?\n\nConsider people, each of whom initially knows one secret (with no two people knowing the same secret). Each day, two people chosen uniformly at random meet and exchange all the secrets they know. What is the expected number of days until everyone knows everyone else’s secrets? Intuitively, the answer is , because any given person has to wait days between meetings, and at each meeting, the number of secrets he knows approximately doubles (or towards the end, the number of secrets he doesn’t know is approximately halved). Replacing people by qubits and meetings by CNOT gates, one can see why a ‘phase transition’ from a sparse to a dense tableau might occur after random unitary gates are applied. However, this argument does not pin down the proportionality constant , so that is what we varied in the experiment.\n\nThe results of the experiment are presented in Figure 2. When , the time per measurement appears to grow roughly linearly in , whereas when (meaning that the number of unitary gates has only doubled), the time per measurement appears to grow roughly quadratically, so that running the simulations took hours of computing time 444Based on our heuristic analysis, we conjecture that for intermediate , the time per measurement grows as for some . However, we do not have enough data to confirm or refute this conjecture. Thus, Figure 2 gives striking evidence for a “phase transition” in simulation time, as increasing the number of unitary gates by only a constant factor shifts us from a regime of simple states that are easy to measure, to a regime of complicated states that are hard to measure. This result demonstrates that CHP’s performance depends strongly on the circuit being simulated. Without knowing what sort of tableaus a circuit will produce, all we can say is that the time per measurement will be somewhere between linear and quadratic in .", null, "Figure 2: Average time needed to simulate a measurement after applying βnlog2n unitary gates to n qubits, on a 650MHz Pentium III with 256MB RAM.\n\n## V Complexity of Simulating Stabilizer Circuits\n\nThe Gottesman-Knill theorem shows that stabilizer circuits are not universal for quantum computation, unless quantum computers can be simulated efficiently by classical ones. To a computer scientist, this theorem immediately raises a question: where do stabilizer circuits sit in the hierarchy of computational complexity theory? In this section we resolve that question, by proving that the problem of simulating stabilizer circuits is complete for a classical complexity class known as (pronounced “parity-L”) 555See www.complexityzoo.com for definitions of and several hundred other complexity classes. The usual definition of is as the class of all problems that are solvable by a nondeterministic logarithmic-space Turing machine, that accepts if and only if the total number of accepting paths is odd. But there is an alternate definition that is probably more intuitive to non-computer-scientists. This is that is the class of problems that reduce to simulating a polynomial-size CNOT circuit, i.e. a circuit composed entirely of NOT and CNOT gates, acting on the initial state . (It is easy to show that the two definitions are equivalent, but this would require us first to explain what the usual definition means!)\n\nFrom the second definition, it is clear that ; in other words, any problem reducible to simulating CNOT circuits is also solvable in polynomial time on a classical computer. But this raises a question: what do we mean by “reducible”? Problem is reducible to problem if any instance of problem can be transformed into an instance of problem ; this means that problem is “harder” than problem in the sense that the ability to answer an arbitrary instance of problem implies the ability to answer an arbitrary instance of problem (but not necessarily vice-versa).\n\nWe must, however, insist that the reduction transforming instances of problem into instances of problem not be too difficult to perform. Otherwise, we could reduce hard problems to easy ones by doing all the difficult work in the reduction itself. In the case of , we cannot mean “reducible in polynomial time,” which is a common restriction, since then the reduction would be at least as powerful as the problem it reduces to! Instead we require the reduction to be performed in the complexity class , or logarithmic space—that is, by a Turing machine that is given a read-only input of size , and a write-only output tape, but only bits of read/write memory. The reduction works as follows: first specifies a CNOT circuit on its output tape; then an “oracle” tells the circuit’s output (which we can take to be, say, the value of the first qubit after the circuit is applied), then specifies another CNOT circuit on its output tape, and so on. A useful result of Hertrampf, Reith, and Vollmer hrv says that this seemingly powerful kind of reduction, in which can make multiple calls to the CNOT oracle, is actually no more powerful than the kind with only one oracle call. (In complexity language, what hrv showed is that : any problem in with oracle is also in itself.)\n\nIt is conjectured that ; in other words, that an oracle for simulating CNOT circuits would let an machine compute more functions than it could otherwise. Intuitively, this is because writing down the intermediate states of such a circuit requires more than a logarithmic number of read/write bits. Indeed, contains some surprisingly “hard” problems, such as inverting matrices over damm . On the other hand, it is also conjectured that , meaning that even with an oracle for simulating CNOT circuits, an machine could not simulate more general circuits with AND and OR gates. As usual in complexity theory, neither conjecture has been proved.\n\nNow define the Gottesman-Knill problem as follows. We are given a stabilizer circuit as a sequence of gates of the form , , , or , where are indices of qubits. The problem is to decide whether qubit will be with certainty after is applied to the initial state . (If not, then qubit will be with probability either or .)\n\nSince stabilizer circuits are a generalization of CNOT circuits, it is obvious that Gottesman-Knill is -hard (i.e. any problem can be reduced to it). Our result says that Gottesman-Knill is in . Intuitively, this means that any stabilizer circuit can be simulated efficiently using CNOT gates alone—the additional availability of Hadamard and phase gates gives stabilizer circuits at most a polynomial advantage. In our view, this surprising fact helps to explain the Gottesman-Knill theorem, by providing strong evidence that stabilizer circuits are not even universal for classical computation (assuming, of course, that classical postprocessing is forbidden).\n\n###### Theorem 4\n\nGottesman-Knill is in .\n\nProof. We will show how to solve Gottesman-Knill using a logarithmic-space machine with an oracle for simulating CNOT circuits. By the result of Hertrampf, Reith, and Vollmer hrv described above, this will suffice to prove the theorem.\n\nBy the principle of deferred measurement, we can assume that the stabilizer circuit has only a single measurement gate at the end (say of qubit ), with all other measurements replaced by CNOT’s into ancilla qubits. In the tableau algorithm of Section III, let be the values of the variables after gates of have been applied. Then will simulate by computing these values. The first task of is to decide whether the measurement has a determinate outcome—or equivalently, whether for every , where is the number of unitary gates. Observe that in the CNOT, Hadamard, and phase procedures, every update to an or variable replaces it by the sum modulo of one or two other or variables. Also, iterating over all and takes only bits of memory. Therefore, despite its memory restriction, can easily write on its output tape a description of a CNOT circuit that simulates the tableau algorithm using bits (the ’s being omitted), and that returns for any desired . Then to decide whether the measurement outcome is determinate, simply iterates over all from to .\n\nThe hard part is to decide whether or is measured in case the measurement outcome is determinate, for this problem involves the variables, which do not evolve in a linear way as the ’s and ’s do. Even worse, it involves the complicated-looking and nonlinear procedure. Fortunately, though, it turns out that the measurement outcome can be computed by keeping track of a single complex number . This is a product of phases of the form or , and therefore takes only bits to specify. Furthermore, although the “obvious” ways to compute use more than bits of memory, can get around that by making liberal use of the oracle.\n\nFirst computes what would be if the CNOT, Hadamard, and phase procedures did not modify the ’s. Let be a Pauli matrix with a phase of or , which therefore takes bits to specify. Also, let be the Pauli matrix represented by the bits in the usual way: , , , . Then the procedure is as follows.\n\nfor to\n\nfor to\n\nif then\n\nnext\n\nmultiply by the phase of ( or )\n\nnext\n\nThe “answer” is if and if (note that will never be at the end). However, also needs to account for the ’s, as follows.\n\nfor to\n\nif\n\nfor to\n\nif gate is a Hadamard or phase on\n\nif then\n\nend if\n\nif gate is a CNOT from to\n\nif then\n\nend if\n\nnext\n\nend if\n\nnext\n\nThe measurement outcome, , is then if and if . As described above, the machine needs only bits to keep track of the loop indices , and additional bits to keep track of other variables. Its correctness follows straightforwardly from the correctness of the tableau algorithm.\n\nFor a problem to be -complete simply means that it is -hard and in . Thus, a corollary of Theorem 4 is that Gottesman-Knill is -complete.\n\n## Vi Canonical Form\n\nHaving studied the simulation of stabilizer circuits, in this section we turn our attention to manipulating those circuits. This task is of direct relevance to quantum computer architecture: because the effects of decoherence build up over time, it is imperative (even more so than for classical circuits) to minimize the number of gates as well as wires and other resources. Even if fault-tolerant techniques will eventually be used to tame decoherence, there remains the bootstrapping problem of building the fault-tolerance hardware! In that regard we should point out that fault-tolerance hardware is likely to consist mainly of CNOT, Hadamard, and phase gates, since the known fault-tolerant constructions (for example, that of Aharonov and Ben-Or ab ) are based on stabilizer codes.\n\nAlthough there has been some previous work on synthesizing CNOT circuits iky ; pmh ; mn and general classical reversible circuits spmh ; lckl , to our knowledge there has not been work on synthesizing stabilizer circuits. In this section we prove a canonical form theorem that is extremely useful for stabilizer circuit synthesis. The theorem says that given any circuit consisting of CNOT, Hadamard, and phase gates, there exists an equivalent circuit that applies a round of Hadamard gates only, then a round of CNOT gates only, and so on in the sequence H-C-P-C-P-C-H-P-C-P-C. One easy corollary of the theorem is that any tableau satisfying the commutativity conditions of Proposition 3 can be generated by some stabilizer circuit. Another corollary is that any unitary stabilizer circuit has an equivalent circuit with only gates.\n\nGiven two -qubit unitary stabilizer circuits , we say that and are equivalent if for all stabilizer states , where is the final state when is applied to 666The reason we restrict attention to unitary circuits is simply that, if measurements are included, then it is unclear what it even means for two circuits to be equivalent. For example, does deferring all measurements to the end of a computation preserve equivalence or not?. By linearity, it is easy to see that equivalent stabilizer circuits will behave identically on all states, not just stabilizer states. Furthermore, there exists a one-to-one correspondence between circuits and tableaus:\n\n###### Lemma 5\n\nLet be unitary stabilizer circuits, and let be their respective final tableaus when we run them on the standard initial tableau. Then and are equivalent if and only if .\n\nProof. Clearly if and are equivalent. For the other direction, it suffices to observe that a unitary stabilizer circuit acts linearly on Pauli operators (that is, rows of the tableau): if it maps to and to , then it maps to . Since the rows of the standard initial tableau form a basis for , the lemma follows.\n\nOur proof of the canonical form theorem will use the following two lemmas.\n\n###### Lemma 6\n\nGiven an -qubit stabilizer state, it is always possible to apply Hadamard gates to a subset of the qubits so as to make the matrix have full rank (or equivalently, make all basis states have nonzero amplitude).\n\nProof. We can always perform row additions on the stabilizer matrix without changing the state that it represents. Suppose the matrix has rank ; then by Gaussian elimination, we can put the stabilizer matrix in the form\n\nwhere is and has rank . Then since the rows are linearly independent, must have rank ; therefore it has an submatrix of full rank. Let us permute the columns of the and matrices simultaneously to obtain\n\nand then perform Gaussian elimination on the bottom rows to obtain\n\nNow commutativity relations imply\n\n (A1A2)(DTI)=0\n\nand therefore . Notice that this implies that the matrix has full rank, since otherwise the matrix would have column rank less than . So performing Hadamards on the rightmost qubits yields a state\n\nwhose matrix has full rank.\n\n###### Lemma 7\n\nFor any symmetric matrix , there exists a diagonal matrix such that , with some invertible binary matrix.\n\nProof. We will let be a lower-triangular matrix with s all along the diagonal:\n\n Mii=1 (1) Mij=0 i\n\nSuch an is always invertible. Then there exists a diagonal such that if and only if\n\n Aij=∑kMikMjk (3)\n\nfor all pairs with . (We pick appropriately to satisfy the equations for automatically, and both sides of the equation are symmetric, covering the cases with .)\n\nWe will perform induction on and to solve for the undetermined elements of . For the base case, we know that . We will determine for by supposing we have already determined for either , or , . We consider equation (3) for and note that unless . Then\n\n Aij=∑k\n\nBy the induction hypothesis, we have already determined in the sum both (since ) and (since and ), so this equation uniquely determines . We can thus find a unique that satisfies (3) for all .\n\nSay a unitary stabilizer circuit is in canonical form if it consists of rounds in the sequence H-C-P-C-P-C-H-P-C-P-C.\n\n###### Theorem 8\n\nAny unitary stabilizer circuit has an equivalent circuit in canonical form.\n\nProof. Divide a tableau into four matrices , , , and , containing the destabilizer bits, destabilizer bits, stabilizer bits, and stabilizer bits respectively:\n\n(We can ignore the phase bits .) Since unitary circuits are reversible, by Lemma 5 it suffices to show how to obtain the standard initial tableau starting from an arbitrary by applying CNOT, Hadamard, and phase gates 777Actually, this gives the canonical form for the inverse of the circuit, but of course the same argument holds for the inverse circuit too, which is also a stabilizer circuit. We cannot use row additions, since although they leave states invariant they do not in general leave circuits invariant.\n\nThe procedure is as follows.\n\n(1) Use Hadamards to make have full rank (this is possible by Lemma 6).\n\n(2) Use CNOT’s to perform Gaussian elimination on , producing\n\n(3) Commutativity of the stabilizer implies that is symmetric, therefore is symmetric, and we can apply phase gates to add a diagonal matrix to and use Lemma 7 to convert to the form for some invertible .\n\n(4) Use CNOT’s to produce\n\nNote that when we map to , we also map to .\n\n(5) Apply phases to all qubits to obtain\n\nSince is full rank, there exists some subset of qubits such that applying two phases in succession to every will preserve the above tableau, but set . Apply two phases to every .\n\n(6) Use CNOT’s to perform Gaussian elimination on , producing\n\nBy commutativity relations, , therefore .\n\n(8) Now commutativity of the destabilizer implies that is symmetric, therefore we can again use phase gates and Lemma 7 to make for some invertible .\n\n(9) Use CNOT’s to produce\n\n(10) Use phases to produce\n\nthen by commutativity relations, . Next apply two phases each to some subset of qubits in order to preserve the above tableau, but set .\n\n(11) Use CNOT’s to produce\n\nSince Theorem 8 relied only on a tableau satisfying the commutativity conditions, not on its being generated by some stabilizer circuit, an immediate corollary is that any tableau satisfying the conditions is generated by some stabilizer circuit. We can also use Theorem 8 to answer the following question: how many gates are needed for an -qubit stabilizer circuit in the worst case? Cleve and Gottesman cg showed that gates suffice for the special case of state preparation, and Gottesman gottesman3 and Dehaene and De Moor dm showed that gates suffice for stabilizer circuits more generally; even these results were not obvious a priori. However, with the help of our canonical form theorem we can show a stronger upper bound.\n\n###### Corollary 9\n\nAny unitary stabilizer circuit has an equivalent circuit with only gates.\n\nProof. Patel, Markov, and Hayes pmh showed that any CNOT circuit has an equivalent CNOT circuit with only gates. So given a stabilizer circuit , first put into canonical form, then minimize the CNOT segments. Clearly the Hadamard and Phase segments require only gates each.\n\nCorollary 9 is easily seen to be optimal by a Shannon counting argument: there are distinct stabilizer circuits on qubits, but at most with gates.\n\nA final remark: as noted by Moore and Nilsson mn , any CNOT circuit has an equivalent CNOT circuit with gates and parallel depth . Thus, using the same idea as in Corollary 9, we obtain that any unitary stabilizer circuit has an equivalent stabilizer circuit with gates and parallel depth . (Moore and Nilsson showed this for the special case of stabilizer circuits composed of CNOT and Hadamard gates only.)\n\n## Vii Beyond Stabilizer Circuits\n\nIn this section, we discuss generalizations of stabilizer circuits that are still efficiently simulable. The first (easy) generalization, in Section VII.1, is to allow the quantum computer to be in a mixed rather than a pure state. Mixed states could be simulated by simply purifying the state, and then simulating the purification, but we present an alternative and slightly more efficient strategy.\n\nThe second generalization, in Section VII.2, is to initial states other than the computational basis state. Taken to an extreme, one could even have noncomputable initial states. When combined with arbitrary quantum circuits, such quantum advice is very powerful, although its exact power (relative to classical advice) is unknown aaronson . We consider a more modest situation, in which the initial state may include specific ancilla states, consisting of at most qubits each. The initial state is therefore a tensor product of blocks of qubits. Given an initial state of this form and general stabilizer circuits, including measurements and classical feedback based on measurement outcomes, universal quantum computation is again possible shor2 ; gc . However, we show that an efficient classical simulation exists, provided only a few measurements are allowed.\n\nThe final generalization, in Section VII.3, is to circuits containing a few non-stabilizer gates. The qualifier “few” is essential here, since it is known that unitary stabilizer circuits plus any additional gate yields a universal set of quantum gates nrs ; solovay . The running time of our simulation procedure is polynomial in , the number of qubits, but is exponential in the , the number of non-stabilizer gates.\n\n### vii.1 Mixed States\n\nWe first present the simulation for mixed states. We allow only stabilizer mixed states—that is, states that are uniform distributions over all states in a subspace (or equivalently, all stabilizer states in the subspace) with a given stabilizer of generators. Such mixed states can always be written as the partial trace of a pure stabilizer state, which immediately provides one way of simulating them.\n\nIt will be useful to see how to write the density matrix of the mixed state in terms of the stabilizer. The operator , when is a Pauli operator, is a projection onto the eigenspace of . Therefore, if the stabilizer of a pure state has generators , then the density matrix for that state is\n\n ρ=12nn∏i=1(I+Mi).\n\nThe density matrix for a stabilizer mixed state with stabilizer generated by is\n\n ρ=12rr∏i=1(I+Mi).\n\nTo perform our simulation, we find a collection of operators and that commute with both the stabilizer and the destabilizer. We can choose them so that for , but . This can be done by solving a set of linear equations, which in practice takes time . If we start with an initial mixed state, we will assume it is of the form (so on the first qubits and the completely mixed state on the last qubits). In that case, we choose and .\n\nWe could purify this state by adding and to the stabilizer and and to the destabilizer for . Then we could simulate the system by just simulating the evolution of this pure state through the circuit; the extra qubits are never altered.\n\nA more economical simulation is possible, however, by just keeping track of the original -generator stabilizer and destabilizer, plus the operators and . Formally, this allows us to maintain a complete tableau and generalize the tableau algorithm from Section III. We place the generators of the stabilizer as rows of the tableau, and the corresponding elements of the destabilizer as rows . The new operators and () become rows and , respectively. Let if and if . Then we have that rows and commute unless , in which case and anticommute.\n\nWe can keep track of this new kind of tableau in much the same way as the old kind. Unitary operations transform the new rows the same way as rows of the stabilizer or destabilizer. For example, to perform a CNOT from control qubit to target qubit , set and , for all .\n\nMeasurement of qubit is slightly more complex than before. There are now three cases:\n\nCase I: for some . In this case anticommutes with an element of the stabilizer, and the measurement outcome is random. We update as before, for all rows of the tableau.\n\nCase II: for all . In this case is in the stabilizer. The measurement outcome is determinate, and we can predict the result as before, by calling to add up rows for those with .\n\nCase III: for all , but for some or . In this case commutes with all elements of the stabilizer but is not itself in the stabilizer. We get a random measurement result, but a slightly different transformation of the stabilizer than in Case I. Observe that row anticommutes with . This row takes the role of row from Case I, and the row takes the role of row . Update as before with this modification. Then swap rows and and rows and . Finally, increase to : the stabilizer has gained a new generator.\n\nAnother operation that we might want to apply is discarding the qubit , which has the effect of performing a partial trace over that qubit in the density matrix. Again, this can be done by simply keeping the qubit in our simulation and not using it in future operations. Here is an alternative: put the stabilizer in a form such that there is at most one generator with an on qubit , and at most one with a on qubit . Then drop those two generators (or one, if there is only one total). The remaining generators describe the stabilizer of the reduced mixed state. We also must put the and operators in a form where they have no entries in the discarded location, while preserving the structure of the tableau (namely, the commutation relations of Proposition 3). This can also be done in time , but we omit the details, as they are rather involved.\n\n### vii.2 Non-Stabilizer Initial States\n\nWe now show how to simulate a stabilizer circuit where the initial state is more general, involving non-stabilizer initial states. We allow any number of ancillas in arbitrary states, but the overall ancilla state must be a tensor product of blocks of at most qubits each. An arbitrary stabilizer circuit is then applied to this state. We allow measurements, but only of them in total throughout the computation. We do allow classical operations conditioned on the outcomes of measurements, so we also allow polynomial-time classical computation during the circuit.\n\nLet the initial state have density matrix : a tensor product of blocks of at most qubits each. Without loss of generality, we first apply the unitary stabilizer circuit , followed by the measurement (that is, a measurement of the first qubit in the standard basis). We then apply the stabilizer circuit , followed by measurement on the second qubit, and so on up to .\n\nWe can calculate the probability of obtaining outcome for the first measurement as follows:\n\n p(0) =Tr[(I+Z1)U1ρU†1]/2 =Tr[(I+U†1Z1U1)ρ]/2 =1/2+Tr[(U†1Z1U1)ρ]/2.\n\nBut is a stabilizer operation, so is a Pauli matrix, and is therefore a tensor product operation. We also know is a tensor product of blocks of at most qubits, and the trace of a tensor product is the product of the traces. Let and where ranges over the blocks. Then\n\n p(0)=12+m∏j=1Tr(Pjρj).\n\nSince and are both -dimensional matrices, each can be computed in time .\n\nBy flipping an appropriately biased coin, Alice can generate an outcome of the first measurement according to the correct probabilities. Conditioned on this outcome (say of ), the state of the system is\n\n (I+Z1)U1ρU†1(1+Z1)4p(0) .\n\nAfter the next stabilizer circuit , the state is\n\n U2(I+Z1)U1ρU†1(1+Z1)U†24p(0) .\n\nThe probability of obtaining outcome for the second measurement, conditioned on the outcome of the first measurement being , is then\n\nBy expanding out the terms, and then commuting and past and , we can write this as\n\n 8∑i=1m∏j=1Tr(P(2)ijρij).\n\nEach term can again be computed in time .\n\nSimilarly, the probability of any particular sequence of measurement outcomes can be written as a sum\n\n p(m1m2⋯md)=22d−1∑i=1m∏j=1Tr(P(d)ijρij),\n\nwhere each trace can be computed in time . It follows that the probabilities of the two outcomes of the measurement can be computed in time" ]	[ null, "https://media.arxiv-vanity.com/render-output/2994480/x1.png", null, "https://media.arxiv-vanity.com/render-output/2994480/x2.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92280614,"math_prob":0.9803029,"size":50386,"snap":"2020-24-2020-29","text_gpt3_token_len":10621,"char_repetition_ratio":0.16940574,"word_repetition_ratio":0.048440453,"special_character_ratio":0.19868614,"punctuation_ratio":0.11133646,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9921488,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-02T08:46:38Z\",\"WARC-Record-ID\":\"<urn:uuid:ee454ae4-e131-4854-89db-b8af8b7486d1>\",\"Content-Length\":\"1049277\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fb5946b1-3d4b-4578-a3a2-723dab147fe7>\",\"WARC-Concurrent-To\":\"<urn:uuid:ed1ba319-bc9a-4684-a4f4-f9f82eeb8e98>\",\"WARC-IP-Address\":\"172.67.158.169\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/quant-ph/0406196/\",\"WARC-Payload-Digest\":\"sha1:BBJMQVRSV7ZNKWHIOXN2DZA7CH6NY6OY\",\"WARC-Block-Digest\":\"sha1:A6DPIMKE7Z722UDCDSKGZOBX3XAXQUZP\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347423915.42_warc_CC-MAIN-20200602064854-20200602094854-00497.warc.gz\"}"}
https://www.ias.ac.in/listing/bibliography/pram/R._Misra	[ "• R Misra\n\nArticles written in Pramana – Journal of Physics\n\n• Vibrational dynamics of the orgganometallic compound triarylorganoantimony (V) SbPh3 [O_{2}CC(OH)Ph_{2}]_{2}\n\nA complete normal coordinate analysis was performed for five-coordinate non-rigid triarylantimony diester SbPh3(O2CR2), known to be a bioactive molecule, using Wilson G-D matrix method and Urey Bradley force field. The study of vibrational dynamics was performed using the concept of group frequencies and band intensities.\n\n• Efficient use of correlation entropy for analysing time series data\n\nThe correlation dimension $D_{2}$ and correlation entropy $K_{2}$ are both important quantifiers in nonlinear time series analysis. However, use of $D_{2}$ has been more common compared to $K_{2}$ as a discriminating measure. One reason for this is that $D_{2}$ is a static measure and can be easily evaluated from a time series. However, in many cases, especially those involving coloured noise, $K_{2}$ is regarded as a more useful measure. Here we present an efficient algorithmic scheme to compute $K_{2}$ directly from a time series data and show that $K_{2}$ can be used as a more effective measure compared to $D_{2}$ for analysing practical time series involving coloured noise.\n\n• # Editorial Note on Continuous Article Publication\n\nPosted on July 25, 2019" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91227174,"math_prob":0.94913226,"size":1548,"snap":"2023-14-2023-23","text_gpt3_token_len":385,"char_repetition_ratio":0.10427461,"word_repetition_ratio":0.0,"special_character_ratio":0.2370801,"punctuation_ratio":0.051282052,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9900938,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-24T04:04:29Z\",\"WARC-Record-ID\":\"<urn:uuid:05adeddf-a9f6-4816-8d73-ba545912fb8d>\",\"Content-Length\":\"27025\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ea7192be-3d26-452c-8408-6ec2cadd9f44>\",\"WARC-Concurrent-To\":\"<urn:uuid:6b9bad12-7425-45f1-920c-3246f5f56825>\",\"WARC-IP-Address\":\"52.66.166.230\",\"WARC-Target-URI\":\"https://www.ias.ac.in/listing/bibliography/pram/R._Misra\",\"WARC-Payload-Digest\":\"sha1:D3ESZPJZ4VYLYTJ7LPY5HGTEKS7TH37S\",\"WARC-Block-Digest\":\"sha1:DW6XG4POQ7ADCRF6HY4HRPJAJ7XBGKET\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945242.64_warc_CC-MAIN-20230324020038-20230324050038-00314.warc.gz\"}"}
https://la.mathworks.com/help/econ/what-is-a-conditional-variance-model.html	[ "## Conditional Variance Models\n\n### General Conditional Variance Model Definition\n\nConsider the time series\n\n`${y}_{t}=\\mu +{\\epsilon }_{t},$`\n\nwhere ${\\epsilon }_{t}={\\sigma }_{t}{z}_{t}$. Here, zt is an independent and identically distributed series of standardized random variables. Econometrics Toolbox™ supports standardized Gaussian and standardized Student’s t innovation distributions. The constant term, $\\mu$, is a mean offset.\n\nA conditional variance model specifies the dynamic evolution of the innovation variance,\n\n`${\\sigma }_{t}^{2}=Var\\left({\\epsilon }_{t}\|{H}_{t-1}\\right),$`\n\nwhere Ht–1 is the history of the process. The history includes:\n\n• Past variances, ${\\sigma }_{1}^{2},{\\sigma }_{2}^{2},\\dots ,{\\sigma }_{t-1}^{2}$\n\n• Past innovations, ${\\epsilon }_{1},{\\epsilon }_{2},\\dots ,{\\epsilon }_{t-1}$\n\nConditional variance models are appropriate for time series that do not exhibit significant autocorrelation, but are serially dependent. The innovation series ${\\epsilon }_{t}={\\sigma }_{t}{z}_{t}$ is uncorrelated, because:\n\n• E(εt) = 0.\n\n• E(εtεt–h) = 0 for all t and $h\\ne 0.$\n\nHowever, if ${\\sigma }_{t}^{2}$ depends on ${\\sigma }_{t-1}^{2}$, for example, then εt depends on εt–1, even though they are uncorrelated. This kind of dependence exhibits itself as autocorrelation in the squared innovation series, ${\\epsilon }_{t}^{2}.$\n\nTip\n\nFor modeling time series that are both autocorrelated and serially dependent, you can consider using a composite conditional mean and variance model.\n\nTwo characteristics of financial time series that conditional variance models address are:\n\n• Volatility clustering. Volatility is the conditional standard deviation of a time series. Autocorrelation in the conditional variance process results in volatility clustering. The GARCH model and its variants model autoregression in the variance series.\n\n• Leverage effects. The volatility of some time series responds more to large decreases than to large increases. This asymmetric clustering behavior is known as the leverage effect. The EGARCH and GJR models have leverage terms to model this asymmetry.\n\n### GARCH Model\n\nThe generalized autoregressive conditional heteroscedastic (GARCH) model is an extension of Engle’s ARCH model for variance heteroscedasticity . If a series exhibits volatility clustering, this suggests that past variances might be predictive of the current variance.\n\nThe GARCH(P,Q) model is an autoregressive moving average model for conditional variances, with P GARCH coefficients associated with lagged variances, and Q ARCH coefficients associated with lagged squared innovations. The form of the GARCH(P,Q) model in Econometrics Toolbox is\n\n`${y}_{t}=\\mu +{\\epsilon }_{t},$`\n\nwhere${\\epsilon }_{t}={\\sigma }_{t}{z}_{t}$ and\n\n`${\\sigma }_{t}^{2}=\\kappa +{\\gamma }_{1}{\\sigma }_{t-1}^{2}+\\dots +{\\gamma }_{P}{\\sigma }_{t-P}^{2}+{\\alpha }_{1}{\\epsilon }_{t-1}^{2}+\\dots +{\\alpha }_{Q}{\\epsilon }_{t-Q}^{2}.$`\n\nNote\n\nThe `Constant` property of a `garch` model corresponds to κ, and the `Offset` property corresponds to μ.\n\nFor stationarity and positivity, the GARCH model has the following constraints:\n\n• $\\kappa >0$\n\n• ${\\gamma }_{i}\\ge 0,\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}{\\alpha }_{j}\\ge 0$\n\n• ${\\sum }_{i=1}^{P}{\\gamma }_{i}+{\\sum }_{j=1}^{Q}{\\alpha }_{j}<1$\n\nTo specify Engle’s original ARCH(Q) model, use the equivalent GARCH(0,Q) specification.\n\n### EGARCH Model\n\nThe exponential GARCH (EGARCH) model is a GARCH variant that models the logarithm of the conditional variance process. In addition to modeling the logarithm, the EGARCH model has additional leverage terms to capture asymmetry in volatility clustering.\n\nThe EGARCH(P,Q) model has P GARCH coefficients associated with lagged log variance terms, Q ARCH coefficients associated with the magnitude of lagged standardized innovations, and Q leverage coefficients associated with signed, lagged standardized innovations. The form of the EGARCH(P,Q) model in Econometrics Toolbox is\n\n`${y}_{t}=\\mu +{\\epsilon }_{t},$`\n\nwhere ${\\epsilon }_{t}={\\sigma }_{t}{z}_{t}$ and\n\n`$\\mathrm{log}{\\sigma }_{t}^{2}=\\kappa +\\sum _{i=1}^{P}{\\gamma }_{i}\\mathrm{log}{\\sigma }_{t-i}^{2}+\\sum _{j=1}^{Q}{\\alpha }_{j}\\left[\\frac{\|{\\epsilon }_{t-j}\|}{{\\sigma }_{t-j}}-E\\left\\{\\frac{\|{\\epsilon }_{t-j}\|}{{\\sigma }_{t-j}}\\right\\}\\right]+\\sum _{j=1}^{Q}{\\xi }_{j}\\left(\\frac{{\\epsilon }_{t-j}}{{\\sigma }_{t-j}}\\right).$`\n\nNote\n\nThe `Constant` property of an `egarch` model corresponds to κ, and the `Offset` property corresponds to μ.\n\nThe form of the expected value terms associated with ARCH coefficients in the EGARCH equation depends on the distribution of zt:\n\n• If the innovation distribution is Gaussian, then\n\n`$E\\left\\{\\frac{\|{\\epsilon }_{t-j}\|}{{\\sigma }_{t-j}}\\right\\}=E\\left\\{\|{z}_{t-j}\|\\right\\}=\\sqrt{\\frac{2}{\\pi }}.$`\n\n• If the innovation distribution is Student’s t with ν > 2 degrees of freedom, then\n\n`$E\\left\\{\\frac{\|{\\epsilon }_{t-j}\|}{{\\sigma }_{t-j}}\\right\\}=E\\left\\{\|{z}_{t-j}\|\\right\\}=\\sqrt{\\frac{\\nu -2}{\\pi }}\\frac{\\Gamma \\left(\\frac{\\nu -1}{2}\\right)}{\\Gamma \\left(\\frac{\\nu }{2}\\right)}.$`\n\nThe toolbox treats the EGARCH(P,Q) model as an ARMA model for$\\mathrm{log}{\\sigma }_{t}^{2}.$ Thus, to ensure stationarity, all roots of the GARCH coefficient polynomial,$\\left(1-{\\gamma }_{1}L-\\dots -{\\gamma }_{P}{L}^{P}\\right)$, must lie outside the unit circle.\n\nThe EGARCH model is unique from the GARCH and GJR models because it models the logarithm of the variance. By modeling the logarithm, positivity constraints on the model parameters are relaxed. However, forecasts of conditional variances from an EGARCH model are biased, because by Jensen’s inequality,\n\n`$E\\left({\\sigma }_{t}^{2}\\right)\\ge \\mathrm{exp}\\left\\{E\\left(\\mathrm{log}{\\sigma }_{t}^{2}\\right)\\right\\}.$`\n\nAn EGARCH(1,1) specification will be complex enough for most applications. For an EGARCH(1,1) model, the GARCH and ARCH coefficients are expected to be positive, and the leverage coefficient is expected to be negative; large unanticipated downward shocks should increase the variance. If you get signs opposite to those expected, you might encounter difficulties inferring volatility sequences and forecasting (a negative ARCH coefficient can be particularly problematic). In this case, an EGARCH model might not be the best choice for your application.\n\n### GJR Model\n\nThe GJR model is a GARCH variant that includes leverage terms for modeling asymmetric volatility clustering. In the GJR formulation, large negative changes are more likely to be clustered than positive changes. The GJR model is named for Glosten, Jagannathan, and Runkle . Close similarities exist between the GJR model and the threshold GARCH (TGARCH) model—a GJR model is a recursive equation for the variance process, and a TGARCH is the same recursion applied to the standard deviation process.\n\nThe GJR(P,Q) model has P GARCH coefficients associated with lagged variances, Q ARCH coefficients associated with lagged squared innovations, and Q leverage coefficients associated with the square of negative lagged innovations. The form of the GJR(P,Q) model in Econometrics Toolbox is\n\n`${y}_{t}=\\mu +{\\epsilon }_{t},$`\n\nwhere${\\epsilon }_{t}={\\sigma }_{t}{z}_{t}$ and\n\n`${\\sigma }_{t}^{2}=\\kappa +\\sum _{i=1}^{P}{\\gamma }_{i}{\\sigma }_{t-i}^{2}+\\sum _{j=1}^{Q}{\\alpha }_{j}{\\epsilon }_{t-j}^{2}+\\sum _{j=1}^{Q}{\\xi }_{j}I\\left[{\\epsilon }_{t-j}<0\\right]{\\epsilon }_{t-j}^{2}.$`\n\nThe indicator function $I\\left[{\\epsilon }_{t-j}<0\\right]$ equals 1 if ${\\epsilon }_{t-j}<0$, and 0 otherwise. Thus, the leverage coefficients are applied to negative innovations, giving negative changes additional weight.\n\nNote\n\nThe `Constant` property of a `gjr` model corresponds to κ, and the `Offset` property corresponds to μ.\n\nFor stationarity and positivity, the GJR model has the following constraints:\n\n• $\\kappa >0$\n\n• ${\\gamma }_{i}\\ge 0,\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}{\\alpha }_{j}\\ge 0$\n\n• ${\\alpha }_{j}+{\\xi }_{j}\\ge 0$\n\n• ${\\sum }_{i=1}^{P}{\\gamma }_{i}+{\\sum }_{j=1}^{Q}{\\alpha }_{j}+\\frac{1}{2}{\\sum }_{j=1}^{Q}{\\xi }_{j}<1$\n\nThe GARCH model is nested in the GJR model. If all leverage coefficients are zero, then the GJR model reduces to the GARCH model. This means you can test a GARCH model against a GJR model using the likelihood ratio test.\n\n Engle, Robert F. “Autoregressive Conditional Heteroskedasticity with Estimates of the Variance of United Kingdom Inflation.” Econometrica. Vol. 50, 1982, pp. 987–1007.\n\n Glosten, L. R., R. Jagannathan, and D. E. Runkle. “On the Relation between the Expected Value and the Volatility of the Nominal Excess Return on Stocks.” The Journal of Finance. 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https://naima.readthedocs.io/en/latest/api-models.html	[ "# Models¶\n\n## API¶\n\nclass naima.models.Bremsstrahlung(particle_distribution, n0=<Quantity 1. 1 / cm3>, *kwargs)[source]\n\nBremsstrahlung radiation on a completely ionised gas.\n\nThis class uses the cross-section approximation of Baring, M.G., Ellison, D.C., Reynolds, S.P., Grenier, I.A., & Goret, P. 1999, Astrophysical Journal, 513, 311.\n\nThe default weights are assuming a completely ionised target gas with ISM abundances. If pure electron-electron bremsstrahlung is desired, n0 can be set to the electron density, weight_ep to 0 and weight_ee to 1.\n\nParameters\nn0Quantity float\n\nTotal ion number density.\n\nOther Parameters\nweight_eefloat\n\nWeight of electron-electron bremsstrahlung. Defined as $$\\sum_i Z_i X_i$$, default is 1.088.\n\nweight_epfloat\n\nWeight of electron-proton bremsstrahlung. Defined as $$\\sum_i Z_i^2 X_i$$, default is 1.263.\n\nproperty We\n\nTotal energy in electrons used for the radiative calculation\n\ncompute_We(Eemin=None, Eemax=None)\n\nTotal energy in electrons between energies Eemin and Eemax\n\nParameters\nEeminQuantity float, optional\n\nMinimum electron energy for energy content calculation.\n\nEemaxQuantity float, optional\n\nMaximum electron energy for energy content calculation.\n\nflux(energy, args, kwargs)\n\nDifferential flux at a given distance from the source.\n\nParameters\nphoton_energyQuantity float or array\n\nPhoton energy array.\n\ndistanceQuantity float, optional\n\nDistance to the source. If set to 0, the intrinsic differential luminosity will be returned. Default is 1 kpc.\n\nsed(photon_energy, distance=<Quantity 1. kpc>)\n\nSpectral energy distribution at a given distance from the source.\n\nParameters\nphoton_energyQuantity float or array\n\nPhoton energy array.\n\ndistanceQuantity float, optional\n\nDistance to the source. If set to 0, the intrinsic luminosity will be returned. Default is 1 kpc.\n\nset_We(We, Eemin=None, Eemax=None, amplitude_name=None)\n\nNormalize particle distribution so that the total energy in electrons between Eemin and Eemax is We\n\nParameters\nWeQuantity float\n\nDesired energy in electrons.\n\nEeminQuantity float, optional\n\nMinimum electron energy for energy content calculation.\n\nEemaxQuantity float, optional\n\nMaximum electron energy for energy content calculation.\n\namplitude_namestr, optional\n\nName of the amplitude parameter of the particle distribution. It must be accesible as an attribute of the distribution function. Defaults to amplitude.\n\nclass naima.models.BrokenPowerLaw(amplitude, e_0, e_break, alpha_1, alpha_2)[source]\n\nOne dimensional power law model with a break.\n\nParameters\namplitudefloat\n\nModel amplitude at the break energy\n\ne_0Quantity float\n\nReference point\n\ne_breakQuantity float\n\nBreak energy\n\nalpha_1float\n\nPower law index for x < x_break\n\nalpha_2float\n\nPower law index for x > x_break\n\nNotes\n\nModel formula (with $$A$$ for amplitude, $$E_0$$ for e_0, $$\\alpha_1$$ for alpha_1 and $$\\alpha_2$$ for alpha_2):\n\n$\\begin{split}f(E) = \\left \\{ \\begin{array}{ll} A (E / E_0) ^ {-\\alpha_1} & : E < E_{break} \\\\ A (E_{break}/E_0) ^ {\\alpha_2-\\alpha_1} (E / E_0) ^ {-\\alpha_2} & : E > E_{break} \\\\ \\end{array} \\right.\\end{split}$\nstatic eval(e, amplitude, e_0, e_break, alpha_1, alpha_2)[source]\n\nOne dimensional broken power law model function\n\nclass naima.models.EblAbsorptionModel(redshift, ebl_absorption_model='Dominguez')[source]\n\nA TableModel containing the different absorption values from a specific model.\n\nIt returns dimensionless opacity values, that could be multiplied to any model.\n\nParameters\nredshiftfloat\n\nRedshift considered for the absorption evaluation.\n\nebl_absorption_model{‘Dominguez’}\n\nName of the EBL absorption model to use (Dominguez by default).\n\nNotes\n\nDominguez model refers to the Dominguez 2011 EBL model. Current implementation does NOT perform an interpolation in the redshift, so it just uses the closest z value from the finely binned tau_dominguez11.npz file (delta_z=0.01).\n\nclass naima.models.ExponentialCutoffBrokenPowerLaw(amplitude, e_0, e_break, alpha_1, alpha_2, e_cutoff, beta=1.0)[source]\n\nOne dimensional power law model with a break.\n\nParameters\namplitudefloat\n\nModel amplitude at the break point\n\ne_0Quantity float\n\nReference point\n\ne_breakQuantity float\n\nBreak energy\n\nalpha_1float\n\nPower law index for x < x_break\n\nalpha_2float\n\nPower law index for x > x_break\n\ne_cutoffQuantity float\n\nExponential Cutoff energy\n\nbetafloat, optional\n\nExponential cutoff rapidity. Default is 1.\n\nNotes\n\nModel formula (with $$A$$ for amplitude, $$E_0$$ for e_0, $$\\alpha_1$$ for alpha_1, $$\\alpha_2$$ for alpha_2, $$E_{cutoff}$$ for e_cutoff, and $$\\beta$$ for beta):\n\n$\\begin{split}f(E) = \\exp(-(E / E_{cutoff})^\\beta)\\left \\{ \\begin{array}{ll} A (E / E_0) ^ {-\\alpha_1} & : E < E_{break} \\\\ A (E_{break}/E_0) ^ {\\alpha_2-\\alpha_1} (E / E_0) ^ {-\\alpha_2} & : E > E_{break} \\\\ \\end{array} \\right.\\end{split}$\nstatic eval(e, amplitude, e_0, e_break, alpha_1, alpha_2, e_cutoff, beta)[source]\n\nOne dimensional broken power law model function\n\nclass naima.models.ExponentialCutoffPowerLaw(amplitude, e_0, alpha, e_cutoff, beta=1.0)[source]\n\nOne dimensional power law model with an exponential cutoff.\n\nParameters\namplitudefloat\n\nModel amplitude\n\ne_0Quantity float\n\nReference point\n\nalphafloat\n\nPower law index\n\ne_cutoffQuantity float\n\nCutoff point\n\nbetafloat\n\nCutoff exponent\n\nNotes\n\nModel formula (with $$A$$ for amplitude, $$\\alpha$$ for alpha, and $$\\beta$$ for beta):\n\n$f(E) = A (E / E_0) ^ {-\\alpha} \\exp (- (E / E_{cutoff}) ^ \\beta)$\nstatic eval(e, amplitude, e_0, alpha, e_cutoff, beta)[source]\n\nOne dimensional power law with an exponential cutoff model function\n\nclass naima.models.InverseCompton(particle_distribution, seed_photon_fields=['CMB'], kwargs)[source]\n\nInverse Compton emission from an electron population.\n\nIf you use this class in your research, please consult and cite Khangulyan, D., Aharonian, F.A., & Kelner, S.R. 2014, Astrophysical Journal, 783, 100\n\nParameters\nparticle_distributionfunction\n\nParticle distribution function, taking electron energies as a Quantity array or float, and returning the particle energy density in units of number of electrons per unit energy as a Quantity array or float.\n\nseed_photon_fieldsstring or iterable of strings (optional)\n\nA list of gray-body or non-thermal seed photon fields to use for IC calculation. Each of the items of the iterable can be either:\n\n• A string equal to CMB (default), NIR, or FIR, for which radiation fields with temperatures of 2.72 K, 30 K, and 3000 K, and energy densities of 0.261, 0.5, and 1 eV/cm³ will be used (these are the GALPROP values for a location at a distance of 6.5 kpc from the galactic center).\n\n• A list of length three (isotropic source) or four (anisotropic source) composed of:\n\n1. A name for the seed photon field.\n\n2. Its temperature (thermal source) or energy (monochromatic or non-thermal source) as a Quantity instance.\n\n3. Its photon field energy density as a Quantity instance.\n\n4. Optional: The angle between the seed photon direction and the scattered photon direction as a Quantity float instance.\n\nOther Parameters\nEeminQuantity float instance, optional\n\nMinimum electron energy for the electron distribution. Default is 1 GeV.\n\nEemaxQuantity float instance, optional\n\nMaximum electron energy for the electron distribution. Default is 510 TeV.\n\nnEedscalar\n\nNumber of points per decade in energy for the electron energy and distribution arrays. Default is 300.\n\nproperty We\n\nTotal energy in electrons used for the radiative calculation\n\ncompute_We(Eemin=None, Eemax=None)\n\nTotal energy in electrons between energies Eemin and Eemax\n\nParameters\nEeminQuantity float, optional\n\nMinimum electron energy for energy content calculation.\n\nEemaxQuantity float, optional\n\nMaximum electron energy for energy content calculation.\n\nflux(photon_energy, distance=<Quantity 1. kpc>, seed=None)[source]\n\nDifferential flux at a given distance from the source from a single seed photon field\n\nParameters\nphoton_energyQuantity float or array\n\nPhoton energy array.\n\ndistanceQuantity float, optional\n\nDistance to the source. If set to 0, the intrinsic luminosity will be returned. Default is 1 kpc.\n\nseedint, str or None\n\nNumber or name of seed photon field for which the IC contribution is required. If set to None it will return the sum of all contributions (default).\n\nsed(photon_energy, distance=<Quantity 1. kpc>, seed=None)[source]\n\nSpectral energy distribution at a given distance from the source\n\nParameters\nphoton_energyQuantity float or array\n\nPhoton energy array.\n\ndistanceQuantity float, optional\n\nDistance to the source. If set to 0, the intrinsic luminosity will be returned. Default is 1 kpc.\n\nseedint, str or None\n\nNumber or name of seed photon field for which the IC contribution is required. If set to None it will return the sum of all contributions (default).\n\nset_We(We, Eemin=None, Eemax=None, amplitude_name=None)\n\nNormalize particle distribution so that the total energy in electrons between Eemin and Eemax is We\n\nParameters\nWeQuantity float\n\nDesired energy in electrons.\n\nEeminQuantity float, optional\n\nMinimum electron energy for energy content calculation.\n\nEemaxQuantity float, optional\n\nMaximum electron energy for energy content calculation.\n\namplitude_namestr, optional\n\nName of the amplitude parameter of the particle distribution. It must be accesible as an attribute of the distribution function. Defaults to amplitude.\n\nclass naima.models.LogParabola(amplitude, e_0, alpha, beta)[source]\n\nOne dimensional log parabola model (sometimes called curved power law).\n\nParameters\namplitudefloat\n\nModel amplitude\n\ne_0Quantity float\n\nReference point\n\nalphafloat\n\nPower law index\n\nbetafloat\n\nPower law curvature\n\nNotes\n\nModel formula (with $$A$$ for amplitude and $$\\alpha$$ for alpha and $$\\beta$$ for beta):\n\n$f(e) = A \\left(\\frac{E}{E_{0}}\\right)^ {- \\alpha - \\beta \\log{\\left (\\frac{E}{E_{0}} \\right )}}$\nstatic eval(e, amplitude, e_0, alpha, beta)[source]\n\nOne dimenional log parabola model function\n\nclass naima.models.PionDecay(particle_distribution, nh=<Quantity 1. 1 / cm3>, nuclear_enhancement=True, *kwargs)[source]\n\nPion decay gamma-ray emission from a proton population.\n\nCompute gamma-ray spectrum arising from the interaction of a relativistic proton distribution with stationary target protons using the parametrization of Kafexhiu et al. (2014).\n\nIf you use this class in your research, please consult and cite Kafexhiu, E., Aharonian, F., Taylor, A.M., & Vila, G.S. 2014, Physical Review D, 90, 123014.\n\nParameters\nparticle_distributionfunction\n\nParticle distribution function, taking proton energies as a Quantity array or float, and returning the particle energy density in units of number of protons per unit energy as a Quantity array or float.\n\nnhQuantity\n\nNumber density of the target protons. Default is $$1 \\mathrm{cm}^{-3}$$.\n\nnuclear_enhancementbool\n\nWhether to apply the energy-dependent nuclear enhancement factor considering a target gas with local ISM abundances. See Section IV of Kafexhiu et al. (2014) for details. Here the proton-nucleus inelastic cross section of Sihver et al. (1993, PhysRevC 47, 1225) is used.\n\nOther Parameters\nEpminQuantity float\n\nMinimum proton energy for the proton distribution. Default is 1.22 GeV, the dynamical threshold for pion production in pp interactions.\n\nEpmaxQuantity float\n\nMinimum proton energy for the proton distribution. Default is 10 PeV.\n\nnEpdscalar\n\nNumber of points per decade in energy for the proton energy and distribution arrays. Default is 100.\n\nhiEmodelstr\n\nMonte Carlo model to use for computation of high-energy differential cross section. Can be one of Geant4, Pythia8, SIBYLL, or QGSJET. See Kafexhiu et al. (2014) for details. Default is Pythia8.\n\nuseLUTbool\n\nWhether to use a lookup table for the differential cross section. The only lookup table packaged with naima is for the Pythia 8 model and ISM nuclear enhancement factor.\n\nproperty Wp\n\nTotal energy in protons\n\ncompute_Wp(Epmin=None, Epmax=None)\n\nTotal energy in protons between energies Epmin and Epmax\n\nParameters\nEpminQuantity float, optional\n\nMinimum proton energy for energy content calculation.\n\nEpmaxQuantity float, optional\n\nMaximum proton energy for energy content calculation.\n\nflux(energy, args, kwargs)\n\nDifferential flux at a given distance from the source.\n\nParameters\nphoton_energyQuantity float or array\n\nPhoton energy array.\n\ndistanceQuantity float, optional\n\nDistance to the source. If set to 0, the intrinsic differential luminosity will be returned. Default is 1 kpc.\n\nsed(photon_energy, distance=<Quantity 1. kpc>)\n\nSpectral energy distribution at a given distance from the source.\n\nParameters\nphoton_energyQuantity float or array\n\nPhoton energy array.\n\ndistanceQuantity float, optional\n\nDistance to the source. If set to 0, the intrinsic luminosity will be returned. Default is 1 kpc.\n\nset_Wp(Wp, Epmin=None, Epmax=None, amplitude_name=None)\n\nNormalize particle distribution so that the total energy in protons between Epmin and Epmax is Wp\n\nParameters\nWpQuantity float\n\nDesired energy in protons.\n\nEpminQuantity float, optional\n\nMinimum proton energy for energy content calculation.\n\nEpmaxQuantity float, optional\n\nMaximum proton energy for energy content calculation.\n\namplitude_namestr, optional\n\nName of the amplitude parameter of the particle distribution. It must be accesible as an attribute of the distribution function. Defaults to amplitude.\n\nclass naima.models.PowerLaw(amplitude, e_0, alpha)[source]\n\nOne dimensional power law model.\n\nParameters\namplitudefloat\n\nModel amplitude.\n\ne_0Quantity float\n\nReference energy\n\nalphafloat\n\nPower law index\n\nNotes\n\nModel formula (with $$A$$ for amplitude, $$\\alpha$$ for alpha):\n\n$f(E) = A (E / E_0) ^ {-\\alpha}$\nstatic eval(e, amplitude, e_0, alpha)[source]\n\nOne dimensional power law model function\n\nclass naima.models.Synchrotron(particle_distribution, B=<Quantity 3.24e-06 G>, kwargs)[source]\n\nSynchrotron emission from an electron population.\n\nThis class uses the approximation of the synchrotron emissivity in a random magnetic field of Aharonian, Kelner, and Prosekin 2010, PhysRev D 82, 3002 (arXiv:1006.1045).\n\nParameters\nparticle_distributionfunction\n\nParticle distribution function, taking electron energies as a Quantity array or float, and returning the particle energy density in units of number of electrons per unit energy as a Quantity array or float.\n\nBQuantity float instance, optional\n\nIsotropic magnetic field strength. Default: equipartition with CMB (3.24e-6 G)\n\nOther Parameters\nEeminQuantity float instance, optional\n\nMinimum electron energy for the electron distribution. Default is 1 GeV.\n\nEemaxQuantity float instance, optional\n\nMaximum electron energy for the electron distribution. Default is 510 TeV.\n\nnEedscalar\n\nNumber of points per decade in energy for the electron energy and distribution arrays. Default is 100.\n\nproperty We\n\nTotal energy in electrons used for the radiative calculation\n\ncompute_We(Eemin=None, Eemax=None)\n\nTotal energy in electrons between energies Eemin and Eemax\n\nParameters\nEeminQuantity float, optional\n\nMinimum electron energy for energy content calculation.\n\nEemaxQuantity float, optional\n\nMaximum electron energy for energy content calculation.\n\nflux(energy, args, *kwargs)\n\nDifferential flux at a given distance from the source.\n\nParameters\nphoton_energyQuantity float or array\n\nPhoton energy array.\n\ndistanceQuantity float, optional\n\nDistance to the source. If set to 0, the intrinsic differential luminosity will be returned. Default is 1 kpc.\n\nsed(photon_energy, distance=<Quantity 1. kpc>)\n\nSpectral energy distribution at a given distance from the source.\n\nParameters\nphoton_energyQuantity float or array\n\nPhoton energy array.\n\ndistanceQuantity float, optional\n\nDistance to the source. If set to 0, the intrinsic luminosity will be returned. Default is 1 kpc.\n\nset_We(We, Eemin=None, Eemax=None, amplitude_name=None)\n\nNormalize particle distribution so that the total energy in electrons between Eemin and Eemax is We\n\nParameters\nWeQuantity float\n\nDesired energy in electrons.\n\nEeminQuantity float, optional\n\nMinimum electron energy for energy content calculation.\n\nEemaxQuantity float, optional\n\nMaximum electron energy for energy content calculation.\n\namplitude_namestr, optional\n\nName of the amplitude parameter of the particle distribution. It must be accesible as an attribute of the distribution function. Defaults to amplitude.\n\nclass naima.models.TableModel(energy, values, amplitude=1)[source]\n\nA model generated from a table of energy and value arrays.\n\nThe units returned will be the units of the values array provided at initialization. The model will return values interpolated in log-space, returning 0 for energies outside of the limits of the provided energy array.\n\nParameters\nenergyQuantity array\n\nArray of energies at which the model values are given\n\nvaluesarray\n\nArray with the values of the model at energies energy.\n\namplitudefloat\n\nModel amplitude that is multiplied to the supplied arrays. Defaults to 1." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6004161,"math_prob":0.97815496,"size":14178,"snap":"2023-40-2023-50","text_gpt3_token_len":3758,"char_repetition_ratio":0.16868915,"word_repetition_ratio":0.4607356,"special_character_ratio":0.2390323,"punctuation_ratio":0.1496228,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9951927,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-11T23:03:40Z\",\"WARC-Record-ID\":\"<urn:uuid:7b39869d-1c95-43b9-b764-a44968bd7557>\",\"Content-Length\":\"91426\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ff149a9f-2d35-4630-82a6-04891d15d357>\",\"WARC-Concurrent-To\":\"<urn:uuid:48a0d554-9690-455b-879e-a3beafcabba5>\",\"WARC-IP-Address\":\"104.17.32.82\",\"WARC-Target-URI\":\"https://naima.readthedocs.io/en/latest/api-models.html\",\"WARC-Payload-Digest\":\"sha1:6ZCWTJFDBIHXUFKAKNQOJSTHYSCPKQB6\",\"WARC-Block-Digest\":\"sha1:TJKAVCCPDXJVYAA5DMH37M3WRJAJU3QF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679518883.99_warc_CC-MAIN-20231211210408-20231212000408-00409.warc.gz\"}"}
https://crypto.stackexchange.com/questions/11338/why-does-merkles-puzzle-requires-eve-quadratic-complexity-of-effort-to-break-th	[ "# Why does Merkle's Puzzle requires Eve quadratic complexity of effort to break the system?\n\nThe way Applied cryptography 2ED explains the puzzle is as follows (I paraphrase it):\n\n1. Bob generates 2^20 messages of the form x,y where x is a puzzle number and y is the secret key. Both x and y are different for each of the one million message. Encrypt each message using symmetric cipher with a different 20-bit key. Send all messages to Alice.\n\n2. Alice picks one at random and brute force it. She should be able to recover x and y.\n\n3. Alice encrypts her message to Bob by using y she just recovered (using symmetric cipher). Her message will contain x, the puzzle number.\n\n4. Bob looks up the secret key for puzzle x and decrypts the message.\n\nAccording to the book and many Internet sources, Eve would have to do theta(n^2) work to brute force the communication.\n\nThe book doesn't seem to say much about how x is included in Alice's response to Bob. It must be not be encrypted right? Because Bob has to do a O(1) look up. That is, Alice would have to send x+E(private message, y) to Bob.\n\nThen why can't Eve just wait for Alice's response to Bob and compute that puzzle?\n\nThis is the brute force algorithm I think would satisfy 2^n if x is encrypted in Alice's response as well.\n\nfor y in 2^20:\nfor x in 2^20:\nc = symmetric_enc(y, x)\nif c == c_from_alice:\nreturn yes, c, x, y\nreturn no, none, none, none\n\n\nBut if x is encrypted in Alice response, then Bob have to brute force like Eve too, no?\n\n$x$ is not encrypted in Alice’s response. The point is that, even though Eve knows the puzzle number $x$, she has no way to tell which of the puzzles Bob sent is the one with this number, unless she solves all of them (or at any rate half of them, on average).\n• Thanks. If x is encrypted in Alice's response, how is Bob able to tell which x in constant time? Now that I think about it, Bob might only need to do n encryption to find out which x. In other words, since Bob has a table with x and key as columns, he can do 2^20 encryptions to find out. So this is the difference and my assumption that Bob does only O(1) is wrong right? Otherwise, Eve can do O(1) as well. Am I correct? Does x really have to be encrypted in Alice's response? Oct 28, 2013 at 10:57\n• @CppLearner Sorry, I should have been clearer: $x$ is not encrypted in Alice’s response. I have edited the answer to make it clearer, I hope. Oct 28, 2013 at 11:55" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9237668,"math_prob":0.88119096,"size":1374,"snap":"2022-27-2022-33","text_gpt3_token_len":349,"char_repetition_ratio":0.12189781,"word_repetition_ratio":0.008,"special_character_ratio":0.24235807,"punctuation_ratio":0.117056854,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98789984,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-30T13:02:55Z\",\"WARC-Record-ID\":\"<urn:uuid:18526bd1-8954-42c3-8fb9-6330e8e0accc>\",\"Content-Length\":\"232153\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:43e461b0-177a-4ef0-a785-9b6abdfed546>\",\"WARC-Concurrent-To\":\"<urn:uuid:01280b58-7a50-49a4-944e-01e00c70ebdb>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://crypto.stackexchange.com/questions/11338/why-does-merkles-puzzle-requires-eve-quadratic-complexity-of-effort-to-break-th\",\"WARC-Payload-Digest\":\"sha1:F7IBBOCYUAU7L3ZLEHYISJZ5DS72H7LQ\",\"WARC-Block-Digest\":\"sha1:ONTO5TWBAJS26SPS46C7VDUO4AKMWSW3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103821173.44_warc_CC-MAIN-20220630122857-20220630152857-00484.warc.gz\"}"}
https://www.lumoslearning.com/llwp/resources/applistings.html?iid=575826804	[ "## Equation Solvers - Lumos Educational App Store", null, "Price -\\$3.99\n\n#### DESCRIPTION:\n\nEquation Solver 4-in-1 is the ultimate app that helps you solve linear equations in one variable, quadratic equations with real and complex roots, cubic and quartic equations. It's intuitive and easy to use. Just type in any equation you want to solve and the equation calculator will show you the result. Everything you need to solve equations or check results! Features: • Linear Equation Solver. • Quadratic Equation Solver. • Cubic Equation Solver. • Quartic Equation Solver. • Solves equations with fractions and parentheses. Note: the app doesn’t yet support equations with a v\n\n#### OVERVIEW:\n\nEquation Solvers is a free educational mobile app By Intemodino Group s.r.o..It helps students practice the following standards .\n\nThis page not only allows students and teachers download Equation Solvers but also find engaging Sample Questions, Videos, Pins, Worksheets, Books related to the following topics.\n\nDeveloper URL: https://intemodino.com\n\nSoftware Version: 4.0.3\n\nCategory: Education\n\nRelease Date: 2012-11-11T06:29:34Z", null, "", null, "", null, "", null, "#### Are you the Developer?\n\nRate this App?\n0\n\n0 Ratings & 0 Reviews\n\n5\n0\n0\n4\n0\n0\n3\n0\n0\n2\n0\n0\n1\n0\n0", null, "EdSearch WebSearch", null, "" ]	[ null, "https://is1-ssl.mzstatic.com/image/thumb/Purple118/v4/a8/d5/87/a8d587f7-16fa-fc1e-6c56-e4f44835dbd8/source/100x100bb.jpg", null, "https://is5-ssl.mzstatic.com/image/thumb/Purple118/v4/3f/29/52/3f295283-b777-c9af-613a-00699054f946/pr_source.png/392x696bb.png", null, "https://is5-ssl.mzstatic.com/image/thumb/Purple118/v4/35/66/96/35669627-9ea4-0758-3b3a-ce0745ae5ddc/pr_source.png/392x696bb.png", null, "https://is5-ssl.mzstatic.com/image/thumb/Purple128/v4/36/f3/8b/36f38b2c-e404-ef2e-aeba-2f9c889796b7/pr_source.png/392x696bb.png", null, "https://is4-ssl.mzstatic.com/image/thumb/Purple118/v4/05/20/60/05206086-b75b-48e0-ba78-6bd6dda69182/pr_source.png/392x696bb.png", null, "https://statc.lumoslearning.com/llwp/wp-content/uploads/2018/01/edSearch_Logo_LowRes.png", null, "https://googleads.g.doubleclick.net/pagead/viewthroughconversion/1066999506/", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8534884,"math_prob":0.94096774,"size":1106,"snap":"2019-51-2020-05","text_gpt3_token_len":248,"char_repetition_ratio":0.1814882,"word_repetition_ratio":0.0,"special_character_ratio":0.21338156,"punctuation_ratio":0.17391305,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98057294,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-05T20:22:55Z\",\"WARC-Record-ID\":\"<urn:uuid:64170fa2-2db1-4785-bd16-db8746b9e37a>\",\"Content-Length\":\"151686\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ecdc4eb6-05fd-4d7f-810b-4e3b9c5147c5>\",\"WARC-Concurrent-To\":\"<urn:uuid:a6bb226a-61bb-4f6f-a96a-4eae35382b30>\",\"WARC-IP-Address\":\"3.223.110.75\",\"WARC-Target-URI\":\"https://www.lumoslearning.com/llwp/resources/applistings.html?iid=575826804\",\"WARC-Payload-Digest\":\"sha1:F6A6EOM6QBCIAMY64DYBMH6MYNA73FFR\",\"WARC-Block-Digest\":\"sha1:OL2FCP6JUHFZCHA53ZPWRI4MW7MTJDF7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540482038.36_warc_CC-MAIN-20191205190939-20191205214939-00263.warc.gz\"}"}
https://physicsforme.com/2014/08/12/fields-medals-2014/	[ "# Fields Medals 2014", null, "The Fields Medal is widely considered to be the most prestigious award in mathematics, and the 2014 Fields Medalists were just announced.\n\nThe Fields Medal is given to between two and four mathematicians under the age of 40 at the International Congress of Mathematicians, held once every four years. According to the International Mathematical Union, the body that awards the Fields Medal, the award intends “to recognize outstanding mathematical achievement for existing work and for the promise of future achievement.”\n\nIt is usually awarded for mathematical research that solves or extends problems that have vexed mathematicians for decades or centuries, or research that greatly expands on or even creates new areas of mathematical thought.\n\nHere are the four 2014 Fields Medalists:\n\nArtur Avila\n\nArtur Avila is a Brazilian mathematician who has contributed to a number of fields. Some of his most notable research is in the study of chaos theory and dynamical systems. These areas seek to understand the behavior of systems that evolve over time in which very small changes in initial conditions can lead to wildly varying outcomes, such as weather patterns, as typified in the classic example of a butterfly’s wings flapping leading to a change in weather hundreds of miles away.\n\nOne of Avila’s major contributions to this field of study was in clarifying that a certain broad class of dynamical systems fall into one of two categories. They either eventually evolve into a stable state, or fall into a chaotic stochastic state, in which their behavior can be described probabilistically.\n\nManjul Bhargava\n\nManjul Bhargava‘s research is focused on number theory and algebra. One of the basic subjects in algebraic number theory is the behavior of polynomials with integer coefficients, like 3×2 + 4xy -5y2.\n\nCarl Friederich Gauss, one of the greatest mathematicians of the late eighteenth and early nineteenth centuries, developed a powerful tool for analyzing polynomials like the one above, where the variables are all raised to at most the second power.\n\nBhargava, by intensely studying Gauss’ work and adding to it an impressive level of geometric and algebraic insight, was able to extend Gauss’ tool to higher degree polynomials, in which we raise the variables to higher powers than two. This work vastly expands the ability of number theorists to study these fundamental mathematical objects.\n\nMartin Hairer\n\nMartin Hairer researches stochastic partial differential equations. Differential equations show up throughout mathematics, physics, and engineering. They describe processes that change over time, like the movement of a shell shot from a cannon, or the price of a stock or bond.\n\nDifferential equations come in a variety of flavors. Ordinary differential equations are equations that only have one variable involved. The motion of a cannonball, for example, can be modeled with a simple ordinary differential equation in which the only variable is the time since the cannon was fired.\n\nPartial differential equations involve processes that depend on multiple variables. In many physical settings, both time and the current position of an object are needed to determine the future trajectory of the object. These describe a much wider variety of processes in the world, and are generally much harder to work with than one-variable ordinary equations.\n\nDifferential equations can also be either deterministic or stochastic. The cannonball’s movement, or the movement of a satellite orbiting earth, are deterministic: outside of measurement error, once we’ve solved the equation, we have no doubt about where the cannonball or satellite will be at a given point in time. Stochastic equations have a random element involved. The motion of sugar grains stirred in a cup of coffee, or a stock’s price at a given moment in time are both best described by models that have an element of noise or randomness.\n\nStochastic partial differential equations — equations that have multiple input variables and random noise elements — have traditionally been very difficult for mathematicians to work with. Hairer developed a new theoretical framework that makes these equations far more tractable, opening the door to being able to solve a number of equations with both large amounts of mathematical interest in their own rights, and with powerful applications in the sciences and engineering.\n\nMaryam Mirzakhani\n\nMaryam Mirzakhani‘s work focuses on the geometry of Riemann surfaces. Riemann surfaces are a classic type of non-Euclidean geometry: while a Riemann surface still has two dimensions like a plane, and we can still define lines, angles, and curves on the surface, the way that the measurement of angles and distances will come out can be very different from what happens on a normal Euclidean plane.\n\nA basic example of this that we’ve looked at before is the Riemann sphere: a version of a sphere in which we still have a notion of lines and angles, but where strange things can happen, like triangles with three ninety degree angles.\n\nRiemann surfaces can get far more complicated than the Riemann sphere, and one of the major research areas in the study of these surfaces is how one Riemann surface can be smoothly deformed or smushed into another surface. These deformations themselves have their own strange geometries, called “moduli spaces”, and Mirzakhani has contributed several interesting results in understanding these mysterious spaces.\n\nRead also: Avila, Bhargava, Hairer, Mirzakhani by Terence Tao\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]	[ null, "https://physicsforme.files.wordpress.com/2014/08/fieldsmedalfrontandback.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9390653,"math_prob":0.94025475,"size":5601,"snap":"2020-34-2020-40","text_gpt3_token_len":1113,"char_repetition_ratio":0.11381097,"word_repetition_ratio":0.0,"special_character_ratio":0.18175326,"punctuation_ratio":0.08784474,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97868043,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-25T23:43:49Z\",\"WARC-Record-ID\":\"<urn:uuid:d54a0596-5b2c-489c-8c22-ec4d346093a7>\",\"Content-Length\":\"118477\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef91b5dc-5ee6-4309-9baa-b339faf63f7c>\",\"WARC-Concurrent-To\":\"<urn:uuid:7904d59c-7127-4693-af5f-19797697346f>\",\"WARC-IP-Address\":\"192.0.78.24\",\"WARC-Target-URI\":\"https://physicsforme.com/2014/08/12/fields-medals-2014/\",\"WARC-Payload-Digest\":\"sha1:VFAREG3RSA5V4V2TGNKMH4YTCOFZ2KBU\",\"WARC-Block-Digest\":\"sha1:BMNMA5I6LH5EPDRNDAZJP35X3NEA4ZID\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400228998.45_warc_CC-MAIN-20200925213517-20200926003517-00302.warc.gz\"}"}
https://wpcalc.com/en/total-asset-turnover/	[ "Asset turnover is a financial ratio which calculates the company’s efficacy at using its assets in generating sales or revenue.\n\n### Calculator of Total Asset Turnover\n\n Currency = Sales revenue = Beginning assets = Ending assets = Total asset turnover =\n\nFormula of Total Asset Turnover\n\nTotal Asset Turnover = Net sales revenue / Average total assets Average total assets = (Beginning assets + Ending assets) / 2\n\n### Example of Total Asset Turnover\n\nIn 2012 and 2013 a textile company had Rs.960000 and Rs.1690000 in assets respectively. The company generated revenue of Rs.1360000. Calculate the total asset turnover ratio for the company.\n\nGiven,\n\n• Beginning assets = Rs.960000\n• Ending assets = Rs.1690000\n• Sales revenue = Rs.1360000\n\nSolution:\nStep 1 :\n\nAverage total assets = (Beginning assets + Ending assets) / 2\n= (960000 + 1690000) / 2\n= 2650000 / 2\n= 1325000 Rs\n\nStep 2 :\n\nTotal Asset Turnover = Net sales revenue / Average total assets\n= 1360000 / 1325000\n= 1.026", null, "" ]	[ null, "https://secure.gravatar.com/avatar/", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79973453,"math_prob":0.99369895,"size":910,"snap":"2019-51-2020-05","text_gpt3_token_len":244,"char_repetition_ratio":0.17108168,"word_repetition_ratio":0.18300654,"special_character_ratio":0.33516484,"punctuation_ratio":0.103225805,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99990463,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-12T05:34:13Z\",\"WARC-Record-ID\":\"<urn:uuid:86ebef2b-f713-4283-8c6c-c4ac53ae40c3>\",\"Content-Length\":\"45112\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3f91ec0c-d963-4e2f-a351-82fd405d986b>\",\"WARC-Concurrent-To\":\"<urn:uuid:b5d2902a-a047-4ba3-88d2-16f86a83aa25>\",\"WARC-IP-Address\":\"89.184.79.27\",\"WARC-Target-URI\":\"https://wpcalc.com/en/total-asset-turnover/\",\"WARC-Payload-Digest\":\"sha1:YDDZMTSTSN4L7ZYMMC5UEIQOQC22QS4E\",\"WARC-Block-Digest\":\"sha1:M6KE3GH3KYNONOX25ED5MTEMCSVR67OF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540537212.96_warc_CC-MAIN-20191212051311-20191212075311-00320.warc.gz\"}"}
https://osdata.com/programming/calculations/subtration.html	[ "", null, "music", null, "subtraction\n\nsummary\n\nThis subchapter looks at subtraction.\n\nfree computer programming text book project\n\nIf you like the idea of this project,", null, "stub section\n\nThis subchapter is a stub section. It will be filled in with instructional material later. For now it serves the purpose of a place holder for the order of instruction.\n\nProfessors are invited to give feedback on both the proposed contents and the propsed order of this text book. Send commentary to Milo, PO Box 1361, Tustin, California, 92781, USA.\n\nsubtraction\n\nThis subchapter looks at subtraction.\n\nnumber systems\n\nHere is a little bit of college level work on subtraction.\n\nA system of numbers os considered to be closed under an operation if it reproduces itself.\n\nS is a set of numbers. S is closed with respect to subtraction if for any two numbers a and b in the set S, the difference a - b is also a number in the set S. A set closed with respect to subtraction is called a modul.\n\nThe set of all integers {0, ±1, ±2, ±3, …} is a modul. The set of natural numbers {1, 2, 3, …} is not a modul.\n\nThe set of all even integers {0, ±2, ±4, ±6, …} is a modul.\n\nThe set of all rational numbers is a modul.\n\nThe set of all real numbers is a modul.\n\nThe set of all complex numbers is a modul.\n\nThe set of all purely imagiunary numbers ib is a modul.\n\nA theorem regarding moduls:\n\n(a) A modul always contains the number zero (0).\n\n(b) If a modul contains the number a then it also contains the number -a.\n\n(c) A modul is always closed with respect to addition.\n\nAnother theorem regarding moduls:\n\nAny modul M containing only integers includes all multiples of the greatest common divisor of the numbers in M.\n\nassembly language instructions\n\nFor most processors, integer arithmetic is faster than floating point arithmetic. This can be reversed in special cases such digital signal processors.\n\nOn many processors, floating point arithmetic is in an optional unit or optional coprocessor rather than being included on the main processor. This allows the manufacturer to charge less for the business machines that don’t need floating point arithmetic.\n\nThe basic four integer arithmetic operations are addition, subtraction, multiplication, and division. Arithmetic operations can be signed or unsigned (unsigned is useful for effective address computations). Some older processors don’t include hardware multiplication and division. Some processors don’t include actual multiplication or division hardware, instead looking up the answer in a massive table of results embedded in the processor.\n\nA specialized, but common, form of subtraction is an decrement instruction, which subtracts one from the contents of a register or memory location. For address computations, “decrement” may mean the subtraction of a constant other than one. Some processors have “short” or “quick” subtraction instructions that extend decrement to include a small range of values.\n\n• SUB Subtract; DEC VAX; signed subtraction of scalar quantities (8, 16, or 32 bit integer or 32, 64, or 128 bit floating point) in general purpose registers or memory, available in two operand (first operand subtracted from second operand with result replacing second operand) and three operand (first operand subtracted from second operand with result placed in third operand) (SUBB2 integer subtract byte 2 operand, SUBB3 integer subtract byte 3 operand, SUBW2 integer subtract word 2 operand, SUBW3 integer subtract word 3 operand, SUBL2 integer subtract long 2 operand, SUBL3 integer subtract long 3 operand) (SUBF2 subtract float 2 operand, SUBF3 subtract float 3 operand, SUBD2 subtract double float 2 operand, SUBD3 subtract double float 3 operand, SUBG2 subtract G float 2 operand, SUBG3 subtract G float 3 operand, SUBH2 subtract H float 2 operand, SUBH3 subtract H float 3 operand); clears or sets flags\n• SUB Subtract Integers; Intel 80x86; integer subtraction of the contents of a register or memory (8, 16, or 32 bits) from a memory location or a register; sets or clear flags\n• SUB Subtract; Motorola 680x0, Motorola 68300; signed subtract of the contents of a data register (8, 16, or 32 bits) from a memory location or subtracts the contents of a memory location (8, 16, or 32 bits) from a data register; sets or clear flags\n• SUB Subtract; MIX; subtract word or partial word field contents of memory from A-register (accumulator), overflow toggle possibly set\n• SR Subtract Register; IBM 360/370; RR format; signed subtract of the contents of a general purpose register (32 bits) from a general purpose register (32 bits); register to register only; sets or clears flags\n• S Subtract; IBM 360/370; RX format; signed subtract of the contents of a memory location (32 bits) from a general purpose register (32 bits); main memory to register only; sets or clears flags\n• SH Subtract Half-word; IBM 360/370; RX format; signed subtract of the contents of a memory location (16 bits) from a general purpose register (low order 16 bits); main memory to register only; sets or clears flags\n• SUBA Subtract Address; Motorola 680x0, Motorola 68300; unsigned subtract of the contents of a memory location or register (16 or 32 bits) from an address register; does not modify flags\n• SUBI Subtract Immediate; Motorola 680x0, Motorola 68300; signed subtract of immediate data (8, 16, or 32 bits) from a register or memory location; sets or clears flags\n• SUBQ Subtract Quick; Motorola 680x0, Motorola 68300; signed subtract of an immediate value of 1 to 8 inclusive from a register or memory lcoation; sets or clears flags for data registers and memory locations, does not modify flags for an address register\n• DEC Decrement; DEC VAX; decrements the integer contents of a general purpose register or contents of memory (DECB byte, DECW word, DECL longword); equivalent to SUBx2 #1, sum, but shorter and executes faster; clears or sets flags\n• DEC Decrement by 1; Intel 80x86; decrements the contents of a register or memory (8, 16, or 32 bits); sets or clear flags (does not modify carry flag)\n• SBWC Subtract With Carry; DEC VAX; integer subtraction (32 bit) in general purpose registers or memory, first operand and the C (carry) flag subtracted from second operand with result replacing second operand; used for extended precision subtraction; clears or sets flags\n• SBB Subtract Integers with Borrow; Intel 80x86; integer subtraction of the contents of a register or memory (8, 16, or 32 bits) and carry flag from a memory location or a register; sets or clear flags\n• SUBX Subtract Extended; Motorola 680x0, Motorola 68300; (signed subtract of a data register [8, 16, or 32 bits] and the extend bit from a data register) or (signed subtract of the contents of memory location [8, 16, or 32 bits] and the extend bit from the contents of another memory location while predecrementing both the source and destination address pointer registers), used to implement multi-precision integer arithmetic; sets or clears flags\n\nfree music player coding example\n\nCoding example: I am making heavily documented and explained open source code for a method to play music for free — almost any song, no subscription fees, no download costs, no advertisements, all completely legal. This is done by building a front-end to YouTube (which checks the copyright permissions for you).\n\nView music player in action: www.musicinpublic.com/.\n\nCreate your own copy from the original source code/ (presented for learning programming).\n\nview text bookHTML file\n\nBecause I no longer have the computer and software to make PDFs, the book is available as an HTML file, which you can convert into a PDF.\n\n Tweets by @osdata\n\nfree computer programming text book project\n\nBuilding a free downloadable text book on computer programming for university, college, community college, and high school classes in computer programming.\n\nIf you like the idea of this project,\n\nsend donations to:\nMilo\nPO Box 1361\nTustin, California 92781\n\nSupporting the entire project:\n\nIf you have a business or organization that can support the entire cost of this project, please contact Pr Ntr Kmt (my church)\n\nSome or all of the material on this web page appears in the", null, "", null, "This web site handcrafted on Macintosh", null, "computers using Tom Bender’s Tex-Edit Plus", null, "and served using FreeBSD", null, ".", null, "†UNIX used as a generic term unless specifically used as a trademark (such as in the phrase “UNIX certified”). UNIX is a registered trademark in the United States and other countries, licensed exclusively through X/Open Company Ltd.\n\nNames and logos of various OSs are trademarks of their respective owners." ]	[ null, "https://osdata.com/ads/tsosdragon.jpg", null, "https://osdata.com/pict/oslogo.gif", null, "http://www.google.com/logos/Logo_25wht.gif", null, "http://www.google.com/logos/Logo_25wht.gif", null, "https://osdata.com/pict/mademac.gif", null, "https://osdata.com/pict/mac.gif", null, "https://osdata.com/pict/texedit.gif", null, "https://osdata.com/pict/freebsdm.gif", null, "https://osdata.com/pict/bvbstar.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8911496,"math_prob":0.732467,"size":4785,"snap":"2022-05-2022-21","text_gpt3_token_len":987,"char_repetition_ratio":0.11796695,"word_repetition_ratio":0.08108108,"special_character_ratio":0.20041797,"punctuation_ratio":0.111363634,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9547404,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[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\":\"2022-01-20T10:25:28Z\",\"WARC-Record-ID\":\"<urn:uuid:0b77259e-6bf4-4935-851e-b276065e09bd>\",\"Content-Length\":\"24055\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3d809cf4-a3e1-47ea-a11f-31c91aa9caba>\",\"WARC-Concurrent-To\":\"<urn:uuid:0feb108f-70f3-419e-97ea-284f525d5953>\",\"WARC-IP-Address\":\"162.214.69.47\",\"WARC-Target-URI\":\"https://osdata.com/programming/calculations/subtration.html\",\"WARC-Payload-Digest\":\"sha1:SXP3BEMEJZI4UBUAZI6DD6O5H5E7VUWS\",\"WARC-Block-Digest\":\"sha1:5IU7A7ABHXDOY3DK5FGCAYOEJWN2F4UT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301737.47_warc_CC-MAIN-20220120100127-20220120130127-00641.warc.gz\"}"}
https://eng.libretexts.org/Courses/Delta_College/C___Programming_I_(McClanahan)/14%3A_Overloading_in_C/14.02%3A_Overload_a_Function/14.2.2%3A_Function_overloading_and_const_keyword	[ "$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\\| #1 \\\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\\| #1 \\\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\n#include<iostream>\nusing namespace std;\n\nclass Test\n{\nprotected:\nint num1;\npublic:\nTest (int i):num1(i) { }\nvoid fun() const\n{\ncout << \"fun() const called \" << endl;\n}\nvoid fun()\n{\ncout << \"fun() called \" << endl;\n}\n};\n\nint main()\n{\nTest t1 (10);\nconst Test t2 (20);\nt1.fun();\nt2.fun();\nreturn 0;\n}\n\n\nOutput: The above program compiles and runs fine, and produces following output.\n\nfun() called\nfun() const called\n\nThe two methods ‘void fun() const’ and ‘void fun()’ have same signature except that one is const and other is not. Also, if we take a closer look at the output, we observe that, ‘const void fun()’ is called on const object and ‘void fun()’ is called on non-const object.\n\nC++ allows member methods to be overloaded on the basis of const type. Overloading on the basis of const type can be useful when a function return reference or pointer. We can make one function const, that returns a const reference or const pointer, other non-const function, that returns non-const reference or pointer.\n\nRules related to const parameters are interesting. Let us first take a look at following two examples. The program 1 fails in compilation, but program 2 compiles and runs fine.\n\n// PROGRAM 1 (Fails in compilation)\n#include<iostream>\nusing namespace std;\n\nvoid fun(const int Num1)\n{\ncout << \"fun(const int) called \" << Num1;\n}\nvoid fun(int Num1)\n{\ncout << \"fun(int ) called \" << Num1;\n}\nint main()\n{\nconst int myNum = 10;\nfun(myNum);\nreturn 0;\n}\n\n\nOutput is:\n\nerror: redefinition of ‘void fun(int)’\nvoid fun(int Num2)\n^~~\n\nThen the second piece of code:\n\n// PROGRAM 1 (Fails in compilation)\n#include<iostream>\nusing namespace std;\n\nvoid fun(const int i)\n{\ncout << \"fun(const int) called \";\n}\nvoid fun(int i)\n{\ncout << \"fun(int ) called \" ;\n}\nint main()\n{\nconst int i = 10;\nfun(i);\nreturn 0;\n}\n\n\nwhich produces the outpu:\n\nconst fun() GeeksforGeeks\n\n\nC++ allows functions to be overloaded on the basis of const-ness of parameters only if the const parameter is a reference or a pointer. That is why the program 1 failed in compilation, but the program 2 worked fine. This rule actually makes sense. In program 1, the parameter ‘myNum’ is passed by value, so ‘Num1’ in fun() is a copy of ‘myNum’ in main(). Hence fun() cannot modify ‘myNum’ of main(). Therefore, it doesn’t matter whether ‘Num1’ is received as a const parameter or normal parameter. When we pass by reference or pointer, we can modify the value referred or pointed, so we can have two versions of a function, one which can modify the referred or pointed value, other which can not." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7616951,"math_prob":0.9956131,"size":2762,"snap":"2022-27-2022-33","text_gpt3_token_len":715,"char_repetition_ratio":0.13669325,"word_repetition_ratio":0.09166667,"special_character_ratio":0.28167993,"punctuation_ratio":0.13271028,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9684679,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-04T21:51:12Z\",\"WARC-Record-ID\":\"<urn:uuid:24501824-f3af-4647-afb3-1603bc278d82>\",\"Content-Length\":\"101037\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8e5bfe3f-800d-43fb-8dc7-3a122f7cde81>\",\"WARC-Concurrent-To\":\"<urn:uuid:3b63e8be-8ae3-4b99-ad90-9e316035b181>\",\"WARC-IP-Address\":\"13.249.39.44\",\"WARC-Target-URI\":\"https://eng.libretexts.org/Courses/Delta_College/C___Programming_I_(McClanahan)/14%3A_Overloading_in_C/14.02%3A_Overload_a_Function/14.2.2%3A_Function_overloading_and_const_keyword\",\"WARC-Payload-Digest\":\"sha1:XXHYPANZROVMQZV4A7TJO4I2DKMFGUR3\",\"WARC-Block-Digest\":\"sha1:3SQBFEUPAMVFQNEQIKSLVIV2EHGLNXUI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104496688.78_warc_CC-MAIN-20220704202455-20220704232455-00032.warc.gz\"}"}
https://developers.arcgis.com/kotlin/api-reference/arcgis-maps-kotlin/com.arcgismaps.geometry/-spatial-reference/index.html	[ "# SpatialReference\n\nThe spatial reference specifies how geometry coordinates relate to real-world space. Instances of this class represent a specific coordinate system identified by a well-known ID (WKID) number or well-known text (WKT) string. There are two broad classes of coordinate systems:\n\n• Geographic coordinate systems use a 3-dimensional spherical surface to define locations on the earth.\n\n• Projected coordinate systems use a flat, 2-dimensional surface. See [https://developers.arcgis.com/documentation/spatial-references/] for more information about spatial references.\n\nSpatialReference ensures that you can accurately view, query, and analyze the layers of a GeoModel.\n\nThe spatial reference value is available from a map or scene after loading has completed, and is immutable. If you want to set this value for a new map or scene, use the ArcGISMap.ArcGISMap(SpatialReference) or ArcGISScene.ArcGISScene(SceneViewTilingScheme) constructors.\n\n200.1.0\n\n## Constructors\n\nCreates a spatial reference based on WKID. The method will create a spatial reference that has only horizontal coordinate system and does not have vertical coordinate system associated with it.\n\nfun SpatialReference(wkid: Int, verticalWkid: Int)\n\nCreates a spatial reference based on WKID for the horizontal coordinate system and vertical coordinate system. The method creates a spatial reference that has both horizontal and vertical coordinate systems. When the vertical WKID is 0, the method is equivalent to calling SpatialReference.SpatialReference(Int), and does not define a vertical coordinate system part.\n\nCreates a spatial reference based on well-known text.\n\nobject Companion\n\n## Properties\n\nIf the given spatial reference is a projected coordinate system, then this returns the geographic coordinate system of that system. If the spatial reference is a projected coordinate system, then a spatial reference object representing the underlying geographic coordinate system is returned. Every projected coordinate system has an underlying geographic coordinate system. If the spatial reference is a geographic coordinate system, then a reference to itself is returned. If the spatial reference is a local spatial reference, a null is returned with an error.\n\nTrue if spatial reference has a vertical coordinate system set; false otherwise. A spatial reference can optionally include a definition for a vertical coordinate system (VCS), which can be used to interpret the z-values of a geometry. A VCS defines linear units of measure, the origin of z-values, and whether z-values are \"positive up\" (representing heights above a surface) or \"positive down\" (indicating that values are depths below a surface).\n\nTrue if spatial reference is a Geographic Coordinate System.\n\nTrue if coordinate system is horizontally pannable.\n\nTrue if spatial reference is a Projected Coordinate System.\n\nThe unit of measure for the horizontal coordinate system of this spatial reference.\n\nThe unit of measure for the vertical coordinate system of this spatial reference. Is null if SpatialReference.hasVertical is false.\n\nThe well-known ID for the vertical coordinate system (VCS), or 0 if the spatial reference has no VCS or has a custom VCS.\n\nval wkid: Int\n\nThe well-known ID for the horizontal coordinate system, or 0 if the spatial reference has a custom horizontal coordinate system.\n\nThe well-known text for the horizontal and vertical coordinate system.\n\n## Functions\n\nopen operator override fun equals(other: Any?): Boolean" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82315713,"math_prob":0.83642983,"size":3134,"snap":"2023-40-2023-50","text_gpt3_token_len":596,"char_repetition_ratio":0.2370607,"word_repetition_ratio":0.04845815,"special_character_ratio":0.17772815,"punctuation_ratio":0.09756097,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9555883,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T23:37:48Z\",\"WARC-Record-ID\":\"<urn:uuid:2cba322c-814e-4d1b-8f79-58975e2b376f>\",\"Content-Length\":\"41321\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5898805e-94fe-4334-b881-53612695db80>\",\"WARC-Concurrent-To\":\"<urn:uuid:b9381c3b-c45c-4821-9a1c-1d70365d5c6c>\",\"WARC-IP-Address\":\"18.160.18.26\",\"WARC-Target-URI\":\"https://developers.arcgis.com/kotlin/api-reference/arcgis-maps-kotlin/com.arcgismaps.geometry/-spatial-reference/index.html\",\"WARC-Payload-Digest\":\"sha1:BNXMOC3FSC64J3QJ6VA5XZYTYYXODNTD\",\"WARC-Block-Digest\":\"sha1:GIM3F4VEELPQPSUX3JCSBSZQOZWFB5NB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506539.13_warc_CC-MAIN-20230923231031-20230924021031-00158.warc.gz\"}"}
https://calculatorsonline.org/rounding-numbers/what-is-845-rounded-to-the-nearest-whole	[ "# 845 rounded to the nearest whole\n\nHere you will see step by step solution to round of the 845 to the nearest whole. What is 845 rounded to the nearest whole? 845 rounded to the nearest whole is 845, check the explanation that how to rounding the 845 to nearest whole.\n\n## Answer: Rounded 845 to nearest whole is\n\n= 845\n\n### How to round 845 to the nearest whole?\n\nTo round of the number 845 simply find the digit at whole place, then look to the right digit next to the whole place, if this number is greater than or equal to 5 (5, 6, 7, 8, 9) round up or if number is less then 5 (0, 1, 2, 3, 4) round down whole(5) number. Remove all the digits to the right of the rounding digit.\n\n#### Solution for decimal number 845 to the nearest whole\n\nGiven number is => 845\n\n• Number at whole place is = 5\n• 845 = 5\n• Now we need to find the digit at the right side of whole place = 0\n• 0 is smaller than 5, whole place digit 5 we don't need to change it. It will remain unchanged.\n• In 845, after rounding digit 5, remove all the numbers in the right side, then rewrite the numbers.\n• Final Conclusion: 845\n\nHence, the 845 to the nearest whole is 845." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86989665,"math_prob":0.9956066,"size":1119,"snap":"2022-40-2023-06","text_gpt3_token_len":307,"char_repetition_ratio":0.22690582,"word_repetition_ratio":0.04888889,"special_character_ratio":0.30741733,"punctuation_ratio":0.1124498,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9603134,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-03T07:12:05Z\",\"WARC-Record-ID\":\"<urn:uuid:6989d9d0-5c7d-4c97-8f61-9a5ab2404f7b>\",\"Content-Length\":\"18095\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ee52b7b7-e071-4ed1-b489-a2de9a13554d>\",\"WARC-Concurrent-To\":\"<urn:uuid:1d4deb13-9f4c-491e-bed4-0159e82d8c3b>\",\"WARC-IP-Address\":\"172.67.209.140\",\"WARC-Target-URI\":\"https://calculatorsonline.org/rounding-numbers/what-is-845-rounded-to-the-nearest-whole\",\"WARC-Payload-Digest\":\"sha1:IJHXBFIHRTDGJ346DY3HBUZZCB2EB2LA\",\"WARC-Block-Digest\":\"sha1:Y32E6ZWBT4NSNYCZNVCMGZYAMXVS2DUA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500044.16_warc_CC-MAIN-20230203055519-20230203085519-00689.warc.gz\"}"}
https://hq.holyfiregames.com/hq/topic/d551be-magnification-of-concave-lens-formula	[ "A magnification of 2 indicates the image is twice the size of the object and a magnification of 1 indicates an image size being the same as the object size. The sign conventions for the given quantities in the lens equation and magnification equations are as follows: f is + if the lens is a double convex lens (converging lens) f is - if the lens is a double concave lens (diverging lens) d i is + if the image is a real image and located on the opposite side of the lens. On the other hand, the magnification m is negative when the image formed is real and inverted. An object 5 cm high is held 25 cm away from a converging lens of focal length 20 cm. u = object distance Please note that the magnification formula is applicable both in convex lenses and concave lenses. Code to add this calci to your website . On the other hand, the magnification m is negative when the image formed is real and inverted. Calculate the position of the images formed by the fol-lowing concave lenses. In the case of a concave lens, it is always positive. Magnification of Convex Lens: It is a ratio between the image height and object height. (click on the green letters for the solutions). If the magnification is positive, then the image is upright compared to the object (virtual image). An expression showing the relation between object distance, image distance and focal length of a mirror is called mirror formula. Example Problem #1 A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a concave mirror having a focal length of 15.2 cm. Learn more about Reflection of Light here. Section 3: Concave Lenses 14 Here are some exercises with concave lenses. Q. Please note that the magnification formula is applicable both in convex lenses and concave lenses. Using lens formula the equation for magnification can also be obtained as . Therefore, the relationship between the object distance, the image distance and the focal length of a lens is given by the $$\\text{Lens Formula: }\\frac{1}{u}+\\frac{1}{v}=\\frac{1}{f}$$ The lens formula may be applied to convex lenses as well as concave lenses provided the ‘real is positive’ sign convention is followed. (a)For an object with u= 12cmif the focal length is 4cm Solution: As the lens here is concave, the focal length (f) = -10 cm. Hence, the expression for magnification (m) becomes: m = h’/h = -v/u. The focal length of the mirror is f. a. Determine the image distance and the image size. Be careful with the sign of the focal length in the lens formula! $$\\frac{1}{m(m+1)}$$ b. Formula : Where, f - Focal length, d i - Image distance, d 0 - Object distance. Is lens formula applicable only for convex lens? Assumptions and Sign conventions Draw the ray diagram and find the position, size and nature of the image formed. If the magnification is positive, then the image is upright compared to the object (virtual image). The magnification ($$M$$) of the image formed can be calculated using the following formula. Example 2: The distance of an object of height 6 cm from a concave lens is 20 cm. The two formulas given above are together referred to as the thin lens formula. We also have another formula for magnification in lenses Magnification = v/u where v is image distance u is object distance Note: - If magnification (m) is positive, It means image formed is virtual and erect If magnification (m) is negative, It means image formed is real and inverted Questions Example 10.3 - A concave lens has focal length of 15 cm. As a demonstration of the effectiveness of the mirror equation and magnification equation, consider the following example problem and its solution. Physics Grade XI Reference Note: Mirror formula for concave mirror when real image is formed and for convex mirror. What will be the distance of the object, when a concave mirror produces an image of magnification m? Online physics calculator that calculates the concave mirror equation from the given values of object distance (do), the image distance (di), and the focal length (f). If ‘ $$i$$ ’ is positive, the image is upright and if ‘ $$i$$ ’ is negative, the image is inverted. Exercise 2. Magnification of Convex Lens: It is a ratio between the image height and object height. Object distance (o) = + 20 cm Now by putting the values of (f) and (o) in 1/v+ 1/o = 1/f, we will get, 1/v = 1/f - 1/o Where, $$v$$ is the object height $$u$$ is the image height. At what distance should the object from the lens be placed so … And the magnification m is positive when the image formed is virtual and erect. A magnification of 2 indicates the image is twice the size of the object and a magnification of 1 indicates an image size being the same as the object size. Lens formula is applicable for convex as well as concave lenses. In the case of a concave lens, it is always positive. It is an equation that relates the focal length, image distance, and object distance for a spherical mirror. If its focal length is 10 cm, calculate the size and position of the image formed. Solved Example for You. m = h 2 /h 1 = v//u = (f-v)/f = f/(f+u) This equation is valid for both convex and concave lenses and for real and virtual images. You can try the following sample problem using this. It is given as, \\frac {1} {i} + \\frac {1} {o} = \\frac {1} {f} i= distance of the image from the lens. Define lens formula. A concave lens of focal length 15 cm forms an image 10 cm from the lens. If magnification is negative then the image is inverted as … These lenses have negligible thickness. Using lens formula the equation for magnification can also be obtained as m = h2/h1 = v//u = (f-v)/f = f/ (f+u) This equation is valid for both convex and concave lenses and for real and virtual images. Table shows the sign convention for the values of object distance, image … And the magnification m is positive when the image formed is virtual and erect. The magnification is negative for real image and positive for virtual image." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9050861,"math_prob":0.9981354,"size":5948,"snap":"2021-21-2021-25","text_gpt3_token_len":1406,"char_repetition_ratio":0.20777254,"word_repetition_ratio":0.2236842,"special_character_ratio":0.23839946,"punctuation_ratio":0.08966695,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996923,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-14T04:50:24Z\",\"WARC-Record-ID\":\"<urn:uuid:ca481768-542a-4a96-9f84-c50a1b0cf100>\",\"Content-Length\":\"16453\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1e4bd611-62f2-4cfe-9f41-e7aeaf955fc3>\",\"WARC-Concurrent-To\":\"<urn:uuid:79873004-83eb-420e-a118-fcc6ff192443>\",\"WARC-IP-Address\":\"104.21.31.145\",\"WARC-Target-URI\":\"https://hq.holyfiregames.com/hq/topic/d551be-magnification-of-concave-lens-formula\",\"WARC-Payload-Digest\":\"sha1:WP5P7TOWUVYJMN3BCDIPCXAUSS7WIDDE\",\"WARC-Block-Digest\":\"sha1:4NCSELMQ4IIEMYAPIXBFJIDNGH36VYMN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991737.39_warc_CC-MAIN-20210514025740-20210514055740-00304.warc.gz\"}"}