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https://ch.mathworks.com/matlabcentral/cody/problems/43977-converting-binary-to-decimals/solutions/1692291	[ "Cody\n\n# Problem 43977. Converting binary to decimals\n\nSolution 1692291\n\nSubmitted on 14 Dec 2018 by HH\n• Size: 9\n• This is the leading solution.\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\nd = ('010111'); y_correct = 23; assert(isequal(bin_to_dec(d),y_correct))\n\n[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In unix (line 32) In bin_to_dec (line 2) In ScoringEngineTestPoint1 (line 3) In solutionTest (line 3)] ans = 1\n\n2 Pass\nd = ('110000'); y_correct = 48; assert(isequal(bin_to_dec(d),y_correct))\n\n[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In unix (line 32) In bin_to_dec (line 2) In ScoringEngineTestPoint2 (line 3) In solutionTest (line 5)] ans = 1\n\n3 Pass\nd = ('1000110'); y_correct = 70; assert(isequal(bin_to_dec(d),y_correct))\n\n[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In unix (line 32) In bin_to_dec (line 2) In ScoringEngineTestPoint3 (line 3) In solutionTest (line 7)] ans = 1" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.66905594,"math_prob":0.9425149,"size":1300,"snap":"2019-51-2020-05","text_gpt3_token_len":373,"char_repetition_ratio":0.13271604,"word_repetition_ratio":0.44927537,"special_character_ratio":0.3123077,"punctuation_ratio":0.0995671,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96720415,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-21T22:48:15Z\",\"WARC-Record-ID\":\"<urn:uuid:c4de0bbc-3931-4ff2-959b-d3adcaeb56d5>\",\"Content-Length\":\"72358\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a1a88233-0ce6-445d-9dd5-845e01661171>\",\"WARC-Concurrent-To\":\"<urn:uuid:f0e0d6c5-8073-4df3-9783-027694233ff3>\",\"WARC-IP-Address\":\"104.96.217.125\",\"WARC-Target-URI\":\"https://ch.mathworks.com/matlabcentral/cody/problems/43977-converting-binary-to-decimals/solutions/1692291\",\"WARC-Payload-Digest\":\"sha1:WFDQYNXSX5IX6PEWDX2MIEWKOG3AB43Z\",\"WARC-Block-Digest\":\"sha1:EZ2UJHBPOI666UDDNDZ7XYZ3QJOJ6TV7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250606226.29_warc_CC-MAIN-20200121222429-20200122011429-00316.warc.gz\"}"}
https://rpgmaker.net/forums/topics/24439/?post=875370	[ "", null, "# [RM2K3] IS THERE A WAY TO CHECK REQUIRED EXP FOR NEXT LEVEL?\n\n## Posts\n\nPages: 1\nSo Rm2k3 lets you store the EXP of a character in a variable, but only the current EXP, not the EXP needed for the next level. So if I want to manually track that, and my max lvl is 99, do I have to make 99 conditional branches for every character that check their levels and enters a value inside a variable?\n\nThe only way to do that is to manually set variables to the EXP required I entered in the Database. If I change my mind later on and change EXP values required for my characters, I'm looking at updating hundreds of conditional branches.\n\nSo yeah. Any neat little trick to do this?\nI'd suggest looking up the RM2k3 EXP algorithm and implement that in event code. Take a character's level (and if each character has different EXP curve parameters the ID too), calculate the EXP needed to reach character level+1, then subtract their current EXP. It'd be better than the hell of 99 conditional branches. idk the algorithm off the top of my head and my google-fu can't find it but it's hopefully in the help documentation.\nIDK if my solution is the roughly the same as GRS's. But when setting a character's level it'll set their EXP to what's required. You could duplicate a hero with the same exp table and have their level always = the real characters level but +1 and grab their exp from there.\nIt's definitely more convenient than mine, that's a clever trick!\nI don't really understand what both of you just said, but I'll look through the help file\nTo expand on Darken's solution: We have Hero Higgins who at some point we want to find the EXP to next level on. To do so we make a clone of HH in the database with the exact same EXP curve, we'll call it Clone Higgins. Then when you want to find HH's EXP to their next level you take Clone Higgins and set their level to HH's level+1. RM2k3 will give Clone Higgins the amount of EXP needed to reach that level which you can then assign to a variable. You take Clone Higgin's EXP and subtract Hero Higgin's current EXP and that will give you the EXP HH needs to reach the next level.\n\nExample with numbers!\n\nHero Higgins is lv10. Our level curve is 10,000 EXP to reach level 10, 14,350 EXP to reach lv11 (not real RM2k3 exp numbers). Hero Higgins has 11,840 EXP right now. We want to find out the EXP remaining:\n\n1) Assign Hero Higgin's current EXP to variable 1. V1 is now 11,840.\n2) Set Clone Higgin's level to 11. The safest way is to reduce Clone Higgin's level by 99 then increase it by Hero Higgin's current level. So something like:\n`VARIABLE: V3 = Hero Higgin's LevelCHANGE LEVEL: Clone Higgins -99 (this will set it to level 1, 0 exp)CHANGE LEVEL: Clone Higgins + VARIABLE: V3(this will add Hero Higgin's level 10, to Clone Higgin's lv1 giving him level 11)`\n3) Assign Clone Higgin's current EXP to variable 2. Since Clone Higgins level was set to lv11 in step 2 his current EXP would be 14,350, the EXP required to reach lv11 with our curve.\n4) Subtract V2 by V1. 14,350 - 11,840: 2,510. This is Hero Higgin's EXP needed to reach level 11, available to do what you want with it.\n\nI can throw together a demo project tonight too if it'll help.\nAh. I see what you're saying.\n\nI have to say, Rm2k3's way of handling EXP and level ups is pretty peculiar. Most RPGs of the SNES era reset your EXP to zero when you reach a new level, and the EXP required for the next level is higher than the previous one, but not a number added on top of the amount needed for the previous one. Unlike Rm2k3, which doesn't reset your EXP to zero, but lets you keep the EXP you needed for the level up as your current EXP, and just adds more on top.\n\nSo what I kinda wanna do is make a common event that checks if a character's level increased, and reset their EXP to zero. But it only needs to do it once, at the moment the level is increased. Then I'd use my own EXP values instead of Rm2k3's. Do you think that's feasible, or is this fighting the engine too much? And how could I pull this off without once again checking every level?\nPersonally trying to make a custom EXP curve in 2k3 sounds like hell but all those dynRPG(?) plugins might've made it more convenient. I haven't used 2k/3 for ages though so I can't offer any real support on this though. Hopefully some 2k/3 power users can help, or maybe a new thread about trying to create a custom EXP curve might get them to pop in?\nTo me, resetting the experience every level seems like it'd be pretty tedious. To your point about SNES RPGs doing this, I don't know if that's even true. A quick glance at all the major final fantasy games of the SNES era keeps a cumulative total of EXP, and then tells you how much is needed for the next level. BUT! Regardless, here is how you calculate experience needed.\n\nEvery experience curve in RM2k3 has three properties: Base Level, Acceleration, and Extra Value. (WARNING: in the official translation, the values are mislabeled and Extra should be labeled Acceleration! These formulas below are written assuming the 2nd value is Acceleration and the 3rd value is Extra.)\n\nSo, given your experience curve has:\nB = Base Value\nA = Acceleration\nX = Extra value\n\nExperience needed to get from level L to L+1:\nB + X + A(L)\n\nTotal experience needed to reach level L:\n(L - 1)(B + X) + A( L(L-1) / 2 )\n\nExamples..\nGiven B=25, X=15, A=8\nExperience needed to get from 5 to 6:\nB + E + A(L) = 25 + 15 + 8(5) = 40 + 40 = 80\n\nTotal experience needed to reach level 6:\n(L-1)(B+X) + A( L(L-1) / 2)\n= (6-1)(25+15) + 8( 6(5) / 2 )\n= 5(40) + 8(30/2)\n= 200 + 8(15)\n= 200 + 120\n= 320\n\nIf you want to know exactly how much EXP is needed to get to the next level:\nGiven the hero's level is 6, we want to target level 7.\nTake the hero's EXP (assign it to a variable) and subtract it from the Total Experience needed to reach level 7.\n\nSo say the hero has 350 EXP.\nGiven the same base, acceleration, and extra value parameters..\nTotal exp needed to reach level 7:\n(L-1)(B+X) + A(L(L-1) / 2)\n= 6(40) + 8(7(6) / 2)\n= 240 + 8(42 / 2)\n= 240 + 8(21)\n= 240 + 168\n= 408\nNeeded for next level: 408 - 350 = 58\n\nIt's a nontrivial amount of calculation to do in RM2k3, but simple enough to code up in a common event and reference whenever you need it. Hope this helps you out!\nIn my project I use this method to get the experience.\nTalk to the boy to see the experience and with the girl to get a quantity.\nhttp://www.mediafire.com/file/31w9wv2cil8h9vx/hero+exp.rar\nauthor=narcodis\nTo me, resetting the experience every level seems like it'd be pretty tedious. To your point about SNES RPGs doing this, I don't know if that's even true. A quick glance at all the major final fantasy games of the SNES era keeps a cumulative total of EXP, and then tells you how much is needed for the next level. BUT! Regardless, here is how you calculate experience needed.\n\nEvery experience curve in RM2k3 has three properties: Base Level, Acceleration, and Extra Value. (WARNING: in the official translation, the values are mislabeled and Extra should be labeled Acceleration! These formulas below are written assuming the 2nd value is Acceleration and the 3rd value is Extra.)\n\nSo, given your experience curve has:\nB = Base Value\nA = Acceleration\nX = Extra value\n\nExperience needed to get from level L to L+1:\nB + X + A(L)\n\nTotal experience needed to reach level L:\n(L - 1)(B + X) + A( L(L-1) / 2 )\n\nExamples..\nGiven B=25, X=15, A=8\nExperience needed to get from 5 to 6:\nB + E + A(L) = 25 + 15 + 8(5) = 40 + 40 = 80\n\nTotal experience needed to reach level 6:\n(L-1)(B+X) + A( L(L-1) / 2)\n= (6-1)(25+15) + 8( 6(5) / 2 )\n= 5(40) + 8(30/2)\n= 200 + 8(15)\n= 200 + 120\n= 320\n\nIf you want to know exactly how much EXP is needed to get to the next level:\nGiven the hero's level is 6, we want to target level 7.\nTake the hero's EXP (assign it to a variable) and subtract it from the Total Experience needed to reach level 7.\n\nSo say the hero has 350 EXP.\nGiven the same base, acceleration, and extra value parameters..\nTotal exp needed to reach level 7:\n(L-1)(B+X) + A(L(L-1) / 2)\n= 6(40) + 8(7(6) / 2)\n= 240 + 8(42 / 2)\n= 240 + 8(21)\n= 240 + 168\n= 408\nNeeded for next level: 408 - 350 = 58\n\nIt's a nontrivial amount of calculation to do in RM2k3, but simple enough to code up in a common event and reference whenever you need it. Hope this helps you out!\n\nI think you're right actually, for some reason I assumed EXP was reset to 0 with every level.\n\nI'm really bad at the math stuff, but why not just swap those labels around to suit the official translation? Seems less confusing to go off of that. The rest of your post looks like Einstein's calculations to me, but I'll try to make it work.\nPages: 1" ]	[ null, "https://www.facebook.com/tr", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91310364,"math_prob":0.84332794,"size":12041,"snap":"2020-24-2020-29","text_gpt3_token_len":3399,"char_repetition_ratio":0.13641272,"word_repetition_ratio":0.73195875,"special_character_ratio":0.29806495,"punctuation_ratio":0.10354119,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97459924,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-27T05:53:33Z\",\"WARC-Record-ID\":\"<urn:uuid:7652bfe9-5a22-44f5-adc1-39cc8b616b37>\",\"Content-Length\":\"41946\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5149a53a-f75d-4860-ad4e-cf407406a040>\",\"WARC-Concurrent-To\":\"<urn:uuid:87bd1eb2-0145-432c-96d2-bb81d51a20cd>\",\"WARC-IP-Address\":\"23.239.86.130\",\"WARC-Target-URI\":\"https://rpgmaker.net/forums/topics/24439/?post=875370\",\"WARC-Payload-Digest\":\"sha1:DL7EVN23WZAANLPKL3G6XZCS6R2ZQZFF\",\"WARC-Block-Digest\":\"sha1:NZASNMPS7UWCOVWQE4ZFEE37PS7MSQXB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347392141.7_warc_CC-MAIN-20200527044512-20200527074512-00398.warc.gz\"}"}
https://dotnettutorials.net/lesson/while-loop-in-c/	[ "# While Loop in C\n\n## While Loop in C Language with Examples\n\nIn this article, I am going to discuss While Loop in C Language with Definitions, Syntax, Flow Charts, and Examples. Please read our previous articles, where we discussed Switch Statements in C Language with Examples. Loop Control Statements are very, very important for logical programming. The Looping Statements are also called Iteration Statements. So, we can use the word Looping and Iteration and the meanings are the same. Before understanding the while loop in C, let us first understand what is a loop, why we need the loop, and the different types of loops available in the C program.\n\n##### What is looping?\n\nThe process of repeatedly executing a statement or group of statements until the condition is satisfied is called looping. In this case, when the condition becomes false the execution of the loops terminates. The way it repeats the execution of the statements will form a circle that’s why iteration statements are called loops.\n\n##### Why do we need looping?\n\nThe basic purpose of the loop is code repetition. In implementation whenever the repetitions are required, then in place of writing the statements, again and again, we need to go for looping.\n\n##### Iteration or looping Statements:\n\nIteration statements create loops in the program. It repeats the same code fragment several times until a specified condition is satisfied is called iteration. Iteration statements execute the same set of instructions until a termination condition is met. C provides the following loop for iteration statements:\n\n1. while loop\n2. for loop\n3. do-while loop\n##### What is While Loop in C Language:\n\nA loop is nothing but executes a block of instructions repeatedly as long as the condition is true. How many times it will repeat means as long as the given condition is true. When the condition fails, it will terminate the loop execution.\n\nA while loop is used for executing a statement repeatedly until a given condition returns false. Here, statements may be a single statement or a block of statements. The condition may be any expression, and true is any nonzero value. The loop iterates while the condition is true. If you see the syntax and flow chart parallelly, then you will get more clarity of the while loop.\n\n###### While Loop Syntax in C Language:\n\nFollowing is the syntax to use the while loop in C Programming Language.", null, "While we are working with a while loop first we need to check the condition, if the condition is true then the control will pass within the body if it is false the control pass outside the body.\n\nWhen we are working with an iteration statement after execution of the body control will be passed back to the condition, and until the condition become false. If the condition is not false then we will get an infinite loop.\n\nIt is something similar to the if condition, just condition, and statements, but the flow is different from the if condition. How it is different lets us understand through the flow-chart.\n\n##### Flow Chart of While Loop in C Language:\n\nThe following diagram shows the flow chart of the while loop.", null, "The flow chart will start. The start is represented by the oval symbol. Then it will check the condition. As discussed earlier, every condition is having two outputs i.e. true and false. If it is true what will happen and it is false what will happen, we need to check.\n\nSuppose, the condition is true, then what all statements defined inside the block (within the while loop block) will execute. After execution of statements, will it end? NO, it will not end. After execution of statements, once again it will go and check the condition. It will repeat the same process as long as the given condition is true. Suppose, the condition is false, then it will come to end. This is the execution flow of a while loop.\n\n##### Program to understand While loop in C Language:\n```#include<stdio.h>\nint main()\n{\nint a=1;\nwhile(a<=4)\n{\nprintf(\"%d\", a);\na++;\n}\nreturn 0;\n}\n```\n\nOutput: 1234\n\nThe variable a is initialized with value 1 and then it has been tested for the condition. If the condition returns true then the statements inside the body of the while loop are executed else control comes out of the loop. The value of a is incremented using the ++ operator then it has been tested again for the loop condition.\n\n##### Example to Print no 1 to 5 using a while loop in C Language\n```#include <stdio.h>\nint main()\n{\nint i = 1;\nwhile(i <= 5)\n{\nprintf(\"%d \", i);\ni = i + 1;\n}\nreturn 0;\n}```\n\nOutput: 1 2 3 4 5\n\n##### Example: Print the nos 10, 9, 8…. 1 using while loop\n```#include <stdio.h>\nint main()\n{\nint i;\ni = 10;\nwhile(i >= 1)\n{\nprintf(\"%d \", i);\ni = i - 1;\n}\nreturn 0;\n}```\n\nOutput: 10 9 8 7 6 5 4 3 2 1\n\nExample: Print the numbers in the following format up to a given number and that number is entered from the keyboard.\n\n2 4 6 8 …………………….. up to that given number\n\n```#include <stdio.h>\nint main()\n{\nint i, n;\nprintf(\"Enter a Number : \");\nscanf(\"%d\", &n);\ni = 2;\nwhile(i <= n)\n{\nprintf(\"%d \", i);\ni = i + 2;\n}\nreturn 0;\n}```\n###### Output:", null, "###### Program to Enter a number and print the Fibonacci series up to that number using a while loop in C Language.\n```#include <stdio.h>\nint main ()\n{\nint i, n, j, k;\nprintf (\"Enter a Number : \");\nscanf (\"%d\", &n);\n\ni = 0;\nj = 1;\nprintf (\"%d %d \", i, j);\nk = i + j;\nwhile (k <= n)\n{\nprintf (\" %d\", k);\ni = j;\nj = k;\nk = i + j;\n}\n\nreturn 0;\n}```\n###### Output:", null, "Example: Enter a number from the keyboard and then calculate the no of digits and sum of digits of that number using a while loop.\n\n```#include <stdio.h>\nint main()\n{\nint no, nd, sd, rem;\nprintf(\"Enter a Number : \");\nscanf(\"%d\", &no);\nnd = 0;\nsd = 0;\nwhile (no > 0)\n{\nrem = no % 10;\nnd = nd + 1;\nsd = sd + rem;\nno = no / 10;\n}\nprintf(\"The number of digit is %d\", nd);\nprintf(\"\\nThe sum of digit is %d\", sd);\nreturn 0;\n}```\n###### Output:", null, "##### What is the pre-checking process or entry controlled loop?\n\nThe pre-checking process means before the evaluation of the statement block conditional part will be executed. When we are working with a while loop always pre-checking process will occur. The loop in which before executing the body of the loop if the condition is tested first then it is called an entry controlled loop.\n\nWhile loop is an example of entry controlled loop because in the while loop before executing the body first condition is evaluated if the condition is true then the body will be executed otherwise the body will be skipped." ]	[ null, "https://dotnettutorials.net/wp-content/uploads/2021/07/word-image-194.png", null, "https://dotnettutorials.net/wp-content/uploads/2021/07/word-image-195.png", null, "https://dotnettutorials.net/wp-content/plugins/wp-fastest-cache-premium/pro/images/blank.gif", null, "https://dotnettutorials.net/wp-content/plugins/wp-fastest-cache-premium/pro/images/blank.gif", null, "https://dotnettutorials.net/wp-content/plugins/wp-fastest-cache-premium/pro/images/blank.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8756127,"math_prob":0.9087377,"size":6717,"snap":"2022-27-2022-33","text_gpt3_token_len":1571,"char_repetition_ratio":0.17354387,"word_repetition_ratio":0.06526995,"special_character_ratio":0.2523448,"punctuation_ratio":0.13711414,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97383046,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,3,null,3,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-26T21:04:44Z\",\"WARC-Record-ID\":\"<urn:uuid:7e6f26e1-b692-4044-85d1-5504b067ffbc>\",\"Content-Length\":\"105390\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9f2ea043-89a1-4182-ba0c-1dffe2423b1d>\",\"WARC-Concurrent-To\":\"<urn:uuid:e5748948-60f2-48b9-b545-64d5df749830>\",\"WARC-IP-Address\":\"35.209.228.125\",\"WARC-Target-URI\":\"https://dotnettutorials.net/lesson/while-loop-in-c/\",\"WARC-Payload-Digest\":\"sha1:S47IRX4VFY66Y4S7TOPSCUA4L6U2HBUG\",\"WARC-Block-Digest\":\"sha1:XEPYDZRIVIGOSJ5QP4XJEUM3VX6D7OA4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103271864.14_warc_CC-MAIN-20220626192142-20220626222142-00527.warc.gz\"}"}
https://www.hindawi.com/journals/aaa/2013/838302/	[ "#### Abstract\n\nA nonlinear generalization of the famous Camassa-Holm model is investigated. Provided that initial value and satisfies an associated sign condition, it is shown that there exists a unique global weak solution to the equation in space in the sense of distribution, and .\n\n#### 1. Introduction\n\nIn recent years, a lot of works have been carried out to investigate the Camassa-Holm equation , which is a completely integrable equation. In fact, the Camassa-Holm equation arises as a model describing the unidirectional propagation of shallow water waves over a flat bottom . The equation was originally derived much earlier as a bi-Hamiltonian generalization of the Korteweg-de Vries equation (see ). Johnson , Constantin and Lannes derived models which include the Camassa-Holm equation (1). It has been found that (1) conforms with many conservation laws (see [6, 7]) and possesses smooth solitary wave solutions if [3, 8] or peakons if [3, 9]. Equation (1) is also regarded as a model of the geodesic flow for the right invariant metric on the Bott-Virasoro group if and on the diffeomorphism group if (see ). The well-posedness of local strong solutions for generalized forms of (1) has been given in . The sharpest results for the global existence and blow-up solutions are found in Bressan and Constantin [18, 19].\n\nRecently, Li et al. studied the following generalized Camassa-Holm equation: where is a natural number. Obviously, (2) reduces to (1) if . The authors applied the pseudoparabolic regularization technique to build the local well-posedness for (2) in Sobolev space with via a limiting procedure. Provided that the initial value satisfies a sign condition and , it is shown that there exists a unique global strong solution for (2) in space . However, the existence and uniqueness of the global weak solution for (2) is not investigated in .\n\nThe objective of this paper is to establish the well-posedness of global weak solutions for (2). Using the estimates in with , which are derived from the equation itself, we prove that there exists a unique global weak solution to (2) in space with if , and satisfies an associated sign condition.\n\nThe structure of this paper is as follows. The main result is given in Section 2. Several lemmas are given in Section 3. Section 4 establishes the proof of the main result.\n\n#### 2. Main Results\n\nFirstly, we give some notations.\n\nThe space of all infinitely differentiable functions with compact support in is denoted by . is the space of all measurable functions such that . We define with the standard norm . For any real number , we let denote the Sobolev space with the norm defined by where .\n\nFor and nonnegative number , let denote the Frechet space of all continuous -valued functions on . We set .\n\nDefining and letting with and (convolution of and ), we know that for any with . Notation (or equivalently ) means that (or equivalently ) for an arbitrary sufficiently small .\n\nFor the equivalent form of (2), we consider its Cauchy problem\n\nDefinition 1. A function is called a global weak solution to problem (5) if for every , , and all , it holds that with .\n\nNow, we give the main result of this work.\n\nTheorem 2. Let , , , and (or equivalently , ). Then, problem (5) has a unique global weak solution in the sense of distribution, and .\n\n#### 3. Several Lemmas\n\nLemma 3 (see ). Let with . Then, the Cauchy problem (5) has a unique solution where depends on .\n\nLemma 4 (see ). Let , , and (or equivalently , . Then, problem (5) has a unique solution satisfying\n\nUsing the first equation of system (5) derives from which one has the conservation law\n\nLemma 5 (see ). Let , and the function is a solution of problem (5) and the initial data . Then, the following inequality holds:\nFor , there is a constant such that\nFor , there is a constant such that\n\nFor (2), consider the problem\n\nLemma 6 (see ). Let , , and let be the maximal existence time of the solution to problem (5). Then, problem (14) has a unique solution . Moreover, the map is an increasing diffeomorphism of with for .\n\nDifferentiating (14) with respect to yields which leads to\n\nThe next lemma is reminiscent of a strong invariance property of the Camassa-Holm equation (the conservation of momentum ).\n\nLemma 7 (see ). Let with , and let be the maximal existence time of the problem (5). It holds that where and .\n\nLemma 8. If , , such that , (or equivalently, ), then the solution of problem (5) satisfies\n\nProof. Using , it follows from Lemma 7 that . Letting , we have from which we obtain On the other hand, we have The inequalities (19), (20), and (21) derive that inequality (18) is valid. Similarly, if , , we still know that (18) is valid.\n\nLemma 9. For , , it holds that where is a constant independent of .\n\nThe proof of this lemma can be found in Lai and Wu .\n\nFrom Lemma 3, it derives that the Cauchy problem has a unique solution depending on the parameter . We write to represent the solution of problem (23). Using Lemma 3 derives that since .\n\nLemma 10. Provided that , , , and (or equivalently , ), then there exists a constant independent of such that the solution of problem (23) satisfies\n\nProof. Using identity (10) and Lemma 9, if with , we have where is independent of .\nFrom Lemma 8, we have which completes the proof.\n\nLemma 11. For any , with , it holds that\n\nThe proof of this lemma can be found in .\n\n#### 4. Existence and Uniqueness of Global Weak Solution\n\nProvided that , for problem (23), applying Lemmas 5, 9, and 10, and the Gronwall’s inequality, we obtain the inequalities where , and is a constant independent of . It follows from the Aubin’s compactness theorem that there is a subsequence of , denoted by , such that and their temporal derivatives are weakly convergent to a function and its derivative in and , respectively, where is an arbitrary fixed positive number. Moreover, for any real number , is convergent to the function strongly in the space for and converges to strongly in the space for .\n\n##### 4.1. The Proof of Existence for Global Weak Solution\n\nFor an arbitrary fixed , from Lemma 10, we know that is bounded in the space . Thus, the sequences , , , and are weakly convergent to , , , and in for any , separately. Using , we know that satisfies the equation with and . Since is a separable Banach space and is a bounded sequence in the dual space of , there exists a subsequence of , still denoted by , weakly star convergent to a function in . As weakly converges to in , it results that almost everywhere. Thus, we obtain . Since is an arbitrary number, we complete the global existence of weak solutions to problem (5).\n\nProof of Uniqueness. Suppose that there exist two global weak solutions and to problem (5) with the same initial value , , we consider its associated regularized problem (23). Letting , from Lemma 10, we get and which is independent of . Still denoting , and , it holds that\nMultiplying both sides of (30) by , we get Using , , , , we have Applying Lemma 11 repeatedly, we have For , using Lemma 11 derives Using (32)–(34), we get Applying results in . Consequently, we know that the global weak solution is unique.\n\n#### Acknowledgment\n\nThis work is supported by the Fundamental Research Funds for the Central Universities (JBK120504)." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8544092,"math_prob":0.99231327,"size":12408,"snap":"2022-40-2023-06","text_gpt3_token_len":3278,"char_repetition_ratio":0.1435021,"word_repetition_ratio":0.16378482,"special_character_ratio":0.2689394,"punctuation_ratio":0.2004,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99780583,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-07T22:27:47Z\",\"WARC-Record-ID\":\"<urn:uuid:c313b862-122b-47ee-b9c0-91b8040d1ed9>\",\"Content-Length\":\"1049037\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:21645f5d-e3d3-4fa6-8c35-a60ffa71154e>\",\"WARC-Concurrent-To\":\"<urn:uuid:43a3ea12-6cc4-40f4-807b-1118149d987b>\",\"WARC-IP-Address\":\"18.160.46.10\",\"WARC-Target-URI\":\"https://www.hindawi.com/journals/aaa/2013/838302/\",\"WARC-Payload-Digest\":\"sha1:TMHQUQRTIF3WYIJID3GLJQWB4ZLRLUTM\",\"WARC-Block-Digest\":\"sha1:TWGK5JX3AMTMB6Y6HSLGGCINHYREHGLS\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030338280.51_warc_CC-MAIN-20221007210452-20221008000452-00610.warc.gz\"}"}
http://grr.blahnet.com/home/software/microsoft/excel	[ "Home‎ > ‎Software‎ > ‎Microsoft‎ > ‎\n\n## Excel calculated value in a text string\n\n`=\"Buy In \\$50 Pot - Total Pot Winning pot \" &TEXT(COUNTA(A3:A21)*50,\"\\$0.00\")`\n\n## Calculate for non blank values\n\n`=IF(ISBLANK(D3),\"\",(\\$C3/D3)-1)`\n\n## Convert minutes to Hours and minutes\n\nFormat the cell as time format\n=H13/1440\n\n## Macro to hide columns\n\n`Sub Button2_Click()`\n` Dim cl As Range, rTest As Range`\n` `\n` Set rTest = Range(\"d2\", Range(\"d2\").End(xlToRight))`\n` For Each cl In rTest`\n` If Not cl.Value = \"blah\" Then`\n` cl.EntireColumn.Hidden = True`\n` End If`\n` Next cl`\n`End Sub`\n\n## Substring\n\nServername\n`Datacenter =MID(A2,17,3)`\n`Product Code =MID(A2,8,3)`\n`Swimlane =MID(A2,8,3) `\n`Type =MID(A2,11,2)`\n`Environment =MID(A2,3,4)`\n`Tier =MID(A2,2,1)`" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.539059,"math_prob":0.9605897,"size":643,"snap":"2023-14-2023-23","text_gpt3_token_len":231,"char_repetition_ratio":0.114241004,"word_repetition_ratio":0.0,"special_character_ratio":0.3437014,"punctuation_ratio":0.15862069,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99573636,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-07T14:54:22Z\",\"WARC-Record-ID\":\"<urn:uuid:0532cfe0-a26c-40a7-bc26-77a732ef348f>\",\"Content-Length\":\"51905\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:61ccd181-f52c-4589-a884-65e060341840>\",\"WARC-Concurrent-To\":\"<urn:uuid:9ce7e7b2-d9b4-4241-8beb-b8ed7874d2c6>\",\"WARC-IP-Address\":\"172.253.122.121\",\"WARC-Target-URI\":\"http://grr.blahnet.com/home/software/microsoft/excel\",\"WARC-Payload-Digest\":\"sha1:F5A64Y54TKDHIRG3REM7CSVAHGGCVFA6\",\"WARC-Block-Digest\":\"sha1:EU35GT4QRODSA2XK67ADYWCUJGV43MYN\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224653930.47_warc_CC-MAIN-20230607143116-20230607173116-00438.warc.gz\"}"}
https://studyforce.com/verify-trigonometric-identities/	[ "# How to Verify Trigonometric Identities\n\nProving identities is a major part of any trigonometry course. This involves getting one of the equation to be identical to the other side. Try using one of the following strategies to begin, and then use others if necessary to continue the verification process.\n\n• First, start with what appears to be the more difficult side of the given identity.\n• Try to transform this side so that it eventually matches identically to the other side.\n• Don’t hesitate to start over and work with the other side of the identity if you are having trouble.\n\n1. Look for ways to use known identities such as the reciprocal identities, quotient identities, and even/odd properties. If the identity includes a squared trigonometric expression. try using a variation of a Pythagorean identity.\n\n2. Try rewriting each trigonometric expression in terms of sines and cosines.\n\n3. Factor out a greatest common factor and use algebraic factoring techniques such as factoring the difference of two squares or the sum/difference of two cubes.\n\n4. If a single term appears in the denominator of a quotient, try separating the quotient in two or more quotients:\n\n$$\\frac{A+B}{C} = \\frac{A}{C}+\\frac{B}{C}$$\n\n5. If there are two or more fractional expressions, try combining the expressions using a common denominator:\n\n$$\\frac{A}{C}+\\frac{B}{D} = \\frac{AD+BC}{CD}$$\n\n6. If the numerator or denominator of one or more quotients contains an expression of the form $B + \\sqrt{C}$, try multiplying the numerator and denominator by its conjugate $B – \\sqrt{C}$:\n\n$$\\frac{A}{B+\\sqrt{C}}$$ $$= \\frac{A}{B+ \\sqrt{C}} \\cdot \\color{red} \\frac{B- \\sqrt{C}}{B-\\sqrt{C}}$$ $$=\\frac{A(B-\\sqrt{C})}{B^2-C}$$\n\nThe idea here is that multiplying by the denominator’s conjugate, you force the denominator to become radical-less. This can lead to further manipulation of the terms." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84108996,"math_prob":0.9952826,"size":1823,"snap":"2021-31-2021-39","text_gpt3_token_len":454,"char_repetition_ratio":0.13743815,"word_repetition_ratio":0.0,"special_character_ratio":0.24190894,"punctuation_ratio":0.0872093,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99980074,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-17T07:26:42Z\",\"WARC-Record-ID\":\"<urn:uuid:5857f0ee-d32d-429d-b6d4-fffda0ce9459>\",\"Content-Length\":\"41377\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3cef1d9b-cc3f-4db7-9414-617549c7d0de>\",\"WARC-Concurrent-To\":\"<urn:uuid:4ca10cde-1e1a-436e-ac0e-e91eb884fd7d>\",\"WARC-IP-Address\":\"54.39.48.233\",\"WARC-Target-URI\":\"https://studyforce.com/verify-trigonometric-identities/\",\"WARC-Payload-Digest\":\"sha1:PZHIWN7ETVCMJVKCUWVAF6C6WXXHLS2F\",\"WARC-Block-Digest\":\"sha1:ZYVWGIEURK7M66PLJ2HXQILSC2NPXIP5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780055601.25_warc_CC-MAIN-20210917055515-20210917085515-00208.warc.gz\"}"}
https://oeis.org/A156921	[ "The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.", null, "Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)\n A156921 FP1 polynomials related to the generating functions of the right hand columns of the A156920 triangle. 14\n 1, 1, 1, 1, -6, 1, 7, -79, 119, 126, -270, 1, 28, -515, 1654, 8689, -65864, 142371, -82242, -99090, 113400, 1, 86, -2255, 5784, 300930, -3904584, 20663714, -41517272, -80232259, 657717054 (list; graph; refs; listen; history; text; internal format)\n OFFSET 0,5 COMMENTS The FP1 polynomials appear in the numerators of the GF1 o.g.f.s. of the right hand columns of A156920. The FP1 can be calculated with the formula for the RHC sequence, see A156920, and the formula for the general structure of the generating function GF1, see below. An appropriate name for the FP1 polynomials seems to be the flower polynomials of the first kind because the zero patterns of these polynomials look like flowers. The zero patterns of the FP2, see A156925, and the FP1 resemble each other closely. A Maple program that generates for a right hand column with a certain RHCnr its GF1 and FP1 can be found below. RHCnr stands for right hand column number and starts from 1. LINKS FORMULA G.f.: GF1(z;RHCnr) := FP1(z;RHCnr)/product((1-(2m-1)z)^(RHCnr+1-m),m=1..RHCnr) Row sums (n) = (-1)^(1+(n+1)(n+2)/2)A098695(n). EXAMPLE The first few rows of the \"triangle\" of the coefficients of the FP1 polynomials. In the columns the coefficients of the powers of z^m, m=0,1,2,... , appear. [1, 1, -6] [1, 7, -79, 119, 126, -270] [1, 28, -515, 1654, 8689, -65864, 142371, -82242, -99090, 113400] Matrix of the coefficients of the FP1 polynomials. The coefficients in the columns of this matrix are the powers of z^m, m=0,1,2,.. . [1, 0 ,0, 0, 0, 0, 0, 0, 0, 0] [1, 0 ,0, 0, 0, 0, 0, 0, 0, 0] [1, 1, -6, 0 ,0, 0, 0, 0, 0, 0] [1, 7, -79, 119, 126, -270, 0, 0, 0, 0] [1, 28, -515, 1654, 8689, -65864, 142371, -82242, -99090, 113400] The first few FP1 polynomials are: FP1(z; RHCnr=1) = 1 FP1(z; RHCnr=2) = 1 FP1(z; RHCnr =3) = 1+z-6z^2 Some GF1(z;RHCnr) are: GF1(z;RHCnr= 3) = (1+z-6z^2)/((1-5z)(1-3z)^2(1-z)^3) GF1(z;RHCnr= 4) = (1+7z-79z^2+119z^3+126z^4-270z^5)/((1-7z)(1-5z)^2(1-3z)^3(1-z)^4) MAPLE RHCnr:=4: if RHCnr=1 then RHCmax :=1; else RHCmax:=(RHCnr-1)(RHCnr)/2 end if: RHCend:=RHCnr+RHCmax: for k from RHCnr to RHCend do for n from 0 to k do S2[k, n]:=sum((-1)^(n+i)binomial(n, i)i^k/n!, i=0..n) end do: G(k, x):= sum(S2[k, p]((2p)!/p!) x^p/(1-4x)^(p+1), p=0..k)/(((-1)^(k+1)2x)/(-1+4x)^(k+1)): fx:=simplify(G(k, x)): nmax:=degree(fx); RHC[k-RHCnr+1]:= coeff(fx, x, k-RHCnr)/2^(k-RHCnr) end do: a:=n-> RHC[n]: seq(a(n), n=1..RHCend-RHCnr+1); for nx from 0 to RHCmax do num:=sort(sum(A[t]z^t, t=0..RHCmax)); nom:=Product((1-(2u-1)z)^(RHCnr-u+1), u=1..RHCnr): RHCa:= series(num/nom, z, nx+1); y:=coeff(RHCa, z, nx)-A[nx]; x:=RHC[nx+1]; A[nx]:=x-y; end do: FP1[RHCnr]:=sort(num, z, ascending); GenFun[RHCnr] :=FP1[RHCnr]/product((1-(2m-1)z)^(RHCnr-m+1), m=1..RHCnr); CROSSREFS Cf. A156920, A156925, A156927, A156933. For the first few GF1's see A000340, A156922, A156923, A156924. The number of FP1 terms follow the triangular numbers A000217, with quite surprisingly one exception here a(0)=1. Abs(Row sums (n)) = A098695(n). For the polynomials in the denominators of the GF1(z;RHCnr) see A157702. Sequence in context: A204205 A143019 A337369 * A094214 A001622 A186099 Adjacent sequences: A156918 A156919 A156920 * A156922 A156923 A156924 KEYWORD easy,sign,tabf,uned,changed AUTHOR Johannes W. Meijer, Feb 20 2009 STATUS approved\n\nLookup \| Welcome \| Wiki \| Register \| Music \| Plot 2 \| Demos \| Index \| Browse \| More \| WebCam\nContribute new seq. or comment \| Format \| Style Sheet \| Transforms \| Superseeker \| Recent\nThe OEIS Community \| Maintained by The OEIS Foundation Inc.\n\nLast modified April 20 23:46 EDT 2021. Contains 343143 sequences. (Running on oeis4.)" ]	[ null, "https://oeis.org/banner2021.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5658153,"math_prob":0.9971606,"size":3581,"snap":"2021-04-2021-17","text_gpt3_token_len":1535,"char_repetition_ratio":0.13950238,"word_repetition_ratio":0.08941606,"special_character_ratio":0.5071209,"punctuation_ratio":0.25726143,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99742997,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-21T05:25:04Z\",\"WARC-Record-ID\":\"<urn:uuid:4bcb19ec-7507-4a35-a117-7dd50390df31>\",\"Content-Length\":\"23253\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a17f6486-e995-4086-8eb9-3815fda616fe>\",\"WARC-Concurrent-To\":\"<urn:uuid:c1bf9c45-077c-400f-a6d5-2a512faeef41>\",\"WARC-IP-Address\":\"104.239.138.29\",\"WARC-Target-URI\":\"https://oeis.org/A156921\",\"WARC-Payload-Digest\":\"sha1:7PRSKYARIRG6U4NCL4G2QFCQ4NAMWAWE\",\"WARC-Block-Digest\":\"sha1:W3QRODYOUSVC72IKMJJKPUM6GIDU6UM4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039508673.81_warc_CC-MAIN-20210421035139-20210421065139-00286.warc.gz\"}"}
https://space.stackexchange.com/questions/2880/are-there-any-natural-circular-orbits	[ "# Are there any natural circular orbits?\n\nI just saw How does orbital eccentricity affect positions of Lagrange points $L_4$ and $L_5$? and it questions the difference between circular and elliptical orbits.\n\nWe know the Moon does not have a circular orbit. Do we know of any natural bodies that are actually in a circular orbit? Is there any possible way for a circular orbit to occur naturally?\n\n• How do you define circular? Perfectly 100% circle according to x squared plus y squared equals value? Then I think that the answer will be \"No\". How much variation will you accept? – Hennes Nov 23 '13 at 15:22\n• – Everyone Nov 23 '13 at 17:27\n• Having conveyed that, I wonder whether A*.SE is possibly a better fit for the question – Everyone Nov 23 '13 at 17:28\n\nIf my understanding in correct, the orbit is perfectly circular if the dimensionless orbital parameter of eccentricity (e) is zero. However, I'm not sure how the inclination works with this, and that comes down to the question \"circular in what plane?\" Anyway, I present to you:\n\nAsteroid 113474 (2002 ST57).\n\nhttp://ssd.jpl.nasa.gov/sbdb.cgi?sstr=113474\n\nElement Value Uncertainty (1-sigma) Units\ne 9.338379075815981E-5 1.3966e-07", null, "Sadly, however, the uncertainty is sufficiently lower than the value itself to show that it does, in fact, have some eccentricity. If you are satisfied with any object that could theoretically be circular with our current understanding then that's easy. There are lots of objects with an uncertainty enough such that a perfectly circular orbit is possible. Actually, this turns up a huge number of candidates due to the simple fact that we don't know the the orbital parameters in the first place!\n\nI argue that this above orbit is pretty darn circular. If you want a theoretically perfect circle, then the answer is probably \"no\". If you want further navel-gazing, I will question whether the concept of a circle is valid in non-Euclidean geometry. Since we have the spacetime distortion of Jupiter thrown in there, perhaps circles don't exist?\n\nIn case it wasn't obvious, my example was possible because I used a large sample set and just searched for a candidate that fit my criteria\n\nNonetheless, there is a very good reason that things are circular-ish. Our solar system's plane exists for very good physical reasons, which is that a collapsing gas cloud has an angular momentum value that it can't get rid of, and can't be held in a single body by gravity. Then, the planets are relatively circular, because if they were not, the solar system would not be as stable. Orbital resonances (which is what the long term dynamics depend upon) apply to relatively circular orbits, and like to keep things relatively circular. There are certain to be some exoplanet systems that more strongly favored circularizing the orbits. Because of that, I'm sure that one day we'll discover some exquisitely circular orbit. But that still doesn't make it a literal circle by the theoretical standard.\n\n• A little late, but I just read Deimos also has a close to circular orbit with an eccentricity of 0.00033 – Everyone Oct 10 '14 at 16:24" ]	[ null, "https://i.stack.imgur.com/aEScE.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.94857365,"math_prob":0.8959346,"size":2214,"snap":"2019-51-2020-05","text_gpt3_token_len":484,"char_repetition_ratio":0.10995475,"word_repetition_ratio":0.0,"special_character_ratio":0.21725383,"punctuation_ratio":0.11374407,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9520596,"pos_list":[0,1,2],"im_url_duplicate_count":[null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-24T23:23:16Z\",\"WARC-Record-ID\":\"<urn:uuid:a8323916-d548-4f00-9292-1a8a0660b458>\",\"Content-Length\":\"143680\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a34f4135-0e9d-4b45-8720-c2f54a7ae57d>\",\"WARC-Concurrent-To\":\"<urn:uuid:90f39935-b0b5-46c4-b68e-9858e1b6407e>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://space.stackexchange.com/questions/2880/are-there-any-natural-circular-orbits\",\"WARC-Payload-Digest\":\"sha1:CYLEARUBVVMULPEWWFGQ5XRHXOZEGEBP\",\"WARC-Block-Digest\":\"sha1:HTSZIOHD52CABYBHIF6AT2KY2WELVUZ4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250626449.79_warc_CC-MAIN-20200124221147-20200125010147-00010.warc.gz\"}"}
https://yetanothermathblog.com/tag/quartic-graph/	[ "# Quartic graphs with 12 vertices\n\nThis is a continuation of the post A table of small quartic graphs. As with that post, it’s modeled on the handy wikipedia page Table of simple cubic graphs.\n\nAccording to SageMath computations, there are 1544 connected, 4-regular graphs. Exactly 2 of these are symmetric (ie, arc transitive), also vertex-transitive and edge-transitive. Exactly 8 of these are vertex-transitive but not edge-transitive. None are distance regular.\n\nExample 1: The first example of such a symmetric graph is the circulant graph with parameters (12, [1,5]), depicted below. It is bipartite, has girth 4, and its automorphism group has order 768, being generated by", null, "$(9,11), (5,6), (4,8), (2,10), (1,2)(5,9)(6,11)(7,10), (1,7), (0,1)(2,5)(3,7)(4,9)(6,10)(8,11)$.\n\nExample 2: The second example of such a symmetric graph is the cuboctahedral graph, depicted below. It has girth 3, chromatic number 3, and its automorphism group has order 48, being generated by", null, "$(1,10)(2,7)(3,6)(4,8)(9,11), (1,11)(3,4)(6,8)(9,10), (0,1,9)(2,8,10)(3,7,11)(4,5,6)$." ]	[ null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93896013,"math_prob":0.99848264,"size":884,"snap":"2023-14-2023-23","text_gpt3_token_len":212,"char_repetition_ratio":0.125,"word_repetition_ratio":0.08510638,"special_character_ratio":0.23190045,"punctuation_ratio":0.15730338,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9940644,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-01T16:24:33Z\",\"WARC-Record-ID\":\"<urn:uuid:b26bb631-6789-44a5-84e7-cb1c9f921329>\",\"Content-Length\":\"75208\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:55233fd0-766c-4ad8-b6da-96a998e5ec3a>\",\"WARC-Concurrent-To\":\"<urn:uuid:d6e36054-e90e-41f8-aba9-c44154ecd277>\",\"WARC-IP-Address\":\"192.0.78.24\",\"WARC-Target-URI\":\"https://yetanothermathblog.com/tag/quartic-graph/\",\"WARC-Payload-Digest\":\"sha1:A6Y5TQINENCLQVI3FL4NIGX3BPQBCSTE\",\"WARC-Block-Digest\":\"sha1:NWLEPTDRD2STT5MATYMIIPMMQLC7PVWJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224647895.20_warc_CC-MAIN-20230601143134-20230601173134-00657.warc.gz\"}"}
https://www.faceprep.in/print-the-armstrong-numbers-between-the-two-intervals/	[ "# Print the Armstrong numbers between the two intervals \| faceprep\n\nProgram to print all the Armstrong numbers between the two intervals is discussed here. A number is said to be an Armstrong number when the sum of nth power of digit of the number is equal to the number itself.", null, "## Algorithm to print Armstrong numbers between two intervals\n\n1. Input the start and end values.\n2. Repeat from i = start_value to end_value.\n3. Repeat until (temp != 0)\n4. remainder = temp % 10\n5. result = resut + pow(remainder,n)\n6. temp = temp/10\n7. if (result == number)\n8. Print the number\n9. Repeat steps from 2 to 8 until the end_value is encountered.\n\n## Program to print the Armstrong numbers between the two intervals\n\n// C program to print the Armstrong numbers between the two intervals\n\n#include <stdio.h>\n#include <math.h>\n\nint main()\n{\nint start, end, i, temp1, temp2, remainder, n = 0, result = 0;\n\nprintf(“Enter start value and end value : “);\nscanf(“%d %d”, &start, &end);\nprintf(“\\nArmstrong numbers between %d an %d are: “, start, end);\n\nfor(i = start + 1; i < end; ++i)\n{\ntemp2 = i;\ntemp1 = i;\n\nwhile (temp1 != 0)\n{\ntemp1 /= 10;\n++n;\n}\n\nwhile (temp2 != 0)\n{\nremainder = temp2 % 10;\nresult += pow(remainder, n);\ntemp2 /= 10;\n}\n\nif (result == i) {\nprintf(“%d “, i);\n}\n\nn = 0;\nresult = 0;\n\n}\nprintf(“\\n”);\nreturn 0;\n}\n\n// C++ program to print the Armstrong numbers between the two intervals\n\n#include <iostream>\n#include <math.h>\nusing namespace std;\n\nint main()\n{\nint start, end, i, temp1, temp2, remainder, n = 0, result = 0;\n\ncout << “Enter start value and end value : “;\ncin >> start >> end;\ncout << “\\nArmstrong numbers between ” << start << ” and ” << end << ” are : “;\n\nfor(i = start + 1; i < end; ++i)\n{\ntemp2 = i;\ntemp1 = i;\n\nwhile (temp1 != 0)\n{\ntemp1 /= 10;\n++n;\n}\n\nwhile (temp2 != 0)\n{\nremainder = temp2 % 10;\nresult += pow(remainder, n);\ntemp2 /= 10;\n}\n\nif (result == i) {\ncout << i << ” “;\n}\n\nn = 0;\nresult = 0;\n\n}\ncout << endl;\nreturn 0;\n}\n\n//Java program to print the Armstrong numbers between the two intervals\n\nimport java.util.*;\npublic class sum_of_primes {\n\npublic static void main(String[] args)\n{\nScanner sc = new Scanner(System.in);\nSystem.out.print(“Enter start and end values: “);\nint start = sc.nextInt();\nint end = sc.nextInt();\nint i, temp1, temp2, remainder, n = 0, result = 0;\n\nfor(i = start + 1; i < end; ++i)\n{\ntemp2 = i;\ntemp1 = i;\n\nwhile (temp1 != 0)\n{\ntemp1 /= 10;\n++n;\n}\n\nwhile (temp2 != 0)\n{\nremainder = temp2 % 10;\nresult += Math.pow(remainder, n);\ntemp2 /= 10;\n}\n\nif (result == i) {\nSystem.out.print(i + ” “);\n}\n\nn = 0;\nresult = 0;\n\n}\n\n}\n\n}\n\n# Python program to print the Armstrong numbers between the two intervals\n\nimport math\nstart = int(input(“Enter start value : “))\nend = int(input(“Enter end value : “))\nresult = 0\nn = 0\nfor i in range(start+1 ,end,1):\ntemp2 = i\ntemp1 = i\n\nwhile (temp1 != 0):\ntemp1 = int(temp1/10)\nn = n + 1\n\nwhile (temp2 != 0):\nremainder = temp2 % 10\nresult = result + math.pow(remainder, n)\ntemp2 = int(temp2/10)\n\nif (result == i):\nprint(i,” “)\n\nn = 0\nresult = 0\n\nOutput:", null, "Time complexity: O(n)" ]	[ null, "https://i1.faceprep.in/Companies-1/given-number-is-armstrong-or-not.png", null, "https://i1.faceprep.in/Companies-1/armstrong numbers between the two intervals.PNG", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.624161,"math_prob":0.9997028,"size":2934,"snap":"2020-10-2020-16","text_gpt3_token_len":926,"char_repetition_ratio":0.16177474,"word_repetition_ratio":0.35888502,"special_character_ratio":0.38718474,"punctuation_ratio":0.2158516,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997712,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-27T11:53:06Z\",\"WARC-Record-ID\":\"<urn:uuid:6ee34ba2-19ba-4853-90da-ca0b0358fd31>\",\"Content-Length\":\"82015\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b3d60490-69a3-4990-bd72-c2feec16ae16>\",\"WARC-Concurrent-To\":\"<urn:uuid:c59a2f32-af3c-4453-be06-4bb3c1e9ff86>\",\"WARC-IP-Address\":\"52.66.100.35\",\"WARC-Target-URI\":\"https://www.faceprep.in/print-the-armstrong-numbers-between-the-two-intervals/\",\"WARC-Payload-Digest\":\"sha1:FODMONGG3B3CHYM74SIYBQVH4VF7J424\",\"WARC-Block-Digest\":\"sha1:CPLJMDR3L6XSUGDWBE3HNASG72Q5MHYI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146681.47_warc_CC-MAIN-20200227094720-20200227124720-00537.warc.gz\"}"}
https://lc101-advent.herokuapp.com/challenges/3	[ "# Advent of Code, Day 3: wishlist snooping\n\nSome kids are very secretive, and encrypt their wishlists. Luckily, they don't know much about cryptography and only ever use caesar ciphers.\n\nCaesar ciphers are susceptible to frequency analysis attacks. In English, the most common letters are `e`, then `t`, then `a`. If a letter to Santa is encrypted using the `m` as the key, it's pretty likely that the most common letters will be `q`, then `f`, then `m`.\n\nWe have a plaintext frequency table. (You can get this off of wikipedia, or just use your `frequency_table` on a book from Project Gutenberg. It looks something like this:\n\n```Letter: A B C D E F ... Z\nFrequency: .08167 .01492 .02782 .04253 .12702 .02228 .00074\n```\n\nTo figure out the encryption key, we will guess each one in turn, and compute the frequency table of the message under this decryption key. That looks something like this:\n\n```Letter: A B C D E F ... Z\nFrequency: .02228 .02015 .06094 .06966 .00153 .00772 .12702\n```\n\nWe need to compare each of these 26 tables to the normal English-language table to figure out which key gives us output that looks most like English. There are a number of good ways to do that, but here's one:\n\n```def distance(freq_table1, freq_table2):\n\"\"\"\nCompute the sum-of-squares distance between\ntwo letter-frequency distributions\n\"\"\"\n# dist= 0\n# For every letter a..z:\n# dist += ((freq of letter in freq_table1) - (freq of letter in freq_table2))**2\n# return dist\n```\n\nhere are some example wishlists to try out:\n\n• ```Jkgx Ygtzg,\nGrr O xkgrre xkgrre cgtz lux inxoyzsgy oy g zaxzrk. O'bk grcgey cgtzkj g zaxzrk hkigayk zaxzrky gxk znk iuurkyz.\n\nZngtq eua,\nRubk,\nKsore\n```\n(This one is pretty easy to do by hand because I left in the punctuation, capitalization and spaces)\n• `xyulmuhnucfceysioxisiofceygycqiofxfceyuxmgulcijfyumyhyrnsyulcqcffacpysiogihysbiqxiymnbunmiohxjlynnswiifcnxiymnigynbunmnbyxyuf`\n• `uvrijrekrpfligifsrscpaljkjfdvgvijfeyzivukfivrukyzjzwefkkyvezjkreutfiivtkvuznzccjkzcckvccpflznrekrgcrpjkrkzfezdknvcmvpvrijslkzdnizkzexkyzjsvtrljvzsvczvmvkyvivzjtyizjkdrjdrxztjfdvnyviv`\n\nWhen you've got a solution, submit the plaintext of the following letter: `ghduvdqwdlzrxogolnhdwrbsxsdwrbnlwwhqdwrbexqqbdqgdwrbkruvhdqgdovrdvpdoosodqhwkdwlfrxogulghlqlzrxogolnhdfrorulqjerrnwrr`" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7064741,"math_prob":0.7059282,"size":2216,"snap":"2020-45-2020-50","text_gpt3_token_len":732,"char_repetition_ratio":0.09358047,"word_repetition_ratio":0.06329114,"special_character_ratio":0.22698556,"punctuation_ratio":0.17021276,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96657526,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-30T16:55:03Z\",\"WARC-Record-ID\":\"<urn:uuid:a0698d25-1bbf-4f27-825b-23622f7e9683>\",\"Content-Length\":\"4367\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e12736ee-0fe7-4d82-9d68-d1ff8110929b>\",\"WARC-Concurrent-To\":\"<urn:uuid:8eb18205-d0ad-44f3-bbd0-ee9e8db6b073>\",\"WARC-IP-Address\":\"52.73.243.231\",\"WARC-Target-URI\":\"https://lc101-advent.herokuapp.com/challenges/3\",\"WARC-Payload-Digest\":\"sha1:FBFXAG2WIANHCEFA2BQEG5LOYIQIAJEY\",\"WARC-Block-Digest\":\"sha1:MALARW6UZDH5OMXIASIP6RAJMSAAPHJZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107911027.72_warc_CC-MAIN-20201030153002-20201030183002-00599.warc.gz\"}"}
https://listserv.uni-heidelberg.de/cgi-bin/wa?A2=9710&L=LATEX-L&D=0&H=N&P=1698098	[ "", null, "## LATEX-L Archives\n\n##### Mailing list for the LaTeX3 project\n Options: Use Forum View Use Monospaced Font Show Text Part by Default Show All Mail Headers Message: [<< First] [< Prev] [Next >] [Last >>] Topic: [<< First] [< Prev] [Next >] [Last >>] Author: [<< First] [< Prev] [Next >] [Last >>]\n\n Subject: Re: LPPL under review at savannah.gnu.org From: \"C.M. Connelly\" <[log in to unmask]> Reply To: Mailing list for the LaTeX3 project <[log in to unmask]> Date: Tue, 9 Jul 2002 13:40:25 -0700 Content-Type: text/plain Parts/Attachments: text/plain (48 lines)\n```\"MS\" == Martin Schroeder <[log in to unmask]>\n\nLD> I think the LPPL is trying to define and enforce a\nLD> distribution policy within the license. This is a strange\nLD> idea. Imagine what mess it would be if the Linux kernel\nLD> imposed the same restrictions on system calls ?-) Instead\nLD> a specification was issued\nLD> (http://www.opengroup.org/onlinepubs/007908799/) to\nLD> encourage the necessary standardization and\nLD> uniformity. Defining a standard interface and behaviour is\nLD> a complex matter that can hardly be implemented by a\n\nMS> \"The Single UNIX® Specification, Version 2\" -- which I\nMS> find irrelevant here.\n\nYes, the specification is irrelevant, but Loic's point was that a\nlicense cannot force standardization; that job has to be left to a\ngroup of interested parties who draft a standards document that\ndefine what bits make up a complete system, how they interact,\ntheir interface, what sort of output they produce, and so on.\n\nIn other words, he wasn't suggesting you look at the document\nbecause it would tell you anything about LaTeX, he was suggesting\nthat the process that produced the document is worth looking at.\n(And perhaps that the document itself might serve as a model for a\nLaTeX standard.)\n\nThe basic idea here is that if you want to keep LaTeX pure, you\ncan take one of two approaches:\n\n1. Impose a restrictive, non-free license that prevents\nmodification of key components\n\n2. Develop an open standard that defines the behavior of the\nsystem in a testable way\n\nClaire\n\n+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+\nMan cannot be civilised, or be kept civilised by what he does in his\nspare time; only by what he does as his work.\nW.R. Lethaby\n+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+" ]	[ null, "https://listserv.uni-heidelberg.de/archives/images/default_archive_home_64x64.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9065225,"math_prob":0.8086681,"size":1953,"snap":"2022-27-2022-33","text_gpt3_token_len":539,"char_repetition_ratio":0.19548486,"word_repetition_ratio":0.0,"special_character_ratio":0.31797236,"punctuation_ratio":0.104477614,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98460597,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-03T21:10:01Z\",\"WARC-Record-ID\":\"<urn:uuid:b131c2ff-2162-4cb3-9d55-f53c5d643290>\",\"Content-Length\":\"78700\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:594d7ee4-f266-49b4-843b-e32d50db7e3a>\",\"WARC-Concurrent-To\":\"<urn:uuid:3181de65-11a4-4981-9126-d64b36832e16>\",\"WARC-IP-Address\":\"129.206.100.94\",\"WARC-Target-URI\":\"https://listserv.uni-heidelberg.de/cgi-bin/wa?A2=9710&L=LATEX-L&D=0&H=N&P=1698098\",\"WARC-Payload-Digest\":\"sha1:MZPK66KN4IQFC3WCVLGVHVYBT5Q56KXF\",\"WARC-Block-Digest\":\"sha1:X35BW7D7JC5R5KLPABFR4XJS4W5SZQ6C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104249664.70_warc_CC-MAIN-20220703195118-20220703225118-00024.warc.gz\"}"}
http://www.dailymathguide.com/2020/05/unlike-terms.html	[ "UNLIKE TERMS - Daily Math Guide\n\n## DMG\n\nHome » , » UNLIKE TERMS\n\n## THE UNLIKE TERMS\n\nUnlike term is the opposite of like terms. The unlike term(s) is an expression composed of numerical or literal coefficients. It is said that when we identify Algebraic terms, it should be a variable or a constant separated by the \"+\" or \"-\" sign. Either of them. In this article, we are looking for literal coefficients, or variables, which are not the same nor similar. By definition, it is a term or term in a series of Algebraic expressions that have no literal coefficients the same, or they are the dissimilar literal coefficients in Algebraic expression. It is also a term that has not the same literal coefficients including exponents. It is said to be that the unlike terms cannot be added nor subtracted because the said operation only happened in the Algebraic expressions with like terms. Like and unlike terms are totally opposite.\n\n### Sample Step Guide on How to Do It\n\nIllustration:\n\n, look at the terms of the given expression. They are obviously dissimilar. The first term is composed of 2xy and only y for the second. In this series of expressions, we cannot perform its indicated operation, hence there's no term similar. It will remain as is to its position.\n- this is obviously an, unlike terms. This expression is three terms with no term similar The literal coefficients are xy and -3y thus, xy and -3y are unlike terms or dissimilar terms. The \"4\" is also part of the series of expression but it's a constant.\n-this is another example of expression which of, unlike terms. Notice that \"x\" is present to the first two terms but they don't have similar powers or exponents. So this means that the first term has \"three- x's\" and the second term has \"two-x's\". Hence, they are not the same.\n\n-this is an expression with \"like terms\" or similar terms. The literal coefficients of the first and second terms are x3 and x3. This is so because its powers are the same which is 3. Therefore this is \"not\" an, unlike terms as an example.\n, this expression is also an example of \"like terms\". Their literal coefficient is the same in terms of variables and powers. There are three terms in this expression with one constant which is -7.\n\n### Example Problems with Solutions\n\nIdentify if the given is similar (like) terms or dissimilar (unlike) terms.\n\n-like terms\n\n-unlike terms\n\n-unlike terms\n\n-like terms\n\n-unlike terms\n\n-like terms\n\n-unlike terms\n\n-like terms\n\n-like terms\n\n-like terms\n\n### Problems with Partial Solutions\n\nWrite \"like\" if the given expression has like terms and can be simplified, otherwise \"unlike\". Write your answer by clicking the blank after each series of expressions.\n\n= _________\n\n= _________\n\n= _________\n\n= _________\n\n= _________\n\n= _________\n\n= _________\n\n= _________\n\n### Simple Quizzes\n\nWrite \"like\" if the given expression has like terms, otherwise \"unlike\".\n\n## Click for a PDF worksheet!\n\nRelated References: like terms, unlike and like terms\n\nCheck our Select button to Share at the bottom for more services!\n\nSelect button to Share :" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6188156,"math_prob":0.885204,"size":8577,"snap":"2023-40-2023-50","text_gpt3_token_len":2271,"char_repetition_ratio":0.28741398,"word_repetition_ratio":0.5533981,"special_character_ratio":0.2583654,"punctuation_ratio":0.06426932,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97584903,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-06T01:24:24Z\",\"WARC-Record-ID\":\"<urn:uuid:95e7686b-b115-418c-b620-57d43f1d5405>\",\"Content-Length\":\"130849\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3bbb469e-1f2f-4715-9e79-f75e360b3726>\",\"WARC-Concurrent-To\":\"<urn:uuid:1d1506ec-dc55-41e5-9ccb-49b271b9329c>\",\"WARC-IP-Address\":\"172.253.63.121\",\"WARC-Target-URI\":\"http://www.dailymathguide.com/2020/05/unlike-terms.html\",\"WARC-Payload-Digest\":\"sha1:ANI22MDXLR7PGSLRGCN3HBFUXSMNXQEN\",\"WARC-Block-Digest\":\"sha1:HEYGCH55CB6KMMPGE5DUI7NIZBKKGTT3\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100575.30_warc_CC-MAIN-20231206000253-20231206030253-00703.warc.gz\"}"}
http://cybergis.cigi.uiuc.edu/cyberGISwiki/doku.php/ct?do=diff&rev2%5B0%5D=1442431771&rev2%5B1%5D=&difftype=sidebyside	[ "", null, "", null, "# Differences\n\nThis shows you the differences between two versions of the page.\n\nLink to this comparison view\n\nct [2015/09/16 14:29]\nyanliu\nct [2015/09/16 14:36] (current)\nyanliu\nLine 7: Line 7:\n* [[ct/pgap\|PGAP]] * [[ct/pgap\|PGAP]]\n* [[ct/pabm\|Parallel Agent-Based Modeling]] * [[ct/pabm\|Parallel Agent-Based Modeling]]\n- * [[ct/sptw\|Simple Parallel Tiff Writer]]\n* [[ct/gkde\|Parallel Kernel Density Estimation]] * [[ct/gkde\|Parallel Kernel Density Estimation]]\n* [[ct/mapalgebra\|Parallel Map Algebra]] * [[ct/mapalgebra\|Parallel Map Algebra]]\n+ * [[ct/sptw\|Simple Parallel Tiff Writer]]\n\n====Introduciton==== ====Introduciton====\nLine 24: Line 24:\n- [[ct/ppysal\|Parallel PySAL]]. The Parallel PySAL library provides a set of scalable [[http://pysal.org\|PySAL]] functions. Currently, parallel PySAL components implemented using the multiprocessing python library. The Fisher-Jenks classification algorithm is integrated in the toolkit; - [[ct/ppysal\|Parallel PySAL]]. The Parallel PySAL library provides a set of scalable [[http://pysal.org\|PySAL]] functions. Currently, parallel PySAL components implemented using the multiprocessing python library. The Fisher-Jenks classification algorithm is integrated in the toolkit;\n- [[ct/prasterblaster\|pRasterBlaster]]. pRasterBlaster is a high-performance map reprojection software contributed by the Center of Excellence for Geospatial Information Science ([[http://cegis.usgs.gov\|CEGIS]]) within the [[http://usgs.gov\|U.S. Geological Survey]]; - [[ct/prasterblaster\|pRasterBlaster]]. pRasterBlaster is a high-performance map reprojection software contributed by the Center of Excellence for Geospatial Information Science ([[http://cegis.usgs.gov\|CEGIS]]) within the [[http://usgs.gov\|U.S. Geological Survey]];\n- - [[ct/pgap\|PGAP]]. PGAP is a scalable Parallel Genetical Algorithm (PGA) solver for the Generalized Assignment Problem (GAP). This code provides an efficient PGA implementation for combinatorial optimization problem-solving and scaled up to 262K processor cores on BlueWaters with marginal communcation cost.+ - [[ct/pgap\|PGAP]]. PGAP is a scalable Parallel Genetical Algorithm (PGA) solver for the Generalized Assignment Problem (GAP). This code provides an efficient PGA implementation for combinatorial optimization problem-solving and scaled up to 262K processor cores on BlueWaters with marginal communication cost;\n- [[ct/pabm\|Parallel Agent-Based Modeling (PABM)]]. PABM is an illustrative software for scalable spatially explicit agent-based modeling (ABM); - [[ct/pabm\|Parallel Agent-Based Modeling (PABM)]]. PABM is an illustrative software for scalable spatially explicit agent-based modeling (ABM);\n+ - [[ct/gkde\|Parallel Kernel Density Estimation]] A multi-GPU code for data-intensive kernel density estimation (KDE). Spatial computational domain is first estimated for KDE. Then the study area is decompose into sub-regions through an adaptive partitioning approach. Each sub-region is processed by a GPU node. These GPU nodes are communicated through massage passing interface (MPI);\n+ - [[ct/mapalgebra\|Parallel Map Algebra]] This package contains a parallel map algebra code using CUDA, MPI, and Parallel I/O. It is extracted from a parallel geospatial programming models training package;\n+ - [[ct/sptw\|Simple Parallel Tiff Writer]] This is a parallel IO code for writing GeoTIFF file on a parallel file system.\n\nLine 46: Line 49:", null, "" ]	[ null, "http://cybergis.cigi.uiuc.edu/cyberGISwiki/lib/tpl/cybergis/images/dark_logo-version3.png", null, "http://cybergis.cigi.uiuc.edu/cyberGISwiki/lib/tpl/cybergis/images/text.png", null, "http://cybergis.cigi.uiuc.edu/cyberGISwiki/lib/exe/indexer.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6980654,"math_prob":0.55189836,"size":4120,"snap":"2019-13-2019-22","text_gpt3_token_len":1060,"char_repetition_ratio":0.12827988,"word_repetition_ratio":0.5083682,"special_character_ratio":0.24538834,"punctuation_ratio":0.1303681,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96298367,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,6,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-25T20:44:54Z\",\"WARC-Record-ID\":\"<urn:uuid:68729725-1d54-402d-9a53-11bda38f5ac9>\",\"Content-Length\":\"21203\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aa5f499d-28b3-47ca-b79d-6af96508d6e2>\",\"WARC-Concurrent-To\":\"<urn:uuid:23b9a9bc-58ed-4ceb-8874-feaaa59087f3>\",\"WARC-IP-Address\":\"192.17.12.194\",\"WARC-Target-URI\":\"http://cybergis.cigi.uiuc.edu/cyberGISwiki/doku.php/ct?do=diff&rev2%5B0%5D=1442431771&rev2%5B1%5D=&difftype=sidebyside\",\"WARC-Payload-Digest\":\"sha1:AKL25QRKSSKMN7QFPGCGNZMHYWWCYSSX\",\"WARC-Block-Digest\":\"sha1:I4P2VXE6KDARIVFHHORL7527H22MAH3Z\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912204300.90_warc_CC-MAIN-20190325194225-20190325220225-00404.warc.gz\"}"}
https://www.hindawi.com/journals/isrn/2011/312789/	[ "Research Article \| Open Access\n\nVolume 2011 \|Article ID 312789 \| 11 pages \| https://doi.org/10.5402/2011/312789\n\n# Bivariate Poincaré Series for the Algebra of Covariants of a Binary Form\n\nAccepted17 May 2011\nPublished29 Jun 2011\n\n#### Abstract\n\nA formula for computation of the bivariate Poincaré series for the algebra of covariants of binary -form is found.\n\n#### 1. Introduction\n\nLet be the complex vector space of binary forms of degree endowed with the natural action of the special linear group . Consider the corresponding action of the group on the coordinate rings and . Denote by and by the subalgebras of -invariant polynomial functions. In the language of classical invariant theory, the algebras and are called the algebra of invariants and the algebra of covariants for the binary form of degree , respectively. The algebra is a finitely generated bigraded algebra: where each subspace of covariants of degree and order is finite dimensional. The formal power series , is called the bivariate Poincaré series of the algebra of covariants . It is clear that the series is the Poincaré series of the algebra and the series is the Poincaré series of the algebra with respect to the usual grading of the algebras under degree. The finitely generation of the algebra of covariants implies that its bivariate Poincaré series is the power series expansion of a rational function of two variables . We consider here the problem of computing efficiently this rational function.\n\nCalculating the Poincaré series of the algebras of invariants and covariants was an important object of research in invariant theory in the 19th century. For the cases , the series were calculated by Sylvester, see in [1, 2] the big tables of , named them as generating functions for covariants, reduced form. All those calculations are correct up to .\n\nRelatively recently, Springer found an explicit formula for computing the Poincaré series of the algebra of invariants . In the paper we have proved a Cayley-Sylvester-type formula for calculating of and a Springer-type formula for calculation of . By using the formula, the bivariate Poincaré series is calculated for .\n\n#### 2. Cayley-Sylvester-Type Formula for dim(𝒞𝑑)𝑖,𝑗\n\nTo begin, we give a proof of the Cayley-Sylvester-type formula for the dimension of the graded component .\n\nLet and be standard irreducible representation of the Lie algebra . The basis elements ,, of the algebra act on by the derivations : The action of is extended to an action on the symmetrical algebra in the natural way.\n\nLet be the maximal unipotent subalgebra of . The algebra , defined by is called the algebra of semi-invariants of the binary form of degree . For any element , a natural number is called the order of the element if the number is the smallest natural number such that It is clear that any semi-invariant of order is the highest weight vector for an irreducible -module of the dimension in .\n\nThe classical theorem of Roberts implies an isomorphism of the algebra of covariants and the algebra of semi-invariants. Furthermore, the order is preserved through the isomorphism. Thus, it is enough to compute the Poincaré series of the algebra .\n\nThe algebra is -graded and each is a completely reducible representation of the Lie algebra . Thus, the following decomposition holds here is the multiplicity of the representation in the decomposition of . On the other hand, the multiplicity of the representation is equal to the number of linearly independent homogeneous semi-invariants of degree and order for the binary -form. This argument proves the following.\n\nLemma 2.1.\n\nThe set of weights (eigenvalues of the operator ) of a representation denote by , in particular, .\n\nA formal sum is called the character of a representation , here denotes the multiplicity of the weight . Since, the multiplicity of any weight of the irreducible representation is equal to 1, we have\n\nThe character of the representation equals (see ), where is the complete symmetrical function\n\nBy replacing with , , we obtain the specialized expression for : here is the number nonnegative integer solutions of the equation on the assumption that . In particular, the coefficient of (the multiplicity of zero weight ) is equal to , and the coefficient of is equal to .\n\nOn the other hand, the decomposition (*) implies the equality for the characters: We can summarize what we have shown so far in the following.\n\nTheorem 2.2.\n\nProof. The weight appears once in any representation , for . Therefore Similarly, Thus, By using Lemma 2.1 we obtain\n\nFor another proof of the formula see .\n\nNote that the original Cayley-Sylvester formula is Also, in we proved that Here are the components of standard grading of the algebras , under degree.\n\n#### 3. Calculation of dim(𝐶𝑑)𝑖,𝑗\n\nIt is well known that the number of nonnegative integer solutions of the following system is given by the coefficient of of the generating function We will use the notation to denote the coefficient of in the series expansion of . Thus It is clear that\n\nSimilarly, the number of nonnegative integer solutions of the following system\n\nequals\n\nTherefore, Thus, the following statement holds.\n\nTheorem 3.1. The number of linearly independent covariants of degree and order for the binary d- form is given by the formula\nIt is clear that By using the decomposition where is the -binomial coefficient one obtains the well-known formula for instance, see .\n\n#### 4. Explicit Formula for 𝒫𝑑(𝑧,𝑡)\n\nLet us prove Springer-type formula for the bivariate Poincaré series of the algebra covariants of the binary -form. Consider the -algebra of formal power series. For an integer define the -linear function in the following way\n\nThe main idea of the ensuing calculations is that the Poincaré series can be expressed in terms of function . The following simple but important statement holds.\n\nLemma 4.1.\n\nProof. Theorem 2.2 implies that . Then\n\nLet be a -linear function defined by\n\nfor . Note that and , . Also, put for . It is clear that if , .\n\nIn important special cases, calculating the functions can be reduced to calculating the functions . The following statements hold.\n\nLemma 4.2. (i) For holds .\n(ii) For and for holds\n\nProof. (i) The statement follows from the linearity of the function and from the following simple observation: for and .\n(ii) Let . Then for we have\n\nNow we can present Springer-type formula for calculating of the bivariate Poincaré series .\n\nTheorem 4.3. here is -shifted factorial.\n\nProof. Consider the partial fraction decomposition of the rational function : It is easy to see, that Using the above Lemmas we obtain\n\nCorollary 4.4. A denominator of the bivariate Poincaré series , can be written in the form where are the degrees of elements of homogeneous system of parameters for the algebra of invariants .\n\nProof. The formula of Theorem 4.3 implies that the bivariate Poincaré series has the form for some polynomials . Thus, the Poincaré series for the algebra of invariants has the form The algebra of invariants is Cohen-Macaulay, and its transcendence degree for equals , see . Therefore it has a homogeneous system of parameters and the denominator of its Poincaré series can be written in the following way , where are the degrees of elements of this homogeneous system of parameters.\n\n#### 5. Examples\n\nFor direct computations we use the following technical lemma.\n\nLemma 5.1. For one has here , and are natural numbers.\n\nProof. Taking into account Lemma 4.2 we get In a similar fashion we prove the general case.\n\nBy using Lemma 5.1 the bivariate Poincaré series for are found. All these results agree with Sylvester's calculations up to , see [1, 2].\n\nBelow is the list of several series:\n\n1. J. J. Sylvester and F. Franklin, “Tables of the generating functions and groundforms for the binary quantics of the first ten orders,” American Journal of Mathematics, vol. 2, no. 3, pp. 223–251, 1879. View at: Publisher Site \| Google Scholar\n2. J. J. Sylvester, “Tables of the generating functions and groundforms of the binary duodecimic, with some general remarks, and tables of the irreducible syzygies of certain quantics,” American Journal of Mathematics, vol. 4, no. 1–4, pp. 41–61, 1881. View at: Publisher Site \| Google Scholar\n3. T. A. Springer, “On the invariant theory of SU(2),” Indagationes Mathematicae, vol. 42, no. 3, pp. 339–345, 1980. View at: Google Scholar \| Zentralblatt MATH\n4. M. Roberts, “The covariants of a binary quantic of the n-th degree,” The Quarterly Journal of Mathematics, vol. 4, pp. 168–178, 1861. View at: Google Scholar\n5. W. Fulton and J. Harris, Representation Theory, vol. 129 of Graduate Texts in Mathematics, Springer, New York, NY, USA, 1991.\n6. L. Bedratyuk, “The Poincaré series for the algebra of covariants of a binary form,” International Journal of Algebra, vol. 4, no. 25, pp. 1201–1207, 2010. 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https://www.arxiv-vanity.com/papers/hep-th/9808014/	[ "Adv. Theor. Math. Phys. 2 (1998) 1249-1286\n\nBranes at conical singularities\n\nand holography\n\nB.S. Acharya, J. M. Figueroa-O’Farrill, C.M. Hull, B. Spence\n\nDepartment of Physics\n\nQueen Mary and Wesfield College\n\nLondon E1 4NS, UK\n\nfootnotetext: e-print archive: http://xxx.lanl.gov/abs/hep-th/9705006\n\nAbstract For supergravity solutions which are the product of an anti-de Sitter space with an Einstein space , we study the relation between the amount of supersymmetry preserved and the geometry of . Depending on the dimension and the amount of supersymmetry, the following geometries for are possible, in addition to the maximally supersymmetric spherical geometry: Einstein-Sasaki in dimension , -Sasaki in dimension , -dimensional manifolds of weak holonomy and -dimensional nearly Kähler manifolds. Many new examples of such manifolds are presented which are not homogeneous and hence have escaped earlier classification efforts. String or theory in these vacua are conjectured to be dual to superconformal field theories. The brane solutions interpolating between these anti-de Sitter near-horizon geometries and the product of Minkowski space with a cone over lead to an interpretation of the dual superconformal field theory as the world-volume theory for branes at a conical singularity (cone branes). We propose a description of those field theories whose associated cones are obtained by (hyper-)Kähler quotients.\n\n## 1 Introduction and Motivation\n\nMaldacena has conjectured that the ’t Hooft large limit of supersymmetric Yang–Mills with gauge group is dual to type IIB superstring theory on , where the subscript indicates that the sizes of the spaces grow with . A more precise version of the conjecture was formulated in [46, 71] where a simple recipe was given for relating gauge theory correlators to string theory S-matrix elements, and these are given in terms of classical supergravity in the large limit.\n\nThis conjecture was motivated by considering parallel -branes for large and taking a limit in which the gauge theory on the brane decouples from the physics of the bulk . When is small (with the string coupling), the system is well-described by -dimensional super Yang-Mills theory with gauge group, while if is large, the system is described by IIB string theory in the near-horizon geometry, which is . These are then dual descriptions of the same system, leading to the conjectured equivalence. String loop corrections correspond to corrections in the gauge theory so that in the large limit, one can use classical supergravity theory in the background.\n\nThis can be generalised to any -brane of superstring theory or theory and this leads to a relation between the worldvolume theory with gauge symmetry and the string or theory in the space-time which arises in the near-horizon limit of the -brane spacetime . Of particular interest are those cases—the -brane and the and -branes—in which the near-horizon geometry is a supersymmetric anti-de Sitter space solution of the form () and the worldvolume theory is a superconformal field theory. In these cases, there is a holographic duality between the string or theory in anti-de Sitter space and the superconformal field theory (which may be thought of as being at the boundary of the anti-de Sitter space [59, 71]). The superconformal symmetry in dimensions is identified with the anti-de Sitter supersymmetry in dimensions.\n\nAn interesting extension of these ideas is to explore -branes which exhibit near-horizon geometries of the form , where is an Einstein manifold. If the geometry is supersymmetric, then the string or theory in that background is expected to be holographically dual to a superconformal field theory. These vacua are not maximally supersymmetric unless is a round sphere or the real projective space, which is the near-horizon geometry of branes on an orientifold , and as a result the dual field theories have less than maximal supersymmetry. Many such cases have been studied already, particularly in the context of type IIB superstring theory, but also in theory . A simple modification is to let be a finite quotient of the sphere [54, 58, 63, 6, 2, 36, 26]. Alternatively, if we recall that is a circle bundle over , one can replace with other Kähler–Einstein -folds. This was done in the type IIB context (i.e., ) in , generalising . Another generalisation is to replace with another homogeneous space . Homogeneous vacua of supergravity theories were studied intensively in the early days of Kaluza–Klein supergravity (see, e.g., ). There is a complete classification in dimension seven and a partial list in dimension five . One such five-dimensional example is , whose dual conformal field theory was recently discussed by Klebanov and Witten (see also ). They interpreted the vacuum in terms of the near-horizon geometry of parallel -branes at a conical singularity of a Calabi–Yau threefold. Similar ideas have been considered in .\n\nThe purpose of the present paper, which subsumes , is to study those theory or superstring vacua of the form that preserve some supersymmetry. Our aim will be to understand the constraints supersymmetry imposes on the geometry of the Einstein manifolds , and then to use the geometry of these manifolds in order to identify the superconformal symmetries which are required under the adS/CFT correspondence. This is certainly likely to be useful in understanding generic features of this correspondence in supersymmetric cases.\n\nAny solution of the form is the near-horizon geometry of a -brane solution which will be supersymmetric if the near-horizon geometry is, and which interpolates between and a vacuum where is a cone over (defined below) and is ()-dimensional Minkowski space. The case in which (with ) is a coset space was considered in , but we will consider general Einstein spaces which in addition preserve some supersymmetries. The cone is Ricci-flat and the number of supersymmetries of the vacuum depends on the number of covariantly constant spinors. This is determined by the holonomy group of , and such holonomies have been classified (see, e.g., ). The number of supersymmetries on depends on the number of Killing spinors on , but these all arise from covariantly constant spinors on , leading to a characterisation of supersymmetric anti-de Sitter solutions.\n\nWe will show that for those vacua which are supersymmetric, the geometry of is highly constrained. Depending on , the possible geometries of are as follows: nearly Kähler for , weak holonomy for , Einstein–Sasaki for , and -Sasaki for ; this gives many supersymmetric compactifications of string and theory which have not been considered previously.\n\nGiven a -brane solution of the above type, the interpolating solution argument then leads to a conjectured duality between the string or theory in a supersymmetric background of the form and a -dimensional superconformal field theory on the worldvolume of coincident -branes located at the conical singularity of the vacuum. The detailed description of the dual theories will be left to future investigations, and in this paper we will focus instead on generic features of the field theories which follow from the common properties of each of the geometries listed above.\n\nThis paper is organised as follows. In Section 2 we set the notation concerning the supersymmetric branes which we will study in this paper and we review the simple solutions with spherical near-horizon geometries. In Section 3 we discuss the solutions which can be interpreted as branes sitting at a conical singularity in a Ricci-flat manifold , and we characterise the supersymmetric solutions in terms of the holonomy of . In Section 4 we discuss their near-horizon geometries in detail. Section 5 contains many examples including those in the early Kaluza–Klein literature as well as some more recent ones which have hereto not been considered in relation with supergravity. As an application of our results, in Section 6 we describe how the near-horizon geometry induces the superconformal symmetry of the dual theory. Finally in Section 7 we offer some conclusions.\n\n## 2 Supersymmetric Branes and their Near-Horizon Geometries\n\nIn this section we review the near-horizon geometries of the elementary brane solutions of string and theory. This section serves mostly to establish the notation.\n\n### 2.1 M-branes\n\nEleven-dimensional supergravity consists of the following fields [62, 19]: a Lorentzian metric , a closed -form with a quantised flux and a gravitino . By a supersymmetric vacuum we will mean any solution of the equations of motion with for which the supersymmetry variation , regarded as a linear equation on the spinor parameter , has nontrivial solutions. The eleven-dimensional spinorial representation is -dimensional and real, so there at most linearly independent solutions. An important physical invariant of a supersymmetric vacuum is the fraction of the supersymmetry that the solution preserves. For example, and the flat metric on eleven-dimensional Minkowski spacetime is a supersymmetric vacuum with ; that is, it is maximally supersymmetric. Other maximally supersymmetric vacua are and with and having quantised flux on the and , respectively.\n\nEleven-dimensional supergravity has four types of elementary solutions with : the -wave and the Kaluza–Klein monopole [43, 68, 52], and the elementary brane solutions: the -brane and the -brane . There should also be an -brane solution [52, 5]. In what follows we will focus on the - and -branes.\n\n#### 2.1.1 The M2-brane\n\nThe following background describes a number of parallel -branes :\n\n g =H−23g2+1+H13g8 (1) F =±dvol2+1∧dH−1 ,\n\nwhere and are the metric and volume form on the three-dimen-sional Minkowskian worldvolume of the branes; is the metric on the eight-dimensional euclidean space transverse to the branes; and is a harmonic function on . For example, if we demand that depend only on the transverse radial coordinate on , we then find that\n\n H(r)=1+a6r6 ;a6≡25π2Nℓ6p (2)\n\nis the only solution with . This corresponds to coincident membranes at . Here is the eleven-dimensional Planck length.\n\nOther choices of are possible. For example one can consider multicentre generalisations, where is an arbitrary harmonic function on with pointlike singularities and suitable asymptotic behaviour , say, as . This corresponds to parallel branes localised at the singularities of . More generally, we can take to be invariant under some subgroup of isometries of . These solutions correspond to branes which are ‘delocalised’ or smeared on the -orbits.\n\nWe can easily determine the fraction of the supersymmetry which the above solution preserves . The supersymmetry variation of the gravitino in a bosonic background is given by\n\n δεΨM=∇Mε−1288(ΓMPQRS−8δMPΓQRS)FPQRSε , (3)\n\nwhere is the spin connection. In the -brane Ansatz given by (1) and (2), equation (3) is solved by\n\n ε=H−16ε∞ ,\n\nwhere, choosing to be the directions tangent to the worldvolume of the brane, obeys\n\n Γ012ε∞=ε∞ .\n\nBecause squares to the identity and is traceless, we see that . Nevertheless, the -brane interpolates between two maximally supersymmetric solutions: flat Minkowski space infinitely far away from the brane, and near the brane horizon [35, 21].\n\nIndeed, notice that the metric on the transverse space is given by\n\n g8=dr2+r2gS , (4)\n\nwhere is the metric on the unit sphere . In the near-horizon region ,\n\n H(r)∼a6r6 ,\n\nso that the membrane metric becomes\n\n g=a−4r4g2+1+a2r−2dr2+a2gS .\n\nThe last term is the metric on a round of radius ; whereas the first two combine to produce the metric on -dimensional anti-de Sitter spacetime with “radius” :\n\nwith .\n\n#### 2.1.2 The M5-brane\n\nThe -brane is the magnetic dual to the -brane. The background describing a number of parallel -branes is given by :\n\n g =H−13g5+1+H23g5 (6) F =±3⋆5dH ,\n\nwhere and are the metric and volume form on the six-dimensional Minkowskian worldvolume of the branes; and are the metric and Hodge operator on the five-dimensional euclidean space transverse to the branes; and is a harmonic function on . For example, if we demand that depend only on the transverse radial coordinate on , we then find that\n\n H(r)=1+a3r3 ;a3≡πNℓ3p (7)\n\nis the only solution with . This corresponds to coincident fivebranes at .\n\nAs for the membrane solution, multicentre and ‘delocalised’ fivebrane solutions also exist.\n\nThe fraction of the supersymmetry which is preserved can be computed as before . The gravitino shift equation in the bosonic background given by above is solved by\n\n ε=H−112ε∞ ,\n\nwhere, choosing to be the directions tangent to the worldvolume of the fivebrane, obeys\n\n Γ012345ε∞=ε∞ .\n\nsquares to the identity and is traceless, so that again . Nevertheless, as for the -brane, the -brane also interpolates between two maximally supersymmetric solutions: flat Minkowski space infinitely far away from the brane, and near the brane horizon [35, 21].\n\nIndeed,\n\n g5=dr2+r2gS , (8)\n\nwhere is now the metric on the unit sphere . In the near-horizon region ,\n\n H(r)∼a3r3 ,\n\nso that the fivebrane metric becomes\n\n g∼a−1rg5+1+a2r−2dr2+a2gS .\n\nThe last term is the metric on a round of radius , whereas the first two combine to produce the metric on -dimensional anti-de Sitter spacetime with “radius” , analogous to (5),\n\nwith now .\n\n### 2.2 String Branes\n\nThe near-horizon geometries of -branes in type II string theory are not of the form\n\nwith the exception of the -brane of type IIB string theory, for which the near-horizon geometry is . For other -branes () the near-horizon geometry is conformal to (9), the conformal factor being nontrivial, and either singular for or zero for as .\n\nGeometries of the form\n\nfor some space also arise, and compactifying on leads to a geometry. For example, arises from a D1-brane lying inside a D5-brane , while is the near-horizon geometry for the extreme Reissner-Nordström black hole.\n\n#### 2.2.1 The D3-brane in Type IIB\n\nThe metric for the -brane of type IIB is given by ,\n\n g=H−12g3+1+H12g6 , (11)\n\nwhere is the metric on the Minkowski worldvolume of the brane and is the euclidean metric on the transverse . The self-dual -form has (quantised) flux on the unit transverse five-sphere and the dilaton is constant. is again harmonic in and the unique spherically symmetric solution with is\n\n H(r)=1+a4r4 ;a4≡4πgNℓ4s\n\nwhere is the string coupling constant, given by the exponential of the constant dilaton, and is the string length. The solution corresponds to parallel -branes at . The ten-dimensional Planck length is , so that can be rewritten as\n\n a4=4πNℓ4p . (12)\n\nThe near-horizon analysis is similar to that for the and branes above, and the near-horizon geometry is\n\nwhere the anti-de Sitter and sphere radii are now equal, .\n\n## 3 Branes at Conical Singularities\n\nAll the branes considered in the previous section interpolate between flat Minkowski spacetime asymptotically far away from the brane and near the brane horizon. Both of these vacua are maximally supersymmetric and the branes themselves preserve of the supersymmetry. In this section we discuss the brane solutions of that interpolate between near-horizon geometries of the form and an asymptotic geometry of the form , where is a -dimensional Einstein manifold and is the cone over . This can be interpreted as a set of coincident branes at a conical singularity in a Ricci-flat manifold. These brane solutions interpolate between vacua which are not maximally supersymmetric and the brane solutions themselves will preserve a smaller fraction of the supersymmetry. In this section we will characterise those branes with in terms of the holonomy group of the transverse space.\n\n### 3.1 Cone Branes\n\nThe -brane solutions above all have metrics of the form\n\n g=H−αgp+1+H2βgE , (13)\n\nwhere\n\n H(r)=1+(ar)β , (14)\n\nis the Minkowski metric on the worldvolume of the -brane, and where is the flat Euclidean metric on . For the -brane, -brane, and -brane, respectively, and . The Euclidean metric can be written as\n\n gE=dr2+r2gS (15)\n\nwhere is the round metric on the unit sphere and is a radial coordinate. In the near-horizon limit in which the constant term in can be dropped, the metric (13) becomes\n\n g∼(ra)αβgp+1+a2r−2dr2+a2gS ,\n\nwhich is the metric on , as can be seen by changing variables to , where the anti-de Sitter radius is .\n\nReplacing the sphere by any other -dimensional Einstein manifold with the same cosmological constant gives another solution of the field equations on , and we will be interested in those choices of that give spontaneous compactifications to anti-de Sitter space that preserve some nonzero fraction of the supersymmetry. Note that as is a complete Einstein space with positive cosmological constant, it follows by Myer’s Theorem that it is compact (see, e.g., ). There is a brane solution of the form \n\n g=H−αgp+1+H2βgC , (16)\n\nwhere is again given by (14) and the metric is given by replacing the spherical metric with the metric of in (15) to give\n\n gC=dr2+r2gX .\n\nThe transverse space to the brane is now the metric cone over with topology , with the open half-line and metric . Since is Einstein with , it follows that is Ricci-flat; however unlike the case of the sphere, where is actually flat, is now metrically singular at the apex of the cone .\n\nIn the near-horizon region , the constant term in can be dropped and the metric becomes approximately\n\n g∼(ar)αβgp+1+a2r−2dr2+a2gC ,\n\nwhich is the metric on . For large , and the metric becomes\n\n g∼gp+1+gC ,\n\non the product of ()-dimensional Minkowski space with the cone . We interpret these solutions as describing coincident branes located at the conical singularity of , or cone branes for short. Note that whereas the solution has a conical singularity at , the brane metric (16) approaches the metric of as , which is non-singular at . However is a horizon for the brane and the solution can be continued through the horizon. In general there will be a singularity inside the horizon.\n\nThese solutions can in principle be generalised by replacing with more general harmonic functions on the cone .\n\n### 3.2 Supersymmetry and Holonomy\n\nIn this subsection we wish to describe the amount of supersymmetry preserved by the solutions and , and the brane solution (16) that interpolates between them. In each case, the number of supersymmetries is the number of linearly independent spinors such that the supersymmetry variation of the gravitini vanish in this background: . On , such spinors are of the form where is a Killing spinor on and is a Killing spinor on , satisfying\n\n ∇(X)Mψ=ϵ12ΓMψ (17)\n\nwhere is either or , depending on the field . For simplicity, and also for ease of comparison with the mathematical literature on Killing spinors, we have rescaled the metric on the Einstein manifold in such a way that the coefficient on the right-hand side of equation (17) is . The number of supersymmetries of the solution is given by where is the number of Killing spinors on with , is the number of Killing spinors on and for theory and for Type II strings.\n\nIf the dimension of is even, then there are as many solutions with as with , whereas if is odd, then changing the sign of corresponds to reversing the orientation of . Thus for odd-dimensional , the number of solutions depends on the orientation. For example, for , if there are Killing spinors with one choice of orientation of , there will be no Killing spinors with the opposite choice of orientation, unless is the round -sphere, in which case with either choice of orientation . We will see below that this is very easy to understand in terms of the holonomy of the cone .\n\nOn , the supersymmetries are of the form where and are covariantly constant spinors on Minkowski space and the cone , respectively. In particular, satisfies\n\n ∇(C)ψ=0 . (18)\n\nThe number of supersymmetries of the solution is then given by where is the number of covariantly constant spinors on the cone and is the number of parallel spinors on .\n\nAs shown in , there is a one to one correspondence between Killing spinors on satisfying (17) and covariantly constant spinors on satisfying (18). Each covariantly constant spinor on restricts to a Killing spinor on satisfying (17) for a particular orientation of . If is odd, so that is even, then the sign of is also correlated to the chirality of the parallel spinor on . If is even, so that is odd, then the sign of depends on which one of the two irreducible representations of the Clifford algebra we use to embed the unique spinor representation of .\n\nIn the round -sphere compactification of theory, there are solutions of (17) with and solutions with , of which only have the right sign to be Killing spinors, while has covariantly constant spinors. In this case, , , and , so there are supersymmetries on both and . For other , there are only solutions of (17) with one particular orientation , and parallel spinors on restrict to spinors satisfying (17) with that choice of orientation. This is because the spinors left invariant by any of the possible holonomy groups which act irreducibly in eight dimensions all have the same chirality. On the other hand, as will be seen shortly, when is an Einstein manifold admitting Killing spinors and has dimension , both orientations give the same numbers of supersymmetries. For Type IIB compactifications on -manifolds, this conflicts with a claim made in . At any rate, we are interested in the supersymmetric cases, so we choose .\n\nFor -, - and -branes, = respectively, whereas = . Using our result for the numbers of asymptotic and near-horizon supersymmetries we find that for - and -branes the number of supersymmetries of the near-horizon geometry is twice the number of supersymmetries for the asymptotic conical geometry , unless is a round sphere, in which case the number of supersymmetries is the same in both cases. Applying this to the -brane gives the same number of supersymmetries asymptotically and near the horizon, but in this case, as we shall see, the only smooth spaces that admit Killing spinors are and . In each of these cases, the number of supersymmetries can be further reduced by orbifolding. In particular, for the -brane with non-spherical near-horizon geometry, the asymptotic space is actually an orbifold and in this case the near-horizon limit has twice the number of supersymmetries as the asymptotic region.\n\nAs an example, consider the squashed -sphere of , which has , and , so that or for the squashed -sphere compactification, depending on the orientation. The cone has one parallel spinor, so the near-horizon geometry has or , while the asymptotic conical geometry has .\n\nIn summary, the amount of supersymmetry on the brane and in the near-horizon geometry is determined by the number of parallel spinors on the cone , and this can now be analysed group-theoretically.\n\nAssume that the base of the cone, , is simply-connected. Its cone will be simply-connected also, since and are homotopy equivalent. Simply-connected manifolds admitting parallel spinors are classified by their holonomy group . Because is Ricci-flat, we know that it cannot be a locally symmetric space. Moreover, by a theorem of Gallot , the holonomy group acts on irreducibly unless is flat, in which case is the round sphere. Therefore, we need only consider irreducible holonomy groups of manifolds which are not locally symmetric. In other words, those in Berger’s table (see, e.g., [7, 66]).\n\nOf those, the ones which admit parallel spinors are given in the following table, which also lists the number (or in even dimension) of linearly independent parallel spinors.\n\nIf , and hence , were not simply-connected, then we could still use part of the above analysis. Gallot’s theorem still applies and the existence of parallel spinors constrains the restricted holonomy group of the manifold to be contained in the above Table. However a spinor which is invariant under the restricted holonomy group need not be parallel, because it may still transform nontrivially under parallel transport along noncontractible loops. Therefore the number of parallel spinors in will be at most the number shown in Table 1.\n\n## 4 Near-horizon Geometries of Cone Branes\n\nIn this section we will discuss the relation between the geometry of and the holonomy of its cone . We will do this in some generality, and give the relation between the geometry of and the number of supersymmetries preserved by a solution, when one exists, and this characterises the near-horizon geometries of the supersymmetric cone branes discussed in the previous section.\n\nOn every cone there is a privileged vector field , called the Euler vector, which generates the rescaling diffeomorphisms. Moreover on any manifold of reduced holonomy, the holonomy principle guarantees the existence of certain parallel tensors, corresponding to the singlets under the holonomy group in the tensor products of the holonomy representation. Therefore on a cone of reduced holonomy we will be able to build interesting geometric structures out of these parallel tensors and the Euler vector. These structures will in turn induce interesting geometric structures on , which we identify with . These geometric structures are summarised in the following table, which also lists the numbers of Killing spinors; that is, solutions of equation (17), with the number of solutions with .\n\nWe now proceed to describe these geometries in detail.\n\n### 4.1 Cones with Spin7 Holonomy\n\nAny eight-dimensional manifold with holonomy possesses a parallel self-dual -form , known as the Cayley form. Contracting the Euler vector into the Cayley form gives a -form on which restricts to a -form on :\n\n ϕ(u,v,w)≡Ω(ξ,u,v,w) ,\n\nfor any vectors tangent to . In fact, one can write the Cayley form restricted to as\n\n Ω=dr∧ϕ+⋆ϕ ,\n\nwith the Hodge operator on . Notice that is a -form on , and by its restriction to we simply mean evaluating it at on vector fields tangent to . In other words, acts both on vectors tangent and normal to . From the fact that is parallel in , it follows that, in , obeys\n\n ∇ϕ=⋆ϕ .\n\nThis condition says that has weak holonomy , and we say that defines a nearly parallel structure. In fact, as proven in a manifold has weak holonomy if and only if its metric cone has holonomy contained in . In the early Kaluza–Klein literature it would have been said that has Weyl holonomy, but we will not follow this nomenclature.\n\n### 4.2 Cones with G2 holonomy\n\nOn any seven-dimensional manifold with holonomy there is a parallel -form , known as the associative form. Contracting the Euler vector into the associative form yields a -form on which restricts to a -form on :\n\n ω(u,v)≡Φ(ξ,u,v) ,\n\nfor any vectors tangent to . We can define a -tensor on as follows:\n\n ⟨J(u),v⟩≡ω(u,v) .\n\nIt is possible to show that is an orthogonal almost complex structure:\n\n J2=−1and⟨J(u),J(v)⟩=⟨u,v⟩ ,\n\nwhence is an almost hermitian manifold. From the fact that is parallel on , it follows that on ,\n\n ∇vJ(v)=0for any v;\n\nalthough . Moreover is not integrable. This means that is a non-Kähler nearly Kähler manifold .\n\nIn fact, the converse is also true and a six-dimensional manifold is non-Kähler nearly Kähler if and only if its cone has holonomy contained in . A different proof that a six-dimensional manifold admits Killing spinors if and only if it is non-Kähler nearly Kähler appeared earlier in .\n\nNearly Kähler -manifolds share many properties with Calabi–Yau -folds. For example, there is a natural -form defined by contracting the Euler vector with the coassociative -form on the cone. In particular, nearly Kähler -manifolds have vanishing first Chern class . For example, is nearly Kähler, but cannot be Kähler because is trivial.\n\n#### 4.2.1 Scholium on Almost Hermitian Manifolds\n\nThe notion of nearly Kähler manifolds should not be confused with the notion of an almost Kähler manifold, which simply means an almost hermitian manifold whose associated -form is closed; or in other words, a symplectic manifold with a compatible almost complex structure. Unfortunately, given the many different generalisations of the notion of Kähler manifolds, there is a large possibility for confusion, so we digress momentarily to settle the notation once and for all. As shown in , there are 16 classes of almost hermitian manifolds, the class of Kähler manifolds being but one of them. The classes are defined in the following way.\n\nOne can define almost hermitian geometry in terms of -structures. Just like a Riemannian metric on an -dimensional manifold allows us to reduce the structure group of the tangent bundle from to , an almost hermitian structure allows us to reduce the group further to . This means that tensor bundles can be consistently decomposed into -irreducible sub-bundles. Let be the associated -form. Its covariant derivative is a section through a sub-bundle , with the cotangent bundle of . is not irreducible under , but rather decomposes into four irreducible components . There are therefore sixteen -invariant sub-bundles (not necessarily irreducible) in : , , (),…, . The sixteen classes of hermitian manifolds correspond to these sixteen sub-bundles: a manifold belonging to the class corresponding to the sub-bundle of to which belongs.\n\nClearly corresponds to the class of Kähler manifolds; but there are other classes of manifolds which are close to Kähler in some sense. The class of non-Kähler nearly Kähler manifolds can be shown to be the one for which . The class of almost Kähler manifolds consists of those for which . To make matters even more confusing there exist also semi-Kähler manifolds for which and quasi-Kähler manifolds which have a more complicated definition which we will not need. For details the reader is referred to .\n\n### 4.3 Calabi–Yau Cones\n\nThe class of Calabi–Yau -folds is the class of compact Ricci-flat Kähler -folds. Let us concentrate first on the Kähler condition. Any Kähler manifold has a parallel -form, the Kähler form, and a corresponding orthogonal complex structure which is also parallel. Together with the Euler vector, we can therefore build two objects: a -form obtained by contracting the Euler vector into the Kähler form, and a vector obtained by acting with the complex structure on the Euler vector. The vector is clearly tangent to and restricts there to a vector , whereas the -form restricts to a -form on which is naturally dual to :\n\n χ=I(ξ)andθ=⟨χ,−⟩ .\n\nIt follows from the definition that has unit norm and is a Killing vector. Furthermore, the covariant derivative of defines a -tensor in ,\n\n T(v)=∇vχ , (19)\n\nwhich, because is parallel in , satisfies the following identity:\n\n ∇uT(v)=θ(v)u−⟨u,v⟩χ . (20)\n\nThe triple defines a Sasaki structure on . More precisely, a triple on an odd-dimensional Riemannian manifold , where is a unit norm Killing vector, its dual -form, and obeys equation (20), is called a Sasaki structure on , and is called Sasakian. Equivalently , a manifold is Sasakian if and only if its metric cone is Kähler.\n\nIf in addition is Ricci-flat (hence Calabi–Yau), then is Einstein. We say that is Sasaki–Einstein. Therefore a manifold is Sasaki–Einstein if and only if its metric cone is Calabi–Yau [29, 30, 4].\n\nThis means, in particular, that a Sasaki–Einstein manifold of dimension also possesses two distinguished -forms obtained by contracting the Euler vector into the real and imaginary parts of the complex volume ()-form on .\n\nThe Killing vector in a -dimensional Sasakian manifold defines a foliation whose leaves are the integral curves of . The manifold is called regular if these curves are closed and have the same length. If this is the case, defines a action on and can be understood as a circle bundle over the orbit space, a -dimensional manifold , which can be shown to be Kähler. Moreover if is Sasaki–Einstein then is Kähler–Einstein. Regularity is a very stringent condition which is not satisfied by most Sasaki(–Einstein) manifolds. When the integral curves of are closed but of different lengths, the orbit space is an orbifold which is everywhere smooth except at a finite number of points [29, 10].\n\n### 4.4 Hyperkähler Cones\n\nIn a hyperkähler manifold we have a parallel quaternionic structure consisting of three orthogonal complex structures satisfying the quaternion algebra, as well as their corresponding Kähler forms: . From the discussion above, on a cone each complex structure gives rise to one Sasaki structure on . Moreover the fact that the three complex structures on satisfy the quaternion algebra means that the Killing vectors in the three Sasaki structures are orthonormal and obey an Lie algebra. Three Sasaki structures satisfying these conditions define a -Sasaki structure on (see for the latest word on -Sasakian manifolds). Equivalently, one can prove the converse and show that a manifold is -Sasakian if and only if its cone is hyperkähler [30, 4, 11].\n\nThe three Killing vectors define a foliation of the -Sasakian manifold . We say that is regular if the foliation fibres. This means that is an or bundle over a quaternionic Kähler space . Equivalently, is a circle bundle over the twistor space of . If is not regular, but the Killing vectors are complete, then the orbit space defines a quaternionic Kähler orbifold .\n\n## 5 Examples\n\nIn this section we list the known near-horizon geometries for different types of cone branes. The homogeneous examples are all known from the early days of Kaluza–Klein supergravity, but we do list some non-homogeneous examples as well. For each type of cone brane we discuss the possible smooth near-horizon geometries and where applicable we discuss orbifolds.\n\n### 5.1 M2 Cone Branes\n\nThe transverse space to an cone brane is the metric cone over a -dimensional manifold . From Table 2 one can read off the following possibilities for simply-connected , which are listed in Table 3. In the non-simply-connected case the number of Killing spinors is at most the one shown in the table.\n\nBesides the sphere and its quotients, we have the following types of near-horizon geometries.\n\n#### 5.1.1 X7 is 3-Sasakian\n\nWe must distinguish between regular and non-regular -Sasakian -manifolds. The regular manifolds were classified in [30, 11] and they all happen to be homogeneous spaces. Therefore they have been known since the early days of Kaluza–Klein supergravity . In the present context they have been discussed in . Apart from and , the only other homogeneous regular example is , which is called in and in .\n\nOn the other hand there are an infinite number of different non-regular -Sasakian -manifolds. The topology of a seven-dimensional -Sasakian manifold is highly constrained. First of all, Bochner’s theorem implies that the first Betti number of a -Sasakian manifold vanishes: . In seven dimensions, by Poincaré duality. Furthermore, as shown in [32, 31], , so that the only nonzero Betti numbers are . Recently an infinite family with arbitrary has been constructed in [13, 9]. This implies that there exist -Sasakian -manifolds of every possible rational homotopy type allowed. These examples are toric; that is, they admit an action of a torus preserving the -Sasakian structure. They are constructed via a -Sasakian quotient akin to the hyperkähler quotient . Indeed, the two quotients are related via the cone construction. In other words, suppose is a -Sasakian manifold and is its hyperkähler cone. Then if admits a triholomorphic action commuting with the Euler vector, then the hyperkähler quotient of is a cone over the -Sasaki quotient of . The cones over the toric -Sasakian manifolds in [13, 9] form a subclass of the toric hyperkähler manifolds studied in and which have been considered in the context of intersecting branes in .\n\nAll -Sasakian manifolds have an infinitesimal isometry. If the Killing vectors are complete, they integrate to an action of or . In the regular case, the orbit space is smooth; otherwise not. In any case, the infinitesimal isometry is a generic feature of these manifolds and one which will play a role in Section 6, when we discuss the dual field theories.\n\n#### 5.1.2 X7 is Sasaki–Einstein\n\nThere are many known Sasaki–Einstein -manifolds. The regular examples can be understood as bundles over Kähler–Einstein -folds . They have not been classified. The known results are summarised in the following table .\n\nThe first four examples are homogeneous and have therefore appeared in the early Kaluza–Klein literature. The first three are listed using the nomenclature of . In [24, 31] these spaces are called , and , respectively. Here is the Stiefel manifold of orthonormal oriented -frames in and is the Grassmannian of oriented -planes in . is the complex flag manifold in consisting of pairs where is a complex plane in and is a complex line. The last class of examples does not consist of homogeneous manifolds: are the del Pezzo surfaces consisting of blowing up points in general position on .\n\nIn addition, there are two infinite families of homogeneous non-regular Sasaki–Einstein -manifolds: the of [73, 15] and the of . Of course they are mentioned in .\n\n#### 5.1.3 X7 has Weak G2 Holonomy\n\nThe canonical example of a weak holonomy manifold which is not Sasaki–Einstein is the squashed -sphere, which is a homogenous space\n. It turns out that this generalises, and every -Sasakian -manifold can be squashed to a manifold with weak holonomy [32, 31]. The squashing is done as follows. A -Sasakian manifold is foliated by the action of the Sasakian Killing vectors. At any point , the Killing vectors are tangent to the unique leaf of the foliation passing through . The tangent space to at has an orthogonal decomposition . The squashing of the metric is done by introducing a parameter and rescaling the metric on the leaves of the foliation by . We can then compute the Ricci tensor as a function of and notice that there are two values for which it is Einstein: one is the original -Sasakian metric, and the other gives rise to a weak holonomy structure.\n\nBy squashing the infinite toric family of [13, 9] in this way, one obtains a huge number of weak holonomy manifolds which are not Sasaki–Einstein. These spaces are not homogeneous and hence have not been considered before in the Kaluza–Klein supergravity context. There are examples with arbitrary and all other Betti numbers vanishing, since squashing does not change the topology.\n\nAnother infinite dimensional family of weak holonomy manifolds consists of the Aloff–Wallach spaces , with defined by\n\n eiθ↦⎛⎜⎝eikθeiℓθe−i(k+ℓ)θ⎞⎟⎠ .\n\nThis family is remarkable because it contains an infinite number of distinct homotopy types and even exotic pairs, i.e., homeomorphic non-diffeomorphic pairs . For or these spaces admit metrics with weak holonomy. The Aloff–Wallach spaces agree, allowing for some redundancy in the notation, with the spaces of . Finally, the other weak holonomy manifold from the early Kaluza–Klein literature is the homogeneous space of .\n\nIn summary, all known examples are homogeneous, hence previously known in the Kaluza–Klein context, except for the squashed toric -Sasakian manifolds.\n\n### 5.2 M5 Cone Branes\n\nThe transverse space to an cone brane is a five-dimensional cone over a four-dimensional manifold . As proven in , any complete connected -dimensional spin manifold, with , admitting real Killing spinors is locally isometric to the round -sphere. Therefore the only possible smooth near-horizon geometry for an cone brane is either spherical or elliptical ; these describe the near-horizon geometry of branes at a point in or on the orientifold . More generally we can consider orbifolds of which are of the form . Such an orbifold will preserve supersymmetry if is an ADE subgroup of the hyperkähler which acts on . This clearly induces a near-horizon geometry of the form which has two singular points induced from the orbifold fixed points of acting on . These orbifolds preserve half of the supersymmetries obtained in the maximally supersymmetric case.\n\n### 5.3 D3 Cone Branes\n\nThe transverse space to a cone brane is a six-dimensional cone over a five-dimensional manifold . From Table 2 we see that there are two simply-connected possibilities. This means is a sphere or a Sasaki–Einstein manifold. Non-simply connected examples, e.g., or more generally , can be obtained by taking quotients. There are an infinite number of smooth quotients of the sphere which possess Killing spinors. These can be determined as follows .\n\nIn his solution of the spherical space problem, Wolf classified all the discrete subgroups , for which is smooth . Given one such subgroup, the spin structures on are in one-to-one correspondence with the lifts of to an isomorphic subgroup ; that is, to a which is mapped to isomorphically under the covering map . Finally, for with spin structure given by , the Killing spinors in are precisely the -invariant spinors in . The results are as follows.\n\nThere are two families of subgroups of for which is smooth: and described below.\n\nLet , where be the cyclic subgroup of of order generated by the following element:\n\n ⎛⎜ ⎜ ⎜ ⎜⎝R(1n)R(an)R(bn)⎞⎟ ⎟ ⎟ ⎟⎠with(a,n)=(b,n)=11≤a,b\n\nwhere denotes the rotation matrix\n\n R(θ)=(−cos2πθsin2πθ−sin2πθcos2πθ) ,\n\nand denotes the greatest common divisor, so that means that and are coprime.\n\nSimilarly let , with and , denote the subgroup of generated by elements and subject to the relations: and . The generators can be written as\n\nwhere is a identity matrix, and where and and is the identity matrix of order 2. This subgroup has order .\n\nIt is not hard to show that has precisely one spin structure when is odd, and none for even. Similarly has precisely one spin structure when both and are odd, and none otherwise.\n\nComputing the -invariant spinors for each of the cases above is an easy matter. One sees that is of type whenever any one of the following equalities is satisfied:\n\n a+b±1=nora−b=±1 ,\n\nand has no Killing spinors otherwise. Similarly," ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9190416,"math_prob":0.97227454,"size":41255,"snap":"2021-31-2021-39","text_gpt3_token_len":9404,"char_repetition_ratio":0.16886863,"word_repetition_ratio":0.06734604,"special_character_ratio":0.20782936,"punctuation_ratio":0.100743294,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9902422,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-21T01:44:18Z\",\"WARC-Record-ID\":\"<urn:uuid:d8de6f2d-53e6-4a18-9273-a059f58ef68e>\",\"Content-Length\":\"1049554\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:de437762-3774-4749-b366-ac9420427999>\",\"WARC-Concurrent-To\":\"<urn:uuid:cdfe9991-a54e-4b8a-853f-1a09bb4e10f0>\",\"WARC-IP-Address\":\"172.67.158.169\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/hep-th/9808014/\",\"WARC-Payload-Digest\":\"sha1:ZL7AWMTVBFRVYCQM5RR2NPLXCJXGUM4X\",\"WARC-Block-Digest\":\"sha1:5N3WSSTACYJSNENOBZNQYLYPNS5QN6YW\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057131.88_warc_CC-MAIN-20210921011047-20210921041047-00477.warc.gz\"}"}
https://mathexpressionsanswerkey.com/math-expressions-grade-4-unit-5-lesson-8-answer-key/	[ "# Math Expressions Grade 4 Unit 5 Lesson 8 Answer Key Focus on Mathematical Practices\n\n## Math Expressions Common Core Grade 4 Unit 5 Lesson 8 Answer Key Focus on Mathematical Practices\n\nMath Expressions Grade 4 Unit 5 Lesson 8 Homework\n\nMath Expressions Grade 4 Focus on Mathematical Practices Lesson 8 Pdf Answer Key Question 1.\nYonni has a 5 gallon fish tank. He needs to change the water in the fish tank. How many cups of water will Yonni need to replace all the water in the fish tank?\nYonni needs 80 cups of water to replace all the water in the fish tank.\n\nExplanation:\nYonni has a 5 gallon fish tank\nIf 1 gallon is equal to 16 cups then 5 gallons will be 5 x 16 = 80 cups.", null, "Therefore, Yonni needs 80 cups of water to replace all the water in the fish tank.\n\nMath Expressions Grade 4 Focus on Mathematical Practices Pdf Answer Key Question 2.\nBarry is building a fence around his backyard. The backyard is in the shape of a rectangle and the longest side of the yard is 20 meters. The fence will have a perimeter of 60 meters. How many meters long is the short side of the backyard?\nThe length of the shortest side of the backyard is 10 meters\n\nExplanation:\nTo find the width, multiply the length given by 2 and subtract the result from the perimeter.Now you will have the total length for the remaining 2 sides.The remaining number divided by 2 is the width.\nMultiply length of the longest side 20 by 2\n20 x 2 = 40\nSubtract 40 from the perimeter 60\n60-40= 20\nDivide the number 20 by 2\n20 didided by 2 = 10\nTherefore, the length of the shortest side of the backyard is 10 meters.\n\nMath Expressions Grade 4 Lesson 8 Pdf Answer Key Unit 5 Question 3.\nYesi’s dog weighed 5 pounds when she got him. He now weighs 45 pounds. How much weight did Yesi’s dog gain, in ounces?\nThe weight Yesi’s dog gained is 640 ounces\n\nExplanation:\nYesi’s dog weighed 5 pounds when she got him. He now weighs 45 pounds\nThe weight of now is 45 – 5 = 40 pounds\nIf 1 pound is 16 ounces then 40 pounds will be 40 x 16 = 640 ounces", null, "Therefore, the weight Yesi’s dog gained is 640 ounces.\n\nMath Expressions Grade 4 Focus on Mathematical Practices Pdf Question 4.\nFiona’s family ¡s replacing the carpet in their living room. The living room is in the shape of a square. The length of one wall is 16 feet. How many square feet of carpet does Fiona’s family need to buy to replace their old carpet?\nFiona’s family need to buy 256 square feet of carpet to replace the old carpet.\n\nExplanation:\nTo know the number of square feet of carpet the family need to buy we need to find the area of their square shaped living room\nLength of the living room is 16 feet\nArea of a square is l x l\n16 x 16 = 256", null, "Therefore, Fiona’s family need to buy 256 square feet of carpet to replace the old carpet.\n\nTrevon drew the two rectangles below. He wanted to know the difference between the areas of the two rectangles. What is the difference between the two areas?", null, "The difference between the areas of the two rectangles is 60 dm", null, "Explanation:\nTrevon drew the two rectangles\nArea of a rectangle is l x w\nThe length and width of recatngle 1 are l= 16dm and w = 9 dm\nArea = l x w = 16 x 9 = 144\nThe length and width of recatngle 2 are l= 12dm and w = 7 dm\nArea = l x w = 12 x 7 = 84\nThe difference of areas of 2 rectangles is 144 – 84 = 60 dm.\n\nMath Expressions Grade 4 Unit 5 Lesson 8 Remembering\n\nSolve. Then explain the meaning of the remainder.\n\nFocus on Mathematical Practices Grade 5 Answer Key Unit 5 Question 1.\nThere are 43 students at a musical performance. Each row in the auditorium has 8 seats.If the students fill seats row by row from front to back, how many people are in the last row?\n3 people are in the last row\n\nExplanation:\nThere are 43 students at a musical performance. Each row in the auditorium has 8 seats\nIf the students fill seats row by row from front to back then\nAfter 1 st row filled by 8 students their will be 43-8=35 students left\n35-8=27\n27-8=19\n19-8=11\n11-8=3\nSimilarly after 5 ows getting filled by 8 students their will be 3 students left in the last rwo.\n\nWrite whether each number is prime or composite.\n\n49\n49 is not a prime number\n\nExplanation:\n49 is not a prime number because it can be divided not only by itself and 1 but also by 7.\n\nQuestion 3.\n31\n31 is a prime number\n\nExplanation:\n31 is a prime number as it can be divided exactly only by self and 1.\n\nQuestion 4.\n17\n17 is a prime number\n\nExplanation:\n17 is a prime number as it can be divided exactly only by self and 1.\n\nQuestion 5.\nThe perimeter of a postage stamp is 90 millimeters. The longer side of the stamp is 25 millimeters. What is the length of the shorter side?\nThe length of the shortest side of the postage stamp is 20 millimeters\n\nExplanation:\nTo find the width, multiply the length given by 2 and subtract the result from the perimeter.Now you will have the total length for the remaining 2 sides.The remaining number divided by 2 is the width.\nMultiply length of the longest side 25 by 2\n25 x 2 = 50\nSubtract 300 from the perimeter 960\n90-50= 40\nDivide the number 40 by 2\n40 didided by 2 = 20", null, "Therefore, the length of the shortest side of the postage stamp is 20 millimeters.\n\nQuestion 6.\nStretch Your Thinking The directions for lemonade say to put 2 cups of the liquid concentrate into 1 gallon of water. If Olivia only wants to make 1 pint of lemonade, how many fluid ounces of concentrate should she use? Explain." ]	[ null, "https://mathexpressionsanswerkey.com/wp-content/uploads/2021/09/Math-Expressions-Grade-4-Unit-5-Lesson-8-Answer-Key-5.png", null, "https://mathexpressionsanswerkey.com/wp-content/uploads/2021/09/Math-Expressions-Grade-4-Unit-5-Lesson-8-Answer-Key-3.png", null, "https://mathexpressionsanswerkey.com/wp-content/uploads/2021/09/Math-Expressions-Grade-4-Unit-5-Lesson-8-Answer-Key-4.png", null, "https://mathexpressionsanswerkey.com/wp-content/uploads/2021/09/Math-Expressions-Grade-4-Unit-5-Lesson-8-Answer-Key-1.png", null, "https://mathexpressionsanswerkey.com/wp-content/uploads/2021/09/Math-Expressions-Grade-4-Unit-5-Lesson-8-Answer-Key-2.png", null, "https://mathexpressionsanswerkey.com/wp-content/uploads/2021/09/Math-Expressions-Grade-4-Unit-5-Lesson-8-Answer-Key-1-1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90143234,"math_prob":0.9683671,"size":6219,"snap":"2022-27-2022-33","text_gpt3_token_len":1617,"char_repetition_ratio":0.114239745,"word_repetition_ratio":0.33220625,"special_character_ratio":0.26193923,"punctuation_ratio":0.07868602,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9977121,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-30T16:29:16Z\",\"WARC-Record-ID\":\"<urn:uuid:0d5793f8-0cb1-4eda-ba77-7a42fb5b6599>\",\"Content-Length\":\"59610\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d15856ee-99ab-48e3-9dff-bf50eca0ec3b>\",\"WARC-Concurrent-To\":\"<urn:uuid:d4f71b62-6a10-4a5d-8661-4078c3beb309>\",\"WARC-IP-Address\":\"164.90.151.146\",\"WARC-Target-URI\":\"https://mathexpressionsanswerkey.com/math-expressions-grade-4-unit-5-lesson-8-answer-key/\",\"WARC-Payload-Digest\":\"sha1:PBKBACJW6E7JIJF7N66L2S5JTFYQKIHR\",\"WARC-Block-Digest\":\"sha1:7VPOZWNMUSABYYFNT5G4TWFQCDQRKXXS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103850139.45_warc_CC-MAIN-20220630153307-20220630183307-00398.warc.gz\"}"}
https://www.cut-the-knot.org/Curriculum/Algebra/GregBrockman/GregBrockmanDucciSequences.shtml	[ "# Ducci Sequences\n\nBy Greg Brockman\n\n### Some Background\n\nThis applet demonstrates the behavior of arbitrary vector-length Ducci sequences over the real numbers. (For the case of integers, see Integer Iterations on a Circle.) So just what does that mean? Well, first of all, we start with any list of n numbers, or a vector of length n. Next, we build a new vector by replacing each number by the absolute value of the difference between it and its neighbor to the right (we think about these vectors as being cyclic). We can repeat this process, and the sequence of vectors thus obtained is called a Ducci sequence.\n\nMore formally, let f: Rn → Rn be defined by\n\nf(ν1, ..., νn) = (\|ν1 - ν2\|, \|ν2 - ν3\|, ..., \|νn - ν1\|).\n\nThen the Ducci sequence of an n-vector ν is defined as {f i(ν)}, where i = 1, 2, 3, ... (f i is the iterated f. For example, f 2(ν) = f(f(ν)).)\n\nNow, since when generating successive elements in a Ducci sequence, we wrap a vector as if cyclic, it can be helpful to think of a vector in a Ducci sequence as simply some numbers arranged around a circle. I have taken this approach in this applet.\n\n### Instructions\n\nSo here's how to run the applet: First of all, click \"Add Circle\" to begin. There are now two modes: Edit and Run. By default, you start out in Edit mode.\n\n#### Options for Edit Mode\n\n• Add Slot (green button): Each new circle you add corresponds to increasing the length of the corresponding vector by 1.\n\n• You can select a given circle by clicking on it. If it's already selected, then nothing happens.\n\n• To add a value to the selected circle, enter values into the text boxes at the top. Values in the selected circle will automatically update, or you can press \"Enter\" if you wish.\n\n• You can remove the selected circle by pushing the \"Del\" key.\n\n• When ready, click \"Play Ducci!\" to enter Run mode.", null, "#### Options for Run Mode\n\n• Next Step calculates the next vector in the Ducci sequence. If animation is enabled, then you get a cool movie effect.\n\n• The Automate button automatically runs Next Step multiple times.\n\n• If you want to skip to a certain generation number (for example, maybe you want to find the 507th vector in the Ducci sequence under consideration), you can use the \"See generation number\" box. Just type in the number you want and press enter.\n\n• The buttons at top are fairly self-explanatory: Reset Values takes you back to the 0th vector (the starting one); Edit Values switches you back to Edit Mode; and Create New Game completely deletes all information and starts you with a fresh vector.", null, "#### Common Options\n\n• Toggling Display Exact/Show Approx. changes whether the exact values of numbers in circles are displayed or whether a decimal approximation is shown.\n\n##### Hint\n• If you aren't sure what a button does, you can always hold your mouse over it and wait for the tooltip.\n\n##### Note\n• The Ducci sequences are calculated using 300 digit precision on π, e, and 2. Thus, after about 1000 generations, your typical Ducci sequence incorrectly starts behaving like a homogeneous one (for the meaning of this term, see my paper) rather than a heterogeneous one. The program automatically shows a warning after 1000 generations.\n\n• The color of a circle indicates its magnitude relative to the maximum element in the current vector. This color is blue for the maximum element and becomes (linearly) darker as the element's magnitude shrinks.\n\n### Some Suggestions of What To Do\n\nMy paper discusses under what conditions Ducci sequences of odd length vectors start repeating. You can use the applet to examine the behavior of various Ducci sequences. What happens if you start out with an irrational, but then replace it by a good rational approximation to it? Can you get any non-repeating sequences that don't go to the zero vector (the answer to this question is actually unknown; such sequences exist in general, but the known ones cannot be built using the given input method)?\n\nSome cool phenomena, connected with papers before mine, show up when you work with vectors of length 4. Can you find a vector that never actually hits the zero vector? If not, can you build any that take a long time to get there? (Spoiler—using the given input tools, you won't be able to build a vector that never actually hits the zero vector. But given sufficient cleverness, it's possible to built vectors that take arbitrarily long to get there.)\n\nYou should also take a look at the exact values, and see as they behave as you iterate. How quickly do the coefficients on the various numbers grow? Under what conditions must they grow without bound?\n\nOn the whole, there is an essentially unlimited spectrum of questions that one can ask. I recommend using this applet to get an intuition of how Ducci sequences over the real numbers behave.", null, "" ]	[ null, "https://www.cut-the-knot.org/gifs/tbow_sh.gif", null, "https://www.cut-the-knot.org/gifs/tbow_sh.gif", null, "https://www.cut-the-knot.org/gifs/tbow_sh.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9028614,"math_prob":0.8748944,"size":5231,"snap":"2019-51-2020-05","text_gpt3_token_len":1170,"char_repetition_ratio":0.12033671,"word_repetition_ratio":0.011086474,"special_character_ratio":0.22079909,"punctuation_ratio":0.114035085,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9621435,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-18T10:43:43Z\",\"WARC-Record-ID\":\"<urn:uuid:c408ae25-82b6-4a14-95e5-7d4de257ec1b>\",\"Content-Length\":\"16072\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5ba12e78-b11b-4276-903e-5c393fbe3628>\",\"WARC-Concurrent-To\":\"<urn:uuid:e31040ce-5154-4324-94eb-1044480a43a8>\",\"WARC-IP-Address\":\"107.180.50.227\",\"WARC-Target-URI\":\"https://www.cut-the-knot.org/Curriculum/Algebra/GregBrockman/GregBrockmanDucciSequences.shtml\",\"WARC-Payload-Digest\":\"sha1:JBRZISGL42YNSIVRXYYFQU7KGH4FE4SG\",\"WARC-Block-Digest\":\"sha1:MPQIXQZ63LCJSXXV2CJ6SZHIGK3467CH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250592394.9_warc_CC-MAIN-20200118081234-20200118105234-00177.warc.gz\"}"}
https://www.khanacademy.org/math/algebra2/rational-expressions-equations-and-functions/simplify-rational-expressions/v/simplifying-rational-expressions-w-common-monomial-factors	[ "Main content\n\n# Simplifying rational expressions: common monomial factors\n\nCCSS Math: HSA.APR.D.6\n\n## Video transcript\n\n- [Voiceover] So I have a rational expression here and my goal is to simplify it, but while I simplify it, I wanna make the simplified expression be algebraically equivalent. So, if there are certain x values that would make this thing undefined, that I have to restrict my simplified expression by those x values. So, you could pause this video and take a go at it and then we'll do it together. All right. So let's just think real quick. What x values would make this expression undefined? Well, it's undefined if we try to divide by zero. So if x is zero, 14 times zero is zero, it's going to be undefined. And so, we could say this is, we could say x does not, cannot be equal to zero. For any other x, we can evaluate this expression. Now let's actually try to simplify it. So, when you look at the numerator and the denominator, every term we see is divisible by x and every term is divisible by seven. So it looks like we can factor seven x out of the numerator and seven x out of the denominator. So the numerator, we can rewrite as seven x times, we factor seven x out of 14 x squared, we're gonna be left with two x and then you factor seven x out of seven x, you're gonna be left with a one. And one way to think about it is we did the distributive property in reverse, if we were to do it again, seven x times two x is 14 x squared, seven x times one is seven x. All right. And now let's factor seven x out of the denominator. So, 14 x could be rewritten as seven x times two, times two, and remember, I wanna keep this algebraically equivalent, so I wanna keep the constraint that x cannot be equal to zero. And so we divide the numerator and the denominator by seven x or one way to think about it, you can divide seven x by seven x and just get one and we are left with two x plus one over two. Now this was the original expression, x could take on any value, but if we want it to be algebraically equivalent to our original expression, it has to have the same constraints on. So we're going to have to say x, x does not equal zero, and it's a very subtle but really important thing. For example, if you defined a function by this right over here, the domain of the function could not include zero. And so, if you simplified how you defined that function to this, if you want that function to be the same, it needs to have the same domain. It has to be defined for the same inputs. And so that's why we're putting the exact same constraints for them to be equivalent. If you got rid of this constraint, these two would be equivalent everywhere except for x equal zero. This one would have been defined for x equal zero, this one would not have, and so they wouldn't have been algebraically equivalent. This makes them algebraically equivalent. And of course you can write this in different ways. You can divide each of these terms by two, if you like. So you could divide two x by two and get x and then divide one by two and get 1/2. Once again, we would wanna keep the x cannot be equal to zero. Let's do another one of these. So it's a slightly hairier expression, but let's do the same drill. See if we could simplify it. But as we simplify it, we really wanna be conscientious of restricting the Zs here so that we get an algebraically equivalent expression. So, let's think about where this is undefined. We can think about where it's undefined by factoring the denominator here. Actually, let me just... So, this is going to be equal to, actually, let me just... Let me just do the first step where I could say, \"Well, what's a common factor \"in the numerator and the denominator?\" Every term here is divided by z squared and every term is also divided by 17, so it looks like 17 z squared can be factored out. So, 17 z squared can be factored out of the numerator and then we would be left with, we factor out of 17 z squared out of 17 z to the third and we're gonna be left with just a z. 17 z squared, you factor 17 z squared out of that and you're gonna be left with just a one. And once again, you can distribute the 17 z squared, multiply it times z, you get 17 z to the third. 17 z squared times one is 17 z squared. All right. All of that is going to be over, we wanna factor out of 17 z squared out of the denominator, 17 z squared times, and so 34 z to the third divided by 17 z squared, 34 divided by 17 is two, and z to the third divided by z squared is z and then we have minus 51 divided 17 is three and z squared divided by z squared is just one. So we'll just leave it like that. And so over here, it becomes a little bit clear of, well one, how we're gonna simplify it. We're just gonna divided 17 z squared by 17 z squared, but let's be careful to restrict the domain here. So, we can tell if z is equal to zero, then this, 17 z squared is gonna be equal to zero and we make the denominator equal to zero and we could see that even looking here. So we could say z cannot be equal to zero. Now what else? Z cannot be equal to whatever makes this two z minus three equals zero. So let's think about what makes two z minus three equal to zero. Two z minus three is equal to zero. You can add three to both sides and you would get two z is equal to three. Divide both sides by two and you would get z is equal to 3/2. So z cannot be equal to zero and z cannot be equal to 3/2. So that's how we're gonna restrict our domain. But now let's simplify. So, if we simplify it, these two cancel out and we are going to be left with, we're gonna be left with z plus one over two z minus three and we wanna keep that constraint. z cannot be equal to zero. And actually, this second constraint is redundant because we still have the two z minus three here. If someone were to just look at this expression, like, well, the denominator can't be equal to zero and so, z cannot be equal to 3/2. So this is still, if we just left it the way it is, we don't even have to rewrite this, that would be redundant since looking at this expression is clearly not defined for z equals 3/2. So, there you go. And so I'm gonna ask you for what values does this expression not defined, well then you would also include that. Actually, let me just write it. It doesn't hurt to be redundant. z does not equal 3/2. But this constraint right here is really important because it's not obvious by looking at this expression. This expression by itself would be defined for z equals zero, but if we want it to be algebraically equivalent to this one and that one, it has to be constrained in the same way." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.958932,"math_prob":0.99381304,"size":6607,"snap":"2019-13-2019-22","text_gpt3_token_len":1636,"char_repetition_ratio":0.13327275,"word_repetition_ratio":0.058551617,"special_character_ratio":0.24004844,"punctuation_ratio":0.11359026,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99946195,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-24T09:01:29Z\",\"WARC-Record-ID\":\"<urn:uuid:0a4a8d4a-427f-41f8-8dc7-23346779b02a>\",\"Content-Length\":\"171590\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:72fd671f-3eea-4d83-8932-fed613453e4e>\",\"WARC-Concurrent-To\":\"<urn:uuid:d07d2eb2-c163-4ea1-8074-10d08d1d40b5>\",\"WARC-IP-Address\":\"172.217.164.147\",\"WARC-Target-URI\":\"https://www.khanacademy.org/math/algebra2/rational-expressions-equations-and-functions/simplify-rational-expressions/v/simplifying-rational-expressions-w-common-monomial-factors\",\"WARC-Payload-Digest\":\"sha1:MKIB4P7QNXJ3RHXDVNUF6RYV2QMJSH5R\",\"WARC-Block-Digest\":\"sha1:IKKQBBU7FZOB43BIQ56O5ZNASXALXTNE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232257601.8_warc_CC-MAIN-20190524084432-20190524110432-00189.warc.gz\"}"}
https://www.tutorialathome.in/wbut-mca-sem-i-c-language/sum-digit-until-single-digits-2011	[ "Sum of its Individual Digit Until Single Digits - WBUT MCA 2011\n18-11-2019 170 times", null, "### Write a program to accept a number and find sum of its individual digits repeatedly till the result is a single digit. [WBUT MCA 2011]\n\n```#include <stdio.h>\nint main()\n{\nint no,i,sum;\nprintf(\"Enter A Number : \");\nscanf(\"%d\",&no);\ndo{\nsum=0;\nwhile(no>0)\n{\nsum += no%10;\nno = no/10;\n}\nno=sum;\n}while(sum>9);\nprintf(\"One Digit Sum is %d\",sum);\n}\n```", null, "" ]	[ null, "https://www.tutorialathome.in/postimg/2019/11/sum-of-digit-upto-single-digit.jpg", null, "https://www.tutorialathome.in/wbut-mca-sem-i-c-language/adsimg/internship.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.57362694,"math_prob":0.96078146,"size":424,"snap":"2020-24-2020-29","text_gpt3_token_len":133,"char_repetition_ratio":0.0952381,"word_repetition_ratio":0.0,"special_character_ratio":0.370283,"punctuation_ratio":0.16494845,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9539497,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-15T02:59:39Z\",\"WARC-Record-ID\":\"<urn:uuid:5b12fdb3-f896-420d-989a-b4fff52802b7>\",\"Content-Length\":\"66236\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:57505275-e6ae-4e1e-a546-cdc68376fa17>\",\"WARC-Concurrent-To\":\"<urn:uuid:be012def-11ab-43b6-ae69-78353179f95a>\",\"WARC-IP-Address\":\"103.118.16.62\",\"WARC-Target-URI\":\"https://www.tutorialathome.in/wbut-mca-sem-i-c-language/sum-digit-until-single-digits-2011\",\"WARC-Payload-Digest\":\"sha1:QKB343JNR5HOSG2U3K5ZPFWHGBZIFH5X\",\"WARC-Block-Digest\":\"sha1:ZYSW2XZYYZHXKI26RQG7U2LFDFWUK4PG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593657154789.95_warc_CC-MAIN-20200715003838-20200715033838-00350.warc.gz\"}"}
http://www.neverendingbooks.org/noncommutative-geometry-2	[ "", null, "Again I\nspend the whole morning preparing my talks for tomorrow in the master\nclass. Here is an outline of what I will cover :\n– examples of\nnoncommutative points and curves. Grothendieck’s characterization of\ncommutative regular algebras by the lifting property and a proof that\nthis lifting property in the category alg of all l-algebras is\nequivalent to being a noncommutative curve (using the construction of a\ngeneric square-zero extension).\n– definition of the affine\nscheme rep(n,A) of all n-dimensional representations (as always,\nl is still arbitrary) and a proof that these schemes are smooth\nusing the universal property of k(rep(n,A)) (via generic\nmatrices).\n– whereas rep(n,A) is smooth it is in general\na disjoint union of its irreducible components and one can use the\nsum-map to define a semigroup structure on these components when\nl is algebraically closed. I’ll give some examples of this\nsemigroup and outline how the construction can be extended over\narbitrary basefields (via a cocommutative coalgebra).\n\ndefinition of the Euler-form on rep A, all finite dimensional\nrepresentations. Outline of the main steps involved in showing that the\nEuler-form defines a bilinear form on the connected component semigroup\nwhen l is algebraically closed (using Jordan-Holder sequences and\nupper-semicontinuity results).\n\nAfter tomorrow’s\nlectures I hope you are prepared for the mini-course by Markus Reineke on non-commutative Hilbert schemes\nnext week.\n\nPublished in featured" ]	[ null, "http://www.neverendingbooks.org/DATA/nog.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8450646,"math_prob":0.73664194,"size":1473,"snap":"2022-40-2023-06","text_gpt3_token_len":332,"char_repetition_ratio":0.09938734,"word_repetition_ratio":0.0,"special_character_ratio":0.18669382,"punctuation_ratio":0.05859375,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9548606,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-01T16:00:59Z\",\"WARC-Record-ID\":\"<urn:uuid:ef3a9b73-2721-4488-9587-6928070e2abd>\",\"Content-Length\":\"40588\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1f3fb795-99fe-43a5-8766-d7471f41ec46>\",\"WARC-Concurrent-To\":\"<urn:uuid:8d54bbb6-499f-464b-9f67-4a74e295f4f6>\",\"WARC-IP-Address\":\"143.129.73.7\",\"WARC-Target-URI\":\"http://www.neverendingbooks.org/noncommutative-geometry-2\",\"WARC-Payload-Digest\":\"sha1:H3RBNHWI2ZV6AQCOWNMI2VB2KABI4SAW\",\"WARC-Block-Digest\":\"sha1:HAXTOL3XEVFSBNFRCRALUKAID4TIPNUT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499946.80_warc_CC-MAIN-20230201144459-20230201174459-00509.warc.gz\"}"}
http://davidson16807.github.io/mental-slide-rule/slide.rule.html	[ "# Visualizing Math with a Mental Slide Rule\n\nSlide rules are really neat.\n\nThey give you a great intuitive understanding of mathematics. If you never used them before, they're easy enough. Here's a primer. If you haven't noticed, you can perform addition and subtraction using two ordinary rulers. There's obviously nothing magical here: you just mete out two distances, stacking them end to end, then measure their sum.\n\nDrag the bottom ruler\n\nSlide rules work the same way, only with multiplication. This is made possible because slide rules place numbers on a logarithmic scale. An ordinary ruler just uses a linear scale. Slide rules work on the principle that log(ab) = log(a) + log(b).\n\nDrag the bottom ruler\n\nSlide rules aren't normally used for mental math. This is because the markers on a slide rule indicate linear increments. Line spacing always changes, so unless you have a photographic memory, it is very difficult to visualize the addition of two distances.\n\nIf line spacing were constant though, we could multiply just by visualizing the slide rule.\n\nIt is possible to create a slide rule where line spacing is constant, but our lines could no longer represent linear increments. They would instead represent logarithmic increments. Every increment would represent a multiplication by some constant number.\n\nLet's try to design this sort of slide rule. There are many values we could assign to its increments, but there are constraints we should consider. The scale must approximate the form a^(1//n), a^(2//n), … a^(n//n). This is necessary to keep line spacing constant. Here, a is the base of our log scale. For us humans, it makes sense to set a = 10, since anything else will reduce the usability of our slide rule. If a != 10, our slide rule would not repeat once at every power of 10 and we'd likely have to convert number bases before calculations.\n\nWith these constraints in mind, we can browse from a pool of candidate log scales. Any viable log scale is of the form 10^(1//n), 10^(2//n), … 10^(n//n), where n is the number of increments on the scale. n is also the number of increments we'll need to memorize, so n should be small enough to remember but large enough to offer precision.\n\nWhen school children memorize their multiplication tables, they generally memorize no more than 32 combinations, so we should be able to start simple, with n<32. Unlike the school children, though, every number we memorize goes further. Multiplication tables have a space complexity of O(N^2), where N is the number of digits that can be multiplied. That means if we want to add just one extra number to our multiplication table, the number of combinations we have to memorize increases exponentially. Not only that, but the multiplication table only works well for multiplication. On the other hand, our slide rule only has a space complexity of O(N). That means if we want to add one extra number to our ruler, we only have to remember one extra value. And as we will see, memorizing that value will allow us to do way more than multiplication.\n\nIt turns out n = 10 is a rather good start, and if you want to learn more you can set n to some other multiple of 10.\n\nDrag the bottom ruler\n\nQuite a few increments on this scale are conveniently close to whole numbers, those being 2, 4, 5, and 8. This probably has to do with the fact 10 is divisible by 2 and 5, but this is just a guess. Other numbers resemble rational fractions of 10, those being 1.25 and 2.5. There are 3 increments on this scale that correspond to neither whole numbers nor rational fractions, but we find these numbers are very close to pi and its multiples, ¹/₂pi and 2 pi. It's fortunate that multiplication by pi is so common that most normal slide rules already come with a special notch for pi. Using n = 10 has an added benefit: it leverages our existing knowledge of addition in base 10.\n\nBy visualizing the slide rule, we can divide and multiply just as easily as we add and subtract.\n\nBecause there are 10 numbers on our scale, our answers will be accurate to within ~10% the actual answer, assuming we round correctly. We can achieve greater accuracy by memorizing additional numbers.\n\nWe note that multiplication will be just as easy as single digit addition, but in fact it may be even easier. This is because there are additional mneumonics that help us visualize the scale. We will go through some mneumonics now.\n\nThe scale repeats at every power of 10, just as with arabic numerals, so we can visualize the scale as a sort of clock. Just like a clock, the numbers \"roll over\" if they go above or below the limits of the scale - 2 hours past 11:00 is 1:00.\n\nDrag the inner wheel\n\nThe clock is divided into two halves. The halves are marked by two values. The first value is 1, which is multiplicative unity. The second is sqrt(10), which is close to pi. There are four \"steps\" within each half. There are a total of 10 steps. The halves have a symmetry about them, so that the values on one half bear the same mantissa as the inverse of values on the other half. On the left half, we have 4, 5, 2 pi, and 8. On the right half, we have the mantissae of ¹/₄, ¹/₅ (or 2), ¹/₂pi (very close to pi/₂), and ¹/₈.\n\nIf we want to multiply two numbers, we count distance clockwise from the first number. The distance is determined by the second number - how far away that second number is from 1 (that is, the start of the scale)\n\nWhat's 158 2?\n\n1.58 (¹/₂pi) is the 2nd step from 1, so we find 2 on the clock and count 2 steps clockwise: 2, 2.5, pi. The mantissa is near the same as pi's, so the answer is close to 314.\n\nYou can also go the other way around: start at 1.58 and count 3 steps clockwise\nI find it's easier to visualize smaller distances, so in this case I start from 2 and count 2 steps clockwise.\nWhat's 158 * 8?\n\n1.58 (¹/₂pi) is the 2nd step from 1. 8 is the 9th step on the scale. Therefore, we find 2 on the clock and count 8 steps clockwise. The mantissa is about 1.25, so the answer is close to 1250.\n\nThere's a better way to visualize it though. Short distances are easier to visualize than large distances. 8 is much closer to 1 in the counter-clockwise direction, so it's easier to start at 1.58 and look 1 step counter-clockwise.\n\nIf we want to divide two numbers, we do the same thing, only counter-clockwise.\n\nWhat's 1.58 // 2?\n\nTake ¹/₂pi and count 3 steps counter-clockwise: ¹/₂pi, ¹/₈, 1, 8. The mantissa is 8, so the answer is close to 0.8. The answer is infact 0.79.\n\nSince these numbers are close together, an easier way is to count the steps from 1.58 to 2. The two numbers are seperated only by a single step, so the answer must be 1 step counter-clockwise from 1.\nWhat's 2 // 1.58?\n\nOur answer here is merely the inverse of the last. The two numbers are seperated only by a single step, so the answer must be 1 step clockwise from 1. This is a general rule of thumb: if the denominator is larger, count the steps going down; if the denominator is smaller, count the steps going up.\n\nWhat's 1.58 // 8?\n\nJust as with multiplying by 8, there are 2 ways to go about division. There are 9 clockwise steps between 1 and 8, and that means to divide we can count 9 steps counter-clockwise.\n\nBut if the number is greater than sqrt(10), it make sense to consider the other direction. There's only 1 counter-clockwise step between 1 and 8, and for division that means we can count 1 step clockwise.\n\nTo divide by large mantissae, it's often easier to multiply by the inverse of the number. Finding the inverse is easy - it's just the value on the opposite side of the clock. Let's divide 1.58 by 8. ¹/₈ is the first step on the clock, so the answer is one step clockwise from ¹/₂pi. The mantissa is 2. The answer is near 0.2. The actual answer is 0.19625.\n\nWhat do we do if a number is between steps?\n\nThere's a few things we can do in that case. We can of course memorize additional numbers on the scale, but memory only takes us so far. Oftentimes, the best solution is interpolation - we estimate where the number lies between steps, then tack that on to an approximation that uses one of the two nearest values. If an answer is between steps, we interpolate between the two nearest values.\n\nA more precise solution is to find the correction. Finding the correction in the log scale is just the same as finding the correction in arabic numerals - take the difference between the number and the approximation, multiply it seperately, then tack it on to the rest of your answer. You should be able to get an accurate number much faster than you would by multiplying with arabic numerals, though often you'll find the first order approximations are good enough.\n\nThis all sounds like stuff we can solve with other mental math. Oftentimes, I still find it easier to calculate in more traditional ways, particularly when dealing with simple whole numbers. However, the method does help us to make intuitive sense of the calculations. It also helps us focus on what's most important when calculating problems in the wild: first get the order of magnitude right, then get a good approximation for significant figures. Only then do we refine as needed. In lots of everyday problems, the order of magnitude will just \"feel\" right, so only the mantissa needs to be considered.\n\n## There are other things we can do with this visualization, too.\n\nFor example, what's log_10(1.58)?\n\nEasy - 0.2. ¹/₂pi is the 2nd step, and each of the 10 steps indicate a log_10 increment of 0.1. This is due to the way we designed our scale - I tricked you into learning the log_10 table.\n\nThis brings us to our next point:\n\nVisualizing the slide rule is very similar to memorizing the log_10 table\n\nBoth are equally valid strategies. The difference lies with the part of the brain your using. If I memorized the log_10 table and was asked to multiply two numbers, I'd map the two numbers to their log_10 values, add the log_10 values together, then map the sum back to get the answer. With the log_10 table, I'm using rote memorization along with existing skillsets in addition. If I memorized the slide rule, I'd find the two numbers on the slide rule, visualize the addition, and estimate the order of magnitude. There's still rote memorization to a certain extent, but now I'm also using more spatial reasoning, intuition, and visual memory.\n\nThe good news is this: if you visualize the slide rule, the log_10 values come easy and you can still fall back on using the log_10 table if you find it useful. I find using the log_10 table useful for calculating the powers and roots of large numbers.\n\nWhat's log_10(158)?\n\nEasy - 2.2. It's the same mantissa, but now it's on the order of 10^2, so it's 0.2 + 2 = 2.2.\n\nWhat's the square of 1.58?\n\nWe know that's the 2nd step, so the answer must be double the distance, on the 4th step. The answer is near 2.5. The exact answer to 4 figures is 2.465.\n\nWhat's the square root of 1.58?\n\nAgain, we know that's the 2nd step, so the answer is half that, on the 1st step. The answer is near 1.25. The exact answer to 4 figures is 1.253.\n\nWhat's the cube of 1.58?\n\nThat's on the 6th step, just a step past pi, near 4, and in fact it is 3.87.\n\nThe fourth power of 1.58?\n\nThat's on the 8th step, 2 steps shy of 10, near 2 pi, and in fact it is 6.076.\n\nThe fifth power?\n\nNear 10, and in fact it is 9.54.\n\nWe start to notice our answers become less accurate at higher powers. This is because the minor inaccuracies we started out with are amplified by repeated multiplication. This is a concern common to all slide rules.\n\nOkay, we've shown what this can do. Now we want greater resolution. There are a number of integers missing on the scale.\n\nWhat if I want to multiply by 9?\n\nWell, 9 is the squareroot of 81. Look at the slide rule and we know log_10(81) ~ 1.9, so log_10(9) ~~ 1.9/2, or 0.95. To be exact, it is 0.954. It's just over half past the last step.\n\nWhat if I want to multiply by 7?\n\nWell, 7 is the squareroot of 49. We know log_10(49) ~~ 1.7, so log_10(7) ~~ 0.85. It is in fact 0.845. There's another way to figure out: 72 = 89, so 7 is half step shy of 8.\n\nWhat if I want to multiply by 3?\n\nWe know 3^2 = 9, and log_10(9) ~~ 0.95, so log_10(3) ~~ 0.475. That's quarter step shy of a half circle. It is in fact 0.477.\n\n6 is merely 23, so log_10(6) ~~ log_10(2)+log_10(3) ~~ 0.3 + 0.475 ~~ 0.775. 6 is quarter step shy of 2pi.\n\nIf we find the answers to the questions above, we will immediately know log_10 of ¹/₉, ¹/₇, ¹/₃, and ¹/₆ - we need only look at the opposite side of the clock!\n\nIt's particularly useful knowing positions for ¹/₉. We know ¹/₉ = 1.111…, so 1/2 step from any whole number is approximately that number repeated. Let's look at the whole numbers.\n\n 10^0.0 ~~ 1, so 10^0.05 ~~ 1.111… ~~ ¹/₉ 10^0.3 ~~ 2, so 10^0.35 ~~ 2.222… 10^0.475 ~~3, so 10^0.525 ~~ 3.333… ~~ ¹/₃ 10^0.6 ~~ 4, so 10^0.65 ~~ 4.444… 10^0.7 ~~ 5, so 10^0.75 ~~ 5.555… 10^0.775 ~~6, so 10^0.825 ~~ 6.666… ~~ ²/₃ 10^0.85 ~~ 7, so 10^0.9 ~~ 7.777… ~~ 8 10^0.9 ~~ 8, so 10^0.95 ~~ 8.888… ~~ 9 10^0.95 ~~ 9, so 10^1.0 ~~ 9.999… = 10 (by definition)\n\nThe actual numbers will be given shortly, but the above explanation is a useful mneumonic for memorizing the precise values of half steps. Once you memorize the precise values you can improve your accuracy to 6%.\n\nYou know what? Knowing the position for ¹/₉ is so useful, why don't we take it further? If we memorize all the quarter steps from 1 to 1.58, the values for all the other powers of two will fall into place. Once we memorize all the quarter steps, our accuracy will improve to 3%.\n\n1 2 4 8\n1.0 2.1 4.2 8.41\n1.1 2.2 4.4 8.91\n1.1 2.3 4.7 9.44\n1.2 2.5 5 1\n1.3 2.6 5.31\n1.4 2.8 5.62 1.41 is the squareroot of 2\n1.5 3 5.96\n1.58 3.16 6.31 all these are near multiples of pi\n\nFor the rest of the numbers, I use seperate mneumonics. They are as follows:\n\nThe quarter steps past 1pi and 2pi are all at or near noteworthy whole numbers and fractions:\n\n 3.35 6.68 around multiples of ¹/₃ 3.54 7.08 around multiples of ⁷/₂ 3.76 7.5 around multiples of ³/₈\n\nThe quarter steps past 1.58 all end in 8. The 10th's place is sequential:\n\n 1.58 1.68 1.78 1.88\n\nThe scale appears as follows:\n\nWhen I first memorized the scale, I was driving alone on a long car ride. There wasn't a lot I could do that was productive without becoming distacted, so I'd recite the scale to myself, occasionally referring to the copy of the slide rule above. By the end of the trip, I could recite the thing forwards and backwards. I was also able to recite the inverse of every quarter step, the log of every quarter step, and the multiples of every quarter step from 1 to 1.25. I recommend doing something similar for anyone interested in learning the system.\n\n## With greater precision, we find we can solve more complex problems.\n\nWhat's the double log of 158?\n\nFrom our previous answer we found log_10(158) = 2.2. The answer here is log_10(2.2), which is about 3.5. The exact answer to 4 digits is 0.342.\n\nWhat's log_2(158)?\n\nWe know log_10(2) = 0.3 and log_10(158) = 2.2. Find a half circle past 2 and the answer is a quarter step past that. It's near 7.25. It is in fact 7.304.\n\nBut there's an easier way! We know 2^3 = 8 and 2^4 = 16, so 10 is somewhere between them. Every time we multiply by 2, we take 3 steps on the scale, and 8 is 1 step from 10, so 2^3.33 = 10. 158 is close to 1016, so the log must be very to 3.33+4, or 7.33.\n\nBut that's not enough. What can we do to visualize log_2? There is in fact an easy way to do so: the distances on any log scale represent the exact same thing, but the units we use to measure distance will differ. We can think of the log base as this unit of measure. On our base 10 scale, we indicate distance relative to the number \"10\", and due to the design of our clock, this is the distance covered by one full turn. On a base 2 scale, we indicate distance relative to 2, which is the third step on our clock. This means the \"1\" appears on the third step clockwise from the top, and everything else is scaled according to that.\n\nWhat's the natural log of 158?\n\nFirst we have to find log_10(e).\n\nWe'll do the same thing we did for log_2. e is between the 4th and 5th step, so log_10(e) is 0.45. To convert the log base we know we must divide by log_10(e).\n\nWe know log_10(158) = 2.2, so ln(158) is just 2.2 // 4.5, which is 3.5 steps clockwise of 3.5 steps, 2 steps past pi, or 5. When we run it through a calculator, we find the exact answer to 4 figures is 5.056 - not too shabby.\n\nNatural log is so common it might serve us just to remember that number - if we want the natural log of a number, just take log_10 and find the number 3.5 steps clockwise from that (well, 3.6 if you must be exact).\n\nHowever, as with log₂, there is another way. If we remember ln(10) = 2.3, then we know 158 is 22.3 + ln(1.58). 1.58 is on the 2nd step, and 2.71 maps to ~4.5, so what's 2//4.5? 2 is 2 steps from pi, so 2//pi is 2 steps from 1, and 2//4.5 is 1.5 steps past that, around 4.44. 22.3 + 0.444 = 5.0.\n\nWhat's e^2.5?\n\ne is between 4.25 steps and 4.5 steps. We'll interpolate that to 4.4 steps. That means e^2.5 is near 2.54.4 steps. Ok, so what's 2.54.4? 4.4 is a 1.5 steps past pi. 2.5 is 1 step short of pi, so 2.5pi is 1 step short of 1, and our answer is just 1.5 steps past that, on the first half step, or 1.12. So what's on the 1.12th step? That's around 1.25. Our intuition tells us it's on the order of ten, so we guess 12.5. The exact answer is 12.18\n\nWhat's 1.58 tetrated by 2, i.e. 1.58^1.58?\n\nThis is where our \"intuitive understanding\" comes in handy. What happens when you calculate an exponent like 1.58^2?\n\nWe find the distance covered by 1.58 (finding the logarithm), add two of those distances together (finding the sum), then answer the number that corresponds to that distance (finding the power of 10). In other words, we understand that all the following are equivalent:\n\n log_10(1.58^2) log_10(1.58 * 1.58) log_10(1.58) + log_10(1.58) 2log_10(1.58)\n\nAs a result, we know log_10(1.58^1.58) = 1.58log_10(1.58). We are asked to find 1.58^1.58, which we know is 10^(1.58 log_10(1.58)).\n\nThe exponent is close to 1.58 * 0.2, or 0.316. We know 10^0.316 is a little over halfway between 2 and 2.12, so we interpolate between the two and answer 2.06. As it turns out, that's only off by 1 out of 20000.\n\n## We also find ourselves with better math reasoning skills.\n\nWhat's 40 rounded to the nearest order of magnitude?\n\nWe normally round on a linear scale. On a linear scale, we round up if the number is 5 or greater; we round down if otherwise. This is because 5 is halfway between 0 and 10 on a linear scale. But what if we're doing a rough order of magnitude calculation? The halfway mark between 0 and 10 isn't 5 on a log scale.\n\nThe halfway mark on our scale is closer to pi (3.16 to be precise). We therefore know to round up for numbers bigger than 3.16, and round down for numbers less than 3.16. 40 is therefore closer to the order of 100 than the order of 10.\n\nI have a set of numbers from 0 to 10: 0.2, 7.8, 6.9, 9.8, 5.4, 8.7, 2.7, 7.5, 7.1, 8.5. Are these more likely to be uniformly distributed or log-uniformly distributed?\n\nWhat would happen if we place the mantissa for those numbers on our clock? Most of numbers fit on the left side of the clock. This indicates the set is more likely a log distribution.\n\nWhat about another set? 5.7, 1.9, 4.2, 3.4, 7.1, 2.0, 1.2, 3.8, 1.0, 4.9\n\nSome of the numbers appear to the left of the clock, but it's definitely not so skewed to that side. This suggests it's more likely a log-based distribution.\n\nWhat's the frequency of the musical note, \"Middle E\"?\n\nMiddle C has a frequency of 261 Hz. E is 4 notes from C. An octave represents a doubling in frequency, so an octave higher than C is 2612 = 522 Hz. There's 12 notes in the octave including flats and sharps. The frequency of these notes are all evenly distributed across the log scale.\n\nThis means the frequency of every note corresponds to a multiplication by 2^(1//12). How many quarter steps are there between 1 and 2?\n\nThat's right, I tricked you into learning musical ratios, too. Every note represents a multiplication by one quarter step. E is 4 notes from C, so we look 4 quarter steps from 2.6: 3.3. We answer 330 Hz. Wikipedia tells me the correct answer is 329.6 Hz.\n\nThere are 8 full notes in an octave, so that means every full note has 12/8 = 1.5 quarter steps.\n\n## And we find we can solve real world problems.\n\n### Calculating arrival times\n\nLet's say you're driving on a highway at 65mph. The road sign says there's 50 miles left to your destination. At what time do you expect to arrive?\n\nMost people just round their speed to 60mph. That way, they know they travel about a mile a minute, so a 50 mile journey might take 50 minutes. That's a good start, but we know we can do better. 65 is a little less than 6.66 on the clock, which is a half step past 6. A half step shy of 5 is 4.47, so we know that's the mantissa for our answer. At this point, you can intuit that there's 45 minutes left to your journey, since that's the only order of magnitude that makes sense.\n\nOn further reflection you find 65 is almost halfway between the two nearest quarter steps: 6.30 and 6.66. You find the answer must then lie between 45 and 47. You interpolate and answer 46 minutes, which is in fact correct.\n\nUsing the mental slide rule is also much more versatile, since the method works with other speeds too. What if we were going 70mph?\n\n7.08 is 3 quarter steps past 6, so the answer is 3 quarter steps shy of 5: 4.22. There's around 42 minutes left to your journey.\n\nYour check comes to $8.47. You think 17% is reasonable for a tip. Your check is three quarters shy of 1. The tip therefore is three quarters shy of 17%, which is a quarter step past ¹/₂pi. The tip is a half step shy of ¹/₂pi, or$1.41. Actual answer is $1.44. ### Converting inches to mm You only have a imperial ruler. An object measures ³/₈ inch. What's that in mm?. A common rule of thumb is that there's 25.4 millimeters in an inch, around ¹/₄. 3/(84) = 3/32. The difference between 3 and 32 is around a quarter step, so the answer is a quarter step counter-clockwise of 1: 9.44. Our intuition tells us that's also the correct order of magnitude, so 9.44 millimeters it is. This is probably good enough for most people. Want the correction? The correction is ³/₈ * 0.4, which is the same mantissa as 4³/₈ or 12/8. The mantissa is a step behind three quarters, or a quarter shy of 1 - 9.44. We intuit the correction is 0.0944, so a closer answer is 9.44+0.0944, or 9.53. The correct answer is 9.525. ### Calculating sales tax$3.50 for milk - what's the total with tax?\nLet's say sales tax is 106%.\n\nEvery quarter step on the log scale represents a multiplication by 106%, so sales tax is super easy. Just look a quarter step past 3.50. It's around \\$3.75 with tax.\n\n### Trigonometry\n\nWhat's the width of a 19\" 4:3 monitor?\n\nThink back to the pythagorean theorem. What's the width:diagonal ratio on a 4:3 monitor?\n\nThe diagonal has a square of 16+9 = 25, so without even consulting our mental slide rule we know the ratio is 4:5, or 0.8.\n\n4:5 is a step shy of 1. 1.9 is a quarter shy of 2, so 1.90.8 is a quarter shy of 1.58 - 1.5. 15\" inches is our answer. The exact answer is 15.2.\n\nWhat's the height?\n\nThe ratio is 3:5, now. That's a 2 and a quarter shy of 1. 2.25 shy of 19 is 2.5 shy of 3 steps, or a half step past 1 - 1.12. 11.2\" is our answer. The exact answer is 11.4\".\n\nWhat's the angle of the diagonal?\n\nIf we normalize the width we have ⁴/₅ or 0.8, so we want to find acos(0.8).\n\nAt times like this I'd like to remind you of something:\n\nyou don't have to get there, you just need to get closer.\n\nYou won't need the taylor series to solve this one, but you will have to think for yourself.\n\nWe know cosine starts at 1 and goes to 0 once it reaches 90°. It repeats after that, so we only care how it looks from -90° to +90°. We want a function that resembles this.\n\nIt kind of resembles a parabola, like y=1-x^2 where x is measured as a fraction of a right angle.\n\nWe know 0.8 = 1-x^2, so 1-0.8 = x^2 = 0.2. We need the squareroot of 0.2.\n\n0.2 is 7 steps backwards from 1, so our answer lies 3.5 steps backwards from 1. 0.44 is our answer expressed as a fraction of a right angle. We multiply by 90 to get the degrees - this is a half step shy of 4.4, so we answer 40°. The exact answer is 39.9°. Not too shabby.\n\nObviously, y=1-x^2 is not a perfect estimate. You could remember the taylor series, but that takes way more figuring. For mental math, it's much more effective to remember a better fitting exponent. After some trial and error, I've found y=1-x^1.75 provides much more accurate results.\n\nIf you need a more concise mental algorithm, it is as follows. To find cos(theta), where theta is an angle expressed in degrees, you must perform the following:\n\n• find the conterclockwise distance from 1 to theta (find 1-log_10(theta))\n• subtract 0.5 from the distance (x=theta/90)\n• double the distance and subtract ⅒ of it (x^1.8)\n• find the number that appears at that distance and subtract it from 1 (1-x^1.8)\n\nHow to find 1-cos(40°)\n\nWhat's e^(ipi/5)?\n\nRemember euler's theorem: e^(theta i) = cos(theta) + sin(theta)i. It's OK to write down your answers.\n\nRemember: sine is just cosine rotated 90°.\n\npi/5 is 2/5 of a right angle. we need cos and sin of that.\n\n2/5 is 0.4. That's 4 steps counter-clockwise from 1, so 0.4^1.75 is 7 steps counter-clockwise from 1, or 0.2. 1-0.2=0.8, so that's cosine, AKA the real component of the complex number.\n\nFor sine, we just do the same thing for 1-0.4, or 0.6. 0.6 is 2.25 steps counter-clockwise, so 0.6^1.75 is ~4 steps counter-clockwise, or 4. 1-0.4=0.6, so that's the imaginary component.\n\nOur guess is 0.8+0.6i. We run it through the computer and find the correct answer is 0.81+0.59i. Noice.\n\nAnd now, your final exam: I have an object at coordinate (3.3, 2.8) and I rotate it pi//5 radians clockwise. Where is it located now?\n\nWe know from the previous question the rotation is (3.3+2.8i)(0.81+0.59i). It's OK to write down your answers.\n\nThe neat thing with slide rules is you can set them to multiply by a certain number and from then on you know at a glance what all the answers are for when multiplying by that number.\n\nStart working with the numbers that are closest to 1 on the scale, since these are the easiest when visualizing multiplication and division.\n\n0.81 is 1 step counter-clockwise, so look 1 step counter-clockwise from 3.3 and 2.8: we have 2.66 and 2.2i.\n\n0.59 is 2.25 steps counter-clockwise, so look that distance from 3.3 and 2.8: we have 2.0i and 1.68ii.\n\nThe answer is 2.66 + 2.2i + 2.0i - 1.68, so the answer is 1 + 4.2i.\n\nThe exact answer is 1.02 + 4.20i.\n\n## Beyond quarter steps.\n\nMake sure you start small. Start by remembering the half steps. Recite the scale forwards and backwards. Try your hand at multiplying, dividing, and exponentiating random numbers. I personally train with a scripting language like python or R, but a couple d10 or d20 dice work, too.\n\nOnce you have the basics down, you can memorize smaller increments to achieve greater accuracy.\n\nFirst, you have to decide which increment you want to memorize. When you're memorizing a scale, it's important to start with an increment that accomplishes two things: 1.) it's easy to remember the scale, and 2.) it works well in base 10. A multiple like 1.1 is a great example. It's easy to calculate multiples of 1.1. For whole numbers, you just repeat the number twice, so for example 3 * 1.1 = 3.3. For multiple digits, the first and last digit stay the same, and every other number gets added to it's neighbor, so for example 1.25 * 1.1 is:\n\n{:(\\ \\ \\ 1250),(+ 0125),(_),(\\ \\ \\ 1375):}\n\nSo if we're memorizing the scale and we forget what an increment on our scale is, we can just multiply by some adjacent number that we do know. This won't excuse you from memorizing the thing, but it can help you while you memorize it.\n\nWe know log_10(1.1) = 0.4, so we can easily memorize a scale where every log increment is 0.2 since this would mean memorizing every fifth of a step. Since log_10(1.05) = 0.2, we know ever fifth step will represent a multiplication by 1.05.\n\nMemorizing the first few steps of any scale will be critical, because any other step in the scale can be derived by multiplying one of the first steps and some whole number. Let's take a look at the first few fifth step increments\n\n 1.047 1.096 1.148 1.202 1.259\n\nWhat sense can we make of them? Well, the trailing numbers are all very close to multiples of 0.05.\n\n 1.047 ~~ 1 + 0.05 * 1 1.096 ~~ 1 + 0.05 * 2 1.148 ~~ 1 + 0.05 * 3 1.202 ~~ 1 + 0.05 * 4 1.259 ~~ 1 + 0.05 * 5\n\nIn other words, it's almost like we're adding 0.05 to each number. But that's not completely the case, because 1.259 rounds up to 1.26, and that's 1.20 + 0.06. By the time we've reached the first full step, it's like we're adding 0.06. Let's look at the next few fifth steps.\n\n 1.318 ~~ 1.259 + 0.06 1.380 ~~ 1.318 + 0.06 1.445 ~~ 1.380 + 0.06 1.514 ~~ 1.445 + 0.06 1.585 ~~ 1.514 + 0.06\n\nBut that's not completely correct either, because by the time we reach 1.445 it's like we're adding by 0.07, and by the time we reach past 1.58, it's like we're adding by 0.08:\n\n 1.660 ~~ 1.58 + 0.08 1.738 ~~ 1.66 + 0.08 1.820 ~~ 1.74 + 0.08 1.905 ~~ 1.82 + 0.08 1.995 ~~ 1.91 + 0.08\n\nAnd of course the increment keeps increasing. The good news is this: it doesn't matter for our purposes. The error remains in line with the precision we want to accomplish. For us the mneumonic works just fine:\n\n• In the first full step, the fifths steps increment by ~0.05\n• In the second full step, the fifths steps increment by ~0.06\n• In the third full step, the fifths steps increment by ~0.08\n\nNow, let's say you do want to memorize the fifths scale. You memorize the scale, paying particular attention to the steps from 1 to 1.25. As you did with the half steps, you ought to recite the scale and try your hand at some random problems. If you memorized the quarter steps, the fifth steps should come easier - the ¹/₅ and ⁴/₅ steps are only slightly off from the ¹/₄ and ³/₄ steps.\n\nIf we wish to go further, we can even remember every tenth of a step (multiplication by ~1.025). The tenth scale includes both the fifth step scale and the half step scale, so by the time you remember the half and fifth steps, you'll already have at least 6 of the 10 substeps memorized for each full step of the tenths scale.\n\nThe Tenths Scale:\n\n1 2 3 4 5 6 8\n1.025 2.05 3.075 4.1 5.125 6.15 8.2\n1.05 2.1 3.15 4.2 5.25 6.3 8.4\n1.075 2.15 3.225 4.3 5.375 6.45 8.6\n1.1 2.2 3.3 4.4 5.5 6.6 8.8\n1.125 2.25 3.375 4.5 5.625 6.75 9.0\n1.15 2.3 3.45 4.6 5.75 6.9 9.2\n1.175 2.35 3.525 4.7 7.07 9.4\n1.2 2.4 3.6 4.8 7.2 9.6\n1.225 2.45 3.675 4.9 7.43 9.8\n1.26 2.5 3.75 7.6\n1.292 2.57\n1.32 2.63\n1.347 2.69\n1.38 2.75\n1.41 2.82\n1.45 2.88\n1.481 2.95\n1.51\n1.55\n1.58\n1.629\n1.66\n1.70\n1.73\n1.78\n1.82\n1.86\n1.90\n1.95" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92206806,"math_prob":0.9846899,"size":28298,"snap":"2023-14-2023-23","text_gpt3_token_len":8743,"char_repetition_ratio":0.1394642,"word_repetition_ratio":0.025004772,"special_character_ratio":0.33401653,"punctuation_ratio":0.16164424,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9970474,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-09T16:51:23Z\",\"WARC-Record-ID\":\"<urn:uuid:2372ee7c-5903-43b7-98ba-7cbedf5e723f>\",\"Content-Length\":\"57519\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9212c7dd-7aa9-4e8f-974d-7ea6929c3a25>\",\"WARC-Concurrent-To\":\"<urn:uuid:72895e9c-cfe0-4c44-ab4f-0849ec136e3a>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"http://davidson16807.github.io/mental-slide-rule/slide.rule.html\",\"WARC-Payload-Digest\":\"sha1:AJL7WAZ5RVR6DTBEXE75L54PHAXHWQGK\",\"WARC-Block-Digest\":\"sha1:WSEKPNVGXOVBNE5C62CVJTLTBVD47MDU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224656788.77_warc_CC-MAIN-20230609164851-20230609194851-00332.warc.gz\"}"}
https://www.wolfram.com/wolfram-u/courses/finance/financial-time-series-processing-fin027/	[ "", null, "# Financial Time Series Processing\n\nEstimated Time: 18 min\n\nCourse Level: Beginner\n\n## Summary\n\nFinancial value over time can be represented as time series objects in the Wolfram Language. This video class demonstrates how to import financial data as time series and how to inspect and specify its properties, such as temporal regularity, resampling method and window length. It explains how time series objects can be used as input to other built-in functions and also directly used for arithmetic calculations. Examples illustrate computing asset returns using log and ratios and visualizing stock trading data along with buy and sell signals.\n\nFeatured Products & Technologies: Wolfram Language, Wolfram Notebooks\n\n### You'll Learn To\n\n• Construct financial time series with ordered time-value pairs\n• Access financial data from the Wolfram Knowledgebase\n• Alter the sampling frequency and viewing window of a time series object\n• Resample data using interpolation\n• Calculate the relative value of investments" ]	[ null, "https://www.wolfram.com/wolfram-u/wp-content/themes/wolfram-u/assets/img/WolframU.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8317826,"math_prob":0.7534776,"size":1137,"snap":"2023-40-2023-50","text_gpt3_token_len":215,"char_repetition_ratio":0.123565756,"word_repetition_ratio":0.0,"special_character_ratio":0.17502199,"punctuation_ratio":0.057803467,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9591235,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T12:22:42Z\",\"WARC-Record-ID\":\"<urn:uuid:87f17acf-604f-4aa6-8cc2-1badb7354a00>\",\"Content-Length\":\"30019\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:774b7e29-daaa-4570-98b1-ddb852c6e39d>\",\"WARC-Concurrent-To\":\"<urn:uuid:bdfed2e3-085d-4374-90b0-9ceee9e6a5e0>\",\"WARC-IP-Address\":\"140.177.9.134\",\"WARC-Target-URI\":\"https://www.wolfram.com/wolfram-u/courses/finance/financial-time-series-processing-fin027/\",\"WARC-Payload-Digest\":\"sha1:FO6WSQEYHZQAYZRODITVVXMOQU2FEA3G\",\"WARC-Block-Digest\":\"sha1:PJFVM53NSCJWYQO26Q4Y3PQYXAEEQX42\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100529.8_warc_CC-MAIN-20231204115419-20231204145419-00199.warc.gz\"}"}
https://forum.defold.com/t/count-object-rotation-turns/72884	[ "", null, "# Count object rotation turns\n\nThis would be a long post with question in the end.\n\nHi I need to rotate object via player input (mouse for now) and count rotation turns (or spins). Object could be rotated in any direction - clockwise (cw) or counterclockwise (ccw).\n\nFor base for movement I’ve taken this example - britzl/defold-orthographic. Shout out to author, his examples always helps me a lot =)\n\nIn update() function of the hitman/player script I add something like this to count rotation turns when player hold “mouse button 1”:\n\n``````-- [hitman.script](https://github.com/britzl/defold-orthographic/blob/master/example/shared/objects/hitman.script)\nlocal function process_spin(self, angle)\n-- transform angle from rad to degree and normilize to 0 - 360\nangle = angle * 180/3.14159\nif angle < 0 then\nangle = 360 + angle\nend\n\n-- set initial values on grab - when the player click mouse button 1\nif not self.grab then\nself.grab = true\nself.initial_angle = angle\nend\n\n-- get the current angle of the object\nlocal current_angle = angle\n\n-- calculate the difference between the current and initial angles\nlocal angle_diff = current_angle - self.initial_angle\nangle_diff = math.floor(angle_diff)\n\n-- normalize the angle difference to be between -180 and 180 degrees\nif angle_diff > 180 then\nangle_diff = angle_diff - 360\nelseif angle_diff < -180 then\nangle_diff = angle_diff + 360\nend\n\n-- check and set rotation direction: cw or ccw\nif self.rotation_direction == 0 then\nif angle_diff < 0 then\nself.rotation_direction = 1\nelseif angle_diff > 0 then\nself.rotation_direction = -1\nend\nend\n\n-- normilize rotation to 0 - 360 degree\nif angle_diff < 0 then\nangle_diff = 360 + angle_diff\nend\n\n-- count turns\nif (self.rotation_direction == 1 and angle_diff <= 5 and angle_diff > 0)\nor (self.rotation_direction == -1 and angle_diff >= 355) then\nself.turn_count = self.turn_count + 1\nself.initial_angle = angle\nend\nend\n\nfunction update(self, dt)\n...\n-- line 27\n\nif self.input[hash('touch')] then\nprocess_spin(self, angle)\nelseif self.grab then\nself.grab = false\nself.last_diff = 0\nself.rotation_direction = 0\nend\n...\nend\n``````\n\nAnd this code works at first. The bug is when player change rotation direction while holding “mouse button 1”. So I add check for rotation direction in process_spin function (don’t like how it looks, need to be cleared):\n\n``````-- check change in dirrection\n-- this if statement changed a lot, but always has some bugs...\nif self.last_diff ~= 0 and (self.last_diff - angle_diff) ~=0\nand ((angle_diff ~= 0 and self.rotation_direction == -1) or (self.rotation_direction == 1)) then\nlocal new_direction = 0\nif (self.last_diff - angle_diff) > 0 then\nnew_direction = 1\nelseif (self.last_diff - angle_diff) < 0 then\nnew_direction = -1\nend\n\nif new_direction ~= self.rotation_direction then\nself.grab = false\nself.rotation_direction = 0\nself.initial_angle = angle\nself.last_diff = 0\nend\n\nif self.last_diff ~= angle_diff then\nself.last_diff = angle_diff\nend\n``````\n\nAnd this part gave me a lot of issues, when player change direction on early start around 0 degree. When one start rotate in one direction from `angle_diff` = 0 to `angle_diff` = 1 degree and drastically changes the direction to, say, `angle_diff` = 358 degree. Code go crazy on this 1 - 358 degree change, because it looks like very quick full circle =)\n\nAfter some playing with if statement I thought that I’m going a wrong way and I need to deep dive into quaternion topic (never used them before). Long story short I came up with this code:\n\n``````-- [hitman.script](https://github.com/britzl/defold-orthographic/blob/master/example/shared/objects/hitman.script)\n-- line 24\nlocal rotation = vmath.quat_rotation_z(angle)\nlocal rot_diff = vmath.conj(go.get_rotation()) * rotation\nlocal direction = math.floor(rot_diff.z * 100)\n-- direction < 0 - clockwise; direction > 0 - counterclockwise.\ngo.set_rotation(rotation)\n``````\n\nThis code seems to be working, but I’m not sure that it is optimal for such a task. Quat conjugate and multiply could be heavy to calculate… Have not done performance testing yet.\n\nMaybe I could get rotation direction in more easy or efficient way? In fact when I set new rotation `go.set_rotation(rotation)` object rotate in right direction, “closest direction”. So engine already know it would be cw or ccw. Maybe I can get this info without additional calculations?\n\n3 Likes\n\nSorry, I was not clear of my goal.\nGame Object will be moved and rotated by player input. Now it is WASD + mouse, but I want to add controller (gamepad) support in future. Think of it like: you are rotating object around it’s axis using mouse. Here is GIF of how it looks for now:", null, "Well, the example logic of @britzl’s code still appplies imho.\nKeeping a separate value with the accumulated angle is the way I would go too.\nYes, you would have to modify it to handle the case of rotating it both ways.\n\n1 Like\n\nYeah, in my example I rotated on any action press but that can easily be changed:\n\n``````function on_input(self, action_id, action)\nif action_id == hash(\"right\") then\nif action.pressed then\nself.accumulated_rotation = 0\nself.rotation_direction = 1\nelseif action.released then\nself.rotation_direction = 0\nend\nelseif action_id == hash(\"left\") then\nif action.pressed then\nself.accumulated_rotation = 0\nself.rotation_direction = -1\nelseif action.released then\nself.rotation_direction = 0\nend\nend\nend\n``````\n\nI’m rotating not by click, but by the mouse move, as in https://github.com/britzl/defold-orthographic/blob/master/example/shared/objects/hitman.script example.\nI’m going to rework my script. I’ll post here the result.\n\n1 Like\n\nIf you want the rotation of a GO1 to be controlled by the location of another GO2 (which might be controlled by an input action) then you could attach a collision object to GO1, and on_message it. This will do rotation:\n\n``````local p = message.other_position\nlocal t = go.get_position()\nlocal angle = math.atan2(p.y- t.y, p.x - t.x)\nlocal rot = vmath.quat_rotation_z(angle)\ngo.set_rotation(rot)\n``````\n\nJust needs some tweaking to do the counting.\nEdit: Argg, just looked at the soln below, clearly I don’t know what I am talking about!\n\nIf you want the rotation of a GO1 to be controlled\n\nI’ve already done it, via example that I mentioned in first post. My initial question was exactly about rotation counting", null, "I haven’t managed to perform rotation+count as were advised in previous posts. And come up with something that I don’t like, and has bugs )\nI’ve prepared my demo here:\n\nUse Mouse to rotate Game Object around it’s axis. Use Left mouse button to start rotation count.\n\nBUT code breaks down to count rotations when you start to use WASD to move object and Mouse to rotate simultaneously. I need another way of finding rotation direction.\n\n1 Like\n\nI’ve reworked it", null, "Demo in github repo.\nUse Mouse to rotate Game Object around it’s axis. Use Left mouse button to start rotation count. Use WASD to move object.\nThanks everyone for tips", null, "Would appreciate if someone has advises how to improve my code.\n\n1 Like" ]	[ null, "https://forum.defold.com/uploads/default/original/3X/3/9/397d7884aa8851a1697ec548fb01325643938a16.png", null, "https://forum.defold.com/uploads/default/original/3X/d/1/d1e35d3d674d146197a1e826b92ab51a3788390a.gif", null, "https://forum.defold.com/images/emoji/twitter/slight_smile.png", null, "https://forum.defold.com/images/emoji/twitter/slight_smile.png", null, "https://forum.defold.com/images/emoji/twitter/handshake.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.75112,"math_prob":0.82021916,"size":4272,"snap":"2023-14-2023-23","text_gpt3_token_len":1077,"char_repetition_ratio":0.20688847,"word_repetition_ratio":0.003030303,"special_character_ratio":0.2661517,"punctuation_ratio":0.11914324,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99039555,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,9,null,1,null,null,null,null,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-29T20:33:21Z\",\"WARC-Record-ID\":\"<urn:uuid:d2298e79-8668-4f6a-ac10-c3fd26c4eebb>\",\"Content-Length\":\"40694\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:119aa9ef-d624-41fd-9109-a983385c7a9d>\",\"WARC-Concurrent-To\":\"<urn:uuid:34f9b69e-228c-4e68-93fc-73782530b6f9>\",\"WARC-IP-Address\":\"52.210.249.1\",\"WARC-Target-URI\":\"https://forum.defold.com/t/count-object-rotation-turns/72884\",\"WARC-Payload-Digest\":\"sha1:GAAIVF5COROHVV55XNZMNHMXHRMMNTHM\",\"WARC-Block-Digest\":\"sha1:BDWJISYOHOQCUDHT7QR2XZN7T3L652GB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949025.18_warc_CC-MAIN-20230329182643-20230329212643-00489.warc.gz\"}"}
https://answers.everydaycalculation.com/multiply-fractions/2-4-times-21-8	[ "# Answers\n\nSolutions by everydaycalculation.com\n\n## Multiply 2/4 with 21/8\n\n1st number: 2/4, 2nd number: 2 5/8\n\nThis multiplication involving fractions can also be rephrased as \"What is 2/4 of 2 5/8?\"\n\n2/4 × 21/8 is 21/16.\n\n#### Steps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 2/4 × 21/8 = 2 × 21/4 × 8 = 42/32\n3. After reducing the fraction, the answer is 21/16\n4. In mixed form: 15/16\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad\n\n#### Multiply Fractions Calculator\n\n×\n\n© everydaycalculation.com" ]	[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7813366,"math_prob":0.97037274,"size":464,"snap":"2021-21-2021-25","text_gpt3_token_len":204,"char_repetition_ratio":0.17826086,"word_repetition_ratio":0.0,"special_character_ratio":0.47413793,"punctuation_ratio":0.075630255,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98020357,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-12T17:26:21Z\",\"WARC-Record-ID\":\"<urn:uuid:2f3be669-8b31-4a0f-9b5b-1ed95bfee97b>\",\"Content-Length\":\"7812\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b6cc8d77-d233-4344-8ffb-803357e5ac58>\",\"WARC-Concurrent-To\":\"<urn:uuid:0d3a8fb7-1da0-4368-96cf-0d3e3e481a4c>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/multiply-fractions/2-4-times-21-8\",\"WARC-Payload-Digest\":\"sha1:G6TJPQZFR24SGVBHB6M237UHQU3TONVI\",\"WARC-Block-Digest\":\"sha1:OZRPKUN5TH6KNWGBEI5MN7AOOLFDY2UF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487586239.2_warc_CC-MAIN-20210612162957-20210612192957-00111.warc.gz\"}"}
https://socratic.org/questions/a-triangle-has-sides-a-b-and-c-if-the-angle-between-sides-a-and-b-is-pi-8-the-an-1	[ "# A triangle has sides A,B, and C. If the angle between sides A and B is (pi)/8, the angle between sides B and C is pi/6, and the length of B is 13, what is the area of the triangle?\n\nApr 23, 2018\n\n$\\textrm{a r e a} \\approx 20.3798$\n\n#### Explanation:\n\nFirst of all, let's rewrite in standard notation. It really helps to do these if we always label the triangle consistently.\n\nThe triangle has sides $a , b , c$ (small letters) where $b = 13$ with opposing angles $C = \\frac{\\pi}{8} = {22.5}^{\\circ}$, $A = \\frac{\\pi}{6} = {30}^{\\circ}$.\n\nThe remaining angle is\n\n$B = \\pi - A - C = \\pi - \\frac{\\pi}{8} - \\frac{\\pi}{6} = \\frac{17 \\pi}{24} = {127.5}^{\\circ}$\n\nThe Law of Sines says\n\n$\\frac{b}{\\sin} B = \\frac{c}{\\sin} C$\n\n$c = \\frac{b \\sin C}{\\sin} B = \\frac{13 \\sin \\left({22.5}^{\\circ}\\right)}{\\sin} \\left({127.5}^{\\circ}\\right)$\n\nNow the area is\n\n$\\textrm{a r e a} = \\frac{1}{2} b c \\sin A = \\frac{{\\left(13\\right)}^{2} \\sin \\left({30}^{\\circ}\\right) \\sin \\left({22.5}^{\\circ}\\right)}{2 \\setminus \\sin {127.5}^{\\circ}}$\n\nI'm tempted to work out an exact answer, because those sines are all expressible using the usual operations including square root. But it's late, and the calculator says\n\n$\\textrm{a r e a} \\approx 20.3798$\n\nJun 18, 2018\n\ncolor(maroon)(A_t == 20.37 \" sq units\"\n\n#### Explanation:\n\n$b = 13 , \\hat{A} = \\frac{\\pi}{6} , \\hat{C} = \\frac{\\pi}{8} , \\hat{B} = \\pi - \\frac{\\pi}{6} - \\frac{\\pi}{8} = \\frac{17 \\pi}{24}$\n\nApplying the Law of sines,\n\n$a = \\frac{b \\sin A}{\\sin} B = \\frac{13 \\cdot \\sin \\left(\\frac{\\pi}{6}\\right)}{\\sin} \\left(\\frac{17 \\pi}{24}\\right) \\approx 8.19$\n\nFormula for area of \" Delta, A_t = (1/2) * a b * sin C\n\ncolor(maroon)(A_t = (1/2) * 8.19 * 13 * sin ( pi/8) = 20.37 \" sq units\"#" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89677984,"math_prob":0.9999962,"size":730,"snap":"2023-40-2023-50","text_gpt3_token_len":182,"char_repetition_ratio":0.12672177,"word_repetition_ratio":0.0,"special_character_ratio":0.24657534,"punctuation_ratio":0.11038961,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000079,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-01T19:54:22Z\",\"WARC-Record-ID\":\"<urn:uuid:d4576133-ad8c-4c5b-855f-c4794eeba524>\",\"Content-Length\":\"37277\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ea4159a0-a30f-4d02-9b56-9d0e41b0229c>\",\"WARC-Concurrent-To\":\"<urn:uuid:aa6b124b-61ef-49b3-b042-a5efb29c9024>\",\"WARC-IP-Address\":\"216.239.32.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/a-triangle-has-sides-a-b-and-c-if-the-angle-between-sides-a-and-b-is-pi-8-the-an-1\",\"WARC-Payload-Digest\":\"sha1:6ZF3TNBCQOZMV4DNXHPUCK733EKWQ245\",\"WARC-Block-Digest\":\"sha1:3A5RXZPQ4ERQKGBF5ZLWKP65VIDW34LU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100304.52_warc_CC-MAIN-20231201183432-20231201213432-00375.warc.gz\"}"}
https://mathoverflow.net/questions/204443/fractional-sobolev-spaces-and-interpolation-in-unbounded-lipschitz-domains	[ "# Fractional Sobolev spaces and interpolation in unbounded Lipschitz domains\n\nI am not really familiar with the topic, thus I am looking for some references about the following problem.\n\nLet $s>0$ be a positive real and let $p\\in(1,+\\infty)$. We define the Bessel Potential spaces on $\\mathbb{R}^n$ via the Fourier transform $\\mathcal{F}$ as follows. $$H^{s,p}(\\mathbb{R}^n)=\\{f\\in L^p(\\mathbb{R^n}):\\mathcal{F}^{-1}(1+\|\\xi\|^2)^{\\frac{s}{2}}\\mathcal{F}f\\in L^p(\\mathbb{R^n})\\}.$$\n\nIn the case of $s\\in\\mathbb{Z}^+$ it holds that $H^{s,p}(\\mathbb{R}^n)=W^{s,p}(\\mathbb{R}^n)$ where the latter is the standard Sobolev space.\n\nMoreover, the Bessel potential spaces can be obtained via complex interpolation from the standard Sobolev spaces of integer order. Therefore the spaces $H^{s,p}(\\mathbb{R}^n)$ are also referred as Fractional Sobolev Space.\n\nLet now $\\Omega\\subset \\mathbb{R}^n$ an open subset (bounded or unbounded), we can define $$H^{s,p}(\\Omega)=\\{f\\in L^p(\\Omega): \\exists g\\in H^{s,p}(\\mathbb{R}^n): g\\big\|_{\\Omega}=f \\}$$\n\nwith the norm $\\\|f\\\|_{H^{s,p}(\\Omega)}=\\inf\\{\\\|g\\\|_{H^{s,p}(\\mathbb{R}^n)}:g\\big\|_{\\Omega}=f\\}$.\n\nTherefore, my question is. Let $k$ be in $\\mathbb{Z}^+$. Are the spaces $H^{s,p}(\\Omega)$ interpolation spaces between $W^{k,p}(\\Omega)$ and $W^{k+1.p}(\\Omega)$ if $k<s<k+1$ when $\\Omega$ is an unbounded Lipschitz domain with noncompact boundary?\n\nIn particular, if $T$ is bounded linear operator $T: W^{k,p}(\\Omega)\\to W^{k,p}(\\Omega)$ and $T:W^{k+1,p}(\\Omega)\\to W^{k+1,p}(\\Omega)$, do we have the boundedness $T:H^{s,p}(\\Omega)\\to H^{s,p}(\\Omega)$ for every $s\\in[k,k+1]?$\n\nCan you please suggest me some references?\n\nYou cannot expect this without imposing some condition at infinity. Otherwise, $\\Omega$ could have a tentacle\" which stretches to infinity and thins rapidly. In this fashion, you can have functions in $W^{k,p}(\\Omega)$ which do not have an extension in $W^{k,p}(R^n)$.\n\n• I expected that some regularity is needed. Can you suggest me some detailed references please?\n– Pit\nMay 1, 2015 at 15:23\n\nHere are two very good references:\n\nR. A. Adams, Sobolev Spaces, Pure and Applied Mathematics, A Series of Monographs and Textbooks, 65 (Academic Press, New York, 1975).\n\nH. Triebel, Interpolation Theory, Function Spaces, Di erential Operators (North-Holland, Amsterdam, 1978).\n\nBest regards." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.65530264,"math_prob":0.99975246,"size":1571,"snap":"2022-40-2023-06","text_gpt3_token_len":578,"char_repetition_ratio":0.19719209,"word_repetition_ratio":0.0,"special_character_ratio":0.34818587,"punctuation_ratio":0.1253482,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999881,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-31T07:18:48Z\",\"WARC-Record-ID\":\"<urn:uuid:2c2f6e29-56e8-4fa4-b7e6-072fa7d67ba2>\",\"Content-Length\":\"114489\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:329f39c9-dee6-4d22-9322-11a6328e4b45>\",\"WARC-Concurrent-To\":\"<urn:uuid:a99738a7-979c-4175-be7a-eabdeb12ceda>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/204443/fractional-sobolev-spaces-and-interpolation-in-unbounded-lipschitz-domains\",\"WARC-Payload-Digest\":\"sha1:DPQTYV3QE56EZUUTYMZZGHONUENZTSYU\",\"WARC-Block-Digest\":\"sha1:M3G25NCJ4XQ55NEAUK3IQ5SESL3NATWH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499845.10_warc_CC-MAIN-20230131055533-20230131085533-00699.warc.gz\"}"}
http://intermath.org/transformations-of-data/	[ "# Transformations of Data\n\nCompute the mean, median, and mode for any set of data.\n\nWhat happens to the mean, median, and mode if you add 0.7 to each element of the data? What happens to these values if you subtract 0.5 from each element of the data?\n\nIf the mean, median, and mode of the data set abcd are 16, 11 and 4, what are the mean, median, and mode of the data set a + xb + xc + xd + x, where x is any real number?", null, "Extension\nDescribe what happens to the mean, median, and mode of a data set when you multiply each data point by a number k." ]	[ null, "http://intermath.org/wp-content/uploads/2018/10/angle.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8526503,"math_prob":0.98364496,"size":530,"snap":"2021-43-2021-49","text_gpt3_token_len":150,"char_repetition_ratio":0.17490494,"word_repetition_ratio":0.17857143,"special_character_ratio":0.29056603,"punctuation_ratio":0.18571429,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9936885,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-28T05:43:15Z\",\"WARC-Record-ID\":\"<urn:uuid:f2b2e4fb-a9c5-475d-8ea1-952aa05099ba>\",\"Content-Length\":\"47502\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:82da681a-1de3-4bf3-9dd8-64dcefb0e4d2>\",\"WARC-Concurrent-To\":\"<urn:uuid:d79f56cf-e4f0-4faf-bc2b-ca5538598829>\",\"WARC-IP-Address\":\"198.71.233.87\",\"WARC-Target-URI\":\"http://intermath.org/transformations-of-data/\",\"WARC-Payload-Digest\":\"sha1:XSCMSD22UMX6HWGDRTEXVYVF6XD4VGL5\",\"WARC-Block-Digest\":\"sha1:3RETFDF5R5ZUS5S7EMM6DBGENIDOAYQ3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323588257.34_warc_CC-MAIN-20211028034828-20211028064828-00536.warc.gz\"}"}
https://techcommunity.microsoft.com/t5/excel/freeze-excel-formula-when-criteria-met/td-p/3702169	[ "", null, "SOLVED\n\n# Freeze Excel formula when criteria met\n\nHello Guys\n\nMight be easy but i have a headache here\n\nPreface\n\nI want the \"Days in transit\" Colum to stop updating the formula when I select the Yes answer in the \"Arrived at destination\" Colum currently the formula on \"Days in transit\" is \"=TODAY()-G2\"", null, "i have tried =IF(L3=\"yes\",\"\",\"=TODAY()-G2\")\n\nbut i don't think the formula is correct\n\nI'm out of ideas\n\n5 Replies\n\n# Re: Freeze Excel formula when criteria met\n\n=IF(L2=\"yes\",\"\",TODAY()-G2)\n\nMaybe like this.\n\n# Re: Freeze Excel formula when criteria met\n\nthank you that helped\n\nNow I just need it to stop counting when yes is selected in \"Arrived at destination\"\n\nthe no logic is working correctly so its something i need to put in the middle i believe\n\n=IF(L3=\"yes\",\"__not sure what to put here__\",TODAY()-G3)\n\n# Re: Freeze Excel formula when criteria met\n\nIf possible please update date of arrival in column and rephrase the formula as below\nIF \"date of arrival\" column >0 then, date arrival - G2 else TODAY()-G2\nbest response confirmed by simon_70 (Occasional Contributor)\nSolution\n\n# Re: Freeze Excel formula when criteria met\n\nInstead of No/Yes in the Arrived at destination column, I'd leave it blank initially, and enter the date of arrival when the item has arrived. You can then use\n\n=IF(L2=\"\",TODAY(),L2)-G2\n\n# Re: Freeze Excel formula when criteria met\n\nhey this worked thank you all" ]	[ null, "https://www.facebook.com/tr", null, "https://techcommunity.microsoft.com/t5/image/serverpage/image-id/427942i65DDB24EB6ABC9AC/image-size/large", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88948125,"math_prob":0.76159745,"size":1134,"snap":"2022-40-2023-06","text_gpt3_token_len":392,"char_repetition_ratio":0.20088495,"word_repetition_ratio":0.17156863,"special_character_ratio":0.3888889,"punctuation_ratio":0.09505703,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95149255,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-02T09:38:41Z\",\"WARC-Record-ID\":\"<urn:uuid:fb16a38b-660f-41c4-a526-f06fe56c0bb1>\",\"Content-Length\":\"380594\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2a63b3e8-8fee-4aab-847f-78a8718c16fe>\",\"WARC-Concurrent-To\":\"<urn:uuid:a50b7563-8ca7-4c3b-9846-88e2b4cd8e60>\",\"WARC-IP-Address\":\"23.218.122.245\",\"WARC-Target-URI\":\"https://techcommunity.microsoft.com/t5/excel/freeze-excel-formula-when-criteria-met/td-p/3702169\",\"WARC-Payload-Digest\":\"sha1:4QNQ75PMF7UMMGP2Q4ZGZB2S7FIRSBE2\",\"WARC-Block-Digest\":\"sha1:JNRS6K626RQJCWWAJFCXE2A6NWJ7OQXX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499967.46_warc_CC-MAIN-20230202070522-20230202100522-00486.warc.gz\"}"}
https://www.smartvidya.co.in/2020/09/solved-question-paper-for-lecturer-in_87.html	[ "### Solved Question Paper for Lecturer in Electrical Engineering in DTTE, GNCTD conducted by UPSC\n\n11. For any medium, electric flux density D is related to electric field intensity E by the equation:", null, "12. In any linear resistive network, the voltage across or the current through any resistor or source may be calculated by adding algebraically all the individual voltages or currents caused by the separate independent sources acting alone, with all other independent voltage sources replaced by short circuits and all other independent current sources replaced by open circuits. Which one of the following theorems states the above definition?\n(a) Maximum power transfer theorem\n(b) Superposition theorem\n(c) Maxwell’s theorem\n(d) Source transformation theorem\n\n13. Consider the following circuit:", null, "14. If two sinusoidal waves whose phases are to be compared:\n1. Both be written as sine waves or both cosine waves\n2. Both be written with positive amplitudes\n3. Each has the same frequency\n4. One wave should be sine and the other cosine wave\n\nWhich of the above statements are correct?\n(a) 1, 2 and 4 only\n(b) 1, 3 and 4 only\n(c) 1, 2 and 3 only\n(d) 2, 3 and 4 only\n\n15. The drawback of 𝑚-derived filter is removed by connecting number of sections in addition to prototype and 𝑚-derived sections with terminating:\n(a) Full sections\n(b) One-fourth sections\n(c) Half sections\n(d) Square of three-fourth sections\n\n16. The hybrid parameter of the open circuit reverse voltage gain is indicated as:", null, "17. Two identical 750 turn coils 𝐴 and 𝐵 lie in parallel planes. A current changing at the rate of 1500 𝐴/𝑠 in 𝐴 induces an emf of 11.25 𝑉 in 𝐵. If the self-inductance of each coil is 15 mH, then the percentage of the flux produced in coil 𝐴 which links the turns of 𝐵 will be:\n(a) 40 %\n(b) 50 %\n(c) 60 %\n(d) 70 %\n\n18. What is the capacitor voltage that is associated with the current for the wave form shown when the value of the capacitance is 5 𝜇F?", null, "(a) 𝑣(𝑡)=1\n(b) 𝑣(𝑡)=2\n(c) 𝑣(𝑡)=4\n(d) 𝑣(𝑡)=8\n\n19. Current measured during a test is 30.4 𝐴, flowing in a resistor of 0.105 Ω . It is noted that the ammeter reading is low by 1.2 % and the marked resistance is high by 0.3 %. The true power will be nearly:\n(a) 79.5 𝑊\n(b) 89.3 𝑊\n(c) 99.1 𝑊\n(d) 109.0 𝑊\n\n20. Schering Bridge is used for the measurement of:\n(a) Capacitors\n(b) Inductors\n(c) Q-of inductors\n(d) Medium value resistors\n\nShare:\n\n## Featured Post\n\n### UPSC Civil Service Preliminary Paper-1 Previous Year Solved Question Papers\n\nCivil Service Preliminary Paper-1 Previous Year Solved Questions for the year 2019 Civil Service Preliminary Paper-1 Previous Year Solved Qu...", null, "Name\n\nEmail \n\nMessage " ]	[ null, 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ouD62sW2GFyDElnCD7EAyd9gag2Jh1WsdOsdIfQYwCs+boGLeRmEaZuYLN9aBFaGWPrXMP+WCUsZqct1jIzEWGALUHLK3YsStj9Jx4e2mEh6MIjnFuWNO7rz3B62IHnmjryeOdrt8XFQG5BykYRctUqCZKs38NhVppODNWOPtox9wGvXXCc7kNQqwjSfTME9OTmQTY7DNUnlnXXweOMtBz6fU2Bj/C5otmbtQO91QMCGnJjwJ8uhyMkwpw23wg71FlDEOkNj8TDwIbzGwjj1x6ah7Jsg0CR3GJnWOnKFAuHvPdhjdsi2SQrjIz2ORKhNN820nT5LsaAl5ajC9lz3zAjt7kDFN6hHwZDoEDuhISmoTipV2JvvgFv902QGeN2u9p44fwtJIKNLIOJXV4QdsbcZFJ9DTtALXL/hNBbgX7PtGFShwqERgi1Baz3CrnRfLC4fwyaPSWDYzA9xHt1AtNba97rw2wyuQoiMBoccQgeJ55BZm0didxaWHr4USE32pkmcmOja55fnuCw3YBj8rCk5LaPCj3I2CDLM0TBGhxnBV740/h+1qHJ5jaCNOKDPLlTApo0EZLf9NwmCb47A3KX4sDkxI4+LW9JM3OnzBxq99zGij6f1SN3pUY3VRfJrfcbm94a5fdK8wOUZ1qdlcidMaYhuTVmSZan4WqaZNBV9jUKYMVsaS+EWJeA6e2nEK75ql6TUcyQKf3cKQJ5cgMS0ERefG7RVpef8Pvv/wOFjzdSrW95t1PRmmaGk2pbPSt4h4zF04TSCE3HDxRyY8OayC3Zme5vED4VN1NdTv8JO90G95Tcprw7lqxxdtiBTfUUWxOYZjdNnq1lE7mt8HT4UCE3QKmfmy/Nsz0VrHG3EJ6yDVGX+Bvs/H+H6FzQ2wCV7NgOJWuaVeq71pkbuFOuKHvT4ui9/hAokBtQw8o74XLlSw3MfbzyTNq8iQjdZtNeoZS08iqAKHVwv0AcfYG7wAOXbaRqtAU4ByLdDIBAkdxQZW8J+4DhAHt/I1baXFj7ezhIWjxfdzR3dNown3K8O1unZR14A/7+Du6juDifGK/SVLIFAQ25AaCXXYEuTFtrW1CiR7NEQE9uwI8M9JPGepLOdT93PSkp9fQQMJCbVTQh5cq7haNuglhJFjHUr+b4JU4lASnRhBGwkJvVOvmGcq14RarJ8wLx40fwiNjV4KJUnSNQQm5WXpuv38MFfUTbOfY9ZyjazbvKtte6TUfQ4neWbqsRvWfRRfYVyD1MRaiUjhHg6wpX4O2/J5/2XX5AeL1pcIKUWLiTOgbferDq8kutjmUX2RG5+8F15FyRkF0QMFmxnaOni8g9Mg17K567c9/C9vDU0i+PJyHM7fQCACJ3b+waO2PhDMBVZOOKc4V6no8QSgt2hcPpK2QxRhIi9xio913m+TvsvTew9v/DPqE/n+BBbClwHBfWhQOYcNV2A/7to/7jb/wCi8fM+Recwm3y0bhlu3Df4mP+RG5EYjF/xVaHsg+x+QTzKt0nk+xPV3ZCin1GZnub2fYf4N3hG9x5V/CP9Vu4CY9w3P8C1m9PB8KayD0Q0MMVg5NJ2e3HXHkfwN/hoUgijfwxN98NqX4zKiaTstuPa/t3sIv+zEQq7QRZ0iGviNxDoj1UWew8w1vbmTJCu8t2OP+SX9HcbJUjfoTb1N5mOyIDCNXNY/zd6W2KI3IPRbhJlSO+7k/t4io7QU0C4EjRhdvRVEaz34nczXCb+VtCc/MviV7hfLwFz7UcqGSVVjMKWNMP95DIPRzWkyqJfzWVLs1bvCGltU4+m5vCBFKtKpFbRYTuF4MAkXsxTUmCqAgQuVVE6H4xCBC5F9OUJIiKAJFbRYTuF4MAkXsxTUmCqAgQuVVE6H4xCBC5F9OUJIiKAJFbRYTuF4MAkXsxTUmCqAgQuVVE6H4xCBC5F9OUJIiKAJFbRYTuF4MAkXsxTUmCqAgQuVVE6H4xCBC5F9OUJIiKAJFbRYTuF4MAkXsxTUmCqAgQuVVE6H4xCBC5F9OUJIiKAJFbRYTuF4MAkXsxTUmCqAgQuVVE6N6CwLwiLBC5LU1Jj2QE8GSp7AzCqUdYIHLL7UfXFgTmF2GByG1pTnokIzC/CAtEbrn96NqOwMwiLFQgd8VQfezY3PsQ7vhJ/eI428sRbrcf4MEYE96OJT2dCgIVIizwqioTzvW4cXRKyJ3EVdl4H+ExfrEgndhjDsZfkQ41v8QPsF2/BT88JiHjLLlM59ErxOFNFr8xlctJfnM9CO6+2ENyTEKYEjkUuYyHWVY6XB6PQfZg9xjDBSM3+Ad4HgkLC7mf4bi/gU0hToqlpjz2ISOAchA5j6z1JouJaMliOo9EYzECyIe08xHqE/hrFxxnM4NOK8mxCiB6VRH+E6LgLTiOGlVBTlchwgJvexeyYFDqGeByfsNcG8iNk4cdRHXivpvIzU7ofwrh2t10EDpuGGBY5OTnwxZcxwFdTJhkRGIEfwuBHEJjqOpVLscuB8vmNQpgpSokJf+yCAuX0x42uTySo4112ClZ93arJ7fQtNfhj3oxDC3kBpAPPL/0JlB3GYv6OrI2knPH54pml5NM4hrraZIDgBOzLEBUiSxJB7mBME6GBm0AqZI8un6sITfaTg3CQFjJLUA0kqVr0drmh/OINQTRWZtZoq2YGZY1qjbhqD+iHMV4N11WK8FCjGI8VOAVODxyw3iKrEjuSpMHHgkIojAAz00mWS6bdB6/gM/vFZsbURR5Oy21BGbX618e3csBa13TzmzuAL3WsUrmKEffHfASw+PO42Ycj2lZFgOzSt1bpimQO+mB5iGMl4cz4dSL8gLx40e49jz41UZuGEaLtMQkEZHbkCUmR0qcOuRObNHMsyQ8MIrnQv+8/giRji7qZJKbnr/C/vRXF3BNMg+F3Gif2YYwMbt2f4PwSXYPZhOXgrckFR3zL+k8afqxLjJZjO4xVjUeok7jHbJWe0hyZ3LkJnaoZecwglqxtD9UyF2uWZOJgp6cqZbIzZrlCiC5SzSi/Moo18KNZXWP4RxiyjY3ymEYHdJQfaOA3HuheXKjTVxKToNmT21Qg80tuddyvuPexaxZQCkOLL9MK05WllQOxWziv68kn7QdH72JZOgwJeaVvaRun9Ykd4lmr0PuKQ+JKIe1joiFfhTrtpka5maUg5lGa/APf1TK+O9B7lQTLFtzJz5bBxzLsJ2YZ3Xt7Upc6ixRKoe8wtpZ7tPPKK+5S70ZJdoKNUWpWTNlmxvnBTaNjEvWTaLuDjWhRDkcsE6Kp8/RxjVUyI2AGDQzrjIalqShBrlzs/fG1e/jRezAJgzE1gRmWzZapBiK3GVyiC0RqwaLdX3A3kOeCrnRA2DRWimBi3sqyofqcsB7kLFelqnvWjcpxq2fjNhbCKe8ldcqB5sP/4Dw+iq/KaweUpNPXSA3oF1ttNNeID68hzXTXK7Y3sgXLP8Jt5/epSuWbBJS0M6Yt3WiNiZmklbOLXq8QBx9EXvVHeCrsdYtwGPKwMo2ySHqhX7u2WyFaIZnkdxQZW8Ja+wQAu9K7Hm+Ai/4CqenB/Bddv0ZwvsIYmVbQaXVz2ZydPCWpJW17qwN+Ps7uI/iepvJOqhZvSzK5JBdeIqLsF5Bk0+tITcANN0VaBVX2POT1drWytPDGSKgJzcOa4Ul9uYSJvb4nPZzN5eV3pwGAgZys8ol5snKu4VjnQ8WdHLxLZDrmX2JoxOEfpsTAhZyMzGSbyjXLTwDl/gb7Dwi9pxIsZS6lpCbiVnx63cdIvzr9x2Ek5+E6SpPvxUREFxIHQnMY5adQFVMP+4vFcg9bgWp9AkhwNc4rjLzkvvKN5X3qAwtCZF7aMRnW57YBTkjbxeRe7ZkG6Hi3EX8djYnGBC5R+DIfIsUDgZc5DKuYk9DQiL3NNph+rXg7tw3sDZ++Jtty3D9B3g6heLgovrffXYFBpG7KyQXnU+F1eXn3yF49wWiuxtw/rGGzU0IP4972OT26AwLEpF7WLxnWhpOJmW3Hzv08gP4u0fpDMgKnWBABIjcA4I966LYGYm3frIbVOwIDUJ1cxzbq35V+dvMvvEgcveN8N8pf+4HN33kMTwQRO7hMV9sicmWZt1HHuOITOQeB/cFlirOAh9xAqmCSuRWEaH7xSBA5F5MU5IgKgJEbhURul8MAkTuxTQlCaIiQORWEaH7xSBA5F5MU5IgKgJEbhURul8MAv8PJkzQQFlvIrYAAAAASUVORK5CYII=", null, 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", null, 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", null, 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https://community.qualtrics.com/XMcommunity/discussion/4445/answer-choice-randomization-problem-with-randomized-subset	[ "Have you already seen Work:New.0 ? Tell us in a few words what you've learned from the video on the Community!\n\n# Answer Choice Randomization: Problem with randomized subset\n\nHi,\nI have the following problem:\nI want to include 15 single-choice answers (3 groups / 5 answers possibilities per group) in one question and the respondent should get only 1 randomly chosen answer from each group - in total 3 answers.\nI tried with the 'Advanced Randomization' function, but I just managed to create one 'Random Subset'. So I need to create several 'Random Subsets' in one question.\nDo you have any idea how to create more than one subset or how to solve the problem in another way, so that the respondents just get one answer from each group? I struggle with the problem for hours. So I really hope you can help me.\n\nJennifer\n\n•", null, "Raleigh, NC Wizard ✭✭✭✭✭\n\nTo do it without JavaScript, you can use survey flow randomizers to randomly pick one choice from each group by setting flags for the choices to be shown. For example, let's say Group 1 is made up of Choices 1 through 5. Create a Group 1 randomizer with five embedded data blocks under it and randomly choose 1 block. The blocks would set a 'display' flag for each choice (e.g., choice1_flag = 1). Create a similar randomizer for each group. Then in your question add display logic to all your choices to only display the choice if the flag is set to 1.\n\n• You will probably have to do some javascript to achieve this.\n\nCreate a question with all 15 answers to use as a base.\n\nIn the javascript, make 3 random numbers to choose one from each group. Then hide the radio buttons for the non selected answers.\n\nNote that this will not change the order of the buttons, it will only hide the buttons. If you want a random order, you will have to randomize the order of all 15 answers and then use the random number to select a pointer to the answer location.\n\n• Unfortunately I'm not able to write a javascript code :-(\n\nSeriously though, it is very easy to do this in Qualtrics. If you search the forum, there was a recent post on hiding the radio buttons. There are also many posts on randomizing.\n\nOff the top of my head, here is a starting point:\nvar grp1 = [1, 4, 7, 10, 13]; // answer group 1 locations\nvar grp2 = [2, 5, 8, 11, 14];\nvar grp3 = [3, 6, 9, 12, 15];\n\nvar pick1 = Math.floor((Math.random() * 5)); // random number between 0 and 4\nvar pick2 = Math.floor((Math.random() * 5));\nvar pick3 = Math.floor((Math.random() * 5));\n\nfor (var i = 0; i < grp1.length; i++) {\nif (i <> pick1) {" ]	[ null, "https://us.v-cdn.net/6030293/uploads/userpics/486/nLFBSWRAZPTWT.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8674745,"math_prob":0.84242743,"size":2635,"snap":"2022-27-2022-33","text_gpt3_token_len":680,"char_repetition_ratio":0.12010642,"word_repetition_ratio":0.0,"special_character_ratio":0.2690702,"punctuation_ratio":0.13357401,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97947246,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-16T04:52:27Z\",\"WARC-Record-ID\":\"<urn:uuid:be34e3f5-ae0a-4336-be66-d21c3891d87a>\",\"Content-Length\":\"722077\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:493da456-2bd3-4700-a760-2d1496f2f88e>\",\"WARC-Concurrent-To\":\"<urn:uuid:415ce2f3-5c45-4c9d-8b00-835d0b455552>\",\"WARC-IP-Address\":\"162.159.128.79\",\"WARC-Target-URI\":\"https://community.qualtrics.com/XMcommunity/discussion/4445/answer-choice-randomization-problem-with-randomized-subset\",\"WARC-Payload-Digest\":\"sha1:7WR2JTXX3M5TX2ONZVFCQCZC2B3H3DR6\",\"WARC-Block-Digest\":\"sha1:CVEO3CFXUNVOJGL6B4UPTMC7A6LTVCLL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572220.19_warc_CC-MAIN-20220816030218-20220816060218-00051.warc.gz\"}"}
https://zilongshanren.com/post/2016-03-13-undertand-javascript-this-keyword/	[ "1. This 绑定的内容与函数无关，而与函数的执行环境有关。\n2. 函数的 This 绑定的内容可以通过 bind，apply 和 call 函数来动态进行修改。\n3. 巧用闭包可以消除不必要的 This 动态绑定，提高代码的可读性。\n\n## This 绑定内容与函数无关，而与执行环境有关\n\n```var name = \"zilong\";\n\nvar myfunc = function() {\nconsole.log(this.name);\n}\n\nmyfunc();\n```\n\n```[[scope chain]] = [\n{\nActive Object {\narguments: ...\nthis: [global Object],\n...\n},\nglobal Object: {\nname: 'zilong'\n...\n}\n}\n]\n```\n\n```var name = \"zilong\";\nvar sex = \"man\";\n\nvar myfunc = function(name) {\nthis.name = name;\n}\n\nmyfunc.prototype.print = function() {\nconsole.log(this.name);\nconsole.log(this.sex);\nconsole.log(sex);\n}\n\nvar obj = new myfunc(\"hello\");\nobj.print();\n```\n\nRESULTS:\n\n```hello\nundefined\nman\n```\n\n```[[scope chain]] = [\n{\nActive Object {\narguments: ...\nthis: obj,\n...\n},\nglobal Object: {\nname: 'zilong',\nsex: 'man',\n...\n}\n}\n]\n```\n\n## This 绑定的内容可以被动态修改\n\n```var name = \"zilong\";\nvar sex = \"man\";\n\nvar myfunc = function(name) {\nthis.name = name;\n}\n\nmyfunc.prototype.print = function() {\nconsole.log(this.name);\nconsole.log(this.sex);\nconsole.log(sex);\n}.bind(this);\n\nvar obj = new myfunc(\"hello\");\nobj.print();\n```\n\nRESULTS:\n\n```zilong\nman\nman\n```\n\n```var name = \"zilong\";\nvar sex = \"man\";\n\nvar myfunc = function(name) {\nthis.name = name;\n}\n\nmyfunc.prototype.print = function() {\nconsole.log(this.name);\nconsole.log(this.sex);\nconsole.log(sex);\n};\n\nvar obj = new myfunc(\"hello\");\nmyfunc.prototype.print.call(obj, \"hello\");\n```\n\nRESULTS:\n\n```hello\nundefined\nman\n```\n\n``` var obj = new myfunc(\"hello\");\nmyfunc.prototype.print.call(this, \"hello\");\n// 下面的 window 和 this 是等价的。\n// myfunc.prototype.print.call(window, \"hello\");\n```\n\nRESULTS:\n\n```zilongshanren\nman\nman\n```\n\n## 巧用闭包消除 This 动态绑定，提高代码可读性\n\n```var a = 10;\nvar obj = {\na : 1,\nb : 2,\n\nsum : function() {\nreturn this.a + a;\n}.bind(this);\n\n}\n}\n\nconsole.log(obj.sum());\n```\n\nRESULTS:\n\n```3\n```\n\n```var a = 10;\nvar obj = {\na : 1,\nb : 2,\n\nsum : function() {\nvar self = this;\n\nreturn self.a + a;\n};\n\n}\n}\n\nconsole.log(obj.sum());\n```\n\nRESULTS:\n\n```3\n```" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.6963116,"math_prob":0.9125112,"size":3112,"snap":"2022-05-2022-21","text_gpt3_token_len":1481,"char_repetition_ratio":0.15122265,"word_repetition_ratio":0.3097561,"special_character_ratio":0.28631106,"punctuation_ratio":0.24630542,"nsfw_num_words":3,"has_unicode_error":false,"math_prob_llama3":0.9574816,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-20T01:07:57Z\",\"WARC-Record-ID\":\"<urn:uuid:8ae8669e-988b-41a0-a43e-1ce4bab14ae2>\",\"Content-Length\":\"25402\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5d99ed23-730a-41d5-9b7b-8b75619a6a66>\",\"WARC-Concurrent-To\":\"<urn:uuid:9e7b5def-2b03-4d89-85b2-4ddce4395d3b>\",\"WARC-IP-Address\":\"1.15.88.122\",\"WARC-Target-URI\":\"https://zilongshanren.com/post/2016-03-13-undertand-javascript-this-keyword/\",\"WARC-Payload-Digest\":\"sha1:CPGVCRDIH6DS3SICTZZUL4HRC4HYDFUJ\",\"WARC-Block-Digest\":\"sha1:I44KA5D3IWZSGI5S6GXUXXAP73HQ2SE3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662530553.34_warc_CC-MAIN-20220519235259-20220520025259-00609.warc.gz\"}"}
https://discourse.matplotlib.org/t/debugging-process-gets-stuck-with-matpltlib-after-show/13612	[ "", null, "# debugging process gets stuck with matpltlib after show()\n\nHi all,\n\nI am new born in Python ( 1 week old)\n\nCan you pls help to understand the basic concept of\nmatpltlib interacting with Python\n\nthe mutter is:\nduring debugging the debug processes stacks when fig is created\nfor example, in code\n\nimport matplotlib.pyplot as plt\nfrom pylab import *\nx= 23;\ny = 111111;\nprint(23456)\nplt.plot(range(10))\nplot([1,2,3])\nshow()\nprint(11111111)\na=888\n\nit is impossible after show() to continue debug in any IDE for example Wingwar\nor pythonxy\nas stated in\nBeginning Python Visualization - Crafting Visual Transformation Scripts (2009)\npage 187\n\nNote If you’re not using matplotlib interactively in Python, be sure\nto call the function show() after all\ngraphs have been generated, as it enters a user interface main loop\nthat will stop execution of the rest of\nyour code. The reason behind this behavior is that matplotlib is\ndesigned to be embedded in a GUI as well.\nIn Windows, if you’re working from interactive Python, you need only\nissue show() once; close the figures\n(or figures) to return to the shell. Subsequent plots will be drawn\nautomatically without issuing show(), and\nyou’ll be able to plot graphs interactively.\n\nI tried the code\n\nas suggested in\nhttp://matplotlib.sourceforge.net/users/shell.html\n\ncode\ntaken from people\nfrom wingware\nhttp://www.wingware.com/doc/howtos/matplotlib\n\nt = Timer(0, show)\nt.start()\n\nbut still debugging process gets\nstuck…\n\nimport\nmatplotlib as mpl\n\nfrom\npylab import plot,show,close,ion\n\nx\n= range(10)\n\nplot(x)\n\n‘show()’\n\nfrom\n\nt\n= Timer(0, show)\n\nt.start()\n\n‘ion()\nthe same result with or not’\n\na\n= 1222233\n\ny\n= [2, 8, 3, 9, 4]\n\nplot(y)\n\nzz=\n12346\n\nprint(44444)\n\nBest Regards\n\nSandy\n\n···" ]	[ null, "https://discourse.matplotlib.org/uploads/default/original/1X/226e21a6b2d1e1ae3ffbb5d50c20c828692648a1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6602115,"math_prob":0.7409335,"size":1782,"snap":"2021-43-2021-49","text_gpt3_token_len":476,"char_repetition_ratio":0.105174355,"word_repetition_ratio":0.036101084,"special_character_ratio":0.26094276,"punctuation_ratio":0.11494253,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95296615,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-22T17:05:09Z\",\"WARC-Record-ID\":\"<urn:uuid:70dde085-14fb-4789-9b6d-8d4f331dfd2a>\",\"Content-Length\":\"14606\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0f9d9245-4f6e-4276-ab43-e47182c04525>\",\"WARC-Concurrent-To\":\"<urn:uuid:0374f317-5089-41c9-be15-2a49fb18e36e>\",\"WARC-IP-Address\":\"165.22.13.220\",\"WARC-Target-URI\":\"https://discourse.matplotlib.org/t/debugging-process-gets-stuck-with-matpltlib-after-show/13612\",\"WARC-Payload-Digest\":\"sha1:UER7ZL6KKQB7VGNLG2Q63LDXSSXFUQJA\",\"WARC-Block-Digest\":\"sha1:4CYJFGDKI2IWZEDCMKSAMPB4LOVKBALN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585516.51_warc_CC-MAIN-20211022145907-20211022175907-00674.warc.gz\"}"}
http://www.expertsmind.com/questions/properties-of-t-distribution-30135289.aspx	[ "## Properties of t distribution, Mathematics\n\nAssignment Help:\n\nProperties of t distribution\n\n1. The t distribution ranges from - ∞ to ∞ first as does the general distribution\n\n2. The t distribution as the standard general distribution is bell shaped and symmetrical around mean zero\n\n3. The shapes of the t distribution changes like the number of degrees of freedom changes\n\n4. The t distribution is more platykurtic that the normal distribution\n\n5. The t distribution has a greater dispersion than the standard general distribution. Like n gets larger the t distribution approaches the general distribution while n = 30 the difference is extremely small\n\nRelation among the t distribution and standard general distribution is displayed in the given diagram", null, "Note that the t distribution has different shapes depending on the size of the sample. When the sample is rather small the height of the t distribution is shorter than the general distribution and the tails are wider.\n\n#### Derive a linear system - gauss jordan elimination, Suppose that, on a certa...\n\nSuppose that, on a certain day, 495 passengers want to fly from Honolulu (HNL) to New York (JFK); 605 passengers want to fly from HNL to Los Angeles (LAX); and 1100 passengers want\n\nCalculate 50%\n\n#### Develop a linear program, The production manager of Koulder Refrigerators m...\n\nThe production manager of Koulder Refrigerators must decide how many refrigerators to produce in each of the next four months to meet demand at the lowest overall cost. There is a\n\n#### Applications of rational numbers, Kaylee makes 56 packages in seven hours T...\n\nKaylee makes 56 packages in seven hours Taylor makes 20% more packages in nine hours who makes more packages per hour\n\nThe power\n\n#### Linear equations, Linear Equations We'll begin the solving portion of ...\n\nLinear Equations We'll begin the solving portion of this chapter by solving linear equations. Standard form of a linear equation: A linear equation is any equation whi\n\n#### Real number, if HCFof 657 and 963 is expressable in the form of 657x+963x-1...\n\nif HCFof 657 and 963 is expressable in the form of 657x+963x-15findx\n\n#### Find the number of zeros of the polynomial, Find the number of zeros of the...\n\nFind the number of zeros of the polynomial from the graph given. (Ans:1)\n\n#### Linear programming problem., Ask question #Minimum 100 words acca paper mil...\n\nAsk question #Minimum 100 words acca paper mill produces two grades of paper viz.,xand y.Bacause of raw material restrictions, it cannot produce more than 400 tones of grade x pape\n\n#### Simpson rule - approximating definite integrals, Simpson's Rule - Approxima...\n\nSimpson's Rule - Approximating Definite Integrals This is the last method we're going to take a look at and in this case we will once again divide up the interval [a, b] int", null, "", null, "" ]	[ null, "http://www.expertsmind.com/CMSImages/1649_Properties%20of%20t%20distribution.png", null, "http://www.expertsmind.com/questions/CaptchaImage.axd", null, "http://www.expertsmind.com/prostyles/images/3.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88508487,"math_prob":0.9781627,"size":2671,"snap":"2021-31-2021-39","text_gpt3_token_len":594,"char_repetition_ratio":0.15373078,"word_repetition_ratio":0.004576659,"special_character_ratio":0.22051667,"punctuation_ratio":0.079918034,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9931388,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,1,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-21T04:25:52Z\",\"WARC-Record-ID\":\"<urn:uuid:8f99abab-03e4-4501-be3a-b8347894fb6f>\",\"Content-Length\":\"66367\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:feeb197a-716c-4970-95d2-4294c4ef6c4c>\",\"WARC-Concurrent-To\":\"<urn:uuid:6e85c554-be85-4b73-b771-6d6d684b99ae>\",\"WARC-IP-Address\":\"198.38.85.49\",\"WARC-Target-URI\":\"http://www.expertsmind.com/questions/properties-of-t-distribution-30135289.aspx\",\"WARC-Payload-Digest\":\"sha1:QXYGPUSMUUOAIUW2VRUZHNJYBJNDJR7O\",\"WARC-Block-Digest\":\"sha1:5YL2SYTBJCSMYZUA5CJP4WRGMTX3QS2W\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057158.19_warc_CC-MAIN-20210921041059-20210921071059-00642.warc.gz\"}"}
https://people.utm.my/alihassan/biography/	[ "# Biography\n\nProfessor Dr. Ali H. M. Murid received his B.Sc. (Mathematics) and M.Sc. (Applied Mathematics) from Iowa State University, USA in 1983 and 1986 respectively, Diploma of Education from UTM in 1988, and Ph.D from UTM in 1997. His Ph.D thesis was on boundary integral equation formulations for numerical conformal mapping written under the supervisions of Allahyarham Prof. Dr. Mohd. Rashidi Md. Razali (UTM) and Prof. Dr. M. Zuhair Nashed (formerly at University of Delaware, USA). He joined UTM in 1988 and has taught several courses at both undergraduate and graduate levels. The undergraduate courses that he has taught are Calculus, Calculus of Several Variables, Vector Calculus, Diﬀerential Equations, Numerical Methods, Functions of Complex Variables, and Mathematical Modelling. For graduate courses, he has taught Applied and Computational Complex Analysis, Mathematical Methods, Numerical Integral Equations, and Advanced Engineering Mathematics. He was the Chairperson for the Mathematics Program at the Ibnu Sina Institute for Fundamental Science Studies (now known as Ibnu Sina Institute for Scientific and Industrial Research) for two years from June 2009 until May 2011. He was a research fellow at UTM Centre for Industrial and Applied Mathematics (UTM-CIAM) from 2013-2018. His research interests are applied and computational complex analysis and mathematical modeling of industrial problems." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9434018,"math_prob":0.50363815,"size":1408,"snap":"2019-26-2019-30","text_gpt3_token_len":311,"char_repetition_ratio":0.11894587,"word_repetition_ratio":0.0,"special_character_ratio":0.20738636,"punctuation_ratio":0.15261044,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95803124,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-16T18:52:52Z\",\"WARC-Record-ID\":\"<urn:uuid:dc72aacc-f859-4c73-ba12-bf09decf1d53>\",\"Content-Length\":\"36041\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:71e87b47-ce60-454d-b5e1-66a8454879cb>\",\"WARC-Concurrent-To\":\"<urn:uuid:a27ea6dd-21ec-471c-9bc6-3671da52f181>\",\"WARC-IP-Address\":\"161.139.21.59\",\"WARC-Target-URI\":\"https://people.utm.my/alihassan/biography/\",\"WARC-Payload-Digest\":\"sha1:YIEK7Q6VUWJA7MGP6KHAUYWPT3PX3GDF\",\"WARC-Block-Digest\":\"sha1:XFXCDPMUOKODDFGBBAFNGUKIU7LXGAPL\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195524685.42_warc_CC-MAIN-20190716180842-20190716202842-00251.warc.gz\"}"}
https://dividedby.org/0-divided-by-50	[ "Home » Divided by 50 » 0 Divided by 50\n\n# 0 Divided by 50\n\nWelcome to 0 divided by 50, our post which explains the division of zero by fifty to you. 🙂\n\nThe number 0 is called the numerator or dividend, and the number 50 is called the denominator or divisor.\n\nThe quotient of 0 and 50, the ratio of 0 and 50, as well as the fraction of 0 and 50 all mean (almost) the same:\n\n0 divided by 50, often written as 0/50.\n\nRead on to find the result of 0 divided by 50 in decimal notation, along with its properties.\n\nNominator (Dividend)\nDenominator (Divisor)\nShow Steps\n0\n=\n0 Remainder 0\n\n## What is 0 Divided by 50?\n\nWe provide you with the result of the division 0 by 50 straightaway:\n\n0 divided by 50 = 0\n\nThe result of 0/50 is an integer, which is a number that can be written without decimal places.\n• 0 divided by 50 in decimal = 0\n• 0 divided by 50 in fraction = 0/50\n• 0 divided by 50 in percentage = 0%\n\nNote that you may use our state-of-the-art calculator above to obtain the quotient of any two integers or decimals, including 0 and 50, of course.\n\nRepetends, if any, are denoted in ().\n\nThe conversion is done automatically once the nominator, e.g. 0, and the denominator, e.g. 50, have been inserted.\n\nNo need to press the button, unless you want to start over.\n\nGive it a try now with a similar division by 50.\n\n## What is the Quotient and Remainder of 0 Divided by 50?\n\nHere we provide you with the result of the division with remainder, also known as Euclidean division, including the terms in a nutshell:\n\nThe quotient and remainder of 0 divided by 50 = 0 R 0\n\nThe quotient (integer division) of 0/50 equals 0; the remainder (“left over”) is 0.\n\n0 is the dividend, and 50 is the divisor.\n\nIn the next section of this post you can find the frequently asked questions in the context of zero over fifty, followed by the summary of our information.\n\n## Zero Divided by Fifty\n\nYou already know what 0 / 50 is, but you may also be interested in learning what other visitors have been searching for when coming to this page.\n\nThe FAQs include, for example:\n\n• What is 0 divided by 50?\n• How much is 0 divided by 50?\n• What does 0 divided by 50 equal?\n\nIf you have read our article up to this line, then we take it for granted that you can answer these FAQs and similar questions about the ratio of 0 and 50.\n\nObserve that you may also locate many calculations such as 0 ÷ 50 using the search form in the sidebar.\n\nThe result page lists all entries which are relevant to your query.\n\nGive the search box a go now, inserting, for instance, zero divided by fifty, or what’s 0 over 50 in decimal, just to name a few potential search terms.\n\nFurther information, such as how to solve the division of zero by fifty, can be found in our article Divided by, along with links to further readings.\n\n## Conclusion\n\nTo sum up, 0/50 = 0. It is a whole number with no fractional part.\n\nAs division with remainder the result of 0 ÷ 50 = 0 R 0.\n\nFor questions and comments about the division of 0 by 50 fill in the comment form at the bottom, or get in touch by email using a meaningful subject line.\n\nIf our content has been helpful to you, then you might also be interested in the Remainder of 2 Divided by 50.\n\nPlease push the sharing buttons to let your friends know about the quotient of 0 and 50, and make sure to place a bookmark in your browser.\n\nThanks for visiting our article explaining the division of 0 by 50." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9555617,"math_prob":0.9021801,"size":3069,"snap":"2022-05-2022-21","text_gpt3_token_len":781,"char_repetition_ratio":0.15040784,"word_repetition_ratio":0.04890388,"special_character_ratio":0.26425546,"punctuation_ratio":0.11111111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9964054,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-27T15:26:52Z\",\"WARC-Record-ID\":\"<urn:uuid:67e17a50-0285-4084-9c9a-9546b557382a>\",\"Content-Length\":\"103548\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2431ee50-01a3-47ec-ac3f-29a1318c4ee4>\",\"WARC-Concurrent-To\":\"<urn:uuid:7023dba4-ba3f-4a2e-b2f1-121f5dcafab0>\",\"WARC-IP-Address\":\"104.21.26.145\",\"WARC-Target-URI\":\"https://dividedby.org/0-divided-by-50\",\"WARC-Payload-Digest\":\"sha1:42PDNI3SEQTM5E3U2Y66KOHYWGTW2LH5\",\"WARC-Block-Digest\":\"sha1:OK5XGX3EM5LVWELB4NRQ64Q76CVDU2QO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662658761.95_warc_CC-MAIN-20220527142854-20220527172854-00555.warc.gz\"}"}
https://cs.stackexchange.com/questions/88898/determine-if-a-line-going-through-n-rectangles-exists	[ "Determine if a line going through n rectangles exists\n\nI guess this is kinda like asking if 3 (or more) rectangles are collinear.\n\nThe question is, given n rectangles on the 2D plane (given as the rectangle's top-left corner coordinates and it's width/height), does there exist a line that goes through all of them?\n\nNotes:\n\n• The rectangles' sides are perpendicular or parallel to the axes. So basically, none of the rectangle sides are slanted.\n• Rectangles themselves can intersect\n• Does not actually need to produce the line equation\n\nI've been thinking through this for a good portion of today, and can't figure out (or find online) an answer. I keep finding edge cases and can't figure out a guaranteed method. The line of best fit for the mid-point or the corners of all n rectangles seem like a good heuristic for a line, should one exist, but I don't think it always works.\n\nThe corners of the rectangles seem like good starting points to determine the extreme slopes of a potential line, but it's still unclear to me how i'd use them to determine if a line exists for all n rectangles. I also know how I'd do this intuitively IRL, but not sure how to turn it into an algorithm.\n\nVisual Example for 4 rectangles:", null, "• One necessary but not sufficient condition is that every rect (short for iso-oriented rectangle) must overlap the convex hull of \"opposite\" rects. For each dimension, one such pair is one rect with minimum and one with maximum coordinate value. – greybeard Mar 5 '18 at 8:47\n• If only I understood the paragraph Stabbing line segments in $ℝ^2$ in Hohmeyer, M.E.; Teller, S.J.: Stabbing isothetic boxes and rectangles in O(n log n) time in Computational Geometry: Theory and Applications 2 (1992) pp. 201-207. – greybeard Mar 6 '18 at 19:17\n\nThere is a line through the four rectangles if and only if you can pick one edge from each rectangle such that the line intersects each of those four edges -- or equivalently, if there is a point on each of those edges such that the resulting four points are colinear. So a plausible algorithm is to enumerate all $4^4$ ways to choose one edge from each rectangle; then check whether there exists a line that goes through those four edges.\n\nHow do we do the latter test? Let's suppose the endpoints of the first edge are the two points $P',P''$ (i.e., these are two adjacent corners of the first rectangle). Then any point $P$ on that edge can be expressed as $P = (1-\\alpha) P' + \\alpha P''$ for some $0 \\le \\alpha \\le 1$. Similarly, we can express any point $Q$ on the second edge as $Q = (1-\\beta) Q' + \\beta Q''$ where $Q',Q''$ are the endpoints of the second edge, and similarly for the two remaining points $R,S$ on the two remaining edges.\n\nWe can express algebraically the condition that $P,Q,R$ are colinear (on the same line). Let $P_x$ be the x-coordinate of the point $P$, and $P_y$ be its y-coordinate. Then $P,Q,R$ are colinear iff\n\n$$(Q_x-P_x)(R_y-P_y) = (Q_y-P_y)(R_x-P_x).$$\n\nWe can plug in $Q_x = (1-\\beta)Q'_x + \\beta Q''_x$, and so on, to get an equation in terms of the three $\\alpha,\\beta,\\gamma$ unknowns. Similarly, the condition that $P,Q,S$ are colinear can be expressed as an equation in terms of $\\alpha,\\beta,\\delta$. Now there exists a linear through these four edges iff there exists a common solution to these two equations such that $0 \\le \\alpha,\\beta,\\gamma,\\delta \\le 1$. This reduces the problem to solving two quadratic equations over four real unknowns. Now you can apply standard methods for solving such equations to determine whether a simultaneous solution exists.\n\n• There is a line through the four rectangles if and only if you can pick one edge from each rectangle such that the line goes through those four edges the rectangles' sides are perpendicular or parallel to the axes, I can't see a similar condition for the line. – greybeard Mar 5 '18 at 6:16\n• @greybeard, thanks for the comment. In retrospect, my writing was unclear. I didn't mean that the line has to be horizontal or vertical. Rather, I meant that the line intersects each of those four edges somewhere. I've edited the question to try to improve the wording. I hope it makes sense now. Thanks again for the feedback! – D.W. Mar 5 '18 at 7:28\n• (I read more into line goes through[…] edges than warranted. You seem to have posted in a hurry: straightforward (but not necessarily) algorithm?) – greybeard Mar 5 '18 at 8:40" ]	[ null, "https://i.stack.imgur.com/BEbcF.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9426079,"math_prob":0.99165505,"size":1161,"snap":"2019-43-2019-47","text_gpt3_token_len":258,"char_repetition_ratio":0.13137424,"word_repetition_ratio":0.0,"special_character_ratio":0.21705426,"punctuation_ratio":0.083333336,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99939156,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-20T07:19:46Z\",\"WARC-Record-ID\":\"<urn:uuid:fa6c98a7-77fc-44f5-bf31-76539ac894c9>\",\"Content-Length\":\"142326\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:085bfb44-6f62-4bf4-8a42-b82840833be4>\",\"WARC-Concurrent-To\":\"<urn:uuid:387b956a-6014-4f46-bcd2-42ba2d58d379>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://cs.stackexchange.com/questions/88898/determine-if-a-line-going-through-n-rectangles-exists\",\"WARC-Payload-Digest\":\"sha1:SPVHMBRUOZWTNQ23SYIIEUV2SJ5OZTMP\",\"WARC-Block-Digest\":\"sha1:BZKHG7F2KEC3UZRY65VO45OJDC7J67F2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986703625.46_warc_CC-MAIN-20191020053545-20191020081045-00091.warc.gz\"}"}
https://docs.xilinx.com/r/1.4.1-English/ug1414-vitis-ai/vai_q_pytorch-QAT	[ "# vai_q_pytorch QAT - 1.4.1 English\n\n## Vitis AI User Guide (UG1414)\n\nDocument ID\nUG1414\nRelease Date\n2021-12-13\nVersion\n1.4.1 English\n\nAssuming that there is a pre-defined model architecture, use the following steps to do quantization aware training. Take the ResNet18 model from Torchvision as an example. The complete model definition is here.\n\n1. Check if there are non-module operations to be quantized. ResNet18 uses `‘+’` to add two tensors. Replace them with `pytorch_nndct.nn.modules.functional.Add`.\n2. Check if there are modules to be called multiple times. Usually such modules have no weights; the most common one is the `torch.nn.ReLu` module. Define multiple such modules and then call them separately in a forward pass. The revised definition that meets the requirements is as follows:\n``````class BasicBlock(nn.Module):\nexpansion = 1\n\ndef __init__(self,\ninplanes,\nplanes,\nstride=1,\ndownsample=None,\ngroups=1,\nbase_width=64,\ndilation=1,\nnorm_layer=None):\nsuper(BasicBlock, self).__init__()\nif norm_layer is None:\nnorm_layer = nn.BatchNorm2d\nif groups != 1 or base_width != 64:\nraise ValueError('BasicBlock only supports groups=1 and base_width=64')\nif dilation > 1:\nraise NotImplementedError(\"Dilation > 1 not supported in BasicBlock\")\n# Both self.conv1 and self.downsample layers downsample the input when stride != 1\nself.conv1 = conv3x3(inplanes, planes, stride)\nself.bn1 = norm_layer(planes)\nself.relu1 = nn.ReLU(inplace=True)\nself.conv2 = conv3x3(planes, planes)\nself.bn2 = norm_layer(planes)\nself.downsample = downsample\nself.stride = stride\n\n# Use a functional module to replace ‘+’\n\nself.relu2 = nn.ReLU(inplace=True)\n\ndef forward(self, x):\nidentity = x\n\nout = self.conv1(x)\nout = self.bn1(out)\nout = self.relu1(out)\n\nout = self.conv2(out)\nout = self.bn2(out)\n\nif self.downsample is not None:\nidentity = self.downsample(x)\n\n# Use function module instead of ‘+’\n# out += identity\nout = self.relu2(out)\n\nreturn out\n``````\n3. Insert `QuantStub` and `DeQuantStub`.\n\nUse `QuantStub` to quantize the inputs of the network and `DeQuantStub` to de-quantize the outputs of the network. Any sub-network from `QuantStub` to `DeQuantStub` in a forward pass will be quantized. Multiple QuantStub-DeQuantStub pairs are allowed.\n\n``````class ResNet(nn.Module):\n\ndef __init__(self,\nblock,\nlayers,\nnum_classes=1000,\nzero_init_residual=False,\ngroups=1,\nwidth_per_group=64,\nreplace_stride_with_dilation=None,\nnorm_layer=None):\nsuper(ResNet, self).__init__()\nif norm_layer is None:\nnorm_layer = nn.BatchNorm2d\nself._norm_layer = norm_layer\n\nself.inplanes = 64\nself.dilation = 1\nif replace_stride_with_dilation is None:\n# each element in the tuple indicates if we should replace\n# the 2x2 stride with a dilated convolution instead\nreplace_stride_with_dilation = [False, False, False]\nif len(replace_stride_with_dilation) != 3:\nraise ValueError(\n\"replace_stride_with_dilation should be None \"\n\"or a 3-element tuple, got {}\".format(replace_stride_with_dilation))\nself.groups = groups\nself.base_width = width_per_group\nself.conv1 = nn.Conv2d(\n3, self.inplanes, kernel_size=7, stride=2, padding=3, bias=False)\nself.bn1 = norm_layer(self.inplanes)\nself.relu = nn.ReLU(inplace=True)\nself.layer1 = self._make_layer(block, 64, layers)\nself.layer2 = self._make_layer(\nblock, 128, layers, stride=2, dilate=replace_stride_with_dilation)\nself.layer3 = self._make_layer(\nblock, 256, layers, stride=2, dilate=replace_stride_with_dilation)\nself.layer4 = self._make_layer(\nblock, 512, layers, stride=2, dilate=replace_stride_with_dilation)\nself.fc = nn.Linear(512 * block.expansion, num_classes)\n\nself.quant_stub = nndct_nn.QuantStub()\nself.dequant_stub = nndct_nn.DeQuantStub()\n\nfor m in self.modules():\nif isinstance(m, nn.Conv2d):\nnn.init.kaiming_normal_(m.weight, mode='fan_out', nonlinearity='relu')\nelif isinstance(m, (nn.BatchNorm2d, nn.GroupNorm)):\nnn.init.constant_(m.weight, 1)\nnn.init.constant_(m.bias, 0)\n\n# Zero-initialize the last BN in each residual branch,\n# so that the residual branch starts with zeros, and each residual block behaves like an identity.\n# This improves the model by 0.2~0.3% according to https://arxiv.org/abs/1706.02677\nif zero_init_residual:\nfor m in self.modules():\nif isinstance(m, Bottleneck):\nnn.init.constant_(m.bn3.weight, 0)\nelif isinstance(m, BasicBlock):\nnn.init.constant_(m.bn2.weight, 0)\n\ndef forward(self, x):\nx = self.quant_stub(x)\n\nx = self.conv1(x)\nx = self.bn1(x)\nx = self.relu(x)\nx = self.maxpool(x)\n\nx = self.layer1(x)\nx = self.layer2(x)\nx = self.layer3(x)\nx = self.layer4(x)\n\nx = self.avgpool(x)\nx = torch.flatten(x, 1)\nx = self.fc(x)\nx = self.dequant_stub(x)\nreturn x\n``````\n4. Use QAT APIs to create the quantizer and train the model.\n``````def _resnet(arch, block, layers, pretrained, progress, kwargs):\nmodel = ResNet(block, layers, kwargs)\nif pretrained:\nreturn model\n\ndef resnet18(pretrained=False, progress=True, kwargs):\nr\"\"\"ResNet-18 model from\n`\"Deep Residual Learning for Image Recognition\" <https://arxiv.org/pdf/1512.03385.pdf>'_\n\nArgs:\npretrained (bool): If True, returns a model pre-trained on ImageNet\nprogress (bool): If True, displays a progress bar of the download to stderr\n\"\"\"\nreturn _resnet('resnet18', BasicBlock, [2, 2, 2, 2], pretrained, progress,\nkwargs)\n\nmodel = resnet18(pretrained=True)\n\n# Generate dummy inputs.\ninput = torch.randn([batch_size, 3, 224, 224], dtype=torch.float32)\n\n# Create a quantizer\nquantizer = torch_quantizer(quant_mode = 'calib',\nmodule = model,\ninput_args = input,\nbitwidth = 8,\nqat_proc = True)\nquantized_model = quantizer.quant_model\nquantized_model.parameters(),\nlr,\nweight_decay=weight_decay)\n\n# Use the optimizer to train the model, just like a normal float model.\n…\n``````\n5. Convert the trained model to a deployable model.\n\nAfter training, dump the quantized model to xmodel. (```batch size=1``` is must for compilation of xmodel).\n\n``````# vai_q_pytorch interface function: deploy the trained model and convert xmodel\n# need at least 1 iteration of inference with batch_size=1\nquantizer.deploy(quantized_model)\ndeployable_model = quantizer.deploy_model\nval_dataset2 = torch.utils.data.Subset(val_dataset, list(range(1)))" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.55043685,"math_prob":0.98484457,"size":6499,"snap":"2023-40-2023-50","text_gpt3_token_len":1715,"char_repetition_ratio":0.13841416,"word_repetition_ratio":0.018445322,"special_character_ratio":0.2703493,"punctuation_ratio":0.2359736,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99526983,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-06T16:59:28Z\",\"WARC-Record-ID\":\"<urn:uuid:eafaa378-0a3a-48fc-88d8-6bc40bc8729d>\",\"Content-Length\":\"48133\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4190353d-a228-483e-ae3e-14a6d885aa37>\",\"WARC-Concurrent-To\":\"<urn:uuid:2ee5adcf-02e2-4af4-92d2-b04216e74b56>\",\"WARC-IP-Address\":\"18.218.43.160\",\"WARC-Target-URI\":\"https://docs.xilinx.com/r/1.4.1-English/ug1414-vitis-ai/vai_q_pytorch-QAT\",\"WARC-Payload-Digest\":\"sha1:VGQUVKAZFDCUA2NFTXEEZJBT5QBF5LU4\",\"WARC-Block-Digest\":\"sha1:WH4DNKJGQPSU5RPFLPEOYVOCRDQXBUA2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100602.36_warc_CC-MAIN-20231206162528-20231206192528-00663.warc.gz\"}"}
https://www.climbwork.cz/raymond/2021-02-1183/	[ " ball mill grinding media calculation formula\n\n## ball mill grinding media calculation formula\n\n•", null, "### Calculate Top Ball Size of Grinding Media Equation &\n\n04/04/2018· The development of the ball mill during the twentieth century has been described as the most significant development in the machinery for performing the grinding of ores (Lynch and Rowland 2006). A key part of the implementation of ball mills\n\n•", null, "### Calculate and Select Ball Mill Ball Size for Optimum Grinding\n\n08/04/2021· Calculation of top size grinding media AZZARONI’s Formula; I attach Fred Bond’s first empirical equation for sizing grinding balls for ball mills as well as a few other links to good related articles on the topic. OPTIMAL BALL DIAMETER IN A MILL ; Grinding\n\n•", null, "### Bond formula for the grinding balls size calculation\n\n19/10/2017· The grinding balls diameter determined by the Bond formula has a recommendatory character and serves as a starting point for calculating the necessary proportion grinding media feeding a new mill. More precisely adjust the ball load in the mill can only by industrial test performing. During the industrial tests necessary to accurately monitor the grinding quality, mill\n\n•", null, "### Ball charges calculators thecementgrindingoffice\n\n- Ball charges: This calculator gives the surface and the average weight of the ball charges. It gives also a rough interpretation of the ball charge efficiency: Ball top size (bond formula): calculation of the top size grinding media (balls or cylpebs):-Modification of the Ball Charge: This calculator analyses the granulometry of the material inside the mill and proposes a modification of\n\n•", null, "### How to Size a Ball Mill -Design Calculator & Formula\n\n08/04/2018· 2) Ball milling a ball mill with a diameter of 2.44 meters, inside new liners, grinding wet in open circuit. When the grinding conditions differ from these specified conditions, efficiency factors (Rowland and Kjos, 1978) have to be used in conjunction with equation 1. In general, therefore, the required mill power is calculated using the\n\n•", null, "### Mill Steel Charge Volume Calculation\n\n19/03/2017· We can calculate the steel charge volume of a ball or rod mill and express it as the % of the volume within the liners that is filled with grinding media. While the mill is stopped, the charge volume can be gotten by measuring the diameter inside the liners and the distance from the top of the charge to the top of the mill. The % loading or change volume can then be read off the graph below or\n\n•", null, "### How To Calculate The Media In Ball Mill Ball Mill\n\nCalculate media charge in a ball mill xls dbm crus ball mill charge calculation in cement ball mill grinding media calculation the ball charge mill consists handbook for dry process plants 187 learn more technical notes 8 grinding r p king media or charge in the mill and dm is the diameter of e is a function of the ball media size distribution . Get Price; How Calculate The Vare In Ball Mill\n\n•", null, "### Page 1 Ball Milling Theory freeshell.org\n\ninvolve grinding). With Lloyd's ball milling book having sold over 2000 copies, there are probably over 1000 home built ball mills operating in just America alone. This article borrows from Lloyd's research, which was obtained from the commercial ball milling industry, and explains some of the key design criteria for making your own ball mill. This is a good starting point for anyone\n\n•", null, "### Ball Mill Critical Speed Mineral Processing & Metallurgy\n\n17/03/2017· A Ball Mill Critical Speed (actually ball, rod, AG or SAG) is the speed at which the centrifugal forces equal gravitational forces at the mill shell’s inside surface and no balls will fall from its position onto the shell. The imagery below helps explain what goes on inside a mill as speed varies. Use our online formula. The mill speed is typically defined as the percent of the Theoretical\n\n•", null, "### Ball Mill Grinding Media Calculation Ball Mill\n\nBall mill grinding media calculation formula metallurgy apr 05 2018 cement ball mill grinding media calculationrrcser cement ball mill grinding media calculation a ball mill is a type of grinder used to grind and blend materials for what is the gradient of overflow ball mills calculation for ball mill drive asinagpurcirclein. Get Details Ball Mill Calculations Grinding Media Filling Degree\n\n•", null, "### Calculation of Grinding Balls Surface Area and Volume\n\nNecessity of such calculations may arise when choosing a container for reloading grinding media into a ball mill;when selecting the optimum characteristics of grinding ballsto suit specific grinding requirements, etc. Most peoplecould say that this task\n\n•", null, "### How to Size a Ball Mill -Design Calculator & Formula\n\n08/04/2018· 2) Ball milling a ball mill with a diameter of 2.44 meters, inside new liners, grinding wet in open circuit. When the grinding conditions differ from these specified conditions, efficiency factors (Rowland and Kjos, 1978) have to be used in conjunction with equation 1. In general, therefore, the required mill\n\n•", null, "### Ball Mill Parameter Selection & Calculation Power\n\n30/08/2019· 1 Calculation of ball mill capacity. The production capacity of the ball mill is determined by the amount of material required to be ground, and it must have a certain margin when designing and selecting. There are many factors affecting the production capacity of the ball mill, in addition to the nature of the material (grain size, hardness, density, temperature and humidity), the degree of\n\n•", null, "### The grinding balls bulk weight in fully unloaded mill\n\n11/04/2017· Method with complete grinding media discharge from the mill inner drum. Method without unloading grinding balls. Calculations by the first method are most accurate, but require a lot of labor costs and time. In this article, we will consider the technique for determining the grinding balls bulk weight in fully unloaded mill. This method used in the mills repair (armor plates replacement). The\n\n•", null, "### Ball Mill Calculations Ball Mill\n\nBall Mill Motor Power Draw Sizing And Design Formula. The following equation is used to determine the power that wet grinding overflow ball mills should draw for mills larger than 33 meters 10 feet diameter inside liners the top size of the balls used affects the power drawn by the mill this is called the ball size factor s rod and ball mills by ca rowland and dm kjos allischalmers\n\n•", null, "### Ball Mill Grinding Media Calculation Ball Mill\n\nTo calculate grinding media charge for continuous type ball mill m 0000676 x d2 x l example to calculate grinding media charge for a 180 cm dia x 180 cm long batch type ball mill with duralox 50 mm thick bricks formula to be used is m 0000929 x d2 x l d 180 10 170 cms. Get Details.\n\n•", null, "### The working principle of ball mill Meetyou Carbide\n\n22/05/2019· Therefore, in the rolling ball mill, the grinding efficiency is increased as the ball diameter decreases. It has been proven that the highest grinding efficiency can be obtained with a small ball of .mm diameter. However, the ball diameter is too small to wear too fast, and it is also difficult to discharge due to the small gap of the ball. Therefore, the ball used in wet-grinding the mixture\n\n•", null, "### Ball Mill Speed Calculation Formula For Wet Grinding\n\nBall Mill Grinding Media Calculation. To calculate grinding media charge for continuous type ball mill m 0000676 x d2 x l example to calculate grinding media charge for a 180 cm dia x 180 cm long batch type ball mill with duralox 50 mm thick bricks formula to\n\n•", null, "### Ball mill Wikipedia\n\nA ball mill, a type of grinder, is a cylindrical device used in grinding (or mixing) materials like ores, chemicals, ceramic raw materials and paints.Ball mills rotate around a horizontal axis, partially filled with the material to be ground plus the grinding medium. Different materials are used as media, including ceramic balls, flint pebbles, and stainless steel balls.\n\n•", null, "### bead mill grinding media calculation\n\ncalculate ball mill grinding media in cement . May 19, 2014 calculate ball mill grinding media in cement, Links: goo.gl/DII9h4 More details : goo.gl/N1nfWU (Hot!!!) Zenith is quite Get Price. Energy distribution and particle trajectories in a grinding chamber of Sep 19, 2002 The calculations are restricted to a stationary operation of a stirred ball mill in the absence of" ]	[ null, "https://www.climbwork.cz/pic/2015/116.jpg", null, "https://www.climbwork.cz/pic/2015/165.jpg", null, "https://www.climbwork.cz/pic/2015/137.jpg", null, "https://www.climbwork.cz/pic/2015/150.jpg", null, "https://www.climbwork.cz/pic/2015/51.jpg", null, "https://www.climbwork.cz/pic/2015/29.jpg", null, "https://www.climbwork.cz/pic/2015/86.jpg", null, "https://www.climbwork.cz/pic/2015/190.jpg", null, "https://www.climbwork.cz/pic/2015/199.jpg", null, "https://www.climbwork.cz/pic/2015/135.jpg", null, "https://www.climbwork.cz/pic/2015/53.jpg", null, "https://www.climbwork.cz/pic/2015/6.jpg", null, "https://www.climbwork.cz/pic/2015/51.jpg", null, "https://www.climbwork.cz/pic/2015/39.jpg", null, "https://www.climbwork.cz/pic/2015/111.jpg", null, "https://www.climbwork.cz/pic/2015/158.jpg", null, "https://www.climbwork.cz/pic/2015/21.jpg", null, "https://www.climbwork.cz/pic/2015/139.jpg", null, "https://www.climbwork.cz/pic/2015/127.jpg", null, "https://www.climbwork.cz/pic/2015/87.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86074924,"math_prob":0.94956404,"size":7340,"snap":"2021-31-2021-39","text_gpt3_token_len":1551,"char_repetition_ratio":0.19833697,"word_repetition_ratio":0.14995858,"special_character_ratio":0.21321526,"punctuation_ratio":0.0761194,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9510161,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40],"im_url_duplicate_count":[null,2,null,1,null,1,null,3,null,3,null,1,null,1,null,1,null,1,null,2,null,1,null,1,null,3,null,1,null,1,null,1,null,3,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-18T22:39:39Z\",\"WARC-Record-ID\":\"<urn:uuid:aee21e41-993a-4360-b115-b5ad2a11ec3e>\",\"Content-Length\":\"23768\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d62992d9-e56a-41be-9df8-f22d99bc9775>\",\"WARC-Concurrent-To\":\"<urn:uuid:edd72296-71f1-48b8-b08a-5125808ec9da>\",\"WARC-IP-Address\":\"172.67.135.73\",\"WARC-Target-URI\":\"https://www.climbwork.cz/raymond/2021-02-1183/\",\"WARC-Payload-Digest\":\"sha1:3J7ZZFUXDKKKOZRX27YXY6MIIXECULPT\",\"WARC-Block-Digest\":\"sha1:RSQZ7NJ4YIAW5G7ICRAOV7D2KTPWYIED\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056578.5_warc_CC-MAIN-20210918214805-20210919004805-00179.warc.gz\"}"}
http://forums.wolfram.com/mathgroup/archive/2007/Oct/msg00598.html	[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: Recursion limit: a good idea to ignore?\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg82228] Re: Recursion limit: a good idea to ignore?\n• From: Jon Harrop <jon at ffconsultancy.com>\n• Date: Tue, 16 Oct 2007 03:16:45 -0400 (EDT)\n• References: <feutsm\\$st\\$1@smc.vnet.net>\n\n```Will Robertson wrote:\n> I've written a potentially inefficient algorithm to combine multiple\n> joined polygons into a single one. On large plots I get the warning\n> \"\\$RecursionLimit::reclim: Recursion depth of 256 exceeded.\" and a\n> bunch of errors because of this.\n>\n> Now, should I try and re-write my program (the speed of it is not\n> entirely critical) to avoid recursion or is it \"okay\" to locally\n> remove the recursion limit for such code?\n\nDeep recursion is often a bad idea. In Mathematica code, it is an even worse\nidea that usual.\n\nThe following excerpt demonstrates unbounded recursion that eventually\nresults in a Segmentation Fault (on AMD64 Debian Linux):\n\n\\$ MathKernel\nMathematica 5.1 for Linux x86 (64 bit)\n-- Motif graphics initialized --\n\nIn:= f := 0\n\nIn:= f[n_] := 1+f[n-1]\n\nIn:= f\n\n\\$RecursionLimit::reclim: Recursion depth of 256 exceeded.\n\nOut= 255 + Hold[f[746 - 1]]\n\nIn:= \\$RecursionLimit=Infinity\n\nOut= Infinity\n\nIn:= f\n\nOut= 1000\n\nIn:= f\n\nOut= 10000\n\nIn:= f\nSegmentation fault\n\nSo recursing too deep (applying your function to a large problem) can cause\nthe whole Mathematica kernel to die when it overruns the system stack.\n\nThe term rewriter at the core of Mathematica is a very powerful and\ngeneral-purpose evaluator that requires much bigger stack frames when it\nrecurses compared to most other languages. So Mathematica code is more\nprone to this problem than most other languages.\n\nFor example, the equivalent OCaml code:\n\n\\$ ocamlnat\nObjective Caml version 3.10.0 - native toplevel\n\n# let rec f = function\n\| 0 -> 0\n\| n -> 1 + f(n - 1);;\nval f : int -> int = <fun>\n# f 1000;;\n- : int = 1000\n# f 10000;;\n- : int = 10000\n# f 100000;;\n- : int = 100000\n# f 1000000;;\nStack overflow during evaluation (looping recursion?).\n#\n\nNote that OCaml recursed ~10x deeper before failing and failed with a nice\ncatchable exception.\n\nIn the language zoo, there are some notable exceptions that do not suffer\nfrom this problem. The SML/NJ compiler is one of the most famous examples.\nBefore compiling code, it rearranges all function calls into continuation\npassing style (CPS) so that they consume heap space rather than the (far\nmore limited) system stack space.\n\nI have not studied Mathematica's implementation of tail recursion in detail\nso I cannot say whether or not performing such a transformation will help\n\nThe solution to your problem is to avoid recursive function calls that\nconsume stack space. This typically entails either rewriting the offending\nfunction in tail recursive form, or rewriting it in imperative form.\n\n--\nDr Jon D Harrop, Flying Frog Consultancy\nhttp://www.ffconsultancy.com/products/?u\n\n```\n\n• Prev by Date: Re: Convert to Text Display has no keyboard short cut\n• Next by Date: Re: Run notebook so graphs are displayed and not their code\n• Previous by thread: Re: Recursion limit: a good idea to ignore?\n• Next by thread: Re: Recursion limit: a good idea to ignore?" ]	[ null, "http://forums.wolfram.com/mathgroup/images/head_mathgroup.gif", null, "http://forums.wolfram.com/mathgroup/images/head_archive.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/2.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/0.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/0.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/7.gif", null, "http://forums.wolfram.com/mathgroup/images/search_archive.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.819393,"math_prob":0.77954304,"size":3036,"snap":"2019-26-2019-30","text_gpt3_token_len":791,"char_repetition_ratio":0.105540894,"word_repetition_ratio":0.016129032,"special_character_ratio":0.28985506,"punctuation_ratio":0.14821124,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9526161,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-18T16:21:09Z\",\"WARC-Record-ID\":\"<urn:uuid:0048c486-dd50-4ac7-8f0e-b2367ec1eabe>\",\"Content-Length\":\"44311\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fc3716a9-1580-43ac-b560-a58da52d7ed2>\",\"WARC-Concurrent-To\":\"<urn:uuid:7128dfe9-0a38-4772-ba5b-3ad3dc882774>\",\"WARC-IP-Address\":\"140.177.205.73\",\"WARC-Target-URI\":\"http://forums.wolfram.com/mathgroup/archive/2007/Oct/msg00598.html\",\"WARC-Payload-Digest\":\"sha1:MBQ6JBN4AR6YUB46LNNNNLYM5XBQDZOB\",\"WARC-Block-Digest\":\"sha1:ZFDBVNWGED4GMPLRNTOKWF4TRKHCMSPQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525659.27_warc_CC-MAIN-20190718145614-20190718171614-00541.warc.gz\"}"}