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https://www.w3resource.com/java-exercises/datetime/java-datetime-exercise-31.php	[ " Java exercises: Compute the difference between two dates (Hours, minutes, milli, seconds and nano) - w3resource", null, "# Java DateTime, Calendar Exercises: Compute the difference between two dates (Hours, minutes, milli, seconds and nano)\n\n## Java DateTime, Calendar: Exercise-31 with Solution\n\nWrite a Java program to compute the difference between two dates (Hours, minutes, milli, seconds and nano).\n\nSample Solution:\n\nJava Code:\n\n``````import java.time.;\nimport java.util.;\n\npublic class Exercise31 {\npublic static void main(String[] args)\n{\nLocalDateTime dateTime = LocalDateTime.of(2016, 9, 16, 0, 0);\nLocalDateTime dateTime2 = LocalDateTime.now();\nint diffInNano = java.time.Duration.between(dateTime, dateTime2).getNano();\nlong diffInSeconds = java.time.Duration.between(dateTime, dateTime2).getSeconds();\nlong diffInMilli = java.time.Duration.between(dateTime, dateTime2).toMillis();\nlong diffInMinutes = java.time.Duration.between(dateTime, dateTime2).toMinutes();\nlong diffInHours = java.time.Duration.between(dateTime, dateTime2).toHours();\nSystem.out.printf(\"\\nDifference is %d Hours, %d Minutes, %d Milli, %d Seconds and %d Nano\\n\\n\",\ndiffInHours, diffInMinutes, diffInMilli, diffInSeconds, diffInNano );\n\n}\n}\n```\n```\n\nSample Output:\n\n```Difference is 6686 Hours, 401180 Minutes, 24070844780 Milli, 24070844 Seconds and 780000000 Nano\n```\n\nN.B.: The result may varry for your system date and time.\n\nFlowchart:", null, "Java Code Editor:\n\nImprove this sample solution and post your code through Disqus\n\nWhat is the difficulty level of this exercise?\n\n\n\nNew Content: Composer: Dependency manager for PHP, R Programming" ]	[ null, "https://www.w3resource.com/w3r_images/java-exercises.png", null, "https://www.w3resource.com/w3r_images/java-datetime-exercise-flowchart-31.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7707749,"math_prob":0.8725169,"size":1609,"snap":"2019-26-2019-30","text_gpt3_token_len":374,"char_repetition_ratio":0.1576324,"word_repetition_ratio":0.13541667,"special_character_ratio":0.24860162,"punctuation_ratio":0.28387097,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97850394,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-23T15:39:49Z\",\"WARC-Record-ID\":\"<urn:uuid:e1f3edf7-62cb-4708-aa85-b9d7d7c0a3d5>\",\"Content-Length\":\"105568\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0e45ea64-7d34-4997-b802-0f812b61200b>\",\"WARC-Concurrent-To\":\"<urn:uuid:bff08968-6248-43f6-b1c5-683e724ed1f6>\",\"WARC-IP-Address\":\"104.25.131.109\",\"WARC-Target-URI\":\"https://www.w3resource.com/java-exercises/datetime/java-datetime-exercise-31.php\",\"WARC-Payload-Digest\":\"sha1:GX7URM7BBYB7FQM3WOOZ466OCLT53OOD\",\"WARC-Block-Digest\":\"sha1:TDLPGNZ5CVWZ5V36V6KADJW2K6ZYZK3Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195529480.89_warc_CC-MAIN-20190723151547-20190723173547-00454.warc.gz\"}"}
https://www.projecteuclid.org/euclid.dmj/1461252850	[ "Duke Mathematical Journal\n\nGamma classes and quantum cohomology of Fano manifolds: Gamma conjectures\n\nAbstract\n\nWe propose Gamma conjectures for Fano manifolds which can be thought of as a square root of the index theorem. Studying the exponential asymptotics of solutions to the quantum differential equation, we associate a principal asymptotic class $A_{F}$ to a Fano manifold $F$. We say that $F$ satisfies Gamma conjecture I if $A_{F}$ equals the Gamma class $\\widehat{\\Gamma}_{F}$. When the quantum cohomology of $F$ is semisimple, we say that $F$ satisfies Gamma conjecture II if the columns of the central connection matrix of the quantum cohomology are formed by $\\widehat{\\Gamma}_{F}\\operatorname{Ch}(E_{i})$ for an exceptional collection $\\{E_{i}\\}$ in the derived category of coherent sheaves $\\mathcal{D}^{b}_{\\mathrm{coh}}(F)$. Gamma conjecture II refines a part of a conjecture by Dubrovin. We prove Gamma conjectures for projective spaces and Grassmannians.\n\nArticle information\n\nSource\nDuke Math. J., Volume 165, Number 11 (2016), 2005-2077.\n\nDates\nReceived: 18 June 2014\nRevised: 6 September 2015\nFirst available in Project Euclid: 21 April 2016\n\nPermanent link to this document\nhttps://projecteuclid.org/euclid.dmj/1461252850\n\nDigital Object Identifier\ndoi:10.1215/00127094-3476593\n\nMathematical Reviews number (MathSciNet)\nMR3536989\n\nZentralblatt MATH identifier\n1350.14041\n\nCitation\n\nGalkin, Sergey; Golyshev, Vasily; Iritani, Hiroshi. Gamma classes and quantum cohomology of Fano manifolds: Gamma conjectures. Duke Math. J. 165 (2016), no. 11, 2005--2077. doi:10.1215/00127094-3476593. https://projecteuclid.org/euclid.dmj/1461252850" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5178158,"math_prob":0.8497016,"size":19417,"snap":"2019-43-2019-47","text_gpt3_token_len":6392,"char_repetition_ratio":0.20120537,"word_repetition_ratio":0.07619048,"special_character_ratio":0.3395993,"punctuation_ratio":0.28061742,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99819064,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-19T21:12:24Z\",\"WARC-Record-ID\":\"<urn:uuid:925d5e32-2f55-4437-a64a-663d78719350>\",\"Content-Length\":\"60717\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1b65e627-0d20-416b-90b3-0068089a98f1>\",\"WARC-Concurrent-To\":\"<urn:uuid:7c2e637a-13c2-4209-a8bc-13211245402f>\",\"WARC-IP-Address\":\"132.236.27.47\",\"WARC-Target-URI\":\"https://www.projecteuclid.org/euclid.dmj/1461252850\",\"WARC-Payload-Digest\":\"sha1:TSXG54O66AR3GCV3REFHZCRWVG7GQHCN\",\"WARC-Block-Digest\":\"sha1:UVDNO7VXYBPKLFT7AIGHZFQNRUFJQGHF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986697760.44_warc_CC-MAIN-20191019191828-20191019215328-00023.warc.gz\"}"}
https://open.bccampus.ca/browse-our-collection/find-open-textbooks/?uuid=c732fe64-79c3-4638-aba0-b2ca258244a1&contributor=&keyword=&subject=Math	[ "## Elementary Differential Equations with Boundary Value Problems\n\nSeptember 30, 2014 \| Updated: May 10, 2019\nAuthor: William F. Trench, Trinity University\n\nElementary Differential Equations with Boundary Value Problems is written for students in science, engineering, and mathematics who have completed calculus through partial differentiation. If your syllabus includes Chapter 10 (Linear Systems of Differential Equations), your students should have some preparation in linear algebra. In writing this book, the author aimed to make the language easy to understand. As an elementary text, this book is written in an informal but mathematically accurate way, illustrated by appropriate graphics. The author has tried to formulate mathematical concepts succinctly in language that students can understand and has minimized the number of explicitly stated theorems and defonitions, preferring to deal with concepts in a more conversational way, copiously illustrated by 299 completely worked out examples. Where appropriate, concepts and results are depicted in 188 figures.This text also includes 2041 numbered exercises, many with several parts. They range in difficulty from routine to very challenging. •\n\nSubject Areas\nMath/Stats, Math - General\n\nOriginal source\ndigitalcommons.trinity.edu\n\nTell us you are using this Open Textbook\n\nSupport for adapting an open textbook\n\nNeed help?\n\n#### Get This Book\n\nSelect a file format", null, "" ]	[ null, "https://i.creativecommons.org/l/by-nc-sa/4.0/88x31.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92427534,"math_prob":0.41897148,"size":1912,"snap":"2021-31-2021-39","text_gpt3_token_len":372,"char_repetition_ratio":0.09067086,"word_repetition_ratio":0.041666668,"special_character_ratio":0.17730126,"punctuation_ratio":0.1221865,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9605712,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-17T23:24:36Z\",\"WARC-Record-ID\":\"<urn:uuid:4935fd1d-f4a6-46c0-992d-53b306a89da2>\",\"Content-Length\":\"92961\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f69bff63-b0bc-4765-8dab-45422159e0c0>\",\"WARC-Concurrent-To\":\"<urn:uuid:5e5bb4d1-592c-4272-8c82-6b5b2243b5bc>\",\"WARC-IP-Address\":\"204.239.18.16\",\"WARC-Target-URI\":\"https://open.bccampus.ca/browse-our-collection/find-open-textbooks/?uuid=c732fe64-79c3-4638-aba0-b2ca258244a1&contributor=&keyword=&subject=Math\",\"WARC-Payload-Digest\":\"sha1:SHNHTIMVUXGGOGRLQYZCMDDWVSVZZOX5\",\"WARC-Block-Digest\":\"sha1:JMQUFE6INZ2ZNMF5SXKRQGEHC6PUJAMI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780055808.78_warc_CC-MAIN-20210917212307-20210918002307-00533.warc.gz\"}"}
https://www.hackerearth.com/practice/basic-programming/complexity-analysis/time-and-space-complexity/tutorial/	[ "Basic Programming\nTopics:\nTime and Space Complexity\n• Input/Output\n• Complexity Analysis\n• Time and Space Complexity\n• Implementation\n• Operators\n• Bit Manipulation\n• Recursion\n\n# Time and Space Complexity\n\nSometimes, there are more than one way to solve a problem. We need to learn how to compare the performance different algorithms and choose the best one to solve a particular problem. While analyzing an algorithm, we mostly consider time complexity and space complexity. Time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function of the length of the input. Similarly, Space complexity of an algorithm quantifies the amount of space or memory taken by an algorithm to run as a function of the length of the input.\n\nTime and space complexity depends on lots of things like hardware, operating system, processors, etc. However, we don't consider any of these factors while analyzing the algorithm. We will only consider the execution time of an algorithm.\n\nLets start with a simple example. Suppose you are given an array $A$ and an integer $x$ and you have to find if $x$ exists in array $A$.\n\nSimple solution to this problem is traverse the whole array $A$ and check if the any element is equal to $x$.\n\nfor i : 1 to length of A\nif A[i] is equal to x\nreturn TRUE\nreturn FALSE\n\n\nEach of the operation in computer take approximately constant time. Let each operation takes $c$ time. The number of lines of code executed is actually depends on the value of $x$. During analyses of algorithm, mostly we will consider worst case scenario, i.e., when $x$ is not present in the array $A$. In the worst case, the if condition will run $N$ times where $N$ is the length of the array $A$. So in the worst case, total execution time will be $(N * c + c)$. $N * c$ for the if condition and $c$ for the return statement ( ignoring some operations like assignment of $i$ ).\n\nAs we can see that the total time depends on the length of the array $A$. If the length of the array will increase the time of execution will also increase.\n\nOrder of growth is how the time of execution depends on the length of the input. In the above example, we can clearly see that the time of execution is linearly depends on the length of the array. Order of growth will help us to compute the running time with ease. We will ignore the lower order terms, since the lower order terms are relatively insignificant for large input. We use different notation to describe limiting behavior of a function.\n\n$O$-notation:\nTo denote asymptotic upper bound, we use $O$-notation. For a given function $g(n)$, we denote by $O(g(n))$ (pronounced “big-oh of g of n”) the set of functions:\n$O(g(n)) =$ { $f(n)$ : there exist positive constants $c$ and $n_0$ such that $0 \\le f (n) \\le c * g(n)$ for all $n \\ge n_0$ }\n\n$\\Omega$-notation:\nTo denote asymptotic lower bound, we use $\\Omega$-notation. For a given function $g(n)$, we denote by $\\Omega(g(n))$ (pronounced “big-omega of g of n”) the set of functions:\n$\\Omega(g(n)) =$ { $f(n)$ : there exist positive constants $c$ and $n_0$ such that $0 \\le c * g(n) \\le f(n)$ for all $n \\ge n_0$ }\n\n$\\Theta$-notation:\nTo denote asymptotic tight bound, we use $\\Theta$-notation. For a given function $g(n)$, we denote by $\\Theta(g(n))$ (pronounced “big-theta of g of n”) the set of functions:\n$\\Theta(g(n)) =$ { $f (n)$ : there exist positive constants $c_1,\\;c_2$ and $n_0$ such that $0 \\le c_1 * g(n) \\le f (n) \\le c_2 * g(n)$ for all $n \\gt n_0$ }", null, "Time complexity notations\n\nWhile analysing an algorithm, we mostly consider $O$-notation because it will give us an upper limit of the execution time i.e. the execution time in the worst case.\n\nTo compute $O$-notation we will ignore the lower order terms, since the lower order terms are relatively insignificant for large input.\nLet $f(N) = 2 * N^2 + 3 * N + 5$\n$O(f(N)) = O(2 * N^2 + 3 * N + 5) = O(N^2)$\n\nLets consider some example:\n\n1.\n\nint count = 0;\nfor (int i = 0; i < N; i++)\nfor (int j = 0; j < i; j++)\ncount++;\n\n\nLets see how many times count++ will run.\n\nWhen $i = 0$, it will run $0$ times.\nWhen $i = 1$, it will run $1$ times.\nWhen $i = 2$, it will run $2$ times and so on.\n\nTotal number of times count++ will run is $0 + 1 + 2 + ... + (N-1) = \\frac{N * (N-1)}{2}$. So the time complexity will be $O(N^2)$.\n\n2.\n\nint count = 0;\nfor (int i = N; i > 0; i /= 2)\nfor (int j = 0; j < i; j++)\ncount++;\n\nThis is a tricky case. In the first look, it seems like the complexity is $O(N * logN)$. $N$ for the $j's$ loop and $logN$ for $i's$ loop. But its wrong. Lets see why.\n\nThink about how many times count++ will run.\n\nWhen $i = N$, it will run $N$ times.\nWhen $i = N / 2$, it will run $N / 2$ times.\nWhen $i = N / 4$, it will run $N / 4$ times and so on.\n\nTotal number of times count++ will run is $N + N / 2 + N / 4 + ... + 1 = 2 * N$. So the time complexity will be $O(N)$.\n\nThe table below is to help you understand the growth of several common time complexities, and thus help you judge if your algorithm is fast enough to get an Accepted ( assuming the algorithm is correct ).\n\nLength of Input (N) Worst Accepted Algorithm\n$\\le [10..11]$ $O(N!), O(N^6)$\n$\\le [15..18]$ $O(2^N * N^2)$\n$\\le [18..22]$ $O(2^N * N)$\n$\\le 100$ $O(N^4)$\n$\\le 400$ $O(N^3)$\n$\\le 2K$ $O(N^2 * logN)$\n$\\le 10K$ $O(N^2)$\n$\\le 1M$ $O(N * logN)$\n$\\le 100M$ $O(N), O(logN), O(1)$\n\nGo to next tutorial\n\nContributed by: Akash Sharma" ]	[ null, "https://he-s3.s3.amazonaws.com/media/uploads/a1dddec.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82132775,"math_prob":0.9999503,"size":5326,"snap":"2019-43-2019-47","text_gpt3_token_len":1582,"char_repetition_ratio":0.12514092,"word_repetition_ratio":0.19081633,"special_character_ratio":0.35035673,"punctuation_ratio":0.11281589,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000052,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-12T00:56:47Z\",\"WARC-Record-ID\":\"<urn:uuid:379e12ed-eb12-4d1f-bce1-df641e9bc8b9>\",\"Content-Length\":\"93266\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:223ac813-ccab-4216-8967-9ed6d5a6bfa5>\",\"WARC-Concurrent-To\":\"<urn:uuid:022989c8-4669-4e33-9f25-39d9bdac1e00>\",\"WARC-IP-Address\":\"54.169.44.24\",\"WARC-Target-URI\":\"https://www.hackerearth.com/practice/basic-programming/complexity-analysis/time-and-space-complexity/tutorial/\",\"WARC-Payload-Digest\":\"sha1:DUKHRJVCHF2E5NSTH2W5MSRXKMLH63HT\",\"WARC-Block-Digest\":\"sha1:NBBS4QXAAV5OLQB3EI77AHPYP4QCIA5J\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496664469.42_warc_CC-MAIN-20191112001515-20191112025515-00279.warc.gz\"}"}
https://answers.everydaycalculation.com/add-fractions/12-5-plus-12-6	[ "Solutions by everydaycalculation.com\n\nAdd 12/5 and 12/6\n\n1st number: 2 2/5, 2nd number: 2 0/6\n\n12/5 + 12/6 is 22/5.\n\nSteps for adding fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 5 and 6 is 30\n2. For the 1st fraction, since 5 × 6 = 30,\n12/5 = 12 × 6/5 × 6 = 72/30\n3. Likewise, for the 2nd fraction, since 6 × 5 = 30,\n12/6 = 12 × 5/6 × 5 = 60/30\n4. Add the two fractions:\n72/30 + 60/30 = 72 + 60/30 = 132/30\n5. After reducing the fraction, the answer is 22/5\n6. In mixed form: 42/5\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad" ]	[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6368537,"math_prob":0.9992146,"size":408,"snap":"2019-43-2019-47","text_gpt3_token_len":194,"char_repetition_ratio":0.23019803,"word_repetition_ratio":0.0,"special_character_ratio":0.5588235,"punctuation_ratio":0.07272727,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99705034,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-19T20:39:09Z\",\"WARC-Record-ID\":\"<urn:uuid:dbe473f6-e23c-49a0-b8c9-6254b47e2ab3>\",\"Content-Length\":\"8684\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5f8a3cfc-a470-4935-a885-f803598bd4e0>\",\"WARC-Concurrent-To\":\"<urn:uuid:707737b8-93e6-4296-baef-ab38d957ade8>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/add-fractions/12-5-plus-12-6\",\"WARC-Payload-Digest\":\"sha1:F2BBLHLYIOOE7CHHVRQ6IIOPIT36SEP2\",\"WARC-Block-Digest\":\"sha1:MYPL24BQVWTNKQ45IVPXDVPIXBEPVYP4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986697760.44_warc_CC-MAIN-20191019191828-20191019215328-00336.warc.gz\"}"}
https://analyticsindiamag.com/hands-on-to-reagent-end-to-end-platform-for-applied-reinforcement-learning/	[ "###### Hands-on to ReAgent: End-to-End Platform for Applied Reinforcement Learning", null, "# Hands-on to ReAgent: End-to-End Platform for Applied Reinforcement Learning", null, "Facebook ReAgent, previously known as Horizon is an end-to-end platform for using applied Reinforcement Learning in order to solve industrial problems. The main purpose of this framework is to make the development & experimentation of deep reinforcement algorithms fast. ReAgent is built on Python. It uses PyTorch framework for data modelling\n\nAnd training and TorchScript for serving. ReAgent holds different algorithms for data preprocessing, feature engineering, model training & evaluation and lastly for optimized serving. ReAgent was first presented in the research paperHorizon: Facebook’s Open Source Applied Reinforcement Learning Platform by Jason Gauci, Edoardo Conti, Yitao Liang, Kittipat Virochsiri, Yuchen He, Zachary Kaden, Vivek Narayanan, Xiaohui Ye, Zhengxing Chen, Scott Fujimoto.\n\n`Register for Data & Analytics Conclave>>`\n\nThe key features of Facebook’s ReAgent are :\n\n• Capable of handling Large-dimension datasets.\n• Provides optimized algorithms for data preprocessing, training, etc.\n• Gives a highly efficient production environment for model serving.\n\nAlgorithms Supported by ReAgent\n\nWorkflow of ReAgent\n\nThe image below shows the overall workflow of ReAgent for decision making and reasoning. It starts its decision-making process by using predefined rules and then with the help of feedback, multi-armed bandits, those decisions are optimized and at last, contextual feedback trains the contextual bandits and reinforcement learning models. These trained models are then deployed via TorchScript library.\n\nDependencies of ReAgent Platform\n\nPython >=3.7\n\nInstallation\n\nReAgent can be installed using the docker image and manually. In this case, we are cloning the GitHub repository and installing all the required dependencies of ReAgent via pip.\n\n``` %%bash\n%cd /content/ReAgent/ ```\n\nThen install the required python packages:\n\n`!pip install -r requirements.txt `\n\n``` !pip install pytorch_lightning\n!pip install \".[gym]\" ```\n\nYou can check the detailed-installation process here and usage of ReAgent is discussed here.\n\nTo know more about ReAgent Serving Platform(RASP), you can check this documentation.\n\nDemo – Reinforcement Learning on CartPole Problem\n\n###### Yandex Upgrades Machine Learning Library CatBoost\n\nThis demo explains the usage of ReAgent on CartPole reinforcement learning. The code below uses the OpenAI Gym environment.\n\n1. Import all the required modules and packages.\n``` from reagent.gym.envs.gym import Gym\nimport pandas as pd\nfrom matplotlib import pyplot as plt\nimport seaborn as sns\nimport numpy as np\nimport torch\nimport torch.nn.functional as F\nimport tqdm.autonotebook as tqdm ```\n1. Define the environment by passing CartPole to the Gym class.\n``` env = Gym('CartPole-v0')\ndef reset_env(env, seed):\nnp.random.seed(seed)\nenv.seed(seed)\nenv.action_space.seed(seed)\ntorch.manual_seed(seed)\nenv.reset()\nreset_env(env, seed=0) ```\n1. Next, is to create a policy which contains a simple scorer or Multilayer perceptron network and softmax sampler.\n```from reagent.net_builder.discrete_dqn.fully_connected import FullyConnected\nfrom reagent.gym.utils import build_normalizer\nnorm = build_normalizer(env)\nnet_builder = FullyConnected(sizes=, activations=[\"linear\"])\ncartpole_scorer = net_builder.build_q_network(\nstate_feature_config=None,\nstate_normalization_data=norm['state'],\noutput_dim=len(norm['action'].dense_normalization_parameters)) ```\n\nThe idea behind the policy is that agents will simply execute this cart pole environment.\n\n``` from reagent.gym.policies.policy import Policy\nfrom reagent.gym.policies.samplers.discrete_sampler import SoftmaxActionSampler\nfrom reagent.gym.agents.agent import Agent\npolicy = Policy(scorer=cartpole_scorer, sampler=SoftmaxActionSampler())\nagent = Agent.create_for_env(env, policy) ```\n1. Now, create a trainer that takes reinforcement learning algorithms to train. The following trainer can be created from the commands below:\n``` from reagent.training.reinforce import (\nReinforce, ReinforceParams\n)\nfrom reagent.optimizer.union import classes\ntrainer = Reinforce(policy, ReinforceParams(\ngamma=0.99,\n)) ```\n1. After creating a Trainer, start the training of the model by creating a function that changes the observed transitions into a training batch and then evaluating the reward for all episodes via RL interaction loop. The code for it is shown below:\n``` import reagent.types as rlt\ndef to_train_batch(trajectory):\nstate=rlt.FeatureData(torch.from_numpy(np.stack(trajectory.observation)).float()),\naction=F.one_hot(torch.from_numpy(np.stack(trajectory.action)), 2),\nreward=torch.tensor(trajectory.reward),\nlog_prob=torch.tensor(trajectory.log_prob)\n) ```\n\nRun agent on the environment and record the rewards.\n\n``` from reagent.gym.runners.gymrunner import evaluate_for_n_episodes\neval_rewards = evaluate_for_n_episodes(100, env, agent, 500, num_processes=20)\nStart the loop\nnum_episodes = 200\nreward_min = 20\nmax_steps = 200\nreward_decay = 0.8\ntrain_rewards = []\nrunning_reward = reward_min\nfrom reagent.gym.runners.gymrunner import run_episode\nwith tqdm.trange(num_episodes, unit=\" epoch\") as t:\nfor i in t:\ntrajectory = run_episode(env, agent, max_steps=max_steps, mdp_id=i)\nbatch = to_train_batch(trajectory)\ntrainer.train(batch)\nep_reward = trajectory.calculate_cumulative_reward(1.0)\nrunning_reward = reward_decay\nrunning_reward += (1 - reward_decay) ep_reward\ntrain_rewards.append(ep_reward)\nt.set_postfix(reward=running_reward) ```\n1. Finally, print all the rewards on each training episode in a form of the graph as shown below.\n\nConclusion\n\nIn this write-up, we have discussed ReAgent Platform aka Horizon and its demo with an example of CartPole Problem with reinforcement learning.\n\nNote : The codes mentioned above are not suitable for colab, due to some dependency issues. The following is the Jupyter Notebook file, to reproduce the above experiment.\n\nOfficial Codes, Docs & Tutorials are available at:\n\nWhat Do You Think?\n\n`Join our Telegram Group. Be part of an engaging community`" ]	[ null, "https://analyticsindiamag.com/wp-content/uploads/2021/09/Data-Science-AIM-Banner-2022-V4.gif", null, "data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.74262875,"math_prob":0.610258,"size":6494,"snap":"2021-43-2021-49","text_gpt3_token_len":1418,"char_repetition_ratio":0.117103234,"word_repetition_ratio":0.0,"special_character_ratio":0.20095472,"punctuation_ratio":0.16313364,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9532594,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-16T18:08:35Z\",\"WARC-Record-ID\":\"<urn:uuid:ecebdc1e-3f5c-42ae-b10b-adc69987f6a6>\",\"Content-Length\":\"145890\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fe9dbc3b-fb07-4807-b6fc-04f09e50c0d9>\",\"WARC-Concurrent-To\":\"<urn:uuid:c23935a5-dbdc-42f6-be48-e7d026988479>\",\"WARC-IP-Address\":\"104.26.14.105\",\"WARC-Target-URI\":\"https://analyticsindiamag.com/hands-on-to-reagent-end-to-end-platform-for-applied-reinforcement-learning/\",\"WARC-Payload-Digest\":\"sha1:UJVDEKHIFOTIXT5ZBTS2P6KSXBJHPJI7\",\"WARC-Block-Digest\":\"sha1:N4K6VRIEIICMMJFKZS6K2PXJD3WNUCFQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323584913.24_warc_CC-MAIN-20211016170013-20211016200013-00535.warc.gz\"}"}
https://achemmic.com/tag/tazarotene-ic50/	[ "## Heterosis also known as the hybrid energy occurs when the suggest\n\nHeterosis also known as the hybrid energy occurs when the suggest phenotype of hybrid off-spring is better than that of their two inbred parents. studying expression info for each gene can produce prejudiced and varying estimates and unreliable exams of heterosis highly. To deal with these disadvantages Tazarotene IC50 we produce a hierarchical style to acquire information throughout genes. Applying our building framework all of us derive scientific Bayes estimators and a great inference technique to identify gene expression heterosis. Simulation effects show which our proposed technique outperforms the greater traditional technique used to discover gene phrase heterosis. This information has ancillary Sagopilone material on line. = you 2 as well as the offspring (= 3). Allow = you … means the total range of genes beneath study. All of us use to represent the suggest expression standard of gene of genotype sama dengan = minutes {= ? (+ exhibits HPH LPH or Tazarotene IC50 MPH if and only if > 0 > 0 or ≠ 0 respectively. Past work on estimating gene expression heterosis using microarray data (Swanson-Wagner et al. 2006 Wang et al. 2006 Bassene et al. 2010 has used separate estimates for each gene obtained by replacing population means (= 1 2 3 = 1 ··· and are problematic because they are biased and tend to underestimate and (see Appendix A). Though the sample average estimator of is unbiased with only a few observations for each gene in a typical microarray experiment the sample average estimators of may each be highly variable. Because high-throughput technologies measure expression of hundreds of thousands of genes simultaneously we can utilize information across genes to improve estimation and testing of gene expression heterosis for each individual gene. For gene = (? = ? (+ ? \|helps to develop statistical inferences for RICTOR all three types of gene expression heterosis. We model and equal to the absolute value of a draw from a normal distribution with probability 1 ? and and draw inferences about gene expression heterosis from estimates of these posteriors. We compare the empirical Bayes method with the sample average method through simulation studies where datasets were generated based on real heterosis microarray experiments or hypothetical probability models. Simulation studies show that the empirical Bayes Sagopilone estimators of have smaller mean square errors (MSEs) than the sample average estimators that have been used previously. Furthermore the empirical Bayes estimators of and are less biased than the sample average estimators and the inferences we draw using our empirical Bayes approach are superior to traditional approaches for detecting all forms of heterosis. The remainder of the paper proceeds as follows. Section 2 presents the proposed hierarchical model in full detail. Section 3 derives Tazarotene IC50 the empirical Bayes inference and estimators strategy based on the framework constructed in section 2. Section 4 summarizes analysis results of two real experiments. Section 5 presents results of several simulation studies. Section 6 summarizes our work. R code and C code for the analysis of real experiments in section 4 the simulation studies in section 5 and the implementation Tazarotene IC50 of all our algorithms is available upon request. 2 HIERARCHICAL GENE EXPRESSION HETEROSIS MODEL Let denote the normalized log-scale gene expression Tazarotene IC50 measurement for genotype = 1 ··· is the total number of replicates for genotype (= 1 2 3 = 1 ··· by = min{ sama dengan is believed Sagopilone by sama dengan? (+ succumbed (3) uses Smyth (2004). Tazarotene IC50 The blend model for the purpose of in (1) models the cases wherever Sagopilone parental means are even and wherever parental means differ correspondingly. The hyperparameter specifies the proportion of genes which might be expressed among two father and mother equally. Likewise the blend model intended for in (2) describes the cases where mean gene expression in the offspring is equal or not to the average of two parental means. When necessary the model (1)–(3) may be modified as needed to better capture the features of a given dataset. For example the mixture model could include more than one normal distribution component intended for or ≡ (≡ and are the natural sample average estimators of and – given and – are is a two-component mixture distribution where each component density is itself an infinite mixture of normal distributions with." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8870676,"math_prob":0.9381193,"size":4499,"snap":"2021-43-2021-49","text_gpt3_token_len":911,"char_repetition_ratio":0.1250278,"word_repetition_ratio":0.011220196,"special_character_ratio":0.19493221,"punctuation_ratio":0.053424656,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9708389,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-22T00:23:09Z\",\"WARC-Record-ID\":\"<urn:uuid:2b9afd0c-608c-4864-a956-6b9cb4d78c86>\",\"Content-Length\":\"30984\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:482798a4-0996-4147-81af-0726c8aad669>\",\"WARC-Concurrent-To\":\"<urn:uuid:d3642df3-8f58-4574-a144-7ef40b2ec655>\",\"WARC-IP-Address\":\"136.0.111.15\",\"WARC-Target-URI\":\"https://achemmic.com/tag/tazarotene-ic50/\",\"WARC-Payload-Digest\":\"sha1:YUAVEY7GBSBCFD7CUA3BIEEYDQIGFLS2\",\"WARC-Block-Digest\":\"sha1:HBC6EUPBMWBDWJUYPWNQVMDG4PKZA75I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585449.31_warc_CC-MAIN-20211021230549-20211022020549-00638.warc.gz\"}"}
https://www.analyticsvidhya.com/blog/2021/04/simple-understanding-and-implementation-of-knn-algorithm/	[ "Sai Patwardhan — April 21, 2021\nThis article was published as a part of the Data Science Blogathon.\n\n## Overview:\n\nK Nearest Neighbor (KNN) is intuitive to understand and an easy to implement the algorithm. Beginners can master this algorithm even in the early phases of their Machine Learning studies.\n\nThis KNN article is to:\n\n· Understand K Nearest Neighbor (KNN) algorithm representation and prediction.\n\n· Understand how to choose K value and distance metric.\n\n· Required data preparation methods and Pros and cons of the KNN algorithm.\n\n· Pseudocode and Python implementation.\n\n## Introduction:\n\nK Nearest Neighbor algorithm falls under the Supervised Learning category and is used for classification (most commonly) and regression. It is a versatile algorithm also used for imputing missing values and resampling datasets. As the name (K Nearest Neighbor) suggests it considers K Nearest Neighbors (Data points) to predict the class or continuous value for the new Datapoint.\n\nThe algorithm’s learning is:\n\n1. Instance-based learning: Here we do not learn weights from training data to predict output (as in model-based algorithms) but use entire training instances to predict output for unseen data.\n\n2. Lazy Learning: Model is not learned using training data prior and the learning process is postponed to a time when prediction is requested on the new instance.\n\n3. Non -Parametric: In KNN, there is no predefined form of the mapping function.\n\n## How does KNN Work?\n\n1. #### Principle:\n\nConsider the following figure. Let us say we have plotted data points from our training set on a two-dimensional feature space. As shown, we have a total of 6 data points (3 red and 3 blue). Red data points belong to ‘class1’ and blue data points belong to ‘class2’. And yellow data point in a feature space represents the new point for which a class is to be predicted. Obviously, we say it belongs to ‘class1’ (red points)\n\nWhy?\n\nBecause its nearest neighbors belong to that class!\n\nYes, this is the principle behind K Nearest Neighbors. Here, nearest neighbors are those data points that have minimum distance in feature space from our new data point. And K is the number of such data points we consider in our implementation of the algorithm. Therefore, distance metric and K value are two important considerations while using the KNN algorithm. Euclidean distance is the most popular distance metric. You can also use Hamming distance, Manhattan distance, Minkowski distance as per your need. For predicting class/ continuous value for a new data point, it considers all the data points in the training dataset. Finds new data point’s ‘K’ Nearest Neighbors (Data points) from feature space and their class labels or continuous values.\n\nThen:\n\nFor classification: A class label assigned to the majority of K Nearest Neighbors from the training dataset is considered as a predicted class for the new data point.\n\nFor regression: Mean or median of continuous values assigned to K Nearest Neighbors from training dataset is a predicted continuous value for our new data point\n\n2. #### Model Representation\n\nHere, we do not learn weights and store them, instead, the entire training dataset is stored in the memory. Therefore, model representation for KNN is the entire training dataset.\n\n## How to choose the value for K?\n\nK is a crucial parameter in the KNN algorithm. Some suggestions for choosing K Value are:\n\n1. Using error curves: The figure below shows error curves for different values of K for training and test data.\n\nAt low K values, there is overfitting of data/high variance. Therefore test error is high and train error is low. At K=1 in train data, the error is always zero, because the nearest neighbor to that point is that point itself. Therefore though training error is low test error is high at lower K values. This is called overfitting. As we increase the value for K, the test error is reduced.\n\nBut after a certain K value, bias/ underfitting is introduced and test error goes high. So we can say initially test data error is high(due to variance) then it goes low and stabilizes and with further increase in K value, it again increases(due to bias). The K value when test error stabilizes and is low is considered as optimal value for K. From the above error curve we can choose K=8 for our KNN algorithm implementation.\n\n2. Also, domain knowledge is very useful in choosing the K value.\n\n3. K value should be odd while considering binary(two-class) classification.\n\n## Required Data Preparation:\n\n1. Data Scaling: To locate the data point in multidimensional feature space, it would be helpful if all features are on the same scale. Hence normalization or standardization of data will help.\n\n2. Dimensionality Reduction: KNN may not work well if there are too many features. Hence dimensionality reduction techniques like feature selection, principal component analysis can be implemented.\n\n3. Missing value treatment: If out of M features one feature data is missing for a particular example in the training set then we cannot locate or calculate distance from that point. Therefore deleting that row or imputation is required.\n\n## Python implementation:\n\nImplementation of the K Nearest Neighbor algorithm using Python’s scikit-learn library:\n\n### Step 1: Get and prepare data\n\n```import pandas as pd\nimport numpy as np\nimport matplotlib.pyplot as plt\nfrom sklearn.datasets import make_classification\nfrom sklearn.model_selection import train_test_split\nfrom sklearn.preprocessing import StandardScaler\nfrom sklearn.neighbors import KNeighborsClassifier\nfrom sklearn import metrics```\n\nAfter loading important libraries, we create our data using sklearn.datasets with 200 samples, 8 features, and 2 classes. Then data is split into the train(80%) and test(20%) data and scaled using StandardScaler.\n\n```X,Y=make_classification(n_samples= 200,n_features=8,n_informative=8,n_redundant=0,n_repeated=0,n_classes=2,random_state=14)\nX_train, X_test, y_train, y_test= train_test_split(X, Y, test_size= 0.2,random_state=32)\nsc= StandardScaler()\nsc.fit(X_train)\nX_train= sc.transform(X_train)\nsc.fit(X_test)\nX_test= sc.transform(X_test)\nX.shape```\n`(200, 8)`\n\n### Step 2: Find the value for K\n\nFor choosing the K value, we use error curves and K value with optimal variance, and bias error is chosen as K value for prediction purposes. With the error curve plotted below, we choose K=7 for the prediction\n\n```error1= []\nerror2= []\nfor k in range(1,15):\nknn= KNeighborsClassifier(n_neighbors=k)\nknn.fit(X_train,y_train)\ny_pred1= knn.predict(X_train)\nerror1.append(np.mean(y_train!= y_pred1))\ny_pred2= knn.predict(X_test)\nerror2.append(np.mean(y_test!= y_pred2))\n# plt.figure(figsize(10,5))\nplt.plot(range(1,15),error1,label=\"train\")\nplt.plot(range(1,15),error2,label=\"test\")\nplt.xlabel('k Value')\nplt.ylabel('Error')\nplt.legend()```\n\n### Step 3: Predict:\n\nIn step 2, we have chosen the K value to be 7. Now we substitute that value and get the accuracy score = 0.9 for the test data.\n\n```knn= KNeighborsClassifier(n_neighbors=7)\nknn.fit(X_train,y_train)\ny_pred= knn.predict(X_test)\nmetrics.accuracy_score(y_test,y_pred)```\n`0.9`\n\n## Pseudocode for K Nearest Neighbor (classification):\n\nThis is pseudocode for implementing the KNN algorithm from scratch:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8210587,"math_prob":0.9748358,"size":7889,"snap":"2021-43-2021-49","text_gpt3_token_len":1748,"char_repetition_ratio":0.14533925,"word_repetition_ratio":0.0033641716,"special_character_ratio":0.21751806,"punctuation_ratio":0.13951936,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99785817,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-01T08:35:40Z\",\"WARC-Record-ID\":\"<urn:uuid:ae654de9-aadf-4b54-859d-1249761acf2b>\",\"Content-Length\":\"135287\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0dbc0f65-c8bb-4a29-a729-616cbb92a593>\",\"WARC-Concurrent-To\":\"<urn:uuid:0f78bd41-4205-435d-a006-361ca58b2524>\",\"WARC-IP-Address\":\"172.67.38.119\",\"WARC-Target-URI\":\"https://www.analyticsvidhya.com/blog/2021/04/simple-understanding-and-implementation-of-knn-algorithm/\",\"WARC-Payload-Digest\":\"sha1:H2KYF4Q6NFRKGWWVITT3WUMJINS7RN7L\",\"WARC-Block-Digest\":\"sha1:JFWPKXIMZ7B4P2WG3F2TWD6WAMBMPALR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964359976.94_warc_CC-MAIN-20211201083001-20211201113001-00079.warc.gz\"}"}
https://arxiv.org/abs/1609.03895	[ "# Title:Lepton mass and mixing in a simple extension of the Standard Model based on T7 flavor symmetry\n\nAbstract: A simple Standard Model Extension based on $T_7$ flavor symmetry which accommodates lepton mass and mixing with non-zero $\\theta_{13}$ and CP violation phase is proposed. At the tree- level, the realistic lepton mass and mixing pattern is derived through the spontaneous symmetry breaking by just one vacuum expectation value ($v$) which is the same as in the Standard Model. Neutrinos get small masses from one $SU(2)_L$ doublet and two $SU(2)_L$ singlets in which one being in $\\underline{1}$ and the two others in $\\underline{3}$ and $\\underline{3}^*$ under $T_7$ , respectively. The model also gives a remarkable prediction of Dirac CP violation $\\delta_{CP}=172.598^\\circ$ in both normal and inverted hierarchies which is still missing in the neutrino mixing matrix.\n Comments: 23 pages, 6 figures Subjects: High Energy Physics - Phenomenology (hep-ph) Journal reference: Physics of Atomic Nuclei, Vol. 82, No 2, (2019), pp. 168--182 DOI: 10.1134/S1063778819020133 Cite as: arXiv:1609.03895 [hep-ph] (or arXiv:1609.03895v1 [hep-ph] for this version)\n\n## Submission history\n\nFrom: Vo Van Vien [view email]\n[v1] Tue, 13 Sep 2016 15:25:13 UTC (34 KB)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8758997,"math_prob":0.88050586,"size":1130,"snap":"2020-10-2020-16","text_gpt3_token_len":296,"char_repetition_ratio":0.09502664,"word_repetition_ratio":0.046783626,"special_character_ratio":0.2778761,"punctuation_ratio":0.08530806,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9730454,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-26T14:07:27Z\",\"WARC-Record-ID\":\"<urn:uuid:e4e4a27c-2748-4a5a-a3a5-3111340eb289>\",\"Content-Length\":\"20852\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f94f817d-1f48-4284-aef9-46acf0286f0c>\",\"WARC-Concurrent-To\":\"<urn:uuid:dbb61ee3-17eb-4e83-b760-cfe75a8b75c1>\",\"WARC-IP-Address\":\"128.84.21.199\",\"WARC-Target-URI\":\"https://arxiv.org/abs/1609.03895\",\"WARC-Payload-Digest\":\"sha1:F7YMLCABTJCPUGXC2ZIXSFYMUBWOGAO3\",\"WARC-Block-Digest\":\"sha1:I4DSGIEKTU45DJ23Z2OAXLWKAU2ZKWIX\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146342.41_warc_CC-MAIN-20200226115522-20200226145522-00131.warc.gz\"}"}
https://www.daniweb.com/programming/software-development/threads/398790/basic-broken-code-emptying-form-fields-return-problem	[ "Hi guys,\n\nQuestion 1. I'm a pretty rubbish coder, haven't really done much of it in the couple of years since my degree, at work they've recently expressed an interest in getting me back into it. My problem is that i feel as though I've been taught the syntax (which i totally understand) I can easily write up a a for loop or whatever, my problem is knowing how to apply that. So i guess I would define it as knowing the words of another language but not having any idea how to speak fluently lol. Any ideas where i can start getting some help ? useful links would be appreciated, preferably anything that helped you guys that is more about how to code rather than syntax (i plan to spend more time on here etc).\n\nQuestion 2, which is sort of an example of my problem (see below), I actually broke this when trying to fix it (trying to add in the return 1,2,3 for the different possibilities), but basically I'm currently struggling with using forms. Using 2 random numbers the user must enter the correct answer for multiplying the random numbers. How can I empty the fields and basically reset the form after a correct answer ?\n\n``````Public Class MathLoader\n\nDim generator As New Random\n\nDim a1 As Integer = generator.Next(1, 100)\nDim b1 As Integer = generator.Next(1, 100)\n\nPrivate Sub Form1_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load\n\nEnd Sub\n\nPrivate Function correctCheck() As Integer\n\nDim z1 As Double = a1 * b1\nDim x1 As Integer = 0\n\nIf TextBox1.Text = \"\" Then\nLabel3.Text = \"Please enter a value\"\nReturn 2\n\nElseIf TextBox1.Text = z1 Then\n\nLabel3.Text = \"Correct !\"\nans1.Text = z1\nBeep()\nReturn 1\n\nElse\nLabel3.Text = \"Wrong...\"\nReturn 3\n\nEnd If\n\nEnd Function\n\nPrivate Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click\n'Try\n\nran1.Text = a1\nran2.Text = b1\n\nIf ans1.Text = \"\" Then\n\nElseIf ans1.Text <> \"\" Then\nFor index = 1 To 3\nIf Me.correctCheck() = 1 Then\nMessageBox.Show(\"Complete\")\nElseIf Me.correctCheck() = 2 Then\nMessageBox.Show(\"2\")\nElseIf Me.correctCheck() = 3 Then\nMessageBox.Show(\"3\")\nElse\nMessageBox.Show(\"4\")\nEnd If\nNext\nMessageBox.Show(\"You're out of guesses\")\n\nEnd If\n\n'Catch e As Exception\n' MessageBox.Show(e.ToString)\n\n'End Try\n\nEnd Sub\n\nEnd Class\n``````\n\nI learned how to code by writing code. I refer to the msdn library because it is the instructions on how to use the language. Your pretty much just going to have to dive right into it. It will work itself out naturally. When you get stuck find the answer …\n\n## All 2 Replies", null, "I learned how to code by writing code. I refer to the msdn library because it is the instructions on how to use the language. Your pretty much just going to have to dive right into it. It will work itself out naturally. When you get stuck find the answer and keep going. Can't find the answer post it here.\n\nYour above code could look like below. Probably not going to do you any good other than seeing some other code thats not yours. You should look at the variable names though. One thing you can start right now is naming your variables with a meaningful name. You will end up doing it eventually if you like to learn the hard way. You could also start now as your beginning.\n\n``````Public Class Form3\n\nPrivate Sub Form3_Load(ByVal sender As Object, ByVal e As System.EventArgs) Handles Me.Load\nEnd Sub\n\nPrivate Sub ButtonGuessAnswer_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles ButtonGuessAnswer.Click\n'Use static to keep track of how many tries.\nStatic NumberOfGuesses As Integer\n\n'Should be self explanatory\nMessageBox.Show(\"Wrong.\")\nElse\nDim Number1 As Double = CDbl(LabelRandomNumber1.Text)\nDim Number2 As Double = CDbl(LabelRandomNumber2.Text)\n\nIf CDbl(TextBoxAnswer.Text) = Number1 * Number2 Then\nMessageBox.Show(\"Correct.\")\n'Got it right. Start over.\nNumberOfGuesses = 0\nElse\nMessageBox.Show(\"Wrong.\")\nEnd If\nEnd If\n\nNumberOfGuesses += 1\n\nIf NumberOfGuesses = 3 Then\n'Tried three times and couldn't get it.\n'reset the number of guesses and reset\n'the display\nNumberOfGuesses = 0\nEnd If\nEnd Sub\n\n'Resets the screen with new values\nDim Generator As New Random\n\nLabelRandomNumber1.Text = Generator.Next(1, 100)\nLabelRandomNumber2.Text = Generator.Next(1, 100)" ]	[ null, "https://static.daniweb.com/connect/images/anonymous.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81975764,"math_prob":0.74385905,"size":2298,"snap":"2021-04-2021-17","text_gpt3_token_len":577,"char_repetition_ratio":0.1102877,"word_repetition_ratio":0.02631579,"special_character_ratio":0.25500435,"punctuation_ratio":0.12253829,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9646731,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-17T09:13:14Z\",\"WARC-Record-ID\":\"<urn:uuid:c51c78bd-bb0c-4574-b25d-2bcbb9ca3833>\",\"Content-Length\":\"66181\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:50ee2093-686d-451e-8582-aea3796c7b1f>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c5e649d-7e84-41d6-be3b-f49d2212fab8>\",\"WARC-IP-Address\":\"104.22.4.5\",\"WARC-Target-URI\":\"https://www.daniweb.com/programming/software-development/threads/398790/basic-broken-code-emptying-form-fields-return-problem\",\"WARC-Payload-Digest\":\"sha1:NJPXFHZGXMOX557H67UFGVYRU2ULZMBQ\",\"WARC-Block-Digest\":\"sha1:5O6EJXR3MFBK7JTFGM3MYVN6RTZPZWQT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703511903.11_warc_CC-MAIN-20210117081748-20210117111748-00750.warc.gz\"}"}
https://www.numbers.education/41367.html	[ "Is 41367 a prime number? What are the divisors of 41367?\n\n## Parity of 41 367\n\n41 367 is an odd number, because it is not evenly divisible by 2.\n\nFind out more:\n\n## Is 41 367 a perfect square number?\n\nA number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 41 367 is about 203.389.\n\nThus, the square root of 41 367 is not an integer, and therefore 41 367 is not a square number.\n\n## What is the square number of 41 367?\n\nThe square of a number (here 41 367) is the result of the product of this number (41 367) by itself (i.e., 41 367 × 41 367); the square of 41 367 is sometimes called \"raising 41 367 to the power 2\", or \"41 367 squared\".\n\nThe square of 41 367 is 1 711 228 689 because 41 367 × 41 367 = 41 3672 = 1 711 228 689.\n\nAs a consequence, 41 367 is the square root of 1 711 228 689.\n\n## Number of digits of 41 367\n\n41 367 is a number with 5 digits.\n\n## What are the multiples of 41 367?\n\nThe multiples of 41 367 are all integers evenly divisible by 41 367, that is all numbers such that the remainder of the division by 41 367 is zero. There are infinitely many multiples of 41 367. The smallest multiples of 41 367 are:\n\n## Numbers near 41 367\n\n### Nearest numbers from 41 367\n\nFind out whether some integer is a prime number" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8574098,"math_prob":0.998147,"size":370,"snap":"2021-04-2021-17","text_gpt3_token_len":110,"char_repetition_ratio":0.19125684,"word_repetition_ratio":0.0,"special_character_ratio":0.35675675,"punctuation_ratio":0.16304348,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99876475,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-16T10:49:16Z\",\"WARC-Record-ID\":\"<urn:uuid:1c38311f-34b5-49e1-8af5-33cb357dc125>\",\"Content-Length\":\"18950\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a4c34a99-2c10-4727-aa50-9f7f679deb09>\",\"WARC-Concurrent-To\":\"<urn:uuid:d706426b-ff1e-4a98-a06c-428e4732ad4b>\",\"WARC-IP-Address\":\"213.186.33.19\",\"WARC-Target-URI\":\"https://www.numbers.education/41367.html\",\"WARC-Payload-Digest\":\"sha1:QVCQNY4JD3BG6EL32DCFTII3P5DXXDBR\",\"WARC-Block-Digest\":\"sha1:KYUAVL2YKY6CQ7SUG44NAHAM6R6HWDHE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038056325.1_warc_CC-MAIN-20210416100222-20210416130222-00084.warc.gz\"}"}
https://www.teachoo.com/subjects/cbse-maths/class-10th/ch15-10th-probability/	[ "# Chapter 15 Class 10 Probability\n\nGet NCERT Solutions for Chapter 15 Class 10 free at teachoo. Solutions to all exercise questions, examples and optional is available with detailed explanations.\n\nIn Class 9, we studied about Empirical or Experimental Probability.\n\nIn this chapter, we will study\n\n• Theoretical Probability, that is, P(E) = Number of outcomes with E / Total possible outcomes\n• Probability of complementary event, i.e., P(not E)\n• Probability of Impossible and sure events\n• Probability of questions where die is thrown twice\n• Probability of card questions\n• Finding Probability using distance and area\n\nClick on an exercise or a topic link below to get started." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8672455,"math_prob":0.79611504,"size":318,"snap":"2020-45-2020-50","text_gpt3_token_len":66,"char_repetition_ratio":0.11464968,"word_repetition_ratio":0.0,"special_character_ratio":0.1981132,"punctuation_ratio":0.118644066,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99241173,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-02T05:55:02Z\",\"WARC-Record-ID\":\"<urn:uuid:58f8d4ed-fb25-45da-b84e-8f726e1668ec>\",\"Content-Length\":\"36994\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7dfac3ec-3b24-48a2-af40-fbc441979f36>\",\"WARC-Concurrent-To\":\"<urn:uuid:ec752f81-d9ca-4efb-8f3a-b148e64e73e6>\",\"WARC-IP-Address\":\"52.23.32.39\",\"WARC-Target-URI\":\"https://www.teachoo.com/subjects/cbse-maths/class-10th/ch15-10th-probability/\",\"WARC-Payload-Digest\":\"sha1:JEBVUXNC6LBXF2NQIAJKVQAXELLI4QOI\",\"WARC-Block-Digest\":\"sha1:GLZ7OGMNH6XURORBREMIV7G2WSOH4VH6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141692985.63_warc_CC-MAIN-20201202052413-20201202082413-00321.warc.gz\"}"}
https://amses-journal.springeropen.com/articles/10.1186/s40323-020-00159-0	[ "# Cross section shape optimization of wire strands subjected to purely tensile loads using a reduced helical model\n\n## Abstract\n\nThis paper introduces a shape optimization of wire strands subjected to tensile loads. The structural analysis relies on a recently developed reduced helical finite element model characterized by an extreme computational efficacy while accounting for complex geometries of the wires. The model is extended to consider interactions between components and its applicability is demonstrated by comparison with analytical and finite element models. The reduced model is exploited in a design optimization identifying the optimal shape of a 1 + 6 strand by means of a genetic algorithm. A novel geometrical parametrization is applied and different objectives, such as stress concentration and area minimization, and constraints, corresponding to operational limitations and requirements, are analyzed. The optimal shape is finally identified and its performance improvements are compared and discussed against the reference strand. Operational benefits include lower stress concentration and higher load at plastification initiation.\n\n## Introduction\n\nWire ropes are basic structural elements in engineering and construction. Thanks to their complex hierarchical composition, wire ropes design permits to achieve a response tailored for specific applications and load cases. They are used as a structural link in bridges and cranes, for lifting objects or as tracks in cable-ways. They offer high longitudinal stiffness, while keeping a low transversal bending stiffness. This allows for easy storage, movement and deployment, thanks to the use of drums, sheaves and pulleys .\n\nEven though many different designs have been proposed throughout history [2, 3], the general composition of a rope (see Fig. 1) has not changed. The basic element is the helical wire, arranged in a bundle to form a strand. The obtained strands can be themselves helically arranged to obtain the stranded rope. Compared to fibre materials used in fibre ropes (that have been in use for millennia ), the use of structural materials allow for an increased load carrying capability.\n\nThe mining field in the 19th century was a driving industry for the development of wire ropes: the aim was to replace the employed metal chain—characterized by small damage tolerance—with an element that would be comparable in structural response. This is achieved by using rope designs, where multiple load paths architecture provides time for servicing and replacing the damaged component, avoiding catastrophic failure .\n\nOut of the vast number of rope applications, this work focuses on those where loads are mainly tensile, in which case stationary ropes are employed . They are utilized, for example, in cable-stayed bridge, in cable-ways and in cranes as guy lines for booms suspension (see Fig. 2). They can also be found in civil constructions in the form of pre-stressed concrete strands . Stationary ropes are usually multi-layered strands, having therefore each component in a single helical configuration, as opposed to stranded ropes described above.\n\nRopes have many geometrical parameters in defining the overall response of the strand . While at the level of the strand and the rope there are numerous combinations of parameters (number of wires, layout, lay-factor), the very basic component, the wire, offers most often only its diameter as degree of freedom, due to the ease and lower costs in manufacturing circular wires. As a consequence, strands present local stress concentrations due to the radial pressure concentration at the wires contact locations. Departing from the geometrical constraint of round wires, the aforementioned design drawbacks are mitigated permitting to achieve optimized overall operational characteristics.\n\nFor the case of tensile-dominated applications where single strands are used, the focus will be on providing examples of how the geometry could be optimized for minimum weight or minimum stress concentration, while satisfying application-dependent requirements such as limit load, axial stiffness, axial load at plastification or bending stiffness. 1 + 6 strands are basic, yet among the most used, strands. It has been chosen as reference strand in this work for its relatively straightforward geometry. Accordingly, a novel geometrical parametrization for the 1 + 6 wire strands is proposed.\n\nFor the analysis to come, a reduced model is employed. Rope theory literature has been developed since the 1860s and a plethora of models have been proposed. Complexity of analytical models for wire strands span from the simple assumption of helical springs in parallel to a more refined curved beam theory, mainly based on Love’s theory or on general theory of rods by Green and Laws , accounting for bending and torsion. Besides, finite element (FE) models have also being employed to model more complex phenomena as residual stress after manufacturing , contact and friction and electromagnetic interactions within power cables . Reduced helical models [7, 13,14,15] have been introduced in more recent years and utilize the concept of helical symmetry to reduce the computational domain, have also being successfully used in various fields. The computational efficiency of reduced models and their ability to model complex geometries permit to challenge the limitation of purely circular wires and to propose an alternative approach to the strand design, by means of a shape optimization. In order to permit such a procedure, part of the considered domain needs to be modified to allow for the contact definition.\n\nThis paper is structured as follows: “Modeling techniques comparison” presents the modeling technique used and how it stands against alternative techniques; in “Optimization procedure” section the optimization framework is introduced and the selection of objectives and constraints is discussed; “Results” section contains the discussion on the performance benefits of the optimal shape compared to the reference and a sensitivity analysis carried on the resulting strand; finally, conclusions are drawn in “Conclusions” section.\n\n## Modeling techniques comparison\n\n### Reduced helical model\n\nWhen a helical structure is deformed uniformly along its entire length, the state variables (strains and stresses) are uniform along helical lines. Its overall response can be exactly analysed by taking a representative two-dimensional surface. This is a property called translational invariance , and it is exploited to derive a reduced finite element model whose formulation is similar in idea to the generalized plane strain elements . Other models have been proposed that use this same property, such as those by Zubov , Treyssede , Frikha et al. and Karathanasopoulos and Kress . Differently from the aforementioned models, the one used in this work has been derived within the finite strain framework, therefore being able to better describe the wire motions. Additionally, it was developed for complex geometries and interactions on the transverse cross section.\n\nThe reduced model permits to have a complex geometry, while keeping a low number of elements. This allows fine meshes and local strains and stresses to be studied, without the need of a volumetric FE and very computationally expensive simulations. On the other hand though, it is limited by its derivation assumption: only uniform loadcases can be studied, such as axial elongation and twist, radial compaction and thermal expansion . Accordingly, any load case—which determines that each transverse cross section of the structure behaves identically—can be considered.\n\n### Requirements on modeling approaches\n\nFor our optimization, four requirements are essential to be satisfied by the chosen modeling technique. An analytical model as found in Feyrer , and two three-dimensional FE models (based on either on solid volumetric or beam elements) are compared to the reduced model.\n\nAxial response As the axial elongation is the load case to optimize for, our model needs to be able to fully capture the interaction between wires, including stiffening due to contact among wires and material plasticity. Figure 3 shows how all models are able to predict the overall axial behaviour.\n\nComputational efficiency A main focus when approaching an optimization routine is to assure that the core simulation—that computes the objective value—is as efficient as possible, as it is run multiple times. Therefore, in Fig. 4 a comparison between solution times to quantify the speed of each model is shown. Apart from the analytical model, the beam and reduced models are comparable in solving the analysis, with the solid FE being significantly slower.\n\nComplex geometries With the goal of setting up a shape optimization, the chosen model will need to be able to fully describe the geometry of the strand (and in particular of the outer wire). Solid and reduced FE models are the only ones that satisfy this requirement, because both the analytical and the beam FE models rely on a narrow database of cross sections for contact definition.\n\nBending response A calculation of the bending response is also required in the optimization routine, to constrain the strand flexibility. Solid and beam FE models and analytical models can directly describe such a load case. The reduced model, on the other hand, because the transversal slices would not behave independently from their axial location, is inherently not capable of modeling bending.\n\nTable 1 highlights how the reduced model stands out against the alternative modeling approaches.\n\n### Extension of the reduced helical model to account for contact\n\nBecause the influence of contact between wires is important to fully characterize the stress state within the strand, an extension of the model found in was required (Fig. 5b). The model was originally developed for the analysis of a single constituent, either free helices or solid regions (e.g. solid cylinder with inclusions). Strands have instead distinct components that are free to rotate and move relative to each other. Therefore, an interaction law needs to be introduced. Instead of simply merging the contact points , the current work uses a contact law with exponential pressure-overclosure behaviour.\n\nIn order to use the contact definitions already available in Abaqus, a geometrical expedient is introduced. Since each component is locally planar and there is a relative out-of-plane rotation, in order to enable a three-dimensional contact, an auxiliary master surface must be defined. This allows the interaction to actually represent a surface-to-surface contact rather then a line-to-line one, that would eventually create an artificial—localized-kink. This surface is obtained by extruding the nodes of the inner core perpendicularly to the reference plane. These nodes are then connected by shell elements and rigidly constrained to the corresponding parent nodes to guarantee the helical symmetry. Figure 5b shows such contact surface, with highlighted the nodes connected to the corresponding master node lying on the reference cross section.\n\n### Approximation of the bending stiffness\n\nAs suggested in the work by Foti , the bending of a strand exhibits two distinctive extremes.\n\n• Stick phase, where the bending curvature is low enough that the friction between components prevents them from sliding relatively to each other. All wires form a cross section with connected elements associated with high bending stiffness.\n\n• Slip phase, curvatures are high enough that friction can be ignored and each component is assumed to freely bend about its neutral plane, determining an overall reduction in bending stiffness.\n\nThe two values of stiffnesses, both in stick and in slip phase, are well approximated by the bending stiffness of the straight rod having the same transverse cross section.\n\n\\begin{aligned} K_{stick} = E_{0} I_{0} + \\sum \\limits _{i=1}^6 E_{i} {\\tilde{I}}_{i} \\end{aligned}\n(1)\n\\begin{aligned} K_{slip} = E_{0} I_{0} + \\sum \\limits _{i=1}^6 E_{i} I_{i} \\end{aligned}\n(2)\n\nwhere E is the Young modulus, I is the moment of inertia of the each wire with respect to its own neutral plane and $${\\tilde{I}}$$ is the moment of inertia with respect to the strand neutral plane. Subscript 0 refers to the core wire, while values of $$i>0$$ refer to the outer wires ($$i=1 \\cdots 6$$).\n\nThis approximation allows us to consider bending without involving more complex models. Figure 6 shows how the analytically computed stiffness values match the results obtained by Foti . However, the ability to characterize the transition between the two phases (that depends on the friction coefficient $$\\mu$$) is not maintained.\n\nThe axial force applied to the strand also influences the bending response , due to the increased friction at the contact between wires when the strand is elongated. Considering the fact that, for the applications considered in this work, axial forces are high and curvatures are low, the stick phase stiffness $$K_{stick}$$ will be considered.\n\n### Material model\n\nThroughout all simulations presented here the material model is an elastic-ideally plastic constitutive law. Figure 7 shows the stress-strain curve corresponding to the material parameters as in Table 2. This choice of constitutive law allows to model failure by a limit load analysis. The material of the analysed structure is replaced by an ideally plastic material with lower yield stress. This makes the limit load, i.e. the maximum load the structure can sustain before plastic collapse, representative of the breaking load.\n\n## Optimization procedure\n\n### Objectives\n\nThe aim is to obtain wire shapes which reduce local stress concentration and therefore reduce plastification, fatigue damages, thereby extending life time. In addition, lightweight design increases structural efficiency and decreases material costs. As a result, it has been chosen to consider two objectives.\n\nThe first is stress concentration minimization, defined as\n\n\\begin{aligned} \\gamma = \\max \\left( \\frac{\\sigma _{VM}^{max}}{\\sigma _{VM}^n}\\right) \\end{aligned}\n(3)\n\nwhere $$\\sigma _{VM}^{max}$$ is the largest Von Mises stress acting in the cross section (located at the wire-to-wire contact point) and $$\\sigma _{VM}^n$$ is the nominal value at the center of the core wire, i.e. the tensile stress occurring as a result of the applied deformation. Because of the nonlinear local response, the stress concentration at contact point varies with the applied load history. In particular, it will reach its maximum value $$\\gamma$$ at the initiation of plastification (Fig. 10).\n\nThe second objective is area minimization, that, at constant lay-length, directly translates into weight reduction. It is considered as the effective area covered by the material in the transverse cross section. Due to the choice of the ideally plastic constitutive law, when a limit load is given, the minimum value of the area is bounded by the yield stress.\n\n### Constraints\n\nOptimization procedures need to have constraints that avoid infeasible solutions to be accepted. For instance, simplifying a rope structure to a single isotropic rod would prevent any stress concentration, therefore minimizing the objective. In such a case though, the rope would lose the favourable bending flexibility and damage tolerance, thereby not fullfilling fundamental requirements of rope structures. Such characteristics are main factors in the selection of ropes in an application and need to be maintained. While the damage tolerance is kept by solely considering a shape optimization (that keeps the multi-component nature of the strand, contrarily to a topology optimization), the bending stiffness is taken as inequality constraint, where the upper bound is defined by the bending stiffness $$K_{stick}$$ of the reference strand.\n\nAdditionally, each application sets a maximum load the rope is required to carry. The breaking load of the selected rope needs to be higher than such value. Therefore, because the optimal shape needs to satisfy the same requirements as its respective initial geometry, the breaking load is considered as a constraint as well.\n\n### Geometrical setup\n\nFigure 8 shows the geometrical parametrization used in the considered procedure. The optimization aims at a wide variability, while keeping the number of design parameters reasonably low. It presents a straight core wire and 6 helical wires around it. The analysis considers constant the number of wires and the lay-length (i.e. the axial length corresponding to a full turn of an outer wire).\n\nFigure 8b shows the degrees of freedom that our shape parametrization has. Besides the total strand wire radius R and the outer wire diameter d, the shape is parametrized by the use of two auxiliary circles that can be moved and scaled on the cross section. These fillets bring in a total of 3 parameters ($$\\rho _1$$, $$r_1$$ and $$r_2$$).\n\nTo fully define the geometry, the following geometrical constraints are imposed as well:\n\n• Minimum interwire distance (gap) is set to be at the mirror plane (highlighted point 1 in Fig. 8b), to allow for the contact initiation;\n\n• Concave outer shape, with a curvature corresponding to the radius of the strand, R (point 2);\n\n• Flat outer wire surface (point 3) with given angular distance $$\\Omega$$, that permits relative movement between adjacent outer wires without contact.\n\nIn the case which reducing the concentration at the contact point is our objective, the optimal shape would morph into a shape that allows a surface-to-surface contact. Doing so would provide a larger area for radial force transmission and thus reduce the localization. We though encode the geometry so to have the contact surface to be concave or convex, in order not to restrict the design space. Figure 8c shows potential candidates satisfying the geometrical constraints.\n\n### Optimization routine\n\nBecause of the complex geometry and the geometrical constraints to be considered, a genetic algorithm has been chosen to find a global minimum of the considered problem. A pool of 100 different feasible geometries —based on the parametrization—has been created as initial population. The optimization is allowed to have up to 100 generations, with Matlab default values for mutation and crossover .\n\nEach optimization has either area minimization or stress concentration minimization as single objective, as discussed in Section 3.1.\n\nConstraints are enforced by a multiplicative penalty factor as follows:\n\n\\begin{aligned} {\\tilde{f}} = f \\prod _{i=1}^{n} \\left( 1+\\frac{\|g_i-{\\hat{g}}_i\|}{{\\hat{g}}_i} \\right) \\prod _{j=1}^{m} \\left( 1+\\frac{\\mathrm {max}(0,h_j-{\\hat{h}}_j)}{{\\hat{h}}_j} \\right) \\end{aligned}\n\nwhere $$g_i$$ and $$h_j$$ are the current values of the n equality constraint functions and m disequality constraint functions. $${\\hat{g}}_i$$ and $${\\hat{h}}_j$$ are the given constraint values and $${\\tilde{f}}$$ is the objective value of the constrained problem.\n\n## Results\n\nThe reference strand taken as initial design is characterized by a 1 + 6 layout, with a core diameter of 2.50 mm and an outer wire diameter of 2.25 mm. Additional constant geometry properties, as the number of wires n, lay-factor LL, gap g and angular distance $$\\Omega$$ are listed in Table 3. Material properties used correspond to an ideally plastic law, as discussed in “Material Model” section. The strand is extended axially to a nominal axial strain of $$1\\%$$. The objective is stress concentration minimization and the selected constraints are the limit load and bending flexibility of the reference strand.\n\nThe optimization procedure is coordinated by the built-in Matlab R2018a Optimization Toolbox , while each simulation is solved by Abaqus 6.14 with a custom subroutine (called User Element) developed in a previous work .\n\nThe optimal shape is shown in Fig. 9 on the right, where the contour of the Von Mises stress concentration, i.e. the stress normalized with the nominal value measured at the center of the core wire, is displayed. The increment plotted refers to the nominal strain at which the plastification starts, that corresponds (as shown in Fig. 10) to the highest stress concentration within the loading history. The pressure distribution associated with the surface-to-surface contact results in the reduction of stress concentration into a more homogeneous field in the new geometry. The reference strand has a maximum local Von Mises stress of $$148\\%$$ higher than the nominal stress (corresponding to a concentration $$\\gamma = 2.48$$), while the optimal presents only a $$4\\%$$ higher Von Mises stress ($$\\gamma = 1.04$$). The initiation of plastification happens therefore at significantly larger strains, as shown in Fig. 11, where the accumulated equivalent plastic strain at the location of plastic initiation is plotted against the loading history. Delayed plastification means as well that the axial load that the structure can bear without having any local defects is increased, as it is highlighted in Fig. 12 by the value $$F_p^{opt}$$ ($$34.8\\,\\mathrm {kN}$$) being 3.74 times higher than $$F_p^{ref}$$ ($$9.3\\,\\mathrm {kN}$$).\n\nContrarily to Fig. 11, the curves in Fig. 12 do not show any effect of the early local plastification. This is due to the considerably small area affected by this phenomenon, making its contribution to the axial force is negligible. Figure 12 shows the force-strain curves and it can be seen that the limit load is kept as required by the constraint, with the optimal strand being more compliant than the reference strand by less than $$2.5\\%$$. As for the bending flexibility, it showed the same value as the reference, with less than $$0.1\\%$$ variation.\n\nFigure 13 presents another optimal shape, obtained with an alternative choice of objectives and constraints. When axial force at plastic initiation $$F_p$$ is considered as the only constraint, the required transverse surface is allowed to decrease, because it is not bounded by the limit load requirement. If area minimization is therefore considered, the resulting shape shown in Fig. 13b is obtained, with the area value of $$10.8\\, \\text {mm}^2$$, corresponding to the $$37\\%$$ of the reference strand area ($$A = 28.9\\,\\text {mm}^2)$$. A delayed plastification‘ start can provide margin for reducing the safety factor and consequently the required weight.\n\n### Sensitivity analysis\n\nAny production process is subjected to tolerances, and thus it could be expected that the manufactured optimal wire would not match perfectly the computed one. To study this sensitivity, it has been chosen to slightly vary a parameter $$\\rho _1$$, that determines the inner surface curvature of the outer wire. Figure 14 shows the maximum stress concentration measured when adding a small perturbation $$\\Delta \\rho _1$$ to the optimal $${\\hat{\\rho }}_1$$. In both directions (whose corresponding geometries are illustrated in Fig. 14) there is a detrimental effect, due to the reintroduction of local concentration, losing partially the benefit of the optimal shape. Values are though significantly lower than the reference strand value ($$\\gamma = 2.48$$). In particular, the results show that a larger $$\\rho _1$$ ($$\\Delta \\rho _1 > 0$$, corresponding to a smaller curvature of the inner contact surface) is safer, as $$\\gamma$$ increases less for such values.\n\n## Conclusions\n\nThe reduced helical model capability to resolve local stresses has been proven essential to allow for the proposed optimization. It computes stress concentrations without recurring to a solid FE model, that would have rendered the entire routine computationally very expensive. Within the limitations set by the reduced helical model assumptions, the applicability and potential of the chosen approach was demonstrated by showing that a optimized design of the strand, and in particular of the outer wires, was found.\n\nSuch optimization framework complements the state-of-the-art design of strands, since an optimal cross section—providing beneficial characteristics—could be tailored to each application. The strand manufacturer would need to have ways to produce the resulting geometry by successive drawing through custom made dies. While this surely increases the complexity and cost of the strand manufaCturing processes, it is feasible, as non-circular wires have already been used in full-locked spiral rope . Compared to the compaction of strands—the process of radially compressing a strand that originally had round wires—the approach proposed in this work reduce dirt infiltration and determines a better contact pressure distribution, without introducing unwanted pre-stresses. This analysis can directly be extended to more complex geometries such as multi-layer strands and it could also be coupled with other models to analyse the next hierarchical level, the wire rope. For instance, the reduced model could compute the homogenized properties of the wire strand to be used in a beam model, that would effectively simulate a stranded wire rope.\n\n## Availability of data and materials\n\nData available on request from the authors. Plot data is found in the additional material “Plot_data.xlsx”.\n\n## References\n\n1. Cardou A, Jolicoeur C. Mechanical models of helical strands. Appl Mech Rev. 1997;50(1):1.\n\n2. Verreet R. Die Geschichte des Drahtseiles. Drahtwelt. 1989;75(6):100–6.\n\n3. Sayenga D. The Birth and Evaluation of the American Wire Rope Industry. First Annual Wire Rope Proceedings. Pullman, Washington 99164: Engineering Extension Service. Washington: Washington State University; 1980.\n\n4. Costello GA. Theory of wire rope., Mechanical engineering seriesNew York: Springer; 1997.\n\n5. Feyrer K. Wire ropes: tension, endurance, reliability, vol. 14. Berlin: Springer; 2015.\n\n6. Raoof M. Wire recovery length in a helical strand under axial-fatigue loading. Int J Fatig. 1991;13(2):127–32.\n\n7. Filotto FM, Kress G. Nonlinear planar model for helical structures. Comput Struct. 2019;224:106111.\n\n8. Love AEH. Treatise on mathematical theory of elasticity. A treatise on the mathematical theory of elasticity. 4th edn. 1944; p. 643.\n\n9. Green AE, Laws N. A General Theory of Rods. Mech Gener Cont. 1968;293:49–56.\n\n10. Frigerio M, Buehlmann PB, Buchheim J, Holdsworth SR, Dinser S, Franck CM, et al. Analysis of the tensile response of a stranded conductor using a 3D finite element model. Int J Mech Sci. 2016;106:176–83.\n\n11. Xiang L, Wang HY, Chen Y, Guan YJ, Dai LH. Elastic-plastic modeling of metallic strands and wire ropes under axial tension and torsion loads. Int J Solids Struct. 2017;129:103–18.\n\n12. Del-Pino-López JC, Hatlo M, Cruz-Romero P. On simplified 3D finite element simulations of three-core armored power cables. Energies. 2018;11:11.\n\n13. Treyssède F. Elastic waves in helical waveguides. Wave Motion. 2008;45(4):457–70.\n\n14. Frikha A, Cartraud P, Treyssède F. Mechanical modeling of helical structures accounting for translational invariance. Part 1: static behavior. Int J Solids Struct. 2013;50(9):1373–82.\n\n15. Karathanasopoulos N, Kress G. Two dimensional modeling of helical structures, an application to simple strands. Comput Struct. 2016;174:79–84. https://doi.org/10.1016/j.compstruc.2015.08.016.\n\n16. Cheng AHD, Rencis JJ, Abousleiman Y. Generalized plane strain elasticity problems. Trans Model Simul. 1995;10:167–74.\n\n17. Zubov LM. Exact nonlinear theory of tension and torsion of helical springs. Doklady Phys. 2002;47(8):623–6.\n\n18. Foti F, Martinelli L. An analytical approach to model the hysteretic bending behavior of spiral strands. Appl Math Modell. 2016;40(13–14):6451–67.\n\n19. The MathWorks I. Global Optimization Toolbox User’s Guide; 2018.\n\n20. Puzzi S, Carpinteri A. A double-multiplicative dynamic penalty approach for constrained evolutionary optimization. Struct Multidiscip Optimiz. 2008;35(5):431–45.\n\n21. Dassault Systèmes. Abaqus 6.14 Online Documentation; 2014.\n\n22. Bergen Cable Technology I. Cable 101. https://bergencable.com/cable-101.\n\n23. Wikipedia. Chords Bridge. https://en.wikipedia.org/wiki/Chords_Bridge.\n\n24. Kobelco. Kobelco Construction Machinery Europe B.V. https://www.kobelco-europe.com.\n\nNot applicable.\n\n## Funding\n\nThe authors acknowledge the support of the Swiss National Science Foundation (project No. 159583 and Grant No. 200020_1595831).\n\n## Author information\n\nAuthors\n\n### Contributions\n\nFMF performed the optimization simulations and analyses and wrote the paper. FR created the beam and solid models used in the present work and reviewed the paper. GK commented and reviewed the paper. All authors read and approved the final manuscript.\n\n### Corresponding author\n\nCorrespondence to Francesco Maria Filotto.\n\n## Ethics declarations\n\n### Competing interests\n\nThe authors declare that they have no competing interests.", null, "" ]	[ null, "https://amses-journal.springeropen.com/track/article/10.1186/s40323-020-00159-0", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9080596,"math_prob":0.95265114,"size":29553,"snap":"2022-05-2022-21","text_gpt3_token_len":6148,"char_repetition_ratio":0.13577448,"word_repetition_ratio":0.010585586,"special_character_ratio":0.2082699,"punctuation_ratio":0.119885825,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9781507,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-20T14:06:33Z\",\"WARC-Record-ID\":\"<urn:uuid:dc0f73a4-c9cf-4837-89ab-40e824990211>\",\"Content-Length\":\"274323\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4c99807f-357a-479f-a60b-0c82997eaa98>\",\"WARC-Concurrent-To\":\"<urn:uuid:09debd49-68e9-4d6b-ade4-ab9cc06effe5>\",\"WARC-IP-Address\":\"146.75.36.95\",\"WARC-Target-URI\":\"https://amses-journal.springeropen.com/articles/10.1186/s40323-020-00159-0\",\"WARC-Payload-Digest\":\"sha1:5APU6XFCHVBTS4O6CVVDPAO47TUO7DSE\",\"WARC-Block-Digest\":\"sha1:C2FTZMOK7CL55FC6UP7D6Y72ZNTSO6CM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662532032.9_warc_CC-MAIN-20220520124557-20220520154557-00364.warc.gz\"}"}
https://cre8math.com/2018/03/17/the-geometry-of-polynomials/	[ "# The Geometry of Polynomials\n\nI recently needed to make a short demo lecture, and I thought I’d share it with you. I’m sure I’m not the first one to notice this, but I hadn’t seen it before and I thought it was an interesting way to look at the behavior of polynomials where they cross the x-axis.\n\nThe idea is to give a geometrical meaning to an algebraic procedure: factoring polynomials. What is the geometry of the different factors of a polynomial?\n\nLet’s look at an example in some detail:", null, "$f(x)=2(x-4)(x-1)^2.$", null, "Now let’s start looking at the behavior near the roots of this polynomial.", null, "Near", null, "$x=1,$ the graph of the cubic looks like a parabola — and that may not be so surprising given that the factor", null, "$(x-1)$ occurs quadratically.", null, "And near", null, "$x=4,$ the graph passes through the x-axis like a line — and we see a linear factor of", null, "$(x-4)$ in our polynomial.\n\nBut which parabola, and which line? It’s actually pretty easy to figure out. Here is an annotated slide which illustrates the idea.", null, "All you need to do is set aside the quadratic factor of", null, "$(x-1)^2,$ and substitute the root,", null, "$x=1,$ in the remaining terms of the polynomial, then simplify. In this example, we see that the cubic behaves like the parabola", null, "$y=-6(x-1)^2$ near the root", null, "$x=1.$ Note the scales on the axes; if they were the same, the parabola would have appeared much narrower.\n\nWe perform a similar calculation at the root", null, "$x=4.$", null, "Just isolate the linear factor", null, "$(x-4),$ substitute", null, "$x=4$ in the remaining terms of the polynomial, and then simplify. Thus, the line", null, "$y=18(x-4)$ best describes the behavior of the graph of the polynomial as it passes through the x-axis. Again, note the scale on the axes.\n\nWe can actually use this idea to help us sketch graphs of polynomials when they’re in factored form. Consider the polynomial", null, "$f(x)=x(x+1)^2(x-2)^3.$ Begin by sketching the three approximations near the roots of the polynomial. This slide also shows the calculation for the cubic approximation.", null, "Now you can begin sketching the graph, starting from the left, being careful to closely follow the parabola as you bounce off the x-axis at", null, "$x=-1.$", null, "Continue, following the red line as you pass through the origin, and then the cubic as you pass through", null, "$x=2.$ Of course you’d need to plot a few points to know just where to start and end; this just shows how you would use the approximations near the roots to help you sketch a graph of a polynomial.", null, "Why does this work? It is not difficult to see, but here we need a little calculus. Let’s look, in general, at the behavior of", null, "$f(x)=p(x)(x-a)^n$ near the root", null, "$x=a.$ Given what we’ve just been observing, we’d guess that the best approximation near", null, "$x=a$ would just be", null, "$y=p(a)(x-a)^n.$\n\nJust what does “best approximation” mean? One way to think about approximating, calculuswise, is matching derivatives — just think of Maclaurin or Taylor series. My claim is that the first", null, "$n$ derivatives of", null, "$f(x)=p(x)(x-a)^n$ and", null, "$y=p(a)(x-a)^n$ match at", null, "$x=a.$\n\nFirst, observe that the first", null, "$n-1$ derivatives of both of these functions at", null, "$x=a$ must be 0. This is because", null, "$(x-a)$ will always be a factor — since at most", null, "$n-1$ derivatives are taken, there is no way for the", null, "$(x-a)^n$ term to completely “disappear.”\n\nBut what happens when the", null, "$n$th derivative is taken? Clearly, the", null, "$n$th derivative of", null, "$p(a)(x-a)^n$ at", null, "$x=a$ is just", null, "$n!p(a).$ What about the", null, "$n$th derivative of", null, "$f(x)=p(x)(x-a)^n$?\n\nThinking about the product rule in general, we see that the form of the", null, "$n$th derivative must be", null, "$f^{(n)}(x)=n!p(x)+ (x-a)(\\text{terms involving derivatives of } p(x)).$ When a derivative of", null, "$p(x)$ is taken, that means one factor of", null, "$(x-a)$ survives.\n\nSo when we take", null, "$f^{(n)}(a),$ we also get", null, "$n!p(a).$ This makes the", null, "$n$th derivatives match as well. And since the first", null, "$n$ derivatives of", null, "$p(x)(x-a)^n$ and", null, "$p(a)(x-a)^n$ match, we see that", null, "$p(a)(x-a)^n$ is the best", null, "$n$th degree approximation near the root", null, "$x=a.$\n\nI might call this observation the geometry of polynomials. Well, perhaps not the entire geometry of polynomials…. But I find that any time algebra can be illustrated graphically, students’ understanding gets just a little deeper.\n\nThose who have been reading my blog for a while will be unsurprised at my geometrical approach to algebra (or my geometrical approach to anything, for that matter). Of course a lot of algebra was invented just to describe geometry — take the Cartesian coordinate plane, for instance. So it’s time for algebra to reclaim its geometrical heritage. I shall continue to be part of this important endeavor, for however long it takes….", null, "### Vince Matsko\n\nMathematician, educator, consultant, artist, puzzle designer, programmer, blogger, etc., etc. @cre8math\n\n## 6 thoughts on “The Geometry of Polynomials”\n\n1.", null, "William Meisel says:\n\nVince – this is really interesting and I am surprised that I have never come across it anywhere before.\nDoes a similar thing work for functions of two variables? I’m thinking of things that are particularly tough to sketch by hand, like Descarte’s Folium (x^3 + y^3 = 3xy). I have read articles that suggest using something called Newton’s Polygon to sketch these, but I will admit I have not been able to follow the steps that clearly. (My intuition is that this method will not generalize to higher dimensions, because I guess you would have to use partial derivatives, but I am often wrong.)\n\nLike\n\n2.", null, "Vince Matsko says:\n\nWilliam – yes, I got it to work with the Folium of Descartes! And in several other cases that I tried as well. Too much to go into here, but I plan to write a more detailed blog post on what I found out in a few weeks (this weekend I’ll be writing about the Bay Area Mathematical Artists). Sorry to keep you in suspense….but thanks for the inspiration!\n\nLiked by 1 person\n\n3.", null, "Rachel Campos says:\n\nWell written post.\n\nLike" ]	[ null, "https://s0.wp.com/latex.php", null, "https://vincematsko.files.wordpress.com/2018/03/poly0b.png", null, "https://vincematsko.files.wordpress.com/2018/03/poly0c.png", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://vincematsko.files.wordpress.com/2018/03/poly0d.png", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://vincematsko.files.wordpress.com/2018/03/day137poly1.png", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://vincematsko.files.wordpress.com/2018/03/day137poly2.png", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://vincematsko.files.wordpress.com/2018/03/day137poly3.png", null, "https://s0.wp.com/latex.php", null, "https://vincematsko.files.wordpress.com/2018/03/poly1d.png", null, "https://s0.wp.com/latex.php", null, "https://vincematsko.files.wordpress.com/2018/03/poly1f.png", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://2.gravatar.com/avatar/27c6b760e350d92f9864bdedf4dc755d", null, "https://0.gravatar.com/avatar/cdb944b28cbd9b46801bfd351b55c087", null, "https://2.gravatar.com/avatar/27c6b760e350d92f9864bdedf4dc755d", null, "https://1.gravatar.com/avatar/7e8814c77793c3318957b3974db8c27a", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9595482,"math_prob":0.99262357,"size":5069,"snap":"2020-45-2020-50","text_gpt3_token_len":1087,"char_repetition_ratio":0.13030602,"word_repetition_ratio":0.011086474,"special_character_ratio":0.20694417,"punctuation_ratio":0.08730159,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996388,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122],"im_url_duplicate_count":[null,null,null,8,null,8,null,null,null,null,null,8,null,null,null,null,null,8,null,null,null,null,null,null,null,null,null,null,null,8,null,null,null,null,null,null,null,null,null,9,null,null,null,8,null,null,null,8,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-23T22:37:49Z\",\"WARC-Record-ID\":\"<urn:uuid:8bc79dcf-db98-4ec2-8a62-c5baebf501fe>\",\"Content-Length\":\"113952\",\"Content-Type\":\"application/http; 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https://stats.stackexchange.com/questions/560688/why-is-my-highest-coefficient-not-significant-but-lower-ones-are-poisson-regre	[ "# Why is my highest coefficient not significant but lower ones are? (Poisson regression)\n\nI'm looking to see if income level affects the number of pets a person has, controlling for number of children. I have run a GLM (Poisson family) model, and found that if a person is categorised as 'Low' income, their number of expected pets decreases by 21.3%. I also found that for every child a person has, their number of expected pets increases by 4.03%.\n\nHowever, the 21.3% decrease is not significant, and the 4.03% increase is significant. I'm wondering if this is incorrect/whether I have done something wrong? I am aware I may be confused about how statistical significance works, so just want to check whether the below makes sense!\n\nThe R outputs for estimates and p-values look like this:\n\n Estimate Pr(>\|z\|)\nLevel_of_IncomeLow: -0.239043 0.0830 .\nNum_of_children 0.039527 7.54e-10 ***\n\n\nMy calculations:\n\nExponent of -0.239043: 0.787381\n1-0.787381=0.212619\n= 21.3% decrease\n\nExponent of 0.039527: 1.040319\n= 4.03% increase\n\n\nEssentially, is it okay that the larger effect is not statistically significant, but the smaller effect is? It just feels wrong!\n\n• Significance of a test depends on the ratio of the coefficient to its standard error (which you should also report in your analyses). The standard error is determined (i) the sample size, (ii) the variance of the residuals of the model, and (iii) the variation in the independent variables. Components (i) and (ii) affect the tests for both coefficients. However, it's possible that for (iii) there is more \"relative\" variation in the number of children in your sample than the level of income. Jan 16, 2022 at 21:51\n• This means that it is easier to \"detect\" an effect. There could still be an income effect but the sample size is too low to detect it. Jan 16, 2022 at 21:51\n• The fact that it depends on a ratio is extremely important. For example, if you rescale the variable, it affects both the numerator and denominator, and doesn't alter the significance. This makes significance a measure that doesn't depend on the units of measurement. Jan 16, 2022 at 21:54\n• Thank you for all the helpful answers! I will also add standard error into my report Jan 16, 2022 at 23:32\n• You can't compare the size of coefficients that are in different units. There's no basis on which to call one coefficient \"larger\" or \"smaller\" when you're comparing incommensurable units. Jan 17, 2022 at 0:46\n\nlow (vs. high) has a restricted range (i.e., only 2 values). Number of children presumably has a much wider range. The narrower the range of an $$X$$ variable, the less power there is to test it. That's a general principle. Specific to this example, it sounds like the underlying variable (income) was subjected to a median split, or something similar. If so, you should be aware that categorizing a continuous variable has long been considered poor statistical practice.\n\nSignificance is not purely about the magnitude of an effect. For instance, if the parameter/effect is $$1$$ light-year or $$9.46×10^{12}$$ km, then it is different figures but the same distance.\n\n(You might argue that this example is bad because I changed the units, but how are you gonna make sure that the units match when you compare 'pets per unit of income' with 'pets per number of kids'. When you change the currency then you change the effect size.)\n\nSignificance is an expression of the magnitude of an observed effect size in terms of how likely or unlikely it is to have an at least as strong deviation from the value predicted by some null hypothesis, given that the null hypothesis is true.\n\nWhen this probability is low then you might consider the effect as an anomaly from the point of view of the null hypothesis. If you detect an effect with a small magnitude to be an anomaly, but an effect with a large magnitude is not, then it means that the experiment has different sensitivity for the two effects.\n\nis it okay that the larger effect is not statistically significant, but the smaller effect is? It just feels wrong!\n\nIt depends on the relative effect size. Relative to the error of the estimate and this might be different for the two coefficients.\n\nBelow you see an example where the estimated intercept of 1.724 is different from 0 but it is not more significant than the estimated slope of 0.095 being different from 0.\n\nThe slope is a smaller effect size if you compare the plain figures like 0.095 < 1.724. But the slope has a much smaller variation from sample to sample, and such a small effect is significant.", null, "" ]	[ null, "https://i.stack.imgur.com/0XtCS.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91200876,"math_prob":0.9321947,"size":1057,"snap":"2023-14-2023-23","text_gpt3_token_len":290,"char_repetition_ratio":0.10161444,"word_repetition_ratio":0.012121212,"special_character_ratio":0.307474,"punctuation_ratio":0.17256637,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99199045,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-27T19:56:33Z\",\"WARC-Record-ID\":\"<urn:uuid:062a11b8-cc47-4cd0-8d94-79b0b93817a7>\",\"Content-Length\":\"167111\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:77bc0e1f-e38a-428d-bb69-363a6f77e3fd>\",\"WARC-Concurrent-To\":\"<urn:uuid:17002d2d-0227-4fc8-a9b3-aeb94bfa69b7>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/560688/why-is-my-highest-coefficient-not-significant-but-lower-ones-are-poisson-regre\",\"WARC-Payload-Digest\":\"sha1:W4OEA6TVWJIOQ35J7PAT6VR6GITBAEYC\",\"WARC-Block-Digest\":\"sha1:N35JYWUDWBDZHHTSSXOV3YE3FNBBX265\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296948684.19_warc_CC-MAIN-20230327185741-20230327215741-00560.warc.gz\"}"}
https://simple.m.wikipedia.org/wiki/Torque	[ "# Torque\n\ntendency of a force to rotate an object\n\nIn physics, torque is the tendency of a force to turn or twist. If a force is used to begin to spin an object, or to stop an object from spinning, a torque is made.\n\nThe force applied to a lever, multiplied by the distance from the lever's fulcrum, multiplied again by the sine of the angle created, is described as torque. This is also known as \"r cross f,\" or \"force times fulcrum distance times sine theta.\"\n\n## Fulcrum\n\nFulcrum is the axis of rotation or point of support on which a lever turns in raising or moving something.\n\n## Equation\n\nThe equation for torque is:\n\n${\\boldsymbol {\\tau }}=\\mathbf {r} \\times \\mathbf {F} \\,\\!$\n\nwhere F is the net force vector and r is the vector from the axis of rotation to the point where the force is acting. The Greek letter Tau is used to represent torque.\n\nThe units of torque are force multiplied by distance. The SI unit of torque is the newton-metre. The most common English unit is the foot-pound." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8505344,"math_prob":0.98042774,"size":1427,"snap":"2023-14-2023-23","text_gpt3_token_len":343,"char_repetition_ratio":0.12508784,"word_repetition_ratio":0.0,"special_character_ratio":0.23686054,"punctuation_ratio":0.10469314,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99955446,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-30T09:18:45Z\",\"WARC-Record-ID\":\"<urn:uuid:f0f2a30f-745b-488c-9155-24064448c478>\",\"Content-Length\":\"60683\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e541b948-09ab-4316-bf00-422cee67610e>\",\"WARC-Concurrent-To\":\"<urn:uuid:e94a9ced-d2f3-49be-9f5b-cc83b760aed1>\",\"WARC-IP-Address\":\"208.80.154.224\",\"WARC-Target-URI\":\"https://simple.m.wikipedia.org/wiki/Torque\",\"WARC-Payload-Digest\":\"sha1:5DQNDGZQCFPXYBSZRFFQPLQYHTY6OFTN\",\"WARC-Block-Digest\":\"sha1:F6EEWUTZ274ERYFAFBCOHE3V3DEBQTCI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224645417.33_warc_CC-MAIN-20230530063958-20230530093958-00442.warc.gz\"}"}
https://www.millisecond.com/support/docs/current/html/language/attributes/position.htm	[ "", null, "", null, "Inquisit Language Reference\n\n# position attribute\n\nThe position attribute controls the screen location at which stimuli are presented.\n\n## Member of\n\n<button> <caption> <checkboxes> <clock> <dropdown> <image> <likert> <listbox> <openended> <picture> <radiobuttons> <shape> <slider> <slidertrial> <text> <textbox> <video>\n\n## Syntax\n\n/ position = (x value, y value)\n\n## Parameters\n\n x value A value or property indicating the horizontal screen coordinate in pixels, percent(default), or points. If x is set to null and y is specified, x is set to the pixel value of y. y value A value or property indicating the vertical screen coordinate in pixels, percent(default), or points. If y is set to null and x is specified, y is set to the pixel value of x.\n\n## Remarks\n\nHorizontal position is relative to the left side. 0% is the left edge of the screen and 100% is the right edge. Vertical position is relative to the top of the screen, with 0% placing the stimulus at the top and 100% placing it at the bottom edge. Percentages may be specified as decimals (e.g., 52.968) for increased precision. The default position is the middle of the screen (50%, 50%).\n\n## Examples\n\nThe following sets the position of the text to the lower left corner of the screen:\n\n``<text sometext>/ items = (\"ipsum\")/ position = (0, 100)</text>``\n\nThe following sets the position of the text to the middle of the screen with 800 X 600 resolution:\n\n``<text sometext>/ items = (\"ipsum\")/ position = (400px, 300px)</text>``\n\nThe following sets the position based on the trial number:\n\n``<text sometext>/ items = (\"ipsum\")/ position = (trial.mytrial.currenttrialnumber * 5, trial.mytrial.currenttrialnumber * 5)</text>``" ]	[ null, "https://www.facebook.com/tr", null, "https://www.millisecond.com/support/docs/current/html/stylesheets/images/up.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7054482,"math_prob":0.96307707,"size":1560,"snap":"2022-27-2022-33","text_gpt3_token_len":401,"char_repetition_ratio":0.15167095,"word_repetition_ratio":0.22134387,"special_character_ratio":0.27051282,"punctuation_ratio":0.113074206,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9505468,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-17T08:29:39Z\",\"WARC-Record-ID\":\"<urn:uuid:19ca24f9-ea09-4728-9022-9d6c0293e9b7>\",\"Content-Length\":\"7511\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7debe4ec-2075-44e7-b2d4-3db09488f327>\",\"WARC-Concurrent-To\":\"<urn:uuid:2536910b-0666-42bc-baa7-9204e6ac9826>\",\"WARC-IP-Address\":\"44.242.33.6\",\"WARC-Target-URI\":\"https://www.millisecond.com/support/docs/current/html/language/attributes/position.htm\",\"WARC-Payload-Digest\":\"sha1:BIDUF644PHJSJGE74C3KB47D4O6LAGSB\",\"WARC-Block-Digest\":\"sha1:LXF7UXPC2LCA4M657PD74MTIG55FBTIL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572870.85_warc_CC-MAIN-20220817062258-20220817092258-00249.warc.gz\"}"}
https://nbviewer.ipython.org/github/gpeyre/numerical-tours/blob/master/matlab/sparsity_7_sudoku.ipynb	[ "# Sudoku using POCS and Sparsity¶\n\nImportant: Please read the installation page for details about how to install the toolboxes. $\\newcommand{\\dotp}{\\langle #1, #2 \\rangle}$ $\\newcommand{\\enscond}{\\lbrace #1, #2 \\rbrace}$ $\\newcommand{\\pd}{ \\frac{ \\partial #1}{\\partial #2} }$ $\\newcommand{\\umin}{\\underset{#1}{\\min}\\;}$ $\\newcommand{\\umax}{\\underset{#1}{\\max}\\;}$ $\\newcommand{\\umin}{\\underset{#1}{\\min}\\;}$ $\\newcommand{\\uargmin}{\\underset{#1}{argmin}\\;}$ $\\newcommand{\\norm}{\\\|#1\\\|}$ $\\newcommand{\\abs}{\\left\|#1\\right\|}$ $\\newcommand{\\choice}{ \\left\\{ \\begin{array}{l} #1 \\end{array} \\right. }$ $\\newcommand{\\pa}{\\left(#1\\right)}$ $\\newcommand{\\diag}{{diag}\\left( #1 \\right)}$ $\\newcommand{\\qandq}{\\quad\\text{and}\\quad}$ $\\newcommand{\\qwhereq}{\\quad\\text{where}\\quad}$ $\\newcommand{\\qifq}{ \\quad \\text{if} \\quad }$ $\\newcommand{\\qarrq}{ \\quad \\Longrightarrow \\quad }$ $\\newcommand{\\ZZ}{\\mathbb{Z}}$ $\\newcommand{\\CC}{\\mathbb{C}}$ $\\newcommand{\\RR}{\\mathbb{R}}$ $\\newcommand{\\EE}{\\mathbb{E}}$ $\\newcommand{\\Zz}{\\mathcal{Z}}$ $\\newcommand{\\Ww}{\\mathcal{W}}$ $\\newcommand{\\Vv}{\\mathcal{V}}$ $\\newcommand{\\Nn}{\\mathcal{N}}$ $\\newcommand{\\NN}{\\mathcal{N}}$ $\\newcommand{\\Hh}{\\mathcal{H}}$ $\\newcommand{\\Bb}{\\mathcal{B}}$ $\\newcommand{\\Ee}{\\mathcal{E}}$ $\\newcommand{\\Cc}{\\mathcal{C}}$ $\\newcommand{\\Gg}{\\mathcal{G}}$ $\\newcommand{\\Ss}{\\mathcal{S}}$ $\\newcommand{\\Pp}{\\mathcal{P}}$ $\\newcommand{\\Ff}{\\mathcal{F}}$ $\\newcommand{\\Xx}{\\mathcal{X}}$ $\\newcommand{\\Mm}{\\mathcal{M}}$ $\\newcommand{\\Ii}{\\mathcal{I}}$ $\\newcommand{\\Dd}{\\mathcal{D}}$ $\\newcommand{\\Ll}{\\mathcal{L}}$ $\\newcommand{\\Tt}{\\mathcal{T}}$ $\\newcommand{\\si}{\\sigma}$ $\\newcommand{\\al}{\\alpha}$ $\\newcommand{\\la}{\\lambda}$ $\\newcommand{\\ga}{\\gamma}$ $\\newcommand{\\Ga}{\\Gamma}$ $\\newcommand{\\La}{\\Lambda}$ $\\newcommand{\\si}{\\sigma}$ $\\newcommand{\\Si}{\\Sigma}$ $\\newcommand{\\be}{\\beta}$ $\\newcommand{\\de}{\\delta}$ $\\newcommand{\\De}{\\Delta}$ $\\newcommand{\\phi}{\\varphi}$ $\\newcommand{\\th}{\\theta}$ $\\newcommand{\\om}{\\omega}$ $\\newcommand{\\Om}{\\Omega}$\n\nThis numerical tour explores the use of numerical schemes to solve the Sudoku game.\n\nThis tour was written by <http://lcav.epfl.ch/~lu/ Yue M. Lu> and <http://www.ceremade.dauphine.fr/~peyre/ Gabriel Peyr >.\n\nThe idea of encoding the Sudoku rule using a higer dimensional lifting, linear constraints and binary constraint is explained in:\n\nAndrew C. Bartlett, Amy N. Langville, An Integer Programming Model for the Sudoku Problem, The Journal of Online Mathematics and Its Applications, Volume 8. May 2008.\n\nThe idea of removing the binary constraint and using sparsity constraint is exposed in:\n\nP. Babu, K. Pelckmans, P. Stoica, and J. Li, Linear Systems, Sparse Solutions, and Sudoku, IEEE Signal Processing Letters, vol. 17, no. 1, pp. 40-42, 2010.\n\nThis tour elaborarates on these two ideas. In particular it explains why $L^1$ minimization is equivalent to a POCS (projection on convex sets) method to find a feasible point inside a convex polytope.\n\nIn :\naddpath('toolbox_signal')\n\n\n## Game Encoding and Decoding¶\n\nThe basic idea is to use a higher dimensional space of size \|(n,n,n)\| to represent a Sudoku matrix of size \|(n,n)\|. In this space, the arrays are constrained to have binary entries.\n\nSize of the Sudoku. This number must be a square.\n\nIn :\nn = 9;\n\n\nCreate a random integer matrix with entries in 1...9.\n\nIn :\nx = floor(rand(n)n)+1;\n\n\nComparison matrix used for encoding to binary format.\n\nIn :\nU = repmat( reshape(1:n, [1 1 n]), n,n );\n\n\nEncoding in binary format.\n\nIn :\nencode = @(x)double( repmat(x, [1 1 n])==U );\nX = encode(x);\n\n\nThe resulting matrix has binary entries and has size \|(n,n,n)\|. One has \|x(i,j)=k\| if \|X(i,j,k)=1\| and \|X(i,j,l)=0\| for \|l~=k\|.\n\nDecoding from binary format. One use a \|min\| to be able to recover even if the matrix \|X\| is not binary (because of computation errors).\n\nIn :\n[tmp,x1] = min( abs(X-1), [], 3 );\n\n\nShow that decoding is ok.\n\nIn :\ndisp(['Should be 0: ' num2str(norm(x-x1,'fro')) '.']);\n\nShould be 0: 0.\n\n\n## Encoding Constraints¶\n\nFor \|X\| to be a valid encoded matrix, it should be binary and satifies that each \|X(i,j,:)\| contains only a single \|1\|, which can be represented using a linear contraint \|AencX(:)=1\|.\n\nNow we construct one encoding constraint.\n\nIn :\ni = 4; j = 4;\nZ = zeros(n,n,n);\nZ(i,j,:) = 1;\n\n\nAdd this constraint to the encoding constraint matrix.\n\nIn :\nAenc = [];\nAenc(end+1,:) = Z(:)';\n\n\nShow that constraint is satisfied.\n\nIn :\ndisp(['Should be 1: ' num2str(AencX(:)) '.']);\n\nShould be 1: 1.\n\n\nExercise 1\n\nBuild the encoding matrix \|Aenc\|. Display it.\n\nIn :\nexo1()", null, "In :\n%% Insert your code here.\n\n\nShow that constraint \|AencX(:)=1\| is satisfied.\n\nIn :\ndisp(['Should be 0: ' num2str(norm(AencX(:)-1)) '.']);\n\nShould be 0: 0.\n\n\n## Sudoku Rules Constraints¶\n\nIn a Sudoku valid matrix \|x\|, each column, row and sub-square of \|x\| should contains all the values in 0...n. This can be encoded on the high dimensional \|X\| using linear constraints \|ArowX=1\|, \|AcolX=1\| and \|AblockX=1\|.\n\nA valid Sudoku matrix.\n\nIn :\nx = [8 1 9 6 7 4 3 2 5;\n5 6 3 2 8 1 9 4 7;\n7 4 2 5 9 3 6 8 1;\n6 3 8 9 4 5 1 7 2;\n9 7 1 3 2 8 4 5 6;\n2 5 4 1 6 7 8 9 3;\n1 8 5 7 3 9 2 6 4;\n3 9 6 4 5 2 7 1 8;\n4 2 7 8 1 6 5 3 9]\n\nx =\n\n8 1 9 6 7 4 3 2 5\n5 6 3 2 8 1 9 4 7\n7 4 2 5 9 3 6 8 1\n6 3 8 9 4 5 1 7 2\n9 7 1 3 2 8 4 5 6\n2 5 4 1 6 7 8 9 3\n1 8 5 7 3 9 2 6 4\n3 9 6 4 5 2 7 1 8\n4 2 7 8 1 6 5 3 9\n\n\n\nEncode it in binary format.\n\nIn :\nX = encode(x);\n\n\nSelect the index of the entries of a row.\n\nIn :\ni=3; k=5;\nZ = zeros(n,n,n);\nZ(i,:,k) = 1;\n\n\nFill the first entries of the row matrix.\n\nIn :\nArow = [];\nArow(end+1,:) = Z(:)';\n\n\nShow that constraint is satisfied.\n\nIn :\ndisp(['Should be 1: ' num2str(ArowX(:)) '.']);\n\nShould be 1: 1.\n\n\nExercise 2\n\nBuild the full row matrix \|Arow\|. Display it.\n\nIn :\nexo2()", null, "In :\n%% Insert your code here.\n\n\nShow that constraint \|ArowX(:)=1\| is satisfied.\n\nIn :\ndisp(['Should be 0: ' num2str(norm(ArowX(:)-1)) '.']);\n\nShould be 0: 0.\n\n\nExercise 3\n\nBuild the full column matrix \|Acol\|. Display it.\n\nIn :\nexo3()", null, "In :\n%% Insert your code here.\n\n\nShow that constraint \|AcolX(:)=1\| is satisfied.\n\nIn :\ndisp(['Should be 0: ' num2str(norm(AcolX(:)-1)) '.']);\n\nShould be 0: 0.\n\n\nNow we proceed to block constraints.\n\nSize of a block.\n\nIn :\np = sqrt(n);\n\n\nThe upper left square should contain all numbers in \|{1,...,n}\|.\n\nIn :\nk = 1;\nZ = zeros(n,n,n);\nZ(1:p,1:p,k) = 1;\n\n\nAdd it as the first row of the block constraint matrix.\n\nIn :\nAblock = [];\nAblock(end+1,:) = Z(:)';\n\n\nShow that constraint is satisfied.\n\nIn :\ndisp(['Should be 1: ' num2str(AblockX(:)) '.']);\n\nShould be 1: 1.\n\n\nExercise 4\n\nCreate the full block matrix. Display it.\n\nIn :\nexo4()", null, "In :\n%% Insert your code here.\n\n\nShow that constraint \|AblockX(:)=1\| is satisfied.\n\nIn :\ndisp(['Should be 0: ' num2str(norm(AblockX(:)-1)) '.']);\n\nShould be 0: 0.\n\n\n## Inpainting Constraint¶\n\nA Sudoku game asks to fill the missing entries of a partial Sudoku matrix \|x1\| to obtain a full Sudoku matrix \|x\|.\n\nThe fact that for each available entry \|(i,j)\| on must have \|x(i,j)=x1(i,j)\| can be encoded using a linear constraint.\n\nLoad a Sudoku with missing entries, that are represented as 0. This is an easy grid.\n\nIn :\nx1 = [0 1 0 0 0 0 3 0 0;\n0 0 3 0 8 0 0 4 0;\n7 0 2 0 0 3 0 0 1;\n0 3 0 9 4 0 1 0 0;\n9 0 0 0 0 0 0 0 6;\n0 0 4 0 6 7 0 9 0;\n1 0 0 7 0 0 2 0 4;\n0 9 0 0 5 0 7 0 0;\n0 0 7 0 0 0 0 3 0];\n\n\nRetrieve the indexes of the available entries.\n\nIn :\n[I,J] = ind2sub( [n n], find(x1(:)~=0) );\nv = x1(x1(:)~=0);\n\n\nCreate a vector corresponding to the constraint that \|x1(I(i),J(i))==v(i)\|.\n\nIn :\ni = 1;\nZ = zeros(n,n,n);\nZ(I(i), J(i), v(i)) = 1;\n\n\nFill the first entries of the row matrix.\n\nIn :\nAinp = [];\nAinp(end+1,:) = Z(:)';\n\n\nExercise 5\n\nBuild the full inpainting matrix \|Ainp\|. Display it.\n\nIn :\nexo5()", null, "In :\n%% Insert your code here.\n\n\nShow that constraint \|AinpX1(:)=1\| is satisfied.\n\nIn :\nX1 = encode(x1);\ndisp(['Should be 0: ' num2str(norm(AinpX1(:)-1)) '.']);\n\nShould be 0: 0.\n\n\n## Solving the Sudoku by Binary Integer Programming¶\n\nThe whole set of constraints can be written \|AX(:)=1\|, where the matrix \|A\| is defined as a concatenation of all the constraint matrices.\n\nIn :\nA = [Aenc; Arow; Acol; Ablock; Ainp];\n\n\nPre-compute the pseudo-inverse of A.\n\nIn :\npA = pinv(A);\n\n\nIf the Sudoku game has an unique solution \|x\|, then the corresponding lifted vector \|X\| is the only solution to \|AX(:)=1\| under the constraint that \|X\| is binary.\n\nThis is the idea proposed in:\n\nAndrew C. Bartlett, Amy N. Langville, An Integer Programming Model for the Sudoku Problem, The Journal of Online Mathematics and Its Applications, Volume 8. May 2008.\n\nUnfortunately, solving a linear system under binary constraints is difficult, in fact solving a general integer program is known to be NP-hard. It means that such a method is very slow to solve Sudoku for large \|n\|.\n\nOne can use branch-and-bounbs methods to solve the binary integer program, but this might be slow. One can use for instance the command \|bintprog\| of Matlab (optimization toolbox), with an arbitrary objective function (since one wants to solve a feasability problem, no objective is needed).\n\nExercise 6\n\nImplement the Soduku solver using an interger linear programming algorithm.\n\nIn :\nexo6()\n\nIn :\n%% Insert your code here.\n\n\n## Removing the Binary Constraint¶\n\nIf one removes the binary constraint, one simply wants to compute a solution to the linear system \|AX(:)=1\|. But unfortunately it has an infinite number of solutions (and the set of solutions is not bounded).\n\nIt is thus unlikely that chosing a solution at random will work, but let's try it by projecting any vector on the constraint \|AX(:)=1\|.\n\nFirst define the orthogonal projector on the constraint \|{X \\ AX(:)=1}\|.\n\nIn :\nprojector = @(u)reshape( u(:) - pA(Au(:)-1), [n n n]);\n\n\nWe project an arbitrary vector (that does not satisfy the constraint) onto the constraint \|AX(:)=1\|.\n\nIn :\nXproj = projector( zeros(n,n,n) );\n\n\nCheck that \|Xproj\| projects onto itself because it satisfies the constraints.\n\nIn :\nd = projector(Xproj)-Xproj;\ndisp(['Should be 0: ' num2str(norm(d(:), 'fro')) '.']);\n\nShould be 0: 4.1665e-14.\n\n\nPlot the histogrtam of the entries of \|Xproj\|. As you can see, they are not binary, meaning that the binary constraint is violated.\n\nIn :\nclf;\nhist(Xproj(:), 30);\naxis('tight');", null, "It is thus not a solution to the Sudoku problem. We emphasize this by counting the number of violated constraints after decoding / re-encoding.\n\nIn :\n[tmp,xproj] = min( abs(Xproj-1), [], 3 );\nXproj1 = encode(xproj);\ndisp(['Number of violated constraints: ' num2str(sum(AXproj1(:)~=1)) '.']);\n\nNumber of violated constraints: 119.\n\n\n## Solving the Sudoku by Projection on Convex Sets¶\n\nA way to improve the quality of the result is to find a vector that satisfies both \|AX(:)=1\| and \|0<=X<=1\|. Note that this last constraint can be modified to \|X>=0\| because of the fact that the entries of \|X(i,j,:)\| must sum to 1 because of \|AX(:)\|.\n\nA way to find a point inside this polytope \|P = {X \\ AX(:)=1 and X>=0}\| is to start from a random initial guess.\n\nIn :\nXproj = zeros(n,n,n);\n\n\nAnd iteratively project on each constraint. This corresponds to the POCS algorithm to find a feasible point into the (non empty) intersection of convex sets.\n\nIn :\nXproj = max( projector(Xproj),0 );\n\n\nExercise 7\n\nPerform iterative projections (POCS) on the two constraints \|AXproj(:)=1\| and \|Xproj>=0\|. Display the decay of the error \|norm(AXproj(:)-1)\| in logarithmic scale.\n\nIn :\nexo7()", null, "In :\n%% Insert your code here.\n\n\nDisplay the histogram of the recovered values.\n\nIn :\nclf;\nhist(Xproj(:), 30);\naxis('tight');", null, "As you can see, the resulting vector is (nearly, up to convergence errors of the POCS) a binary one, meaning that it is actually the (unique) solution to the Sudoku problem.\n\nWe check this by counting the number of violated constraints after decoding and re-encoding.\n\nIn :\n[tmp,xproj] = min( abs(Xproj-1), [], 3 );\nXproj1 = encode(xproj);\ndisp(['Number of violated constraints: ' num2str(sum(AXproj1(:)~=1)) '.']);\n\nNumber of violated constraints: 0.\n\n\nExercise 8\n\nProve (numerically) that for this grid, the polytope of constraints \|P={X \\ AX(:)=1 and X>=0}\| is actually reduced to a singleton, which is the solution of the Sudoku problem.\n\nIn :\nexo8()\n\nIn :\n%% Insert your code here.\n\n\nUnfortunately, this is not always the case. For more difficult grids, \|P\| might not be reduced to a singleton.\n\nThis is a difficult grid.\n\nIn :\nx1 = [0 0 3 0 0 9 0 8 1;\n0 0 0 2 0 0 0 6 0;\n5 0 0 0 1 0 7 0 0;\n8 9 0 0 0 0 0 0 0;\n0 0 5 6 0 1 2 0 0;\n0 0 0 0 0 0 0 3 7;\n0 0 9 0 2 0 0 0 8;\n0 7 0 0 0 4 0 0 0;\n2 5 0 8 0 0 6 0 0];\n\n\nExercise 9\n\nTry the iterative projection on convexs set (POCS) method on this grid (remember that you need to re-define \|A\| and \|pA\|). What is your conclusion ? ill the constraint matrix OCS heck wether this is a valid solution.\n\nIn :\nexo9()\n\nNumber of violated constraints: 20.", null, "In :\n%% Insert your code here.\n\n\n## Decoding Using L1 Sparsity¶\n\nThe true solution has exactly \|n^2\| non zero entries, while a feasible point within the convex polytope \|P\| is usually not as sparse.\n\nCompute the sparsity of a projected vector.\n\nIn :\nXproj = projector( zeros(n,n,n) );\ndisp(['Sparsity: ' num2str(sum(Xproj(:)~=0)) ' (optimal: ' num2str(nn) ').']);\n\nSparsity: 729 (optimal: 81).\n\n\nOne can prove that any solution to \|AX(:)=1\| has more than \|n^2\| non zeros, and that the true Sudoku solution is the unique solution to \|AX(:)=1\| with \|n^2\| entries.\n\nOne can thus (in principle) solve the Sudoku by finding the solution to \|AX(:)=1\| with minimal L0 norm.\n\nUnfortunately, solving this problem is known to be in some sense NP-hard.\n\nA classical method to approximate the solution to the minimum L0 norm problem it to replace it by a minimum L1 norm solution, which can be computed with polynomial time algorithms.\n\nThis idea is put forward in:\n\nP. Babu, K. Pelckmans, P. Stoica, and J. Li, Linear Systems, Sparse Solutions, and Sudoku, IEEE Signal Processing Letters, vol. 17, no. 1, pp. 40-42, 2010.\n\nThe L1 norm of the Sudoku solution is \|n^2\|. The L1 norm of a projected vector is usually larger.\n\nIn :\ndisp(['L1 norm: ' num2str(norm(Xproj(:),1)) ' (optimal: ' num2str(nn) ').']);\n\nL1 norm: 102.7585 (optimal: 81).\n\n\nUnfortunately, all the vectors in the (bouned) polytope \|AX(:)=1\| and \|X>=0\| has the same L1 norm, equal to \|81\|.\n\nThis shows that the L1 minimization has the same properties as the POCS algorithm. They work if and only if the polytope is reduced to a single point.\n\nNevertheless, one can compute the solution with minimum L1 norm which corresponds to the Basis pursuit problem. This problem is equivalent to a linear program, and can be solved using standard interior points method (other algorithms, such as Douglas-Rachford could be used as well).\n\nDefine a shortcut for the resolution of basis pursuit.\n\nIn :\nsolvel1 = @(A)reshape(perform_solve_bp(A, ones(size(A,1),1), n^3, 30, 0, 1e-10), [n n n]);\n\n\nSolve the L1 minimization.\n\nIn :\nXbp = solvel1(A);\n\n\nCompute the L1 norm of the solution, to check that it is indeed equal to the minimal possible L1 norm \|n^2\|.\n\nIn :\ndisp(['L1 norm: ' num2str(norm(Xbp(:),1)) ' (optimal: ' num2str(nn) ').']);\n\nL1 norm: 81 (optimal: 81).\n\n\nUnfortunately, on this difficult problem, similarely to POCS, the L1 method does not works.\n\nIn :\n[tmp,xbp] = min( abs(Xbp-1), [], 3 );\nXbp1 = encode(xbp);\ndisp(['Number of violated constraints: ' num2str(sum(AXbp1(:)~=1)) '.']);\n\nNumber of violated constraints: 16.\n\n\n## Decoding Using more Aggressive Sparsity¶\n\nSince the L1 norm does not perform better than POCS, it is tempting to use a more agressive sparsity measure, like \|L^alpha\| norm for \|alpha<1\|.\n\nThis leads to non-convex problems, and one can compute a (not necessarily globally optimal) local minimum.\n\nAn algorithm to find a local minimum of the energy is the reweighted L1 minimization, described in:\n\nE. J. Cand s, M. Wakin and S. Boyd, Enhancing sparsity by reweighted l1 minimization, J. Fourier Anal. Appl., 14 877-905.\n\nThis idea is introduced in the paper:\n\nP. Babu, K. Pelckmans, P. Stoica, and J. Li, Linear Systems, Sparse Solutions, and Sudoku, IEEE Signal Processing Letters, vol. 17, no. 1, pp. 40-42, 2010.\n\nAt each iteration of the algorithm, one minimizes\n\n$$min \\sum 1/u_k \|x_k\|$$\n\nsubject to $$A x=1$$\n\nThe weights are then updated as\n\n$$u_k=\|x_k\|^{1-\\alpha} + \\epsilon$$\n\nThe weighted L1 minimization can be recasted as a traditional L1 minimization using a change of variables.\n\n$$min \|y\|_1$$\n\nsubject to $$A diag(u) y=1$$ and $$x_k = y_k u_k$$\n\nSet the target \|alpha\|, that should be in \|0<=alpha<=1\|.\n\nIn :\nalpha = 0;\n\n\nSet the regularisation parameter \|epsilon\|, that avoids a division by zero.\n\nIn :\nepsilon = 0.1;\n\n\nInitial set of weights.\n\nIn :\nu = ones(n,n,n);\n\n\nSolve the weighted L1 minimization.\n\nIn :\nXrw = solvel1( Adiag(u(:)) ) . u;\n\n\nUpdate the weights.\n\nIn :\nu = (abs(Xrw).^(1-alpha)+epsilon);\n\n\nExercise 10\n\nCompute the solution using the reweighted L1 minimization. Track the evolution of the number of invalidated constraints as the algorithm iterates.\n\nIn :\nexo10()\n\nNumber of violated constraints: 0.\n\nIn :\n%% Insert your code here.\n\n\nDisplay the histogram.\n\nIn :\nhist(Xrw(:), 30);", null, "While reweighting L1 works for reasonnably complicated Sudoku, it might fail on very difficult one.\n\nThis is the Al Escargot puzzle, believed to be the hardest Sudoku available.\n\nIn :\nx1 = [1 0 0 0 0 7 0 9 0;\n0 3 0 0 2 0 0 0 8;\n0 0 9 6 0 0 5 0 0;\n0 0 5 3 0 0 9 0 0;\n0 1 0 0 8 0 0 0 2;\n6 0 0 0 0 4 0 0 0;\n3 0 0 0 0 0 0 1 0;\n0 4 0 0 0 0 0 0 7;\n0 0 7 0 0 0 3 0 0];\n\n\nExercise 11\n\nTry reweighted L1 on this puzzle. ill matrix. olve\n\nIn :\nexo11()", null, "In :\n%% Insert your code here.\n\n\nExercise 12\n\nTry other sparsity-enforcing minimization methods, such as Orthogonal Matching Pursuit (OMP), or iterative hard thresholding.\n\nIn :\nexo12()\n\nIn :\n%% Insert your code here.\n\n\nExercise 13\n\nTry the different methods of this tour on a large number of Sudokus.\n\nIn :\nexo13()\n\nIn :\n%% Insert your code here.\n\n\nExercise 14\n\nTry the different methods of this tour on larger Sudokus, for \|n=4,5,6\|.\n\nIn :\nexo14()\n\nIn :\n%% Insert your code here." ]	[ null, "data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAkAAAAGwCAIAAADOgk3lAAAACXBIWXMAAAsSAAALEgHS3X78AAAA IXRFWHRTb2Z0d2FyZQBBcnRpZmV4IEdob3N0c2NyaXB0IDguNTRTRzzSAAAKDElEQVR4nO3c0U4z RxZGURjN+78yc/FH0ciIUDFgf5te6y65OjrVZpcbkte3t7cXAKj5z7MHAIB7CBgASQIGQJKAAZAk YAAkCRgASQIGQJKAAZAkYAAkCRgASQIGQJKAAZAkYAAkCRgASQIGQJKAAZAkYAAkCRgASQIGQJKA AZAkYAAkCRgASQIGQJKAAZAkYAAkCRgASQIGQJKAAZAkYAAkCRgASQIGQJKAAZAkYAAkCRgASQIG QJKAAZAkYAAkCRgASQIGQJKAAZAkYAAkCRgASQIGQJKAAZAkYAAkCRgASQIGQJKAAZAkYAAkCRgA SQIGQJKAAZAkYAAkCRgASQIGQJKAAZAkYAAkCRgASQIGQJKAAZAkYAAkCRgASQIGQJKAAZAkYAAk CRgASQIGQJKAAZAkYAAkCRgASQIGQJKAAZAkYAAkCRgASQIGQJKAAZAkYAAkCRgASQIGQJKAAZAk YAAkCRgASQIGQJKAAZAkYAAkCRgASQIGQJKAAZAkYAAkCRgASQIGQJKAAZAkYAAkCRgASQIGQJKA 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https://www.irrigationbox.co.nz/irrigatable-area-calculator-by-water-supply	[ "# Irrigatable Area\n\nThis calculator finds the land area that can be irrigated with a given flow of water. The minimum system capacity (supply) is the available water from the supply. The water needs is the peak crop water need during a specific time period. The available hours of operation per day is measured as the available hours for irrigation on a worst case day. The system efficiency is based on the irrigation efficiency and distribution uniformity.\n\n## Irrigatable Area As Limited By Water Supply\n\nMinimum System Capacity (Supply):\nWater Needs:\nOperation Hours Per Day:\nSystem Efficiency:\nIrrigated Area:\n\n### The Equation\n\nThis calulator uses this formula to determine the Irrigatable area.", null, "Where:", null, "= Irrigatable Area (sq. ft)", null, "= Minimum system capacity, or supply (gpm)", null, "= Water needs (in/day)", null, "= Operation hours per day (hrs)", null, "= System Efficiency (as a decimal)\n\nReference: Washington State University" ]	[ null, "https://www.irrigationbox.co.nz/Images/uploaded/Calculator-Images/Irrigatable_Area.gif", null, "https://www.irrigationbox.co.nz/Images/uploaded/Calculator-Images/A.gif", null, "https://www.irrigationbox.co.nz/Images/uploaded/Calculator-Images/S.gif", null, "https://www.irrigationbox.co.nz/Images/uploaded/Calculator-Images/Wsubn.gif", null, "https://www.irrigationbox.co.nz/Images/uploaded/Calculator-Images/hrs.gif", null, "https://www.irrigationbox.co.nz/Images/uploaded/Calculator-Images/E.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7855242,"math_prob":0.82860565,"size":274,"snap":"2021-31-2021-39","text_gpt3_token_len":65,"char_repetition_ratio":0.08148148,"word_repetition_ratio":0.0,"special_character_ratio":0.229927,"punctuation_ratio":0.11363637,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97661215,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,1,null,4,null,4,null,1,null,2,null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-03T15:00:18Z\",\"WARC-Record-ID\":\"<urn:uuid:51d8dce4-1914-46ad-9a85-700c95d670ba>\",\"Content-Length\":\"77241\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:751823fb-4d49-4608-97f5-ffc8a285601e>\",\"WARC-Concurrent-To\":\"<urn:uuid:541849d0-dbaa-479d-a0a8-8cb6f3c0076b>\",\"WARC-IP-Address\":\"13.70.72.44\",\"WARC-Target-URI\":\"https://www.irrigationbox.co.nz/irrigatable-area-calculator-by-water-supply\",\"WARC-Payload-Digest\":\"sha1:HGRZRAB2JSAT5L4LQKOPSGUWN6LNP7NA\",\"WARC-Block-Digest\":\"sha1:XOX7K4VATPJ7VP4ICREUIKEVHDQWQIHF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154459.22_warc_CC-MAIN-20210803124251-20210803154251-00494.warc.gz\"}"}
https://downloads.haskell.org/~ghc/6.8.1/docs/html/users_guide/data-type-extensions.html	[ "## 8.4. Extensions to data types and type synonyms\n\n### 8.4.1. Data types with no constructors\n\nWith the `-fglasgow-exts` flag, GHC lets you declare a data type with no constructors. For example:\n\n``` data S -- S :: \ndata T a -- T :: -> \n```\n\nSyntactically, the declaration lacks the \"= constrs\" part. The type can be parameterised over types of any kind, but if the kind is not `` then an explicit kind annotation must be used (see Section 8.7.3, “Explicitly-kinded quantification”).\n\nSuch data types have only one value, namely bottom. Nevertheless, they can be useful when defining \"phantom types\".\n\n### 8.4.2. Infix type constructors, classes, and type variables\n\nGHC allows type constructors, classes, and type variables to be operators, and to be written infix, very much like expressions. More specifically:\n\n• A type constructor or class can be an operator, beginning with a colon; e.g. `::`. The lexical syntax is the same as that for data constructors.\n\n• Data type and type-synonym declarations can be written infix, parenthesised if you want further arguments. E.g.\n\n``` data a :: b = Foo a b\ntype a :+: b = Either a b\nclass a :=: b where ...\n\ndata (a :*: b) x = Baz a b x\ntype (a :++: b) y = Either (a,b) y\n```\n\n• Types, and class constraints, can be written infix. For example\n\n```\tx :: Int :: Bool\nf :: (a :=: b) => a -> b\n```\n\n• A type variable can be an (unqualified) operator e.g. `+`. The lexical syntax is the same as that for variable operators, excluding \"(.)\", \"(!)\", and \"()\". In a binding position, the operator must be parenthesised. For example:\n\n``` type T (+) = Int + Int\nf :: T Either\nf = Left 3\n\nliftA2 :: Arrow (~>)\n=> (a -> b -> c) -> (e ~> a) -> (e ~> b) -> (e ~> c)\nliftA2 = ...\n```\n\n• Back-quotes work as for expressions, both for type constructors and type variables; e.g. `Int `Either` Bool`, or `Int `a` Bool`. Similarly, parentheses work the same; e.g. `(::) Int Bool`.\n\n• Fixities may be declared for type constructors, or classes, just as for data constructors. However, one cannot distinguish between the two in a fixity declaration; a fixity declaration sets the fixity for a data constructor and the corresponding type constructor. For example:\n\n``` infixl 7 T, ::\n```\n\nsets the fixity for both type constructor `T` and data constructor `T`, and similarly for `::`. `Int `a` Bool`.\n\n• Function arrow is `infixr` with fixity 0. (This might change; I'm not sure what it should be.)\n\n### 8.4.3. Liberalised type synonyms\n\nType synonyms are like macros at the type level, and GHC does validity checking on types only after expanding type synonyms. That means that GHC can be very much more liberal about type synonyms than Haskell 98:\n\n• You can write a `forall` (including overloading) in a type synonym, thus:\n\n``` type Discard a = forall b. Show b => a -> b -> (a, String)\n\nf x y = (x, show y)\n\ng :: Discard Int -> (Int,String) -- A rank-2 type\ng f = f 3 True\n```\n\n• You can write an unboxed tuple in a type synonym:\n\n``` type Pr = (# Int, Int #)\n\nh :: Int -> Pr\nh x = (# x, x #)\n```\n\n• You can apply a type synonym to a forall type:\n\n``` type Foo a = a -> a -> Bool\n\nf :: Foo (forall b. b->b)\n```\n\nAfter expanding the synonym, `f` has the legal (in GHC) type:\n\n``` f :: (forall b. b->b) -> (forall b. b->b) -> Bool\n```\n\n• You can apply a type synonym to a partially applied type synonym:\n\n``` type Generic i o = forall x. i x -> o x\ntype Id x = x\n\nfoo :: Generic Id []\n```\n\nAfter expanding the synonym, `foo` has the legal (in GHC) type:\n\n``` foo :: forall x. x -> [x]\n```\n\nGHC currently does kind checking before expanding synonyms (though even that could be changed.)\n\nAfter expanding type synonyms, GHC does validity checking on types, looking for the following mal-formedness which isn't detected simply by kind checking:\n\n• Type constructor applied to a type involving for-alls.\n\n• Unboxed tuple on left of an arrow.\n\n• Partially-applied type synonym.\n\nSo, for example, this will be rejected:\n\n``` type Pr = (# Int, Int #)\n\nh :: Pr -> Int\nh x = ...\n```\n\nbecause GHC does not allow unboxed tuples on the left of a function arrow.\n\n### 8.4.4. Existentially quantified data constructors\n\nThe idea of using existential quantification in data type declarations was suggested by Perry, and implemented in Hope+ (Nigel Perry, The Implementation of Practical Functional Programming Languages, PhD Thesis, University of London, 1991). It was later formalised by Laufer and Odersky (Polymorphic type inference and abstract data types, TOPLAS, 16(5), pp1411-1430, 1994). It's been in Lennart Augustsson's hbc Haskell compiler for several years, and proved very useful. Here's the idea. Consider the declaration:\n\n``` data Foo = forall a. MkFoo a (a -> Bool)\n\| Nil\n```\n\nThe data type `Foo` has two constructors with types:\n\n``` MkFoo :: forall a. a -> (a -> Bool) -> Foo\nNil :: Foo\n```\n\nNotice that the type variable `a` in the type of `MkFoo` does not appear in the data type itself, which is plain `Foo`. For example, the following expression is fine:\n\n``` [MkFoo 3 even, MkFoo 'c' isUpper] :: [Foo]\n```\n\nHere, `(MkFoo 3 even)` packages an integer with a function `even` that maps an integer to `Bool`; and ```MkFoo 'c' isUpper``` packages a character with a compatible function. These two things are each of type `Foo` and can be put in a list.\n\nWhat can we do with a value of type `Foo`?. In particular, what happens when we pattern-match on `MkFoo`?\n\n``` f (MkFoo val fn) = ???\n```\n\nSince all we know about `val` and `fn` is that they are compatible, the only (useful) thing we can do with them is to apply `fn` to `val` to get a boolean. For example:\n\n``` f :: Foo -> Bool\nf (MkFoo val fn) = fn val\n```\n\nWhat this allows us to do is to package heterogenous values together with a bunch of functions that manipulate them, and then treat that collection of packages in a uniform manner. You can express quite a bit of object-oriented-like programming this way.\n\n#### 8.4.4.1. Why existential?\n\nWhat has this to do with existential quantification? Simply that `MkFoo` has the (nearly) isomorphic type\n\n``` MkFoo :: (exists a . (a, a -> Bool)) -> Foo\n```\n\nBut Haskell programmers can safely think of the ordinary universally quantified type given above, thereby avoiding adding a new existential quantification construct.\n\n#### 8.4.4.2. Type classes\n\nAn easy extension is to allow arbitrary contexts before the constructor. For example:\n\n```data Baz = forall a. Eq a => Baz1 a a\n\| forall b. Show b => Baz2 b (b -> b)\n```\n\nThe two constructors have the types you'd expect:\n\n```Baz1 :: forall a. Eq a => a -> a -> Baz\nBaz2 :: forall b. Show b => b -> (b -> b) -> Baz\n```\n\nBut when pattern matching on `Baz1` the matched values can be compared for equality, and when pattern matching on `Baz2` the first matched value can be converted to a string (as well as applying the function to it). So this program is legal:\n\n``` f :: Baz -> String\nf (Baz1 p q) \| p == q = \"Yes\"\n\| otherwise = \"No\"\nf (Baz2 v fn) = show (fn v)\n```\n\nOperationally, in a dictionary-passing implementation, the constructors `Baz1` and `Baz2` must store the dictionaries for `Eq` and `Show` respectively, and extract it on pattern matching.\n\nNotice the way that the syntax fits smoothly with that used for universal quantification earlier.\n\n#### 8.4.4.3. Record Constructors\n\nGHC allows existentials to be used with records syntax as well. For example:\n\n```data Counter a = forall self. NewCounter\n{ _this :: self\n, _inc :: self -> self\n, _display :: self -> IO ()\n, tag :: a\n}\n```\n\nHere `tag` is a public field, with a well-typed selector function `tag :: Counter a -> a`. The `self` type is hidden from the outside; any attempt to apply `_this`, `_inc` or `_display` as functions will raise a compile-time error. In other words, GHC defines a record selector function only for fields whose type does not mention the existentially-quantified variables. (This example used an underscore in the fields for which record selectors will not be defined, but that is only programming style; GHC ignores them.)\n\nTo make use of these hidden fields, we need to create some helper functions:\n\n```inc :: Counter a -> Counter a\ninc (NewCounter x i d t) = NewCounter\n{ _this = i x, _inc = i, _display = d, tag = t }\n\ndisplay :: Counter a -> IO ()\ndisplay NewCounter{ _this = x, _display = d } = d x\n```\n\nNow we can define counters with different underlying implementations:\n\n```counterA :: Counter String\ncounterA = NewCounter\n{ _this = 0, _inc = (1+), _display = print, tag = \"A\" }\n\ncounterB :: Counter String\ncounterB = NewCounter\n{ _this = \"\", _inc = ('#':), _display = putStrLn, tag = \"B\" }\n\nmain = do\ndisplay (inc counterA) -- prints \"1\"\ndisplay (inc (inc counterB)) -- prints \"##\"\n```\n\nAt the moment, record update syntax is only supported for Haskell 98 data types, so the following function does not work:\n\n```-- This is invalid; use explicit NewCounter instead for now\nsetTag :: Counter a -> a -> Counter a\nsetTag obj t = obj{ tag = t }\n```\n\n#### 8.4.4.4. Restrictions\n\nThere are several restrictions on the ways in which existentially-quantified constructors can be use.\n\n• When pattern matching, each pattern match introduces a new, distinct, type for each existential type variable. These types cannot be unified with any other type, nor can they escape from the scope of the pattern match. For example, these fragments are incorrect:\n\n```f1 (MkFoo a f) = a\n```\n\nHere, the type bound by `MkFoo` \"escapes\", because `a` is the result of `f1`. One way to see why this is wrong is to ask what type `f1` has:\n\n``` f1 :: Foo -> a -- Weird!\n```\n\nWhat is this \"`a`\" in the result type? Clearly we don't mean this:\n\n``` f1 :: forall a. Foo -> a -- Wrong!\n```\n\nThe original program is just plain wrong. Here's another sort of error\n\n``` f2 (Baz1 a b) (Baz1 p q) = a==q\n```\n\nIt's ok to say `a==b` or `p==q`, but `a==q` is wrong because it equates the two distinct types arising from the two `Baz1` constructors.\n\n• You can't pattern-match on an existentially quantified constructor in a `let` or `where` group of bindings. So this is illegal:\n\n``` f3 x = a==b where { Baz1 a b = x }\n```\n\nInstead, use a `case` expression:\n\n``` f3 x = case x of Baz1 a b -> a==b\n```\n\nIn general, you can only pattern-match on an existentially-quantified constructor in a `case` expression or in the patterns of a function definition. The reason for this restriction is really an implementation one. Type-checking binding groups is already a nightmare without existentials complicating the picture. Also an existential pattern binding at the top level of a module doesn't make sense, because it's not clear how to prevent the existentially-quantified type \"escaping\". So for now, there's a simple-to-state restriction. We'll see how annoying it is.\n\n• You can't use existential quantification for `newtype` declarations. So this is illegal:\n\n``` newtype T = forall a. Ord a => MkT a\n```\n\nReason: a value of type `T` must be represented as a pair of a dictionary for `Ord t` and a value of type `t`. That contradicts the idea that `newtype` should have no concrete representation. You can get just the same efficiency and effect by using `data` instead of `newtype`. If there is no overloading involved, then there is more of a case for allowing an existentially-quantified `newtype`, because the `data` version does carry an implementation cost, but single-field existentially quantified constructors aren't much use. So the simple restriction (no existential stuff on `newtype`) stands, unless there are convincing reasons to change it.\n\n• You can't use `deriving` to define instances of a data type with existentially quantified data constructors. Reason: in most cases it would not make sense. For example:;\n\n```data T = forall a. MkT [a] deriving( Eq )\n```\n\nTo derive `Eq` in the standard way we would need to have equality between the single component of two `MkT` constructors:\n\n```instance Eq T where\n(MkT a) == (MkT b) = ???\n```\n\nBut `a` and `b` have distinct types, and so can't be compared. It's just about possible to imagine examples in which the derived instance would make sense, but it seems altogether simpler simply to prohibit such declarations. Define your own instances!\n\n### 8.4.5. Declaring data types with explicit constructor signatures\n\nGHC allows you to declare an algebraic data type by giving the type signatures of constructors explicitly. For example:\n\n``` data Maybe a where\nNothing :: Maybe a\nJust :: a -> Maybe a\n```\n\nThe form is called a \"GADT-style declaration\" because Generalised Algebraic Data Types, described in Section 8.4.6, “Generalised Algebraic Data Types (GADTs)”, can only be declared using this form.\n\nNotice that GADT-style syntax generalises existential types (Section 8.4.4, “Existentially quantified data constructors ”). For example, these two declarations are equivalent:\n\n``` data Foo = forall a. MkFoo a (a -> Bool)\ndata Foo' where { MKFoo :: a -> (a->Bool) -> Foo' }\n```\n\nAny data type that can be declared in standard Haskell-98 syntax can also be declared using GADT-style syntax. The choice is largely stylistic, but GADT-style declarations differ in one important respect: they treat class constraints on the data constructors differently. Specifically, if the constructor is given a type-class context, that context is made available by pattern matching. For example:\n\n``` data Set a where\nMkSet :: Eq a => [a] -> Set a\n\nmakeSet :: Eq a => [a] -> Set a\nmakeSet xs = MkSet (nub xs)\n\ninsert :: a -> Set a -> Set a\ninsert a (MkSet as) \| a `elem` as = MkSet as\n\| otherwise = MkSet (a:as)\n```\n\nA use of `MkSet` as a constructor (e.g. in the definition of `makeSet`) gives rise to a `(Eq a)` constraint, as you would expect. The new feature is that pattern-matching on `MkSet` (as in the definition of `insert`) makes available an `(Eq a)` context. In implementation terms, the `MkSet` constructor has a hidden field that stores the `(Eq a)` dictionary that is passed to `MkSet`; so when pattern-matching that dictionary becomes available for the right-hand side of the match. In the example, the equality dictionary is used to satisfy the equality constraint generated by the call to `elem`, so that the type of `insert` itself has no `Eq` constraint.\n\nThis behaviour contrasts with Haskell 98's peculiar treatment of contexts on a data type declaration (Section 4.2.1 of the Haskell 98 Report). In Haskell 98 the definition\n\n``` data Eq a => Set' a = MkSet' [a]\n```\n\ngives `MkSet'` the same type as `MkSet` above. But instead of making available an `(Eq a)` constraint, pattern-matching on `MkSet'` requires an `(Eq a)` constraint! GHC faithfully implements this behaviour, odd though it is. But for GADT-style declarations, GHC's behaviour is much more useful, as well as much more intuitive.\n\nFor example, a possible application of GHC's behaviour is to reify dictionaries:\n\n``` data NumInst a where\nMkNumInst :: Num a => NumInst a\n\nintInst :: NumInst Int\nintInst = MkNumInst\n\nplus :: NumInst a -> a -> a -> a\nplus MkNumInst p q = p + q\n```\n\nHere, a value of type `NumInst a` is equivalent to an explicit `(Num a)` dictionary.\n\nThe rest of this section gives further details about GADT-style data type declarations.\n\n• The result type of each data constructor must begin with the type constructor being defined. If the result type of all constructors has the form `T a1 ... an`, where `a1 ... an` are distinct type variables, then the data type is ordinary; otherwise is a generalised data type (Section 8.4.6, “Generalised Algebraic Data Types (GADTs)”).\n\n• The type signature of each constructor is independent, and is implicitly universally quantified as usual. Different constructors may have different universally-quantified type variables and different type-class constraints. For example, this is fine:\n\n``` data T a where\nT1 :: Eq b => b -> T b\nT2 :: (Show c, Ix c) => c -> [c] -> T c\n```\n\n• Unlike a Haskell-98-style data type declaration, the type variable(s) in the \"`data Set a where`\" header have no scope. Indeed, one can write a kind signature instead:\n\n``` data Set :: * -> * where ...\n```\n\nor even a mixture of the two:\n\n``` data Foo a :: (* -> ) -> where ...\n```\n\nThe type variables (if given) may be explicitly kinded, so we could also write the header for `Foo` like this:\n\n``` data Foo a (b :: * -> *) where ...\n```\n\n• You can use strictness annotations, in the obvious places in the constructor type:\n\n``` data Term a where\nLit :: !Int -> Term Int\nIf :: Term Bool -> !(Term a) -> !(Term a) -> Term a\nPair :: Term a -> Term b -> Term (a,b)\n```\n\n• You can use a `deriving` clause on a GADT-style data type declaration. For example, these two declarations are equivalent\n\n``` data Maybe1 a where {\nNothing1 :: Maybe1 a ;\nJust1 :: a -> Maybe1 a\n} deriving( Eq, Ord )\n\ndata Maybe2 a = Nothing2 \| Just2 a\nderiving( Eq, Ord )\n```\n\n• You can use record syntax on a GADT-style data type declaration:\n\n``` data Person where\nAdult { name :: String, children :: [Person] } :: Person\nChild { name :: String } :: Person\n```\n\nAs usual, for every constructor that has a field `f`, the type of field `f` must be the same (modulo alpha conversion).\n\nAt the moment, record updates are not yet possible with GADT-style declarations, so support is limited to record construction, selection and pattern matching. For example\n\n``` aPerson = Adult { name = \"Fred\", children = [] }\n\nshortName :: Person -> Bool\nhasChildren (Adult { children = kids }) = not (null kids)\nhasChildren (Child {}) = False\n```\n\n• As in the case of existentials declared using the Haskell-98-like record syntax (Section 8.4.4.3, “Record Constructors”), record-selector functions are generated only for those fields that have well-typed selectors. Here is the example of that section, in GADT-style syntax:\n\n```data Counter a where\nNewCounter { _this :: self\n, _inc :: self -> self\n, _display :: self -> IO ()\n, tag :: a\n}\n:: Counter a\n```\n\nAs before, only one selector function is generated here, that for `tag`. Nevertheless, you can still use all the field names in pattern matching and record construction.\n\n### 8.4.6. Generalised Algebraic Data Types (GADTs)\n\nGeneralised Algebraic Data Types generalise ordinary algebraic data types by allowing constructors to have richer return types. Here is an example:\n\n``` data Term a where\nLit :: Int -> Term Int\nSucc :: Term Int -> Term Int\nIsZero :: Term Int -> Term Bool\nIf :: Term Bool -> Term a -> Term a -> Term a\nPair :: Term a -> Term b -> Term (a,b)\n```\n\nNotice that the return type of the constructors is not always `Term a`, as is the case with ordinary data types. This generality allows us to write a well-typed `eval` function for these `Terms`:\n\n``` eval :: Term a -> a\neval (Lit i) \t = i\neval (Succ t) = 1 + eval t\neval (IsZero t) = eval t == 0\neval (If b e1 e2) = if eval b then eval e1 else eval e2\neval (Pair e1 e2) = (eval e1, eval e2)\n```\n\nThe key point about GADTs is that pattern matching causes type refinement. For example, in the right hand side of the equation\n\n``` eval :: Term a -> a\neval (Lit i) = ...\n```\n\nthe type `a` is refined to `Int`. That's the whole point! A precise specification of the type rules is beyond what this user manual aspires to, but the design closely follows that described in the paper Simple unification-based type inference for GADTs, (ICFP 2006). The general principle is this: type refinement is only carried out based on user-supplied type annotations. So if no type signature is supplied for `eval`, no type refinement happens, and lots of obscure error messages will occur. However, the refinement is quite general. For example, if we had:\n\n``` eval :: Term a -> a -> a\neval (Lit i) j = i+j\n```\n\nthe pattern match causes the type `a` to be refined to `Int` (because of the type of the constructor `Lit`), and that refinement also applies to the type of `j`, and the result type of the `case` expression. Hence the addition `i+j` is legal.\n\nThese and many other examples are given in papers by Hongwei Xi, and Tim Sheard. There is a longer introduction on the wiki, and Ralf Hinze's Fun with phantom types also has a number of examples. Note that papers may use different notation to that implemented in GHC.\n\nThe rest of this section outlines the extensions to GHC that support GADTs. The extension is enabled with `-XGADTs`.\n\n• A GADT can only be declared using GADT-style syntax (Section 8.4.5, “Declaring data types with explicit constructor signatures”); the old Haskell-98 syntax for data declarations always declares an ordinary data type. The result type of each constructor must begin with the type constructor being defined, but for a GADT the arguments to the type constructor can be arbitrary monotypes. For example, in the `Term` data type above, the type of each constructor must end with `Term ty`, but the `ty` may not be a type variable (e.g. the `Lit` constructor).\n\n• You cannot use a `deriving` clause for a GADT; only for an ordinary data type.\n\n• As mentioned in Section 8.4.5, “Declaring data types with explicit constructor signatures”, record syntax is supported. For example:\n\n``` data Term a where\nLit { val :: Int } :: Term Int\nSucc { num :: Term Int } :: Term Int\nPred { num :: Term Int } :: Term Int\nIsZero { arg :: Term Int } :: Term Bool\nPair { arg1 :: Term a\n, arg2 :: Term b\n} :: Term (a,b)\nIf { cnd :: Term Bool\n, tru :: Term a\n, fls :: Term a\n} :: Term a\n```\n\nHowever, for GADTs there is the following additional constraint: every constructor that has a field `f` must have the same result type (modulo alpha conversion) Hence, in the above example, we cannot merge the `num` and `arg` fields above into a single name. Although their field types are both `Term Int`, their selector functions actually have different types:\n\n``` num :: Term Int -> Term Int\narg :: Term Bool -> Term Int\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8017702,"math_prob":0.95439124,"size":20383,"snap":"2023-40-2023-50","text_gpt3_token_len":5179,"char_repetition_ratio":0.13935915,"word_repetition_ratio":0.055555556,"special_character_ratio":0.26340577,"punctuation_ratio":0.1612825,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9686569,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-11T05:54:59Z\",\"WARC-Record-ID\":\"<urn:uuid:d621c7b5-672f-4526-9b84-c8bb1f14d490>\",\"Content-Length\":\"35005\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a75f70ae-67f2-4f15-bd48-c55c97906cd8>\",\"WARC-Concurrent-To\":\"<urn:uuid:ab3bde90-27fa-45b9-aa2c-f7357c6c8781>\",\"WARC-IP-Address\":\"146.75.29.175\",\"WARC-Target-URI\":\"https://downloads.haskell.org/~ghc/6.8.1/docs/html/users_guide/data-type-extensions.html\",\"WARC-Payload-Digest\":\"sha1:QLIFJPJ2MUW22UK2PFH24J7MRNYCF3O6\",\"WARC-Block-Digest\":\"sha1:XUZGKJ6HUBDQXS2RC7U346ZKOO47OC6S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679103558.93_warc_CC-MAIN-20231211045204-20231211075204-00684.warc.gz\"}"}
http://essays.grokearth.com/2012/10/golden-ratio-in-nature.html	[ "## Saturday, October 20, 2012\n\n### Golden Ratio in Nature\n\nIn words, the golden ratio is:\nThe ratio of the sum of two quantities over the larger quantity is equal to the ratio of the larger quantity over the smaller one.\nThe golden ratio expressed as an equation is:\n\n(a + b) / a = 1.618 = a / b\n\nFrom the golden ratio, we get a mathematical constant - a number. It is the number 1.6 which, by convention, is represented by the Greek letter φ (pronounced \"fee\").\n\nThe number φ is a mathematical curiosity. Like the mathematical constant π, φ appears everywhere.", null, "Ginkgo Leaf\nThe good, of course, is always beautiful, and the beautiful never lacks proportion.\nPlato\nThe golden ratio is found in the proportions of leaves. I measured the golden ratio in the leaf of a ginkgo tree.\n\nUsing the typographical units of pica, I measured 28 pica from the notch to the base of the stem of a ginkgo leaf. I then measured 17 pica from the top of the stem to the base of the stem (length a).\n\nThe quotient of 28 pica over 17 pica yields the golden ratio 1.6.\n\nScientists have recently observed nanoscopic symmetry. The symmetry they observed had the attributes of the golden ratio, demonstrating this curious proportion at the quantum level.\nThe universe cannot be read until we have learnt the language and become familiar with the characters in which it is written. It is written in mathematical language, and the letters are triangles, circles and other geometrical figures, without which means it is humanly impossible to comprehend a single word.\nGalileo Galilei\nReferences" ]	[ null, "http://4.bp.blogspot.com/-zfFTrdhC39c/UILBQU_hISI/AAAAAAAADPM/1vwDj94wMiI/s200/ginkgoGoldenRatio2.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9014342,"math_prob":0.9574846,"size":1447,"snap":"2022-40-2023-06","text_gpt3_token_len":325,"char_repetition_ratio":0.12820514,"word_repetition_ratio":0.016064256,"special_character_ratio":0.2211472,"punctuation_ratio":0.11785714,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9883596,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-27T23:48:07Z\",\"WARC-Record-ID\":\"<urn:uuid:479db622-c728-4208-a6b9-1bcea7e5a3a8>\",\"Content-Length\":\"288956\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef1eec13-d11e-423e-9431-c770b0f80e52>\",\"WARC-Concurrent-To\":\"<urn:uuid:f1f56eb6-ae39-47b4-b461-92c953896213>\",\"WARC-IP-Address\":\"172.253.115.121\",\"WARC-Target-URI\":\"http://essays.grokearth.com/2012/10/golden-ratio-in-nature.html\",\"WARC-Payload-Digest\":\"sha1:V2BYWEEI6JEY2K2KP4OFCFOU5RDJ4X6R\",\"WARC-Block-Digest\":\"sha1:AWGHMW4POORYYXBGH66STKUWGPECL543\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499468.22_warc_CC-MAIN-20230127231443-20230128021443-00304.warc.gz\"}"}
https://www.mathstrength.org/grade-4	[ "", null, "## Formative Assessment and Bridging activities", null, "These materials are part of an iterative design process and will continue to be refined during the 2021-2022 school year. Feedback is being accepted at the link below.\n*Share Feedback for Grade 4 Modules\n\nNote: Links marked with will open in a new tab\nThe Bridging Standards in bold below are currently live. Others are coming soon!\n\nStandard 4.3c Standard 4.3d\n\nStandard 4.5a\n\nStandard 4.5b\n\nStandard 4.5c\n\nStandard 4.6a\n\nStandard 4.6b Standard 4.8a\n\nStandard 4.8b\n\nStandard 4.8d\n\nStandard 4.10a\n\nStandard 4.14b\n\nStandard 4.15\n\nStandard 4.16\n\n## Standard 4.2a\n\nStandard 4.2a Compare and order fractions and mixed numbers.\n\n(Pull down for more)\n\nUnderstanding the Learning Progression\n\nBig Ideas:\n\n• Initial fraction ideas begin with creating experiences with a variety of visual representations (area models, number lines, set models, etc.) which can be utilized to compare and order fractions. Over time, reasoning strategies and generalizations are developed through modeling and mental imagery. There may be a transition from creating models to applying those generalizations and reasoning strategies to solve problems.\n\n• A variety of reasoning strategies can be applied to compare fractions. Potential strategies include but are not limited to:\n\n• reasoning about relative magnitude ;\n\n• unit fraction reasoning;\n\n• benchmark reasoning ( i.e. I know 4/8 is ½ so ⅝ is a little more); and\n\n• equivalence.\n\n• Behr and Post (1992) indicate that “a child’s understanding of the ordering of two fractions needs to be based on an understanding of the ordering of unit fractions” (1992, p.21).\n\n• The reasoning strategy applied is often impacted by number choice. For example, I might use the benchmark of ½ if I am comparing 3/10 (a little less than ½) and ⅝ (a little more than ½).\n\nImportant Assessment Look-fors:\n\n• Students recognize and utilize benchmark fractions.\n\n• Students use their understanding of the ordering of unit fractions to ordering of two fractions.\n\n• Students recognize and apply their understanding of equivalent fractions.\n\n• Students recognize fractions equivalent to ½.\n\n• Students represent and work with fractions greater than 1.\n\n• Students organize their thinking in a way that helps them make sense of the problem. Student representations support their thinking/reasoning.\n\n• Student strategies make sense for the given set of numbers.\n\n• It is evident in the strategy used that the student understands the relationship between the numerator and the denominator.\n\nPurposeful questions:\n\n• Can you share where you started and why?\n\n• How might you use your understanding of unit fractions to compare and order these fractions?\n\n• How can benchmark be used to help you understand the value of these fractions?\n\n• Is that fraction greater than 1? Less than 1? How do you know?\n\n• What do you know about the value of these numbers?\n\n• What relationships do you see?\n\n• I’m looking at your number line and I’m noticing _____. Tell me more about that...", null, "### Student Strengths\n\nStudents can name, write, represent and compare fractions and mixed numbers represented by a model\n\n### Bridging Concepts\n\nStudents can make generalizations about fractional relationships across representations (i.e., when the numerator is half of the denominator, the fraction is always equal to ½).\n\n### Standard 4.2a\n\nStudents can compare and order fractions and mixed numbers with and without models", null, "", null, "", null, "", null, "", null, "## Standard 4.2b\n\nStandard 4.2b Represent equivalent fractions.\n\n(Pull down for more)\n\nUnderstanding the Learning Progression\n\nBig Ideas:\n\n• When two fractions are equivalent that means there are two ways of describing the same amount by using different sized fractional parts. (Van de Walle et al, 2019)\n\n• A variety of representations and models can be used to identify different names for equivalent fractions: region/area, set and measurement models. (Students should use area representations, strips of paper, tape diagrams, number lines, counters and other manipulatives to reason about equivalence.)\n\n• Intuitive methods using drawings and manipulatives support student understanding. Students can develop an understanding of equivalent fractions and also develop from that understanding a conceptually based algorithm. Delay sharing “a rule.” (Van de Walle et al, 2019)\n\nImportant Assessment Look-fors:\n\n• Evidence that the student is able to identify equivalent fractions modeled using a variety of different models: area, set and/or measurement.\n\n• The student uses strategies like removing lines or adding additional lines to the area and number line models to show how the models are equivalent.\n\n• The student is able to use a variety of strategies and can justify their reasoning as to why fractions are equivalent.\n\n• When the student represents their fraction as models, they represent the whole using the same size and shape model. Click this link for more information about student representations.\n\nPurposeful questions:\n\n• What strategy (or strategies) did you use to determine which fraction models are equivalent?\n\n• Which fraction models are the easiest for you to identify? What made it easy?\n\n• Which fraction models are the hardest for you to identify? What made it difficult?\n\n• What relationships do you see?\n\n• Can you create another equivalent model that is different from ones shown?", null, "### Student Strengths\n\nStudents can name and write fractions and mixed numbers represented by a model. Students can represent fractions and mixed numbers with models and symbols.\n\n### Bridging Concepts\n\nStudents can create a model to represent a fraction. (Area models tend to be easier for students to grasp while set and measurement models tend to be more difficult. Students may need additional support bridging their understanding of area models to help support their understanding of set and measurement models.)\n\n### Standard 4.2b\n\nStudents can represent equivalent fractions.", null, "", null, "", null, "", null, "## Standard 4.2C\n\nStandard 4.2c Identify the division statement that represents a fraction, with models and in context.\n\n(Pull down for more)\n\nUnderstanding the Learning Progression\n\nBig Ideas:\n\n• Fractions have multiple meanings and interpretations. Generally there are five main interpretations: fractions as parts of wholes or parts of sets; fractions as the result of dividing two numbers; fractions as the ratio of two quantities; fractions as operators; and fractions as measures (Behr, Harel, Post, and Lesh 1992; Kieren 1988; Lamon 1999). When a fraction is presented in symbolic form, devoid of context, the intended interpretation of the fraction is not evident. The various interpretations are needed, however, in order to make sense of fraction problems and situations. Students need to explore and understand that fractional parts are equal shares of a whole or set model.\n\n• A fraction can also represent the result obtained when two numbers are divided. This interpretation of fraction is sometimes referred to as the quotient meaning, since the quotient is the answer to a division problem. Chapin and Johnson (2000) give these examples, “the number of gumdrops each child receives when 40 gumdrops are shared among 5 children can be expressed as 40/5 , 8/1 , or 8; when two steaks are shared equally among three people, each person gets 2/ 3 of a steak for dinner. We often express the quotient as a mixed number rather than an improper fraction – 15 feet of rope can be divided to make two jump ropes, each 7 1/ 2 ( 15/2 ) feet long” (p .99– 101).\n\n• When partitioning a whole into more equal shares the parts become smaller. (Teaching Student-Centered Mathematics, Grades 3-5, John Van de Walle)\n\n• When exploring the concept of fractions and connecting it to the division statement, students should be able to identify and recognize that the fraction is the amount that each person would receive when dividing equally.\n\nImportant Assessment Look-fors:\n\n• The student is able to accurately partition models when exploring fair share problems.\n\n• When given context, the student is able to successfully use models to identify the division statement that represents the fraction.\n\n• The student can interpret the fraction and division statement as the amount each person would receive.\n\n• The student can identify the difference between a division statement that represents an improper fraction versus a proper fraction within context. The student can also represent a fair share problem when the numerator is larger or smaller than the denominator.\n\nPurposeful questions:\n\n• Will each person receive more or less than a whole? Explain your reasoning.\n\n• Identify the division statements that represent this fraction as presented in the context.\n\n• How much will each person receive when shared equally?\n\n• Once the students have discovered how much each person will receive, ask the students if each person was to combine their equal share together, what would be the combined total? What do you notice?\n\n• Example: 2 cookies shared with 3 students. Each person would receive ⅔ of a cookie. When combining the equal shares from each of the 3 people, ⅔ + ⅔ + ⅔ = 6/3 or 2. The sum is equivalent to 2 whole cookies or the original amount of cookies that was shared.\n\n• If the dividend is smaller than the divisor, how does this relate to how much each person would receive if the amount was shared equally? Would this be true if the dividend was larger than the divisor?\n\n• Example: 2 pizzas shared with 3 people versus 3 pizzas shared with 2 people.", null, "### Student Strengths\n\nStudents can name and write fractions and mixed numbers represented by a model.\n\n### Bridging Concepts\n\nStudents understand the concept of division and know what the dividend and divisor represents.\n\n### Standard 4.2C\n\nStudents can identify the division statement that represents a fraction, with models and in context.", null, "", null, "", null, "## Standard 4.3a\n\nStandard 4.3a Read, write, represent, and identify decimals expressed through thousandths.\n\n(Pull down for more)\n\nUnderstanding the Learning Progression\n\nBig Ideas:\n\n• The structure of the base-ten number system is based upon a simple pattern of tens, where each place is ten times the value of the place to its right. This is known as a ten-to-one place value relationship (Van de Walle et al., 2019)\n\n• Decimals is another form of writing fractions and the connection between the two is important in understanding the concepts of decimals (i.e. connect 1/10 as 0.1 and 1/100 as 0.01 and 1/1000 as 0.001 - Reading the decimal fractions will help students “hear” the connection).\n\n• Understanding of the base-ten system to the relationship between adjacent places and how numbers compare can help support students round for decimals to thousandths. For example, it is important to deepen understanding and fluency with decimals in the different forms, seeing .57 as 5 tenths and 7 hundredths as well as 57 hundredths (Common Core Standards Writing Team, 2019, p. 64). This ability to rename and decompose decimals can help students round to the nearest whole number, tenth or hundredth.\n\n• The decimal point separates the whole from the fractional part. The place value system extends infinitely in both directions of the decimal point, to very large and very small numbers. Connected uses of decimals in real life using money, metric measurements, batting averages can support student understanding.\n\nImportant Assessment Look-fors:\n\n• The student is able to identify the decimal represented using a variety of different models.\n\n• The student can model a given decimal using base ten blocks when the whole is defined.\n\n• The student is able to read and write decimals, especially with zero placeholders?\n\n• The student is able to represent decimals in a variety of forms?\n\n• When given a decimal, the student can identify the place value position and value of each digit.\n\nPurposeful questions:\n\n• Is the decimal represented more or less than a whole? Explain your answer.\n\n• If the whole changed, how would that affect the decimal represented?\n\n• When modeling with base ten blocks, what would the cube represent if the whole is equal to a rod? Explain your answer.\n\n• How can you represent this decimal in a variety of ways, such as number line, money, or 10-by-10 grid?", null, "### Student Strengths\n\nStudents can identify the ten-to-one relationship within the base-ten system of whole numbers.\nStudents can read and write various amounts of money and recognize that the number before the decimal point represents whole dollars and the amount to the right of the decimal point represent a part of a dollar.\n\n### Bridging Concepts\n\nStudents connect the idea that money is a model or representation of decimals (i.e., that 10 dimes equals a dollar or 100 pennies is equal to a dollar). Students can extend their understanding of ten-to-one place value relationships to decimals.\n\n### Standard 4.3a\n\nRead, write, represent, and identify decimals expressed through thousandths.", null, "", null, "", null, "", null, "## Standard 4.3c\n\nStandard 4.3c Compare and order decimals.\n\n(Pull down for more)\n\nUnderstanding the Learning Progression\n\nBig Ideas:\n\n• This standard builds upon the work students did in previous grades in understanding place value and comparing and ordering whole numbers. In grade 4, in addition to comparing greater numbers, students began relating decimal fractions and decimal numbers and comparing decimals using visual models. The place value understanding that supports the ability to compare decimals also supports the understanding of rounding decimals, which is introduced in grade 5 as SOL 5.1.\n\n• Concepts of whole numbers, fractions, and decimals are connected and applied when comparing and ordering.\n\n• Using manipulatives to construct decimals helps students develop an understanding of the relative size of decimal numbers for comparing and ordering.\n\n• It is important for students to connect decimal number sense concepts such as representations, decimals benchmarks, and/or fractions when comparing and ordering decimals.\n\nImportant Assessment Look-fors:\n\n• The student can compare decimals with different amounts of digits. (Example 0.9 and 0.234)\n\n• The student can justify which decimal is larger or smaller using a variety of strategies that focus on number sense such as models, decimal benchmarks and/or identifying the value of the greatest place value.\n\n• The student can order decimals least to greatest or greatest to least.\n\n• The student can apply a variety of strategies when ordering decimals with similar digits and/or different amounts of digits. (Example: 0.9; 0.901; 0.09; 0.009)\n\nPurposeful questions:\n\n• Identify which decimal is the greatest? Which one is the least? Explain your answer.\n\n• Which decimal(s) can be placed in the space provided so that the decimals are in order from least to greatest? (Example: 0.142; 0.45 ______; 0.8)\n\n• Compare the following decimals using two different strategies to justify which one is greater or least.", null, "### Student Strengths\n\nStudents can compare and order whole numbers with similar numbers of digits and/or smaller numbers.\n\n### Bridging Concepts\n\nStudents use understanding of the ten-to-one base ten relationships to create decimal representations (i.e., base ten blocks, decimal circles/squares, etc.).\n\n### Standard 4.3c\n\nStudents can compare and order decimals.", null, "", null, "", null, "## Standard 4.4a\n\nStandard 4.4a Demonstrate fluency with multiplication facts through 12 x 12, and corresponding division facts.\n\n(Pull down for more)\n\nUnderstanding the Learning Progression\n\nBig Ideas:\n\n• Computational fluency is the ability to think flexibly in order to choose appropriate strategies to solve problems accurately and efficiently (VDOE Grade 4 Curriculum Framework).\n\n• All of the facts are conceptually related so students can figure out new or unknown facts using what they already know (Van de Walle et al, 2018).\n\n• The development of computational fluency relies on quick access to number facts. There are patterns and relationships that exist in the facts. These relationships can be used to learn and retain the facts (VDOE Grade 4 Curriculum Framework).\n\n• Mastering the basic facts is a developmental process. Students move through phases, starting with counting, then more efficient reasoning strategies, and eventually quick recall and mastery. Instruction must help students through these phases without rushing them to know their facts only through memorization (Van de Walle et al, 2018).\n\n• When students struggle with developing basic fact fluency, they may need to return to foundational ideas. Just providing additional drill will not resolve their challenges and can negatively affect their confidence and success in mathematics (Van de Walle et al, 2018).\n\n• In order to develop and use strategies to learn the multiplication facts through the twelves table, students should use concrete materials, a hundreds chart, and mental mathematics. Strategies to learn the multiplication facts include an understanding of multiples, properties of zero and one as factors, commutative property, and related facts (VDOE Grade 4 Curriculum Framework).\n\nImportant Assessment Look-fors:\n\n• The student knows and is able to apply a variety of strategies (i.e., partial products, using friendly numbers, repeated addition, and/or decomposition strategies, recall, etc.).\n\n• The student demonstrates an understanding of the term “product” and uses a strategy to find a product that leads to a correct answer.\n\n• The student identifies multiple number sentences with the same product.\n\n• Student’s work shows understanding of the inverse relationship between multiplication and division.\n\nPurposeful questions:\n\n• What strategies are most efficient for this fact (or set of facts) and how do you know?\n\n• How can knowing one fact help you with another fact (or a fact that you don’t know)?\n\n• What makes one fact related to another fact? How can knowing related facts be helpful?\n\n• What do you know about the relationship between multiplication and division? How can that relationship help you solve problems?\n\n• What are some ways that you can break apart these numbers to make this problem easier?", null, "### Student Strengths\n\nStudents developed an understanding of the meanings of multiplication and division of whole numbers through activities and practical problems involving equal-sized groups, arrays, and length models.\n\n### Bridging Concepts\n\nStudents have developed efficient reasoning strategies that will lead them to quick recall and memorization of facts of 0, 1, 2, 5 and 10.\n\n### Standard 4.4A\n\nThe student will demonstrate fluency with multiplication facts through 12 × 12, and the corresponding division facts.", null, "", null, "", null, "", null, "", null, "## Standard 4.4b\n\nStandard 4.4b Estimate and determine sums, differences, and products of whole numbers.\n\n(Pull down for more)\n\nUnderstanding the Learning Progression\n\nBig Ideas:\n\n• Flexible methods of computation for all four operations involve taking apart (decomposing) and combining (composing) numbers in a variety of ways (Van de Walle et al, 2018).\n\n• Students should explore and apply the properties of addition and multiplication as strategies for solving addition, subtraction, multiplication, and division problems using a variety of representations (e.g., manipulatives, diagrams, and symbols). (VDOE Grade 4 Curriculum Framework)\n\n• Flexible methods for computation require deep understanding of the operations and the properties of operations (commutative property, associative property, and the distributive property). How addition and subtraction, as well as multiplication and division, are related as inverse operations is also critical knowledge (Van de Walle et al, 2018).\n\n• Estimation can be used to determine the approximation for and then to verify the reasonableness of sums, differences, products, and quotients of whole numbers. An estimate is a number that lies within a range of the exact solution, and the estimation strategy used in a particular problem determines how close the number is to the exact solution. (VDOE Grade 4 Curriculum Framework)\n\nImportant Assessment Look-fors:\n\n• Students' work shows that they understand and can apply terms such as estimate, sum, difference and product.\n\n• Students’ can select a strategy that they understand and can apply successfully.\n\n• Students can explain their solution and justify the reasonableness of their answer.\n\n• Student’s work contains a variety of strategies and representations.\n\n• Students are able to use estimation to determine the reasonableness of their answer (i.e., a little more than, a little less than, closer to, etc.).\n\nPurposeful questions:\n\n• Tell me about the operation you decided to use and why it makes sense.\n\n• How did you represent your thinking?\n\n• How do you know __ is closest to __?\n\n• Why did you choose that place value to estimate?", null, "### Student Strengths\n\nThe student will estimate and determine the sum or difference of two whole numbers up to 9,999.\nThe student will represent multiplication and division through 10 × 10, using a variety of approaches and models.\n\n### Bridging Concepts\n\nStudents use their place value understanding and ability to round to estimate sums and differences of two whole numbers.Students use strategies based on place value and properties of the operations to multiply whole numbers.\n\n### Standard 4.4B\n\nThe student will estimate and determine sums, differences, and products of whole numbers.", null, "", null, "", null, "", null, "## Standard 4.4c\n\nStandard 4.4c Estimate and determine quotients of whole numbers, with and without remainders.\n\n(Pull down for more)\n\nUnderstanding the Learning Progression\n\nBig Ideas:\n\n• Flexible methods of computation involve taking apart (decomposing) and combining (composing) numbers in a variety of ways (Van de Walle et al, 2018).\n\n• Students should explore and apply the properties of addition and multiplication as strategies for solving division problems using a variety of representations (e.g., manipulatives, diagrams, and symbols). (VDOE Grade 4 Curriculum Framework)\n\n• Flexible methods for computation require deep understanding of the operations and the properties of operations (commutative property, associative property, and the distributive property). How addition and subtraction, as well as multiplication and division, are related as inverse operations is also critical knowledge (Van de Walle et al, 2018).\n\n• Estimation can be used to determine the approximation for and then to verify the reasonableness of quotients of whole numbers. An estimate is a number that lies within a range of the exact solution, and the estimation strategy used in a particular problem determines how close the number is to the exact solution (VDOE Grade 4 Curriculum Framework).\n\n• Some division situations will produce a remainder, but the remainder will always be less than the divisor. If the remainder is greater than the divisor, that means at least one more can be given to each group (fair sharing) or at least one more group of the given size (the dividend) may be created (Georgia Department of Education Grade 4 Curriculum).\n\nImportant Assessment Look-fors:\n\n• Students’ use methods they understand and can explain.\n\n• Students' work shows that they understand and can apply the term quotient(s).\n\n• Students’ work demonstrates an understanding of place value and identifying related facts that correlate with the problem.\n\n• Students can explain their solution and the reasonableness of their answer.\n\n• Student’s work contains a variety of strategies and representations.\n\n• The student is able to use estimation to determine the reasonableness of their answer (ie, a little more than, a little less than, closer to, etc.).\n\nProbing questions:\n\n• How did you represent your thinking?\n\n• How do you know __ is closest to __?\n\n• Why did you choose that place value for your estimate?\n\n• What is the meaning of a remainder in a division problem?\n\n• What effect does a remainder have on a quotient?\n\n• How are remainders and divisors related?", null, "### Student Strengths\n\nStudents developed an understanding of the meanings of multiplication and division of whole numbers through activities and practical problems involving equal-sized groups, arrays, and length models.\nThe students have worked on fluency of facts for 0, 1, 2, 5, and 10.\n\n### Bridging Concepts\n\nStudents may still be working on developing efficient reasoning strategies that will lead them to quick recall and memorization of facts from 0 - 12.\n\n### Standard 4.4c\n\nThe student will estimate and determine quotients of whole numbers, with and without remainders.", null, "", null, "", null, "", null, "", null, "## Standard 4.4d\n\nStandard 4.4d Create and solve single-step and multistep practical problems involving addition, subtraction, and multiplication of whole numbers and single step practical problems with division.\n\n(Pull down for more)\n\nUnderstanding the Learning Progression\n\nBig Ideas:\n\n• Use reasoning and a variety of strategies that the student understands and is able to explain.\n\n• Makes sense of the problem rather than relying on keywords that may not always be helpful. (See VDOE Standards of Learning- Grade 4 Document p.19).\n\n• Inverse operations are related and can flexibly be used to solve problems.\n\n• Estimation, based on number sense and an understanding of place value, can be used to determine if an answer is reasonable.\n\n• Students will be stronger problem solvers given opportunities to engage with a variety of problem types. For more information about addition/subtraction problem types see the Grade 3 VDOE Standards of Learning Document p. 15 and for multiplication/division problem types see the Grade 4 VDOE Standards of Learning Document pp.20-21.\n\nImportant Assessment Look-fors:\n\n• Students are able to create a problem using the given information.\n\n• Students are able to correctly represent the problem they created and it matches the way they solved it.\n\n• Student work shows that they understand the problem because they have planned an approach to solve the problem and/or a way to represent their thinking.\n\n• Student work shows that they understand what each number in the problem represents.\n\n• Student work shows that they understand what operations can be used, the meaning behind the operation or how the operations are related.\n\nPurposeful questions:\n\n• How did you represent your thinking? How do you know it matches the story?\n\n• Is there another way that you could solve this problem?\n\n• What equation/number sentence matches your thinking?\n\n• What operations did you decide to use? Why?", null, "### Student Strengths\n\nStudents determine the sum or difference of two whole numbers to 4 digits.\nStudents create and solve single-step and multistep practical problems involving sums or differences of two whole numbers, each 9,999 or less.\n\n### Bridging Concepts\n\nStudents use their understanding of multiplication and division to solve a variety of single step contextual problems.\nStudents use place value understanding and properties of operations to solve multiplication problems up to 10 x 10.\n\n### Standard 4.4D\n\nStudents create and solve single-step and multistep practical problems involving addition, subtraction, and multiplication of whole numbers and single step practical problems with division.", null, "", null, "", null, "", null, "", null, "", null, "### Routines:", null, "" ]	[ null, 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http://techtalks.tv/events/13/?page=1	[ "## TechTalks from event: IEEE FOCS 2010\n\n51st Annual IEEE Symposium on Foundations of Computer Science\n\n• Geometric complexity theory (GCT) Authors: Ketan Mulmuley\nGeometric complexity theory (GCT) is an approach towards the fundamental lower bound problems of complexity theory through algebraic geometry and representation theory. This talk will give a brief introduction to GCT without assuming any background in algebraic geometry or representation theory.\n• How to Grow Your Lower Bounds Authors: Mihai Patrascu\nI will survey the state of the art in proving hardness for data structure problems in the cell-probe model of computation.\n• How To Think About Mechanism Design Authors: Tim Roughgarden, Stanford University\nMechanism design studies optimization problems where the underlying data --- such as the value of a good or the cost of performing a task --- is initially unknown to the algorithm designer. Auction settings are canonical examples, where the private data is the willingness to pay the bidders for the goods on sale, and the optimization problem is to allocate the goods to maximize some objective, such as revenue or overall value to society. A \"mechanism\" is a protocol that interacts with participants and determines a solution to the underlying optimization problem. We first explain why designing mechanisms with good game-theoretic properties boils down to algorithm design in a certain \"constrained computational model.\" We differentiate between single-parameter problems, where this computational model is well understood, and multi-parameter problems, where it is not. We then study two specific problem domains: revenue-maximizing auctions, and the recent idea of analyzing auctions via a \"Bayesian thought experiment\"; and welfare-maximizing mechanisms, with an emphasis on black-box reductions that transmute approximation algorithms into truthful approximation mechanisms.\n• Constructive Algorithms for Discrepancy Minimization Authors: Nikhil Bansal\nGiven a set system (V,S), V=[n] and S={S_1,ldots,S_m}, the minimum discrepancy problem is to find a 2-coloring X:V -> {-1,+1}, such that each set is colored as evenly as possible, i.e. find X to minimize max_{j in [m]} \|sum_{i in S_j} X(i)\|. In this paper we give the first polynomial time algorithms for discrepancy minimization, that achieve bounds similar to those known existentially using the so-called Entropy Method. We also give a first approximation-like result for discrepancy. Specifically we give efficient randomized algorithms to: 1. Construct an \\$O(sqrt{n})\\$ discrepancy coloring for general sets systems when \\$m=O(n)\\$, matching the celebrated result of Spencer up to constant factors. Previously, no algorithmic guarantee better than the random coloring bound, i.e. O((n log n)^{1/2}), was known. More generally, for \\$mgeq n\\$, we obtain a discrepancy bound of O(n^{1/2} log (2m/n)). 2. Construct a coloring with discrepancy \\$O(t^{1/2} log n)\\$, if each element lies in at most \\$t\\$ sets. This matches the (non-constructive) result of Srinivasan cite{Sr}. 3. Construct a coloring with discrepancy O(lambdalog(nm)), where lambda is the hereditary discrepancy of the set system. In particular, this implies a logarithmic approximation for hereditary discrepancy. The main idea in our algorithms is to gradually produce a coloring by solving a sequence of semidefinite programs, while using the entropy method to guide the choice of the semidefinite program at each stage.\n• Bounded Independence Fools Degree-2 Threshold Functions Authors: Ilias Diakonikolas and Daniel M. Kane and Jelani Nelson\nLet x be a random vector coming from any k-wise independent distribution over {-1,1}^n. For an n-variate degree-2 polynomial p, we prove that E[sgn(p(x))] is determined up to an additive eps for k = poly(1/eps). This gives a large class of explicit pseudo-random generators against such functions and answers an open question of Diakonikolas et al. (FOCS 2009).\n\nIn the process, we develop a novel analytic technique we dub multivariate FT-mollification. This provides a generic tool to approximate bounded (multivariate) functions by low-degree polynomials (with respect to several different notions of approximation). A univariate version of the method was introduced by Kane et al. (SODA 2010) in the context of streaming algorithms. In this work, we refine it and generalize it to the multivariate setting. We believe that our technique is of independent mathematical interest. To illustrate its generality, we note that it implies a multidimensional generalization of Jackson's classical result in approximation theory due to (Ganzburg 1979).\n\nTo obtain our main result, we combine the FT-mollification technique with several linear algebraic and probabilistic tools. These include the invariance principle of of Mossell, O'Donnell and Oleszkiewicz, anti-concentration bounds for low-degree polynomials, an appropriate decomposition of degree-2 polynomials, and a generalized hyper-contractive inequality for quadratic forms which takes the operator norm of the associated matrix into account. Our analysis is quite modular; it readily adapts to show that intersections of halfspaces and degree-2 threshold functions are fooled by bounded independence. From this it follows that Omega(1/eps^2)-wise independence derandomizes the Goemans-Williamson hyperplane rounding scheme.\n\nOur techniques unify, simplify, and in some cases improve several recent results in the literature concerning threshold functions. For the case of ``regular'' halfspaces we give a simple proof of an optimal independence bound of Th\n\n• The Coin Problem, and Pseudorandomness for Branching Programs / Pseudorandom Generators for Regular Branching Programs Authors: Joshua Brody and Elad Verbin / Mark Braverman and Anup Rao and Ran Raz and Amir Yehudayoff\nThe emph{Coin Problem} is the following problem: a coin is given, which lands on head with probability either \\$1/2 + \beta\\$ or \\$1/2 - \beta\\$. We are given the outcome of \\$n\\$ independent tosses of this coin, and the goal is to guess which way the coin is biased, and to be correct with probability \\$ge 2/3\\$. When our computational model is unrestricted, the majority function is optimal, and succeeds when \\$\beta ge c /sqrt{n}\\$ for a large enough constant \\$c\\$. The coin problem is open and interesting in models that cannot compute the majority function.\n\nIn this paper we study the coin problem in the model of emph{read-once width-\\$w\\$ branching programs}. We prove that in order to succeed in this model, \\$\beta\\$ must be at least \\$1/ (log n)^{Theta(w)}\\$. For constant \\$w\\$ this is tight by considering the recursive tribes function.\n\nWe generalize this to a emph{Dice Problem}, where instead of independent tosses of a coin we are given independent tosses of one of two \\$m\\$-sided dice. We prove that if the distributions are too close, then the dice cannot be distinguished by a small-width read-once branching program.\n\nWe suggest one application for this kind of theorems: we prove that Nisan's Generator fools width-\\$w\\$ read-once emph{permutation} branching programs, using seed length \\$O(w^4 log n log log n + log n log (1/eps))\\$. For \\$w=eps=Theta(1)\\$, this seedlength is \\$O(log n log log n)\\$. The coin theorem and its relatives might have other connections to PRGs. This application is related to the independent, but chronologically-earlier, work of Braverman, Rao, Raz and Yehudayoff (which might be submitted to this FOCS).\n\n• Settling the Polynomial Learnability of Mixtures of Gaussians Authors: Ankur Moitra and Gregory Valiant\nGiven data drawn from a mixture of multivariate Gaussians, a basic problem is to accurately estimate the mixture parameters. We give an algorithm for this problem that has a running time, and data requirement polynomial in the dimension and the inverse of the desired accuracy, with provably minimal assumptions on the Gaussians. As simple consequences of our learning algorithm, we can perform near-optimal clustering of the sample points and density estimation for mixtures of \\$k\\$ Gaussians, efficiently.\n\nThe building blocks of our algorithm are based on the work (Kalai emph{et al}, STOC 2010)~cite{2Gs} that gives an efficient algorithm for learning mixtures of two Gaussians by considering a series of projections down to one dimension, and applying the emph{method of moments} to each univariate projection. A major technical hurdle in~cite{2Gs} is showing that one can efficiently learn emph{univariate} mixtures of two Gaussians. In contrast, because pathological scenarios can arise when considering univariate projections of mixtures of more than two Gaussians, the bulk of the work in this paper concerns how to leverage an algorithm for learning univariate mixtures (of many Gaussians) to yield an efficient algorithm for learning in high dimensions. Our algorithm employs emph{hierarchical clustering} and rescaling, together with delicate methods for backtracking and recovering from failures that can occur in our univariate algorithm.\n\nFinally, while the running time and data requirements of our algorithm depend exponentially on the number of Gaussians in the mixture, we prove that such a dependence in necessary.\n\n• Polynomial Learning of Distribution Families Authors: Mikhail Belkin and Kaushik Sinha\nThe question of polynomial learnability of probability distributions, particularly Gaussian mixture distributions, has recently received significant attention in theoretical computer science and machine learning. However, despite major progress, the general question of polynomial learnability of Gaussian mixture distributions still remained open. The current work resolves the question of polynomial learnability for Gaussian mixtures in high dimension with an arbitrary but fixed number of components.\n\nThe result for Gaussian distributions relies on a very general result of independent interest on learning parameters of distributions belonging to what we call {it polynomial families}. These families are characterized by their moments being polynomial of parameters and, perhaps surprisingly, include almost all common probability distributions as well as their mixtures and products. Using tools from real algebraic geometry, we show that parameters of any distribution belonging to such a family can be learned in polynomial time.\n\nTo estimate parameters of a Gaussian mixture distribution the general results on polynomial families are combined with a certain deterministic dimensionality reduction allowing learning a high-dimensional mixture to be reduced to a polynomial number of parameter estimation problems in low dimension.\n\n• Agnostically learning under permutation invariant distributions Authors: Karl Wimmer\nWe generalize algorithms from computational learning theory that are successful under the uniform distribution on the Boolean hypercube \\${0,1}^n\\$ to algorithms successful on permutation invariant distributions, distributions where the probability mass remains constant upon permutations in the instances. While the tools in our generalization mimic those used for the Boolean hypercube, the fact that permutation invariant distributions are not product distributions presents a significant obstacle.\n\nUnder the uniform distribution, halfspaces can be agnostically learned in polynomial time for constant \\$eps\\$. The main tools used are a theorem of Peres~cite{Peres04} bounding the {it noise sensitivity} of a halfspace, a result of~cite{KOS04} that this theorem this implies Fourier concentration, and a modification of the Low-Degree algorithm of Linial, Mansour, Nisan~cite{LMN:93} made by Kalai et. al.~cite{KKMS08}. These results are extended to arbitrary product distributions in~cite{BOWi08}.\n\nWe prove analogous results for permutation invariant distributions; more generally, we work in the domain of the symmetric group. We define noise sensitivity in this setting, and show that noise sensitivity has a nice combinatorial interpretation in terms of Young tableaux. The main technical innovations involve techniques from the representation theory of the symmetric group, especially the combinatorics of Young tableaux. We show that low noise sensitivity implies concentration on ``simple'' components of the Fourier spectrum, and that this fact will allow us to agnostically learn halfspaces under permutation invariant distributions to constant accuracy in roughly the same time as in the uniform distribution over the Boolean hypercube case.\n\n• Learning Convex Concepts from Gaussian Distributions with PCA Authors: Santosh Vempala\nWe present a new algorithm for learning a convex set in \\$R^n\\$ given labeled examples drawn from any Gaussian distribution. The efficiency of the algorithm depends on the dimension of the {em normal subspace}, the span of vectors orthogonal to supporting hyperplanes of the convex set. The key step of the algorithm uses a Singular Value Decomposition (SVD) to approximate the relevant normal subspace. The complexity of the resulting algorithm is \\$poly(n)2^{ ilde{O}(k)}\\$ for an arbitrary convex set with normal subspace of dimension \\$k\\$. For the important special case of the intersection of \\$k\\$ halfspaces, the complexity is [ poly(n,k,1/eps) + n cdot min , k^{O(log k/eps^4)}, (k/eps)^{O(klog (1/eps))} ] to learn a hypothesis that correctly classifies \\$1-eps\\$ of the unknown Gaussian distribution. This improves substantially on existing algorithms and is the first algorithm to achieve a fixed polynomial dependence on \\$n\\$. The proof of our main result is based on a monotonicity property of Gaussian space." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8510702,"math_prob":0.9573388,"size":17901,"snap":"2019-26-2019-30","text_gpt3_token_len":3762,"char_repetition_ratio":0.12471364,"word_repetition_ratio":0.024806201,"special_character_ratio":0.18708453,"punctuation_ratio":0.07083043,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99331903,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-16T16:47:59Z\",\"WARC-Record-ID\":\"<urn:uuid:73a160af-4226-4403-8b15-b23e6a0e6c6c>\",\"Content-Length\":\"79414\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:72105938-d6bc-4cb9-998c-3aabcfd47152>\",\"WARC-Concurrent-To\":\"<urn:uuid:c9d7d0d1-928f-4e7f-9d98-40d5434e5adb>\",\"WARC-IP-Address\":\"173.230.131.71\",\"WARC-Target-URI\":\"http://techtalks.tv/events/13/?page=1\",\"WARC-Payload-Digest\":\"sha1:76RPHQPRSDNB4745OZ6JAX3MSUU5BQFH\",\"WARC-Block-Digest\":\"sha1:AOMZ7SCWLOHYCBWGW6RIPJPPOF4W6OZX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195524679.39_warc_CC-MAIN-20190716160315-20190716182315-00348.warc.gz\"}"}
https://calculomates.com/en/divisors/of/2629/	[ "# Divisors of 2629\n\n## Divisors of 2629\n\nThe list of all positive divisors (that is, the list of all integers that divide 22) is as follows :\n\nAccordingly:\n\n2629 is multiplo of 1\n\n2629 is multiplo of 11\n\n2629 is multiplo of 239\n\n2629 has 3 positive divisors\n\n## Parity of 2629\n\n2629is an odd number,as it is not divisible by 2\n\n## The factors for 2629\n\nThe factors for 2629 are all the numbers between -2629 and 2629 , which divide 2629 without leaving any remainder. Since 2629 divided by -2629 is an integer, -2629 is a factor of 2629 .\n\nSince 2629 divided by -2629 is a whole number, -2629 is a factor of 2629\n\nSince 2629 divided by -239 is a whole number, -239 is a factor of 2629\n\nSince 2629 divided by -11 is a whole number, -11 is a factor of 2629\n\nSince 2629 divided by -1 is a whole number, -1 is a factor of 2629\n\nSince 2629 divided by 1 is a whole number, 1 is a factor of 2629\n\nSince 2629 divided by 11 is a whole number, 11 is a factor of 2629\n\nSince 2629 divided by 239 is a whole number, 239 is a factor of 2629\n\n## What are the multiples of 2629?\n\nMultiples of 2629 are all integers divisible by 2629 , i.e. the remainder of the full division by 2629 is zero. There are infinite multiples of 2629. The smallest multiples of 2629 are:\n\n0 : in fact, 0 is divisible by any integer, so it is also a multiple of 2629 since 0 × 2629 = 0\n\n2629 : in fact, 2629 is a multiple of itself, since 2629 is divisible by 2629 (it was 2629 / 2629 = 1, so the rest of this division is zero)\n\n5258: in fact, 5258 = 2629 × 2\n\n7887: in fact, 7887 = 2629 × 3\n\n10516: in fact, 10516 = 2629 × 4\n\n13145: in fact, 13145 = 2629 × 5\n\netc.\n\n## Is 2629 a prime number?\n\nIt is possible to determine using mathematical techniques whether an integer is prime or not.\n\nfor 2629, the answer is: No, 2629 is not a prime number.\n\n## How do you determine if a number is prime?\n\nTo know the primality of an integer, we can use several algorithms. The most naive is to try all divisors below the number you want to know if it is prime (in our case 2629). We can already eliminate even numbers bigger than 2 (then 4 , 6 , 8 ...). Besides, we can stop at the square root of the number in question (here 51.274 ). Historically, the Eratosthenes screen (which dates back to Antiquity) uses this technique relatively effectively.\n\nMore modern techniques include the Atkin screen, probabilistic tests, or the cyclotomic test." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9168329,"math_prob":0.9876593,"size":2120,"snap":"2021-04-2021-17","text_gpt3_token_len":642,"char_repetition_ratio":0.20274103,"word_repetition_ratio":0.09178744,"special_character_ratio":0.37216982,"punctuation_ratio":0.13822894,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9992803,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-27T22:26:45Z\",\"WARC-Record-ID\":\"<urn:uuid:2e4b0209-006a-4358-9fa8-62620e978c31>\",\"Content-Length\":\"16159\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d3baf991-3eec-4dad-adef-791f61f97d26>\",\"WARC-Concurrent-To\":\"<urn:uuid:3017ca2f-04ce-467f-919a-4257ca132d41>\",\"WARC-IP-Address\":\"104.21.88.17\",\"WARC-Target-URI\":\"https://calculomates.com/en/divisors/of/2629/\",\"WARC-Payload-Digest\":\"sha1:BWEIXWRYQBZQUOXE2FUVVKFCZRQQ35P4\",\"WARC-Block-Digest\":\"sha1:F3OK6SS2XD325GNECHG6Z77Y5B4V5AK7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610704833804.93_warc_CC-MAIN-20210127214413-20210128004413-00577.warc.gz\"}"}
https://projecteuler.net/problem=502	[ "", null, "## Counting Castles", null, "Published on Saturday, 7th February 2015, 04:00 pm; Solved by 219;\nDifficulty rating: 100%\n\n### Problem 502\n\nWe define a block to be a rectangle with a height of 1 and an integer-valued length. Let a castle be a configuration of stacked blocks.\n\nGiven a game grid that is w units wide and h units tall, a castle is generated according to the following rules:\n\n1. Blocks can be placed on top of other blocks as long as nothing sticks out past the edges or hangs out over open space.\n2. All blocks are aligned/snapped to the grid.\n3. Any two neighboring blocks on the same row have at least one unit of space between them.\n4. The bottom row is occupied by a block of length w.\n5. The maximum achieved height of the entire castle is exactly h.\n6. The castle is made from an even number of blocks.\n\nThe following is a sample castle for w=8 and h=5:", null, "Let F(w,h) represent the number of valid castles, given grid parameters w and h.\n\nFor example, F(4,2) = 10, F(13,10) = 3729050610636, F(10,13) = 37959702514, and F(100,100) mod 1 000 000 007 = 841913936.\n\nFind (F(1012,100) + F(10000,10000) + F(100,1012)) mod 1 000 000 007." ]	[ null, "https://projecteuler.net/images/print_page_logo.png", null, "https://projecteuler.net/images/icon_info.png", null, "https://projecteuler.net/project/images/p502_castles.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8612652,"math_prob":0.9797678,"size":1027,"snap":"2019-51-2020-05","text_gpt3_token_len":291,"char_repetition_ratio":0.1202346,"word_repetition_ratio":0.0,"special_character_ratio":0.34469327,"punctuation_ratio":0.118421055,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9775317,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-12T09:49:49Z\",\"WARC-Record-ID\":\"<urn:uuid:5c1ce188-7c4a-48b4-93ce-7b8248646400>\",\"Content-Length\":\"6012\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cb556af4-00e6-4686-9536-c7250d66032b>\",\"WARC-Concurrent-To\":\"<urn:uuid:b01b35ef-5387-4a29-b730-f6e4b535b8fc>\",\"WARC-IP-Address\":\"31.170.122.77\",\"WARC-Target-URI\":\"https://projecteuler.net/problem=502\",\"WARC-Payload-Digest\":\"sha1:TSKCI6Q4OYXSARQWJJRUAEYA4XB24O7A\",\"WARC-Block-Digest\":\"sha1:HBZDBXB725QJEKKKT3GB2HR3JHDMMM3O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540542644.69_warc_CC-MAIN-20191212074623-20191212102623-00409.warc.gz\"}"}
https://reason.town/neural-network-from-scratch-pytorch/	[ "# How to Build a Neural Network from Scratch with Pytorch\n\nThis blog post will teach you how to build a neural network from scratch with Pytorch. We’ll cover everything from loading data to training and evaluating the network.\n\nCheckout this video:\n\n## Introduction\n\nNeural networks are a powerful machine learning tool that is used for both supervised and unsupervised learning tasks. In this tutorial, we will learn how to build a neural network from scratch using Pytorch, a popular deep learning framework. We will also learn how to train and evaluate our neural network. By the end of this tutorial, you will be able to build your own neural networks and use them for various tasks.\n\n## What is a neural network?\n\nA neural network is a series of layers, or nodes, that process information in a way that is similar to the brain. Each node is connected to several others, and they work together to solve complex problems. Neural networks are used in a variety of fields, including image recognition, natural language processing, and predictive analytics.\n\nBuilding a neural network from scratch can be a daunting task, but Pytorch makes it easy with its pre-built functions and classes. In this tutorial, we’ll show you how to build a simple neural network with Pytorch.\n\nFirst, we need to import the torch library. This will give us access to all of the functions we need to build our network.\n\nNext, we’ll create our first layer. This is called the input layer, and it will be responsible for taking in our data and feeding it into the rest of the network. We can create an input layer with the following code:\n\n## How do neural networks work?\n\nNeural networks are computer systems that are modeled after the human brain. These systems are able to learn and recognize patterns. Neural networks are composed of a series of layers. The first layer is the input layer. The second layer is the hidden layer. The hidden layer is made up of neurons. The third layer is the output layer.\n\nNeural networks are trained by providing them with a set of training data. This data is fed into the input layer. The hidden layer then learns to recognize patterns in the data. The outputlayer produces the results of the neural network.\n\nPytorch is a deep learning framework that provides a way to implement neural networks on a variety of devices including GPUs and CPUs. Pytorch allows for easy and flexible implementation of neural networks.\n\n## What is Pytorch?\n\nPytorch is a Python-based deep learning framework that enables you to build neural networks from scratch. It offers an easy-to-use API and can be used for both research and production. In this tutorial, you will learn how to build a neural network from scratch using Pytorch.\n\n## Why use Pytorch to build a neural network from scratch?\n\nPytorch is a powerful python library that makes it easy to build complex neural networks from scratch. In this tutorial, we will use Pytorch to build a simple neural network from scratch. We will also discuss some of the advantages of using Pytorch over other libraries like Tensorflow. By the end of this tutorial, you will be able to build and train your own neural networks with Pytorch.\n\n## What are the steps to build a neural network from scratch with Pytorch?\n\nThere are a few steps to building a neural network from scratch with Pytorch. First, you need to define the network architecture, which includes the number of layers and the number of neurons in each layer. Next, you need to initialize the weights and biases for each neuron. After that, you need to define the forward propagation function, which takes in an input and produces an output. Finally, you need to define the backpropagation function, which calculates the gradient of the error with respect to the weights and biases.\n\n## Conclusion\n\nIn this tutorial, we’ve seen how to build a neural network from scratch using Pytorch. We’ve covered the basic concepts of neural networks, such as layers and weights, and seen how to implement a simple network using only numpy. Then, we used Pytorch to build a more complex network and train it on a dataset. Finally, we saw how to evaluate the network on new data and make predictions.\n\nIf you’re interested in learning more about neural networks, we recommend checking out the following resources:\n\n– [A Complete Beginner’s Guide to Neural Networks](https://tosche.net/neural-networks/)\n– [How to Build a Neural Network from Scratch with Pytorch](https://medium.com/@jamesloyys/how-to-build-a-neural-network-from-scratch-with-pytorch-84f2e75bb2cf)\n– [https://towardsdatascience.com/build-your-first-neural network from scratch with Python and Keras](https://towardsdatascience.com/build-your first neural network from scratch with Python and Keras)\n\n-Aurélien Géron, “Hands-On Machine Learning with Scikit-Learn, Keras, and TensorFlow: Concepts, Tools, and Techniques to Build Intelligent Systems”, 2nd Edition, O’Reilly Media, Inc., 2019.\n-“Neural Networks and Deep Learning”, Michael Nielsen, Determination Press, 2015." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.888888,"math_prob":0.61088556,"size":5477,"snap":"2023-14-2023-23","text_gpt3_token_len":1163,"char_repetition_ratio":0.19568792,"word_repetition_ratio":0.110593714,"special_character_ratio":0.19883148,"punctuation_ratio":0.11376673,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9688573,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-20T19:16:16Z\",\"WARC-Record-ID\":\"<urn:uuid:41d12956-0651-4fb7-8aa2-2ce5750d0acc>\",\"Content-Length\":\"154380\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ffe0fa2c-1434-4088-b847-e936f097bec9>\",\"WARC-Concurrent-To\":\"<urn:uuid:9cf79c30-24ef-47ee-8d1c-b13a48653a36>\",\"WARC-IP-Address\":\"104.21.38.119\",\"WARC-Target-URI\":\"https://reason.town/neural-network-from-scratch-pytorch/\",\"WARC-Payload-Digest\":\"sha1:3CS7BCDW2OVWVTOP3K7DK3INX744CQPP\",\"WARC-Block-Digest\":\"sha1:NS4CJBPXCWVVBBGWBVCSG6KXHVAJUD4I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943555.25_warc_CC-MAIN-20230320175948-20230320205948-00104.warc.gz\"}"}
https://dontgetserious.com/step-by-step-strategy-for-solving-word-problems/	[ "# Step by Step Strategy for Solving Word Problems\n\n63\n\nSolving word problems can be trickier and more intimidating than solving a math equation. Most students are comfortable working with given numbers, but the simple addition of reading is enough to send a shiver down the spines of most students. However, solving even the most challenging word problems is not difficult provided that you comprehend the mathematical concept behind the question.\n\n## Strategies Used to Solve word problems", null, "The following steps can help equip students with the tools necessary to help them become confident in solving the mystery of word problems.\n\n## 1) Understanding the problem\n\n”A recent Adobe Education Exchange article said that the first step to successfully solve word problems is to read the question carefully”. This helps you understand and figure out the scenario of the problem. Be careful not to jump to any conclusion about solving it until you have a full understanding of the problem. The word problem will provide you with all the necessary information that you need to find a solution. Once you have the bigger picture, it’s time to identify the problem and come up with the measurements for finding the answer.\n\n## 2) Gather information\n\nSometimes curriculum writers add extra information that is not necessary for finding the answer to a math problem. The additional details are meant to train the students to ignore the unnecessary information and stay focused on finding the real problem. You will have to filter through the question to identify the details needed to solve that particular problem. Every situation requires different formats but using a visual representation makes it easier. You can use a list, table, or chart to note down the information given and leave blanks for the information not provided.\n\n## 3) Create the equation\n\nIn this step, you will need to identify the keywords. You can use a pencil to underline or circle the phrases that tell you what you need to find. These terms include words like sum, addition, more than, increased, which all mean to add. Once you know what you need to solve, you can determine the formula, equation, and steps you need to use to come up with the correct answer. The step also helps reinforce thinking to identify the information you don’t need.\n\n## 4) Solve the problem\n\nUsing the information provided and the formula, you can now key in the values to develop an equation that will help you solve the problem to get the unknown variable. Always be keen as you do your calculation to avoid mistakes and ensure that you are using the correct order of operations.\n\nOnce done with the computation, review your answer to ensure that it’s accurate. Using the information given and you can test it to see if it’s within the expected margin of the result. If it is outrageous or unreasonable, review your steps and calculation for errors. Going through the problem carefully will help you figure out where you made a mistake.\n\n## 6) Practice word problems often\n\nThe practice is the key to mastering word problem-solving. When you practice word problems, they will often become easier to solve as you notice similarities and patterns in solving them. These help you gain confidence even when you are dealing with challenging word problems.\n\n## Conclusion\n\nWhile the level of difficulty varies, the above steps are the basic planned approach that can help you answer word problems.\n\n63" ]	[ null, "https://dontgetserious.com/wp-content/uploads/2020/11/Solve-word-problems.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92251337,"math_prob":0.7836634,"size":3370,"snap":"2021-31-2021-39","text_gpt3_token_len":636,"char_repetition_ratio":0.1553773,"word_repetition_ratio":0.0,"special_character_ratio":0.18664688,"punctuation_ratio":0.0762987,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9595149,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-27T09:54:11Z\",\"WARC-Record-ID\":\"<urn:uuid:cff49b6d-7457-470e-9e7d-13742be91831>\",\"Content-Length\":\"90914\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bdc69e9b-d1de-4f6e-baa7-a1e81b6a5bbe>\",\"WARC-Concurrent-To\":\"<urn:uuid:39e74a72-0a2f-4211-a8d8-f0f5734e1f6f>\",\"WARC-IP-Address\":\"104.21.22.225\",\"WARC-Target-URI\":\"https://dontgetserious.com/step-by-step-strategy-for-solving-word-problems/\",\"WARC-Payload-Digest\":\"sha1:QZE7V5DBUDTIFBWZEDPYY7O5Y4GI5ZMU\",\"WARC-Block-Digest\":\"sha1:3OEKAV4CY7F4NAB362PVM7GA2FY5FCEQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780058415.93_warc_CC-MAIN-20210927090448-20210927120448-00536.warc.gz\"}"}
https://cellrank.readthedocs.io/en/latest/pancreas_basic.html	[ "# Pancreas Basics¶\n\nThis tutorial shows how to apply CellRank in order to infer initial or terminal states of a developmental process and how to compute probabilistic fate mappings. The first part of this tutorial is very similar to scVelo’s tutorial on pancreatic endocrinogenesis. The data we use here comes from Bastidas-Ponce et al. (2018). For more info on scVelo, see the documentation or read the article.\n\nThis is the high level mode to interact with CellRank, which is quick and simple but does not offer as many options as the lower level mode, in which you interact directly with our kernels and estimators. If you would like to get to know this more advanced way of interacting with CellRank, see the pancreas advanced tutorial.\n\n## Import packages & data¶\n\nEasiest way to start is to download Miniconda3 along with the environment file found here. To create the environment, run conda create -f environment.yml.\n\n:\n\nimport scvelo as scv\nimport scanpy as sc\nimport cellrank as cr\nimport numpy as np\n\nscv.settings.verbosity = 3\nscv.settings.set_figure_params('scvelo')\ncr.settings.verbosity = 2\n\n\nFirst, we need to get the data. The following commands will download the adata object and save it under datasets/endocrinogenesis_day15.5.h5ad.\n\n:\n\nadata = cr.datasets.pancreas()\n\nAbundance of ['spliced', 'unspliced']: [0.81 0.19]\n\n:\n\nAnnData object with n_obs × n_vars = 2531 × 27998\nobs: 'day', 'proliferation', 'G2M_score', 'S_score', 'phase', 'clusters_coarse', 'clusters', 'clusters_fine', 'louvain_Alpha', 'louvain_Beta', 'palantir_pseudotime'\nvar: 'highly_variable_genes'\nuns: 'clusters_colors', 'clusters_fine_colors', 'day_colors', 'louvain_Alpha_colors', 'louvain_Beta_colors', 'neighbors', 'pca'\nobsm: 'X_pca', 'X_umap'\nlayers: 'spliced', 'unspliced'\nobsp: 'connectivities', 'distances'\n\n\n## Pre-process the data¶\n\nFilter out genes which don’t have enough spliced/unspliced counts, normalize and log transform the data and restrict to the top highly variable genes. Further, compute principal components and moments for velocity estimation. These are standard scanpy/scvelo functions, for more information about them, see the scVelo API.\n\n:\n\nscv.pp.filter_and_normalize(adata, min_shared_counts=20, n_top_genes=2000)\n\nFiltered out 22024 genes that are detected 20 counts (shared).\nNormalized count data: X, spliced, unspliced.\nExctracted 2000 highly variable genes.\nLogarithmized X.\ncomputing moments based on connectivities\n'Ms' and 'Mu', moments of un/spliced abundances (adata.layers)\n\n\n## Run scVelo¶\n\nWe will use the dynamical model from scVelo to estimate the velocities. The first step, estimating the parameters of the dynamical model, may take a while (~10min). To make sure we only have to run this once, we developed a caching extension called scachepy. scachepy does not only work for recover_dynamics, but it can cache the output of almost any scanpy or scvelo function. To install it, simply run\n\npip install git+https://github.com/theislab/scachepy\n\n\nIf you don’t want to install scachepy now, don’t worry, the below cell will run without it as well and this is the only place in this tutorial where we’re using it.\n\n:\n\ntry:\nimport scachepy\nc = scachepy.Cache('../../cached_files/basic_tutorial/')\nexcept ModuleNotFoundError:\nprint(\"You don't seem to have scachepy installed, but that's fine, you just have to be a bit patient (~10min). \")\n\nYou don't seem to have scachepy installed, but that's fine, you just have to be a bit patient (~10min).\nrecovering dynamics\n'fit_pars', fitted parameters for splicing dynamics (adata.var)\n\n\nOnce we have the parameters, we can use these to compute the velocities and the velocity graph. The velocity graph is a weighted graph that specifies how likely two cells are to transition into another, given their velocity vectors and relative positions.\n\n:\n\nscv.tl.velocity(adata, mode='dynamical')\n\ncomputing velocities\n'velocity', velocity vectors for each individual cell (adata.layers)\ncomputing velocity graph\n'velocity_graph', sparse matrix with cosine correlations (adata.uns)\n\n:\n\nscv.pl.velocity_embedding_stream(adata, basis='umap', legend_fontsize=12, title='', smooth=.8, min_mass=4)\n\ncomputing velocity embedding", null, "## Run CellRank¶\n\nCellRank builds on the velocities computed by scVelo to compute initial and terminal states of the dynamical process. It further uses the velocities to calculate how likely each cell is to develop from each initial state or towards each terminal state.\n\n## Identify terminal states¶\n\nTerminal states can be computed by running the following command:\n\n:\n\ncr.tl.terminal_states(adata, cluster_key='clusters', weight_connectivities=0.2)\n\nComputing transition matrix based on logits using 'deterministic' mode\nEstimating softmax_scale using 'deterministic' mode\nSetting softmax_scale=3.7951\nFinish (0:00:09)\nUsing a connectivity kernel with weight 0.2\nComputing transition matrix based on connectivities\nFinish (0:00:00)\nComputing eigendecomposition of the transition matrix\nAdding .eigendecomposition\nadata.uns['eig_fwd']\nFinish (0:00:00)\nComputing Schur decomposition\nUnable to import PETSc or SLEPc.\nYou can install it from: https://slepc4py.readthedocs.io/en/stable/install.html\nDefaulting to method='brandts'.\nAdding .eigendecomposition\nadata.uns['eig_fwd']\n.schur\n.schur_matrix\nFinish (0:00:09)\nComputing 3 macrostates\nINFO: Using pre-computed schur decomposition\nAdding .macrostates_memberships\n.macrostates\n.schur\n.coarse_T\n.coarse_stationary_distribution\nFinish (0:00:00)\nAdding adata.obs['terminal_states_probs']\nadata.obs['terminal_states']\nadata.obsm['macrostates_fwd']\n.terminal_states_probabilities\n.terminal_states\n\n\nThe most important parameters in the above function are:\n\n• estimator: this determines what’s going to behind the scenes to compute the terminal states. Options are cr.tl.estimators.CFLARE (“Clustering and Filtering of Left and Right Eigenvectors”) or cr.tl.estimators.GPCCA (“Generalized Perron Cluster Cluster Analysis”). The latter is the default, it computes terminal states by coarse graining the velocity-derived Markov chain into a set of macrostates that represent the slow-time scale dynamics of the process, i.e. it finds the states that you are unlikely to leave again, once you have entered them.\n\n• cluster_key: takes a key from adata.obs to retrieve pre-computed cluster labels, i.e. ‘clusters’ or ‘louvain’. These labels are then mapped onto the set of terminal states, to associate a name and a color with each state.\n\n• n_states: number of expected terminal states. This parameter is optional - if it’s not provided, this number is estimated from the so-called ‘eigengap heuristic’ of the spectrum of the transition matrix.\n\n• method: This is only relevant for the estimator GPCCA. It determines the way in which we compute and sort the real Schur decomposition. The default, krylov, is an iterative procedure that works with sparse matrices which allows the method to scale to very large cell numbers. It relies on the libraries SLEPSc and PETSc, which you will have to install separately, see our installation instructions. If your dataset is small (<5k cells), and you don’t want to install these at the moment, use method='brandts'. The results will be the same, the difference is that brandts works with dense matrices and won’t scale to very large cells numbers.\n\n• weight_connectivities: additionally to the velocity-based transition probabilities, we use a transition matrix computed on the basis of transcriptomic similarities to make the algorithm more robust. Essentially, we are taking a weighted mean of these two sources of inforamtion, where the weight for transcriptomic similarities is defined by weight_connectivities.\n\nWhen running the above command, CellRank adds a key terminal_states to adata.obs and the result can be plotted as:\n\n:\n\ncr.pl.terminal_states(adata)", null, "## Identify initial states¶\n\nThe same sort of analysis can now be repeated for the initial states, only that we use the function cr.tl.initial_states this time:\n\n:\n\ncr.tl.initial_states(adata, cluster_key='clusters')\n\nComputing transition matrix based on logits using 'deterministic' mode\nEstimating softmax_scale using 'deterministic' mode\nSetting softmax_scale=3.7951\nFinish (0:00:08)\nUsing a connectivity kernel with weight 0.2\nComputing transition matrix based on connectivities\nFinish (0:00:00)\nComputing eigendecomposition of the transition matrix\nAdding .eigendecomposition\nadata.uns['eig_bwd']\nFinish (0:00:00)\nWARNING: For 1 macrostate, stationary distribution is computed\nAdding .macrostates_memberships\n.macrostates\nFinish (0:00:00)\nAdding adata.obs['initial_states_probs']\nadata.obs['initial_states']\nadata.obsm['macrostates_bwd']\n.terminal_states_probabilities\n.terminal_states", null, "We found one initial state, located in the Ngn3 low EP cluster.\n\n## Compute fate maps¶\n\nOnce we know the terminal states, we can compute associated fate maps - for each cell, we ask how likely is the cell to develop towards each of the identified terminal states.\n\n:\n\ncr.tl.lineages(adata)\n\nComputing lineage probabilities towards terminal states\nComputing absorption probabilities\nAdding adata.obsm['to_terminal_states']\nadata.obs['to_terminal_states_dp']\n.absorption_probabilities\n.diff_potential\nFinish (0:00:00)\nAdding lineages to adata.obsm['to_terminal_states']\nFinish (0:00:00)", null, "We can aggregate the above into a single, global fate map where we associate each terminal state with color and use the intensity of that color to show the fate of each individual cell:\n\n:\n\ncr.pl.lineages(adata, same_plot=True)", null, "This shows that the dominant terminal state at E15.5 is Beta, consistent with known biology, see e.g. Bastidas-Ponce et al. (2018).\n\n## Directed PAGA¶\n\nWe can further aggragate the individual fate maps into a cluster-level fate map using an adapted version of PAGA with directed edges. We first compute scVelo’s latent time with CellRank identified root_key and end_key, which are the probabilities of being an initial or a terminal state, respectively.\n\n:\n\nscv.tl.recover_latent_time(adata, root_key='initial_states_probs', end_key='terminal_states_probs')\n\ncomputing latent time using initial_states_probs, terminal_states_probs as prior\n\n\nNext, we can use the inferred pseudotime along with the initial and terminal states probabilities to compute the directed PAGA.\n\n:\n\nscv.tl.paga(adata, groups='clusters', root_key='initial_states_probs', end_key='terminal_states_probs',\nuse_time_prior='velocity_pseudotime')\n\nrunning PAGA using priors: ['velocity_pseudotime', 'initial_states_probs', 'terminal_states_probs']\n\n:\n\ncr.pl.cluster_fates(adata, mode=\"paga_pie\", cluster_key=\"clusters\", basis='umap',\nlegend_kwargs={'loc': 'top right out'}, legend_loc='top left out',\nnode_size_scale=5, edge_width_scale=1, max_edge_width=4, title='directed PAGA')", null, "We use pie charts to show cell fates averaged per cluster. Edges between clusters are given by transcriptomic similarity between the clusters, just as in normal PAGA.\n\nGiven the fate maps/probabilistic trajectories, we can ask interesting questions like:\n\nTo find out more, check out the CellRank API.\n\n## Compute lineage drivers¶\n\nWe can compute the driver genes for all or just the subset of lineages. We can also restric this to some subset of clusters by specifying clusters=... (not shown below).\n\nIn the resulting dataframe, we also see the p-value, the corrected p-value (q-value) and the 95% confidence interval for the correlation statistic.\n\n:\n\ncr.tl.lineage_drivers(adata)\n\nComputing correlations for lineages ['Epsilon' 'Alpha' 'Beta'] restricted to clusters None in layer X with use_raw=False\nAdding .lineage_drivers\nadata.var['to Epsilon corr']\nadata.var['to Alpha corr']\nadata.var['to Beta corr']\nFinish (0:00:00)\n\n:\n\nEpsilon corr Epsilon pval Epsilon qval Epsilon ci low Epsilon ci high Alpha corr Alpha pval Alpha qval Alpha ci low Alpha ci high Beta corr Beta pval Beta qval Beta ci low Beta ci high\nindex\nGhrl 0.802445 0.000000e+00 0.000000e+00 0.788123 0.815898 -0.102534 2.297290e-07 1.708022e-06 -0.140933 -0.063827 -0.409512 4.725654e-106 6.750934e-104 -0.441431 -0.376558\nAnpep 0.456914 7.357066e-136 7.357066e-133 0.425527 0.487202 -0.063882 1.298457e-03 4.477439e-03 -0.102589 -0.024982 -0.228949 1.017801e-31 2.867046e-30 -0.265542 -0.191697\nGm11837 0.449616 6.364047e-131 4.242698e-128 0.417977 0.480167 -0.045986 2.068035e-02 4.854542e-02 -0.084796 -0.007037 -0.238269 2.593631e-34 7.980404e-33 -0.274681 -0.201175\nIrx2 0.399584 1.901441e-100 9.507205e-98 0.366325 0.431823 0.517187 3.359556e-182 3.359556e-179 0.488060 0.545163 -0.640866 0.000000e+00 0.000000e+00 -0.663266 -0.617318\nCcnd2 0.384303 3.214070e-92 1.285628e-89 0.350590 0.417020 0.152544 1.074181e-14 1.732551e-13 0.114262 0.190375 -0.351178 6.074690e-76 4.672839e-74 -0.384874 -0.316547\n... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...\nDlk1 -0.323106 1.064602e-63 1.252473e-61 -0.357567 -0.287767 -0.168021 1.477981e-17 2.869865e-16 -0.205637 -0.129910 0.325836 7.865248e-65 4.494427e-63 0.290563 0.360224\nGng12 -0.342606 4.647161e-72 7.149478e-70 -0.376541 -0.307752 -0.330585 7.894530e-67 9.868163e-65 -0.364847 -0.295428 0.462704 7.128450e-140 2.376150e-137 0.431522 0.492782\nNkx6-1 -0.349302 4.411242e-75 7.352071e-73 -0.383051 -0.314622 -0.318417 8.753797e-62 8.753797e-60 -0.352999 -0.282965 0.457423 3.288706e-136 9.396304e-134 0.426055 0.487693\nNnat -0.357368 7.895266e-79 1.579053e-76 -0.390887 -0.322902 -0.324286 3.462537e-64 4.073572e-62 -0.358716 -0.288976 0.466845 8.497522e-143 3.399009e-140 0.435810 0.496771\nPdx1 -0.370468 3.605169e-85 8.011487e-83 -0.403604 -0.336361 -0.332613 1.076943e-67 1.435924e-65 -0.366821 -0.297507 0.481221 2.648652e-153 1.765768e-150 0.450709 0.510609\n\n2000 rows × 15 columns\n\nAfterwards, we can plot the top 5 driver genes (based on the correlation), e.g. for the Alpha lineage:\n\n:\n\ncr.pl.lineage_drivers(adata, lineage=\"Alpha\", n_genes=5)", null, "" ]	[ null, "https://cellrank.readthedocs.io/en/latest/_images/pancreas_basic_15_1.png", null, "https://cellrank.readthedocs.io/en/latest/_images/pancreas_basic_22_0.png", null, "https://cellrank.readthedocs.io/en/latest/_images/pancreas_basic_25_3.png", null, "https://cellrank.readthedocs.io/en/latest/_images/pancreas_basic_29_1.png", null, "https://cellrank.readthedocs.io/en/latest/_images/pancreas_basic_31_0.png", null, "https://cellrank.readthedocs.io/en/latest/_images/pancreas_basic_38_0.png", null, "https://cellrank.readthedocs.io/en/latest/_images/pancreas_basic_44_0.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6925549,"math_prob":0.7475577,"size":17036,"snap":"2021-04-2021-17","text_gpt3_token_len":4856,"char_repetition_ratio":0.12576327,"word_repetition_ratio":0.053114437,"special_character_ratio":0.32102606,"punctuation_ratio":0.21275946,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9629222,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,2,null,2,null,1,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-26T18:24:27Z\",\"WARC-Record-ID\":\"<urn:uuid:57048424-377c-4568-9036-566b316a3644>\",\"Content-Length\":\"134503\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1fd1596d-043a-42c0-b919-270a9c729f12>\",\"WARC-Concurrent-To\":\"<urn:uuid:2de858aa-f340-4ead-ada5-53e91014e851>\",\"WARC-IP-Address\":\"104.17.32.82\",\"WARC-Target-URI\":\"https://cellrank.readthedocs.io/en/latest/pancreas_basic.html\",\"WARC-Payload-Digest\":\"sha1:4RJIZTGNFJ42GIYWRSKBFHWYFWHOJRTA\",\"WARC-Block-Digest\":\"sha1:HGDQU72VCT3HC4GIUJ7R3ZIFQXKQ3PMJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610704803308.89_warc_CC-MAIN-20210126170854-20210126200854-00796.warc.gz\"}"}
https://number.academy/132036	[ "# Number 132036\n\nNumber 132,036 spell 🔊, write in words: one hundred and thirty-two thousand and thirty-six . Ordinal number 132036th is said 🔊 and write: one hundred and thirty-two thousand and thirty-sixth. Color #132036. The meaning of number 132036 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 132036. What is 132036 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 132036.\n\n## What is 132,036 in other units\n\nThe decimal (Arabic) number 132036 converted to a Roman number is (C)(X)(X)(X)MMXXXVI. Roman and decimal number conversions.\n\n#### Weight conversion\n\n132036 kilograms (kg) = 291086.6 pounds (lbs)\n132036 pounds (lbs) = 59891.1 kilograms (kg)\n\n#### Length conversion\n\n132036 kilometers (km) equals to 82044 miles (mi).\n132036 miles (mi) equals to 212492 kilometers (km).\n132036 meters (m) equals to 433184 feet (ft).\n132036 feet (ft) equals 40246 meters (m).\n132036 centimeters (cm) equals to 51982.7 inches (in).\n132036 inches (in) equals to 335371.4 centimeters (cm).\n\n#### Temperature conversion\n\n132036° Fahrenheit (°F) equals to 73335.6° Celsius (°C)\n132036° Celsius (°C) equals to 237696.8° Fahrenheit (°F)\n\n#### Time conversion\n\n(hours, minutes, seconds, days, weeks)\n132036 seconds equals to 1 day, 12 hours, 40 minutes, 36 seconds\n132036 minutes equals to 3 months, 1 week, 16 hours, 36 minutes\n\n### Codes and images of the number 132036\n\nNumber 132036 morse code: .---- ...-- ..--- ----- ...-- -....\nSign language for number 132036:", null, "", null, "", null, "", null, "", null, "", null, "Number 132036 in braille:", null, "QR code Bar code, type 39", null, "", null, "Images of the number Image (1) of the number Image (2) of the number", null, "", null, "More images, other sizes, codes and colors ...\n\n## Share in social networks", null, "## Mathematics of no. 132036\n\n### Multiplications\n\n#### Multiplication table of 132036\n\n132036 multiplied by two equals 264072 (132036 x 2 = 264072).\n132036 multiplied by three equals 396108 (132036 x 3 = 396108).\n132036 multiplied by four equals 528144 (132036 x 4 = 528144).\n132036 multiplied by five equals 660180 (132036 x 5 = 660180).\n132036 multiplied by six equals 792216 (132036 x 6 = 792216).\n132036 multiplied by seven equals 924252 (132036 x 7 = 924252).\n132036 multiplied by eight equals 1056288 (132036 x 8 = 1056288).\n132036 multiplied by nine equals 1188324 (132036 x 9 = 1188324).\nshow multiplications by 6, 7, 8, 9 ...\n\n### Fractions: decimal fraction and common fraction\n\n#### Fraction table of 132036\n\nHalf of 132036 is 66018 (132036 / 2 = 66018).\nOne third of 132036 is 44012 (132036 / 3 = 44012).\nOne quarter of 132036 is 33009 (132036 / 4 = 33009).\nOne fifth of 132036 is 26407,2 (132036 / 5 = 26407,2 = 26407 1/5).\nOne sixth of 132036 is 22006 (132036 / 6 = 22006).\nOne seventh of 132036 is 18862,2857 (132036 / 7 = 18862,2857 = 18862 2/7).\nOne eighth of 132036 is 16504,5 (132036 / 8 = 16504,5 = 16504 1/2).\nOne ninth of 132036 is 14670,6667 (132036 / 9 = 14670,6667 = 14670 2/3).\nshow fractions by 6, 7, 8, 9 ...\n\n### Calculator\n\n 132036\n\n#### Is Prime?\n\nThe number 132036 is not a prime number. The closest prime numbers are 132019, 132047.\n\n#### Factorization and factors (dividers)\n\nThe prime factors of 132036 are 2 * 2 * 3 * 11003\nThe factors of 132036 are 1 , 2 , 3 , 4 , 6 , 12 , 11003 , 22006 , 33009 , 44012 , 66018 , 132036\nTotal factors 12.\nSum of factors 308112 (176076).\n\n#### Powers\n\nThe second power of 1320362 is 17.433.505.296.\nThe third power of 1320363 is 2.301.850.305.262.656.\n\n#### Roots\n\nThe square root √132036 is 363,367582.\nThe cube root of 3132036 is 50,921062.\n\n#### Logarithms\n\nThe natural logarithm of No. ln 132036 = loge 132036 = 11,79083.\nThe logarithm to base 10 of No. log10 132036 = 5,120692.\nThe Napierian logarithm of No. log1/e 132036 = -11,79083.\n\n### Trigonometric functions\n\nThe cosine of 132036 is 0,413998.\nThe sine of 132036 is 0,910278.\nThe tangent of 132036 is 2,198751.\n\n### Properties of the number 132036\n\nIs a Friedman number: No\nIs a Fibonacci number: No\nIs a Bell number: No\nIs a palindromic number: No\nIs a pentagonal number: No\nIs a perfect number: No\n\n## Number 132036 in Computer Science\n\nCode typeCode value\nPIN 132036 It's recommendable to use 132036 as a password or PIN.\n132036 Number of bytes128.9KB\nCSS Color\n#132036 hexadecimal to red, green and blue (RGB) (19, 32, 54)\nUnix timeUnix time 132036 is equal to Friday Jan. 2, 1970, 12:40:36 p.m. GMT\nIPv4, IPv6Number 132036 internet address in dotted format v4 0.2.3.196, v6 ::2:3c4\n132036 Decimal = 100000001111000100 Binary\n132036 Decimal = 20201010020 Ternary\n132036 Decimal = 401704 Octal\n132036 Decimal = 203C4 Hexadecimal (0x203c4 hex)\n132036 BASE64MTMyMDM2\n132036 MD5604b3c7cafdff80cb186630886d13bb1\n132036 SHA178996f85e3ce8f3f962cdd5be09c8bf9dbb53fbd\n132036 SHA22412bdf339a08576297853773d909556f3bd706acb39371186d0003bed\n132036 SHA384ea72dd285ecef550d64b7843821fecbf56976908dce741d09983dedc113948863476d0ea33b7eb3206e67a545a8ba13e\nMore SHA codes related to the number 132036 ...\n\nIf you know something interesting about the 132036 number that you did not find on this page, do not hesitate to write us here.\n\n## Numerology 132036\n\n### Character frequency in number 132036\n\nCharacter (importance) frequency for numerology.\n Character: Frequency: 1 1 3 2 2 1 0 1 6 1\n\n### Classical numerology\n\nAccording to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 132036, the numbers 1+3+2+0+3+6 = 1+5 = 6 are added and the meaning of the number 6 is sought.\n\n## Interesting facts about the number 132036\n\n### Asteroids\n\n• (132036) 2002 CC125 is asteroid number 132036. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from White Sands Observatory in Socorro on 2/7/2002.\n\n## № 132,036 in other languages\n\nHow to say or write the number one hundred and thirty-two thousand and thirty-six in Spanish, German, French and other languages. The character used as the thousands separator.\n Spanish: 🔊 (número 132.036) ciento treinta y dos mil treinta y seis German: 🔊 (Anzahl 132.036) einhundertzweiunddreißigtausendsechsunddreißig French: 🔊 (nombre 132 036) cent trente-deux mille trente-six Portuguese: 🔊 (número 132 036) cento e trinta e dois mil e trinta e seis Chinese: 🔊 (数 132 036) 十三万二千零三十六 Arabian: 🔊 (عدد 132,036) مائة و اثنان و ثلاثون ألفاً و ستة و ثلاثون Czech: 🔊 (číslo 132 036) sto třicet dva tisíce třicet šest Korean: 🔊 (번호 132,036) 십삼만 이천삼십육 Danish: 🔊 (nummer 132 036) ethundrede og toogtredivetusindseksogtredive Dutch: 🔊 (nummer 132 036) honderdtweeëndertigduizendzesendertig Japanese: 🔊 (数 132,036) 十三万二千三十六 Indonesian: 🔊 (jumlah 132.036) seratus tiga puluh dua ribu tiga puluh enam Italian: 🔊 (numero 132 036) centotrentaduemilatrentasei Norwegian: 🔊 (nummer 132 036) en hundre og tretti-to tusen og tretti-seks Polish: 🔊 (liczba 132 036) sto trzydzieści dwa tysiące trzydzieści sześć Russian: 🔊 (номер 132 036) сто тридцать две тысячи тридцать шесть Turkish: 🔊 (numara 132,036) yüzotuzikibinotuzaltı Thai: 🔊 (จำนวน 132 036) หนึ่งแสนสามหมื่นสองพันสามสิบหก Ukrainian: 🔊 (номер 132 036) сто тридцять двi тисячi тридцять шiсть Vietnamese: 🔊 (con số 132.036) một trăm ba mươi hai nghìn lẻ ba mươi sáu Other languages ...\n\n## News to email\n\nPrivacy Policy.\n\n## Comment\n\nIf you know something interesting about the number 132036 or any natural number (positive integer) please write us here or on facebook." ]	[ null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-1.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-3.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-2.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-0.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-3.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-6.png", null, "https://number.academy/img/braille-132036.svg", null, "https://numero.wiki/img/codigo-qr-132036.png", null, "https://numero.wiki/img/codigo-barra-132036.png", null, "https://numero.wiki/img/a-132036.jpg", null, "https://numero.wiki/img/b-132036.jpg", null, "https://numero.wiki/s/share-desktop.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5324778,"math_prob":0.9660858,"size":7203,"snap":"2022-27-2022-33","text_gpt3_token_len":2612,"char_repetition_ratio":0.1579386,"word_repetition_ratio":0.0088573955,"special_character_ratio":0.43301404,"punctuation_ratio":0.16027088,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99168646,"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],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,1,null,1,null,1,null,1,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-18T09:51:46Z\",\"WARC-Record-ID\":\"<urn:uuid:bcd0ec60-c7cd-48bd-b1a0-d83832909ac3>\",\"Content-Length\":\"41520\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cca9acee-7498-4bb4-bf4b-085df5c4660e>\",\"WARC-Concurrent-To\":\"<urn:uuid:a2971986-12dc-48e2-b95e-4ff97b92bd64>\",\"WARC-IP-Address\":\"162.0.227.212\",\"WARC-Target-URI\":\"https://number.academy/132036\",\"WARC-Payload-Digest\":\"sha1:R52YHHJM7E7Y4TDDE54T5L7HJPTKJXL6\",\"WARC-Block-Digest\":\"sha1:VTYFU4O5T3R46RJKO33T4KPHPZHCAXEO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573193.35_warc_CC-MAIN-20220818094131-20220818124131-00219.warc.gz\"}"}
https://www.sharkbase.ca/shark-functions.html	[ "# LIST OF ALL SHARK FUNCTIONS:\n\nA function is just like an operation; the addition operation (+) requires two arguments (numbers) and returns their sum; a function requires some arguments (0 to 3) and returns some value.\n\nEvery function has a type: numeric, string, or logical, depending on the values it returns.\n\nA function has arguments; the values put into the function between the parentheses, separated by commas. Shark functions have at most three arguments; a few have none. Some arguments may be optional.\n\nThe arguments can be variables or expressions of type string: str var or str exp, or of type numeric: num var or <num exp>. (Expressions are discussed in Section 3.5)\n\n# Within the above list are some specific Mathematical Functions. The mathematical functions may be divided into five groups::\n\n#### Logarithmic functions:\n\nEXP( e to the power <num exp>\nLOG( natural logarithm of <num exp>\nLOG10( base 10 logarithm of <num exp>\nPOW( <num exp1> to the power <num exp2>\nSQRT( square root of <num exp>\n\n#### Trigonometric functions:\n\nSIN( sine of <num exp> in radians\nCOS( cosine of <num exp> in radians\nTAN( tangent of <num exp> in radians\nASIN( arc sine of <num exp>; returns a value in radians between -𝜋/2 and 𝜋/2\nACOS( arc cosine of <num exp>; returns a value in radians between 0 and 𝜋\nATAN( arc tangent of <num exp>; returns a value in radians between -𝜋/2 and 𝜋/2\n\n#### Hyperbolic functions:\n\nSINH( hyperbolic sine of <num exp>\nCOSH( hyperbolic cosine of <num exp>\nTANH( hyperbolic tangent of <num exp>\n\n#### Integer-valued functions:\n\nCEIL( ceiling integer: the integer equal to or just above a given number\nFLOOR( floor integer: the integer equal to or just below a given number\nINT( the integer part of a given number (the fractional part is discarded); useful for currency \"rounding\", for example.\n\nNote that for positive numbers, INT( and FLOOR( return the same result, but for negative numbers, INT( and CEIL( produce the same result. This is because discarding the decimal part of a real number reduces its distance from zero.\n\n#### Real-valued functions:\n\nABS( displays the absolute value of <num exp>\nMOD( evaluates remainder of a division of a <num exp> by a different <num exp>\n\n#### Examples:\n\n```\n1>:PICTURE='999.999999'\n1>? EXP(1)\n2.718282 ;this is the value of e\n1>? LOG(3)\n1.098612\n1>? LOG10(3)\n0.477121\n1>? POW(2,4)\n16.000000\n1>? POW(2,.5)\n1.414214\n1>? SQRT(4)\n2.000000\n1>? SQRT(2)\n1.414214\n1>? SIN(2) ;2 is in radians\n0.909297\n1>? ASIN(1)\n1.570796\n1>? 2ASIN(1)\n3.141593 ; this is, of course, 𝜋; store this to a variable if you need 𝜋\n1>? CEIL(3.14)\n4.000000\n1>? FLOOR(3.14)\n3.000000\n1>? INT(23.45)\n23.000000\n1>? INT(-23.45)\n-23.000000\n1>x=23.45\n1>? INT(10x)\n234.000000\n1>x=23.999\n1>? INT(x)\n23.000000\n1>? INT(3.14)\n3.000000\n1>? FLOOR(-3.14)\n-4.000000\n1>? INT(-3.14) ;note that for negative numbers, INT( and FLOOR( give different values\n-3.000000\n1>? MOD(5,2) ;this is the remainder of 5 divided by 2\n1.000000\n1>? MOD(-3.14, .7)\n-0.340000\n1>? ABS(-3.14)\n3.140000\n1>:PICTURE='9999999.99'\n1>x=15.689\n1>? INT(x100+.5)/100 Note: this is an example of rounding to the nearest 1 cent\n15.69\n1>x=15.244\n1>? INT(x100+.5)/100\n15.24\n```\n\n# Within the above list of all functions are some specific Comparative Functions:\n\n```MASK(: Manipulates bits in a string based on comparison between 2 strings\nMAX(: Compare two expressions of any type and return the larger.\n```\n\n# ALPHABETICAL LIST OF ALL SHARK FUNCTIONS:\n\n## !(\n\nConverts a string to upper case.\n\n```!(<str exp>)\n```\n\n<str exp> is the text to be converted to upper case\n\nType: character\n\nAll lower-case letters in the <str exp> are converted into upper case by the !( function. See also the LOWER( function.\n\nExamples:\n\n```1>a='Aa12b'\n1>? !(a)\nAA12B\n1>? !('David!')\nDAVID!\n```\n\nNote that only the lower-case letters are changed.\n\n## #\n\nGets the current record number.\n\n`#`\n\nType: numeric\n\nThis function returns the record number of the current record of the selected file. Note that ? # displays the current record number in the form specified by the system variable :PICTURE (see Section 2.7). Shark also has a more general form of this function, RECNO(, which allows the user to specify a file other than the selected file.\n\nWhen used with the option RECNO(filenum), it gives the record number of the current record in file filenum.\n\nExamples:\n\n```1>USE employee\n1>? #\n1.00\n1>GO BOTTOM\n1>? #\n6.00\n1>GO TOP\n1>? #\n1.00\n1>SKIP 2\n1>? #\n3.00\n1>? recno(2); selects current record in file #2\n```\n\n## \\$( (substring)\n\nThis is shorthand for the SUBSTR( function.\n\n## ASTERISK ()\n\nDetermines whether a record is deleted. Type: logical\n\nIn the selected file, the current record pointer points at a record. If this record has been marked for deletion (in BROWSE or EDIT, or with the DELETE command), then gives the value T; otherwise, it is false.\n\nExample:\n\n```1>USE employee\n1>DELETE RECORD 2\n1 DELETE(S)\n1>GO 2\n1>? \nT\n```\n\nDo not confuse this function with command (NOTE), or with the * comment marker.\n\n## @( (at)\n\nGets the location of a substring.\n\n```@(find str exp, <str exp> [,start)\n\nfind str exp the string searched for\n<str exp> the text to be searched\n```\n\nOption:\n\nstart byte position at which to start looking for find str exp within the string: <str exp>. If it occurs, the function returns the character position of the first (left-most) substring of <str exp> which is the same as find str exp; if it does not occur, the function returns a 0.\n\n@( and its equivalent, the AT( function, can easily search for successive occurrences of a substring within a string.\n\nConsider the previous paragraph. How many occurences are there of the substring \"string\"? How many of \"s\"? This function can determine the answer as follows if the paragraph is stored as a single string in the variable STR:\n\nSTR='@( and its equivalent, the AT( function, can easily search for successive occurrences of a substring within a string.\n\n```SUBSTR='s'\nCOUNT=0\nSTART=0\nLEN_STR=LEN(STR)\nDO WHILE START<LEN_STR\nSTART=@(SUBSTR,STR,START+1)\nIF START>0\nCOUNT=COUNT+1\nELSE\nBREAK\nENDIF\nENDDO\n```\n\nBe sure that start always advances; if it stays in the same place, it will perform the same operation indefinitely since it will never get to the end.\n\nAmbitious programmers will have noted that, if substr is longer than one byte, start should begin at 1-LEN(SUBSTR) and advance each time by the length. Furthermore, the condition on the DO WHILE line should then be:\n\n```start<=len_str-len_sub\n```\n\n(assuming you've saved the length of the substring in this last variable).\n\nExamples:\n\n```1>greeting='Good morning'\n1>? @('oo', greeting)\n2.00\n1>? @('good',greeting)\n0.00\n```\n\nIn a program:\n\n```IF @(answer,'YNQynq')>0\n```\n\nchecks whether the user response is correct.\n\nThis program segment finds and marks every occurrence of a specified string in a page of text displayed on the screen:\n\n```CLS ;start by erasing the screen\nDO WHILE t ;set up continuous loop -break embedded in segment\nIF row()<mLineNumber ;stop at end of screen\nIF .not. read(mPrintLine,1) ;be sure there's more to display\nBREAK ;if not, break out of loop\nENDIF\n?? mPrintLine ;display the line, no linefeed\nIF mText>' ' ;if looking for input text\nmPrintLine=!(mPrintLine) ;capitalize line from file\nmSearchPos=1 ;start search at 1\nDO WHILE t\nmSearchPos=@(mText,mPrintLine,mSearchPos)\nIF mSearchPos=0 ;no more to find, so...\nBREAK ; get out\nELSE\n* we found something...highlight it and increment search-start position\nCOLOR colorFind,row(),mSearchPos-1,row(),mSearchPos+mTextLen-2\nmSearchPos=mSearchPos+mTextLen\nENDIF\nENDDO\nENDIF\n? ;emit carriage return/linefeed pair\nELSE\nBREAK ;stop when screen is full\nENDIF\nENDDO\n\n```\n\n## ABS(\n\ndisplays the absolute value of a <num exp>\n\nExample:\n\n```1>? abs(\n1>? @('oo', greeting)\n2.00\n1>? @('good',greeting)\n0.00\n```\n\n## ACOS(\n\narc cosine of a <num exp> returns a value in radians between 0 and π\n\n## ASC(\n\nConverts a character to its ASCII number.\n\n```ASC(<str exp>)\n<str exp> the first character of this string is converted\n\nType: numeric\n```\n\nThe characters in the character set used by the computer are numbered from 0 to 255. For the first character of the string <str exp>, ASC( returns the corresponding number. RANK( is a synonym for ASC(. See also the functions CHR(, CTONUM(, and NUMTOC(.\n\nExamples:\n\n```1>? ASC('x')\n120.00\n1>? ASC('xyz')\n120.00\n```\n\nNote that only the first character of the string matters.\n\n## ASIN(\n\narc sine of <num exp> returns a value in radians between -π/2 and π/2\n\n## ATAN(\n\narc tangent of <num exp> returns a value in radians between -π/2 and π/2\n\n## AT(\n\nGets the location of a substring . . . a synonym for @(\n\n```AT(find str exp, <str exp>, [start]\n```\n\nfind str exp the string searched for <str exp> the text to be searched\n\nOption:\n\n```[start] byte position at which to start looking for\n```\n\nType: numeric\n\nThis function finds out whether a string: find str exp occurs in the string: <str exp>. If it occurs, the function returns the character position of the first (left-most) substring of <str exp> which is the same as find str exp; if it does not occur, the function returns a 0.\n\n@( and its equivalent, the AT( function, can easily search for successive occurences of a substring within a string.\n\nConsider the previous paragraph. How many occurrences are there of the suBstring \"string\"? How many of \"s\"? This function can determine the answer as follows if the paragraph is stored as a single string in the variable STR:\n\nExample:\n\n```SUBSTR='s'\nCOUNT=0\nSTART=0\nLEN_STR=len(str)\nDO WHILE start<len_str\nstart=@(substr,str,start+1)\nIF start>0\ncount=count+1\nELSE\nBREAK\nENDIF\nENDDO\n```\n\nBe sure that start always advances; if it stays in the same place, it will perform the same operation indefinitely since it will never get to the end.\n\nAmbitious programmers will have noted that, if SUBSTR is longer than one byte, START should begin at 1-LEN(SUBSTR) and advance each time by the length. Furthermore, the condition on the DO WHILE line should then be\n\n```START<=LEN_STR-LEN_SUB\n```\n\n(assuming you've saved the length of the substring in this last variable).\n\nExamples:\n\n```1>greeting='Good morning'\n1>? @('oo', greeting)\n2.00\n1>? @('good',greeting)\n0.00\n```\n\nIn a program:\n\n```IF @(answer,'YNQynq')>0\n```\n\nchecks whether the user response is correct.\n\nThis program segment finds and marks every occurrence of a specified string in a page of text displayed on the screen:\n\n```CLS ;start by erasing the screen\nDO WHILE t ;break embedded in segment\nIF row()<mLineNumber ;stop at end of screen\nIF .not. read(mPrintLine,1) ;be sure there's more to display\nBREAK ;if not, get out\nENDIF\n?? mPrintLine ;display the line, no linefeed\nIF mText>' ' ;if looking for input text\nmPrintLine=!(mPrintLine) ;capitalize line from file\nmSearchPos=1 ;start search at 1\nDO WHILE t\nmSearchPos=@(mText,mPrintLine,mSearchPos)\nIF mSearchPos=0 ;no more to find, so...\nBREAK ; get out\nELSE\n* we found something...highlight it and increment search-start position\nCOLOR colorFind,row(),mSearchPos-1,row(),\nmSearchPos+mTextLen-2\nmSearchPos=mSearchPos+mTextLen\nENDIF\nENDDO\nENDIF\n? ;emit carriage return/linefeed pair\nELSE\nBREAK ;sop when screen is full\nENDIF\nENDDO\n```\n\n## BIT(\n\nBit-set function determines if a given bit is 0 or 1\n\n```BIT(string,bit position)\n```\n\nstring a string or string variable to test bit position a numeric expression; position of a given bit within string\n\nType: logical\n\nEach character in the ASCII character set is identified by an eight-bit binary number from 0 to 255 inclusive. Each bit may be either 0 or 1; for example, the letter A has a decimal value of 65 and a binary value of 01000001.\n\nWhen used with the SET( and RESET( functions, which turn specific bits to 1 or 0 respectively, the BIT( function can be used to access large amounts of logical data much more compactly than in a set of logical variables. BIT( returns T (true) if the specified bit is set (0), F (false) if not set (0).\n\nNOTE: Bit positions are counted differently than in some other schemes. In these functions, all bits are counted from the left of the string starting at 1, so that each character contains bits numbered as follows: 1. Bits 1 to 8. 2. Bits 9 to 16. 3. Bits 17 to 24. . . . and so on\n\nExample in a program:\n\nTo print the binary value of each character in an input string:\n\n```SET RAW ON ;eliminates spaces between listed output\nDO WHILE t\nACCEPT 'Enter a short string or binary representation: ' TO string\nIF string=' '\nBREAK\nENDIF\n?\nREPEAT LEN(string)8 TIMES VARYING position\n?? IFF(BIT(string,position),'1','0')\nIF MOD(position,8)=0\n?? ' '\nENDIF\nENDREPEAT\nENDDO\n```\n\nNow run the program:\n\n```Enter a short string for binary representation: Bit <--- enter text \"Bit\"\n01000010 01101001 01110100 <--- result displays on screen\nEnter a short string for binary representation: <--- prompt reappears\n```\n\n## BLANK(\n\nCreates a string of blanks or other specified characters.\n\n`BLANK(<num exp>[,charnum)`\n\n<num exp> a number from 0 to 255\n\nOption:\n\n```charnum the ASCII number of the character used to fill the\nblank string; default is 32, the blank character\n```\n\nType: character\n\nThis function creates a string of <num exp> blanks or other specified characters.\n\nWhen charnum is specified, many interesting effects can be created, particularly by using the special pattern characters in the IBM screen character set, 176-178, and the solid block character, 219.\n\nExamples:\n\n```1>name='DAVID'\n1>? name+BLANK(15)+name\nDAVID DAVID\n1>num=23\n1>? name+BLANK(num+5)+name\nDAVID DAVID\n1>? BLANK(20,65)\nAAAAAAAAAAAAAAAAAAAA\n1>? BLANK(20,196)\n--------------------\n```\n\n## BOF\n\nGives the beginning-of-file flag for the currently selected data file.\n\n`BOF`\n\nType: logical\n\nIf the current record pointer is on the first record of the file in use and a SKIP -1 is issued, BOF returns T (true); otherwise it is F (false). Since SKIP -n is treated as n SKIP -1 commands, BOF returns true if SKIP -n goes past the last record.\n\nTo check an open data file other than the currently selected one, use the BOF( function.\n\nExamples:\n\n```1>USE employee\n1>GO 3\n1>SKIP -2\n1>? #\n1.00\n1>? BOF\nF\n1>SKIP -1\n1>? BOF\nT\n1>GO 4\n1>SKIP -5\n1>? #\n1.00\n1>? BOF\nT\n```\n\n## BOF(\n\nGives the beginning-of-file flag for a specified data file.\n\n`BOF(<filenum>)`\n\nOption:\n\n`<filenum> the number of the data file to be checked`\n\nType: logical\n\nFor the data file number specified, if the current record pointer is on the first record and a SKIP -1 is issued, BOF( returns T (true); otherwise it is F (false). Since SKIP -n is treated as n SKIP -1 commands, BOF( returns true if SKIP n goes past the first record.\n\nAn extended form of the BOF function which, since it takes no parameter, works only on the currently selected data file. If no <filenum> is specified, the current file is assumed.\n\nExamples:\n\n```1>USE employee\n1>GO 3\n1>SKIP -2\n1>? #\n1.00\n1>? BOF()\nF\n1>SELECT 2\n2>USE inventry\n2>GO TOP\n2>? #\n1.00\n2>? BOF()\nF\n2>SKIP -1\n2>? #\n1.00\n2>? BOF()\nT\n2>SELECT 1\n1>? BOF(2)\nT\n```\n\n## CEIL(\n\nceiling integer: the integer equal to or just above a given number\n\nRelated functions: CEIL( = the integer equal to or just above a given number; FLOOR( = the integer equal to or just below a given number; INT( = the integer part of a mixed number after discarding the fractional part; MOD( = the balance after truncating a mixed number.\n\n## CEN(\n\nCenters a line of text.\n\n```CEN(<str exp>,<num exp>)\n\n<str exp> the text to be centered\n<num exp> the line width```\n\nType: character\n\nThis function centers (from the present position) the text <str exp> in a line (column) with <num exp> characters.\n\nExamples:\n\n```1>compiler='Shark'\n1>? CEN(compiler,40)\nShark\n1>@ 10,20 SAY CEN('Center this',40)```\n\nNote: the last command centers the text between columns 20 and 60.\n\n## CHR(\n\nConverts an ASCII number to character.\n\n(To convert a numeric value to a string, use STR()\n\n```CHR(<num exp>)\n\n<num exp> a number from 0 to 255```\n\nType: character\n\nThe characters in the character set used by the computer are numbered from 0 to 255; this number for a character is called the ASCII number. For a given <num exp> in this range, CHR(<num exp>) is the corresponding character.\n\nThis function is useful to send control codes to the printer. For instance,\n\n`1>? CHR(27)+CHR(120)+CHR(1)`\n\nputs the Epson LQ-1500 printer into letter quality mode.\n\nThe functions ASC( and RANK( do the reverse. These functions combine nicely. If the memory variable LETTER contains a letter of the alphabet (other than z or Z), then LETTER=CHR(ASC(LETTER)+1) places in LETTER the next letter of the alphabet.\n\nExamples:\n\n1. To send unprintable or other characters to a printer\n\n`1>? CHR(27)`\n\nsends ESC to the printer\n\n```1>letter='C'\n1>? CHR(RANK(letter)+1)```\n\nsends \"D\" to the printer\n\n2. To set a standard IBM or Epson printer into double-wide mode:\n\n`1>SET PRINT ON`\n`1>? CHR(14)+'First line.'`\n\nprints:\n\n`First line. `\n\nNOTE: the character '0' is difficult to send to a printer since it represents the end-of-line instruction. A better method of sending characters is the\n\n`PSTR`\n\ncommand, which permits sending any character to the printer. Example:\n\n`PSTR 27,W,1Bh,41h,3h`\n\n## CLOSE(\n\nCloses a file.\n\n`CLOSE([filenum])`\n\nOption:\n\n`<filenum> the DOS file number (between 1 and 4)`\n\nType: logical\n\nThis function closes the DOS file (in particular, the sequential file) opened with the ROPEN( or WOPEN( function. It returns T if successful, F otherwise. See the functions ROPEN(, WOPEN(, SEEK(, SSEEK(, READ(, WRITE(, GET(, PUT(, IN(, OUT(, and CLOSE).\n\nIf filenum is not specified, filenum=1 is the default.\n\nExample:\n\n```1>ok=ROPEN('a:label.prg',3)\n1>? ok\nT\n1>ok=CLOSE(3)\n```\n\n## COL(\n\nGets print column position.\n\n`COL()`\n\nType: numeric\n\nThis function gives the current column position of the cursor; if the printer is on, it returns the column position of the printer head. See the commands SET PRINT ON and SET FORMAT TO PRINT, and the function ROW(.\n\nExample:\n\n`@ ROW(),COL()+3 SAY 'Hello'`\n\nprints 'Hello' starting three characters to the right of the end of the last printing.\n\n## COS(\n\ncosine of <num exp> in radians\n\n## COSH(\n\nhyperbolic cosine of <num exp>\n\n## CTONUM(\n\nConvert a hexadecimal string into a decimal number.\n\n```CTONUM(type,string exp)\n\ntype the length of the numeric value to be returned\nstring exp the string to be evaluated as a hexadecimal value```\n\nType: numeric\n\nA general conversion function for converting hexadecimal values into decimal numbers. Input can be any length string or string variable up to eight characters as follows:\n\n``` Type String Length Returns\n1 1 byte integer 0 to 255\n2 2 bytes integer -32768 to 32767\n4 4 bytes integer +/- 2 billion\n8 8 bytes a floating point number```\n\nIf string is shorter, conversion still assumes the string is the format of the given width. When type is 1, this function is equivalent to RANK( or ASC(.\n\nThe NUMTOC( and CHR( functions convert numbers into strings.\n\nDo not confuse these function with STR( and VAL(, which convert decimal numbers into their string representations, and vice versa.\n\nExamples:\n\n```1>? CTONUM(1,'a')\n97.00\n1>? CTONUM(2,'ab')\n25185.00\n1>? CTONUM(4,'abc')\n6513249.00\n1>? CTONUM(4,'abcd'); number too large for format in :PICTURE\n```\n\n## DATE(\n\nTo display a date in a specific format, or to update :DATE with the computer's system date.\n\n`DATE([type],<str exp>)`\n\ntype being one of nine type basic date-output formats:\n\n`DATE(1)`\n`DATE(2)`\n`DATE(3), etc`\n\nOption: <str exp> the date to be converted\n\nType: character\n\nThis function has three distinctly different purposes and results:\n\n1. with only type specified in the range 1-9, rewrites the current Shark date in :DATE in the format specified by the type, and returns the result.\n\n2. with two parameters (the format in the range 1-9, and the date) returns the given date in the specified format. :DATE is not affected.\n\n3. with only type specified as zero, updates the Shark date with the computer's current system date, and the computer time with the current system time. The Shark and system dates are always the same when Shark is started, but the Shark does not automatically advance at midnight as the computer's system date should. When Shark can be in use overnight, or even for days at a time, it may be important to ensure that these dates are kept in synchronization. Date(0) returns a string of length zero.\n\nThe type can be given in either of two forms, a name or number (numeric expression) as follows:\n\n```Type Date-output format\n----------- -----------------------------------------------------------------\n0 a string of length zero representing the full current system date\ndisplayed as in Type 4 on this list\n1 or YMD 6-character format without slashes: yymmdd\n2 or MDY 8-character format with slashes: mm/dd/yy\n3 or Char Spelled out: Month dd, yyyy\n4 or Full Spelled out: Weekday, Month dd, yyyy\n5 or Lchar Last day of month spelled out in format 3 (Char)\n6 or DMY 11-byte string in format dd-MMM-yyyy (example 03-NOV-1990)\n7 or Variable formatted without slashes according to SET DATE TO command (See SET DATE TO)\n8 or Long 8-character format without slashes: yyyymmdd\n9 or Last Last day of month in format 1 (YMD) or 8 (Long), depending on whether SET DATE TO command sets year to 'YY' or 'YYYY'\n```\n\nShortcut: When specifying type by name, only the first character is usually required. The exception is for Lchar, Long, and Last, which require two characters to resolve ambiguity. If only one is given, Lchar is assumed.\n\nNote: No name equivalent is provided for type 0, which updates the Shark date from the computer BIOS calendar setting.\n\n```<str exp> must contain the date in one of the following formats:\n\nmmddyy ddmmyy yymmdd mmddyyyy ddmmyyyy yyyymmdd (<-- This last format is the ISO-8601 STANDARD format)\n```\n\nOptionally, a slash, a hyphen, or a space may be used to separate the elements of these formats. For example, YY/MM/DD, YY-MM-DD, DD MM YYYY are all equally valid.\n\nThere should be two digits each for month and the day, and two or four digits for the year. 01 3 92 is not acceptable. If <str exp> is not acceptable, then DATE( returns a string of blanks.\n\nIn the event of ambiguity, dates will be decoded in accordance with the format set in the SET DATE TO command. For example:\n\n```SET DATE TO date is interpreted as Comment\n------------- ----- --------------- --------\n'ddmyy' 11/03/99 March 11, 1999 Difficult to use for commerce\n'mmddyy' 11/03/99 November 3, 1999 \"\n'mmddyy' 11/03/60 November 3, 2060 \"\n'yyyymmdd' 20110419 2011 April 19 ISO-8601 STANDARD FORMAT\n```\n\nThis variety of date formats caters to many varied uses. However, the only reliable date format in business and science is the completely unambiguous ISO8601 'yyyymmdd' format shown above.\n\nSee also the system variable :DATE and the command SET DATE TO.\n\nExamples:\n\n```1>:DATE= '10/05/22'\n1>? :DATE\n10/05/22\n1>? DATE()\n20221005\n1>? :DATE\n20221005\n1>? DATE(1)\n221005\n1>? :DATE\n221005\n1>? DATE(2)\n10/05/22\n1>? :DATE\n10/05/22\n1>? DATE(3)\nOctober 5, 2022\n1>? :DATE\nOctober 5, 2022\n1>? DATE(4)\nSaturday, October 5, 2022\n1>? :DATE\nSaturday, October 5, 2022\n1>? DATE(4,'12/08/21')\nTuesday, August 21, 2012\n1>? :DATE\nSunday, December 8, 2021\n1>? DATE(5,'20-03-27')\nMarch 31, 2020\n1>? DATE(6,'21-03-27')\n27-Mar-2021\n1>? :DATE\nSaturday, October 5, 2022\n1>? :DATE(8)\n20110419\n```\n\nThe date can be recalled with any of the above date() functions, and styled to suit the user. For example, using the PIC( function, put the following in the SHARKNET.CNF file so any Shark program can access it at any time:\n\n```:DATE=DATE(8)\n:UNDOC=PIC(DATE(8),\"XXXX.XX.XX\")\n```\n\nThen the system variable :UNDOC when recalled will display:\n\n```1>? :UNDOC\n1>? 2011.04.19```\n\nThus, a preferred date format can be stored in Shark's .CNF file as a system variable. It's then visible to all applications in Shark:\n\n`:UNDOC=PIC(DATE(8),\"XXXX.XX.XX\")`\n\nNOTE: If the time has passed midnight during the current run of Shark, you can update the Shark date with DATE(0):\n\n```1>? DATE(3)\nSunday, October 5, 2019\n1>? DATE(0)\n1>? DATE(4)\nSunday, October 6, 2019```\n\nNOTE: Formatting the date display has no effect on date calculations.\n\nA NOTE FOR PROGRAMMERS & TECH TYPES:\n\nShark keeps all file dates in pre-Y2K MS-DOS format. Each date component before year 2000 is saved simply as the last 2 digits of each component. The 20th century is epoch \"00\". \"1998\" for example is stored as \"0098\". An old file from \"April 15,1998\" will show its file date as \"4-15-98\". It's actually saying \"04-15-0098\", but omitting the zeros. If you type \"DIR\" or \"DIRF(\" at the prompt, this is the date format you will see.\n\nAfter the epoch rolls over to the next millenium (the 21st century), the \"00\" of the 19th millenium will become \"01\" and Shark will display the DOS date not as \"0120\", but as \"120\", again dropping the leading zero. \"April 15,2000\" will thus display as \"4-15-120\".\n\nShark automates this formatting using the \"SET DATE TO\" instruction, so the display task is automated. However, the date display when listing files still remains in the older MS-DOS display format. Only rarely will the Shark programmer see this.\n\nWhen doing occasional file searches or backups, Shark programmers sometimes work with this display quirk by writing a simple formatting subroutine to always display the dates in a preferred display format when listing files:\n\n``` DISPDATE.PRG\n* A date display subroutine to sort out a file listing display:\n\nif VAL(\\$(dirx(5),7,3))<100\nmdatprefx='19'; set a 20th century date prefix\nMDAT=mdatprefx+\\$(DIRX(5),7,2)+\".\"+IFF(\\$(DIRX(5),1,1)=' ','0'+\\$(DIRX(5),2,1),\\$(DIRX(5),1,2))+\".\"+\\$(DIRX(5),4,2)\nelse\nmdatprefx='20'; set a 21st century date prefix\nMDAT=mdatprefx+\\$(DIRX(5),8,2)+\".\"+IFF(\\$(DIRX(5),1,1)=' ','0'+\\$(DIRX(5),2,1),\\$(DIRX(5),1,2))+\".\"+\\$(DIRX(5),4,2)\n\nendif```\n\nThus you can set your date display as you prefer. The dates in Shark are absolute and are unaffected by any display choices, so your files and dates will always be correct regardless of your display choices. The above subroutine will show dates in a program like this:\n\n```BEFORE:\nVPI.EXE 260326 4-08-91 10:56a\nVPI.MSG 15809 4-08-101 10:57a\nVPI.SGN 1972 4-08-121 10:58a\n\nAFTER displaying file dates with the above subroutine:\nVPI.EXE 260326 1991.04.08 10:56a\nVPI.MSG 15809 2001.04.08 10:57a\nVPI.SGN 1972 2021.04.08 10:58a```\n\nMost of the time, the Shark programmer will deal with preferred formats stored in a system variable such as :UNDOC, and will rarely change date formats once a preferred setup is implemented.\n\n## DAYS(\n\nComputes dates and date differences in days.\n\n```DAYS(<str exp1>,<str exp2>) Example: DAYS(2005-03-18,2005-06-27)\nDAYS(<str exp>,<num exp>) Example: DAYS(2005-03-18,30)\n```\n\nIn the first form: str exp1 and str exp2 are dates\n\nIn the second form: <str exp> is a date and <num exp> is a number\n\nType: numeric/character\n\nIn the first form, DAYS( returns the number of days between two dates. The result is an integer.\n\nIn the second form, DAYS( returns the date (as a string) which is <num exp> days past or before the date <str exp>.\n\nThe string expressions containing dates can be of many different formats (see the DATE( function for more examples):\n\n```\tyyyy/mm/dd 2005/03/18\nyyyy-mm-dd 2005-03-18\nyyyy mm dd 2005 03 18\nmm/dd/yy 03/18/05\nmm-dd-yy 03-18-05\nmm dd yy 03 18 05\n```\n\nThere should be two digits each for yy, mm, and dd, and four digits for yyyy. 01 3 90 is not acceptable (dd contains only 1 digit).\n\nIn the second form, the date is returned in the format set with the SET DATE TO command (default: mmddyyyy). If you wish a different format, use the DATE( function. See also MONTHS( and SET DATE TO.\n\nExamples:\n\n```1>? DAYS('04 06 90','04 29 90')\n23.00\n1>? DAYS('01/01/88','01 23 90')\n753.00\n1>? DAYS('01/01/90','01 23 88')\n-708.00\n1>? DAYS('01/01/91','01 02 91')\n1.00\n1>? DAYS('01/02/91','01 01 91')\n-1.00\n1>? DAYS('04 03 90',30)\n050290\n1>? DAYS('02 03 90',30)\n030590\n1>? DAYS('02 03 90',-3)\n010490\n1>? DAYS('020390',-30)\n010490\n1>monthday='0203'\n1>offset=30\n1>? DAYS(monthday+'90',offset+1)\n030690\n```\n\nNote that the MONTH( and DATE( functions always count the days in leap years:\n\n```\n1>? DAYS('02/28/88','03 01 88') ;leap year\n2.00\n1>? DAYS('02/28/90','03 01 90') ;not a leap year\n1.00\n```\n\nThus, DAYS(2005-03-18,-365) may not give the same result as MONTHS(2005-03-18,-12)\n\nDAYS( and DATE( may be combined to form complex expressions. For instance, the end of the month closest to today in the form set in the SET DATE TO command:\n\n`DATE(7,DAYS(DATE(2),-15))`\n\nThe end of NEXT month:\n\n`DATE(5,DAYS(DATE(2),30))`\n\nSee DATE( and MONTHS( functions, and SET DATE TO command.\n\n## DBF(\n\n``` DBF(type[,<filenum>)\n\ntype the information required from the data file header```\n\nOption:\n\n``` <filenum> the data file number; default is the currently\nselected data file```\n\nType: character/numeric\n\nEach data file has a file head which contains information about its structure, most of which is displayed with the LIST STRUCTURE command.\n\nUsing the DBF(, DBFX( and FLD( functions provides users access to this information in a programmable form suitable for display and use in expressions.\n\nThe type can be given in either of two forms, a name or number (numeric expression) as follows:\n\n```TYPE EXPLANATION RESULT\n1 or (T)ype file type (\"0\", \"1\", \"2\" or \"3\") string\n2 or (N)ame data file name string\n3 or (F)ields number of fields in structure integer\n4 or (R)ecords number of records in file integer\n5 or (I)ndexes number of indexes currently open integer\n6 or (M)aster index number of master index integer\n7 or (Q)ualified fully-qualified file name from\nSharkBase FILES structure if\none exists; otherwise from DOS string\n8 or (E)xpanded fully-qualified file name from DOS string```\n\nShortcut: When specifying type by name, only the first character is required.\n\nExamples:\n\n```1>? DBF(type),DBF(n),DBF(fields),DBF(recs),DBF(indexes),DBF(master)\n0 CUSTOMER.DBF 9.00 4.00 2.00 1.00\n1>? DBF(q)\nC:\\SHARK\\DBF\\CUSTOMER.DBF\n\n1>SET INDEX TO NAME\n1>APPE BLANK\n1>GO TOP <--- takes you to blank (NEWLY APPENDED) record of indexed table\n1>GO BOTT <--- takes you to last record before new record appended to table\n1>GOTO DBF(R) <--- takes you to the last (NEWLY APPENDED) record in table```\n\nSince appended record is blank, it won't show as the last (BOTTOM) record in an indexed table. DBF(R) will take you to the last record in the table.\n\n## DBFX(\n\nGives additional information about an open data file; an extension to the DBF( function.\n\n``` DBF(type[,<filenum>)\n\ntype one of three types of information as listed below```\n\nOption:\n\n``` <filenum> the data file number; default is the currently\nselected data file```\n\nType: logical\n\nThe DBF( and FLD( functions provide information contained in the data file headers.\n\nShark provides this extended function to give three additional types of information in a programmable form suitable for display and use in expressions.\n\nThe type can be given in either of two forms, a name or number (numeric expression) as follows:\n\n```TYPE EXPLANATION\n1 or Filter TRUE if a FILTER is in effect\n2 or Limit TRUE if a LIMIT is in effect\n3 or Relation TRUE if file is related to another\n4 or Write TRUE if current record has changes to be written to disk```\n\nShortcut: When specifying type by name, only the first character is required.\n\nIn a program, the programmer has control over whether filters, limits and relations are established, so the first three options are primarily helpful in Conversational SharkBase, in program debugging and in displaying the current environment for the aid of end-users.\n\nThe fourth option, however, is extremely helpful in programming, especially on networks, for it allows the programmer not only to check whether the user actually wants to save changes made to the record, but to avoid locking and changing records when the user was only looking.\n\nA second valuable use of this feature is to time-stamp changes and leave a record of who made them.\n\nExamples:\n\n```1>USE customer\n1>SET FILTER TO state='CA'\n1>? DBFX(Filter) or DBFX(1)\nT\n1>? DBFX(R) or DBFX(3)\nF```\n\nIn a program segment, starting with the READ command):\n\n```READ\nIF DBFX(w)\n@ 22,0 say CEN('Record changed. Save the Changes (Y/N)?',80)\nCURSOR 23,39\nIF !(CHR(INKEY()))='Y'\nREPLACE changedby WITH DATE(1)+user+TIME(1)\nFLUSH\nELSE\nELSE\nENDIF```\n\nThe same approach is used in programming for SHARE files on a network, but the actual technique is beyond the scope of this section due to the many different ways networking-data-indegrity is implemented by various operating systems. See specific network sections of your SharkBase documentation for more on this topic.\n\n## DELETED(\n\nDetermines whether a record is deleted.\n\n` DELETED(<filenum>)`\n\nOption:\n\n` <filenum> the number of the data file to be checked`\n\nType: logical\n\nIn a specified data file, the current record pointer points at a record. If this record has been marked for deletion (in BROWSE or EDIT, or with the DELETE command), then DELETED( gives the value T; otherwise, it is false.\n\nThis is a more general form of the function, which operates the same way as DELETED( but, because it allows no parameter, works only with the currently selected data file.\n\nExamples:\n```1>? DELETED(4)\nT\n```\n\n## DESCEND(\n\nThe main use of DESCEND( in Shark is to create indexes that are in reverse, or descending, order.\n\nRemember that an index arranges its keys in order of the value of the characters making up the keys. \"A\" comes before \"B\" because the value 65 is smaller than 66.\n\nIn C Language terms, Shark reverses all bits in string; equivalent to NOT of string. Shark uses this function to reverse the order of an indexed file.\n\n`DESCEND(string)`\n``` string the string to be converted; every \"1\" bit is returned\nas \"0\", and every \"0\" bit is returned as \"1\"```\n\nType: character\n\nEach character in the ASCII character set is identified by an eight-bit binary number from 0 to 255 inclusive. Each bit may be either 0 or 1; for example, the letter A has a decimal value of 65 and a binary value of 01000001, while the letter B has a decimal value of 66 and a binary value of 01000010.\n\nThe DESCEND( function returns a \"mirror image\" of the input string . . . i.e., one in which evey bit set to 1 is reset to 0, and every 0 is set to 1.\n\nThe result is familiar to programmers who use C and other low-level languages, who would use the NOT operator to obtain the same result.\n\nThe main use of DESCEND( in Shark is to create indexes that are in reverse, or descending, order.\n\nBut suppose the index expression, instead of being on the key expression itself is on the NOT value of it, as in:\n\n`INDEX ON DESCEND(NAME) TO REVERSE`\n\nNow you have an index called REVERSE.NDX in which B comes before A.\n\nThis function is especially useful in creating reports. For example, suppose you want to report transactions for your customers in reverse date order (latest transactions first). Your index command could be:\n\n`INDEX ON CUST+DESCEND(DATE) TO CUSTREPT`\n\nNote: if you create an index using the DESCEND( function, you will have to use the same function to modify any find strings if you need to use the index to FIND, SEEK, etc.\n\nExample in a report form:\n\n```TITLE - Transactions by Customer, with Dates in Descending Order\nFILE - trans\nINDEX - cust+descend(date) to custrept\nFILE - cust index cust1\nRELATION - custnum to 2\nFIELDS - date(6,date),desc,amount\nPICTURE - ,,9999999.99\nSUBTOTAL - custn\nMESSAGE - 'Customer: '+pic(custnum,'xxx-x-xx -- ')+name#2```\n\nNote: This report form includes a number of features worth noting, including the fact that it creates its own index, sets up a relation to a second file, includes underlined headings and subtotals, and includes one line split into two. In report forms, the comma (,) is a continuation character when it is the last character on a line. See REPORT.\n\n## DIR(\n\nGet file information from disk directory.\n\n``` DIR(filespec) (referred to as \"Form 1\")\nDIR() (referred to as \"Form 2\") ```\n\nOption:\n\n`filespec refers to a string or string expression containing a file name (or skeleton using ? and/or * wildcards), with optional drive and/or path specification (enclosed in single quotes)`\n\nType: character\n\nSearches for specific files specified by the drive path, subdirectory and/or filename and provides specific information on files found.\n\nIf an argument is given as shown in Form 1, it must be a string or string expression naming a file, with * and ? wildcards optional). Returns the first file name found matching filespec. If no match was found, blank is returned.\n\nIf no argument is given as shown in Form 2, the previous filespec is used to find the next matching file. If no more matching file names are found, blank is returned.\n\nDIR( differs from DIRF( in that DIR( requires the user to specify the pathname unless the search is to be confined to the current directory. DIRF( (see below) is usually preferable since it searches the directory as which a current FILES structure points for the file name specified. See DIRF(.\n\nExamples:\n\n`? DIR('c:\\path\\.bak')`\n\nlocates and displays the name of the first file with extension BAK in the subdirectory \\PATH, while\n\n`? DIR('.bak')`\n\nlocates and displays the name of the first file with extension BAK in the current directory.\n\nIf a file is found, entering DIR() with no parameters will locate the next file meeting the filespec. This will continue until the response is a blank, indicating there are no more files meeting the filespec.\n\nOnce a file is identified with DIR(, the DIRX( function may be used to obtain additional information from the DOS directory in a form suitable for use in a program.\n\n## DIRF(\n\nGet file information from disk directory.\n\n``` DIR(filespec) (referred to as \"Form 1\")\nDIR() (referred to as \"Form 2\") ```\n\nOption:\n\n`filespec a string or string expression containing a file name (or skeleton using ? and/or * wildcards), with optional drive and/or path specification (enclosed in single quotes)`\n\nType: character\n\nSearches for specific files specified by the filename and the path supplied from the FILES structure (see FILES...ENDFILES), and provides specific information on files found.\n\nIf an argument is given (Form 1), it must be a string or string expression naming a file, with * and ? wildcards optional). Returns the first file name found matching filespec. If no match was found, blank is returned. If a disk name is supplied, the path supplied by the FILES structure is ignored.\n\nIf no argument is given (Form 2), the previous filespec is used to find the next matching file. If no more matching file names are found, blank is returned.\n\nDIR( differs from DIRF( in that DIR( requires the user to specify the pathname unless the search is to be confined to the current directory. See DIR(.\n\nExamples, assuming the following FILES structure is in effect:\n\n```FILES\n.prg,c:\\shark\\prg\n.frm,c:\\shark\\prg\n.bak,c:\\shark\\prg\n.db,c:\\shark\\data\n.ntx,c:\\shark\\ntx\n.cpl,c:\\shark\\cpl\nENDFILES\nDIRF('.prg')```\n\nlocates the first file with extension PRG in the subdirectory C:\\SHARK\\PRG. NB: this request only LOCATES the file information and doesn't display any result! Read further for instructions on how to DISPLAY needed information.\n\n`DIRF('.bak')`\n\nlocates the first file with extension BAK in the subdirectory C:\\SHARK\\PRG.\n\nNOTE that the above instruction only locates the file and doesn't display any result. To display the found file, you must type:\n\n`? DIRF('.bak')`\n`DIRF('.dbt')`\n\nlocates the first file with extension DBT in the subdirectory C:\\SHARK\\DATA.\n\n`DIRF('.txt')`\n\nlocates the first file with extension TXT in the current directory, since there was no matching filename pattern in the FILES structure.\n\nIf a file is found, entering DIRF() with no parameters will locate the next file meeting the filespec. This will continue until the response is a blank, indicating there are no more files meeting the filespec.\n\nOnce a file is identified with DIRF(, the DIRX( function may be used to obtain additional information from the DOS directory in a form suitable for use in a program.\n\n## DIRX(\n\nObtain additional information about file located with the DIR( or DIRF( function.\n\n``` DIRX(type)\n\ntype the name or number of the information required about a file\n```\n\nType: character/numeric\n\nOnce a file is identified with DIR( or DIRF( additional information can be obtained from the DOS directory in a form suitable for use in a program.\n\nThe type can be given in either of two forms, a name or number (numeric expression) as follows:\n\n```Type Explanation Result\n1 or Name last file name found with DIR( string\n2 or Size size in bytes of last name found with DIR( integer\n3 or Attribute DOS file attribute as follows integer\n1 - directory\n2 - system\n3 - hidden\n5 - normal\n4 or Time time file created or last updated string\n5 or Date date file created or last updated string\n6 directory\n7 sub-directory\n```\n\nShortcut: When specifying type by name, only the first character is required.\n\nThese functions have many uses. Use them to write a program that backs up recently modified files, a program that lists files so the user can pick one, etc.\n\nExamples:\n\n```DIRX(n) returns the filename (e.g. \"Name\")\nDIRX(a) returns 4 if the file is read only (e.g. \"Attribute\")\n```\n\n## EOF(\n\nGives the end-of-file flag for a specified data file.\n\n```\nEOF(<filenum>)```\n\nOption:\n\n` <filenum> the number of the data file to be checked`\n\nType: logical\n\nFor the data file number specified, if the current record pointer is on the last record and a SKIP is issued, EOF( returns T (true); otherwise it is F (false). Since SKIP n is treated as n SKIP commands, EOF( returns true if SKIP n goes past the last record.\n\nAlso, if a LOCATE or CONTINUE command is unsuccessful, or if NEAREST does not find an index key equal to or greater than the FIND string, EOF( returns T.\n\nIf no <filenum> is specified, the current file is assumed.\n\nAn alternate form of the function - EOF - works only on the currently selected data file, since it takes no parameter. See EOF.\n\nExamples:\n\n```1>USE employee\n1>GO 4\n1>SKIP 2\n1>? #\n6.00\n1>? EOF()\nF\n1>SELECT 2\n2>SKIP\n2>? EOF()\nT\n2>GO BOTTOM\n2>SKIP\n2>? #\n6.00\n2>SELECT 1\n1>? EOF(2)\nT\n```\n\n## EOF\n\nGives the end-of-file flag for the currently selected data file.\n\n`EOF`\n\nType: logical\n\nIf the current record pointer is on the last record of the file in use and a SKIP is issued, EOF returns T (true); otherwise it is F (false). Since SKIP n is treated as n SKIP commands, EOF returns true if SKIP n goes past the last record. Also, if a LOCATE or CONTINUE command is unsuccessful, or if NEAREST does not find an index key equal to or greater than the FIND string, EOF returns T.\n\nExamples:\n\n```1>USE employee\n1>GO 4\n1>SKIP 2\n1>? #\n6.00\n1>? EOF\nF\n1>SKIP\n1>? EOF\nT\n1>GO 4\n1>SKIP 3\n1>? #\n6.00\n1>? EOF\nT\n```\n\n## EXP(\n\ne to the power <num exp>\n\n## FIELD(\n\nGet the number of the Get Table entry corresponding to a variable or field name.\n\n```\nFIELD(<name>) ```\n<name> =the name of field or variable in a Get Table\n\nType: numeric\n\nWhile in full-screen editing mode (with READ, BROWSE, EDIT, etc.), each input variable and field is put into a Get Table that can be controlled with an ON FIELD structure.\n\nFIELD( returns the number from 1 to 64 of any editing field on screens created with @ GET and TEXT macros. This function is usually used on an ON FIELD structure to redirect the sequence of data entry.\n\nSee READ and ON FIELD in the Command Reference section.\n\nExample in a program:\n\n```ON FIELD\nFIELD qty\nIF qty<0\n@ 22,0 say CEN('Quantity cannot be negative. Press any key',80)\ncc=INKEY()\nERASE 22,22\n:FIELD=FIELD(qty)\nENDIF\nENDON\n```\n\n## FILE(\n\nVerifies whether a file exists.\n\n``` FILE(<str exp>)\n\n<str exp> = a file name ```\n\nType: logical\n\nThis function looks up the file whose name is given by <str exp>; if the file is found, the function returns T, otherwise it returns F.\n\nIf no extension is given in the file name, DBF is assumed (a data file is looked for).\n\nExamples:\n\n```1>? FILE('employee')\nT\nT\n1>? FILE(mfile)\nF\n1>? FILE('a:'+mfile)\nT\n```\n\n## FLD(\n\nGet information about a field in a data file.\n\n```\nFLD(type,<fieldnum> [,<filenum>])\n\ntype one of the four attributes of a field\n<fieldnum> = the number of the field to be checked```\n\nOption:\n\n` <filenum> the number of the data file to be checked`\n\nType: character/numeric\n\nEach field in a data file has four attributes as shown with the LIST STRUCTURE command: name, type, width and (for numeric variables) number of decimal places. The FLD() function is often used in conjunction with the DBF() function.\n\nThese attributes can be retrieved in a form suitable for use in a program with the FLD() function.\n\nThe type can be given in either of two forms, a name or number (numeric expression) as follows:\n\n```Type Explanation Result\n1 or Name string containing field name string\n2 or Type string containing field type string\n3 or Width number containing width of field integer\n4 or Decimals number of decimal places in field integer\n```\n\nShortcut: When specifying type by name, only the first character is required.\n\nExample in a program:\n\n```REPEAT DBF(records) TIMES varying fldnum\nREPEAT 4 times VARYING type\n?? FLD(type,fldnum)\nENDREPEAT\n?\nENDREPEAT\n```\n\n## FLOOR(\n\nfloor integer: the integer equal to or just below a given number\n\nRelated functions: CEIL( = the integer equal to or just above a given number; FLOOR( = the integer equal to or just below a given number; INT( = the integer part of a mixed number after discarding the fractional part; MOD( = the balance after truncating a mixed number.\n\n## GET(\n\nGets a string from a DOS file.\n\n``` GET(str var,<width num exp>[,filenum])\n\nstr var stores the string\n<width num exp> = the width of the string requested (must be in range 1 to 254)```\n\nOption:\n\n` filenum the DOS file number (between 1 and 4)`\n\nType: logical\n\nThis function imports a string of <width num exp> characters from a DOS file opened with the ROPEN( function; the character number pointer is normally positioned with the SEEK( function.\n\nIf successful in getting all the bytes requested, GET( returns T (true) and sets str var to the string imported from the file. If str var does not exist, GET( will create it.\n\nIf the function is unsuccessful, it returns F (false). This will be the result if the GET( function tries to get data beyond the end of the file. Note, however, that even if GET( returns F, one of more characters may still have been imported from the file; it is wise to check the value and width of str var to ensure part of a file is not lost.\n\nIf filenum is not given, filenum=1 is assumed.\n\nGET( READ(, IN(, and WRAP( are the only functions that change the contents of the memory variable used as an argument.\n\nExample in a program:\n\n```IF ROPEN('test',3)\nDO WHILE GET(string,80,3)\n? string\nENDDO\nENDIF\nok=CLOSE(3)\n```\n\n## IFF(\n\nAllows inline IF...THEN logic in expressions.\n\n``` IFF(cond,<exptrue>,<expfalse>)\n\ncond a logical expression\n<exptrue> the expression to be returned if cond is TRUE\n<expfalse> the expression to be returned if cond is FALSE```\n\nType: Type: character/numeric/logical\n\nThis function returns <exptrue> if cond is true, <expfalse> otherwise. The type of the value returned is the same as the expression selected by the condition. IFF( is very useful in the FIELDS line of reports or in commands such as SUM, AVERAGE, REPLACE, or LIST.\n\nExamples:\n\n```1>? IFF(married,'Married','Single ')\n1>SUM IFF(quant>500, quantprice, 0),IFF(state='NY',1,0)\n```\n\nThe first command prints \"Married\" or \"Single\" according to the value of a logical field named MARRIED. The second command will return the sum of all quantities for transactions where quantity is greater than 500, and a count of all records where STATE='NY', thus combining two separate commands (SUM FOR and COUNT FOR) into one.\n\nCaution: do not use expressions of different types or widths in reports, since this may cause the REPORT command to fail.\n\n## IN(\n\nInputs a single character from a sequential file.\n\n``` IN(str var[,<filenum>])\n\nstr var stores the character```\n\nOption:\n\n` <filenum> the DOS file number (between 1 and 4)`\n\nType: logical\n\nThis function reads the next character of the DOS file (opened with the ROPEN() function into the string variable str var; if str var does not exist, it will be created. str var cannot be a matrix variable.\n\nIf filenum is not given, filenum=1 is assumed. IN( returns T if successful, F otherwise.\n\nThis function is especially useful to communicate over the standard COM1, COM2 devices, for conversion of WordStar or other non-standard files to standard ASCII files, to encrypt/decrypt a file through a translation table.\n\nIN(, GET(, READ(, and WRAP( are the only functions that change the contents of the memory variable used as an argument.\n\nSee the functions OUT(, ROPEN(, WOPEN(, SEEK(, SSEEK(, and CLOSE(.\n\n## IFKEY(\n\nTests if a character is waiting in the keyboard buffer.\n\n` IFKEY()`\n\nType: logical\n\nIt is often useful to test whether a key has been pressed on the keyboard without waiting indefinitely if a key is not pressed.\n\nThe IFKEY( function returns T (true) if a keystroke is waiting in the keyboard buffer, F (false) if not. The keyboard buffer is not affected.\n\nExample in a program, to create a timing loop that ends as soon as any key is pressed:\n\n```start=VAL(TIME(seconds))\nDO WHILE VAL(TIME(seconds))-start<3\nIF IFKEY()\nBREAK\nENDIF\nENDDO\n```\n\n## INKEY(\n\nEvery key on a keyboard has a numeric value. INKEY() waits and gets the numeric value of a key-press.\n\n`INKEY()`\n\nType: numeric\n\nThis function suspends program execution until a key is pressed. It returns a number identifying the key. Nothing is displayed on the screen. Any key can be read (except Alt, Ctrl, and shift which merely affect the characters produced by other keys) including all function keys, editing keys, and alternate keys. (Depending on your computer, function keys may be pre-programmed in a variety of ways. It's even possible that F11 and F12 may not recognized by your computer's BIOS program, and may be ignored by Shark. Although it's possible, it's more likely that your PC will recognize these keys.)\n\nStandard keys are identified with their ASCII number. Other keys return values between 256 and 511.\n\nExamples:\n\n``` Key INKEY()\nCtrl-C 3\nA 65\nAlt-A 285\n<F1> 315\nShift-<F1> 340\nCtrl-<F1> 350\nAlt-<F1> 360\nF12 390\n```\n\nTo find out the number identifying a key, give the command:\n\n```1>? INKEY()\n```\n\nThen (1) press <ENTER>, then (2) the key. The key's character number will be displayed.\n\nUsing INKEY() the user can program his own EDIT, set up cursor controlled menus, and so on.\n\nA simple example showing the many uses of INKEY():\n\nINKEY() makes a more attractive screen than the WAIT command. INKEY() can also be used to make choices:\n\n```DO WHILE t\nCLS\n@ 10,25 SAY \"Hello! Press Any Key\"\nWAIT=INKEY()\nDO CASE\nCLS\n@ 3,30 say \"Some Announcements:\"\nBOX 2,15,4,65\n@ 7,27 say \"Select from the following:\"\n@ 10,20 say \"<--- <R>eturn to Start or <Q>uit --->\"\nCOLOR 112,10,25,10,27\nCOLOR 112,10,50,10,52\nCURSOR 24,79\n:KEY=0\nWAIT=INKEY()\n if <R> or <r>, <R>eturn to opening menu\nCASE :KEY=114 .or. :KEY=82\nLOOP\n* if <Q> or <q>\nCASE :KEY=81 .or. :KEY=113\nCLS\nCANCEL\nOTHERWISE\nCLS 22,22\nRING\n@ 15,20 SAY CEN(\"PLEASE TRY AGAIN!\",40)\nWAIT=INKEY()\nCLS\n@ 10,30 \"Thank you!\"\nENDCASE\nENDDO\n\n```\n\n## INSERT(\n\nOverwrites a string at a given position with another string.\n\n``` INSERT(str expover,<str exp>,<num exp>)\n\nstr expover the string expression to overwrite\n<str exp> the string expression to overwrite with\n<num exp> the position```\n\nType: character\n\nThis function takes the string in str expover and overwrites the string with <str exp> starting at position <num exp>.\n\nExamples:\n\n```1>line=' '\n1>customer='John Smith'\n1>ponumber='32109'\n1>amount='910.56'\n1>line=INSERT(line,customer,1)\n1>? line\nJohn Smith\n1>line=INSERT(line,ponumber,15)\n1>? line\nJohn Smith 32109\n1>line=INSERT(line,amount,25)\n1>? line\nJohn Smith 32109 910.56\n1>line=' c '\n1>newline=INSERT(line,customer,@('c',line))\n1>? newline\nJohn Smith\n```\n\nNote: The last example shows the use of INSERT( with \"templates\". The line variable is the template. The character \"c\" in it designates the place where the customer has to be inserted. Such templates are useful in report generators or for creating screen displays.\n\n## INT(\n\nGives the integer part of a given number (the fractional part is discarded); useful for currency \"rounding\", for example.\n\nRelated functions: CEIL( = the integer equal to or just above a given number; FLOOR( = the integer equal to or just below a given number; INT( = the integer part of a mixed number after discarding the fractional part; MOD( = the balance after truncating a mixed number.\n\nExample:\n\n```1> ? int(3.14159)\n1> 3.00\n```\n\n## LEFT(\n\nGets the left part of a string.\n\n``` LEFT(<str exp>, <num exp>)\n\n<str exp> the string from which the new string is formed\n<num exp> the number of characters to place in the new string\n```\n\nType: character\n\nThis function takes the first <num exp> characters from the string <str exp>. It is equivalent to, but more efficient than:\n\n```1>\\$(<str exp>, 1,<num exp>).\n```\n\nIf <num exp> is greater than the width of <str exp>, this function returns all of <str exp>.\n\nWherever an expression calls for a substring starting at the beginning, use LEFT( instead of \\$( or SUBSTR(.\n\nExample:\n\n```1>a='David Bark'\n1>? LEFT(a,5)\nDavid\n1>? LEFT(a,50)\nDavid Bark\n```\n\n## LEN(\n\nGets the width (length) of a string.\n\n``` LEN(<str exp>)\n\n<str exp> is the string being measured```\n\nType: numeric\n\nThis function returns the width (including trailing blanks) of the string <str exp>.\n\nExamples:\n\n```1>name='David Barberr'\n1>? LEN(name)\n13.00\n1>? LEN(name+' is a nice boy')\n27.00\n```\n\nNote that the width of a string is at least 1!\n\n## LOC(\n\nLOC( Gets the current byte position in a file opened with ROPEN( or WOPEN(.\n\nLOC(<filenum>)\n\nOption:\n\n<filenum> is the number of the sequential or random file, 1 to 4 (default 1)\n\nType: number\n\nWhenever a file is opened with the ROPEN( or WOPEN( function, Shark maintains a pointer at a current position, which is where any PUT( or GET( function would take effect. The position pointer is set with the SEEK( and SSEEK( functions, and reset every time the IN(, OUT, READ(, WRITE(, PUT(, and GET( function is used.\n\nIf filenum is not given, filenum=1 is assumed.\n\nA common use of LOC( is to get the current position before a SEEK( so that the pointer can be reset to the original position after some operation.\n\n## LOG(\n\nLOG( <num exp>)\n\ngives the natural logarithm of <num exp>\n\nType: number\n\nExample:\n\n```1>? log(100)\n1> 4.60\n```\n\n## LOG10(\n\ngives the base 10 logarithm of <num exp>\n\nType: number\n\nExample:\n\n```1>? log10(100)\n1> 2.00\n```\n\n## LOWER(\n\nConverts the string <str exp> to lower case.\n\n` LOWER( <str exp>)`\n<str exp> is the text to be converted to lower case\n\nType: character\n\nAll upper-case letters in the <str exp> are converted into lower case by the LOWER( function. See also the !( and UPPER( functions.\n\nExamples:\n\n```1>a='Aa12b'\n1>? LOWER(a)\naa12b\n1>? LOWER('David!')\ndavid!\n```\n\nNote that only the upper-case letters, A-Z, are changed (to a-z). No other characters are affected.\n\n## LTRIM(\n\nTrims blanks from the left-hand side of a string:\n\nLTRIM(<str exp> ) <str exp> the string to be trimmed\n\nType: character\n\nThis function gets rid of the blanks on the left of a string. See also TRIM( which removes the blanks on the right side of a string.\n\nExamples:\n\n1>a=' David ' 1>? a David 1>? LEN(a) 14.00 1>? LTRIM(a)+' is trimmed on the left' David is trimmed on the left 1>? LEN(LTRIM(a)) 9.00 1>blank=' ' 1>? LEN(LTRIM(blank)) 1.00\n\nNote: LTRIM(blank) is a single blank.\n\n```MASK(<operation>,<string>,<mask>) Manipulates bits in a string based on comparison between 2 strings\n<operation> a number representing one of the three masking operations supported by SharkBase: AND, OR, and XOR\n<string> the string to be modified by the operation\n<mask> the string used to modify the bits in string1 ```\n\nType: character\n\nEach character in the ASCII character set is identified by an eight-bit binary number from 0 to 255 inclusive. Each bit may be either 0 or 1; for example, the letter A has a decimal value of 65 and a binary value of 01000001, while the letter B has a decimal value of 66 and a binary value of 01000010. Although SharkBase has functions to set individual bits to 1 or 0 (SET( and RESET( respectively), MASK( can be used to change any number of bits at once.\n\nThe three MASK( operations are used to return a string that is the result of bitwise operations between two strings, comparing each bit in one string with the bit in the same position in the other string. If the two bits are both 1 (on or true), AND and OR both produce a 1 in that position, while XOR (exclusive OR) returns a 0. If one is a 1 and the other a 0, OR and XOR both produce a 1 and AND produces a 0. Finally, if both are 0, all three produce a 0.\n\nThere is one other operator of this type, known as the bitwise NOT, but it is not implemented through the MASK( function. Since the primary use of the bitwise NOT is to create indexes in decending order, this operation is implemented through the DESCEND( function. See DESCEND(.\n\nThe use of AND, OR and XOR is the same as in C and other low-level languages that permit access to data at the bit level, and will rarely be used by most SharkBase programmers. Those who require these operations already know how to use them and need no additional instructions for SharkBase.\n\nExample in a program:\n\nGiven any screen color (See COLOR and SET COLOR), determine the reverse that SharkBase would use in highlighting input windows during READ operations (useful in using the same color to highlight messages and other text on the screen):\n\n```CLS\nDO WHILE t\nINPUT 'Color.... ' to in\nlow=mod(in,16) ;separate the low-order and high-order bits\nhi=int(in/16)\nout=low16+hi ;the rest of these operations are too complex to explain...just be assured it works\nnew=int(in/16)+16mod(in,16)\n:color=in\n? 'Main Color..',in\n:color=rev\n? 'Reverse.....',rev\nENDDO\n```\n\n## MAX(\n\nGiven any two expressions of the same type, MAX( returns the higher value. It must be remembered that string comparisons are based on the ASCII value of the characters in the two strings. Comparing two logical expressions has no meaning.\n\nYou may find that a comparison between a numeric expression and a numeric field results in an error. This can be avoided by ensuring both arguments are an expression, accomplished most easily by adding 0 to a field.\n\nExamples:\n\n```1>? MAX(123,amount+0) ;adding 0 ensures handling as an expression\n5241.34\n1>? MAX('hello','goodbye')\nhello```\n```MAX(<exp1>,<exp2>) Compare two expressions of any type and return the larger.\n<exp1> any expression\n<exp2> any expression of the same type as <exp1> ```\n\nType: Type: character/numeric/logical\n\n```\n\nchoices the number of choices offered by the menu\nhotkeys one character for each possible option, beginning\nwith zero (normally the exit option)\nwidth the width of the menu lightbar\n```\n\nOption:\n\n`seconds exit MENU( after this many seconds; function returns value of 65`\n\nType: numeric\n\nThe MENU( function pauses program execution and superimposes a movable lightbar (reverse-video line) over a menu of selections previously written to the screen, usually with TEXT.\n\nThe menu lightbar is moved up and down with the up arrow and dn arrow keys. If you press down arrow while on the bottom, the lightbar cycles automatically to the top. Similarly, pressing up arrow while on the top cycles to the bottom.\n\nThe user can select any item by moving the lightbar over it and pressing Enter, or entering its line number as a one-digit number. In either case, the line number selected is returned by the function, and the key pressed stored in the system variable :KEY. Both the function value and :KEY can be tested in a subsequent DO CASE structure to determine the program's next actions. The inkey() function displays the :KEY value:\n\n```CLS\nDO WHILE t\n@ 10,0 SAY \"To display a key code, press any key . . .\"\n? inkey()\nENDDO\n```\n\nIf the user presses 0 or <Home> , MENU( returns zero, although the first line covered by the lightbar is 1. Options over 9 can be accessed only by the lightbar.\n\nAlternately, you may choose to use the <hotkeys> option for this function's first argument. Instead of the number of choices, a string may be supplied. The length of the string determines the number of choices and the letters in the string math the numbers of the choice. For example, if if you supplied a string literal or variable with the value \"Qmdn\" for <hotkeys>, pressing \"Q\" or \"q\" would return zero, \"M\" or \"m\" would return 1, \"D\" or \"d\" would return 2, and \"N\" or \"n\" would return 3. <Hotkeys> may be caps or lower case; MENU( is not case-sensitive.\n\nIf any cursor key except , and <Home> is pressed, MENU( returns the number of the line highlighted by the lightbar, and :KEY contains the key number that would be returned by the INKEY( function. (If SET FUNCTION OFF, all function keys have the same effect as these cursor keys.)\n\nWhile the MENU( function is active, all other typewriter keys are ignored.\n\nNote: If an ON KEY structure is in effect, it is ignored while MENU( is waiting for input.\n\nExamples:\n\n1. This is a program which shows specifically the values returned by Menu( when any key is pressed:\n\n```CLS\n@ 19,46 say 'Returns'\n@ 19,64 say ':KEY'\nDO WHILE t\nCURSOR 5,20\n@ 20,50 say var\n@ 20,65 say :KEY\nENDDO\n```\n\n2. In a simple real-life program:\n\n```CLS\nTEXT\n\n PREPARE SALES TOTALS\n CALCULATE TAX SUMMARY FOR A TAX PERIOD\n PRINT OUT SALES TRANSACTIONS & MARGINS\n REVIEW DEPOSITS JOURNAL\n\nENDTEXT\nCHOICE=0\nCURSOR 8,12 ; positions cursor at row 8, column 12\nCHOICE=MENU(4,1); 1-wide lightbar moves over 4 menu rows, saves value to choice\n\nDO CASE\n * * * **\nCASE CHOICE = 0\n... etc\nCASE CHOICE = 1\n... etc\n```\n\n3. Another real-life program:\n\n```ERASE\nWINDOW 1,2,23,77 double ;draw frame around screen\n@ 1,3 say DATE(full)\n@ 3,3 say CEN(:company,74)\nWINDOW 8,25,22,75 blank ;use WINDOW to position TEXT\nWINDOW\nCURSOR 10,23\nIF ans=0 .or. :key=335\nENDIF\n@ ans+9,23 say CHR(149) ;• character as bullet\nDO CASE\nCASE ans=1\n... etc.\n```\n\n4. Part of the same program using both hotkeys and timeout:\n\n```WINDOW 8,25,22,75 blank ;use WINDOW to position TEXT\nWINDOW\nCURSOR 10,23\nhotkeys='Qeprsmwgf' ;hotkeys can be a string variable\n ; this allows 8 options plus zero (quit)\nIF ans=0 .or. :key=335 .or. ans=65 ;return of 65 means program timed-out\nENDIF\n@ ans+9,23 say CHR(149) ;• character as bullet\nDO CASE\nCASE ans=1\n... etc.\n```\n\n## MIN(\n\nCompare two expressions of any type and return the smaller.\n\n```\nMIN(exp1,exp2)\n\nexp1 any character or numeric expression\nexp2 any expression of the same type as exp1 ```\n\nType: character/numeric\n\nGiven any two expressions of the same type, MIN( returns the lower value. It must be remembered that string comparisons are based on the ASCII value of the characters in the two strings. (Comparing two logical expressions has no meaning.)\n\nYou may find that a comparison between a numeric expression and a numeric field results in an error. This can be avoided by ensuring both arguments are an expression, accomplished most easily by adding 0 to a field.\n\nExamples:\n\n```\n1>? MIN(123,amount+0) ;adding 0 ensures handling as an expression\n123.00\n1>? MIN('Hello','Goodbye')\nGoodbye\n```\n\n## MOD(\n\nThe MODULO function returns the remainder of one number divided by another.\n\n`MOD(<num exp1>,<num exp2>)`\n\nFor example:\n\n```1>? MOD(5,2) gives the remainder of 5 divided by 2, e.g.\n1.000000\n1>? MOD(-3.14, .7)\n-0.340000\n```\n``` MOD(<num exp1>,<num exp2>) <num exp1> modulo <num exp1>: returns 0 if <num exp2> is 0; returns the value num with the same sign\nas <num exp1>, less than <num exp2>, satisfying <num exp1>=i<num exp2>+num for some integer i.\n\n1>? MOD(5,2) this is the remainder of 5 divided by 2\n1.000000\n1>? MOD(-3.14, .7)\n-0.340000\n```\n\nRelated functions: CEIL( = the integer equal to or just above a given number; FLOOR( = the integer equal to or just below a given number; INT( = the integer part of a mixed number after discarding the fractional part; MOD( = the balance after truncating a mixed number.\n\n## MONTHS(\n\nComputes dates and date differences in months.\n\n```\nMONTHS(<date1>,<date2>) or\nMONTHS(<date1>,<num exp>)\n\n<date1> a string expression containing a valid date\n<date2> a string expression containing a valid date\n<num exp> a numeric expression such as a number of months\n```\n\nWhen <date2> is specified, MONTHS( returns number of months between the two dates. When <num exp> is specified, MONTHS( returns date that many months away\n\nMONTHS( computes the difference between the two dates in months, or computes a date a given number of months before or after a specified date. Fractional parts of months are discarded.\n\nIf a computed date is after the last date of the month, the date will be adjusted to the last day of the month. For example, MONTHS('013190',1) results in 022890.\n\nExamples:\n\n```1>? MONTHS('04 06 90','04 29 90')\n0.00\n1>? MONTHS('01/01/90','02/01/90')\n1.00\n1>? MONTHS('02/01/90','01/01/90')\n-1.00\n1>? MONTHS('01/01/90','01/01/92')\n24.00\n1>? MONTHS('02/01/90',10)\n120190\n1>? MONTHS('01/01/90',-6)\n070189\n```\n\nNote that the MONTH( and DATE( functions always count the days in leap years:\n\nThus, DAYS(2005-03-18,-365) may not give the same result as MONTHS(2005-03-18,-12)\n\n## NDX(\n\nGet information on index files in use.\n\n``` NDX(type [,indexnum] [,<filenum>])\n\ntype the name or number of the information required```\n\nOptions:\n\n``` the number of the index being checked (1 to 7);\ndefault is thge master index\n<filenum> the number of the data file to be checked```\n\nType: character/logical\n\nNDX( is used to primarily in programs to get the information on the current environment in a form suitable for use in expressions.\n\nThe type can be given in either of two forms, a name or number (numeric expression) as follows:\n\n```Type Explanation Result\n1 or Name name of index file string\n2 or Key key on which index was created string\n3 or DBF_Name name of data file on which index\nwas created string\n4 or Filter TRUE if filter or FOR clause was\nin effect when index was created logical```\n\nShortcut: When specifying type by name, only the first character is required.\n\nExamples:\n\n```1>? NDX(n),NDX(key),NDX(dbf),NDX(filter)\nCUST1.NDX CUSTNUM CUSTOMER.DBF F\n```\n\n## NTX(\n\n```This is an obsolete function identical to NDX(, which refers to the\ndiscontinued Clipper .ntx file type. The older Clipper NTX indexes are\nno longer used in Shark.\n```\n\n## NUMTOC(\n\n```Convert a decimal number to a hexadecimal string.\n\nNUMTOC(type,number)\n\ntype the length of the string to be created\nnumber the number to be converted\n\nType: character\n\nA general conversion function for converting decimal numbers into\nhexadecimal values. Input can be any number, and the returned string length\ncan be up to eight characters as follows:\n\nThe type can be given in either of two forms, a name or number (numeric\nexpression) as follows:\nType Range of Number String Length Returned\n1 integer 0 to 255 1 byte\n2 integer -32768 to 32767 2 bytes\n4 integer +/- 2 billion 4 bytes\n8 a floating point number 8 bytes\nShortcut: When specifying type by name, only the first character is required.\n\nTypes 1, 2 and 4 return hexadecimal integers. Any fractional parts are ignored.\n\nWhen type is 1, this function is equivalent to CHR(. Values outside\nthe range of 0 to 155 return the modulus of 256 for type 1.\n\nThe CTONUM(, RANK( and ASC( functions convert strings into numbers.\n\nDo not confuse these function with STR( and VAL(, which convert decimal\nnumbers into their string representations, and vice versa.\n\nExamples:\n1>? NUMTOC(1,97)\na\n1>? NUMTOC(2,25185)\nab\n1>? NUMTOC(4,6513249)\nabc\n```\n\n## OUT(\n\n```\nOutputs a single character to a sequential file.\n\nOUT(str var [,<filenum>])\n\nstr var contains the character\n\nOption:\n\n<filenum> the DOS file number (between 1 and 4)\n\nType: logical\n\nThis function outputs the character in str var to the sequential file\n(opened with the WOPEN( function).\n\nIf filenum is not given, filenum=1 is assumed. OUT( returns T if successful, F otherwise.\n\nThis function is especially useful to communicate over the standard COM1,\nCOM2 devices, for conversion of Word Star or other non-standard files to\nstandard ASCII files, to encrypt/decrypt a file through a translation table.\n\nSee the functions IN(, ROPEN(, WOPEN(, CLOSE(, SEEK(, SSEEK(, READ(, and WRITE(.\n```\n\n## PIC(\n\n```\nFormats the PICTURE of a number or string.\n\nPIC(exp,format)\n\nexp is the number or string to be formatted\nformat the format clause\n\nType: character\n\nThis function returns the exp formatted with the format clause\nformat. See the command @ for the description of the format clauses.\nPIC( is especially useful in preparing numeric values for printing.\n\nPIC( always returns a string, even when a number or numeric expression\nis being formatted.\n\nExamples:\n1>number=1123.89\n1>format='9,999.99'\n1>? PIC(number,'9,999.99')\n1,123.89\n1>? PIC(number,format)\n1,123.89\n1>format='9999'\n1>? PIC(number,format)\n1123\n1>format='\\$\\$\\$,\\$\\$\\$.99'\n1>? PIC(number,format)\n\\$1,123.89\n1>format='\\$\\$\\$,\\$\\$\\$.999'\n1>? PIC(number,format)\n\\$1,123.890\n1>string='abcd'\n1>format='xX9!'\n1>? PIC(string,'xx9!')\nabcD\n1>? PIC(string,format)\nabcD\n1>format='X-X-X-X'\n1>? PIC(string,format)\na-b-c-d\n1>SET ZERO ON\n1> ? 0,\"\",str(0,5,2),\"\",pic(0,\"99.99\"),\"\",\"0\"\n0.00 * 0.00 * 0.00 * 0\n1>SET ZERO OFF\n1> ? 0,\"\",str(0,5,2),\"\",pic(0,\"99.99\"),\"\",\"0\"\n 0.00 * * 0\n\n```\n\n## POW(\n\n```<num exp1> to the power <num exp2>\nExample:\n\n1>:PICTURE='999.999999'\n1>? POW(2,4)\n16.00000```\n\n## PRINTER(\n\n```\nThis is an MS-DOS function to test whether a printer is ready to print.\n\nPRINTER(printernum)\n\nOption:\n\nprinternum the number of the LPT port (1 or 2)\n\nType: logical\n\nWhenever a program has to print, it needs a printer turned on and\non-line. When it is unsuccessful in printing, Shark intercepts the customary\nDOS error (the infamous \"Abort, Retry, Ignore?\") and ends execution.\n\nThe PRINTER( function gives programmers a way to ensure the printer is\ncorrectly set up before sending output to the screen. This makes it possible\nto suspend execution under program control, prompt for correction action, or\neven SPOOL the output to a disk file instead of the printer.\n\nExamples in programs:\n\nDO WHILE .NOT. PRINTER()\nWINDOW 10,10,15,69 DOUBLE\n@ 12,10 SAY CEN('Turn on printer and press any key . . .',60)\nRING\nCURSOR 13,39\ncc=INKEY()\nWINDOW\nENDDO\nIF .NOT. PRINTER(2) ;test LPT2\nSPOOL printfil\nENDIF\nIF PRINTER()\nSET PRINT ON\nENDIF\n```\n\n## PUT(\n\n```\nPuts a string into a DOS file.\n\nPUT(<str exp>[,filenum])\n\n<str exp> the string to overwrite with\n\nOption:\n\n<filenum> the DOS file number (between 1 and 4)\n\nType: logical\n\nA DOS file was opened with the WOPEN( function; the character number\npointer was normally positioned with the SEEK( function. This function\noverwrites the file from the character chosen by the character number pointer\nwith the string <str exp>.\n\nIf filenum is not given, filenum=1 is assumed. PUT( returns T if successful, F otherwise.\n\nExamples:\n1>byte=CHR(13)\n1>ok=WOPEN('test',3)\n1>ok=SEEK(5221)\n1>ok=PUT(byte,3)\n```\n\n## RAND(\n\n```\nGives a random number in the range 0<=n<1.\n\nRAND()\n\nOption:\n\nseed a number used to initiate the random series\n\nType: numeric\n\nA series of successive calls to the RAND( function will return a uniform\ndistribution of random numbers.\n\nThe first time RAND( is called, <seed> (any numeric expression) may be\nspecified. All subsequent calls should be without the seed. If no initial\nseed is provided, a random seed is chosen by the program.\n\nRAND( always returns a number equal to or greater than 0 and less than 1.\nIf you need a random series of integers between zero and 5000, use\n5000*RAND().\n\nNote: if you provide the initial seed, every execution of RAND( ) will\nreturn the same series of numbers.\n```\n\n## RANK(\n\n```\nConverts a character to its ASCII number.\n\nRANK(<str exp>)\n\n<str exp> the first character of this string is converted\n\nType: numeric\n\nThe characters in the character set used by the computer are numbered\nfrom 0 to 255. For the first character of the string <str exp>, RANK(\nCTONUM(, and NUMTOC(.\n\nExamples:\n\n1>? RANK('x')\n120.00\n1>? RANK('xyz')\n120.00\n\nNote that only the first character of the string matters.\n```\n\n```\nReads a line of a sequential file.\n\nstr var stores the line read in\n\nOption:\n\n<filenum> the DOS file number (between 1 and 4)\n\nType: logical\n\nThis function reads the next line of the sequential file (opened with the\nROPEN( function) into the string variable str var. If str var does not\nexist, it will be created; str var cannot be a matrix variable.\n\nA line is terminated by the carriage return character (ASCII 13). Since\nthe line is read into a string variable, it cannot be longer than 254\ncharacters.\n\nIf filenum is not given, filenum=1 is assumed. READ( returns T if\nsuccessful, F otherwise.\n\nIn Shark programs, READ( normally appears in an IF or DO WHILE command.\n\nREAD(, IN(, GET(, and WRAP( are the only functions that change the\ncontents of the memory variable used as an argument.\n\nSee the functions WRITE(, ROPEN(, WOPEN(, CLOSE(, IN(, OUT(, and SSEEK(.\n\nExamples:\n\n1. In Conversational Shark:\n\n1>ok=ROPEN('a:label.prg')\n1>? line\n\n2. Two programs to print a text file, TEST (in the second version it is\nassumed that TEST has no more than 20 lines):\nSET WIDTH TO 80\nSET PRINT ON\nIF ROPEN('test')\n? line\nENDDO\nok=CLOSE()\nENDIF\nDIM CHAR 80 matrix\nSET WIDTH TO 80\nSET PRINT ON\nIF ROPEN('test',1)\nREPEAT 20 times VARYING num\nmatrix[num]=input,\nELSE\nBREAK\nENDIF\nENDREPEAT\nIF CLOSE(1)\n? matrix\nENDIF\nENDIF\n```\n\n## RECNO(\n\n```\nGets the current record number in any open data file.\n\nRECNO(<filenum>)\n\nOption:\n\n<filenum> the number of any open data file; default is the selected data file\n\nType: numeric\n\nThis function returns the record number of the current record of any\nspecified data file; if no <filenum> is given as in recno(), returns the record number the\nselected file. Note that ? RECNO() displays the current record number in the\nform specified by the system variable :PICTURE (see Section 2.7).\n\nShark also has a more limited form of this function, #, which applies\nonly to the selected data file.\n\nExamples:\n1>USE employee\n1>USE#2 customer\n1>? RECNO(1)\n1.00\n1>GO BOTTOM\n1>? RECNO(1)\n6.00\n1>GO TOP\n1>? RECNO()\n1.00\n1>SKIP#2 2\n1>? RECNO(2)\n3.00\n```\n\n## REMLIB(\n\nRemoves a library entry.\n\n`REMLIB(volume)`\n\nvolume the number of the library entry to be removed.\n\nType: logical\n\nThis function deletes a library entry. The function accepts the library volume number you wish to delete as its argument and returns T (true) if the delete operation was successful, F (false) if not.\n\nOnce a library entry (volume) is deleted, its space in the library is made available for new text.\n\nLibraries are created with the SET LIBRARY TO command.\n\nExample:\n\n```1>? REMLIB(50)\nT\n```\n\n## REPLACE(\n\nReplaces, in a string expression, all occurrences of a string with another string.\n\n` REPLACE(<str exp>,<str exp1>,<str exp2>) `\n\nReplace in the string expression <str exp> all occurrences of the string <str exp1> with the string <str exp2>\n\nType: character\n\nThis function looks up the first occurrence of . This process continues as long as <str exp1> occurs in <str exp>.\n\nExamples:\n\n1. A field contains a number as right justified characters, padded on the left with blanks. The following REPLACE( changes these numbers to right justified numbers padded on the left with zeros.\n\n```1>number=' 123'\n1>number=REPLACE(number,' ','0')\n1>? number\n00000123\n```\n\n2. In writing checks, dollar amounts may be left padded with dollar signs:\n\n```1>number=' 123.11'\n1>number=REPLACE(number,' ','\\$')\n1>? number\n\\$\\$\\$\\$\\$123.11```\n\n3. Renaming a variable in a program line. The variable OLDN is renamed FIRSTNUMB.\n\n```1>line='newn=oldn+oldn+(oldn/3)'\n1>line=REPLACE(line,'oldn','firstnumb')\n1>? line\nnewn=firstnumb+firstnumb+(firstnumb/3)\n```\n\n## RESET(\n\nSets a bit in a string to 0.\n\n``` RESET(<str exp>,bit position)\n\n<str exp> the string or string expression on which the\nfunction is to act\nbit position the number of the bit, numbered from the left\nstarting at 1, which is to be set to zero ```\n\nType: logical\n\nA bit is any of the eight binary digits in a character's ASCII number representation. Each bit can have only one of two possible values, 0 and 1.\n\nThe SET( and RESET( functions are used to manipulate the bits within a string or string expression. SET( makes a bit 1, and RESET( makes a bit 0. The BIT( function tests the value of a specific bit.\n\nAmong the chief uses for these functions is compression of logical (true/false) data by using just one bit for each data item instead of an entire byte for a logical field or two bytes for a logical variable.\n\nSee the BIT( function for programming examples.\n\nExamples:\n\n```1>str='PS'\n1>? BIT(str,15)\nT\n1>str=RESET(str,15)\n1>? str,BIT(str,15)\nPQ F\n```\n\n## RIGHT(\n\nGets the right-hand part of a string.\n\n` RIGHT(<str exp>, <num exp>) `\n\n<str exp> the string from which the new string is formed <num exp> the number of characters to place in the new string\n\nType: character\n\nThis function returns the last (that is, the rightmost) <num exp> characters from the string <str exp>.\n\nIf <num exp> is greater than the width of <str exp>, this function returns all of <str exp>.\n\nExample:\n\n```1>a='David Bark'\n1>? RIGHT(a,5)\nBark\n1>? RIGHT(a,50)\nDavid Bark\n```\n\n## ROPEN(\n\nOpens a DOS file for reading.\n\n``` ROPEN(<str exp> [, <filenum> ])\n\n<str exp> the file name```\n\nOption:\n\n`<filenum> the DOS file number (between 1 and 4) `\n\nType: logical\n\nThis function opens <str exp>, a DOS file (in particular, a sequential file or input device, such as COM1), for reading only. The current position pointer (see the SEEK( function) is set to the beginning of the file.\n\nIf filenum is not given, filenum=1 is assumed. If no file extension is given, the extension TXT is used.\n\nROPEN( returns T if successful, F otherwise.\n\nSee the functions WOPEN(, CLOSE(, READ(, WRITE(, IN(, OUT(, GET(, PUT(,SEEK(, and SSEEK(.\n\nExamples:\n\n```1>? ROPEN('a:label.prg')\nT```\n\nIn an Shark program, ROPEN( normally appears in an IF command:\n\n```IF ROPEN('file',2)\n? data\nENDDO\nok=CLOSE(2)\nENDIF\n```\n\n## ROW(\n\nGets print row position.\n\n` ROW() `\n\nType: numeric\n\nIn an MS-DOS system, this function gives the current row (line) position of the cursor; if the printer is on, it returns the column position of the printer head. See the commands SET PRINT ON and SET FORMAT TO PRINT, and the function COL(.\n\nExample:\n\n`@ ROW()+1,COL()+3 SAY 'Hello'`\n\nprints 'Hello' starting on the next line three characters to the right of the end of the last printing.\n\n## SEEK(\n\nGoes to a given character number in a plain text file.\n\n` SEEK(<num exp> [,filenum]) `\n\n<num exp> the character number\n\nOption:\n\n` <filenum> a Shark file number (between 1 and 4)`\n\nType: logical\n\nThis function repositions the character number pointer in the text file (opened with the ROPEN( or WOPEN( function) to the value given by <num exp>.\n\nIf filenum is not given, filenum=1 is assumed. If no file extension is given, the extension TXT is used.\n\nIf SEEK( is successful, it returns T (true); otherwise F (false). In a Shark program, SEEK( normally occurs in an IF or DO WHILE command.\n\nOnce the character pointer is properly positioned, use the GET( and PUT( functions to manipulate the characters.\n\nRelated functions: SSEEK(, ROPEN(, WOPEN(, CLOSE(, GET(, and PUT(.\n\nExample:\n\n```1>ok=ROPEN('a:label.prg',4)\n1>ok=SEEK(522,4)\n```\n\n## SET(\n\nSets a bit in a string to 1.\n\n```\nSET(<str exp>,bit position)\n\n<str exp> the string or string expression on which the\nfunction is to act\nbit position the number of the bit, numbered from the left\nstarting at 1, which is to be set to 1 ```\n\nType: logical\n\nA bit is any of the eight binary digits in a character's ASCII number representation. Each bit can have only one of two possible values, 0 and 1.\n\nThe SET( and RESET( functions are used to manipulate the bits within a string or string expression. SET( makes a bit 1, and RESET( makes a bit 0. The BIT( function tests the value of a specific bit.\n\nAmong the chief uses for these functions is compression of logical (true/false) data by using just one bit for each data item instead of an entire byte for a logical field or two bytes for a logical variable.\n\nSee the BIT( function for programming examples.\n\nExamples:\n\n```1>str='PQ'\n1>? BIT(str,15)\nF\n1>str=RESET(str,15)\n1>? str,BIT(str,15)\nPS T\n```\n\n## SIN(\n\nsine of <num exp> in radians\n\n## SINH(\n\nhyperbolic sine of <num exp>\n\n## SQRT(\n\nsquare root of <num exp>\n\n## SPACE(\n\nGets the amount of space left in the data space. SPACE()\n\nType: numeric\n\nThis function returns the available memory in the 64K data space (see Appendix A).\n\nExample:\n\n```1>? SPACE()\n27155.00\n```\n\nAs a general rule, when SPACE is down to less than 500.00, there is not a lot of room left for code & variables!\n\n## SSEEK(\n\nGoes to a given line number in a sequential file.\n\n```\nSSEEK(<num exp> [,<filenum>)\n\n<num exp> the line number ```\n\nOption:\n\n```\n<filenum> the DOS file number (between 1 and 4)```\n\nType: logical\n\nThis function repositions the line number pointer in the sequential file (opened with the ROPEN( function) to the value given by <num exp>.\n\nIf filenum is not given, filenum=1 is assumed. If no file extension is given, the extension TXT is used.\n\nIf filenum is not given, filenum=1 is assumed. If no file extension is given, the extension TXT is used.\n\nSee the functions SEEK(, ROPEN(, WOPEN(, CLOSE(, READ(, IN(, and OUT(.\n\nExample:\n\n```1>ok=ROPEN('a:label.prg',4)\n1>line=''\n1>ok=SSEEK(5,4)\n1>? ok\nT\n1>? line\nGOTO top\n1>ok=SSEEK(900,4)\n1>? ok\nF\n```\n\n## STR(\n\nType: character\n\nConverts a number to a string.\n\n```\nSTR(<num expression>,<width num expression>[,<decimals num expression<])\n\n<num expression> the number to be converted\n<width num expression> the width of the string\n```\n\nExample:\n\n```1> ? STR(1234,4)\n1> 1234\n1> ? STR(1234,12)\n1> 1234\n\n```\n\nOption:\n\n```\n<decimals num expression> the number of decimals\n```\n\nThis function gets a number by evaluating the <num expression>, and converts it into a string of given width. Optionally, the number of decimals can be specified (the default is 0). See also PIC(.\n\nExamples:\n\n```1>x=123.456\n1>? STR(x,8)\n123\n1>? LEN(STR(x,8))\n8.00\n1>? STR(x,10)\n123\n1>? STR(x,10,1); NOTE 'x' is the num expression; '10' is the width of the num expression; and '1' is the number of decimals\n123.4\n```\n\nWhen combined with the VAL( function, STR( is a convenient way of rounding decimal numbers with a given precision.\n\nFor example:\n\n```1>:PICTURE='9999.99999'\n1>a=29.95748\n1>? a\n29.95748\n1>? VAL(STR(a+.005,10,2))\n29.96000\n1>? VAL(STR(a+.00005,10,4))\n29.95750\n```\n\n## SUBSTR(\n\nType: character\n\nGets a substring of a string.\n\n```\nSUBSTR(<str exp>, <start num exp>, <width num exp>)\n\n<str exp> the string from which the new string is formed\n<start num exp> the position from which the new string is taken\n<width num exp> the number of characters to place in the new string\n```\n\nThis function, a synonym for the \\$( function, takes the string in <str exp> from position <start num exp> (fractions are thrown away); the number of characters taken is <width num exp>. (In both numeric expressions, the fractions are disregarded).\n\nExamples:\n\n```1>name='David Barberr'\n1>? SUBSTR(name, 7,3)\nBar\n1>? SUBSTR(name, 7,12)\nBarberr\n1>? LEN(SUBSTR(name,7,12))\n7.00\n```\n\nNote that SUBSTR(name,7,12) is of width 7, not 12; there are only 7 letters left in name from position 7.\n\n```1>s=3\n1>t=1\n1>? SUBSTR(name+name,(s+t)/2,1.9)\na\n```\n\nNote that 1.9 was taken as 1.\n\n## TAN(\n\ntangent of <num exp> in radians\n\n## TANH(\n\nhyperbolic tangent of <num exp>\n\n## TEST(\n\nTests a string whether it is a valid expression.\n\n```TEST(<str exp>)\n\n<str exp> the string to be tested\n\nType: logical```\n\nThis function tests the string in <str exp> as to whether it is a valid expression; in particular, all variables must be defined and must be of the proper type. It returns T if it is, F otherwise. If the test is successful, TYPE( can be used to find the type of the expression.\n\nIn a Shark program, use TEST( to find out whether a selection criteria typed in by the user is correct. Or use it to ensure that a variable exists in a subroutine that may be called from several programs.\n\nExamples:\n\n```1>? TEST('check=0')\nF false because check is not defined\n1>check=0\n1>? TEST('check=0')\nT\n1>? TEST('check=0.or.check=1')\nT\n1>? TEST('check=\"A\".or.check=1')\nF false because \"A\" is of character\ntype\n1>? TEST('num')\nF\n1>num=5\n1>? TEST('num')\nT\n1>? TEST('num+') false because the second operand\nis missing\nF\nIF .NOT. TEST('check')\ncheck=0\nENDIF\n```\n\n## TIME(\n\nGets the system time.\n\n```\nTIME(type)```\n\nOption:\n\n```\ntype one of three types of information requested, as listed below ```\n\nType: character\n\nThis function returns a string containing the current system time, and changes the format of the system variable :TIME to this format. (See the system variable :TIME in Section 2. Recall that :TIME is initialized when Shark starts up, but can be reinitialized with a :TIME= command.)\n\nThe type can be given in either of two forms, a name or number (numeric expression) as follows:\n\n```Type Explanation Example\n1 or HMS Hours,minutes,seconds,hundredths 17:26:36.07\n2 or AMPM Hours,minutes with AM/PM 5:36 pm\n3 or Seconds Seconds since midnight 62804.80```\n\nShortcut: When specifying type by name, only the first character is required. The default parameter is 1, the time in the 24-hour format hh:mm:ss.hh.\n\nExamples:\n\n```1>? :TIME\n15:45:59\n1>? TIME(2)\n3:46 pm\n1>:TIME\n3:46 pm\nstart=VAL(TIME(seconds))\nprogram lines\nfinish=VAL(TIME(seconds))\n? 'Program execution took',start-finish,'seconds.'```\n\nThis displays how long the running of <program lines> took, in seconds.\n\n## TRIM(\n\nTrims blanks from the right-hand side of a string.\n\n```\nTRIM(<str exp>)\n\n<str exp> the string to be trimmed ```\n\nType: character\n\nShark stores strings in fields padded on the right with blanks. In actual use, these blanks may get in the way. TRIM( gets rid of the blanks on the right of a string. TRIM( can be used in the key of an index. See also LTRIM( .\n\nExamples:\n\n```1>a='Vancouver '\n1>b='BC'\n1>? a+','+b\nVancouver ,BC\n1>? LEN(a)\n14.00\n1>? trim(a)+','+b\nVancouver,BC\n1>? LEN(TRIM(a))\n9.00\n1>blank=' '\n1>? LEN(TRIM(blank))\n1.00```\n\nNote: TRIM(blank) is a single blank.\n\n## TYPE(\n\nGets the type of an expression.\n\n` TYPE(exp)`\nexp = any expression\n\nType: character\n\nFor an expression, exp, this function returns the type of the expression as a one character string:\n\n```C for character\nL for logical\nN for numeric\nM for memo\nF for float ;a type used by other xBase languages; treated as N internally\nD for date ;a type used by other xBase languages; treated as C internally\nU Undefined ;created by GLOBAL or VARIABLES command, but not yet given\na value or type```\n\nTo test whether a string is a valid expression, use the TEST( function.\n\nExamples:\n\n```1>a='name'\n1>? TYPE(a)\nC\n1>? TYPE(a+a)\nC\n1>n=12\n1>? TYPE(a+n)\n1. Invalid variable type found when executing an expression.\n1>? TYPE(a<a)\nL\n1>? TYPE(a<a .OR. n<5)\nL\n1>? TYPE(note)\nM\n1>? TYPE(date)\nD\n1>TYPE(n+5/10)\nN\n```\n\n## UPPER(\n\n```\nConverts a string to upper case.\n\nUPPER(<str exp>)\n\n<str exp> the text to be converted to upper case\n\nType: character\n\nAll lower-case letters in the <str exp> are converted into upper case by\nthe UPPER( function, which is a synonym for the !( function. See also the\nLOWER( function.\n\nExamples:\n\n1>a='Aa12b'\n1>? UPPER(a)\nAA12B\n1>? UPPER('David!')\nDAVID!\nNote that only the lower-case letters are changed.\n```\n\n## VAL(\n\n```\nConverts a string to its numeric value.\n\nVAL(<str exp>)\n\n<str exp> the string to be evaluated\n\nType: numeric\n\nThis function takes the string <str exp> which is a number, and returns\nit as a number. If the whole string cannot be interpreted as a number, it\ntakes as much of the front part as it can.\n\nExamples:\n\n1>a='123.23'\n1>? VAL(a)\n123.23\n1>? VAL(123.23)\n1. Invalid variable type found when executing an expression.\n1>? VAL('a12')\n0.00\n1>? VAL('12a')\n12.00\n1>? VAL(DATE(2))\nyields the current month as a number.\n```\n\n## WOPEN(\n\n```\nOpens a DOS file for writing.\n\nWOPEN(<str exp>[,<filenum>])\n\n<str exp> the file name\n\nOption:\n\n<filenum> the DOS file number (between 1 and 4)\n\nType: logical\n\nThis function opens the DOS file, and in particular, the sequential file\n<str exp> (or output device, such as COM1:) for writing; if the file does not\nexist, it will be created. If filenum is not given, filenum=1 is assumed. If\nno file extension is given, the extension TXT is used. It returns T if\nsuccessful, F otherwise.\n\nSee the functions ROPEN(, CLOSE(, GET(, PUT(, SEEK(, SSEEK(, WRITE(,\n\nExample:\n1>ok=WOPEN('a:label.prg')\n1>? ok\nT\n\nIn an Shark program, WOPEN( normally appears in an IF command:\n\nIF WOPEN('file',2)\n\n2. Two programs to print a text file, TEST (in the second version it is\nassumed that TEST has no more than 20 lines):\n\nSET WIDTH TO 80\nSET PRINT ON\nIF ROPEN('test')\n? line\nENDDO\nok=CLOSE()\nENDIF\n```\n\n## WRAP(\n\n```\nWraps text for output.\n\nWRAP(str var, <num exp>)\n\n<str exp> the string to be wrapped\n<num exp> the line width\n\nType: character\n\nThis function returns one line of text with word wrapping from the string\nin str var; str var now contains what is left of the string. In other\nwords, the string returned will contain as many words as will fit in a line\n(the line width is given by <num exp>). If the whole contents of str var is\none line, str var becomes the blank string (of length 1)\n\nWRAP(, IN(, GET(, and READ( are the only functions that change the\ncontents of the memory variable used as an argument.\n\nExamples:\n1>text='This text line is going to be wrapped in a printed line of width 30.'\n1>? text\nThis text line is going to be wrapped in a printed line of width 30.\n1>temp=text\n1>WRAP(temp,30)\nThis text line is going to be\n1>? temp\nwrapped in a printed line of width 30.\n1>WRAP(temp,30)\nwrapped in a printed line of\n1>? temp\nwidth 30.\n1>WRAP(temp,30)\nwidth 30.\n1>? temp\n1>? LEN(temp)\n1.00\n```\n\n## WRITE(\n\n```\nWrites a new line into the sequential file.\n\nWRITE(str var[,<filenum>])\n\nstr var the line to be written\n\nOption:\n\n<filenum> the DOS file number (between 1 and 4)\n\nType: logical\n\nThis function writes (appends) the contents of the string variable\nstr var to the end of a sequential file opened with the WOPEN( function.\nIf filenum is not given, filenum=1 is assumed. WRITE( returns T if\nsuccessful, F otherwise.\nSee the functions ROPEN(, WOPEN(, READ(, IN(, OUT(, GET(, PUT(, and CLOSE(.\n\nExample:\n\n1>ok=WOPEN('customer.frm',2)\n1>ok=WRITE('FIELDS - cust,orderno,amount',2)\n1>ok=CLOSE(2)\ncreates the CUSTOMER.FRM report form file:\nFIELDS - cust,orderno,amount\nNow the command REPORT CUSTOMER will run the report. This example\nillustrates how a program can be written that creates a report form file.\nIn an Shark program, WRITE( normally appears in an IF command:\nIF WRITE('file',2)\n\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7461294,"math_prob":0.89724463,"size":90035,"snap":"2023-40-2023-50","text_gpt3_token_len":24623,"char_repetition_ratio":0.168031,"word_repetition_ratio":0.26417378,"special_character_ratio":0.28135726,"punctuation_ratio":0.14196177,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9548307,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-05T15:57:12Z\",\"WARC-Record-ID\":\"<urn:uuid:428071d9-6140-473b-9c5e-296289164f24>\",\"Content-Length\":\"137328\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:75753a43-7c60-44c7-bb8a-74235fb84d3a>\",\"WARC-Concurrent-To\":\"<urn:uuid:17e56925-1832-43d4-8a69-36d9ea19c8b5>\",\"WARC-IP-Address\":\"23.253.205.12\",\"WARC-Target-URI\":\"https://www.sharkbase.ca/shark-functions.html\",\"WARC-Payload-Digest\":\"sha1:4WNR232WUATJOJOYNPOC7EUVU2V3STRT\",\"WARC-Block-Digest\":\"sha1:B5GOVCJBY6YKIJ65WH7MNTRZ2VGD5HO4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100551.2_warc_CC-MAIN-20231205140836-20231205170836-00124.warc.gz\"}"}
https://answers.everydaycalculation.com/subtract-fractions/50-35-minus-30-10	[ "Solutions by everydaycalculation.com\n\n## Subtract 30/10 from 50/35\n\n1st number: 1 15/35, 2nd number: 3 0/10\n\n50/35 - 30/10 is -11/7.\n\n#### Steps for subtracting fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 35 and 10 is 70\n2. For the 1st fraction, since 35 × 2 = 70,\n50/35 = 50 × 2/35 × 2 = 100/70\n3. Likewise, for the 2nd fraction, since 10 × 7 = 70,\n30/10 = 30 × 7/10 × 7 = 210/70\n4. Subtract the two fractions:\n100/70 - 210/70 = 100 - 210/70 = -110/70\n5. After reducing the fraction, the answer is -11/7\n\n#### Subtract Fractions Calculator\n\n-\n\nUse fraction calculator with our all-in-one calculator app: Download for Android, Download for iOS" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6325994,"math_prob":0.9959787,"size":303,"snap":"2019-35-2019-39","text_gpt3_token_len":116,"char_repetition_ratio":0.18060201,"word_repetition_ratio":0.0,"special_character_ratio":0.46864685,"punctuation_ratio":0.12857144,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99940753,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-15T07:46:36Z\",\"WARC-Record-ID\":\"<urn:uuid:6f896ef7-2e50-438c-ac8d-c0ada81358c8>\",\"Content-Length\":\"7709\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:30cbc126-46a5-4c4b-b6a8-b316c649fde5>\",\"WARC-Concurrent-To\":\"<urn:uuid:f3e4a518-2eb1-4ae0-afab-f0314ffa937f>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/subtract-fractions/50-35-minus-30-10\",\"WARC-Payload-Digest\":\"sha1:T7OKU6YPX5EZJVCY5HBRC3F3RNYR73CI\",\"WARC-Block-Digest\":\"sha1:6TGPZSNYJ7GLPJR4HFGSZPHEQJRHYPCT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514570830.42_warc_CC-MAIN-20190915072355-20190915094355-00095.warc.gz\"}"}
https://discuss.codechef.com/t/codechef-is-broken/75632	[ "", null, "# CodeChef is broken\n\nJust try to run this code in the Problem: HS08TEST\n\nIt gives a runtime error of NZEC but when I submitted it. CodeChef accepted the code. CodeChef is broken. This wasted my 2 days and now I won’t be using this platform.\n\nimport java.util.;\nimport java.lang.\n;\nimport java.io.;\nimport java.text.\n;\n\n/* Name of the class has to be “Main” only if the class is public. */\nclass Codechef\n{\npublic static void main (String[] args) throws java.lang.Exception\n{\nint amount;\ndouble balance;\nScanner scanner = new Scanner(System.in);\namount = scanner.nextInt();\nbalance = scanner.nextDouble();\nDecimalFormat df = new DecimalFormat(\"###.00\");\nif((amount>0 && amount<=2000) && (balance>=0 && balance<=2000)){\nif(amount%5==0 && amount<=balance && (amount+0.5<=balance)){\nbalance = balance-amount;\nbalance = balance-0.50;\nSystem.out.println(df.format(balance));\n}\nelse{\nSystem.out.println(df.format(balance));\n}\n}else{\nSystem.out.println(df.format(balance));\n}\n\n``````}\n``````\n\n}\n\nRunning without input it does give random error sometimes, and that is not a problem ,you have to give custom input before you run on codechef ide.\n\n1 Like\n\nbhai, if you run this code on codechef ide then you have to give custom inputs, otherwise it will give you NZEC, i also wasted so many days figuring this out.\nso, whenever you run any code, and if it requires any input(like amount and balance is required in your code) then do give custom input then run, else directly try to submit.\n\nprovide custom input before running the program." ]	[ null, "https://s3.amazonaws.com/discourseproduction/original/3X/7/f/7ffd6e5e45912aba9f6a1a33447d6baae049de81.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.54297686,"math_prob":0.8135703,"size":968,"snap":"2020-34-2020-40","text_gpt3_token_len":237,"char_repetition_ratio":0.1493776,"word_repetition_ratio":0.0,"special_character_ratio":0.30061984,"punctuation_ratio":0.23232323,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9766128,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-30T16:32:52Z\",\"WARC-Record-ID\":\"<urn:uuid:8da62f43-1098-4b6f-9069-a1decd569044>\",\"Content-Length\":\"18614\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d359378d-bca6-4e69-8909-783bdcc062e4>\",\"WARC-Concurrent-To\":\"<urn:uuid:acef7c43-5ae5-4b53-89ca-0420e3d6692f>\",\"WARC-IP-Address\":\"18.213.158.143\",\"WARC-Target-URI\":\"https://discuss.codechef.com/t/codechef-is-broken/75632\",\"WARC-Payload-Digest\":\"sha1:LKKOGDONRLOHZXTDD66RZRZ6RUGKNP2P\",\"WARC-Block-Digest\":\"sha1:PGCTYNT3VGGAASTV3ZFB7FSVWXUT4X2A\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600402127075.68_warc_CC-MAIN-20200930141310-20200930171310-00396.warc.gz\"}"}
https://www.helpteaching.com/questions/115558/what-fraction-is-equal-to-3x5x4-x2	[ "##### Question Info\n\nThis question is public and is used in 192 tests or worksheets.\n\nType: Multiple-Choice\nCategory: Fractions and Ratios\nScore: 9\nAuthor: jlasbell\n\nView all questions by jlasbell.\n\n# Fractions and Ratios Question\n\nView this question.\n\nAdd this question to a group or test by clicking the appropriate button below.\n\n## Grade 7 Fractions and Ratios\n\nWhat fraction is equal to $((3x)/5)/(x/4 +x/2) ?$\n1. $x^2/5$\n2. $(9x^2)/20$\n3. $4/5$\n4. $9/5$\nYou need to have at least 5 reputation to vote a question down. Learn How To Earn Badges." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7413716,"math_prob":0.68358326,"size":527,"snap":"2019-51-2020-05","text_gpt3_token_len":150,"char_repetition_ratio":0.17590822,"word_repetition_ratio":0.33766234,"special_character_ratio":0.26565465,"punctuation_ratio":0.115384616,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97613543,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-10T22:41:31Z\",\"WARC-Record-ID\":\"<urn:uuid:67b8e8ef-e20b-4914-8c69-af40cac79a89>\",\"Content-Length\":\"16941\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2bf7d9aa-cf9b-4c02-b77b-26c7083a15c1>\",\"WARC-Concurrent-To\":\"<urn:uuid:65dcbe09-1fbc-49d0-8706-83aa0f5e9547>\",\"WARC-IP-Address\":\"52.6.170.220\",\"WARC-Target-URI\":\"https://www.helpteaching.com/questions/115558/what-fraction-is-equal-to-3x5x4-x2\",\"WARC-Payload-Digest\":\"sha1:ESHF6DUO2MLQDLJ435HGUPIDNQZPU4IA\",\"WARC-Block-Digest\":\"sha1:SP3EJLRZ2URO35CGVEVLBQ2AD6N2MYFR\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540529006.88_warc_CC-MAIN-20191210205200-20191210233200-00502.warc.gz\"}"}
https://www.gradesaver.com/textbooks/math/algebra/elementary-linear-algebra-7th-edition/chapter-4-vector-spaces-4-3-subspaces-of-vector-spaces-4-3-exercises-page-167/5	[ "## Elementary Linear Algebra 7th Edition\n\nSince the sum of continuous functions is continuous and the multiplication of a continuous function by a scalar is continuous, then $W$ the set of all functions that are continuous on $[-1,1]$ is a vector subspace of the set of all functions that are integrable on $[-1,1]$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9308023,"math_prob":0.99851465,"size":328,"snap":"2019-51-2020-05","text_gpt3_token_len":80,"char_repetition_ratio":0.19444445,"word_repetition_ratio":0.14545454,"special_character_ratio":0.24085365,"punctuation_ratio":0.07575758,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9830679,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-12T22:03:59Z\",\"WARC-Record-ID\":\"<urn:uuid:ff38646a-4738-47e2-9aad-51bd32468e42>\",\"Content-Length\":\"68308\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cfa03205-a948-4f4a-b73e-83c48a1f7b3d>\",\"WARC-Concurrent-To\":\"<urn:uuid:f64fc12a-2ec8-4a32-a164-b5d234841617>\",\"WARC-IP-Address\":\"52.73.200.166\",\"WARC-Target-URI\":\"https://www.gradesaver.com/textbooks/math/algebra/elementary-linear-algebra-7th-edition/chapter-4-vector-spaces-4-3-subspaces-of-vector-spaces-4-3-exercises-page-167/5\",\"WARC-Payload-Digest\":\"sha1:VNRBX6PELJZAEDLMNXBA7TVD2CUFQM3R\",\"WARC-Block-Digest\":\"sha1:5DGO6LPZNI2DPVQBWST5SRBKVASHDUJC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540547165.98_warc_CC-MAIN-20191212205036-20191212233036-00453.warc.gz\"}"}
https://electricalbaba.com/significance-of-burden-in-potential-transformer/	[ "# Significance of Burden in Potential Transformer\n\nBefore going into the impact of Burden on the performance of a Potential transformer, we will discuss about Burden. The Burden of an Instrument Transformer is the rated Volt-Ampere loading which is permissible without errors exceeding the limits for a particular class of Instrument Transformer.\n\nBasically, there are two Classes of an Instrument Transformer, one is Protection Class and another one is Metering Class. Protection Class (PS) is one which is used for the protection scheme where as Metering Class is used for the purpose of Metering.\n\nNow, we will discuss about errors in a Potential Transformer. There are two types of error in a Potential Transformer, Ratio Error and Phase Angle Error.\n\n### Ratio Error:\n\nTransformation Ratio of a Potential Transformer varies with the operating condition and therefore the voltage induced in the secondary of a Potential Transformer will also vary. The error in secondary voltage of a Potential Transformer is defined as\n\n% Ration Error = (Kn-R)×100/R\n\nWhere Kn = Rated primary voltage / Rated secondary voltage\n\nR = Primary Voltage / Secondary Voltage\n\n### Phase Angle Error:\n\nIn an ideal Potential Transformer, there should not be any phase difference between the primary voltage and the secondary voltage reversed. But in a practical Potential Transformer there exists a phase difference between them.\n\nThe important thing to note that, Ratio error is of importance when we use PT for Protection purpose but when we use PT for Metering then both Ration and Phase Angle Error are important to consider.\n\nNow we are on the stage to discuss the impact of Burden on the performance of a Potential Transformer.\n\n### Impact of Burden on the Performance of PT:\n\nThe effects of Burden on the performance of a Potential Transformer are as follows:\n\nIf we increase the Burden, the secondary as well as primary current will increase but as the primary voltage will remain constant (because primary of a PT is connected to the line), the secondary voltage will decrease. Therefore, the voltage ratio will increase. Thus we see that increasing the Burden, increases the Error in the Ratio. Also, as the power factor of the Burden reduces the transformation ratio of Potential Transformer increases.\n\nPT accuracy performance changes linearly with burden and can be plotted as\n\nHere RCF is Ratio Correction Factor, which is defined as defined as the factor that when multiplied by the potential transformer output will yield the correct result.\n\nMathematically,\n\nRCF = Transformation Ratio/Nominal Ratio =R / Kn\n\nThe transformer nameplates show a “marked ratio” usually an even number, such as 20 to 1. The actual ratio of primary to secondary quantity may be slightly higher or lower than the marked value by an amount 1 called ratio error. For example, if the actual ratio is 20.2 to 1, then the RCF is 1.01 and the ratio error is 1%.\n\nRCF = 20.2/20 = 1.01\n\nSo, % Ratio Error = (RCF-1)×100 = 1%\n\n### 4 thoughts on “Significance of Burden in Potential Transformer”\n\n1.", null, "Thanks so much! The term \"burden\" is not that common in latam, so this was very handy!\n\n2.", null, "RCF IS Kn/R, I THINK, INSTEAD OF R/Kn\n\n3.", null, "4.", null, "" ]	[ null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%2050%2050'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%2050%2050'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%2050%2050'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%2050%2050'%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9037011,"math_prob":0.97085047,"size":3087,"snap":"2023-14-2023-23","text_gpt3_token_len":660,"char_repetition_ratio":0.17061304,"word_repetition_ratio":0.07321773,"special_character_ratio":0.20861678,"punctuation_ratio":0.08900524,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99319094,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-01T23:16:28Z\",\"WARC-Record-ID\":\"<urn:uuid:955fb7ab-283a-420b-b184-75044e93cafc>\",\"Content-Length\":\"68813\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:865fd9d4-9a52-41f7-baca-28c9e189f72f>\",\"WARC-Concurrent-To\":\"<urn:uuid:834db7a7-058d-40cb-8bc3-98ebccdcf302>\",\"WARC-IP-Address\":\"104.21.74.120\",\"WARC-Target-URI\":\"https://electricalbaba.com/significance-of-burden-in-potential-transformer/\",\"WARC-Payload-Digest\":\"sha1:4MJWUG63RQCK76OLKHLYHDCQ26VR6O5Z\",\"WARC-Block-Digest\":\"sha1:33NJ2YCHOL3DDQNSJ3CQ44GKRLANOS2L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224648209.30_warc_CC-MAIN-20230601211701-20230602001701-00480.warc.gz\"}"}
https://mafiadoc.com/measuring-market-power-in-the-greek-food-and-agecon-search_5ba3dd1b097c4749358b46ba.html	[ "## Measuring market power in the Greek food and ... - AgEcon Search\n\nAug 30, 2011 - Abstract: This paper measures the degree of market power of the Greek food and beverages manufacturing industry over the period ...\n\nMeasuring market power in the Greek food and beverages manufacturing industry Anthony N. Rezitis 1 Maria A. Kalantzi 2 1\n\nAssociate Professor, Department of Business Administration of Food and Agricultural Enterprises, University Western Greece, G. Seferi Str. 2, 30 100 Agrinio, Greece, E-mail: [email protected] 2 PhD Candidate, Department of Business Administration of Food and Agricultural Enterprises, University Western Greece, G. Seferi Str. 2, 30 100 Agrinio, Greece, E-mail: [email protected]\n\nPaper prepared for presentation at the EAAE 2011 Congress Change and Uncertainty Challenges for Agriculture, Food and Natural Resources August 30 to September 2, 2011 ETH Zurich, Zurich, Switzerland\n\nCopyright 2011 by [Rezitis Anthony and Kalantzi Maria]. All rights reserved. Readers may make verbatim copies of this document for non-commercial purposes by any means, provided that this copyright notice appears on all such copies. Abstract: This paper measures the degree of market power of the Greek food and beverages manufacturing industry over the period 1983–2007 at the three-digit SIC level. The present study also estimates the “deadweight” loss and the reduction of consumers’ income due to the possible existence of market power in the Greek food and beverages manufacturing industry. Based on Bresnahan’s (1989) conjectural variation model, three different approaches are used to investigate competitive conditions of the Greek food and beverages manufacturing industry. The first approach assesses the extent of market power of the whole industry over the period 1983–2007; the second approach tests the degree of market power in each one of the nine sectors of the industry over the whole period, i.e. 1983–2007; and the third one estimates the extent of market power for the whole Greek food and beverages manufacturing industry for specific sub-periods of the period 1983–2007. The methodology of Dickson and Yu (1989) is adopted to measure the welfare losses. The empirical results indicate the presence of some degree of market power in the whole Greek food and beverages manufacturing industry as well as in each one sector of the industry during the period 1983– 2007 and, as a result, the existence of welfare losses. In addition, the empirical findings support the presence of some degree of market power for each sub-period of the period 1983– 2007 in the whole Greek food and beverages manufacturing industry and the existence of welfare losses. Keywords: Conjectural variation, Greek food and beverages manufacturing industry, Market power; Welfare losses JEL classification: D43, D60, L66, Q10\n\n1\n\n2\n\nvariation model of competition, a manufacturing industry is considered in which firms face a demand function p ≡ p (Y , z ) , with Y = ∑ yi , and i = 1,..., n where p is the output price, yi i\n\nrepresents the quantity supplied by firm i , n is the number of firms and z is a vector of exogenous factors affecting the demand curve. Thus, the profit maximization problem of firm i is given as: (1) max π i = p (Y , z ) ⋅ yi − C ( yi , wi )\n\nwhere C ( yi , wi ) is the cost function of firm i and wi is a vector of input prices of firm i. The first order condition of the profit maximization problem (1) is given as: ∂π i ∂C ∂p ∂Y = p+ ⋅ yi − i = 0 (2) ∂yi ∂Y ∂yi ∂yi Rearranging Eq. (2), the following expression is obtained: ⎛ ∂p Y ⎞ ⎛ ∂Y yi ⎞ p − MCi θ (3) = −⎜ ⋅ ⎟⋅⎜ ⋅ ⎟=− i p h ⎝ ∂Y p ⎠ ⎝ ∂yi Y ⎠\n\nwhere MCi is the marginal cost of firm i, h ≡ ( ∂Y ∂p ) (Y p ) < 0 is the price elasticity of output demand and θ i ≡ ( ∂Y ∂yi ) (Y yi ) is the conjectural variation elasticity of firm i which represents the reaction of total industry to the change of output of firm i and it is a measure of competition. When θi takes the value of zero for all firms, then the industry is under conditions of perfect competition, while the value of one indicates a monopolistic market. Values of θi between zero and one support the presence of Cournot oligopoly in the food and beverages manufacturing industry. According to Bresnahan (1989), multiplying equation (3) by yi Ci , summing over i and rearranging, the following expression, i.e. the supply function, is obtained: f ⎞ ⎛ (4) S y ⎜1 + ⎟ = MC h⎠ ⎝ where S y is the ratio of aggregate revenue to total cost, MC is the industry-level (weighted) f p − MC 1 . According to Cowling and Waterson (1976), the average = h p degree of competition parameter f ( with 0 ≤ f ≤ 1) measures the average deviation of\n\nmarginal cost and −\n\nfirms’ behavior from the monopolistic case and, if properly identified in the estimation process, expresses the true degree of market power exerted by firms, with f = 1 indicating monopolistic market power, 0 < f < 1 Cournot oligopoly, and f = 0 perfect competition. Furthermore, a specified cost function, i.e. C ( yi , wi ) and a cost share equation for labor are added to the empirical system in order to improve the precision of the estimates. Following the methodology of Dickson and Yu (1989), the industry demand curve is h represented by Y = 1 p , where h is the absolute value of the demand elasticity, h. In\n\n( )\n\nε\n\naddition, the weighted industry marginal cost curve MC is presented by Y = MC , where ε is the inverse of the weighted industry marginal cost elasticity. Based on the Lerner\n\n1\n\nL = − f h , where\n\nL is the Lerner index which is the relative mark-up or the price-cost margin. 3\n\n(\n\nindex, L = po − MC\n\n)\n\npo = f h , the oligopoly price ( po ) and the oligopoly output (Yo ) are\n\ngiven as:\n\n⎛ h ⎞ 1ε po = ⎜⎜ ⎟⎟ Yo h f − ⎝ ⎠\n\n(5) hε\n\n⎛ h − f ⎞ h +ε h (6) Yo = 1 po = ⎜⎜ ⎟⎟ h ⎝ ⎠ The net loss of welfare (deadweight loss or Harberger loss) due to the existence of oligopoly is depicted by: 1\n\n1 h WL = ∫ ⎡(1 Y ) − Y 1 ε ⎤dY ⎣ ⎦ Y H\n\n(7)\n\no\n\nThe reduction of consumers’ welfare due to the transfer of income from consumers to the monopolist in the case of an oligopoly (Tullock loss) is presented by: 1 h WLT = WLH + ⎡(1 Yo ) − Yo1 ε ⎤ Yo (8) ⎣ ⎦ 3. Model Formulation and Data Variables The translog specification for the total cost function with two inputs, one output and symmetry and linear homogeneity restrictions imposed is given below: ⎛ C ln ⎜ t ⎜W ⎝ k ,t\n\n⎞ 1 ⎟⎟ = a 0 + a y ln Yt + a yy ln Yt 2 ⎠\n\n⎛W 1 g ll ln ⎜ l ,t ⎜W 2 ⎝ k ,t\n\n(\n\n)\n\n2\n\n⎛W + g ly ln Yt ln ⎜ l ,t ⎜W ⎝ k ,t\n\n2\n\n⎞ ⎛ Wl ,t ⎟⎟ + xt Τ + xty Τ ln Yt + xtl Τ ln ⎜⎜ ⎠ ⎝ W k ,t\n\n⎞ ⎛ Wl ,t ⎟⎟ + a l ln ⎜⎜ ⎠ ⎝ W k ,t\n\n⎞ ⎟⎟ + ⎠\n\n(9)\n\n⎞ ⎟⎟ ⎠\n\nwhere Ct represents the industry-level cost, Wl,t, Wk,t correspond to the two input prices, i.e. labor and capital respectively; Yt corresponds to the output quantity of industry; T is the time trend. The convention for the translog function is followed by letting variables with upper bars, i.e. input prices and output, be normalized by their means to avoid possible multicolinearity. The demand function is denoted by: ⎛ p ⎞ ⎛ ⎞ It ln Yt = a + h ln ⎜ t × 100 ⎟ + z159 ln ⎜ × 100 ⎟ + ⎝ bt ⎠ ⎝ bt × P O Pt ⎠\n\n⎛ ⎛ ⎞⎞ It z s ⎜⎜ D S s × ln ⎜ × 100 ⎟ ⎟⎟ s = 151 ⎝ bt × P O Pt ⎠⎠ ⎝ 158\n\n(10)\n\nwhere h is the industry demand elasticity, pt is the output price, bt is a price deflator, I t is the Gross National Product, POPt is the population of Greece, z159 is the income demand elasticity of sector 159 where sector 159 is the manufacture of beverages as referred to Table 1 and DSs (s=151,…,158) is a dummy variable, which is set to one for the s sector and zero otherwise in order to account for possible differences in the income demand elasticity among the sectors of the food and beverages manufacturing industry. Note that the s sectors are defined in Table 1. Furthermore, zs (s=151,…,158) refers to the change in the income demand elasticity of s sector (s=151,…,158) with respect to the sector 159, i.e. the manufacture of beverages. Applying Shephard’s Lemma to the cost function (9), the cost share equation for labor is given as: ⎛W ⎞ Sl ,t = al + gly ln Y t + gll ln ⎜ l ,t ⎟ + xtlT (11) ⎜W ⎟ ⎝ k ,t ⎠ 4\n\nand the cost share equation for capital is the following: ⎛W ⎞ S k ,t = ak + g ky ln Yt + g kl ln ⎜ l ,t ⎟ + xtk T (12) ⎜W ⎟ ⎝ k ,t ⎠ Note that since the sum of the dependent variables over the two cost share equations (11) and (12) always equals 1 then only one factor share equation is linearly independent. As a result, it is necessary to omit one equation (equation for capital) to avoid singularity of the estimated covariance matrix. Three different approaches are used in the present study to investigate competitive conditions in the Greek food and beverages manufacturing industry. The first approach is based on the estimated system of equations (9), (10), (11) and the following supply function (13) where a unique estimate of the degree of market power ( f15 ) is obtained for the whole industry over the period 1983–2007:\n\n⎛W f ⎞ ⎛ SY ,t ⎜1 + 15 ⎟ = a y + a yy ln Yt + gly ln ⎜ l ,t ⎜W h ⎠ ⎝ ⎝ k ,t where SY ,t\n\n⎞ (13) ⎟⎟ + xtyT ⎠ is the ratio of the industry’s aggregate revenue to total cost and f15 is the\n\nconjectural variation elasticity for the whole food and beverages manufacturing industry. The second approach investigates two related estimated equation systems. The first estimated system includes the equations (9), (10), (11) and the following supply function (14) where the f parameter is allowed to change among the sectors of the food and beverages manufacturing industry while it is unchanged over time: 158 ⎛W ⎞ f f SY ,t + 159 SY ,t + ∑ s ( DS s × SY ,t ) = a y + a yy ln Yt + gly ln ⎜ l ,t ⎟ + xtyT (14) ⎜W ⎟ h s =151 h ⎝ k ,t ⎠ where f159 is the conjectural variation elasticity of sector 159, i.e. the manufacture of beverages and f s ( s = 151,...,158 ) refers to the change in the conjectural variation elasticity of s sector ( s = 151,...,158 ) with respect to sector 159. In addition, DS s\n\n( s = 151,...,158)\n\nis a\n\ndummy variable, which is set to one for the s sector and zero otherwise (Table 1). The empirical results of the aforementioned estimated system, provided in the next section, indicate that all of the sectors have the same conjectural variation elasticity, f, except that of sector 157, i.e. manufacture of prepared animal feeds. Thus, the supply function (14) is respecified and a slightly modified system (second estimated system) is estimated under the second approach. More analytically, the second estimated system of the second approach contains the equations (9), (10), (11) and the following supply function (15) where the estimate of the degree of market power ( f16 ) is the same for all sectors of the food and beverages manufacturing industry except that of sector 157 (manufacture of prepared animal feeds ): ⎛W ⎞ f f SY ,t + 16 SY ,t + 157 ( DS157 × SY ,t ) = a y + a yy ln Yt + gly ln ⎜ l ,t ⎟ + xtyT (15) ⎜W ⎟ h h k t , ⎝ ⎠ where f16 refers to the conjectural variation elasticity which is the same across all sectors of the Greek food and beverages manufacturing industry except that of sector 157, f157 is the change in the conjectural variation elasticity of the sector 157 with respect to the other sectors of manufacturing, i.e. f16 , and DS157 is the dummy variable corresponding to sector 157.\n\n5\n\nThe third approach is based on the estimated system of equations (9), (10), (11) and the following supply function (16) which allows f to change for each sub-period of the period 1983–2007 but to remain the same among sectors: 8 ⎛W ⎞ f f SY ,t + 1 SY ,t + ∑ t ( DTt × SY ,t ) = a y + a yy ln Yt + gly ln ⎜ l ,t ⎟ + xtyT (16) ⎜W ⎟ h t =2 h k t , ⎝ ⎠ where f1 refers to the conjectural variation elasticity of the sub-period t=1, i.e. the years 1983–1985, and DTt ( t = 2,...,8 ) is a dummy variable, which is set to one for the sub-period t\n\n( t = 2,...,8 ) refers to the change of the conjectural ( t = 2,...,8 ) with respect to the sub-period t = 1 , i.e. 1983–\n\nand zero otherwise.2 Note also that ft variation elasticity of sub-period t\n\n1985. The sample comprised annual data for the period 1983–2007 for nine sectors at the three-digit SIC level of the Greek food and beverages manufacturing industry, i.e. SIC: 151159, based on the Statistical Nomenclature of Economic Activity of 2003 (STAKOD_2003). The data used in the estimation was obtained by the Annual National Industrial Survey of the Hellenic Statistical Authority (EL.STAT.) The nine sectors of the Greek food and beverages manufacturing industry are presented in Table 1. 4. Empirical Results The empirical results of the three different approaches used to measure the degree of market power in the Greek food and beverages manufacturing industry are presented in Table 2. The equation systems of the three different approaches are estimated using the non-linear three-stage least square (NL3SLS) estimation technique. The econometric analysis is conducted using Shazam 9.0 software. The empirical findings of the three different approaches are plausible and consistent with economic theory in terms of the signs and the magnitudes of the coefficients and indicate that the translog cost function satisfies the restrictions of monotonicity and concavity at the sample mean (Table 2).3 According to the first approach, the empirical results indicate that all the estimated coefficients of the translog cost function are statistically significant at any conventional level of significance (Table 2, First Approach). Furthermore, the results support that the cost shares of labor ( al = 0.7779 ) and capital ( ak = 0.2222 ) are statistically\n\nsignificant at any conventional level of significance. These labor and capital share estimates are similar to the corresponding mean labor and capital cost shares calculated from the data, i.e. 0.7071 and 0.2929 respectively. The results of the supply function reveal that the conjectural variation elasticity ( f15 = 0.7104 ) is statistically significant at any conventional level of significance and its value is ranged between zero and one. These results imply the presence of some degree of market power in the Greek food and beverages manufacturing industry over the period 1983–2007. The estimated results of the demand function indicate that first, the price elasticity of output demand ( h = -0.7130 ) is statistically significant at any conventional level of significance; second, the income demand elasticity of sector 159 2\n\nIt is noted that the sub-period 2 (t=2) corresponds to the period 1986–1988, the sub-period 3 (t=3) to the period 1989–1991, the sub-period 4 (t=4) to the period 1992–1994, the sub-period 5 (t=5) to the period 1995– 1997, the sub-period 6 (t=6) to the period 1998–2000, the sub-period 7 (t=7) to the period 2001–2003 and the sub-period 8 (t=8 ) to the period 2004–2007. 3 The monotonicity restriction at the sample mean implies that al>0 and ak>0, while the concavity restriction implies that the Hessian matrix is negative semidefinite, i.e. all Allen-Uzawa own-partial elasticities of substitution are negative at the sample mean.\n\n6\n\n( z159 = 0.3969 ) ,\n\ni.e. the manufacture of beverages, is statistically significant; and third the\n\nchange of the income demand elasticity of each one of the sectors of the food and beverages manufacturing industry, zs ( s = 151,...,158 ) , with respect to sector 159, i.e. the manufacture of beverages, is statistically significant at any conventional level of significance. Furthermore, the results regarding scale economies imply the presence of increasing returns to scale. Finally, relative to the whole Greek food and beverages manufacturing industry, the Harberger loss is about €96.53 million in terms of 2007 value added (or 2.74% of the 2007 value added) whereas the Tullock loss is about €176.51 million in terms of 2007 value added (or 5.01% of the 2007 value added) (Table 3). The empirical findings of the first estimated system of the second approach show that most of the estimated parameters of the translog cost function are statistically significant at any conventional level of significance (Table 2, Second Approach, First Estimated System). In addition, the results support that the cost shares of labor ( al = 0.7706 ) and capital\n\n( ak = 0.2294 ) are statistically significant at any conventional level of significance. Moreover, the results of the supply function show that only sector 157, i.e. the manufacture of prepared animal feeds, has a statistically different degree of market power from sector 159, i.e. the manufacture of beverages, since only the change in the conjectural variation elasticity of sector 157 ( f157 ) with respect to sector 159 is statistically significant. These results are reinforced by the fact that the Wald test does not reject the null hypothesis which supports that all sectors, except sector 157, have the same degree of market power as that of sector 159.4 The estimated results of the demand function indicate that first, the price elasticity of output demand ( h = -0.6782 ) is statistically significant at any conventional level of significance; second, the income demand elasticity of sector 159\n\n( z159 = 0.4029 ) ,\n\ni.e. the\n\nmanufacture of beverages, is statistically significant; and third the change in the income demand elasticity of each one of the sectors of the food and beverages manufacturing industry, zs ( s = 151,...,158 ) , with respect to sector 159, i.e. the manufacture of beverages, is statistically significant at any conventional level of significance. Furthermore, the results regarding scale economies imply the presence of increasing returns to scale. According to the empirical results of the second estimated system of the second approach, all the estimated coefficients of the translog cost function are statistically significant at any conventional level of significance (Table 2, Second Approach, Second Estimated System). Furthermore, the results support that the cost shares of labor ( al = 0.7782 ) and capital\n\n( ak = 0.2218)\n\nare statistically significant at any conventional level of\n\nsignificance. Moreover, the results of the supply function reveal that first, the conjectural variation elasticity of all the sectors except sector 157, i.e. f16 , is statistically significant at any conventional level of significance; and second, the change of the conjectural variation elasticity of sector 157 ( f157 = −0.0004 ) with respect to the conjectural variation elasticity of all the other sectors significance.\n\nNote\n\n( f16 = 0.7714 ) that\n\nthe\n\nis statistically significant at any conventional level of\n\nconjectural\n\nvariation\n\nelasticity\n\nof\n\nsector\n\n157\n\nis\n\n4\n\nWald test: the null hypothesis is f 151 = f 152 = f 153 = f 154 = f 155 = f 156 = f 158 = 0 whereas the alternative is that at least one of the aforementioned sectors is different from zero. The Wald test used follows\n\n( ) distribution with 7 degrees of freedom. The t-statistic and the p-value are 4.30 and 0.7452\n\nthe chi-squared χ\n\n2\n\nrespectively.\n\n7\n\n0.7710 ( f16 + f157 = 0.7710 ) . In addition, both conjectural variation elasticities of all sectors of the industry except sector 157 ( f16 = 0.7714 ) and that of sector 157 ( f16 + f157 = 0.7710 ) , i.e. sector 157 the manufacture of prepared animal feeds, are ranged between zero and one which indicates the presence of some degree of market power for each sector of the Greek food and beverages manufacturing industry over the period 1983–2007. The empirical results of the demand equation reveal that, first, the price elasticity of output demand ( h = -0.7742 ) is statistically significant at any conventional level of significance; second, the income demand elasticity of sector 159\n\n( z159 = 0.3701) ,\n\ni.e. the\n\nmanufacture of beverages, is statistically significant; and third the change in the income demand elasticity of each one of the sectors of the food and beverages manufacturing industry, zs ( s = 151,...,158 ) , with respect to sector 159, i.e. the manufacture of beverages, is statistically significant at any conventional level of significance. Furthermore, the results regarding scale economies imply the presence of increasing returns to scale. Finally, the Harberger loss is about €1.47 million in terms of 2007 value added (or 2.43% of the 2007 value added) for sector 157, i.e. the manufacture of prepared animal feeds, and €87.26 million (or 2.52% of the 2007 value added) for all other sectors, i.e. sectors 151–156 & 158–159, whereas the Tullock loss is about €2.58 million in terms of 2007 value added (or 4.25% of the 2007 value added) for sector 157 and €151.31 million (or 4.37% of the 2007 value added) for all other sectors of the Greek food and beverages manufacturing industry, i.e. sectors 151–156 & 158–159, (Table 3). The empirical findings of the third approach (Table 2, Third Approach) show that all the estimated coefficients of the translog cost function are statistically significant at any conventional level of significance. In addition, the cost shares of labor ( al = 0.7817 ) and capital ( ak = 0.2183) are statistically significant at any conventional level of significance. Furthermore, the empirical findings of the supply function indicate that first, the conjectural variation elasticity ( f1 ) of the sub-period t=1, i.e. years 1983–1985, is statistically significant at any conventional level of significance and second, the change of the conjectural variation elasticity of each sub-period ( f t , t = 2,...,8 ) with respect to the conjectural variation elasticity of the sub-period t = 1 , f1 , is statistically significant.5 Moreover, the empirical findings indicate that the conjectural variation elasticity of each sub-period is ranged between zero and one implying the presence of some degree of market power for each sub-period of the period 1983–2007 for the whole Greek food and beverages manufacturing industry. Figure 1 depicts the degree of market power for the whole Greek food and beverages manufacturing industry for each sub-period of the period 1983–2007. According to Figure 1, the sub-period t=1, i.e. the years 1983–1985, presents the highest conjectural variation elasticity ( f1 = 0.6612 ) and as a result the highest degree of market power whereas the sub-period t=8, i.e. the years 2004– 2007, shows the lowest conjectural variation elasticity ( f1 + f8 = 0.6596 ) and as a result the 5\n\nMore specific, the conjectural variation approach of the sub-period 1983–1985 is 0.6612 (f1=0.6612), sub-period 1986–1988 is 0.6609 (f1+f2=0.6609), of the sub-period 1989–1991 is 0.6605 (f1+f3=0.6605), sub-period 1992–1994 is 0.6603 (f1+f4=0.6603), of the sub-period 1995–1997 is 0.6600 (f1+f5=0.6600), sub-period 1998–2000 is 0.6598 (f1+f6=0.6598), of the sub-period 2001–2003 is 0.6600 (f1+f7=0.6600), sub-period 2004–2007 is 0.6596 (f1+f8=0.6596).\n\nof of of of\n\nthe the the the\n\n8\n\nlowest degree of market power. In general, Figure 1 shows that the degree of market power gradually decreases during the sub-periods under consideration except that of the sub-period t=7, i.e. the years 2001–2003, where an increase in the degree of market power occurred ( f1 + f 7 = 0.6600 ) . In particular, while the conjectural variation elasticity gradually decreases and takes the value of 0.6598 ( f1 + f 6 = 0.6598 ) for the sub-period t=6, i.e. the years 1998–2000, it becomes slightly higher and takes the value of 0.6600 ( f1 + f 7 = 0.6600 ) for the sub-period t=7, i.e. the years 2001–2003. Two important events took place which resulted in the gradual decrease in the degree of market power in the Greek food and beverages manufacturing industry during the period 1983–2007 except that of the period 2001–2003, i.e. the sub-period t=1,…, 8 except that of t=7. The first event is the deregulation of international markets and the gradual abolition of protectionism since the mid-1980s which led to imports being gradually increased. The second event is the introduction of research, development and innovation in the Greek food industry through different Developmental Laws and Operational Programmes towards the 1990s which resulted in the increase in the competitiveness of some problematic and smallscale firms in the Greek food and beverages manufacturing industry. However, the degree of market power in the Greek food and beverages manufacturing industry slightly increased during the period 2001–2003, i.e. the sub-period t=7, probably due to the launch of the euro in 2000. The launch of the euro led some small-scale firms to exit the Greek food market since they could not operate in the Single European Market and the European Monetary Union. The empirical findings of the demand equation imply that first, the price elasticity of output demand ( h = -0.6627 ) is statistically significant at any conventional level of significance; second, the income demand elasticity of sector 159\n\n( z159 = 0.3350 ) ,\n\ni.e. the\n\nmanufacture of beverages, is statistically significant, and third the change in the income demand elasticity of each one of the sectors of the food and beverages manufacturing industry, zs ( s = 151,...,158 ) , with respect to sector 159, i.e. the manufacture of beverages, is statistically significant at any conventional level of significance. Furthermore, the results regarding scale economies imply the presence of increasing returns to scale. Finally, the Harberger loss is ranged between 2.71% for the sub-period t=8, i.e. the years 2004–2007, which equals €338.53 million in terms of value added and 3.40% for the sub-period t=1, i.e. the years 1983–1985, which is equal to €27.60 million in terms of value added whereas the Tullock loss is ranged between 5.29% for the sub-period t=8, i.e. the years 2004–2007, which is equal to €660.82 million in terms of value added and 6.34% for the sub–period t=1, i.e. the years 1983–1985, which is equal to €51.47 million in terms of value added (Table 3). 5. Conclusions The objective of this paper has been to measure the degree of market power in the Greek food and beverages manufacturing industry over the period 1983–2007. For this purpose, three different approaches of the Bresnahan’s conjectural variation method are used. The first approach assesses the degree of market power of the whole Greek food and beverages manufacturing industry over the period 1983–2007. The second approach tests the degree of market power in each one of the nine sectors of the industry over the period under consideration, i.e. 1983–2007, and the third one estimates the extent of market power for the whole Greek food and beverages manufacturing industry for specific sub-periods of the period 1983–2007. The empirical results of the first approach imply the presence of imperfect competition in the whole Greek food and beverages manufacturing industry for the period 1983–2007. The\n\n9\n\nempirical findings of the second approach suggest the presence of a non-competitive market structure in each one of the nine sectors of the Greek food and beverages manufacturing industry for the period 1983–2007, with only sector 157, i.e. the manufacture of prepared animal feeds, showing a different degree of market power from all other sectors of the industry, i.e. sectors 151–156 & 157–158. Finally, according to the empirical findings of the third approach, each sub-period of the period 1983–2007, i.e. the sub-period t=1,…,8, appears to operate in conditions of imperfect competition in the whole Greek food and beverages manufacturing industry, with the sub-period t=1, i.e. the years 1983–1985, showing the highest degree of market power whereas the sub-period t=8, i.e. the years 2004–2007, the lowest degree of market power. Two important events took place which resulted in the gradual decrease of the degree of market power in the Greek food industry during the period 1983–2007 except that for the period 2001–2003. The first is the deregulation of international markets and the gradual abolition of protectionism since the mid-1980s and the second is the introduction of research, development and innovation in the Greek food market through different Developmental Laws and Operational Programmes towards the 1990s. However, the degree of market power in the Greek food and beverages manufacturing industry was slightly increased during the period 2001–2003 probably due to the launch of the euro in 2000. Furthermore, the present study estimates the net welfare loss (deadweight loss or Harberger loss) and the reduction of consumers’ welfare due to the transfer of income from consumers to the monopolist (Tullock loss) in the case of imperfect competition. According to the first approach, the empirical results indicate that, for the whole Greek food and beverages manufacturing industry, the Harberger loss is about 2.74% whereas the Tullock loss is about 5.01%. The findings, regarding the second approach, imply that the Harberger and the Tullock losses are 2.43% and 4.25% respectively for sector 157 whereas for all other sectors of the food and beverages manufacturing industry, the Harberger and the Tullock losses are 2.52% and 4.37% respectively. Finally, the empirical results of the third approach reveal that the Harberger loss is ranged between 2.71% for the sub-period 2004–2007 and 3.40% for the sub-period 1983–1985 while the Tullock loss is between 5.29% for the sub-period 2004–2007 and 6.34% for the sub-period 1983–1985. References Anders, M.S. (2008), “Imperfect competition in German food retailing: Evidence from State level Data”, International Atlantic Economic Society, (36), 441-454. Bhuyan, S., and Lopez, R.A. (1997), “Oligopoly Power in the Food and Tobacco Industries”, American Journal of Agricultural Economics, (79), 1035-1043. Bourlakis, C. A. (1992b), “Profits and market power in Greek Manufacturing Industries”, Ph.D. Thesis, School of Economic and Social Studies, University of East Anglia. Bresnahan, T.F.(1982),“The oligopoly solution concept is identified”, Economics Letters, (10), 87-92. Bresnahan, T.F. (1989), “Empirical studies of industries with market power.” Handbook of Industrial Organization, (2), R. Schmalensee and R. D. Willig, ed., 1011-1157. Cowling, K. and Waterson M. (1976), “Price-cost margins and market structure”, Economica, (43), 267-274. Dickson, V.A. and Yu, W. (1989),“Welfare losses in Canadian manufacturing under alternative oligopoly regimes”,International Journal of Industrial Organization,(7), 257-267. Harberger, A.C., (1954), “Monopoly and Resource Allocation”, American Economic Review, (2), 77–87. Lau, L. (1982), “On identifying the degree of competitiveness from industry price and output data”, Economic Letters, (10), 93-99.\n\n10\n\nLopez, A.R., Aziza, M.A. and Liron-Espana, C. (2002), “Market power and/or efficiency: A structural approach”, Review of Industrial Organization, (20), 115-126. Peterson, E.B and Connor, J.M, (1995), “A comparison of oligopoly welfare loss estimates for U.S. food manufacturing”, American Journal of Agricultural Economics, (77), 300-308. Tullock, G. (1967), “The welfare cost of tariffs, monopolies, and theft”, Western Economic Journal, (5), 224-232. Table 1. Classification of sectors SIC\n\nSector description\n\n151 152 153 154 155 156 157 158 159\n\nProduction, processing and preserving of meat and meat products Processing and preserving of fish and fish products Processing and preserving of fruits and vegetables Manufacture of vegetable and animal oils and fats Manufacture of dairy products Manufacture of grain milk products, starches and starch products Manufacture of prepared animal feeds Manufacture of other food products Manufacture of beverages\n\nTable 2. Empirical results of the conjectural variation model of Greek food and beverages manufacturing industry over the period 1983-2007 Coef. Cost a0 ay ayy gly al akα gll gklα gkyα xt xty xtl Supply f15 f16 f151 f152 f153 f154 f155 f156 f157 f158 f159 f1 f2 f3 f4\n\nFirst Approach Estimated System: Equations (9), (10), (11), (13)\n\nSecond Approach First Estimated System: Second Estimated System: Equations (9), (10), (11), Equations (9), (10), (14) (11), (15)\n\nThird Approach Estimated System: Equations (9), (10), (11), (16)\n\n18.5820*(435.310) 0.6812(22.505) 0.1030(8.720) 0.0635(12.216) 0.7779(51.072) 0.2222(14.585) 0.0755(16.165) -0.0755(-16.165) -0.0635(-12.216) -0.0211(-7.832) 0.0048(2.947) -0.0052(-4.633)\n\n18.5790(402.810) 0.6588(15.071) 0.0284 (0.525) 0.0599(11.381) 0.7706(48.463) 0.2294(14.430) 0.0759(16.233) -0.0759(-16.233) -0.0599(-11.381) -0.0186(-5.911) 0.0041(2.478) -0.0049*(-4.351)\n\n18.5850435.53) 0.6791 (22.513) 0.0905* (7.174) 0.0645* (12.565) 0.7782* (50.960) 0.2218* (14.527) 0.0757* (16.181) -0.0757* (-16.181) -0.0645* (-12.565) -0.0209* (-7.737) 0.0048* (2.975) -0.0052* (-4.642)\n\n18.4200* (347.57) 0.4287* (5.898) 0.1041* (8.840) 0.0450* (6.969) 0.7817* (52.457) 0.2183* (14.651) 0.0755* (16.753) -0.0755* (-16.753) -0.0450* (-6.969) -0.0098* (-2.890) 0.0233* (4.770) -0.0066* (-6.304)\n\n0.7104 (3.106) ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─\n\n─ ─ -0.0005 (-1.474) -0.0007 (-1.265) -0.0002 (-0.584) -0.0002 (-0.699) -0.0001 (-0.606) -0.0003 (-1.037) -0.0008(-1.693) -0.0001 (-0.997) 0.6760* (3.158) ─ ─ ─ ─\n\n─ 0.7714* (3.440) ─ ─ ─ ─ ─ ─ -0.0004 (2.130) ─ ─ ─ ─ ─ ─\n\n─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ 0.6612* (2.830) -0.0003* (-1.884) -0.0007 (-2.444) -0.0009 (-2.415)\n\n11\n\nf5 f6 f7 f8 Demand a h z 151 z152 z153 z154 z155 z156 z157 z158 z159 Scale Economies (SCE) b\n\na\n\n─ ─ ─ ─\n\n─ ─ ─ ─\n\n─ ─ ─ ─\n\n-0.0012 -0.0014 -0.0012 -0.0016\n\n19.6640*(7.820) -0.7130(-3.106) -0.1715(-14.612) -0.2941(-24.530) -0.0924(-7.968) -0.2141(-18.443) -0.0759(-6.663) -0.1564(-13.536) -0.2298(-19.516) -0.0373(-3.276) 0.3969(2.005)\n\n19.4330(8.186) -0.6782(-3.157) -0.1710(-13.841) -0.2883(-23.547) -0.0822(-6.681) -0.2157(-17.595) -0.0748(-6.201) -0.1552(-12.736) -0.2351(-19.293) -0.0354(-2.954) 0.4029(2.102)\n\n20.1900 (8.133) -0.7742* (-3.440) -0.1695* (-14.603) -0.2915* (-24.368) -0.0924* (-8.098) -0.212* (-18.432) -0.0758* (-6.774) -0.1553* (-13.624) -0.2364 * (-19.820) -0.0378* (-3.386) 0.3701* (1.881)\n\n19.9770* (7.794) -0.6627* (-2.830) -0.1717* (-14.608) -0.2878* (-23.912) -0.0839* (-7.209) -0.2196* (-18.920) -0.0770* (-6.722) -0.1529* (-13.200) -0.2282* (-19.353) -0.0366* (-3.202) 0.3350* (1.670)\n\n0.2568*(12.463)\n\n0.2876(8.687)\n\n0.2583 (12.582)\n\n0.2583* (12.582)\n\n(-2.421) (-2.391) (-2.216) (-2.273)\n\nak=1-al, gkl = -gll, gky = -gly. The corresponding values in parentheses are “t-ratios” obtained by applying the Wald test\n\n( (\n\nstatistic, b SCE = 1 − ∂ ln Ct Wk ,t\n\n)\n\n)\n\n∂ ln Y = 1 − ( aY + xty 13) at the point of approximation. The corresponding values\n\nin parentheses are “t-ratios” obtained by applying the Wald test statistic. indicates 1% significance levels, indicates 5% significance levels, * indicates 10% significance levels.\n\nTable 3. Estimated Harberger and Tullock losses in the Greek food and beverages manufacturing industry over the period 1983-2007 Approaches\n\nSectors/ Time periods\n\nFirst Approach\n\nHarberger loss a (WLH)\n\nTullock loss a (WLT)\n\n2.74\n\n5.01\n\n151-159\n\nSecond Approach (2ndEstimated System) Third Approach\n\na\n\nHarberger loss c\n\nTullock loss c\n\nHarberger loss d\n\nTullock loss d\n\n41972.69\n\n3523.14\n\n96.53\n\n176.51\n\n151-156& 158-159\n\n2.52\n\n4.37\n\n40819.40\n\n3462.55\n\n87.26\n\n151.31\n\n157\n\n2.43\n\n4.25\n\n1153.29\n\n60.59\n\n1.47\n\n2.58\n\n1983-1985 1986-1988 1989-1991 1992-1994 1995-1997 1998-2000 2001-2003 2004-2007\n\n3.40 3.22 3.00 2.93 2.84 2.78 2.83 2.71\n\n6.34 6.07 5.73 5.63 5.49 5.39 5.48 5.29\n\n811.86 1416.70 2535.86 4624.16 5451.29 6733.66 8176.80 12491.87\n\n─ ─ ─ ─ ─ ─ ─ ─\n\n─ ─ ─ ─ ─ ─ ─ ─\n\n─ ─ ─ ─ ─ ─ ─ ─\n\n27.60 45.62 76.08 135.49 154.82 187.2 231.4 338.53\n\n51.47 85.99 145.31 260.34 299.28 362.94 448.09 660.82\n\nThe estimated Harberger and Tullock losses are ratios, b The value added is in million Euros, c The Harberger and Tullock losses are in terms of 2007 value added and in million Euros, d The Harberger and Tullock losses are in terms of value added and in million Euros.\n\nD E G R E EO FM A R K E TP O W E R\n\nFigure 1. Degree of market power for the whole Greek food and beverages manufacturing industry for specific sub-periods of the period 1983-2007 0.6615 0.6610 0.6605 0.6600 0.6595 0.6590 0.6585 19831985\n\n19861988\n\n19891991\n\n19921994\n\n19951997\n\n19982000\n\n20012003\n\n20042007\n\nSUB-PERIODS\n\n12" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8650327,"math_prob":0.9419974,"size":40318,"snap":"2020-10-2020-16","text_gpt3_token_len":11361,"char_repetition_ratio":0.18854493,"word_repetition_ratio":0.2983524,"special_character_ratio":0.3219902,"punctuation_ratio":0.14734773,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9557162,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-01T15:23:00Z\",\"WARC-Record-ID\":\"<urn:uuid:b9412052-66ef-4a79-bc8e-60328aba7df0>\",\"Content-Length\":\"101726\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:52109dc5-cd52-43d5-996d-ccf4ebf5da3a>\",\"WARC-Concurrent-To\":\"<urn:uuid:a84e66e1-8e82-4b97-afe7-0132965857cb>\",\"WARC-IP-Address\":\"104.27.163.127\",\"WARC-Target-URI\":\"https://mafiadoc.com/measuring-market-power-in-the-greek-food-and-agecon-search_5ba3dd1b097c4749358b46ba.html\",\"WARC-Payload-Digest\":\"sha1:UJL6SCJEEFJZF5F2J7IBDDT22BUQT3OL\",\"WARC-Block-Digest\":\"sha1:2P2ZPDUIQJGHGBD2B6VYYBPZLPA7WNVN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370505731.37_warc_CC-MAIN-20200401130837-20200401160837-00053.warc.gz\"}"}
https://food-le.com/numberle	[ "Rating:\nNumberle\n5\n\n# Numberle\n\nNumberle is a math puzzle game inspired by Wordle, which became popular in early 2022. The main objective of the Numberle game is to correctly estimate the mathematical equation in six attempts. As you add your own equations, colored suggestions will appear to indicate how close you are to solving the puzzle; if all of the rows are highlighted in green, you have won! This game is ideal for brain training while also having fun with your buddies.\n\n## HOW TO PLAY\n\nSimply enter any valid equation to find hints to begin the game. You will have a total of six attempts to estimate the target equation. You can use numbers (0-9) and arithmetic signs (+ - * / =) to calculate.\n\nDetermine the numbers and signs in the equation.\n\nIf any numbers or arithmetic signs appear in the target equation but are in the wrong position, they will be marked in brown. If there are numbers or signs in the equation and they are in the correct place, they will be highlighted in green. The gray tint indicates that certain numbers or signs do not appear in the equation.\n\nMake an attempt to answer the target equation.\n\nTo win the game, you must correctly guess the equation (all spots are green). You may quickly post your results on social media at the end of the game, as well as copy the URL and challenge your friends!\n\nPUZZLE WORD brain skill logic quiz braining seach\nComment (0)", null, "Be the first to comment", null, "" ]	[ null, "https://food-le.com/themes/foodle/resources/images/icons/empty-comment.png", null, "https://food-le.com/game-tracking-views.ajax", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93964535,"math_prob":0.9817204,"size":1284,"snap":"2022-40-2023-06","text_gpt3_token_len":267,"char_repetition_ratio":0.15,"word_repetition_ratio":0.0,"special_character_ratio":0.211838,"punctuation_ratio":0.087649405,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9703908,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,8,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-27T13:57:01Z\",\"WARC-Record-ID\":\"<urn:uuid:540e0abd-de5a-43e9-b406-2fba2cec7720>\",\"Content-Length\":\"80618\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7329e4aa-a1c4-441a-bf06-154eebef9124>\",\"WARC-Concurrent-To\":\"<urn:uuid:7456b1ed-bccd-4077-bb80-d43cb676a5f6>\",\"WARC-IP-Address\":\"104.21.62.86\",\"WARC-Target-URI\":\"https://food-le.com/numberle\",\"WARC-Payload-Digest\":\"sha1:FNWPU2J77LNTRR2NUF3FNSHXZRKNDQK6\",\"WARC-Block-Digest\":\"sha1:WTB2PME7BWA2EZV25PRNDMQYZL457XKZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335034.61_warc_CC-MAIN-20220927131111-20220927161111-00294.warc.gz\"}"}
https://cmc.deusto.eus/category/publications/umberto-biccari/	[ "## Control and Numerical approximation of Fractional Diffusion Equations\n\nUmberto Biccari, Mahamadi Warma, Enrique Zuazua. Control and Numerical approximation of Fractional Diffusion Equations (2022) Handb. Numer. Anal. Elsevier. ISSN:1570-8659, DOI: https://doi.org/10.1016/bs.hna.2021.12.001 Abstract. The aim of this work is to give…\n\n## Internal control for a non-local Schrödinger equation involving the fractional Laplace operator\n\nU. Biccari Internal control for a non-local Schrödinger equation involving the fractional Laplace operator (2021) Abstract: We analyze the interior controllability problem for a nonlocal Schr\\\"odinger equation involving the fractional…\n\n## A stochastic approach to the synchronization of coupled oscillators\n\nUmberto Biccari, Enrique Zuazua. A stochastic approach to the synchronization of coupled oscillators. Frontiers in Energy Research, section Smart Grids. Front. Energy Res. Vol. 8 (2020). DOI: 10.3389/fenrg.2020.00115 Abstract. This paper…\n\n## Controllability of the one-dimensional fractional heat equation under positivity constraints\n\nU. Biccari, M. Warna, E. Zuazua Internal observability for coupled systems of linear partial differential equations. Commun. Pure Appl. Anal., Vol 19. No. 4 (2019), pp. 1949-1978. DOI: 10.3934/cppaa.2020086\n\n## Null-controllability properties of a fractional wave equation with a memory term\n\nU. Biccari, M. Warma Null-controllability properties of a fractional wave equation with a memory term. Evol. Eq. Control The., Vol. 9, No. 2 (2020), pp. 399-430 Abstract: We study the…\n\n## Propagation of one and two-dimensional discrete waves under finite difference approximation\n\nU. Biccari, A. Marica, E. Zuazua Propagation of one and two/dimensional discrete waves under finite difference approximation, Found. Comput. Math., Vol. 20 (2020), pp. 1401-1438. DOI: 10.1007/s10208-020-09445-0 Abstract: We analyze…\n\n## Null controllability of a nonlocal heat equation with an additive integral kernel\n\nU. Biccari, V. Hernández-Santamaría Null controllability of a nonlocal heat equation with an additive integral kernel. SIAM J. Control Optim., Vol. 57, No. 4 (2019), pp. 2924-2938, DOI: 10.1137/18M1218431 Abstract:…\n\n## Null-controllability properties of the wave equation with a second order memory term\n\nU. Biccari, S. Micu Null-controllability properties of the wave equation with a second order memory termJ DIFFER EQUATIONS, Vol. 267, No. 2 (2019), pp. 1376-1422 doi.org/10.1016/j.jde.2019.02.009 Abstract: We study the…\n\n## Dynamics and control for multi-agent networked systems: a finite difference approach\n\nU. Biccari, D. Ko, E. Zuazua Dynamics and control for multi-agent networked systems: a finite difference approach. Math. Models Methods Appl. Sci., Vol. 29, No. 4 (2019), pp. 755–790. DOI:…\n\n## Boundary controllability for a one-dimensional heat equation with a singular inverse-square potential\n\nU. Biccari Boundary controllability for a one-dimensional heat equation with a singular inverse-square potential, Math. Control Relat. F., Vol. 9, No. 1 (2019), pp. 191-219, DOI: 10.3934/mcrf.2019011 Abstract: We analyse…\n\n## The Poisson equation from non-local to local\n\nU. Biccari, V. Hernández-Santamaría The Poisson equation from non-local to local, Electronic Journal of Differential Equations, Vol. 2018 (2018), No. 145, pp. 1-13. DOI: arXiv:1801.09470 Abstract: We analyze the limit…\n\n## Controllability of a one-dimensional fractional heat equation: theoretical and numerical aspects\n\nU. Biccari, V. Hernández-Santamaría Controllability of a one-dimensional fractional heat equation: theoretical and numerical aspects, IMA J. Math. Control Inf., Vol. 36, No. 4 (2019), pp. 1199-1235. DOI: 10.1093/imamci/dny025 Abstract:…\n\n## Local regularity for fractional heat equations\n\nU. Biccari, M. Warma, E. Zuazua Local regularity for fractional heat equations<, Recent Advances in PDEs: Analysis, Numerics and Control, SEMA SIMAI Springer Series, Vol. 17 (2018). DOI: 10.1007/978-3-319-97613-6 Abstract:…\n\n## Local elliptic regularity for the Dirichlet fractional Laplacian\n\nU. Biccari, M. Warma, E. Zuazua Local elliptic regularity for the Dirichlet fractional Laplacian Advanced Nonlinear Studies, Vol. 17, Nr. 2 (2017), pp. 387-409. DOI: 10.1515/ans-2017-0014 Abstract: We analyze the…\n\n## Addendum: Local elliptic regularity for the Dirichlet fractional Laplacian\n\nU. Biccari, M. Warma, E. Zuazua Addendum: Local elliptic regularity for the Dirichlet fractional Laplacian Adv. Nonlinear Stud., Vol. 17, Nr. 4 (2017), pp. 837-839. DOI: 10.1515/ans-2017-6020 Abstract: In ,…\n\n## Null controllability for a heat equation with a singular inverse-square potential involving the distance to the boundary function\n\nU. Biccari, E. Zuazua Null controllability for a heat equation with a singular inverse-square potential involving the distance to the boundary function J. Differential Equations, Vol. 261, No. 5 (2016),…" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6929025,"math_prob":0.6277076,"size":6148,"snap":"2022-05-2022-21","text_gpt3_token_len":1762,"char_repetition_ratio":0.15332031,"word_repetition_ratio":0.044902913,"special_character_ratio":0.27553675,"punctuation_ratio":0.25040388,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9700815,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-27T03:00:50Z\",\"WARC-Record-ID\":\"<urn:uuid:8c0e98fe-b497-44b9-a2c3-04689a9e3d3c>\",\"Content-Length\":\"123492\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:78ebaddc-b4a5-4983-a815-16b06602aff7>\",\"WARC-Concurrent-To\":\"<urn:uuid:ddffc9e6-d92d-4f55-b7e3-24cadf3fa093>\",\"WARC-IP-Address\":\"151.80.161.28\",\"WARC-Target-URI\":\"https://cmc.deusto.eus/category/publications/umberto-biccari/\",\"WARC-Payload-Digest\":\"sha1:76G6DMMXAT44O4VDZAWELSQZA75G7DYF\",\"WARC-Block-Digest\":\"sha1:CYS47BLMYBA4RW6J3QAYSUUCKMXFB7AH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662631064.64_warc_CC-MAIN-20220527015812-20220527045812-00139.warc.gz\"}"}
https://homework.cpm.org/category/CON_FOUND/textbook/a2c/chapter/9/lesson/9.3.2/problem/9-157	[ "", null, "", null, "### Home > A2C > Chapter 9 > Lesson 9.3.2 > Problem9-157\n\n9-157.", null, "Remember the annual compound interest formula.\nA = P(1 + r)t\n\nFirst substitute the values into the equation.\n25000 = 15000(1.08)t\n\nDivide both sides by 15000.\n\n$\\frac{5}{3}=(1.08)^{\\textit{t}}$\n\nRemember the rules of logarithms and take the log of both sides.\n\n$\\text{log}\\left(\\frac{5}{3}\\right)=\\textit{t}(\\text{log}1.08)$\n\nSolve for t.\n\nThe account will be worth 25000 in between 6 and 7 years.\n\nt = 6.64" ]	[ null, "https://homework.cpm.org/dist/7d633b3a30200de4995665c02bdda1b8.png", null, 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https://de.scribd.com/document/347016535/Repeat-Paper-01-Edexcel-Physics	[ "Sie sind auf Seite 1von 6\n\n# FORCES and MOTION\n\n## a. Speed, Velocity and Acceleration\n\na. Speed/ average speed is defined as the distance moved per time, and hence,\nthe equation for speed is :\n\nQ1 1. If S for speed, d for distance travelled and t for time, then show that S=d/t.\n\n## 4. Show that 3.6km/h = m/s.\n\nQ2 Calculate the average speed of a car that travels 2500m in 500 seconds.\n\nQ3 Calculate the average speed of a bus that travels 150km in 2.5 hours.\n\nQ4 How far, in meters, will a train travel in 500 seconds at an average speed of 25\nm/s?\n\n10000 m/s?\n\nkm/h?\n\n30 000 km/h?\n\n## Q8 How long, in seconds, should a person take to travel 2000m at an average\n\nspeed of 4m/s?\n\nQ9 How long, in seconds, should a rocket take to travel 1 000 000m at an average\nspeed of 20 000m/s?\n\n## Sanjaya Perera(BSc(Hons).Maths(Sp)) - 0773 440 468 Page 1\n\nFORCES and MOTION\nRevision Paper 01- Edexcel IGCSE in Physics\n\n## b. Velocity is a physical vector quantity; both magnitude and direction are\n\nneeded to define it.\nVelocity is defined as a measure of the distance an object travels in a stated\ndirection (displacement) in a given length of time, and hence, the equation for\nvelocity is:\n( )\n=\n\nQ10 1. If V for velocity, d for displacement and t for time, then show that V=d/t.\n\n## c. Acceleration is a physical vector quantity; both magnitude and direction are\n\nneeded to define it.\nAcceleration is defined as the rate at which objects change their velocity, and\nhence, the equation for acceleration is:\n\n=\n\nQ11 1. If a for acceleration, u for initial velocity, v for final velocity and t for time\ntaken for change velocity, then show that a=(v-u)/t.\n\n## 3. What is meant by deceleration?\n\nQ12 A car is travelling at 40m/s. It accelerates steadily for 10seconds until velocity\nbecomes 70m/s. What is its acceleration?\n\n## Sanjaya Perera(BSc(Hons).Maths(Sp)) - 0773 440 468 Page 2\n\nFORCES and MOTION\nRevision Paper 01- Edexcel IGCSE in Physics\n\nQ13 Calculate the acceleration of a rocket the moves from rest (0 m/s) to 1000 m/s\nin 20 seconds.\n\nQ14 By how much does the velocity of a cyclist change in 2 seconds when\naccelerating at 5 m/s2?\n\nQ15 What is the final velocity of a rocket after 30 seconds if it accelerates at 50m/s2\nfrom 100 m/s?\n\nQ16 What is the initial velocity of a bus if after 4 seconds of accelerating at 3m/s2 it\nreaches 18 m/s?\n\n## Q17 An object strikes the ground travelling at 20m/s. It is brought to rest in\n\n0.05seconds. What is its acceleration? Explain the negative sign of your\n\n*******************************************************************\n\nb. Distance-time graphs\na. Gradient of the distance (Y-axis) Vs time (X-axis) graph denotes speed.\ni. A straight line sloping upwards means it has a steady speed.\nii. A horizontal line means the object is stopped.\niii. A steeper gradient means a higher speed.\niv. A curved line means the speed is changing.\nv. A straight line sloping downwards means it has a steady speed\nand a steady velocity in the negative direction.\n\nQ1 Using given details, sketch a distance-time graph to show the motion of a car\nbetween two junctions (A to C).\n\n## Start from junction A at t=t0\n\nMoving quickly from A to B and reached to B at t=t1\nStop for a moment and start its journey again from B at t=t2\nMoving slowly from B to junction C and end its journey at t=t3\n\nQ2 Show that the average speed of the above car for its whole journey is\n\n.\n( )\n\n## Sanjaya Perera(BSc(Hons).Maths(Sp)) - 0773 440 468 Page 3\n\nFORCES and MOTION\nRevision Paper 01- Edexcel IGCSE in Physics\n\n## Q3 Following distance-time graph shows a car travelling along a road, starting\n\nfrom the city O to the city G.\nDistance Vs Time\n120\n\n100 E\nDistance/km\n\n80\n\n60 D\nC\n40\n\n20\nA B\n0\nF\nO G\n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20\nTime/h\n\nI. Discuss the travelling behavior of the car from OA, AB, BC, CD,\nDE, EF and FG respectively.\n(Hint: uniform speed, not moving, speed changing, accelerating, positive direction, negative\ndirection, deceleration)\nII. What is the total time that the car is stopped during its travel?\nIII. Compare the speed difference between OA and DE.\nIV. Find the speed of the car between EF.\nV. Find the average speed of the whole journey.\nVI. What can you say about the city O (starting location) and city G (ending\nlocation)?\n\n*****************************************************************\n\nc. Velocity-time graphs\na. Gradient of the velocity (Y-axis) Vs time (X-axis) graph denotes acceleration.\ni. A straight line sloping upwards means it has a uniform\nacceleration.\nii. A horizontal line means the object is moving uniform velocity.\niii. The area under the graph is equal to the travelled distance.\niv. A straight line sloping downwards means it has a uniform\nacceleration in the negative direction/ deceleration.\n\n## Sanjaya Perera(BSc(Hons).Maths(Sp)) - 0773 440 468 Page 4\n\nFORCES and MOTION\nRevision Paper 01- Edexcel IGCSE in Physics\n\n## Q1 Using following details, sketch a velocity-time graph to show the motion of a\n\ncar between two junctions (A to C).\n\n## Start from junction A with the initial speed U at t=0.\n\nVelocity of the car increases until becomes V, within time t and reaches to\nthe junction B.\nThen the car moves with the uniform velocity V during time T and reaches\nto Junction C.\nI. Show that the acceleration (a) of the car within first time t is given by\n( )\n= .\nII. Show that the distance (S) travelled with uniform velocity (V) by the\ncar within time T is given by = .\n\n## Q2 Using following details, sketch a velocity-time graph to show the motion of a\n\nbike rider on a straight road.\n\nVelocity Vs Time\n120\n\n100 F\nVelocity (m/s)\n\n80\n\n60 D\nE\n40\n\n20 C\nB\nG H\n0\nA\n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20\nTime (s)\n\nI. Discuss the travelling behavior of the rider from AB, BC, CD, DE,\nEF, FG and GH respectively.\nII. What is the total time that the rider is stopped during his ride?\nIII. Compare the acceleration difference between AB and EF.\nIV. Find the deceleration of the bike after apply the brake between FG.\nV. How far he rides the bike after apply the brake (FG).\n\n## Sanjaya Perera(BSc(Hons).Maths(Sp)) - 0773 440 468 Page 5\n\nFORCES and MOTION\nRevision Paper 01- Edexcel IGCSE in Physics\n\nVI. How far he rides the bike under uniform velocity {(BC) &(DE)}.\n\n## Q3 A car is travelling at 20m/s. It accelerates uniformly at 3m/s2 for 5 seconds.\n\nI. Draw a velocity-time graph for the car during the period that it is\naccelerating. Include numerical detail on the axes of your graph.\nII. Calculate the distance which car travels while it is accelerating.\n\n## Q4 Plot a velocity-time graph using the data in the following table.\n\nv(m/s) 0.0 2.5 5.0 7.5 10.0 10.0 10.0 10.0 10.0 10.0\nt(s) 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0\nI. The acceleration during the first 4 seconds.\nII. The distance travelled in the last 6 seconds of the motion shown.\n\n## Q5 Consider the following velocity-time graph;\n\n12\n\n10\nvelocity (m/s)\n\n0\n0 10 20 30 40 50 60\ntime (s)\n\n## I. Find the acceleration and deceleration.\n\nII. 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