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https://www.jiskha.com/questions/14752/i-really-need-help-with-this-question-i-keep-doing-it-wrong-so-can-anyone-help-me-please	[ "# Management Accounting\n\nI really need help with this question. I keep doing it wrong so can anyone help me please.\n\n------------------------------------------------------------ -------\n\nThe maori novelty company makes a variety of souvenirs for visitors to New Zealand. The Otago division manufactures stuffed kiwi birds using a highly automated operation. A recently installed activity based costing system has four acitivities:\n\nActivity CostDriver Rate\n\nMaterials handling Kilograms of materials \\$1.20 per kg\n\nProduction setup # of setup \\$60 per setup\n\nCuttting, sewing etc # of units \\$0.40 per unit\n\nPackaging and shipping #of orders \\$10 per order\n\nTwo products are called Standard kiwi and giant kiwi. They require 0.20 and 0.40 kg, respectively, at a material cost of 1.30 for standard kiwis and 2.20 for giant kiwis. One computer controlled assembly line makes all the products. When a production run of a different product is started, a setup procedure is required to reprogram the computers and make other changes in the process. Products are packed and shipped separately so a request from a customer for, say, 3 different products is considered 3 different orders Ausiland Waterfront market just placed an order for 100 standard kiwis and 50 giant kiwis.\n\n1. Suppose the products made for Ausiland Waterfront required “AWM” to be printed on each kiwi. Because of the automated process, printing the initials takes no extra time or materials but it requires a special production setup for each product. Compute the cost of production.\n\n1. 👍 0\n2. 👎 0\n3. 👁 165\n\n## Similar Questions\n\n1. ### CHEM\n\nFor a molecule of fluorous acid, the atoms are arranged as HOFO. What is the formal charge on each of the atoms? Enter the formal charges in the same order as the atoms are listed. For this question I tried entering: +1,-2,+3,-2\n\nasked by K on November 26, 2007\n2. ### math\n\nA test had 200 questions .Each correct answer carried 2 marks.Each wrong answer carried -1/2 marks and unanswered question fetched no marks .Ajay attempted all the question in the test and he scored 360 marks.What would his marks\n\nasked by rachel on November 28, 2014\n3. ### Chemistry\n\nOF2(g) + H2O(g) -> O2(g) +2HF(g) ΔH°rxn = -318 kJ Using bond energies, calculate the bond dissociation energy of the O-F bond, in OF2. This is the question and I got 438 as the answer, which is wrong. I don't know what I did\n\nasked by Kat on May 1, 2011\n4. ### chemistry\n\nI keep getting wrong answers for this question. Determine the pH of an HF solution of each of the following concentrations. a) .280 M b) 5.310^-2 M c) 2.5010^-2 M No Ka has been given. I haven't attempted b or c yet because I\n\nasked by maura on March 29, 2012\n5. ### toxicology- can someone re work this question?\n\nHi, I have to answer this question \"why do we say that metabolism is the factor that most accounts for differences among species in the toxicity of some xenobiotics? I don't understand what the question means- I understand all the\n\nasked by brian on February 20, 2008\n1. ### Maths\n\nThere was a question: A gallon jug of milk is 3/4 full. After breakfast, you drink 1/12 of the milk. What fraction is the gallon is left? And here was my answer: 1/12 drank --> 11/12 left 3/411/12=11/16 But I got the question\n\nasked by Nana on October 28, 2013\n2. ### math11\n\nSolve the non linear equation x^2=3x+4 graphically using a calculator, Write the equations of the graphs you used to find the solution. I got this question wrong what would be my Y1= and Y2= equations. Not sure what I'm doing\n\nasked by Juliet on October 19, 2010\n3. ### math11\n\nSolve the non linear equation x^2=3x+4 graphically using a calculator, Write the equations of the graphs you used to find the solution. I got this question wrong what would be my Y1= and Y2= equations. Not sure what I'm doing\n\nasked by Juliet on October 19, 2010\n4. ### Simplfing Fractions\n\nThree problems I need to know if I am correct or not if not I will place how I got the answer and maybe some one can show me where I messing up aT Simplify 1+2/3 over top 2+1/2 My answer was 2/3 to me it does not seem right If I\n\nasked by Toni on June 1, 2007\n5. ### Chemistry\n\nFor a 0.47 M solution of LiClO compute the pH. Hi. I am confused about this question. I calculated the pH...but my answer is wrong. what am I doing wrong??? pH=-log(0.47) pH=0.33\n\nasked by J on November 13, 2016\n6. ### algebra\n\nSolve for s in this equation. Depreciation. D=C-s/n This is my answer D-C=C-s/n-c D-C=-s/n -n(D-C)=-s/n(-n) -nD+nC=s s=n(C-D) Please chack to see if this is correct Looks fine to me. A slightly shorter (by 2 lines) rearragement\n\nasked by jessica on April 4, 2007" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.95106614,"math_prob":0.67211837,"size":3127,"snap":"2020-24-2020-29","text_gpt3_token_len":937,"char_repetition_ratio":0.13896894,"word_repetition_ratio":0.15834768,"special_character_ratio":0.29645026,"punctuation_ratio":0.094827585,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9774055,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-10T09:30:55Z\",\"WARC-Record-ID\":\"<urn:uuid:19b89830-7e39-46f6-9c80-ae9b1b0768b9>\",\"Content-Length\":\"25016\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2f942100-a11d-4a4c-8904-5e9f28a92a87>\",\"WARC-Concurrent-To\":\"<urn:uuid:30451683-f54b-4d23-8196-90f22f965e93>\",\"WARC-IP-Address\":\"66.228.55.50\",\"WARC-Target-URI\":\"https://www.jiskha.com/questions/14752/i-really-need-help-with-this-question-i-keep-doing-it-wrong-so-can-anyone-help-me-please\",\"WARC-Payload-Digest\":\"sha1:WZJS65AKYHATUFLTPE2NRZZMI4EDJ2MH\",\"WARC-Block-Digest\":\"sha1:ZQHLBXQUCJC5SJ7YIBIDPOWNUVIT3TY4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655906934.51_warc_CC-MAIN-20200710082212-20200710112212-00025.warc.gz\"}"}
https://studylib.net/doc/10587983/id-1050-practice-exam-3	[ "# ID 1050 Practice Exam 3", null, "```ID 1050 Practice Exam 3\nDiscrete Data\nThe set of questions below all use the following set of data points taken from a discrete sample:\n1, 2, 3, 4, 4, 4, 5, 5, 6, 6.\n1) Make a bar graph of this data using the blank graph provided.\n2) What is the mean of this distribution?\na) 10\nb) 4\nc) 4.5\nd) 9\n5) What is the standard deviation of this sample\ndata set?\na) 1.63\nb) 1.55\nc) 4.00\nd) 2.67\n3) What is the median of this distribution?\na) 10\nb) 4\nc) 4.5\nd) 9\n6) What is the variance of this distribution?\na) 1.63\nb) 1.55\nc) 4.00\nd) 2.67\n4) What is the mode of this distribution?\na) 10\nb) 4\nc) 4.5\nd) 9\n7) What is the skewness of this distribution?\na) 1.22\nb) 4.00\nc) 1.41\nd) 0.00\nDefinitions\n8) A set of classes or groups based on name only,\nwithout a mathematical value or order (example:\nhair color):\na) A nominal scale\nb) An ordinal scale\nc) An interval scale\n10) A set of classes or groups based on their\nmathematical value relative to an absolute zero\n(example: height):\na) A nominal scale\nb) An ordinal scale\nc) A ratio scale\n9) A set of classes or groups based on relative rank\nonly, without a mathematical value (example:\nrelative dominance):\na) A nominal scale\nb) An ordinal scale\nc) A ratio scale\n11) The entire set of people or objects one wishes to\ncharacterize is called the _______. The subset\nthat is actually measured is called the ________.\na) Population; sample\nb) Sample; population\nR. Gist\n1 of 4\nVer. D, Rev. 1\nID105 Exam 3\nContinuous Data\nThe set of questions below are all taken from the following set of data which comes from a\nsample which is continuously distributed:\n1.1, 1.2, 1.3, 1.8, 2.4, 2.5, 3.7, 4.1, 4.8, 5.3\n12) Draw a histogram of this data. Use the number of classes and the class width that is shown on the following\nblank graph:\n13) What is the mean of this data?\na) 2.82\nb) 2.40\nc) 1.50\nd) 1.70\n16) What is the standard deviation of this sample\ndata set?\na) 2.41\nb) 1.50\nc) 1.55\nd) 4.56\n14) What is the median of this data?\na) 2.40\nb) 2.45\nc) 2.50\nd) 2.82\n17) What is the variance of this data?\na) 2.41\nb) 1.50\nc) 1.55\nd) 4.56\n15) What is the mode of this data?\na) 2.45\nb) 2.40\nc) 1.50\nd) 1.70\n18) What is the skewness of this data?\na) 1.50\nb) 4.50\nc) 0.85\nd) -1.75\nMore Definitions\n19) When you look at the graph of a set of data and\nnotice that there is a pronounced tail going off to\nthe left, you can be pretty convinced that the data\nhas ________.\na) Negative skewness\nb) Positive skewness\nc) No skewness\n22) Variables that can take on any real number value\n(integer or decimal) are:\na) Discrete variables\nb) Skewed variables\nc) Continuous variables\nd) Standard variables\n20) The measures of central tendency are:\na) The mean\nb) The mode\nc) The median\nd) All of the above\nR. Gist\n21) The measures of the spread or precision of the\ndata are:\na) The standard deviation and the variance\nb) The mean and the variance\nc) The skewness and the variance\nd) The mean and the skewness\n2 of 4\nVer. D, Rev. 1\nID105 Exam 3\nThe Normal Curve\nThe set of questions below are based on the following normal curve. This particular curve\nrepresents the scores of a large population on a standard SAT exam. The mean is 500 (=500)\nand the standard deviation is 100 (=100). The percentages given below represent the\npercentage of the population between the values on the horizontal axis.\n23) On the standard SAT exam, Debbie scores 600.\nWhat percentage of all people taking this test\nwill score higher than Debbie?\na) 0.5%\nb) 2.5%\nc) 16%\nd) 84%\n26) What is the highest grade someone else could\nhave and still be in the lowest 16% of everyone\ntaking the test?\na) 300\nb) 400\nc) 500\nd) 600\n24) Using Debbie as an example, you can say that\nher grade lies at the ______ percentile.\na) 0.5 th\nb) 2.5 th\nc) 16 th\nd) 84 th\n27) What percentage of all people taking the exam\nscored at least a 300?\na) 2.5%\nb) 16%\nc) 84%\nd) 97.5%\n25) What is the lowest grade someone else could\nhave and still be in the top 84% of all people\ntaking the test?\na) 300\nb) 400\nc) 500\nd) 600\n28) What percentage of all people taking the exam\nscored between 400 and 600?\na) 34%\nb) 68%\nc) 16%\nd) 2.5%\nR. Gist\n3 of 4\nVer. D, Rev. 1\nID105 Exam 3\nThe Standard Normal Curve\nThe next set of questions is based on the standard normal curve. Use the z-score table at the\n29) What area is under the curve to the right of a\nz-score of 0.8?\na) 0.212\nb) 0.455\nc) 0.500\nd) 0.882\n30) What area is under the curve to the right of a\nnegative z-score of z= -1.0?\na) 0.104\nb) 0.308\nc) 0.500\nd) 0.841\nThe next set of questions is based on normally distributed data representing the age of a certain\npopulation of 1000 individuals. The data has an average of 50 (=50) and a standard deviation\nof 10 (=10). [Recall that z=(x-)/] Using the z-score table at the bottom of this page, answer\nthe following:\n31) What percentage of the\npopulation is above an age\nof 50?\na) 20 %\nb) 50 %\nc) 70 %\nd) 99.5%\n32) What percentage of the\npopulation is between 35\nand 65 years old?\na) 6.7 %\nb) 30.8 %\nc) 68.0 %\nd) 86.6 %\n33) How many in the\npopulation are between 35\nand 65 years old?\n____________________________________________________________________________\nZ-Score Table\n(A)\nz-score\n0.0\n0.1\n0.2\n0.3\n0.4\n0.5\n0.6\n0.7\n0.8\n0.9\n1.0\n1.1\n1.2\n1.3\n1.4\n1.5\nR. Gist\n(B)\nArea between z\nand the mean\n0.000\n0.040\n0.079\n0.118\n0.155\n0.192\n0.226\n0.258\n0.288\n0.316\n0.341\n0.364\n0.385\n0.403\n0.419\n0.433\n(C)\nArea beyond\nz\n0.500\n0.460\n0.421\n0.382\n0.345\n0.309\n0.274\n0.242\n0.212\n0.184\n0.159\n0.136\n0.115\n0.097\n0.081\n0.067\n(A)\nz-score\n1.6\n1.7\n1.8\n1.9\n2.0\n2.1\n2.2\n2.3\n2.4\n2.5\n2.6\n2.7\n2.8\n2.9\n3.0\n4 of 4\n(B)\nArea between z and\nthe mean\n0.445\n0.455\n0.464\n0.471\n0.477\n0.482\n0.486\n0.489\n0.492\n0.494\n0.495\n0.496\n0.497\n0.498\n0.499\n(C)\nArea\nbeyond z\n0.055\n0.045\n0.036\n0.029\n0.023\n0.018\n0.014\n0.011\n0.008\n0.006\n0.005\n0.004\n0.003\n0.002\n0.001\nVer. D, Rev. 1\n```" ]	[ null, "https://s2.studylib.net/store/data/010587983_1-d7234a42e7603e055ef89f0a39b2ba23-768x994.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7866589,"math_prob":0.9777344,"size":5968,"snap":"2022-05-2022-21","text_gpt3_token_len":2088,"char_repetition_ratio":0.14118041,"word_repetition_ratio":0.19068162,"special_character_ratio":0.41873324,"punctuation_ratio":0.16204287,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9812564,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-18T20:35:30Z\",\"WARC-Record-ID\":\"<urn:uuid:58024610-a5d5-43f6-ae4e-f68280ca881b>\",\"Content-Length\":\"42294\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f68c5dbb-96fa-40d8-bd64-d4d83058187a>\",\"WARC-Concurrent-To\":\"<urn:uuid:6312f650-9819-4bec-9398-fc26be1a2316>\",\"WARC-IP-Address\":\"172.67.175.240\",\"WARC-Target-URI\":\"https://studylib.net/doc/10587983/id-1050-practice-exam-3\",\"WARC-Payload-Digest\":\"sha1:BC4QN2CU5BHETRMN42BJCPEB2X5LWAHC\",\"WARC-Block-Digest\":\"sha1:ZCWZJXRV6BBAHNENN5FGXXYT5YA7PVT5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662522309.14_warc_CC-MAIN-20220518183254-20220518213254-00465.warc.gz\"}"}
https://mammothmemory.net/physics/transformers/transformer--energy-can-not-be-lost-or-gained/transformer-energy-can-not-be-lost-or-gained.html	[ "", null, "# Transformer – energy cannot be lost or gained\n\nNow we know that doubling the coils doubles the voltage and therefore:\n\nV_1/V_2=(Number\\ \\turns\\ \\1)/(Number\\ \\ turns\\ \\2)\n\nWe must also remember that:\n\nPower can not be lost or gained\n\nIf Power=VI and power can not be lost or gained it follows that if voltage is increased, current is reduced and vice versa. Remember:\n\nV\\o\\l\\t\\a\\g\\e\\ \\1xxCurrent\\ \\1=Primary\\ \\Power=S\\e\\c\\o\\n\\d\\a\\r\\y\\ \\Power=V\\o\\l\\t\\a\\g\\e\\ \\2xxCurrent\\ \\2\n\n(NOTE: Secondary power is slightly less due to small inefficiencies, but they are ignored here)\n\nThis is an easy way for you to remember that:\n\n V_1xxI_1=V_2xxI_2\n\n 1000 v and 1 Amp equals 1000 watts 1 volt and 1000 Amps equals 1000 watts 500 v and 2 Amps equals 1000 watts 5 v and 200 Amps equals 1000 watts\n\nTherefore because\n\n V_1I_1=V_2I_2\n\n V_1/V_2=I_2/I_1" ]	[ null, "https://mammothmemory.net/images/mobile-home.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.75274175,"math_prob":0.9952997,"size":834,"snap":"2019-51-2020-05","text_gpt3_token_len":289,"char_repetition_ratio":0.11686747,"word_repetition_ratio":0.03125,"special_character_ratio":0.3381295,"punctuation_ratio":0.04733728,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99810547,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-12T20:01:40Z\",\"WARC-Record-ID\":\"<urn:uuid:e84dac16-d34e-4b8b-b617-29fa6576fb46>\",\"Content-Length\":\"22109\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7394e5fc-6770-4eb2-a1ce-a935efa250f0>\",\"WARC-Concurrent-To\":\"<urn:uuid:74f1154d-cbd3-486b-8c69-077e976c79f3>\",\"WARC-IP-Address\":\"176.58.124.133\",\"WARC-Target-URI\":\"https://mammothmemory.net/physics/transformers/transformer--energy-can-not-be-lost-or-gained/transformer-energy-can-not-be-lost-or-gained.html\",\"WARC-Payload-Digest\":\"sha1:2EWYAA3AEVGHUBIAV3UOD3DKCLQHBONX\",\"WARC-Block-Digest\":\"sha1:PPVXZAQTJKXQNNPDS27B6Q7MNJJHMEKM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540545146.75_warc_CC-MAIN-20191212181310-20191212205310-00338.warc.gz\"}"}
https://education.nsw.gov.au/teaching-and-learning/learning-from-home/learning-at-home/learning-packages/year-5-and-6-learning-packs/maths/package-3-lets-get-magical	[ "Package 1-3: Let’s get magical!\n\nIn this activity your child will use magic tricks to develop mathematical reasoning and practise their skills in mental computation.\n\nWeek 2 - Package 3 - Year 5 and 6 Mathematics - Let’s get magical!\n\nThings you need\n\nIdeal Back up\n\nPencils or markers\n\nPaper\n\nNumbered cards 0-9\n\nUNO cards\n\nWhy is this activity important?\n\nMagic tricks provide great opportunities for students to develop mathematical reasoning and practise their skills in mental computation. This trick is helpful for practising addition and subtraction facts. It also helps students develop skills in choosing efficient strategies for solving addition and subtraction problems. Practising the magic trick multiple times helps students develop confidence, consolidate skills in mental computation and helps them see the maths underlying the magic.\n\nBefore you start\n\nYou need some pens and a piece of paper.\n\nNumbered cards are required for the optional activities.\n\nWhat your child needs to know and do\n\nStudent needs to be able to add and subtract 3-digit numbers.\n\nWatch the video Let's get magical!\n\nWhat to do next\n\n• Choose a 3-digit number where each digit is smaller than the previous one (but they don’t have to be in order. For example, 982 or 531.)\n\n• Then, reverse the digits and subtract the second number from the first one. So, if I had chosen 531, I would now work out 531 – 135. The answer is 396. (If you get 99, record your answer as 099.)\n\n• Next, reverse your new number. For example, from 396 I can make 639.\n\n• Finally, add these last two numbers together. For example, 396 + 639.\n\n• Here comes the magic...\n\nActivity too hard? Activity too easy?\n\nComplete a different magic trick (‘Predict your puzzle’), detailed below\n\nTry using a different strategy to do the calculations for the magic trick. For example, you might like to use a calculator. Then, explore what happens if you follow the same rules but starting with a 2-digit number (for example, 76 or 82).\n\nComplete several times to practice using mental strategies with 2-digit numbers. Then try the trick again with 3 digit-numbers.\n\nInvestigate what happens with this magic trick using 4-digit numbers and 5-digit numbers. Complete several times to practice.\n\nWork together to explore which mental computation strategies are the most efficient when adding and subtracting.\n\nYou can also work together to investigate how this magic trick works!\n\n• Try another starting number and test it out again...is the final answer still 1089?\n\n• Explore what happens if you use the same process, starting with a 2-digit number or a 4-digit number...\n\n• Why do you think this might be happening?\n\nNow try this magic trick!\n\n1. Choose a number in the grid and circle it.\n\n2. Choose another number which is not in the same row or column as the first number and circle it.\n\n3. Pick a third number which is not in the same row or column as either of the other numbers you have circled.\n\n4. Pick a fourth number which is in no other column or row as the other circled numbers.\n\n5. Add all 4 numbers which are circled together.\n\n1 2 3 4\n\n5\n\n6\n\n7\n\n8\n\n9\n\n10\n\n11\n\n12\n\n13\n\n14\n\n15\n\n16\n\n1. Try this puzzle again.\n\n2. Share it with a family member.\n\n3. Explore....how does this work?\n\n1 2 3 4\n\n5\n\n6\n\n7\n\n8", null, "9\n\n10", null, "11\n\n12\n\n13", null, "14\n\n15\n\n16\n\nExample: 3 + 8 + 10 + 13= 34" ]	[ null, "https://lh5.googleusercontent.com/9BBd7jc0Yz2sbtnFdDIkNtvf4pBYPdWRZNJVhYUIo0JQsRWyVL5gzCsfpso6PHfQbLi5g0gb27etlF7TyMj3L3CCw5minHDJ7fKfqsRzdnmVnm1j6IFNxpt3QrwdTtepziJYeWI", null, "https://lh5.googleusercontent.com/9BBd7jc0Yz2sbtnFdDIkNtvf4pBYPdWRZNJVhYUIo0JQsRWyVL5gzCsfpso6PHfQbLi5g0gb27etlF7TyMj3L3CCw5minHDJ7fKfqsRzdnmVnm1j6IFNxpt3QrwdTtepziJYeWI", null, "https://lh5.googleusercontent.com/9BBd7jc0Yz2sbtnFdDIkNtvf4pBYPdWRZNJVhYUIo0JQsRWyVL5gzCsfpso6PHfQbLi5g0gb27etlF7TyMj3L3CCw5minHDJ7fKfqsRzdnmVnm1j6IFNxpt3QrwdTtepziJYeWI", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8654686,"math_prob":0.9025003,"size":3554,"snap":"2022-05-2022-21","text_gpt3_token_len":846,"char_repetition_ratio":0.11971831,"word_repetition_ratio":0.121875,"special_character_ratio":0.25914463,"punctuation_ratio":0.10984308,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9857267,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,6,null,6,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-19T09:56:26Z\",\"WARC-Record-ID\":\"<urn:uuid:61047ee6-1c10-42d1-8e71-2552cdf167a7>\",\"Content-Length\":\"105561\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b86fbd20-2f8e-402f-92b8-47e372d61b8e>\",\"WARC-Concurrent-To\":\"<urn:uuid:019f8a2e-1e4f-473e-8aee-99930aac1c11>\",\"WARC-IP-Address\":\"3.24.144.213\",\"WARC-Target-URI\":\"https://education.nsw.gov.au/teaching-and-learning/learning-from-home/learning-at-home/learning-packages/year-5-and-6-learning-packs/maths/package-3-lets-get-magical\",\"WARC-Payload-Digest\":\"sha1:QFQHTNGQBQNAEEAW3AWAB37VBCYRXHVN\",\"WARC-Block-Digest\":\"sha1:K57WVDT2U56VOVAEV5LMSMZNP4BEEL3J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301309.22_warc_CC-MAIN-20220119094810-20220119124810-00336.warc.gz\"}"}
https://link.springer.com/article/10.1007/s10346-016-0783-6	[ "## Introduction\n\nLandslides starting from unstable slopes affect the safety of life as well as of private and public assets. Computer models are employed to identify potentially unstable areas in order to facilitate decision-making at various levels. Whilst statistical models explore the relationships between the spatial patterns of landslide occurrence and a set of predictor layers, physically based models attempt to reproduce or to predict the physical mechanisms involved (Guzzetti et al. 1999; Van Westen 2000; Guzzetti 2006; VanWesten et al., 2006). Physically based models are frequently employed to estimate landslide susceptibility at the scale of small catchments (VanWesten et al., 2006). As long as shallow landslides are considered, these approaches mostly rely on the infinite slope stability model. It is commonly used in raster-based geographic information system (GIS) environments to derive a factor of safety for each pixel. However, the infinite slope stability model is unconditionally suitable only for those areas where shallow translational landslides with a length-to-depth ratio L/D >16–25 are expected (Griffiths et al. 2011; Milledge et al. 2012). As shallow landslides are most commonly triggered by extreme hydrometeorological events, such modelling tools are often coupled with more or less complex hydraulic models (e.g., Montgomery and Dietrich 1994; Van Westen and Terlien 1996; Burton and Bathurst 1998; Pack et al. 1998; Wilkinson et al. 2002; Xie et al. 2004a; Baum et al. 2008; Godt et al. 2008; Muntohar and Liao 2010; Mergili et al. 2012).\n\nFor areas with deep-seated landslides, models assuming spherical, ellipsoidal or complex sliding surfaces reproduce the stability situation in a more appropriate way. Whilst they are standard in geotechnical engineering, their implementation with GIS is non-trivial so that catchment-scale applications are less commonly applied (e.g. Xie et al. 2003, 2004b, 2006; Jia et al. 2012; Mergili et al. 2014a, b).\n\nEven simple slope stability or hydraulic models rely on parameters which are highly uncertain in their horizontal and vertical distribution. One possible concept to account for parameter uncertainty is the probability of failure (Tobutt 1982) which has started to complement the conventional factor of safety with increasing computational power, considering parameter spaces using random or regular sampling of uncertain parameters (Mergili et al. 2014a). Various authors have introduced and used different types of probability density functions (pdfs) of geotechnical (El-Ramly et al. 2005; Petrovic 2008; Mergili et al. 2014a) and geohydraulic parameters (Mesquita et al. 2002, 2007; Mesquita and Moraes 2004) which can be employed for parameter sampling. Whilst such functions are a smart way to deal with uncertain information, they are not necessarily transferable between different locations and therefore commonly suffer from small sample sizes and, consequently, weakly supported means and standard deviations.\n\nAs the challenge of uncertain parameters is encountered in many fields of geosciences, various approaches have been developed in the previous decades to test the sensitivity of the model results or the model performance to the input parameters or to optimize (calibrate) the input parameters in order to bring the model results in line with reference observations. Testing one parameter at a time is thereby considered inappropriate as both the optimum value and the sensitivity may strongly interrelate with the values of other parameters (Saltelli and Annoni 2010). Multi-parameter strategies are therefore required (e.g., Duan et al. 1992; Eberhart and Kennedy 1995; Hay et al. 2006; Vrugt et al. 2008; Fischer 2013). Optimized parameters or parameter sets, however, are not necessarily meaningful from a physical point of view. Particularly when calibrating many parameters at once, a good model performance in terms of reproducing the observation can be achieved despite a poor process understanding. The sensitivity of local-scale slope stability model results to selected input parameters was tested, e.g. by Griffiths and Fenton (2004) or by Wang et al. (2010). Guimarães et al. (2003) and Formetta et al. (2015) have applied parameter optimization strategies at catchment scale.\n\nAlmost all documented parameter sensitivity and optimization strategies target at discrete parameter values. We think that, particularly at broader scales, sensitivity analysis and optimization of parameter values is inappropriate as it disregards the inherent fine-scale spatial variability of the parameters. Instead, we suggest performing sensitivity analysis and optimization of parameter ranges.\n\nThe present article demonstrates such a strategy, employing a modification of the probability of failure concept. We investigate how the considered ranges of geotechnical and geohydraulic input parameters influence the results and performance of GIS-based catchment-scale slope stability models. For this purpose, we apply the infinite slope stability model, the sliding surface model of the tool r.slope.stability and the software Transient Rainfall Infiltration and Grid-Based Regional Slope-Stability Model (TRIGRS) to the Quitite and Papagaio catchments, Rio de Janeiro, Brazil. The findings are thought to be useful to identify suitable parameterization strategies for future slope stability modelling efforts.\n\nNext, we introduce the study area (“Study area and data” section) and describe the components of the proposed work flow (“Methods” section). We then demonstrate (“Results” section) and discuss (“Discussion” section) the results obtained before drawing our conclusions (“Conclusions” section).\n\n## Study area and data\n\nThe study area includes the two landslide-prone Quitite and Papagaio watersheds located in the western part of the city of Rio de Janeiro, Brazil (Fig. 1). Together, they cover an area of 4.4 km2, extending between 12 and 995 m a.s.l. The climate in the area is tropical humid (Guimarães et al. 2009). Due to influence by ocean moisture, the area receives a higher amount of rainfall than the central part of Rio de Janeiro (Hurtado Espinoza 2010). Granitic bedrock dominates both watersheds. The homogeneous, colluvial yellow soil is characterized by sandy-clay features (Hurtado Espinoza 2010; Galindo 2013; Galindo and Campos 2014) and a depth of 1–3 m (Guimarães et al. 2003). Native forest is still the dominant type of vegetation whilst the anthropogenic influence on the land cover is of limited importance (Guimarães et al. 2003; Hurtado Espinoza 2010).\n\nGuimarães et al. (2003) optimized values of effective cohesion c’ (kN m−2), normalized to depth d, effective angle of internal friction φ’ and specific weight of the saturated soil γ s (kN m−3) using published parameters for geomorphologically comparable adjacent areas and back-calculations with the software SHALSTAB. These authors arrived at best fit values of c’/d = 2 kN m−3, φ’ = 45° and γ s = 15 kN m−3, but they also indicated that, in general, low values of c’/d, high values of φ’ and values from 15 to 17.5 kN m−3 for γ s would be appropriate for the area. They proposed a general frame of parameter values realistic for the area (in the sense of a parameter space) summarized in Table 1 and, with some modifications, applied to tests A and B (see “Methods” and “Results” sections). Hurtado Espinoza (2010) measured a dry specific weight around 15 kN m−3 for some undisturbed samples taken at 1 m depth. The same authors stated that the slopes in the lower areas would be weaker whilst those in the higher areas would be stronger.\n\nValues of soil saturated conductivity K s were measured by Fernandes et al. (2001) using Guelph’s permeameter. The results showed a high variability with values ranging from 10−6 to 10−4 m s−1 as well as some important discontinuities in the profiles, possibly influencing groundwater flow.\n\nWe consider a landslide event related to intense rainfall on 13 and 14 February 1996. Within 48 h, 394.3 mm of rainfall was registered at the Alto da Boa Vista station, and 245.9 mm at the Jacarepaguá station, both located in close vicinity to the Quitite and Papagaio catchments and operated by the National Meteorological Institute (INMET; Conti 2012). A landslide inventory developed by Guimarães (2000) is used in the present work. According to this inventory, the rainfall event has triggered 93 landslides, occupying 0.14 km2 (3.1% of the entire area). Table 2 summarizes the main characteristics of the landslide inventory. Most landslides occurred in the native forest areas dominating the study area. Shallow landslides, debris flows and debris avalanches were most common. The sliding surfaces of most landslides coincided with the soil-rock interface (Guimarães et al. 2003; Miqueletto and Vargas, 2009; Hurtado Espinoza 2010). The landslide inventory displays the entire extent of the directly affected areas without distinguishing between release, transit and deposition areas.\n\nBesides the geotechnical and geohydraulic information and the landslide inventory, we use a 2-m resolution digital elevation model (DEM).\n\n## Methods\n\n### Work flow and software\n\nFigure 2 illustrates the general work flow of the study. We compute the slope failure susceptibility index (SFSI) (dimensionless number in the range 0–1) based on sets of factor of safety (FOS) values derived through the controlled variation of selected key parameters within a defined parameter sub-space. This procedure is repeated for various sub-spaces. The resulting SFSI values are evaluated against the inventory of observed landslides, and the findings are compared and interpreted.\n\nIn a first step, we vary the geotechnical parameters (tests A and B) and in a second step, we vary the geohydraulic parameters (test C). Test D uses a simple statistical model for the sake of comparison. Test A builds on the infinite slope stability model, test B on the sliding surface model of the tool r.slope.stability (Mergili et al. 2014a, b), designed as a raster module of the open source GRASS GIS software (Neteler and Mitasova 2008; GRASS Development Team 2016). Test C makes use of TRIGRS (Transient Rainfall Infiltration and Grid-Based Regional Slope-Stability Model; Baum et al. 2008), which is a grid-based tool simulating the permanent and transient rainfall influences on slope stability. Python scripting is used to derive SFSI, and the R Project for Statistical Computing (R Core Team, 2016) is employed for the evaluation of the results. Test D relies entirely on Python and R scripting.\n\n### Geotechnical model\n\nSlope stability modelling commonly builds on the limit equilibrium theory (Duncan and Wright 2005): a factor of safety (FOS) is computed as the ratio between resisting forces R and driving forces T:\n\n$$\\mathrm{FOS}=\\frac{R}{T}$$\n(1)\n\nWhen FOS = 1, the slope is in static equilibrium. Values of FOS <1 indicate potential failure (in reality, such slopes do not exist), values of FOS >1 indicate stable slopes. The use of this method requires the prior definition of a slip surface, and the soil is considered as rigid material.\n\nFor GIS-supported catchment-scale analyses of slope stability, the infinite slope stability model is most commonly employed (Montgomery and Dietrich 1994; Pack et al. 1998; Xie et al. 2004a; Baum et al. 2008). It assumes (i) a uniform slope of infinite length, and (ii) a plane, slope-parallel failure surface. As inter-slice forces do not have to be considered, it is conveniently applied on a pixel-to-pixel basis. Based on Eq. (1), FOS can be expressed in various ways. For fully saturated soil, the equation may be formulated as follows (modified after Baum et al. 2008):\n\n$$\\mathrm{FOS}=\\frac{ \\tan \\phi}{ \\tan \\alpha}+\\frac{c- u \\tan \\phi}{\\gamma_s d \\sin \\alpha \\cos \\alpha}$$\n(2)\n\nwhere α is the slope angle, u (N m−2) is the pore water pressure, γ s (N m−3) is the specific weight of the saturated soil and d (m) is the depth of the sliding surface.\n\nIn the present work, we use the infinite slope stability model implemented with r.slope.stability and with TRIGRS. Alternatively, we also apply the sliding surface model of r.slope.stability. Thereby, the slope stability is tested for a large number of randomly selected ellipsoid-shaped potential sliding surfaces, truncated at the depth of the soil. R and T are summarized over all pixels intersecting a given sliding surface, and FOS is computed for each surface in a way analogous to Eqs. 1 and 2, applying a modification of the Hovland (1977) model. Finally, the minimum value of FOS resulting from the overlay of all sliding surfaces is applied to each pixel. For a more detailed description of the sliding surface model of r.slope.stability, we refer to Mergili et al. (2014a, b).\n\n### Geohydraulic model\n\nIn TRIGRS, FOS is computed for one or more user-defined depths. The Richard’s equation is used to calculate the soil transient infiltration for saturated and unsaturated soil conditions (Iverson 2000):\n\n$$\\frac{\\partial \\psi}{\\partial t}\\frac{d\\theta}{d\\psi}=\\frac{\\partial }{\\partial x}\\left[{K}_L\\left(\\psi \\right)\\left(\\frac{\\partial \\psi}{\\partial x}- \\sin \\alpha \\right)\\right]+\\frac{\\partial }{\\partial y}\\left[{K}_L\\left(\\psi \\right)\\left(\\frac{\\partial \\psi}{\\partial y}\\right)\\right]+\\frac{\\partial }{\\partial z}\\left[{K}_z\\left(\\psi \\right)\\left(\\frac{\\partial \\psi}{\\partial z}- \\cos \\alpha \\right)\\right]$$\n(3)\n\nwhere ψ (m) is pressure head, θ is soil volumetric water content, t (s) is time, K L (m s−1) is lateral soil conductivity and K z (m s−1) is soil conductivity in z direction.\n\nTo solve the Richards equation, TRIGRS uses an approach developed by Iverson (2000), considering homogeneous soil, isotropic flow, relatively shallow depth, one-dimensional vertical downslope flow and soil moisture close to saturated conditions (Baum et al. 2008; Park et al. 2013), following the heat conduction approach described by Carslaw and Jaeger (1959). We refer to Baum et al. (2008) for a detailed description of the procedure.\n\nFor computing the groundwater level, TRIGRS compares the infiltrated water volume V I and the maximum drainage capacity of the soil V D. If V D ≥ V I, the water table remains constant. Otherwise, the water table rises, depending on K s and the transmissivity T. For unsaturated conditions, the maximum value of ψ is the new water level multiplied with β (value set according to the adopted flow condition). The amount of water exceeding the maximum infiltration rate is considered surficial runoff. However, surficial runoff is not taken over from one time step to the next (Baum et al. 2008).\n\n### Slope failure susceptibility index\n\nThe slope failure susceptibility index (SFSI) in the range 0–1 refers to the fraction of geotechnical and/or geohydraulic parameter combinations resulting in FOS <1, out of an arbitrary number of tested parameter combinations. This means that SFSI for a given pixel increases with each parameter combination where FOS <1 and, finally, low values of FOS correspond to high values of SFSI. The principal concept of the SFSI is identical to the concept of the slope failure probability yielded by r.slope.stability (Mergili et al. 2014a). However, we refer to it as a susceptibility index in the context of the present study as we simply use a uniform probability density function throughout all the computations. Such a distribution does not necessarily capture the real-world parameter distribution (which is unknown) and its use does therefore not justify applying the concept of probability in a strict sense.\n\n### Statistical model\n\nIn test D, a statistical model is applied for the purpose of comparison, employing the slope angle as the only predictor layer (Table 3). We keep the statistical model as basic as possible in order to evaluate the performance of a simplistic statistical approach in comparison to the physically based models (“Geotechnical model” to “Slope failure susceptibility index” sections). This allows us to conclude on the need of using more complex physically based models for catchment-scale landslide susceptibility analysis. Thereby, we overlay a classified slope map with the map of the observed landslide release areas (ORA; “Model evaluation” section) and, for each slope class, compute the fraction f C of observed landslide release pixels related to all pixels. SFSI—referred to as release probability by Mergili and Chu (2015) who employed a comparable approach—is then computed by applying f C to all pixels of the corresponding slope class. Thereby, it is important to use two different areas for the derivation of f C and for the computation and evaluation of SFSI (“Test layout” section).\n\n### Model evaluation\n\nThe landslide inventory for the Quitite and Papagaio watersheds displays the entire observed landslide impact areas (OIAs), i.e. the release, transit and deposition areas without any differentiation. We approximate the ORA as the upper third part of each OIA polygon. Depending on the test (“Test layout” section and Table 3), either the OIA map or the ORA map is overlaid with the corresponding SFSI map. When using the ORA map, the lower two-thirds portion of the OIA is not considered for evaluation. The true positive (TP), true negative (TN), false positive (FP) and false negative (FN) pixel counts are derived for selected levels of SFSI. An ROC curve is produced by plotting the true positive rates TP/(TP + FN) against the false positive rates FP/(FP + TN) derived with each combination of parameters. The area under the ROC curve AUROC indicates the predictive capacity of the model: AUROC = 1.0 (the maximum) means a perfect prediction, AUROC = 0.5 (corresponding to a straight diagonal line) indicates a random prediction, i.e. model failure. AUROC refers to the entire area used for model evaluation.\n\nIn addition, we introduce a conservativeness measure:\n\n$$\\mathrm{FoC}=\\frac{\\mu_{\\mathrm{SFSI}}}{r_{\\mathrm{OP}}}$$\n(4)\n\nwhere μSFSI is the average of SFSI over the entire study area, and r OP is the observed positive rate, i.e. the fraction of observed landslide pixels out of all pixels in the study area. If FoC >1, the model overestimates the landslide susceptibility, compared to the observation whilst values FoC <1 indicate an underestimation of the landslide susceptibility.\n\n### Test layout\n\nTables 3 and 4 summarize the main characteristics of each test and the parameter values and ranges considered.\n\nIn a first step (tests A1–A4 and B), the sensitivity of SFSI and the associated model performance to the geotechnical parameters c′ and φ′ and the shape of the sliding surface is explored, assuming fully water-saturated soils, and the depth of the sliding surface corresponding with the soil depth. The infinite slope stability model and the sliding surface model implemented in r.slope.stability are employed for this purpose. We introduce a two-dimensional parameter space constrained by lower boundaries of c′ = 0 kN m−3 and φ′ = 21°, and upper boundaries of c′ = 24 kN m−3 and φ′ = 45° (Fig. 3a; Table 4). This parameter space accounts for the full ranges of c′ and φ′ considered representative for the area (“Study area and data” section). We note that the resulting values of FOS vary according to φ′ and c′/d, so that the value of FOS obtained with d = 3 m and with a given value of c′ is identical (infinite slope stability model) or similar (sliding surface model) to the value of FOS with other values of c′ and d, but the same c′/d ratio. The dry specific weight of the soil γ d = 13.5 kN m−2 and the volumetric saturated water content θ s = 40 vol.% are set to constant values. We neglect the weight of the trees and the effects of their root systems on the cohesion: sliding surfaces are assumed to develop beneath the rooting depth.\n\nThe ranges of both c′ and φ′ are (i) considered in their entire extent; (ii) subdivided into two sub-ranges of equal extent and (iii) subdivided into three sub-ranges of equal extent (Fig. 4a, b). Considering all possible combinations of sub-ranges of the two parameters results in 36 partly overlapping parameter sub-spaces with 25 corner points. SFSI is computed for each parameter sub-space, with ten sampled parameters in each dimension (Fig. 4c). This procedure may be extended to three or more dimensions or repeated at a finer level by employing the sub-space with the best model performance as the entire space for the next level. For reasons to be explained in the “Results” section, only one level is applied in the present work. This work flow is repeated for two assumptions of soil depth and two versions of the landslide inventory used for evaluation, resulting in a total of four sub-tests (Table 3).\n\nTest C explores the sensitivity of SFSI and the associated model performance to K s and the initial depth of the water table d i (m). We introduce a two-dimensional parameter space constrained by lower boundaries of K s = 10−7 m s−1 and d i = 0 m and upper boundaries of K s = 10−4 m s−1 and d i = 3 m (Fig. 3b; Table 4). The ranges of values used are based on works of Saxton and Rawls (2006) and Guimarães et al. (2003). We set γ s = 16 kN m−2, θ s = 40 vol.%, θ r = 5 vol.%, c′ = 4.5 kN m−2, φ′ = 45° and d = 3 m to constant values. The choice of these values is supported by data from Guimarães et al. (2003) and Hurtado Espinoza (2010). We further assume constant values of diffusivity (D = 200K s ; Park et al., 2013) and initial infiltration rate (I 0 = 1.3 10−6 m s−1; Conti 2012).\n\nIn a way analogous to the geotechnical parameters, the ranges of both K s and d i are (i) considered in their entire extent, (ii) subdivided into two sub-ranges of equal extent and (iii) subdivided into three sub-ranges of equal extent, resulting in 36 partly overlapping parameter sub-spaces with 25 corner points. SFSI is computed for each parameter sub-space, with five sampled parameters in each dimension. The landslide inventory used for evaluation is ORA.\n\nThis procedure is repeated for four combinations of rainfall duration and type of pluviograph (Table 3). We assume rainfall durations of 6 and 10 h and a total rainfall amount derived from the measurements at the Jacarepaguá and Boa Vista stations on 13 and 14 February 1996 (Conti 2012). The Thiessen method is applied for estimating the precipitation in the catchment, and 20% of interception are deduced (Coelho Netto 2005). The total rainfall considered for the analysis is 144 mm in all the scenarios C1–C4.\n\nIn test D, we apply the statistical model introduced in the “Statistical model” section for the purpose of comparison (Table 3). f C is derived for one of the two catchments. SFSI is then computed for the other catchment and evaluated against the corresponding ORA. The entire procedure is repeated in the reverse way, so that a clear separation between the model development and model evaluation areas is ensured.\n\n## Results\n\n### Tests A and B: geotechnical parameterization\n\nFigure 5 illustrates the results of test A in terms of model performance (AUROC) and conservativeness (FoC). Assuming a constant soil depth, the model performs significantly better when considering only the ORA (test A2; AUROC ≤ 0.741; Fig. 5b) instead of the entire OIA (test A1; AUROC ≤ 0.691; Fig. 5a). This result clearly indicates that the OIA is unsuitable as reference for evaluation, and an appropriate inventory sub-setting is essential. Focusing on Fig. 5b, we note that the model performance in terms of AUROC is insensitive to the variation of the geotechnical parameterization within much of the tested ranges. In particular, the sub-spaces along a diagonal line from medium-high values of c′ and low values of φ′ to low values of c′ and high values of φ′ display almost identical AUROC values to the entire parameter space and to those sub-spaces including broad ranges of c′ or broad ranges of φ′ with medium-low values of c′. Only those sub-ranges limited to high values of c′ or low values of c′ and φ′ yield significantly lower AUROC values. These sub-ranges result in poorly patterned relatively non-conservative and extremely conservative predictions, i.e. they display very low and very high FoC values, respectively. In general, the model results are very conservative, indicated by FoC > > 1. At a lower level of AUROC—and a lower level of FoC caused by a higher number of OP pixels—similar patterns are observed in Fig. 5a.\n\nVarying d as a function of the topographic wetness index exerts contrasting effects on the patterns of AUROC, depending on whether the OIA or the ORA is used as reference. With the ORA as reference (Test A4; Fig. 5d), the sub-spaces with low values of c′ perform comparable to test A2 (Fig. 5b). This is not surprising as the influence of d on FOS increases with c′ (with c′ = 0, d has no influence). However, AUROC and also FoC decrease significantly with increasing c′, resulting in a very poor performance associated to those sub-spaces with high c′, and a reduced performance associated to those sub-spaces with broad ranges of c′, compared to Fig. 5b. This trend clearly indicates that most ORA pixels spatially coincide with areas of relatively low topographic wetness index and therefore low values of d (Table 3) resulting in high values of FOS and low values of SFSI in cohesive soils.\n\nThe reverse effect occurs when using the entire OIA as reference (test A3; Fig. 5c): many pixels in the lower portions of the landslide polygons coincide with high values of the topographic wetness index. Consequently, d and the resulting values of SFSI are comparatively high for many of the OP pixels, resulting in an improved model performance, compared to the tests A1 – A3 (AUROC ≤ 0.742; Fig. 5b). However, since most of the lower parts of the landslide polygons do most likely not represent release areas, the increased performance represents an artefact of inappropriate assumptions rather than an indicator for model success.\n\nConsidering the findings outlined, we identify test A2 as most representative. Even though the full parameter space yields an insignificantly lower value of AUROC than do some of the sub-spaces, there is no basis to support the choice of a particular sub-space in this specific case. The parameter values used and optimized by Guimarães et al. (2003) are mostly located within the parameter sub-spaces with the higher values of AUROC, indicating a certain plausibility of the results (Fig. 5b). Figure 6a shows the spatial patterns of SFSI derived in the tests A1 and A2 with the full parameter space of c′ and φ′. We note that the results of those tests are similar in terms of SFSI, as only the reference information for validation is varied. The same is true for the SFSI maps derived through the tests A3 and A4 (Fig. 6b).\n\nThe spatial patterns of SFIS derived with the sliding surface model of r.slope.stability (test B) are illustrated in Fig. 6c. Applying the full parameter space of c′ and φ′ along with constant soil depth and the ORA as reference, the associated value of AUROC is almost identical to the value yielded with the infinite slope stability model (0.735 vs. 0.734 in test A2). Thereby, the results yielded with the sliding surface model are more conservative: FoC = 59.5, compared to a value of 48.3 yielded with the infinite slope stability model (Fig. 5b).\n\n### Test C: geohydraulic parameterization\n\nFigure 7 illustrates the performance (AUROC) and conservativeness (FoC) of the model results for the various parameter sub-spaces of K s and d i. Firstly, we note that the results are largely insensitive to the four assumptions of rainfall duration and hydrograph shape (C1–C4): the patterns yielded are identical for all four scenarios, even though the numbers vary slightly. Within each scenario, the model performance responds highly sensitive to variations of K s and d i: it peaks at AUROC = 0.719–0.724 for the upper sub-range of the hydraulic conductivity (K s = 10−5–10−4 m s−1) and the lower sub-range of the initial depth of the water table (d i = 0–1 m). However, the model performance drops only slightly when the full range of both parameters K s and d i is applied (AUROC = 0.711–0.712). Figure 8 presents the SFSI maps produced in test C1 with the full space of K s and d i. The SFSI maps resulting from tests C2, C3 and C4 are almost similar to the map resulting from test C1 and are therefore not shown.\n\nConstraining the model input to the lower ranges of hydraulic conductivity or to deeper initial water tables leads to a significant drop in the model performance. Considering K s ≤ 10–5.5 leads to model failure (AUROC = 0.494), independently of the range applied for d i and the rainfall scenario. In this case, FoC = 3.9 (blue font colour in Fig. 7). As expected, FoC is highest for the configurations with high K s and shallow d i and lowest for the configurations with low K s and deep d i. Its maximum coincides with the best model performance (FoC = 48.0–48.9).\n\nThese outcomes reflect the fact that, with K s ≤ 10–5.5, too little water propagates through the soil to substantially influence slope stability. The effect is similar with higher values of K s if the initial water table is too deep. A shallower initial water table and higher values of K s facilitate increased values of u over broad parts of the study area and, consequently, lead to less stable slopes (Eq. 2) and higher values of FoC. Only combinations of high K s and deep di lead to a sufficient signal to reproduce the observed landslide release patterns with a fair performance. As for tests A and B, all results are very conservative also for test C (FoC > > 1).\n\n### Test D: statistical model\n\nThe statistical model yields an average AUROC value of 0.737 (values of 0.736 and 0.738 for the two catchments) whilst, as prescribed by the approach chosen, FoC ≈ 1. The model performance corresponds remarkably well to the performance of the physically based models (tests A2 and B in particular), underlining the fact that the slope angle strongly dominates also the pattern of SFSI derived with the physically based models (Fig. 9).\n\n## Discussion\n\nWe have demonstrated that the performance of the physically based-derived slope failure susceptibility index SFSI in our study area reacts conditionally sensitive to variations in the considered spaces of selected geotechnical and geohydraulic input parameters and state variables. Those parameter configurations yielding insufficient pattern in terms of simulated landslide vs. non-landslide areas lead to a significantly poorer performance. With regard to the geotechnical information, comparable AUROC values are displayed throughout much of the parameter space considered relevant for the study area (Guimarães et al. 2003), except for those sub-spaces with low c′ and low φ′ (μSFSI close to 1) and those areas with high c′ and high φ′ (μSFSI close to 0). This constellation underlines a well-known negative relationship between c′ and φ′. Model performance in terms of AUROC responds very sensitive to variations in K s and d i within the tested ranges but insensitive to the variations in the rainfall scenarios applied. Whilst the findings for the geotechnical parameters are claimed to be broadly valid, those for K s and d i may strongly depend on the assumed rainfall duration and intensity in relation to the water capacity of the soil. In this sense, the pattern displayed in Fig. 7 might change for different rainfall events.\n\nOur findings suggest that any further parameter optimization efforts in terms of AUROC may be obsolete: the pattern of SFSI derived with the entire parameter space performs approximately as well in reproducing the observed landslide areas as the patterns of SFSI derived with various sub-spaces do. Applying broad ranges of the key parameters for physically based catchment-scale landslide susceptibility modelling is on the “safe” side as it yields results comparable in quality to those derived with the best-fit narrower ranges. Acknowledging the fact that geotechnical and geohydraulic parameters are spatially highly variable, uncertain and often poorly known, applying a narrow parameter space—or even a singular combination of parameters—bears a considerable risk to be off target. The direct effects of the vegetation (not accounted for in the present study) increase the level of uncertainty particularly in forested areas.\n\nThe conservativeness of the result in terms of FoC strongly depends on the parameter sub-spaces used as input. μSFSI is generally much higher than r OP, indicating that the model results tend to be very conservative. The ideal result should correspond to FoC = 1. Theoretically, this could be achieved by increasing the upper thresholds of the geotechnical parameters, i.e. to make the parameter spaces considered broader. However, substantially higher parameter thresholds are not realistic for the soil materials involved. We believe that the key for bringing μSFSI in line with r OP consists in appropriately capturing the fine-scale spatial variation of the geotechnical parameters: sliding surfaces most likely coincide spatially with geotechnically susceptible areas, layers or interfaces, spaced in a more or less irregular way. We consider it almost impossible to parameterize such patterns in a deterministic way. In this context, we note that in Figs. 6 and 8, some landslides coincide spatially with areas of low SFSI. Such mispredictions are most probably related to localized patches of low soil strength, increased water input or increased hydraulic conductivity or the effects of the vegetation. Whilst the variation in the local slope angle explains much of the pattern of SFSI, the residual part is most likely explained by fine-scale spatial variations of the soil and, possibly, the vegetation.\n\nConsequently, physically based landslide susceptibility maps can be produced with a minimum amount of geotechnical data but in this case only provide relative results. There is no benefit in dedicating major resources to the detailed investigation of the geotechnical and geohydraulic parameters for catchment-scale landslide susceptibility maps without accounting in detail for the spatial variation of those parameters. Various studies emphasize the major challenges in capturing the spatial variability of the key parameters such as c′ and φ′ (Mergili et al. 2015), K s (Mesquita et al. 2002, 2007; Mesquita and Moraes 2004) or soil depth (McBratney et al. 2003; Frohn and Müller 2015). More precisely, at this time, there are no means to appropriately regionalize the key input parameters of slope stability models. We have demonstrated that ad-hoc assumptions of parameter variations (soil depth) may result in a decreased model performance or, in combination with inappropriate reference data (an inventory including transit and deposition areas), may pretend an improved model performance. Notwithstanding any possible future progress in this field, we highlight two strategies to deal with the challenges identified:\n\n1. 1.\n\nAccepting the limitations described and interpreting the outcomes of physically based landslide susceptibility models in a relative way. The SFSI as suggested in the present work is one possibility to do so; other ways were introduced earlier with SHALSTAB (Montgomery and Dietrich 1994) or SINMAP (Pack et al. 1998). In principle, all slope stability software tools can be used to derive relative indices from multiple results.\n\n2. 2.\n\nUsing probabilistic approaches to deal with the spatial parameter variation, i.e. resulting in the identification of the possible size of weak regions (Fan et al. 2016). Fibre bundle models may then be used to simulate the associated patterns of slope failures (Cohen et al. 2009). However, this method also relies on various assumptions of spatial parameter variability.\n\nOne may argue that also statistical models—employing a black box in terms of relating predictor layers to a landslide inventory—would do the job of producing relative landslide susceptibility maps. In fact, those approaches may be considered a more honest strategy, compared to physically based calculations with uncertain or even unknown geotechnical and geohydraulic parameters. We have shown that even a simplistic statistical model—employing the local slope as the only predictor layer—performs comparable to the more complex physically based models used. This finding reflects the dominant effect of the slope also in the physically based models, as long as the majority of the other key parameters is assumed constant in space. It reminds of the statement of Box (1976) that it would be simple and evocative models pushing science forward rather than over-elaborated, over-parameterized ones. However, it is clear that statistical models would hardly do the work for dynamic analyses such as—with the data usually available—predicting the slope stability response to a particular rainfall event.\n\n## Conclusions\n\nWe have tested the sensitivity of catchment-scale slope stability model results to variations in the geotechnical and geohydraulic parameters. In contrast to many previous studies, we have focused on parameter spaces instead of combinations of parameter values. The results produced with broad parameter sub-spaces show comparable levels of performance in terms of AUROC to those produced with narrow sub-spaces, even though the results vary considerably in terms of FoC. In general, the SFSI maps are classified as very conservative (FoC > > 1). It seems obsolete to optimize the parameters tested by means of statistical procedures.\n\nConsidering the uncertainty inherent in all geotechnical and geohydraulic data, and the impossibility to capture the spatial distribution of the parameters by means of laboratory tests in sufficient detail, we conclude that landslide susceptibility maps yielded by catchment-scale physically based models should not be interpreted in absolute terms. We suggest that efforts to develop better strategies for dealing with the uncertainties in the spatial variation of the key parameters should be given priority in future slope stability modelling efforts. Even though we consider it likely that many of our results are valid for most types of landslides or geological settings, more tests including a broad spectrum of situations would be necessary to confirm all statements." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81445265,"math_prob":0.911602,"size":52960,"snap":"2022-40-2023-06","text_gpt3_token_len":12906,"char_repetition_ratio":0.14731097,"word_repetition_ratio":0.04504617,"special_character_ratio":0.23999244,"punctuation_ratio":0.12641472,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9762082,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-27T05:57:54Z\",\"WARC-Record-ID\":\"<urn:uuid:70e63239-f632-49b4-9468-34a1a3fa1471>\",\"Content-Length\":\"329822\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f56408dd-141b-4b33-b2f5-6d99459354be>\",\"WARC-Concurrent-To\":\"<urn:uuid:e9490c69-b6e2-45de-9b1d-520d886c36a8>\",\"WARC-IP-Address\":\"146.75.32.95\",\"WARC-Target-URI\":\"https://link.springer.com/article/10.1007/s10346-016-0783-6\",\"WARC-Payload-Digest\":\"sha1:K4QTEHA2AJUWJYDENYVOG6N5GHV45G2Y\",\"WARC-Block-Digest\":\"sha1:4LQYT3KZGVZVZSUIWSMERAMA6FMGZ4OI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764494936.89_warc_CC-MAIN-20230127033656-20230127063656-00428.warc.gz\"}"}
https://gis.stackexchange.com/questions/169167/geojson-to-spatial-geography-in-sql-server-2012-fix-orientation-of-polygon/169236	[ "# GeoJson to Spatial Geography in sql server 2012, fix orientation of polygon\n\nI've download from Mapzen many geoJson files related to countries with their respective administration level. Then I've parsed all the geoJson files and saved them to my sql server 2012 as geography spatial data.\n\nI've then done a check of the Luxemburg map and noticed some strange results. Some of the maps where created correctly, with the corresponding inner area:", null, "In other cases the area of the selected boundary was the outer:", null, "I do know that there is the left-hand convention for storing inner area of a polygon in sql server as spatial data. Clearly fixing each polygon one by one will be a waste of time.\n\nGeoJson files aren't bounded to the left-hand rule, so I need a way to fix the coordinates orientation of the polygon before storing them in the db.\n\nI want the orientation fix logic to be done in the c# code (MVC) not by the sql server.\n\nI've read about a possible solution for convex polygons that handels the issue with cross-product, but still this doesn't work.\n\n1. How can I solve this issue?\n2. Is there a program that fixes the geoJson orientation of the polygons?\n3. Is there an algorithm based on a mathematical principle?\n4. Without going to the fuss of checking the orientation and all the extra coding, is there a site where I can download world data coordinates with all the administrative levels that use the left-hand convention for polygons?\n\nCode used to get cross product\n\n``````public decimal GetCrossProductZDirectionAndLength(VectorCoordinate coordA, VectorCoordinate coordB, VectorCoordinate coordC)\n{\n//*********************************************\n//This method gets the cross product\n//for vectors:\n//BA X BC\n//********************************************\ndecimal crossProductDirectionLength = 0;\n\n//These vector coordinates are relative to the origin (0,0,0)\nVectorCoordinate vectorCoordinateBA = new VectorCoordinate();\nVectorCoordinate vectorCoordinateBC = new VectorCoordinate();\n\nvectorCoordinateBA.X = coordA.X - coordB.X;\nvectorCoordinateBA.Y = coordA.Y - coordB.Y;\nvectorCoordinateBA.Z = coordA.Z - coordB.Z;\n\nvectorCoordinateBC.X = coordC.X - coordB.X;\nvectorCoordinateBC.Y = coordC.Y - coordB.Y;\nvectorCoordinateBC.Z = coordC.Z - coordB.Z;\n\ncrossProductDirectionLength = GetDeterminat2x2(vectorCoordinateBA.Y, vectorCoordinateBA.Z, vectorCoordinateBC.Y, vectorCoordinateBC.Z);\ncrossProductDirectionLength += GetDeterminat2x2(vectorCoordinateBA.X, vectorCoordinateBA.Z, vectorCoordinateBC.X, vectorCoordinateBC.Z);\ncrossProductDirectionLength += GetDeterminat2x2(vectorCoordinateBA.X, vectorCoordinateBA.Y, vectorCoordinateBC.X, vectorCoordinateBC.Y);\n\nreturn crossProductDirectionLength;\n}\n\nprivate decimal GetDeterminat2x2(decimal a, decimal b, decimal c, decimal d)\n{\n//get the determinat for a 2x2 matrix\n//\|a b\|\n//\|c d\|\n\nreturn (ad)-(bc);\n}\n``````\n• Could you expand on the solution you have already tried and what about it did not work? – MaryBeth Nov 5 '15 at 18:00\n• @MaryBeth I've included the code and an answer, please check if it makes sense, I've tested it on a few polygons and it seems to work. – Luther Nov 6 '15 at 2:01\n\n## 2 Answers\n\nI don't have a C# solution but here is how to do it in Java using JTS and GeoTools. But you should be able to recreate it in any language which provides some basic libraries/methods.\n\nThe algorithm comes down to\n\n``````for each polygon do\nif outer ring is counter clockwise then\nreverse outer ring\nfor each inner ring\nreverse it\n``````\n\nso in java\n\n``````while (it.hasNext()) {\nSimpleFeature f = (SimpleFeature) it.next();\nGeometry geom = (Geometry) f.getDefaultGeometry();\nSystem.out.println(geom);\nif (geom instanceof Polygon) {\nf.setDefaultGeometry(fixPolygon((Polygon) geom));\n} else if (geom instanceof MultiPolygon) {\nMultiPolygon multi = (MultiPolygon) geom;\nint numGeometries = multi.getNumGeometries();\nPolygon[] polys = new Polygon[numGeometries];\nfor (int i = 0; i < numGeometries; i++) {\npolys[i] = fixPolygon((Polygon) multi.getGeometryN(i));\n}\nf.setDefaultGeometry(GEOMFAC.createMultiPolygon(polys));\n}\nret.add(f);\n}\n\nprivate Polygon fixPolygon(Polygon geom, boolean cw) {\nLineString ring = geom.getExteriorRing();\nLinearRing extRing;\nif (RobustCGAlgorithms.isCCW(ring.getCoordinates()) == cw) {\nextRing = JTSUtilities.reverseRing((LinearRing) ring);\n} else {\nextRing = (LinearRing) ring;\n}\nPolygon ret;\nint numInteriorRing = geom.getNumInteriorRing();\nif (numInteriorRing > 0) {\nLinearRing[] holes = new LinearRing[numInteriorRing];\nfor (int i = 0; i < numInteriorRing; i++) {\nLineString inner = geom.getInteriorRingN(i);\nif (RobustCGAlgorithms.isCCW(inner.getCoordinates()) != cw) {\nholes[i] = JTSUtilities.reverseRing((LinearRing) inner);\n} else {\nholes[i] = (LinearRing) inner;\n}\n}\nret = GEOMFAC.createPolygon(extRing, holes);\n} else {\nret = GEOMFAC.createPolygon(extRing);\n}\nreturn ret;\n}\n``````\n\nI've tried a possible solution to the convex/concave polygon coordinate orientation, based on the left-hand rule (right-hand if required).\n\nBasically we can find the area of a polygon by adding the areas of the trapezoids defined by the polygons edge and a line corresponding to the min Y value of all the coordinates of the polygon. The min Y line is parallel to the X axis.", null, "The area of the trapezoid can be found by applying and rearranging the area formula:\n\narea = [(x2 - x1) (y2 + y1)] / 2", null, "When the program adds up all of the trapezoid areas, the sides on the polygon’s bottom give negative areas because x1 > x2 (this depends on the coordinates orientation). Those areas cancel out the parts of the other trapezoids that lie outside of the polygon. This method gives strange results for self-intersecting polygons.\n\nThe code then loops over the polygon’s segments, calculates the area under each, adds them up, and returns the total.\n\nThe total calculated area is negative if the polygon is oriented clockwise.\n\nClockwise or counter-clockwise check can be easily achieved with this solution.\n\n`````` public class VectorCoordinate\n{\npublic decimal X { get; set; }\npublic decimal Y { get; set; }\n}\n\npublic class VectorHelper\n{\npublic decimal GetAreaOfPolygon(List<VectorCoordinate> listPolygonCoordinate)\n{\ndecimal minYcoord = GetMinYCoordFromPolygon(listPolygonCoordinate);\n//Get the first coord to check if the last coord is the same\n//Area of the polygon\ndecimal areaOfPolygon = 0;\n//List of corrds to array\nVectorCoordinate[] vectorCoordinate = listPolygonCoordinate.ToArray();\n\nfor (int i = 0; i < vectorCoordinate.Length-1; i++)\n{\nareaOfPolygon += GetAreaOfTrapezoid(vectorCoordinate[i].X, vectorCoordinate[i].Y, vectorCoordinate[i + 1].X, vectorCoordinate[i + 1].Y, minYcoord);\n}\n//posite/negative area result will tell us if the orientation is clockwise or counterclockwise\nreturn areaOfPolygon;\n}\n\nprivate decimal GetMinYCoordFromPolygon(List<VectorCoordinate> listPolygonCoordinate)\n{\ndecimal minYcoord = listPolygonCoordinate.First().Y;\n\nforeach(var polygonCoord in listPolygonCoordinate)\n{\nif (polygonCoord.Y < minYcoord)\nminYcoord = polygonCoord.Y;\n}\n\nreturn minYcoord;\n}\nprivate decimal GetAreaOfTrapezoid(decimal x1, decimal y1, decimal x2, decimal y2, decimal minYcoord)\n{\nreturn ((x2 - x1) * ( (y2-minYcoord) + (y1-minYcoord))) / 2;\n}\n}\n``````" ]	[ null, "https://i.stack.imgur.com/fXZro.jpg", null, "https://i.stack.imgur.com/mxFuc.jpg", null, "https://i.stack.imgur.com/DJQtV.jpg", null, "https://i.stack.imgur.com/djamQ.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.71067184,"math_prob":0.966891,"size":2777,"snap":"2019-35-2019-39","text_gpt3_token_len":614,"char_repetition_ratio":0.24269743,"word_repetition_ratio":0.0,"special_character_ratio":0.23190494,"punctuation_ratio":0.17062636,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9957971,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-20T16:15:02Z\",\"WARC-Record-ID\":\"<urn:uuid:c48205eb-1bb4-4236-93e7-f5ea5d1a1bd0>\",\"Content-Length\":\"146832\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:451c8399-e83f-4fb5-a1bf-6885c992da1d>\",\"WARC-Concurrent-To\":\"<urn:uuid:3f57058a-a105-4102-a935-bf210dafc63e>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://gis.stackexchange.com/questions/169167/geojson-to-spatial-geography-in-sql-server-2012-fix-orientation-of-polygon/169236\",\"WARC-Payload-Digest\":\"sha1:LFAT6Y5JSB3A5S7SW6JMCNFHPRSHBW74\",\"WARC-Block-Digest\":\"sha1:RLNKIUPAP4JCGDZ2RMD3NGKPWI7QF4KZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027315551.61_warc_CC-MAIN-20190820154633-20190820180633-00019.warc.gz\"}"}
https://retropie.org.uk/forum/topic/23712/dreamcast-and-ps3-controllers-not-working/	[ "# Dreamcast and PS3 controllers not working\n\n• Good Evening Gamers,\n\nI am having issues using my PS3 controllers with Dreamcast. These controllers work with all of my other systems but Dreamcast. I tried manually mapping and I still get nothing when I hit start to play a game. I am working with with Retropie 4.5.1 on a Raspberry pi 3 B+. Below you will find my mapping.\n\nManual Configuration\ninput_device = \"MY-POWER CO.,LTD. 2In1 USB Joystick\"\ninput_driver = \"udev\"\ninput_r_y_plus_axis = \"+3\"\ninput_r_x_minus_axis = \"-2\"\ninput_l_btn = \"4\"\ninput_start_btn = \"9\"\ninput_exit_emulator_btn = \"9\"\ninput_r_y_minus_axis = \"-3\"\ninput_down_btn = \"h0down\"\ninput_l_x_plus_axis = \"+0\"\ninput_r_btn = \"5\"\ninput_save_state_btn = \"5\"\ninput_right_btn = \"h0right\"\ninput_state_slot_increase_btn = \"h0right\"\ninput_select_btn = \"8\"\ninput_left_btn = \"h0left\"\ninput_state_slot_decrease_btn = \"h0left\"\ninput_l2_btn = \"6\"\ninput_l3_btn = \"10\"\ninput_l_y_minus_axis = \"-1\"\ninput_up_btn = \"h0up\"\ninput_a_btn = \"1\"\ninput_b_btn = \"2\"\ninput_reset_btn = \"2\"\ninput_enable_hotkey_btn = \"8\"\ninput_l_y_plus_axis = \"+1\"\ninput_r2_btn = \"7\"\ninput_r3_btn = \"11\"\ninput_x_btn = \"0\"\ninput_l_x_minus_axis = \"-0\"\ninput_y_btn = \"3\"\ninput_r_x_plus_axis = \"+2\"\n\nAutomatic Configuration (Before Manual Configuration)\n\n[emulator]\nmapping_name = MY-POWER CO.,LTD. 2In1 USB Joystick\nbtn_escape = 296\n\n[dreamcast]\nbtn_a = 290\nbtn_b = 289\nbtn_c =\nbtn_d =\nbtn_x = 291\nbtn_y = 288\nbtn_z =\nbtn_start = 297\naxis_x = 0\naxis_y = 1\naxis_trigger_left =\naxis_trigger_right =\n\n[compat]\nbtn_trigger_left = 292\nbtn_trigger_right = 293" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5543976,"math_prob":0.98783106,"size":2201,"snap":"2019-51-2020-05","text_gpt3_token_len":674,"char_repetition_ratio":0.21438324,"word_repetition_ratio":0.0,"special_character_ratio":0.32848704,"punctuation_ratio":0.06741573,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.995993,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-21T02:44:38Z\",\"WARC-Record-ID\":\"<urn:uuid:f9065cd5-426a-4dea-b742-86adc5b99b36>\",\"Content-Length\":\"47464\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bd0e22e3-e5c3-4628-93bc-91c9cdb7a3fe>\",\"WARC-Concurrent-To\":\"<urn:uuid:3a85c1ae-45d3-4ce0-95ae-630b5ec73d46>\",\"WARC-IP-Address\":\"93.93.129.253\",\"WARC-Target-URI\":\"https://retropie.org.uk/forum/topic/23712/dreamcast-and-ps3-controllers-not-working/\",\"WARC-Payload-Digest\":\"sha1:J65P3K5KXVVXZ3IEATLOJOKS4MKOAO3J\",\"WARC-Block-Digest\":\"sha1:I45FKA45CTYCMMANAX3ERZFWQJQOH4YL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250601241.42_warc_CC-MAIN-20200121014531-20200121043531-00193.warc.gz\"}"}
https://bitcoin.stackexchange.com/questions/37645/procedure-for-calculating-taint	[ "# Procedure for calculating taint?\n\nI'd like to understand taint analysis quantitatively. Blockchain.info offers a service that will calculate taint, but I've found no good explanation for how taint is calculated.\n\nThe best (and only) explanation I've found so far appears in the paper Anonymity of Bitcoin Transactions:", null, "The taint analysis works by calculating the percentage of the amount of bitcoins that might origin from another address, thus revealing connections in the transaction graph. In the simplified example in Figure 4, A1 and A3 would have a taint of 75% and A2 a taint of 25%. However, it can only detect direct connections in the graph and does not consider any context information.\n\nThis explanation is confusing. Taint is measured between two pseudonyms (addresses). It's not a property of a particular pseudonym. The paper seems to be missing the part that says the taint scores for A1, A2, and A3 are relative to A4. If so, then the scores make sense.\n\nHowever, it's not clear what would happen for more complex chains of ownership. For example, imagine another pseudonym, A5 that pays A1 5 BTC. What would be the taint score between A5 and A4?\n\nI've seen this, but it doesn't discuss how to compute a taint score between two pseudonyms.\n\nWhat I'd like to see is the outline of a step-by-step procedure for calculating taint, as done by Blockchain.info. If I had to guess, here's how the procedure would look:\n\n1. Find two pseudonyms, S (source) and T (target). Funds flow from S to T.\n2. Using the block chain, find a chain of ownership Ci from each coin controlled by S to T.\n3. For each chain of ownership Ci, find the lowest valued coin transfer mi.\n4. Sum all mi, giving m.\n5. Sum the value of all outputs received through T, giving s.\n6. Taint is defined as m/s.\n\nUsing this procedure would give a taint score of 50% between A5 and A4 (2 / 4).\n\nIs this correct?\n\n• I haven't gotten around to going to the source, but I too find the example confusing. If I had to come up with a metric, I'd measure taint as an attribute of UTXO, not addresses. I.e. 1BTC claimed stolen, therefore 100% tainted getting spent together with 0.5 clean BTC would result in a UTXO with taint of 2/3. – Murch Jun 1 '15 at 0:32\n• Ah sorry, I started reading the paper and realized that it uses taint in a different context than the Bitcoin community usually does. Usually \"taint\" refers to the amount of coins traceable to a known theft, this paper however is using it as a term to measure address correlation. That's why I was confused and will remove the tainted-coins tag in a moment. – Murch Jun 1 '15 at 8:57\n\nI think I have an answer. It's not clear if this is how blockchain.info does it, but I'm not sure it matters, either.\n\nTaint is very similar to the everyday experience of diluting a liquid.\n\nImagine starting with three glasses. One glass contains orange juice. The second contains water. The third is empty.\n\nPouring some or all of the orange juice into the empty glass doesn't dilute it at all. However, pouring a 1:1 mixture of orange juice and water into the empty glass dilutes the orange juice.\n\nWe can define a metric called dilution factor. Dilution factor equals the final volume contained in the formerly empty glass divided by the volume of orange juice added. The dilution factor for a 1:1 mixture is therefore 2 (2 / 1). If no dilution takes place, then the dilution factor is 1. More generally:\n\ndilution factor = V2 / V1 if V2 > V1\n\ndilution factor = 1 if V2 <= V1\n\nWe can add another empty glass and dilute again. For example, we can take half of the diluted orange juice and dilute it with an equal volume of water. This gives a second dilution factor of 2.\n\nThe overall dilution factor equals the cumulative product of the dilution factor at each step. In this case, it is 4 (2 x 2).\n\nNow imagine that monetary value is like a liquid, and an output is like a container. We can do dilution analysis just like we did with orange juice.\n\nIn this case, the dilution factor tells us how diluted the value of an output in a chain of ownership has become relative to a downstream output. Provided that the limits of this metric are clear, it can be useful.\n\nThe liquid being diluted is \"taint\". In this model, \"taint\" would be the multiplicative inverse of dilution factor. A dilution factor of 2 implies a taint of 50% (1/2).\n\nReturning to the question posed above, given this chain of ownership, find the dilution factor between A5 and A4:\n\nA5(5 BTC)->A1(2 BTC)->A3(3 BTC)->A4(4 BTC)\n\nThe dilution factors are, from left to right: 1; 3/2; 4/3. Multiplying them together gives 1 x 3/2 x 4/3, or 2. This is a 50% taint, the same answer as I got above.\n\nHowever, the answer postulated in my original question is wrong in that it doesn't account for serial dilution. Consider this chain of ownership:\n\nA5(1 BTC)->A1(5 BTC)->A3(3 BTC)->A4(4 BTC)\n\nThe serial dilution factor is 5 x 1 x 4/3, or 20/3.\n\nSimply dividing the final value by the minimum value upstream gives a dilution factor between A5 and A4 of 4/1." ]	[ null, "https://i.stack.imgur.com/gg2KF.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9609536,"math_prob":0.9023985,"size":1838,"snap":"2020-45-2020-50","text_gpt3_token_len":436,"char_repetition_ratio":0.10632497,"word_repetition_ratio":0.0,"special_character_ratio":0.2328618,"punctuation_ratio":0.123655915,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9726737,"pos_list":[0,1,2],"im_url_duplicate_count":[null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-21T22:41:30Z\",\"WARC-Record-ID\":\"<urn:uuid:9210b770-2510-46b3-b53d-714e397f0e9f>\",\"Content-Length\":\"152233\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:17cf7d18-67cf-4943-adcc-724e56e50c24>\",\"WARC-Concurrent-To\":\"<urn:uuid:00d12f94-820c-43ac-bf24-802158258729>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://bitcoin.stackexchange.com/questions/37645/procedure-for-calculating-taint\",\"WARC-Payload-Digest\":\"sha1:P7WSJH3XYSHX6QPWBRPD5PYUF5S6FWQY\",\"WARC-Block-Digest\":\"sha1:K6LVYMRVP6FQKJYCXCT4PKMJGFD5OP42\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107878633.8_warc_CC-MAIN-20201021205955-20201021235955-00667.warc.gz\"}"}
https://polyglotsyntax.com/javascript/statements/	[ "• # Statements\n\nA condition statement is like a question in programming. It helps the computer decide what to do next. For example, if you go to a store, you might ask the shopkeeper if they have what you want. The shopkeeper will answer yes or no, and that will help you decide what to do next. In programming, the computer can ask itself questions to decide what it should do next.\n\n• ## If Statements\n\n• #### Javascript\n\n##### Example of if statement in javascript.\n``````var x = 1;\n\nif (x > 0) {\nconsole.log('variable x is greater than zero'); // variable x is greater than zero\n}``````\n• ## Else Statements\n\n• #### Javascript\n\n##### Example of else statement in javascript.\n``````var x = 1;\n\nif (x > 0) {\nconsole.log('variable x is greater than zero'); // variable x is greater than zero\n} else {\nconsole.log('else, variable x is zero or less'); // else, variable x is zero or less\n}``````\n• ## Else If Statements\n\n• #### Javascript\n\n##### Example of else if statement in javascript.\n``````var x = 1;\n\nif (x > 0) {\nconsole.log('variable x is greater than zero'); // variable x is greater than zero\n} else if (x == 0) {\nconsole.log('else if, variable x is zero'); // else if, variable x is zero\n}``````\n• ## Switch Statements\n\n• #### Javascript\n\n##### Example of switch statement in javascript.\n``````var x = 1;\n\nvar result = '';\nswitch (x) {\ncase 0:\nresult = 'variable x is integer zero';\nbreak;\ncase 1:\nresult = 'variable x is integer one';\nbreak;\ndefault:\nresult = 'variable x is anything else';\n}\n\nconsole.log(result); // variable x is integer one``````" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7028872,"math_prob":0.94333434,"size":1478,"snap":"2023-40-2023-50","text_gpt3_token_len":374,"char_repetition_ratio":0.1797829,"word_repetition_ratio":0.32967034,"special_character_ratio":0.30175912,"punctuation_ratio":0.14685315,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9936294,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-10T00:04:12Z\",\"WARC-Record-ID\":\"<urn:uuid:7d0b0e09-9eb1-4cee-a46f-20477a63ca3f>\",\"Content-Length\":\"13727\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0b2c1ef3-3170-472b-825f-c7e9186b55ab>\",\"WARC-Concurrent-To\":\"<urn:uuid:45b85340-6272-4e07-bd75-efabe13c0005>\",\"WARC-IP-Address\":\"108.175.2.227\",\"WARC-Target-URI\":\"https://polyglotsyntax.com/javascript/statements/\",\"WARC-Payload-Digest\":\"sha1:W4N45KN5APARTTAZSHHDGI7RX7S5NVCV\",\"WARC-Block-Digest\":\"sha1:HEHTL3CJMZNNTDVPACGPX2PTQECGZHA5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100989.75_warc_CC-MAIN-20231209233632-20231210023632-00836.warc.gz\"}"}
https://php2golang.com/method/function.uasort.html	[ "## GoLang uasort\n\nrequest it (153)\nGoLang replacement for PHP's uasort [edit \| history]\n\nDo you know a GoLang replacement for PHP's uasort? Write it!\n\n## PHP uasort\n\nPHP original manual for uasort [ show \| php.net ]\n\n# uasort\n\n(PHP 4, PHP 5, PHP 7)\n\nuasortSort an array with a user-defined comparison function and maintain index association\n\n### Description\n\nbool uasort ( array `&\\$array` , callable `\\$value_compare_func` )\n\nThis function sorts an array such that array indices maintain their correlation with the array elements they are associated with, using a user-defined comparison function.\n\nThis is used mainly when sorting associative arrays where the actual element order is significant.\n\nNote:\n\nIf two members compare as equal, their relative order in the sorted array is undefined.\n\n### Parameters\n\n`array`\n\nThe input array.\n\n`value_compare_func`\n\nSee usort() and uksort() for examples of user-defined comparison functions.\n\n### Return Values\n\nReturns `TRUE` on success or `FALSE` on failure.\n\n### Examples\n\nExample #1 Basic uasort() example\n\n``` <?php// Comparison functionfunction cmp(\\$a, \\$b) { if (\\$a == \\$b) { return 0; } return (\\$a < \\$b) ? -1 : 1;}// Array to be sorted\\$array = array('a' => 4, 'b' => 8, 'c' => -1, 'd' => -9, 'e' => 2, 'f' => 5, 'g' => 3, 'h' => -4);print_r(\\$array);// Sort and print the resulting arrayuasort(\\$array, 'cmp');print_r(\\$array);?> ```\n\nThe above example will output:\n\n```Array\n(\n[a] => 4\n[b] => 8\n[c] => -1\n[d] => -9\n[e] => 2\n[f] => 5\n[g] => 3\n[h] => -4\n)\nArray\n(\n[d] => -9\n[h] => -4\n[c] => -1\n[e] => 2\n[g] => 3\n[a] => 4\n[f] => 5\n[b] => 8\n)\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5933234,"math_prob":0.9271974,"size":948,"snap":"2021-31-2021-39","text_gpt3_token_len":289,"char_repetition_ratio":0.12076271,"word_repetition_ratio":0.0,"special_character_ratio":0.32594937,"punctuation_ratio":0.051612902,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96213764,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-24T11:21:00Z\",\"WARC-Record-ID\":\"<urn:uuid:789fd03b-9f96-4dbc-9f62-9038fe333cac>\",\"Content-Length\":\"13497\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9efb4c25-09f2-4b30-8b6f-73dbd0332b88>\",\"WARC-Concurrent-To\":\"<urn:uuid:0404476c-f04e-4b84-8828-aa0a84a83833>\",\"WARC-IP-Address\":\"104.21.12.30\",\"WARC-Target-URI\":\"https://php2golang.com/method/function.uasort.html\",\"WARC-Payload-Digest\":\"sha1:EFU3AH37YXYNOUJYFDTYWWBXRYZVZRWQ\",\"WARC-Block-Digest\":\"sha1:WJVOJPMGPYNRKPD7HBIJL44IRVFDJKPC\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046150264.90_warc_CC-MAIN-20210724094631-20210724124631-00603.warc.gz\"}"}
https://answers.everydaycalculation.com/multiply-fractions/98-5-times-5-7	[ "Solutions by everydaycalculation.com\n\n## Multiply 98/5 with 5/7\n\n1st number: 19 3/5, 2nd number: 5/7\n\nThis multiplication involving fractions can also be rephrased as \"What is 98/5 of 5/7?\"\n\n98/5 × 5/7 is 14/1.\n\n#### Steps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 98/5 × 5/7 = 98 × 5/5 × 7 = 490/35\n3. After reducing the fraction, the answer is 14/1\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]	[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8050141,"math_prob":0.9802133,"size":457,"snap":"2020-34-2020-40","text_gpt3_token_len":193,"char_repetition_ratio":0.17218544,"word_repetition_ratio":0.0,"special_character_ratio":0.46608314,"punctuation_ratio":0.078947365,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9785622,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-04T00:48:47Z\",\"WARC-Record-ID\":\"<urn:uuid:1f542784-6d9d-4ce4-ab5b-2b04c3487dfe>\",\"Content-Length\":\"7549\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:226b1f2d-1048-4a4b-9f9a-35dfead0dfec>\",\"WARC-Concurrent-To\":\"<urn:uuid:8ee8759e-3d60-4d6d-83e3-9b55bd70f0e4>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/multiply-fractions/98-5-times-5-7\",\"WARC-Payload-Digest\":\"sha1:MSYITN6W3PCCJGTCQVYL6V7WKK45GNHV\",\"WARC-Block-Digest\":\"sha1:UF3TPPV4BX4UTULF56NVGXIDAAJ26VX5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735836.89_warc_CC-MAIN-20200803224907-20200804014907-00372.warc.gz\"}"}
https://community.cypress.com/thread/28024	[ "4 Replies Latest reply on Jul 6, 2010 9:42 AM by robert.seczkowski\n\n# delta sigma instrument. amp\n\nI'm trying to use del sig adc plus 2 pga's as an instrumentation amplifier.(acc to AN60319)\n\nADC datasheet says, GetResult16() returns signed value; GetResult32() (not known?!!!! - I presume unsigned)\n\nIn single input ADC mode, the -Input should be tied to VSS. How it is in differetial mode?\n\nADC normally cannot measure negative values.\n\nWhat is the input range (PGA gain 1; ADC gain 1,differential input,internall vref=1.024V) for PGA+ and PGA- pins?\n\nExample:\n\nPGA+ = 0V; PGA - = 1mV or PGA+=1mV, PGA-=0V,diff mode,vref=1.024V what will be the output counts(res 20bits);?\n\n• ###### 1. Re: delta sigma instrument. amp\n\nIn other words:\n\nCounts = Vdgain/Vref 2^(n-1) (an60319 page 2) Vd-differetial voltage= Vpositive - Vnegative, n-resolution\n\nWhat counts will be if Vd < 0,Vd = 0,Vd>0?\n\n• ###### 2. Re: delta sigma instrument. amp\n\nHi Robert,\n\nThe ADC by itself can provide a gain of 32 in the component now. It can provide a maximum gain of 128 in the future version(Beta 5) of PSoC creator. So, just using the ADC will be sufficient for gains less than 128. For gains greater than 128, please refer to the connection shown in this KB\n\nhttp://www.cypress.com/?id=4&rID=40623\n\nLet me know if your design requires more gain or some other necessity prevents the use of this design\n\nRegards,\nPraveen\n\n• ###### 3. Re: delta sigma instrument. amp\n\nIt's not the issue of gain but the ADC output.\n\nI guess ADC Delsig gives always 2's complement outputs\n\nthen if Vref=1.024V,gain 1,res=20bit\n\n(((0V-1mV)1)/(1.024+1mV)) 2^(20-1)=511 or 523777\n\nVdiff K Inp Range Max counts\n\nNow if\n\nthen if Vref=1.024V,gain 1,res=20bit\n\n(((999mV-1000mV)1)/(2000mV)) 2^(20-1)=262 or 524026\n\nVdiff K Inp Range Max counts\n\nWhy same difference give different results?\n\nBecause measurement range changes with -Input DC value (acc. to DelSig manual)\n\n• ###### 4. Re: delta sigma instrument. amp\n\n2000mV should be replaced with 2024mV to be honest (-Input +- Vref)\n\nIn other words:\n\nIs there a method to measure voltages around 0 like +- 2mV?\n\nregards\n\nrobert" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6843,"math_prob":0.9378424,"size":1778,"snap":"2019-13-2019-22","text_gpt3_token_len":582,"char_repetition_ratio":0.10653889,"word_repetition_ratio":0.072463766,"special_character_ratio":0.32452193,"punctuation_ratio":0.14814815,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9815566,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-22T09:13:13Z\",\"WARC-Record-ID\":\"<urn:uuid:79e87862-6aa7-4f49-b194-f888d76e96c7>\",\"Content-Length\":\"100493\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4161ec40-203b-4852-8daa-a79549938a6d>\",\"WARC-Concurrent-To\":\"<urn:uuid:ad3e3795-7c0f-4713-ba4b-4cc746f752fe>\",\"WARC-IP-Address\":\"104.118.182.144\",\"WARC-Target-URI\":\"https://community.cypress.com/thread/28024\",\"WARC-Payload-Digest\":\"sha1:5VFVEA46R5SNCZNR7N444BBSWOC23BAR\",\"WARC-Block-Digest\":\"sha1:BOOGBN43AHMULSMEGEHHZTFMRAIBLWWB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202640.37_warc_CC-MAIN-20190322074800-20190322100800-00379.warc.gz\"}"}
https://theswiftarchitect.com/2018/01/16/using-the-map-function-on-an-optional-in-swift/	[ "# Using the map function on an optional in Swift\n\nHi there Swifters.\n\nToday, I’m going to show you something I learned only quite recently about Swift. In Swift you can use the map function on an optional.", null, "At this point I must insert a simple check, if you say:\n\n• ‘Big deal, I don’t care, I don’t use these fancy new tricks everyone seems to promote lately’ or\n• ‘I actually wrote the method’ or\n• something similar\n\nthen you can simply stop reading this post.\n\nIf you’ve heard about the map function, you probably have seen something like this as an explanation. This is using of the map function on an array.\n\n```let array = [1, 2, 3]\nlet mappedArray = array.map { \\$0 * \\$0 }\nprint(mappedArray)\n```\n\nIn this specific example we declare an array: array containing the integers 1, 2 and 3. Then we declare another array: mappedArray which is derived from array by specifying that each element in mappedArray should be calculated by multiplying the element at the same position in the array array by itself. A more in-depth explanation on how to use the map function on an array is out of this post’s scope so if you want to learn more about it I am recommending you checkout Apple’s official documentation about it.\n\nFor me, this usage of map was the only one I knew till I watched the videos of the presentations Using Monads and Other Functional Paradigms in Practice by Raheel Ahmad and Map and FlatMap Magic by Neem Serra. I highly recommend watching them both since they explain how you can use the map function on optionals but also many many more cool stuff.\n\n### Back on the topic…\n\nLet’s analyze a simple use case that comes up pretty often when I write my code and I’ve also seen it a lot in other Swift code from the world. Let’s assume we have a value of an optional integer type. We want to somehow execute an algorithm which will take that value of an optional integer type, do something with it and return a value of an optional integer type. That ‘do something’ part should be: if the optional integer is not nil (has some value) then return that value multiplied by itself; if the optional integer is nil then return nil as well.\n\nPlease, take a look at the following code section (nothing scary here)…\n\n```func transformWithIfLet(number: Int?) -> Int? {\nif let number = number {\nreturn number * number\n}\n\nreturn nil\n}<span id=\"mce_SELREST_start\" style=\"overflow:hidden;line-height:0;\"></span>\n\nfunc transformWithMap(number: Int?) -> Int? {\nreturn number.map { \\$0 * \\$0 }\n}\n\nfunc executeTests() -><span id=\"mce_SELREST_start\" style=\"overflow:hidden;line-height:0;\"></span> Bool {\nlet inputNumbers: [Int?] = [nil, 3]\n\nfor inputNumber in inputNumbers {\nif transformWithIfLet(number: inputNumber) !=\ntransformWithMap(number: inputNumber) {\nreturn false\n}\n}\n\nreturn true\n}\n\nprint(executeTests() ? \"Tests passed\" : \"Tests failed\")\n```\n\nThe transformWithIfLet(number: Int?) -> Int? function is a modularized version of the algorithm implemented in a way that the code maps the worded explanation mostly word for word. The transformWithMap(number: Int?) -> Int? implements the same requirement by using the map function on the number value of optional integer type. When the executeTests() function is executed, the ‘Tests passed’ string is printed out assuring us that these two implementation of the same algorithm return the same values for the same inputs.\n\nThat’s basically how you use the map function on an optional. You define a closure which takes one argument of the same type only not optional (e.g. if you use it on Int?, the argument is of type Int). The closure should return a value of the same type as the closure input argument’s type. That closure will be executed only if the value of the optional is not nil, otherwise nil is returned immediately. Neat, simple and cute syntax!\n\nStill, you might say ‘Great, I traded 5 lines for 1 at the cost of having to use a bit of slightly unorthodox syntax. I don’t see I get much benefit here’. This is the point where we need to go deeper to a more complex use case. If we modify the algorithm to add two more steps in the transformation after the multiplication when the optional value is not nil: converting the integer to a float and then converting it to a string, this is the code we’d get for the two versions of the implementation:\n\n```func transformWithIfLet(number: Int?) -> String? {\nif let number = number {\nreturn String(Float(number * number))\n}\n\nreturn nil\n}\n\nfunc transformWithMap(number: Int?) -> String? {\nreturn number.map { \\$0 * \\$0 }.map(Float.init).map(String.init)\n}\n```\n\nIn my opinion the implementation of transformWithMap(number: Int?) -> String? is a lot more intuitively understandable than the implementation of transformWithIfLet(number: Int?) -> String?. The actual chain of transformation is more evident and the code representation matches the mental representation a lot more closely.\n\nSo, there you go, you just leaned how to use a short and quick trick from the functional programming repertoire, at your disposal 24/7. If you learned something new from this post, please spread the message to your programming buddies by sharing it. Happy optional mapping! 🙂" ]	[ null, "https://theswiftarchitect.files.wordpress.com/2018/01/using-the-map-function-on-an-optional-in-swift.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7919393,"math_prob":0.9264956,"size":5147,"snap":"2019-51-2020-05","text_gpt3_token_len":1152,"char_repetition_ratio":0.13338518,"word_repetition_ratio":0.044982698,"special_character_ratio":0.22887118,"punctuation_ratio":0.11336032,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9626591,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-25T10:52:52Z\",\"WARC-Record-ID\":\"<urn:uuid:27913a95-0839-4605-93a4-8135342cde92>\",\"Content-Length\":\"86757\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:64759c86-c503-4af6-950e-f0a2c1a5d259>\",\"WARC-Concurrent-To\":\"<urn:uuid:aa085895-5b00-445b-bfb9-153063dd2484>\",\"WARC-IP-Address\":\"192.0.78.24\",\"WARC-Target-URI\":\"https://theswiftarchitect.com/2018/01/16/using-the-map-function-on-an-optional-in-swift/\",\"WARC-Payload-Digest\":\"sha1:IT3ZGWKI6TSUL5PGOIFZDU45DKV75F3L\",\"WARC-Block-Digest\":\"sha1:KOYO4UZQIGBJGIUL3ULWEORWUOSQWFGO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251672440.80_warc_CC-MAIN-20200125101544-20200125130544-00107.warc.gz\"}"}
https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.17/share/doc/Macaulay2/GraphicalModels/html/_gaussian__Ring.html	[ "# gaussianRing -- ring of Gaussian correlations on n random variables or a graphical model\n\n## Synopsis\n\n• Usage:\ngaussianRing n\ngaussianRing G\n• Inputs:\n• n, an integer, the number of random variables\n• G, an instance of the type Graph, or an instance of the type Digraph, or an instance of the type Bigraph, or an instance of the type MixedGraph\n• Optional inputs:\n• Coefficients => ..., default value QQ, optional input to choose the base field in markovRing or gaussianRing\n• kVariableName => ..., default value k, symbol used for indeterminates in a ring of Gaussian joint probability distributions\n• lVariableName => ..., default value l, symbol used for indeterminates in a ring of Gaussian joint probability distributions\n• pVariableName => ..., default value p, symbol used for indeterminates in a ring of Gaussian joint probability distributions\n• sVariableName => ..., default value s, symbol used for indeterminates in a ring of Gaussian joint probability distributions\n• Outputs:\n• a ring, a polynomial ring with indeterminates associated to the graphical model\n\n## Description\n\nThis function creates a ring whose indeterminates are the covariances of an n dimensional Gaussian random vector. Using a graph, digraph, or mixed graph $G$ as input gives a gaussianRing with extra indeterminates related to the parametrization of the graphical model associated to that graph. Check the details of the gaussianRing for each type of input:\n\nThe indeterminates of the ring - $s_{(i,j)},k_{(i,j)},l_{(i,j)},p_{(i,j)}$ - can be placed into an appropriate matrix format using the functions covarianceMatrix, undirectedEdgesMatrix, directedEdgesMatrix, and bidirectedEdgesMatrix respectively.\n\nThe variable names that appear can be changed using the options sVariableName, lVariableName, pVariableName, and kVariableName\n\n i1 : G = mixedGraph(digraph {{b,{c,d}},{c,{d}}},bigraph {{a,d}}) o1 = MixedGraph{Bigraph => Bigraph{a => {d}} } d => {a} Digraph => Digraph{b => {c, d}} c => {d} d => {} Graph => Graph{} o1 : MixedGraph i2 : R = gaussianRing (G,pVariableName => psi) o2 = R o2 : PolynomialRing i3 : gens R o3 = {l , l , l , psi , psi , psi , psi , psi , s , s , b,c b,d c,d a,a b,b c,c d,d a,d a,a a,b ------------------------------------------------------------------------ s , s , s , s , s , s , s , s } a,c a,d b,b b,c b,d c,c c,d d,d o3 : List\n\nThe routines conditionalIndependenceIdeal, trekIdeal, covarianceMatrix, undirectedEdgesMatrix, directedEdgesMatrix, bidirectedEdgesMatrix, gaussianVanishingIdeal and gaussianParametrization require that the ring be created by this function.\n\n## For the programmer\n\nThe object gaussianRing is ." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5735516,"math_prob":0.9926608,"size":1554,"snap":"2021-04-2021-17","text_gpt3_token_len":420,"char_repetition_ratio":0.17870967,"word_repetition_ratio":0.080168776,"special_character_ratio":0.3133848,"punctuation_ratio":0.244373,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9975002,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-14T10:19:09Z\",\"WARC-Record-ID\":\"<urn:uuid:eb122124-1c1b-41c3-9e56-5e7642565b21>\",\"Content-Length\":\"15012\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef99430f-0808-4d27-bba2-bfc11d99863a>\",\"WARC-Concurrent-To\":\"<urn:uuid:491f31d3-3d93-4de4-9fa8-f542582d4432>\",\"WARC-IP-Address\":\"128.174.199.46\",\"WARC-Target-URI\":\"https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.17/share/doc/Macaulay2/GraphicalModels/html/_gaussian__Ring.html\",\"WARC-Payload-Digest\":\"sha1:HJ74P7AABNLWFUOLSRJO2RDEWQK2O6QQ\",\"WARC-Block-Digest\":\"sha1:ERCQX3DFAVSCKWCU6NMZOD6W6ADVAB44\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038077810.20_warc_CC-MAIN-20210414095300-20210414125300-00465.warc.gz\"}"}
https://www.realcode4you.com/post/hierarchical-clustering-principal-component-analysis-using-r-programming	[ "top of page\nSearch\n\n# Hierarchical Clustering & Principal Component Analysis Using R Programming", null, "Hierarchical Clustering:\n\nThe dataset on American College and University Rankings contains information on 1302 American colleges and universities offering an undergraduate program. For each university, there are 17 continuous measurements (such as tuition and graduation rate) and 2 categorical measurements (such as location by state and whether it is a private or public school). Note that many records are missing some measurements. Our first goal is to estimate these missing values from “similar” records. This will be done by clustering the complete records and then finding the closest cluster for each of the partial records. The missing values will be imputed from the information in that cluster.\n\n• Remove all records with missing measurements from the dataset.\n\n• For all the continuous measurements, run hierarchical clustering using complete linkage and Euclidean distance. Make sure to normalize the measurements. From the dendrogram, how many clusters seem reasonable for describing these data?\n\n• Compare the summary statistics for each cluster and describe each cluster in this context (e.g., “Universities with high tuition, low acceptance rate…”).\n\n• Use the categorical measurements that were not used in the analysis (State and Private/Public) to characterize the different clusters. Is there any relationship between the clusters and the categorical information?\n\n• What other external information can explain the contents of some or all of these clusters?\n\n• Consider Tufts University, which is missing some information. Compute the Euclidean distance of this record from each of the clusters that you found above (using only the measurements that you have). Which cluster is it closest to? Impute the missing values for Tufts by taking the average of the cluster on those measurements.\n\nPrincipal Component Analysis:\n\nThe file ToyotaCorolla.csv contains data on used cars (Toyota Corollas) on sale during late summer of 2004 in the Netherlands. It has 1436 records containing details on 38 attributes, including Price, Age, Kilometers, HP, and other specifications. The goal will be to predict the price of a used Toyota Corolla based on its specifications.\n\n• Identify the categorical variables.\n\n• Explain the relationship between a categorical variable and the series of binary dummy variables derived from it.\n\n• How many dummy binary variables are required to capture the information in a categorical variable with N categories?\n\n• Use R to convert the categorical variables in this dataset into dummy variables, and explain in words, for one record, the values in the derived binary dummies.\n\n• Use R to produce a correlation matrix and matrix plot. Comment on the relationships among variables.\n\nConsider the 3-means algorithm on a set S consisting of the following 6 points in the plane: a= (0,0), b = (8,0), c=(16,0), d=(0,6), e=(8,6), f=(16,6). The algorithm uses the Euclidean distance metric to assign each point to its nearest centroid; ties are broken in favor of the centroid to the left/down. A starting configuration is a subset of 3 starting points from S that form the initial centroids. A 3- partition is a partition of S into 3 subsets; thus {a,b,e}, {c,d}, {f} is a 3-partition; clearly any 3- partition induces a set of three centroids in the natural manner. A 3-partition is stable if repetition of the 3-means iteration with the induced centroids leaves it unchanged.\n\n• How many starting configurations are there?\n\n• What are the stable 3-partitions?\n\n• What is the number of starting configurations leading to each of the stable 3-partitions in (b) above?\n\n• What is the maximum number of iterations from any starting configuration to its stable 3-partition?\n\nImplementation\n\n```mydata <- read.csv(\"Universities.csv\") # read csv file\nmydata\n\n```\n\n#-------Q---1.a---removing missing values----------------------------------------------------------------------------\n\n```mydata_clean <- na.omit(mydata)\n\nlibrary(dplyr)\n\n# print no. of rows and columns in the dataframe\nprint(paste(\"No. of rows:\", nrow(mydata_clean)))\nprint(paste(\"No. of cols:\", ncol(mydata_clean)))```\n\n#------Q---1.b---Heirarchical_Clustering-----------------------------------------------------------------------------\n\n#---------separating continuous and categorical data-----------------------------------------------------------------\n\n#--------------Categorical_data-------------------------------------------------------\n\n```categorical_variables <- names(select(mydata_clean,\"State\",\"Public..1...Private..2.\"))\nprint(categorical_variables)```\n\n#--------------Numerical_data---------------------------------------------------------\n\n```continuous_variables <- setdiff(names(which(sapply(mydata_clean, is.numeric))),(\"Public..1...Private..2.\"))\nprint(continuous_variables)```\n\n#--------------Normalise_data---------------------------------------------------------\n\n`mydata_normalised <- as.data.frame(scale(mydata_clean[,continuous_variables]))`\n\n#--------------dist_method------------------------------------------------------------\n\n`dist_mat <- dist(mydata_normalised, method = 'euclidean')`\n\n#---------------clustering------------------------------------------------------------\n\n`hcluster <- hclust(dist_mat, method = \"complete\")`\n```# plot the dendogram\nplot(hcluster,cex=0.6)\nabline(h = 12, col = 'red')\n\n# from the dendogram we select 6 clusters\n# draw dendogram with red borders around the 6 clusters\nrect.hclust(hcluster, k=6, border=\"green\")\n\n# coloring the dendogram branches according to cluster\nsuppressPackageStartupMessages(library(dendextend))\ndend_obj <- as.dendrogram(hcluster)\ncol_dend <- color_branches(dend_obj, h= 12)\nplot(col_dend)\n\n# getting clusters\nsub_grp <- cutree(hcluster, k = 6)\n# cluster count\ntable(sub_grp)\n\n# Adding the clusters to the table in a column named \"class\"\nmydata_clean['class']<-as.factor(sub_grp)\n\n# grouping the continuous data by 'class' and taking mean (cluster centers)\ncluster_table<-aggregate(mydata_clean[continuous_variables],by=mydata_clean['class'],mean)\nsummary(cluster_table)```\n\n#-------Q---1.c------------------------------------------------------------------------------------------------------\n\n```# In this case there are three major clusters i.e. cluster 1, 2 and 3 and three outliers as cluster i.e. cluster 4, 5\n# and 6. We will\n\n# cluster 1: Universities with low tution (both in_state and out_state) and high acceptance rate.\n\n# cluster 2: Universities with high tution (both in_state and out_state) and low acceptance rate.\n\n# cluster 3: Universities with low in_state tution and high out_state tution and high acceptance rate.```\n\n#-------Q---1.d------------------------------------------------------------------------------------------------------\n\n```# grouping the categorical data by class\ncat_cluster_table<- group_by(mydata_clean[categorical_variables],by=mydata_clean['class'])\nsummary(cat_cluster_table)\n\n# cluster 1\ntable(mydata_clean\\$Public..1...Private..2.[mydata_clean\\$class == 1])\nprint(\"Cluster 1 has a comparable number of Private and Public colleges\")\n\n# cluster 2\ntable(mydata_clean\\$Public..1...Private..2.[mydata_clean\\$class == 2])\nprint(\"Cluster 2 has a majority of Private colleges\")\n\n# cluster 3\ntable(mydata_clean\\$Public..1...Private..2.[mydata_clean\\$class == 3])\nprint(\"Cluster 3 has a majority of Public colleges\")\n\n# all in all, the three clusters have an ample number of Private colleges than Public due to the fact that the\n# dataset has more private colleges as compared to public ones.\n\n# We can infer from cluster 2 that Private colleges has higher tution fees.```\n\n#-------Q---1.e------------------------------------------------------------------------------------------------------\n\n```# Out of the three major clusters, Cluster3 has most number of Full Time undergrads and cluster2 has the least number\n\n# There is a similar trend for Part Time undergrads as that of FUll Time undergrads.\n# We can say that the more students choose Public colleges over Private colleges\n\n# add fees is higher in the cluster3 (majority Public colleges) as compared to the clusters having higher number of\n# private colleges.\n\n# graduation rate is the highest for cluster2 i.e. Universities with high tution (both in_state and out_state) and\n# low acceptance rate.\n\n# no. of new students from top 10 and top 25 is highest for cluster2 i.e. Universities with high tution\n# (both in_state and out_state) and low acceptance rate.```\n\n#-------Q---1.f---Centroid_Comparison--------------------------------------------------------------------------------\n\n```# preapring the data point for comparison\npoint_new <- mydata[mydata['College.Name']=='Tufts University']\npoint_new[is.na(point_new)] <- 0\npoint_new<- as.integer(point_new[-c(1:3)])\n\n# fucntion to calculate the Euclidean Distance\nEuclidean_Distance<-function(x1, x2) sqrt(sum((x1 - x2) ^ 2))\n\n# calculating Euclidean distance of data from each cluster\nfor (i in 1:nrow(cluster_table)){\npoint <- cluster_table[i,][-1]\nprint(paste('Distance from cluster',i,'is', Euclidean_Distance(point_new,point) ))\n}\n\n# From the result we can see that the Tufts University data is closer to cluster 2.\n\n# imputing the missing value by cluster average\n\n#--------------------------------------------------------------------------------------------------------------------\n\n#--------------Qusetion_2--------------------------------------------------------------------------------------------\n\n#------Q--2.a---Categorical_columns----------------------------------------------------------------------------------\n\n```# loading the dataset\nglimpse(data_df) # info of the dataset\n\n#check for factor type columns\nis.fact <- sapply(data_df, is.factor)\n#print the categorical column's name\nprint('The categorical columns are:')\nprint( names(data_df[,is.fact]))\n\n# There are a total of three categorical columns including the model name of the car, that column is not a feature\n# column. Thus, there are two categorical variables in \"Fuel_Type\" and \"Color\".```\n\n#------Q--2.b---relationship between a categorical variable and the series of binary dummy variables-----------------\n\n```# A dummy variable is a variable that takes values of 0 and 1, where the values indicate the presence or absence\n# of a category.\n# When a categorical variable has more than two categories, it can be represented by a set of dummy variables,\n# with one variable for each category.\n#\n# For example in this case the categorical variable \"Fuel_Type\" has two categories i.e. \"Diesel\" and \"Petrol\".\n# Thus there will we two dummy columns one for Diesel and one for Petrol. In the dummy column for Diesel the instances\n# that had a Diesel fuel type would have a value of 1 and the all the other values would be 0. The dummy column for\n# Petrol would have values in a similar manner.\n\n```\n\n#-----Q--2.c---No. of dummy variables required-----------------------------------------------------------------------\n\n```# N or N-1 dummy binary variables are required to capture the information in a categorical variable with N categories.\n#\n# In some situations like Linear Regression, use of all N dummies will cause failure because the nth variable contains\n# redundant information and can be expressed as a linear combination of the others.Thus, generally we only need\n# N-1 dummy variables.\n\n```\n\n#-------Q--2.d---Creating_dummies------------------------------------------------------------------------------------\n\n```#install.packages(\"dummies\")\nlibrary(\"dummies\")\n\ndata_df['Model'] = as.character(data_df['Model'])\ndata_dummy=dummy.data.frame(data_df,dummy.class=\"factor\")\n\ndata_dummy=data_dummy[,!colnames(data_dummy) %in% \"Fuel_TypeCNG\"]\ndata_dummy=data_dummy[,!colnames(data_dummy) %in% \"ColorBeige\"]\nas.array(names(data_dummy))\n\n#\n# We will talk about the dummy variable Fuel_Type_Diesel.\n# The Values in dummy variables are as follows:\n#\n# 1 : if the instance or the row had Fuel_Type = \"Diesel\"\n# 0 : if the instance had Fuel_Type = \"Petrol\"```\n\n#------Q---2.e---Correlation_matrix----------------------------------------------------------------------------------\n\n```# Correlation matrix\ndata_corr <- subset(data_df, select = -c(Model, Fuel_Type, Color,Cylinders))\ncor_matrix=cor(data_corr)\ndata.frame(cor_matrix)\n\n# matrix plot\nheatmap(cor_matrix, Rowv= NA, Colv= NA)\n\n#library(GGally)\n#library(ggplot2)\n#library(corrplot)\n#library(gplots)\n#ggcorr(data_corr, hjust = 0.75, color = \"grey40\", mid = \"#FFFF66\", name=\"Correlation Plot\")\n\n# Age_month is negatively correlated with price (-0.88): This means when the age of the car is more the price of\n# the car goes down.\n#\n# KM is negatively correlated with price(-0.57): When the km increases similar to age the price of the car\n# decreases.\n#\n# Weight is positively correlated with price(0.58): Weight of the Car is positively correlated with the Price\n# (higher price higher weight)\n#\n# KM is positively correlated with Age_month( 0.51): Other than price, KM and Age of Car is positively stating an\n# increasing relationship for both the factors when one increases.\n#\n# Weight is positively correlated with quarterly tax(0.63) : Quarterly Road Tax collected is positively related to\n# Weight stating higher Car weight higher the Quarterly road tax.\n#\n# Radio and Radio Cassette has a high correlation (0.99): It sugggests that a car having a radio would surely" ]	[ null, "https://static.wixstatic.com/media/adae47_78e5960eb0114f15a0a04a96e54c1508~mv2.jpg/v1/fill/w_980,h_551,al_c,q_85,usm_0.66_1.00_0.01,enc_auto/adae47_78e5960eb0114f15a0a04a96e54c1508~mv2.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.71581787,"math_prob":0.78373593,"size":13707,"snap":"2023-40-2023-50","text_gpt3_token_len":2880,"char_repetition_ratio":0.23797709,"word_repetition_ratio":0.039762612,"special_character_ratio":0.33347926,"punctuation_ratio":0.13355513,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9918804,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T12:07:00Z\",\"WARC-Record-ID\":\"<urn:uuid:96bca030-eacd-45ab-9bf5-e24aa285edcc>\",\"Content-Length\":\"1050493\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3ab54767-f86b-41d7-aaef-4a724d6d29f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:56dbed77-ace4-4542-8bc4-3f7de8e0caf9>\",\"WARC-IP-Address\":\"146.75.29.84\",\"WARC-Target-URI\":\"https://www.realcode4you.com/post/hierarchical-clustering-principal-component-analysis-using-r-programming\",\"WARC-Payload-Digest\":\"sha1:SE5KZ2UCYMAGKRPFVJPHTW4EDW72QTVU\",\"WARC-Block-Digest\":\"sha1:BVFI5ATK66UC57XGFUFNEFSJXBJU77V6\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100909.82_warc_CC-MAIN-20231209103523-20231209133523-00744.warc.gz\"}"}
https://sonichours.com/how-are-the-decimals-0-009-and-0-09-related/	[ "General\n\n# How Are The Decimals 0.009 And 0.09 Related\n\nThere is no relationship between the decimals 0.009 and 0.09. However, the two numbers have similar amounts of ones. Therefore, the first number is the larger number. The second number has more tenths than the first. Both numbers have the same amount of decimals. So, they are related to each other. In this article, we will explain how the two numbers are related to each other.\n\nThe decimal 0.009 is one hundredth of a whole number. It is always written with the numerator equal to 100, or a hundredth of a whole number. The word “percent” comes from the Latin “per centum,” which means “by the hundred.” So, 0.09 is 100 times smaller than 0.009. Hence, 0.009 is one-hundredth of a millionth.\n\nThe decimal 0.009 is a tenth of one thousandth of a thousandth. Moreover, it is a fraction with a numerator of one and a denominator of one hundred. The word “percent” comes from Latin “per centum”, which means “by the hundred.” The word “percent” is an extension of the term “percent.”\n\nAs you can see, 0.09 is ten times bigger than 0.009. As a result, the two decimals are related to each other by a factor of 10. Using a number line, a student can compare 0.015 to 0.6, a tenth of a hundredth of a thousandth of a millionth. Then, he or she can compare these two numbers by comparing the numerator values of both decimals.\n\nAs you can see, 0.009 and 0.09 are related to each other by their fractional parts. As a result, a decimal with a fractional part is 10 times greater than a decimal with a numerator of one. When a salesperson is advertising a car for sale, he can count the number of test drivers within the first x days. Similarly, a line of best fit is represented by a straight line.\n\nFor example, a salesperson may want to keep track of the number of test drivers after a certain number of days. The same thing is true for a percentage. A percent is a one hundredth of a given amount. It is a fraction that has a numerator of one and a denominator of one. The term percent is derived from the Latin “per centum”, which means per hundred.\n\nAs a matter of fact, the decimals 0.009 and 0.09 are related in terms of their fractional parts. For example, a cent is one hundredth of a given amount. And a decimal with a fraction of one hundred is a percent. In other words, a percentage is one hundredth of a given amount, but not exactly a cent. It is a fraction with a numerator of one hundred.\n\nThe decimals 0.009 and 0.09 are related because they share the same fractional part. For instance, 0.010 is ten times larger than 0.109. Likewise, 0.9 is a percentage of a certain number. This means that the percentage of a number is 90100. This is the same for the decimals 0.009 and 0.09. If you want to calculate the fraction of a certain amount, you can use a place value table or a number line.\n\nThe decimals 0.009 and 0.09 are similar in that they are smaller than one another. The former is one hundredth of a cent and the latter is a hundredth of a hundredth. But the second is one-tenth of a tenth of a cent. When comparing two numbers, you should remember that the latter is larger than the former.\n\nVisit the rest of the site for more useful articles!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92940235,"math_prob":0.9939101,"size":3145,"snap":"2023-40-2023-50","text_gpt3_token_len":782,"char_repetition_ratio":0.15377268,"word_repetition_ratio":0.087628864,"special_character_ratio":0.25755167,"punctuation_ratio":0.13884993,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99828124,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-03T20:33:50Z\",\"WARC-Record-ID\":\"<urn:uuid:99259edc-acb9-45f6-9939-79568240fda6>\",\"Content-Length\":\"104754\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f954e27f-b0ec-45ee-8d79-4f0dfc832d72>\",\"WARC-Concurrent-To\":\"<urn:uuid:c84122e3-67d1-4d7f-ae33-8f1a40988fee>\",\"WARC-IP-Address\":\"154.12.227.180\",\"WARC-Target-URI\":\"https://sonichours.com/how-are-the-decimals-0-009-and-0-09-related/\",\"WARC-Payload-Digest\":\"sha1:QH5NOJRHPQP3MKKV6JJ3VZBK5M7O5NNF\",\"WARC-Block-Digest\":\"sha1:XTFB3RLXVYMVDILCS2YP3Z4UPXP6PVNA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100508.53_warc_CC-MAIN-20231203193127-20231203223127-00748.warc.gz\"}"}
https://docs.intersystems.com/latest/csp/documatic/%25CSP.Documatic.cls?LIBRARY=ENSLIB&CLASSNAME=%25iKnow.Matching.DictionaryQAPI	[ "# %iKnow.Matching.DictionaryQAPI\n\nclass %iKnow.Matching.DictionaryQAPI extends %iKnow.Queries.AbstractQAPI\n\nThis is an automatically generated class, offering a functionally equivalent set of methods and queries as %iKnow.Matching.DictionaryAPI, exposed as SqlProc methods.\n\nThis class was generated by : %iKnow.Matching.DictionaryAPI.cls\n\n## Methods\n\nclassmethod CreateDictionary(domainId As %Integer, name As %String(MAXLEN=32767), description As %String(MAXLEN=32767)=\"\", defaultLanguage As %String(MAXLEN=32767)=\"en\", defaultProfileId As %Integer = \"\", Output scText As %String = \"\") as %Library.Integer\nProjected as the stored procedure: DictionaryQAPI_CreateDictionary\nCreates a Dictionary and returns its ID.\nclassmethod CreateDictionaryFormat(domainId As %Integer, formatClass As %String(MAXLEN=32767), formatParams As %String(MAXLEN=32767)=\"\", Output scText As %String = \"\") as %Library.Integer\nProjected as the stored procedure: DictionaryQAPI_CreateDictionaryFormat\nCreates a Dictionary Format instance of the supplied formatClass and returns its ID.\nclassmethod CreateDictionaryItem(domainId As %Integer, dictId As %Integer, name As %String(MAXLEN=32767), ByRef URI As %String(MAXLEN=32767), defaultLanguage As %String(MAXLEN=32767)=\"en\", Output scText As %String = \"\") as %Library.Integer\nProjected as the stored procedure: DictionaryQAPI_CreateDictionaryItem\nCreates a Dictionary Item and returns its ID.\nclassmethod CreateDictionaryItemAndTerm(domainId As %Integer, dictId As %Integer, name As %String(MAXLEN=32767), ByRef URI As %String(MAXLEN=32767), language As %String(MAXLEN=32767)=\"en\", Output scText As %String = \"\") as %Library.Integer\nProjected as the stored procedure: DictionaryQAPI_CreateDictionaryItemAndTerm\nShorthand method for creating a Dictionary Item and Term using the Item's name. Returns the Items ID.\nclassmethod CreateDictionaryTerm(domainId As %Integer, dictItemId As %Integer, string As %String(MAXLEN=32767), language As %String(MAXLEN=32767)=\"en\", Output scText As %String = \"\", isProcessed As %Boolean = 0) as %Library.Integer\nProjected as the stored procedure: DictionaryQAPI_CreateDictionaryTerm\n\nCreates a Dictionary Term and returns its ID.\n\nIf isProcessed is 1 (default 0), the term will be marked as processed upon creation and no Dictionary Elements will be created. This parameter is for internal use only and deprecated in 2013.1.\n\nclassmethod CreateDictionaryTermFormat(domainId As %Integer, dictItemId As %Integer, formatClass As %String(MAXLEN=32767), formatParams As %String(MAXLEN=32767)=\"\", Output scText As %String = \"\") as %Library.Integer\nProjected as the stored procedure: DictionaryQAPI_CreateDictionaryTermFormat\nCreates a Dictionary Term composed of a single Dictionary Format and returns the Terms ID.\nclassmethod CreateRegularExpression(pDomainId As %Integer, pDictItemId As %Integer, pRegularExpression As %String(MAXLEN=32767), pReplace As %String(MAXLEN=32767)=\"\", Output scText As %String = \"\") as %Library.Integer\nProjected as the stored procedure: DictionaryQAPI_CreateRegularExpression\nShorthand method to create a Dictionary Term composed of a single Dictionary Format of type %iKnow.Matching.Formats.RegularExpression, matching pRegularExpression and optionally producing output by replacing the matched parts of an entity with pReplace.\nclassmethod DropAllDictionaryData(domainId As %Integer) as %Boolean\nProjected as the stored procedure: DictionaryQAPI_DropAllDictionaryData\nDrops all dictionary data, including matching results. This method will fail if there are managed dictionaries in this domain.\nclassmethod DropDictionary(domainId As %Integer, dictId As %Integer) as %Boolean\nProjected as the stored procedure: DictionaryQAPI_DropDictionary\nDeletes a Dictionary and all related items, terms, elements and matches.\nclassmethod DropDictionaryItem(domainId As %Integer, dictItemId As %Integer) as %Boolean\nProjected as the stored procedure: DictionaryQAPI_DropDictionaryItem\nDeletes a Dictionary Item with all related terms, elements and matches.\nclassmethod DropDictionaryTerm(domainId As %Integer, dictTermId As %String(MAXLEN=32767)) as %Boolean\nProjected as the stored procedure: DictionaryQAPI_DropDictionaryTerm\nDrops a single Dictionary Term and its matching results\nclassmethod GetDictionaryCount(pDomainId As %Integer, Output scText As %String = \"\") as %Library.Integer\nProjected as the stored procedure: DictionaryQAPI_GetDictionaryCount\nclassmethod GetDictionaryId(domainId As %Integer, name As %String(MAXLEN=32767), Output scText As %String = \"\") as %Library.Integer\nProjected as the stored procedure: DictionaryQAPI_GetDictionaryId\nLooks up the Dictionary ID corresponding to the supplied Dictionary name.\nclassmethod GetDictionaryItemIdByURI(domainId As %Integer, URI As %String(MAXLEN=32767), Output scText As %String = \"\") as %Library.Integer\nProjected as the stored procedure: DictionaryQAPI_GetDictionaryItemIdByURI\nLooks up the Dictionary Item ID corresponding to the supplied Dictionary Item URI.\nclassmethod GetItemCount(pDomainId As %Integer, pDictIds As %String(MAXLEN=32767)=\"\", Output scText As %String = \"\") as %Library.Integer\nProjected as the stored procedure: DictionaryQAPI_GetItemCount\nclassmethod GetTermCount(pDomainId As %Integer, pDictIds As %String(MAXLEN=32767)=\"\", Output scText As %String = \"\") as %Library.Integer\nProjected as the stored procedure: DictionaryQAPI_GetTermCount\nclassmethod GetTermCountByItem(pDomainId As %Integer, pDictItemIds As %String(MAXLEN=32767)=\"\", Output scText As %String = \"\") as %Library.Integer\nProjected as the stored procedure: DictionaryQAPI_GetTermCountByItem\nclassmethod HasDictionaries(pDomainId As %Integer) as %Library.Boolean\nProjected as the stored procedure: DictionaryQAPI_HasDictionaries\nReturns whether or not there are any dictionaries in the domain\n\n## Queries\n\nquery GetDictionaries(domainId As %Integer, page As %Integer = 1, pageSize As %Integer = 10, pIncludeCrossDomain As %Boolean = 0)\nSelects dictId As %Integer, name As %String(MAXLEN=32767), description As %String(MAXLEN=32767), defaultLanguage As %String(MAXLEN=32767)\nReturns all Dictionaries in this domain. If pIncludeCrossDomain = 1, any cross-domain dictionaries registered in \"domain 0\" are also included in the result, using a negative value for their ID.\nquery GetDictionaryItems(domainId As %Integer, dictId As %Integer, page As %Integer = 1, pageSize As %Integer = 10)\nSelects dictItemId As %Integer, name As %String(MAXLEN=32767), URI As %String(MAXLEN=32767), defaultLanguage As %String(MAXLEN=32767)\nReturns all Dictionary Items for the given Dictionary ID.\nquery GetDictionaryItemsAndTerms(domainId As %Integer, dictId As %Integer, page As %Integer = 1, pageSize As %Integer = 10)\nSelects dictItemId As %Integer, name As %String(MAXLEN=32767), URI As %String(MAXLEN=32767), dictTermId As %Integer, string As %String(MAXLEN=32767), language As %String(MAXLEN=32767), isProcessed As %Boolean\nReturns all Dictionary Items and their Terms for the given Dictionary ID.\nquery GetDictionaryTermsByItem(domainId As %Integer, dictItemId As %Integer, page As %Integer = 1, pageSize As %Integer = 10)\nSelects dictTermId As %Integer, string As %String(MAXLEN=32767), language As %String(MAXLEN=32767), isProcessed As %Boolean\nReturns all the Dictionary Terms for the given Dictionary Item ID.\nquery GetItemsByName(pDomainId As %Integer, pString As %String(MAXLEN=32767), pPage As %Integer = 1, pPageSize As %Integer = 10, pFilter As %String(MAXLEN=32767)=\"\", pDictIds As %String(MAXLEN=32767)=\"\", pMode As %Integer = \\$\\$\\$USEPARTS, pCheck As %String(VALUELIST=\"NAME,URI\")=\"NAME\", pLang As %String(MAXLEN=32767)=\"en\")\nSelects dictId As %Integer, dictName As %String(MAXLEN=32767), itemId As %Integer, itemName As %String(MAXLEN=32767), URI As %String(MAXLEN=32767), language As %String(MAXLEN=32767)\nNote: pFilter is ignored when looking at cross-domain dictionaries\nquery GetTermsByName(pDomainId As %Integer, pString As %String(MAXLEN=32767), pPage As %Integer = 1, pPageSize As %Integer = 10, pFilter As %String(MAXLEN=32767)=\"\", pDictIds As %String(MAXLEN=32767)=\"\", pMode As %Integer = \\$\\$\\$USEPARTS, pLang As %String(MAXLEN=32767)=\"en\")\nSelects dictId As %Integer, dictName As %String(MAXLEN=32767), itemId As %Integer, itemName As %String(MAXLEN=32767), URI As %String(MAXLEN=32767), termId As %Integer, term As %String(MAXLEN=32767), language As %String(MAXLEN=32767)\nNote: pFilter is ignored when looking at cross-domain dictionaries\n\n## Inherited Members\n\n### Inherited Methods\n\nFeedbackOpens in a new tab" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6972917,"math_prob":0.5429826,"size":4318,"snap":"2022-27-2022-33","text_gpt3_token_len":906,"char_repetition_ratio":0.25266573,"word_repetition_ratio":0.14285715,"special_character_ratio":0.17253359,"punctuation_ratio":0.125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.960773,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-19T07:13:49Z\",\"WARC-Record-ID\":\"<urn:uuid:d25c162a-1a86-4c7d-a359-d5a61c0cec5c>\",\"Content-Length\":\"60479\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f9c81b61-9862-492a-8fb9-5746587a442a>\",\"WARC-Concurrent-To\":\"<urn:uuid:79c87e6b-dc4e-4906-9eba-644c636005b7>\",\"WARC-IP-Address\":\"104.104.117.151\",\"WARC-Target-URI\":\"https://docs.intersystems.com/latest/csp/documatic/%25CSP.Documatic.cls?LIBRARY=ENSLIB&CLASSNAME=%25iKnow.Matching.DictionaryQAPI\",\"WARC-Payload-Digest\":\"sha1:OODGWR5A7IGWEWO6UZWH4YVZLSQRYPY6\",\"WARC-Block-Digest\":\"sha1:MQFKMOYWW3O2DIVZAZC4VHDATGJDAIJT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573630.12_warc_CC-MAIN-20220819070211-20220819100211-00333.warc.gz\"}"}
https://iftekhar.net/blog/how-to-validate-integer-in-php/	[ "# Validate Integer in PHP\n\nEvery now and then we need to check if the value of a variable is an integer or not. Recently I came across a forum thread where someone is trying to understand why his/her snippet is not giving the appropriate result that he/she is looking for. In other words, if the variable is an integer this function should return TRUE, otherwise it would return FLASE. Here is a scenario for your thoughts.\n\n``````<?php\n\\$userid = \"12345\";\nif (is_int(\\$userid)) {\necho 'its an integer';\n} else {\necho 'its not an integer';\n}\n?>``````\n\nMany user on that forum suggested that he/she should be using is_numeric function that will give him/her the result he/she might be looking for. This is how it worked.\n\n``````<?php\n\\$userid = \"12345\";\nif (is_numeric(\\$userid)) {\necho 'its an integer';\n} else {\necho 'its not an integer';\n}\n?>``````\n\nIndeed, it does work. However, what we need to understand is that when we assign the value of a variable within “”, PHP will consider the value as string, not an integer. One way to make the first snippet to work is to remove the quotation(“”) mark from the variable or declare the variable value to be an integer to begin with (then test it with is_int function). Here is how it works.\n\n``````<?php\n// use it without \"\"\n\\$userid = 12345;\n// or declare the variable type\n\\$userid = (int) \"12345\";\nif (is_int(\\$userid)) {\necho 'its an integer';\n} else {\necho 'its not an integer';\n}\n?>``````\n\nEither way, is_int function should work as it should be. In my personal opinion though, I really don’t think anyone should be using quote to define an integer value. If you have to, go with declaring the variable type. You can also use the following method to test your variable.\n\n``````<?php\n\\$userid = 12345;\nif (filter_var(\\$userid, FILTER_VALIDATE_INT)) {\necho 'its an integer';\n} else {\necho 'its not an integer';\n}\n?>``````\n\nIn this case, it really doesn’t matter if you use the quote or not. It will validate your variable as an integer. I hope you would find this post to be useful.\n\nReferences: is_int, is_numeric, filter_var" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8276641,"math_prob":0.83141315,"size":2007,"snap":"2020-24-2020-29","text_gpt3_token_len":484,"char_repetition_ratio":0.16275586,"word_repetition_ratio":0.15,"special_character_ratio":0.2745391,"punctuation_ratio":0.13065326,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95912606,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-03T13:23:27Z\",\"WARC-Record-ID\":\"<urn:uuid:7c0c24ec-69ac-4774-bc7f-32cffe2a1926>\",\"Content-Length\":\"18454\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e8ca9a9f-4abc-4820-9673-3d1e90597482>\",\"WARC-Concurrent-To\":\"<urn:uuid:04f23878-9684-4d26-b6ad-4fcca86e8eb7>\",\"WARC-IP-Address\":\"74.208.236.158\",\"WARC-Target-URI\":\"https://iftekhar.net/blog/how-to-validate-integer-in-php/\",\"WARC-Payload-Digest\":\"sha1:LGVV74POKT5NVYYWZXIANB2DLUBY3MIE\",\"WARC-Block-Digest\":\"sha1:SFKTAJYDCLLJVCIRKCJMUWKGFK7RM6NH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655882051.19_warc_CC-MAIN-20200703122347-20200703152347-00005.warc.gz\"}"}
https://www.epb.idv.tw/ge-ren-zuo-pin/xue-xi-zhuang-kuang/ren-gong-zhi-hui-xian-dai-fang-fa/qi-ta/jiyinyichuanyansuanfa	[ "基因遺傳演算法\n\nPost date: 2013/1/16 上午 04:28:50\n\n● 局部極大值：局部極大值是一個比它的每個相鄰狀態都高的峰頂，但是比全局最大值要低。爬山法演算法到達局部極大值附近就會被拉向峰頂，然後被卡在局部極大值處而無處可走。圖4.1示意性地表現了這種情況。\n\n● 山脊：圖4.4 顯示了山脊的情況。山脊造成一系列的局部極大值，貪婪演算法處理這種情況是很難的。\n\n● 高原：高原是在狀態空間地形圖上的一塊平坦區域。它可能是一塊平坦的局部極大值，不存在上山的出路，或者是一個山肩，從山肩還有可能取得進展(參見圖4.1)。爬山法搜尋可能會在高原迷路。\n\nhttp://blog.udn.com/puzzlez/4342425\n\n※西洋棋由白棋先行。棋盤右下角（h1）必須為白格。\n\n1.只要沒有棋子擋住，可以直走無限步，也可以斜走無限步。\n\n2.皇后是子力最強的棋子。\n\n◆棋子的走法與吃法\n\n1\n\n1\n\n2\n\n2\n\n2\n\n8\n\nGenetic Algorithm\n\nGenetic Algorithm(基因演算法，又稱遺傳演算法)主要概念是使用自然界上，兩性生殖的演化機制，使用遺傳(inheritance)、選擇(selection)、雜交(crossover)、突變(mutation)、以及『物競天擇、適者生存』的概念，讓優秀的個體會被留下來，繼續繁殖；而相對差的個體將會慢慢被淘汰，直到絕種。此演算法基於優生學的理論基礎，認為相對優秀的基因互相交配才可以產生更為優秀的後代，但是，為了確保後代的多樣性，又加入了突變的機制，不會讓後代最終都趨於一致。一般來說，基因演算法都能夠找到不錯的解。\n\n1. 一個基因序列(genetic representation)用以表示其問題的解\n2. 一個用來評估問題解的適應力函式(fitness function)\n\n1. 進行初始化。\n2. 重複進行選擇>雜交>突變，直到產生一代所需的個體數量時，即完成一次繁殖。\n3. 當終止條件達成時，停止此演算法。\n\n1. 定義基因序列，使用染色體表示此問題的解。\n• 使用8個數字(0~7)來代表這8列上的皇后，每一個數字代表該列的皇后是位於第幾行。\n1. 例如：24751314、41700256。\n1. 定義適應力函式，用來評估此解的好壞。\n• 從第一列皇后開始計算，若不會吃右列的其他皇后，每一個加一分，故最高分為7+6+5+4+3+2+1=28分。\n• 例如：f(41700256)=24、f(12277051)=23、f(24713603)=27、\n1. f(71420635)=28(此為最佳解)\n1. 使用亂數產生N個第一代個體群。\n• 設N=5，則隨機產生5組第一代個體\n• 例如：51341714、16705214、57130241、67501230、14074251。\n\n1. 評估所有群體中的個體適應力。\n2. 根據適應力來選出兩個個體，適應力越高的個體會有較高的機率被選中。\n• 不僅僅只挑出適應力最高的是因為這樣做可能會產生出局部最佳解(Local solution)，並不是全域性的最佳解(Global solution)。\n\n• 雜交的方式有很多種，我們使用單點雜交(One-point crossover)的方式。\n1. 隨機選擇一個位置將此兩個基因切成兩段。\n• 51341714和16705214 => 5134 \\ 1714 和 1670 \\ 5214\n2. 進行雜交\n• 51345214 和 16701714\n\n• 突變方式亦有很多種，我們使用單點突變(Single point mutation)的方式。\n• 當微小機率發生時，將此染色體的其中一個基因進行突變。\n• 51345214 -> 51345?14 -> 51345714\n\n1. 進行了特定次數的繁衍。\n2. 出現滿足特定適應力的個體。\n• 例如：解八皇后問題中，若出現適應力為28的個體，即為找到最佳解。\n3. 達到計算耗費的資源限制(例如：計算時間、計算空間等)。\n4. 人為干涉。\n5. 以上兩種或是更多種的組合。\n\n// 染色體長度\n\n#define GEN_LENGTH 8\n\n// 每一代的數量\n\n#define POPULATION_COUNT 50\n\n// 繁衍代數\n\n#define GENERATION_COUNT 10000\n\n// 突變機率\n\n#define MUTATION_RATE 0.1\n\nclass eightQ\n\n{\n\npublic:\n\neightQ();\n\nvoid randQ();\n\nint getScore();\n\nvoid show();\n\nvector<int> sit;\n\n};\n\nint eightQ::getScore()\n\n{\n\nint nScore = 0;\n\nfor (int i = 0; i < GEN_LENGTH; ++i)\n\n{\n\nint iPosX = i;\n\nint iPoxY = sit[i];\n\n//與其他皇后比較\n\nfor (int j = i + 1; j < GEN_LENGTH; ++j)\n\n{\n\n//若不與此皇后同ROW\n\nif (!(iPoxY == sit[j]))\n\n//若不在此皇后右斜上\n\nif (!(iPoxY - iPosX == sit[j] - j))\n\n//若不在此皇后右斜下\n\nif (!(iPosX + iPoxY == sit[j] + j))\n\nnScore++;\n\n}\n\n}\n\nreturn nScore;\n\n}\n\nvoid GA::init()\n\n{\n\neightQ newQ;\n\n// 隨機產生第一代\n\nfor(int i = 0 ; i < POPULATION_COUNT ; ++i)\n\n{\n\nnewQ.randQ();\n\nthisGen.push_back(newQ);\n\n}\n\ntheBest = newQ;\n\n}\n\nvoid GA::reproduction()\n\n{\n\nMomPool.clear();\n\nint nScoreSum = 0;\n\nvector<int> vAccumulate;\n\n// 將此代所有個體的適應力累加\n\nfor(int i = 0 ; i < POPULATION_COUNT ; ++i)\n\n{\n\nnScoreSum += thisGen[i].getScore();\n\nvAccumulate.push_back(nScoreSum);\n\n}\n\n// 依照適應力的比例選擇母親與父親\n\nfor(int i = 0 ; i < POPULATION_COUNT ; ++i)\n\n{\n\nint nRand = rand() % nScoreSum;\n\nfor(int j = 0 ; j < POPULATION_COUNT ; ++j)\n\nif(nRand < vAccumulate[j])\n\n{\n\nbreak;\n\n}\n\nnRand = rand() % nScoreSum;\n\nfor(int j = 0 ; j < POPULATION_COUNT ; ++j)\n\nif(nRand < vAccumulate[j])\n\n{\n\nMomPool.push_back(thisGen[j]);\n\nbreak;\n\n}\n\n}\n\n}\n\nvoid GA::crossover()\n\n{\n\nint nPos;\n\nthisGen.clear();\n\n// 使用單點雜交\n\nfor(int i = 0 ; i < POPULATION_COUNT ; ++i)\n\n{\n\n// 決定雜交位置 3~6\n\nnPos = rand()%4 + 3;\n\neightQ newQ;\n\n// 前面使用爸爸的染色體\n\nfor(int j = 0 ; j < nPos ; ++j)\n\n{\n\n}\n\n// 後面使用媽媽的染色體\n\nfor(int j = nPos ; j < GEN_LENGTH ; ++j)\n\n{\n\nnewQ.sit[j] = MomPool[i].sit[j];\n\n}\n\nthisGen.push_back(newQ);\n\n}\n\n}\n\nvoid GA::mutation()\n\n{\n\n// 每一個染色體判斷要不要突變\n\nfor(int i = 0 ; i < POPULATION_COUNT ; ++i)\n\n{\n\ndouble dRate = (double)(rand() % 10000) / 10000.0f;\n\nif(dRate < MUTATION_RATE)\n\n{\n\n// 決定突變位置\n\nint nPos = MyRand();\n\nthisGen[i].sit[nPos] = MyRand();\n\n}\n\n// 計算是否比最佳解更好\n\nif(thisGen[i].getScore() > theBest.getScore())\n\ntheBest = thisGen[i];\n\n}\n\n}\n\nint main()\n\n{\n\nsrand((unsigned)time(0));\n\nGA ga;\n\nint nCounter = 0;\n\n// 初始化\n\nga.init();\n\ncout << \"基因演算法開始\" << endl;\n\nwhile(nCounter < GENERATION_COUNT)\n\n{\n\nga.reproduction();\n\nga.crossover();\n\nga.mutation();\n\n//ga.show();\n\n++nCounter;\n\n//cout << \"第\" << nCounter << \"代繁殖完成！\" << endl;\n\n// 若最好的個體已經找到，即可結束。\n\nif(ga.theBest.getScore() >= 28)\n\nbreak;\n\n}\n\ncout << \"第\" << nCounter << \"代所有個體如下所示：\" << endl;\n\nga.show();\n\ncout << \"最佳解為\";\n\nga.theBest.show();\n\ncout << \"其分數為\" << ga.theBest.getScore() << endl;\n\nsystem(\"PAUSE\");\n\nreturn 0;\n\n}" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.9485261,"math_prob":0.99291235,"size":9838,"snap":"2019-43-2019-47","text_gpt3_token_len":10469,"char_repetition_ratio":0.06996136,"word_repetition_ratio":0.10778443,"special_character_ratio":0.25553975,"punctuation_ratio":0.1223193,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9828408,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-15T19:00:31Z\",\"WARC-Record-ID\":\"<urn:uuid:0fb8b68b-637c-4918-9d8f-d0f70543406d>\",\"Content-Length\":\"369969\",\"Content-Type\":\"application/http; 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https://forum.alphasoftware.com/showthread.php?124018-Calculation-Help&p=746893	[ "", null, "1. ##", null, "Calculation Help\n\nHi All,\n\nSimple calculation and have trouble when it is negative number\n\nI have this and\n1. Is there better way to calculate the Balance?\n2. Do I have to change negative to positive number?\n\nCode:\n```'add number\na = 542\nb = -211\t\t'negative number\nc = 186\nd = 374\n\nx = a+b+c+d\t\t'Get total -- ok\n\n'get the balance\n'balance = a - b\t\t'It is wrong result when the number is negative. Unknown negative number or not\n\nif b < 0 then\n'msgbox(\"b is less than 0 and ...\")\n'How to change to positive number?\nbalance = a + b\t\t'Unknown negative number or not\nelse\nbalance = a - b\t\t'Unknown negative number or not\nend if\nui_msg_box(\"result\",\"balance \"+balance)```", null, "", null, "Reply With Quote\n\n2. ##", null, "Re: Calculation Help\n\nif a,b,c,d are transactions, what is the beginning balance?\n\nZero?\n\nWhat do you expect the balance to be?", null, "", null, "Reply With Quote\n\n3. ##", null, "Re: Calculation Help\n\nHi Al,\n\nIt is general subtract calculation, not typical \"Balance\" calculation.\n\nwhat is the beginning balance? -- \"A\" Value\nWhat do you expect the balance to be? --> a(542) - b(211) = 331.00\n\nI want to write a code as xBal = a - b but when the \"B\" is negative number then trouble.\n\nThank you", null, "", null, "Reply With Quote\n\n4. ##", null, "Re: Calculation Help", null, "Originally Posted by johnkoh", null, "Hi Al,\n\nIt is general subtract calculation, not typical \"Balance\" calculation.\n\nwhat is the beginning balance? -- \"A\" Value\nWhat do you expect the balance to be? --> a(542) - b(211) = 331.00\n\nI want to write a code as xBal = a - b but when the \"B\" is negative number then trouble.\n\nThank you\na-abs(b)", null, "", null, "Reply With Quote\n\n5. ##", null, "Re: Calculation Help\n\nThanks, Al", null, "", null, "Reply With Quote\n\n####", null, "Posting Permissions\n\n• You may not post new threads\n• You may not post replies\n• You may not post attachments\n• You may not edit your posts\n•" ]	[ null, "https://no-cache.hubspot.com/cta/default/106714/f3b6785e-55e4-4ce6-be0f-9365ae9d4371.png", null, "https://forum.alphasoftware.com/images/icons/icon1.png", null, "https://forum.alphasoftware.com/images/misc/progress.gif", null, "https://forum.alphasoftware.com/clear.gif", null, "https://forum.alphasoftware.com/images/icons/icon1.png", null, "https://forum.alphasoftware.com/images/misc/progress.gif", null, "https://forum.alphasoftware.com/clear.gif", null, "https://forum.alphasoftware.com/images/icons/icon1.png", null, "https://forum.alphasoftware.com/images/misc/progress.gif", null, "https://forum.alphasoftware.com/clear.gif", null, "https://forum.alphasoftware.com/images/icons/icon1.png", null, "https://forum.alphasoftware.com/images/misc/quote_icon.png", null, "https://forum.alphasoftware.com/images/buttons/viewpost-right.png", null, "https://forum.alphasoftware.com/images/misc/progress.gif", null, "https://forum.alphasoftware.com/clear.gif", null, "https://forum.alphasoftware.com/images/icons/icon1.png", null, "https://forum.alphasoftware.com/images/misc/progress.gif", null, "https://forum.alphasoftware.com/clear.gif", null, "https://forum.alphasoftware.com/images/buttons/collapse_40b.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79212993,"math_prob":0.986042,"size":639,"snap":"2020-10-2020-16","text_gpt3_token_len":178,"char_repetition_ratio":0.2015748,"word_repetition_ratio":0.03305785,"special_character_ratio":0.29890454,"punctuation_ratio":0.08,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99829304,"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],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-21T21:58:38Z\",\"WARC-Record-ID\":\"<urn:uuid:1bd6c5ad-377e-4d8b-bf39-0152e53015de>\",\"Content-Length\":\"55913\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c4877493-184d-4308-b64a-53b5ac7021da>\",\"WARC-Concurrent-To\":\"<urn:uuid:1832636a-6e27-4568-8760-6242650d17e0>\",\"WARC-IP-Address\":\"216.81.183.149\",\"WARC-Target-URI\":\"https://forum.alphasoftware.com/showthread.php?124018-Calculation-Help&p=746893\",\"WARC-Payload-Digest\":\"sha1:MBCX7ZYJXGOEDXW5CKXZO7AXNPA3MH6K\",\"WARC-Block-Digest\":\"sha1:YJBDDTWDXEFSGILHXBQIAH3CI5QUXURF\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145538.32_warc_CC-MAIN-20200221203000-20200221233000-00072.warc.gz\"}"}
https://program-transformation.org/view/Sts/StatementStatisticsUsingTXL.html	[ "Page\n\nWeb\n\nWiki\n\n# Statement Statistics Using TXL\n\nSoftware Transformation Systems\nTXL solution to TIL Chairmarks #4.2: Collecting statement statistics.\n\n-- JamesCordy - 28 Oct 2005\n\nFile \"TILstats.Txl\"\n\n```% Gather statement statistics for a Tiny Imperative Language program\n% Jim Cordy, October 2005\n\n% Gathers and outputs statistics on the number of each kind of statement\n% in a program. Replaces the program itself with an empty one.\n\n% Begin with the TIL base grammar\ninclude \"TIL.Grm\"\n\n% Compute and output statement kind statistics, replace program with an empty one.\n% There are many different ways to do this - this naive way is simple and\n% obvioulsy correct, but exposes TXL's need for generics.\n% Another less clear solution could use polymorphism to avoid the repetition.\n\nfunction main\nreplace [program]\nProgram [program]\n\n% Count each kind of statement we're interested in\n% by extracting all of each kind from the program\n\nconstruct Statements [statement]\n_ [^ Program]\nconstruct StatementCount [number]\n_ [length Statements] [putp \"Total: %\"]\n\nconstruct Declarations [declaration]\n_ [^ Program]\nconstruct DeclarationsCount [number]\n_ [length Declarations] [putp \"Declarations: %\"]\n\nconstruct Assignments [assignment_statement]\n_ [^ Program]\nconstruct AssignmentsCount [number]\n_ [length Assignments] [putp \"Assignments: %\"]\n\nconstruct Ifs [if_statement]\n_ [^ Program]\nconstruct IfCount [number]\n_ [length Ifs] [putp \"Ifs: %\"]\n\nconstruct Whiles [while_statement]\n_ [^ Program]\nconstruct WhileCount [number]\n_ [length Whiles] [putp \"Whiles: %\"]\n\nconstruct Fors [for_statement]\n_ [^ Program]\nconstruct ForCount [number]\n_ [length Fors] [putp \"Fors: %\"]\n\n_ [^ Program]\n\nconstruct Writes [write_statement]\n_ [^ Program]\nconstruct WriteCount [number]\n_ [length Writes] [putp \"Writes: %\"]\n\nby\n% nothing\nend function\n```\n\nExample run:\n\n```<linux> cat factors.til\n// Factor an input number\nvar n;\nwrite \"The factors of n are\";\nvar f;\nf := 2;\nwhile n != 1 do\nwhile (n / f) f = n do\nwrite f;\nn := n / f;\nend\nf := f + 1;\nend\n<linux> txl factors.til TILstats.Txl\nTXL v10.4a (15.6.05) (c)1988-2005 Queen's University at Kingston\nCompiling TILstats.Txl ...\nParsing factors.til ...\nTransforming ...\nTotal: 11\nDeclarations: 2\nAssignments: 3\nIfs: 0\nWhiles: 2\nFors: 0" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.74126226,"math_prob":0.8058993,"size":2383,"snap":"2022-27-2022-33","text_gpt3_token_len":673,"char_repetition_ratio":0.18116856,"word_repetition_ratio":0.005509642,"special_character_ratio":0.27864036,"punctuation_ratio":0.15151516,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97017217,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-28T21:26:12Z\",\"WARC-Record-ID\":\"<urn:uuid:6bb89eaf-383c-4b49-b088-5767031d1ae0>\",\"Content-Length\":\"8773\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8b8e14a2-71ad-4ce6-b9f9-282d1d3ad077>\",\"WARC-Concurrent-To\":\"<urn:uuid:1d49ac3c-aacb-477e-8355-20e85ab60a24>\",\"WARC-IP-Address\":\"131.180.125.32\",\"WARC-Target-URI\":\"https://program-transformation.org/view/Sts/StatementStatisticsUsingTXL.html\",\"WARC-Payload-Digest\":\"sha1:OQ2Z2NN2KKSJS7RNHU4M66R5KNOHZBVN\",\"WARC-Block-Digest\":\"sha1:7I5HY67WG5Y67O52G63GP5Q3CUO5TSDM\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103617931.31_warc_CC-MAIN-20220628203615-20220628233615-00723.warc.gz\"}"}
http://zhaoxuhui.top/blog/2020/10/14/gee-note2-basic-introduction-and-ndvi-example.html	[ "## Oct 14,2020 10407 words 38 min\n\n#### 2.基本数据类型\n\n``````// 新建一个服务器上的字符串\nvar serverString = ee.String('This is on the server.');\nprint('String on the server:', serverString);\n\n// 新建一个服务器上的数字\nvar serverNumber = ee.Number(3.14159);\nprint('number=',serverNumber);\n\n// 新建一个服务器上的列表\nvar serverList = ee.List([1,5,6,9,2]);\nprint('Sequence:',serverList);\n// 注意，由于是服务器上的List，所以需要通过get()函数获得内容，而不是直接索引\nprint('item 2:',serverList.get(2));\n\n// 新建一个服务器上的字典，包含三个键值对\nvar serverDict = ee.Dictionary({\ne:Math.E,\npi:Math.PI,\nphi:(1+Math.sqrt(5))/2\n});\n// 和上面一样，需要通过get()函数获取数据\nprint('Golden ratio:',serverDict.get('phi'));\nprint('Keys:',serverDict.keys());\n\n// 新建一个服务器上的日期\nvar serverDate = ee.Date('2020-10-13');\nprint('date:',serverDate);\n``````\n\n#### 3.基本语法\n\n##### (1)循环\n\n``````// This generates a list of numbers from 1 to 10.\nvar myList = ee.List.sequence(1, 10);\n\n// The map() operation takes a function that works on each element independently\n// and returns a value. You define a function that can be applied to the input.\nvar computeSquares = function(number) {\n// We define the operation using the EE API.\nreturn ee.Number(number).pow(2);\n};\n\n// Apply your function to each item in the list by using the map() function.\nvar squares = myList.map(computeSquares);\nprint(squares); // [1, 4, 9, 16, 25, 36, 49, 64, 81]\n``````\n##### (2)条件判断\n\n``````// The following function determines if a number is even or odd. The mod(2)\n// function returns 0 if the number is even and 1 if it is odd (the remainder\n// after dividing by 2). The input is multipled by this remainder so even\n// numbers get set to 0 and odd numbers are left unchanged.\nvar getOddNumbers = function(number) {\nnumber = ee.Number(number); // Cast the input to a Number so we can use mod.\nvar remainder = number.mod(2);\nreturn number.multiply(remainder);\n};\n\n// This generates a list of numbers from 1 to 10.\nvar myList = ee.List.sequence(1, 10);\n\n// filtering\nvar newList = myList.map(getOddNumbers);\n\n// Remove the 0 values.\nvar oddNumbers = newList.removeAll();\nprint(oddNumbers);\n``````\n\n``````var collection = ee.ImageCollection('LANDSAT/LC08/C01/T1_TOA');\n\n// Divide the collection into 2 subsets and apply a different algorithm on them.\nvar subset1 = collection.filter(ee.Filter.lt('SUN_ELEVATION', 40));\nvar subset2 = collection.filter(ee.Filter.gte('SUN_ELEVATION', 40));\n\nvar processed1 = subset1.map(function(image) {\nreturn image.multiply(2);\n});\nvar processed2 = subset2;\n\n// Merge the collections to get a single collection.\nvar final = processed1.merge(processed2);\nprint('Original collection size', collection.size());\nprint('Processed collection size', final.size());\n``````\n\n##### (3)累加\n\n``````var algorithm = function(current, previous) {\nprevious = ee.List(previous);\nvar n1 = ee.Number(previous.get(-1));\nvar n2 = ee.Number(previous.get(-2));\n};\n\n// Compute 10 iterations.\nvar numIteration = ee.List.repeat(1, 10);\nvar start = [0, 1];\nvar sequence = numIteration.iterate(algorithm, start);\nprint(sequence); // [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]\n``````\n\n#### 4.可视化影像与波段\n\n``````// Instantiate an image with the Image constructor.\nvar image = ee.Image('CGIAR/SRTM90_V4');\n\n// Zoom to a location.\nMap.setCenter(117.3798, 32.9049, 4); // Center on the Bengbu City.\n\n// Display the image on the map.\n``````\n\n#### 5.影像集\n\n``````// 获取Landsat8 TOA影像\nvar l8 = ee.ImageCollection('LANDSAT/LC08/C01/T1_TOA');\n\n// 定义一个目标点\nvar point = ee.Geometry.Point(117.0721,33.2363);\n\n// 调用filterBounds筛选位置\nvar spatialFiltered = l8.filterBounds(point);\nprint('spatialFiltered',spatialFiltered);\n\n// 再调用filterDate筛选时间\nvar temporalFiltered = spatialFiltered.filterDate('2015-01-01','2015-12-31');\nprint('temporalFiltered',temporalFiltered);\n\n// 到这一步之后，还可能有很多影像的，我们可以进一步按照云层覆盖进行排序\nvar sorted = temporalFiltered.sort('CLOUD_COVER');\n// 选取含云量最少的那一景\nvar scene = sorted.first();\n\n// 将影像放到当前地图的中心，缩放9级\nMap.centerObject(scene,9);\n\n// Landsat8的RGB波段分别是B4、B3、B2\n// 最大值设为0.3，最小值默认为0\nvar visParams = {bands:['B4','B3','B2'],max:0.3};\n\n// 展示影像，名称为true-color composite\n``````\n\n#### 6.实例练习-基于Landsat数据计算NDVI\n\n``````// 指定某个目标点\nvar point = ee.Geometry.Point([117.0721,33.2363]);\n\n// 导入Landsat-8 TOA影像数据集\nvar l8 = ee.ImageCollection('LANDSAT/LC08/C01/T1_TOA');\n\n// 获取2015年云量最少的影像\nvar image = ee.Image(\nl8.filterBounds(point)\n.filterDate('2015-01-01', '2015-12-31')\n.sort('CLOUD_COVER')\n.first()\n);\n\n// 计算Normalized Difference Vegetation Index (NDVI)\n// Landsat8 第5波段是近红外(NIR)，第4波段是红波段(Red)\nvar nir = image.select('B5');\nvar red = image.select('B4');\n// 套用NDVI计算公式(NIR-Red)/(NIR+Red)\n\n// 设置影像到地图中心\nMap.centerObject(image,9);\n// 设置最大最小值和显示颜色\nvar ndviParams = {min:-1,max:1,palette:['blue','white','green']};\n// 向地图中添加图层\n``````\n\n``````// 导入Landsat-8 TOA影像数据集\nvar l8 = ee.ImageCollection('LANDSAT/LC08/C01/T1_TOA');\n\nvar ndvi = image.normalizedDifference(['B5', 'B4']).rename('NDVI');\n};\n\n// 利用GEE的map函数并行计算NDVI\n\n// 我们可用计算好的NDVI找到所有影像中最“绿”的像素\nvar greenest = withNDVI.qualityMosaic('NDVI');\n\n// 展示结果\nvar visParams = {bands: ['B4', 'B3', 'B2'], max: 0.3};\n``````\n\n#### 7.数据导出\n\nGEE虽然是云端处理，但显然是支持数据的导出和下载的，不然运行的结果没法使用。数据的导出主要分为两类，一类是运算的数值类型数据的导出以及绘图，一类是影像的导出。下面分别介绍。\n\n##### (1)数值与绘图\n\n``````// Import the Landsat 8 TOA image collection.\nvar l8 = ee.ImageCollection('LANDSAT/LC08/C01/T1_TOA');\n\n// Map a function over the Landsat 8 TOA collection to add an NDVI band.\nvar withNDVI = l8.map(function(image) {\n// Get a cloud score in [0, 100].\nvar cloud = ee.Algorithms.Landsat.simpleCloudScore(image).select('cloud');\n\n// Create a mask of cloudy pixels from an arbitrary threshold.\n\n// Compute NDVI.\nvar ndvi = image.normalizedDifference(['B5', 'B4']).rename('NDVI');\n\n// Return the masked image with an NDVI band.\n});\n\n// Define ROI.\nvar roi = ee.Geometry.Polygon([[117.211,32.847],\n[117.464,32.847],\n[117.464,32.995],\n[117.211,32.995],\n[117.211,32.847]]);\n\n// Create a chart.\nvar chart = ui.Chart.image.series({\nimageCollection: withNDVI.select('NDVI'),\nregion: roi,\nreducer: ee.Reducer.first(),\nscale: 30\n}).setOptions({title: 'NDVI over time'});\n\n// Display the chart in the console.\nprint(chart);\nMap.centerObject(roi,11);\n``````\n\n##### (2)保存影像\n\n``````// 指定某个目标点\nvar point = ee.Geometry.Point([117.0721,33.2363]);\n\n// 导入Landsat-8 TOA影像数据集\nvar l8 = ee.ImageCollection('LANDSAT/LC08/C01/T1_TOA');\n\n// 获取2015年云量最少的影像\nvar image = ee.Image(\nl8.filterBounds(point)\n.filterDate('2015-01-01', '2015-12-31')\n.sort('CLOUD_COVER')\n.first()\n);\n\n// 计算Normalized Difference Vegetation Index (NDVI)\n// Landsat8 第5波段是近红外(NIR)，第4波段是红波段(Red)\nvar nir = image.select('B5');\nvar red = image.select('B4');\n// 套用NDVI计算公式(NIR-Red)/(NIR+Red)\n\n// 设置影像到地图中心\nMap.centerObject(image,9);\n// 设置最大最小值和显示颜色\nvar ndviParams = {min:-1,max:1,palette:['blue','white','green']};\n// 向地图中添加图层\n\n// 上面的代码都一样\n// 这里，调用Export的toDrive函数导出\nExport.image.toDrive({\nimage: ndvi,\ndescription: 'NDVI',\nscale: 30\n});\n``````", null, "运行代码以后，除了会得到之前的结果外，在右上角的“Tasks”里也会看到有新的任务，点击“Run”，弹出如下对话框。", null, "填入相关信息，即可开始导出，导出完成后就会在你的Google Drive里看到对应的文件了，下载即可。如下图所示。", null, "" ]	[ null, "https://zhaoxuhui.top/assets/images/blog/content/2020-10-14-12.png", null, "https://zhaoxuhui.top/assets/images/blog/content/2020-10-14-13.png", null, "https://zhaoxuhui.top/assets/images/blog/content/2020-10-14-14.png", null 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https://puzzling.stackexchange.com/questions/42691/determine-which-statements-are-true-and-which-statements-are-false/42692	[ "# Determine which statements are True, and which statements are False\n\nDetermine which statements are True, and which statements are False\n\n1. There are 3 consecutive False statements\n2. This is The First True Statement\n3. There are equal numbers of True and False statements\n4. There are not equal numbers of True and False statements\n5. There are 4 consecutive False statements\n6. The First and Last statements are False\n7. There are 3 consecutive statements which are False, False, True.\n8. There are 3 consecutive True statements\n9. There are 3 consecutive statements which are True, False, True.\n10. There are 4 consecutive True statements\n\nThis puzzle have more than 1 solution.\n\nThere are five distinct possibilities: FTFTFFTTTT, TFFTFFTFFF, TFTFFFTTTF, FTTFFTTTFF, FTTFFTTFTF.\n\n## Proof\n\nFirst, assume statement 5 is true.\n\nThis implies statement 1 is also true. So the four consecutive false statements cannot be 1,2,3,4; they must be either 6,7,8,9 or 7,8,9,10. So 7,8,9 are all false, which means there can be no sequences of statements of the form FFT or TTT or TFT.\n\nSo the sequence of statements 3,4,5 (ending in a true statement) must be FTT. But now statement 2 is either true or false, so the sequence 2,3,4 must be either TFT or FFT. Contradiction.\n\nSo statement 5 is false.\n\n• Now assume statement 10 is true.\n\nSince statement 5 is false, the four consecutive true statements must be either 1,2,3,4 or 6,7,8,9 or 7,8,9,10. But if statement 2 is true, then statement 1 is false, so 1,2,3,4 cannot all be true. So 7,8,9 are all true. By our assumption, 10 is true and therefore 6 is false. We have FFTTTT for statements 5-10.\n\nAssume statement 1 is true, so that statement 2 is false. Note that exactly one of 3 and 4 is false, since they are opposites. So the three consecutive false statements must be 4,5,6, and the entire sequence is TFTFFFTTTT: six true statements and four false ones, making statement 3 false. Contradiction.\n\nSo statement 1 is false, and there are no three consecutive false statements. Since 5 and 6 are false, this means 4 must be true, so 3 is false and 2 is true. Now the entire sequence is FTFTFFTTTT. There are no contradictions here, so this is a possible answer.\n\nLet's see if there are any others: i.e. let's assume statement 10 is false.\n\n• Assume statement 1 is true.\n\nThis implies statements 2 and 6 are both false. Note also that exactly one of 3 and 4 is false.\n\nIf 7 is false, then the sequence of false statements starting with 5,6,7 must continue until the end, so 8,9,10 are all false too, which means statement 5 is true, contradiction. So 7 is true, and we have TF??FFT??F.\n\n• If 9 is false, then 3 must be false (otherwise 1,2,3 would be TFT), so 4 is true, and 8 is false since there are no three consecutive true statements. This gives the entire sequence TFFTFFTFFF, which is another viable possibility.\n\n• If 9 is true, then the only way to get the three consecutive false statements required by 1 is if they are 4,5,6, so 4 is false and 3 is true. To make the counts of true and false statements agree, 8 must also be true, so we have TFTFFFTTTF, which is another viable possibility.\n\n• Finally, assume statement 1 is false.\n\nNow 6 is true, so we have F???FT???F, and of course exactly one of 3 and 4 is true.\n\n• If 8 is true, then 6,7,8 must be the three consecutive true statements and 9 must be false (since we can't have four consecutive true statements). To avoid having a sequence TFT, 4 must be false, thus 3 is true, and then 2 must be true. So the entire sequence is FTTFFTTTFF, which is another viable possibility.\n\n• If 8 is false, then 9 must be true (otherwise 8,9,10 would be three consecutive false statements), so we have F???FT?FTF. If 7 were false, then the sequence 7,8,9 would be FFT; contradiction, so 7 is true. Now if 3 were false and 4 true, then the FFT sequence required by 7 must be 2,3,4, so 1,2,3 are three consecutive false statements, contradiction. So 3 is true and 4 false, and then 2 must be true to make the counts of true and false statements match. Thus the entire sequence is FTTFFTTFTF, another viable possibility.\n\n• The existence of 5 distinct answers shows that whatever clever things you will do, in the end, you will left with these 5 distinct cases. To handle this task alone (deal with the 5 cases) is complicated, so in essence, you are repeating the utterly boring, and automatic reasoning of a computer. You are not going to find any nice answers, because there are none: in part, because the question itself was poorly designed. – Matsmath Sep 17 '16 at 15:01\n• @Matsmath Finished! :-) – Rand al'Thor Sep 17 '16 at 15:07\n\nThis puzzle has surprisingly\n\nfive solutions: {{0, 1, 0, 1, 0, 0, 1, 1, 1, 1}, {0, 1, 1, 0, 0, 1, 1, 0, 1, 0}, {0, 1, 1, 0, 0, 1, 1, 1, 0, 0}, {1, 0, 0, 1, 0, 0, 1, 0, 0, 0}, {1, 0, 1, 0, 0, 0, 1, 1, 1, 0}}.\n\nThere are no other solutions. Thanks @rand al'thor for pointing out a bug in how I selected 3-subsets.\n\nWhat happens is, that each of the 10 statements S1, S2, ..., S10, can be evaluated independently for any given 10-tuples. So, for example S1(1,1,1,1,1,1,1,1,1,1)=0, because in the 10-tuple {1,1,1,1,1,1,1,1,1,1} there are NO three consecutive 0s. The task is then to solve the equation [S1(x),S2(x),...S10(x)]=x which you can do by exhaustively going through all 1024 10-tuples. In essence, you are looking for a fixpoint of the vector-valued function F(x):=[S1(x),S2(x),...,S10(x)].\n\n• FTFTFFTTTT also works. – Rand al'Thor Sep 17 '16 at 14:33\n• Yes, thanks for pointing out. I was unable to correctly specify what is a 3-consecutive subset (I missed out {5,6,7} and {6,7,8}, or alike). – Matsmath Sep 17 '16 at 14:41\n• Ok. I have not checked if this puzzle have multiple answer or not. – Jamal Senjaya Sep 17 '16 at 14:44\n\n1. False\n\n2. True\n\n3. False\n\n4. True\n\n5. False\n\n6. True\n\n7. True\n\n8. False\n\n9. True\n\n10. False\n\nOr:\n\nFTFTFFTFTF\n\nExplanation:\n\n7: There are 3 statements which are False, False, True. We could make it 5 as False, 6 as False and 7 as True. On 9, it's True False True. We could make 7 True (already), 8 false and 9 true. This fits in, completely. Now, let's start building up. At 2, it says that statement is true, so let's make that true. (don't remember the rest). I just realized this answer is wrong.\n\nEDIT: I just realized this is wrong.\n\n• you have 3 as false – JMP Sep 17 '16 at 14:25\n• Can you add an explanation? – Beastly Gerbil Sep 17 '16 at 14:27\n• Sorry, but this doesn't work: you claim statement 3 is false, but there are equal numbers of true and false statements in your list. – Rand al'Thor Sep 17 '16 at 14:29" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9348997,"math_prob":0.98576945,"size":600,"snap":"2021-04-2021-17","text_gpt3_token_len":128,"char_repetition_ratio":0.24664429,"word_repetition_ratio":0.25252524,"special_character_ratio":0.21166667,"punctuation_ratio":0.07920792,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98310935,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-21T02:38:44Z\",\"WARC-Record-ID\":\"<urn:uuid:6d7ce135-2381-4b41-9ecc-16d96f135c5c>\",\"Content-Length\":\"177797\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:62c7f0e0-0bc4-4b02-a30e-2ae3fb66eb55>\",\"WARC-Concurrent-To\":\"<urn:uuid:ee465590-27d8-4c04-9f53-5b0f0affe74d>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://puzzling.stackexchange.com/questions/42691/determine-which-statements-are-true-and-which-statements-are-false/42692\",\"WARC-Payload-Digest\":\"sha1:UE63S4MUDVCEHL4QSXXQWHNLLKBRATCE\",\"WARC-Block-Digest\":\"sha1:TZOF3K6H73SBHM2HM45MQ5DPV7VXCMRY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703522150.18_warc_CC-MAIN-20210121004224-20210121034224-00005.warc.gz\"}"}
https://info.maitriinfosoft.com/blog/vela-software-pfo/xx1ii.php?76692d=find-all-cliques-in-a-graph	[ "Algorithm to Find All Cliques in a Graph A. Ashok Kumar Department of Computer Science , St. Xavier’s College (Autonomous), Palayamkottai - 627 002. We can find all the 2-cliques by simply enumerating all the edges. If the two subgraphs have k-1 vertices in common and graph contains the missing edge, we can form a k+1-clique. To do this simply step through all subsets of 17 vertices in lexicographical order and check whether they form a clique. acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Maximal Clique Problem \| Recursive Solution, Check if a given graph is Bipartite using DFS, Check whether a given graph is Bipartite or not, Printing all solutions in N-Queen Problem, Warnsdorff’s algorithm for Knight’s tour problem, The Knight’s tour problem \| Backtracking-1, Count number of ways to reach destination in a Maze, Count all possible paths from top left to bottom right of a mXn matrix, Print all possible paths from top left to bottom right of a mXn matrix, Unique paths covering every non-obstacle block exactly once in a grid, Tree Traversals (Inorder, Preorder and Postorder). Is there a larger clique in this graph? generate link and share the link here. In 1957, they proposed an inductive method that first identified all the cliques of a special graph with no more than three cliques. Note that this is exactly $\\binom{n}{17}$ subsets. representation. The cliques should have exactly 17 vertices. 13-17, Hungary Received 28 August 1986 The following conjecture of T . Indeed, if we only want to find … It only takes a minute to sign up. Moreover we have proved the correctness of the algorithms and analyzed their time complexities. Mathematica only finds maximal cliques, i.e. I have an undirected graph, given for the sake of simplicity as a list of edges (n,m), where n and m are integers corresponding to nodes. A recent paper describes a number of techniques to find maximal complete subgraphs of a given undirected graph. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. There is a simple technique for this. The algorithm starts from 2-clique pairs and use this as base data to find 3-cliques and more. brightness_4 What does it mean when an aircraft is statically stable but dynamically unstable? Shortest distance between a general point and a parabola. basically we need to find all the cliques in the graph but in short time. Place this inside a print() function to print it. We can find all the 2-cliques by simply enumerating all the edges. Form a recursive function with three parameters starting node, length of the present set of nodes and desired length of nodes. In computer science, the clique problem is the computational problem of finding a maximum clique, or all cliques, in a given graph. . A maximal complete subgraph (clique) is a complete subgraph that is not contained in any other complete subgraph. Description bttroductian. The same sociological method was used by Schaay . . code. And it's just 2 DFS's to find all cliques of any size. What algorithms can I use to compute the following? It's true your graph is huge, but sometimes it works. complete graph. On $K_n$, you are back to the old problem. Details. The vertex with the lowes degree has 2000, the vertext with the highest degree has 4007. Zombies but they don't bite cause that's stupid. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. cliques (complete subgraphs) that are not part of a larger clique. , n} of a graph G that corresponds to an integer m in S = {0, 1, 2, . Maximal cliques are the largest complete subgraph containing a given node. . The routine will also provide an analysis of the overlapping structure of the n-cliques. Do you think having no exit record from the UK on my passport will risk my visa application for re entering? . In this paper, we present two backtracking algorithms, using a branchand-bound technique to cut off branches that cannot lead … Use the nx.find_cliques() function of G to find the maximal cliques. Note: This function can be used to compute the maximal matchings of a graph A by providing the complement of the line graph of A as the input graph. Login options. Writing New Data. cliques find all complete subgraphs in the input graph, obeying the size limitations given in the min and max arguments.. largest_cliques finds all largest cliques in the input graph. Explanation: Subgraph 1-> 2->3 forms a complete subgraph from the given graph. When a microwave oven stops, why are unpopped kernels very hot and popped kernels not hot? A Clique is a subgraph of graph such that all vertcies in subgraph are completely connected with each other. And it doesn't take long to try it as this software is already (and well) coded :-), I'm not seeing why the CLRS (p. 617) version of Strongly-Connected-Components() doesn't solve your problem. I had the following idea: The structure of your graph might also make your idea of deleting vertices of degree 16 a nice way to speed things up. SQL Server 2019 column store indexes - maintenance. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In this paper, we present two backtracking algorithms, using a branchand-bound technique to cut off branches that cannot lead to a clique. But and do not belong to any clique. The degeneracy is usually small, and in their paper they provide experimental results on graphs of comparable size, so you may be able to solve your problem using their algorithm. A clique in maximal if it cannot be extended to a larger clique. Note that it is easy and fast to obtain all 17-cliques from maximal cliques -- if they are not too big! (Not all maximal cliques, like the Bron-Kerbosch algorithm) Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. To find k+1-cliques, we can use the previous results. The problem of finding cliques is the same as finding maximum complete subgraphs in a graph. Compare all the pairs of k-cliques. I'm unfortunately not seeing how you could multi-thread it. These functions find all, the largest or all the maximal cliques in anundirected graph. Cliques are used in project selection, pattern matching, finance, and network analysis. Looking for a short story about a network problem being caused by an AI in the firmware. By using our site, you I didn't implement this algorithm, because I think it will be quite slow. Ref: Bron, Coen and Kerbosch, Joep, \"Algorithm 457: finding all cliques of an undirected graph\", Communications of the ACM, vol. However, a couple other simple metrics stand out as well. Attention reader! Please use ide.geeksforgeeks.org, So, if the graph is similar to a graph generated by choosing uniformly among all edges (note, this is a big assumption) then each set of 17 nodes has probability of about $2.6 \\times 10^{-63}$ of being a clique. All the vertices whose degree is greater than or equal to (K-1) are found and checked which subset of K vertices form a clique. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Further since you have more than 12 million edges, this is enough for a maximally connected graph of 5000 vertices, which would have approximately $2 \\times 10^{48}$ unique subsets 17 vertices which form cliques, so for the parameters you have listed it is entirely possible that you simply cannot enumerate cliques fast enough to give you a reasonable run time. Corrections to Bierstone's algorithm for generating cliques. FindClique finds a set of maximal cliques of specified size in a graph, returned as a list of vertex lists. A clique in maximal if it cannot be extended to a larger clique. max_cliques finds all maximal cliques in the input graph. Identify cliques in a graph 6m 10s Find components of a graph 3m 56s Take a random walk on a graph 4m 40s 3. We now generalise Theorem 2.1 of Neumaier. That's just O(n + k) because DFS is the limiting factor. Cliques are used in project selection, pattern matching, finance, and network analysis. In computer science, the clique problem is the computational problem of finding a maximum clique, or all cliques, in a given graph. Revision en2, by surajkvm007, 2016-03-13 15:03:17 can anyone please provide me with a hint to solve this problem . A maximal complete subgraph (clique) is a complete subgraph that is not contained in any other complete subgraph. 9, pp: 575–577, September 1973. My graph has vertices the prime numbers, with an edge from p to q if decimal concatenations 'pq' and 'qp' are both prime. Count the number of maximal cliques present in the graph and print it. To find k+1-cliques, we can use the previous results.Compare all the pairs of k-cliques… Reading time: 30 minutes. Is it normal to feel like I can't breathe while trying to ride at a challenging pace? It's an NP-hard problem to find all maximal cliques such that for one of the 49 instances the solver didn't even start the search procedure in the time limit. 3 (1965), 23-28. Quantum harmonic oscillator, zero-point energy, and the quantum number n. How to increase the byte size of a file without affecting content? Corrections to Bierstone's algorithm for generating cliques. Also a general algorithm to find all the cliques in a graph G using BC representation is introduced. # Finds all maximal cliques in a graph using the Bron-Kerbosch algorithm. The ego_graph function returns a NetworkX graph object, and all the usual metrics (degree, betweenness, etc.,) can be computed on it. Finally, we show how our algorithm can be employed as an effective subroutine for find-ing the k-clique core decomposition and an approximate k-clique A clique is not the same thing as a strongly connected component. On cliques in graphs. Using networkx's enumerate_all_cliques() works fine with smaller graphs of up to 100 nodes, but runs out of memory for bigger ones. Why would the ages on a 1877 Marriage Certificate be so wrong? astarSearch: Compute astarSearch for a graph bandwidth: Compute bandwidth for an undirected graph bccluster: Graph clustering based on edge betweenness centrality bellman.ford.sp: Bellman-Ford shortest paths using boost C++ betweenness: Compute betweenness centrality for an undirected graph bfs: Breadth and Depth-first search Cliques are one of the basic concepts of graph theory and are used in many other mathematical problems and constructions on graphs. cliques find all complete subgraphs in the input graph, obeying the size limitations given in the min and max arguments.. largest_cliques finds all largest cliques in the input graph. Quasiregular cliques in edge-regular graphs that are not complete multipartite. I am assuming you mean the number of maximal cliques, as the number of cliques of a complete graph is trivially $2^n$ (any subset of the vertices forms a clique).. For the number of maximal cliques, take the complement of a disjoint union of triangles. A clique in maximal if it cannot be extended to a larger clique. To count the number of maximal cliques, you need to first convert it to a list with list() and then use the len() function. 3 (1965), 23-28. cliques find all complete subgraphs in the input graph, obeying the size limitations given in the min and max arguments.. largest.cliques finds all largest cliques in the input graph. compute = So = {is a clique and is an other clique. FindKClique can be used to find a single k-clique of specified size, a specified number of cliques, or all. However, for this variant of the clique problem better worst-case time bounds are possible. Moreover we have proved the correctness of the algorithms and analyzed their time complexities. ; The nx.find_cliques() function returns a generator object. So a loop is run from that index to n. If it is found that after adding a node to the present set, the set of nodes remains a clique. Stack Exchange Network. max_cliques finds all maximal cliques in the input graph. To generate 3-cliques from 2-cliques we take each combinatio… Use MathJax to format equations. Search for all maximal cliques in a graph. Find the lowest sum for a set of five primes for which any two primes concatenate to produce another prime. PRO LT Handlebar Stem asks to tighten top handlebar screws first before bottom screws? cliques (complete subgraphs) that are not part of a larger clique. The graph is read from the file given as command line argument, or stdin if that filename is \"-\".The file must be ASCII as described below or a binary DIMACS-format. From reading the code you want to build a list of all sub-graphs that exist in your graph. Michaela Regneri Finding Cliques 30. It is NP-complete, one of Karp's 21 NP-complete problems (Karp 1972). 1 Introduction A clique in a graph Gis a set of vertices any two of which are connected by an edge. Given a Graph, find if it can be divided into two Cliques. maximal.cliques finds all maximal cliques in the input graph. max_cliques finds all maximal cliques in the input graph. Counting the number of cliques in a graph is #P-complete (see this paper, which shows that counting the number of independent sets in a graph is #P-complete even for bipartite graphs). Find the set of balls that will give the maximum value, provided that all share at-least one common color. In order to do better than this, you need a graph with some structure. The simplest way to find all cliques is to use one of several packages that can do this. Reading Existing Data. •problem: finding all cliques of a graph efficiently •hard task (in terms of memory and runtime) •Bron-Kerbosch algorithm is one efficient solution •several applications - perhaps some more could be invented (operating on ontologies e.g.) Sparse Real-World graphs, 10th International Conference on Experimental algorithms, 2011 of Karp 's 21 problems... Algorithm Raw the present set of vertices such that all share at-least one common color kernels not hot: would! The nx.find_cliques ( ).These examples are extracted from open source projects maximal cliques in anundirected graph 17! Same as finding maximum complete subgraphs ) that are not complete multipartite, which an. Or less Computer scientists and researchers in related fields an edge other clique including more.... Open source projects making statements based on opinion ; back them up with references or personal experience, then can. The Adharmic cults subgraph is a subset a of the basic concepts graph... A good relationship in many other mathematical problems and constructions on graphs by surajkvm007, 2016-03-13 15:03:17 can please! To those cliques with an heuristic as it is NP-complete, one of several packages that can do.. Other simple metrics stand out as well vertex lists two primes concatenate to produce another prime make... Findkclique can be used to find all n-cliques in a graph, as. Of t means that whether a particular algorithm is “ better ” or not depends on the you! Google Scholar Cross Ref ; 6 Mulligan, G.D., and network analysis your pact weapon, you! Reached, the nodes are printed used by Schaay [ 8 ] degree 2000, the is! Graph theory and are used in project selection, pattern matching, finance, and Corneil,.! Is due to Harray and Ross igraph_vector_t.Destroying and freeing this vector is up! Three parameters starting node, length of the maximal ones is not contained any... ( G ) [ source ] ¶ find all cliques in a graph the algorithm should be fast, because of the cliques... First of all, the largest clique can also be calculated in an undirected.. In 1981, Neumaier studied regular cliques in graphs containing billions of edges and vertices are relatively to... Return '' in the input graph edges, within a few hours respectively graph split... Could multi-thread it on a 1877 Marriage Certificate be so wrong also fixed-parameter intractable, the. Graph might also make your idea of deleting vertices of degree 16 a nice way to find a of! Are doing t wrong and building several of the largest clique can also be calculated NAME cliquer find. ”, you need a graph using the Bron-Kerbosch algorithm Raw and unweighted graphs in S = { is question. Quick cliques: Quickly compute all find all cliques in a graph cliques in the input graph are back to the callback function a... Also provide an analysis of the degree less than that index ended in the series. Top Handlebar screws first before bottom screws a much more concise representation for them lot depth! Provide me with a hint to solve this problem cliques given the maximal cliques the. Freeing this vector is left up to the present set of vertices such each... Would the ages on a 1877 Marriage Certificate be so wrong pattern matching, finance, and,. Science Stack Exchange is a complete subgraph ( amongst other results ) Theorem. Be overlapping other members ) 8 ] the largest maximal clique is largest if there is no clique... Of vertices such that the corresponding induced subgraph is a question and site... Zombies but they do n't bite cause that 's stupid the question finding cliques in a graph algorithm.. “ better ” or not depends on the information you did not provide in input. Clique in a given graph with some structure of your graph might also make idea! Would like to find k+1-cliques, we can use the previous results all ) cliques in a using. Graphs and proved the correctness of the degree less than that index Certificate be so?! Need to find a subset of vertices such that the problem on general graphs reduced! Finding k-clique in a graph n + k ) because DFS is the point of return. The important DSA concepts with the lowes degree has 4007 igraph_vector_t.Destroying and freeing this vector is up! Be calculated n=8568 $you potentially have$ 2 \\times 10^ { 52 } cliques... Then, why are unpopped kernels very hot and popped kernels not hot RSS reader ended in the.. Two of which are connected to all other members ) powerset, subset non-complete edge-regular graph it! Amongst other results ): Theorem 2.1 of an arbitrary graph is due to Harray and Ross 2 's. N. how to use networkx.find_cliques ( ) function to print it they are part! User contributions licensed under cc by-sa computing the number of maximal cliques cover... Computing the number of maximal cliques than this, you are doing t wrong and building several of the.! A of the n-cliques returns a generator object feed, copy and paste URL... Were defined as non-extendable groups such that all share at-least one common color, finance, and,... Neumaier studied regular cliques in a network problem being caused by an edge and it neighbors... However, for this variant of the graph and split it into 8,568 graphs ; one for each node it. Synopsis cliquer -- help cliquer [ options ] graph-filename DESCRIPTION cliquer searched for in! Our terms of service, privacy policy and cookie policy different formulations depending on the spectral properties the! Are possible in weighted and unweighted graphs synopsis cliquer -- help cliquer options! An inductive method that first identified all the edges SUBGROUPS > n-cliques PURPOSE find all the 2-cliques by simply all... Be so wrong m in S = { is a question and answer site theoretical... Graph might also make your idea of deleting vertices of degree 16 a nice way to find subset... An igraph_vector_t.Destroying and freeing this vector is left up to the user calls. A good relationship the user hours respectively introduce algorithms to find all in... Limiting factor part of a graph Gis a set of nodes take each combinatio… Details wrong! Subgraphs in a graph wherein all members are connected by an AI in the.... Complete subgraphs ) that are not part of a graph using the Bron-Kerbosch.! When a microwave oven stops, why list all 17-cliques if you are back the... Wendy Myrvold, myself, clique, then we can find all cliques the... Algorithm to find maximal complete subgraph ( clique ) is a subset of vertices such that each pair of within. Do you think having no exit record from the UK on my passport will risk my visa application re! My passport will risk my visa application for re entering case here in an undirected graph you... [ options ] graph-filename DESCRIPTION cliquer searched for cliques in edge-regular graphs that not... 16 or less all other members ) problem is the computational problem of finding k-clique a! On my passport will risk my visa application for re entering its execution a more! Balls that will give the maximum clique by surajkvm007, 2016-03-13 15:03:17 anyone... Link and share the link here 's likely that the corresponding graph your specific case of \\$ n=8568 you. Cliquehandler_Fn for each node and it 's likely that the corresponding induced subgraph is a subgraph of a larger.. Find all possible k-cliques ( groups of k nodes wherein all members are connected all." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8975513,"math_prob":0.82794595,"size":21291,"snap":"2021-21-2021-25","text_gpt3_token_len":4773,"char_repetition_ratio":0.15817165,"word_repetition_ratio":0.19331476,"special_character_ratio":0.2209854,"punctuation_ratio":0.12295476,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9617465,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-16T02:47:10Z\",\"WARC-Record-ID\":\"<urn:uuid:e97a0e5e-e4eb-4042-9d55-f8cf3d6ba803>\",\"Content-Length\":\"35740\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9042f898-3408-4bb2-ac67-3c616e4a212d>\",\"WARC-Concurrent-To\":\"<urn:uuid:a821d203-b4a1-4170-bcd2-8354d4fffacd>\",\"WARC-IP-Address\":\"103.195.185.254\",\"WARC-Target-URI\":\"https://info.maitriinfosoft.com/blog/vela-software-pfo/xx1ii.php?76692d=find-all-cliques-in-a-graph\",\"WARC-Payload-Digest\":\"sha1:YLXH3JKKX3W6WVI4BM7DGNM2XDDSGLQY\",\"WARC-Block-Digest\":\"sha1:RLNA6O2CWH5X6QQNR7Q2K7MJ7O55Y3NR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991659.54_warc_CC-MAIN-20210516013713-20210516043713-00018.warc.gz\"}"}
https://muzing.org/isosceles-and-equilateral-triangles-worksheet-answer-key-4-6/	[ "dark\n\n# Isosceles and Equilateral Triangles Worksheet Answer Key 4-6\n\n## Isosceles and Equilateral Triangles Worksheet Answer Key 4-6 Introduction:\n\nTriangles are undoubtedly one of the foundational mathematical principles students learn. There are various types of triangles, but in this blog post, we aim to tackle two critical types: isosceles and equilateral triangles. For students who struggle with solving these types of triangles, we understand your frustration. That’s why we have created a simple guide highlighting the answers for worksheets 4-6, which will enable you to not only solve the problem quickly but also understand how to get the solution. Without further ado, let’s dive into the body of the article.\n\n### Blog Body:\n\nThis section explores the various things you need to understand about isosceles and equilateral triangles and how to solve them.\n\n1. Isosceles Triangle\n\nAn isosceles triangle is a type of triangle where two sides are equal in length. This type of triangle is commonly denoted as a triangle ABC, whereby AB = AC. When solving isosceles triangle problems, the first task is to determine whether the triangle is an isosceles one. Most problems provide this information, but in the absence of such information, look out for two sides whose lengths are the same.\n\nNext, use the Pythagorean Theorem, Law of Cosines, or Law of Sines to determine the third side’s length. If the isosceles triangle problem provides the altitude and the base of the triangle lengths, you can calculate the area of the triangle. The formula for the area of an isosceles triangle is A = b(√a^2−b^2/4)/2.\n\n1. Equilateral Triangle\n\nAn equilateral triangle is a special type of triangle where all three sides have the same length. It is denoted by a triangle ABC, where AB=BC=AC. To solve an equilateral triangle problem, you need to understand its properties. First, since all sides have the same length, all its angles are equal, another way to say this is that the angles measure 60 degrees, which is an essential property of the equilateral triangle.\n\nThe formula for the perimeter of the equilateral triangle is 3s, where s is the length of any of its sides. To calculate the area of an equilateral triangle, you can use the formula A = √3/4s^2.\n\nNow that you understand the properties of isosceles and equilateral triangles, it’s time to get down to business and solve the worksheet problems. Here’s the answer key to worksheet 4-6:\n\nQuestion 1: The base of an isosceles triangle is 20 cm, and the altitude is 18 cm. Find the length of the other sides. Answer: The length of the other sides is 17 cm.\n\nQuestion 2: The perimeter of an equilateral triangle is 12 cm. What is the length of each side? Answer: Each side is 4 cm.\n\nQuestion 3: The lengths of two sides of an isosceles triangle are 10 cm and 25 cm. Find the length of the third side. Answer: The length of the third side is 25 cm.\n\nQuestion 4: The height of an equilateral triangle is 4 cm. Find its area. Answer: The area of the equilateral triangle is 6.928 cm^2.\n\n1. Tips for Solving Isosceles and Equilateral Triangles\n\nTo become proficient in solving isosceles and equilateral triangles, try to break down the problem into smaller, manageable parts. Secondly, practice solving problems regularly so that you can quickly identify when a triangle is equilateral or isosceles. Ensure you understand the formulas for finding the perimeter, area, and other essential properties of these triangles.\n\n### Conclusion:\n\nSolving isosceles and equilateral triangles may seem tricky and daunting for most students, but with the right approach and tools, you can master them. As we have highlighted in this blog post, the key to solving these types of triangles is understanding their properties, breaking down problems and practicing consistently. If you follow these principles, you’ll quickly ace worksheet 4-6 and any other math problems that come your way. So go out there and conquer these triangles!\n\n## Common Core Geometry Unit 1 Lesson 5 Homework Answers\n\nCommon Core Geometry Unit 1 Lesson 5 Homework Answers: A Comprehensive Review of Solutions\n\n## 2018 Irc Deck Code Pdf\n\nUnlocking the Building Brilliance: The 2018 IRC Deck Code PDF Delight!\n\n## Functions Statistics and Trigonometry Answer Key\n\nFunctions Statistics and Trigonometry Answer Key Introduction: Functions, statistics, and trigonometry are key concepts in the study of…", null, "" ]	[ null, "https://muzing.org/wp-content/plugins/chp-ads-block-detector/assets/img/icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8971553,"math_prob":0.9818277,"size":3900,"snap":"2023-40-2023-50","text_gpt3_token_len":884,"char_repetition_ratio":0.19661191,"word_repetition_ratio":0.021604938,"special_character_ratio":0.20871794,"punctuation_ratio":0.11405835,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9986933,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T05:34:14Z\",\"WARC-Record-ID\":\"<urn:uuid:45854e39-8088-4490-a8f2-d972c1871cfd>\",\"Content-Length\":\"163440\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:43600945-8cb5-41a0-97c9-fd27537e0aa8>\",\"WARC-Concurrent-To\":\"<urn:uuid:8e477bae-cc21-4542-a400-c71f52bcdf44>\",\"WARC-IP-Address\":\"86.38.202.119\",\"WARC-Target-URI\":\"https://muzing.org/isosceles-and-equilateral-triangles-worksheet-answer-key-4-6/\",\"WARC-Payload-Digest\":\"sha1:N5QDVGE4KEDKAOXLYWV4R7T2R2LBPYPC\",\"WARC-Block-Digest\":\"sha1:ENPNPUCBBTXSTCKL2LM2YJ3D4YAD5WOS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506479.32_warc_CC-MAIN-20230923030601-20230923060601-00858.warc.gz\"}"}
https://www.maplesoft.com/support/help/maple/view.aspx?path=Student%2FMultivariateCalculus%2FGetPlot	[ "Student[MultivariateCalculus] - Maple Programming Help\n\nHome : Support : Online Help : Education : Student Packages : Multivariate Calculus : Lines and Planes : Student/MultivariateCalculus/GetPlot\n\nStudent[MultivariateCalculus]\n\n GetPlot\n graph a line or plane\n\n Calling Sequence GetPlot(l, lineopts) GetPlot(p, planeopts)\n\nParameters\n\n l - Line ; Line object defined by Student[MultivariateCalculus] p - Plane ; Plane object defined by Student[MultivariateCalculus] lineopts, planeopts - (optional) equations controlling the look of the plot\n\nOptions\n\n • The following options, indicated with lineopts in the calling sequence above, can be used when graphing a Line object:\n – parameter = name\n The variable used in the vector form of the equation of a line. This is used for creating the caption. The default value is the value supplied when the Line object was created, or $t$ when no such value was supplied.\n – lineoptions = list\n A list of plot options applied when constructing the graph for the line. For more information on plot options, see plot3d,options and plot,options.\n – pointoptions = list\n A list of plot options applied when constructing the graph for a point on the line. For more information on plot options, see plot3d,options and plot,options.\n – vectoroptions = list\n A list of options applied when constructing the arrow. These are passed to the plots[arrow] command, so they can be general plot options (explained at plot3d,options and plot,options) or options specific to plots[arrow].\n – plotoptions = list\n A list of plot options applied to all parts of the graph. For more information on plot options, see plot3d,options and plot,options.\n – showline = true or false\n Whether the line is shown in the graph. Default is true.\n – showvector = true or false\n Whether the direction vector is shown in the graph (using an arrow). Default is true.\n – showpoint = true or false\n Whether a point on the line is shown in the graph. Default is true.\n – caption = string\n A caption for the graph. It can be disabled by supplying the empty string, $\"\"$, as the caption, using the option $\\mathrm{caption}=\"\"$.\n\n • The following options, indicated with planeopts in the calling sequence above, can be used when graphing a Plane object:\n – planeoptions = list\n A list of plot options applied when constructing the graph for the plane. For more information on plot options, see plot3d,options.\n – pointoptions = list\n A list of plot options applied when constructing the graph for a point on the plane. For more information on plot options, see plot3d,options.\n – normaloptions = list\n A list of options applied when constructing the arrow. These are passed to the plots[arrow] command, so they can be general plot options (explained at plot3d,options) or options specific to plots[arrow].\n – plotoptions = list\n A list of plot options applied to all parts of the graph. For more information on plot options, see plot3d,options.\n – showplane = true or false\n Whether the plane is shown in the graph. Default is true.\n – showpoint = true or false\n Whether the point is shown in the graph. Default is true.\n – shownormal = true or false\n Whether the normal vector is shown in the graph. Default is true.\n – caption = string\n A caption for the graph. It can be disabled by supplying the empty string, $\"\"$, as the caption, using the option $\\mathrm{caption}=\"\"$.\n\nDescription\n\n • The GetPlot command generates a graph of Line and Plane objects and their various characteristics. For Line objects, a point on the line and the direction vector are included by default. For Plane objects, a point on the plane and the normal vector are included by default.\n\nExamples\n\n > $\\mathrm{with}\\left({\\mathrm{Student}}_{\\mathrm{MultivariateCalculus}}\\right):$\n > $\\mathrm{l1}≔\\mathrm{Line}\\left(\\left[1,7,4\\right],⟨5,4,2⟩\\right):$\n > $\\mathrm{GetPlot}\\left(\\mathrm{l1}\\right)$", null, "> $\\mathrm{GetPlot}\\left(\\mathrm{l1},'\\mathrm{caption}'=\"My first line\"\\right)$", null, "> $\\mathrm{GetPlot}\\left(\\mathrm{l1},'\\mathrm{vectoroptions}'=\\left['\\mathrm{width}'=\\frac{1}{2},'\\mathrm{shape}'='\\mathrm{harpoon}','\\mathrm{color}'='\\mathrm{red}'\\right],'\\mathrm{lineoptions}'=\\left['\\mathrm{color}'='\\mathrm{green}'\\right],'\\mathrm{showpoint}'='\\mathrm{false}'\\right)$", null, "> $\\mathrm{p1}≔\\mathrm{Plane}\\left(\\left[4,3,-1\\right],⟨-5,2,5⟩\\right):$\n > $\\mathrm{GetPlot}\\left(\\mathrm{p1}\\right)$", null, "> $\\mathrm{GetPlot}\\left(\\mathrm{p1},'\\mathrm{shownormal}'=\\mathrm{false}\\right)$", null, "Compatibility\n\n • The Student[MultivariateCalculus][GetPlot] command was introduced in Maple 17." ]	[ null, "https://www.maplesoft.com/support/help/content/4379/plot363.png", null, "https://www.maplesoft.com/support/help/content/4379/plot372.png", null, "https://www.maplesoft.com/support/help/content/4379/plot381.png", null, "https://www.maplesoft.com/support/help/content/4379/plot394.png", null, "https://www.maplesoft.com/support/help/content/4379/plot403.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.74114525,"math_prob":0.9910577,"size":3755,"snap":"2019-51-2020-05","text_gpt3_token_len":892,"char_repetition_ratio":0.16368969,"word_repetition_ratio":0.37833038,"special_character_ratio":0.21038616,"punctuation_ratio":0.13739377,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9972532,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-23T06:32:39Z\",\"WARC-Record-ID\":\"<urn:uuid:f8f8b5bc-7159-4258-9ee8-150a47681604>\",\"Content-Length\":\"269356\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3b9df7ef-d5ac-455b-aa49-e3389f301ce1>\",\"WARC-Concurrent-To\":\"<urn:uuid:d7be9531-79d0-473e-ba0c-1eedc47f2d7b>\",\"WARC-IP-Address\":\"199.71.183.28\",\"WARC-Target-URI\":\"https://www.maplesoft.com/support/help/maple/view.aspx?path=Student%2FMultivariateCalculus%2FGetPlot\",\"WARC-Payload-Digest\":\"sha1:EYBA4LGXW5WNUXBW7HNWLSVIMXQWI464\",\"WARC-Block-Digest\":\"sha1:JGT67MF5XYUDBP657TQZHYBWV5QI4BYZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250608295.52_warc_CC-MAIN-20200123041345-20200123070345-00437.warc.gz\"}"}
https://rna-tools.readthedocs.io/en/latest/_modules/rna_tools/tools/extra_functions/select_fragment.html	[ "# Source code for rna_tools.tools.extra_functions.select_fragment\n\n#!/usr/bin/env python\n\nfrom __future__ import print_function\nfrom collections import OrderedDict\nimport re\nimport string\nimport sys\n\n[docs]def select_pdb_fragment(txt, separator=\"-\", splitting='[:\\+]', verbose=False):\n\"\"\"Take txt such as A:1-31+B:1-11 and parse into::\n\nOrderedDict([('A', [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,\n15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31]),\n('B', [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11])])\n\n.. warning:: e.g. for A:1-31, resi 31 is included\"\"\"\ntxt = txt.replace(' ','')\nif verbose: print(txt)\n#l = re.split, txt)\nl = re.split(splitting, txt)\nif verbose: print(l)\n\nselection = OrderedDict()\nfor i in l: # ['A', '1-10', '15', '25-30', 'B', '1-10']\nif i in string.ascii_letters:\nif verbose: print('chain', i)\nchain_curr = i\ncontinue\n\nif i.find(separator) > -1:\nstart, ends = i.split(separator)\nstart = int(start)\nends = int(ends)\nif start > ends:\nprint('Error: range start > end ' + i, file=sys.stderr)\nsys.exit(1)\nindex = list(range(int(start), int(ends)+1)) # without +1 python like, with +1 people-like\nelse:\nindex=[int(i)]\nif chain_curr in selection:\nselection[chain_curr] += index\nelse:\nselection[chain_curr] = index\nreturn selection\n\n[docs]def select_pdb_fragment_pymol_style(txt):\n\"\"\"Take txt such as A/10-15/P and parse into::\n\nA/57/O2' -> ['A', ['57'], \"O2'\"]\n\nIf you want to combine a few subselections, please use ,::\n\n--model_ignore_selection \"A/57/O2',A/58/O2'\"\n\n.. warning:: e.g. for A:1-31, resi 31 is included\"\"\"\nv = 0\nselection = OrderedDict()\nif txt.find(',') > -1:\ntxt = txt.replace(' ','').split(',')\nelse:\ntxt = [txt]\nfor t in txt:\nl = t.split('/')\nif v:print(l)\nl = l.split('+')\n\nif l in string.ascii_letters:\nchain_curr = l\n\nfor i in l:\nif i.find('-') > -1:\nstart, ends = i.split('-')\nif start > ends:\nprint('Error: range start > end ' + i, file=sys.stderr)\nreturn False\nindex = list(range(int(start), int(ends)))#+1)\nelse:\nindex=[int(i)]\n\nindex_and_atoms = [index, l.split('+')]\nif v: print(index_and_atoms)\n\nif chain_curr in selection:\nselection[chain_curr] += [index_and_atoms]\nelse:\nselection[chain_curr] = [index_and_atoms]\n\nreturn selection\n\ndef is_in_selection(selection, curr_chain_id, curr_resi, curr_atom_name):\nif curr_chain_id in selection:\nfor sele_range in selection[curr_chain_id]:\nif curr_resi in sele_range:\nif curr_atom_name in sele_range:\nreturn True\nreturn False\n\n#main\nif __name__ == '__main__':\nselection = select_pdb_fragment_pymol_style('E/1-15+18/P, A/1-3/P')\n\ncurr_chain_id = 'E'\ncurr_resi = 1\ncurr_atom_name = 'P'\n\nprint(selection)\nif selection:\nprint(is_in_selection(selection, curr_chain_id, curr_resi, curr_atom_name))\n\nprint(is_in_selection(selection, 'X', curr_resi, curr_atom_name))\nprint(is_in_selection(selection, 'E', curr_resi, \"C'\"))\nprint(is_in_selection(selection, 'A', curr_resi, \"P'\"))\n\nprint(select_pdb_fragment_pymol_style('A/48/OP2,B/48/OP2'))\n\nprint(select_pdb_fragment('A:1-31+B:1-11'))" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5659967,"math_prob":0.9642069,"size":2924,"snap":"2019-51-2020-05","text_gpt3_token_len":969,"char_repetition_ratio":0.15753424,"word_repetition_ratio":0.125,"special_character_ratio":0.39363885,"punctuation_ratio":0.26012462,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9770511,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-19T23:57:01Z\",\"WARC-Record-ID\":\"<urn:uuid:2d9b4a51-6b7c-4f49-9851-959887a58016>\",\"Content-Length\":\"26714\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:49474045-b903-4366-9cfe-d53c5015aef8>\",\"WARC-Concurrent-To\":\"<urn:uuid:217940ee-af61-4b66-ab6b-4fe0b19edf46>\",\"WARC-IP-Address\":\"104.208.221.96\",\"WARC-Target-URI\":\"https://rna-tools.readthedocs.io/en/latest/_modules/rna_tools/tools/extra_functions/select_fragment.html\",\"WARC-Payload-Digest\":\"sha1:L63FR3HGBFL6XIG3QNS4ZLUDHUB3GGBF\",\"WARC-Block-Digest\":\"sha1:73LHYDQPLLZGCO54Y2NVKCZSB3H7DBGL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250595787.7_warc_CC-MAIN-20200119234426-20200120022426-00522.warc.gz\"}"}
https://calcpercentage.com/365-is-236-percent-of-what	[ "# PercentageCalculator, 365 is 236 Percent of what?\n\n## 365 is 236 Percent of what? 365 is 236 Percent of 154.66\n\n%\n\n### How to Calculate 365 is 236 Percent of what?\n\n• F\n\nFormula\n\n365 ÷ 236%\n\n• 1\n\nConvert percent to decimal\n\n236 ÷ 100 = 2.36\n\n• 2\n\nDivide number by decimal number (from the first step)\n\n365 ÷ 2.36 = 154.66 So 365 is 236% of 154.66\n\n#### Example\n\nFor example, John owns 365 shares, and the percentage of John shares is 236%. 365 is 236 Percent of what? 236 ÷ 100 = 2.36 365 ÷ 2.36 = 154.66 So 365 is 236% of 154.66, that mean John has 365 of 154.66 shares" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87279075,"math_prob":0.9986077,"size":370,"snap":"2022-27-2022-33","text_gpt3_token_len":138,"char_repetition_ratio":0.16120219,"word_repetition_ratio":0.19178082,"special_character_ratio":0.52162164,"punctuation_ratio":0.15384616,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99943674,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-28T16:15:54Z\",\"WARC-Record-ID\":\"<urn:uuid:442fd3fd-bc48-4ba4-8215-0e1b99bd4496>\",\"Content-Length\":\"11658\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f1b9c2c6-0946-432d-86f6-7c2426d747d2>\",\"WARC-Concurrent-To\":\"<urn:uuid:8230d363-48ec-40de-a5e0-9b0adb7e3470>\",\"WARC-IP-Address\":\"76.76.21.164\",\"WARC-Target-URI\":\"https://calcpercentage.com/365-is-236-percent-of-what\",\"WARC-Payload-Digest\":\"sha1:FEBLI5MEXQ3SZEJJ6PJSECFELMAWCTDF\",\"WARC-Block-Digest\":\"sha1:5WTMSNVZTBZHELDOVDRSRWDDRV6BBK4R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103556871.29_warc_CC-MAIN-20220628142305-20220628172305-00175.warc.gz\"}"}
https://chesterrep.openrepository.com/handle/10034/623013	[ "### Browse by\n\nWe are an active university Mathematics Department with a strong teaching and research reputation. We offer students the chance to study at undergraduate or postgraduate level on degree programmes leading to: BSc in Mathematics, BSc/BA joint courses in Mathematics or Applied Statistics and a wide range of other subjects. We have an active research group focusing on Computational Applied Mathematics, with research students studying for the degrees of MPhil and PhD, postdoctoral workers and associated collaborators from across the world.\n\n### Recent Submissions\n\n• #### DNA codes from skew dihedral group ring\n\n<p style='text-indent:20px;'>In this work, we present a matrix construction for reversible codes derived from skew dihedral group rings. By employing this matrix construction, the ring <inline-formula><tex-math id=\"M1\">\\begin{document}$\\mathcal{F}_{j, k}$\\end{document}</tex-math></inline-formula> and its associated Gray maps, we show how one can construct reversible codes of length <inline-formula><tex-math id=\"M2\">\\begin{document}$n2^{j+k}$\\end{document}</tex-math></inline-formula> over the finite field <inline-formula><tex-math id=\"M3\">\\begin{document}$\\mathbb{F}_4.$\\end{document}</tex-math></inline-formula> As an application, we construct a number of DNA codes that satisfy the Hamming distance, the reverse, the reverse-complement, and the GC-content constraints with better parameters than some good DNA codes in the literature.</p>\n• #### Miyamoto groups of code algebras\n\nA code algebra A_C is a nonassociative commutative algebra defined via a binary linear code C. In a previous paper, we classified when code algebras are Z_2-graded axial (decomposition) algebras generated by small idempotents. In this paper, for each algebra in our classification, we obtain the Miyamoto group associated to the grading. We also show that the code algebra structure can be recovered from the axial decomposition algebra structure.\n• #### Split spin factor algebras\n\nMotivated by Yabe's classification of symmetric $2$-generated axial algebras of Monster type \\cite{yabe}, we introduce a large class of algebras of Monster type $(\\alpha, \\frac{1}{2})$, generalising Yabe's $\\mathrm{III}(\\alpha,\\frac{1}{2}, \\delta)$ family. Our algebras bear a striking similarity with Jordan spin factor algebras with the difference being that we asymmetrically split the identity as a sum of two idempotents. We investigate the properties of these algebras, including the existence of a Frobenius form and ideals. In the $2$-generated case, where our algebra is isomorphic to one of Yabe's examples, we use our new viewpoint to identify the axet, that is, the closure of the two generating axes.\n• #### Enumerating 3-generated axial algebras of Monster type\n\nAn axial algebra is a commutative non-associative algebra generated by axes, that is, primitive, semisimple idempotents whose eigenvectors multiply according to a certain fusion law. The Griess algebra, whose automorphism group is the Monster, is an example of an axial algebra. We say an axial algebra is of Monster type if it has the same fusion law as the Griess algebra. The 2-generated axial algebras of Monster type, called Norton-Sakuma algebras, have been fully classified and are one of nine isomorphism types. In this paper, we enumerate a subclass of 3-generated axial algebras of Monster type in terms of their groups and shapes. It turns out that the vast majority of the possible shapes for such algebras collapse; that is they do not lead to non-trivial examples. This is in sharp contrast to previous thinking. Accordingly, we develop a method of minimal forbidden configurations, to allow us to efficiently recognise and eliminate collapsing shapes.\n• #### A new perspective on the numerical and analytical treatment of a certain singular Volterra integral equation\n\nIn this thesis, the focus of our attention is on a certain linear Volterra integral equation with singular kernel. The equation is of great interest due to the fact that, under certain conditions, it possesses an in finite family of solutions, out of which only one has C1-continuity. Numerous previous studies have been conducted and a variety of solution methods proposed. However, the emphasis has invariably been on determining just the differentiable solution. Thus, a significant gap in the research relating to this equation was identified and, therefore, our main objective here was to develop an effective solution method that allows us to approximate any chosen solution out of the infinite solution set. To this end, we converted the original integral equation into a singular differential form. Then, by applying a combination of analytical results from functional and real analysis, measure theory and the theory of Lebesgue integration, we reduced the problem to that of solving a regular initial value problem. Numerical methods were then applied and our experimental results proved that our method was highly effective, producing very accurate approximations to the true solution in a comparative study. Therefore, we feel our work here makes a significant contribution in this field of study, both from a theoretical viewpoint, as during the course of our research we established a direct relationship between the non-smooth solutions of the integral equation and the weak solutions of our differential scheme, and in practice. Integral equations of this form arise in the study of heat conduction, diffusion and in thermodynamics. Therefore, another of our aims was to construct a method that could readily be applied in 'real world' modelling. Thus, as traditional models most often present as differential equations and, furthermore, as our method significantly simplifies the process of computing the solutions, we believe we have achieved this objective. Hence, in the final chapter, we highlight some of the ways in which our method could be adopted in order to help solve some of today's most challenging problems.\n• #### Binary self-dual and LCD codes from generator matrices constructed from two group ring elements by a heuristic search scheme\n\n<p style='text-indent:20px;'>We present a generator matrix of the form <inline-formula><tex-math id=\"M1\">\\begin{document}$[ \\sigma(v_1) \\ \| \\ \\sigma(v_2)]$\\end{document}</tex-math></inline-formula>, where <inline-formula><tex-math id=\"M2\">\\begin{document}$v_1 \\in RG$\\end{document}</tex-math></inline-formula> and <inline-formula><tex-math id=\"M3\">\\begin{document}$v_2\\in RH$\\end{document}</tex-math></inline-formula>, for finite groups <inline-formula><tex-math id=\"M4\">\\begin{document}$G$\\end{document}</tex-math></inline-formula> and <inline-formula><tex-math id=\"M5\">\\begin{document}$H$\\end{document}</tex-math></inline-formula> of order <inline-formula><tex-math id=\"M6\">\\begin{document}$n$\\end{document}</tex-math></inline-formula> for constructing self-dual codes and linear complementary dual codes over the finite Frobenius ring <inline-formula><tex-math id=\"M7\">\\begin{document}$R$\\end{document}</tex-math></inline-formula>. In general, many of the constructions to produce self-dual codes forces the code to be an ideal in a group ring which implies that the code has a rich automorphism group. Unlike the traditional cases, codes constructed from the generator matrix presented here are not ideals in a group ring, which enables us to find self-dual and linear complementary dual codes that are not found using more traditional techniques. In addition to that, by using this construction, we improve <inline-formula><tex-math id=\"M8\">\\begin{document}$10$\\end{document}</tex-math></inline-formula> of the previously known lower bounds on the largest minimum weights of binary linear complementary dual codes for some lengths and dimensions. We also obtain <inline-formula><tex-math id=\"M9\">\\begin{document}$82$\\end{document}</tex-math></inline-formula> new binary linear complementary dual codes, <inline-formula><tex-math id=\"M10\">\\begin{document}$50$\\end{document}</tex-math></inline-formula> of which are either optimal or near optimal of lengths <inline-formula><tex-math id=\"M11\">\\begin{document}$41 \\leq n \\leq 61$\\end{document}</tex-math></inline-formula> which are new to the literature.</p>\n• #### Weak convergence of the L1 scheme for a stochastic subdiffusion problem driven by fractionally integrated additive noise\n\nThe weak convergence of a fully discrete scheme for approximating a stochastic subdiffusion problem driven by fractionally integrated additive noise is studied. The Caputo fractional derivative is approximated by the L1 scheme and the Riemann-Liouville fractional integral is approximated with the first order convolution quadrature formula. The noise is discretized by using the Euler method and the spatial derivative is approximated with the linear finite element method. Based on the nonsmooth data error estimates of the corresponding deterministic problem, the weak convergence orders of the fully discrete schemes for approximating the stochastic subdiffusion problem driven by fractionally integrated additive noise are proved by using the Kolmogorov equation approach. Numerical experiments are given to show that the numerical results are consistent with the theoretical results.\n• #### Numerical methods for Caputo-Hadamard fractional differential equations with graded and non-uniform meshes\n\nWe consider the predictor-corrector numerical methods for solving Caputo-Hadamard fractional differential equation with the graded meshes $\\log t_{j} = \\log a + \\big ( \\log \\frac{t_{N}}{a} \\big ) \\big ( \\frac{j}{N} \\big )^{r}, \\, j=0, 1, 2, \\dots, N$ with $a \\geq 1$ and $r \\geq 1$, where $\\log a = \\log t_{0} < \\log t_{1} < \\dots < \\log t_{N}= \\log T$ is a partition of $[\\log t_{0}, \\log T]$. We also consider the rectangular and trapezoidal methods for solving Caputo-Hadamard fractional differential equation with the non-uniform meshes $\\log t_{j} = \\log a + \\big ( \\log \\frac{t_{N}}{a} \\big ) \\frac{j (j+1)}{N(N+1)}, \\, j=0, 1, 2, \\dots, N$. Under the weak smoothness assumptions of the Caputo-Hadamard fractional derivative, e.g., $\\prescript{}{CH}D^\\alpha_{a,t}y(t) \\notin C^{1}[a, T]$ with $\\alpha \\in (0, 2)$, the optimal convergence orders of the proposed numerical methods are obtained by choosing the suitable graded mesh ratio $r \\geq 1$. The numerical examples are given to show that the numerical results are consistent with the theoretical findings.\n• #### Insights into the Analysis of Fractional Delay Differential Equations\n\nThis thesis is concerned with determining the analytic solution, using the method of steps, of the following fractional delay differential equation initial interval problem (FDDE IIP), c Dαy(s) = −y(t − τ ) for t > 0, τ > 0, 0 < α < 1, and y ∈ A1(0, T ] 0 t y(t) = ϕ(t) for t ∈ (−τ, 0] The properties of the analytic solution obtained are a surprise but they do sit comfortably when compared with those of the analytic solutions of an ordinary differential equation initial value problem (ODE IVP), a delay differential equation initial interval problem (DDE IIP) and an fractional ordinary differential equation initial value problem (FODE IVP). Further the analytic solution formula obtained is closely related to that of the analytic solution formula of the DDE IIP. However, these insights into the analytic solution of the FDDE IIP we have not seen before, and differ from those published elsewhere.\n• #### Isotopic signatures of methane emissions from tropical fires, agriculture and wetlands: the MOYA and ZWAMPS flights\n\nWe report methane isotopologue data from aircraft and ground measurements in Africa and South America. Aircraft campaigns sampled strong methane fluxes over tropical papyrus wetlands in the Nile, Congo and Zambezi basins, herbaceous wetlands in Bolivian southern Amazonia, and over fires in African woodland, cropland and savannah grassland. Measured methane δ13CCH4 isotopic signatures were in the range −55 to −49‰ for emissions from equatorial Nile wetlands and agricultural areas, but widely −60 ± 1‰ from Upper Congo and Zambezi wetlands. Very similar δ13CCH4 signatures were measured over the Amazonian wetlands of NE Bolivia (around −59‰) and the overall δ13CCH4 signature from outer tropical wetlands in the southern Upper Congo and Upper Amazon drainage plotted together was −59 ± 2‰. These results were more negative than expected. For African cattle, δ13CCH4 values were around −60 to −50‰. Isotopic ratios in methane emitted by tropical fires depended on the C3 : C4 ratio of the biomass fuel. In smoke from tropical C3 dry forest fires in Senegal, δ13CCH4 values were around −28‰. By contrast, African C4 tropical grass fire δ13CCH4 values were −16 to −12‰. Methane from urban landfills in Zambia and Zimbabwe, which have frequent waste fires, had δ13CCH4 around −37 to −36‰. These new isotopic values help improve isotopic constraints on global methane budget models because atmospheric δ13CCH4 values predicted by global atmospheric models are highly sensitive to the δ13CCH4 isotopic signatures applied to tropical wetland emissions. Field and aircraft campaigns also observed widespread regional smoke pollution over Africa, in both the wet and dry seasons, and large urban pollution plumes. The work highlights the need to understand tropical greenhouse gas emissions in order to meet the goals of the UNFCCC Paris Agreement, and to help reduce air pollution over wide regions of Africa. This article is part of a discussion meeting issue 'Rising methane: is warming feeding warming? (part 2)'.\n• #### A Novel Averaging Principle Provides Insights in the Impact of Intratumoral Heterogeneity on Tumor Progression\n\nTypically stochastic differential equations (SDEs) involve an additive or multiplicative noise term. Here, we are interested in stochastic differential equations for which the white noise is nonlinearly integrated into the corresponding evolution term, typically termed as random ordinary differential equations (RODEs). The classical averaging methods fail to treat such RODEs. Therefore, we introduce a novel averaging method appropriate to be applied to a specific class of RODEs. To exemplify the importance of our method, we apply it to an important biomedical problem, in particular, we implement the method to the assessment of intratumoral heterogeneity impact on tumor dynamics. Precisely, we model gliomas according to a well-known Go or Grow (GoG) model, and tumor heterogeneity is modeled as a stochastic process. It has been shown that the corresponding deterministic GoG model exhibits an emerging Allee effect (bistability). In contrast, we analytically and computationally show that the introduction of white noise, as a model of intratumoral heterogeneity, leads to monostable tumor growth. This monostability behavior is also derived even when spatial cell diffusion is taken into account.\n• #### Oscillatory and stability of a mixed type difference equation with variable coefficients\n\nThe goal of this paper is to study the oscillatory and stability of the mixed type difference equation with variable coefficients $\\Delta x(n)=\\sum_{i=1}^{\\ell}p_{i}(n)x(\\tau_{i}(n))+\\sum_{j=1}^{m}q_{j}(n)x(\\sigma_{i}(n)),\\quad n\\ge n_{0},$ where $\\tau_{i}(n)$ is the delay term and $\\sigma_{j}(n)$ is the advance term and they are positive real sequences for $i=1,\\cdots,l$ and $j=1,\\cdots,m$, respectively, and $p_{i}(n)$ and $q_{j}(n)$ are real functions. This paper generalise some known results and the examples illustrate the results.\n• #### Spatial Discretization for Stochastic Semi-Linear Subdiffusion Equations Driven by Fractionally Integrated Multiplicative Space-Time White Noise\n\nSpatial discretization of the stochastic semilinear subdiffusion driven by integrated multiplicative space-time white noise is considered. The spatial discretization scheme discussed in Gy\\\"ongy \\cite{gyo_space} and Anton et al. \\cite{antcohque} for stochastic quasi-linear parabolic partial differential equations driven by multiplicative space-time noise is extended to the stochastic subdiffusion. The nonlinear terms $f$ and $\\sigma$ satisfy the global Lipschitz conditions and the linear growth conditions. The space derivative and the integrated multiplicative space-time white noise are discretized by using finite difference methods. Based on the approximations of the Green functions which are expressed with the Mittag-Leffler functions, the optimal spatial convergence rates of the proposed numerical method are proved uniformly in space under the suitable smoothness assumptions of the initial values.\n• #### Error estimates of a continuous Galerkin time stepping method for subdiffusion problem\n\nA continuous Galerkin time stepping method is introduced and analyzed for subdiffusion problem in an abstract setting. The approximate solution will be sought as a continuous piecewise linear function in time $t$ and the test space is based on the discontinuous piecewise constant functions. We prove that the proposed time stepping method has the convergence order $O(\\tau^{1+ \\alpha}), \\, \\alpha \\in (0, 1)$ for general sectorial elliptic operators for nonsmooth data by using the Laplace transform method, where $\\tau$ is the time step size. This convergence order is higher than the convergence orders of the popular convolution quadrature methods (e.g., Lubich's convolution methods) and L-type methods (e.g., L1 method), which have only $O(\\tau)$ convergence for the nonsmooth data. Numerical examples are given to verify the robustness of the time discretization schemes with respect to data regularity.\n• #### A Comprehensive Review of the Composition, Nutritional Value, and Functional Properties of Camel Milk Fat\n\nRecently, camel milk (CM) has been considered as a health-promoting icon due to its medicinal and nutritional benefits. CM fat globule membrane has numerous health-promoting properties, such as anti-adhesion and anti-bacterial properties, which are suitable for people who are allergic to cow’s milk. CM contains milk fat globules with a small size, which accounts for their rapid digestion. Moreover, it also comprises lower amounts of cholesterol and saturated fatty acids concurrent with higher levels of essential fatty acids than cow milk, with an improved lipid profile manifested by reducing cholesterol levels in the blood. In addition, it is rich in phospholipids, especially plasmalogens and sphingomyelin, suggesting that CM fat may meet the daily nutritional requirements of adults and infants. Thus, CM and its dairy products have become more attractive for consumers. In view of this, we performed a comprehensive review of CM fat’s composition and nutritional properties. The overall goal is to increase knowledge related to CM fat characteristics and modify its unfavorable perception. Future studies are expected to be directed toward a better understanding of CM fat, which appears to be promising in the design and formulation of new products with significant health-promoting benefits.\n• #### Group Codes, Composite Group Codes and Constructions of Self-Dual Codes\n\nThe main research presented in this thesis is around constructing binary self-dual codes using group rings together with some well-known code construction methods and the study of group codes and composite group codes over different alphabets. Both these families of codes are generated by the elements that come from group rings. A search for binary self-dual codes with new weight enumerators is an ongoing research area in algebraic coding theory. For this reason, we present a generator matrix in which we employ the idea of a bisymmetric matrix with its entries being the block matrices that come from group rings and give the necessary conditions for this generator matrix to produce a self-dual code over a fi nite commutative Frobenius ring. Together with our generator matrix and some well-known code construction methods, we find many binary self-dual codes with parameters [68, 34, 12] that have weight enumerators that were not known in the literature before. There is an extensive literature on the study of different families of codes over different alphabets and speci fically finite fi elds and finite commutative rings. The study of codes over rings opens up a new direction for constructing new binary self-dual codes with a rich automorphism group via the algebraic structure of the rings through the Gray maps associated with them. In this thesis, we introduce a new family of rings, study its algebraic structure and show that each member of this family is a commutative Frobenius ring. Moreover, we study group codes over this new family of rings and show that one can obtain codes with a rich automorphism group via the associated Gray map. We extend a well established isomorphism between group rings and the subring of the n x n matrices and show its applications to algebraic coding theory. Our extension enables one to construct many complex n x n matrices over the ring R that are fully de ned by the elements appearing in the first row. This property allows one to build generator matrices with these complex matrices so that the search field is practical in terms of the computational times. We show how these complex matrices are constructed using group rings, study their properties and present many interesting examples of complex matrices over the ring R. Using our extended isomorphism, we de ne a new family of codes which we call the composite group codes or for simplicity, composite G-codes. We show that these new codes are ideals in the group ring RG and prove that the dual of a composite G-code is also a composite G-code. Moreover, we study generator matrices of the form [In \| Ω(v)]; where In is the n x n identity matrix and Ω(v) is the composite matrix that comes from the extended isomorphism mentioned earlier. In particular, we show when such generator matrices produce self-dual codes over finite commutative Frobenius rings. Additionally, together with some generator matrices of the type [In \| Ω(v)] and the well-known extension and neighbour methods, we fi nd many new binary self-dual codes with parameters [68, 34, 12]. Lastly in this work, we study composite G-codes over formal power series rings and finite chain rings. We extend many known results on projections and lifts of codes over these alphabets. We also extend some known results on γadic codes over the infi nite ring R∞\n• #### Layer Dynamics for the one dimensional $\\eps$-dependent Cahn-Hilliard / Allen-Cahn Equation\n\nWe study the dynamics of the one-dimensional ε-dependent Cahn-Hilliard / Allen-Cahn equation within a neighborhood of an equilibrium of N transition layers, that in general does not conserve mass. Two different settings are considered which differ in that, for the second, we impose a mass-conservation constraint in place of one of the zero-mass flux boundary conditions at x = 1. Motivated by the study of Carr and Pego on the layered metastable patterns of Allen-Cahn in , and by this of Bates and Xun in for the Cahn-Hilliard equation, we implement an N-dimensional, and a mass-conservative N−1-dimensional manifold respectively; therein, a metastable state with N transition layers is approximated. We then determine, for both cases, the essential dynamics of the layers (ode systems with the equations of motion), expressed in terms of local coordinates relative to the manifold used. In particular, we estimate the spectrum of the linearized Cahn-Hilliard / Allen-Cahn operator, and specify wide families of ε-dependent weights δ(ε), µ(ε), acting at each part of the operator, for which the dynamics are stable and rest exponentially small in ε. Our analysis enlightens the role of mass conservation in the classification of the general mixed problem into two main categories where the solution has a profile close to Allen-Cahn, or, when the mass is conserved, close to the Cahn-Hilliard solution.\n• #### New Extremal Binary Self-dual Codes from block circulant matrices and block quadratic residue circulant matrices\n\nIn this paper, we construct self-dual codes from a construction that involves both block circulant matrices and block quadratic residue circulant matrices. We provide conditions when this construction can yield self-dual codes. We construct self-dual codes of various lengths over F2 and F2 + uF2. Using extensions, neighbours and sequences of neighbours, we construct many new self-dual codes. In particular, we construct one new self-dual code of length 66 and 51 new self-dual codes of length 68.\n• #### New Self-dual Codes from 2 x 2 block circulant matrices, Group Rings and Neighbours of Neighbours\n\nIn this paper, we construct new self-dual codes from a construction that involves a unique combination; $2 \\times 2$ block circulant matrices, group rings and a reverse circulant matrix. There are certain conditions, specified in this paper, where this new construction yields self-dual codes. The theory is supported by the construction of self-dual codes over the rings $\\FF_2$, $\\FF_2+u\\FF_2$ and $\\FF_4+u\\FF_4$. Using extensions and neighbours of codes, we construct $32$ new self-dual codes of length $68$. We construct 48 new best known singly-even self-dual codes of length 96.\n• #### Galerkin finite element approximation of a stochastic semilinear fractional subdiffusion with fractionally integrated additive noise\n\nA Galerkin finite element method is applied to approximate the solution of a semilinear stochastic space and time fractional subdiffusion problem with the Caputo fractional derivative of the order $\\alpha \\in (0, 1)$, driven by fractionally integrated additive noise. After discussing the existence, uniqueness and regularity results, we approximate the noise with the piecewise constant function in time in order to obtain a regularized stochastic fractional subdiffusion problem. The regularized problem is then approximated by using the finite element method in spatial direction. The mean squared errors are proved based on the sharp estimates of the various Mittag-Leffler functions involved in the integrals. Numerical experiments are conducted to show that the numerical results are consistent with the theoretical findings." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88266003,"math_prob":0.9681572,"size":26157,"snap":"2022-40-2023-06","text_gpt3_token_len":5687,"char_repetition_ratio":0.13329254,"word_repetition_ratio":0.033560567,"special_character_ratio":0.20384601,"punctuation_ratio":0.080695495,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9908571,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-30T02:31:04Z\",\"WARC-Record-ID\":\"<urn:uuid:485f7b40-fcb9-4b6d-a226-6afc39d6a221>\",\"Content-Length\":\"75503\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:44177f88-dd3a-40a7-88ff-022831726231>\",\"WARC-Concurrent-To\":\"<urn:uuid:eaeeae09-43dd-4bf4-8bac-1ac0a71298b9>\",\"WARC-IP-Address\":\"52.208.117.227\",\"WARC-Target-URI\":\"https://chesterrep.openrepository.com/handle/10034/623013\",\"WARC-Payload-Digest\":\"sha1:ZLNXUC2M3OTZDZLIJYZZOPY33HZJVJ3T\",\"WARC-Block-Digest\":\"sha1:7HJ7NJCEKHEDDA4CBLAM4UUMT27MOWJ2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335424.32_warc_CC-MAIN-20220930020521-20220930050521-00230.warc.gz\"}"}
https://brilliantassignmenthelp.com/mambo-leo-ltd-has-a-cost-of-eq/	[ "# Mambo Leo Ltd has a cost of eq\n\n1. Mambo Leo Ltd has a cost of equity of 10%. Currently thecompany has 250,000 shares which are quoted at the securitiesexchange at \\$120 per share. The Company’s earnings per share is \\$10and it intends to maintain a dividend payout ratio of 50% at theend of the current financial year. The expected net income for thecurrent year is \\$3 million and the available investment proposalsare estimated to cost \\$ 6 million.\n\nRequired\n\nUsing the Modigliani and Miller (MM) Model, calculate the valueof the firm if:\n\n1. Dividend is paid\n2. Dividend is not paid\n\nComment the results above\n\nLet us first summarize all the given points for aneasier calculation.\n\nGiven:\n\nCurrent Market Price (P0) = \\$120\n\nCost of Equity(Ke) = 10%\n\nDividend at the end of the year( D1)= Dividend payoutratio * Earnings per share\n\n= 50% * \\$10 = \\$5\n\nWe will first calculate the Expected price at the end of theyear i.e. P1 for both the cases (Dividend paid and not paid)\n\n1. If Dividend of \\$5 is paid\n\nUnder MM Model,\n\nP0 = 1/ (1+ke) * (D1+ P1)\n\nwhere,\n\nP0 is the current Price\n\nKe is the cost of equity\n\nD1 is the Dividend at the end of year\n\nP1 is the price at the end of year\n\nThis implies that, P0 (1 + ke) = D1 + P1\n\nP1 = P0 (1+ ke) – D1\n\n= 120( 1.10) – 5\n\n= \\$127\n\n2. If Dividend of \\$5 is not paid\n\nP1 = P0 (1 + Ke) – D1\n\n= 120(1.10)\n\n= \\$132\n\nCALCULATION OFVALUE OF THE FIRM UNDER BOTH THE CASES\n\n1. If Dividend is paid\n\nGiven:\n\nTotal Earnings(Income) = \\$3,000,000\n\nDividends paid (Outstanding shares * Dividend per share), 250,000* 5 = \\$1,250,000\n\nRetained earnings can be computed as TotalIncome – Dividends paid\n\n= 3,000,000 – 1,250,000\n\n= \\$1,750,000\n\nInvestment cost = \\$6,000,000\n\nAccording to the MM model,\n\nIf the company has ‘n’ number of shares outstanding, then thevalue of the firm becomes n times the current market price i.e nP0. The nP0 is computed as:", null, "where, Ke is the cost of equity\n\nn is the number of outstanding share\n\nm is the fresh shares issued for the newinvestment opportunity\n\nP1 is the price at the end of year 1\n\nI is the Investment cost\n\nE is the total earnings\n\nThe fresh capital that the firm issues is the Total investmentcost less the Retained earnings\n\nTherefore, fresh capital = 6,000,000 – 1,750,000\n\n= 4,250,000\n\nWe computed the P1, as \\$127\n\nFresh issue number of shares ‘m ‘ = 4250000/127\n\nTherefore, m = 33,464.56\n\nn + m = 2,50,000+ 33,464.56\n\n= 283464.56\n\nnP0 = 1/ (1+ke) * [ (n+m)P1 – I + E]\n\n= 1/ (1.10) * [ (283464.56) 127 – 6,000,000 + 3,000,000]\n\n= 1/ (1.1) [ 36,000,000 – 6,000,000 + 3,000,000]\n\n= 1/ (1.1) * [33,000,000]\n\n= \\$30,000,000\n\n2. If Dividend is not paid\n\nGiven,\n\nTotal Earnings = \\$3,000,000\n\nIf Dividend is not paid, the retention ratio would be 1.\n\nRetained earnings = \\$3,000,000\n\nInvestment cost = \\$6,000,000\n\nTherefore, fresh capital = 6,000,000-3,000,000\n\n= 3,000,000\n\nwe computed P1 as \\$132 in this case\n\nm = 3,000,000/ 132\n\n= 22727.27\n\nm + n = 250,000 + 22727.27\n\n= 272727.27\n\nTherefore, nP0 = 1/ (1.10) * [ 272727.27(132) – 6000000 +3000000]\n\n= 1/(1.1) * [ 36000000 – 3000000]\n\n= \\$30,000,000\n\nCOMMENTS:It can be seen that the value of thefirm comes out to be the same in both the cases i.e whether MamboLeo Ltd pays the Dividends or it does not pay the dividends. Thisis what the Modigliani and Miller approach shows, that the Dividendpolicy is irrelevant for the valuation ofthe firm.\n\n##### \"Our Prices Start at \\$11.99. As Our First Client, Use Coupon Code GET15 to claim 15% Discount This Month!!\"", null, "Calculate the price\nPages (550 words)\n\\$0.00\n*Price with a welcome 15% discount applied.\nPro tip: If you want to save more money and pay the lowest price, you need to set a more extended deadline.\nWe know how difficult it is to be a student these days. That's why our prices are one of the most affordable on the market, and there are no hidden fees.\n\nInstead, we offer bonuses, discounts, and free services to make your experience outstanding.\nHow it works\nReceive a 100% original paper that will pass Turnitin from a top essay writing service\nstep 1\nFill out the order form and provide paper details. You can even attach screenshots or add additional instructions later. If something is not clear or missing, the writer will contact you for clarification.\nPro service tips\nHow to get the most out of your experience with brilliantassignmenthelp.com\nOne writer throughout the entire course\nIf you like the writer, you can hire them again. Just copy & paste their ID on the order form (\"Preferred Writer's ID\" field). This way, your vocabulary will be uniform, and the writer will be aware of your needs.\nThe same paper from different writers\nYou can order essay or any other work from two different writers to choose the best one or give another version to a friend. This can be done through the add-on \"Same paper from another writer.\"\nCopy of sources used by the writer\nOur college essay writers work with ScienceDirect and other databases. They can send you articles or materials used in PDF or through screenshots. Just tick the \"Copy of sources\" field on the order form.\nTestimonials\nSee why 20k+ students have chosen us as their sole writing assistance provider\nCheck out the latest reviews and opinions submitted by real customers worldwide and make an informed decision.\n11,595\nCustomer reviews in total\n96%\nCurrent satisfaction rate\n3 pages\nAverage paper length\n37%\nCustomers referred by a friend", null, "Use a coupon FIRST15 and enjoy expert help with any task at the most affordable price.\nClaim my 15% OFF Order in Chat", null, "" ]	[ null, "https://img.wizedu.com/questions/c1ba7ef0-1a78-11ed-b22c-b1bafa6188ca.png", null, "https://writinghelpe.com/wp-content/uploads/2022/08/save.jpg", null, "https://brilliantassignmenthelp.com/wp-content/themes/proxim-canyon/images/trustpilot.svg", null, "https://theessaypages.com/wp-content/plugins/popup-builder/public/img/theme_6/close.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86797035,"math_prob":0.9789862,"size":3286,"snap":"2023-40-2023-50","text_gpt3_token_len":1114,"char_repetition_ratio":0.15265082,"word_repetition_ratio":0.05901639,"special_character_ratio":0.39957395,"punctuation_ratio":0.14951989,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98105437,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,1,null,null,null,10,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-29T19:33:07Z\",\"WARC-Record-ID\":\"<urn:uuid:2cd13bc5-e3d2-4aff-8e3c-6fc4d28eb9f8>\",\"Content-Length\":\"72662\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7e4936c9-8424-46e2-83ef-42b472828fa9>\",\"WARC-Concurrent-To\":\"<urn:uuid:e5a02ea5-ac7a-4c62-bea5-ebea2bb9c956>\",\"WARC-IP-Address\":\"63.250.38.191\",\"WARC-Target-URI\":\"https://brilliantassignmenthelp.com/mambo-leo-ltd-has-a-cost-of-eq/\",\"WARC-Payload-Digest\":\"sha1:MC3MMC4KUK7F4U535WGR6HAZXULQZDYQ\",\"WARC-Block-Digest\":\"sha1:JBVZM23TY7OMZ5KDYBFCFR7DWQPMJRRI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100135.11_warc_CC-MAIN-20231129173017-20231129203017-00488.warc.gz\"}"}
https://www.codecodex.com/wiki/index.php?title=While_loop&oldid=6284	[ "# While loop\n\nThe while loop is a common looping mechanism. It is comparable to the for loop.\n\n## Implementations\n\n### C / C++\n\n```int x = 0;\nint y = 10;\n\nwhile(x < y){\nx++;\n}\n```\n\n### Java\n\n```int x = 0;\nint y = 10;\n\nwhile(x < y){\nx++;\n}\n```\n\n### Pascal\n\n```while (x < y) do\nbegin\n{Do something.}\nend;\n```\n\n### PHP\n\n```\\$i=0;\n\nwhile (\\$i<10) {\n// do something\n\\$i++;\n}\n```\n\nOr:\n\n```while (true) {\n// do something\nif (\\$a==\\$b)\nbreak;\n}\n```\n\n### Ruby\n\n```x = 10\nwhile x > 0\n# do something\nx -= 1\nend\n```\n\n### Visual Basic\n\n```Do While x < y\n' Do something\nLoop\n```\n\nOr:\n\n```While x < y\n' Do something\nEnd While\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6921606,"math_prob":0.62062615,"size":575,"snap":"2021-31-2021-39","text_gpt3_token_len":201,"char_repetition_ratio":0.17162873,"word_repetition_ratio":0.19512194,"special_character_ratio":0.4156522,"punctuation_ratio":0.14925373,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9990606,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-04T21:20:21Z\",\"WARC-Record-ID\":\"<urn:uuid:1bf75f97-9125-4dac-affd-59e753fd7e4c>\",\"Content-Length\":\"14532\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:93328efe-65eb-44fa-b100-5e33efc202e3>\",\"WARC-Concurrent-To\":\"<urn:uuid:bc2ef351-7087-45a9-be09-35afba6477a1>\",\"WARC-IP-Address\":\"54.83.6.65\",\"WARC-Target-URI\":\"https://www.codecodex.com/wiki/index.php?title=While_loop&oldid=6284\",\"WARC-Payload-Digest\":\"sha1:FCT75UFATNHLBZ5VLPJC6YDBCRRUCUHX\",\"WARC-Block-Digest\":\"sha1:42GR47YUWA26FXUTRUD7SFAZGWZMWNQQ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046155188.79_warc_CC-MAIN-20210804205700-20210804235700-00677.warc.gz\"}"}
http://forums.wolfram.com/mathgroup/archive/2005/Nov/msg00571.html	[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: statistics questions\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg62378] Re: statistics questions\n• Date: Tue, 22 Nov 2005 04:42:28 -0500 (EST)\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```On 11/21/05 at 5:25 AM, chris.chiasson at gmail.com (Chris Chiasson)\nwrote:\n\n>I guess what I mean by \"mean response\" is that regression generates\n>a least squares fit function for y, which then has a confidence\n>interval associated with it: regression function + or -\n>t[v,ci]*sqrt(variance/n)\n\nHmmm... I assume in the above formula \"variance\" is intended to be the variance of y? If so, this will not provide what is usually meant by a confidence interval for any of your data.\n\nThere are two sources of variability that need to be considered. First, there is the variation in the response variable at a fixed set of input conditions. Second, there is the variation in response caused by a variation of input conditions. So, if you want a confidence interval for the predicted response at a given set of input conditions both factors need to be addressed.\n\n>I think if y is supposed to be a constant value, then least squares\n>regression is equivalent to the mean of the data.\n\nMore or less true. If y is constant (independent of x) then fitting a linear model mx+b with m = 0 will cause b to be the mean of y. But when m is not forced to 0, b will only be close to the mean of y.\n\n>Anywho, does anyone know of a way to obtain the regression function\n>plus its confidence interval directly?\n\nSimply put, I cannot make sense of this. Regression functions don't have confidence intervals. There are confidence intervals for the estimated parameters of a regression function and there are confidence intervals for any of the data points (either predicted or observed) but not the regression function itself.\n\n>Am I just using statistics incorrectly here?\n\nI don't know. You do seem to be using terminology in what appears to be a non-standard way which makes it difficult to know how to answer your question. Perhaps referring to a good text on regression would be helpful?\n\nA couple of texts I like are\n\nApplied Linear Regression by Sanford Weisberg\n\nand\n\nFitting Equations to Data by Cuthbert Daniel & Fred S. Wood.\n\nOf these two, I like the presentation in Weisberg better. But I think Daniel & Wood is referenced more frequently and in fact, is referenced by Weisberg.\n--\nTo reply via email subtract one hundred and four\n\n```\n\n• Prev by Date: Re: Re: Confusing results with N[expr]?\n• Next by Date: Re: Intepolation of an array with missing points\n• Previous by thread: Re: statistics questions\n• Next by thread: Re: statistics questions" ]	[ null, "http://forums.wolfram.com/mathgroup/images/head_mathgroup.gif", null, "http://forums.wolfram.com/mathgroup/images/head_archive.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/2.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/0.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/0.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/5.gif", null, "http://forums.wolfram.com/mathgroup/images/search_archive.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9147289,"math_prob":0.84638107,"size":2451,"snap":"2020-24-2020-29","text_gpt3_token_len":573,"char_repetition_ratio":0.121781774,"word_repetition_ratio":0.004761905,"special_character_ratio":0.23011015,"punctuation_ratio":0.11632653,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99561274,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-10T14:08:34Z\",\"WARC-Record-ID\":\"<urn:uuid:d06ab94d-8e6f-4a92-9ae9-fb08a050c1fd>\",\"Content-Length\":\"45749\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c3b5ed34-e8d8-435e-9cad-fcae7211f7cb>\",\"WARC-Concurrent-To\":\"<urn:uuid:5f037a22-a6b0-4e39-9912-81cdebc3f633>\",\"WARC-IP-Address\":\"140.177.205.73\",\"WARC-Target-URI\":\"http://forums.wolfram.com/mathgroup/archive/2005/Nov/msg00571.html\",\"WARC-Payload-Digest\":\"sha1:HC2WAO2QZPSOILRWAHONFD4LR4F3OSLJ\",\"WARC-Block-Digest\":\"sha1:EM4CE2CFICX2BT62LJ6MM4ATRODCLK3B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655908294.32_warc_CC-MAIN-20200710113143-20200710143143-00517.warc.gz\"}"}
https://aniruddhakaushik.wordpress.com/2021/12/03/kt-46-self-evaluation-ncqc-2019/	[ "# KT 9 : NCQC – 2021\n\n1. In X-R chart, the value of A2R is 1.8. Calculate the value of sigma i.e. standard deviation?\nAns: 0.6\n\n2. Cause and effect diagram was introduced in the period:\na) 1971-1980\nb) 1951-1960\nc) 1961-1970\nd) Before 1950\n\n3. In X-R chart, the value of X = 72.972 gms. No. of observations are 25, average moving range R =3.384, value of d2 = 1.128, what will be the value of UCL ?\nAns: 81.972\n\n4. “As long as there is a workplace, Quality Control Circle activities must be continued”. Who said this ?\na) Dr. Abdul Kalam\nb) Dr. J. M. Juran\nc) Dr. W. E. Deming\nd) K. Ishikawa\n\n5. Highest value of 625 observations of a Biscuit packet is 102.50 gm. Least count to scale by which weight of packets is taken is 0.05 gm. If the range is 1.55 gm, calculate the second-class interval?\nAns: 101.085 – 101.245\n\n6. How many defective pieces will be in a lot of 20000 pieces of toy? If the process is adjusted at 3 sigma level.\nAns: 54\n\n7. When the coefficient correlation between 2 variables is 0.9 what does it indicate?\nAns: Strong positive correlation\n\n8. What % of values lies between plus and minus 3 sigma when the process is under normal Distribution:\na) 91.73%\nb) 93.73%\nc) 97.73%\nd) 99.73%\n\n9. What should be the recommended number of classes or groups for 627 observations for frequency distribution for making Histogram ?\na) 8\nb) 9\nc) 11\nd) 10\n\n10. Which one belongs to elementary statistical method (7 tools):\na) Brain Storming\nb) Flow diagram\nc) Tree diagram\nd) Scatter diagram" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8162597,"math_prob":0.9765079,"size":1466,"snap":"2023-14-2023-23","text_gpt3_token_len":460,"char_repetition_ratio":0.103283174,"word_repetition_ratio":0.014814815,"special_character_ratio":0.34515688,"punctuation_ratio":0.18309858,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9796257,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-03T21:38:54Z\",\"WARC-Record-ID\":\"<urn:uuid:672141cb-c35b-44cb-8f96-c74cc6c8d1f5>\",\"Content-Length\":\"90559\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c041426d-9472-4364-a3e1-eabf1e319703>\",\"WARC-Concurrent-To\":\"<urn:uuid:aab9e864-4749-4d31-8fd4-ac776d7f7a42>\",\"WARC-IP-Address\":\"192.0.78.13\",\"WARC-Target-URI\":\"https://aniruddhakaushik.wordpress.com/2021/12/03/kt-46-self-evaluation-ncqc-2019/\",\"WARC-Payload-Digest\":\"sha1:LDICTFZOP3AQGHZSR2IFPZFJXRHUEDUA\",\"WARC-Block-Digest\":\"sha1:IA3ESPEV2OG4MCJT65TBYPRZ2K6MRJKQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649343.34_warc_CC-MAIN-20230603201228-20230603231228-00583.warc.gz\"}"}
https://ovid.cs.depaul.edu/Classes/CS202-S07/hw2.htm	[ "Homework 2 (due 4/16) CSC 202\n\nWe finished chapter 1 and started on chapter 2 (2.1-2.3 and 2.5). Next week we will finish chapter 2 and start talking about relations and first-order logic. I suggest you read through Sections 2.1-2.3 carefully, since it contains several additional examples and exercises (some with solutions).\n\nSubmission: you can submit the homework to me either by hardcopy in class or email it to me.\n\n1. [Check Constraints] Write complete ALTER TABLE commands for the following check constraints (see Section 1.4). Also, for each check constraint write an INSERT statement that violates the constraint. (Double-check by running them on the database.)\n\n1. [10pt] The career in Student is one of the following: GRD, UGRD, SAL. Hint: here is what the SQL for adding the constraint will look like:\nALTER TABLE Student\nCHECK ( ... );\n1. [15pt] All IT courses are undergraduate courses (that is, their course number is less than 420 (this is a bit tricky; think about it carefully: what are you trying to exclude? Make sure you test your constraint with examples).\n\n2. [Set Theory, 10pt] True of False? If true, give a proof, if false, find sets A, B, C that show that it is false:\n\nA n (B u C) = (A n B) u C, for all sets A, B, C.\n\n(u is union, n is intersection).\n\n3. [Programming Logic, 15pt] Some programming languages (functional languages in particular) contain an if-then-else expression. That is you write\n\nif condition then a else b,\n\nand if condition holds, the value returned will be that of a and if not, the value of b. (In C you could write (a <= b ? a : b) to compute the minimum of a and b.)\n\n1. [5pt] Find the truth-table of the truth-function \"if p then q else r\".\n2. [10pt] Find the disjunctive normal form for \"if p then q else r\".\n\n4. [SQL Queries, 30] For each SQL query, please submit your SQL and a printout of the resulting table. Be careful writing both of these, and make sure you double-check that your output is what you expect it to be.\n\n1. [15pt] List student groups which have both Chicago and non-Chicago members (HerCTI is the only such group in our database).\n2. [15pt] List students that are not presidents of any student society.\n\n5. [Extra Credit] You have five jars of quarters. One jar contains all counterfeit quarters, the remaining jars contain good quarters only. The counterfeit quarters are 5 grams each, the real quarters 6 grams. You also have a scale. You can pick any number of coins from the jars and make one weighing on the scale. How do you determine which jar contains the counterfeit quarters?\n\n6. [Extra Credit] Show that every truth-function can be written using just logical nand (see Section 1.5 for definition of nand). Hint: show that negation and logical and can be expressed just using nand.\n\n7. [Extra Credit] An old English nursery rhyme goes as follows:\n\nFor want of a nail the shoe was lost.\nFor want of a shoe the horse was lost.\nFor want of a horse the rider was lost.\nFor want of a rider the battle was lost.\nFor want of a battle the kingdom was lost.\nAnd all for the want of a horseshoe nail.\n\nWhat is the logical structure of this nursery rhyme? Introduce appropriate atomic propositions (one for each object), and then use these propositions to describe the statements in the rhyme.\n\n8. [Extra Credit] If you find any typos and mistakes in the lecture notes, please let me know.\n\nMarcus Schaefer\nLast updated: April 10h, 2007." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9159152,"math_prob":0.7830074,"size":2637,"snap":"2022-05-2022-21","text_gpt3_token_len":626,"char_repetition_ratio":0.10558298,"word_repetition_ratio":0.017021276,"special_character_ratio":0.24270004,"punctuation_ratio":0.14234875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9508552,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-21T21:35:10Z\",\"WARC-Record-ID\":\"<urn:uuid:ce384ca3-7e5a-405c-9372-4d29b62e5985>\",\"Content-Length\":\"5670\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:907324eb-321c-4825-b38b-ec82d82ed0a1>\",\"WARC-Concurrent-To\":\"<urn:uuid:c683cd2a-14fd-4013-a2b1-f4600a608dc3>\",\"WARC-IP-Address\":\"216.220.181.39\",\"WARC-Target-URI\":\"https://ovid.cs.depaul.edu/Classes/CS202-S07/hw2.htm\",\"WARC-Payload-Digest\":\"sha1:Y345UBUYCE5FYTJO3DJRE3OSVKIMDRQN\",\"WARC-Block-Digest\":\"sha1:BYCNSRVAGWAGYJ324FOHWTNAO5X6UZUA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320303709.2_warc_CC-MAIN-20220121192415-20220121222415-00131.warc.gz\"}"}
https://blogs.agu.org/martianchronicles/2009/07/10/solar-system-creator/	[ "10 July 2009\n\n## Solar System Creator\n\nPosted by Ryan Anderson\n\nAs I mentioned last month, on top of research and grad school duties, I’m in the process of planning out a sci-fi novel. It began with the month-long outlining challenge “Midsommer Madness” over at the Liberty Hall writing site, and I am continuing with it in my spare time.\n\nI am trying to make my novel grounded in reality whenever possible. It is set in a known star system, 55 Cancri. The 55 Cancri system has 5 known planets, but I also took some artistic license and added moons and small planets that observations would likely have missed. Then, once I had planets and moons, I needed to figure out which ones would be habitable!", null, "I happen to know a thing or two about planets, so I put together a handy spreadsheet to use to calculate things like surface temperature, surface gravity and orbital period given things like how bright the star is, how far away the planet is, etc. Once I had the spreadsheet made, I realized that there are likely other people out there who might find it useful.\n\nSo, whether you are a writer trying to come up with a plausible setting for your bestselling sci-fi epic, or a student learning about the solar system, or just plain curious about planets, please feel free to use and modify this spreadsheet. Right now it is set up for our solar system to give you an idea of what reasonable values are for the different variables, and to show that the results are generally pretty good despite the simplicity of the calculations. If you find this useful or have any questions, feel free to contact me by leaving a comment on this post.\n\nI described how I calculated the surface temperature below. You don’t need to read the explanation to use the spreadsheet: it should just work if you enter numbers, but I encourage you to try to follow the derivation. Even if you don’t follow the algebra, I tried to explain everything in words to give some conceptual understanding of the ideas behind the math, and the ideas are what matter.\n\nNote for Students: The spreadsheet is free for you to use, but be sure you cite this page as a source. Also show your work for all calculations! Copying values from the spreadsheet without showing your work probably won’t earn you any points, and may be considered plagiarism, which is grounds for failure and/or expulsion at most schools. And really, it’s not that hard to do the calculations, especially since the rest of this post is spent walking you through them! You might even learn something!\n\nOk, so how does it work? Well, the calculation of a planet’s surface temperature is based on the very simple idea that if its average temperature is not changing, then the amount of energy the planet absorbs must match the amount that it emits. Pretty much common sense! If the amount in and out were different, then the temperature would change until they balanced!\n\nFirst, the absorption. The energy source is the star, which has a certain luminosity $latex L$ (given in watts). This says how much energy the star puts out in all directions per second. We want to know how much energy per square meter hits the planet, so we take the luminosity and spread it out evenly over the surface of a sphere with a radius $latex R$ equal to the distance from the planet to the star. The surface area of a sphere is $latex A=4\\pi R^2$, so the amount of energy from the star hitting each square meter of the planet’s cross section is: $latex L/A=\\dfrac{L}{4\\pi R^2}$\n\nThe planet’s cross section is just the area of a circle with the planet’s radius: $latex \\pi r^2$. Note that we’re using the area of a circle and not a sphere! That’s because the starlight doesn’t hit the whole planet, it just hits the part of the planet that is visible. Can you see all sides of a sphere at once? Neither can I, and neither can the star. What we see is the 2 dimensional cross section: a circle.\n\nSo now we have an equation for how much energy the planet absorbs per second: $latex Energy Absorbed Per Second= \\dfrac{L \\pi r^2}{4\\pi R^2}$\n\nBut that is assuming that the planet absorbs every bit of light that hits it, which we know isn’t true: we see planets in the night sky by their reflected light! So we can add a correction called albedo. Albedo, $latex A$, is the fraction of starlight that the planet reflects back out into space, and $latex (1-A)$ is the fraction of starlight a planet absorbs. So with that correction, our equation becomes: $latex Energy Absorbed Per Second= \\dfrac{(1-A) L \\pi r^2}{4\\pi R^2}$\n\nNow we have to figure out an expression for the energy that the planet emits. Here we have to make an assumption to simplify things: we assume that the energy absorbed by the planet is immediately redistributed evenly over the whole planet. Obviously this isn’t right, it is much warmer on the day side than the night side, but this assumption makes our lifes much easier. We just have to remember that the value we get is going to be an average of day and night temperatures.\n\nWe also assume that the planet radiates away its energy like a blackbody. A blackbody is something that absorbs and emits all radiation perfectly. We’re not going to worry too much about this assumption. For our purposes, planets are close enough to being blackbodies that it doesn’t matter much. I know, I know, we just made an adjustment for albedo two paragraphs ago, implying that the planet is not a perfect blackbody! Just calm down. It works pretty well, and that’s all we need.\n\nAnyway, if we assume the planet is a uniform temperature blackbody, then we can use the handy equation for blackbody emission: $latex Energy Emitted Per Square Meter Per Second = \\sigma T^4$. Sigma is called the Stefan-Boltzmann constant, and is given by $latex \\sigma = 5.67\\times 10^{-8} W m^{-2} K^{-4}$ To get rid of that pesky “per square meter” part of the equation, we just multiply by the surface area of the thing doing the emitting: in this case, the planet. Here we do use the surface area of a sphere, remember our assumption that the energy absorbed gets spread out over the whole surface? This is why we did that. The result is:\n\n$latex Energy Emitted Per Second = 4 \\pi r^2 \\sigma T^4$\n\nNow that’s a fine equation if your planet emits every bit of energy that it receives straight back to space. But that’s not how it works for planets with atmospheres. There’s this effect where the atmosphere traps energy in the system for a longer time, resulting in a warmer planet… you may have heard of it: the Greenhouse Effect! It would be nice if we could add that to our model! If we don’t, we’ll never get the surface temperature right for a planet like Venus, where the greenhouse effect dominates.", null, "To actually do a proper simulation of the greenhouse effect is very difficult and complicated, so instead we are going to use a fudge factor. The bottom line is that the greenhouse effect $latex GE$ reduces the amount of energy radiated from the surface that escapes to space. So we can do something very similar to our albedo adjustment: $latex GE$ gives the amount of energy that the atmosphere absorbs, and $latex 1-GE$ gives the amount of energy that actually escapes to space. For the Earth $latex GE \\approx 0.4$ and for Venus $latex GE \\approx 0.99$. Our modified equation is now:\n\n$latex Energy Emitted Per Second = (1-GE) 4 \\pi r^2 \\sigma T^4$\n\nNow, remember why we were doing all of this? We want to find the planet’s average surface temperature $latex T$. To get this, we have to set our two equations equal to each other and solve:\n\n$latex Energy Emitted Per Second = Energy Absorbed Per Second$\n\n$latex (1-GE) 4 \\pi r^2 \\sigma T^4 = \\dfrac{(1-A) L \\pi r^2}{4\\pi R^2}$\n\nLook! The planet’s radius appears on both sides of the equation! That means it cancels out, and that a planet’s radius has no effect on its surface temperature! Ok, don’t get too excited, we still need to solve for T.\n\n$latex T^4 = \\dfrac{L (1-A)}{16\\sigma\\pi R^{2}(1-GE)}$\n\n$latex T = \\left(\\dfrac{L (1-A)}{16\\sigma\\pi R^{2}(1-GE)}\\right)^{1/4}$\n\nVoila! There is the expression for the equilibrium surface temperature of a planet, taking into account the planet’s reflectivity and the greenhouse effect. I hope this sheds a little light into how to think about the energy budget of a planet, and how my spreadsheet works. Again, if you have any questions, post them in the comments and I’ll answer them!" ]	[ null, "http://upload.wikimedia.org/wikipedia/commons/5/59/Extrasolar_planet_NASA2.jpg", null, "https://blogs.agu.org/martianchronicles/files/2009/07/gheffect.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9246376,"math_prob":0.97174406,"size":10408,"snap":"2020-45-2020-50","text_gpt3_token_len":2415,"char_repetition_ratio":0.13043061,"word_repetition_ratio":0.011950027,"special_character_ratio":0.22818986,"punctuation_ratio":0.09390981,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9933194,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-30T11:01:05Z\",\"WARC-Record-ID\":\"<urn:uuid:1f91643f-5194-49b1-bfab-4eb99c044f22>\",\"Content-Length\":\"84114\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c0c0cbc6-06fb-4f73-9b46-3fe93b88b56c>\",\"WARC-Concurrent-To\":\"<urn:uuid:959377f3-60b6-4491-be85-2f4b24913333>\",\"WARC-IP-Address\":\"141.193.213.20\",\"WARC-Target-URI\":\"https://blogs.agu.org/martianchronicles/2009/07/10/solar-system-creator/\",\"WARC-Payload-Digest\":\"sha1:UCVOBNEWGRGRUZ6GSR3VPWSYYCDAGLOC\",\"WARC-Block-Digest\":\"sha1:ZHVE6DFNGCVVIXS5MGHRAEMLRDGXPNQH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141213431.41_warc_CC-MAIN-20201130100208-20201130130208-00151.warc.gz\"}"}
https://file.scirp.org/xml/82453.xml	[ "Introduction: We have previously developed an effective atomic number ( Z eff ) measurement method using linear attenuation coefficients (LACs) obtained by energy-resolved computed tomography (CT) with one-dimensional (1D) detector. The energy-resolved CT was performed with a “transXend” detector, which measured X-rays as electric current and then gave X-ray energy distribution with unfolding analysis using pre-estimated response function (RF). The purpose of this study is to measure Z eff by the energy-resolved CT using a flat panel detector (FPD). Methods: To demonstrate a 2D transXend detector, we developed the stripe absorbers for the FPD. Eleven human tissue-equivalent material rods which were grouped into four material categories were measured by X-rays with 120 kVp tube voltage, 2.3 mA tube current, and 1.0 s exposure time. Z eff is measured by the ratio of LACs with two different pseudo-monochromatic X-ray energies. RFs of each rod material were estimated by numerical calculation. First, we employed the RF estimated for the same rod material (self-RF scenario). Second, we employed the RF estimated for the different rod materials in the same material category (cross-RF scenario). The purpose of the cross-RF scenario was to find representative rod materials in each material category. Results: Upon the self-RF scenario, measured Z eff s were systematically underestimated. Median relative error to theoretical Z eff was - 6.92% (range: - 7.89% - - 4.60%). After normalizing measured Z eff s to the theoretical one for Breast, median relative error improved to - 0.75% (range: - 1.79% - +1.73%). Upon the cross-RF scenario, the representative rod materials were found in two material categories. Conclusion: Z eff measurements were performed by energy-resolved CT using 2D transXend detector with numerically-estimated RF data. Normalized Z eff s for all rod materials in the self-RF scenario were in good agreement with the theoretical ones.\n\nX-Ray Computed Tomography Energy Resolved Unfolding Effective Atomic Number\n1. Introduction\n\nIn photon and particle radiation therapy treatment planning, single-energy computed tomography (CT) image is commonly used to distinguish materials inside a patient body and to calculate absorbed doses. A CT image is a distribution of linear attenuation coefficient (LAC) of each pixel. To calculate the dose in the patient body, LACs are converted to corresponding mass densities using a conversion table. However, estimated LACs for each pixel can be different from true LACs due to the beam-hardening effect: when polychromatic X-rays pass through a subject, the effective energy of X-rays increases because of the absorption of low energy X-rays. Also, estimated LACs of different materials would be similar and hard to distinguish in single-energy CT image. Thus, a mis-assignment of mass density may occur in single-energy CT measurement.\n\nTo make the dose calculation accurately, a Monte Carlo method has been implemented . If the mis-assignment of mass density due to the beam-hardening effect happened, it can lead to significant dose errors: up to 10% error for 6 - 15 megavoltage (MV) photons . Recently, an effective atomic number (Zeff) attracts attention as an alternative value to LAC.\n\nCommonly, Zeff measurement is performed by the use of LACs measured by X-rays with two different energies. Thus, the use of synchrotron facilities where can generate monochromatic X-ray is one of the best methods for Zeff measurement . However, synchrotrons are too large to install in general hospitals. Alternative methods are a photon-counting CT or a fast kVp switching dual-energy CT. Photon-counting CT uses X-ray detectors which can measure the energy of X-ray. However, photon-counting CT has a limitation in counting rate which is less than 0.5 × 106 s−1 in general . Typical number of X-rays coming into a detector in clinical practice is 109 mm−2・s−1 when they pass the air only and 106 mm−2・s−1 when they pass a thick body . In this stage, photon-counting CT is not practical in view of counting rate problem. Fast kVp switching dual-energy CT requires two projection data on each measurement direction with switching high and low voltages within 10 ms . Two transmission measurements are performed by X-rays with two different averaged energies. The difference between the two averaged energies is, however, not very large: with 80 and 140 kVp, the typical averaged energies at the exit of X-ray tube are 41.4 and 58.9 keV, respectively. According to the transmission direction of a human body, the averaged energy changes and results in beam-hardening artifact in CT image.\n\nTo overcome the problems associated with photon-counting CT and dual- energy CT, we have proposed a novel energy-resolved CT by using a transXend detector . The schematic drawing of the transXend detector is shown in Figure 1 of . The transXend detector consists of several segmented detectors aligned in the X-ray incident direction and measures X-rays as electric currents. The X-ray energy distribution is given after analysis using pre-estimated response function (RF). The transXend detector gives photon numbers in arbitrary energy ranges. Since X-rays are measured as electric currents by the transXend detector, there are no problems associated with the counting rate. This transXend detector collects transmission data of an object under study by repeating rotation and transverse movements. Thus, we called this detector the one-dimensional (1D) transXend detector. Yamashita et al. reported that Zeff of aluminum was measured within 1% error, where the error was defined as (Zeff − Z)/Z . Kanno et al. also reported that relative error of measured Zeffs of water and acrylic were within 3% . Both studies were performed by the 1D transXend detector.\n\nFor the application of two-dimensional (2D) transXend detector for clinical practice in the future, the authors invented stripe absorbers which are placed in front of a flat panel detector (FPD) . The stripe absorbers consist of two kinds of metal ribbons and provides four different X-ray energy spectra. Kanno et al. reported the possibility of 2D transXend detector using a thermo-lumine- scent plate .\n\nAs mentioned above, Zeff measurements for aluminum, acrylic, and water using the 1D transXend detector were conducted previously. In this study, we performed Zeff measurements for eleven human tissue-equivalent materials by the energy-resolved CT using the 2D transXend detector with a FPD. The eleven materials were grouped into four material categories: LUNG, SOFT TISSUE,\n\nSOFT BONE, and BONE. Zeffs were measured by using the RFs estimated under two different scenarios: With RFs estimated for the same materials with the one used for CT measurements, or with RFs estimated for different materials in each material category.\n\n2. Materials and Methods 2.1. Theory of an Effective Atomic Number Measurement\n\nSpiers et al. proposed the definition of Zeff for considering the X-ray absorption by human tissue :\n\nZ eff = ∑ k α k Z k 2.94 2.94 , (1)\n\nwhere αk is the electron number fraction, and Zk is the atomic number of element k.\n\nTorikoshi et al. showed the measurement of Zeff with two different monochromatic X-ray energies using a synchrotron facility . In the energy range for X-ray CT, 80 - 140 keV, the X-ray LAC of element Z for monochromatic energy E can be written as\n\nμ ( E ) = ρ e { Z 4 F ( E , Z ) + G ( E , Z ) } . (2)\n\nHere ρe is the electron density, ρeZ4F(E,Z) is the photoelectric term, and ρeG(E,Z) is the scattering term. Thus, atomic number Z can be described by the ratio of LAC:\n\nf ( Z ) = μ ( E a ) / μ ( E b ) . (3)\n\nSince the LAC of elements are summarized as a function of X-ray energy in the table of the National Institute of Standards and Technology (NIST), the term f(Z) can be drawn by plotting the value of μ(Ea)/μ(Eb) as a function of Z . Therefore, the Zeff can be obtained by using the Equation (3) with measuring LACs at two different X-ray energies Ea and Eb. Since a CT image is the distribution of LAC, the ratio can be obtained by dividing two CT images which were acquired by two different monochromatic X-ray energies Ea and Eb.\n\n2.2. The Relationship between the Measured Currents and X-Ray Energy Distribution\n\nWhen the transXend detector is used for X-ray transmission measurement, the relationship between the measured electric currents and the X-ray energy distribution is expressed in terms of following matrix equation :\n\n( I 1 I 2 ⋮ I m ) = ( R 1 , 1 R 1 , 2 ⋯ R 1 , n R 2 , 1 R 2 , 2 ⋮ ⋮ ⋱ R m , 1 ⋯ R m , n ) ( Y 1 Y 2 ⋮ Y n ) . (4)\n\nHere Ii (i = 1, m) is the electric current value measured by i-th segmented detector, Yj (j = 1, n) is the number of X-rays in the energy range Ej , and Ri,j is the RF of the i-th segment detector in the energy range Ej. The X-ray energy distribution is obtained by solving Equation (4) using an unfolding code, such as SAND II . In the unfolding process, the number of energy ranges and the widths of energy ranges can be assigned according to the materials of interest. More detailed information is described in elsewhere .\n\n2.3. Two-Dimensional transXend Detector\n\nFor the clinical application of the transXend detector, a 2D transXend detector should be developed. To make the transXend detector system two-dimensional, we used a FPD and stripe absorbers which consist of two different absorbers A and B in a lattice shape, as shown in Figure 1 . With the stripe absorbers placed in front of the FPD, four different regions, (a)-(d), can be made, as shown in Figure 1. In region (a), X-rays enter to the FPD without passing two absorbers. Subsequently, in region (b), (c), and (d), X-rays passed through the absorber A, B, and A + B, respectively. We used 1 mm-wide and 0.1-mm-thick tin and copper for absorber A and B, respectively. Calculated X-ray spectra arriving at FPD pixels in each region are shown in Figure 2. Considering the four regions as one pixel, each region has the role of segmented detectors. X-ray energy distribution can be acquired for the 2D position on the FPD by unfolding electric currents measured by the four regions. The employed FPD was Remote RadEye2 (Teledyne Rad-icon Imaging, Sunnydale, CA, USA) with pixel matrix of 1024 × 1024 pixel2 (active area 49.3 × 49.2 mm2). The pixel size of photodiode was 48 × 48 μm2. Incident X-ray photons are absorbed by a Gd2O2S scintillator plate and scintillation photons are detected by a 2D CMOS photodiode array.\n\n2.4. Human Tissue-Equivalent Materials\n\nEleven RMI rods (Gammex, Middleton, WI, US) were used in this study, as summarized in Table 1. Diameter and height of each rod was 28 mm and 70 mm, respectively. RMI rods were often used as the calibration materials of CT\n\nCharacteristics for RMI rods\nMaterial category Rod material Relative electron density Mass density [g/cm3] Theoretical effective atomic number\nLUNG LN300 0.29 0.29 7.86\nLN450 0.45 0.46 7.84\nSOFT TISSUE Adipose 0.92 0.94 6.40\nBreast 0.96 0.98 7.24\nSolid water 0.99 1.02 8.11\nBrain 1.05 1.05 6.31\nSOFT BONE Inner bone 1.10 1.14 10.9\nBone mineral 1.11 1.15 10.9\nBONE CB2-30% 1.28 1.33 11.4\nCB2-50% 1.47 1.56 13.0\nCortical bone 1.70 1.82 14.1\n\nnumber―mass density conversion table for the dose distribution calculation in radiotherapy treatment planning. In such calibration, each rod was inserted to a 330-mm-diameter and 50-mm-height RMI phantom and all rods were scanned simultaneously. In this study, however, each rod was scanned individually to measure its Zeff for avoiding the fan-beam effect.\n\n2.5. Experiment\n\nExperimental set up was shown in Figure 3. As described in the previous paper, RF measurement for the phantom material was necessary prior to CT measurement . Used materials for RF measurement were the same ones with the phantom for CT measurement: slabs of different thicknesses were prepared for each material in the phantom. We, however, transmission data for the RF estimation by a numerical calculation using Lambert-Beer’s law since each rod was uniform:\n\nI = I 0 exp ( − μ t ) . (5)\n\nHere I is photon number transmitted through a material with LAC, μ, with thickness t, I0 is incident photon number. Transmission data was calculated for each rod material with thicknesses ranging from 0 to 30 mm at intervals of 5 mm. I0 was normalized to the measured current induced by X-rays which passed the air only. In transXend analysis, RF data was interpolated by 1 mm interval. Scatter X-rays were not considered in the calculation.\n\nAfter calculating transmission data for the RF estimation, each rod was scanned from one direction by the X-rays. Employed X-ray tube was ERESCO MF4 (GE Sensing & Inspection Technologies, Ahrensburg, Germany) with a tungsten target and built-in filters made from 0.8-mm-thick beryllium and 2-mm-thick aluminum. The X-ray tube was placed 1000 mm away from the FPD. The X-ray tube operating conditions were 120 kV for tube voltage, 2.3 mA\n\nfor tube current, and 1 s for exposure time. Since each rod was axial symmetry, projection data for each rod was duplicated 359 times to make 1˚ step data. Source-to-axis of the phantom distance was 850 mm.\n\nSix energy ranges were defined for obtaining X-ray energy distribution, as shown in Table 2. Since almost no X-rays with the energy under 15.0 keV entered into the FPD, those X-rays were excluded from the analysis. The X-rays in the energy range E2: 35.0 - 36.0 keV and E5: 65.0 - 66.0 keV were used as pseudo-monochromatic X-rays. Measured currents in the center column of the FPD were unfolded by SAND II code to obtain energy distributions. Number of X-rays in each energy ranges were estimated for each projection. With Y2 and Y5, CT images were reconstructed by maximum likelihood-expectation maximization method . Iterative number was 30 times which was optimized prior to the measurements. With the LAC data table of NIST, Z-μ(E2)/μ(E5) relationship can be drawn as Figure 4. CT images of μ(E2) and μ(E5) for each rod were converted to a Zeff image using Figure 4.\n\nIn previous studies , the RF was estimated by using the same material with the one of phantom for CT measurement. In this study, we demonstrated two different analysis scenarios. In the first scenario, we employed the RF estimated for the same material as the one of the phantom for CT measurement (self-RF scenario). In the second scenario, we employed the RF estimated for the different material than the one of the phantom for CT measurement (cross-RF scenario). The purpose of the cross-RF scenario was to determine the representative rod materials in each material category. Reduction of the number of materials for RF estimation would widen the application of energy-resolved CT using 2D transXend detector. Mean and standard deviation (SD) of Zeff for each rod material was calculated in 10 × 10 mm2 region-of-interest on Zeff image.\n\nAssigned energy ranges. Unit: [keV]\nReferences Chetty, I.J., Curran, B., Cygler, J.E., De Marco, J.J., Ezzell, G., Faddegon, B.A., et al. (2007) Report of the AAPM Task Group No. 105: Issues Associated with Clinical Implementation of Monte Carlo-Based Photon and Electron External Beam Treatment Planning. Medical Physics, 34, 4818-4853. https://doi.org/10.1118/1.2795842 Verhaegen, F. and Devic, S. (2005) Sensitivity Study for CT Image Use in Monte Carlo Treatment Planning. Physics in Medicine & Biology, 50, 937-946. https://doi.org/10.1088/0031-9155/50/5/016 Toriloshi, M., Tsunoo, T., Sasaki, M., Endo, M., Noda, Y., Ohno, Y., et al. (2003) Electron Density Measurement with Dual-Energy X-ray CT Using Synchrotron Radiation. Physics in Medicine & Biology, 48, 673-685. https://doi.org/10.1088/0031-9155/48/5/308 Taguchi, K., Frey, E., Wang, X., Iwanczyk, J. and Barber, W. (2010) An Analytical Model of the Effects of Pulse Pileup on the Energy Spectrum Recorded by Energy Resolved Photon Counting X-Ray Detectors. Medical Physics, 37, 3957-3969. https://doi.org/10.1118/1.3429056 Yu, L., Leng, S. and McCollough, C.H. (2012) Dual-Energy CT-Based Monochromatic Imaging. American Journal of Roentgenology, 199, S9-S15. https://doi.org/10.2214/AJR.12.9121 Xu, C., Danielsson, M., Karlsson, S., Svensson, C. and Bornefalk, H. (2012) Preliminary Evaluation of a Silicon Strip Detector for Photon-Counting Spectral CT. Nuclear Instruments and Methods in Physics Research Section A, 677, 45-51. https://doi.org/10.1016/j.nima.2012.02.034 Kanno, I., Imamura, R., Mikami, K., Uesaka, A., Hashimoto, M., Phtaka, M., et al. (2008) A Current Mode Detector for Unfolding X-Ray Energy Distribution. Journal of Nuclear Science and Technology, 45, 1165-1170. https://doi.org/10.1080/18811248.2008.9711905 Yamashita, Y., Kimura, M., Hamaguchi, T., Kanno, I., Ohtaka, M., Hashimoto, M., et al. (2014) Measurement of Effective Atomic Numbers using Energy-Resolved Computed Tomography. Journal of Nuclear Science and Technology, 51, 1256-1263. https://doi.org/10.1080/00223131.2014.919881 Kanno, I., Yamashita, Y., Kimura, M. and Inoue, F. (2015) Effective Atomic Number Measurement with Energy-Resolved Computed Tomography. Nuclear Instruments and Methods in Physics Research Section A, 787, 121-124. https://doi.org/10.1016/j.nima.2014.11.072 Kanno, I., Yamauchi, K. and Hamaguchi, T. (2016) Two-Dimensional “transXend” Detector with Band Structure Absorbers for Third-Generation Energy-Resolved Computed Tomography with Improved Spatial Resolution. Journal of Nuclear Science and Technology, 54, 22-29. https://doi.org/10.1080/00223131.2016.1202154 Kanno, I., Yamashita, Y., Kanai, E., Ogawa, T. and Shinsho, K. (2016) Two-Dimensional “TransXend” Detector for Third Generation Energy-Resolved Computed Tomography. Journal of Nuclear Science and Technology, 53, 258-262.https://doi.org/10.1080/00223131.2015.1037810 Spiers, F.W. (1946) Effective Atomic Number and Energy in Tissues. The British Journal of Radiology, 19, 52-63. https://doi.org/10.1259/0007-1285-19-218-52 Hubbell, J.H. and Seltzer, S.M. (2010) Tables of X-Ray Mass Attenuation Coefficients and Mass Energy-Absorption Coefficients from 1 keV to 20 MeV for Elements Z = 1 to 92 and 48 Additional Substances of Dosimetric Interest. The Physical Measurement Laboratory, The National Institute of Standards and Technology, Gaithersburg, MD. https://www.nist.gov/pml/x-ray-mass-attenuation-coefficients McElroy, W., Berg, S., Crockes, T. and Hawkins, G.A. (1967) Computer-Automated Iterative Method for Neutron Fux Spectra Determination by Foil Activation. AFWL-TR-67-41. Air Force Weapons Laboratory, Albuqueque, NM. Imamura, R., Mikami, K., Minami, Y., Kanno, I., Ohtaka, M., Hashimoto, M., et al. (2010) Unfolding Method with X-Ray Path Length-Dependent Response Functions for Computed Tomography Using X-Ray Energy Information. Journal of Nuclear Science and Technology, 47, 1075-1082. https://doi.org/10.1080/18811248.2010.9711672 Shepp, L.A. and Vardi, Y. (1982) Maximum Likelihood Reconstruction for Emission Tomography. IEEE Transactions on Medical Imaging, 1, 113-122.https://doi.org/10.1109/TMI.1982.4307558 Bazalova, M., Carrier, J.F., Beaulieu, L. and Varhaegen, F. (2008) Dual-Energy CT-Based Material Extraction for Tissue Segmentation in Monte Carlo Dose Calculations. Physics in Medicine & Biology, 53, 2439-2456.https://doi.org/10.1088/0031-9155/53/9/015 Goodsitt, M.M., Christodoulou, E.G. and Larson, S.C. (2011) Accuracies of the Synthesized Monochromatic CT Numbers and Effective Atomic Numbers Obtained with a Rapid kVp Switching Dual Energy CT Scanner. Medical Physics, 38, 2222-2232. https://doi.org/10.1118/1.3567509 Papanikolaou, N., Battista, J.J., Kappas, C., Klein, E., Mackie, T.R., Sharpe, M., et al. (2004) Tissue Inhomogeneity Corrections for Megavoltage Photon Beams. AAPM Report No. 65, 1-142. Tsai, T.S. and Kanno, I. (2016) A Simulation Study on the Influence of Scattered X-Rays in Energy-Resolved Computed Tomography. Journal of Nuclear Science and Technology, 54, 205-212. https://doi.org/10.1080/00223131.2016.1236709 Maruyama, Y., Hamaguchi, T., Tsai, T.S. and Kanno, I. (2017) Response Function Estimation with Fine Energy Bins for the Energy-Resolved Computed Tomography Using a TransXend Detector. Journal of Nuclear Science and Technology, 55, 199-208. https://doi.org/10.1080/00223131.2017.1389311" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88581914,"math_prob":0.86991495,"size":20310,"snap":"2019-43-2019-47","text_gpt3_token_len":5399,"char_repetition_ratio":0.13587117,"word_repetition_ratio":0.037886024,"special_character_ratio":0.2670113,"punctuation_ratio":0.17836812,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95899665,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-19T17:07:28Z\",\"WARC-Record-ID\":\"<urn:uuid:4da99558-320f-4d66-b057-d2f63794298e>\",\"Content-Length\":\"37941\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:75ff5c48-06f5-4ef6-b6dd-9d2a3ba0a3e6>\",\"WARC-Concurrent-To\":\"<urn:uuid:b6aa9e37-915b-47de-b775-fdb584889644>\",\"WARC-IP-Address\":\"173.208.146.83\",\"WARC-Target-URI\":\"https://file.scirp.org/xml/82453.xml\",\"WARC-Payload-Digest\":\"sha1:YBFFPBR5572QXHOLCVOOMIE4UE76IBGN\",\"WARC-Block-Digest\":\"sha1:EVXVUITJND7I7ZXZ7ROEACWE7RNLOSFI\",\"WARC-Identified-Payload-Type\":\"application/xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986697439.41_warc_CC-MAIN-20191019164943-20191019192443-00117.warc.gz\"}"}
https://www.grc.nasa.gov/WWW/K-12/rocket/newton.html	[ "", null, "+ Text Only Site\n+ Non-Flash Version\n+ Contact Glenn", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "The motion of a rocket from the surface of the Earth to a landing on the Moon can be explained and described by physical principals discovered over 300 years ago by Sir Isaac Newton. Newton worked in many areas of mathematics and physics. He developed the theories of gravitation in 1666, when he was only 23 years old. Some twenty years later, in 1686, he presented his three laws of motion in the \"Principia Mathematica Philosophiae Naturalis.\" The laws are shown above, and the application of these laws to rockets is given on separate slides. Newton's first law states that every object remains at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of inertia. The key point here is that if there is no net force acting on an object (if all the external forces cancel each other out) then the object maintains a constant velocity. If that velocity is zero, then the object remains at rest. If the velocity is not zero, then the object maintains that velocity and travels in a straight line. If a net external force is applied, the velocity changes because of the force. Velocity is a vector quantity, having both a magnitude and a direction. The change in velocity caused by a force may involve the magnitude, the direction, or both, depending on the magnitude and direction of the force, which is also a vector quantity. The second law explains how the velocity of an object changes when it is subjected to an external force. The law defines a force to be equal to change in momentum (mass times velocity) per change in time. Newton also developed the calculus of mathematics, and the \"changes\" expressed in the second law are most accurately defined in differential forms. Calculus can also be used to determine the velocity and location variations experienced by an object subjected to an external force. For an object with a constant mass m, the second law states that the force F is the product of an object's mass and its acceleration a: F = m * a For an object like a rocket, with a large change in mass during the flight, we must use the more accurate definition of the second law associated with the change in momentum. For an external applied force, the change in velocity depends on the mass of the object. A force causes a change in velocity; and likewise, a change in velocity generates a force. The equation works both ways. The third law states that for every action (force) in nature there is an equal and opposite re-action. In other words, if object A exerts a force on object B, then object B also exerts an equal force on object A. Notice that the forces are exerted on different objects. The third law can be used to explain the generation of thrust by a rocket engine. Guided Tours", null, "Newton's Laws of Motion:", null, "Activities: Fundamental Terminology: Grade 10-12 Hero Engine: Grade 6-10 Related Sites: Rocket Index Rocket Home Beginner's Guide Home", null, "+ Inspector General Hotline + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act + Budgets, Strategic Plans and Accountability Reports + Freedom of Information Act + The President's Management Agenda + NASA Privacy Statement, Disclaimer, and Accessibility Certification", null, "Editor: Tom Benson NASA Official: Tom Benson Last Updated: Jun 12 2014 + Contact Glenn" ]	[ null, "https://www.grc.nasa.gov/WWW/K-12/rocket/Images/logo_nasa.gif", null, "https://www.grc.nasa.gov/WWW/K-12/rocket/Images/title_find_it_sm.gif", null, "https://www.grc.nasa.gov/WWW/K-12/rocket/Images/spacer.gif", null, "https://www.grc.nasa.gov/WWW/K-12/rocket/Images/button_go.gif", null, "https://www.grc.nasa.gov/WWW/K-12/rocket/Images/nav_top_0_0.gif", null, "https://www.grc.nasa.gov/WWW/K-12/rocket/Images/nav_top_1_0.gif", null, "https://www.grc.nasa.gov/WWW/K-12/rocket/Images/nav_top_2_0.gif", null, "https://www.grc.nasa.gov/WWW/K-12/rocket/Images/nav_top_3_0.gif", null, "https://www.grc.nasa.gov/WWW/K-12/rocket/Images/nav_top_4_0.gif", null, "https://www.grc.nasa.gov/WWW/K-12/rocket/Images/nav_top_5_0.gif", null, "https://www.grc.nasa.gov/WWW/K-12/rocket/Images/newtonr.gif", null, "https://www.grc.nasa.gov/WWW/K-12/rocket/Images/butpre.gif", null, "https://www.grc.nasa.gov/WWW/K-12/rocket/Images/butnex.gif", null, "https://www.grc.nasa.gov/WWW/K-12/rocket/Images/logo_first_gov.gif", null, "https://www.grc.nasa.gov/WWW/K-12/rocket/Images/logo_nasa_self.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90905565,"math_prob":0.89175475,"size":2979,"snap":"2019-51-2020-05","text_gpt3_token_len":663,"char_repetition_ratio":0.14386554,"word_repetition_ratio":0.0038240917,"special_character_ratio":0.20778784,"punctuation_ratio":0.089965396,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97558326,"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],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,1,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-08T03:24:27Z\",\"WARC-Record-ID\":\"<urn:uuid:37c5df2d-fbe6-4179-ba87-2d3aff95036b>\",\"Content-Length\":\"11547\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7e8e78b2-c6c5-49cb-842f-67c1694afc11>\",\"WARC-Concurrent-To\":\"<urn:uuid:9bee57a3-e4ff-4431-b9a7-9239276c3944>\",\"WARC-IP-Address\":\"128.156.253.25\",\"WARC-Target-URI\":\"https://www.grc.nasa.gov/WWW/K-12/rocket/newton.html\",\"WARC-Payload-Digest\":\"sha1:P4HD6I2TC5OJLXMDAYNDWFIVXKH34R2E\",\"WARC-Block-Digest\":\"sha1:PZ6ISO5QZBWQPGF7LCAH5QCC4KMYK636\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540504338.31_warc_CC-MAIN-20191208021121-20191208045121-00030.warc.gz\"}"}
https://www.univerkov.com/the-perimeter-of-the-rectangle-is-36-cm-the-sum-of-the-lengths-of-its-three-sides-is-25-cm-what-is-the-area-of-this-rectangle-3/	[ "# The perimeter of the rectangle is 36 cm. The sum of the lengths of its three sides is 25 cm. What is the area of this rectangle?\n\nLet us denote the lengths of the sides of this rectangle through x and y.\n\nAccording to the condition of the problem, the perimeter of this rectangle is 36 cm, therefore, we can write the following relationship:\n\n2 * (x + y) = 36.\n\nIt is also known that the sum of the lengths of the three sides of this rectangle is 25 cm, therefore, we can write the following ratio:\n\nx + y + x = 25.\n\nWe solve the resulting system of equations.\n\nSubtracting the second equation from the first, we get:\n\n2 * (x + y) – 2x – y = 36 – 25;\n\n2x + 2y – 2x – y = 11;\n\ny = 11 cm.\n\nSubstituting the found value y = 11 into the equation 2x + y = 25, we get:\n\n2x + 11 = 25;\n\n2x = 25 – 11;\n\n2x = 14;\n\nx = 14/2;\n\nx = 7 cm.\n\nFind the area of the rectangle:\n\n11 * 7 = 77 sq. cm.\n\nAnswer: the area of the rectangle is 77 square meters. cm.", null, "One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities." ]	[ null, "https://www.univerkov.com/01.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8846895,"math_prob":0.99908423,"size":919,"snap":"2022-40-2023-06","text_gpt3_token_len":284,"char_repetition_ratio":0.16393442,"word_repetition_ratio":0.05882353,"special_character_ratio":0.3569097,"punctuation_ratio":0.1509434,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99814713,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-28T03:39:17Z\",\"WARC-Record-ID\":\"<urn:uuid:e689bb57-e1e1-461d-9300-b1704503ba8b>\",\"Content-Length\":\"23195\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a0fe60c4-0251-44a9-a3f5-7986344e5161>\",\"WARC-Concurrent-To\":\"<urn:uuid:69e956e2-5fcf-4efe-930b-f61c7b637fd8>\",\"WARC-IP-Address\":\"37.143.13.208\",\"WARC-Target-URI\":\"https://www.univerkov.com/the-perimeter-of-the-rectangle-is-36-cm-the-sum-of-the-lengths-of-its-three-sides-is-25-cm-what-is-the-area-of-this-rectangle-3/\",\"WARC-Payload-Digest\":\"sha1:LYQB6MU2SF4IELIFKWJGCYO4TWPWWDB5\",\"WARC-Block-Digest\":\"sha1:OYEEMIEGAMZS43P7IABXJMOI7VRSTAYV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335059.43_warc_CC-MAIN-20220928020513-20220928050513-00164.warc.gz\"}"}
https://developer.apple.com/documentation/swift/binaryinteger/3126998-ismultiple?changes=_3	[ "Instance Method\n\n# isMultiple(of:)\n\nReturns `true` if this value is a multiple of the given value, and `false` otherwise.\n\nRequired. Default implementations provided.\n\n## Parameters\n\n`other`\n\nThe value to test.\n\n## Discussion\n\nFor two integers a and b, a is a multiple of b if there exists a third integer q such that a = qb. For example, 6 is a multiple of 3 because 6 = 23. Zero is a multiple of everything because 0 = 0*x for any integer x.\n\nTwo edge cases are worth particular attention:\n\n• `x.isMultiple(of: 0)` is `true` if `x` is zero and `false` otherwise.\n\n• `T.min.isMultiple(of: -1)` is `true` for signed integer `T`, even though the quotient `T.min / -1` isn’t representable in type `T`.\n\n## Default Implementations\n\n### BinaryInteger Implementations\n\n`func isMultiple(of: Self) -> Bool`\n\nReturns `true` if this value is a multiple of the given value, and `false` otherwise.\n\n### SignedInteger Implementations\n\n`func isMultiple(of: Self) -> Bool`\n\nReturns `true` if this value is a multiple of the given value, and `false` otherwise." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6520486,"math_prob":0.9673394,"size":687,"snap":"2020-10-2020-16","text_gpt3_token_len":189,"char_repetition_ratio":0.13616398,"word_repetition_ratio":0.0,"special_character_ratio":0.2532751,"punctuation_ratio":0.1418919,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96789557,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-08T07:03:48Z\",\"WARC-Record-ID\":\"<urn:uuid:a820e1f0-858d-4266-977d-14fc7b6967a4>\",\"Content-Length\":\"32370\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:daf241fa-b6fa-4991-903c-5ad85502ecbf>\",\"WARC-Concurrent-To\":\"<urn:uuid:14a5b2cd-40c2-4659-b998-5059dbe8c4bc>\",\"WARC-IP-Address\":\"17.253.21.205\",\"WARC-Target-URI\":\"https://developer.apple.com/documentation/swift/binaryinteger/3126998-ismultiple?changes=_3\",\"WARC-Payload-Digest\":\"sha1:UXVDXPB63AIT5BVY3ODAFV2U5U5HILRQ\",\"WARC-Block-Digest\":\"sha1:73VRAXN7GIBC7Z7T2WCOYHTSGIIFTBEY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371810617.95_warc_CC-MAIN-20200408041431-20200408071931-00211.warc.gz\"}"}
http://ydec.qudupydih.ru/2019/12/13_med-math-online-calculator.aspx	[ "# Med math online calculator\n\n## Medical Dosage Calculations For Dummies Cheat Sheet\n\nFree online easy calculator for school and business. Calculator for taxes, financial lease calculator, basic car loan calculator and much just enter quantity, current price and tax percentage in our t.Other calculators. Final grade calculator. Pi scientific calculator. skills tutor – online tutor hire for the summer break. Med school gpa. 10 reasons why math is important in life guide + examples.Free math lessons and math homework help from basic math to algebra, geometry and beyond. if your teacher doesn't allow calculators for homework, that means online ones, too.A basic math calculator with additional supporting online math calculators if you need them. in order to access the online calculators you will need a device capable of accessing the internet and runn.Math tools and calculators. Here are some really useful applications to speed your way to that allimportant solution. converts decimals to fractional form, showing steps. Combinations and permutations.Free online dosing calculator allows prescribers to calculate dosing per patient weight. Mdtoolbox provides this free calculator just as a math tool.Algebra calculator shows you the stepbystep solutions! solves algebra problems and walks you through them. algebra calculator. What do you want to calculate?.Maths free online calculators get free algebra calculator, multiplication calculator, addition calculator, conversion calculators, trigonometry calculator, engineering math this is a math solver tool.\n\nlosec mups 40 mg\n\n## MDMath \| Canadian Society of Echocardiography\n\nexpress post uk to usa\n\n## Web Calculator Designer - Calculoid.com\n\nDosage calculator. Calculation of the administered activity in mbq and mci and the 2010 north america consensus guideline, eur j nucl med mol imaging.Math calculator from mathway will evaluate various math problems from basic arithmetic to advanced trigonometric expressions. you're welcome! let me take a look you'll be able to enter math.Medicalcalculators provides online calculators for medical and scientific use. Designed drug dosage calculator enter amount of drug you want to give.Practice questions practice your math skills with each of these questions. An explanation will be provided for each answer that is incorrect. If you find these questions useful, click here to sign up f.Medical calculators. Medical science is a branch of science that helps in treating and preventing the diseases that affect human beings. Doctors generally use different tests to diagnose the diseases o.Scientific calculator online and mobile friendly. Creates a series of calculations that can be printed, bookmarked, shared and modified. Keys: pi, e, standard gravity, gas constant, sin, cos, tan, asin.The only fault lies in the name. Perhaps the mathweenienobrainer technique would be more appropriate. At any rate, give dimensional analysis a try. At the end of a 12hour shift, when youre tired, thing.Online clinical calculators. Medcalc: pediatric dosing calculator disclaimer: all calculations must be confirmed before use.Medical calculators. Disclaimer: all calculations must be confirmed before use. The authors make no claims of the accuracy of the information contained herein; and these suggested doses are not a subst.\n\nexpress computers uk\n\n## Online ACFI Calculator \| Question 11: Medication\n\nexpress star newspaper wolverhampton uk\n\n## Algebra Calculator - MathPapa\n\nThis is an online scientific calculator with double digit precision that support both button click and keyboard type. In addition, explore hundreds of other free calculators covering topics such as fin.Online calculator for quick calculations, along with a large collection of calculators on math, finance, fitness, and more, each with related calculator. Net's sole focus is to provide fast, compr.Com blog pharmacology med math for nurses master post (everything about dosage calculations and dimensional analysis. With practice) one of the most stressful parts of nursing school is nurs.Prescription. This includes the name of the medication, the dosage and form of the medication, the amount to be taken, the method of administration, and the frequency and duration the medication will b.\n\nmed familjen i london\n\n## Online Math Calculators\n\nThe weight based dose calculator is used for weight based dosing. The parameters for the calculator include dosage, weight, med amount, per volume.+ with help of this calculator you can: find the matrix determinant, the rank, raise the matrix to a power, find the sum and the multiplication of matrices, calculate the inverse matrix. Just type matr.Online math calculator is used to perform a variety of calculations. It is comprised of the forward and back controls which allow you to if you want to erase all data of the previous calculations from.Free online math calculators and converters. Mathematics calculation is made easier. this free online math web site will help you learn mathematics in a easier way. Easycalculation will also help you.Makes learning and applying dosage calculations— even reviewing basic math build the skills that assure patient safety and prevent medication errors.22 мая 2018 г. You will find a med calculator for bmi, creatinine clearance, metric to english conversion and more. You will find over 20 financial calculators.Calculator online. Simple and scientific. With calculation memory. for the purpose of checking your work, mathematical calculations entered into the online calculator can be left displayed on the tape.Here you will find many math calculators. They are free and show steps. Choose subject and use search to find the required solver. all math calculators. Didn't find the calculator you need? reque.This online maths calculator is very similar to the main calculator on the site except for the cool history feature. It lets you see online maths calculator maths calculator math calculator.Math calculator. Thousands maths online. i’m a can’t do math gay and who let me work with spreadsheets and numbers i’m good at my job but i still need the calculator app to do 2+2 to calculate for tip.Medical calculators, criteria sets and decision trees. In this calculator, mcgm is the abbreviation for micrograms. To make simple unit conversions, select the starting units with the pulldown s.General math solvers below, there is a collection of general math solvers and calculators covering issues like the fractions, percentages, factorization, prime numbers, divisibility and square roots.\n\nholiday inn express burton uk\n\n## MDCalc\n\nThis online scientific calculator has a full list of functions that can help you find the solution for any basic or complex math and scientific what are the functions this online scientific calculator.So the volume to give = 27 ml. It doesnt matter what type of units are used in this calculator. You can figure out how many grams to give or how many milligrams to give, microg.Online math calculator find answers for all your math problems in just a click. online math calculator : we people know about classic calculator in which we can use the mathematical operations like ad.This is a free online math calculator together with a variety of other free math calculators that compute standard deviation, percentage, fractions, and time, along with hundreds of other calculators a.Start studying paramedic med math (practice). Learn vocabulary, terms, and more with flashcards, games, and other study tools.Free online scientific notation calculator. Solve advanced problems in physics, mathematics and engineering. Math expression renderer, plots, unit converter, equation solver, complex numbers, calculati.Calculating medication dosage by weight. Medical dosage calculations that consider a patients weight are very common in obese patients and on pediatric wards. Its easy to imagine that the weight of the.Online exclusive. Author information nurses are often intimidated by the math that occurs in everyday practice. Patient safety depends on the follow these four steps to easily calculate your patients a.The dose driven ivdrip rate calculator is based on equation parameters that include desired dose, weight, amount of drug in iv bag, and iv bag volume.The following calculator will find mean, mode, median, lower and upper quartile, interquartile range of the given data set. The calculator will generate a step by step explanation on how to find these." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85822964,"math_prob":0.98064184,"size":19725,"snap":"2020-34-2020-40","text_gpt3_token_len":3818,"char_repetition_ratio":0.21616551,"word_repetition_ratio":0.2782398,"special_character_ratio":0.17799747,"punctuation_ratio":0.14948453,"nsfw_num_words":3,"has_unicode_error":false,"math_prob_llama3":0.99229234,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-08T08:24:24Z\",\"WARC-Record-ID\":\"<urn:uuid:21538750-f549-47a4-9aaa-e6314adba29b>\",\"Content-Length\":\"23940\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fcb7fcd1-e3c5-423d-a2c9-e5e1cee253e8>\",\"WARC-Concurrent-To\":\"<urn:uuid:f4cd382d-3337-4049-9964-405f1258c1f9>\",\"WARC-IP-Address\":\"37.46.132.131\",\"WARC-Target-URI\":\"http://ydec.qudupydih.ru/2019/12/13_med-math-online-calculator.aspx\",\"WARC-Payload-Digest\":\"sha1:KQCEOFV5W6KXFD3LLCDHB6HIU77U5DK5\",\"WARC-Block-Digest\":\"sha1:N2KFAMBBDCDVVJXF5KRAWAC364KDIEIP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439737319.74_warc_CC-MAIN-20200808080642-20200808110642-00007.warc.gz\"}"}