{"group":{"id":1,"name":"Community","lockable":false,"created_at":"2012-01-18T18:02:15.000Z","updated_at":"2026-04-06T14:01:22.000Z","description":"Problems submitted by members of the MATLAB Central community.","is_default":true,"created_by":161519,"badge_id":null,"featured":false,"trending":false,"solution_count_in_trending_period":0,"trending_last_calculated":"2026-04-06T00:00:00.000Z","image_id":null,"published":true,"community_created":false,"status_id":2,"is_default_group_for_player":false,"deleted_by":null,"deleted_at":null,"restored_by":null,"restored_at":null,"description_opc":null,"description_html":null,"published_at":null},"problems":[{"id":179,"title":"Monte-Carlo integration","description":"Write a function that estimates a d-dimensional integral to at least 1% relative precision.\r\n\r\nInputs:\r\n\r\n* d: positive integer. The dimension of the integral.\r\n* fun: function handle. The function accepts a row-vector of length d as an argument and returns a real scalar as a result.\r\n\r\nOutput:\r\n\r\n* I: is the integral over fun from 0 to 1 in each direction.\r\n\r\n     1     1        1            \r\n     /     /        /           \r\n I = |dx_1 |dx_2 ...| dx_d  fun([x_1,x_2,...,x_d])\r\n     /     /        /            \r\n     0     0        0            \r\n\r\nExample:\r\n\r\n  fun = @(x) x(1)*x(2)\r\n  d = 2\r\n\r\nThe result should be 0.25. An output I=0.2501 would be acceptable, because the relative deviation would be abs(0.25-0.2501)/0.25 which is smaller than 1%.\r\n\r\nThe functions in the test-suite are all positive and generally 'well behaved', i.e. not fluctuating too much. Some of the tests hav a relatively large d.","description_html":"\u003cp\u003eWrite a function that estimates a d-dimensional integral to at least 1% relative precision.\u003c/p\u003e\u003cp\u003eInputs:\u003c/p\u003e\u003cul\u003e\u003cli\u003ed: positive integer. The dimension of the integral.\u003c/li\u003e\u003cli\u003efun: function handle. The function accepts a row-vector of length d as an argument and returns a real scalar as a result.\u003c/li\u003e\u003c/ul\u003e\u003cp\u003eOutput:\u003c/p\u003e\u003cul\u003e\u003cli\u003eI: is the integral over fun from 0 to 1 in each direction.\u003c/li\u003e\u003c/ul\u003e\u003cpre\u003e     1     1        1            \r\n     /     /        /           \r\n I = |dx_1 |dx_2 ...| dx_d  fun([x_1,x_2,...,x_d])\r\n     /     /        /            \r\n     0     0        0            \u003c/pre\u003e\u003cp\u003eExample:\u003c/p\u003e\u003cpre class=\"language-matlab\"\u003efun = @(x) x(1)*x(2)\r\nd = 2\r\n\u003c/pre\u003e\u003cp\u003eThe result should be 0.25. An output I=0.2501 would be acceptable, because the relative deviation would be abs(0.25-0.2501)/0.25 which is smaller than 1%.\u003c/p\u003e\u003cp\u003eThe functions in the test-suite are all positive and generally 'well behaved', i.e. not fluctuating too much. Some of the tests hav a relatively large d.\u003c/p\u003e","function_template":"function I = dquad(d,fun)\r\n  I=fun(0.5*ones(1,d));\r\nend","test_suite":"%% 1d integral: integrate x^2 from 0 to 1\r\nfun = @(x) x^2;\r\nassert(abs((dquad(1,fun) - 1/3)*3)\u003c0.01)\r\n\r\n%% 2d integral from the example\r\nfun = @(x) x(1)*x(2);\r\nassert(abs((dquad(2,fun) - 0.25)*4)\u003c0.01)\r\n\r\n%% constant in most dimensions\r\nfun = @(x) 1+sin(x(1));\r\nassert(abs((dquad(50,fun) -  1.45969769)/1.45969769)\u003c0.01)\r\n\r\n%% volume of d-dimensional 2^d box with a spherical hole, d between 5 and 10\r\nd = randi([5 10],1)\r\nr = rand*0.8\r\nfun = @(x) 2^d*(norm(x)\u003er);\r\ndball = exp(d/2*log(pi)+d*log(r)-gammaln(d/2+1));\r\nassert(abs((dquad(d,fun) - 2^d+dball)/(2^d-dball))\u003c0.01)\r\n","published":true,"deleted":false,"likes_count":7,"comments_count":8,"created_by":203,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":153,"test_suite_updated_at":"2012-01-31T21:50:54.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2012-01-30T15:04:16.000Z","updated_at":"2026-04-03T20:58:12.000Z","published_at":"2012-02-01T19:02:46.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"targetMode\":\"\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"targetMode\":\"\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\\n\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eWrite a function that estimates a d-dimensional integral to at least 1% relative precision.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eInputs:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003ed: positive integer. The dimension of the integral.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003efun: function handle. The function accepts a row-vector of length d as an argument and returns a real scalar as a result.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eOutput:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eI: is the integral over fun from 0 to 1 in each direction.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"code\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c![CDATA[     1     1        1            \\n     /     /        /           \\n I = |dx_1 |dx_2 ...| dx_d  fun([x_1,x_2,...,x_d])\\n     /     /        /            \\n     0     0        0]]\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eExample:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"code\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c![CDATA[fun = @(x) x(1)*x(2)\\nd = 2]]\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThe result should be 0.25. An output I=0.2501 would be acceptable, because the relative deviation would be abs(0.25-0.2501)/0.25 which is smaller than 1%.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThe functions in the test-suite are all positive and generally 'well behaved', i.e. not fluctuating too much. Some of the tests hav a relatively large d.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"}],"problem_search":{"errors":[],"problems":[{"id":179,"title":"Monte-Carlo integration","description":"Write a function that estimates a d-dimensional integral to at least 1% relative precision.\r\n\r\nInputs:\r\n\r\n* d: positive integer. The dimension of the integral.\r\n* fun: function handle. The function accepts a row-vector of length d as an argument and returns a real scalar as a result.\r\n\r\nOutput:\r\n\r\n* I: is the integral over fun from 0 to 1 in each direction.\r\n\r\n     1     1        1            \r\n     /     /        /           \r\n I = |dx_1 |dx_2 ...| dx_d  fun([x_1,x_2,...,x_d])\r\n     /     /        /            \r\n     0     0        0            \r\n\r\nExample:\r\n\r\n  fun = @(x) x(1)*x(2)\r\n  d = 2\r\n\r\nThe result should be 0.25. An output I=0.2501 would be acceptable, because the relative deviation would be abs(0.25-0.2501)/0.25 which is smaller than 1%.\r\n\r\nThe functions in the test-suite are all positive and generally 'well behaved', i.e. not fluctuating too much. Some of the tests hav a relatively large d.","description_html":"\u003cp\u003eWrite a function that estimates a d-dimensional integral to at least 1% relative precision.\u003c/p\u003e\u003cp\u003eInputs:\u003c/p\u003e\u003cul\u003e\u003cli\u003ed: positive integer. The dimension of the integral.\u003c/li\u003e\u003cli\u003efun: function handle. The function accepts a row-vector of length d as an argument and returns a real scalar as a result.\u003c/li\u003e\u003c/ul\u003e\u003cp\u003eOutput:\u003c/p\u003e\u003cul\u003e\u003cli\u003eI: is the integral over fun from 0 to 1 in each direction.\u003c/li\u003e\u003c/ul\u003e\u003cpre\u003e     1     1        1            \r\n     /     /        /           \r\n I = |dx_1 |dx_2 ...| dx_d  fun([x_1,x_2,...,x_d])\r\n     /     /        /            \r\n     0     0        0            \u003c/pre\u003e\u003cp\u003eExample:\u003c/p\u003e\u003cpre class=\"language-matlab\"\u003efun = @(x) x(1)*x(2)\r\nd = 2\r\n\u003c/pre\u003e\u003cp\u003eThe result should be 0.25. An output I=0.2501 would be acceptable, because the relative deviation would be abs(0.25-0.2501)/0.25 which is smaller than 1%.\u003c/p\u003e\u003cp\u003eThe functions in the test-suite are all positive and generally 'well behaved', i.e. not fluctuating too much. Some of the tests hav a relatively large d.\u003c/p\u003e","function_template":"function I = dquad(d,fun)\r\n  I=fun(0.5*ones(1,d));\r\nend","test_suite":"%% 1d integral: integrate x^2 from 0 to 1\r\nfun = @(x) x^2;\r\nassert(abs((dquad(1,fun) - 1/3)*3)\u003c0.01)\r\n\r\n%% 2d integral from the example\r\nfun = @(x) x(1)*x(2);\r\nassert(abs((dquad(2,fun) - 0.25)*4)\u003c0.01)\r\n\r\n%% constant in most dimensions\r\nfun = @(x) 1+sin(x(1));\r\nassert(abs((dquad(50,fun) -  1.45969769)/1.45969769)\u003c0.01)\r\n\r\n%% volume of d-dimensional 2^d box with a spherical hole, d between 5 and 10\r\nd = randi([5 10],1)\r\nr = rand*0.8\r\nfun = @(x) 2^d*(norm(x)\u003er);\r\ndball = exp(d/2*log(pi)+d*log(r)-gammaln(d/2+1));\r\nassert(abs((dquad(d,fun) - 2^d+dball)/(2^d-dball))\u003c0.01)\r\n","published":true,"deleted":false,"likes_count":7,"comments_count":8,"created_by":203,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":153,"test_suite_updated_at":"2012-01-31T21:50:54.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2012-01-30T15:04:16.000Z","updated_at":"2026-04-03T20:58:12.000Z","published_at":"2012-02-01T19:02:46.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"targetMode\":\"\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"targetMode\":\"\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\\n\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eWrite a function that estimates a d-dimensional integral to at least 1% relative precision.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eInputs:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003ed: positive integer. The dimension of the integral.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003efun: function handle. The function accepts a row-vector of length d as an argument and returns a real scalar as a result.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eOutput:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"ListParagraph\\\"/\u003e\u003cw:numPr\u003e\u003cw:numId w:val=\\\"1\\\"/\u003e\u003c/w:numPr\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eI: is the integral over fun from 0 to 1 in each direction.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"code\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c![CDATA[     1     1        1            \\n     /     /        /           \\n I = |dx_1 |dx_2 ...| dx_d  fun([x_1,x_2,...,x_d])\\n     /     /        /            \\n     0     0        0]]\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eExample:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"code\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c![CDATA[fun = @(x) x(1)*x(2)\\nd = 2]]\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThe result should be 0.25. 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