{"group":{"id":1,"name":"Community","lockable":false,"created_at":"2012-01-18T18:02:15.000Z","updated_at":"2026-05-26T00:16:20.000Z","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":"2026-05-26T00:00:00.000Z","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":2409,"title":"Find the Connecting Path","description":"Here is a matrix x\r\n\r\n x = [7 6 8 5 7 2 4 5 1 3 0 0 0\r\n      7 7 7 7 7 0 0 0 0 0 0 0 0\r\n      0 0 0 5 5 5 5 5 0 0 0 0 0\r\n      0 0 0 1 1 1 1 1 1 0 0 0 0\r\n      0 0 0 0 0 2 2 2 2 2 2 2 0\r\n      0 6 6 6 6 6 0 0 0 0 0 0 0\r\n      0 3 3 3 3 3 3 3 3 3 0 0 0\r\n      0 0 0 0 0 0 0 0 0 2 2 2 0\r\n      0 0 0 0 0 0 4 4 4 4 4 4 4\r\n      0 0 8 8 8 8 8 0 0 0 0 0 0\r\n      0 0 4 4 4 4 4 4 4 4 4 4 4\r\n      0 3 4 1 7 6 8 0 0 2 9 0 0]\r\n\r\nIf n = 7 then the output matrix y should be\r\n\r\n y = [1 0 0 0 1 0 0 0 0 0 0 0 0\r\n      1 1 1 1 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0] \r\n\r\nAll the places of n = 7 will become 1 and horizontally they are connected by a path given by 1's which is represented in the output y.\r\n\r\nIf n = 9 then y(12,11) = 1 and others are 0's.\r\n\r\nOutput should display path as per given n.","description_html":"\u003cp\u003eHere is a matrix x\u003c/p\u003e\u003cpre\u003e x = [7 6 8 5 7 2 4 5 1 3 0 0 0\r\n      7 7 7 7 7 0 0 0 0 0 0 0 0\r\n      0 0 0 5 5 5 5 5 0 0 0 0 0\r\n      0 0 0 1 1 1 1 1 1 0 0 0 0\r\n      0 0 0 0 0 2 2 2 2 2 2 2 0\r\n      0 6 6 6 6 6 0 0 0 0 0 0 0\r\n      0 3 3 3 3 3 3 3 3 3 0 0 0\r\n      0 0 0 0 0 0 0 0 0 2 2 2 0\r\n      0 0 0 0 0 0 4 4 4 4 4 4 4\r\n      0 0 8 8 8 8 8 0 0 0 0 0 0\r\n      0 0 4 4 4 4 4 4 4 4 4 4 4\r\n      0 3 4 1 7 6 8 0 0 2 9 0 0]\u003c/pre\u003e\u003cp\u003eIf n = 7 then the output matrix y should be\u003c/p\u003e\u003cpre\u003e y = [1 0 0 0 1 0 0 0 0 0 0 0 0\r\n      1 1 1 1 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0] \u003c/pre\u003e\u003cp\u003eAll the places of n = 7 will become 1 and horizontally they are connected by a path given by 1's which is represented in the output y.\u003c/p\u003e\u003cp\u003eIf n = 9 then y(12,11) = 1 and others are 0's.\u003c/p\u003e\u003cp\u003eOutput should display path as per given n.\u003c/p\u003e","function_template":"function y = your_fcn_name(x,n)\r\n  y = x;\r\nend","test_suite":"%%\r\nx = [7 6 8 5 7 2 4 5 1 3 0 0 0;7 7 7 7 7 0 0 0 0 0 0 0 0;0 0 0 5 5 5 5 5 0 0 0 0 0;0 0 0 1 1 1 1 1 1 0 0 0 0;0 0 0 0 0 2 2 2 2 2 2 2 0;0 6 6 6 6 6 0 0 0 0 0 0 0;0 3 3 3 3 3 3 3 3 3 0 0 0;0 0 0 0 0 0 0 0 0 2 2 2 0;0 0 0 0 0 0 4 4 4 4 4 4 4;0 0 8 8 8 8 8 0 0 0 0 0 0;0 0 4 4 4 4 4 4 4 4 4 4 4;0 3 4 1 7 6 8 0 0 2 9 0 0];\r\nn = 7;\r\ny_correct =[1 0 0 0 1 0 0 0 0 0 0 0 0;1 1 1 1 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0];\r\nassert(isequal(your_fcn_name(x,n),y_correct))\r\n\r\n%%\r\nx = [7 6 8 5 7 2 4 5 1 3 0 0 0;7 7 7 7 7 0 0 0 0 0 0 0 0;0 0 0 5 5 5 5 5 0 0 0 0 0;0 0 0 1 1 1 1 1 1 0 0 0 0;0 0 0 0 0 2 2 2 2 2 2 2 0;0 6 6 6 6 6 0 0 0 0 0 0 0;0 3 3 3 3 3 3 3 3 3 0 0 0;0 0 0 0 0 0 0 0 0 2 2 2 0;0 0 0 0 0 0 4 4 4 4 4 4 4;0 0 8 8 8 8 8 0 0 0 0 0 0;0 0 4 4 4 4 4 4 4 4 4 4 4;0 3 4 1 7 6 8 0 0 2 9 0 0];\r\nn = 4;\r\ny_correct =[0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 1 1 1 1 1 1 1;0 0 0 0 0 0 0 0 0 0 0 0 1;0 0 1 1 1 1 1 1 1 1 1 1 1;0 0 1 0 0 0 0 0 0 0 0 0 0]\r\nassert(isequal(your_fcn_name(x,n),y_correct))","published":true,"deleted":false,"likes_count":0,"comments_count":6,"created_by":20855,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":17,"test_suite_updated_at":"2014-07-28T04:59:47.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2014-07-09T12:12:25.000Z","updated_at":"2026-05-28T02:54:46.000Z","published_at":"2014-07-09T12:12:35.000Z","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\u003eHere is a matrix x\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[ x = [7 6 8 5 7 2 4 5 1 3 0 0 0\\n      7 7 7 7 7 0 0 0 0 0 0 0 0\\n      0 0 0 5 5 5 5 5 0 0 0 0 0\\n      0 0 0 1 1 1 1 1 1 0 0 0 0\\n      0 0 0 0 0 2 2 2 2 2 2 2 0\\n      0 6 6 6 6 6 0 0 0 0 0 0 0\\n      0 3 3 3 3 3 3 3 3 3 0 0 0\\n      0 0 0 0 0 0 0 0 0 2 2 2 0\\n      0 0 0 0 0 0 4 4 4 4 4 4 4\\n      0 0 8 8 8 8 8 0 0 0 0 0 0\\n      0 0 4 4 4 4 4 4 4 4 4 4 4\\n      0 3 4 1 7 6 8 0 0 2 9 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\u003eIf n = 7 then the output matrix y should be\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[ y = [1 0 0 0 1 0 0 0 0 0 0 0 0\\n      1 1 1 1 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 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\u003eAll the places of n = 7 will become 1 and horizontally they are connected by a path given by 1's which is represented in the output y.\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\u003eIf n = 9 then y(12,11) = 1 and others are 0's.\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 should display path as per given n.\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\"}]}"},{"id":2565,"title":"Determine the Result of a Move in Reversi","description":"Note: This is closely related to \u003chttp://www.mathworks.com/matlabcentral/cody/problems/2538-find-the-next-legal-move-in-reversi Problem 2538\u003e, Find the Next Legal Move in Reversi.\r\n\r\n\u003chttp://en.wikipedia.org/wiki/Reversi Reversi\u003e, also known as Othello, is a game in which reversible white/black chips are placed on a grid. The goal is to have the most pieces once the board is full (i.e. there are no more legal moves). A move, to be legal, must flip at least one opponent's chip by flanking it. \"Flanking\" occurs by surrounding a line of opposing chips with two of your own. It can occur on straight lines and diagonals.\r\n\r\nWe represent a small 4x4 board this way, with 0 for empty squares, 1 for black chips, and 2 for white chips, \r\n\r\n [ 0 0 0 0\r\n   0 1 2 0\r\n   0 2 1 0 \r\n   0 0 0 0 ]\r\n\r\nNow assume black now moves where the asterisk (*) is placed. This is column-wise matrix index position 14.\r\n\r\n [ 0 0 0 0\r\n   0 1 2 *\r\n   0 2 1 0 \r\n   0 0 0 0 ]\r\n\r\nYour Cody challenge is to determine the state of the board once the appropriate chips have been flipped over. In this case the correct answer would be\r\n\r\n [ 0 0 0 0\r\n   0 1 1 1\r\n   0 2 1 0 \r\n   0 0 0 0 ]\r\n\r\nExample\r\n\r\nFor the inputs boardIn, side, and move\r\n\r\n boardIn = ...\r\n   [ 0 0 0 0 0\r\n     0 1 1 2 2 \r\n     0 2 1 1 0\r\n     0 0 2 1 0 \r\n     0 0 0 0 0 ]\r\n side = 1 (black)\r\n move = 9 (matrix index of black's next move)\r\n\r\nThe correct answer is\r\n\r\n boardOut = ...\r\n   [ 0 0 0 0 0\r\n     0 1 1 2 2 \r\n     0 1 1 1 0\r\n     0 1 1 1 0 \r\n     0 0 0 0 0 ]\r\n\r\n\r\n","description_html":"\u003cp\u003eNote: This is closely related to \u003ca href = \"http://www.mathworks.com/matlabcentral/cody/problems/2538-find-the-next-legal-move-in-reversi\"\u003eProblem 2538\u003c/a\u003e, Find the Next Legal Move in Reversi.\u003c/p\u003e\u003cp\u003e\u003ca href = \"http://en.wikipedia.org/wiki/Reversi\"\u003eReversi\u003c/a\u003e, also known as Othello, is a game in which reversible white/black chips are placed on a grid. The goal is to have the most pieces once the board is full (i.e. there are no more legal moves). A move, to be legal, must flip at least one opponent's chip by flanking it. \"Flanking\" occurs by surrounding a line of opposing chips with two of your own. It can occur on straight lines and diagonals.\u003c/p\u003e\u003cp\u003eWe represent a small 4x4 board this way, with 0 for empty squares, 1 for black chips, and 2 for white chips,\u003c/p\u003e\u003cpre\u003e [ 0 0 0 0\r\n   0 1 2 0\r\n   0 2 1 0 \r\n   0 0 0 0 ]\u003c/pre\u003e\u003cp\u003eNow assume black now moves where the asterisk (*) is placed. This is column-wise matrix index position 14.\u003c/p\u003e\u003cpre\u003e [ 0 0 0 0\r\n   0 1 2 *\r\n   0 2 1 0 \r\n   0 0 0 0 ]\u003c/pre\u003e\u003cp\u003eYour Cody challenge is to determine the state of the board once the appropriate chips have been flipped over. In this case the correct answer would be\u003c/p\u003e\u003cpre\u003e [ 0 0 0 0\r\n   0 1 1 1\r\n   0 2 1 0 \r\n   0 0 0 0 ]\u003c/pre\u003e\u003cp\u003eExample\u003c/p\u003e\u003cp\u003eFor the inputs boardIn, side, and move\u003c/p\u003e\u003cpre\u003e boardIn = ...\r\n   [ 0 0 0 0 0\r\n     0 1 1 2 2 \r\n     0 2 1 1 0\r\n     0 0 2 1 0 \r\n     0 0 0 0 0 ]\r\n side = 1 (black)\r\n move = 9 (matrix index of black's next move)\u003c/pre\u003e\u003cp\u003eThe correct answer is\u003c/p\u003e\u003cpre\u003e boardOut = ...\r\n   [ 0 0 0 0 0\r\n     0 1 1 2 2 \r\n     0 1 1 1 0\r\n     0 1 1 1 0 \r\n     0 0 0 0 0 ]\u003c/pre\u003e","function_template":"function boardOut = boardResult(boardIn,side,move)\r\n  boardOut = boardIn;\r\nend","test_suite":"%%\r\nboardIn = ...\r\n [ 0 0 0 0\r\n   0 1 2 0\r\n   0 2 1 0 \r\n   0 0 0 0 ];\r\nside = 1;\r\nmove = 14;\r\nboardOut = ...\r\n [ 0 0 0 0\r\n   0 1 1 1\r\n   0 2 1 0 \r\n   0 0 0 0 ];\r\nassert(isequal(boardOut,boardResult(boardIn,side,move)))\r\n\r\n%%\r\nboardIn = ...\r\n [ 0 0 0 0\r\n   0 1 1 1\r\n   0 2 1 0 \r\n   0 0 0 0 ];\r\nside = 2;\r\nmove = 5;\r\nboardOut = ...\r\n [ 0 2 0 0\r\n   0 2 1 1\r\n   0 2 1 0 \r\n   0 0 0 0 ];\r\nassert(isequal(boardOut,boardResult(boardIn,side,move)))\r\n\r\n%%\r\nboardIn = ...\r\n [ 0 2 0 0\r\n   0 2 1 1\r\n   0 2 1 0 \r\n   0 0 0 0 ];\r\nside = 1;\r\nmove = 4;\r\nboardOut = ...\r\n [ 0 2 0 0\r\n   0 2 1 1\r\n   0 1 1 0 \r\n   1 0 0 0 ];\r\nassert(isequal(boardOut,boardResult(boardIn,side,move)))\r\n\r\n%%\r\nboardIn = ...\r\n [ 0 0 0 0 0\r\n   0 1 1 2 2 \r\n   0 2 1 1 0\r\n   0 0 2 1 0 \r\n   0 0 0 0 0 ];\r\nside = 1;\r\nmove = 9;\r\nboardOut = ...\r\n [ 0 0 0 0 0\r\n   0 1 1 2 2 \r\n   0 1 1 1 0\r\n   0 1 1 1 0 \r\n   0 0 0 0 0 ]\r\nassert(isequal(boardOut,boardResult(boardIn,side,move)))\r\n\r\n","published":true,"deleted":false,"likes_count":2,"comments_count":0,"created_by":7,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":22,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2014-09-09T16:32:03.000Z","updated_at":"2026-05-25T01:53:55.000Z","published_at":"2014-09-09T16:36:06.000Z","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\u003eNote: This is closely related to\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:hyperlink w:docLocation=\\\"http://www.mathworks.com/matlabcentral/cody/problems/2538-find-the-next-legal-move-in-reversi\\\"\u003e\u003cw:r\u003e\u003cw:t\u003eProblem 2538\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\u003e\u003cw:r\u003e\u003cw:t\u003e, Find the Next Legal Move in Reversi.\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:hyperlink w:docLocation=\\\"http://en.wikipedia.org/wiki/Reversi\\\"\u003e\u003cw:r\u003e\u003cw:t\u003eReversi\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\u003e\u003cw:r\u003e\u003cw:t\u003e, also known as Othello, is a game in which reversible white/black chips are placed on a grid. The goal is to have the most pieces once the board is full (i.e. there are no more legal moves). A move, to be legal, must flip at least one opponent's chip by flanking it. \\\"Flanking\\\" occurs by surrounding a line of opposing chips with two of your own. It can occur on straight lines and diagonals.\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\u003eWe represent a small 4x4 board this way, with 0 for empty squares, 1 for black chips, and 2 for white chips,\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[ [ 0 0 0 0\\n   0 1 2 0\\n   0 2 1 0 \\n   0 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\u003eNow assume black now moves where the asterisk \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e(*\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e) is placed. This is column-wise matrix index position 14.\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[ [ 0 0 0 0\\n   0 1 2 *\\n   0 2 1 0 \\n   0 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\u003eYour Cody challenge is to determine the state of the board once the appropriate chips have been flipped over. In this case the correct answer would be\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[ [ 0 0 0 0\\n   0 1 1 1\\n   0 2 1 0 \\n   0 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=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eFor the inputs boardIn, side, and move\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[ boardIn = ...\\n   [ 0 0 0 0 0\\n     0 1 1 2 2 \\n     0 2 1 1 0\\n     0 0 2 1 0 \\n     0 0 0 0 0 ]\\n side = 1 (black)\\n move = 9 (matrix index of black's next move)]]\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 correct answer is\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[ boardOut = ...\\n   [ 0 0 0 0 0\\n     0 1 1 2 2 \\n     0 1 1 1 0\\n     0 1 1 1 0 \\n     0 0 0 0 0 ]]]\u003e\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":{"problems":[{"id":2409,"title":"Find the Connecting Path","description":"Here is a matrix x\r\n\r\n x = [7 6 8 5 7 2 4 5 1 3 0 0 0\r\n      7 7 7 7 7 0 0 0 0 0 0 0 0\r\n      0 0 0 5 5 5 5 5 0 0 0 0 0\r\n      0 0 0 1 1 1 1 1 1 0 0 0 0\r\n      0 0 0 0 0 2 2 2 2 2 2 2 0\r\n      0 6 6 6 6 6 0 0 0 0 0 0 0\r\n      0 3 3 3 3 3 3 3 3 3 0 0 0\r\n      0 0 0 0 0 0 0 0 0 2 2 2 0\r\n      0 0 0 0 0 0 4 4 4 4 4 4 4\r\n      0 0 8 8 8 8 8 0 0 0 0 0 0\r\n      0 0 4 4 4 4 4 4 4 4 4 4 4\r\n      0 3 4 1 7 6 8 0 0 2 9 0 0]\r\n\r\nIf n = 7 then the output matrix y should be\r\n\r\n y = [1 0 0 0 1 0 0 0 0 0 0 0 0\r\n      1 1 1 1 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0] \r\n\r\nAll the places of n = 7 will become 1 and horizontally they are connected by a path given by 1's which is represented in the output y.\r\n\r\nIf n = 9 then y(12,11) = 1 and others are 0's.\r\n\r\nOutput should display path as per given n.","description_html":"\u003cp\u003eHere is a matrix x\u003c/p\u003e\u003cpre\u003e x = [7 6 8 5 7 2 4 5 1 3 0 0 0\r\n      7 7 7 7 7 0 0 0 0 0 0 0 0\r\n      0 0 0 5 5 5 5 5 0 0 0 0 0\r\n      0 0 0 1 1 1 1 1 1 0 0 0 0\r\n      0 0 0 0 0 2 2 2 2 2 2 2 0\r\n      0 6 6 6 6 6 0 0 0 0 0 0 0\r\n      0 3 3 3 3 3 3 3 3 3 0 0 0\r\n      0 0 0 0 0 0 0 0 0 2 2 2 0\r\n      0 0 0 0 0 0 4 4 4 4 4 4 4\r\n      0 0 8 8 8 8 8 0 0 0 0 0 0\r\n      0 0 4 4 4 4 4 4 4 4 4 4 4\r\n      0 3 4 1 7 6 8 0 0 2 9 0 0]\u003c/pre\u003e\u003cp\u003eIf n = 7 then the output matrix y should be\u003c/p\u003e\u003cpre\u003e y = [1 0 0 0 1 0 0 0 0 0 0 0 0\r\n      1 1 1 1 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0\r\n      0 0 0 0 1 0 0 0 0 0 0 0 0] \u003c/pre\u003e\u003cp\u003eAll the places of n = 7 will become 1 and horizontally they are connected by a path given by 1's which is represented in the output y.\u003c/p\u003e\u003cp\u003eIf n = 9 then y(12,11) = 1 and others are 0's.\u003c/p\u003e\u003cp\u003eOutput should display path as per given n.\u003c/p\u003e","function_template":"function y = your_fcn_name(x,n)\r\n  y = x;\r\nend","test_suite":"%%\r\nx = [7 6 8 5 7 2 4 5 1 3 0 0 0;7 7 7 7 7 0 0 0 0 0 0 0 0;0 0 0 5 5 5 5 5 0 0 0 0 0;0 0 0 1 1 1 1 1 1 0 0 0 0;0 0 0 0 0 2 2 2 2 2 2 2 0;0 6 6 6 6 6 0 0 0 0 0 0 0;0 3 3 3 3 3 3 3 3 3 0 0 0;0 0 0 0 0 0 0 0 0 2 2 2 0;0 0 0 0 0 0 4 4 4 4 4 4 4;0 0 8 8 8 8 8 0 0 0 0 0 0;0 0 4 4 4 4 4 4 4 4 4 4 4;0 3 4 1 7 6 8 0 0 2 9 0 0];\r\nn = 7;\r\ny_correct =[1 0 0 0 1 0 0 0 0 0 0 0 0;1 1 1 1 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0];\r\nassert(isequal(your_fcn_name(x,n),y_correct))\r\n\r\n%%\r\nx = [7 6 8 5 7 2 4 5 1 3 0 0 0;7 7 7 7 7 0 0 0 0 0 0 0 0;0 0 0 5 5 5 5 5 0 0 0 0 0;0 0 0 1 1 1 1 1 1 0 0 0 0;0 0 0 0 0 2 2 2 2 2 2 2 0;0 6 6 6 6 6 0 0 0 0 0 0 0;0 3 3 3 3 3 3 3 3 3 0 0 0;0 0 0 0 0 0 0 0 0 2 2 2 0;0 0 0 0 0 0 4 4 4 4 4 4 4;0 0 8 8 8 8 8 0 0 0 0 0 0;0 0 4 4 4 4 4 4 4 4 4 4 4;0 3 4 1 7 6 8 0 0 2 9 0 0];\r\nn = 4;\r\ny_correct =[0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 1 0 0 0 0 0 0;0 0 0 0 0 0 1 1 1 1 1 1 1;0 0 0 0 0 0 0 0 0 0 0 0 1;0 0 1 1 1 1 1 1 1 1 1 1 1;0 0 1 0 0 0 0 0 0 0 0 0 0]\r\nassert(isequal(your_fcn_name(x,n),y_correct))","published":true,"deleted":false,"likes_count":0,"comments_count":6,"created_by":20855,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":17,"test_suite_updated_at":"2014-07-28T04:59:47.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2014-07-09T12:12:25.000Z","updated_at":"2026-05-28T02:54:46.000Z","published_at":"2014-07-09T12:12:35.000Z","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\u003eHere is a matrix x\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[ x = [7 6 8 5 7 2 4 5 1 3 0 0 0\\n      7 7 7 7 7 0 0 0 0 0 0 0 0\\n      0 0 0 5 5 5 5 5 0 0 0 0 0\\n      0 0 0 1 1 1 1 1 1 0 0 0 0\\n      0 0 0 0 0 2 2 2 2 2 2 2 0\\n      0 6 6 6 6 6 0 0 0 0 0 0 0\\n      0 3 3 3 3 3 3 3 3 3 0 0 0\\n      0 0 0 0 0 0 0 0 0 2 2 2 0\\n      0 0 0 0 0 0 4 4 4 4 4 4 4\\n      0 0 8 8 8 8 8 0 0 0 0 0 0\\n      0 0 4 4 4 4 4 4 4 4 4 4 4\\n      0 3 4 1 7 6 8 0 0 2 9 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\u003eIf n = 7 then the output matrix y should be\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[ y = [1 0 0 0 1 0 0 0 0 0 0 0 0\\n      1 1 1 1 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 0 0 0\\n      0 0 0 0 1 0 0 0 0 0 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\u003eAll the places of n = 7 will become 1 and horizontally they are connected by a path given by 1's which is represented in the output y.\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\u003eIf n = 9 then y(12,11) = 1 and others are 0's.\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 should display path as per given n.\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\"}]}"},{"id":2565,"title":"Determine the Result of a Move in Reversi","description":"Note: This is closely related to \u003chttp://www.mathworks.com/matlabcentral/cody/problems/2538-find-the-next-legal-move-in-reversi Problem 2538\u003e, Find the Next Legal Move in Reversi.\r\n\r\n\u003chttp://en.wikipedia.org/wiki/Reversi Reversi\u003e, also known as Othello, is a game in which reversible white/black chips are placed on a grid. The goal is to have the most pieces once the board is full (i.e. there are no more legal moves). A move, to be legal, must flip at least one opponent's chip by flanking it. \"Flanking\" occurs by surrounding a line of opposing chips with two of your own. It can occur on straight lines and diagonals.\r\n\r\nWe represent a small 4x4 board this way, with 0 for empty squares, 1 for black chips, and 2 for white chips, \r\n\r\n [ 0 0 0 0\r\n   0 1 2 0\r\n   0 2 1 0 \r\n   0 0 0 0 ]\r\n\r\nNow assume black now moves where the asterisk (*) is placed. This is column-wise matrix index position 14.\r\n\r\n [ 0 0 0 0\r\n   0 1 2 *\r\n   0 2 1 0 \r\n   0 0 0 0 ]\r\n\r\nYour Cody challenge is to determine the state of the board once the appropriate chips have been flipped over. In this case the correct answer would be\r\n\r\n [ 0 0 0 0\r\n   0 1 1 1\r\n   0 2 1 0 \r\n   0 0 0 0 ]\r\n\r\nExample\r\n\r\nFor the inputs boardIn, side, and move\r\n\r\n boardIn = ...\r\n   [ 0 0 0 0 0\r\n     0 1 1 2 2 \r\n     0 2 1 1 0\r\n     0 0 2 1 0 \r\n     0 0 0 0 0 ]\r\n side = 1 (black)\r\n move = 9 (matrix index of black's next move)\r\n\r\nThe correct answer is\r\n\r\n boardOut = ...\r\n   [ 0 0 0 0 0\r\n     0 1 1 2 2 \r\n     0 1 1 1 0\r\n     0 1 1 1 0 \r\n     0 0 0 0 0 ]\r\n\r\n\r\n","description_html":"\u003cp\u003eNote: This is closely related to \u003ca href = \"http://www.mathworks.com/matlabcentral/cody/problems/2538-find-the-next-legal-move-in-reversi\"\u003eProblem 2538\u003c/a\u003e, Find the Next Legal Move in Reversi.\u003c/p\u003e\u003cp\u003e\u003ca href = \"http://en.wikipedia.org/wiki/Reversi\"\u003eReversi\u003c/a\u003e, also known as Othello, is a game in which reversible white/black chips are placed on a grid. The goal is to have the most pieces once the board is full (i.e. there are no more legal moves). A move, to be legal, must flip at least one opponent's chip by flanking it. \"Flanking\" occurs by surrounding a line of opposing chips with two of your own. It can occur on straight lines and diagonals.\u003c/p\u003e\u003cp\u003eWe represent a small 4x4 board this way, with 0 for empty squares, 1 for black chips, and 2 for white chips,\u003c/p\u003e\u003cpre\u003e [ 0 0 0 0\r\n   0 1 2 0\r\n   0 2 1 0 \r\n   0 0 0 0 ]\u003c/pre\u003e\u003cp\u003eNow assume black now moves where the asterisk (*) is placed. This is column-wise matrix index position 14.\u003c/p\u003e\u003cpre\u003e [ 0 0 0 0\r\n   0 1 2 *\r\n   0 2 1 0 \r\n   0 0 0 0 ]\u003c/pre\u003e\u003cp\u003eYour Cody challenge is to determine the state of the board once the appropriate chips have been flipped over. In this case the correct answer would be\u003c/p\u003e\u003cpre\u003e [ 0 0 0 0\r\n   0 1 1 1\r\n   0 2 1 0 \r\n   0 0 0 0 ]\u003c/pre\u003e\u003cp\u003eExample\u003c/p\u003e\u003cp\u003eFor the inputs boardIn, side, and move\u003c/p\u003e\u003cpre\u003e boardIn = ...\r\n   [ 0 0 0 0 0\r\n     0 1 1 2 2 \r\n     0 2 1 1 0\r\n     0 0 2 1 0 \r\n     0 0 0 0 0 ]\r\n side = 1 (black)\r\n move = 9 (matrix index of black's next move)\u003c/pre\u003e\u003cp\u003eThe correct answer is\u003c/p\u003e\u003cpre\u003e boardOut = ...\r\n   [ 0 0 0 0 0\r\n     0 1 1 2 2 \r\n     0 1 1 1 0\r\n     0 1 1 1 0 \r\n     0 0 0 0 0 ]\u003c/pre\u003e","function_template":"function boardOut = boardResult(boardIn,side,move)\r\n  boardOut = boardIn;\r\nend","test_suite":"%%\r\nboardIn = ...\r\n [ 0 0 0 0\r\n   0 1 2 0\r\n   0 2 1 0 \r\n   0 0 0 0 ];\r\nside = 1;\r\nmove = 14;\r\nboardOut = ...\r\n [ 0 0 0 0\r\n   0 1 1 1\r\n   0 2 1 0 \r\n   0 0 0 0 ];\r\nassert(isequal(boardOut,boardResult(boardIn,side,move)))\r\n\r\n%%\r\nboardIn = ...\r\n [ 0 0 0 0\r\n   0 1 1 1\r\n   0 2 1 0 \r\n   0 0 0 0 ];\r\nside = 2;\r\nmove = 5;\r\nboardOut = ...\r\n [ 0 2 0 0\r\n   0 2 1 1\r\n   0 2 1 0 \r\n   0 0 0 0 ];\r\nassert(isequal(boardOut,boardResult(boardIn,side,move)))\r\n\r\n%%\r\nboardIn = ...\r\n [ 0 2 0 0\r\n   0 2 1 1\r\n   0 2 1 0 \r\n   0 0 0 0 ];\r\nside = 1;\r\nmove = 4;\r\nboardOut = ...\r\n [ 0 2 0 0\r\n   0 2 1 1\r\n   0 1 1 0 \r\n   1 0 0 0 ];\r\nassert(isequal(boardOut,boardResult(boardIn,side,move)))\r\n\r\n%%\r\nboardIn = ...\r\n [ 0 0 0 0 0\r\n   0 1 1 2 2 \r\n   0 2 1 1 0\r\n   0 0 2 1 0 \r\n   0 0 0 0 0 ];\r\nside = 1;\r\nmove = 9;\r\nboardOut = ...\r\n [ 0 0 0 0 0\r\n   0 1 1 2 2 \r\n   0 1 1 1 0\r\n   0 1 1 1 0 \r\n   0 0 0 0 0 ]\r\nassert(isequal(boardOut,boardResult(boardIn,side,move)))\r\n\r\n","published":true,"deleted":false,"likes_count":2,"comments_count":0,"created_by":7,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":22,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2014-09-09T16:32:03.000Z","updated_at":"2026-05-25T01:53:55.000Z","published_at":"2014-09-09T16:36:06.000Z","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\u003eNote: This is closely related to\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:hyperlink w:docLocation=\\\"http://www.mathworks.com/matlabcentral/cody/problems/2538-find-the-next-legal-move-in-reversi\\\"\u003e\u003cw:r\u003e\u003cw:t\u003eProblem 2538\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\u003e\u003cw:r\u003e\u003cw:t\u003e, Find the Next Legal Move in Reversi.\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:hyperlink w:docLocation=\\\"http://en.wikipedia.org/wiki/Reversi\\\"\u003e\u003cw:r\u003e\u003cw:t\u003eReversi\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\u003e\u003cw:r\u003e\u003cw:t\u003e, also known as Othello, is a game in which reversible white/black chips are placed on a grid. The goal is to have the most pieces once the board is full (i.e. there are no more legal moves). A move, to be legal, must flip at least one opponent's chip by flanking it. \\\"Flanking\\\" occurs by surrounding a line of opposing chips with two of your own. It can occur on straight lines and diagonals.\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\u003eWe represent a small 4x4 board this way, with 0 for empty squares, 1 for black chips, and 2 for white chips,\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[ [ 0 0 0 0\\n   0 1 2 0\\n   0 2 1 0 \\n   0 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\u003eNow assume black now moves where the asterisk \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e(*\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e) is placed. This is column-wise matrix index position 14.\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[ [ 0 0 0 0\\n   0 1 2 *\\n   0 2 1 0 \\n   0 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\u003eYour Cody challenge is to determine the state of the board once the appropriate chips have been flipped over. In this case the correct answer would be\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[ [ 0 0 0 0\\n   0 1 1 1\\n   0 2 1 0 \\n   0 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=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eFor the inputs boardIn, side, and move\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[ boardIn = ...\\n   [ 0 0 0 0 0\\n     0 1 1 2 2 \\n     0 2 1 1 0\\n     0 0 2 1 0 \\n     0 0 0 0 0 ]\\n side = 1 (black)\\n move = 9 (matrix index of black's next move)]]\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 correct answer is\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[ boardOut = ...\\n   [ 0 0 0 0 0\\n     0 1 1 2 2 \\n     0 1 1 1 0\\n     0 1 1 1 0 \\n     0 0 0 0 0 ]]]\u003e\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\"}]}"}],"errors":[],"facets":[[{"value":"Board Games II","count":1,"selected":false}],[{"value":"medium","count":2,"selected":false}]],"term":"tag:\"reversi\"","page":1,"per_page":50,"sort":"map(difficulty_value,0,0,999) asc"}}