{"group":{"id":1,"name":"Community","lockable":false,"created_at":"2012-01-18T18:02:15.000Z","updated_at":"2026-04-16T00:12:35.000Z","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":"2026-04-16T00:00:00.000Z","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":1182,"title":"Hangman (easy)","description":"What is the best letter to start a \u003chttp://en.wikipedia.org/wiki/Hangman_(game) hangman\u003e game with?\r\n\r\nYou are given a cell array with all valid N-letter words. Your output should be the letter that has the highest chance of occurring (at least once) within any randomly chosen word in this dictionary. \r\n\r\nYou can assume that words will always be in all capital letters, and the cell array will always be a row.\r\n\r\n*Example:*\r\n\r\nwords={'AAA','BED','BEG','BAD'};\r\n\r\nYou should return letter='B';\r\n\r\nnote: Letter 'B' occurrs in _three_ different words. Letter 'A', while occurring four times (counting repetitions), only appears in _two_ different words. \r\n\r\n*Follow-up:* \r\n\r\nIf you are going to be losing hours of sleep over the issue of whether choosing the letter with the highest chance of occurring within any randomly chosen word is actually the _best_ 'simple' strategy in a hangman game, then the next problem in this series - \u003chttp://www.mathworks.com/matlabcentral/cody/problems/1184-hangman-strategy Hangman (strategy)\u003e - is for you. Go ahead and test this or a different strategy there, and the contest machinery will score it based on its performance in a series of simulated hangman games. ","description_html":"\u003cp\u003eWhat is the best letter to start a \u003ca href=\"http://en.wikipedia.org/wiki/Hangman_(game)\"\u003ehangman\u003c/a\u003e game with?\u003c/p\u003e\u003cp\u003eYou are given a cell array with all valid N-letter words. Your output should be the letter that has the highest chance of occurring (at least once) within any randomly chosen word in this dictionary.\u003c/p\u003e\u003cp\u003eYou can assume that words will always be in all capital letters, and the cell array will always be a row.\u003c/p\u003e\u003cp\u003e\u003cb\u003eExample:\u003c/b\u003e\u003c/p\u003e\u003cp\u003ewords={'AAA','BED','BEG','BAD'};\u003c/p\u003e\u003cp\u003eYou should return letter='B';\u003c/p\u003e\u003cp\u003enote: Letter 'B' occurrs in \u003ci\u003ethree\u003c/i\u003e different words. Letter 'A', while occurring four times (counting repetitions), only appears in \u003ci\u003etwo\u003c/i\u003e different words.\u003c/p\u003e\u003cp\u003e\u003cb\u003eFollow-up:\u003c/b\u003e\u003c/p\u003e\u003cp\u003eIf you are going to be losing hours of sleep over the issue of whether choosing the letter with the highest chance of occurring within any randomly chosen word is actually the \u003ci\u003ebest\u003c/i\u003e 'simple' strategy in a hangman game, then the next problem in this series - \u003ca href=\"http://www.mathworks.com/matlabcentral/cody/problems/1184-hangman-strategy\"\u003eHangman (strategy)\u003c/a\u003e - is for you. Go ahead and test this or a different strategy there, and the contest machinery will score it based on its performance in a series of simulated hangman games.\u003c/p\u003e","function_template":"function letter = hangman(words)\r\n  letter='S';\r\nend","test_suite":"%%\r\nwords={'AAA','BED','BEG','BAD'};\r\nassert(isequal(hangman(words),'B'));\r\n\r\n%%\r\nwords={'BUZZ','COZY','DOZE','FUZZ','GAZE','HAZE','JAZZ','LAZY','SIZE','ZERO','ZONE'};\r\nassert(isequal(hangman(words),'Z'));\r\n\r\n%%\r\nrng default;\r\nwords=unique(char('A'+randi(26,[100,3])-1),'rows');\r\nassert(isequal(sum(any(words==hangman(cellstr(words)'),2)),max(arrayfun(@(x)sum(any(words==x,2)),'A':'Z'))));\r\n\r\n%%\r\nrng default;\r\nwords=unique(char('A'+randi(26,[200,4])-1),'rows');\r\nassert(isequal(sum(any(words==hangman(cellstr(words)'),2)),max(arrayfun(@(x)sum(any(words==x,2)),'A':'Z'))));\r\n\r\n%%\r\nrng default;\r\nwords=unique(char('A'+randi(26,[500,5])-1),'rows');\r\nassert(isequal(sum(any(words==hangman(cellstr(words)'),2)),max(arrayfun(@(x)sum(any(words==x,2)),'A':'Z'))));\r\n","published":true,"deleted":false,"likes_count":4,"comments_count":0,"created_by":43,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":52,"test_suite_updated_at":"2013-01-08T05:17:12.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2013-01-07T03:59:12.000Z","updated_at":"2025-12-15T20:07:42.000Z","published_at":"2013-01-07T04:03:43.000Z","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\u003eWhat is the best letter to start a\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://en.wikipedia.org/wiki/Hangman_(game)\\\"\u003e\u003cw:r\u003e\u003cw:t\u003ehangman\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\u003e\u003cw:r\u003e\u003cw:t\u003e game with?\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\u003eYou are given a cell array with all valid N-letter words. Your output should be the letter that has the highest chance of occurring (at least once) within any randomly chosen word in this dictionary.\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\u003eYou can assume that words will always be in all capital letters, and the cell array will always be a row.\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:rPr\u003e\u003cw:b/\u003e\u003c/w:rPr\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\u003ewords={'AAA','BED','BEG','BAD'};\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\u003eYou should return letter='B';\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\u003enote: Letter 'B' occurrs in\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:rPr\u003e\u003cw:i/\u003e\u003c/w:rPr\u003e\u003cw:t\u003ethree\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e different words. Letter 'A', while occurring four times (counting repetitions), only appears in\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:rPr\u003e\u003cw:i/\u003e\u003c/w:rPr\u003e\u003cw:t\u003etwo\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e different words.\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:rPr\u003e\u003cw:b/\u003e\u003c/w:rPr\u003e\u003cw:t\u003eFollow-up:\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 you are going to be losing hours of sleep over the issue of whether choosing the letter with the highest chance of occurring within any randomly chosen word is actually the\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:rPr\u003e\u003cw:i/\u003e\u003c/w:rPr\u003e\u003cw:t\u003ebest\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e 'simple' strategy in a hangman game, then the next problem in this series -\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/1184-hangman-strategy\\\"\u003e\u003cw:r\u003e\u003cw:t\u003eHangman (strategy)\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\u003e\u003cw:r\u003e\u003cw:t\u003e - is for you. Go ahead and test this or a different strategy there, and the contest machinery will score it based on its performance in a series of simulated hangman games.\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":1182,"title":"Hangman (easy)","description":"What is the best letter to start a \u003chttp://en.wikipedia.org/wiki/Hangman_(game) hangman\u003e game with?\r\n\r\nYou are given a cell array with all valid N-letter words. Your output should be the letter that has the highest chance of occurring (at least once) within any randomly chosen word in this dictionary. \r\n\r\nYou can assume that words will always be in all capital letters, and the cell array will always be a row.\r\n\r\n*Example:*\r\n\r\nwords={'AAA','BED','BEG','BAD'};\r\n\r\nYou should return letter='B';\r\n\r\nnote: Letter 'B' occurrs in _three_ different words. Letter 'A', while occurring four times (counting repetitions), only appears in _two_ different words. \r\n\r\n*Follow-up:* \r\n\r\nIf you are going to be losing hours of sleep over the issue of whether choosing the letter with the highest chance of occurring within any randomly chosen word is actually the _best_ 'simple' strategy in a hangman game, then the next problem in this series - \u003chttp://www.mathworks.com/matlabcentral/cody/problems/1184-hangman-strategy Hangman (strategy)\u003e - is for you. Go ahead and test this or a different strategy there, and the contest machinery will score it based on its performance in a series of simulated hangman games. ","description_html":"\u003cp\u003eWhat is the best letter to start a \u003ca href=\"http://en.wikipedia.org/wiki/Hangman_(game)\"\u003ehangman\u003c/a\u003e game with?\u003c/p\u003e\u003cp\u003eYou are given a cell array with all valid N-letter words. Your output should be the letter that has the highest chance of occurring (at least once) within any randomly chosen word in this dictionary.\u003c/p\u003e\u003cp\u003eYou can assume that words will always be in all capital letters, and the cell array will always be a row.\u003c/p\u003e\u003cp\u003e\u003cb\u003eExample:\u003c/b\u003e\u003c/p\u003e\u003cp\u003ewords={'AAA','BED','BEG','BAD'};\u003c/p\u003e\u003cp\u003eYou should return letter='B';\u003c/p\u003e\u003cp\u003enote: Letter 'B' occurrs in \u003ci\u003ethree\u003c/i\u003e different words. Letter 'A', while occurring four times (counting repetitions), only appears in \u003ci\u003etwo\u003c/i\u003e different words.\u003c/p\u003e\u003cp\u003e\u003cb\u003eFollow-up:\u003c/b\u003e\u003c/p\u003e\u003cp\u003eIf you are going to be losing hours of sleep over the issue of whether choosing the letter with the highest chance of occurring within any randomly chosen word is actually the \u003ci\u003ebest\u003c/i\u003e 'simple' strategy in a hangman game, then the next problem in this series - \u003ca href=\"http://www.mathworks.com/matlabcentral/cody/problems/1184-hangman-strategy\"\u003eHangman (strategy)\u003c/a\u003e - is for you. Go ahead and test this or a different strategy there, and the contest machinery will score it based on its performance in a series of simulated hangman games.\u003c/p\u003e","function_template":"function letter = hangman(words)\r\n  letter='S';\r\nend","test_suite":"%%\r\nwords={'AAA','BED','BEG','BAD'};\r\nassert(isequal(hangman(words),'B'));\r\n\r\n%%\r\nwords={'BUZZ','COZY','DOZE','FUZZ','GAZE','HAZE','JAZZ','LAZY','SIZE','ZERO','ZONE'};\r\nassert(isequal(hangman(words),'Z'));\r\n\r\n%%\r\nrng default;\r\nwords=unique(char('A'+randi(26,[100,3])-1),'rows');\r\nassert(isequal(sum(any(words==hangman(cellstr(words)'),2)),max(arrayfun(@(x)sum(any(words==x,2)),'A':'Z'))));\r\n\r\n%%\r\nrng default;\r\nwords=unique(char('A'+randi(26,[200,4])-1),'rows');\r\nassert(isequal(sum(any(words==hangman(cellstr(words)'),2)),max(arrayfun(@(x)sum(any(words==x,2)),'A':'Z'))));\r\n\r\n%%\r\nrng default;\r\nwords=unique(char('A'+randi(26,[500,5])-1),'rows');\r\nassert(isequal(sum(any(words==hangman(cellstr(words)'),2)),max(arrayfun(@(x)sum(any(words==x,2)),'A':'Z'))));\r\n","published":true,"deleted":false,"likes_count":4,"comments_count":0,"created_by":43,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":52,"test_suite_updated_at":"2013-01-08T05:17:12.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2013-01-07T03:59:12.000Z","updated_at":"2025-12-15T20:07:42.000Z","published_at":"2013-01-07T04:03:43.000Z","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\u003eWhat is the best letter to start a\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://en.wikipedia.org/wiki/Hangman_(game)\\\"\u003e\u003cw:r\u003e\u003cw:t\u003ehangman\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\u003e\u003cw:r\u003e\u003cw:t\u003e game with?\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\u003eYou are given a cell array with all valid N-letter words. Your output should be the letter that has the highest chance of occurring (at least once) within any randomly chosen word in this dictionary.\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\u003eYou can assume that words will always be in all capital letters, and the cell array will always be a row.\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:rPr\u003e\u003cw:b/\u003e\u003c/w:rPr\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\u003ewords={'AAA','BED','BEG','BAD'};\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\u003eYou should return letter='B';\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\u003enote: Letter 'B' occurrs in\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:rPr\u003e\u003cw:i/\u003e\u003c/w:rPr\u003e\u003cw:t\u003ethree\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e different words. Letter 'A', while occurring four times (counting repetitions), only appears in\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:rPr\u003e\u003cw:i/\u003e\u003c/w:rPr\u003e\u003cw:t\u003etwo\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e different words.\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:rPr\u003e\u003cw:b/\u003e\u003c/w:rPr\u003e\u003cw:t\u003eFollow-up:\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 you are going to be losing hours of sleep over the issue of whether choosing the letter with the highest chance of occurring within any randomly chosen word is actually the\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:rPr\u003e\u003cw:i/\u003e\u003c/w:rPr\u003e\u003cw:t\u003ebest\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e 'simple' strategy in a hangman game, then the next problem in this series -\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/1184-hangman-strategy\\\"\u003e\u003cw:r\u003e\u003cw:t\u003eHangman (strategy)\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\u003e\u003cw:r\u003e\u003cw:t\u003e - is for you. Go ahead and test this or a different strategy there, and the contest machinery will score it based on its performance in a series of simulated hangman games.\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 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