{"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":1098,"title":"USC Fall 2012 ACM : Code Word Minimum Flipped Bits","description":"This Challenge is to solve Question A of the \u003chttp://contest.usc.edu/index.php/Fall12/Home USC ACM Fall 2012 Contest\u003e.\r\n\r\nGiven an array M of Valid binary codewords(m codewords of width n) and a Received Corrupted(?) codeword of width n, determine the minimum number of flipped bits in the Received codeword to generate a valid codeword.\r\n\r\n\r\nInput: [ M, v ]\r\n\r\nOutput: e, minimum number of error(flipped) bits\r\n.\r\n\r\nFrom full \u003chttp://contest.usc.edu/index.php/Fall12/Home?action=download\u0026upname=codes.in.txt USC data file\u003e\r\n\r\nInput: [0 0 0; 1 1 1; 1 1 0], [0 1 0]\r\n\r\nOutput: 1 as [0 1 0] can convert to [0 0 0] or [1 1 0] with a single flip\r\n\r\nMatlab one-liner?\r\n\r\nThe Winning C solution - not much help:\r\n\r\n  #include \u003ciostream\u003e\r\n  #include \u003cstdio.h\u003e\r\n  #include \u003cstring\u003e\r\n  using namespace std;\r\n  int main () {\r\n  \tfreopen(\"codes.in\", \"r\", stdin);\r\n  \tint K,n,b;\r\n  \tcin \u003e\u003e K;\r\n  \tfor (int i = 1; i \u003c K + 1; ++i) {\r\n  \t\tcin \u003e\u003e n \u003e\u003e b;\r\n  \t\tstring m[1000], r;\r\n  \t\tfor (int j = 0; j \u003c n; ++j)\r\n  \t\t\tcin \u003e\u003e m[j];\r\n  \t\tcin \u003e\u003e r;\r\n% Process Start\t\t\r\n  \t\tint f = b;\r\n  \t\tfor (int j = 0; j \u003c n; ++j) {\r\n  \t\t\tint d = b;\r\n  \t\t\tfor (int k = 0; k \u003c b; ++k) {\r\n  \t\t\t\tif (m[j][k] == r[k])\r\n  \t\t\t\t\t--d;\r\n  \t\t\t}\r\n  \t\t\tf = ((f \u003c= d) ? f : d);\r\n  \t\t}\r\n % Process End \r\n  \t\tprintf(\"Data Set %d:\\n\", i);\r\n  \t\tprintf(\"%d\\n\\n\", f);\r\n  \t}\r\n  \treturn 0;\r\n  }\r\n  ","description_html":"\u003cp\u003eThis Challenge is to solve Question A of the \u003ca href=\"http://contest.usc.edu/index.php/Fall12/Home\"\u003eUSC ACM Fall 2012 Contest\u003c/a\u003e.\u003c/p\u003e\u003cp\u003eGiven an array M of Valid binary codewords(m codewords of width n) and a Received Corrupted(?) codeword of width n, determine the minimum number of flipped bits in the Received codeword to generate a valid codeword.\u003c/p\u003e\u003cp\u003eInput: [ M, v ]\u003c/p\u003e\u003cp\u003eOutput: e, minimum number of error(flipped) bits\r\n.\u003c/p\u003e\u003cp\u003eFrom full \u003ca href=\"http://contest.usc.edu/index.php/Fall12/Home?action=download\u0026amp;upname=codes.in.txt\"\u003eUSC data file\u003c/a\u003e\u003c/p\u003e\u003cp\u003eInput: [0 0 0; 1 1 1; 1 1 0], [0 1 0]\u003c/p\u003e\u003cp\u003eOutput: 1 as [0 1 0] can convert to [0 0 0] or [1 1 0] with a single flip\u003c/p\u003e\u003cp\u003eMatlab one-liner?\u003c/p\u003e\u003cp\u003eThe Winning C solution - not much help:\u003c/p\u003e\u003cpre class=\"language-matlab\"\u003e#include \u0026lt;iostream\u003e\r\n#include \u0026lt;stdio.h\u003e\r\n#include \u0026lt;string\u003e\r\nusing namespace std;\r\nint main () {\r\n\tfreopen(\"codes.in\", \"r\", stdin);\r\n\tint K,n,b;\r\n\tcin \u003e\u003e K;\r\n\tfor (int i = 1; i \u0026lt; K + 1; ++i) {\r\n\t\tcin \u003e\u003e n \u003e\u003e b;\r\n\t\tstring m[1000], r;\r\n\t\tfor (int j = 0; j \u0026lt; n; ++j)\r\n\t\t\tcin \u003e\u003e m[j];\r\n\t\tcin \u003e\u003e r;\r\n% Process Start\t\t\r\n\t\tint f = b;\r\n\t\tfor (int j = 0; j \u0026lt; n; ++j) {\r\n\t\t\tint d = b;\r\n\t\t\tfor (int k = 0; k \u0026lt; b; ++k) {\r\n\t\t\t\tif (m[j][k] == r[k])\r\n\t\t\t\t\t--d;\r\n\t\t\t}\r\n\t\t\tf = ((f \u0026lt;= d) ? f : d);\r\n\t\t}\r\n% Process End \r\n\t\tprintf(\"Data Set %d:\\n\", i);\r\n\t\tprintf(\"%d\\n\\n\", f);\r\n\t}\r\n\treturn 0;\r\n}\r\n\u003c/pre\u003e","function_template":"function f = USC_No_1(M,v)\r\n  f=0;\r\nend","test_suite":"%%\r\ntic\r\nurlfn='http://contest.usc.edu/index.php/Fall12/Home?action=download\u0026upname=codes.in.txt';\r\nurlwrite(urlfn,'codesin_A.txt'); % Load file from USC\r\ntoc\r\n%%\r\n flip_correct=[1 0 0 54 1 29 37 32 33 32 39 8 0 0 36 36 35];\r\n\r\n fid=fopen('codesin_A.txt','r');\r\n\r\n qty=fscanf(fid,'%i',1);\r\n for ptr=1:qty\r\n  nr=fscanf(fid,'%i',1);\r\n  nc=fscanf(fid,'%i',1);\r\n \r\n  A=zeros(nr,nc);\r\n  for i=1:nr\r\n   strv=fscanf(fid,'%s',1); % Reads a line of text\r\n   A(i,:)=strv-'0'; % vectorize the string\r\n  end\r\n \r\n  strv=fscanf(fid,'%s',1);\r\n  v=strv-'0';\r\n\r\n  USC_flips = USC_No_1(A,v);\r\n\r\n  assert(isequal(USC_flips,flip_correct(ptr)))\r\n \r\n end\r\n fclose(fid);\r\n \r\ntoc\r\n","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":3097,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":20,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2012-12-06T02:14:30.000Z","updated_at":"2026-05-26T13:48:08.000Z","published_at":"2012-12-06T02:40:27.000Z","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\u003eThis Challenge is to solve Question A of the\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://contest.usc.edu/index.php/Fall12/Home\\\"\u003e\u003cw:r\u003e\u003cw:t\u003eUSC ACM Fall 2012 Contest\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\u003e\u003cw:r\u003e\u003cw:t\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\u003eGiven an array M of Valid binary codewords(m codewords of width n) and a Received Corrupted(?) codeword of width n, determine the minimum number of flipped bits in the Received codeword to generate a valid codeword.\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\u003eInput: [ M, v ]\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: e, minimum number of error(flipped) bits .\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\u003eFrom full\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://contest.usc.edu/index.php/Fall12/Home?action=download\u0026amp;upname=codes.in.txt\\\"\u003e\u003cw:r\u003e\u003cw:t\u003eUSC data file\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\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\u003eInput: [0 0 0; 1 1 1; 1 1 0], [0 1 0]\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: 1 as [0 1 0] can convert to [0 0 0] or [1 1 0] with a single flip\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\u003eMatlab one-liner?\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 Winning C solution - not much help:\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[#include \u003ciostream\u003e\\n#include \u003cstdio.h\u003e\\n#include \u003cstring\u003e\\nusing namespace std;\\nint main () {\\n  freopen(\\\"codes.in\\\", \\\"r\\\", stdin);\\n  int K,n,b;\\n  cin \u003e\u003e K;\\n  for (int i = 1; i \u003c K + 1; ++i) {\\n    cin \u003e\u003e n \u003e\u003e b;\\n    string m[1000], r;\\n    for (int j = 0; j \u003c n; ++j)\\n      cin \u003e\u003e m[j];\\n    cin \u003e\u003e r;\\n% Process Start    \\n    int f = b;\\n    for (int j = 0; j \u003c n; ++j) {\\n      int d = b;\\n      for (int k = 0; k \u003c b; ++k) {\\n        if (m[j][k] == r[k])\\n          --d;\\n      }\\n      f = ((f \u003c= d) ? f : d);\\n    }\\n% Process End \\n    printf(\\\"Data Set %d:\\\\n\\\", i);\\n    printf(\\\"%d\\\\n\\\\n\\\", f);\\n  }\\n  return 0;\\n}]]\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":1098,"title":"USC Fall 2012 ACM : Code Word Minimum Flipped Bits","description":"This Challenge is to solve Question A of the \u003chttp://contest.usc.edu/index.php/Fall12/Home USC ACM Fall 2012 Contest\u003e.\r\n\r\nGiven an array M of Valid binary codewords(m codewords of width n) and a Received Corrupted(?) codeword of width n, determine the minimum number of flipped bits in the Received codeword to generate a valid codeword.\r\n\r\n\r\nInput: [ M, v ]\r\n\r\nOutput: e, minimum number of error(flipped) bits\r\n.\r\n\r\nFrom full \u003chttp://contest.usc.edu/index.php/Fall12/Home?action=download\u0026upname=codes.in.txt USC data file\u003e\r\n\r\nInput: [0 0 0; 1 1 1; 1 1 0], [0 1 0]\r\n\r\nOutput: 1 as [0 1 0] can convert to [0 0 0] or [1 1 0] with a single flip\r\n\r\nMatlab one-liner?\r\n\r\nThe Winning C solution - not much help:\r\n\r\n  #include \u003ciostream\u003e\r\n  #include \u003cstdio.h\u003e\r\n  #include \u003cstring\u003e\r\n  using namespace std;\r\n  int main () {\r\n  \tfreopen(\"codes.in\", \"r\", stdin);\r\n  \tint K,n,b;\r\n  \tcin \u003e\u003e K;\r\n  \tfor (int i = 1; i \u003c K + 1; ++i) {\r\n  \t\tcin \u003e\u003e n \u003e\u003e b;\r\n  \t\tstring m[1000], r;\r\n  \t\tfor (int j = 0; j \u003c n; ++j)\r\n  \t\t\tcin \u003e\u003e m[j];\r\n  \t\tcin \u003e\u003e r;\r\n% Process Start\t\t\r\n  \t\tint f = b;\r\n  \t\tfor (int j = 0; j \u003c n; ++j) {\r\n  \t\t\tint d = b;\r\n  \t\t\tfor (int k = 0; k \u003c b; ++k) {\r\n  \t\t\t\tif (m[j][k] == r[k])\r\n  \t\t\t\t\t--d;\r\n  \t\t\t}\r\n  \t\t\tf = ((f \u003c= d) ? f : d);\r\n  \t\t}\r\n % Process End \r\n  \t\tprintf(\"Data Set %d:\\n\", i);\r\n  \t\tprintf(\"%d\\n\\n\", f);\r\n  \t}\r\n  \treturn 0;\r\n  }\r\n  ","description_html":"\u003cp\u003eThis Challenge is to solve Question A of the \u003ca href=\"http://contest.usc.edu/index.php/Fall12/Home\"\u003eUSC ACM Fall 2012 Contest\u003c/a\u003e.\u003c/p\u003e\u003cp\u003eGiven an array M of Valid binary codewords(m codewords of width n) and a Received Corrupted(?) codeword of width n, determine the minimum number of flipped bits in the Received codeword to generate a valid codeword.\u003c/p\u003e\u003cp\u003eInput: [ M, v ]\u003c/p\u003e\u003cp\u003eOutput: e, minimum number of error(flipped) bits\r\n.\u003c/p\u003e\u003cp\u003eFrom full \u003ca href=\"http://contest.usc.edu/index.php/Fall12/Home?action=download\u0026amp;upname=codes.in.txt\"\u003eUSC data file\u003c/a\u003e\u003c/p\u003e\u003cp\u003eInput: [0 0 0; 1 1 1; 1 1 0], [0 1 0]\u003c/p\u003e\u003cp\u003eOutput: 1 as [0 1 0] can convert to [0 0 0] or [1 1 0] with a single flip\u003c/p\u003e\u003cp\u003eMatlab one-liner?\u003c/p\u003e\u003cp\u003eThe Winning C solution - not much help:\u003c/p\u003e\u003cpre class=\"language-matlab\"\u003e#include \u0026lt;iostream\u003e\r\n#include \u0026lt;stdio.h\u003e\r\n#include \u0026lt;string\u003e\r\nusing namespace std;\r\nint main () {\r\n\tfreopen(\"codes.in\", \"r\", stdin);\r\n\tint K,n,b;\r\n\tcin \u003e\u003e K;\r\n\tfor (int i = 1; i \u0026lt; K + 1; ++i) {\r\n\t\tcin \u003e\u003e n \u003e\u003e b;\r\n\t\tstring m[1000], r;\r\n\t\tfor (int j = 0; j \u0026lt; n; ++j)\r\n\t\t\tcin \u003e\u003e m[j];\r\n\t\tcin \u003e\u003e r;\r\n% Process Start\t\t\r\n\t\tint f = b;\r\n\t\tfor (int j = 0; j \u0026lt; n; ++j) {\r\n\t\t\tint d = b;\r\n\t\t\tfor (int k = 0; k \u0026lt; b; ++k) {\r\n\t\t\t\tif (m[j][k] == r[k])\r\n\t\t\t\t\t--d;\r\n\t\t\t}\r\n\t\t\tf = ((f \u0026lt;= d) ? f : d);\r\n\t\t}\r\n% Process End \r\n\t\tprintf(\"Data Set %d:\\n\", i);\r\n\t\tprintf(\"%d\\n\\n\", f);\r\n\t}\r\n\treturn 0;\r\n}\r\n\u003c/pre\u003e","function_template":"function f = USC_No_1(M,v)\r\n  f=0;\r\nend","test_suite":"%%\r\ntic\r\nurlfn='http://contest.usc.edu/index.php/Fall12/Home?action=download\u0026upname=codes.in.txt';\r\nurlwrite(urlfn,'codesin_A.txt'); % Load file from USC\r\ntoc\r\n%%\r\n flip_correct=[1 0 0 54 1 29 37 32 33 32 39 8 0 0 36 36 35];\r\n\r\n fid=fopen('codesin_A.txt','r');\r\n\r\n qty=fscanf(fid,'%i',1);\r\n for ptr=1:qty\r\n  nr=fscanf(fid,'%i',1);\r\n  nc=fscanf(fid,'%i',1);\r\n \r\n  A=zeros(nr,nc);\r\n  for i=1:nr\r\n   strv=fscanf(fid,'%s',1); % Reads a line of text\r\n   A(i,:)=strv-'0'; % vectorize the string\r\n  end\r\n \r\n  strv=fscanf(fid,'%s',1);\r\n  v=strv-'0';\r\n\r\n  USC_flips = USC_No_1(A,v);\r\n\r\n  assert(isequal(USC_flips,flip_correct(ptr)))\r\n \r\n end\r\n fclose(fid);\r\n \r\ntoc\r\n","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":3097,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":20,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":1,"created_at":"2012-12-06T02:14:30.000Z","updated_at":"2026-05-26T13:48:08.000Z","published_at":"2012-12-06T02:40:27.000Z","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\u003eThis Challenge is to solve Question A of the\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://contest.usc.edu/index.php/Fall12/Home\\\"\u003e\u003cw:r\u003e\u003cw:t\u003eUSC ACM Fall 2012 Contest\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\u003e\u003cw:r\u003e\u003cw:t\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\u003eGiven an array M of Valid binary codewords(m codewords of width n) and a Received Corrupted(?) codeword of width n, determine the minimum number of flipped bits in the Received codeword to generate a valid codeword.\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\u003eInput: [ M, v ]\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: e, minimum number of error(flipped) bits .\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\u003eFrom full\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://contest.usc.edu/index.php/Fall12/Home?action=download\u0026amp;upname=codes.in.txt\\\"\u003e\u003cw:r\u003e\u003cw:t\u003eUSC data file\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\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\u003eInput: [0 0 0; 1 1 1; 1 1 0], [0 1 0]\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: 1 as [0 1 0] can convert to [0 0 0] or [1 1 0] with a single flip\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\u003eMatlab one-liner?\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 Winning C solution - not much help:\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[#include \u003ciostream\u003e\\n#include \u003cstdio.h\u003e\\n#include \u003cstring\u003e\\nusing namespace std;\\nint main () {\\n  freopen(\\\"codes.in\\\", \\\"r\\\", stdin);\\n  int K,n,b;\\n  cin \u003e\u003e K;\\n  for (int i = 1; i \u003c K + 1; ++i) {\\n    cin \u003e\u003e n \u003e\u003e b;\\n    string m[1000], r;\\n    for (int j = 0; j \u003c n; ++j)\\n      cin \u003e\u003e m[j];\\n    cin \u003e\u003e r;\\n% Process Start    \\n    int f = b;\\n    for (int j = 0; j \u003c n; ++j) {\\n      int d = b;\\n      for (int k = 0; k \u003c b; ++k) {\\n        if (m[j][k] == r[k])\\n          --d;\\n      }\\n      f = ((f \u003c= d) ? f : d);\\n    }\\n% Process End \\n    printf(\\\"Data Set %d:\\\\n\\\", i);\\n    printf(\\\"%d\\\\n\\\\n\\\", f);\\n  }\\n  return 0;\\n}]]\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":"easy","count":1,"selected":false}]],"term":"tag:\"one-line\"","page":1,"per_page":50,"sort":"map(difficulty_value,0,0,999) asc"}}