{"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":663,"title":"Crypto Addition - v01","description":"Crypto addition is solving for the numeric values of the characters given in an addition equation:\r\n\r\n   abc       756\r\n  +def       342\r\n  ----      ----\r\n  ghij      1098\r\n\r\n\r\nInput [ 'abc' , 'def' , 'ghij' ] Output [ 756 342 1098 ] or [ ] if no solution  \r\n\r\nAll characters have unique values. \r\n\r\nLeading zeros are not allowed.\r\n\r\nThis Crypto Addition v01 is limited to two addends and a sum.\r\n\r\nThe output is the number values [anum bnum cnum] given [astr, bstr,cstr].\r\n\r\nIf there is no solution the output should be [ ], empty set.\r\n\r\nI am planning v02 thru v05 with additional complications.\r\n\r\nv02 will have embedded numbers.\r\n\r\nv03 will allow up to ten addends vs the current two.\r\n\r\nA solid recursive routine can crack each test case in under 75 msec. ","description_html":"\u003cp\u003eCrypto addition is solving for the numeric values of the characters given in an addition equation:\u003c/p\u003e\u003cpre\u003e   abc       756\r\n  +def       342\r\n  ----      ----\r\n  ghij      1098\u003c/pre\u003e\u003cp\u003eInput [ 'abc' , 'def' , 'ghij' ] Output [ 756 342 1098 ] or [ ] if no solution\u003c/p\u003e\u003cp\u003eAll characters have unique values.\u003c/p\u003e\u003cp\u003eLeading zeros are not allowed.\u003c/p\u003e\u003cp\u003eThis Crypto Addition v01 is limited to two addends and a sum.\u003c/p\u003e\u003cp\u003eThe output is the number values [anum bnum cnum] given [astr, bstr,cstr].\u003c/p\u003e\u003cp\u003eIf there is no solution the output should be [ ], empty set.\u003c/p\u003e\u003cp\u003eI am planning v02 thru v05 with additional complications.\u003c/p\u003e\u003cp\u003ev02 will have embedded numbers.\u003c/p\u003e\u003cp\u003ev03 will allow up to ten addends vs the current two.\u003c/p\u003e\u003cp\u003eA solid recursive routine can crack each test case in under 75 msec.\u003c/p\u003e","function_template":"function y = crypto_add(astr,bstr,cstr)\r\n\r\n a=double(astr); % Gives ascii values in a vector\r\n b=double(bstr);\r\n c=double(cstr);\r\n\r\n y = [a b c]; % or y=[];\r\nend","test_suite":"%%\r\n    astr='a'; % []\r\n    bstr='a';\r\n    cstr='aa';\r\n  y_correct = [];\r\nassert(isequal(crypto_add(astr,bstr,cstr),y_correct))\r\n%%\r\n    astr='ab'; % []\r\n    bstr='ab';\r\n    cstr='ab';\r\n  y_correct = [];\r\nassert(isequal(crypto_add(astr,bstr,cstr),y_correct))\r\n%%\r\n   astr='abcdefghij'; %[]\r\n   bstr='abcdefghij';\r\n   cstr='aabcdefghij';\r\n  y_correct = [];\r\nassert(isequal(crypto_add(astr,bstr,cstr),y_correct))\r\n%%\r\n   astr='coca'; % 8186\r\n   bstr='cola'; % 8106\r\n   cstr='oasis'; % 16292\r\ny_correct = [8186 8106 16292];\r\nassert(isequal(crypto_add(astr,bstr,cstr),y_correct))\r\n%%\r\n   astr='xxx';% 999\r\n   bstr='b';  %   1\r\n   cstr='baaa'; % 1000\r\ny_correct = [999 1 1000];\r\nassert(isequal(crypto_add(astr,bstr,cstr),y_correct))\r\n%%\r\n   astr='ma';  % 89 9 98\r\n   bstr='a';\r\n   cstr='am'\r\ny_correct = [89 9 98];\r\nassert(isequal(crypto_add(astr,bstr,cstr),y_correct))\r\n%%\r\n   astr='bmtran'; % 951740 or  651740\r\n   bstr='winner'; % 630087 y   930087\r\n   cstr='tmeteor';% 1581827   1581827\r\ny_correct1 = [951740 630087 1581827];\r\ny_correct2 = [651740 930087 1581827];\r\ny_out=crypto_add(astr,bstr,cstr)\r\nsolved= isequal(y_out,y_correct1) || isequal(y_out,y_correct2)\r\nassert(solved)\r\n%%\r\n% For those that like to hard code answers\r\nasum=0;\r\nwhile asum\u003c1234567890\r\n avec=randperm(10)-1\r\n asum=polyval(avec,10);\r\nend\r\n\r\nbsum=0;\r\nwhile bsum\u003c1234567890\r\n bvec=randperm(10)-1\r\n bsum=polyval(bvec,10);\r\nend\r\n\r\ncsum=asum+bsum;\r\n%csum=csum+20000000000; %To create []\r\ncvec=int2str(csum)-'0' % separate puzzle question\r\n\r\nalpha_vec=randperm(10); % randomize the alphabet\r\nalpha='abcdefghij';\r\n\r\nastr='';\r\nfor i=1:length(avec)\r\n astr=[astr alpha(alpha_vec(avec(i)+1))];\r\nend\r\n\r\nbstr='';\r\nfor i=1:length(bvec)\r\n bstr=[bstr alpha(alpha_vec(bvec(i)+1))];\r\nend\r\n\r\ncstr='';\r\nfor i=1:length(cvec)\r\n cstr=[cstr alpha(alpha_vec(cvec(i)+1))];\r\nend\r\nastr\r\nbstr\r\ncstr\r\ny_correct = [asum bsum csum];\r\nassert(isequal(crypto_add(astr,bstr,cstr),y_correct))\r\n%%\r\n% For those that like to hard code answers\r\nasum=0;\r\nwhile asum\u003c1234567890\r\n avec=randperm(10)-1\r\n asum=polyval(avec,10);\r\nend\r\n\r\nbsum=0;\r\nwhile bsum\u003c1234567890\r\n bvec=randperm(10)-1\r\n bsum=polyval(bvec,10);\r\nend\r\n\r\ncsum=asum+bsum;\r\ncsum=csum+20000000000; %To create []\r\ncvec=int2str(csum)-'0' % separate puzzle question\r\n\r\nalpha_vec=randperm(10); % randomize the alphabet\r\nalpha='abcdefghij';\r\n\r\nastr='';\r\nfor i=1:length(avec)\r\n astr=[astr alpha(alpha_vec(avec(i)+1))];\r\nend\r\n\r\nbstr='';\r\nfor i=1:length(bvec)\r\n bstr=[bstr alpha(alpha_vec(bvec(i)+1))];\r\nend\r\n\r\ncstr='';\r\nfor i=1:length(cvec)\r\n cstr=[cstr alpha(alpha_vec(cvec(i)+1))];\r\nend\r\n\r\ny_correct = [];\r\nassert(isequal(crypto_add(astr,bstr,cstr),y_correct))\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":6,"test_suite_updated_at":"2012-05-05T22:30:58.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2012-05-04T20:33:57.000Z","updated_at":"2012-05-05T22:42:29.000Z","published_at":"2012-05-04T22:11:29.000Z","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\u003eCrypto addition is solving for the numeric values of the characters given in an addition equation:\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[   abc       756\\n  +def       342\\n  ----      ----\\n  ghij      1098]]\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\u003eInput [ 'abc' , 'def' , 'ghij' ] Output [ 756 342 1098 ] or [ ] if no solution\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 characters have unique values.\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\u003eLeading zeros are not allowed.\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\u003eThis Crypto Addition v01 is limited to two addends and a sum.\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 output is the number values [anum bnum cnum] given [astr, bstr,cstr].\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 there is no solution the output should be [ ], empty set.\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\u003eI am planning v02 thru v05 with additional complications.\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\u003ev02 will have embedded numbers.\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\u003ev03 will allow up to ten addends vs the current two.\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\u003eA solid recursive routine can crack each test case in under 75 msec.\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":663,"title":"Crypto Addition - v01","description":"Crypto addition is solving for the numeric values of the characters given in an addition equation:\r\n\r\n   abc       756\r\n  +def       342\r\n  ----      ----\r\n  ghij      1098\r\n\r\n\r\nInput [ 'abc' , 'def' , 'ghij' ] Output [ 756 342 1098 ] or [ ] if no solution  \r\n\r\nAll characters have unique values. \r\n\r\nLeading zeros are not allowed.\r\n\r\nThis Crypto Addition v01 is limited to two addends and a sum.\r\n\r\nThe output is the number values [anum bnum cnum] given [astr, bstr,cstr].\r\n\r\nIf there is no solution the output should be [ ], empty set.\r\n\r\nI am planning v02 thru v05 with additional complications.\r\n\r\nv02 will have embedded numbers.\r\n\r\nv03 will allow up to ten addends vs the current two.\r\n\r\nA solid recursive routine can crack each test case in under 75 msec. ","description_html":"\u003cp\u003eCrypto addition is solving for the numeric values of the characters given in an addition equation:\u003c/p\u003e\u003cpre\u003e   abc       756\r\n  +def       342\r\n  ----      ----\r\n  ghij      1098\u003c/pre\u003e\u003cp\u003eInput [ 'abc' , 'def' , 'ghij' ] Output [ 756 342 1098 ] or [ ] if no solution\u003c/p\u003e\u003cp\u003eAll characters have unique values.\u003c/p\u003e\u003cp\u003eLeading zeros are not allowed.\u003c/p\u003e\u003cp\u003eThis Crypto Addition v01 is limited to two addends and a sum.\u003c/p\u003e\u003cp\u003eThe output is the number values [anum bnum cnum] given [astr, bstr,cstr].\u003c/p\u003e\u003cp\u003eIf there is no solution the output should be [ ], empty set.\u003c/p\u003e\u003cp\u003eI am planning v02 thru v05 with additional complications.\u003c/p\u003e\u003cp\u003ev02 will have embedded numbers.\u003c/p\u003e\u003cp\u003ev03 will allow up to ten addends vs the current two.\u003c/p\u003e\u003cp\u003eA solid recursive routine can crack each test case in under 75 msec.\u003c/p\u003e","function_template":"function y = crypto_add(astr,bstr,cstr)\r\n\r\n a=double(astr); % Gives ascii values in a vector\r\n b=double(bstr);\r\n c=double(cstr);\r\n\r\n y = [a b c]; % or y=[];\r\nend","test_suite":"%%\r\n    astr='a'; % []\r\n    bstr='a';\r\n    cstr='aa';\r\n  y_correct = [];\r\nassert(isequal(crypto_add(astr,bstr,cstr),y_correct))\r\n%%\r\n    astr='ab'; % []\r\n    bstr='ab';\r\n    cstr='ab';\r\n  y_correct = [];\r\nassert(isequal(crypto_add(astr,bstr,cstr),y_correct))\r\n%%\r\n   astr='abcdefghij'; %[]\r\n   bstr='abcdefghij';\r\n   cstr='aabcdefghij';\r\n  y_correct = [];\r\nassert(isequal(crypto_add(astr,bstr,cstr),y_correct))\r\n%%\r\n   astr='coca'; % 8186\r\n   bstr='cola'; % 8106\r\n   cstr='oasis'; % 16292\r\ny_correct = [8186 8106 16292];\r\nassert(isequal(crypto_add(astr,bstr,cstr),y_correct))\r\n%%\r\n   astr='xxx';% 999\r\n   bstr='b';  %   1\r\n   cstr='baaa'; % 1000\r\ny_correct = [999 1 1000];\r\nassert(isequal(crypto_add(astr,bstr,cstr),y_correct))\r\n%%\r\n   astr='ma';  % 89 9 98\r\n   bstr='a';\r\n   cstr='am'\r\ny_correct = [89 9 98];\r\nassert(isequal(crypto_add(astr,bstr,cstr),y_correct))\r\n%%\r\n   astr='bmtran'; % 951740 or  651740\r\n   bstr='winner'; % 630087 y   930087\r\n   cstr='tmeteor';% 1581827   1581827\r\ny_correct1 = [951740 630087 1581827];\r\ny_correct2 = [651740 930087 1581827];\r\ny_out=crypto_add(astr,bstr,cstr)\r\nsolved= isequal(y_out,y_correct1) || isequal(y_out,y_correct2)\r\nassert(solved)\r\n%%\r\n% For those that like to hard code answers\r\nasum=0;\r\nwhile asum\u003c1234567890\r\n avec=randperm(10)-1\r\n asum=polyval(avec,10);\r\nend\r\n\r\nbsum=0;\r\nwhile bsum\u003c1234567890\r\n bvec=randperm(10)-1\r\n bsum=polyval(bvec,10);\r\nend\r\n\r\ncsum=asum+bsum;\r\n%csum=csum+20000000000; %To create []\r\ncvec=int2str(csum)-'0' % separate puzzle question\r\n\r\nalpha_vec=randperm(10); % randomize the alphabet\r\nalpha='abcdefghij';\r\n\r\nastr='';\r\nfor i=1:length(avec)\r\n astr=[astr alpha(alpha_vec(avec(i)+1))];\r\nend\r\n\r\nbstr='';\r\nfor i=1:length(bvec)\r\n bstr=[bstr alpha(alpha_vec(bvec(i)+1))];\r\nend\r\n\r\ncstr='';\r\nfor i=1:length(cvec)\r\n cstr=[cstr alpha(alpha_vec(cvec(i)+1))];\r\nend\r\nastr\r\nbstr\r\ncstr\r\ny_correct = [asum bsum csum];\r\nassert(isequal(crypto_add(astr,bstr,cstr),y_correct))\r\n%%\r\n% For those that like to hard code answers\r\nasum=0;\r\nwhile asum\u003c1234567890\r\n avec=randperm(10)-1\r\n asum=polyval(avec,10);\r\nend\r\n\r\nbsum=0;\r\nwhile bsum\u003c1234567890\r\n bvec=randperm(10)-1\r\n bsum=polyval(bvec,10);\r\nend\r\n\r\ncsum=asum+bsum;\r\ncsum=csum+20000000000; %To create []\r\ncvec=int2str(csum)-'0' % separate puzzle question\r\n\r\nalpha_vec=randperm(10); % randomize the alphabet\r\nalpha='abcdefghij';\r\n\r\nastr='';\r\nfor i=1:length(avec)\r\n astr=[astr alpha(alpha_vec(avec(i)+1))];\r\nend\r\n\r\nbstr='';\r\nfor i=1:length(bvec)\r\n bstr=[bstr alpha(alpha_vec(bvec(i)+1))];\r\nend\r\n\r\ncstr='';\r\nfor i=1:length(cvec)\r\n cstr=[cstr alpha(alpha_vec(cvec(i)+1))];\r\nend\r\n\r\ny_correct = [];\r\nassert(isequal(crypto_add(astr,bstr,cstr),y_correct))\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":6,"test_suite_updated_at":"2012-05-05T22:30:58.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2012-05-04T20:33:57.000Z","updated_at":"2012-05-05T22:42:29.000Z","published_at":"2012-05-04T22:11:29.000Z","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\u003eCrypto addition is solving for the numeric values of the characters given in an addition equation:\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[   abc       756\\n  +def       342\\n  ----      ----\\n  ghij      1098]]\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\u003eInput [ 'abc' , 'def' , 'ghij' ] Output [ 756 342 1098 ] or [ ] if no solution\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 characters have unique values.\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\u003eLeading zeros are not allowed.\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\u003eThis Crypto Addition v01 is limited to two addends and a sum.\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 output is the number values [anum bnum cnum] given [astr, bstr,cstr].\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 there is no solution the output should be [ ], empty set.\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\u003eI am planning v02 thru v05 with additional complications.\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\u003ev02 will have embedded numbers.\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\u003ev03 will allow up to ten addends vs the current two.\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\u003eA solid recursive routine can crack each test case in under 75 msec.\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\"}]}"}],"term":"tag:\"recursion crypto vector string math\"","current_player_id":null,"fields":[{"name":"page","type":"integer","callback":null,"default":1,"directive":null,"facet":null,"facet_method":"and","operator":null,"param":null,"static":null,"prepend":true},{"name":"per_page","type":"integer","callback":null,"default":50,"directive":null,"facet":null,"facet_method":"and","operator":null,"param":null,"static":null,"prepend":true},{"name":"sort","type":"string","callback":null,"default":null,"directive":null,"facet":null,"facet_method":"and","operator":null,"param":null,"static":null,"prepend":true},{"name":"body","type":"text","callback":null,"default":"*:*","directive":null,"facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":false},{"name":"group","type":"string","callback":null,"default":null,"directive":"group","facet":true,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"difficulty_rating_bin","type":"string","callback":null,"default":null,"directive":"difficulty_rating_bin","facet":true,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"id","type":"integer","callback":null,"default":null,"directive":"id","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"tag","type":"string","callback":null,"default":null,"directive":"tag","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"product","type":"string","callback":null,"default":null,"directive":"product","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"created_at","type":"timeframe","callback":{},"default":null,"directive":"created_at","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"profile_id","type":"integer","callback":null,"default":null,"directive":"author_id","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"created_by","type":"string","callback":null,"default":null,"directive":"author","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"player_id","type":"integer","callback":null,"default":null,"directive":"solver_id","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"player","type":"string","callback":null,"default":null,"directive":"solver","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"solvers_count","type":"integer","callback":null,"default":null,"directive":"solvers_count","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"comments_count","type":"integer","callback":null,"default":null,"directive":"comments_count","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"likes_count","type":"integer","callback":null,"default":null,"directive":"likes_count","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"leader_id","type":"integer","callback":null,"default":null,"directive":"leader_id","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true},{"name":"leading_solution","type":"integer","callback":null,"default":null,"directive":"leading_solution","facet":null,"facet_method":"and","operator":null,"param":"term","static":null,"prepend":true}],"filters":[{"name":"asset_type","type":"string","callback":null,"default":null,"directive":null,"facet":null,"facet_method":"and","operator":null,"param":null,"static":"\"cody:problem\"","prepend":true},{"name":"profile_id","type":"integer","callback":{},"default":null,"directive":null,"facet":null,"facet_method":"and","operator":null,"param":"author_id","static":null,"prepend":true}],"query":{"params":{"per_page":50,"term":"tag:\"recursion crypto vector string math\"","current_player":null,"sort":"map(difficulty_value,0,0,999) asc"},"parser":"MathWorks::Search::Solr::QueryParser","directives":{"term":{"directives":{"tag":[["tag:\"recursion crypto vector string math\"","","\"","recursion crypto vector string math","\""]]}}},"facets":{"#\u003cMathWorks::Search::Field:0x00007f3034aa6cc8\u003e":null,"#\u003cMathWorks::Search::Field:0x00007f3034aa6c28\u003e":null},"filters":{"#\u003cMathWorks::Search::Field:0x00007f3034aa6368\u003e":"\"cody:problem\""},"fields":{"#\u003cMathWorks::Search::Field:0x00007f3034aa6f48\u003e":1,"#\u003cMathWorks::Search::Field:0x00007f3034aa6ea8\u003e":50,"#\u003cMathWorks::Search::Field:0x00007f3034aa6e08\u003e":"map(difficulty_value,0,0,999) asc","#\u003cMathWorks::Search::Field:0x00007f3034aa6d68\u003e":"tag:\"recursion crypto vector string math\""},"user_query":{"#\u003cMathWorks::Search::Field:0x00007f3034aa6d68\u003e":"tag:\"recursion crypto vector string math\""},"queried_facets":{}},"query_backend":{"connection":{"configuration":{"index_url":"http://index-op-v2/solr/","query_url":"http://search-op-v2/solr/","direct_access_index_urls":["http://index-op-v2/solr/"],"direct_access_query_urls":["http://search-op-v2/solr/"],"timeout":10,"vhost":"search","exchange":"search.topic","heartbeat":30,"pre_index_mode":false,"host":"rabbitmq-eks","port":5672,"username":"cody-search","password":"78X075ddcV44","virtual_host":"search","indexer":"amqp","http_logging":"true","core":"cody"},"query_connection":{"uri":"http://search-op-v2/solr/cody/","proxy":null,"connection":{"parallel_manager":null,"headers":{"User-Agent":"Faraday v1.0.1"},"params":{},"options":{"params_encoder":"Faraday::FlatParamsEncoder","proxy":null,"bind":null,"timeout":null,"open_timeout":null,"read_timeout":null,"write_timeout":null,"boundary":null,"oauth":null,"context":null,"on_data":null},"ssl":{"verify":true,"ca_file":null,"ca_path":null,"verify_mode":null,"cert_store":null,"client_cert":null,"client_key":null,"certificate":null,"private_key":null,"verify_depth":null,"version":null,"min_version":null,"max_version":null},"default_parallel_manager":null,"builder":{"adapter":{"name":"Faraday::Adapter::NetHttp","args":[],"block":null},"handlers":[{"name":"Faraday::Response::RaiseError","args":[],"block":null}],"app":{"app":{"ssl_cert_store":{"verify_callback":null,"error":null,"error_string":null,"chain":null,"time":null},"app":{},"connection_options":{},"config_block":null}}},"url_prefix":"http://search-op-v2/solr/cody/","manual_proxy":false,"proxy":null},"update_format":"RSolr::JSON::Generator","update_path":"update","options":{"url":"http://search-op-v2/solr/cody"}}},"query":{"params":{"per_page":50,"term":"tag:\"recursion crypto vector string math\"","current_player":null,"sort":"map(difficulty_value,0,0,999) asc"},"parser":"MathWorks::Search::Solr::QueryParser","directives":{"term":{"directives":{"tag":[["tag:\"recursion crypto vector string math\"","","\"","recursion crypto vector string math","\""]]}}},"facets":{"#\u003cMathWorks::Search::Field:0x00007f3034aa6cc8\u003e":null,"#\u003cMathWorks::Search::Field:0x00007f3034aa6c28\u003e":null},"filters":{"#\u003cMathWorks::Search::Field:0x00007f3034aa6368\u003e":"\"cody:problem\""},"fields":{"#\u003cMathWorks::Search::Field:0x00007f3034aa6f48\u003e":1,"#\u003cMathWorks::Search::Field:0x00007f3034aa6ea8\u003e":50,"#\u003cMathWorks::Search::Field:0x00007f3034aa6e08\u003e":"map(difficulty_value,0,0,999) asc","#\u003cMathWorks::Search::Field:0x00007f3034aa6d68\u003e":"tag:\"recursion crypto vector string math\""},"user_query":{"#\u003cMathWorks::Search::Field:0x00007f3034aa6d68\u003e":"tag:\"recursion crypto vector string math\""},"queried_facets":{}},"options":{"fields":["id","difficulty_rating"]},"join":" "},"results":[{"id":663,"difficulty_rating":"unrated"}]}}