Pattern match question

*To*: mathgroup at smc.vnet.net*Subject*: [mg68385] Pattern match question*From*: glassymeow at yahoo.com*Date*: Thu, 3 Aug 2006 06:07:13 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

hi txt = "ZACCZBNRCSAACXBXX"; letters = "ABC"; i want to find the first occurrences of any of the six combinations of the letters of the set "ABC" Globally, and without overlap option. and the space between letters does not important. in the above txt string the result must be: Out[]:= ACCZB CSAACXB i wish a solution using mathematica regular expressions. the Regex pattern (A|B|C).*?(A|B|C).*?(A|B|C) will give the out: ACC , BNRCSA , ACXB because it considers the permutations and not the combinations the following is an old fashion program which will emulate the human pencil and paper method, will solve the problem, but i am sure there are a better solutions. txt = "ZACCZBNRCSAACXBXX"; letters = "ABC"; ptrnLtrs = ""; (* make a string of 26 zero's as the number of the alphbet*) For[i = 1, i <= 26, ptrnLtrs = StringJoin[ptrnLtrs, "0"]; i++] (* replace every letter of the pattern letters *) (* with a corresponding 1 in the string of the zero's *) For[i = 1, i <= StringLength[letters], num = ToCharacterCode[StringTake[letters, {i, i}]]; num = num - 64; ptrnLtrs = StringReplacePart[ptrnLtrs, "1", Flatten[{num, num}]]; i++]; (* the procedural pattern match *) ptrnLtrsBak = ptrnLtrs; y = 0; (* backup for the ptrnLtrs *) beginFlag = 0; result = ""; lst = {}; For[i = 1, i <= StringLength[txt], OneLetter = StringTake[txt, {i, i}]; If[beginFlag == 0 && StringCases[letters, OneLetter] == {}, Goto[jmp]]; num = ToCharacterCode[StringTake[txt, {i, i}]] - 64; If[StringTake[ptrnLtrs, num] == "1", result = StringJoin[result, OneLetter]; ptrnLtrs = StringReplacePart[ptrnLtrs, "0", Flatten[{num, num}]]; , result = StringJoin[result, OneLetter];]; beginFlag = 1; If[ToExpression[ptrnLtrs] == 0, ptrnLtrs = ptrnLtrsBak; Print[result]; result = ""; beginFlag = 0;]; Label[jmp]; i++] Out[]:= ACCZB CSAACXB regards peter glassy