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How to find a position of a substring within a string with fuzzy match


I have come across a problem of matching a string in an OCR recognized text and find the position of it considering there can be arbitrary tolerance of wrong, missing or extra characters. The result should be a best match position, possibly (not necessarily) with length of matching substring.

For example:

String: 9912, 1.What is your name?
Substring: 1. What is your name?
Tolerance: 1
Result: match on character 7

String: Where is our caat if any?
Substring: your cat
Tolerance: 2
Result: match on character 10

String: Tolerance is t0o h1gh.
Substring: Tolerance is too high;
Tolerance: 1
Result: no match

I have tried to adapt Levenstein algorithm, but it doesn't work properly for substrings and doesn't return position.

Algorithm in Delphi would be preferred, yet any implementation or pseudo logic would do.


Solution

  • Here's a recursive implementation that works, but might not be fast enough. The worst case scenario is when a match can't be found, and all but the last char in "What" gets matched at every index in Where. In that case the algorithm will make Length(What)-1 + Tolerance comparasions for each char in Where, plus one recursive call per Tolerance. Since both Tolerance and the length of What are constnats, I'd say the algorithm is O(n). It's performance will degrade linearly with the length of both "What" and "Where".

    function BrouteFindFirst(What, Where:string; Tolerance:Integer; out AtIndex, OfLength:Integer):Boolean;
      var i:Integer;
          aLen:Integer;
          WhatLen, WhereLen:Integer;
    
        function BrouteCompare(wherePos, whatPos, Tolerance:Integer; out Len:Integer):Boolean;
        var aLen:Integer;
            aRecursiveLen:Integer;
        begin
          // Skip perfect match characters
          aLen := 0;
          while (whatPos <= WhatLen) and (wherePos <= WhereLen) and (What[whatPos] = Where[wherePos]) do
          begin
            Inc(aLen);
            Inc(wherePos);
            Inc(whatPos);
          end;
          // Did we find a match?
          if (whatPos > WhatLen) then
            begin
              Result := True;
              Len := aLen;
            end
          else if Tolerance = 0 then
            Result := False // No match and no more "wild cards"
          else
            begin
              // We'll make an recursive call to BrouteCompare, allowing for some tolerance in the string
              // matching algorithm.
              Dec(Tolerance); // use up one "wildcard"
              Inc(whatPos); // consider the current char matched
              if BrouteCompare(wherePos, whatPos, Tolerance, aRecursiveLen) then
                begin
                  Len := aLen + aRecursiveLen;
                  Result := True;
                end
              else if BrouteCompare(wherePos + 1, whatPos, Tolerance, aRecursiveLen) then
                begin
                  Len := aLen + aRecursiveLen;
                  Result := True;
                end
              else
                Result := False; // no luck!
            end;
        end;
    
      begin
    
        WhatLen := Length(What);
        WhereLen := Length(Where);
    
        for i:=1 to Length(Where) do
        begin
          if BrouteCompare(i, 1, Tolerance, aLen) then
          begin
            AtIndex := i;
            OfLength := aLen;
            Result := True;
            Exit;
          end;
        end;
    
        // No match found!
        Result := False;
    
      end;
    

    I've used the following code to test the function:

    procedure TForm18.Button1Click(Sender: TObject);
    var AtIndex, OfLength:Integer;
    begin
      if BrouteFindFirst(Edit2.Text, Edit1.Text, ComboBox1.ItemIndex, AtIndex, OfLength) then
        Label3.Caption := 'Found @' + IntToStr(AtIndex) + ', of length ' + IntToStr(OfLength)
      else
        Label3.Caption := 'Not found';
    end;
    

    For case:

    String: Where is our caat if any?
    Substring: your cat
    Tolerance: 2
    Result: match on character 10
    

    it shows a match on character 9, of length 6. For the other two examples it gives the expected result.