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Building a Levenstein Automata for fast string searching

Revision en2, by intptr, 2015-10-03 17:36:52

Paul Masurel offers a great article about the Levenstein automata : link

He offers an algorithm to compute if the levenstein distance between two strings is less than or equal to D.

The naive algorithm is fairly straightforward :

def levenshtein(s1, s2, D=2):
    Returns True iff the two string
    s1 and s2 is lesser or equal to D
    if D == -1:
        return False
    if len(s1) < len(s2):
        return levenshtein(s2, s1)
    if len(s2) == 0:
        return len(s1) <= D
    return (levenshtein(s1[1:], s2[1:], D-1)   # substitution\
        or levenshtein(s1, s2[1:], D-1)       # insertion\
        or levenshtein(s1[1:], s2, D-1)       # deletion\
        or (
            # character match
            (s1[0] == s2[0]) and \
            levenshtein(s1[1:], s2[1:], D)

This basically tries all possibilities which is with exponential complexity if we don't save any of the calls.

He offers a better algorithm but for the last week I have been trying to understand how it works, here it is :

def levenshtein(s1, s2, D=2): """ Returns True iff the edit distance between the two strings s1 and s2 is lesser or equal to D """ if len(s1) == 0: return len(s2) <= D if len(s2) == 0: return len(s1) <= D # assuming s1[0] is NOT used to build s2, if D > 0: if levenshtein(s1[1:], s2, D - 1): # deletion return True if levenshtein(s1[1:], s2[1:], D - 1): # substitution return True # assuming s1[0] is used to build s2 for d in range(min(D+1, len(s2))): # d is the position where s1[0] # might be used. # it is also the number of character # that are required to be inserted before # using s1[d]. if s1[0] == s2[d]: if levenshtein(s1[1:], s2[d+1:], D - d): return True return False

It would be great if anyone can shed some light on how this works! Thanks!


  Rev. Lang. By When Δ Comment
en2 English intptr 2015-10-03 17:36:52 108
en1 English intptr 2015-10-03 17:36:07 2089 Initial revision (published)