### intptr's blog

By intptr, history, 4 years ago, , 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 == s2) 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 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 is used to build s2
for d in range(min(D+1, len(s2))):
# d is the position where s1
# might be used.
# it is also the number of character
# that are required to be inserted before
# using s1[d].
if s1 == 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! Comments (1)
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