Levenshtein distance

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In information theory and computer science, the Levenshtein distance or edit distance between two strings is given by the minimum number of operations needed to transform one string into the other, where an operation is an insertion, deletion, or substitution of a single character. It is named after Vladimir Levenshtein, who considered this distance in 1965. It is useful in applications that need to determine how similar two strings are, such as spell checkers.

For example, the Levenshtein distance between "kitten" and "sitting" is 3, since these three edits change one into the other, and there is no way to do it with fewer than three edits:

kitten → sitten (substitution of 'k' for 's')
sitten → sittin (substitution of 'e' for 'i')
sittin → sitting (insert 'g' at the end)

It can be considered a generalization of the Hamming distance, which is used for strings of the same length and only considers substitution edits. There are also further generalizations of the Levenshtein distance that consider, for example, exchanging two characters as an operation, like in the Damerau-Levenshtein distance algorithm.

1 The algorithm
2 See also
3 External links

[edit] The algorithm

A commonly-used bottom-up dynamic programming algorithm for computing the Levenshtein distance involves the use of an (n + 1) × (m + 1) matrix, where n and m are the lengths of the two strings. Here is pseudocode for a function LevenshteinDistance that takes two strings, s of length m, and t of length n, and computes the Levenshtein distance between them:

int LevenshteinDistance(char s[1..m], char t[1..n])
   // d is a table with m+1 rows and n+1 columns
   declare int d[0..m, 0..n]
 
   for i from 0 to m
       d[i, 0] := i
   for j from 1 to n
       d[0, j] := j
 
   for i from 1 to m
       for j from 1 to n
           if s[i] = t[j] then cost := 0
                          else cost := 1
           d[i, j] := minimum(
                                d[i-1, j] + 1,     // deletion
                                d[i, j-1] + 1,     // insertion
                                d[i-1, j-1] + cost   // substitution
                            )
 
   return d[m, n]

Two examples of the resulting matrix (the minimum steps to be taken are highlighted):

		k	i	t	t	e	n
	0	1	2	3	4	5	6
s	1	1	2	3	4	5	6
i	2	2	1	2	3	4	5
t	3	3	2	1	2	3	4
t	4	4	3	2	1	2	3
i	5	5	4	3	2	2	3
n	6	6	5	4	3	3	2
g	7	7	6	5	4	4	3

		S	a	t	u	r	d	a	y
	0	1	2	3	4	5	6	7	8
S	1	0	1	2	3	4	5	6	7
u	2	1	1	2	2	3	4	5	6
n	3	2	2	2	3	3	4	5	6
d	4	3	3	3	3	4	3	4	5
a	5	4	3	4	4	4	4	3	4
y	6	5	4	4	5	5	5	4	3

The invariant maintained throughout the algorithm is that we can transform the initial segment s[1..i] into t[1..j] using a minimum of d[i,j] operations. At the end, the bottom-right element of the array contains the answer. This algorithm is essentially part of a solution to the Longest common subsequence problem (LCS), in the particular case of 2 input lists.

[edit] Possible improvements

Possible improvements to this algorithm include:

We can adapt the algorithm to use less space, O(m) instead of O(mn), since it only requires that the previous row and current row be stored at any one time.
We can store the number of insertions, deletions, and substitutions separately, or even the positions at which they occur, which is always j.
We can give different penalty costs to insertion, deletion and substitution.
The initialization of d[i,0] can be moved inside the main outer loop.
This algorithm parallelizes poorly, due to a large number of data dependencies. However, all the costs can be computed in parallel, and the algorithm can be adapted to perform the minimum function in phases to eliminate dependencies.

[edit] Proof of correctness

As mentioned earlier, the invariant is that we can transform the initial segment s[1..i] into t[1..j] using a minimum of d[i,j] operations. This invariant holds since:

It is initially true on row and column 0 because s[1..i] can be transformed into the empty string t[1..0] by simply dropping all i characters. Similarly, we can transform s[1..0] to t[1..j] by simply adding all j characters.
The minimum is taken over three distances, each of which is feasible:
- If we can transform s[1..i] to t[1..j-1] in k operations, then we can simply add t[j] afterwards to get t[1..j] in k+1 operations.
- If we can transform s[1..i-1] to t[1..j] in k operations, then we can do the same operations on s[1..i] and then remove the original s[i] at the end in k+1 operations.
- If we can transform s[1..i-1] to t[1..j-1] in k operations, we can do the same to s[1..i] and then do a substitution of t[j] for the original s[i] at the end if necessary, requiring k+cost operations.
The operations required to transform s[1..n] into t[1..m] is of course the number required to transform all of s into all of t, and so d[n,m] holds our result.

This proof fails to validate that the number placed in d[i,j] is in fact minimal; this is more difficult to show, and involves an argument by contradiction in which we assume d[i,j] is smaller than the minimum of the three, and use this to show one of the three is not minimal.

[edit] Upper and lower bounds

The Levenshtein distance has several simple upper and lower bounds that are useful in applications which compute many of them and compare them. These include:

It is always at least the difference of the sizes of the two strings.
It is at most the length of the longer string.
It is zero if and only if the strings are identical.
If the strings are the same size, the Hamming distance is an upper bound on the Levenshtein distance.
If the strings are called s and t, the number of characters (not counting duplicates) found in s but not in t is a lower bound.