2 #include "levenshtein.h"
5 * This function implements the Damerau-Levenshtein algorithm to
6 * calculate a distance between strings.
8 * Basically, it says how many letters need to be swapped, substituted,
9 * deleted from, or added to string1, at least, to get string2.
11 * The idea is to build a distance matrix for the substrings of both
12 * strings. To avoid a large space complexity, only the last three rows
13 * are kept in memory (if swaps had the same or higher cost as one deletion
14 * plus one insertion, only two rows would be needed).
16 * At any stage, "i + 1" denotes the length of the current substring of
17 * string1 that the distance is calculated for.
19 * row2 holds the current row, row1 the previous row (i.e. for the substring
20 * of string1 of length "i"), and row0 the row before that.
22 * In other words, at the start of the big loop, row2[j + 1] contains the
23 * Damerau-Levenshtein distance between the substring of string1 of length
24 * "i" and the substring of string2 of length "j + 1".
26 * All the big loop does is determine the partial minimum-cost paths.
28 * It does so by calculating the costs of the path ending in characters
29 * i (in string1) and j (in string2), respectively, given that the last
30 * operation is a substition, a swap, a deletion, or an insertion.
32 * This implementation allows the costs to be weighted:
35 * - s (as in "Substitution")
36 * - a (for insertion, AKA "Add")
37 * - d (as in "Deletion")
39 * Note that this algorithm calculates a distance _iff_ d == a.
41 int levenshtein(const char *string1
, const char *string2
,
42 int w
, int s
, int a
, int d
)
44 int len1
= strlen(string1
), len2
= strlen(string2
);
45 int *row0
= malloc(sizeof(int) * (len2
+ 1));
46 int *row1
= malloc(sizeof(int) * (len2
+ 1));
47 int *row2
= malloc(sizeof(int) * (len2
+ 1));
50 for (j
= 0; j
<= len2
; j
++)
52 for (i
= 0; i
< len1
; i
++) {
55 row2
[0] = (i
+ 1) * d
;
56 for (j
= 0; j
< len2
; j
++) {
58 row2
[j
+ 1] = row1
[j
] + s
* (string1
[i
] != string2
[j
]);
60 if (i
> 0 && j
> 0 && string1
[i
- 1] == string2
[j
] &&
61 string1
[i
] == string2
[j
- 1] &&
62 row2
[j
+ 1] > row0
[j
- 1] + w
)
63 row2
[j
+ 1] = row0
[j
- 1] + w
;
65 if (row2
[j
+ 1] > row1
[j
+ 1] + d
)
66 row2
[j
+ 1] = row1
[j
+ 1] + d
;
68 if (row2
[j
+ 1] > row2
[j
] + a
)
69 row2
[j
+ 1] = row2
[j
] + a
;
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