stash: remove now superfluos help for "stash push"
[git.git] / levenshtein.c
blobd2632690d5107b53ee8a7ac4832cd85eb8c7bfc1
1 #include "cache.h"
2 #include "levenshtein.h"
4 /*
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 substitution, a swap, a deletion, or an insertion.
32 * This implementation allows the costs to be weighted:
34 * - w (as in "sWap")
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, *row1, *row2;
46 int i, j;
48 ALLOC_ARRAY(row0, len2 + 1);
49 ALLOC_ARRAY(row1, len2 + 1);
50 ALLOC_ARRAY(row2, len2 + 1);
52 for (j = 0; j <= len2; j++)
53 row1[j] = j * a;
54 for (i = 0; i < len1; i++) {
55 int *dummy;
57 row2[0] = (i + 1) * d;
58 for (j = 0; j < len2; j++) {
59 /* substitution */
60 row2[j + 1] = row1[j] + s * (string1[i] != string2[j]);
61 /* swap */
62 if (i > 0 && j > 0 && string1[i - 1] == string2[j] &&
63 string1[i] == string2[j - 1] &&
64 row2[j + 1] > row0[j - 1] + w)
65 row2[j + 1] = row0[j - 1] + w;
66 /* deletion */
67 if (row2[j + 1] > row1[j + 1] + d)
68 row2[j + 1] = row1[j + 1] + d;
69 /* insertion */
70 if (row2[j + 1] > row2[j] + a)
71 row2[j + 1] = row2[j] + a;
74 dummy = row0;
75 row0 = row1;
76 row1 = row2;
77 row2 = dummy;
80 i = row1[len2];
81 free(row0);
82 free(row1);
83 free(row2);
85 return i;