1 // SPDX-License-Identifier: GPL-2.0
2 #include "levenshtein.h"
3 #include <errno.h>
4 #include <stdlib.h>
5 #include <string.h>
6
7 /*
8 * This function implements the Damerau-Levenshtein algorithm to
9 * calculate a distance between strings.
10 *
11 * Basically, it says how many letters need to be swapped, substituted,
12 * deleted from, or added to string1, at least, to get string2.
13 *
14 * The idea is to build a distance matrix for the substrings of both
15 * strings. To avoid a large space complexity, only the last three rows
16 * are kept in memory (if swaps had the same or higher cost as one deletion
17 * plus one insertion, only two rows would be needed).
18 *
19 * At any stage, "i + 1" denotes the length of the current substring of
20 * string1 that the distance is calculated for.
21 *
22 * row2 holds the current row, row1 the previous row (i.e. for the substring
23 * of string1 of length "i"), and row0 the row before that.
24 *
25 * In other words, at the start of the big loop, row2[j + 1] contains the
26 * Damerau-Levenshtein distance between the substring of string1 of length
27 * "i" and the substring of string2 of length "j + 1".
28 *
29 * All the big loop does is determine the partial minimum-cost paths.
30 *
31 * It does so by calculating the costs of the path ending in characters
32 * i (in string1) and j (in string2), respectively, given that the last
33 * operation is a substition, a swap, a deletion, or an insertion.
34 *
35 * This implementation allows the costs to be weighted:
36 *
37 * - w (as in "sWap")
38 * - s (as in "Substitution")
39 * - a (for insertion, AKA "Add")
40 * - d (as in "Deletion")
41 *
42 * Note that this algorithm calculates a distance _iff_ d == a.
43 */
44 int levenshtein(const char *string1, const char *string2,
45 int w, int s, int a, int d)
46 {
47 int len1 = strlen(string1), len2 = strlen(string2);
48 int *row0 = malloc(sizeof(int) * (len2 + 1));
49 int *row1 = malloc(sizeof(int) * (len2 + 1));
50 int *row2 = malloc(sizeof(int) * (len2 + 1));
51 int i, j;
52
53 for (j = 0; j <= len2; j++)
54 row1[j] = j * a;
55 for (i = 0; i < len1; i++) {
56 int *dummy;
57
58 row2[0] = (i + 1) * d;
59 for (j = 0; j < len2; j++) {
60 /* substitution */
61 row2[j + 1] = row1[j] + s * (string1[i] != string2[j]);
62 /* swap */
63 if (i > 0 && j > 0 && string1[i - 1] == string2[j] &&
64 string1[i] == string2[j - 1] &&
65 row2[j + 1] > row0[j - 1] + w)
66 row2[j + 1] = row0[j - 1] + w;
67 /* deletion */
68 if (row2[j + 1] > row1[j + 1] + d)
69 row2[j + 1] = row1[j + 1] + d;
70 /* insertion */
71 if (row2[j + 1] > row2[j] + a)
72 row2[j + 1] = row2[j] + a;
73 }
74
75 dummy = row0;
76 row0 = row1;
77 row1 = row2;
78 row2 = dummy;
79 }
80
81 i = row1[len2];
82 free(row0);
83 free(row1);
84 free(row2);
85
86 return i;
87 }