Root/tools/perf/util/levenshtein.c

1#include "cache.h"
2#include "levenshtein.h"
3
4/*
5 * This function implements the Damerau-Levenshtein algorithm to
6 * calculate a distance between strings.
7 *
8 * Basically, it says how many letters need to be swapped, substituted,
9 * deleted from, or added to string1, at least, to get string2.
10 *
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).
15 *
16 * At any stage, "i + 1" denotes the length of the current substring of
17 * string1 that the distance is calculated for.
18 *
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.
21 *
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".
25 *
26 * All the big loop does is determine the partial minimum-cost paths.
27 *
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.
31 *
32 * This implementation allows the costs to be weighted:
33 *
34 * - w (as in "sWap")
35 * - s (as in "Substitution")
36 * - a (for insertion, AKA "Add")
37 * - d (as in "Deletion")
38 *
39 * Note that this algorithm calculates a distance _iff_ d == a.
40 */
41int levenshtein(const char *string1, const char *string2,
42        int w, int s, int a, int d)
43{
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));
48    int i, j;
49
50    for (j = 0; j <= len2; j++)
51        row1[j] = j * a;
52    for (i = 0; i < len1; i++) {
53        int *dummy;
54
55        row2[0] = (i + 1) * d;
56        for (j = 0; j < len2; j++) {
57            /* substitution */
58            row2[j + 1] = row1[j] + s * (string1[i] != string2[j]);
59            /* swap */
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;
64            /* deletion */
65            if (row2[j + 1] > row1[j + 1] + d)
66                row2[j + 1] = row1[j + 1] + d;
67            /* insertion */
68            if (row2[j + 1] > row2[j] + a)
69                row2[j + 1] = row2[j] + a;
70        }
71
72        dummy = row0;
73        row0 = row1;
74        row1 = row2;
75        row2 = dummy;
76    }
77
78    i = row1[len2];
79    free(row0);
80    free(row1);
81    free(row2);
82
83    return i;
84}
85

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