xref: /openbmc/linux/tools/perf/util/levenshtein.c (revision 4d39c89f)
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 substitution, 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  */
levenshtein(const char * string1,const char * string2,int w,int s,int a,int d)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 }
88