Linux Kernel  3.7.1
 All Data Structures Namespaces Files Functions Variables Typedefs Enumerations Enumerator Macros Groups Pages
levenshtein.c
Go to the documentation of this file.
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  */
41 int 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 }