tools/perf/util/levenshtein.c

   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 }