LXR linux/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 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 */
  44int 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