LXR linux/tools/perf/util/levenshtein.c

   1#include "levenshtein.h"
   2#include <errno.h>
   3#include <stdlib.h>
   4#include <string.h>
   5
   6/*
   7 * This function implements the Damerau-Levenshtein algorithm to
   8 * calculate a distance between strings.
   9 *
  10 * Basically, it says how many letters need to be swapped, substituted,
  11 * deleted from, or added to string1, at least, to get string2.
  12 *
  13 * The idea is to build a distance matrix for the substrings of both
  14 * strings.  To avoid a large space complexity, only the last three rows
  15 * are kept in memory (if swaps had the same or higher cost as one deletion
  16 * plus one insertion, only two rows would be needed).
  17 *
  18 * At any stage, "i + 1" denotes the length of the current substring of
  19 * string1 that the distance is calculated for.
  20 *
  21 * row2 holds the current row, row1 the previous row (i.e. for the substring
  22 * of string1 of length "i"), and row0 the row before that.
  23 *
  24 * In other words, at the start of the big loop, row2[j + 1] contains the
  25 * Damerau-Levenshtein distance between the substring of string1 of length
  26 * "i" and the substring of string2 of length "j + 1".
  27 *
  28 * All the big loop does is determine the partial minimum-cost paths.
  29 *
  30 * It does so by calculating the costs of the path ending in characters
  31 * i (in string1) and j (in string2), respectively, given that the last
  32 * operation is a substition, a swap, a deletion, or an insertion.
  33 *
  34 * This implementation allows the costs to be weighted:
  35 *
  36 * - w (as in "sWap")
  37 * - s (as in "Substitution")
  38 * - a (for insertion, AKA "Add")
  39 * - d (as in "Deletion")
  40 *
  41 * Note that this algorithm calculates a distance _iff_ d == a.
  42 */
  43int levenshtein(const char *string1, const char *string2,
  44                int w, int s, int a, int d)
  45{
  46        int len1 = strlen(string1), len2 = strlen(string2);
  47        int *row0 = malloc(sizeof(int) * (len2 + 1));
  48        int *row1 = malloc(sizeof(int) * (len2 + 1));
  49        int *row2 = malloc(sizeof(int) * (len2 + 1));
  50        int i, j;
  51
  52        for (j = 0; j <= len2; j++)
  53                row1[j] = j * a;
  54        for (i = 0; i < len1; i++) {
  55                int *dummy;
  56
  57                row2[0] = (i + 1) * d;
  58                for (j = 0; j < len2; j++) {
  59                        /* substitution */
  60                        row2[j + 1] = row1[j] + s * (string1[i] != string2[j]);
  61                        /* swap */
  62                        if (i > 0 && j > 0 && string1[i - 1] == string2[j] &&
  63                                        string1[i] == string2[j - 1] &&
  64                                        row2[j + 1] > row0[j - 1] + w)
  65                                row2[j + 1] = row0[j - 1] + w;
  66                        /* deletion */
  67                        if (row2[j + 1] > row1[j + 1] + d)
  68                                row2[j + 1] = row1[j + 1] + d;
  69                        /* insertion */
  70                        if (row2[j + 1] > row2[j] + a)
  71                                row2[j + 1] = row2[j] + a;
  72                }
  73
  74                dummy = row0;
  75                row0 = row1;
  76                row1 = row2;
  77                row2 = dummy;
  78        }
  79
  80        i = row1[len2];
  81        free(row0);
  82        free(row1);
  83        free(row2);
  84
  85        return i;
  86}
  87