87 lines
2.6 KiB
C
87 lines
2.6 KiB
C
#include "levenshtein.h"
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#include <errno.h>
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#include <stdlib.h>
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#include <string.h>
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/*
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* This function implements the Damerau-Levenshtein algorithm to
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* calculate a distance between strings.
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*
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* Basically, it says how many letters need to be swapped, substituted,
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* deleted from, or added to string1, at least, to get string2.
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*
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* The idea is to build a distance matrix for the substrings of both
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* strings. To avoid a large space complexity, only the last three rows
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* are kept in memory (if swaps had the same or higher cost as one deletion
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* plus one insertion, only two rows would be needed).
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*
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* At any stage, "i + 1" denotes the length of the current substring of
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* string1 that the distance is calculated for.
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*
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* row2 holds the current row, row1 the previous row (i.e. for the substring
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* of string1 of length "i"), and row0 the row before that.
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*
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* In other words, at the start of the big loop, row2[j + 1] contains the
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* Damerau-Levenshtein distance between the substring of string1 of length
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* "i" and the substring of string2 of length "j + 1".
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*
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* All the big loop does is determine the partial minimum-cost paths.
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*
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* It does so by calculating the costs of the path ending in characters
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* i (in string1) and j (in string2), respectively, given that the last
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* operation is a substition, a swap, a deletion, or an insertion.
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*
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* This implementation allows the costs to be weighted:
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*
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* - w (as in "sWap")
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* - s (as in "Substitution")
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* - a (for insertion, AKA "Add")
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* - d (as in "Deletion")
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*
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* Note that this algorithm calculates a distance _iff_ d == a.
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*/
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int levenshtein(const char *string1, const char *string2,
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int w, int s, int a, int d)
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{
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int len1 = strlen(string1), len2 = strlen(string2);
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int *row0 = malloc(sizeof(int) * (len2 + 1));
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int *row1 = malloc(sizeof(int) * (len2 + 1));
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int *row2 = malloc(sizeof(int) * (len2 + 1));
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int i, j;
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for (j = 0; j <= len2; j++)
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row1[j] = j * a;
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for (i = 0; i < len1; i++) {
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int *dummy;
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row2[0] = (i + 1) * d;
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for (j = 0; j < len2; j++) {
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/* substitution */
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row2[j + 1] = row1[j] + s * (string1[i] != string2[j]);
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/* swap */
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if (i > 0 && j > 0 && string1[i - 1] == string2[j] &&
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string1[i] == string2[j - 1] &&
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row2[j + 1] > row0[j - 1] + w)
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row2[j + 1] = row0[j - 1] + w;
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/* deletion */
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if (row2[j + 1] > row1[j + 1] + d)
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row2[j + 1] = row1[j + 1] + d;
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/* insertion */
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if (row2[j + 1] > row2[j] + a)
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row2[j + 1] = row2[j] + a;
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}
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dummy = row0;
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row0 = row1;
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row1 = row2;
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row2 = dummy;
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}
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i = row1[len2];
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free(row0);
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free(row1);
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free(row2);
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return i;
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}
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