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1// SPDX-License-Identifier: GPL-2.02#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 to9 * 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 both15 * strings. To avoid a large space complexity, only the last three rows16 * are kept in memory (if swaps had the same or higher cost as one deletion17 * plus one insertion, only two rows would be needed).18 *19 * At any stage, "i + 1" denotes the length of the current substring of20 * string1 that the distance is calculated for.21 *22 * row2 holds the current row, row1 the previous row (i.e. for the substring23 * 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 the26 * Damerau-Levenshtein distance between the substring of string1 of length27 * "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 characters32 * i (in string1) and j (in string2), respectively, given that the last33 * 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