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 } 88