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 */ 43 int 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