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