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 substitution, 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 */ 41intlevenshtein(const char*string1,const char*string2, 42int w,int s,int a,int d) 43{ 44int len1 =strlen(string1), len2 =strlen(string2); 45int*row0, *row1, *row2; 46int i, j; 47 48ALLOC_ARRAY(row0, len2 +1); 49ALLOC_ARRAY(row1, len2 +1); 50ALLOC_ARRAY(row2, len2 +1); 51 52for(j =0; j <= len2; j++) 53 row1[j] = j * a; 54for(i =0; i < len1; i++) { 55int*dummy; 56 57 row2[0] = (i +1) * d; 58for(j =0; j < len2; j++) { 59/* substitution */ 60 row2[j +1] = row1[j] + s * (string1[i] != string2[j]); 61/* swap */ 62if(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 */ 67if(row2[j +1] > row1[j +1] + d) 68 row2[j +1] = row1[j +1] + d; 69/* insertion */ 70if(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]; 81free(row0); 82free(row1); 83free(row2); 84 85return i; 86}