Commit ef4daa74 by Junio C Hamano

### Merge branch 'st/levenshtein'

```* st/levenshtein:
Document levenshtein.c
Fix deletion of last character in levenshtein distance```
parents cab7d7d8 850fb6ff
 #include "cache.h" #include "levenshtein.h" /* * This function implements the Damerau-Levenshtein algorithm to * calculate a distance between strings. * * Basically, it says how many letters need to be swapped, substituted, * deleted from, or added to string1, at least, to get string2. * * The idea is to build a distance matrix for the substrings of both * strings. To avoid a large space complexity, only the last three rows * are kept in memory (if swaps had the same or higher cost as one deletion * plus one insertion, only two rows would be needed). * * At any stage, "i + 1" denotes the length of the current substring of * string1 that the distance is calculated for. * * row2 holds the current row, row1 the previous row (i.e. for the substring * of string1 of length "i"), and row0 the row before that. * * In other words, at the start of the big loop, row2[j + 1] contains the * Damerau-Levenshtein distance between the substring of string1 of length * "i" and the substring of string2 of length "j + 1". * * All the big loop does is determine the partial minimum-cost paths. * * It does so by calculating the costs of the path ending in characters * i (in string1) and j (in string2), respectively, given that the last * operation is a substition, a swap, a deletion, or an insertion. * * This implementation allows the costs to be weighted: * * - w (as in "sWap") * - s (as in "Substitution") * - a (for insertion, AKA "Add") * - d (as in "Deletion") * * Note that this algorithm calculates a distance _iff_ d == a. */ int levenshtein(const char *string1, const char *string2, int w, int s, int a, int d) { ... ... @@ -25,7 +62,7 @@ int levenshtein(const char *string1, const char *string2, row2[j + 1] > row0[j - 1] + w) row2[j + 1] = row0[j - 1] + w; /* deletion */ if (j + 1 < len2 && row2[j + 1] > row1[j + 1] + d) if (row2[j + 1] > row1[j + 1] + d) row2[j + 1] = row1[j + 1] + d; /* insertion */ if (row2[j + 1] > row2[j] + a) ... ...
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