Merge branch 'st/levenshtein'

* st/levenshtein:
  Document levenshtein.c
  Fix deletion of last character in levenshtein distance

Author: gitster

levenshtein.c

@@ -1,6 +1,43 @@
 #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)