+ * 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:

+ * - 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)

row2[j + 1] > row0[j - 1] + w)

row2[j + 1] = row0[j - 1] + w;

- 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;

if (row2[j + 1] > row2[j] + a)