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+// SPDX-License-Identifier: GPL-2.0
+#include "levenshtein.h"
+#include <errno.h>
+#include <stdlib.h>
+#include <string.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 substitution, 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)
+{
+ int len1 = strlen(string1), len2 = strlen(string2);
+ int *row0 = malloc(sizeof(int) * (len2 + 1));
+ int *row1 = malloc(sizeof(int) * (len2 + 1));
+ int *row2 = malloc(sizeof(int) * (len2 + 1));
+ int i, j;
+
+ for (j = 0; j <= len2; j++)
+ row1[j] = j * a;
+ for (i = 0; i < len1; i++) {
+ int *dummy;
+
+ row2[0] = (i + 1) * d;
+ for (j = 0; j < len2; j++) {
+ /* substitution */
+ row2[j + 1] = row1[j] + s * (string1[i] != string2[j]);
+ /* swap */
+ if (i > 0 && j > 0 && string1[i - 1] == string2[j] &&
+ string1[i] == string2[j - 1] &&
+ row2[j + 1] > row0[j - 1] + w)
+ row2[j + 1] = row0[j - 1] + w;
+ /* deletion */
+ if (row2[j + 1] > row1[j + 1] + d)
+ row2[j + 1] = row1[j + 1] + d;
+ /* insertion */
+ if (row2[j + 1] > row2[j] + a)
+ row2[j + 1] = row2[j] + a;
+ }
+
+ dummy = row0;
+ row0 = row1;
+ row1 = row2;
+ row2 = dummy;
+ }
+
+ i = row1[len2];
+ free(row0);
+ free(row1);
+ free(row2);
+
+ return i;
+}