In levenshtein_internal(), describe algorithm a bit more clearly.

diff --git a/contrib/fuzzystrmatch/fuzzystrmatch.c b/contrib/fuzzystrmatch/fuzzystrmatch.c
index 63921d92717b3333d353ed1329fab42e9f8bcef1..01084da4072c1513192b12099a286e3b794abc48 100644 (file)
--- a/contrib/fuzzystrmatch/fuzzystrmatch.c
+++ b/contrib/fuzzystrmatch/fuzzystrmatch.c
@@ -277,15 +277,25 @@ levenshtein_internal(text *s, text *t,
        ++n;
 
        /*
-        * Instead of building an (m+1)x(n+1) array, we'll use two different
-        * arrays of size m+1 for storing accumulated values. At each step one
-        * represents the "previous" row and one is the "current" row of the
-        * notional large array.
+        * One way to compute Levenshtein distance is to incrementally construct
+        * an (m+1)x(n+1) matrix where cell (i, j) represents the minimum number
+        * of operations required to transform the first i characters of s into
+        * the first j characters of t.  The last column of the final row is the
+        * answer.
+        *
+        * We use that algorithm here with some modification.  In lieu of holding
+        * the entire array in memory at once, we'll just use two arrays of size
+        * m+1 for storing accumulated values. At each step one array represents
+        * the "previous" row and one is the "current" row of the notional large
+        * array.
         */
        prev = (int *) palloc(2 * m * sizeof(int));
        curr = prev + m;
 
-       /* Initialize the "previous" row to 0..cols */
+       /*
+        * To transform the first i characters of s into the first 0 characters
+        * of t, we must perform i deletions.
+        */
        for (i = 0; i < m; i++)
                prev[i] = i * del_c;
 
@@ -297,8 +307,8 @@ levenshtein_internal(text *s, text *t,
                int                     y_char_len = n != t_bytes + 1 ? pg_mblen(y) : 1;
 
                /*
-                * First cell must increment sequentially, as we're on the j'th row of
-                * the (m+1)x(n+1) array.
+                * To transform the first 0 characters of s into the first j
+                * characters of t, we must perform j insertions.
                 */
                curr[0] = j * ins_c;