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Fix levenshtein with costs. The previous code multiplied by the cost in only

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3 of the 7 relevant locations.

Marcin Mank, slightly adjusted by me.
This commit is contained in:
Robert Haas 2009-12-10 01:54:17 +00:00
parent 03d7b0647f
commit da07641481
1 changed files with 7 additions and 7 deletions
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14
contrib/fuzzystrmatch/fuzzystrmatch.c
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@ -5,7 +5,7 @@
*
* Joe Conway <mail@joeconway.com>
*
* $PostgreSQL: pgsql/contrib/fuzzystrmatch/fuzzystrmatch.c,v 1.30 2009/06/11 14:48:51 momjian Exp $
* $PostgreSQL: pgsql/contrib/fuzzystrmatch/fuzzystrmatch.c,v 1.31 2009/12/10 01:54:17 rhaas Exp $
* Copyright (c) 2001-2009, PostgreSQL Global Development Group
* ALL RIGHTS RESERVED;
*
@ -207,13 +207,13 @@ levenshtein_internal(const char *s, const char *t,
n = strlen(t);
/*
* If either m or n is 0, the answer is the other value. This makes sense
* since it would take that many insertions to build a matching string
* We can transform an empty s into t with n insertions, or a non-empty t
* into an empty s with m deletions.
*/
if (!m)
return n;
return n * ins_c;
if (!n)
return m;
return m * del_c;
/*
* For security concerns, restrict excessive CPU+RAM usage. (This
@ -241,7 +241,7 @@ levenshtein_internal(const char *s, const char *t,
/* Initialize the "previous" row to 0..cols */
for (i = 0; i < m; i++)
prev[i] = i;
prev[i] = i * del_c;
/* Loop through rows of the notional array */
for (y = t, j = 1; j < n; y++, j++)
@ -252,7 +252,7 @@ levenshtein_internal(const char *s, const char *t,
* First cell must increment sequentially, as we're on the j'th row of
* the (m+1)x(n+1) array.
*/
curr[0] = j;
curr[0] = j * ins_c;
for (x = s, i = 1; i < m; x++, i++)
{