LUCENE-1183: Optimized TRStringDistance class (in contrib/spell) that uses less memory than the previous version

git-svn-id: https://svn.apache.org/repos/asf/lucene/java/trunk@659016 13f79535-47bb-0310-9956-ffa450edef68
This commit is contained in:
Otis Gospodnetic 2008-05-22 06:24:55 +00:00
parent a379a67875
commit f68c9544de
2 changed files with 66 additions and 89 deletions

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@ -174,6 +174,9 @@ Optimizations
runtime type checking for possible speedup of binaryValue().
(Eks Dev via Mike McCandless)
5. LUCENE-1183: Optimized TRStringDistance class (in contrib/spell) that uses
less memory than the previous version. (Cédrik LIME via Otis Gospodnetic)
Documentation
1. LUCENE-1236: Added some clarifying remarks to EdgeNGram*.java (Hiroaki Kawai via Grant Ingersoll)

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@ -1,6 +1,5 @@
package org.apache.lucene.search.spell;
/**
* Licensed to the Apache Software Foundation (ASF) under one or more
* contributor license agreements. See the NOTICE file distributed with
@ -19,13 +18,16 @@ package org.apache.lucene.search.spell;
*/
/**
* Edit distance class
* Edit distance class.
* Note: this class is not thread-safe.
*/
final class TRStringDistance {
final char[] sa;
final int n;
final int[][][] cache=new int[30][][];
int p[]; //'previous' cost array, horizontally
int d[]; // cost array, horizontally
int _d[]; //placeholder to assist in swapping p and d
/**
@ -33,101 +35,73 @@ final class TRStringDistance {
* In one benchmark times were 5.3sec using ctr vs 8.5sec w/ static method, thus 37% faster.
*/
public TRStringDistance (String target) {
sa=target.toCharArray();
n=sa.length;
sa = target.toCharArray();
n = sa.length;
p = new int[n+1]; //'previous' cost array, horizontally
d = new int[n+1]; // cost array, horizontally
}
//*****************************
// Compute Levenshtein distance
//*****************************
public final int getDistance (String other) {
int d[][]; // matrix
int cost; // cost
// Compute Levenshtein distance: see org.apache.commons.lang.StringUtils#getLevenshteinDistance(String, String)
//*****************************
public final int getDistance (String other) {
/*
The difference between this impl. and the previous is that, rather
than creating and retaining a matrix of size s.length()+1 by t.length()+1,
we maintain two single-dimensional arrays of length s.length()+1. The first, d,
is the 'current working' distance array that maintains the newest distance cost
counts as we iterate through the characters of String s. Each time we increment
the index of String t we are comparing, d is copied to p, the second int[]. Doing so
allows us to retain the previous cost counts as required by the algorithm (taking
the minimum of the cost count to the left, up one, and diagonally up and to the left
of the current cost count being calculated). (Note that the arrays aren't really
copied anymore, just switched...this is clearly much better than cloning an array
or doing a System.arraycopy() each time through the outer loop.)
// Step 1
final char[] ta=other.toCharArray();
final int m=ta.length;
if (n==0) {
return m;
}
if (m==0) {
return n;
}
Effectively, the difference between the two implementations is this one does not
cause an out of memory condition when calculating the LD over two very large strings.
*/
if (m>=cache.length) {
d=form(n, m);
}
else if (cache[m]!=null) {
d=cache[m];
}
else {
d=cache[m]=form(n, m);
// Step 3
}
for (int i=1; i<=n; i++) {
final char s_i=sa[i-1];
// Step 4
for (int j=1; j<=m; j++) {
final char t_j=ta[j-1];
// Step 5
if (s_i==t_j) { // same
cost=0;
}
else { // not a match
cost=1;
// Step 6
}
d[i][j]=min3(d[i-1][j]+1, d[i][j-1]+1, d[i-1][j-1]+cost);
}
}
// Step 7
return d[n][m];
}
/**
*
*/
private static int[][] form (int n, int m) {
int[][] d=new int[n+1][m+1];
// Step 2
for (int i=0; i<=n; i++) {
d[i][0]=i;
final int m = other.length();
if (n == 0) {
return m;
} else if (m == 0) {
return n;
}
for (int j=0; j<=m; j++) {
d[0][j]=j;
// indexes into strings s and t
int i; // iterates through s
int j; // iterates through t
char t_j; // jth character of t
int cost; // cost
for (i = 0; i<=n; i++) {
p[i] = i;
}
return d;
for (j = 1; j<=m; j++) {
t_j = other.charAt(j-1);
d[0] = j;
for (i=1; i<=n; i++) {
cost = sa[i-1]==t_j ? 0 : 1;
// minimum of cell to the left+1, to the top+1, diagonally left and up +cost
d[i] = Math.min(Math.min(d[i-1]+1, p[i]+1), p[i-1]+cost);
}
// copy current distance counts to 'previous row' distance counts
_d = p;
p = d;
d = _d;
}
// our last action in the above loop was to switch d and p, so p now
// actually has the most recent cost counts
return p[n];
}
//****************************
// Get minimum of three values
//****************************
private static int min3 (int a, int b, int c) {
int mi=a;
if (b<mi) {
mi=b;
}
if (c<mi) {
mi=c;
}
return mi;
}
}