/*
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* CDDL HEADER START
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*
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* The contents of this file are subject to the terms of the
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* Common Development and Distribution License, Version 1.0 only
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* (the "License"). You may not use this file except in compliance
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* with the License.
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*
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* You can obtain a copy of the license at legal-notices/CDDLv1_0.txt
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* or http://forgerock.org/license/CDDLv1.0.html.
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* See the License for the specific language governing permissions
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* and limitations under the License.
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*
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* When distributing Covered Code, include this CDDL HEADER in each
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* file and include the License file at legal-notices/CDDLv1_0.txt.
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* If applicable, add the following below this CDDL HEADER, with the
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* fields enclosed by brackets "[]" replaced with your own identifying
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* information:
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* Portions Copyright [yyyy] [name of copyright owner]
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*
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* CDDL HEADER END
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*
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*
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* Copyright 2008 Sun Microsystems, Inc.
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* Portions Copyright 2014 ForgeRock AS
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*/
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package org.opends.server.util;
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import static org.forgerock.util.Reject.*;
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/**
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* This class provides an implementation of the Levenshtein distance algorithm,
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* which may be used to determine the minimum number of changes required to
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* transform one string into another. For the purpose of this algorithm, a
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* change is defined as replacing one character with another, inserting a new
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* character, or removing an existing character.
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* <BR><BR>
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* The basic algorithm works as follows for a source string S of length X and
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* a target string T of length Y:
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* <OL>
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* <LI>Create a matrix M with dimensions of X+1, Y+1.</LI>
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* <LI>Fill the first row with sequentially-increasing values 0 through
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* X.</LI>
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* <LI>Fill the first column with sequentially-increasing values 0 through
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* Y.</LI>
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* <LI>Create a nested loop iterating over the characters in the strings. In
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* the outer loop, iterate through characters in S from 0 to X-1 using an
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* iteration counter I. In the inner loop, iterate through the characters
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* in T from 0 to Y-1 using an iterator counter J. Calculate the
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* following three things and place the smallest value in the matrix in
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* row I+1 column J+1:
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* <UL>
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* <LI>One more than the value in the matrix position immediately to the
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* left (i.e., 1 + M[I][J+1]).</LI>
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* <LI>One more than the value in the matrix position immediately above
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* (i.e., 1 + M[I+1][J]).</LI>
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* <LI>Define a value V to be zero if S[I] is the same as T[J], or one if
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* they are different. Add that value to the value in the matrix
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* position immediately above and to the left (i.e.,
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* V + M[I][J]).</LI>
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* </UL>
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* </LI>
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* <LI>The Levenshtein distance value, which is the least number of changes
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* needed to transform the source string into the target string, will be
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* the value in the bottom right corner of the matrix (i.e.,
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* M[X][Y]).</LI>
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* </OL>
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* <BR><BR>
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* Note that this is a completely "clean room" implementation, developed from a
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* description of the algorithm, rather than copying an existing implementation.
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* Doing it in this way eliminates copyright and licensing concerns associated
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* with using an existing implementation.
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*/
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@org.opends.server.types.PublicAPI(
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stability=org.opends.server.types.StabilityLevel.UNCOMMITTED,
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mayInstantiate=false,
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mayExtend=false,
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mayInvoke=true)
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public final class LevenshteinDistance
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{
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/**
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* Calculates the Levenshtein distance between the provided string values.
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*
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* @param source The source string to compare. It must not be {@code null}.
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* @param target The target string to compare. It must not be {@code null}.
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*
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* @return The minimum number of changes required to turn the source string
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* into the target string.
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*/
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public static int calculate(String source, String target)
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{
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ifNull(source, target);
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// sl == source length; tl == target length
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int sl = source.length();
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int tl = target.length();
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// If either of the lengths is zero, then the distance is the length of the
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// other string.
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if (sl == 0)
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{
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return tl;
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}
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else if (tl == 0)
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{
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return sl;
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}
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// w == matrix width; h == matrix height
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int w = sl+1;
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int h = tl+1;
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// m == matrix array
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// Create the array and fill it with values 0..sl in the first dimension and
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// 0..tl in the second dimension.
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int[][] m = new int[w][h];
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for (int i=0; i < w; i++)
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{
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m[i][0] = i;
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}
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for (int i=1; i < h; i++)
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{
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m[0][i] = i;
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}
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for (int i=0,x=1; i < sl; i++,x++)
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{
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char s = source.charAt(i);
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for (int j=0,y=1; j < tl; j++,y++)
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{
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char t = target.charAt(j);
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// Figure out what to put in the appropriate matrix slot. It should be
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// the lowest of:
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// - One more than the value to the left
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// - One more than the value to the top
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// - If the characters are equal, the value to the upper left, otherwise
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// one more than the value to the upper left.
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m[x][y] = Math.min(Math.min((m[i][y] + 1), (m[x][j] + 1)),
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(m[i][j] + ((s == t) ? 0 : 1)));
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}
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}
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// The Levenshtein distance should now be the value in the lower right
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// corner of the matrix.
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return m[sl][tl];
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}
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}
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