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/*
 * Licensed to the Apache Software Foundation (ASF) under one or more
 * contributor license agreements.  See the NOTICE file distributed with
 * this work for additional information regarding copyright ownership.
 * The ASF licenses this file to You under the Apache License, Version 2.0
 * (the "License"); you may not use this file except in compliance with
 * the License.  You may obtain a copy of the License at
 *
 *      http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */
package org.apache.commons.text.similarity;

import java.util.Arrays;

/**
 * An algorithm for measuring the difference between two character sequences.
 *
 * 

* This is the number of changes needed to change one sequence into another, * where each change is a single character modification (deletion, insertion * or substitution). *

* *

* This code has been adapted from Apache Commons Lang 3.3. *

* * @since 1.0 */ public class LevenshteinDistance implements EditDistance { /** * Default instance. */ private static final LevenshteinDistance DEFAULT_INSTANCE = new LevenshteinDistance(); /** * Threshold. */ private final Integer threshold; /** *

* This returns the default instance that uses a version * of the algorithm that does not use a threshold parameter. *

* * @see LevenshteinDistance#getDefaultInstance() */ public LevenshteinDistance() { this(null); } /** *

* If the threshold is not null, distance calculations will be limited to a maximum length. * If the threshold is null, the unlimited version of the algorithm will be used. *

* * @param threshold * If this is null then distances calculations will not be limited. * This may not be negative. */ public LevenshteinDistance(final Integer threshold) { if (threshold != null && threshold < 0) { throw new IllegalArgumentException("Threshold must not be negative"); } this.threshold = threshold; } /** *

Find the Levenshtein distance between two Strings.

* *

A higher score indicates a greater distance.

* *

The previous implementation of the Levenshtein distance algorithm * was from http://www.merriampark.com/ld.htm

* *

Chas Emerick has written an implementation in Java, which avoids an OutOfMemoryError * which can occur when my Java implementation is used with very large strings.
* This implementation of the Levenshtein distance algorithm * is from http://www.merriampark.com/ldjava.htm

* *
     * distance.apply(null, *)             = IllegalArgumentException
     * distance.apply(*, null)             = IllegalArgumentException
     * distance.apply("","")               = 0
     * distance.apply("","a")              = 1
     * distance.apply("aaapppp", "")       = 7
     * distance.apply("frog", "fog")       = 1
     * distance.apply("fly", "ant")        = 3
     * distance.apply("elephant", "hippo") = 7
     * distance.apply("hippo", "elephant") = 7
     * distance.apply("hippo", "zzzzzzzz") = 8
     * distance.apply("hello", "hallo")    = 1
     * 
* * @param left the first string, must not be null * @param right the second string, must not be null * @return result distance, or -1 * @throws IllegalArgumentException if either String input {@code null} */ @Override public Integer apply(final CharSequence left, final CharSequence right) { if (threshold != null) { return limitedCompare(left, right, threshold); } return unlimitedCompare(left, right); } /** * Gets the default instance. * * @return the default instance */ public static LevenshteinDistance getDefaultInstance() { return DEFAULT_INSTANCE; } /** * Gets the distance threshold. * * @return the distance threshold */ public Integer getThreshold() { return threshold; } /** * Find the Levenshtein distance between two CharSequences if it's less than or * equal to a given threshold. * *

* This implementation follows from Algorithms on Strings, Trees and * Sequences by Dan Gusfield and Chas Emerick's implementation of the * Levenshtein distance algorithm from http://www.merriampark.com/ld.htm *

* *
     * limitedCompare(null, *, *)             = IllegalArgumentException
     * limitedCompare(*, null, *)             = IllegalArgumentException
     * limitedCompare(*, *, -1)               = IllegalArgumentException
     * limitedCompare("","", 0)               = 0
     * limitedCompare("aaapppp", "", 8)       = 7
     * limitedCompare("aaapppp", "", 7)       = 7
     * limitedCompare("aaapppp", "", 6))      = -1
     * limitedCompare("elephant", "hippo", 7) = 7
     * limitedCompare("elephant", "hippo", 6) = -1
     * limitedCompare("hippo", "elephant", 7) = 7
     * limitedCompare("hippo", "elephant", 6) = -1
     * 
* * @param left the first CharSequence, must not be null * @param right the second CharSequence, must not be null * @param threshold the target threshold, must not be negative * @return result distance, or -1 */ private static int limitedCompare(CharSequence left, CharSequence right, final int threshold) { // NOPMD if (left == null || right == null) { throw new IllegalArgumentException("CharSequences must not be null"); } if (threshold < 0) { throw new IllegalArgumentException("Threshold must not be negative"); } /* * This implementation only computes the distance if it's less than or * equal to the threshold value, returning -1 if it's greater. The * advantage is performance: unbounded distance is O(nm), but a bound of * k allows us to reduce it to O(km) time by only computing a diagonal * stripe of width 2k + 1 of the cost table. It is also possible to use * this to compute the unbounded Levenshtein distance by starting the * threshold at 1 and doubling each time until the distance is found; * this is O(dm), where d is the distance. * * One subtlety comes from needing to ignore entries on the border of * our stripe eg. p[] = |#|#|#|* d[] = *|#|#|#| We must ignore the entry * to the left of the leftmost member We must ignore the entry above the * rightmost member * * Another subtlety comes from our stripe running off the matrix if the * strings aren't of the same size. Since string s is always swapped to * be the shorter of the two, the stripe will always run off to the * upper right instead of the lower left of the matrix. * * As a concrete example, suppose s is of length 5, t is of length 7, * and our threshold is 1. In this case we're going to walk a stripe of * length 3. The matrix would look like so: * *
         *    1 2 3 4 5
         * 1 |#|#| | | |
         * 2 |#|#|#| | |
         * 3 | |#|#|#| |
         * 4 | | |#|#|#|
         * 5 | | | |#|#|
         * 6 | | | | |#|
         * 7 | | | | | |
         * 
* * Note how the stripe leads off the table as there is no possible way * to turn a string of length 5 into one of length 7 in edit distance of * 1. * * Additionally, this implementation decreases memory usage by using two * single-dimensional arrays and swapping them back and forth instead of * allocating an entire n by m matrix. This requires a few minor * changes, such as immediately returning when it's detected that the * stripe has run off the matrix and initially filling the arrays with * large values so that entries we don't compute are ignored. * * See Algorithms on Strings, Trees and Sequences by Dan Gusfield for * some discussion. */ int n = left.length(); // length of left int m = right.length(); // length of right // if one string is empty, the edit distance is necessarily the length // of the other if (n == 0) { return m <= threshold ? m : -1; } else if (m == 0) { return n <= threshold ? n : -1; } if (n > m) { // swap the two strings to consume less memory final CharSequence tmp = left; left = right; right = tmp; n = m; m = right.length(); } int[] p = new int[n + 1]; // 'previous' cost array, horizontally int[] d = new int[n + 1]; // cost array, horizontally int[] tempD; // placeholder to assist in swapping p and d // fill in starting table values final int boundary = Math.min(n, threshold) + 1; for (int i = 0; i < boundary; i++) { p[i] = i; } // these fills ensure that the value above the rightmost entry of our // stripe will be ignored in following loop iterations Arrays.fill(p, boundary, p.length, Integer.MAX_VALUE); Arrays.fill(d, Integer.MAX_VALUE); // iterates through t for (int j = 1; j <= m; j++) { final char rightJ = right.charAt(j - 1); // jth character of right d[0] = j; // compute stripe indices, constrain to array size final int min = Math.max(1, j - threshold); final int max = j > Integer.MAX_VALUE - threshold ? n : Math.min( n, j + threshold); // the stripe may lead off of the table if s and t are of different // sizes if (min > max) { return -1; } // ignore entry left of leftmost if (min > 1) { d[min - 1] = Integer.MAX_VALUE; } // iterates through [min, max] in s for (int i = min; i <= max; i++) { if (left.charAt(i - 1) == rightJ) { // diagonally left and up d[i] = p[i - 1]; } else { // 1 + minimum of cell to the left, to the top, diagonally // left and up d[i] = 1 + Math.min(Math.min(d[i - 1], p[i]), p[i - 1]); } } // copy current distance counts to 'previous row' distance counts tempD = p; p = d; d = tempD; } // if p[n] is greater than the threshold, there's no guarantee on it // being the correct // distance if (p[n] <= threshold) { return p[n]; } return -1; } /** *

Find the Levenshtein distance between two Strings.

* *

A higher score indicates a greater distance.

* *

The previous implementation of the Levenshtein distance algorithm * was from * https://web.archive.org/web/20120526085419/http://www.merriampark.com/ldjava.htm

* *

This implementation only need one single-dimensional arrays of length s.length() + 1

* *
     * unlimitedCompare(null, *)             = IllegalArgumentException
     * unlimitedCompare(*, null)             = IllegalArgumentException
     * unlimitedCompare("","")               = 0
     * unlimitedCompare("","a")              = 1
     * unlimitedCompare("aaapppp", "")       = 7
     * unlimitedCompare("frog", "fog")       = 1
     * unlimitedCompare("fly", "ant")        = 3
     * unlimitedCompare("elephant", "hippo") = 7
     * unlimitedCompare("hippo", "elephant") = 7
     * unlimitedCompare("hippo", "zzzzzzzz") = 8
     * unlimitedCompare("hello", "hallo")    = 1
     * 
* * @param left the first CharSequence, must not be null * @param right the second CharSequence, must not be null * @return result distance, or -1 * @throws IllegalArgumentException if either CharSequence input is {@code null} */ private static int unlimitedCompare(CharSequence left, CharSequence right) { if (left == null || right == null) { throw new IllegalArgumentException("CharSequences must not be null"); } /* This implementation use two variable to record the previous cost counts, So this implementation use less memory than previous impl. */ int n = left.length(); // length of left int m = right.length(); // length of right if (n == 0) { return m; } else if (m == 0) { return n; } if (n > m) { // swap the input strings to consume less memory final CharSequence tmp = left; left = right; right = tmp; n = m; m = right.length(); } final int[] p = new int[n + 1]; // indexes into strings left and right int i; // iterates through left int j; // iterates through right int upperLeft; int upper; char rightJ; // jth character of right int cost; // cost for (i = 0; i <= n; i++) { p[i] = i; } for (j = 1; j <= m; j++) { upperLeft = p[0]; rightJ = right.charAt(j - 1); p[0] = j; for (i = 1; i <= n; i++) { upper = p[i]; cost = left.charAt(i - 1) == rightJ ? 0 : 1; // minimum of cell to the left+1, to the top+1, diagonally left and up +cost p[i] = Math.min(Math.min(p[i - 1] + 1, p[i] + 1), upperLeft + cost); upperLeft = upper; } } return p[n]; } }




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