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/****************************************************************************
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**
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** Copyright (C) 2009 Nokia Corporation and/or its subsidiary(-ies).
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** All rights reserved.
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** Contact: Nokia Corporation (qt-info@nokia.com)
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**
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** This file is part of the Qt Linguist of the Qt Toolkit.
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**
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** $QT_BEGIN_LICENSE:LGPL$
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** No Commercial Usage
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** This file contains pre-release code and may not be distributed.
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** You may use this file in accordance with the terms and conditions
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** contained in the Technology Preview License Agreement accompanying
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** this package.
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**
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** GNU Lesser General Public License Usage
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** Alternatively, this file may be used under the terms of the GNU Lesser
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** General Public License version 2.1 as published by the Free Software
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** Foundation and appearing in the file LICENSE.LGPL included in the
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** packaging of this file. Please review the following information to
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** ensure the GNU Lesser General Public License version 2.1 requirements
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** will be met: http://www.gnu.org/licenses/old-licenses/lgpl-2.1.html.
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**
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** In addition, as a special exception, Nokia gives you certain additional
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** rights. These rights are described in the Nokia Qt LGPL Exception
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** version 1.1, included in the file LGPL_EXCEPTION.txt in this package.
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**
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** If you have questions regarding the use of this file, please contact
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** Nokia at qt-info@nokia.com.
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**
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**
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**
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**
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**
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**
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**
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**
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** $QT_END_LICENSE$
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**
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****************************************************************************/
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#include "simtexth.h"
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#include "translator.h"
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#include <QtCore/QByteArray>
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#include <QtCore/QString>
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#include <QtCore/QList>
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QT_BEGIN_NAMESPACE
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typedef QList<TranslatorMessage> TML;
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/*
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How similar are two texts? The approach used here relies on co-occurrence
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matrices and is very efficient.
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Let's see with an example: how similar are "here" and "hither"? The
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co-occurrence matrix M for "here" is M[h,e] = 1, M[e,r] = 1, M[r,e] = 1, and 0
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elsewhere; the matrix N for "hither" is N[h,i] = 1, N[i,t] = 1, ...,
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N[h,e] = 1, N[e,r] = 1, and 0 elsewhere. The union U of both matrices is the
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matrix U[i,j] = max { M[i,j], N[i,j] }, and the intersection V is
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V[i,j] = min { M[i,j], N[i,j] }. The score for a pair of texts is
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score = (sum of V[i,j] over all i, j) / (sum of U[i,j] over all i, j),
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a formula suggested by Arnt Gulbrandsen. Here we have
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score = 2 / 6,
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or one third.
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The implementation differs from this in a few details. Most importantly,
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repetitions are ignored; for input "xxx", M[x,x] equals 1, not 2.
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*/
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/*
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Every character is assigned to one of 20 buckets so that the co-occurrence
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matrix requires only 20 * 20 = 400 bits, not 256 * 256 = 65536 bits or even
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more if we want the whole Unicode. Which character falls in which bucket is
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arbitrary.
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The second half of the table is a replica of the first half, because of
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laziness.
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*/
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static const int indexOf[256] = {
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0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
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0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
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// ! " # $ % & ' ( ) * + , - . /
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0, 2, 6, 7, 10, 12, 15, 19, 2, 6, 7, 10, 12, 15, 19, 0,
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// 0 1 2 3 4 5 6 7 8 9 : ; < = > ?
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1, 3, 4, 5, 8, 9, 11, 13, 14, 16, 2, 6, 7, 10, 12, 15,
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// @ A B C D E F G H I J K L M N O
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0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 6, 10, 11, 12, 13, 14,
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// P Q R S T U V W X Y Z [ \ ] ^ _
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15, 12, 16, 17, 18, 19, 2, 10, 15, 7, 19, 2, 6, 7, 10, 0,
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// ` a b c d e f g h i j k l m n o
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0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 6, 10, 11, 12, 13, 14,
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// p q r s t u v w x y z { | } ~
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15, 12, 16, 17, 18, 19, 2, 10, 15, 7, 19, 2, 6, 7, 10, 0,
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0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
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0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
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0, 2, 6, 7, 10, 12, 15, 19, 2, 6, 7, 10, 12, 15, 19, 0,
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1, 3, 4, 5, 8, 9, 11, 13, 14, 16, 2, 6, 7, 10, 12, 15,
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0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 6, 10, 11, 12, 13, 14,
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15, 12, 16, 17, 18, 19, 2, 10, 15, 7, 19, 2, 6, 7, 10, 0,
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0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 6, 10, 11, 12, 13, 14,
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15, 12, 16, 17, 18, 19, 2, 10, 15, 7, 19, 2, 6, 7, 10, 0
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};
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/*
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The entry bitCount[i] (for i between 0 and 255) is the number of bits used to
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represent i in binary.
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*/
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static const int bitCount[256] = {
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0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4,
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1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
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1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
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2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
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1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
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2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
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2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
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3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
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1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
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2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
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2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
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3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
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2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
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3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
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3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
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4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8
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};
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struct CoMatrix
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{
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/*
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The matrix has 20 * 20 = 400 entries. This requires 50 bytes, or 13
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words. Some operations are performed on words for more efficiency.
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*/
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union {
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quint8 b[52];
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quint32 w[13];
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};
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CoMatrix() { memset( b, 0, 52 ); }
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CoMatrix(const QString &str)
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{
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QByteArray ba = str.toUtf8();
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const char *text = ba.constData();
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char c = '\0', d;
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memset( b, 0, 52 );
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/*
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The Knuth books are not in the office only for show; they help make
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loops 30% faster and 20% as readable.
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*/
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while ( (d = *text) != '\0' ) {
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setCoOccurence( c, d );
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if ( (c = *++text) != '\0' ) {
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setCoOccurence( d, c );
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text++;
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}
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}
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}
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void setCoOccurence( char c, char d ) {
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int k = indexOf[(uchar) c] + 20 * indexOf[(uchar) d];
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b[k >> 3] |= (1 << (k & 0x7));
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}
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int worth() const {
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int w = 0;
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for ( int i = 0; i < 50; i++ )
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w += bitCount[b[i]];
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return w;
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}
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};
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static inline CoMatrix reunion(const CoMatrix &m, const CoMatrix &n)
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{
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CoMatrix p;
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for (int i = 0; i < 13; ++i)
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p.w[i] = m.w[i] | n.w[i];
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return p;
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}
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static inline CoMatrix intersection(const CoMatrix &m, const CoMatrix &n)
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{
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CoMatrix p;
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for (int i = 0; i < 13; ++i)
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p.w[i] = m.w[i] & n.w[i];
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return p;
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}
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StringSimilarityMatcher::StringSimilarityMatcher(const QString &stringToMatch)
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{
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m_cm = new CoMatrix(stringToMatch);
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m_length = stringToMatch.length();
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}
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int StringSimilarityMatcher::getSimilarityScore(const QString &strCandidate)
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{
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CoMatrix cmTarget(strCandidate);
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int delta = qAbs(m_length - strCandidate.size());
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int score = ( (intersection(*m_cm, cmTarget).worth() + 1) << 10 ) /
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( reunion(*m_cm, cmTarget).worth() + (delta << 1) + 1 );
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return score;
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}
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StringSimilarityMatcher::~StringSimilarityMatcher()
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{
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delete m_cm;
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}
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/**
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* Checks how similar two strings are.
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* The return value is the score, and a higher score is more similar
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* than one with a low score.
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* Linguist considers a score over 190 to be a good match.
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* \sa StringSimilarityMatcher
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*/
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int getSimilarityScore(const QString &str1, const QString &str2)
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{
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CoMatrix cmTarget(str2);
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CoMatrix cm(str1);
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int delta = qAbs(str1.size() - str2.size());
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int score = ( (intersection(cm, cmTarget).worth() + 1) << 10 )
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/ ( reunion(cm, cmTarget).worth() + (delta << 1) + 1 );
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return score;
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}
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CandidateList similarTextHeuristicCandidates(const Translator *tor,
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const QString &text, int maxCandidates)
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{
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QList<int> scores;
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CandidateList candidates;
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TML all = tor->translatedMessages();
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foreach (const TranslatorMessage &mtm, all) {
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if (mtm.type() == TranslatorMessage::Unfinished
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|| mtm.translation().isEmpty())
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continue;
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QString s = mtm.sourceText();
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int score = getSimilarityScore(s, text);
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if (candidates.size() == maxCandidates && score > scores[maxCandidates - 1] )
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candidates.removeLast();
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if (candidates.size() < maxCandidates && score >= textSimilarityThreshold) {
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Candidate cand( s, mtm.translation() );
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int i;
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for (i = 0; i < candidates.size(); i++) {
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if (score >= scores.at(i)) {
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if (score == scores.at(i)) {
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if (candidates.at(i) == cand)
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goto continue_outer_loop;
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} else {
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break;
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}
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}
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}
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scores.insert(i, score);
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candidates.insert(i, cand);
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}
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continue_outer_loop:
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;
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}
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return candidates;
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}
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QT_END_NAMESPACE
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