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All rights reserved. ** ** This file is part of TQt Linguist. ** ** This file may be used under the terms of the GNU General ** Public License versions 2.0 or 3.0 as published by the Free ** Software Foundation and appearing in the files LICENSE.GPL2 ** and LICENSE.GPL3 included in the packaging of this file. ** Alternatively you may (at your option) use any later version ** of the GNU General Public License if such license has been ** publicly approved by Trolltech ASA (or its successors, if any) ** and the KDE Free TQt Foundation. ** ** Please review the following information to ensure GNU General ** Public Licensing requirements will be met: ** http://trolltech.com/products/qt/licenses/licensing/opensource/. ** If you are unsure which license is appropriate for your use, please ** review the following information: ** http://trolltech.com/products/qt/licenses/licensing/licensingoverview ** or contact the sales department at sales@trolltech.com. ** ** Licensees holding valid TQt Commercial licenses may use this file in ** accordance with the TQt Commercial License Agreement provided with ** the Software. ** ** This file is provided "AS IS" with NO WARRANTY OF ANY KIND, ** INCLUDING THE WARRANTIES OF DESIGN, MERCHANTABILITY AND FITNESS FOR ** A PARTICULAR PURPOSE. Trolltech reserves all rights not granted ** herein. ** **********************************************************************/ #include "simtexth.h" #include #include #include #include #include #include #include #include typedef TQValueList TML; /* How similar are two texts? The approach used here relies on co-occurrence matrices and is very efficient. Let's see with an example: how similar are "here" and "hither"? The co-occurrence matrix M for "here" is M[h,e] = 1, M[e,r] = 1, M[r,e] = 1, and 0 elsewhere; the matrix N for "hither" is N[h,i] = 1, N[i,t] = 1, ..., N[h,e] = 1, N[e,r] = 1, and 0 elsewhere. The union U of both matrices is the matrix U[i,j] = max { M[i,j], N[i,j] }, and the intersection V is V[i,j] = min { M[i,j], N[i,j] }. The score for a pair of texts is score = (sum of V[i,j] over all i, j) / (sum of U[i,j] over all i, j), a formula suggested by Arnt Gulbrandsen. Here we have score = 2 / 6, or one third. The implementation differs from this in a few details. Most importantly, repetitions are ignored; for input "xxx", M[x,x] equals 1, not 2. */ /* Every character is assigned to one of 20 buckets so that the co-occurrence matrix requires only 20 * 20 = 400 bits, not 256 * 256 = 65536 bits or even more if we want the whole Unicode. Which character falls in which bucket is arbitrary. The second half of the table is a replica of the first half, because of laziness. */ static const int indexOf[256] = { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, // ! " # $ % & ' ( ) * + , - . / 0, 2, 6, 7, 10, 12, 15, 19, 2, 6, 7, 10, 12, 15, 19, 0, // 0 1 2 3 4 5 6 7 8 9 : ; < = > ? 1, 3, 4, 5, 8, 9, 11, 13, 14, 16, 2, 6, 7, 10, 12, 15, // @ A B C D E F G H I J K L M N O 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 6, 10, 11, 12, 13, 14, // P Q R S T U V W X Y Z [ \ ] ^ _ 15, 12, 16, 17, 18, 19, 2, 10, 15, 7, 19, 2, 6, 7, 10, 0, // ` a b c d e f g h i j k l m n o 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 6, 10, 11, 12, 13, 14, // p q r s t u v w x y z { | } ~ 15, 12, 16, 17, 18, 19, 2, 10, 15, 7, 19, 2, 6, 7, 10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 6, 7, 10, 12, 15, 19, 2, 6, 7, 10, 12, 15, 19, 0, 1, 3, 4, 5, 8, 9, 11, 13, 14, 16, 2, 6, 7, 10, 12, 15, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 6, 10, 11, 12, 13, 14, 15, 12, 16, 17, 18, 19, 2, 10, 15, 7, 19, 2, 6, 7, 10, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 6, 10, 11, 12, 13, 14, 15, 12, 16, 17, 18, 19, 2, 10, 15, 7, 19, 2, 6, 7, 10, 0 }; /* The entry bitCount[i] (for i between 0 and 255) is the number of bits used to represent i in binary. */ static const int bitCount[256] = { 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8 }; struct CoMatrix { /* The matrix has 20 * 20 = 400 entries. This requires 50 bytes, or 13 words. Some operations are performed on words for more efficiency. */ union { TQ_UINT8 b[52]; TQ_UINT32 w[13]; }; CoMatrix() { memset( b, 0, 52 ); } CoMatrix( const char *text ) { char c = '\0', d; memset( b, 0, 52 ); /* The Knuth books are not in the office only for show; they help make loops 30% faster and 20% as readable. */ while ( (d = *text) != '\0' ) { setCoocc( c, d ); if ( (c = *++text) != '\0' ) { setCoocc( d, c ); text++; } } } void setCoocc( char c, char d ) { int k = indexOf[(uchar) c] + 20 * indexOf[(uchar) d]; b[k >> 3] |= k & 0x7; } int worth() const { int w = 0; for ( int i = 0; i < 50; i++ ) w += bitCount[b[i]]; return w; } }; static inline CoMatrix reunion( const CoMatrix& m, const CoMatrix& n ) { CoMatrix p; for ( int i = 0; i < 13; i++ ) p.w[i] = m.w[i] | n.w[i]; return p; } static inline CoMatrix intersection( const CoMatrix& m, const CoMatrix& n ) { CoMatrix p; for ( int i = 0; i < 13; i++ ) p.w[i] = m.w[i] & n.w[i]; return p; } CandidateList similarTextHeuristicCandidates( const MetaTranslator *tor, const char *text, int maxCandidates ) { TQValueList scores; CandidateList candidates; CoMatrix cmTarget( text ); int targetLen = tqstrlen( text ); TML all = tor->translatedMessages(); TML::Iterator it; for ( it = all.begin(); it != all.end(); ++it ) { if ( (*it).type() == MetaTranslatorMessage::Unfinished || (*it).translation().isEmpty() ) continue; TQString s = tor->toUnicode( (*it).sourceText(), (*it).utf8() ); CoMatrix cm( s.latin1() ); int delta = TQABS( (int) s.length() - targetLen ); /* Here is the score formula. A comment above contains a discussion on a similar (but simpler) formula. */ int score = ( (intersection(cm, cmTarget).worth() + 1) << 10 ) / ( reunion(cm, cmTarget).worth() + (delta << 1) + 1 ); if ( (int) candidates.count() == maxCandidates && score > scores[maxCandidates - 1] ) candidates.remove( candidates.last() ); if ( (int) candidates.count() < maxCandidates && score >= 190 ) { Candidate cand( s, (*it).translation() ); int i; for ( i = 0; i < (int) candidates.count(); i++ ) { if ( score >= scores[i] ) { if ( score == scores[i] ) { if ( candidates[i] == cand ) goto continue_outer_loop; } else { break; } } } scores.insert( scores.at(i), score ); candidates.insert( candidates.at(i), cand ); } continue_outer_loop: ; } return candidates; }