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-/**********************************************************************
-** Copyright (C) 2000-2008 Trolltech ASA. 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 <metatranslator.h>
-
-#include <tqcstring.h>
-#include <tqdict.h>
-#include <tqmap.h>
-#include <tqstring.h>
-#include <tqstringlist.h>
-#include <tqvaluelist.h>
-
-#include <string.h>
-
-typedef TQValueList<MetaTranslatorMessage> 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<int> 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;
-}