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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 tqcontains 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;
+}