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#include "simtexth.h"
#include "translator.h"
#include <QtCore/QByteArray>
#include <QtCore/QString>
#include <QtCore/QList>
QT_BEGIN_NAMESPACE
typedef QList<TranslatorMessage> 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 {
quint8 b[52];
quint32 w[13];
};
CoMatrix() { memset( b, 0, 52 ); }
CoMatrix(const QString &str)
{
QByteArray ba = str.toUtf8();
const char *text = ba.constData();
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' ) {
setCoOccurence( c, d );
if ( (c = *++text) != '\0' ) {
setCoOccurence( d, c );
text++;
}
}
}
void setCoOccurence( char c, char d ) {
int k = indexOf[(uchar) c] + 20 * indexOf[(uchar) d];
b[k >> 3] |= (1 << (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;
}
StringSimilarityMatcher::StringSimilarityMatcher(const QString &stringToMatch)
{
m_cm = new CoMatrix(stringToMatch);
m_length = stringToMatch.length();
}
int StringSimilarityMatcher::getSimilarityScore(const QString &strCandidate)
{
CoMatrix cmTarget(strCandidate);
int delta = qAbs(m_length - strCandidate.size());
int score = ( (intersection(*m_cm, cmTarget).worth() + 1) << 10 ) /
( reunion(*m_cm, cmTarget).worth() + (delta << 1) + 1 );
return score;
}
StringSimilarityMatcher::~StringSimilarityMatcher()
{
delete m_cm;
}
/**
* Checks how similar two strings are.
* The return value is the score, and a higher score is more similar
* than one with a low score.
* Linguist considers a score over 190 to be a good match.
* \sa StringSimilarityMatcher
*/
int getSimilarityScore(const QString &str1, const QString &str2)
{
CoMatrix cmTarget(str2);
CoMatrix cm(str1);
int delta = qAbs(str1.size() - str2.size());
int score = ( (intersection(cm, cmTarget).worth() + 1) << 10 )
/ ( reunion(cm, cmTarget).worth() + (delta << 1) + 1 );
return score;
}
CandidateList similarTextHeuristicCandidates(const Translator *tor,
const QString &text, int maxCandidates)
{
QList<int> scores;
CandidateList candidates;
TML all = tor->translatedMessages();
foreach (const TranslatorMessage &mtm, all) {
if (mtm.type() == TranslatorMessage::Unfinished
|| mtm.translation().isEmpty())
continue;
QString s = mtm.sourceText();
int score = getSimilarityScore(s, text);
if (candidates.size() == maxCandidates && score > scores[maxCandidates - 1] )
candidates.removeLast();
if (candidates.size() < maxCandidates && score >= textSimilarityThreshold) {
Candidate cand( s, mtm.translation() );
int i;
for (i = 0; i < candidates.size(); i++) {
if (score >= scores.at(i)) {
if (score == scores.at(i)) {
if (candidates.at(i) == cand)
goto continue_outer_loop;
} else {
break;
}
}
}
scores.insert(i, score);
candidates.insert(i, cand);
}
continue_outer_loop:
;
}
return candidates;
}
QT_END_NAMESPACE