/**********************************************************************
**   Copyright (C) 2000 Trolltech AS.  All rights reserved.
**
**   numberh.cpp
**
**   This file is part of Qt Linguist.
**
**   See the file LICENSE included in the distribution for the usage
**   and distribution terms.
**
**   The file is provided AS IS with NO WARRANTY OF ANY KIND,
**   INCLUDING THE WARRANTY OF DESIGN, MERCHANTABILITY AND FITNESS FOR
**   A PARTICULAR PURPOSE.
**
**********************************************************************/

#include <qarray.h>
#include <qcstring.h>
#include <qmap.h>
#include <qstringlist.h>

#include <ctype.h>
#include <metatranslator.h>

typedef QMap<QCString, MetaTranslatorMessage> TMM;
typedef QValueList<MetaTranslatorMessage> TML;

static bool isDigitFriendly( int c )
{
    return ispunct( c ) || isspace( c );
}

static int numberLength( const char *s )
{
    int i = 0;

    if ( isdigit(s[0]) ) {
	do {
	    i++;
	} while ( isdigit(s[i]) ||
		  (isDigitFriendly(s[i]) &&
		   (isdigit(s[i + 1]) ||
		    (isDigitFriendly(s[i + 1]) && isdigit(s[i + 2])))) );
    }
    return i;
}

static QCString zeroKey( const char *key )
{
    QCString zeroed( strlen(key) + 1 );
    char *z = zeroed.data();
    int i = 0, j = 0;
    int len;

    while ( key[i] != '\0' ) {
	len = numberLength( key + i );
	if ( len > 0 ) {
	    i += len;
	    z[j++] = '0';
	} else {
	    z[j++] = key[i++];
	}
    }
    z[j] = '\0';
    return zeroed;
}

static QString translationAttempt( const QString& oldTranslation,
				   const char *oldSource,
				   const char *newSource )
{
    int p = zeroKey( oldSource ).contains( '0' );
    QString attempt;
    QStringList oldNumbers;
    QStringList newNumbers;
    QArray<bool> met( p );
    QArray<int> matchedYet( p );
    int i, j;
    int k = 0, ell, best;
    int m, n;
    int pass;

    /*
      This algorithm is hard to follow, so we'll consider an example all along:
      oldTranslation is "XeT 3.0", oldSource is "TeX 3.0" and newSource is
      "XeT 3.1".

      First, we set up two tables: oldNumbers and newNumbers.  In our example,
      oldNumber[0] is "3.0" and newNumber[0] is "3.1".
    */
    for ( i = 0, j = 0; oldSource[i] != '\0'; i++, j++ ) {
	m = numberLength( oldSource + i );
	n = numberLength( newSource + j );
	if ( m > 0 ) {
	    oldNumbers.append( QCString(oldSource + i, m + 1) );
	    newNumbers.append( QCString(newSource + j, n + 1) );
	    i += m;
	    j += n;
	    met[k] = FALSE;
	    matchedYet[k] = 0;
	    k++;
	}
    }

    /*
      We now go over the old translation, "XeT 3.0", one letter at a time,
      looking for numbers found in oldNumbers.  Whenever such a number is met,
      it is replaced with its newNumber equivalent.  In our example, the "3.0"
      of "XeT 3.0" becomes "3.1".
    */
    for ( i = 0; i < (int) oldTranslation.length(); i++ ) {
	attempt += oldTranslation[i];
	for ( k = 0; k < p; k++ ) {
	    if ( oldTranslation[i] == oldNumbers[k][matchedYet[k]] )
		matchedYet[k]++;
	    else
		matchedYet[k] = 0;
	}

	/*
	  Let's find out if the last character ended a match.  We make two
	  passes over the data.  In the first pass, we try to match only
	  numbers that weren't matched yet; if that fails, the second pass does
	  the trick.  This is useful in some suspicious cases, flagged below.
	*/
	for ( pass = 0; pass < 2; pass++ ) {
	    best = p;
	    for ( k = 0; k < p; k++ ) {
		if ( (!met[k] || pass > 0) &&
		     matchedYet[k] == (int) oldNumbers[k].length() &&
		     numberLength(oldTranslation.latin1() + (i + 1) -
				  matchedYet[k]) == matchedYet[k] ) {
		    if ( best == p || matchedYet[k] > matchedYet[best] )
			best = k;
		}
	    }
	    if ( best != p ) {
		attempt.truncate( attempt.length() - matchedYet[best] );
		attempt += newNumbers[best];
		met[best] = TRUE;
		for ( k = 0; k < p; k++ )
		    matchedYet[k] = 0;
		break;
	    }
	}
    }

    /*
      We flag two kinds of suspicious cases.  They are identified as such with
      comments such as "{2000?}" at the end.

      Example of the first kind: old source text "TeX 3.0" translated as
      "XeT 2.0" is flagged: "TeX 2.0 {3.0?}", no matter what the new text is.
    */
    for ( k = 0; k < p; k++ ) {
	if ( !met[k] )
	    attempt += QString( " {" ) + newNumbers[k] + QString( "?}" );
    }

    /*
      Example of the second kind: "1 of 1" translated as "1 af 1", with new
      source text "1 of 2", generates "1 af 2 {1 or 2?}" because it's not clear
      which of "1 af 2" and "2 af 1" is right.
    */
    for ( k = 0; k < p; k++ ) {
	for ( ell = 0; ell < p; ell++ ) {
	    if ( k != ell && oldNumbers[k] == oldNumbers[ell] &&
		    newNumbers[k] < newNumbers[ell] )
		attempt += QString( " {" ) + newNumbers[k] + QString( " or " ) +
			   newNumbers[ell] + QString( "?}" );
	}
    }
    return attempt;
}

/*
  Augments a MetaTranslator with translations easily derived from similar
  existing (likely obsolete) translations.

  For example, if "TeX 3.0" is translated as "XeT 3.0" and "TeX 3.1" has no
  translation, "XeT 3.1" is added to the translator and is marked Unfinished.
*/

void applyNumberHeuristic( MetaTranslator *tor )
{
    TMM translated, untranslated;
    TMM::Iterator t, u;
    TML all = tor->messages();
    TML::Iterator it;

    for ( it = all.begin(); it != all.end(); ++it ) {
	if ( (*it).type() == MetaTranslatorMessage::Unfinished )
	    untranslated.insert( zeroKey((*it).sourceText()), *it );
	else if ( !(*it).translation().isEmpty() )
	    translated.insert( zeroKey((*it).sourceText()), *it );
    }

    for ( u = untranslated.begin(); u != untranslated.end(); ++u ) {
	t = translated.find( u.key() );
	if ( t != translated.end() ) {
	    MetaTranslatorMessage m( *u );
	    m.setTranslation( translationAttempt((*t).translation(),
						 (*t).sourceText(),
						 (*u).sourceText()) );
	    tor->insert( m );
	}
    }
}
