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#include "simtexth.h"

#include <metatranslator.h>

#include <qmap.h>
#include <qstring.h>
#include <qstringlist.h>
#include <qlist.h>

#include <string.h>

typedef QList<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 {
    Q_UINT8 b[52];
    Q_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 )
{
    QList<int> scores;
    CandidateList candidates;
    CoMatrix cmTarget( text );
    int targetLen = qstrlen( text );

    TML all = tor->translatedMessages();

    foreach ( MetaTranslatorMessage mtm, all ) {
    if ( mtm.type() == MetaTranslatorMessage::Unfinished ||
         mtm.translation().isEmpty() )
        continue;

    QString s = tor->toUnicode( mtm.sourceText(), mtm.utf8() );
    CoMatrix cm( s.latin1() );
    int delta = qAbs( (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.removeAt( candidates.size()-1 );
    if ( (int) candidates.count() < maxCandidates && score >= 190 ) {
        Candidate cand( s, mtm.translation() );

        int i;
        for ( i = 0; i < (int) 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;
}
