goto encode_remaining; // With pawns we have finished special treatments
}
- // In positions withouth pawns, we further flip the squares to ensure leading
+ // In positions without pawns, we further flip the squares to ensure leading
// piece is below RANK_5.
if (rank_of(squares[0]) > RANK_4)
for (int i = 0; i < size; ++i)
// Rs "together" in 62 * 61 / 2 ways (we divide by 2 because rooks can be
// swapped and still get the same position.)
//
- // In case we have at least 3 unique pieces (inlcuded kings) we encode them
+ // In case we have at least 3 unique pieces (included kings) we encode them
// together.
if (entry->hasUniquePieces) {
+ (squares[1] - adjust1)) * 62
+ squares[2] - adjust2;
- // First piece is on a1-h8 diagonal, second below: map this occurence to
+ // First piece is on a1-h8 diagonal, second below: map this occurrence to
// 6 to differentiate from the above case, rank_of() maps a1-d4 diagonal
// to 0...3 and finally MapB1H1H7[] maps the b1-h1-h7 triangle to 0..27.
else if (off_A1H8(squares[1]))
idx *= d->groupIdx[0];
Square* groupSq = squares + d->groupLen[0];
- // Encode remainig pawns then pieces according to square, in ascending order
+ // Encode remaining pawns then pieces according to square, in ascending order
bool remainingPawns = entry->hasPawns && entry->pawnCount[1];
while (d->groupLen[++next])
// Group together pieces that will be encoded together. The general rule is that
// a group contains pieces of same type and color. The exception is the leading
-// group that, in case of positions withouth pawns, can be formed by 3 different
+// group that, in case of positions without pawns, can be formed by 3 different
// pieces (default) or by the king pair when there is not a unique piece apart
// from the kings. When there are pawns, pawns are always first in pieces[].
//
//
// This ensures unique encoding for the whole position. The order of the
// groups is a per-table parameter and could not follow the canonical leading
- // pawns/pieces -> remainig pawns -> remaining pieces. In particular the
+ // pawns/pieces -> remaining pawns -> remaining pieces. In particular the
// first group is at order[0] position and the remaining pawns, when present,
// are at order[1] position.
bool pp = e.hasPawns && e.pawnCount[1]; // Pawns on both sides
d->groupIdx[1] = idx;
idx *= Binomial[d->groupLen[1]][48 - d->groupLen[0]];
}
- else // Remainig pieces
+ else // Remaining pieces
{
d->groupIdx[next] = idx;
idx *= Binomial[d->groupLen[next]][freeSquares];
d->groupIdx[n] = idx;
}
-// In Recursive Pairing each symbol represents a pair of childern symbols. So
+// In Recursive Pairing each symbol represents a pair of children symbols. So
// read d->btree[] symbols data and expand each one in his left and right child
// symbol until reaching the leafs that represent the symbol value.
uint8_t set_symlen(PairsData* d, Sym s, std::vector<bool>& visited) {
for (auto p : bothOnDiagonal)
MapKK[p.first][p.second] = code++;
- // Binomial[] stores the Binomial Coefficents using Pascal rule. There
+ // Binomial[] stores the Binomial Coefficients using Pascal rule. There
// are Binomial[k][n] ways to choose k elements from a set of n elements.
Binomial[0][0] = 1;
for (int leadPawnsCnt = 1; leadPawnsCnt <= 5; ++leadPawnsCnt)
for (File f = FILE_A; f <= FILE_D; ++f)
{
- // Restart the index at every file because TB table is splitted
+ // Restart the index at every file because TB table is split
// by file, so we can reuse the same index for different files.
int idx = 0;