Calculating TO

[grant]

Hi

If my rook can check the king, then by calculation, the TO square is

(oppKing >> 3) + (from & 7) and/or
(from >> 3) + (oppKing & 7)

If my bishop can check the king, is there a formula for calculating the TO squares?

Grant

[Sven]

What exactly do you mean by "TO square": the square from which the rook can give check to the king?

And what is your board representation (it does not quite look like "0x88")?

Sven

[grant]

Sven

The TO square means the square/s that the rook moves to to check the king.
My board representation is A8=0, B8=1 ... H1=63, and I have a bitboard engine.

I just thought of trying a quick calculation to get my TO square in my generate checks function, which also does not require magics.

Grant

[Sven]

In case of bitboards, why don't you do an intersection (using precalculated tables) of all_bishop_targets[from] & all_bishop_targets[opp_king] that results in up to two squares, provided the bishop is not on the same diagonal as the opp_king already?

Same for rook: all_rook_targets[from] & all_rook_targets[opp_king], although I don't know whether the formula you gave wouldn't be slightly faster in the rook case.

For queen the same but there you may get up to four squares.

Sven

[Zach Wegner]

For queen the same but there you may get up to four squares.

That's actually 12.

[Edmund]

Sven

The TO square means the square/s that the rook moves to to check the king.
My board representation is A8=0, B8=1 ... H1=63, and I have a bitboard engine.

I just thought of trying a quick calculation to get my TO square in my generate checks function, which also does not require magics.

Grant

assuming the definition:
#define getDIA(sq) (7 - ((sq)&7) + ((sq)>>3))
#define getADIA(sq) (((sq)&7) + ((sq)>>3))

now getting the intersection of these two values is the problem:
one possible solution would be:
adia+3.5*(adia+dia-7)

but sofar I didn't come up with a solution to find out whether the two are intersecting at all

[Edmund]

but sofar I didn't come up with a solution to find out whether the two are intersecting at all

to test whether they are intersecting:
(dia + adia - 7) = even
and
(dia + adia - 7) / 2 <= dia


maybe you can find an easier way Grant

[grant]

Sven

That is exactly what I was trying to avoid.

For rooks giving check, I simply lookup the bitboard containing all the squares the rook has to pass over to capture the king (if you see what I mean), and AND it with ALLOccupied. If the result is zero, a non-capture check is possible and the TO square can be calculated (as above).

Same with bishops but at the moment I have to use intersecting bitboards to get my TO squares.

Grant

[Edmund]

so in short:
TO = (9*(sqA>>3 + sqA&7)+7*(sqB>>3 - sqB&7))/2

but be aware that TO has to be checked for 0 <= TO <= 63
and sqA and sqB must be on equal colored squares ((sqA+sqB) & 1) == 0

[grant]

Thanks Edmund

The rook calculation above should have read

(sqA & 56) + (sqB & 7) and/or
((sqB & 56) + (sqA & 7)

Oops.

Grant