Author: Ralf Elvsén
Date: 17:46:04 01/28/02
Go up one level in this thread
On January 28, 2002 at 13:26:02, Sune Fischer wrote:
>On January 28, 2002 at 13:04:05, Sune Fischer wrote:
>
>This is what I get for half the rook, similar stuff is done for the backwards
>attacks using the reversed occupied bitboard.
>
>BITBOARD RookAttacksForward(BOARD &occupied, char square)
>{
> BITBOARD a,x,y;
>
> x=occupied>>square; // shift board so square ends in lower left corner
> y=x&RANK1; // mask out ewerything but the first rank
> a=y^(y-2); // get the attacked bits on this rank
> y=x&FILE1; // now mask so we get a clean first file
> y=y^(y-2); // get the attacked squares
> a=a|(y&FILE1); // add the attacked bits after masking with the file again
> a=a<<square); // shift the attacked bits back (for ease of readability)
>
> return a;
>}
>
>
>..untested, but something along those lines.
>Not entirely sure how to do it for the bishop, I guess we just shift to the
>lower right corner when attacking north-west and lower left when attacking
>north-east.
>
>But what happens if x=0?
>Will y then be 0 or something strange, the bitboard is unsigned so it should
>remain 0 right?
>
>-S.
I think one should change
y = x&RANK1
to
y = x|0x80
That way the most significant bit in the rank will be set and when
you do the x^(x-2)-thing you will get a correct answer.
I don't understand how you can use the same "occupied" to get
the file attacks but it's late here... :)
x will never be zero since there is slider at "square".
Ralf
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