Author: Ralf Elvsén
Date: 18:00:46 01/28/02
Go up one level in this thread
On January 28, 2002 at 20:46:04, Ralf Elvsén wrote:
>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
The above is not correct. The most significant bit in the rank
is not nr 7 any longer. I should know better then improvise
bit manipulation 3 am, sorry :)
Ralf
>
>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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