Author: Ralf Elvsén
Date: 03:10:53 01/29/02
Go up one level in this thread
On January 29, 2002 at 04:10:07, Sune Fischer wrote:
>On January 28, 2002 at 21:00:46, Ralf Elvsén wrote:
>
>>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".
>
>
>the idea is this:
>
>occupied = 110110011000111
>x=8
>occupied^(occypied-x*2)= 000000001111100
>
>which are the attacked squares after you mask it in the direction you need.
>
>There are some wrap-around problems for the diagonals and for the rank (the file
>is no problem).
>Example: when a rook on b1 is shifted to a1, the square a2 becomes h1 (a first
>rank square), this is bad because a rook on b1 can't attack a piece on a2 and
>this bit will be set if a2 is occupied and the rest of first rank is empty.
>So I think my above code was not completely correct,
>one more masking at the beginning or end is needed to fix that problem.
Yes, that was what I was trying to do. If (in this case) h1 is set you
don't have to mask after the occ^(occ-2) -thing. I think the easiest way
maybe is.
x = (occ | 0x8080808080808080) >> square
a = (x^(x-2)) >> square
>
>Anyway, it is easier to test and debug than rotated and it may even be faster >on some machines where memory bandwith is a huge bottleneck.
Yes, I think it can be just fine for move generation. But I'm not so
sure how it fits in regarding the other things now handled by e.g.
rotated bitboards.
Ralf
>
>>>Ralf
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