Author: Gerd Isenberg
Date: 01:55:25 04/09/04
Go up one level in this thread
On April 09, 2004 at 02:30:36, rasjid chan wrote:
>
>
>I have a set of simple macros for bitboard operations and they are meant for
>those with a new chess program and like to implement fast complete
>attacks by bishops and rooks. It is basically just three #define and all sliding
>attacks would then be available for use in evaluations and move ordering. Of
>course it is only when we don't yet count instruction cycles,
>but then after things are moving we could EASILY upgrade to
>rotated bitboards.
>
>The only requirement is that the "all" bits must be available. Most chess
>programmers sooner or later will have to use some bitboard operations as
>there seem no escape for the sliding pieces and it is usually necessary to
>incrementally update the "all" bits after makemove()and unmake().
>
>These bitboard operations just make use of simple direct intuitive
>operations.If we have a single bit x, then x - 1 divides the board into
>upper half and lower half and add to it some bitwise operations which
>everyone say is fast. The only memory access is U64 diagonal[2][15]
>and it is much better than U64 attack[64][64].
>For those not using 0x88, they just need some simple editing.
>
>If anyone has something faster or can make them faster, please post.
>
>Best Regards
>Rasjid
>
>
>typedef unsigned _int64 U64
>
>#define GetFile64(sq) ((sq) & 7)
>#define GetRank64(sq) ((sq) >> 3)
>#define iGetFile(sq88) ((sq88) & 7)
>#define iGetRank(sq88) ((sq88) >> 4)
>#define sq88Bit(sq88) ((U64)1 << (((sq88) + ((sq88) & 7)) >> 1))
>#define sq64Bit(sq64) (((U64)1) << (sq64))
>#define sq88RankBits(sq88) ((U64)0xff << ((sq88) >> 1 & 0370))
>#define sq88FileBits(sq88) ((U64) 0x0101010101010101 << ((sq88) & 7))
>
>//*** global variables
>U64 all, diagonal[2][15];
>
>
>//*** Start - bitboard attack operations for sliding pieces
>
>/*
> This test that the in-between arc (x, y) is clear of any bits.
> path is the complete bit path along x, y across the board
> */
>
>#define brq_path_clear(x, y, path)
> ((x) <= (y)
> ? ((x) == (y) || isqBit(y) - 1 & ~(isqBit(x) - 1 | isqBit(x))
> & (path) & all ? 0 : 1)
> : (isqBit(x) - 1 & ~(isqBit(y) - 1 | isqBit(y))
> & (path) & all ? 0 : 1))
>
>/*
> given any two sq88 x, y , this test if they attack along
> diagonals.
> */
>
>#define bstyle_attack(x, y)
> (iGetRank(x) + iGetFile(x) == iGetRank(y) + iGetFile(y)
> ? (brq_path_clear(x, y, diagonal[1][iGetRank(x)
> + iGetFile(x)]) ? 1 : 0)
> : (iGetRank(x) - iGetFile(x) == iGetRank(y) - iGetFile(y)
> ? (brq_path_clear(x, y, diagonal[0][7 + iGetRank(x) - iGetFile(x)])
> ? 1 : 0) : 0))
>
>/*
> given any two sq88 x, y , this test if they attack along
> rank or file.
> */
>
>#define rstyle_attack(x, y)
> (iGetRank(x) == iGetRank(y)
> ? (brq_path_clear(x, y, isqRankBits(x)) ? 1 : 0)
> : (iGetFile(x) == iGetFile(y)
> ? (brq_path_clear(x, y, isqFileBits(x)) ? 1 : 0) : 0))
>
>
>//*** End - bitboard attack operations for sliding pieces
<init code snipped>
Hi Rasjid,
Well, i have some problems to read/understand your macros ;-(
Didn't you need some backslashes to connect all that lines to _one_ macro line?
As far as i understand they act like boolean functions, whether two squares are
connected on an appropriate ray, or whether y is attacked by a bishop/rook on x
(or vice versa).
So you do something like ...
bool isAttacked(BitBoard all, int targetSq, int sourceSq)
{
// precondition: both squares share a common ray
BitBoard inbetween = bbSquaresInbetween[targetSq][sourceSq];
return (inbetween & all) == 0;
}
... but you want to get rid of the huge 64*64 array and do some computation...
Where do you need that macros? To check the validity of a move (e.g. from
hashtable) or to genetate moves or attacks with it, as your subject title
suggests? It it would be interesting to see the usage of bstyle_attack(x,y) and
rstyle_attack(x,y) macros. How do you compute sliding attacks with this macros?
I am not quite sure, but my feeling with your macros is, that they are a bit to
complicated and therefore difficult to read for me ;-)
Note that those boolean statements
booleanExpression ? true : false
booleanExpression ? 1 : 0
booleanExpression
are equivalent.
Expressions with relational operators (==,!=,...) are boolean expressions with
the value range {0,1}, as well a boolean operators (&&,||,!).
Cheers,
Gerd
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