Author: Robert Hyatt
Date: 08:01:28 11/18/02
Go up one level in this thread
On November 18, 2002 at 10:20:49, Vincent Diepeveen wrote:
>On November 17, 2002 at 13:30:22, Eugene Nalimov wrote:
>
>6 compares times 0.5 cycles ==> 3 cycles
>5 shiftlefts times 2 cycles ==> 10 cycles
>
>So it is like 13 cycles to get a single bit out of a bitboard in
>this C code is very slow on the P4. 13 cycles is hell of a lot,
>i can generate several moves in non-bitboards within that time!
Unless something has changed, the P4 is clocked 2x internally. A shift should
take 1/2 clock cycle if I am reading their manual correctly...
That turns this into 5.5 cycles which seems more reasonable. No reason a shiftl
would be that much slower than a compare... And your counting doesn't make any
sense to me anyway as 6 compares by themselves won't do a thing. it would
require
jumps or cmov's or something...
>
>>On November 17, 2002 at 13:25:55, Vincent Diepeveen wrote:
>>
>>>On November 17, 2002 at 12:01:02, Gerd Isenberg wrote:
>>>
>>>>On November 17, 2002 at 10:53:05, Joel wrote:
>>>>
>>>>>Hey All,
>>>>>
>>>>>I am a 2nd year Uni student from Australia who has recently gotten into chess
>>>>>programming. My first attempt was a simple array-based alpha-beta variant which
>>>>>struggled to search more than 6 levels deep in most positions! I think that
>>>>>might have something to do with the fact that there was no move ordering,
>>>>>transposition table, an expensive evaluation function, no killer moves and weak
>>>>>coding :)
>>>>>
>>>>>I have been working on my second attempt for some time now. It uses Bitboards. I
>>>>>have a few questions regarding move generation.
>>>>>
>>>>>It seems to me that the performance of the Bitboard approach relies somewhat
>>>>>heavily on how fast you can retrieve the position of a 1 bit within a 64-bit
>>>>>unsigned integer. I looked for sometime on the Internet for some kind of
>>>>>magical, hacky solution to this dilemna, and the best I could find was this (b &
>>>>>-b) trick which I used in a debatedly clever way. I was just wondering if there
>>>>>is any approach significantly better than the one which I will outline below:
>>>>>
>>>>>1. (b & -b) to clear all 1 bit's except for one.
>>>>>2. get this value, mod it by 67 (which has the property that every possible
>>>>> value returned is unique, thus i can hash to the position of the bit in the
>>>>> 64 bit integer.)
>>>>>
>>>>>I am no expert, but it doesn't seem too ineffecient to me. Any problems?
>>>>>
>>>>>Also, if there are any improvements, I would prefer to find out about the ones
>>>>>which do not involve assembly coding - I do not want to make my program too
>>>>>dependant on any one CPU architecture at this stage.
>>>>>
>>>>>Thanks for your time,
>>>>>Joel
>>>>
>>>>Hi Joel,
>>>>
>>>>nice idea with the mod 67 to get unique bitvalues. But i fear a 64 bit div/mod
>>>>operation is too slow, even with 64bit-processors. If you don't want to use
>>>>assembler eg. intels x86 bsf (bit scan foreward), i think using a lookup table
>>>>indexed by the byte or 16-bit word is most common.
>>>>
>>>>I use this approach:
>>>>
>>>>#define LOWBOARD(bb) (*((UINT32*)&(bb)))
>>>>#define HIGHBOARD(bb) (*(((UINT32*)&(bb))+1))
>>>>
>>>>// precondition: bb not null
>>>>__forceinline UINT BitSearch(BitBoard bb)
>>>>{
>>>> ASSERT(bb != 0);
>>>>#ifdef _M_IX86
>>>> __asm
>>>> {
>>>> bsf eax,[bb+4]
>>>> xor eax, 32
>>>> bsf eax,[bb]
>>>> }
>>>>#else
>>>> BitBoard lsbb = bb & (-(__int64)bb);
>>>> UINT32 lsb = LOWBOARD(lsbb) | HIGHBOARD(lsbb);
>>>> return ((((((((((HIGHBOARD(lsbb)!=0) <<1)
>>>> +((lsb & 0xffff0000)!=0))<<1)
>>>> +((lsb & 0xff00ff00)!=0))<<1)
>>>> +((lsb & 0xf0f0f0f0)!=0))<<1)
>>>> +((lsb & 0xcccccccc)!=0))<<1)
>>>> +((lsb & 0xaaaaaaaa)!=0);
>>>>#endif
>>>>}
>>>>
>>>>Regards,
>>>>Gerd
>>>
>>>Isn't shifting on the P4 *WAY* slower than his mod 67 idea is?
>>
>>No.
>>
>>Thanks,
>>Eugene
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