Author: Dieter Buerssner
Date: 08:51:20 05/28/04
Go up one level in this thread
On May 28, 2004 at 11:43:31, Gerd Isenberg wrote:
>On May 28, 2004 at 11:13:09, Dieter Buerssner wrote:
>
>>On May 28, 2004 at 10:56:59, Gerd Isenberg wrote:
>>
>>>int popCount(BitBoard bb)
>>>{
>>> unsigned int l = low(bb);
>>> unsigned int h = high(bb);
>>> l -= (l >> 1) & 0x55555555; // count 2-bit groups
>>> h -= (h >> 1) & 0x55555555; // count 2-bit groups
>>> l = (l & 0x33333333) + ((l >> 2) & 0x33333333); // count 4-bit groups
>>> h = (h & 0x33333333) + ((h >> 2) & 0x33333333); // count 4-bit groups
>>
>>l += h; // and do the next stage on l only
>>
>>> l = (l + (l >> 4)) & 0x0f0f0f0f; // count 8-bits groups
>>> h = (h + (h >> 4)) & 0x0f0f0f0f; // count 8-bits groups
>>> return ((h+l) * 0x01010101) >> 24; // sum 8-bit results
>>>}
>>
>>Gerd, I think, you can combine the 2 halves already one stage earlier. Or did I
>>miscaclulate this? I thought, the individual counters cannot overflow at that
>>stage already. See also my post
>>http://chessprogramming.org/cccsearch/ccc.php?art_id=349781
>>
>>For me, adding two counter combining stages was faster, than the multiplication.
>I wrote it from scratch to verify the "packing" trick.
>But it seems your add is a bit too early, since 8+8 may overflow to 0x10.
I think, the maximum of the individual counters will be 4 after the
"0x33333333-stage". Not? I should try it out, instead of thinking about it ...
Cheers,
Dieter
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