Author: Flemming Rodler
Date: 19:10:19 04/19/00
Dear all,
Thanks for all the post I got to my thread "Bit counting below". For the problem
of counting the number of bits in a 64 bit variable I implemented the 3 schemes
suggested in the answers plus another scheme that I found on the web. Here are
my findings.
Test computer: AMD K6-II 350 Mhz.
Test: I timed how long each scheme took by calling them 2^20 and 4*2^20 times.
This was done for different number bits set to one.
The different schemes are listed at the end of this post as Method 1,..
Result: for 2^20 repeatition.
#bits on: 1 2 3 4 5 6 7 8
----------------------------------------------------------------------
time in sec. Method 1: 0.07 0.10 0.12 0.15 0.17 0.19 0.22 0.24
Method 2: 0.15 regardless of #bits
Method 3: 0.20
Method 4: 0.20
Result: for 4*2^20 repeatition.
#bits on: 1 2 3 4 5 6 7 8
----------------------------------------------------------------------
time in sec. Method 1: 0.30 0.40 0.49 0.59 0.69 0.78 0.87 0.98
Method 2: 0.59 regardless of #bits
Method 3: 0.77
Method 4: 0.77
So we see that 4 bits is the break-even point for Method 1 and Method 2.
I do not know if it has any practical influence which methods you use in a chess
program but I might try it out with Crafty which I believe uses method 1 on a
PC. It would be interesting to change it to method 2 and see if there is a
significant change in the number of nodes searched per second. If so then it
might be good to use different versions when counting pawns and f.ex. queens.
I hope you will find the info useful and if you have any ideas of how to make
them even faster please let me know so I can update the test results.
Best regards
Flemming
Here are the 4 used methods. They were all compiled with gcc and -O3 turned onI
do not who to give credit for the various methods. Sorry for that. The first two
methods I have from the previous thread of posts. The third method I have from
the DarkThought webpages and the fourth method I got from the link supplied in
the previous posts.
Method 1:
int bitcount1(register unsigned64 b) {
register int c=0;
for(;b;c++,b &= b-1);
return c;
}
Method 2:
static char nbits[256] =
{
0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8,
};
int bitcount2(unsigned64 b)
{
register unsigned long *n = (unsigned long *) &b;
return
nbits[(unsigned char) n[0]] +
nbits[(unsigned char) (n[0] >> 8)] +
nbits[(unsigned char) (n[0] >> 16)] +
nbits[(unsigned char) (n[0] >> 24)] +
nbits[(unsigned char) n[1]] +
nbits[(unsigned char) (n[1] >> 8)] +
nbits[(unsigned char) (n[1] >> 16)] +
nbits[(unsigned char) (n[1] >> 24)];
}
Method 3: I added the register, but it not make any difference
int bitcount3(register unsigned64 b) {
register unsigned32 n;
register const unsigned64 a = b - ((b >> 1) & m1);
register const unsigned64 c = (a & m2) + ((a >> 2) & m2);
n = ((unsigned32) c) + ((unsigned32) (c >> 32));
n = (n & 0x0F0F0F0F) + ((n >> 4) & 0x0F0F0F0F);
n = (n & 0xFFFF) + (n >> 16);
n = (n & 0xFF) + (n >> 8);
return n;
}
Method 4:
int bitcount4(register unsigned64 b)
{
register unsigned long i1,i2;
i1 = ((b & 0xAAAAAAAAL) >> 1) + (b & 0x55555555L);
i1 = ((i1 & 0xCCCCCCCCL) >> 2) + (i1 & 0x33333333L);
i1 = ((i1 & 0xF0F0F0F0L) >> 4) + (i1 & 0x0F0F0F0FL);
i1 = ((i1 & 0xFF00FF00L) >> 8) + (i1 & 0x00FF00FFL);
i1 = ((i1 & 0xFFFF0000L) >> 16) + (i1 & 0x0000FFFFL);
i2 = (((b>>32) & 0xAAAAAAAAL) >> 1) + ((b>>32) & 0x55555555L);
i2 = ((i2 & 0xCCCCCCCCL) >> 2) + (i2 & 0x33333333L);
i2 = ((i2 & 0xF0F0F0F0L) >> 4) + (i2 & 0x0F0F0F0FL);
i2 = ((i2 & 0xFF00FF00L) >> 8) + (i2 & 0x00FF00FFL);
i2 = ((i2 & 0xFFFF0000L) >> 16) + (i2 & 0x0000FFFFL);
return (int)(i1+i2);
}
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