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Subject: Re: Bit counting revisited

Author: Robert Hyatt

Date: 19:39:37 04/19/00

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On April 19, 2000 at 22:10:19, Flemming Rodler wrote:

>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.
>

Not worth testing. I did this exhaustively when I chose the bit counting
code.  I don't count things with several bits set very often...  by far
most of the counts are < 4, with the majority == 0...  I don't count pawns
using the bitmap, this is computed incrementally as part of the material
signature...




>
>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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