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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