Author: Matt Taylor
Date: 03:35:19 01/22/03
Go up one level in this thread
On January 22, 2003 at 06:09:22, Dann Corbit wrote:
>On January 22, 2003 at 05:33:20, Matt Taylor wrote:
>
>>On January 22, 2003 at 04:27:16, Dann Corbit wrote:
>>
>>>On January 22, 2003 at 03:29:05, Matt Taylor wrote:
>>>
>>>>On January 21, 2003 at 17:03:33, Dann Corbit wrote:
>>>>
>>>>>On January 21, 2003 at 15:48:57, Sander de Zoete wrote:
>>>>>
>>>>>>The following instruction I found for the new architecture of Intel Chips
>>>>>>
>>>>>>BSWAP—Byte Swap
>>>>>>
>>>>>>Description
>>>>>>Reverses the byte order of a 32-bit (destination) register:
>>>>>>
>>>>>>Operation
>>>>>>TEMP ¬ DEST
>>>>>>DEST(7..0) ¬ TEMP(31..24)
>>>>>>DEST(15..8) ¬ TEMP(23..16)
>>>>>>DEST(23..16) ¬ TEMP(15..8)
>>>>>>DEST(31..24) ¬ TEMP(7..0)
>>>>>>Flags Affected
>>>>>>None.
>>>>>>Opcode Instruction Description
>>>>>>0F C8+ rd BSWAP r32 Reverses the byte order of a 32-bit register.
>>>>>>
>>>>>>It is only valid for 486 architecture.
>>>>>>
>>>>>>If this instruction can be used, it should be very easy to reverse Reverse
>>>>>>BitBoards into Forward Boards again. Saving a lot of updating in make and unmake
>>>>>>move and also be much faster using the bitboards for generating attack
>>>>>>board for evaluation purposes, incheck checks etc.
>>>>>>
>>>>>>The thing left to do is to reverse the bits per byte.
>>>>>>
>>>>>>I use the following trick, but it doesn't seem to work voor the value 9. Let me
>>>>>>show you what I mean: (and ofcourse, can you help me?)
>>>>>>
>>>>>>In C code, it looks like this:
>>>>>>
>>>>>>typedef unsigned __int64 BITBOARD;
>>>>>>
>>>>>>void ReverseBitsPerByte(BITBOARD bitboard)
>>>>>>{
>>>>>>// 1. Load the constant, k = 5555555555555555h
>>>>>>BITBOARD k = 0x5555555555555555;
>>>>>>
>>>>>>// 2. x = [(x shl 1) and k] or [(x and k) shr 1] result is: EFCDAB8967452301
>>>>>>bitboard = ((bitboard<<1) & k) | ((bitboard & k)>>1 );
>>>>>>}
>>>>>>
>>>>>>Initial bitboard:
>>>>>>11111110
>>>>>>11011100
>>>>>>10111010
>>>>>>10011000
>>>>>>01110110
>>>>>>01010100
>>>>>>00110010
>>>>>>00010000
>>>>>>
>>>>>>Result of ReverseBitsPerByte(bitboard):
>>>>>>01111111
>>>>>>00111011
>>>>>>01011101
>>>>>>00011000<--what the ^*&* goes wrong here? Should be 00011001.
>>>>>>01101110
>>>>>>00101010
>>>>>>01001100
>>>>>>00001000
>>>>>>
>>>>>>Thanks for any help or suggestions.
>>>>>
>>>><snip code>
>>>>
>>>>These are all basically the same algorithm. It's going to take a while to
>>>>rearrange everything when you do it bit-by-bit. Sander has the right idea by
>>>>first swapping bytes and then trying to reverse the bits within the bytes (8*2
>>>>ops + the swap as opposed to 64*2 ops).
>>>
>>>The algorithms don't perform identically.
>>>They don't require 64 ops.
>>>Once you get the "Sander" algorithm completed, compare the two (and also the
>>>COBRA algorithm).
>>
>>Yes, breverse5 is a divide-and-conquer algorithm. The others all iterate 64
>>times. Assuming 2 ops/iteration was a gross underestimate as they use heavy
>>masking and shifting.
>>
>>2 bswaps (and probably a mov, too) will reverse the bytes; afterward, the bits
>>can be reversed as well.
>>
>>Not sure when or if I can have a look at COBRA and think about the algorithm --
>>I've got more than enough mandatory work right now.
>
>The COBRA algorithm is designed for big bit sequences. I don't know how well it
>adapts to 64 bits. I have only just downloaded the paper.
>
>This algorithm:
>
>typedef unsigned long ling Bitboard;
>
>Bitboard breverse5(Bitboard n)
>{
> int i = 64;
> Bitboard m = -1;
> while (i /= 2)
> m ^= m << i, n = n >> i & m | (n & m) << i;
> return n;
>}
>
>iterates 6 times. I suspect it will be faster than something table driven.
>Perhaps also faster than 8 mask and shift operations.
>
>I'll be surprised if anything can beat it by 20%.
Is that a challenge? :-)
Grr...I have so many things I need to get done...and now you go and entice me to
do more optimization!
-Matt
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