Author: Dann Corbit
Date: 01:27:16 01/22/03
Go up one level in this thread
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).
This page took 0.01 seconds to execute
Last modified: Thu, 15 Apr 21 08:11:13 -0700
Current Computer Chess Club Forums at Talkchess. This site by Sean Mintz.