Author: Dan Honeycutt
Date: 11:41:36 02/17/04
Go up one level in this thread
On February 17, 2004 at 13:27:33, Robert Hyatt wrote: >On February 17, 2004 at 12:19:06, Sune Fischer wrote: > >>On February 17, 2004 at 11:49:33, Robert Hyatt wrote: >> >>>On February 17, 2004 at 10:48:35, Sune Fischer wrote: >>> >>>>On February 16, 2004 at 22:41:00, Robert Hyatt wrote: >>>> >>>>>This will make bitmaps insanely difficult to visualize. Remember that bit 0 is >>>>>the LSB (or rightmost bit). That means your chess board is going to look like >>>>>this when you display a bitboard as a hex number: >>>>> >>>>> >>>>> >>>>> h1 g1 f1 e1 d1 c1 b1 a1 >>>>> >>>>>Because the rightmost 8 bits would be displayed in that order. >>>> >>>>Correct. >>>> >>>>> That could drive >>>>>someone to drink drain cleaner. I might be more tempted to make bit 0 either >>>>>square h8, or h1 instead, so that things are not impossible to debug... >>>> >>>>Why on earth would you want to do something horrible like that? >>>> >>> >>> >>>When I display a 64 bit word, it pops out in hex. I want some simple way to >>>translate that 64 bit image into what it represents about each square, so that I >>>can do it mentally. Currently take the 64 bit word in groups of 8 bits. >>>leftmost 8 bits is rank 1, next 8 bits is rank 2, all easy to visualize. >> >>Of course I want the same, that is why I don't get your 'backwards' way of doing >>it. >> >>bitboard b=..; >> >>rank 1: b&0xff >>rank 2: (b>>8)&0xff >>etc. >> >>I think this orientation is a lot easier mentally due to it corresponding well >>with a coordinate system, everything begins at the lower left corner! >> You never begin a coordinate system out at h1 for instance, that would make the >>x-axis negative :-) >> >>Ok, perhaps I am too math-impaired to accept anything else, it simply _must_ be >>this way for me, going against 15 years of school is too much to handle for me >>:) >> >>>I want >>>to renumber the bits, but also make the 64 bit value something I can look at and >>>then visualize without having to "flip" the board as might happen if I just >>>change the square numbers and leave the bit/square correspondence alone. >> >>You have to flip the board by your method AFAICT. >>At least I don't see how it maps easily starting at some obscure location like >>h1, do you go up or down from there or just wizz around randomly? >>How can normality be reconstituted? >>;-) > >I only want 8 consecutive bits to represent 8 consecutive squares, both going >left-to-right. > >Right now I do that. the left-most 8 bits of a word are rank 1, next 8 are rank >2, etc. in an 8 bit chunk, left bit = a file, right bit = h file. I can deal >with that. > >Uri's suggestion inverted that so that left bit = right file, etc, and that was >what I was saying I could not mentally deal with. It makes a lot of sense for >a1 (first square) to either be bit 0 (most logical) or bit 63. At present, the >way I number things, it is (to me) a1=0, but the way BSF/BSR numbers bits, I >have a1=63 and I have to re-map it to my numbering scheme with the 63-x idea. > >I plan on getting rid of that at some point, but there will be some pain >involved, obviously. > >> >>>I have to compute square = 63 - bsf(board) >>> >>>_that_ subtraction. And that needs a register. >>> >>>IE optimal (on opteron): >>> >>>bsfq %rax, %rbx ; result in %rbx >>> >>>done. result = %rbx >>> >>>Currently (on opteron): >>> >>>movq $63, %rbx >>>bsfq %rax, %rcx >>>subq %rcx, %rbx ; result 63-bsf in %rbx >>> >>>That transformation is done everywhere I currently use FirstOne()/LastOne(). >>> >>>Because I numbered the bits to avoid that on the Cray. >> >>Yes I understood this was some old reminiscence from the Cray, what I didn't >>know was that you actually liked the transformation :) >> >>-S. > > >I don't like it. I only do it because bsf/bsr are much faster than any other >approach to extract a bit. Since this seems to be the standard now for >numbering bits, I'm going to change what I do to dump the transformation. I assume this would be for 64 bit hardware. In my fledgling attempt at a bitboard program (using x86 32 bit which is all I know) I have bit 0 = square 0 = a1. in FirstOne() if bsf delivers a bit on the low dword I'm out in one instruction. but if I have to go to the high dword, or in both cases with LastOne() I have to add 32. so it seems like 25% of the time I save an an instruction versus Crafty, but 75% of the time they are the same. Or am I missing something? Dan H.
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