Author: Robert Hyatt
Date: 10:27:33 02/17/04
Go up one level in this thread
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.
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