Author: Dan Honeycutt
Date: 15:08:16 08/26/05
Go up one level in this thread
On August 26, 2005 at 17:30:55, Dan Honeycutt wrote: >On August 26, 2005 at 17:01:16, Robert Hyatt wrote: > >>On August 26, 2005 at 16:55:28, Dan Honeycutt wrote: >> >>>On August 26, 2005 at 16:36:47, Dan Honeycutt wrote: >>> >>>>On August 26, 2005 at 16:15:47, Robert Hyatt wrote: >>>> >>>>>On August 26, 2005 at 16:11:53, Dan Honeycutt wrote: >>>>> >>>>>>Run the lowest bits down the a or h file and you can do pawn captures without >>>>>>having to mask those files. >>>>>> >>>>>>Best >>>>>>Dan H. >>>>> >>>>> >>>>>How would you avoid masking? Otherwise the pawn at h2 would still wrap to a3 on >>>>>a shift... >>>> >>>>h8 = bit 0 >>>>h7 = bit 1 >>>>: >>>>a1 = bit 63 >>>> >>>>If I ran that the right way, the out of bounds bits are lost and 0's are shifted >>>>in on the other side. >>>> >>>>Best >>>>Dan H. >>> >>>Looking at it, I think that's right. Take white pawns on g4 and h4 with >>>captures to the right. g4 is bit 12 and h4 is bit 4. My shift is 9 left, so g4 >>>lands on h5 (bit 3) and h4 is discarded. >>> >>>Best >>>Dan H. >> >> >>The problem is that all 64 squares are represented by one bit. So no matter >>what you do, a pawn on the edge is going to wrap around to the other side of the >>board... Because all 64 bits are shifted at once... >> >>Are we talking about the same thing here??? > >Before After (shifting all 64 bits 9 left) >sq bit sq bit >h8 0 gone >h7 1 gone >h6 2 gone >h5 3 gone >h4 4 gone >h3 5 gone >: >g4 12 h5 3 >: >a1 63 b2 54 > (bits 55 to 63 - squares b1 and the a file - are all 0) > >So my white pawn on g4 who I want to capture on a5 is in the correct bit 3. My >white pawn on h4 (who I do not want to capture on a5 or any other square) has >been shifted out. I take the After bitmap, "and" it with the black pieces and >I've got my pawn captures (half of them) with no need to mask the h file. > >Best >Dan H. I have to leave, but I'll give the numerical values in case that might make it more clear. With white pawns on g4 (bit 12) and h4 (bit 4) my white pawn bitmap has a value of 2^12 + 2^4 = 4096 + 16 = 4112. My 9 shift divides by 2^9, or 512. Thus 4112 \ 512 = 8 so bit 3 is set. If the black piece bitmap has bit 3 set, I have a capture. Without the pawn on h4 I get 4096 \ 512 = 8 which is the same result which is what I want since white pawns on the h file cannot capture to the right. Best Dan H.
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