Author: Robert Hyatt
Date: 21:50:45 05/06/04
Go up one level in this thread
On May 07, 2004 at 00:24:42, Russell Reagan wrote: >On May 06, 2004 at 20:05:03, Robert Hyatt wrote: > >>I don't follow. I currently use 6-bit indices _everywhere_. Because the two >>endpoints are moot. We're not changing that. Just squeezing out the bit for >>the square where the piece in question sits... it is meaningless and could be a >>zero or 1 and produce the same attacks... > >He is saying you can't always get correct attacks from 5 bits. Another way of >putting it is that you can't always squeeze out the square that the attacker >sits on, because it may have already been squeezed out. I realized that after Gerd pointed it out... (see the other sub-thread). I already use just 6 bits... and was not thinking about the case where the piece sits on the end of the rank/file/diagonal... > >Let's say you start with all 8 bits for the 1st rank. You squeeze out the end >bits, squares a1 and h1. You want to get attacks for a rook on a1. You can't >squeeze out the square the attacker is on (a1), because you already squeezed it >out. > >Put a rook on a1 and try to get correct horizontal attacks from 5 bits in all >cases for the 1st rank. Take a simple example, 5 bits = 00000. What is the >attack bit pattern for the 1st rank? Here is the board. You get 5 bits worth of >rank state info, plus the implied rook on a1. > > abcdefgh >8 -------- >7 -------- >6 -------- >5 -------- >4 -------- >3 -------- >2 -------- >1 R-----?? > ^^^^^ > 12345 <-- You only get 5 bits > >What is the correct attack bitboard for this position? > >- represents a 0 bit ># represents a 1 bit > >Choice A: >-------- >-------- >-------- >-------- >-------- >-------- >-------- >-####### > > >Choice B: >-------- >-------- >-------- >-------- >-------- >-------- >-------- >-######- > >Choice C: >Impossible to know without more information.
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