Author: Vincent Diepeveen
Date: 06:50:27 05/25/02
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On May 25, 2002 at 09:32:58, Russell Reagan wrote: A few rude things can be done pretty quick with bitboards at 64 bits processors, which is what Bob aimed at a year or 10 ago, but as soon as you want to evaluate things in detail, then please remember what bitboards are: they provide 1 bit of info a bitboard about a square. That's very little info. So obviously when you require many details about a certain square, which makes other methods galaxies faster than bitboards. There are even several ways to do it faster. >I hear about how great bitboards are because they [insert one of the many >reasons here]. I'm interested in two reasons in particular, because I do not see >how these reasons work. > >The first thing I'm curious about is computing whether or not a square is >attacked by a certain side. This is supposed to cost about a couple of array >lookups and an AND operation, right? Look up the white king bitboard, look up >the black pieces attack bitboards, AND them together, and you can tell if the >king is in check. That's what I've always heard anyway. Clearly this is much >faster than ray tracing the board looking for opposing pieces, but I don't see >how this bitboard method would even work. For example, for this to work, you >have to have a valid "black pieces attack" bitboard. To compute this you could >OR together all of the attack bitboards for each piece. The problem is that you >have pseudo-attacks. So this computation above where bitboards calculate whether >or not a square was attacked really only says "this square MIGHT be attacked", >right? At which point, you would have to do the ray tracing anyway, and it's not >really any faster, right? I'd like some clarification on this. > >The other thing I recall Bob saying was that using bitboards you can calculate >mobility for "free". Again, this seems like you could calculate pseudo-mobility, >but to calculate actual mobility, you'd have to do ray tracing just like you >would with another board representation scheme. I'd appreciate some explaination >of this also. > >Thanks, >Russell
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