Author: Andrew Dados
Date: 13:37:19 01/29/02
Go up one level in this thread
On January 29, 2002 at 14:37:27, Les Fernandez wrote: >On January 29, 2002 at 14:03:24, Robert Hyatt wrote: > >>On January 29, 2002 at 10:47:40, Les Fernandez wrote: >> >>>On January 29, 2002 at 09:57:30, Robert Hyatt wrote: >>> >>>>On January 28, 2002 at 17:48:48, Les Fernandez wrote: >>>> >>>>>Just curious to know what is the best being done now for storing a chess >>>>>position. In the worse case scenario (castling right exists) it takes me 162 >>>>>bits to store 32 pieces, color, location, side to move, castling, enpassant, >>>>>promotion, ce and pv. Now if castling rights do not exist then the worse case >>>>>scenario for the above is 81 bits. Much further reduction in bits/position is >>>>>possible but at the moment I am interested in the above. >>>>> >>>>>Thanks in advance, >>>>> >>>>>Les >>>> >>>> >>>>How can you store a complete chess position without castling rights in 81 >>>>bits? castling rights are certainly not 81 bits of information to get you >>>>up to 162. >>>> >>>>Something is wrong. >>> >>>Hi Bob, >>> >>>The point is that in cases where castling is not available there exists a >>>minimum of 4 rotations that can be applied to a board, top-bot and left-right. >>>The method I discuss to store a position requires 9 bits to store piece, color >>>and x and y location. 32x9=288 and then 36 bits are needed to store the other >>>pertinent things about the position. (288+36)/4=81 bits/position on avg. just >>>as Uri explained here (thx Uri sorry Bob if I was not clear). I thought the >>>scheme I used to store a piece using only 9 bits was different and would >>>appreciate feedback here. >>> >>>Thanks, >>> >>>Les >> >> >>I think your math is wrong. If you need 288+36 bits to store one position, you >>need 288+36-2 bits to store 1/4 of that number of positions. This is log math >>and you _subtract_ exponents, not divide. >> >>ie 2^32 / 4 == 2^30, _not_ 2^8 >> >>That is where I am getting lost in your description... > >Hi Bob, > >First let me say my intention is not to play with semantics as the other >responder eludes too, I am not here to play. Basically I went on the basis of >the following. Lets say I have a Key position (3 pieces) and that I need 63 >bits to store "all" the information about this position. Now if from this >Binary key I can extract say 44 other positions of equivalent nature then I only >need to use 63 bits of space to store the key position but I actully have access >to 44 positions that require no storage at all and yet have the same solution. > >I hope I am saying it right <s>, > >Thanks Bob for the input Than for any database application you'll need to do a 'search' for position, killing your performance. You may need to compare hundreds of entries, or use huge space for indexing. -Andrew-
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