Author: Les Fernandez
Date: 11:37:27 01/29/02
Go up one level in this thread
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
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