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Subject: Re: Question about Bit storage

Author: Uri Blass

Date: 01:57:47 01/30/02

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On January 30, 2002 at 01:56:33, Dann Corbit wrote:

>On January 30, 2002 at 00:03:19, Robert Hyatt wrote:
>>On January 29, 2002 at 13:58:20, Dann Corbit wrote:
>>>No.  His notion is that if you mirror using every symmetry, the total number of
>>>those positions (including ALL reflections) would be less than 2^81 in that
>>>category.
>>
>>OK.  You are a math guy.  If you allow for 8 symmetries, which is false for
>>positions with pawns, you reduce the number of bits by a factor of 8, which
>>is 3 bits.  That is the mistake that is being made here, unless I misunderstand
>>something seriously.  IE for king vs king, allowing _all_ possible permutations
>>even with two kings on one square, you get 64^2 positions, which is
>>2^12.  If you take into account 8 symmetries, you reduce that to 2^9 positions,
>>not 2^(12/8)...
>
>Let's suppose that you need 170 bits to encode a chess position.  Now, with that
>position [for instance], you may have automatically stored 50 permutation of it.
> The net number of bits needed to store each of those 50 positions is 170/50 =
>3.4 bits.

you said
those positions (including ALL reflections) would be less than 2^81 in that
category.

Based on the same wrong logic in this case you can say that
those positions(including ALL reflections) would be less than 2^3.4 in that
category.

It is clearly wrong.




<snipped>
>>How can there be more than 8 "reflections"?  you can find symmetry along thhe
>>vertical center, horizontal center, and the two diagonals.
>
>Well, they are actually more than just reflections.

They are not the same position and I do not see how the fact that there are a
lot of positions that you can get the same position from them help you
practically to generate smaller tablebases.

I am sure that it may be possible to compress the data of tablebases but the
information does not tell me how to do it.

Uri



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