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Subject: Re: Question about Bit storage

Author: Dann Corbit

Date: 11:39:10 01/30/02

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On January 30, 2002 at 04:57:47, Uri Blass wrote:

>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.

Les made an example.  Note that he is not talking about all board positions, but
a subset of them.

><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.

They are not tablebase files in the standard sense.  They are "At least as good
as this" tables.  So, if Les can show a mate in 8, there *might* be a shorter
mate, but it will be no worse than a mate in 8.

The reason that the tables are much smaller is not that it takes less bits to
encode a position.  It takes (as you have seen) the same number of bits.  But
only 1/50th (or whatever) of the positions are needed to be stored.  *THIS* is
the thing that reduces the size of the files by a constant, large factor.

An interesting thing about this approach is that it can use *any* analyzed
position and create dozens (or potentially even hundreds) of possible solution
moves.

Now, the format Les showed is not a good one when the board is full of pieces.
That much is clear.  But it isn't too bad when there are 3 or 4 pieces on the
board.

>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.

I have not gone over the math in Les' document yet, and I am not totally sure
what he is trying to say about 2^81.  But I do know that his method will
compress better than anything else that is available.



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