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Subject: Re: Counting & Encoding Any Chess Position in 157 bits

Author: Ratko V Tomic

Date: 10:43:26 11/10/99

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> I did not think to pick a uniform odds for counts for pieces.
>
> It is possible to calculate the probability of every material
> configuration and choose 1000 random material configuration...

Well, that's point  was making -- to get these probabilities you
will have to calculate how many placement permutations each MC state
has, which amounts to exactly the calculations you and I had done.
These approximate probabilities will then carry the sampling bias
of exactly the same magnitude as the level of approximation used to
obtain them. Any further sampling experiments, biased this way
(since our methods were approximate), will then carry that bias
in any figures they produce, in addition to the regular
statistical uncertainties of finite samples.

In any case, from the experimenting with this code, I think that
the figure I got is at most 20% above the actual number of legal
positions (some minor improvements could be obtained via bishop
optimization and some simple types of the wrong side check
estimates). But even the a global constraint strenghtening done
to get from your to my figure, which removes entire classes of MC
states, including all the placements for such MC states, (as
opposed to checks which only remove a small fraction of placements
within a given MC state), produced only 33 percent reduction in
the upper bound. So, all the other stuff which could practically be
estimated may likely amount to merely another few percent reduction.




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