Author: S. Loinjak
Date: 10:41:39 01/08/03
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Thank you very much for the reply and escpecially for the hint about Petrovich's calculation (he seems to be a famous retrograde analysis composer). Does someone know how Petrovich came to his result and what kind of positions he regarded as being different. Has he taken into account the right to move, castling rights, e.p. rights and all possible material distributions (max. 8 promoted pawns per color). Let me give the following specification: N = number of legal positions in chess -------------------------------------------------- positions are different when they would generate different sets of legal moves so even optically identical positions must be counted up to 192 times (if we cannot prove that some of the special cases are impossible) -- color to move (counts twice) -- castling (max. 16 combinations per position) -- e.p.rights (max. 6 combinations) (example for max. en passant diversity: pawns on 4th rank: wbwwbwbw ==> either one or none of the 5 white pawns moved to the 4th rank a ply before, all could be captured by an opposite pawn ) -------------------------------------------------- Based on the technical level of the year 1978 when Petrovichs got his result N <= 2*10^43 I'd say this must have been a comparably short calculation. And I really doubt although I don't know the method he used that he has taken really everything into account (especially the underpromotions which can easily push up the number by at least 1000). I should not forget to mention Christoph Fieberg's result of N being approximately 3.02*10^46 published in CSS 2002/feb (CSS = Computer ,Schach und Spiele - in German). Although this result does not take into acount the color to move, the castling and the e.p. rights (this would push the estimation to roughly 6.0*10^48) it might be one of the rare calculations where really all underpromotions are counted correctly. My own computation to prove N < 1.0*10^46 runs for almost an hour on a 2 GHz PC - I use 256 MB for lookup tables for combinatoric values (products of binominal coefficients, results of certain local optimization problems) and so nothing is calculated twice. Originally I though it must be easy to prove that N < 1.0*10^43 but meanwhile I think I'm quite near to the real number of the legal positions and I doubt there are less than 1.0*10^44 - I hope I'm wrong so chess could be easier to solve then it seems to me now. Sini
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