Author: Francesco Di Tolla
Date: 08:34:47 06/23/99
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>Volunteers for doing ply 10 or verify ply 9 :-P ? Maybe we can calculate this >distributedly... If you put a file on ftp site with the final leave positions :-) If the trend is preserved (i.e. that there are 25 moves more or less each step) this would imply that to get to 40 full moves (ply 80) it would take 25^80, i.e nearly 10^111, and considering that you have to calculate previous nodes to, you have to consider the sum from 1 to n=10^111 which is 0.5*n(n-1) which brings to more than 10^200 even if only 1% of what you see is not repetition! This is a very rough estimation, based on unsupported hypotesis, but I would be surprised that the real number would not be in the order of 10^n with more than 100... And sombody wants to solve chess? regards Franz
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