Author: Dann Corbit
Date: 10:32:42 01/13/05
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On January 13, 2005 at 11:12:09, Dieter Buerssner wrote: >On January 12, 2005 at 20:33:25, Dann Corbit wrote: > >>If there were 10^120 in the full tree, then about 10^60 would be in the solution >>tree. >> >>It can be less than that. But it cannot be more. > >10^120 nodes makes a rather short game of chess. > >Effort to slove chess depends on many things. Say you have to look at an average >game length of 50 moves only, normal perfectly ordered alpha-beta will look at >around sqrt(35^100) nodes. This is a very huge number, much huger than the >number of different chess positions. Having really huge hash (for all positions) >would of course make the effort smaller. > >I guess, that I won't live anymore, when something like 10^50 bytes of memory is >available for computers. > >One other number. The mass of the earth is 6*10^24 kg. It is mainly made of iron >(the core). But say it was out of oxygen (which is lighter, so we overestimate >the number of atoms). The atomic mass of oxygen is 16 g/mol. Avogadro's constant >is 6.022e23 mol^-1. The number of atoms in the earth is: > > 6e24 kg / 0.016 kg/mol * 6e23 mol^-1 = 5.76 * 10^46 Imagine if instead of the normal means of game tree search, we have an oracle of pv positions for any given node. Since there are only 10e46 or so distinct board positions, we should certainly not have to contain a list of 10^120th nodes. We will also not need to consider those positions which do not occur in practice in a well played game such as KQQQQQQk positions which are won anyway. Anytime we find a losing move, we get to trim that branch or pronounce it dead. There may be as little as 10^30th nodes in the real, final, perfectly ordered solution tree. Nobody knows. In any case, there is no reason to search/solve any positions that are lost. And there is certainly no reason to search/solve positions more than once.
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