Author: Graham Laight
Date: 02:31:05 05/10/01
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I apologise if you've already explained how it's calculated - but I missed it. Would you mind explaining in a straightforward way how you calculated this upper bound, please? It intuitively sounds right to me - I'd just like to know how you calculated it! By the way - as I've said before (excuse the repetition!), my own view on the solving of chess is that, as the standard of chess continues to rise, the proportion of draws will also continue to rise. Eventually, you'll get to a point where almost all games are drawn. At this point, people will (pragmatically) conclude that chess is a draw - and the interest in continuing to drive up the standards of chess programs will disappear. -g On May 10, 2001 at 00:35:50, Uri Blass wrote: >10^50 is an upper bound of the number of the legal positions(2 positions are the >same if the pieces are in the same places and both sides have the same right to >castle and the same rights to capture by the en passant rule). >After less than 10^50 moves the same position happens twice > >If you can mate in 10^50 moves the game contains a repetition of the same >position twice. >If you can win with repetition you have a shorter win without repetition. > >It means that the problem of finding if a position is a draw or a forced mate is >solvable in a finite time even without the 50 move rule > >Uri
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