Author: Dann Corbit
Date: 11:02:05 12/12/02
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On December 12, 2002 at 10:53:35, Heiner Marxen wrote: >On December 11, 2002 at 16:39:53, Dann Corbit wrote: [snip] >A proof can sometimes be much shorter than the explicit optimal tree. >You just give a tree generation recipe like the above (Bb8, Kg8-h8-g8 for >black, white unrestricted). Such a tree can be sufficient for the proof, >but is never explicitly unfolded. How can you create this list without the search? >Our proof does not inspect all the nodes of the tree, but rather a general >property of all the resulting nodes (positions) of the tree. >The length of such a proof is independant (!) from the size of the tree, >and that size may be quite large. Will it not have been necessary to inspect at least a subset of size equal to the square root of all the nodes of the tree to find it (the list)? Actually, I have been thinking about it carefully, and clearly, I was wrong in total. Uri is right -- it could be a mate in 15. It seems highly improbable, but it is possible that there is some forced sequence that leads to a mate down all paths. In this case, the tree terminates in only a few plies. It is still necessary to search the square root of the nodes of the tree to form the solution, but the tree could *conceivably* be very short if we find the right pathway. In such a case, chess could be solved tomorrow. It seems unlikely but still conceivable. [snip]
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