Author: Dieter Buerssner
Date: 11:17:01 03/29/04
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On March 29, 2004 at 13:55:22, Gerd Isenberg wrote: >Node = 2 * moves^(depth/2) * (1/MoveOrdering)^(depth/2) >Help me! What's the minimum of MoveOrdering to get a minimax tree? Good question. Answer for the easier formula without factor of 2 above: MoveOrdering=1/moves. Does the minimax tree have a factor of 2, too? If not, and if I calculated correctly, MO = log_depth(4)*1/moves. I think, it shows the the formula cannot be generally valid, although it might give a good estimate for typical cases. It is clear, that the formula can predict more nodes than the minimax tree for very bad move ordering. Regards, Dieter
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