Author: Dieter Buerssner
Date: 12:46:01 03/29/04
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On March 29, 2004 at 15:22:50, martin fierz wrote: >On March 29, 2004 at 14:30:17, Dieter Buerssner wrote: > >>This has not so much to do with your question, but I doubt the last part of your >>sentence. I believe, it will be impossible to become as bad as the minimax tree, >>even when by purporse ordering the moves "perfectly wrong". You will still have >>plenty of situations, where many different moves give a cutoff. In a previous >>experiment, I got 50% beta cutoffs for the first move, when randomizing the move >>order. Note that this is far away, from a minimax tree (where you would need >>100% beta cutoffs in the last tried moves - there are in general many more move, >>you try before). >hmm, everybody writes this that making A/B MO as bad as possible you return to >minimax. somewhere below gerd just made the same point as you did here. and if >two experienced programmers like you say so, i am of course afraid to contradict >you :-) >but i have to contradict you all the same. in a perfectly misordered tree you >will *never* fail high. which also means that the case that you and gerd were >thinking of never happens. martin, I might have not thought well enough about it. But some things, that might happen: 2 moves have exactly identical minimax score somewhere inside the tree (I think this will happen quite often like piece1x... opponetx... piece2x and piece2x... opponetx... piece1x). I think, you cannot avoid to get cutoffs. But, I have to agree, I thought of it too naively first. Cheers, Dieter
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