Author: martin fierz
Date: 15:08:10 03/29/04
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On March 29, 2004 at 15:46:01, Dieter Buerssner wrote: >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 hi dieter, i see your point though: if you have an evaluation function which always returns 0, then you get cutoffs everywhere. in this sense, i think that a tree is only maximally disordered if you have a >, not a >= in the values of siblings. cheers martin
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