Author: martin fierz
Date: 12:22:50 03/29/04
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On March 29, 2004 at 14:30:17, Dieter Buerssner wrote: >On March 29, 2004 at 10:17:18, martin fierz wrote: > >>aloha! >> >>i was discussing this somewhere in a thread, but thought i'd like to make this >>question more visible in the hope of getting a good answer: >> >>everybody knows that with plain alpha-beta, a fixed number of moves N per node, >>and perfect move ordering a search to depth D needs >> >>nodes(depth) = sqrt(N)^(D/2) nodes. >> >>with absolutely imperfect move ordering it needs >> >>nodes(depth) = N^(D) nodes. > >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). > >Regards, >Dieter hi dieter, 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. cheers martin
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