Author: Uri Blass
Date: 12:43:38 03/29/04
Go up one level in this thread
On March 29, 2004 at 15:22:50, martin fierz wrote: >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 I think that it is impossible never to fail high. What do you do when all legal moves cause a fail high? Uri
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