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Subject: Re: move ordering and node count

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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