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Subject: Re: move ordering and node count

Author: Gerd Isenberg

Date: 10:55:22 03/29/04

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On March 29, 2004 at 12:24:42, Steve Maughan wrote:

>After hacking around in Excel for a couple of minutes I came up with this
>formula.  Of course it's an engineers fomula and not a mathematicials formula
>i.e. it looks about right and I'm not really interested in the proof :-))
>
>Node = 2 * sqrt(moves ^ depth) * sqrt((1 / MoveOrdering) ^ depth)
>
>It's based on the thought that when the MoveOrdering stat falls we need to
>search more than one move at a theoretical cuttoff node i.e. alternative depths.
> Also the effect is going to be multiplicative and as the MoveOrdering
>approaches 1 the nodes approaches the theoretical minimum.
>
>Thoughts?

Interesting!

Node = 2 * moves^(depth/2) * (1/MoveOrdering)^(depth/2)
Help me! What's the minimum of MoveOrdering to get a minimax tree?



>
>Interestingly this means that the difference between a move ordering stat of 0.9
>and 0.95 for a 10 ply search is 31% more nodes.  At a 15 ply search this
>increases to 50% more nodes.  So improving the move ordering has more effect at
>slower time controls - I think we knew this anyway.

Yes, but not so nice to calculate.

Cheers,
Gerd

>
>Regards,
>
>Steve



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