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Subject: Re: Eval() And Node Count

Author: Steffen Jakob

Date: 06:44:43 01/20/06

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On January 20, 2006 at 08:53:48, Michael Neish wrote:

>
>Hi, I just tried a little experiment with my as yet simple program.  I added a
>mobility function for the Queen as the only way of scoring this piece, i.e., no
>piece-square tables.
>
>When I ran the program, I found that it increased the required number of nodes
>to reach the same depth in a given position, by roughly 50% to ply 11, which I
>found curious.
>
>Thinking about it, adding this new feature could be resulting in a greater
>number of positions that score roughly equally, hence search needs to sift
>through more nodes to find the best one.  The opposite is true: when there's a
>really obvious best move like moving one's Queen away from a Pawn that's
>attacking it, search tends to go deeper faster since it can quickly eliminate
>bad branches.
>
>Have I got the correct interpretation?
>
>If so, would adding more refinements to Eval() result in an even greater number
>of nodes required to the same depth, or not necessarily?  If it would, what can
>one do to prevent it?  Could one, say, design Eval() so that position scores
>don't cluster around the zero mark, but are instead more spread out so that
>search doesn't get bogged down trying to distinguish between positions that are
>just a point different?  Easier said than done, maybe.

If you want to experiment with this idea you can reduce the value space of your
evalation function by something like this:

Value Engine::FilterValue(Value v) const {
    return v - (v % M_eval_filter);
}

First you evaluate the position and after this you filter the computed score.
E.g. if M_eval_filter == 5 you only get scores which can be divided by 5.

Best wishes,
Steffen.

P.S.: I prefer to use M_eval_filter == 1 in my engine. ;-)

>
>Properly done, though, I wonder if it might be a viable strategy for reducing
>the size of the tree.
>
>Any comments?
>
>Cheers,
>
>Mike.



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