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Subject: Re: How to evaluate KQ vs KR?

Author: Dieter Buerssner

Date: 12:42:20 05/05/04

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On May 05, 2004 at 10:45:31, Renze Steenhuisen wrote:

>On May 04, 2004 at 16:46:14, Dieter Buerssner wrote:
>
>>Ernst A. Heinz suggests (w = Q/winning side, l = losing side)
>>
>>score = 400 + 1/8 * dist(K_l,R) - 1/8 * edge_dist(K_l)
>>            - 1/8 * corner_dist(K_l) - 1/8 * dist(K_l, K_w)
>>
>
>We might have been reading different articles then but I read:

I am confused, now. What was wrong with the formula I gave?

>[KQ]l[KR]w = 8 + 1/8*dist([K]l,[Q]l)
>               - 1/4*edge_dist([K]l)
>               - 1/16*dist([K]l,[K]w)
>
>[KQ]w[KR]l = 4 + 1/8*dist([K]l,[R]l)
>               - 1/8*edge_dist([K]l)
>               - 1/8*dist([K]l,[K]w)
>               - 1/8*corner_dist([K]l)

What is strange, is that the winning R gets a higher score, as the winning Q. Or
Do I misinterprete the formalism used? Perhaps it is a typo in the paper.
Actually exchanging 4 and 8 would make more sense. The only difference I see to
the formula I gave, is that I used 400 instead of 4. I assumed, that it is meant
like this in the paper (although not explicetly mentionend, at least I don't see
it when I fast read over that part of the paper). But perhaps I am even more
confused, than I think :-)

Tord already mentioned, that 4 looks too low. But when the position is already
on the board, it won't matter at all. It will be enough, when all scores are
positive (when the pos is won), and that they will be lower than scores for KQK
(otherwise the engine might refuse to take the R). The scores should fit
together with other scores of course. Probably one might want to have lower
scores for KQKR than for KBNK, etc.

Regards,
Dieter



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