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Subject: Re: Christmas Tactical Test - results for F8, S7, F7, S6.02 & Ruffian (LONG)

Author: Steve Maughan

Date: 06:13:17 12/24/02

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Dieter,

>With your formula, a problem might be, that it has a time domain.

Yes it is and I see that as a benefit!  If you set the half life to a low value
and steadily increase it you will see the total score, expressed as a
percentage, increase.  This reflects that given more time, programs do play
stronger moves.

>Assume I
>change a version, and test again. All will be the same, but one solution time
>goes from 30 seconds to 60 seconds. This will lose about 2 points. Another
>solution goes from 480 to 180 seconds. This will gain about 1.2 points. My
>feeling would say, the later version is not worse, but with the given formula,
>it is. When just comparing the sum of the logarithms and the number of solutions
>stays the same, to me, the comparision would look more natural. In the example
>above, the later version would win. I see however, that your formula also has
>advantages. It handles the cases of different number of solutions better. To me
>it looks, like it would work best, when the typical solution times are not too
>far from the half-live time.

I have also applied this type of formula to positional test suites.  As an
example suppose that a position had a list of possible moves with some form of
positional score associated with them e.g.

Ng5 = 10 points
Bxe2 = 7 points
Qh2 = 6 points
Qh3 = 6 points
g4 = 4 points
Nd2 = 1 points

Suppose that a program considered g4 to be the best for the first 45 seconds,
then switched to Qh3 and finally switched to Bxe2 after five minutes.  The score
for this position would be calculated as:

Score 1 (g4)  = 4 * (exp(0  *ln(0.5) / HL) - exp(45 * ln(0.5) / HL)
Score 2 (Qh3) = 6 * (exp(45 *ln(0.5) / HL) - exp(300* ln(0.5) / HL)
Score 3 (Bxe2)= 7 * (exp(300*ln(0.5) / HL)

Position Score = Score 1 + Score 2 + Score 3;

It works well - all you need is a set of good quality positions and assocaited
positional moves.

Regards,

Steve



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