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Jesús Muñoz

Joined: 13 Jul 2011
Posts: 694

Post subject: Some explanations.    Posted: Sun Jan 22, 2012 4:09 pm

Hello:

In the output of GNU appears sdn, which is by definition (sd)·sqrt(n) for each sample. I said in my first post that I would use (sdn_i)² for averaging the standard deviation, but actually I am using (n_i)·(sd_i)² for two reasons:

a) More accurate results: I assume that both m_i and sd_i are rounded, but also sdn_i is rounded; so I suppose that the output of (n_i)·(sd_i)² is a little more exact than the output of (sdn_i)².

b) Save time: for each sample, I do three 'copy & paste' (m_i, sd_i and n_i) from a notepad to Excel instead of four (m_i, sd_i, n_i and sdn_i). I save N 'copy & paste' in N samples.

So, the method of averaging the data is now fully explained. I have just realized that Peter Österlund did a similar thing with Perft(13) averaging lots of samples, but he averaged around 180,000 samples! His relative error with the true Perft(13) value was ~ 0.000449%! I have done 54 until the moment, the 0.03%, and with much larger standard deviations (that is why I compare <m>/<sd>, for viewing the quality of the average standard deviation against the average mean). Something is something...

I see that GNU gives a confidence interval with 99% confidence. I knew that this level of confidence is reached between m ± (2.57)sd and m ± (2.58)sd in a normal distribution; so, m ± k·sd has 99% confidence, where k ~ 2.575829303 according to Derive 6. I can give now a minimum value and a maximum value for Perft(14) estimate:

 Code: Averages after N = 54 MonteCarlo perft samples:   ~ 61,881,120,367,616,500,000 ~     10,495,306,430,406,300 (Minimum value with 99% confidence) ~ - (2.575829303) ~ 61,854,086,249,769,100,000 (Maximum value with 99% confidence) ~ + (2.575829303) ~ 61,908,154,485,463,900,000 / ~ 5,896.076 ~ 501,279,504.33

I remember that Excel also rounds its results. In view of the evolution of my tiny experiment, I expect that <sd> will be around 1.05e+16 (with the criterion of stop GNU at more less 5.0128e+8 nodes).

I have done a little search of Perft(14) estimates of people of this forum (except Mr. Labelle), and here are the estimates that I consider more accurate:

 Quote: 61,803,489,628,662,504,195 by Joshua Haglund. 6.187e+19 by François Labelle. 61,886,459,822,115,294,738 by myself. 6.188925e+19 by H.G.Muller. 6.19009592e+19 by Reinhard Scharnagl.

Each person uses his own estimate method: this is the reason why I give the links. The more different estimates (Joshua's and Reinhard's) differ in less than 0.16%!

Regards from Spain.

Ajedrecista.
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 Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First
Subject Author Date/Time
Jesús Muñoz Sun Jan 15, 2012 12:27 pm
Daniel Shawul Sun Jan 15, 2012 1:31 pm
Jesús Muñoz Mon Jan 16, 2012 8:35 pm
Jesús Muñoz Fri Jan 20, 2012 3:27 pm
Some explanations. Jesús Muñoz Sun Jan 22, 2012 4:09 pm
Jesús Muñoz Thu Jan 26, 2012 4:44 pm
Jesús Muñoz Fri Feb 17, 2012 8:00 pm
Jesús Muñoz Fri Mar 02, 2012 5:26 pm
Jesús Muñoz Fri Mar 09, 2012 3:50 pm
Jesús Muñoz Fri Mar 16, 2012 4:57 pm
Jesús Muñoz Fri Mar 23, 2012 3:22 pm
Jesús Muñoz Fri Mar 30, 2012 3:11 pm
Jesús Muñoz Tue Apr 10, 2012 7:21 pm
Peter Österlund Tue Apr 10, 2012 9:24 pm

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