I ask/answer various persons in the same post:
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Why do not compute first Perft(14), Perft(15) and Perft(16)? It looks very strange for me. I also did not know that you have written a multi-core perft utility for Nebiyu. Could you provide the CPU time and the real-life time of the alleged calculation, please? Thanks in advance.Daniel Shawul wrote:I have computed perft(17) on a few million core-hours access I had to a supercomputer at Aragonne labs. I used close to 700000 cores for computation so the computational power is really big. Those guys will not allow you to waste their resources without purpose and benefit to the society, so I had to write a proposal and take training to get that done. Anyway I am not sure if everything went well as intended, so I can not be sure if the result will hold when I do the verification later. Without further adieu, here is the result I got for perft(17) that is not verified yet...drum rolls please...
2172314159265358979323846
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Please take a look here:vittyvirus wrote:What program did you use? And by the way, your calculation suggests perft(17) had a branching factor of almost 27, so it seems somewhat correct.
Statistics on chess games
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Perft estimates
n perft(n)
13 1.9810e+18
14 6.187e+19
15 2.015e+21
16 6.507e+22
17 2.175e+24
18 7.214e+25
19 2.462e+27
20 8.350e+28There are other estimates here in TalkChess:
Re: Perft(20) summary of estimates
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perft(17)= ca.2.172185e+24 (74.900 sec) 30.983098 (Müller, 2011/07/10; the last number is the branching factor).
Perft(17) ~ 2.1729e+24 (Muñoz, 2011/08/04; myself with my hocus-pocus method of logarithms, polynomial adjusts, Lagrange polinomials, etc., explained in some old posts by myself).
perft(17) estimate (1,000,000 random walks in about 78 seconds): 2.1782e+24 (Schüle using Österlund's method, 2011/08/09).I discovered by chance a fairly good approximation to pi some years ago (probably when I was 14 or 15) with only two 3's and two 5's:vittyvirus wrote:Well, if it's not already getting off topic, I've an good approximation for pi:Code: Select all
426880 * Sqrt(10005) pi ~= -------------------- 13591409
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3*[3 + sqrt(5)]/5 ~ 3.14164079
|3*[3 + sqrt(5)]/(5*pi) - 1| ~ 1/(65269.17)
|3*[3 + sqrt(5)]/(5*pi) - 1|*100 ~ 0.001532% of error.------------------------
The end of the joke.zullil wrote:http://en.wikipedia.org/wiki/April_Fools%27_Day
Día de los Santos Inocentes
You can click on the English article, of course!
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Regards from Spain.
Ajedrecista.

