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Perft(13).

#1
Hi everybody!

Perft(n) is the number of possible games in chess, checkers, reversi... up to n plies (for example: e4, e5; Nf3, Nc6; Bb5 are 5 plies). Perft(12) is the highest perft value calculated to date, but a
guy in TalkChess Forum is calculating Perft(13) and he expects to finishing the calculations around the end of this year.

http://talkchess.com/forum/viewtopic.php?topic_view=threads&p=410442&t=39389

Some people are doing their estimations and I also did my own! Here is the document I have just upoladed to Megaupload server:

http://www.megaupload.com/?d=ZUJWWQWF

I hope my estimation will be true. In fact, all estimations I have seen are near mine, but with minor changes. Keep in mind that Perft(13) should be around 2·10^(18), so this is why the exact
calculation of this number will took more than half a year (maybe).

If you LiChess players want to do a poll, this is the thread! How much can we approach to the true value?

Regards from Spain.

Ajedrecista.
#2
Yeah! That's crazy. I won't make an estimation, but I'm very excited about this stuff. The estimations are definitely more interresting than the exact number :)
#3
Hi everybody!

Maybe there will not be any poll about Perft(13) in LiChess, but Mr. Steven Edwards (the person who is computing it) has just started one at TalkChess forum!

http://talkchess.com/forum/viewtopic.php?topic_view=threads&p=414187&t=39678

So far, his estimation fits in my interval! I have seen another estimate in this thread:

http://talkchess.com/forum/viewtopic.php?topic_view=threads&p=413613&t=39630

This last thread cares about estimations on Perft(20), far beyond to calculate it within decades! Here, Uri Blass has proposed a nice method for estimating any Perft(n) (of course, n>12), but when n
grows, estimations became less accurate (obviously). I post this last thread because Sven Schüle applies Blass' method and gives results around 1.98·10^(18) for Perft(13), which do no fit in my
interval by few (around 1%).

Of course, Blass' method is more difficult of implement than mine but is also far better; I am a mere afficionado. I did estimation intervals based on trends, but give results only for odd numbers
(i.e. n=13, 15, 17...), that is somewhere incomplete. For even number of plies (n=14, 16, 18...) I can only get upper bounds, which is useful, but complete intervals are more useful and can only be
provided for odd number of plies using my 'method'... unless someone may improve it!

According with Mr. Edwards, the exact value of Perft(13) will be known 'probably sometime in early 2012'. It is interesting how hardware has evolutioned for making this possible!

Please comment on it.

Regards from Spain.

Ajedrecista.
#4
Hi again.

Paul Byrne, the first person that calculated Perft(12) back in 2006, is going to calculate Perft(13). He must have very strong hardware and very good software for it, because he thinks that Perft(13)
count could take around a month! This is very few time.

http://talkchess.com/forum/viewtopic.php?topic_view=threads&p=414253&t=39678

In the same thread, there are some people that are doing great work and implement various methods for estimating Perft(13)... they are dazzling! Too much better than my 'method', if it could be named
in this way. I look that most of their estimates are around 1.98·10^(18) and do not fit in my interval :( In fact, there are less than five estimates that fit in it, which is not bad IMO.

Well, I did my best with my own estimate... who knows if there will be surprises? But I am looking at their methods and are very smart... so I start thinking that maybe my 'method' is wrong and crack
somewhere; it is true that trends I looked at are not telltale with 100% confidence.

But if I fail my estimate, maybe the error should be less than 1.5%, which is not a tragedy, chiefly if you consider that I am a mere amateur.

Thanks for reading it!

Regards from Spain.

Ajedrecista.
#6
Hi again.

I checked my estimation and now I have updated it:

http://www.megaupload.com/?d=RH1CQTPD

The new interval does not fit in the first one of some days ago, but I am now closer than before. The fact is that I saw a pattern in my first estimate, but did not work in it for some reason; in this
second estimate (I hope it will be the last one by me) I think I get much better results. Also I manage to narrow the interval width from ±0.2% to ±0.013% (more less), which is very good indeed.

With this second estimate, there are more bets than fit in my interval (bets of the Perft(13) poll on TalkChess, links in former posts in this topic). And more important: most bets (I think all the
serious ones there, like Peter Österlund's one) differ from mine less than 0.1% now!

It seems that my first estimate was wrong... let's hope the second one is right! At least I am now among the main estimated result: around 1.98·10^(18) or 1.981·10^(18). Thanks for downloading my
'works' (really it is a hobby that take me few hours) and read the whole thread.

Regards from Spain.

Ajedrecista.
#7
Hi again:

Sorry for bumping this topic. There is a typo in both PDF's I uploaded: 'alpha(n) > beta (n)' is wrong; right is: '|alpha(n)| > |beta(n)|' (absolute values). But this typo does not change my final
estimate: Perft(13) ~ 1.980465·10^(18) ± 0.013%.

I will not upload another PDF for this typo (two are enough, IMO). I include this correction in a Notepad in the same folder as PDF's.

Thanks for read and sorry for the typo. Bye!

Regards from Spain.

Ajedrecista.
Anonymous
#8
Hi!

I found a perft counter that is really fast, although it can not compete with Edwards one.

www.zipproth.de/jetchess/

It is easy to use... and free, of course! Enjoy.

Bye!
Anonymous
#9
Hi!

Perft(12) value of games starting with 1.- a3 is now calculated: 54,239,338,583,061,004. There are other 19 values which are unknown at this momment; when all 20 values of Perft(12) for each initial
move will be known, then the sum will be Perft(13).

So far this number was among estimates of everybody (between 5.4e+16 < value < 5.6e+16).

Bye!

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