[Retros] Happy New Hebrew Year

[andrew buchanan]

Hi Noam,
That's really delightful. The stipulation form is basically about turning endgames into enumeration problems, and I think that will play to your strengths as an endgamer. My position was certainly clotted, and I didn't need to mention dead position.
So your problem is:https://www.janko.at/Retros/d.php?ff=2k5/P7/1P6/p7/P5pp/4p1P1/4P1KP/n2B2b1How many "best plays" in 17.0 moves with Black moving first, ignoring triple repetition?"How many best plays" is a felicitous phrasing.
So Black must play Kb7 very soon, because otherwise White will promote and win. But 1. Kb7? gxh4! 2. Bf2 h5 3. Be1/Bh4 f6 4. Bc3/Bf6 Kg3 (4... h7? 5. Be5!) and Black has no defensive line. bPg4 falls, and Black cannot take wPh2 without allowing wPh6 to promote, as wK can cover e5.
Thus play must begin 1. h3! Kf1/Kxg1/Kh1 2. Kb7 Kxg1/Kf1-Kh1/Kxg1. Critically, not 2... Kxe1? Bf2+! 3. Kf1 and is trapped. So after White's second turn there are 0,1,2,1 ways for wK to have reached e1,f1,g1,h1. This adds to 4, which is L_3 = 3rd Lucas Number. . (see http://oeis.org/A000204.) turn White now just moves along the axis, while Black vibrates between a8 & b7. The total number of ways for wK in turn n is L_(n+1) so indeed turn 17 we hit the magic value 5778.
I wonder only if it might be possible to lighten the pawns on the right hand side. Perhaps you tried remove the g-paws before. At the very least try to remove bPh4 & move sPg3 to h3.
All the best,Andrew

      From: Noam D. Elkies <elkies at math.harvard.edu>
 To: retros at janko.at; andrew at anselan.com 
Cc: kleeman12 at gmail.com
 Sent: Thursday, September 21, 2017 11:48 PM
 Subject: Re: [Retros] Happy New Hebrew Year
   
andrew buchanan <andrew at anselan.com> writes:

> Today is the first day of the new Hebrew Year, 5778,
> a mathematically auspicious number!

Shanah Tovah to all who celebrate Hebrew holidays and/or
auspicious numbers :-)

> In an entirely unrelated matter :D here is a chess position. 
> http://www.janko.at/Retros/d.php?ff=8/8/8/1p6/kPp5/BpPp2p1/pBRP2P1/B1B1K3
> Assuming optimal play, how many different possible sequences are there
> of the next 18.0 moves? (Shorter games don't count. Ignore the rules
> for drawing by repetition & dead position.)

With such a fakakte stipulation :-) one shouldn't have to also resort to
three promoted Bishops.  I think this position works in 17.0 (BTM)
and without having to keep track of the current convention for
"dead reckoning":

https://www.janko.at/Retros/d.php?ff=2k5/P7/1P6/p7/P5pp/4p1P1/4P1KP/n2B2b1

NDE


   
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