[Retros] Challenge
andrew buchanan
andrew at anselan.com
Sun Nov 29 19:38:35 EST 2009
(1) Wow good improvements, Noam! Fortunately for my bruised ego, I managed to find:
k7/2K5/8/8/8/8/1Rp5/8
"#n" for all n
Is this solution unique up to symmetry for KRkp? [Guess yes] Is there an even more economical solution? [Guess no]
(2) And just to get the ball rolling on the original challenge, a few moments spent pushing pieces round a virtual Nalimov board reveals:
4k3/8/5K2/8/8/6N1/3Qpb2/8
"#1", "#2", "#3", "#4", "#5"
Surely this can easily be beaten?
Regards,
Andy.
----- Original Message ----
From: Noam Elkies <elkies at math.harvard.edu>
To: andrew at anselan.com; retros at janko.at
Sent: Sun, November 29, 2009 8:40:10 PM
Subject: Re: [Retros] Challenge
I wrote:
< This may be minimal in material, but for piece count one can do better:
< Kf6,Qh3(or h7) / Kf8,Qd7.
Likewise with R vs. R (or Q), e.g. Kg6,Rf7/Kh8,Q(R)c7, or even R vs. B,
Kg6,Re7/Kh8,Be2. Not sure if R/N or R/P is possible.
NDE
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