[Retros] Challenge

[andrew buchanan]

(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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