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Miscellaneous Problems

Solution

L. A. Garaza
Thèmes 64, Jan. 1962
3rd Prize

[r3k3/1p1p4/1pnB1P2/p2P4/1N2P1P1/Pp5B/3P3P/n3K2R]

12+9. (Maximummer) Mate in 2.
(a) diagram; (b) Pd5 -> c3; (c) Pd5 -> c2

 

(I have been largely inspired from the lengthy analysis devoted to this problem in Jean Bertin's Histoires extraordinaires)

Analysis: The black Ps are responsible for the capture of the 4 missing White pieces. Thus the Na1 must have reached a1 through c2 or b3. If it visited c2, Wh. can't castle anymore. Could it have never visited c2 under the Maximummer condition?

Let's assume the N never visited c2. Then, when the blP from e7 reached b3 the N was already on a1, thus it could have gone to b3 in place of the blP. This would have been a longer move, and it would have forbidden the blP to reach b3 under the Maximummer condition. So that it seems the N had to go to a1 after the blP reached b3...

In fact there is a (tricky) way out. If the blP went to b3 with c4xb3 e.p., answering a check that a b4 pawn just gave to the bl. K, then Na1-b3 was not a legal alternative to the P move. But then, at that time the bl. K was on a5 or c5, and cannot castle anymore.

The conclusion is that White and Black castlings are mutually exclusive. (A complete demonstration of this still requires two proof games respecting the Maximummer condition and reaching the diagram position without destroying Black, resp. Wh., castling right. Such games do exist. See [Ber88], p.120)

Solution: The try 1. Rf1? fails on 1 ... O-O-O, which is the longest legal move. If instead Wh. plays the legal 1. O-O!, then the priority rule for mutually exclusive castlings forbids 1 ... O-O-O? and only leaves 1 ... Rd8 as longest move. Wh. concludes with 2. f7 mate.

 


In the (b) twin position, Wh. can't castle. Indeed, the b2-b4+ c4xb3 e.p. trick could be used to explain how the Na1 never visited c2, but then the wQR could not get out and be captured without disturbing the wK. By default, Bl. may castle.

Solution: 1. Nb5! O-O-O (longest) 2. Nxb6 mate.


In the (c) twin position, the Na1 could not visit b3, so that the b2-b4+ c4xb3 trick is mandatory and Bl. can't castle. Nor can Wh. because the wQR had to disturb the wK.

Solution: 1. Tf1! Rd8 (longest) 2. f7 mate.