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Chess

William Hartston
Friday 18 April 1997 23:02 BST
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Being by inclination a player rather than a problemist, I generally find mate-in-three problems too difficult. I can solve a mate-in-two, if only by looking at every possibility, but three-movers are too much of a strain. Unless, of course, I cheat by looking up the answer and then appreciating the ingenuity of the composer - which is what I did with the mate-in-three above. Composed by A Lobusov, it won first prize in a composing tournament in the Soviet Union in 1983.

When you first look at the position, it seems quite clear that the key to the action will lie on the e-file. At some stage, the knight will move from e6 to deliver a fatal discovered check. The only trouble is that any move of the knight will leave the pawn on f4 unprotected and let the black king escape with Kxf4. We could defend the pawn with 1.g3, but that would let the king escape to f3. No, it's clear that what we have to do is persuade Black to play Rxf4, blocking that square for his king. So all we need is a first move that contains a threat that can only be met by Rxf4. But why should Black possibly want to play Rxf4?

You will never solve the problem at all if you think in this apparently logical manner. The key is to stop yourself thinking about the e-file, which has no part in the solution.

The answer begins with the unlikely looking 1.Rd8! carrying the remarkable threat of 2.Bd5+! Kxd5 3.Nf6 mate. Black has three possible defences:

a) 1...Ne3 covers the d5 square but takes away a possible flight square from the black king. White mates with 2.Nf6+ Nxf6 3.Bxd3.

b) 1...d2 creates room to answer 2.Bd5+ with Kd3, but White instead plays 2.Bd3+! Kxd3 3.Nxc5 mate when the pawn on d2 prevents the king from running away.

c) 1...Rxf4 also prevents the main threat (2.Bd5+ Kxd5 3.Nf6 Ke5) but now, because f4 is occupied, White has 2.Nxc5+ Nxc5 3.Bd5 mate.

Quite apart from the beautifully hidden key, the subsequent white moves in the above variations also exhibit a cyclical effect: Bd5 and Nf6, then Nf6 and Bd3, then Bd3 and Nc5, and finally Nc5 and Bd5.

No, I would not have solved it had I not been able to read the answer.

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