Chess

William Hartston
Sunday 03 May 1998 18:02 EDT
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The above diagrams look like a bad case of double vision - the one on the right is identical to the other, but with every man nudged one square to the right. They are a typically ingenious pair of problems by that master of puzzles, Sam Loyd. In either case, it is White to play and mate in four.

In the first position (and, of course, the second too) Black's king is completely surrounded, but clearing away the clutter and getting a piece through to deliver mate is not easy. 1.Ne4+ Kxe2 2.Ng3+ Kxf3 fails to mate within the specified time, as does 1.Bxe3+ Kxe3 2.Re4+ Kd2.

The solution is surprising: the only move to force mate in four is 1.b8=N!! when Black is remarkably helpless against the threat of 2.Nxd7 3.Nc5 and 4.N5e4 mate. 1...d5 is met by 2.Nc6 dxc4 3.Ne4+ Kxe2 4.Nd4 mate.

So what is the difference in the second position? Why doesn't the analogous solution work? The answer is delightfully hidden: after 1.c8=N Black plays 1...Rxh1! 2.Nxe7 g1=B! when 3.Nd5 leaves Black stalemated.

The solution this time is 1.Nf4+ Kxf2 2.Nxh3+ and now 2...Kxg3 is met by 3.Nf5+ Kxh3 4.Bg4 mate, while 2...Ke2 allows 3.c8=Q! followed by Qa6 mate - both mates impossible in the first position.

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