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The Independent Culture
A FORTNIGHT ago, I reviewed a delightful collection of the great Sam Loyd's problems, The Puzzle King (Pickard and Sons, listed at $22.95; it's about pounds 17.50 in this country) - which I picked up in New York on the way to Las Vegas.

Coming home last Friday, I was flicking through this on the first leg from Vegas to New York in a somewhat desultory way, when I came across a pair of quite superb construction tasks which held my interest, not only for the remainder of that flight and much of my stay overnight at an airport hotel, but also for a good portion of the flight the next day from JFK to Heathrow.

a) In how many moves can the position illustrated be attained in actual play?

b)Kf1 vs Kf8

c)Kf2 vs Kf7 - seems impossible.

Both sides have 15 pieces to capture, and either White or Black - or both - must make two non-captures in their first three moves. So the absolute shortest conceivable is after White's 17th move, while in practice the tasks seem possible only after Black's 17th.

Eventually, after perhaps five hours, I solved a):

1 c4 d5 2 cxd5 Qxd5 3 Qc2 Qxa2 4 Qxh7 Qxb1 5 Rxa7 Qxb2 6 Qxg8 Rxh2 7 Rxb7 Rxh1 8 Rxc7 Rxg1 9 Qxf7+ Kxf7 10 Rxc8 Rxg2 11 Rxb8 Rxb8 12 Bxb2 Rxf2 13 Bxg7 Rxe2+ 14 Bxe2 Rb2 15 Bxf8 Rxd2 16 Bxe7 Rxe2+ 17 Kxe2 Kxe7

Interestingly, though, Sam Loyd's original solution is quite different:

1 c4 d5 2 cxd5 Qxd5 3 Qc2 Qxg2 4 Qxc7 Qxg1 5 Qxb7 Qxh2 6 Qxb8 Qe5 7 Qxc8+ Rxc8 8 Rxh7 Qxb2 9 Rxh8 Qxa2 10 Rxg8 Qxd2+ 11 Kxd2 Rxc1 12 Rxg7 Rxb1 13 Rxf7 Rxf1 14 Rxf8+ Kxf8 15 Rxa7 Rxf2 16 Rxe7 Rxe2+ 17 Kxe2 Kxe7

b) Of course, I did this much faster:

1 e4 d5 2 exd5 Qxd5 3 Qh5 Qxa2 4 Qxh7 Qxb1 5 Qxg8 Qxb2 6 Qxf7+ Kxf7 7 Rxa7 Qxc2 8 Rxb7 Qxd2+ 9 Bxd2 g5 10 Bxg5 Rxh2 11 Rxc7 Rxh1 12 Rxc8 Rxg1 13 Rxb8 Rxg2 14 Rxa8 Rxf2 15 Bxe7 Rxf1+ 16 Kxf1 Kxe7 17 Rxf8 Kxf8

But Sam Loyd's solution was:

1 e4 d5 2 exd5 Qxd5 3 Qh5 Qxa2 4 Qxh7 Qxb1 5 Qxg7 Rxh2 6 Rxa7 Rxg2 7 Rxb7 Rxg1 8 Rxc7 Qxb2 9 Rxc8+ Kd7 10 Rxb8 Rxb8 11 Bxb2 Rxb2 12 Rxg1 Rxc2 13 Qxf7 Rxd2 14 Qxe7+ Kxe7 15 Rxg8 Rxf2 16 Rxf8 Rxf1+ 17 Kxf1 Kxf8

c) This is my position, which I don't think is soluble. But I did find another solution to a) in trying to solve it:

1 c4 d5 2 cxd5 Qxd5 3 Qc2 Qxa2 4 Qxh7 Qxb1 5 Qxg8 Rxh2 6 Qxg7 Bxg7 7 Rxa7 Bxb2 8 Rxa8 Bxc1 9 Rxb8 Bxd2+ 10 Kxd2 Rxg2 11 Rxb7 Rxf2 12 Rxc7 Rxe2+ 13 Nxe2 Qxf1 14 Rxf1 Bg4 15 Rxf7 Bxe2 16 Kxe2 Kxf7 17 Rxe7+ Kxe7