The presence of two German guest participants, however, suggested this may be a purely British phenomenon. Both Arno Zude and Michael Pfannkuche are practically unknown as players, yet Zude won the overall competition, a point and a half ahead of Mestel, and Pfannkuche finished in fifth place.
He is, incidentally, the reigning problem-solving world champion, thus adding support to the theory that the English do not win things because their names are insufficiently romantic. Could an Englishman called 'Pancake' ever win anything?
The competition itself, which was sponsored by East-West Consultants, began with a set of three 'White to play and mate in two' positions for which the solvers were given 20 minutes. The next round, lasting 50 minutes, comprised a three-mover and a four-mover, while rounds three, four and five featured an endgame study and more exotic beasts such as self- mates (where White forces Black to mate him) and help-mates (where the two sides conspire to reach a mating position).
The two-mover round is usually seen by experienced solvers as an easy warm-up, but this time several fell at the earliest hurdle.
ch09out-harts-nws The diagram position was the one that caused the trouble. It came at the end of the set, when many solvers, pushed for time, rushed into the wrong answer. The first thing to note is that White has a mate ready in reply to almost every Black move: if the rook moves, then Re5 is mate, since c4 is now covered by the white bishop; if the bishop moves, then Rd4 is mate (e6 is protected from h6); cxd2 is met by c4 mate; knight moves are met by Nxc3 mate; c5 allows Bb7; d6 is mated by Nc7 or Nf6; and dxe6 again allows Rd4. The only move not allowed for is 1 . . . fxe4.
The next thing to see is that the white queen does not seem to play a part in any of these mates, except occasionally to protect the rook on e4. So all we need is a queen move that rules out the defence with fxe4.
So we may now conclude that the answer is 1. Qh5, stopping fxe4, meeting Kxe4 with 2. Qh1, and cheerfully ignoring Bxh5 because 2. Rd4 is mate. Many solvers did indeed conclude that 1. Qh5 was the answer, and they were wrong. After 1. Qh5 cxd2, White has no mate. 2. c4 fails to the reply Kxe4.
Since this does not work, White needs a queen move that provides mates after fxe4 and, if he renounces protection of the rook on e4, after cxd2. What should make the answer easier to find is that the mate after cxd2 is unchnaghed from that provided in the starting position. The solution is 1. Qb1] when 1 . . . fxe4 is met by 2. Qb3 and 1 . . . cxd2 2. c4 leaves the rook protected from another direction.