Tuesday 28 February 1995
The contest comprised five rounds. After easing into the competition with three mate-in-two problems to solve in 20 minutes, contestants were then allowed 50 minutes for a 3-mover and a 5-mover, then 35 minutes for an endgame study, 30 minutes for two self-mates (where White plays first and forces Black to give mate), and finishing with two helpmates (both sides conspire to reach a mating position in the specified number of moves).
Friedgood dropped only one point, omitting one variation from the complete solution of one of the self-mates. For the lesser contestants, one of the two-movers produced unexpected difficulties.
The diagram position - White to play and mate in two - was composed by F Salazar in 1969. The difficulty lies in the number of tempting tries for White and the ingenuity of Black's defences.
x , , ,
,Hn , V
, V B ,
Z , ,H,
, , , X
With Black's king almost surrounded, White has only to engineer a check and it will be mate. Even 1.Qxg2 (far too crude to be the intended solution) almost works with its threats of 2.Ng4 or 2.Qg5, but 1...Qg8! denies White a mate at move two.
After a brief look at 1.Qh2 (which would be a fine idea if it actually threatened anything) we soon home in on the idea of Rf5, which would be mate but for the reply Kxd4. That thought produces a string of candidate moves: e3, Be3, Bc3, Qa1, Nc2, Nb3, Nxb5, Ne6 and Ndf5 all threaten Rf5 mate but only one of them works.
Black's stock defensive move is Bd3, which works against 1.e3 or 1.Be3. If you discover why it does not work against some of the other moves, you'll be well on the way to a solution. Otherwise you'd best wait until tomorrow.
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