Chess
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Your support makes all the difference.White to play and mate in three - Bo Lindgren, Dagens Nyheter 1997
- round 2 of the WCSC
ALTHOUGH YOU might imagine that their field is primarily aesthetic, chess problem and study composers are in fact highly competitive; and there is an annual World Congress of Problem Solving.
This year's, the 42nd, took place in Netanya in Israel from 23 to 30 October, with the highlight of the programme being the 23rd World Chess Solving Championship (WCSC), fought out on 26 and 27 October.
This consists of six timed rounds, each of which features a different type of composition. Three problems must be solved in each round, and all variations must be given for each problem in order for a team to score maximum points. Teams of three compete, with the two highest scores for each round being combined to make the team total. All scores also count for the individual title.
In order to make the event fair, it is of course essential that the competitors haven't seen the compositions previously, so the majority of those chosen by the event director Bo Lindgren were composed by himself (though possibly, like the one above, already published in obscure sources).
These formed a particularly difficult set that left even ex-world champions reeling, well behind the winners Ofer Comay (Israel), first on 76/90, ahead of Sergey Rumyantsev (Russia) on 75 and Jorma Paavilainen (Finland) on 73.
Indeed, while last year's champion, Georgy Evseev of Russia, didn't compete and the 1996 world champion Noam Elkies finished 27th, the best result of recent winners was made by the 1997 champion Jonathan Mestel, who came in 13th with 60.5/90.
Together with Michael McDowell who was 19th, on 58.5, and Graham Lee, who was 32nd of the 68 competitors, on 46.5, this put Great Britain sixth out of the 21 participating countries, behind the winners Russia, Germany, Israel, Yugoslavia and Finland.
The solution:
1 Bh3! clears the rank for the queen to threaten 2 Qe8+ Re5 3 Qxe5 mate. Black can defend by moving the f7 pawn to allow the bishop to interpose at e6.
But if 1 ...f6, 2 Nc4! and 3 Nd2, mate cannot be prevented, because the unpinned bishop at d4 can no longer check on g7.
And if 1 ...f5 2 Nb7! and 3 Nc5 mate, because the unpinned rook can't check at h5.
Some would-be solvers fell for the trap 1 Bd7?, which fails because after 1 ...f6 2 Nc4 Black can play 2 ...Be6! and White cannot guard f5. Meanwhile 1 Bg4? fails because of 1 ...f5 2 Nb7 fxg4!.
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