Stay up to date with notifications from The Independent

Notifications can be managed in browser preferences.

CHESS / Silliest scenario in the world

William Hartston
Tuesday 21 June 1994 23:02 BST
Comments

THE RESULTS of the PCA/Intel world championship quarter- finals in New York bring closer the scenario of ultimate silliness which has been on the agenda since last year's schism. Here is the position:

The Intel-sponsored PCA world championship is now reduced to four players competing for the right to challenge Garry Kasparov next year. The semi-finals, to be played in September, will be Adams-Anand and Short-Kamsky. However much one might hope for English victories, the prediction on form must be for a final between Anand and Kamsky.

Next month, however, these two will meet in a Fide world title qualifier. (The defending champion, Karpov, joins in at the semi- final stage, incidentally.) So what happens if Anand beats Kamsky and goes on to take the Fide world championship from Karpov, while in the parallel universe of the PCA, Kamsky beats Anand and then bops Kasparov?

Oddly enough, such confusion may be exactly what the chess world needs to bring it back to its senses. A unification decider between the two would be difficult to avoid, and some formula would be agreed between the PCA and Fide to run a single world championship in future. With Florencio Campomanes reportedly not seeking re-election as Fide President, there are signs of peace at last in the long-running chess wars.

Here's another problem which should be simpler to solve:

ch22out-harts-nws

Composed by H Lange, it's White to play and mate in four. At first sight, it looks as though it should be easy. If it were Black's move the game would be over at once: 1 . . . Kf5 2. Bh7 mate or 1 . . . Kh5 2. Bf7 mate, but it's not easy for White to lose a move. 1. Kg4 is stalemate; 1. d6 Kf5 lets the black king escape to e6 after Bh7+; and 1. Be6 Kh7 2. Bf5+ Kg8 also fails to mate in the stipulated number of moves. After fiddling around with the pieces a little, we begin to realise that two bishops and a pawn can't mate on their own. Some subtle king moves are therefore needed. But 1. Kh3 Kf5 or 1. Kf3 Kh5 leave the black king rushing towards the open spaces via f4 or h4.

The answer, as one might have guessed at the start, involves reaching the diagram position with Black to move. The mates are already there; why spoil them? And the way to do it is to start with 1. Kg2] After 1 . . . Kf5 2. Kf3 Kg6 3. Kg3] or 1 . . . Kh5 2. Kh3 Kg6 3. Kg3] it is mate next move.

Join our commenting forum

Join thought-provoking conversations, follow other Independent readers and see their replies

Comments

Thank you for registering

Please refresh the page or navigate to another page on the site to be automatically logged inPlease refresh your browser to be logged in