Chess

William Hartston
Friday 07 March 1997 19:02 EST
Comments

Your support helps us to tell the story

From reproductive rights to climate change to Big Tech, The Independent is on the ground when the story is developing. Whether it's investigating the financials of Elon Musk's pro-Trump PAC or producing our latest documentary, 'The A Word', which shines a light on the American women fighting for reproductive rights, we know how important it is to parse out the facts from the messaging.

At such a critical moment in US history, we need reporters on the ground. Your donation allows us to keep sending journalists to speak to both sides of the story.

The Independent is trusted by Americans across the entire political spectrum. And unlike many other quality news outlets, we choose not to lock Americans out of our reporting and analysis with paywalls. We believe quality journalism should be available to everyone, paid for by those who can afford it.

Your support makes all the difference.

Joseph Ney Babson (1852-1929) bequeathed to the world of chess problemists a challenge - known as the Babson Task - that was to keep them occupied for more than 50 years after his death. The task was to compose a problem in which White and Black both have a pawn on the seventh rank, and in reply to four variations in which the black pawn promotes to queen, rook, bishop and knight, White must promote to the same piece in each case to force mate.

Babson had done it in a non-standard problem - a self-mate where White had to force Black to give mate - but composers despaired over attempts to do it in the standard, White to play and mate in n, format.

For fifty years the Babson task was thought impossible, but in 1983 the Russian Leonid Jarosh cracked it. The diagram shows his extraordinary composition. It is White to play and mate in four.

The first move is 1.a7! setting up the mechanism. After 1...axb1(Q) White plays 2.axb8(Q)! Qxb2 (giving the king an escape square on d3) 3.Qxb3! Qxa1 4.Rxf4 mate.

After 1...axb1(R), 2.axb8(Q)? Rxb2 3.Qxb3 is stalemate; but White plays 2.axb8(R)! Rxb2 3.Rxb3 Kxc4 4.Qa4 mate.

Still more tricky is 1...axb1(B), when Black introduces the idea of Be4 into the defence. The only way to defeat it is 2.axb8(B)! Be4 3.Bxf4! Bxa8 4.Be5 (or Be3) mate.

The last thematic variation is 1...axb1(N) 2.axb8(N)! Nxd2 3.Qc1 Ne4 (otherwise 4.Rxf4 will be mate) 4.Nc6 mate.

Finally, and somewhat messily, we have to fill in the details of the mates if Black does not promote his pawn at the first move: 1...Qxd8+ 2.Kg7! Qc7 3.d8=Q+ Qxd8 4.Rxf4; or 1...Qe5 2.Bxe7 Qd6 3.Nxd6 axb1(Q) 4.Bxf6 mate; or 1...Qxa8 2.Rxf4+ Qe4 3.a8=Q Qxf4 4.Qd5 mate; or 1...Qd6 2.Re1 Qe5 (to stop Re4 mate) 3.Nxe5 fxe5 4.Re4 mate.

The magnitude of this achievement (and the fact that Yarosh went on to publish a slightly simpler version soon after) must make this a candidate for the Greatest Problem of All Time award.

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