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Self-mate in 9

G Broecker,

London Chess Fortnightly 1892

IN THIS problem White is required to force Black to mate him in nine moves. Not much to do with normal chess, you may think, and I can hardly demur too strongly. But before the solution at the end, here is a digression about analogous ideas in actual play.

Unless it is particularly, as we say, "sharp", there is generally a choice of two or three reasonable moves in a middlegame or opening position. Of course, there are many forcing situations in contradiction to this: tactical sequences in which you must recapture, parry a check, prevent a direct attack on the king or perhaps meet some more strategic threat (which will normally involve changing the pawn structure in some way - for example, quite often it's absolutely essential to block a pawn's advance to prevent critical lines being opened).

But the principle remains that at these relatively early stages, when there are plenty of pieces on the board, you often have quite a broad choice: a choice which, however, tends to narrow as the endgame approaches.

This may seem odd - surely when there's more room there will be more reasonable squares for your pieces; but the thing is that any weaknesses which require defending - or of the opponent's to attack - can only be protected/ attacked by the small number of men now available.

This restriction of choice finds its purest form in king and pawn endings, in which often there will be a single good move. And especially when the pawn structure is fixed, the kings may have to dance around in exceedingly complex ways in order to gain or deny entry to each other.

Here, for the kings read bishops and rook. In order to solve the problem, we need to consider where the rook needs to be with Black to move, depending on the number of squares between the bishops.

With no squares in between, the obvious case is Bc6 vs Bb7, when it must be on h8.

With one square between, eg, Bd5 vs Bb7, White should play 1 Rc8! Bc6 2 Rh8!

Similarly, with two squares between it should be on d8, three e8 and four f8.

This gives the solution:

1 Rf8! Bc6 2 Re8 Bd5 3 Rd8 Be4 4 Rc8 Bf3 5 Rh8 Be4 6 Bf3 Bd5 7 Be4 Bc6 8 Bd5 Bb7 9 Bc6 Bxc6 mate.

Another line goes:

1 Rf8 Bc6 2 Re8 Bb7 3 Bf3 Bc6 4 Rd8 Bb7 5 Be4 Bc6 6 Rc8 Bb7 7 Bd5 Bc6 8 Rh8 Bb7 9 Bc6 Bxc6 mate.

jspeelman@compuserve.com

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