How to lose a move December 30, 2005
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Selfmates and rook promotions

We need to come back to the subject of selfmates once more. Dr John Nunn, our resident problem expert and general spoilsport in these matters, drew attention to the fact that the intended solution to my beloved Hasselkus problem which I presented yesterday is unfortunately flawed.

E. Hasselkus, Schach-Express 1948

Selfmate in seven moves

The author gave 1.g8R+ Kf6 2.f8R+ Ke6 3.e8R+ Kd6 4.d8R+ Kc6 5.c8R+ Kb6 6.b8R+ Ka6 7.Ra7+ Qxa7# – with six underpromotions to rooks. However on move four there is an alternative: 4.d8Q+ Kc6 5.Qd7+ Kb6 6.b8Q+ Ka6 7.Qa7+ Qxa7#. That in itself spoils the problem. But there is a more serious fault: a completely different solution was found, with two queen and two rook promotions: 1.g8Q+ Kf6 2.Qg7+ Ke6 3.e8Q+ Kd6 4.d8R+ Kxc7 5.Qd7+ Kb6 6.b8R+ Ka6 7.Qa7+ Qxa7#. Oops.

But there is also positive news from the Doctor. In 1966 the Swedish composer Bo Lindgren presented an eightfold rook underpromotion in a selfmate.

Bo Lindgren, Feenschach 1966

Selfmate in ten moves

The problem won a second prize for the following amazing solution: 1.a8R+ Na7 2.Rxa7+ Kb5 3.b8R+ Kc6 4.c8R+ Kd6 5.d8R+ Kxe6 6.Rc6+ Rxc6 7.e8R+ Kxf6 8.f8R+ Kg6 9.g8R+ Kxh6 10.h8R+ Bh7#. "I believe this setting is sound," writes John.

Click here to replay both problems in JavaScript
[You can click on the moves to follow the notation]

How to lose a move

We return to our subject of Very Long Problems and the amazing convolutions one has to go through just in order to lose a single move. The great master of the art was Karl Fabel, and here is another gem from his problem factory.

Karl Fabel, Die Welt 1947

How many moves to mate?
wKc1,Nb4,d1,Pb2,c3,c5,e3,f2/bKa1,Bd5,Pb3,c4,c6,e4,e5,e6,f3,g4

Once again we are going to give you a lot of help with this problem. For that reason we do not specify the number of moves required to mate, but ask you to work out this number, using the pointers we give you.

Looking at the above position we note that if it were Black to move then we would have a forced mate: 1...g3 (the only legal move Black has) 2.fxg3 f2 3.Nc2+ bxc2 4.Nxf2 Ka2 5.Kxc2 Ka1 6.g4 Ka2 7.g5 Ka1 8.g6 Ka2 9.g7 Ka1 10.g8Q Ka2 11.Qa8 mate.

So all White has to do is to lose a move. White would like to move his king out to g1 and triangulate there (Kg1-h1-h2-g1) and return to c1. Black would be in zugzwang and would have to play the fatal ...g3. But the plan does not work so easily. As soon as White moves his king away Black follows with his king, moving to b1, c1 and even capturing the immobilised knight on d1.

So White has to first prevent the black king from moving to b1 while his own king triangulates. This can be achieved with the help of the other knight, the one that is now on b4. Here we somewhat abruptly end the generous assistance we have been offering. You are on your own now. Try to find a sequence of moves to force mate, and count the number of moves required to do so.

Frederic Friedel