December 31, 2005
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First a comment to yesterday's column, which people seem to be reading quite carefully. We published an eight-fold rook underpromotion problem by Swedish composer Bo Lindgren, which John Nunn assumed was sound – unlike my own beloved Hasselkus example, which I have been using for years. But, sigh, even Lindgren appears to be flawed.

Bo Lindgren, Feenschach 1966

Selfmate in ten moves

The solution intended by the author is: 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#. But Arie Haenel of Jerusalem sent us an alternate solution, with two queen and two rook promotions. One that in fact forces Black to mate in nine moves: 1.a8Q+ Na7 2.Qxa7+ Kb5 3.b8R+ Kc6 4.c8R+ Kd6 5.d8Q+ Kxe6 6.Qdd7+ Kxf6 7.Qd6+ Kxg7 8.Rg8+ Kxh7 9.Qg6+ Bxg6#. This appears to be correct – another famous problem bites the dust.

Back to the subject of very long direct mate. Here's another lovely lose-a-move problem by the great magician of unorthodox chess puzzles.

Karl Fabel, Caissa 1948

How many moves to mate?
wKf7,Nc1,d3,Rd1,Pa2,e6,g6,h5/bKf1,Qg1,Ne1,h1,Bf2,f8,Rg2,h8,Pa3,a7,e7,g3,g7,h2,h6

Once again we are assuming that there are not too many people out there who will get out a board and solve this puzzle during afternoon tea, so we will not be specifying how many moves to mate. That is for you to find, after making liberal use of the help we will provide.

First of all we notice in the above position that Black can only move one of the pieces in the cluster at the bottom right. It is the bishop on f2. But if he does so he is immediately mated by Rxe1. On the other hand, White cannot move any of his pieces either without letting Black escape from the box at the bottom right – except for the white king.

The second thing we notice is that Black cannot move his rook on h8, as it will be captured, and then he will be forced to move the bishop and be mated. Since it is White to move that is not currently an acute problem. But if White succeeds in losing a move the danger would be serious.

The best way for White to lose the move is to triangulate with his king. But to do so the king has to move to a black square, and if it is not careful it will receive a check from the bishop on f2. For instance, after 1.Ke8 Rg8 2.Kd7 Rh8 3.Kd8?? Black would have 3...Bb6+ and then 4.Ba5 or Re2, opening up the position for an easy win. No, the white king must avoid a bishop check at any cost.

There is a black square that White can safely tread upon: b8. So his stategy will be to move his king to b7, using only white squares on the way, then move it to b8, and return to f7 by the same route (Kf7-e8-d7-c8-b7-b8-c8-d7-e8-f7). After that it would be Black to move and his rook would be in zugzwang.

But Black has the a-pawn. Ah, we begin to understand: after the above tempo-losing manoeuvre Black will move the a-pawn, and White will keep repeating the manoeuvre until the a-pawn has no more moves. Then Black will be in final zugzwang, his rook will be captured and he will be mated.

There are still a couple of subtleties to be attended to, but from here on you are on your own. You will notice that we have provided the ASCII description of the problem so that you can use Fritz to help you analyse it. Please do so, as this will be part of the final problem competition that will be presented to you in a couple of days. You will need to tell us exactly how many moves White needs to mate, assuming best defence by the opponent.

Frederic Friedel