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Time to return to the selfmate, and this week, a long one! However, fear not! Many
solvers naturally assume that the longer a solution is the harder it will be to
find, but, in fact, this is not necessarily the case. The requirements of soundness
and uniqueness of solution mean that in a long problem White generally has to have
a great deal of control, with Black usually reduced to only a single legal move at
each point, which can often make such problems easier to solve than a hard 3-mover.
For those who nevertheless would like a clue, it is in the next paragraph, so, as
they say on the TV News every Saturday night, “If you don't want to know the
scores, look away now!”
Clue: If, in the above position, the white bishop stood on c6, rather than d5, then
White could selfmate in three, by 1.Kh6 Kf5 2.Qd7+ Re6+ 3.Kh5 Bg4. So White's task
in the diagram is to find a (10-move) manoeuvre, which brings about the diagram
position, but with the WB on c6, rather than d5.
In order to bring about the required shift of the WB’s position, without allowing Black's
pieces out or stalemating him, White’s own minor pieces and king have to engage in
a careful dance. The solution is 1.Bg8! Kf3 2.Nd2+ Kg4 3.Kf6 Kh5 (this is why
1.Bf7? would not have sufficed at move 1) 4.Bf7+ Kg4 5.Nc4 Kf3 6.Bd5+ Kg4 7.Bc6 Kh5
8.Be8+ Kg4 9.Kg6 Kf3 10.Bc6+ Kg4 Mission accomplished! Now the mate follows:
11.Kh6 Kf5 12.Qd7+ Re6+ 13.Kh5 Bg4#.
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