It is white to move, but the goal is to force black to mate white in 9 moves. So it doesn't matter that black can mate in 1, he must be forced to play the mate in 1. Solution follows in a few hours, good luck, it's mindfucking. I cannot remember the composer, but it's definetly not me. He must be a honoured man (or woman)
Selfmate in 9
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The answer can be found by logic. We must keep our rook on the eight rank to prevent the black dark-squared bishop from moving.
If we get a position where there is no squares between the lightsqared bishops and with white to move, Rh8! would win. Lets assume this:
So by logic, black is actually threathning Bf3!, where white can not play Rh8, because it is already there. So if there is one square between the bishops, where should the rook then be? On c8!
So, what happens if there are two squares between the bishops? The rooks must belong to d8 then.
So when there are three squares between the bishops, the rook belongs on e8, and when there are 4 squares (as there are in the puzzle), the rook obviously belongs on f8!. Notice that g8 always is a bad square, because the rook can interpose on g2, no matter where the bishops are.
So ready to see the solution?
ViktorHNielsen, that was the most amazing work of explanation of a position I've ever seen on any chess site.
Haha!!
