Well, it is unfortunate that you edited your puzzle after I already found a solution. So I'll post one of the solutions to the original puzzle, and if anyone wants to have a go at the edited position he should have a clear idea for the strategy:
As per requested by @Dorian Fusco, here is what my original solution was before. @BoredAtWork, amazingly, and to my surprise, found an alternate way! Congratulations to them!
My Original Solution:
Now do you see why I considered this to be a tricky question? ;D
I spot a
after which Black changes tack:
What a slaughterhouse indeed! Well done!
As requested: an explanation of why I chose the moves I did. With such a terrifyingly long selfmate I thought that for the setter to have composed it at all, the solution must surely be a simple one, with Black always making checks which White can parry only in one way. I ...
Here is a new record of 16 moves that is a slight modification of a position made by Bernd Schwarzkopf and Karl Scherer in Feensnach 1980 on page 13. A PDF can be found here.
FEN: QKN4r/nRRRRRRr/kq6/p1q1R3/1P1q1b2/4q3/5q2/1r4q1 w KQkq - 0 1
In this position that I composed long ago (this position is legal), every move is completely forced and it is checkmate in
!8 moves (15 ply):
1. Qxc7+ Qxc7+ 2. Qxc7+ Qxc7+ 3. Qxc7+ Qxc7+ 4. Bxc7+ Kb7+ 5. Bb8+ Qxe7+ 6. Qxe7+ Qxe7+ 7. Qxe7+ Qxe7+ 8. Rxe7#
The last episode was difficult, with a few twists and turns, but this selfmate left little doubt: the answer is 18 moves.
It's clear that we must manipulate Black's pawn structure somehow, because there's no way a mate can come out of the current structure. The fact that such a manipulation must be forced, i.e. Black's only legal move, ...
I've looked at this problem for so long that I begin to think perhaps the composer made a slight error in the composition. So I think it would be smart to ask the composer if my try is what he had in mind before I continue cracking my brain. My try is as follows: 1.Bh5+ Rg4 Then the next 4 black pawn moves are forced (irrespective of the order in which they ...
To submit something that no one has yet suggested, Rb2..Nc7# is valid but not mandatory for black because the option of Bxf7 is always in play unless black's king is placed into check.
The solution hinges on F7 because outside of A1, it is the subject square of which Black has the most options.
A1 tells us two things when combined with F7:
You must check ...