The Corner Parade

Here is a problem that I made awhile ago on the Matplus forums. It’s probably cooked. Good luck solving!

It’s White to selfmate themselves in 10 moves.

It's hard to say if this is dualfree but the variations I checked appear to be OK. Hard to analyze what happens when white does not start with Ng6+. White gains a tempo but black gains a choice for his knight (g6, f7) and there many things to try.

But I already know you have been quite lucky that one white attempt fails to stalemate. That luck is well deserved as all composers go through 10 occasions of bad luck for every single instance of good luck!

So this is what you escaped from: the first 4 moves the same as in your solution, then: 5. Qf2 Ke4 6. Ka3 Ke5 7. Nd2 (stalemate) Ke4 (legal if it were not stalemate) 8. Nb3 Ke5 9. Qf6+ Ke4 10. Qc2+ Nxc2#.

Imagine being white.

OK, I think I found a dual - in exactly 10 moves as well. Black can play his knight earlier but white always replies Qde6+ which only leads to a change in move order. Note that this solution neither requires Nf8 nor Ra1 in the diagram.

That’s no surprise of course. Thanks for the effort. A simple fix is adding a Black queen on g6 for 1. Nxg6+.

I don't think you really want an extra black queen as it takes away the charm of the lonely knight. But of course you want to keep your knight on h8 to demonstrate its corner-to-corner journey. Not easy to correct it seems.

There are two other duals near the end of your solution - 9. Bg4+ and 9. Bf1+ - but I estimate that is solvable by placing Bh3 on c8 or d7.

When I think of it, could you by any chance be extraordinary lucky to solve all duals by merely placing Bh3 on d7? It prevents my dual solution as well since the essential move Qd7 is now impossible!