The Problemist, 1992 (amended)
Mate in 2
A two-mover featuring the fairy form Checkless Chess, whose sole difference from orthodox chess is that a check is only legal if it gives mate.
My apologies to solvers. The diagram posted yesterday has no solution. I took the problem from a database and assumed it had been tested. I have added a white rook at h6, which makes the problem sound and preserves the clever idea, though I am not sure the composer would approve of the drastic means.
White would like to play Qe3 mate, but at the moment the move is illegal, since Black can reply Qxe3 mate (2...Qd4 is illegal because of 3.Qxd4 mate). The key 1.f7 gives the white king a flight and threatens 2.Qe3, since 2...Qxe3 is now only check, and therefore illegal. 2.f8Q is not a threat, since 2...Re7 would be mate. Black can defend by moving his bishop, as the white king will then be unable to move off the rank because of the check from the g5 rook. 1...Ba2 etc. allows 2.Qd6, which is mate because 2...Qxd6 is illegal, since 3.Ke4 is mate! In similar fashion 1...Bh1 etc. allows 2.Qd4, since 2...Qxd4? is illegal because of 3.Ke6 mate. 1...Qxd3 allows 2.f8Q (or B), since 2...Re7 is no longer mate because of the unguard of f4. Finally there is one ‘orthodox’ variation, 1...b4 2.Sa4.
Without the h6 rook the problem has no solution, since 1.f7? is met by 1...Rxg8 mate. The rook can stand on h7, h6, h5, h4 or h1, but I cannot see a more economical way of making the problem sound.