This problem combines logic and chess in an interesting manner:
Is it possible to design a position so that it can be proved that White
has a mate in two moves, yet that it is impossible to exhibit the mate?
And I don't mean a chessboard with pieces weighing several tonnes,
just the chess position (if possible)...
No. If it is a stalemate in one there can't be mate in two, at least not
in a chessproblem where white (and black) is supposed to play the
best possible move.
I'll reveal the answer later this evening.
a) If black's last move was 1...e5, then White can take en passant
and after black castles (else 2.g8=Q mate), white has 2.b7 mate!
b) If black's last move was something over than 1..e5, then his last
move must have been made by King or rook, making castling illegal!
Therefore white can play 1.Ke6 and mate next move.
As there is insufficient information to know which condition (a or b) is
true, then the mate in two is proved but cannot be exhibited. A superb
A superb problem by Raymond Smullyan.
Smullyan wrote two chesslogic books:
- The chess mysteries of Sherlock Holmes (containing this problem)
- The chess mysteries of the Arabian Knights
Regretfully I do not own The Arabian Knights. If anyone knows how I
can get a copy of that I'll be delighted!
Acolyte (and other mathlovers),
You must read these books.
For you (and others of course) I will present my next problem: "the
so what you're trying to say is that mate is possible either way, but
you don't know which one you can play because you didn't see the
previous move? Is that what is meant by "proved" (you know that you
can mate whatever) but cannot be exhibited (you can't do it because
you don't have enogh info).
Sorry - got punched in the head earlier and am not thinking straight