> email@example.com (Bill Taylor) wrote in message news:<firstname.lastname@example.org>... > > It is an old theorem that in Hex, once the board has been completely > > filled in with two colours, there *must* be a winning path for one > > or other of them. > > > > Now, I can prove this easily enough mathematically, but I'm wondering if > > there is a simple proof, or proof outline, that would be understandable > > and reasonably convincing to the intelligent layman. > > > > Can anyone help out please? >
Neither of the proofs (which are basically the same) posted so far is correct. Both would apparently conclude that a winning path would be formed on a squared board, whereas this is not the case - a squared board could end in a draw.
An actual proof must use the hex nature of the board or, alternatively, that 3 cells meet at each vertex. A proof is given in Cameron Browne's book Hex Strategy, but whether it would convince an intelligent layman is not clear.
Maybe a simpler proof could be achieved by induction?