email@example.com (Bill Taylor) wrote in 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? >
Here's what I'm thinking...
Suppose you have red going top-bottom and blue going left-right. If red does not win, then there must be a path that divides the board into to pieces, a top and a bottom piece. Red cannot be in any position in that path, so it must be blue. Thus blue wins. Rotate pi/2 and switch colors. QED
-- Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University