"Steven Meyers" <firstname.lastname@example.org> wrote in message news://email@example.com... > firstname.lastname@example.org (Bill Taylor) wrote in message news:<email@example.com>... > > 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. > Check http://web.cs.ualberta.ca/~javhar/hex/hex-yproof.html for a > simple proof using the game of Y. Nice proof and a nice page. Another way is to consider not the cells, but just the boundary between red and blue when the board is full. A component of the boundary must be either a closed curve (in which case we can contract it to a point, losing nothing) or a path from an edge to the same edge (contract again) or a path from one edge to another. In the last case, consider an enpoint of a boundary component which is closer than any other to an obtuse corner of board. According to where the other endpoint of that component is, we can determine who has won the game.
Here's a related result in topology. Let T be the Euclidean triangle with vertices (1,0,0) (0,1,0) (0,0,1) in homogeneous coordinates. Let f be any continuous mapping T to T, and write f(x,y,z) = (A(x,y,z), B(x,y,z), C(x,y,z)) Let S be the (closed) subset of T consisting of the points (x,y,z) such that _at least two_ of these inequalities hold: A(x,y,z) >= x B(x,y,z) >= y C(x,y,z) >= z. Then S has a connected component which meets all three sides of T. LH