
Re: Hex Win Proof?
Posted:
Mar 22, 2004 5:08 AM


In article <w565cxi4n1.fsf@pc032.diku.dk>, Torben ÃÂÃÂgidius Mogensen <torbenm@diku.dk> wrote:
> w.taylor@math.canterbury.ac.nz (Bill Taylor) writes: > > > 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. > > The easy part is proving that both players can't win at the same time: > > Assume that there is a white path connecting top and bottom and a > black path connecting left to right. These must intersect, but on a > hex board two paths can only intersect if they share a hex. [snipped]
Why must they intersect? That would seem to be the core of the proof.
The way I would show they intersect is to first show that a path separates the board. But this is something that takes some work to show.

