Date: Mar 22, 2004 5:08 AM
Author: Chan-Ho Suh
Subject: Re: Hex Win Proof?

In article <w565cxi4n1.fsf@pc-032.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.