```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 ÃÂÃÂgidius 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 pathseparates the board.  But this is something that takes some work toshow.
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