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