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Re: Hex Win Proof?
Posted:
Mar 19, 2004 5:14 PM
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> w.taylor@math.canterbury.ac.nz (Bill Taylor) wrote in message news:<716e06f5.0403181938.72a82f90@posting.google.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.
> >
> > Can anyone help out please?
>
Neither of the proofs (which are basically the same) posted so far is
correct. Both would apparently conclude that a winning path would be
formed on a squared board, whereas this is not the case - a squared
board could end in a draw.
An actual proof must use the hex nature of the board or,
alternatively, that 3 cells meet at each vertex. A proof is given in
Cameron Browne's book Hex Strategy, but whether it would convince an
intelligent layman is not clear.
Maybe a simpler proof could be achieved by induction?
Jonathan Welton
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