Date: Mar 20, 2004 4:35 AM
Author: Chan-Ho Suh
Subject: Re: Hex Win Proof?
In article <Xns94B1B0DB1B331webmastertbrauchcom@126.96.36.199>, Tim
Brauch <RnEeMwOs.pVoEst@tbrauch.cNOoSPAMm> wrote:
> firstname.lastname@example.org (Jonathan Welton) wrote in
> > 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
> I wasn't assuming a square board, I was imagining the board set up like
> a parallelogram. At least, that is how I orientate the board when I
> play. Then red goes top to bottom and blue goes left to right (red and
> blue because the board I made uses poker chips).
What Jonathan is trying to point out is that you aren't using the fact
that there are hexagons. If you took a checkerboard and squished it to
form a parallelogram (with angles not 90 degrees), then you would have
a board where every piece of the board looked like a little
parallelogram (instead of a hexagon). Clearly we can color this
checkerboard without a winning path by the usual checkerboard coloring.
[I'm not considering two parallelograms that touch only in a corner to
be part of a path]
Your proof attempt makes no use of the specifics of the Hex board, and
so would apply to any board like the one above.