Date: Mar 20, 2004 4:35 AM
Author: Chan-Ho Suh
Subject: Re: Hex Win Proof?

In article <Xns94B1B0DB1B331webmastertbrauchcom@63.223.5.95>, Tim

Brauch <RnEeMwOs.pVoEst@tbrauch.cNOoSPAMm> wrote:

> j_welton@hotmail.com (Jonathan Welton) wrote in

> news://3dfcbf81.0403191414.33bb386a@posting.google.com:

>

> > 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.

.