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

In article <Xns94B1B0DB1B331webmastertbrauchcom@63.223.5.95>, TimBrauch <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 factthat there are hexagons.  If you took a checkerboard and squished it toform a parallelogram (with angles not 90 degrees), then you would havea board where every piece of the board looked like a littleparallelogram (instead of a hexagon).  Clearly we can color thischeckerboard without a winning path by the usual checkerboard coloring. [I'm not considering two parallelograms that touch only in a corner tobe part of a path]Your proof attempt makes no use of the specifics of the Hex board, andso would apply to any board like the one above.  .
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