In article <Xns94B1B0DB1B331webmastertbrauchcom@220.127.116.11>, Tim Brauch <RnEeMwOs.pVoEst@tbrauch.cNOoSPAMm> wrote:
> firstname.lastname@example.org (Jonathan Welton) wrote in > news://email@example.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. .