Date: Mar 21, 2004 3:43 PM Author: Arthur J. O'Dwyer Subject: Re: Hex Win Proof?

On Sat, 20 Mar 2004, Chan-Ho Suh wrote:

>

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

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

> > >

> > > 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?

Almost certainly not. Induction doesn't usually lead to clear proofs,

the way "pouring water" does. And in this case, it won't lead to a

short-but-obscure proof either: you'll get a long-and-obscure proof.

> > 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.

Both Tim's proof and Brian's proof assume that the reader knows what

a hexagon is; how many sides it has; and how it differs from a square.

I don't see what you're objecting to.

> Your proof attempt makes no use of the specifics of the Hex board, and

> so would apply to any board like the one above.

But it *does* make use of the specific topology of the Hex board.

If it didn't, then (as you noted) it would prove a falsehood. Since

it does not prove a falsehood (which is impossible; falsehoods cannot

be proven), it must use the topology of the board.

Q.E.D.

Bad analogy: I can prove that the speed at which a tennis ball hits

my hand going down is the same speed at which it left my hand going

up, using a simple argument from conservation of energy. You object:

oh, but that proof is flawed, because what if there was a rocket engine

attached to the tennis ball? I respond: any fool can tell that there

is *not* a rocket engine attached to the tennis ball, so why would you

even think that could be a problem?

-Arthur