Date: Mar 23, 2004 3:49 AM Author: Torben Mogensen Subject: Re: Hex Win Proof?

Chan-Ho Suh <suh@math.ucdavis.nospam.edu> writes:

> In article <w5ad28sxnd.fsf@pc-032.diku.dk>, Torben ÃÂÃÂgidius Mogensen

> <torbenm@diku.dk> wrote:

>

> > Chan-Ho Suh <suh@math.ucdavis.nospam.edu> writes:

> >

> > > In article <w565cxi4n1.fsf@pc-032.diku.dk>, Torben ÃÂÃÂgidius Mogensen

> > > <torbenm@diku.dk> wrote:

> > >

> > > > Assume that there is a white path connecting top and bottom and a

> > > > black path connecting left to right. These must intersect, but on a

> > > > hex board two paths can only intersect if they share a hex.

> > > [snipped]

> > >

> > > Why must they intersect? That would seem to be the core of the proof.

> >

> > If you have a rectangle (skewed or not), a curve that connects top and

> > bottom must intersect a curve that connects left and right (though

> > this can be at the end-point of one or both curves). This is true for

> > any two continuous curves, not just for paths through hexes. I

> > suppose you can be deliberately obtuse and require a proof for this,

> > but I think an "intelligent layman" would accept this without proof.

> > After all, what was asked for was a proof that would be

> > "understandable and reasonably convincing".

> > [snipped]

>

> That's a good point, about what's "reasonably convincing" to the

> layman. I was just thinking the other day that probably none of the

> proofs (or attempted proofs) in this discussion is needed to be

> "reasonably convincing" to the layman.

>

> Anyway, it appears to me that the discussion has turned basically on

> how to prove these results (barring considerations of

> laymen-convincing).

>

> But regardless of all that, my point was that you purported to give a

> proof of a result by reducing it to something which is more or less the

> the crux of the matter, and then saying it was obvious.

IMHO, the fact that the paths must intersect is obvious to a layman,

but the fact that a path must exist is not. Hence, it is the latter

part that needs to be detailed.

> It's not a matter of being deliberately obtuse. Showing the result

> for continuous curves is much harder than anything required for the

> Hex proof.

Indeed, as it requires a rigorous definition of what it means to be

continuous. However, a layman has a good intuition about what it means

for a curve to be continuous and what the consequences of that is.

Basically, a non-mathematician will think of a continuous curve as

"something that can be drawn without lifting the pencil", and that is

good enough for this purpose.

> Of course, in this case, we are in a discrete setting and so we

> don't have to work so hard, but I felt that it needed to be pointed

> out to readers that this is something that needs to be demonstrated

> rigorously.

Indeed, if we want to prove it to a mathematician who does not already

accept the intermediate value theorem (of which the intersection

property is a simple consequence).

> It seems your goal was to give a pseudo-proof, convincing to the person

> on the street; on the other hand, you combined that with a fairly

> rigorous proof in the rest of your post. I think it's perfectly

> natural for me to critique your post based on the standards you follow

> in most of your post.

As I said above, I can rely on a laymans intuition about continuous

curves, but there is no such intuition to rely on for the second part

of the proof, which hence needs to be more detailed/rigorous.

Torben