
Re: Hex Win Proof?
Posted:
Mar 22, 2004 2:42 PM


In article <w5ad28sxnd.fsf@pc032.diku.dk>, Torben ÃÂÃÂgidius Mogensen <torbenm@diku.dk> wrote:
> ChanHo Suh <suh@math.ucdavis.nospam.edu> writes: > > > In article <w565cxi4n1.fsf@pc032.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 endpoint 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 laymenconvincing).
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. 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. 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.
It seems your goal was to give a pseudoproof, 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.

