
Re: Hex Win Proof?
Posted:
Mar 23, 2004 3:49 AM


ChanHo Suh <suh@math.ucdavis.nospam.edu> writes:
> 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.
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 nonmathematician 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 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.
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

