
Re: Hex Win Proof?
Posted:
Mar 22, 2004 12:47 PM


w.taylor@math.canterbury.ac.nz (Bill Taylor) wrote in message news:<716e06f5.0403181938.72a82f90@posting.google.com>... > It is an old theorem that in Hex, once the board has been completely > filled in with two colours, there *must* be a winning path for one > or other of them. > > Now, I can prove this easily enough mathematically, but I'm wondering if > there is a simple proof, or proof outline, that would be understandable > and reasonably convincing to the intelligent layman. > > Can anyone help out please? > >
More generally, suppose a square is partitioned into two subsets A and B. Then must one of A or B have a connected subset which has nonempty intersection with opposite sides? The hex win solution would follow from this if the answer is yes, since two hexagons either do not intersect or intersect in a common side.
Conjecture: no in general, but yes if A and B satisfy nice properties.
Andrzej
 > Bill Taylor W.Taylor@math.canterbury.ac.nz >  > The empty board waits. > Stones cascade down onto it! > The game is over. > 

