Date: Mar 20, 2004 4:48 AM
Author: Chan-Ho Suh
Subject: Re: Hex Win Proof?

In article <>, Jonathan
Welton <> wrote:

> > (Bill Taylor) wrote in message
> > news:<>...

> > > 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?

> >
> 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?

Im not sure what proof you are thinking of, but you make it sound like
it's rather complicated.

Here's a proof I think is pretty simple and straightforward (but as
we've some evidence of, not simple to come up with!); you'll have to
tell me if this is the same as Browne's proof. I think the following
proof is understandable to the so-called intelligent layman:

Suppose we've filled up the board with black and white colors.
The boundaries of the hexagons form a graph with edges and vertices.
We can look at the edges that separate hexagons of different colors.
Because of the way the hexagons meet (three to a vertex), there is
always a unique way to continue one of these special edges to another
edge that separates different colors.

In other words, we can't have more than two of these color separating
edges meeting at a common vertex.

So the boundary between the colors consists of either loops (that don't
intersect each other or themselves) and/or paths (that don't intersect
themselves) from a spot on the boundary of the board to another spot on
the boundary of the board.

Now we're not done since we haven't shown the boundary between colors
has at least one of these paths, and furthermore, even if we knew that,
we would have to show there's a path connecting one side of the board
to the opposite side. Then we could conclude there's a chain of one
color joining opposite sides, i.e. there is a winning path.

Here's a simply way to accomplish both these goals. At each corner of
the board, there is a hexagon that is basically shared by each side.
We extend the graph (given by taking the boundaries of all the hexagons
on the board) by adding an edge to each corner hexagon: there's only
one way of doing this without favoring either side. We also pretend
that each side of the board has an extra row of hexagons already
colored by one color. So if one side belongs to white, then the extra
row added adjacently is colored black, and vice versa. Note that the
edge we added at each corner separates these newly added rows and so is
one of the color separating edges of our graph.

Now we've killed two birds with one stone. Since the boundary between
colors now has edges that can't close up to be loops, we must have
paths going from one side to one side (possibly the same side). By
checking the possibilities of where a path starting at a corner can end
up, we can easily see we have a winning path.

I struggled a little in explaining the modification of the board; I can
only wish I was adept in ASCII since with pictures it's pretty simple.
Hopefully people can figure it out.

Final observation, it still remains to show there is *only* one winner!
This is harder (basically one needs to prove a piecewise-linear version
of the Jordan separation theorem), so I'll stop here. In any case, it
doesn't seem the OP was interested in more than showing there is at
least one winner.