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Re: Hex Win Proof?
Posted:
Mar 20, 2004 3:06 AM
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"Steven Meyers" <swmeyers@fuse.net> wrote in message
news://b1d340a7.0403191753.4511693e@posting.google.com...
> 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.
> Check http://web.cs.ualberta.ca/~javhar/hex/hex-yproof.html for a
> simple proof using the game of Y.
Nice proof and a nice page.
Another way is to consider not the cells, but just the boundary
between red and blue when the board is full. A component of
the boundary must be either a closed curve (in which case we
can contract it to a point, losing nothing) or a path from an edge
to the same edge (contract again) or a path from one edge to
another. In the last case, consider an enpoint of a boundary
component which is closer than any other to an obtuse corner
of board. According to where the other endpoint of that
component is, we can determine who has won the game.
Here's a related result in topology. Let T be the Euclidean
triangle with vertices (1,0,0) (0,1,0) (0,0,1) in homogeneous
coordinates. Let f be any continuous mapping T to T, and write
f(x,y,z) = (A(x,y,z), B(x,y,z), C(x,y,z))
Let S be the (closed) subset of T consisting of the points (x,y,z)
such that _at least two_ of these inequalities hold:
A(x,y,z) >= x
B(x,y,z) >= y
C(x,y,z) >= z.
Then S has a connected component which meets all three
sides of T.
LH
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