Date: Mar 20, 2004 2:33 AM Author: Tim Smith Subject: Re: Hex Win Proof?

In article <716e06f5.0403181938.72a82f90@posting.google.com>, Bill Taylor

wrote:

> 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.

The following has a bunch of cases, but I think the intelligent layman can

follow them.

Let's assume Red has top/bottom, and Blue has the sides. We want to show

that if Blue does not have a win when the board is full, the Red must have

one.

There must be at least one Red hex on the top, and at least one on the

bottom, or else Blue would have a win along the top or bottom. Pick a Red

hex on top, U, and a Red hex on the bottom, D, and consider any path between

them. Go along that path from bottom to top.

If Red doesn't have a win, that path must run into a Blue hex. Let S be the

set of all Blue hexes connected to that Blue hex. Call the Red hex that ran

into S A. As you continue along the path, you must eventually reach a Red

hex that is not surrounded by S. Call that Red hex B.

If S does not touch any edge, then there is an all Red path from A to B.

Just go around the edge of S in either direction from A to B. If S touches

one edge, there is still a Red path from A to B.

Thus, we cannot reroute around S only if S connects to two or more edges.

Let's consider the cases:

1. S connects to the top and bottom. Then the Red hexes adjacent to at

least one of its sides form a win for Red, unless S connects is connected to

opposite corners, in which case S would be a win for Blue.

2. S connects to left and right. S is a win for Blue.

3. S connects to the bottom and a side. WLOG we can assume S is connected

to the left side. If the connection to the bottom is to the left of D, then

there is a path from A to B around the boundary of S. If the connection to

the bottom is to the right of D, then there is a connection from B to the

bottom along the boundary of S (unless S connects to the bottom at the

corner, in which case S is a win for Blue).

4. S connects to the top and a side. Similar to case #3. If the top

connection is on the same side of U as the side connection, then there is a

connection from A to B. Otherwise, there is a connection from A to the top.

In all these cases, we are able to reroute the path such that it is still

connected to the top and the bottom, but avoids S. Hence, if Blue does not

have a win, there is a win for Red, and this procedure will even find such a

win.

--

--Tim Smith