In article <firstname.lastname@example.org>, 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.