The box below forms a basis for a game between two players. The idea is that the two players take turns shading in one of the six rectangles (numbered 1 through 6) with one of two colors- say red or blue.

+-+---+------------+
| |   |            |
| | 2 |     3      |
| |   |            |
| +---+--------+---+
|1|            |   |
| |     6      | 4 |
| |            +---+
| |            | 5 |
+-+------------+---+

Either player can use either color on any turn. It is illegal to shade a rectangle with a color that has already been given to a neighboring rectangle. If you don't have a legal move at your turn, you lose the game.

Prove that for each opening move by the first player, the second player can always win.

Since 6 borders all the others, you don't want to go there first and once you've played anywhere else you can't choose 6.

By symmetry there are only 3 opening moves: 1, 2 or 3.

If player A chooses 1, Player B can choose 5, then A may choose 3 or 4, then B chooses 2 and there is no option for A.

If player A chooses 2, Player B can choose 4, then A must choose 5, then B must choose 1 and there is no option for A.

If player A chooses 3, Player B may choose 1 or 5, then A must choose the other, then either 4 or 2 is available to B and there is no option for A.

I could imagine larger boards for this game.   In general it would appear the game favors player B.  I wonder if there is a small board where the first player would win.


Edit:  The above is incorrect.  I didn't read the rules carefully and assumed each player was bound to a single color.

Edited on May 3, 2013, 11:08 am

Posted by Jer on 2013-05-02 13:52:11