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 Colorful Win (Posted on 2013-05-02)
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.

 No Solution Yet Submitted by K Sengupta No Rating

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 Another try Comment 4 of 4 |
Call the players A and B for convenience.
By color symmetry, although A can use either color, I will call the first color used Red.
By board symmetry, a first move of 1 is the same as a first move of 5 and 2 is the same as 4.

So we need to consider four of the 12 possible opening moves:
A plays red on 1,2,3or6

Case A plays red 1:
B plays 5 blue.  6 becomes unavailable.
If A plays 2 blue or 4 red, B will be left with the final move of playing the opposite color in either of the remaining spaces.
If A plays 3 in blue or red, B will be left with the final move of playing the opposite color in either of the remaining spaces.

Case A plays 2 red.
B plays 4 blue.  6 becomes unavailable.
The are only two possible moves left: 1 blue or 5 red
when A chooses one B gets the other.

Case A plays 3 red.
B plays 6 blue.
There are only two possible moves left: 1 red or 5 red.
When A chooses one B gets the other.

Case A plays 6 red.
B plays 3 blue.  (This becomes the same as above, with opposite colors.)
There are only two possible moves left: 1 blue or 5 blue.
When A chooses one B gets the other.

 Posted by Jer on 2013-05-03 11:21:21

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