A 3x3 array of counters is laid out. Players take turns removing counters. The rule for removing counters is to pick a row or column and take any 1,2 or 3 from it. Whoever removes the last counter wins.

Does the first or second player have a winning strategy?
What is this strategy?

The patterns into which the hopeful winner should leave the board fall into 14 categories. The 78 such unplayable tableaux (excluding the full and empty boards) are rotations/reflections of these 14:

2-diagonal:
|   |
|  *|
| * |

4-rectangle:
|   |
|* *|
|* *|

perpendicular sticks:
|  *|
|  *|
|** |

right triangle:
|  *|
| **|
|***|

J-hook:
|  *|
|* *|
|***|

Y:
|* *|
| * |
| * |

baby carriage:
|  *|
|***|
| **|

h or 4:
|*  |
|***|
|* *|

diamond:
| * |
|* *|
| * |

fish:
| * |
|***|
| **|

1:
|** |
| * |
|***|

parentheses:
| **|
|* *|
|** |

Casseopia+:
| **|
|** |
|* *|

right angle with dot:
|* *|
|  *|
|***|

This last suggests a further simplification: the right angle of 5, supplemented by a single piece left in any of the four remaining spots is something one can try for to give the opponent an unplayable position, so that also encompasses the J-hook and the right triangle.

The diamond is two 2-diagonals, as is the perpendicular sticks.

The parentheses is three 2-diagonals.

Posted by Charlie on 2009-06-27 13:48:38