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3x3 Nim (Posted on 2009-06-25) Difficulty: 3 of 5
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?

See The Solution Submitted by Jer    
Rating: 4.0000 (7 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution simplification | Comment 22 of 26 |

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
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