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 3x3 Nim (Posted on 2009-06-25)
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)

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 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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