Postby **ConMan** » Thu Dec 18, 2008 6:06 am UTC

Ok, hopefully the necro justifies the double post, because I'm about to post Barbara's winning strategy for the 3x3 case.

First, let's simplify things by naming the matrix and its entries:

[math]M=\begin{bmatrix}

m_{11} & m_{12} & m_{13} \\

m_{21} & m_{22} & m_{23} \\

m_{31} & m_{32} & m_{33}

\end{bmatrix}[/math]

WLOG, Alan's first move will be either m_{11} = 0 or m_{11} = 1.

Case 1: m_{11} = 0

Barbara plays m_{12} = 0. If Alan plays any move other than placing a non-zero in m_{13}, Barbara plays m_{13} = 0 and wins. So, let Alan play WLOG m_{13} = 1. Notice that now [imath]\det M = m_{21}m_{32} - m_{22}m_{31}[/imath], so Barbara's next two moves, no matter what Alan plays, will be m_{21} = 0 followed by either m_{22} = 0 or m_{31} = 0.

Case 2: m_{11} = 1

In this case, Barbara plays m_{22} = 0. Let's now break Alan's second move down to a few sub-cases written in their most general terms (I put Alan's non-zero moves as 1, but the strategy holds no matter what he chooses).

Case 2.1: m_{12} = 0, m_{21} = 0, m_{23} = 0, m_{32} = 0

Clearly, in all these cases, Alan moved has lined two zeroes up, so Barbara just has to put the third zero in the row or column.

Case 2.2: m_{13} = 0 (equivalent to m_{31} = 0)

Barbara plays m_{33} = 0, forcing Alan to play m_{23} = 1 to avoid an all-zero 3rd column. Barbara then plays m_{32} = 0, and Alan can't block both the 2nd column and the 3rd row from becoming all-zero at Barbara's next turn.

Case 2.3: m_{33} = 0

Barbara plays m_{13} = 0, and it continues as for Case 2.2.

Case 2.4: m_{12} = 1 (equivalent to m_{21} = 1)

Barbara plays m_{23} = 0. To keep the 2nd row from being all zeroes, Alan must play m_{21} = 1. Barbara responds with m_{33} = 0, forcing Alan to play m_{13} = 0. At this point, [imath]\det M = m_{32}[/imath], so Barbara just sets this to zero to win.

Case 2.5: m_{13} = 1 (equivalent to m_{31} = 1

Barbara plays m_{32} = 0, and Alan must avoid the zero 2nd column by playing m_{12} = 1. Now [imath]\det M = m_{23}m_{31} - m_{21}m_{33}[/imath], so like in Case 1 Barbara just zeros the two terms out one at a time.

Case 2.6: m_{23} = 1 (equivalent to m_{23} = 1)

Barbara plays m_{32} = 0, and like Case 2.5 Alan must play m_{12} = 1. This time, [imath]\det M = m_{31} - m_{21}m_{33}[/imath], so Barbara plays m_{31} = 0 and then one of m_{21} = 0 or m_{33} = 0.

Case 2.7: m_{33} = 1

Barbara plays m_{12} = 0 and Alan must respond with m_{32} = 1, and this turns into a rearrangement of the second half of Case 2.6.

And that's it. If I've missed a case, or given incorrect reasoning, feel free to correct me.

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