A related post:

• Detecting patterns of black and white stones on a 2D board

I'll try to apply the same method here.

Your data

I'll start by assigning your data to m:

m = {{-1, -1, -1, -1, -1, -1, -1, -1, -1, -1}, {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1}, {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1}, {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1}, {-1, 0, 0, 0, 0, 0, 0, 0, 0, -2}, {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1}, {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1}, {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1}, {-1, 0, 0, 0, 0, 0, 0, 0, 0, -1}, {-1, -1, -1, -1, -1, -1, -1, -1, -1, -1}}; 

Two-dimensional replacement rules

Using CellularAutomaton:

p = Partition[#, 3] & /@ Permutations[Table[1Permutations[Table[_, {8}]~Append~-2]; p[[All, 2, 2]] = 0; rules = Append[Thread[(Delete[p, 5] /. {1 -> _}) -> 1], ArrayPad[{{x_}}, 1, _] :> x]; CellularAutomaton[rules, m, 1][[2]] // MatrixForm 

$\left( \begin{array}{cccccccccc} -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -2 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ \end{array} \right)$