How many ways are there to fill a 3 × 3 grid with 0s and 1s?

Question

Extra conditions that put a formal solution out of my reach: the centre cell must contain a $0$, and two grids are equal if they have a symmetry, e.g. $$\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 1\end{array} \right) = \left( \begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right) = \left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 0\end{array} \right) $$ For context, this question is part of an investigation into the number of possible checkmate patterns in chess.

Answer 1

Use Burnside's lemma. The number of symmetries of the matrix is eight:

• the identity, leaving $2^8$ admissible matrices unchanged (the centre cell being fixed)
• two 90° rotations leaving $2^2$ matrices unchanged each
• a 180° rotation leaving $2^4$ matrices unchanged
• four reflections leaving $2^5$ matrices unchanged each

So the number of possible matrices up to symmetry is $$\frac{2^8+2\cdot2^2+2^4+4\cdot2^5}8=32+1+2+16=51$$