# Genies' chess on a 10×10 board

The work of Hearth Taxel revealed some other results related to genies' chess. For example, there is an arrangement $$A$$ of pawns on a 10×10 board such that no 3×3 submatrix is empty and

• $$A$$ is symmetric about the main diagonal, which is all pawns
• Removing any pawn from $$A$$ leaves at least one empty submatrix, i.e. it is minimal
• Without the diagonal pawns, $$A$$ is the adjacency matrix of a symmetric graph, where there are automorphisms taking any vertex/edge to any other vertex/edge. In particular this implies that the row and column sums of $$A$$ are all equal

Can you find a solution?

• Does a submatrix have to be a connected subgrid? I mean, does it have to be the intersection of three consecutive rows with three consecutive columns, or can it be the intersection of any three rows and any three columns? Commented Dec 20, 2021 at 12:13
• @Ankoganit See the first problem for what submatrix means: ranks/files don't have to be consecutive. Commented Dec 20, 2021 at 12:22

If the said pawn is on the diagonal, corresponding a to a vertex $$v$$ on the "outer pentagon", take the submatrix corresponding to $$\{v, p, q \}$$ and $$\{ v, p', q' \}$$, where $$p,q$$ are the vertices "opposite" to $$v$$ in the outer pentagon, $$v'$$ is the vertex on the "inner pentagon" connected to $$v$$, and $$\{ p', q' \}$$ are the vertex "opposite" to $$v'$$ in the "inner pentagon". If it's not on the diagonal, since the Petersen graph is 3-arc transitive (in particular: it acts transitively on the edges, so let that said pawn correspond to $$\overline{vv'}$$. Take $$\{v, p, q \}$$ and $$\{v', p', q' \}$$ to be the rows and columns of the submatrix.