A Transformation of a cross-shaped grid filled with 1s (Proof of impossibility?) Consider a cross-shaped grid of size 7 as it shows on the figure (compared to one of size 3). Each cell contains a 1. Le'ts define a transformation $\pi$ of the grid as follows: take any 3 sized sub-cross of the grid and multiply all the cells inside by $-1$.

How many $\pi$ transformations are required to transform a cross-shaped grid of size 2017 that contains a 1 in each cell into a grid that contains $-1$ in every cell?
Any ideas on how to proceed? I was trying to solve the particular case for 7 but even for that I found it quite hard.

For each tile that is not at the edge we can apply the transformation $\pi$ centered at that tile; I'll call such a tile 'transformed'. Because transforming a tile twice is the same as not transforming it at all, it suffices to consider which tiles to transform once, and which not to transform.
Consider one edge of the diamond, say the top left edge. It contains $1009$ tiles, and the next 'row' of adjacent tiles contains $1008$ tiles. For the two tiles at the ends of the edge (i.e. at the corners) to be flipped, we must transform the unique adjacent tile to each corner. These are the first and last tiles of the row of $1008$ adjacent tiles. Every other tile on the edge has precisely two tiles adjacent to it; to flip such a tile we must transform precisely one of these two adjacent tiles. This holds for every tile on the edge, and so the tiles of the adjacent row alternate between having to be transformed and not transformed. But the first and last tile, i.e. the $1$st and $1008$th tile, are both transformed, a contradiction.
This shows that it is impossible to flip all tiles by means of these transformations.
