Let there be a $12×12$ table of white squares. We draw $10$ squares in black. If a white square has $2$ black neighbours, then we draw it in black. We say that $2$ squares are neighbours if they have a common edge.
Does there exist an arrangement of the $10$ black squares such that all the squares will ultimately be black?
My intuition says that it's impossible. I divided the table into 9 $4×4$ squares and tried to use the Pigeonhole Principle, but I am stuck.
Let $P$ be a perimeter of area colored black.
We see that $P\leq 40$ at the begining and that at each step it does not increase*.
So since the whole table has perimeter $48$ it is impossible.
*Since each square has $4$ sides and at least $2$ sides are also sides of two neighbour black cells we must delete these two but we can add at most $2$ others.