Given a square of NxN fields, some of the fields are initially painted to black while the other fields are left empty. Then the table begins its life: every empty field that has at least two black neighbours (by the four edges) is painted to black automatically. Obviously, the full table can be auto-painted if it is allowed to use enough initial fields. It is just one thought further to see that N initial fields are enough for this (e.g. a diagonal of the square is painted initially).
Prove that less than N initial fields are not enough to auto-paint the full table. Try to find an elegant (i.e. short) solution.