I encountered this problem earlier today.
Suppose we have an 8x8 chessboard, and 10 rooks. We fill in some of the squares on the board such that it's impossible to place the 10 rooks on empty squares so that every empty square on the board either contains a rook or is being attacked by a rook (Rooks cannot attack through a filled in square). Find the minimum number of squares we must fill in.
It seems like the answer is 5 (as stated below in the answers) Is there a nice combinatorial proof to this result? Can we generalize this to other numbers of rooks (e.g. 16)?