Consider an infinite chessboard. Each square contains either a 1 or an X in some pattern. (X can be any real number but for a given board, all the X's are the same.)
Each square with an X on it has weight equal to zero.
Each square with a 1 on it has a weight of 1 + N*X where N is the total number of X's on the 8 surrounding squares.
For a given value of X, find a way of tiling the board with the highest average weight per square.
Inspired by various Tower Defense games.
I have been playing with this, and for tilings that involve X's, I haven't been able to improve on alternating stripes (or columns) of X's and 1's. The tiling below gives an average weight of (6X+1)/2 per square, and all other tilings involving X's seem to average out to less.
..1 1 1 1 ..
..X X X X..
..1 1 1 1 ..
..X X X X.. etc.
But this is not always the best. There is one tiling that does not involve X's, and that is
..1 1 1 1..
..1 1 1 1..
..1 1 1 1..
..1 1 1 1..
That gives a weight of 1 per square
And 1 is qreater than (6X+1)/2 when X < 1/6.
So, until proven otherwise, I believe the answer is alternating vertical or horizontal strips if X >= 1/6, and all 1's otherwise.
Final Answer? (spoiler)
