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Boosted board (Posted on 2012-02-07) Difficulty: 3 of 5
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.

See The Solution Submitted by Jer    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Final Answer ?? (spoiler) | Comment 9 of 11 |
(In reply to Final Answer ?? (spoiler) by Steve Herman)

You found all that I have except for the one case that doesn't really matter (it is tied for the best tiling at only 1 point).  As you point out, we may have missed one, but I don't know how to exhaustively try every possibility.

You do have an algebra error:
(8x+3)/4 > (6x+1)/2
x < 1/4 (you have 1/2)

  Posted by Jer on 2012-02-09 21:27:41

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