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 Boosted board (Posted on 2012-02-07)
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)

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 X = 1 (spoiler) | Comment 3 of 11 |
For X = 1, the best I have been able to come up with is solid rows of X's alternating with solid rows of 1's.

i.e.,

..1 1 1 1 ..
..X X X X..
..1 1 1 1 ..
..X X X X..   etc.

In an infinite board,because each 1 is surrounded by 6 X's this gives weights of
..7 7 7 7..
..0 0 0 0..
..7 7 7 7..
..0 0 0 0..
for an average weight per square of 3.5  (7/2).

I imagine that this arrangement will be the best for all values of X greater than 1, leading to an average weight per square of (6X + 1)/2.

If X <= 1/8, then the best arrangement is clearly all 1's, leading to an average grid weight of 1

I have not explored values of X between 1/8 and 1.

 Posted by Steve Herman on 2012-02-08 09:23:44

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