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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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 re: Shaky ground? | Comment 5 of 11 |
(In reply to Shaky ground? by brianjn)

By the way, brianjn, I don't think that your calculation of average grid weight is quite right, perhaps because your boards are finite.

For instance, if X = 1, then grid pattern (f) is better than grid pattern (e).

In grid pattern (e), with infinite tiling, one out of every 4 squares has a weight of 9.  Average weight = 2.25.

In grid pattern (f), with infinite tiling, one out of every 3 squares has a weight of 7.  Average weight = 2.333, so (f) is better.

Neither, of course, is as good as my tiling, where one out of every 2 squares has a weight of 7, for an average weight of 3.5.

 Posted by Steve Herman on 2012-02-08 20:25:54

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