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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.

  Submitted by Jer    
Rating: 4.0000 (1 votes)
Solution: (Hide)
For X ≤ 1/8 all 1's is best. It always has an average weight of 1.

The following pattern has an average weight of (8X+8)/9 which is tied for the best only when X=1/8 weight of 1 (it is never the sole maximum)


From 1/8 to 1/4 the following pattern is the highest with a weight of (3+8X)/4

Above 1/4 this pattern of alternating strips of 1's and X's is the best with an average weight of (1+6X)/2:

At X=1/8 precisely the first 3 are tied, at X=1/4 precisely the last two are tied.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re(3): Final Answer ?? (spoiler)broll2012-02-10 09:16:59
re(2): Final Answer ?? (spoiler)Steve Herman2012-02-09 22:31:31
re: Final Answer ?? (spoiler)Jer2012-02-09 21:27:41
SolutionFinal Answer ?? (spoiler)Steve Herman2012-02-09 16:50:03
re: Final Answer? (spoiler)Jer2012-02-09 10:47:34
re(2): Shaky ground? You misunderstand.brianjn2012-02-08 21:01:28
re: Shaky ground?Steve Herman2012-02-08 20:25:54
Final Answer? (spoiler)Steve Herman2012-02-08 19:36:15
Some ThoughtsX = 1 (spoiler)Steve Herman2012-02-08 09:23:44
re: Shaky ground? You misunderstand.Jer2012-02-08 09:02:35
Shaky ground?brianjn2012-02-08 05:51:45
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