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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)

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Shaky ground? | Comment 1 of 11
I want each '1' to be surrounded by as many 'X' as possible, but to also have them shared with at least another '1' if possible.  The best sharing  of a particular cell is 4 but that is not optimum for this problem.

While my logic is slightly flawed, in each case below I am allowing extrapolation of one cell beyond the grid in recognition that the plane is infinite while I my thoughts are within finite limits.  

I am allowing myself a 5x5 grid in which I distribute the '1's.  I am also considering the 7x7 grid in which that resides.


a)               b)               c)    
   1 . 1 . 1      1 . 1 . 1       1 . . . 1
   . . . . .         . 1 . 1 .       . . . . .
   1 . 1 . 1      1 . 1 . 1       . 1 . . .
   . . . . .         . 1 . 1 .       . . . . .
   1 . 1 . 1      1 . 1 . 1       1 . . . 1
      X=8            X=4           X=8


d)               e)              f)
   1 . . . 1       1 . . . 1       1 . . 1
   . 1 . . .        . . 1 . .        . 1 . . 1
   . . 1 . .        1 . . . 1       . . 1 . .
   . . . 1 .        . . 1 . .        . . . 1 .
   1 . . . 1       1 . . . 1        . 1 . . 1
      X=6           X=8            X=6

g)              h)
   1 . . . 1       1 . . 1 .
    . . . . .        . . . . .
    . . . . .        . . . . .
   1 . . . 1        . 1 . . 1
    . . . . .        . . . . .
      x=8           x=8

Pattern No.1   X val    Grid Wt  Grid Wt/25  Grid Wt/49
  a)      9       8      9 +  9*8      81/25           81/49
  b)    13       4     13 +13*4      65/25           65/49
  c)      5       8      5 +  5*8      45/25           45/49
  d)      7       6      7 +  7*6      49/25           45/49
  e)      8       8      8 +  8*8      72/25           72/49*
  f)       9       6      9 +  9*6      63/25           63/49
  g)      4       8      4 +  4*8      36/25           36/49
  h)      4       8      4 +  4*8      36/25           36/49

Now, if my analysis of this situation is correct e) is the solution although 72/49 will not be the exact average.
  Posted by brianjn on 2012-02-08 05:51:45
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