All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > General
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.)
Final Answer? (spoiler) | Comment 4 of 11 |

I have been playing with this, and for tilings that involve X's, I haven't been able to improve on alternating stripes (or columns) of X's and 1's.  The tiling below gives an average weight of (6X+1)/2 per square, and all other tilings involving X's seem to average out to less.

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

But this is not always the best.  There is one tiling that does not involve X's, and that is

..1 1 1 1.. 
..1 1 1 1..
..1 1 1 1.. 
..1 1 1 1.. 

That gives a weight of 1 per square

And 1 is qreater than (6X+1)/2 when X < 1/6.

So, until proven otherwise, I believe the answer is alternating vertical or horizontal strips if X >= 1/6, and all 1's otherwise.




  Posted by Steve Herman on 2012-02-08 19:36:15
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information