Assume you have a checker board with 7 rows and infinite columns. You can place checkers on only the first 2 rows initially (number these -1 and 0). Then you may jump other checkers up, down, right, and left but not diagonally. The goal is to get as high a row as possible. For example you can get to the second level with four checkers like this:
Level Setup Turn 1 Turn 2 Turn 3
------ ------- ------- ------- -------
2 ····· ····· ····· ···a·
1 ····· ···d· ···d· ·····
0 ·abc· ·ab·· ···a· ·····
-1 ···d· ····· ····· ·····
It turns out you need at least 2 checkers to get to level 1, 4 to get to level 2, 8 to get to level 3, and 20 to get to level 4.
Prove the least number of jumps it would take to get to level 5, and how you would do it.
Note: You may place the initial checkers anywhere you wish, as necessary.
(In reply to
Level 5 Solution by Brian Smith)
A solid approach, weighting each square to "conserve energy" on a jump. When t is the golden ratio, you have the nice fact that
t^(-n) +t^(-n-1)=t^(-n-2)
And I totally agree that results on a half plane of checkers trumps values gained from seven columns.
However, I do have a question; how do we know that nothing good can come from jumping down or away from the "center"? We can jump down if we open up a lower row early, and we might have a strategy of always moving left in the infinite plane case, always gaining fresh checkers. Make sense?
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Posted by owl
on 2004-10-24 14:28:47 |
question to Level 5 Solution
