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.
So to get to the "next" level (call it "N") (after 1) you need to have an arrangement similar to the one shown in "Turn 1":
Level N:.. ......
Level N-1: ...a..
Level N-2: .bc...
B would then jump over C then A to get to level N.
(B cannot start out sitting below A, since A had to jump over something to have gotten to level N-1, and that checker is now gone)
So it seems that the lowest theoretical "cost" of moving a checker to N is equal to the cost of N-1 plus twice the cost of N-2.
From the problem though, it seems there is some additional expense after level 3 (otherwise the cost for N=4 would be 16, not 20.)
|
|
Posted by levik
on 2003-06-24 11:43:38 |
Thoughts...
