There are 10 coins laid out on a table in a straight line

* * * * * * * * * *
1 2 3 4 5 6 7 8 9 10

The goal is to get 5 stacks of 2 coins each in only 5 jumps.
A coin must jump over exactly 2 other coins and land on a third.
Coins may jump in either direction.

Example:
Jump 1: Jump coin 10 on top of coin 7.
Jump 2: Jump coin 8 on top of coin 6. This is possible since coin 7 and 10 now form a stack of two coins.

Here is the answer (of course, there are other symmetric solutions), followed by how I found it:

Step 1:
Jump 5 onto 2
Jump 7 onto 10
(since these jumps don’t cross each other they can be done in either order)

Step 2: Jump 3 onto 8

Step 3:
Jump 1 onto 4, or Jump 4 onto 1 (doesn’t matter)
Jump 6 onto 9, or Jump 9 onto 6 (doesn’t matter)
(since these jumps don’t cross each other they can be done in either order)

Here’s how I got there:

First, as I was playing around, I saw that if I got two adjacent stacks of 2 coins early on, this divided the jumping into two regions. The coins on the left were now blocked from jumping on the coins on the right because they’d have to jump over 4 coins which is not allowed. I quickly saw that if I divided the coins into two jumping regions this way that I was in trouble and got stuck in traps.

I also found that if I had exactly 6 coins in a row (so I had two stacks of 2 coins either on each end or both on the same end) I ran into trouble again. There were several choices for jumps, but I would always get stuck.

So then I decided to focus on how I wanted things to look near the end. Well, if I had an adjacent group of coins whose heights were 1-2-1, that would be good. I could then jump one of the singles over the double onto the other single. And if I had two groups like this 1-2-1 arrangement, that would be even better. But that only accounts for 8 coins.

Ok, so somehow I want to get something like the following arrangements of stack heights:
1-2-1 – 1-2-1 – 2
1-2-1 – 2 – 1-2-1
2 – 1-2-1 – 1-2-1

Of course, the first and third are symmetric, but anyways…I couldn’t find a good way to get the second arrangement, so I tried for the other kind. And that’s how I approached my solution.

Posted by nikki on 2004-10-18 12:36:53