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

Home > General
Jumping Coins (Posted on 2004-10-18) Difficulty: 2 of 5
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.

See The Solution Submitted by Brian Smith    
Rating: 3.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Working backwards | Comment 3 of 8 |

Starting with 5 piles of 2, I first tried 22222 -> 1212121, but realized that one side or the other there'd be a single coin left unable to jump.

But 22222 -> 1211212 (going in opposite directions from the second and third stack) and then to 11111212 and from there to 111111112 and then 1111111111.  Then place them in reverse order:

1111111111
111111 112
111 11 212
121 1 212
22 2 2 2

The last is two steps combined into one (1->3 and 9->5).  Also, instead of 9->5, we could have done 5->9.

Overall this is 7->10, 4->8, 6->2, 1->3 and (9->5 or 5->9), and the last two moves could be done in either order (that is, the 9->5 or 5->9 could come before the 1->3).

Edited on October 18, 2004, 3:54 pm
  Posted by Charlie on 2004-10-18 15:49:22

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