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Author: LordFoolman Big gold star, 5000 posts Old School Fool Add to my Favorite Fools Ignore this person (you won't see their posts anymore) Number: of 5858  
Subject: Re: Jealous Neighbors Date: 8/14/2014 4:35 PM
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I don't have the official answer to the problem. I should have it by Sunday when it's revealed.

But like both of you, I tried to solve it and couldn't find an answer assuming single occupancy for each spot. I would assume that's what the author meant. What makes it impossible, the way I looked at it, is that there is an odd number of neighbors. You can think of the moves as "pairings". For each neighbor that moves, there is necessary one that has to leave for another spot. That's a pair. This makes it necessary that the total number of neighbors must be even. Odd x even = even. Even x even = even. Given that there is 25, this isn't possible. Maybe this is the correct answer to the puzzle.

The only other possibility is that the author of the puzzle is allowing the neighbors to move more than once, and if allowing for that, what is the minimum number of moves? Given this possibility, I would say rotate the outer ring of the array clock or counter clockwise. Same with inner ring that surrounds the center neighbor. That produces 24 moves. Now swap spots between center and one of its neighbors top/bottom etc. that produces 26 moves. One neighbor moves twice, but all others once.
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