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Nine trolls (Posted on 2015-07-23) Difficulty: 2 of 5
Nine trolls are placed in the cells of a three-by-three square.
The trolls in neighboring cells shake hands with each other.
Later they re-arrange themselves in the square and the neighbors greet each other once more.
Then they repeat it again for the 3rd time.

Prove (or provide a counterexample) that there is at least one pair of trolls who didn’t greet each other.

Based on a problem in Russian "Kvantik",2012

No Solution Yet Submitted by Ady TZIDON    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 3 of 9 |
As was pointed out there the three rounds give the opportunity for 36 greetings and there must be 36 unique pairwise greetings.

Each troll must greet 8 others total.

Each round consists of:
1 center troll who greets 4 others,
4 edge trolls who greet 3 others, and
4 corner trolls who greet 2 others.

Over the course of 3 rounds, there are 3 centers, 12 edges, and 12 corners available.

Each troll's greetings after 3 rounds must sum to 8.  The only possibilities are
4+2+2
3+3+2

Three different trolls must use a center on one round and corners on the other two rounds.

The problem is these three trolls can never meet.  Every pairwise meeting involves one edge troll.



Edited on July 23, 2015, 8:45 pm
  Posted by Jer on 2015-07-23 10:25:35

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