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

Being that there are exactly C(9,2)=36 possible pairs and 12*3=36 handshakes that take place, the statement to be proved is equivalent to saying there must be at least one pair of trolls that shake hands more than once, thereby depriving some other pair of a handshake.

Posted by Charlie on 2015-07-23 08:27:41