*Every state A has n incoming edges (self edges included). (3)*
Here's an example: Let A be 0 1 2 1 0 0.

We can reach this state from 3 0 1 0 0 0 applying a move on cup 1, 2 0 1 1 0 0 applying a move on cup 1, 1 0 2 1 0 0 applying a move on cup 1, 0 1 2 1 0 0 applying a move on cup 5 (self edge), 0 1 2 1 0 0 applying a move on cup 6 (self edge)

We can reach this state from 0 1 2 1 0 0 applying a move on cup 1 (self edge), 3 0 1 0 0 0 applying a move on cup 1, 2 0 1 1 0 0 applying a move on cup 1, 1 0 2 1 0 0 applying a move on cup 1, 0 1 2 1 0 0 applying a move on cup 5 (self edge), 0 1 2 1 0 0 applying a move on cup 6 (self edge)

From (2), (3) the graph is eulerian, from (1) the graph is also connected. It follows that we have a cycle that goes through all edges and nodes. Which means we can reach any state from any other state.

From (2), (3) the graph is Eulerian, from (1) the graph is also connected. It follows that we have an *Eulerian cycle* that goes through all edges and nodes. Which means we can reach any state from any other state.

Thanks *Andrei Dragus* for the neat problem.