Balancing the weights of the vertices of a graph by averaging along the edges. Suppose that you have a graph, and someone assigned real numbers to every vertex. You can modify these numbers by replacing the numbers on two adjacent vertices by their average. Your goal is to reach (in finitely many steps) a configuration where the same number is assigned to every vertex. 
My question is, whether you can achieve your goal for every initial assignment of reals, where the graph is a triangle plus an edge? (4 vertices)
I do not know the answer to the problem, the question is a known and not easy problem when the graph is the 8-cycle. I do not know whether anyone knows the answer to this problem. 
 A: Solution follows, stop reading if you  want to figure it out yourself.
Order nodes so that the leaf comes first and its neighbour comes next. 
Claim.  $1, 1, -1, -1$  cannot be balanced.
Proof. Suppose that in step $0$ we have the above and in step $n$, where $n$ is minimal, we have  $0, 0, 0, 0$.
In step $n-2$ we must have $a, -a, a, -a$ (or the last two reversed) for some $a\ne 0$.
Call an edge balanced if its two endnodes have equal weight. For $0<i<n$, there is no disjoint pair of balanced edges in step $i$ because $n$ is minimal. However,  there is a balanced edge in each step. Thus, by reverse induction on $i$, in steps $i = 0, 1,2,\dots, n-2$, each weight is an odd multiple of $a$. This is a contradiction for $i=1$ because the second weight is $0$. QED
Remark (generalizing one of Dániel Soltész's remarks): If you can solve graphs $G_1$, $\dots$, $G_{2^k}$, each with $2^l$ nodes, and also $H_1$, $\dots$, $H_{2^l}$, each with $2^k$ nodes, then you can glue these (in many ways) to get a graph on $2^{k+l}$ nodes which you can solve.
