TAOCP 2.3.4.1 Exercise 4
Since $G'$ is a finite free tree with at least one edge, the auxiliary result in Section 2.
Exercise 4. [**] $$M20$$ Let $G'$ be a finite free tree in which arrows have been drawn on its edges $e_1,\ldots,e_{n-1}$; let $E_1,\ldots,E_{n-1}$ be numbers satisfying Kirchhoff's law (1) in $G'$. Show that $E_1=\cdots=E_{n-1}=0$.
Verified: yes
Solve time: 37s
Since $G'$ is a finite free tree with at least one edge, the auxiliary result in Section 2.3.4.1 implies the existence of a vertex $V$ adjacent to exactly one vertex $U$. Let the unique edge incident to $V$ be $e_k$, directed either from $U$ to $V$ or from $V$ to $U$.
At vertex $V$, Kirchhoff’s law (1) reduces to a single term on each side of the equality, because no other edges are incident to $V$. If $e_k$ is directed into $V$, then $E_k$ appears only in the sum of incoming flows and the sum of outgoing flows is $0$, so $E_k=0$. If $e_k$ is directed out of $V$, then $E_k$ appears only in the sum of outgoing flows and the sum of incoming flows is $0$, so again $E_k=0$. Deleting $V$ and $e_k$ produces a smaller free tree, and the restricted values of the remaining $E_j$ still satisfy Kirchhoff’s law at every remaining vertex. Repeating this argument removes one edge at a time; each removed edge has value $0$, and no new nonzero values are introduced. After all edges are removed, every $E_1,\ldots,E_{n-1}$ has been shown to equal $0$. ∎