Number of spanning arborescences does not depend on $i$.

Question

Let $G$ be a directed graph with vertices $x_1,x_2,\dotsc, x_n$ for which a directed Eulerian circuit exists. A spanning arborescence with root $x_i$ is a spanning tree $T$ of $G$, with root $x_i$, such that for all $j\ne i$ there is a directed path from $x_j$ to $x_i$ in $T$. Show that the number of spanning arborescences of $G$ with root $x_i$ does not depend on $i$.

This question has already been posted here but no answer was received. It come from the book A Course in Combinatorics (Problem 2C), where the hint says the following "lemma":

Let $A_i$ be the number of spanning arborescences with $x_i$ as a root and denote by $r_i$ the number of edges out of $x_i$. Then the number of different Eulerian circuits in $G$ is $$A_i\prod_{j=1}^{n}(r_j-1)!.$$

The hint also sketched the proof: To do this, take $i=1$ and consider some spanning arborescence with $x_1$ as a root. Fix some edge out of $x_1$ as the first edge of an Eulerian circuit. Number the other edges out of $x_1$ arbitrarily. For $i=2,3,\ldots,n$, number the edges out of $x_i$ arbitrarily but such that the edge of the spanning arborescence has number $r_i$. This numbering leads in a natural way to the construction of an Eulerian circuit and this process can be reversed.

My attemp and questions:

1. In the step numbering the edges out of $x_i$, how can I be sure that there is precisely one edge in the spanning arborescence? In other words, why is the edge numbered $r_i$ unique?
2. After numbering, how does it lead to a construction? Here is what I tried: Arriving at a vertex $x_i$, we choose the edge out of $x_i$ with the least number associated to it to be the next edge in our construction. Since $\deg^+(x)=\deg^-(x)$ for all $x\in V(G)$, we must stop at $x_1$. Consider the unused edges (if they exist at all), we see that they must form simple closed paths (for the same reason, i.e. $\deg^+(x)=\deg^-(x)$). Combine all these paths and we get the Eulerian circuit. But how do I prove that the combination does not affect the numbering and the order in which the edges appear in the Eulerian circuit?
3. If the above has been proved, then the total number of numberings is $A_1 r_1!\prod_{j=2}^{n}(r_j-1)!$. But each Eulerian circuit corresponds to $r_1$ numberings, since given any Eulerian circuit, we can produce a numbering starting from $x_1$ and any of the $r_1$ edges out of $x_1$. But I don't see how I can construct a bijection between the set of numberings and the set of Eulerian circuits because we have to divide by $r_1$. And I don't know the point of using the spanning arborescence either.

My thoughts are messy, so ignore them if they don't make any sense. I hope to get a detailed answer (rather than another hint or something). Thanks in advance!

Answer 1

1. An arborescence is a tree with the extra condition that all edges are directed to the root. So from $x_i$ there is a unique path to the root, and the first edge of that path is directed out of $x_i$. Every other edge of the arborescence meeting $x_i$ is not part of a path from $x_i$ to the root (because the path is unique), so the path from this edge to the root goes through $x_i$, i.e. this edge must be directed towards $x_i$.

2. The point is that this doesn't produce several closed circuits that we have to combine, but a single closed circuit. The construction here is that you only choose the edge of the arborescence leaving $x_i$ if you've already visited $x_i$ the maximum number of times. If $x_j\to x_i$ is an edge of the arborescence, then to visit $x_i$ the maximum number of times you must use that edge, so you must first visit $x_j$ the maximum number of times. Thus you can prove by induction (on the length of the path $x_i\to x_1$ in the arborescence) that to cover all the edges leading to $x_1$ you must visit $x_i$ the maximum number of times for every $x_i$. Thus your trail doesn't close off prematurely and you have a single Euler circuit.

3. It depends what you mean by an Euler circuit. We need to regard two circuits as different if they visit edges in a different order, even if one can be obtained from another by starting somewhere in the middle and wrapping round to the start. (You might think of these as circuits which are equivalent but not the same, then each equivalence class corresponds to $r_1$ different Euler circuits starting at $x_1$, but that doesn't matter since we are counting circuits not equivalence classes, and there is a bijection between circuits and orderings.)