# Equivalence of Definitions of Arborescence

## Contents

## Theorem

Let $G = \struct {V, A}$ be a directed graph.

Let $r \in V$.

The following definitions of the concept of **Arborescence** are equivalent:

### Definition 1

$G$ is an **arborescence of root $r$** if and only if:

- For each $v \in V$ there is exactly one directed walk from $r$ to $v$.

### Definition 2

$G$ is an **arborescence of root $r$** if and only if:

- $(1): \quad$ $G$ is an orientation of a tree
- $(2): \quad$ For each $v \in V$, $v$ is reachable from $r$.

### Definition 3

$G$ is an **arborescence of root $r$** if and only if:

- $(1): \quad$ Each vertex $v \ne r$ is the final vertex of exactly one arc.
- $(2): \quad$ $r$ is not the final vertex of any arc.
- $(3): \quad$ For each $v \in V$ such that $v \ne r$ there is a directed walk from $r$ to $v$.

## Proof

### Definition 1 implies Definition 3

Let $G$ be an $r$-arborescence by definition 1.

Let $v \in V$ such that $v \ne r$.

Then there is exactly one directed walk $w$ from $r$ to $v$.

Since $v \ne r$, either:

- $w = \tuple {r, v}$

or:

- $\exists m \in V: w = \tuple {r, \ldots, m, v}$

Thus $v$ is the final vertex of the arc $r v$ or the arc $m v$.

Aiming for a contradiction, suppose that $v$ is the final vertex of distinct arc $x v$ and $y v$.

Then there exist directed walks $w_1$ and $w_2$ from $r$ to $x$ and $r$ to $y$ respectively.

But appending $v$ to $w_1$ and to $w_2$ yields distinct directed walks from $r$ to $v$.

This contradicts the fact that there is exactly one such directed walk.

Thus $v$ is the final vertex of exactly one arc.

$\Box$

Aiming for a contradiction, suppose that $r$ is the final vertex of an arc $x r$.

By Definition 1, there is a directed walk $w$ from $r$ to $x$.

But then $w$ appended to $w$ is a directed walk from $r$ to $x$ which is not equal to $w$.

This contradicts definition 1.

Thus we conclude that $r$ is not the final vertex of any arc.

It follows immediately from definition 1 that there is a directed walk from $r$ to each vertex $v \ne r$.

Thus $G$ is an $r$-arborescence by Definition 3.

$\Box$

### Definition 3 implies Definition 1

Suppose that $G$ is an $r$-arborescence by Definition 3.

Let $v \in V$.

We must show that there is a unique directed walk from $r$ to $v$.

If $v = r$, then $\tuple r$ is a directed walk from $r$ to $v$.

Since $r$ is not the final vertex of any arc, $\tuple r$ is the only such directed walk.

If $v \ne r$, then there exists some directed walk $w$ from $r$ to $v$.

Suppose that $z$ is a directed walk from $r$ to $v$.

Since $v \ne r$, $v$ is the final vertex of exactly one arc $x v$.

If $x = r$, then $z$ must end with $\tuple {r, v}$.

But $r$ is not the final vertex of any arc.

So in fact:

- $z = \tuple {r, v}$

If $x \ne r$, then:

- $z = \tuple {r, \ldots, x, v}$

Continuing inductively from $x$ proves that $z = w$.

Thus for each $v \in V$ there is exactly one directed walk from $r$ to $v$.

So $G$ is an $r$-arborescence by Definition 1.

$\Box$

### Definitions 1 and 3 imply Definition 2

Let $G$ be an $r$-arborescence by Definition 1.

From the above, $G$ is then also an $r$-arborescence by Definition 3.

Let $T = \struct {V, E}$ be the simple graph corresponding to $G$.

That is, for $x, y \in V$, let $\set {x, y} \in E$ if and only if $\tuple {x, y} \in A$ or $\tuple {y, x} \in A$.

We will show that $G$ is an orientation of $T$ and that $T$ is a tree.

By Definition 1, there is exactly one directed walk from $r$ to each vertex $v$.

Let $x, y \in V$ and suppose for the sake of contradiction that $\tuple {x, y} \in A$ and $\tuple {y, x} \in A$.

Let $w_x$ be the directed walk from $r$ to $x$.

Then appending $y$ to $w_x$ yields a directed walk $w_x + y$ from $r$ to $y$.

But then appending $x$ to $w_x + y$ yields a directed walk $w_x + y + x$ from $r$ to $x$, contradicting the fact that $w_x$ is unique.

Thus $A$ is asymmetric.

So $G$ is an orientation of $T$.

We must now show that $T$ is a tree.

By Equivalence of Definitions of Reachable, each vertex $v$ of $G$ is reachable from $r$.

Let $x, y \in V$.

By Definition 1, there is exactly one directed walk from $r$ to $x$ and exactly one directed walk from $r$ to $y$.

Thus there is a walk $w_x$ in $T$ from $r$ to $x$ and a walk $w_y$ from $r$ to $y$.

Reversing $w_x$ and then appending $w_y$ to it (eliding the duplicate $r$) yields a walk from $x$ to $y$.

Thus $T$ is connected.

Next we show that $G$ has no directed cycle.

Aiming for a contradiction, suppose that $x_0, x_1, \dots, x_n$ is a directed walk with $n \ge 2$ and $x_0 = x_n$.

By Definition 1, there is a unique directed walk $w$ from $r$ to $x_0$.

But then $w + \tuple {x_1, \dots, x_n}$ is another directed walk from $r$ to $x_0$, contradicting uniqueness.

Thus $G$ has no directed cycles.

Aiming for a contradiction, suppose that $T$ has a cycle $\tuple {x_0, x_1, \dots, x_n}$, where $x_0 = x_n$ and $n \ge 2$.

Since $G$ has no directed cycles, the arcs corresponding to the edges in this cycle cannot all go in the same direction around the cycle.

But this implies that some vertex of $x_0, x_1, \dots, x_n$ is the final vertex of two different arcs, contradicting Definition 3.

Thus $T$ has no cycles.

As $T$ is connected, it is a tree.

Thus we have shown that $G$ is an orientation of $T$ and that $T$ is a tree.

$\Box$

### Definition 2 implies Definition 1

Let $G$ be an $r$-arborescence by Definition 2.

That is, let $G$ be an orientation of a tree $T$ and that every vertex of $G$ is reachable from $r$.

Let $v \in V$.

By the definition of reachable, there exists a directed walk $w$ from $r$ to $v$.

We must show that $w$ is the only directed walk from $r$ to $v$.

Aiming for a contradiction, suppose that $z$ is a directed walk from $r$ to $v$ and $z \ne v$.

Then either $z$ extends $w$, $w$ extends $z$, or there is some $k$ such that $w_k \ne z_k$.

First suppose that $z$ extends $w$.

So $w = \tuple {x_0, \ldots, x_n}$ and $z = \tuple {x_0, \ldots, x_n, \ldots, x_m}$.

Then $p = \left({x_n, \dots, x_m}\right)$ is a directed walk from $v$ to $v$ with more than one vertex.

By Directed Circuit in Simple Digraph forms Circuit, the vertices of $p$ form a circuit, contradicting the fact that $T$ is a tree.

Thus there exists $k$ such that $w_k \ne z_k$.

It follows from the Well-Ordering Principle that there must be some smallest $k$ such that $w_k \ne z_k$.

Since $w$ and $z$ both end at $v$, it follows from the Well-Ordering Principle that:

- there must be some smallest $n > k$ such that there exists $m > k$ such that $w_n = z_m$
- and that there exists a smallest such $m$.

Then $\tuple {w_{k - 1}, w_k, \ldots, w_n, z_{m - 1}, \dots, z_k, z_{k - 1} }$ forms a cycle, contradicting the fact that $T$ is a tree.

$\blacksquare$