Answer
Showing that a subgraph T = (V , F) of the graph G = (V ,E) is an arborescence of G rooted at r if and only
if T contains r, T has no simple circuits, and for every
vertex v ∈ V other than r, deg−
(v) = 1 in T .
Work Step by Step
Because paths in trees are unique, an arborescence T of a directed graph G is just a subgraph of G that is a tree rooted at r,
--containing all the vertices ofG, with all the edges directed away from the root.
- Thus the in-degree of each vertex other than r is 1. For
the converse,
- it is enough to showthat for each v ∈ V there is a unique directed path from r to v.
- Because the in-degree of each vertex other than r is 1, we can follow the edges of T backwards from v.
This path can never return to a previously visited vertex,
because that would create a simple circuit.
- Therefore the path must eventually stop, and it can stop only at r, whose
in-degree is not necessarily 1.
-Following this path forward gives the path from r to v required by the definition of arborescence.
