Find root of tree in Erlang / Elixir digraph

Question

I have the following super simple digraph tree (code is Elixir):

digraph = :digraph.new()
coords = [{0.0, 0.0}, {1.0, 0.0}, {1.0, 1.0}]
[v0, v1, v2] = (for c <- coords, do:   :digraph.add_vertex(digraph, c))
:digraph.add_edge(digraph, v0, v1)
:digraph.add_edge(digraph, v1, v2)

We can see it is a tree:

iex(82)> :digraph_utils.is_tree(digraph)
true

But how do I find the root of vertex of the tree in an efficient way? Currently I'm doing this:

iex(83)> Enum.filter(:digraph.vertices(digraph), fn x -> :digraph.in_degree(digraph, x) == 0 end)
[{0.0, 0.0}]

Is this the most efficient way of finding the root vertex, that is, by finding the vertex with no in-edges?

Answer 1

There's a function called :digraph_utils.arborescence_root/1 that might do the trick.

For your example, it would return the following:

iex(7)> :digraph_utils.arborescence_root(digraph)
{:yes, {0.0, 0.0}}

Notice if we add another edge making a cycle from first to last vertex, then it will loose its root:

iex(8)> :digraph.add_edge(digraph, v2, v0)
iex(9)> :digraph_utils.arborescence_root(digraph)
:no

In terms of efficiency, this built-in function does the same as your code, so I guess the answer to your question is no, there is not a more efficient way of doing it.

I hope this helps :)