[erlang-questions] graphs and trees

French, Mike Mike.French@REDACTED
Thu Dec 20 10:34:22 CET 2007

A graph is a tree if it's connected and nEdges = nVertices - 1
For a simple test, but not necessarily most efficient, use
digraph:no_edges/1 and digraph:no_vertices/1 then see if 
digraph_utils:components/1 returns a single element list:

 ( digraph:no_edges(G) == digraph:no_vertices(G)-1 ) and 
 ( length(digraph_utils:components(G)) == 1 )

Of course this does not guarantee that the edges are 
directed from the root outward to the leaves of the tree.


-----Original Message-----
From: erlang-questions-bounces@REDACTED
[mailto:erlang-questions-bounces@REDACTED]On Behalf Of mats cronqvist
Sent: 20 December 2007 08:51
To: erlang-questions@REDACTED
Subject: [erlang-questions] graphs and trees

  so i've made a directad acyclic graph by calling various functions in
the digraph module. i theory, the resulting graph should be a tree. is
there some snazzy graph theory trick to show that the graph is indeed a
  disclaimer; i am by training a physicist, and as such uncomfortable
with all data structures more complicated than the fixed-size array. so
no big words please :>


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