
Boost Users : 
Subject: Re: [Boostusers] [BGL] question about connected_components()
From: Jeremiah Willcock (jewillco_at_[hidden])
Date: 20120125 16:27:45
On Wed, 25 Jan 2012, Andriy Gapon wrote:
> on 24/01/2012 23:03 Jeremiah Willcock said the following:
>> In order to do that, the easiest way is to create a new undirected graph from
>> your original graph, adding each directed edge as an undirected edge in the new
>> graph. If you have very large graphs, there are (more complicated) ways to do
>> that without extra memory utilization, but this approach involves the least
>> effort and works unless your graphs are very large or you are doing this
>> operation many times.
>
> My graph is actually a quite large (directed) adjacency_matrix.
> Actually, I need to run connected_components() on a filtered_graph of the said
> graph (based on edge weight). So I think that I found a way to achieve what I
> wanted, but I would like to doublecheck my approach with you.
>
> I arranged my filter to be such that each edge of the filtered graph has an
> opposite edge  for each (u, v) there is a (v, u). And then I use exactly the
> same approach as boost::connected_components(), but without any limitations on a
> directionality of a graph.
If you have both directions of each edge, the normal connected_components
code (possibly with the directionality check commented out) should work
just fine.
> That is, I also use depth_first_search().
> I think that since depth_first_search() colors only the vertices, then it should
> work on my graph in the same way as it would work on an undirected graph.
Yes, I think so.
> And thus it should find the weekly connected components that I want.
> What do you think?
That's the way I would recommend doing it.
> P.S.
> It seems that the connected_components documentation mentions only the first
> version of the connected_components function.
Having two versions is an implementation detail; the second one implements
the "= all defaults" part of the signature given in the documentation.
 Jeremiah Willcock
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