Re: [Boost-users] [BGL] Using copy_graph(g_in, g_out) with filtered_graph (and a special predicate) as input yields an unexpectedly empty graph

Subject: Re: [Boost-users] [BGL] Using copy_graph(g_in, g_out) with filtered_graph (and a special predicate) as input yields an unexpectedly empty graph
From: Cedric Laczny (cedric.laczny_at_[hidden])
Date: 2011-05-13 02:29:10

On Wednesday, 11. May 2011 06:00:54 Jeremiah Willcock wrote:
> On Thu, 5 May 2011, Cedric Laczny wrote:
> > Hi,
> > 
> > in order to select a subgraph based on some predicates, I use
> > filtered_graph() with an appropriate predicate. To get the resulting
> > subgraph (strictly speaking it is a new graph), I simply use
> > copy_graph(fg, new_g). In a special situation, this predicate simply
> > filters on the name and type of a vertex (set via vertex_properties).
> > This is working as expected. The more, I am working with multipartite
> > graphs, so there are no edges between 2 vertices of the same type.
> > When I now define a predicate that will only let vertices of one special
> > type to be visible and I use copy_graph() on this filtered_graph, the
> > resulting graph has no edges (which is ok) but it also has no vertices.
> > I would not expect the latter to happen. In turn, when I let another
> > type of vertices to be visible, the resulting graph is again as
> > expected.
> > Is there perhaps some obvious speciality of copy_graph() I am missing?
> 
> If you manually iterate through vertices() of your filtered_graph, does
> that return the vertices you expect?  It looks like copy_graph (as opposed
> to copy_component) just iterates through vertices and edges, and so
> unreachable vertices should still be copied.
> 

Ok, thank you for the confirmation. I just had a deeper look at the code again 
and it turned out that it was not caused by copy_graph().
Thanks again.

> -- Jeremiah Willcock
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Best,

Cedric