Re: [Boost-users] [BGL] Separating twin vertices

Subject: Re: [Boost-users] [BGL] Separating twin vertices
From: Shaun Jackman (sjackman_at_[hidden])
Date: 2011-03-17 20:49:56

On Thu, 2010-08-12 at 22:18 -0700, Jeremiah Willcock wrote:
> On Thu, 12 Aug 2010, Shaun Jackman wrote:
> 
> > Hi,
> >
> > I have a directed graph where every vertex has a twin vertex. It is
> > known ahead of time which pairs of vertices are twins. I want to remove
> > as few vertices as possible from the graph such that no vertex and its
> > twin is in the same connected component. Could someone suggest an
> > appropriate algorithm?
> 
> Connected component or strongly connected component?  I'm assuming for a 
> directed graph that you mean SCC.  When you remove a vertex, do you also 
> remove its twin?  Is there a relationship between an edge from A -> B and 
> from twin(A) -> twin(B) (or twin(B) -> twin(A)), in the sense that those 
> edges would need to be removed in pairs?  That would be implied by needing 
> to remove a vertex and its twin at the same time.  Do you need to remove 
> entire vertices, or would removing just edges be OK?  Also, do you need an 
> exact solution or would an approximation be acceptable if an exact 
> algorithm was too inefficient?

Hi Jeremiah,

Yes, strongly connected component.
Yes, when you remove a vertex you must also remove its twin.
An edge A -> B implies the existence of a twin edge twin(B) -> twin(A),
and in removing an edge, you must also remove its twin.
Entire vertices need to be removed (not just edges).
An approximate solution would be sufficient.

After some reading of Wikipedia, I found that my graph looks remarkably
similar to an implication graph
http://en.wikipedia.org/wiki/Implication_graph

and my problem sounds remarkably similar to the 2-satisfiability problem
http://en.wikipedia.org/wiki/2-satisfiability

So, armed with this new vocabulary, I'm looking to remove the fewest
number of vertices (and their twins) from an implication graph that is
not 2-satisfiable, to make a new graph that is 2-satisfiable.

Cheers,
Shaun