
Hi,
I am having some problems with understanding the results of the betweenness centrality calculation for my network (a directed graph). I have several nodes with very high outdegree, but zero indegree, or vice versa and these are receiving very high scores
for betweeness centrality. As they can only be on the beginning or end of a path I would expect them to score zero in a directed network. Am I misunderstanding this?
Cheers, Alex


Jan 8, 2013 at 6:32 AM
Edited Jan 8, 2013 at 6:32 AM

Alex:
Yes, you are misunderstanding this. If a vertex has very high outdegree (or indegree, for that matter), then it is on the shortest paths between many vertex pairs, namely each pair of its many adjacent vertices. If A is connected to B, C, and
D, for example, then A is on the shortest paths between B and C, C and D, and B and D.
A vertex's betweenness centrality is determined by the number of shortest paths that include the vertex, so your highoutdegree vertex has a high betweenness centrality.
Please see the
http://en.wikipedia.org/wiki/Centrality#Betweenness_centrality article for information on the betweenness centrality algorithm.
 Tony



I forgot to mention an important fact that might better answer your question: The direction of individual edges is not taken into account when the shortest paths are determined.
 Tony



Hi Tony,
yes, thanks! The second answer was the one that I needed. I had previously found an answer on the forum that said that directionality was taken onto account on calculating betweenness centrality
(discussion:261244 Edge Weights and Betweenness). I think it would be useful to flag this up in the documentation.
Cheers,
Alex



PS Do any of the other centrality measures take directionality into account? I need to gain a measure of the relative importance of my nodes as links in a metabolic/ecological network so directionality is key.
Thanks again



Alex:
You're confusing two different things here. The direction of individual edges is NOT taken into account when the shortest paths are determined. However, the directionality of the
graph IS taken into account in the final step of the betweenness centrality calculations, where NodeXL has to decide whether to divide the result by 2 based on the graph's directionality. (I would copy over the relevant equations from
http://en.wikipedia.org/wiki/Centrality#Betweenness_centrality to show you what I mean, but I'm sure the equation formatting wouldn't paste properly.)
So my statements here and at
http://nodexl.codeplex.com/discussions/261244 are both correct.
 Tony


Jan 10, 2013 at 3:40 AM
Edited Jan 10, 2013 at 3:41 AM

Alex:
The graph's directedness is ignored when calculating PageRank, closeness centrality and eigenvector centrality. It is taken into account, obviously, when calculating degree, indegree and outdegree, which are also considered a type of centrality (http://en.wikipedia.org/wiki/Centrality).
 Tony



Tony,
I need some clarification about computing other basic centrality measures except for indegree and outdegree.
You are saying the graph's direction is ignored when measures are computed, even though a graph is directed, right?
I guess now I understand NodeXL outputs are different from other tools' outputs.
Is there any specific reasons why NodeXL ignores the direction of edges?
If you take edges' direction into account, the number of shortest paths between nodes would be slightly different from one without edge's directions.
I am just wondering the rationale behind this computation.
Jinie


Oct 10, 2013 at 12:26 AM
Edited Oct 10, 2013 at 2:13 AM

Jinie:
I said that the graph's directedness is ignored when calculating PageRank, closeness centrality and eigenvector centrality.
NodeXL uses a graph library called SNAP ( http://snap.stanford.edu/) to calculate these. In the particular version of SNAP that NodeXL uses, the graph's directedness is not taken into account for these calculations.
I can't say why; I don't even know if directedness should be used. The best way to get more information about this is to contact the SNAP team directly; they should be able to answer you more knowledgeably than I can.
 Tony

