Re: [igraph] Circular Reingold Tilford on tree is not planar!?

[Tamás Nepusz]

it should be planar, AFAIK. If it is not planar, that means that the
original non-circular one is not planar

No, it isn't necesarily planar. A while ago I found a graph for which the RT layout was planar but the circular RT layout wasn't, and I also thought it was a bug, but it isn't. Try this graph for instance (given as an edge list):

0-1, 0-2, 0-3, 0-4, 0-5, 0-6, 0-7, 0-8, 0-9, 4-10, 9-10

When you use vertex 0 as root, the RT layout is planar but the circular RT layout isn't.

-- 

T.

 

, the circular one is just the
transformation of that. Can you send us (in private if it's big, to
Tamas and me) an example? I think this is a bug.

Best,
Gabor

On Fri, Jun 24, 2011 at 9:55 AM, Robert Belleman <address@hidden> wrote:

Hi,

I am applying the circular Reingold Tilford layout algorithm on a tree
consisting of 23k nodes and the same number of edges. I was expecting this
algorithm to produce a planar layout but it doesn't.

Can anybody enlighten me?

Thanks,
-- Rob

--
Robert Belleman, PhD, Computational Science, Faculty of Science
Universiteit van Amsterdam, Science Park 904, 1098 XH Amsterdam
The Netherlands.    T: +31 20 525 7272/7463  F: +31 20 525 7490
W: http://www.science.uva.nl/~robbel/    E: address@hidden

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Gabor Csardi <address@hidden>     MTA KFKI RMKI

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