Re: dual graph
From: Dmitry A. Kazakov <mailbox_at_dmitry-kazakov.de>
Date: Mon, 26 Jun 2006 20:09:30 +0200
Message-ID: <xnhqm4tazibp$.m1c6iamjjjl0$.dlg_at_40tude.net>
>> Thanks to everybody for clearing up the matter of "planar".
>>
>> Now, I'm interested in a a subset of the planar graphs.
>>
>> It's the graphs that don't divide the space into more than one region.
>>
>> It seems to me that it's appropriate to call such graphs "hierarchical
>> graphs", or simply "hierarchies". But I'm not sure. Can anyone clear this
>> up?
Date: Mon, 26 Jun 2006 20:09:30 +0200
Message-ID: <xnhqm4tazibp$.m1c6iamjjjl0$.dlg_at_40tude.net>
On Mon, 26 Jun 2006 08:55:02 -0600, Chris Smith wrote:
> David Cressey <dcressey_at_verizon.net> wrote:
>> Thanks to everybody for clearing up the matter of "planar".
>>
>> Now, I'm interested in a a subset of the planar graphs.
>>
>> It's the graphs that don't divide the space into more than one region.
>>
>> It seems to me that it's appropriate to call such graphs "hierarchical
>> graphs", or simply "hierarchies". But I'm not sure. Can anyone clear this
>> up?
> > Well, such a graph has no cycles, so it is a forest. If it's connected, > then it's a tree. I don't know where Dmitry got "ordered".
G is ordered if DAG. Its transitive closure G* is a strict order [on nodes].
[ OK, planarity is lost, but it is not required for hierarchies either. ]
> Any graph > (planar or not, cycles or not, doesn't matter) may be made into an > ordered graph simply by defining an order for its nodes.
If you can order the set of nodes. But that order is not necessary G*.
-- Regards, Dmitry A. Kazakov http://www.dmitry-kazakov.deReceived on Mon Jun 26 2006 - 20:09:30 CEST