Re: dual graph
From: Dmitry A. Kazakov <mailbox_at_dmitry-kazakov.de>
Date: Tue, 27 Jun 2006 09:16:14 +0200
Message-ID: <7fngvin8f2no.wbi23fndg7aq$.dlg_at_40tude.net>
>
> Of course I can order the set of nodes. I just start somewhere, and
> start counting, like so... 1, 2, 3, 4, and so on.
> Who cares if the
> order I come up with is equivalent to the transitive closure of a
> directed graph?
Date: Tue, 27 Jun 2006 09:16:14 +0200
Message-ID: <7fngvin8f2no.wbi23fndg7aq$.dlg_at_40tude.net>
On Mon, 26 Jun 2006 13:33:51 -0600, Chris Smith wrote:
> You've really lost me now. The question was what to call a graph whose
> dual has only one node (i.e., the edges don't divide the plane). The
> correct answer is "forest" (or "tree", if we assume connectedness).
> "Ordered" is decidedly not a correct answer.
OK, I thought the question was about hierarchies. My mistake.
>>> 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*.
>
> Of course I can order the set of nodes. I just start somewhere, and
> start counting, like so... 1, 2, 3, 4, and so on.
> Who cares if the
> order I come up with is equivalent to the transitive closure of a
> directed graph?
The hierarchy does.
-- Regards, Dmitry A. Kazakov http://www.dmitry-kazakov.deReceived on Tue Jun 27 2006 - 09:16:14 CEST