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RE: After XQuery, are we done?

• To: "Burak Emir" <Burak.Emir@e...>,"Elliotte Harold" <elharo@m...>
• Subject: RE: After XQuery, are we done?
• From: "Hunsberger, Peter" <Peter.Hunsberger@S...>
• Date: Tue, 26 Oct 2004 10:11:10 -0500
• Cc: "XML Developers List" <xml-dev@l...>
• Thread-index: AcS7bKLTpBFgbeYmTVO6f+YyQAz2QgAAKotQ
• Thread-topic: After XQuery, are we done?

Burak Emir <Burak.Emir@e...> writes:
> 
> Elliotte Harold wrote:
> 
> > Hunsberger, Peter wrote:
> >
> >
> >> I don't get the distinction.  As soon as you've got a graph you've 
> >> got a tree (or perhaps many trees).
> >>
> >
> > Not necessarily. All trees are graphs but not all graphs are trees.
> > For instance a pure tree can't represent a cycle but a graph can. 
> > XML's rule that a node can only have one parent is not a limit of 
> > graphs in general.
> >
> I think he meant a spanning tree, i.e. one that has all the nodes but 
> all edges. A graph can have many spanning trees.
> 
> This works for undirected edges. For directed, there might not be a 
> spanning tree in the mathematical sense, but you can still
> get one if you reverse the arrows, like in this one
> 
> o -> o <- o

Nothing that sophisticated, I just meant that any tree can be contained
in a graph.  Guess I should have wrote something more like "As soon as
you've got a tree there is a graph that can contain it (and perhaps many
other trees)".

Which still leaves the original question; once you've got a way of
managing and manipulate graphs, why would you need a way to distinguish
trees?  What does recognizing the special case get you?

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