
Re: An Argument for the Subroot Node
Posted:
Mar 11, 2008 11:40 AM


On Mar 11, 5:42 pm, Tony Orlow <t...@lightlink.com> wrote: > G. Frege wrote: > > On Mon, 10 Mar 2008 23:44:16 0500, Tony Orlow <t...@lightlink.com> > > wrote: > > >> So, in conclusion, the "complete" binary tree includes not only a root > >> node, but a subroot node, which is its own parent [...]. > > > Such a structure (graph) is not a /tree/ any more (since a tree does not > > have cycles _by definition_). > > > See: > >http://en.wikipedia.org/wiki/Tree_(graph_theory) > > > F. > > "In graph theory, a tree is a graph in which any two vertices are > connected by exactly one path." > ************************************************************
Where did you get that definition from? According to it, the perimeter of a triangle is a tree...which, of course, it is not.
A tree is a connected graph without any cycles. Period. This is not an "alternative" definition to the nonsense written above. This is THE definition of a tree in graph theory and it is not equivalent to what you wrote above.
Regards Tonio
Ps. By the way, are you trying to make some point by tagging a binary ""tree"" the way you did? Where are you heading?

