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Topic: An Argument for the Subroot Node
Replies: 28   Last Post: Mar 14, 2008 8:32 PM

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Tony Orlow

Posts: 3,142
Registered: 8/23/06
Re: An Argument for the Subroot Node
Posted: Mar 14, 2008 12:21 AM
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briggs@encompasserve.org wrote:
> In article <r0bdt314kmk4gc4pkjbau9j6vf4n99r48q@4ax.com>, G. Frege <nomail@invalid> writes:
>> On Tue, 11 Mar 2008 16:51:03 +0100, G. Frege <nomail@invalid> wrote:
>>
>> Oopos... typo/thinko...
>>

>>>> According to it, the perimeter of a triangle is a tree...
>>>> which, of course, it is not.
>>>>

>>> Wait a second. If the nodes of your triangle were, say a, b, c then
>>> (say) a, b would be connected by more than one path:
>>>
>>> p_1 = ((a,b))
>>>
>>> p_2 = ((a,b), (b,c), (c,a), (a,b))
>>>

>> No?
> Why not:
>
> p_3 = ((a,c),(c,b))
>
> Now the example doesn't imply that cycles are the only way to get multiple
> paths.
>
> You'd better lay out your definition of "path".
>
> It appears that your current working definition of path is along
> the lines of:
>
> An "edge" is an ordered pair of vertexes that are directly connected.
>
> A "path" is a sequence of edges such that the terminal vertex of
> each edge in the sequence is the initial vertex of the next edge
> (if any) in the sequence. In addition, the initial vertex
> of an edge in the sequence may not appear as the terminal vertex
> of the next edge in the sequence. (No direct back-tracking).
>
> It seems fairly easy to prove that:
>
> 1. If a graph has a cycle, it will have two distinct nodes with
> more than one path between them.
>
> 2. If a graph has two distinct nodes with more than one path between
> them, it will contain a cycle.
>
> 3. If a graph is connected, there will be at least one path between
> any pair of distinct nodes.
>
> 4. If a graph has at least one path between any pair of distinct nodes,
> it is connected.


Does this say anything about a node with a single edge between it and
itself? It seems that doesn't quite qualify as a cycle, eh?

:) Tony


Date Subject Author
3/11/08
Read An Argument for the Subroot Node
Tony Orlow
3/11/08
Read Re: An Argument for the Subroot Node
Virgil
3/11/08
Read Re: An Argument for the Subroot Node
G. Frege
3/11/08
Read Re: An Argument for the Subroot Node
Tony Orlow
3/11/08
Read Re: An Argument for the Subroot Node
Virgil
3/11/08
Read Re: An Argument for the Subroot Node
Tony Orlow
3/11/08
Read Re: An Argument for the Subroot Node
Virgil
3/11/08
Read Re: An Argument for the Subroot Node
G. Frege
3/11/08
Read Re: An Argument for the Subroot Node
J. Antonio Perez M.
3/11/08
Read Re: An Argument for the Subroot Node
G. Frege
3/11/08
Read Re: An Argument for the Subroot Node
G. Frege
3/11/08
Read Re: An Argument for the Subroot Node
Tony Orlow
3/11/08
Read Re: An Argument for the Subroot Node
briggs@encompasserve.org
3/14/08
Read Re: An Argument for the Subroot Node
Tony Orlow
3/14/08
Read Re: An Argument for the Subroot Node
David R Tribble
3/14/08
Read Re: An Argument for the Subroot Node
LudovicoVan
3/11/08
Read Re: An Argument for the Subroot Node
Tony Orlow
3/11/08
Read Re: An Argument for the Subroot Node
J. Antonio Perez M.
3/11/08
Read Re: An Argument for the Subroot Node
Tony Orlow
3/11/08
Read Re: An Argument for the Subroot Node
Tony Orlow
3/11/08
Read Re: An Argument for the Subroot Node
Tony Orlow
3/11/08
Read Re: An Argument for the Subroot Node
LudovicoVan
3/11/08
Read Re: An Argument for the Subroot Node
LudovicoVan
3/14/08
Read Re: An Argument for the Subroot Node
Tony Orlow
3/14/08
Read Re: An Argument for the Subroot Node
LudovicoVan
3/14/08
Read Re: An Argument for the Subroot Node
LudovicoVan
3/11/08
Read Re: An Argument for the Subroot Node
David R Tribble
3/11/08
Read Re: An Argument for the Subroot Node
LudovicoVan
3/11/08
Read Re: An Argument for the Subroot Node
LudovicoVan

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