Geoff,
How did you arrive at the V(V+1)/2 figure?
On May 15, 1:21 am, Geoffrey Summerhayes <[EMAIL PROTECTED]> wrote:
> On May 14, 12:41 pm, "Bruno Avila" <[EMAIL PROTECTED]> wrote:
>
> > (reformatting last email)
>
> > These are good questions. In my problem, a binary tree is different if
> > the set of nodes are different. For example:
>
> > a We have 9 different binary trees: {a}, {b}, {c},
> > \ {a,b}, {b,c}, {b,d}, {c,d}, {a,b,c}, {a,b,d}. It does
> > b not matter the number of different binary trees
> > / \ that can be formed by each of these sets. The
> > d - c set {b,c,d} can not be considered because it would
> > for a cycle, so it would not be a tree.
>
> > My initial question was not well specified. Sorry.
>
> That's an unusual set of conditions for a subgraph.
>
> Ok, that would make the minimum V(V+1)/2, which
> would be exact for a linear graph a-b-c-d-... as well as any
> complete one (more than two nodes automatically have a
> cycle).
>
> Maximum might be a ring structure, that would be V*(V-1)
> when V>=3. No guarantee on that one though.
>
> ---
> Geoff
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