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 --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---