Graph Problem: (n+1) Knots & Edges - How Many Possible Graphs?

  • Thread starter Thread starter pixel01
  • Start date Start date
  • Tags Tags
    Graph
AI Thread Summary
The discussion revolves around calculating the total number of possible graphs formed by (n+1) knots, specifically without loops and allowing for tree structures. Participants highlight that the number of combinations of edges connecting these knots exceeds simple r-combinations due to the inclusion of various graph types. The complexity increases with larger values of n, as both tree and non-tree structures must be considered. References to graph theory concepts and enumeration methods are provided to support the analysis. The conversation emphasizes the need for a deeper understanding of graph theory to arrive at an accurate solution.
pixel01
Messages
688
Reaction score
1
Hello all,

I have got this graph problem. There are (n+1) knots which are numbered as A1, A2...An,An+1. We can connect those knots with lines (edges). The question is : How many possible graphs are there totally?
Notes: A graph can be a tree of any kinds, but muts not contain a loop.

Thanks
 
Last edited:
Physics news on Phys.org
EnumaElish said:

Thanks for your ansser.
The number of combinations out of n elements,without repetition is just double the number of graphs of the same type. For n relatively large, we can have graphs without tree, and graphs with tree structure as well, so the final answer must be higher than just the number of combinations out of n elements...
 
Thanks Enuma. I have done it!
Great chance to revise the graph theory.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
Back
Top