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Given $n$ distinct nodes $1,2...n$, I wish to find the number of connected graphs with these $n$ nodes. I have seen the previous question : How to calculate the number of possible connected simple graphs with $n$ labelled vertices. But in my case multi-edges are allowed.

By multi-edges I mean, if I have $n$ nodes a multi edge is a subset of vertices connecting all those vertices. For example in case of $3$ nodes $\{1,2,3\}$ is a multi-edge connecting all the three edges. For example following are the few ( out of $96$ different ways ) ways of connecting three nodes:

  1. Select edge $\{1,2,3\}$.
  2. Select edges $\{1,2,3\}$ , $\{1,3\}$ , $\{2,3\}$ and $\{1,2\}$.
  3. Select edges $\{1,2,3\}$ and $\{1,3\}$.
  4. Select edges $\{1,3\}$ and $\{2,3\}$ and so on.

I am just looking for hints not solutions. As this is part of question from Project Euler question: Power set of power sets.

A friend of mine gave the above method of multi-edges as a hint to me and told me I had to modify the recurrence in the question: How to calculate the number of possible connected simple graphs with $n$ labelled vertices. But I am unable to figure it out and just need a new direction for thinking.

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advocateofnone
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  • I think your terminology is a little different from normal. You say that {1, 2, 3} is a "multi-edge connecting all the three edges" - this seems quite different from either the usual definition of a multi-edge (eg: [(1, 2), (1, 2)] where vertices 1 and 2 are connected multiple times) and more similar to an edge in a hypergraph (where more than 2 vertices can be part of an edge). – gilleain May 10 '16 at 11:47
  • Ok, after reading that Project Euler question it looks like the graphs are actually simple (not multi-) graphs. It says "Two vertices Y1 and Y2 are connected if Y1 ∩ Y2 ≠ ∅." - this is a binary relation, so you either have an edge between Y1 and Y2 (if their intersections are non-empty) OR no edge. – gilleain May 10 '16 at 11:52
  • @gilleain But if you solve that way I can have a graph of size $2^n$ for which the recurrence given can't be calculated as the vlaue $2^n$ is quite large, even OEIS lists only first 50 values for the sequence. Thus we have to look at the graph having $n$ nodes instead of $2^n$ nodes an allow multi-edges. – advocateofnone May 10 '16 at 12:05
  • ok, I agree that $2^n$ will be quite large if n is $10^4$, but surely you need C(n, k) which is only defined on simple graphs on X in R(n)? Or do you have some other value you are computing on your graphs that maps to C(n, k)? – gilleain May 10 '16 at 12:11
  • @gilleain Surely $C(n,k)$ is for simple graphs ( not having multi edges ), but as my question suggets I am looking for a way I could modify $C(n,k)$ so that it takes into account the multi edges. – advocateofnone May 10 '16 at 12:18
  • Well, no, your question does not suggest this, I think. Perhaps you could edit your question to make this clear. Describe the graphs from the Project Euler question and how you are going to map these to your (hyper)graphs and then ask how C(n, k) in the simple graph relates to the connected components (or whatever) of the hypergraphs. – gilleain May 10 '16 at 12:29
  • @gilleain Will do so. I will format the question and include all the details thank you. – advocateofnone May 10 '16 at 12:31

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