I am looking for advice on what would be a reasonable or useful generalization of vertex- and edge-connectivity to the graphs with 0 and 1 vertices (null graph and singleton graph).

Motivation: Creating a software package that computes these quantities, as well as checks connectivity and biconnectivity.

It is often practically useful for such software not to explicitly fail for similar edge cases, but return a result anyway. For example, in Mathematica Min[{}] and Max[{}] are Infinity and -Infinity respectively, Total[{}] is 0, etc.

Similar questions have been discussed here before: Is the empty graph connected?

Specific issues:

  • What is the edge and vertex connectivity of the graph with no vertices, $K_0$?

  • What is the edge and vertex connectivity of the graph with one vertex, $K_1$?

  • Is $K_2$ bi-connected? It is often considered to be, even though its vertex connectivity is 1.

  • Similarly, is $K_3$ tri-connected, etc.?

I am looking for reasons why a particular choice would be useful or consistent with various theorems, other definitions, etc.


1 Answer 1


The standard definition of vertex connectivity is the minimum number of vertices you need to remove to reduce the graph in question to either a disconnected graph or a singleton graph. The second case is needed if $G$ is complete, and it follows that the vertex connectivity of $K_n$ is $n-1$. This is generally the right choice in this case so that results about connectivity apply even if $G$ is complete - for example, the fact that $\delta(G)\geq\lambda(G)\geq\kappa(G)$, where $\lambda,\kappa$ are edge and vertex connectivity respectively. Following this, you would have $\kappa(K_1)=0$. This means that it is no longer true to say $G$ is connected if and only if $\kappa(G)>0$, but you can't have everything.

Normally you don't hear the definition of edge connectivity given in the same way, because any graph with more than one vertex can be reduced to a disconnected graph by removing (possibly $0$) edges. But IMO it would make sense to again add "or to a singleton graph" to deal with this case, meaning that $\lambda(K_1)=0$ also. (I am less sure how standard this is and don't currently have textbooks to hand to check.)

Personally, I take the view that it is much better to define "graph" to require a non-empty vertex set. I can't think of a good reason for needing "$K_0$" to be a graph, and deciding that it isn't avoids some problems with other definitions - of course, these are not insurmountable by any means, but why cause yourself trouble? In my experience this is a fairly common position to take.

  • $\begingroup$ What is your view on whether $K_2$ is biconnected, $K_3$ triconnected, etc.? $\endgroup$
    – Szabolcs
    Commented Apr 19, 2019 at 18:51
  • 1
    $\begingroup$ I think that the best reason to consider the null graph $K_0$ to be a graph is when dealing with exponential generating functions. In particular, the identity $G(x) = e^{C(x)}$ (where $G(x)$ is the egf of all labeled graphs and $C(x)$ the egf of connected graphs) only holds if we consider $K_0$ to be a graph, but not a connected graph. (For similar reasons, we should consider $K_0$ to be a forest, but not a tree.) $\endgroup$ Commented Apr 20, 2019 at 4:55
  • $\begingroup$ @Szabolcs I don't really have an opinion, since I don't tend to use those words. I would say $K_2$ isn't $2$-connected (since $\kappa(K_2)<2$), but since $K_1$ is connected without being $1$-connected, perhaps it makes sense for $K_2$ to be biconnected without being $2$-connected? $\endgroup$ Commented Apr 20, 2019 at 8:18

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