Question

Suppose we are talking about graphs with $n$ labeled vertices. Which graphs are more common: connected or non connected?

Answer 1

Connectedness wins, since the complement of any disconnected graph is connected.

EDIT: Perhaps you'd like a proof of this. Let G be a disconnected graph, G' its complement. If v and u are in different components of G, then certainly they're connected by an edge in G'. And if they're in the same component of G, then there's some w in another component (since G was disconnected), so v-w-u is a path in G'.

Answer 2

I like Jonah Ostroff short and sweet proof, but the key to it lies in the fact that there is not a bijection between the set $S_1$ of connected graphs and the set $S_2$ of disconnected graphs over $n$ labeled vertices for $n \ge 4$, as follows:

• the complement of each disconnected graph is a connected graph (which Ostroff points out)

• the complement of a connected graph can also be a connected graph

• thus the cardinality of the set of connected graphs must be larger than the cardinality of the disconnected graphs, because while there is a one-to-one mapping of each disconnected graph onto a connected graph, there exist connected graphs which do not map to a disconnected graph

For example, for $n=4$:

Take the $12$ possible un-drected Hamiltonian paths of length $4$ on a graph over four labeled vertices.

The complement of each of these paths is also a hamiltonian path.

Since we know that the complement of a disconnected graph is obviously connected for $n>3$, then the number of connected graphs is at least equal to the number of disconnected graphs. Hoewever, since for $n>3$, the complements of at least some of the connected graphs are also connected graphs, that means that there must be more connected graphs than there are unconnected graphs.

The $12$ Hamiltonian paths are those connected graphs over $4$ vertices whose complements are also connect: thus the remaining $2^6 - 12 = 52$ graphs are divided into pairs of complement graphs which are connected and disconnected,

yielding a total of $26$ disconnected graphs, and $26+12=38$ connected graphs over the set of $64$ labeled graphs over $4$ labeled vertices.

The path graphs of length $n$ on the set of $n$ vertices are the canonical example of connected graphs whose complements are also connected graphs (for $n>3$).

Answer 3

For large $n$, not only are the vast majority of graphs on $n$ vertices connected, the vast majority have diameter 2. That is, any two vertices have a neighbor in common. (The standard reference for properties of most graphs on $n$ vertices, for large $n$, is the book "Random Graphs" by Bela Bollobas.)

Answer 4

Connectedness wins by a knockout: the proportion of disconnected graphs is about $n2^{-n+1}$. See Flajolet, Sedgewick "Analytic Combinatorics", p. 138.

Answer 5

I like Jonah Ostroff's proof, but here is an inductive proof (for the heck of it).

Let $c(n)$ and $d(n)$ respectively denote the number of connected and disconnected graph on $n$ vertices.

Evidently, $g(n):=c(n)+d(n)$ is the number of graphs on $n$ vertices. As Jonah Ostroff points out $c(4)=38$ and $d(4)=26$.

So, inductively assume that $c(n) > d(n)$, let $G$ be a graph with vertex set $[n]$ and consider a new vertex $n+1$. If $G$ is connected, then adding any non-empty subset of edges incident to $n+1$ maintains connectivity. On the other hand, if $G$ is disconnected, then adding all edges incident to $n+1$ results in a connected graph.

Therefore,

\[ c(n+1) \geq (2^{n}-1)c(n)+d(n) = (2^n-2)c(n) + g(n). \]

By induction, we have $c(n) > g(n)/2$. Substituting yields

\[ c(n+1) > 2^{n-1} g(n)=g(n+1)/2. \]