Are there more connected or disconnected graphs on n vertices? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-22T18:51:33Zhttp://mathoverflow.net/feeds/question/44877http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/44877/are-there-more-connected-or-disconnected-graphs-on-n-verticesAre there more connected or disconnected graphs on n vertices?Leonid Petrov2010-11-04T21:16:29Z2010-11-06T10:34:00Z
<p>Suppose we are talking about graphs with $n$ labeled vertices. Which graphs are more common: connected or non connected?</p>
http://mathoverflow.net/questions/44877/are-there-more-connected-or-disconnected-graphs-on-n-vertices/44884#44884Answer by Tony Huynh for Are there more connected or disconnected graphs on n vertices?Tony Huynh2010-11-04T22:27:57Z2010-11-05T11:24:26Z<p>I like Jonah Ostroff's proof, but here is an inductive proof (for the heck of it).</p>
<p>Let $c(n)$ and $d(n)$ respectively denote the number of connected and disconnected graph on $n$ vertices. </p>
<p>Evidently, $g(n):=c(n)+d(n)$ is the number of graphs on $n$ vertices. As Jonah Ostroff points out $c(4)=38$ and $d(4)=26$.</p>
<p>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.</p>
<p>Therefore,</p>
<p>\[
c(n+1) \geq (2^{n}-1)c(n)+d(n) = (2^n-2)c(n) + g(n).
\]</p>
<p>By induction, we have $c(n) > g(n)/2$. Substituting yields</p>
<p>\[
c(n+1) > 2^{n-1} g(n)=g(n+1)/2.
\]</p>
http://mathoverflow.net/questions/44877/are-there-more-connected-or-disconnected-graphs-on-n-vertices/44889#44889Answer by Jonah Ostroff for Are there more connected or disconnected graphs on n vertices?Jonah Ostroff2010-11-04T22:53:32Z2010-11-04T22:53:32Z<p>Connectedness wins, since the complement of any disconnected graph is connected.</p>
<p>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'.</p>
http://mathoverflow.net/questions/44877/are-there-more-connected-or-disconnected-graphs-on-n-vertices/44890#44890Answer by zhoraster for Are there more connected or disconnected graphs on n vertices?zhoraster2010-11-04T23:01:26Z2010-11-04T23:01:26Z<p>Connectedness wins by a knockout: the proportion of disconnected graphs is about $n2^{-n+1}$. See Flajolet, Sedgewick "Analytic Combinatorics", p. 138.</p>
http://mathoverflow.net/questions/44877/are-there-more-connected-or-disconnected-graphs-on-n-vertices/44908#44908Answer by Andreas Blass for Are there more connected or disconnected graphs on n vertices?Andreas Blass2010-11-05T01:28:45Z2010-11-05T01:28:45Z<p>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.)</p>
http://mathoverflow.net/questions/44877/are-there-more-connected-or-disconnected-graphs-on-n-vertices/45029#45029Answer by sleepless in beantown for Are there more connected or disconnected graphs on n vertices?sleepless in beantown2010-11-06T04:57:40Z2010-11-06T10:34:00Z<p>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:</p>
<ul>
<li><p>the complement of each disconnected graph is a connected graph (which Ostroff points out)</p></li>
<li><p><strong>the complement of a connected graph can also be a connected graph</strong></p></li>
<li><p>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</p></li>
</ul>
<p>For example, for $n=4$:</p>
<p>Take the $12$ possible un-drected Hamiltonian paths of length $4$ on a graph over four labeled vertices. </p>
<p>The complement of each of these paths is also a hamiltonian path.</p>
<p>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.</p>
<p>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, </p>
<p>yielding a total of $26$ disconnected graphs, and $26+12=38$ connected graphs
over the set of $64$ labeled graphs over $4$ labeled vertices.</p>
<p>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 $n>3$).</p>