Consider a undirected densely connected (every vertex has $>\Theta(1)$ incident edges) graph $G$. Denote its vertices set as $\mathbf{V}$, number of vertices as $n$.

A connected dominating set $\mathbf{D}\subseteq \mathbf{V}$ is a subset of vertices with two properties:

  • Any node in $\mathbf{D}$ can reach any other node in $\mathbf{D}$ by a path that stays entirely within $\mathbf{D}$. That is, $\mathbf{D}$ induces a connected subgraph of $G$.

  • Every vertex in $G$ either belongs to $\mathbf{D}$ or is adjacent to a vertex in $\mathbf{D}$. That is, $\mathbf{D}$ is a dominating set of $G$.

My question is:

  • In such a densely connected graph, there is always a connected dominating set $\mathbf{D}$ such that $|\mathbf{V}\backslash\mathbf{D}|=\Theta(n)$ ?

  • (Actually, I only expect there are $\Theta(n)$ vertices in $\mathbf{V}\backslash\mathbf{D}$ adjacent to $\mathbf{D}$, it is not necessary that $\mathbf{D}$ is a dominating set and every vertex in $\mathbf{V}\backslash\mathbf{D}$ is adjacent to $\mathbf{D}$. I do not know what's the name of such a set $\mathbf{D}$, so I use connected dominating set.)

I am aware of that it is NP-complete to test whether there exists a connected dominating set with size less than a given threshold. But I am not sure whether it is still hard in my setting.

Here, $\Theta$ is the Big-theta notation in computational complexity. I use $> \Theta(1)$ to denote 'larger than any constant number'.

Any proof or counter-example is welcomed !

  • 1
    $\begingroup$ FWIW, the standard notation for your $> \Theta(1)$ is $\omega(1)$. $\endgroup$
    – Neal Young
    Nov 2, 2022 at 17:52
  • $\begingroup$ So, to clarify what you are looking for, you just want any subtree $T$ that has $\Omega(n)$ neighbors that aren't in $T$? (And finding this is equivalent to finding a subtree $T'$ that has $\Omega(n)$ leaves, right? Because by removing the leaves from $T'$ you would get $T$, or, similarly, by adding the neighbors to $T$ you could get $T'$.) $\endgroup$
    – Neal Young
    Nov 2, 2022 at 18:18
  • $\begingroup$ You are right, this is exactly what I expect. $\endgroup$ Nov 3, 2022 at 1:21

1 Answer 1


Yes, there is.

Lemma 1. Let $G$ be any $n$-vertex connected graph in which $\Omega(n)$ vertices have degree 3 or more. Then $G$ contains a connected dominating set $D$ whose complement has size $\Omega(n)$.

Before we give the proof, note that this answer strengthens this result, showing that it suffices for $G$ to meet the following condition: the number of edges in $G$ such that both endpoints have degree at most two is $O(1)$ times the number of vertices in $G$ of degree three or more.

Proof of Lemma 1. Consider any spanning tree $T$ of $G$. Let $n_1$ and $n_2$ be the numbers of leaves and degree-2 vertices in $T$. Let $n^+_3$ be the number of vertices of degree three or more in $T$.

Let $G_2$ be the subgraph of $T$ consisting of the $n_2$ vertices of degree 2 in $T$ and the edges in $T$ between these vertices. (Note that $G_2$ is a subgraph of $T$, and has maximum degree 2, so is a collection of paths.) We claim that the number of edges in $G_2$ is at least $n - 4n_1$, and the number of vertices in $G_2$ is at least $n-2n_1$.

To show the claim, consider the tree $T'$ obtained from $T$ by splicing out all degree-2 vertices (that is, for each maximal path $v_1, v_2, \ldots, v_p$ in $T$ such that each intermediate vertex $v_i$ for $2\le i \le p-1$ has degree two in $T$, remove the path edges and the intermediate vertices, replacing them all by the edge $(v_1, v_p)$, which might not be in $G$). It is well known (and easy to show) that in $T'$ (as in any tree where each non-leaf node has degree 3 or more) the number of edges is at most twice the number of leaf nodes, i.e., at most $2n_1$. And for every edge in $T'$, $G_2$ is missing at most two edges from $T$, so $G_2$ is missing at most $4n_1$ edges of $T$. Likewise, $T'$ has at most $2n_1$ non-leaf nodes, to $G_2$ is missing at most $2n_1$ nodes of $T$. This shows the claim.

It follows that $G_2$ has at most $4n_1$ more nodes than edges, so consists of at most $4n_1$ paths (counting each isolated vertex in $G_2$ as a path). It follows that at least $n-10 n_1$ nodes in $G_2$ have two neighbors in $G_2$. That is, back in $T$, at least $n-10 n_1$ nodes in $T$ have the following property: they have degree 2 in $T$, and both of their neighbors have degree 2 in $T$. Call such a node typical.

Now fix $T$ to be a spanning tree in $G$ with a maximum number of leaves.

Let $u$ be a typical vertex in $G$ and consider any edge $(u,w)$ from $u$ that is not in $T$. Consider adding the edge $(u, w)$ to $T$, creating a cycle, and removing one of the other edges incident to $u$ to break the cycle (so the result is again a spanning tree). This makes one of $u$'s neighbors into a leaf, so by the choice of $T$ the vertex $w$ must have been a leaf in $T$ (otherwise the operation would increase the number of leaves). So, for every typical vertex $u$, each edge from $u$ except the two in $T$ goes to a leaf in $T$.

Assume for contradiction that $n_1 = o(n)$. Then there are at least $n - 10 n_1 = n - o(n)$ typical nodes in $G$, and at least $\Omega(n)$ nodes in $G$ have degree at least three. It follows that at least one of the $n_1$ leaves, say $u$, has two edges to typical vertices, say $v_1$ and $v_2$. But then by adding edges $(u, v_1)$ and $(u, v_2)$ to $T$, and removing one edge from $v_1$ and one from $v_2$ (thereby making a neighbor of each a leaf in the new tree), we can create a spanning tree with more leaves than $T$, contradicting the choice of $T$. So $n_1 = \Omega(n)$.

The non-leaf vertices of $T$ form a connected dominating set, whose complement (the leaves) has size $n_1 = \Omega(n)$. $~~~~~\Box$

  • $\begingroup$ I have a question regarding this claim 'It follows that at least $n-8n_1$ nodes in $G_2$ have two neighbors in $G_2$'. $G_2$ only contains the $n_2$ vertices of degree $2$ in $T$, so, it should be 'at least $n_2 - 8n_1$ nodes' ? $\endgroup$ Nov 3, 2022 at 11:16
  • $\begingroup$ If this claim indeed contains a mistake, the proof above can only show $n_1>=n_2 /9$. However, it can be fixed and Lemma-1 is correct. By definition, we have $n_1+n_2+n_3=n-1$. So, at least one of $n_1$, $n_2$, $n_3$ is $\Theta(n)$. If it is $n_1$, then Lemma-1 is true. If it is $n_2$, since $n_1>=n_2/9$, Lemma-1 is also true. If it is $n_3$, as it is well-known that $n_1 \geq n_3+2$, we also have Lemma-1 is true. $\endgroup$ Nov 3, 2022 at 11:50
  • $\begingroup$ Yes, my edit introduced a gap in the proof. Namely, by the argument about $T'$, there are at least $n-2n_1$ vertices in $G_2$, that is $n_2 \ge n-2n_1$. So at least $n-2 n_1 - 8 n_1 = n - 10n_1$ nodes in $G_2$ are typical (have two neighbors in $G_2$). I'll correct the gap. $\endgroup$
    – Neal Young
    Nov 3, 2022 at 11:55

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