I want to find the smallest possible function $k(n,m)$ such that for any graph $G$ with $n$ vertices and $m$ edges, there exists $n$ vertex sets $S_1,S_2,...,S_n\subseteq V$ each with size $k(n,m)$ and every edge $(u,v)$ has $u,v$ both contained in some $S_i$. In other words, $n$ complete graphs suffice to cover all edges.

The first question is whether $k(n,m)$ can be $n^{1/2-\epsilon}$ when $m$ is sub-quadratic. The question is interesting because $n$ complete subgraph each with size $O(n^{1/2})$ suffice to cover any graph.

I searched for clique edge covering but most of the results only consider the case where cliques cannot cover non-edge. I wonder if there exists any similar research in the setting when cliques can cover non-edge.

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    $\begingroup$ Just a suggestion: if I understand the problem correctly, then "subgraphs" in the title is confusing since the sets $S_i$ are not necessarily cliques in $G$? $\endgroup$ Commented Feb 7, 2023 at 7:47
  • $\begingroup$ @ChristianKomusiewicz: Indeed -- but it's about covering the graph, so I don't find it so confusing that the complete graphs used for this purpose can contain edges not part of the original graph. $\endgroup$
    – a3nm
    Commented Feb 7, 2023 at 15:18
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    $\begingroup$ It's just that when one reads "Cover ... with complete subgraphs" one thinks that all the cliques must be subgraphs of the input graph. Writing "Cover a graph with complete graphs" or "Cover a graph with cliques" would be more fitting in my opinion. $\endgroup$ Commented Feb 8, 2023 at 7:46

1 Answer 1


Here are asymptotic bounds for $k(n, m)$ that are tight up to a logarithmic factor. Note the threshold around $m = \Theta(n^{3/2})$:

Theorem 1. $~~~~\frac{1}{21}\min(\lceil\sqrt n\rceil, \lceil m/(n\log n)\rceil) ~\le~ k(n, m) ~\le ~\min(\sqrt {n}, 2\lceil m/n\rceil).$

Here's the proof. We show the upper bounds (Lemma 1) then the lower bounds (Lemma 2).

Lemma 1. $k(n, m) \le \min(\sqrt {n}, 2\lceil m/n\rceil)$

Proof. First we show $k(n, m) \le \sqrt n$. Partition the vertex greedily into some $p$ parts of size at most $\sqrt n/2$. Then, for each of the ${p\choose 2} \le n$ unordered pairs $\{A, B\}$ of distinct parts, create a vertex set $S_i$ consisting of $A\cup B$ (having size at most $\sqrt n$). There are at most $n$ such pairs $\{A, B\}$, and every edge has both endpoints in $A\cup B$ for some pair. This shows $k(n, m) \le \sqrt n$.

To finish we show $k(n, m) \le 2\lceil m/n\rceil$. Greedily partition the edge set $E$ into $n$ sets $E_1, \ldots, E_n$ such that each contains at most $\lceil m/n\rceil$ edges. For each $i\in [n]$, let $S_i$ be the set of vertices used in edge set $E_i$. Then $|S_i| \le 2 |E_i| \le 2\lceil m/n\rceil$, as desired.$~~~~\Box$

Here is the lower bound.

Lemma 2. $k(n, m) \ge \min(\sqrt n, \lceil m/(n\ln n)\rceil)/21$

Proof. Let $m' = n^{3/2}\ln n$. The lower bound (of $\sqrt n/21$) for $m\ge m'$ follows from the lower bound for $m=m'$. And a lower bound of $1$ holds trivially for all positive $m \le n\ln n$. So assume without loss of generality that $n\ln n \le m \le n^{3/2}\ln n$.

Let $k=k(n, m)$ and $p = m/n^2$. Let $G=(V, E)$ be a random graph where each edge is independently present with probability $p$.

Claim 1: With probability $1-o(1)$ $G$ has the following two properties:

  1. $m/3 = p n^2/3 \le |E| \le p n^2 = m$

  2. For every vertex subset $S\subseteq V$ of size $k$, we have $|E_S| \le 7 k \ln n$, where $E_S$ denotes the set of edges with both endpoints in $S$.

Suppose the claim is true. Let $G$ have these two properties. By definition of $k$ there are $n$ $k$-vertex induced subgraphs in $G$ that collectively contain all of $G$'s edges. By Property 2, each of these $n$ subgraphs contains at most $7 k \ln n$ edges from $S$, so collectively they contain at most $7 n k \ln n$ edges. But they contain $E$, and (by Property 1) $|E|\ge m/3$. So $7 n k \ln n \ge m/3$. Simplifying implies $k \ge m/(21 n \ln n)$, as desired.

To complete the proof we show the claim. The expected number of edges in $G$ is ${n\choose 2}p \sim n^2p/2$. By a standard Chernoff bound the probability that $n^2p/3 \le |E| \le n^2 p$ fails to hold is at most $2\exp(-\Omega(n^2 p))$, which (by the choice of $p$) is $o(1)$. So Property 1 fails with probability $o(1)$.

The number of $k$-vertex subgraphs of $G$ is ${n\choose k} \le n^k$. For each, (using $k \le n^{1/2}$ and $m\le n^{3/2}\ln n$) its expected number of edges is ${k \choose 2} p \le 3.5 k \ln n$, so by a standard Chernoff bound the probability that the number is more than $2\times 3.5 k\ln n$ is at most $\exp(-3.5 k\ln (n)/3) \le \exp(-1.1 k \ln n)$. So by the naive union bound the probability that Property 2 fails is at most $n^k \exp(-1.1 k\ln n) = \exp(-0.1 k\ln n) = o(1).$ So Property 2 fails with probability $o(1)$.

By the naive union bound, the probability that either Property 1 or Property 2 fails is $o(1)$. This proves the claim, and the lemma. $~~~~\Box$


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