Let $G=(V,E)$ be an undirected graph with $|E|=m$ edges.

Given that any clique of size $\ge 2$ can be identified with its corresponding edges, and at most every subset $S\subseteq E$ creates a clique, $G$ can have at most $|P(E)|= 2^m$ cliques of size $\ge 2$.

However, based on the intuition that it's better to arrange the edges into bigger cliques, I'd assume arranging the $m$ edges into one big clique of size $c$ is optimal (whenever it's possible). If $c$ is the size of that clique, we have $m=\binom{c}2 \rightarrow c\le 4\sqrt m$, and I'd therefore assume that one can get an upper bound of the form $2^{4\sqrt m}$.

How would one show this upper bound?

Noting in above simple proof that only subsets $S$ with size $\left|S\right|=\binom{i}2$ for some $i\in \mathbb N$ can ever be edge sets of a clique doesn't seem to help.

  • 1
    $\begingroup$ I would try to show that a maximum number of cliques can be achieved when the $m$ edges are incident on as few vertices as possible (which is a clique when $m$ is a triangular number). I would try to do this by showing that, given an arbitrary $m$-edge graph, if there is a vertex $v$ all of whose incident edges can be moved to other pairs of vertices that are already both of nonzero degree (note there is a wrinkle for neighbours of $v$), then doing so (which reduces the total number of vertices incident on any edge, bringing us closer to a clique) never reduces the number of cliques. $\endgroup$ Commented Aug 19, 2021 at 15:46
  • $\begingroup$ You can take a look at similar proofs of Turán's theorem for inspiration. $\endgroup$ Commented Aug 20, 2021 at 3:39

1 Answer 1


In this paper with DOI 10.1007/s00373-007-0738-8 we have the following theorem:

Let $n$ and $m$ be non-negative integers such that $m ≤ \binom n2$. Let $d$ and $l$ be the unique integers such that $m =\binom d2+l$, where $d ≥ 1$ and $0 ≤ l ≤ d − 1$.
Then the maximum number of cliques in an $(n, m)$-graph equals $ 2^d + 2^l + n − d − 1$.

Here, an $(n,m)$-graph is a graph with $n$ nodes and $m$ edges.

This graph has $n$ 1-cliques and $1$ 0-clique.

Therefore its number of cliques of size $\ge 2$ is $$ 2^d + 2^l − d − 2 \le 2^{d+1} − d − 2\le 2^{d+1} $$ and $d$ is the biggest integer with $m\ge \binom{d}2$.

So we further have $$ d\le \frac{\sqrt{8·m+1}+1}2 $$

Assuming $m\ge 1$, we have $$ \frac{\sqrt{8·m+1}+1}2\le \frac{\sqrt{9m}+1}2 = \frac{3\sqrt m+1}2\le 2\sqrt m $$

And with that, we get that the number of cliques of size $\ge 2$ is upper bounded by $$ 2^{2\sqrt m+1}=2\cdot 4^\sqrt m $$


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