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Recall that $G$ has a clique of size $k$ if it has a complete sub graph consisting of $k$ vertices. Define CLIQUE as the decision problem

$$\{ \langle G, c \rangle \mid G \text{ has a clique of size } c\}$$

Define the problem $\sqrt{n}$-CLIQUE as follows:

$$\{\langle G \rangle \mid G \text{ has a clique of at least size } \sqrt{n}\}$$

where $n$ is the number of vertices of $G$. It is easy to reduce $\sqrt{n}$-CLIQUE to CLIQUE. How can we go the other way and thereby show that $\sqrt{n}$-CLIQUE is NP-complete?

Idea: If $c \geq \sqrt{n}$, we can add dummy vertices to $G$ until $c = \sqrt{n}$. What do we do if $c < \sqrt{n}$? It seems we need to be able to remove vertices without disturbing the size of the largest clique. My idea in this case to remove a vertex if it has less than $c-1$ edges. Obviously the new graph has a clique of size $c$ if and only if the original graph does. But what happens if we can't remove enough vertices? Can we conclude that if a graph has $n > c^2$ vertices each with degree $\geq c$ then a clique exists?

Raphael
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MT_
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2 Answers2

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When $c > \sqrt{n}$, you add an independent set of size $m$ so that $c = \sqrt{n+m}$ (i.e., you need $m = c^2-n$).

When $c < \sqrt{n}$, try doing the same, increasing both $n$ and $c$ at the same time, by adding a complement of an independent set. (You might need to add a few isolated vertices as well.)

Yuval Filmus
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Below is a different reduction from CLIQUE to $\sqrt{n}$-CLIQUE which is essentially an extension of the reduction from CLIQUE to $\frac{n}{2}$-CLIQUE.

Given an instance $(G=(V,E),k)$ of CLIQUE, we construct an instance $G'=(V',E')$ of $\sqrt{n}$-CLIQUE as follows. Suppose $V=\{v_1,...,v_n\}$, then $G'$ is defined as follows $$V'=V\cup\{u_1,...,u_{n-k}\}\cup\{w_1,...,w_{n^2-2n+k}\}$$ $$E'=E\cup\{\{u_i,u_j\}\mid \forall i,j\in[n-k]\}\cup\{\{v_{i},u_j\}\mid \forall i\in[n],\forall j\in[n-k]\}$$ Intuitively, we are constructing $G'$ to have one copy of $G$ which is fully connected to a $K_{n-k}$ clique, and we pad the number of vertices to $n^2$ using isolated vertices.

Claim: $(G=(V,E),k)\in$ CLIQUE if and only if $G'\in\sqrt{n}$-CLIQUE.

proof. $\Longrightarrow$ If $(G=(V,E),k)\in$ CLIQUE, then the copy of $G$ in $G'$ has a clique of size $\geq k$. Since $G$ in $G'$ is fully connected to $K_{n-k}$, $G'$ will have a clique of size $\geq (n-k)+k=n$.

$\Longleftarrow$ If $G'\in\sqrt{n}$-CLIQUE, then suppose for contradiction that $G$ has all cliques of size $<k$. This implies that $G'$ has all cliques of size $<(n-k)+k=n$. A contradiction. Thus $G$ has a clique of size $\geq k$.

Finally, it is easy to construct $G'$ in poly$(n)$-time. So we are done.

Tom Finet
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