Instance: An undirected graph $G$ and a positive integer $k$

Question: Does $G$ contain a vertex cover of size $\leq k$ or a clique of size $\geq k$?

Obviously, this problem is solved by polynomial reduction, but is it from Clique or Vertex Cover? And how?

I've tried to reduce from both problems, but seem to get stuck. If I reduce from Clique, for example, It seems I'd need to guarantee that there isn't a Vertex Cover, but it's not clear to me how to do that. It seems reducing from Vertex Cover is more promising, but I run into the same issue


1 Answer 1


You can easily reduce from clique as follows.

First, notice that the clique problem remains NP-hard even if we restrict $k$ to lie in $3 \leq k \leq n$ (because outside this range the problem is trivially solvable in polynomial time).

Given a graph $G$ on $n$ vertices and $3\leq k \leq n$, construct a graph $G'$ by taking the disjoint union of $G$ with a perfect matching on $2(n+1)$ additional vertices.

Now, $G$ has a clique of size $\geq k$ if and only if $G'$ has a clique of size $\geq k$. This is because if we find a clique of size $\geq k\geq 3$ in $G$ then it cannot lie in the matching part, as this has no cliques of size greater than $2$. Moreover, $G'$ has no vertex cover of size $\leq k$, as you need at least $n+1$ vertices to cover the matching. Thus, $G$ has a clique of size $k$ if and only if $G'$ has a clique of size $\geq k$ or a vertex cover of size $\leq k$ (this second condition always being false).


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