I am trying to show that our problem is NP-Complete by reducing the known problem CLIQUE to our problem.

Regular CLIQUE problem:

Input: An undirected graph $G$ and a positive integer $K$.

Goal: Does $G$ have a clique of size at least $K$?

Our problem:

Input: An undirected graph $G$ and a positive integer $K$.

Goal: Is there a group of size $K$ of super connectors in $G$?

A super connector is a node with at least $C$ (i.e., a fixed number of) edges. A group $S$ of super connectors is a set of super connectors all connected to each other.

A group of super connectors is thereby a clique and a return value of TRUE from our problem will result in TRUE in the CLIQUE problem as well.

My idea to solve this is that I modify the graph by removing all nodes that aren't super connectors. We are then left with a graph where every node is super connectors and can then input the graph in the CLIQUE problem to see if there exists a clique and thereby get a solution for our problem. By modifying the graph like this both of the problems will give the same outputs with the same inputs. Am I on the right path or completely wrong?

  • $\begingroup$ Have you tried fixing the value of $C$? $\endgroup$ May 21, 2019 at 15:23
  • $\begingroup$ The task I am working on actually states that C is fixed, but I am not sure how that is different from getting C as an input. $\endgroup$
    – Klasj
    May 21, 2019 at 15:29
  • $\begingroup$ That's different from your post. Which question do you want answered? $\endgroup$ May 21, 2019 at 15:30
  • $\begingroup$ Hint: all vertices in a $k$-clique have degree at least $k-1$. $\endgroup$ May 21, 2019 at 15:30
  • $\begingroup$ I edited the question with C being fixed. $\endgroup$
    – Klasj
    May 21, 2019 at 15:34

1 Answer 1


The simplest NP-hardness reduction is as follows. Suppose we are given an instance $(G,k)$ of CLIQUE. If $k \leq c$ then we solve CLIQUE by brute force in time $O(n^c)$. Otherwise, $G$ has a $k$-clique iff it has a $k$-clique consisting of super-connectors. This is because any vertex in a $k$-clique must have degree at least $k-1 \geq c$, and so is a super-concentrator.


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