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Input: A graph$G\left(V,E\right)$ .

Here,V is the set of vertices and E is the set of edges.

Output: A clique and independent set, both of size K , if exist.

To prove: The given problem in NP- complete.

To prove that the given problem is in NP, verifier takes the graph G, K and set S and checks that if size of subset S is greater than or equal to and not in. It also checks if every edge of belongs to set E of the graph G.

This is verified in polynomial time. Thus, this problem is in NP.

3-SAT is an NP-complete problem, it is reduced to clique and thus to independent set problem as well.

Consider the example:

$\left(\stackrel{炉}{x}鈭�y鈭�\stackrel{炉}{z}\right)\left(x鈭�\stackrel{炉}{y}鈭�z\right)\left(x鈭�y鈭�z\right)$

In these clauses, see them as the vertices of the graph. Choose one literal from each clause and store that to a set S . The set is the independent set of a graph. Compare if S is equal to K or not. This is done in polynomial time.

Similarly, independent set $\stackrel{炉}{G}$of which is the clique of G is also calculated in polynomial time. Thus, the problem is reducible from 3-SAT, it is proved that it is at least hard as 3-SAT and hence it is in NP-Hard.

Since, given problem is in NP and NP-Hard, it is NP-Complete.