**Prerequisite:** NP-Completeness

A clique is a subgraph of a graph such that all the vertices in this subgraph are connected with each other that is the subgraph is a complete graph. The Maximal Clique Problem is to find the maximum sized clique of a given graph G, that is a complete graph which is a subgraph of G and contains the maximum number of vertices. This is an optimization problem. Correspondingly, the Clique Decision Problem is to find if a clique of size k exists in the given graph or not.

To prove that a problem is NP-Complete, we have to show that it belongs to both NP and NP-Hard Classes. (Since NP-Complete problems are NP-Hard problems which also belong to NP)

**The Clique Decision Problem belongs to NP** – If a problem belongs to the NP class, then it should have polynomial-time verifiability, that is given a certificate, we should be able to verify in polynomial time if it is a solution to the problem.

**Proof:**

__Certificate__– Let the certificate be a set S consisting of nodes in the clique and S is a subgraph of G.__Verification__– We have to check if there exists a clique of size k in the graph. Hence, verifying if number of nodes in S equals k, takes O(1) time. Verifying whether each vertex has an out-degree of (k-1) takes O(k^{2}) time. (Since in a complete graph, each vertex is connected to every other vertex through an edge. Hence the total number of edges in a complete graph =^{k}C_{2}= k*(k-1)/2 ). Therefore, to check if the graph formed by the k nodes in S is complete or not, it takes O(k^{2}) = O(n^{2}) time (since k<=n, where n is number of vertices in G).

Therefore, the Clique Decision Problem has polynomial time verifiability and hence belongs to the NP Class.

**The Clique Decision Problem belongs to NP-Hard **– A problem L belongs to NP-Hard if every NP problem is reducible to L in polynomial time. Now, let the Clique Decision Problem by C. To prove that C is NP-Hard, we take an already known NP-Hard problem, say S, and reduce it to C for a particular instance. If this reduction can be done in polynomial time, then C is also an NP-Hard problem. The Boolean Satisfiability Problem (S) is an NP-Complete problem as proved by the Cook’s theorem. Therefore, every problem in NP can be reduced to S in polynomial time. Thus, if S is reducible to C in polynomial time, every NP problem can be reduced to C in polynomial time, thereby proving C to be NP-Hard.

__Proof that the Boolean Satisfiability problem reduces to the Clique Decision Problem__

Let the boolean expression be – F = (x_{1} v x_{2}) ^ (x_{1}‘ v x_{2}‘) ^ (x_{1} v x_{3}) where x_{1}, x_{2}, x_{3} are the variables, ‘^’ denotes logical ‘and’, ‘v’ denotes logical ‘or’ and x’ denotes the complement of x. Let the expression within each parentheses be a clause. Hence we have three clauses – C_{1}, C_{2} and C_{3}. Consider the vertices as – <x_{1}, 1>; <x_{2}, 1>; <x_{1}’, 2>; <x_{2}’, 2>; <x_{1}, 3>; <x_{3}, 3> where the second term in each vertex denotes the clause number they belong to. We connect these vertices such that –

- No two vertices belonging to the same clause are connected.
- No variable is connected to its complement.

Thus, the graph G (V, E) is constructed such that – V = { <a, i> | a belongs to C_{i} } and E = { ( <a, i>, <b, j> ) | i is not equal to j ; b is not equal to a’ } Consider the subgraph of G with the vertices <x_{2}, 1>; <x_{1}’, 2>; <x_{3}, 3>. It forms a clique of size 3 (Depicted by dotted line in above figure) . Corresponding to this, for the assignment – <x_{1}, x_{2}, x_{3}> = <0, 1, 1> F evaluates to true. Therefore, if we have k clauses in our satisfiability expression, we get a max clique of size k and for the corresponding assignment of values, the satisfiability expression evaluates to true. Hence, for a particular instance, the satisfiability problem is reduced to the clique decision problem.

Therefore, the Clique Decision Problem is NP-Hard.

__Conclusion__

The Clique Decision Problem is NP and NP-Hard. Therefore, the Clique decision problem is NP-Complete.

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