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550 Chap. 17 Limits to ComputationExample 17.2 In this example, we make use of a simple conversion betweentwo graph problems.Theorem 17.2 VERTEX COVER is N P-complete.Proof: Prove that VERTEX COVER is in N P: Simply guess a subsetof the graph and determine in polynomial time whether that subset is in facta vertex cover of size k or less.Prove that VERTEX COVER is N P-hard: We will assume that K-CLIQUE is already known to be N P-complete. (We will see this proof inthe next example. For now, just accept that it is true.)Given that K-CLIQUE is N P-complete, we need to find a polynomialtimetransformation from the input to K-CLIQUE to the input to VERTEXCOVER, and another polynomial-time transformation from the output forVERTEX COVER to the output for K-CLIQUE. This turns out to be asimple matter, given the following observation. Consider a graph G anda vertex cover S on G. Denote by S ′ the set of vertices in G but not in S.There can be no edge connecting any two vertices in S ′ because, if therewere, then S would not be a vertex cover. Denote by G ′ the inverse graphfor G, that is, the graph formed from the edges not in G. If S is of sizek, then S ′ forms a clique of size n − k in graph G ′ . Thus, we can reduceK-CLIQUE to VERTEX COVER simply by converting graph G to G ′ , andasking if G ′ has a VERTEX COVER of size n − k or smaller. If YES, thenthere is a clique in G of size k; if NO then there is not.✷Example 17.3 So far, our N P-completeness proofs have involved transformationsbetween inputs of the same “type,” such as from a Boolean expressionto a Boolean expression or from a graph to a graph. Sometimes anN P-completeness proof involves a transformation between types of inputs,as shown next.Theorem 17.3 K-CLIQUE is N P-complete.Proof: K-CLIQUE is in N P, because we can just guess a collection of kvertices and test in polynomial time if it is a clique. Now we show that K-CLIQUE is N P-hard by using a reduction from SAT. An instance of SATis a Boolean expressionB = C 1 · C 2 · ... · C mwhose clauses we will describe by the notationC i = y[i, 1] + y[i, 2] + ... + y[i, k i ]