4. NP-completeness. Let X be a finite set and F a family of subsets of X such th

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4. NP-completeness. Let X be a finite set and F a family of subsets of X such that every element of X appears in at least one subset in F. We say that a subset C of F is a set cover for X if X -SEc S (that is, the union of the sets in C is X). The cardinality of a set cover C is the number of elements in C. (Note that an element of C is a subset of X.) The Set Cover problem is: Given an in- stance (X,F), and a nonnegative integer k, decide if X has a set cover C in F of cardinality k. For example, if X = { 1 , 2, 3, 4, 5), F= { {1),(2),(3),(25), (1 ,3,5).( 1,2,5),(24.5),{ 1.45)), and k = 2, then the answer is Yes because X has a set cover C in F of cardinality 2, namely C = {(1 , 3, 5), {2, 4, 5)). Show that the Set Cover problem is NP-complete. (Hint. Reduce from Vertex Cover.)

Explanation / Answer

Vertex Cover

Given a Graph G (V,E) and kN, give a Vertex Cover G = (V,K) which is a subset W of V, such that  {i,j} E either and i W or j W.

We shall reduce Set Cover to Vertex Cover and since Vertex Cover is already NP-Complete, we would conclude that Set Cover is NP-Complete as well.

Consider a Graph G = <V,E> with E edges and V vertices.

Now, if we have a set cover C which satisfies the problem <=>, (V-C) is a vertex cover.

1) If C is a Set Cover, there is no edge e = (u,v) in the graph G such that u,v C.

Therefore, for any edge e = (u,v), at least one of u,v must be in (V-C).

=> (V-C) is a vertex cover.

2) If (V-C) is a vertex cover then, between any (u,v) C, if there exists another edge e, none of the endpoints of e should be in (V-C) (It would be against the definition of Vertex Cover). Hence, no pair can be connected to C.

=> C is a Set Cover.

From (1) and (2), Set Cover is a NP-Complete Problem.

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