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Question: In an undirected graph G=(V,E), we say DVis a dominating set if every vV is either in D or adjacent to at least one member of D. In the DOMINATING SET problem, the input is a graph and a budget , and the aim is to find a dominating set in the graph of size at most , if one exists. Prove that this problem is NP-complete.

Short Answer

Expert verified

Any vertex cover problem can be reduced to the dominance set problem.

Step by step solution

01

Define vertex cover problem

The vertex Cover problem for an undirected graphG(V,E) , is said to beSV vertex cover, if and only if any edge of graph G is adjacent to at least one vertex in S. It is NP-complete problem.

Any vertex cover problem is reduced to the dominating set problem

02

Prove that the graph  as NP-Complete problem

Consider an undirected graph G, for any side uv of the graph G , add a point t so that the two vertices u and v of the side are in phase with t respectively. Get a new undirected graph G'.

If the original undirected G graph has a vertex cover S , and S satisfies|S|b , then can also be used as a dominating set of graphs G' that satisfies the conditions.

Assume is not the dominating set of the graph G', that is, there is a point tV(G')so that tSand t is not adjacent to any vertex in s . Since the graph g is connected, there must be edges (u,t) satisfying uS.

The edge ( u,t ) bis an auxiliary edge added to the original graph.

When the edge corresponding to tis u,v, is a cover of G, and the vertex u of the u,vedge does not belong toS, then vertex vmust belong to S.

If is adjacent tov , then it Sis proved that if is a vertex cover of the undirected graphG, and |S|b, thenS is the dominating set of the graphG' .

Therefore, any vertex cover problem can be reduced to the dominance set problem.

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Most popular questions from this chapter

On page 266we saw that 3SATremainsNP-complete even when restricted to formulas in which each literal appears at most twice.

(a)Show that if each literal appears at mostonce,then the problem is solvable in polynomial time.

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