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Lecture 24 Vertex Cover and Hamiltonian Cycle. Vertex Cover. Given a graph G=(V,E), find a minimum subset C of vertices such that every edge is incident to a vertex in C. Decision Version. Given a graph G=(V,E) and positive integer k < |V|, is there a vertex cover C of size at most k?.
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Vertex Cover • Given a graph G=(V,E), find a minimum subset C of vertices such that every edge is incident to a vertex in C.
Decision Version • Given a graph G=(V,E) and positive integer k < |V|, is there a vertex cover C of size at most k?.
Vertex-Cover is NP-complete Proof.
2-approximation • The vertex set of a maximal matching gives 2-approximation, i.e., approx / opt < 2
ρ-Approximation • ρ-approximation is apolynomial-timeapproximation satisfying: 1 < approx(input)/opt(input) <ρ for MIN or 1 < opt(input)/approx(input) <ρ for MAX
Max Independent Set • An independent set is a vertex subset such that no edge exists between any two in the subset. • A vertex subset is a vertex cover if and only if its complement is an independent set. • Given a graph, find a maximum independent set.
Complexity and Approximation • Max Independent Set is NP-hard. • For any constant c>1, there does not exist a polynomial-time c-approximation unless NP=P.
Max Clique • A clique is a vertex subset which induces a complete subgraph. • A vertex subset is a clique of graph G if and only if it is an independent set of the complement of G. • :Max Clique: Given a graph, find a maximum clique.
Complexity and Approximation • Max Clique is NP-hard. • For any constant c>1, there does not exist a polynomial-time c-approximation unless NP=P.
Hamiltonian Cycle • Given a graph G, does G contain a Hamiltonian cycle? • Hamiltonian cycle is a cycle passing every vertex exactly once.
Traveling Salesman • Given n cities with a distance table, find a minimum total-distance tour to visit each city exactly once.
HC <m TSP p • From a given graph G, we need to construct (n cities, a distance table, k).
Quiz Sample • Is Traveling Salesman Problem NP-complete? • Answer: No • Because NP contains only decision problems, TSP does not belong to NP.
Quiz Sample • TSP is NP-hard. Does this mean that for any A in NP, A<m TSP? • Answer: No! • It is in wide sense that if TSP can be solved in p.-t., then every problem in NP can be solved in p.-t.. p
Special Case Theorem • Traveling around a minimum spanning tree is a 2-approximation.
Theorem • Minimum spanning tree + minimum-length perfect matching on odd vertices is 1.5-approximation
Minimum perfect matching on odd vertices has weight at most 0.5 opt.