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?.

Vertex-Cover is NP-complete Proof.

2-approximation The vertex set of a maximal matching gives 2-approximation, i.e., approx / opt < 2

Performance Ratio

ρ-Approximation ρ-approximation is a polynomial-time approximation 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.

HC is NP-complete

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.