1 Introduction. 1. Prove the problem lies in the class NP. 2. Find an NP-complete problem that reduces to it.

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1 1 Introduction There are hundreds of NP-complete problems. For a recent selection see csc.liv.ac.uk/ ped/teachadmin/comp202/annotated_np.html Also, see the book M. R. Garey and D. S. Johnson. Computers and intractability. A Guide to the theory of NP-completeness. WH Freeman and Company, New York, Historically, graph theoretic problems were among the earliest to be proved NP-complete. We ll start by looking at one of the very first such problems: the Vertex Cover problem. All NP-completeness results are proved in the same way 1. Prove the problem lies in the class NP 2. Find an NP-complete problem that reduces to it. Figure 1: Motivation Example: Using vertex cover algorithm to simulate the propagation of worms on computer networks and design optimal strategies for protecting the network against virus attacks. The set {2, 4, 5} is a minimum vertex cover in this computer network. Figure from 1

2 2 The Vertex Cover problem Definition 2.1 (Vertex Cover). Given a graph (V, E) and an integer k does there exist a set C of k vertices such that every edge has one (or both) endpoints in the set C? The set C is called a vertex cover. Theorem 2.2. The vertex cover problem is NP-complete. Proof. The first part of the proof is usually easiest. 2.1 Vertex cover in N P It is fairly obvious that the vertex cover problem is in N P. Simply choose a set of k vertices non-deterministically then check if all edges are covered by those vertices (clearly polynomial). 2.2 Reduce 3-SAT to VC For the second step we must find some suitable NP-complete problem at this stage in our experience there isn t much choice!! This means: find a function f that maps instances of 3-SAT into instances of VC such that Given an arbitrary instance I of 3-SAT f(i) will be an instance of VC, i.e. f(i) will be a graph G together with an integer k, i.e. f(i) = (G, k) f(i) can be computed in polynomial time in I f(i) will be a positive instance if and only if I is a positive instance (i.e. G will have a vertex cover of k vertices if and only if the clauses of I are satisfiable) We now define the function f by a construction that, given I, produces G and k 2

3 Suppose the variables of I are x 1, x 2,..., x n and there are m clauses in the boolean formula: (u 11 u 12 u 13 ) (u 21 u 22 u 23 )... (u m1 u m2 u m3 ) The variables u ij are literals and could be any one of x k. The i, j labels refer to the particular position of the literal in the boolean formula Making G There is one vertex for every x i, one vertex for every x i, and one vertex for every u ij. There is an edge from vertex x i to vertex x i (all values of i). There is an edge between vertex u ij and vertex u ik (all values of i and all values j, k with j k). There is an edge from x i to u jk exactly when x i = u jk, and an edge from x i to u jk exactly when x i = u jk. See Figure 2 for an example of G Making k Define k = n + 2m. All this can be done in time polynomial in the size of I (in fact in time O(mn) which is quadratic in I ) The clauses are satisfiable = there is a vertex cover Suppose we have a satisfying assignment. We find the vertex cover as follows: if x i = T put vertex x i in the vertex cover if x i = F put vertex x i in the vertex cover At this point we have used up n vertices and have covered the n edges from x i to x i (and some others too). 3

4 x 1 x 1 x 2 x 2 x 3 x 3 x 4 x 4 u 12 u 22 u 11 u 13 u 21 u 23 k = n + 2m = = 8 Figure 2: Reduction of (x 1 x 2 x 3 ) ( x 1 x 2 x 4 ) For each triple of vertices u j1, u j2, u j3 we have three edges leading to an x or a x. At least one of these will already be covered; the one corresponding to the true value in the clause. To cover the two remaining edges of the three, and the edges of the triangle u j1, u j2, u j3 we put an appropriate two vertices of the triangle in the vertex cover There is a vertex cover = the clauses are satisfiable Suppose we have a vertex cover. From the form of the graph the cover must have exactly one vertex from each pair {x i, x i }. That s n of the vertices. The other 2m vertices must be used 2 per triangle to cover the triangle edges. We consider the truth assignment Define x i = T if vertex x i is in the cover. 4

5 x 1 x 1 x 2 x 2 x 3 x 3 x 4 x 4 u 12 u 22 u 11 u 13 u 21 u 23 k = n + 2m = = 8 Figure 3: Vertex cover corresponding to x 1 = T, x 2 = F, x 3 = T, x 4 = F Define x i = F if vertex x i is in the cover. Consider the clause corresponding to the vertices u j1, u j2, u j3 and the vertex of this trio not in the vertex cover. That vertex is connected to a vertex x i or x i that must be in the cover which therefore has the value T; so the clause contains a true value. 3 Question Again When faced with a problem, why do we need so much effort to determine whether it is NP-hard (or not)? 5

6 4 Tutorial Problems Question 1 Consider the conjunction of clauses (x 1 x 2 x 3 ) ( x 1 x 2 ) (x 1 x 2 x 3 ) Give the graph constructed in the reduction of 3-SAT to VC. Find a vertex cover of this graph with at most 9 vertices and read off a satisfying assignment. Question 2 Consider the following two problems: UHAM: an instance of this problem is an undirected graph. The decision question is whether G has a Hamiltonian circuit (a sequence of edges that visits all the vertices once only, returning to the initial vertex). DHAM: an instance of this problem is a directed graph. The decision question is whether G has a Hamiltonian circuit (a sequence of directed edges that visits all the vertices once only, returning to the initial vertex). 1. Prove that both these problems are in NP. 2. Find a polynomial time reduction of UHAM to DHAM. 3. Find a polynomial time reduction of DHAM to UHAM. Notes: The first of these is easy no hints, except to remind you that to be in NP is simply to have a polynomial time guess and check algorithm. For the second and third you have to take an instance of one and, in polynomial time, produce an instance of the other with the same answer. The second is easy no further hints at all! 6

7 The third is not too bad: think about replacing a vertex with three vertices joined in a path. Actually, both these problems are N P -complete (there is a polynomial time reduction of 3-SAT to DHAM in the textbook). Here s a similar looking problem: Question 3 EulerPath: Given a graph, find a sequence of edges giving you a path through the graph that visits every edge exactly once, returning to the start. What are your thoughts? In P? In N P? N P -complete? 7