Basic Approximation algorithms

Transcription

1 Approximation slides Basic Approximation algorithms Guy Kortsarz

2 Approximation slides 2 A ρ approximation algorithm for problems that we can not solve exactly Given an NP-hard question finding the optimum seems impossible (that is in polynomial time). Given a candidate solution for an NPC problem its possible to verify in poly time if the solution is valid. Difficulty: To many candidates solutions Let ρ > be some number. Let throughout opt(i) be the optimum value of the solution for I and OPT(I) the solution itself. An ρ approximation algorithm for minimization problem runs in polynomial time and returns for any instance I a solution of cost at most ρ opt(i) for any instance I. For maximization, at least opt(i)/ρ, for any instance.

3 Approximation slides 3 TSP TSP: A metric is a distance between points so that dist(a,b) dist(ac)+dist(c,b). Input: A complete graph with vertices u,u 2...,u n, and a metrics dist(u,v) N+ for every pair u,v. Required: An ordering of the vertices on a circle denoted by {v,v 2,...,v n,v } with v = s Objective function: Minimize n i= dist(v i,v i+ ) Here v n+ = v. Number of candidate solution n!.

4 Approximation slides 4 Finding a lower bound; The metric TSP example The optimum TSP cycle P is an edge plus a spanning tree. Thus, the minimum cost spanning tree T has cost c(t ) c(p ) We use the idea of shortcuts Paths and walks. Path means simple (no vertex appears more than once) but walk may have the same vertices and edges many times. If we are given a cycle walk. Claim: Given an Euler cycle walk that contains all the edges cost c, can be turned into a Hamiltonian path with no larger cost. there exists a TSP path of cost at most c Pf: Shortcut

5 Approximation slides 5 TSP, Cont. Getting an Euler cycle: requires all degrees even Double every edge in MST T Find an Euler cycle (all degrees are even so its possible). Shortcut. T * a 4 a 4 b 3 c 4 2 d b 3 3 c 2 2 d e 2 f e 2 2 f c a b a c d c f c e c c a b d f e

6 Approximation slides 6 Steiner tree Input: A graph G(V,E) a cost function w : E + and a subset S V of terminals Required: A subtree T of G that contains all of S and has minimum cost It is easy to see that we may assume that w is a metric: add/replace the cost function for every e = (u,v). Note that if uv are not neighbors, we can add a new edge uv with a cost tat equals the minimum cost path between u to v path. Clearly this does not increase the optimum. Hence we assume the graph is complete. Thus, if we restrict ourself to G(S) spanned by the terminals S it is also a metric and there exists an Hamiltonian cycle in G(S) of cost 2 opt. Indeed, after doubling the edges and geting an Euler path its possible to shortcut into a path that only contains S. Since we short cut to the cycle it still has cost at most 2 opt.

7 Approximation slides 7 Not knowing the graph in advance: An On-Line approximation for Steiner tree Vertices appear one by one. Say u is the next one. When a vertex appears we get the distances (recall that the graph is a metric) to already added vertices. Let U be vertices that already appear. Since its a metric, we know the distance dist(u,u). Add a shortest path (in the partial graph) between u and U. For simplicity we assume that the terminals are s,s 2,...,s k and P i is the shortest path to the partial graph. Also assume without loss of generality that c(p i ) c(p i+ ). We now show that this algorithm has O(logn) ratio

8 Approximation slides 8 The main claim Claim: the distance of s j+ to the partial graph is at most 2opt/j. Proof: Otherwise we have j vertices s 2,...,s j+. so that c(p i ) > 2opt/j. Say that s i arives before s p for any k,p j. We know that when s p arrives its distance to the partial graph is more than 2opt/j. Thus dist(s p,s k ) > 2opt/j. Since this applies to every two terminals, its means that for every pair of s 2,...,s j,s j+ their distance is more than 2opt/j

9 Approximation slides 9 Getting a contradiction A minimum spanning tree on the above vertices will have j edges each of cost strictly larger than 2opt/j. By the above, the cost of any MST on s 2,...,s j+ more than 2 opt. But above we saw that there is an Hamiltonian cycle of cost at most 2 opt. This is a contradiction.

10 Approximation slides 0 The harmonic number Let k be the number of terminals. Define the nth harmonic number as H k = +/2+/3+...+/(k )+/k. It is known that H k = lnk +Θ(). The Θ() is less than. As vertex s j+ adds 2opt/j to the cost, clearly the cost is at most 2 opt H k. Thus up to low order terms the approximation is 2 lnk On Line ratio.

11 Approximation slides Improving the ratio to 3/2 for Metric TSP Idea due to Christophedes Any TSP solution is a Hamilton path; A simple path that contains all vertices. Any TSP tour of even size decomposes into two perfect matchings

12 Approximation slides 2. Find MST T(V,E ) The algorithm 2. Let X V be the vertices of odd degree in T 3. Compute the graph G X = (X,X X). Find a minimum cost perfect matching M in G X 4. Find an Euler cycle in T M 5. Shortcut Analysis: First, shortcut the tour to X Let M,M 2 be the two perfect matchings of P on X c(p) = c(m )+C(M 2 ) 2c(M) Thus c(t M) 3opt/2 Hence 3/2 ratio Its possible to show that 3/2 is also an upper bound on the integrality gap. In for general metric the best known integrality gap is 4/3.

13 Approximation slides 3 Finding lower bound: The uniform cost vertex-cover example Figure : VC M The size of any matching lower bounds the minimum VC A maximal matching gives a size 2 M vertex-cover Hence, ratio 2.

14 Approximation slides 4 A lower bound is not always required: certificate of failure Say that for input I we want to find a feasible solution minimizing some integral function µ(i). Let opt be opt = mini µ(i). For an integer x say that we have a procedure P(I,x) that has one of the following inputs:. Either it determines that x < opt and then returns F alse 2. Or, it returns a solution S(I) False of cost at most µ(i) ρ x P(I, x) called a certificate of failure procedure (Hochbaum and Shmoys). Claim: The above procedure can be used as an oracle to produce a ρ approximation algorithm.

15 Approximation slides 5 Binary search Assume the costs are integral. lb min, ub max /* lb,ub some lower and upper bounds over minimum and maximum possible value of µ(i) */ While max min > do (a) x (lb+up)/2 (b) If P(I,lb) is false, then lb x (c) Else, ub x 4. Return ub Since opt > lb and opt integer, opt ub. So, ρ ratio

16 Approximation slides 6 Running time Claim: If, P(I,x) is polynomial in I and ub lb always O(exp( I )) then polynomial We now use this to give ratio 2 approximation for the k center problem The input a complete graph on the vertices {,...,n}. A bound k on the number of centers. Every i,j have distance d ij. We assume the triangle inequality (otherwise, no approximation possible).

17 Approximation slides 7 A 2 approximation for undirected k center The algorithm. It is a certificate of failure algorithm with (I,x). Due to Hochbaum and Shmoys.. S 2. V V 3. While S is not a legal solution, do (a) Add to S and arbitrary vertex i V (b) Delete from V all vertices j so that l ij 2 x 4. If S > k return False 5. Else, return S

18 Approximation slides 8 The two properties 2x 2x 2x 2x Figure 2: An illustration of the algorithm in the special case of points in the plane. The distance between any two centers is more than 2x All centers have pairwise distance larger than 2x. Consider our centers that are non-centers in OPT

19 Approximation slides 9 Analysis cont. Let j and p be two centers in our algoritm but not in OPT If q OPT covers both j and p and opt x then by triangle inequality: d j,p d j,q +d p,q 2x. This is a contradiction. If j e.g. computed before p then p is removed So, since S > k the optimum would have more than k, contradiction. Thus S > k implies that there could not be a k subset that covers all of S. In other words, x < opt. This proves that the failure certificate is correct. Remark: It can be that x < opt but the procedure succeeds

20 Approximation slides 20 The two properties. Continued If returns a solution then by construction the solution has radios 2x. Hence ρ = 2 We can use binary search, as opt max distance. Thus, 2 approximation Also better than 2 is as hard as solving: The dominating set problem: Input: G(V,E) and k Question: Is there a dominating set of size at most k, namely, a subset U V, U k so that U N(U) = V? This problem is NPC.

21 Approximation slides 2 Why a ratio better than 2 is not possible Give edges of G costs and non-edges in V V \E costs 2 This implies the triangle inequality holds There is a dominating set of size k if and only if there is a k center solution of size k Approximating within 2 ǫ implies in the case of a yes instance an optimal solution Figure 3: Edges are given cost. Non-edges are gives cost 2. Not all non-edges are shown