Approximation Algorithms

Approximation Algorithms Load Balancing k-center selection Pricing Method Vertex Cover Set Cover Bin Packing TSP

Approximation Algorithms Q. Suppose I need to solve an NP-hard problem. What should I do? A. Theory says you're unlikely to find a poly-time algorithm. Must sacrifice one of three desired features. • Solve problem to optimality. • Solve problem in poly-time. • Solve arbitrary instances of the problem. -approximation algorithm. • Guaranteed to run in poly-time. • Guaranteed to solve arbitrary instance of the problem • Guaranteed to find solution within ratio  of true optimum. Challenge. Need to prove a solution's value is close to optimum, without even knowing what optimum value is!

11.1 Load Balancing

Load Balancing Input. m identical machines; n jobs, job j has processing time tj. • Job j must run contiguously on one machine. • A machine can process at most one job at a time. Def. Let J(i) be the subset of jobs assigned to machine i. The load of machine i is Li = j  J(i) tj. Def. The makespan is the maximum load on any machine L = maxi Li. Load balancing. Assign each job to a machine to minimize makespan.

Load Balancing: List Scheduling List-scheduling algorithm. • Consider n jobs in some fixed order. • Assign job j to machine whose load is smallest so far. Implementation.O(n log n) using a priority queue. List-Scheduling(m, n, t1,t2,…,tn) { for i = 1 to m { Li  0 J(i)  } for j = 1 to n { i = argmink Lk J(i) J(i) {j} Li  Li + tj } } load on machine i jobs assigned to machine i machine i has smallest load assign job j to machine i update load of machine i

Load Balancing: List Scheduling Analysis Theorem. [Graham, 1966]Greedy algorithm is a 2-approximation. • First worst-case analysis of an approximation algorithm. • Need to compare resulting solution with optimal makespan L*. Lemma 1. The optimal makespan L*  maxj tj. Pf. Some machine must process the most time-consuming job. ▪ Lemma 2. The optimal makespan Pf. • The total processing time is j tj . • One of m machines must do at least a 1/m fraction of total work. ▪

Load Balancing: List Scheduling Analysis Theorem. Greedy algorithm is a 2-approximation. Pf. Consider load Li of bottleneck machine i. • Let j be last job scheduled on machine i. • When job j assigned to machine i, i had smallest load. Its load before assignment is Li - tj  Li - tj  Lk for all 1  k  m. blue jobs scheduled before j machine i j 0 Li - tj L = Li

Load Balancing: List Scheduling Analysis Theorem. Greedy algorithm is a 2-approximation. Pf. Consider load Li of bottleneck machine i. • Let j be last job scheduled on machine i. • When job j assigned to machine i, i had smallest load. Its load before assignment is Li - tj  Li - tj  Lk for all 1  k  m. • Sum inequalities over all k and divide by m: • (correct the second eqn. to j) • Now ▪ Lemma 1 Lemma 2

Load Balancing: List Scheduling Analysis Q. Is our analysis tight? A. Essentially yes. Ex: m machines, m(m-1) jobs length 1 jobs, one job of length m machine 2 idle machine 3 idle machine 4 idle m = 10 machine 5 idle machine 6 idle machine 7 idle machine 8 idle machine 9 idle machine 10 idle list scheduling makespan = 19

Load Balancing: List Scheduling Analysis Q. Is our analysis tight? A. Essentially yes. Ex: m machines, m(m-1) jobs length 1 jobs, one job of length m m = 10 optimal makespan = 10

Load Balancing: LPT Rule Longest processing time (LPT). Sort n jobs in descending order of processing time, and then run list scheduling algorithm. LPT-List-Scheduling(m, n, t1,t2,…,tn) { Sort jobs so that t1 ≥ t2 ≥…≥ tn for i = 1 to m { Li  0 J(i)  } for j = 1 to n { i = argmink Lk J(i) J(i) {j} Li  Li + tj } } load on machine i jobs assigned to machine i machine i has smallest load assign job j to machine i update load of machine i

Load Balancing: LPT Rule Observation. If at most m jobs, then list-scheduling is optimal. Pf. Each job put on its own machine. ▪ Lemma 3. If there are more than m jobs, L*  2tm+1. Pf. • Consider first m+1 jobs t1, …, tm+1. • Since the ti's are in descending order, each takes at least tm+1 time. • There are m+1 jobs and m machines, so by pigeonhole principle, at least one machine gets two jobs. ▪ • tj <= t(m+1) <= ½ L* Theorem. LPT rule is a 3/2 approximation algorithm. Pf. Same basic approach as for list scheduling. ▪ Lemma 3( by observation, can assume number of jobs > m )

Load Balancing: LPT Rule Q. Is our 3/2 analysis tight? A. No. Theorem. [Graham, 1969]LPT rule is a 4/3-approximation. Pf. More sophisticated analysis of same algorithm. Q. Is Graham's 4/3 analysis tight? A. Essentially yes. Ex: m machines, n = 2m+1 jobs, 2 jobs of length m+1, m+2, …, 2m-1 and one job of length m.

11.2 Center Selection

r(C) center Center Selection Problem Input. Set of n sites s1, …, sn. Center selection problem. Select k centers C so that maximum distance from a site to nearest center is minimized. k = 4 site

Center Selection Problem Input. Set of n sites s1, …, sn. Center selection problem. Select k centers C so that maximum distance from a site to nearest center is minimized. Notation. • dist(x, y) = distance between x and y. • dist(si, C) = min c  C dist(si, c) = distance from si to closest center. • r(C) = maxi dist(si, C) = smallest covering radius. Goal. Find set of centers C that minimizes r(C), subject to |C| = k. Distance function properties. • dist(x, x) = 0 (identity) • dist(x, y) = dist(y, x) (symmetry) • dist(x, y)  dist(x, z) + dist(z, y)(triangle inequality)

Center Selection Example Ex: each site is a point in the plane, a center can be any point in the plane, dist(x, y) = Euclidean distance. Remark: search can be infinite! r(C) center site

Greedy Algorithm: A False Start Greedy algorithm. Put the first center at the best possible location for a single center, and then keep adding centers so as to reduce the covering radius each time by as much as possible. Remark: arbitrarily bad! greedy center 1 center k = 2 centers site

Center Selection: Greedy Algorithm Greedy algorithm. Repeatedly choose the next center to be the site farthest from any existing center. Observation. Upon termination all centers in C are pairwise at least r(C) apart. Pf. By construction of algorithm. Greedy-Center-Selection(k, n, s1,s2,…,sn) { C =  repeat k times { Select a site si with maximum dist(si, C) Add si to C } return C } site farthest from any center

Center Selection: Analysis of Greedy Algorithm Theorem. Let C* be an optimal set of centers. Then r(C)  2r(C*). Pf. (by contradiction) Assume r(C*) < ½ r(C). • For each site ci in C, consider ball of radius ½ r(C) around it. • Exactly one ci* in each ball; let ci be the site paired with ci*. • Consider any site s and its closest center ci* in C*. • dist(s, C)  dist(s, ci)  dist(s, ci*) + dist(ci*, ci)  2r(C*). • Thus r(C)  2r(C*). ▪ -inequality  r(C*) since ci* is closest center ½r(C) ½r(C) ci ½r(C) C* ci* sites s

Center Selection Theorem. Let C* be an optimal set of centers. Then r(C)  2r(C*). Theorem. Greedy algorithm is a 2-approximation for center selection problem. Remark. Greedy algorithm always places centers at sites, but is still within a factor of 2 of best solution that is allowed to place centers anywhere. Question. Is there hope of a 3/2-approximation? 4/3? e.g., points in the plane Theorem. Unless P = NP, there no -approximation for center-selectionproblem for any  < 2.

11.4 The Pricing Method: Vertex Cover

Weighted Vertex Cover Weighted vertex cover. Given a graph G with vertex weights, find a vertex cover of minimum weight. 2 4 2 4 2 9 2 9 weight = 9 weight = 2 + 2 + 4

Weighted Vertex Cover Pricing method. Each edge must be covered by some vertex i. Edge e pays price pe 0 to use vertex i. Fairness. Edges incident to vertex i should pay  wiin total. Claim. For any vertex cover S and any fair prices pe: epe  w(S). Proof. ▪ 2 4 2 9 each edge e covered byat least one node in S sum fairness inequalitiesfor each node in S

Pricing Method Pricing method. Set prices and find vertex cover simultaneously. Weighted-Vertex-Cover-Approx(G, w) { foreach e in E pe = 0 while( edge i-j such that neither i nor j are tight) select such an edge e increase pe without violating fairness } S  set of all tight nodes return S }

Pricing Method price of edge a-b vertex weight Figure 11.8

Pricing Method: Analysis Theorem. Pricing method is a 2-approximation. Pf. • Algorithm terminates since at least one new node becomes tight after each iteration of while loop. • Let S = set of all tight nodes upon termination of algorithm. S is a vertex cover: if some edge i-j is uncovered, then neither i nor j is tight. But then while loop would not terminate. • Let S* be optimal vertex cover. We show w(S)  2w(S*). S  V,prices  0 fairness lemma all nodes in S are tight each edge counted twice

Extra Slides

Load Balancing on 2 Machines Claim. Load balancing is hard even if only 2 machines. Pf. NUMBER-PARTITIONING PLOAD-BALANCE. NP-complete by Exercise 8.26 a b c d e f g length of job f Machine 1 a d f machine 1 yes machine 2 b Machine 2 c e g 0 Time L

Center Selection: Hardness of Approximation Theorem. Unless P = NP, there is no -approximation algorithm formetric k-center problem for any  < 2. Pf. We show how we could use a (2 - ) approximation algorithm for k-center to solve DOMINATING-SET in poly-time. • Let G = (V, E), k be an instance of DOMINATING-SET. • Construct instance G' of k-center with sites V and distances • d(u, v) = 2 if (u, v)  E • d(u, v) = 1 if (u, v)  E • Note that G' satisfies the triangle inequality. • Claim: G has dominating set of size k iff there exists k centers C* with r(C*) = 1. • Thus, if G has a dominating set of size k, a (2 - )-approximation algorithm on G' must find a solution C* with r(C*) = 1 since it cannot use any edge of distance 2. see Exercise 8.29

Vertex Cover Approximation A vertex cover is a subset of vertices such that every edge in the graph is incident to at least one of these vertices. The vertex cover optimization problem is to ﬁnd a vertex cover of minimum size. For a good strategy, a heuristic is needed

Vertex Cover Consider an arbitrary edge (u, v) in the graph. One of its two vertices must be in the cover, but we do not know which one. The idea of this heuristic is to simply put both vertices into the vertex cover. Then we remove all edges that are incident to u and v (since they are now all covered), and recurse on the remaining edges. For every one vertex that must be in the cover, we put two into our cover, so it is easy to see that the cover we generate is at most twice the size of the optimum cover.

Proof of aprroximation ratio Claim :approx VC yields a factor-2 approximation Proof: Consider the set C output by ApproxVC. Let C*be the optimum VC. Let A be the set of edges selected by the line marked with “(*)” in the ﬁgure. Observe that the size of C is exactly 2|A|because we add two vertices for each such edge. However note that in the optimum VC one of these two vertices must have been added to the VC, and thus the size of C*is at least |A|. Thus we have: |C| ---- = |A| <= |C*| 2 Therefore : |C| ---- <= 2 |C*|

Example

Approximate VC Algorithm : Naive Approach ApproxVC{ C = empty-set while (E is nonempty) do { (*) let (u,v) be any edge of E addboth u and v to C remove from E all edges incident to either u or v } return C; } Can weimprove on it ? Why not considerverticeswithhigherdegreesfirst (GreedyStrategy)

Greedy VC Greedy Approximation for VC GreedyVC(G=(V,E)) { C = empty-set; while (E is nonempty) do { let u be the vertex of maximum degree in G; addu to C; remove from E all edges incident to u; } return C; } For the example, it yields the optimum solution

Greedy VC Example Can weproveGreedy VC outperformstheotherone ? NO ! It can evenperformpoorlythan it. However, it should also be pointed out that the vertex cover constructed by the greedy heuristic is (for typical graphs) smaller than that one computed by the 2-for-1 heuristic, so it would probably be wise to run both algorithms and take the better of the two.

Third Attempt: Use Matching A matching is a subset of edges that have no vertices in common A matching is maximal if no more edges can be added to it. Maximal matchings will help us ﬁnd good vertex covers, and moreover,they are easy to generate: repeatedly pick edges that are disjoint from the ones chosen already, until this is no longer possible. Any vertex cover of a graph G must be at least as large as the number of edges in any matching in G; that is, any matching provides a lower bound on OPT. This is simply because each edge of the matching must be covered by one of its endpoints in any vertex cover!

Example • Figure below shows how to convert from Maximal Matching to Vertex Cover • a) A matching b) Completion to MaxMatch c) Its VC

Vertex Cover from Matching • let S be a set that contains both endpoints of each edge in a maximal matching M. • Then S must be a vertex cover—if it isn’t, that is, if it doesn’t touch some edge e ∪ E, then M could not possibly be maximal since we could still add e to it. But our cover S has 2|M| vertices • We know that any vertex cover must have size at least |M|. Algorithm : Find a maximal matching M ∞ E Return S = {all endpoints of edges in M}

Vertex cover from Matching • This simple procedure always returns a vertex cover whose size is at most twice optimal! • In summary, even though we have no way of ﬁnding the best vertex cover, we can easily ﬁnd another structure, a maximal matching, with two key properties: 1. Its size gives us a lower bound on the optimal vertex cover. 2. It can be used to build a vertex cover, whose size can be related to that of the optimal cover using property 1. Alpha <= 2

Set Cover Problem Revisited • Given a pair (X,F) where X ={x1,x2,...,xm} is a ﬁnite set (a domain of elements) and F ={S1,S2,...,Sn} is a family of subsets of X, such that every element of X belongs to at least one set of F. • For C ∈ F. (This is a collection of sets over X.) We say that C covers the domain if every element of X is in some set of C • The problem is to ﬁnd the minimum-sized subset C of F that covers X.

Set Cover • Vertex Cover is a type of set cover problem. The domain to be covered are the edges, and each vertex covers the subset of incident edges. • Decision-problem formulation of set cover (“does there exist a set cover of size at most k?”) is NP-complete • There is a factor-2 approximation for the vertex cover problem, but it cannot be applied to generate a factor2 approximation for set cover.

Set Cover • It is known that there is no constant factor approximation to the set cover problem • There is however the greedy heuristic, which achieves an approximation bound of ln m, where m = |X|, the size of the underlying domain, we will leave the proof. • A simple greedy approach to set cover works by at each stage selecting the set that covers the greatest number of “uncovered” elements

Set Cover : The Approx. Algorithm Greedy-Set-Cover(X, F) { U = X // U are the items to be covered C = empty // C will be the sets in the cover while (U is nonempty) { // there is someone left to cover • select S in F that covers the most elements of U • addS to C • U=U-S } return C }

Set Cover : Bad Example • The optimal set cover consists ofsets S5 and S6, each of size 16. Initially all three sets S1, S5, and S6 have 16 elements. If ties are broken in theworst possible way, the greedy algorithm will first select set S1. We remove all the covered elements. Now S2,S5 and S6 all cover 8 of the remaining elements. Again, if we choose poorly, S2 is chosen. The pattern repeats,choosing S3 (size 4), S4 (size 2) and finally S5 and S6 (each of size 1).

Bin Packing • Bin packing is another well-known NP-complete problem, which is a variant of the knapsack problem • Given a set of n objects, where si denotes the size of the ith object (0 < si < 1.for simplification) , put objects into bins • Size of a bin is 1 at max. • Use fewest bins as possible • Ex : Fit object sto a truck etc.

Bin Packing : Example

Bin Packing : Approximation Factor • Theorem: The first-fit heuristic achieves a ratio bound of 2. • Proof: Consider an instance {s1,...,sn} of the bin packingproblem. Let S =Σ i si denote the sum of all the object sizes. Let b*denote the optimal number of bins, and bff denote the number of bins used by first-fit. • b*>= S : since no bin can hold a total capacity of more than 1 unit, and evenif we were to fill each bin exactly to its capacity, we would need at least S bins

Bin Packing : Analysis • We claim that bff <= 2S. • To see this, let ti denote the total size of the objects that first-fit puts intobin i. • Consider bins i and i + 1 filled by first-fit. Assume that indexing is cyclical, so if i is the last index(i = bff ) then i+1 = 1. • We claim that ti + ti +1> = 1. If not, then the contents of bins i and i + 1 couldboth be put into the same bin, and hence first-fit would never have started to fill the second bin, preferringto keep everything in the first bin. Thus we have: • bff Σi=1 ( ti + ti+1) > = bff

Approximation Algorithms