# Chapter 11. Approximation Algorithms. Slides by Kevin Wayne Pearson-Addison Wesley. All rights reserved.

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1 Chapter 11 Approximation Algorithms Slides by Kevin Wayne Pearson-Addison Wesley. All rights reserved. 1

2 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. ρ-approximation algorithm. Guaranteed to run in poly-time. Guaranteed to solve arbitrary instance of the problem Guaranteed to find solution within ratio ρ of true optimum. Challenge. Need to prove a solution's value is close to optimum, without even knowing what optimum value is! 2

3 11.1 Load Balancing

4 Load Balancing Input. m identical machines; n jobs, job j has processing time t j. 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 L i = Σ j J(i) t j. Def. The makespan is the maximum load on any machine L = max i L i. Load balancing. Assign each job to a machine to minimize makespan. 4

5 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. List-Scheduling(m, n, t 1,t 2,,t n ) { for i = 1 to m { L i 0 load on machine i J(i) φ } jobs assigned to machine i } for j = 1 to n { i = argmin k L k J(i) J(i) {j} L i L i + t j } machine i has smallest load assign job j to machine i update load of machine i Implementation. O(n log m) using a priority queue. 5

6 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* max j t j. Pf. Some machine must process the most time-consuming job. Lemma 2. The optimal makespan Pf. The total processing time is Σ j t j. L * m 1 j t j. One of m machines must do at least a 1/m fraction of total work. 6

7 Load Balancing: List Scheduling Analysis Theorem. Greedy algorithm is a 2-approximation. Pf. Consider load L i 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 L i - t L j i - t j L k for all 1 k m. blue jobs scheduled before j machine i j 0 L i - t j L = L i 7

8 Load Balancing: List Scheduling Analysis Theorem. Greedy algorithm is a 2-approximation. Pf. Consider load L i 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 L i - t L j i - t j L k for all 1 k m. Sum inequalities over all k and divide by m: L i t j 1 m k L k = 1 m k t k L * Now L i = (L i t j ) t j 2L *. { L* Lemma 1 L* Lemma 2 8

9 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 9

10 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 10

11 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, t 1,t 2,,t n ) { Sort jobs so that t 1 t 2 t n for i = 1 to m { L i 0 J(i) φ } load on machine i jobs assigned to machine i } for j = 1 to n { i = argmin k L k J(i) J(i) {j} L i L i + t j } machine i has smallest load assign job j to machine i update load of machine i 11

12 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* 2 t m+1. Pf. Consider first m+1 jobs t 1,, t m+1. Since the t i 's are in descending order, each takes at least t m+1 time. There are m+1 jobs and m machines, so by pigeonhole principle, at least one machine gets two jobs. Theorem. LPT rule is a 3/2 approximation algorithm. Pf. Same basic approach as for list scheduling. L i = (L i t j ) t j { 2 3 L *. L* 1 2 L* Lemma 3 ( by observation, can assume number of jobs > m ) 12

13 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. 13

14 11.2 Center Selection

15 Center Selection Problem Input. Set of n sites s 1,, s n. Center selection problem. Select k centers C so that maximum distance from a site to nearest center is minimized. k = 4 r(c) center site 15

16 Center Selection Problem Input. Set of n sites s 1,, s n. 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(s i, C) = min c C dist(s i, c) = distance from s i to closest center. r(c) = max i dist(s i, 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) 16

17 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 17

18 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 k = 2 centers center site 18

19 Center Selection: Greedy Algorithm Greedy algorithm. Repeatedly choose the next center to be the site farthest from any existing center. Greedy-Center-Selection(k, n, s 1,s 2,,s n ) { } C = φ repeat k times { Select a site s i with maximum dist(s i, C) Add s i to C } site farthest from any center return C Observation. Upon termination all centers in C are pairwise at least r(c) apart. Pf. By construction of algorithm. 19

20 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 c i in C, consider ball of radius ½ r(c) around it. Exactly one c i * in each ball; let c i be the site paired with c i *. Consider any site s and its closest center c i * in C*. dist(s, C) dist(s, c i ) dist(s, c i *) + dist(c i *, c i ) 2r(C*). Thus r(c) 2r(C*). -inequality r(c*) since c i * is closest center ½ r(c) ½ r(c) C* sites ½ r(c) s c i c i * 20

21 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. e.g., points in the plane Question. Is there hope of a 3/2-approximation? 4/3? Theorem. Unless P = NP, there no ρ-approximation for center-selection problem for any ρ < 2. 21

22 11.4 The Pricing Method: Vertex Cover

23 Weighted Vertex Cover Weighted vertex cover. Given a graph G with vertex weights, find a vertex cover of minimum weight weight = weight = 9 23

24 Weighted Vertex Cover Pricing method. Each edge must be covered by some vertex i. Edge e pays price p e 0 to use vertex i. Fairness. Edges incident to vertex i should pay w i in total. 2 4 for each vertex i : e= ( i, j) p e w i 2 9 Claim. For any vertex cover S and any fair prices p e : e p e w(s). Proof. e E p e pe wi = i S e= ( i, j) i S w( S). each edge e covered by at least one node in S sum fairness inequalities for each node in S 24

25 Pricing Method Pricing method. Set prices and find vertex cover simultaneously. Weighted-Vertex-Cover-Approx(G, w) { foreach e in E p e = 0 e = ( i, j) p = e w i while ( edge i-j such that neither i nor j are tight) select such an edge e increase p e without violating fairness } } S set of all tight nodes return S 25

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

27 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*). w(s) = w i = i S i S p e e=(i, j) i V p e e=(i, j) = 2 p e 2w(S*). e E all nodes in S are tight S V, prices 0 each edge counted twice fairness lemma 27

28 11.6 LP Rounding: Vertex Cover

29 Weighted Vertex Cover Weighted vertex cover. Given an undirected graph G = (V, E) with vertex weights w i 0, find a minimum weight subset of nodes S such that every edge is incident to at least one vertex in S. 10 A 6F 9 16 B 7G C H 9 23 D I 33 7 E 10 J 32 total weight = 55 29

30 Weighted Vertex Cover: IP Formulation Weighted vertex cover. Given an undirected graph G = (V, E) with vertex weights w i 0, find a minimum weight subset of nodes S such that every edge is incident to at least one vertex in S. Integer programming formulation. Model inclusion of each vertex i using a 0/1 variable x i. x i = 0 if vertex i is not in vertex cover 1 if vertex i is in vertex cover Vertex covers in 1-1 correspondence with 0/1 assignments: S = {i V : x i = 1} Objective function: maximize Σ i w i x i. Must take either i or j: x i + x j 1. 30

31 Weighted Vertex Cover: IP Formulation Weighted vertex cover. Integer programming formulation. (ILP) min w i x i i V s. t. x i + x j 1 (i, j) E x i {0,1} i V Observation. If x* is optimal solution to (ILP), then S = {i V : x* i = 1} is a min weight vertex cover. 31

32 Integer Programming INTEGER-PROGRAMMING. Given integers a ij and b i, find integers x j that satisfy: max c t x s. t. Ax b x integral n a ij x j b i 1 i m j=1 x j 0 1 j n x j integral 1 j n Observation. Vertex cover formulation proves that integer programming is NP-hard search problem. even if all coefficients are 0/1 and at most two variables per inequality 32

33 Linear Programming Linear programming. Max/min linear objective function subject to linear inequalities. Input: integers c j, b i, a ij. Output: real numbers x j. (P) max c t x s. t. Ax b x 0 (P) max c j x j n j=1 n s. t. a ij x j b i 1 i m j=1 x j 0 1 j n Linear. No x 2, xy, arccos(x), x(1-x), etc. Simplex algorithm. [Dantzig 1947] Can solve LP in practice. Ellipsoid algorithm. [Khachian 1979] Can solve LP in poly-time. 33

34 LP Feasible Region LP geometry in 2D. x 1 = 0 x 2 = 0 2x 1 + x 2 = 6 x 1 + 2x 2 = 6 34

35 Weighted Vertex Cover: LP Relaxation Weighted vertex cover. Linear programming formulation. (LP) min w i x i i V s. t. x i + x j 1 (i, j) E x i 0 i V Observation. Optimal value of (LP) is optimal value of (ILP). Pf. LP has fewer constraints. Note. LP is not equivalent to vertex cover. ½ ½ Q. How can solving LP help us find a small vertex cover? A. Solve LP and round fractional values. ½ 35

36 Weighted Vertex Cover Theorem. If x* is optimal solution to (LP), then S = {i V : x* i ½} is a vertex cover whose weight is at most twice the min possible weight. Pf. [S is a vertex cover] Consider an edge (i, j) E. Since x* i + x* j 1, either x* i ½ or x* j ½ (i, j) covered. Pf. [S has desired cost] Let S* be optimal vertex cover. Then w i i S* * w i x i 1 w 2 i i S i S LP is a relaxation x* i ½ 36

37 Weighted Vertex Cover Theorem. 2-approximation algorithm for weighted vertex cover. Theorem. [Dinur-Safra 2001] If P NP, then no ρ-approximation for ρ < , even with unit weights Open research problem. Close the gap. 37

39 Generalized Load Balancing Input. Set of m machines M; set of n jobs J. Job j must run contiguously on an authorized machine in M j M. Job j has processing time t j. Each 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 L i = Σ j J(i) t j. Def. The makespan is the maximum load on any machine = max i L i. Generalized load balancing. Assign each job to an authorized machine to minimize makespan. 39

40 Generalized Load Balancing: Integer Linear Program and Relaxation ILP formulation. x ij = time machine i spends processing job j. (IP) min L s. t. x i j = t j for all j J i L for all i M j x i j x i j {0, t j } for all j J and i M j x i j = 0 for all j J and i M j LP relaxation. (LP) min L s. t. x i j = t j for all j J i L for all i M j x i j x i j 0 for all j J and i M j x i j = 0 for all j J and i M j 40

41 Generalized Load Balancing: Lower Bounds Lemma 1. Let L be the optimal value to the LP. Then, the optimal makespan L* L. Pf. LP has fewer constraints than IP formulation. Lemma 2. The optimal makespan L* max j t j. Pf. Some machine must process the most time-consuming job. 41

42 Generalized Load Balancing: Structure of LP Solution Lemma 3. Let x be solution to LP. Let G(x) be the graph with an edge from machine i to job j if x ij > 0. Then G(x) is acyclic. Pf. (deferred) can transform x into another LP solution where G(x) is acyclic if LP solver doesn't return such an x x ij > 0 G(x) acyclic job machine G(x) cyclic 42

43 Generalized Load Balancing: Rounding Rounded solution. Find LP solution x where G(x) is a forest. Root forest G(x) at some arbitrary machine node r. If job j is a leaf node, assign j to its parent machine i. If job j is not a leaf node, assign j to one of its children. Lemma 4. Rounded solution only assigns jobs to authorized machines. Pf. If job j is assigned to machine i, then x ij > 0. LP solution can only assign positive value to authorized machines. job machine 43

44 Generalized Load Balancing: Analysis Lemma 5. If job j is a leaf node and machine i = parent(j), then x ij = t j. Pf. Since i is a leaf, x ij = 0 for all j parent(i). LP constraint guarantees Σ i x ij = t j. Lemma 6. At most one non-leaf job is assigned to a machine. Pf. The only possible non-leaf job assigned to machine i is parent(i). job machine 44

45 Generalized Load Balancing: Analysis Theorem. Rounded solution is a 2-approximation. Pf. Let J(i) be the jobs assigned to machine i. By Lemma 6, the load L i on machine i has two components: leaf nodes Lemma 5 LP Lemma 1 (LP is a relaxation) t j j J(i) j is a leaf = x ij j J(i) j is a leaf Lemma 2 x ij L L * j J optimal value of LP parent(i) t parent(i) L * Thus, the overall load L i 2L*. 45

46 Generalized Load Balancing: Flow Formulation Flow formulation of LP. x i j = t j for all j J i x i j L for all i M j x i j 0 for all j J and i M j x i j = 0 for all j J and i M j Observation. Solution to feasible flow problem with value L are in oneto-one correspondence with LP solutions of value L. 46

47 Generalized Load Balancing: Structure of Solution Lemma 3. Let (x, L) be solution to LP. Let G(x) be the graph with an edge from machine i to job j if x ij > 0. We can find another solution (x', L) such that G(x') is acyclic. Pf. Let C be a cycle in G(x). Augment flow along the cycle C. flow conservation maintained At least one edge from C is removed (and none are added). Repeat until G(x') is acyclic G(x) augment along C G(x') 47

48 Conclusions Running time. The bottleneck operation in our 2-approximation is solving one LP with mn + 1 variables. Remark. Can solve LP using flow techniques on a graph with m+n+1 nodes: given L, find feasible flow if it exists. Binary search to find L*. Extensions: unrelated parallel machines. [Lenstra-Shmoys-Tardos 1990] Job j takes t ij time if processed on machine i. 2-approximation algorithm via LP rounding. No 3/2-approximation algorithm unless P = NP. 48

49 11.8 Knapsack Problem

50 Polynomial Time Approximation Scheme PTAS. (1 + ε)-approximation algorithm for any constant ε > 0. Load balancing. [Hochbaum-Shmoys 1987] Euclidean TSP. [Arora 1996] Consequence. PTAS produces arbitrarily high quality solution, but trades off accuracy for time. This section. PTAS for knapsack problem via rounding and scaling. 50

51 Knapsack Problem Knapsack problem. Given n objects and a "knapsack." Item i has value v > 0 and weighs w i i > 0. Knapsack can carry weight up to W. Goal: fill knapsack so as to maximize total value. we'll assume w i W Ex: { 3, 4 } has value 40. Item Value Weight W =

52 Knapsack is NP-Complete KNAPSACK: Given a finite set X, nonnegative weights w i, nonnegative values v i, a weight limit W, and a target value V, is there a subset S X such that: W SUBSET-SUM: Given a finite set X, nonnegative values u i, and an integer U, is there a subset S X whose elements sum to exactly U? i S w i V i S v i Claim. SUBSET-SUM P KNAPSACK. Pf. Given instance (u 1,, u n, U) of SUBSET-SUM, create KNAPSACK instance: v i = w i = u i V = W = U u i i S u i i S U U 52

53 Knapsack Problem: Dynamic Programming 1 Def. OPT(i, w) = max value subset of items 1,..., i with weight limit w. Case 1: OPT does not select item i. OPT selects best of 1,, i 1 using up to weight limit w Case 2: OPT selects item i. new weight limit = w w i OPT selects best of 1,, i 1 using up to weight limit w w i OPT(i, w) = 0 if i = 0 OPT(i 1, w) if w i > w max{ OPT(i 1, w), v i + OPT(i 1, w w i )} otherwise Running time. O(n W). W = weight limit. Not polynomial in input size! 53

54 Knapsack Problem: Dynamic Programming II Def. OPT(i, v) = min weight subset of items 1,, i that yields value exactly v. Case 1: OPT does not select item i. OPT selects best of 1,, i-1 that achieves exactly value v Case 2: OPT selects item i. consumes weight w i, new value needed = v v i OPT selects best of 1,, i-1 that achieves exactly value v OPT(i, v) = 0 if v = 0 if i = 0, v > 0 OPT(i 1, v) if v i > v min{ OPT(i 1, v), w i + OPT(i 1, v v i )} otherwise V* n v max Running time. O(n V*) = O(n 2 v max ). V* = optimal value = maximum v such that OPT(n, v) W. Not polynomial in input size! 54

55 Knapsack: FPTAS Intuition for approximation algorithm. Round all values up to lie in smaller range. Run dynamic programming algorithm on rounded instance. Return optimal items in rounded instance. Item Value Weight 1 134, , ,810, ,217, ,343,199 7 Item Value Weight W = 11 W = 11 original instance rounded instance 55

56 Knapsack: FPTAS Knapsack FPTAS. Round up all values: v i = v i θ θ, v ˆ i = v i θ v max = largest value in original instance ε = precision parameter θ = scaling factor = ε v max / n Observation. Optimal solution to problems with or are equivalent. Intuition. close to v so optimal solution using is nearly optimal; v ˆ v v small and integral so dynamic programming algorithm is fast. v v ˆ Running time. O(n 3 / ε). Dynamic program II running time is O(n 2 v ˆ max ), where v ˆ max = v max θ = n ε 56

57 Knapsack: FPTAS Knapsack FPTAS. Round up all values: v i = v i θ θ Theorem. If S is solution found by our algorithm and S* is any other feasible solution then (1+ε) v i v i i S Pf. Let S* be any feasible solution satisfying weight constraint. i S* v i i S* v i i S* always round up v i i S (v i + θ) i S solve rounded instance optimally never round up by more than θ v i + nθ i S S n DP alg can take v max (1+ε) v i i S n θ = ε v max, v max Σ i S v i 57

58 Extra Slides

59 Load Balancing on 2 Machines Claim. Load balancing is hard even if only 2 machines. Pf. NUMBER-PARTITIONING P LOAD-BALANCE. NP-complete by Exercise 8.26 a b c d e f g length of job f machine 1 machine 2 a d Machine 1 f b c Machine e2 g yes 0 Time L 59

60 Center Selection: Hardness of Approximation Theorem. Unless P = NP, there is no ρ-approximation algorithm for metric k-center problem for any ρ < 2. Pf. We show how we could use a (2 - ε) approximation algorithm for k- center to solve DOMINATING-SET in poly-time. Let G = (V, E), k be an instance of DOMINATING-SET. see Exercise 8.29 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 - ε)-approximation algorithm on G' must find a solution C* with r(c*) = 1 since it cannot use any edge of distance 2. 60

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