# Approximation Algorithms

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1 Approximation Algorithms Group Members: 1. Geng Xue (A R) 2. Cai Jingli (A B) 3. Xing Zhe (A W) 4. Zhu Xiaolu (A W) 5. Wang Zixiao (A X) 6. Jiao Qing (A R) 7. Zhang Hao (A U) 1

2 Introduction Geng Xue 2

3 Optimization problem Find the minimum/maximum of Vertex Cover : A minimum set of vertices that covers all the edges in a graph. 3

4 NP optimization problem The hardness of NP optimization is NP hard. Can t be solved in polynomial time. 4

5 How to solve? 5

6 How to solve? 6

7 How to solve? 7

8 How to solve? 8

9 How to evaluate? Mine is better! 9

10 How to evaluate? Exact optimal value of optimization problem. relative performance guarantee. Sub-optimal solution : ρ OPT (ρ is factor). The closer factor ρ is to 1, The better the algorithm is. 10

11 Set cover It leads to the development of fundamental techniques for the entire approximation algorithms field. Due to set cover, many of the basic algorithm design techniques can be explained with great ease. 11

12 What is set cover? Definition: Given a universe U of n elements, a collection of subsets of U, S = {S 1, S 2,, S k }, and a cost function c: S Q +, find a minimum cost subcollection of S that covers all elements of U. 12

13 What is set cover? Example: S Sets Cost S 1 {1,2} 1 S 2 {3,4} 1 S 3 {1,3} 2 S 4 {5,6} 1 S 5 {1,5} 3 S 6 {4,6} 1 Universal Set : U = {1,2,3,4,5,6} Find a sub-collection of S that covers all the elements of U with minimum cost. Solution? S 1 S 2 S 4 Cost = = 3 13

14 Approximation algorithms to set cover problem Combinatorial algorithms Linear programming based Algorithms (LP-based) 14

15 Combinatorial & LP-based Combinatorial algorithms Greedy algorithms Layering algorithms LP-based algorithms Dual Fitting Rounding Primal Dual Schema 15

16 Greedy set cover algorithm Cai Jingli 16

17 Greedy set cover algorithm Where C is the set of elements already covered at the beginning of an iteration and α is the average cost at which it covers new elements. 17

18 Example S1 1 S S2 subset cost S1 = {1, 2} 1 S2 = {2, 3, 4} 2 S3 = {3, 4} 3 U 6 5 S4 S4 = {5, 6} 3 Iteration 1: C = {} α1 = 1 / 2 = 0.5 α2 = 2 / 3 = 0.67 α3 = 3 / 2 = 1.5 α4 = 3 / 2 = 1.5 C = {1, 2} Price(1) = 0.5 Price(2) = 0.5 Iteration 2: C = {1, 2} α2 = 2 / 2 = 1 α3 = 3 / 2 = 1.5 α4 = 3 / 2 = 1.5 C = {1, 2, 3, 4} Price(3) = 1 Price(4) = 1 Iteration 3: C = {1, 2, 3, 4 } α3 = 3 / 0 = infinite α4 = 3 / 2 = 1.5 C = {1, 2, 3, 4, 5, 6} Price(5) = 1.5 Price(6) =

19 Example S1 1 S S2 subset cost S1 = {1, 2} 1 S2 = {2, 3, 4} 2 S3 = {3, 4} 3 U 6 5 S4 S4 = {5, 6} 3 The picked sets = {S1, S2, S4} Total cost = Cost(S1)+Cost(S2)+Cost(S4) = Price(1) + Price(2) + + Price(6) = = 6 19

20 Theorem The greedy algorithm is an H n factor approximation algorithm for the minimum set cover problem, where H n = n i Cost(Si) H n OPT 20

21 Proof 1) We know price e = e U cost of the greedy algorithm = c S 1 + c S c S m. 2) We will show price(e k ) OPT n k+1, where e k is the kth element covered. If 2) is proved then theorem is proved also. i Cost(Si) = price e e U n k=1 OPT n k + 1 = OPT = OPT ( n ) = H n OPT n k=1 1 n k

22 price(e k ) OPT n k + 1 Say the optimal sets are O 1, O 2,, O p, so p i=1 OPT = c O 1 + c O c O p = c O i Now, assume the greedy algorithm has covered the elements in C so far. U C O 1 U C = O i U C i=1 price e k = α c(o i) p O i (U C) this because of greedy algorithm + + O p U C (2), i=1,,p. we know 22

23 price(e k ) p OPT n k OPT = c O i i=1 p 2 U C O i U C i=1 3 c O i α O i U C So OPT = p i=1 c O i α O i U C U C = n k + 1 price e k = α OPT n k + 1 p i=1 α U C 23

24 Shortest Superstring Problem (SSP) Given a finite alphabet Σ, and set of n strings, S = {s 1,, s n } Σ +. Find a shortest string s that contains each s i as a substring. Without loss of generality, we may assume that no string s i is a substring of another string s j, j i. 24

25 Application: Shotgun sequencing 25

26 Approximating SSP Using Set Cover 26

27 Set Cover Based Algorithm Set Cover Problem: Choose some sets that cover all elements with least cost Elements The input strings Subsets σ ijk = string obtained by overlapping input strings s i and s j, with k letters. β = S σ ijk, all i, j, k Let π β set(π) = {s S s is a substr. of π} 27

28 Set Cover Based Algorithm Set Cover Problem: Choose some sets that cover all elements with least cost Elements The input strings Subsets σ ijk = string obtained by overlapping input strings s i and s j, with k letters. β = S σ ijk, all i, j, k Let π β set(π) = {s S s is a substr. of π} Cost of a subset set(π) is π A solution to SSP is the concatenation of π obtained from SCP 28

29 Example S = {CATGC, CTAAGT, GCTA, TTCA, ATGCATC} π Set Cost CATGC CTAAGT CATGCTAAGT CATGC, CTAAGT, GCTA 10 CATGC..... GCTA CATGCTA CATGC, GCTA CATGC ATGCATC.... ATGCATCATGC CATGC, ATGCATC 11 CTAAGT TTCA CTAAGTTCA CTAAGT, TTCA 9 ATGCATC CTAAGT ATGCATCTAAGT CTAAGT, ATGCATC 12 29

30 GCTA ATGCATC GCTATGCATC GCTA, ATGCATC 10 TTCA ATGCATC TTCATGCATC TTCA, ATGCATC, CATGC 10 GCTA.... CTAAGT GCTAAGT GCTA, CTAAGT 7 TTCA..... CATGC TTCATGC CATGC, TTCA 7 CATGC.... ATGCATC CATGCATC CATGC, ATGCATC CATGC CATGC 5 CTAAGT CTAAGT 6 GCTA GCTA 4 TTCA TTCA 4 ATGCATC ATGCATC

31 Lemma OPT OPT SCA 2H n OPT 31

32 Layering Technique Xing Zhe 32

33 Layering Technique Set Cover Problem Greedy: H n Combinatorial algorithms Layering? 33

34 Vertex Cover Problem Definition: Given a graph G = (V, E), and a weight function w: V Q + assigning weights to the vertices, find a minimum weighted subset of vertices C V, such that C covers all edges in E, i.e., every edge e i E is incident to at least one vertex in C. 34

35 Vertex Cover Problem Example: all vertices of unit weight

36 Vertex Cover Problem Example: all vertices of unit weight Total cost = 5 36

37 Vertex Cover Problem Example: all vertices of unit weight Total cost = 3 37

38 Vertex Cover Problem Example: all vertices of unit weight Total cost = 3 OPT 38

39 Vertex Cover Problem Vertex cover is a special case of set cover. a b c d e f Edges are elements, vertices are subsets Universe U = {a, b, c, d, e, f, g, h} A collection of subsets S = S 1, S 2, S 3, S 4, S 5, S 6, S 7 g h S 1 = {a, b} S 2 = {b, c, d} S 3 = {c, e, g, h} S 4 = {a} S 5 = {d, e, f} S 6 = {f, g} S 7 = {h} Vertex cover problem is a set cover problem with f = 2 39

40 Vertex Cover Problem Vertex cover is a special case of set cover. a b c d e f Edges are elements, vertices are subsets Universe U = {a, b, c, d, e, f, g, h} A collection of subsets S = S 1, S 2, S 3, S 4, S 5, S 6, S 7 g h S 1 = {a, b} S 2 = {b, c, d} S 3 = {c, e, g, h} S 4 = {a} S 5 = {d, e, f} S 6 = {f, g} S 7 = {h} Vertex cover problem is a set cover problem with f = 2 40

41 Vertex Cover Problem Vertex cover is a special case of set cover. a b c d e f Edges are elements, vertices are subsets Universe U = {a, b, c, d, e, f, g, h} A collection of subsets S = S 1, S 2, S 3, S 4, S 5, S 6, S 7 g h S 1 = {a, b} S 2 = {b, c, d} S 3 = {c, e, g, h} S 4 = {a} S 5 = {d, e, f} S 6 = {f, g} S 7 = {h} Vertex cover problem is a set cover problem with f = 2 41

42 Vertex Cover Problem Vertex cover is a special case of set cover. a b c d e f Edges are elements, vertices are subsets Universe U = {a, b, c, d, e, f, g, h} A collection of subsets S = S 1, S 2, S 3, S 4, S 5, S 6, S 7 g h S 1 = {a, b} S 2 = {b, c, d} S 3 = {c, e, g, h} S 4 = {a} S 5 = {d, e, f} S 6 = {f, g} S 7 = {h} Vertex cover problem is a set cover problem with f = 2 42

43 Notation 1) frequency 2) f : the frequency of the most frequent element. Layering algorithm is an f factor approximation for the minimum set cover problem. 43

44 Notation 1) frequency 2) f : the frequency of the most frequent element. Layering In the vertex algorithm cover problem, is an f factor f =? approximation for the minimum set cover problem. 44

45 Vertex Cover Problem Vertex cover is a special case of set cover. a b c d e f Edges are elements, vertices are subsets Universe U = {a, b, c, d, e, f, g, h} A collection of subsets S = S 1, S 2, S 3, S 4, S 5, S 6, S 7 g h S 1 = {a, b} S 2 = {b, c, d} S 3 = {c, e, g, h} S 4 = {a} S 5 = {d, e, f} S 6 = {f, g} S 7 = {h} Vertex cover problem is a set cover problem with f = 2 45

46 Vertex Cover Problem Vertex cover is a special case of set cover. a b c d e f Edges are elements, vertices are subsets Universe U = {a, b, c, d, e, f, g, h} A collection of subsets S = S 1, S 2, S 3, S 4, S 5, S 6, S 7 g h S 1 = {a, b} S 2 = {b, c, d} S 3 = {c, e, g, h} S 4 = {a} S 5 = {d, e, f} S 6 = {f, g} S 7 = {h} Vertex cover problem is a set cover problem with f = 2 46

47 Approximation factor 1) frequency 2) f : the frequency of the most frequent element. In the vertex cover problem, f = 2 Layering Algorithm Set Cover Problem Vertex Cover Problem factor = f factor = 2 47

48 Approximation factor 1) frequency 2) f : the frequency of the most frequent element. In the vertex cover problem, f = 2 Layering Algorithm Set Cover Problem Vertex Cover Problem factor = f factor = 2 48

49 Vertex Cover Problem arbitrary weight function: w: V Q + degree-weighted function: c > 0 s.t. v V, w v = c deg (v) 49

50 Vertex Cover Problem arbitrary weight function: w: V Q + degree-weighted function: c > 0 s.t. v V, w v = c deg (v) 50

51 Degree weighted function Lemma: If w: V Q + is a degree-weighted function. Then w V 2 OPT 51

52 Proof: Lemma: If w: V Q + is a degree-weighted function. Then w V 2 OPT degree-weighted function, w v = c deg (v) C is the optimal vertex cover in G, deg v E v C OPT = w C = c deg v c E v C (1) in worst case, we pick all vertices. handshaking lemma, deg v = 2 E v V w V = v V w v = v V c deg v = c v V deg v = c 2 E (2) from (1) and (2), w V 2 OPT 52

53 Proof: Lemma: If w: V Q + is a degree-weighted function. Then w V 2 OPT degree-weighted function, w v = c deg (v) C is the optimal vertex cover in G, deg v E v C OPT = w C = c deg v c E v C (1) in worst case, we pick all vertices. handshaking lemma, deg v = 2 E v V w V = v V w v = v V c deg v = c v V deg v = c 2 E (2) from (1) and (2), w V 2 OPT 53

54 Proof: Lemma: If w: V Q + is a degree-weighted function. Then w V 2 OPT degree-weighted function, w v = c deg (v) C is the optimal vertex cover in G, deg v E v C OPT = w C = c deg v c E v C (1) in worst case, we pick all vertices. handshaking lemma, deg v = 2 E v V w V = v V w v = v V c deg v = c v V deg v = c 2 E (2) from (1) and (2), w V 2 OPT 54

55 Proof: Lemma: If w: V Q + is a degree-weighted function. Then w V 2 OPT degree-weighted function, w v = c deg (v) C is the optimal vertex cover in G, deg v E v C OPT = w C = c deg v c E v C (1) in worst case, we pick all vertices. handshaking lemma, deg v = 2 E v V w V = v V w v = v V c deg v = c v V deg v = c 2 E (2) from (1) and (2), w V 2 OPT 55

56 Proof: Lemma: If w: V Q + is a degree-weighted function. Then w V 2 OPT degree-weighted function, w v = c deg (v) C is the optimal vertex cover in G, deg v E v C OPT = w C = c deg v c E v C (1) in worst case, we pick all vertices. handshaking lemma, deg v = 2 E v V w V = v V w v = v V c deg v = c v V deg v = c 2 E (2) from (1) and (2), w V 2 OPT 56

57 Proof: Lemma: If w: V Q + is a degree-weighted function. Then w V 2 OPT degree-weighted function, w v = c deg (v) C is the optimal vertex cover in G, deg v E v C OPT = w C = c deg v c E v C (1) in worst case, we pick all vertices. handshaking lemma, deg v = 2 E v V w V = v V w v = v V c deg v = c v V deg v = c 2 E (2) from (1) and (2), w V 2 OPT 57

58 Proof: Lemma: If w: V Q + is a degree-weighted function. Then w V 2 OPT degree-weighted function, w v = c deg (v) C is the optimal vertex cover in G, deg v E v C OPT = w C = c deg v c E v C (1) in worst case, we pick all vertices. handshaking lemma, deg v = 2 E v V w V = v V w v = v V c deg v = c v V deg v = c 2 E (2) from (1) and (2), w V 2 OPT 58

59 Layering algorithm Basic idea: arbitrary weight function several degree-weighted functions nice property (factor = 2) holds in each layer 59

60 Layering algorithm 1) remove all degree zero vertices, say this set is D 0 2) compute c = min {w v deg(v)} 3) compute degree-weighted function t(v) = c deg (v) 4) compute residual weight function w v = w v t(v) 5) let W 0 = v w v = 0}, pick zero residual vertices into the cover set 6) let G 1 be the graph induced on V (D 0 W 0 ) 7) repeat the entire process on G 1 w.r.t. the residual weight function, until all vertices are of degree zero 60

61 Layering algorithm zero residual weight non-zero residual weight zero degree 61

62 Layering algorithm 62

63 Layering algorithm 63

64 Layering algorithm 64

65 Layering algorithm The vertex cover chosen is C = W 0 W 1 W k 1 Clearly, V C = D 0 D 1 D k 65

66 Layering algorithm Example: all vertices of unit weight: w(v) =

67 Iteration D 0 = { 8 } 67

68 Iteration D 0 = { 8 } compute c = min {w v deg(v)} = 0.25 degree-weighted function: t v = c deg v = deg v 68

69 0.5 Iteration D 0 = { 8 } compute c = min {w v deg(v)} = 0.25 degree-weighted function: t v = c deg v = deg v compute residual weight: w (v) = w(v) t(v) 69

70 0.5 Iteration D 0 = { 8 } compute c = min {w v deg(v)} = 0.25 degree-weighted function: t v = c deg v = deg v compute residual weight: w (v) = w(v) t(v) W 0 = { 3 } 70

71 Iteration D 0 = { 8 } compute c = min {w v deg(v)} = 0.25 degree-weighted function: t v = c deg v = deg v compute residual weight: w (v) = w(v) t(v) W 0 = { 3 } remove D 0 and W 0 71

72 Iteration D 1 = { 7 } 72

73 Iteration D 1 = { 7 } compute c = min {w v deg(v)} = degree-weighted function: t v = c deg v = deg v 73

74 Iteration D 1 = { 7 } compute c = min {w v deg(v)} = degree-weighted function: t v = c deg v = deg v compute residual weight: w (v) = w(v) t(v) 74

75 Iteration D 1 = { 7 } compute c = min {w v deg(v)} = degree-weighted function: t v = c deg v = deg v compute residual weight: w (v) = w(v) t(v) W 1 = { 2, 5 } 75

76 Iteration D 1 = { 7 } compute c = min {w v deg(v)} = degree-weighted function: t v = c deg v = deg v compute residual weight: w (v) = w(v) t(v) W 1 = { 2, 5 } remove D 1 and W 1 76

77 Iteration D 2 = { 6 } 77

78 Iteration D 2 = { 6 } compute c = min {w v deg(v)} = 0.25 degree-weighted function: t v = c deg v = deg v 78

79 Iteration D 2 = { 6 } compute c = min {w v deg(v)} = 0.25 degree-weighted function: t v = c deg v = deg v compute residual weight: w (v) = w(v) t(v) 79

80 Iteration D 2 = { 6 } compute c = min {w v deg(v)} = 0.25 degree-weighted function: t v = c deg v = deg v compute residual weight: w (v) = w(v) t(v) W 2 = { 1 } 80

81 Iteration D 2 = { 6 } compute c = min {w v deg(v)} = 0.25 degree-weighted function: t v = c deg v = deg v compute residual weight: w (v) = w(v) t(v) W 2 = { 1 } remove D 2 and W 2 81

82 Layering algorithm all vertices of unit weight: w(v) = Vertex cover C = W 0 W 1 W 2 = {1, 2, 3, 5} Total cost: w C = 4 82

83 Layering algorithm all vertices of unit weight: w(v) = Optimal vertex cover C = {1, 3, 5} Optimal cost: OPT = w C = 3 83

84 Layering algorithm all vertices of unit weight: w(v) = Vertex cover C = W 0 W 1 W 2 = {1, 2, 3, 5} Total cost: w C = 4 Optimal cost: OPT = w C = 3 w C < 2 OPT 84

85 Approximation factor The layering algorithm for vertex cover problem (assuming arbitrary vertex weights) achieves an approximation guarantee of factor 2. 85

86 Approximation factor The layering algorithm for vertex cover problem (assuming arbitrary vertex weights) achieves an approximation guarantee of factor 2. We need to prove: 1) set C is a vertex cover for G 2) w C 2 OPT 86

87 Proof 1) set C is a vertex cover for G layering algorithm terminates until all nodes are zero degree. all edges have been already covered. 87

88 Proof 2) w C 2 OPT a vertex v C, if v W j, its weight can be decomposed as w v = i j t i (v) a vertex v V C, if v D j, a lower bound on its weight is given by w v i j t i (v) Let C be the optimal vertex cover, in each layer i, C G i is a vertex cover for G i recall the lemma : if w: V Q + is a degree-weighted function, then w V by lemma, t i (C G i ) 2 t i (C G i ) 2 OPT k 1 k 1 w C = i=0 t i C G i 2 i=0 t i C G i 2 w(c ) 88

89 Summary Layering Algorithm Vertex Cover Problem factor = 2 Set Cover Problem factor = f 89

90 Introduction to LP-Duality Zhu Xiaolu 90

91 Introduction to LP-Duality Linear Programming (LP) LP-Duality Theorem a) Weak duality theorem b) LP-duality theorem c) Complementary slackness conditions 91

92 Why use LP? obtaining approximation algorithms using LP Rounding Techniques to solve the linear program and convert the fractional solution into an integral solution. Primal-dual Schema to use the dual of the LP-relaxation in the design of the algorithm. analyzing combinatorially obtained approximation algorithms LP-duality theory is useful using the method of dual fitting 92

93 What is LP? Linear programming: the problem of optimizing (i.e., minimizing or maximizing) a linear function subject to linear inequality constraints. Minimize 7X 1 + X 2 + 5X 3 Subject to X 1 X 2 + 3X X 1 + 2X 2 X 3 6 X 1, X 2, X 3 0 Objective function: The linear function that we want to optimize. Feasible solution: An assignment of values to the variables that satisfies the inequalities. E.g. X=(2,1,3) Cost: The value that the objective function gives to an assignment. E.g =30 93

94 What is LP-Duality? Minimize 7X 1 + X 2 + 5X 3 Subject to X 1 X 2 + 3X X 1 + 2X 2 X 3 6 X 1, X 2, X 3 0 Abstract the constraints: a b c d Find multipliers: y 1, y 2 0 y 1 a y 1 b y 2 c y 2 d Objective function y 1 a + y 2 c y 1 b + y 2 d (1) As large as possible 7X 1 + X 2 + 5X 3 1 X 1 X 2 + 3X X 1 + 2X 2 X 3 = 6X 1 + X 2 + 2X 3 = X 1 + X 2 + 5X 3 2 X 1 X 2 + 3X X 1 + 2X 2 X 3 = 7X 1 + 5X 3 = Range of the objective function Min 94

95 What is LP-Duality? The primal program The dual program Minimize 7X 1 + X 2 + 5X 3 Subject to X 1 X 2 + 3X X 1 + 2X 2 X 3 6 X 1, X 2, X 3 0 Maximize 10Y 1 + 6Y 2 Subject to Y 1 + 5Y 2 7 Y 1 + 2Y 2 1 3Y 1 Y 2 5 Y 1, Y 2 0 n Minimize j=1 c j x j n Subject to j=1 a ij x j b i, i = 1,, m x j 0, j = 1,, n a ij, b i, c j are given rational numbers m Maximize i=1 b i y i m Subject to i=1 a ij y i c j, j = 1,, n y i 0, i = 1,, m a ij, b i, c j are given rational numbers 95

96 Minimize c 1 X c n X n Subject to a 1,1 X a 1,n X n b 1 a m,1 X a m,n X n b m X 1,, X n 0 Minimize 7X 1 + X 2 + 5X 3 Subject to X 1 X 2 + 3X X 1 + 2X 2 X 3 6 X 1, X 2, X 3 0 Y 1 (a 1,1 X a 1,n X n ) + +Y m (a m,1 X a m,n X n ) Y 1 b Y m b m (1) a 1,1 Y a m,1 Y m X (a 1,n Y a m,n Y m ) X n Y 1 b Y m b m (2) Assume: a 1,1 Y a m,1 Y m c 1 a 1,n Y a m,n Y m c n c 1 X c n X n a 1,1 Y a m,1 Y m X (a 1,n Y a m,n Y m ) X n b 1 Y b m Y m (3) Maximize b 1 Y b m Y m Subject to a 1,1 Y a m,1 Y m c 1 a 1,n Y a m,n Y m c n Y 1,, Y m 0 Maximize 10Y 1 + 6Y 2 Subject to Y 1 + 5Y 2 7 Y 1 + 2Y 2 1 3Y 1 Y 2 5 Y 1, Y

97 Weak duality theorem If X =(X 1,,X n ) and Y =(Y 1,,Y m ) are feasible solutions for the primal and dual program, respectively, then m i=1 b i y i n j=1 c j x j 97

98 Weak duality theorem - proof n Minimize j=1 c j x j n Subject to j=1 a ij x j b i, i = 1,, m x j 0, j = 1,, n a ij, b i, c j are given rational numbers m Maximize i=1 b i y i m Subject to i=1 a ij y i c j, j = 1,, n y i 0, i = 1,, m a ij, b i, c j are given rational numbers Since y is dual feasible and x j s are nonnegative, Since x is primal feasible and y j s are nonnegative, Obviously, So, we proof that n j=1 m c j x j a ij y i n j=1 i=1 i=1 j=1 n m n j=1 c j x j b i y i j=1 i=1 m i=1 m x j a ij x j y i b i y i i=1 j=1 i=1 n m m n a ij y i x j = a ij x j y i 98

99 LP-duality theorem If X = (X 1,,X n ) and Y = Y 1,,Y m are optimal solutions for the primal and dual programs, respectively, then n j=1 c j x j m = b i y i i=1 99

100 Complementary slackness conditions Let X and Y be primal and dual feasible solutions, respectively. Then, X and Y are both optimal iff all of the following conditions are satisfied: Primal complementary slackness conditions For each 1 j n: either x j = 0 or m i=1 a ij y i = c j ; And Dual complementary slackness conditions For each 1 i m: either y i = 0 or n j=1 a ij x j = b i. 100

101 Complementary slackness conditions -proof Proof how these two conditions comes up From the proof of weak duality theorem: n n m c j x j a ij y i j=1 j=1 i=1 m i=1 n m x j a ij x j y i b i y i j=1 i=1 (1) (2) By the LP-duality theorem, x and y are both optimal solutions iff: n j=1 c j x j m = b i y i i=1 These happens iff (1) and (2) hold with equality: n c j x j = n m a ij y i x j = m n a ij x j y i = m b i y i j=1 j=1 i=1 i=1 j=1 i=1 101

102 n c j x j = n m a ij y i x j = m n a ij x j y i = m b i y i j=1 j=1 i=1 i=1 j=1 i=1 n m (c j a ij y i j=1 i=1 )x j = 0 Same Process Dual complementary slackness conditions For each 1 j n If x m j > 0 then c j i=1 a ij y i = 0 m Ifc j i=1 a ij y i > 0 then x j = 0 Primal complementary slackness conditions For each 1 j n: either x j = 0 or a ij y i = c j m i=1 102

103 Set Cover via Dual Fitting Wang Zixiao 103

104 Set Cover via Dual Fitting Dual Fitting: help analyze combinatorial algorithms using LP-duality theory Analysis of the greedy algorithm for the set cover problem Give a lower bound 104

105 Dual Fitting Minimization problem: combinatorial algorithm Linear programming relaxation, Dual Primal is fully paid for by the dual Infeasible Feasible (shrunk with factor) Factor Approximation guarantee 105

106 Formulation U: universe S: {S1, S2,, Sk} C: S > Q + Goal:sub-collection with minimum cost 106

107 LP-Relaxation Motivation: NP-hard Polynomial Integer program Fractional program Letting x S : 0 x S 107

108 LP Dual Problem 108

109 LP Dual Problem OPT: cost of optimal integral set cover OPTf: cost of optimal fractional set cover 109

110 Solution ye = price(e)? U = {1, 2, 3} S = {{1, 2, 3}, {1, 3}} C({1, 2, 3}) = 3 C({1, 3}) = 1 Not feasible! Iteration 1: {1, 3} chosen price(1) = price(3) = 0.5 Iteration 2: {1, 2, 3} chosen price(2) = 3 price(1) + price(2) + price(3) = 4 > C({1, 2, 3}) violation 110

111 Solution y e = price(e) H n The vector y defined above is a feasible solution for the dual program There is no set S in S overpacked 111

112 Proof Consider a set S S S = k Consider the iteration for ei: e1, e2,, ei,, ek Select S: price e i C(S) k i + 1 Select S : price e i < C(S) k i + 1 Number the elements in the order of being covered: e1, e2,, ek At least k-i+1 elements uncovered in S price e i C(S) k i + 1 y ei = price(e i) H n C S H n 1 k i + 1 k i=1 y ei C(S) H n 1 k + 1 k = H k H n C(S) C(S) 112

113 The approximation of guarantee of the greedy set cover algorithm is H n Proof: The cost of the set cover picked is: e U price e = H n ( y e ) e U H n OPT f H n OPT 113

114 Constrained Set Multicover Constrained Set Multicover: Each element e has to be covered r e times Greedy algorithm: Pick the most cost-effective set until all elements have been covered by required times 114

115 Analysis of the greedy algorithm 115

116 Analysis of the greedy algorithm 116

117 Analysis of the greedy alrogithm 117

118 Analysis of the greedy algorithm 118

119 Analysis of the greedy algorithm 119

120 Rounding Applied to LP to solve Set Cover Jiao Qing 120

121 Why rounding? Previous slides show LP-relaxation for the set cover problem, but for set cover real applications, they usually need integral solution. Next section will introduce two rounding algorithms and their approximation factor to OPT f and OPT. 121

122 Rounding Algorithms A simple rounding algorithm Randomized rounding 122

123 A simple rounding algorithm Algorithm Description 1. Find an optimal solution to the LP-relaxation. 2. Pick all sets S for which x s 1 in this solution. f Apparently, its algorithm complexity equals to LP problem. 123

124 A simple rounding algorithm Proving that it solves set cover problem. 1. Remember the LP-relaxation: 2. Let C be the collection of picked sets. For e U, e is in at most f sets, one of these sets must has x s 1, therefore, this simple algorithm f at least solve the set cover problem. 124

125 A simple rounding algorithm Firstly we prove that it solves set cover problem. Then we estimate its approximation factor to OPT f and OPT is f. 125

126 Approximation Factor Estimating its approximation factor 1. The rounding process increases x s by a factor of at most f, and further it reduces the number of sets. 2. Thus it gives a desired approximation guarantee of f. 126

127 Randomized Rounding This rounding views the LP-relaxation coefficient x s of sets as probabilities. Using probabilities theory it proves this rounding method has approximation factor O(log n). 127

128 Randomized Rounding Algorithm Description: 1. Find an optimal solution to the LP-relaxation. 2. Independently picks v log n subsets of full set S. 3. Get these subsets union C, check whether C is a valid set cover and has cost OPT f 4v log n. If not, repeat the step 2 and 3 again. 4. The expected number of repetitions needed at most 2. Apparently, its algorithm complexity equals to LP problem. 128

129 Proof Approximation Factor Next,we prove its approximate factor is O(log n). 1. For each set S, its probability of being picked is p s (equals to x s coefficient). 2. Let C be one collection of sets picked. The expected cost of C is: E cost C = p r s is picked c s = p s c s = OPT f s s s s 129

130 Proof Approximation Factor 3. Let us compute the probability that an element a is covered by C. Suppose that a occurs in k sets of S, with the probabilities associated with these sets be p 1,... p k. Because their sum greater than 1. Using elementary calculus, we get: p r a is covered by C k and thus, k 1 1 e p r a is not covered by C 1 e where e is the base of natural logarithms. Hence each element is covered with constant probability by C. 130

131 Proof Approximation Factor 4. And then considering their union C, we get: p r a is not covered by c 5. Summing over all elements a U, we get p r c is not a valid set cover = n 1 e v log n 1 4n With n e v log n 1 4n 6. With: E cost C E cost c i i =OPT f v log n = OPT f v log n Applying Markov s Inequality with t=opt f 4v log n, we get: p r cost C OPT f 4v log n

132 Proof Approximation Factor 7. With p r c is not a valid set cover 1 4 p r cost C OPT f 4v log n 1 4 Hence, p r C is a valid set cover and has cost OPT f 4v log n

133 Approximation Factor The chance one could find a set cover and has a cost smaller than O(log n)opt f is bigger than 50%, and the expected number of iteration is two. Thus this randomized rounding algorithm provides a factor O(log n) approximation, with a high probability guarantee. 133

134 Summary Those two algorithm provide us with factor f and O(log n) approximation, which remind us the factor of greedy and layering algorithms. Even through LP-relaxation, right now we could only find algorithm with factor of f and O log n. 134

135 Set Cover via the Primal-dual Schema Zhang Hao 135

136 Primal-dual Schema? A broad outline for the algorithm The details have to be designed individually to specific problems Good approximation factors and good running times 136

137 Primal-dual Program(Recall) The primal program The dual program n Minimize j=1 c j x j n Subject to j=1 a ij x j b i, i = 1,, m x j 0, j = 1,, n a ij, b i, c j are given rational numbers m Maximize i=1 b i y i m Subject to i=1 a ij y i c j, j = 1,, n y i 0, i = 1,, m a ij, b i, c j are given rational numbers 137

138 Standard Complementary slackness conditions (Recall) Let x and y be primal and dual feasible solutions, respectively. Then, x and y are both optimal iff all of the following conditions are satisfied: Primal complementary slackness conditions For each 1 j n: either x j = 0 or m i=1 a ij y i = c j ; And Dual complementary slackness conditions For each 1 i m: either y i = 0 or n j=1 a ij x j = b i. 138

139 Relaxed Complementary slackness conditions Let x and y be primal and dual feasible solutions, respectively. Primal complementary slackness conditions For each 1 j n: either x j = 0 or c i m α i=1 a ij y i c i ; And Dual complementary slackness conditions n For each 1 i m: either y i = 0 or b i j=1 a ij x j βb i. α, β => The optimality of x and y solutions If α = β = 1 => standard complementary slackness conditions 139

140 Overview of the Schema Pick a primal infeasible solution x, and a dual feasible solution y, such that the slackness conditions are satisfied for chosen α and β (usually x = 0, y = 0). Iteratively improve the feasibility of x (integrally) and the optimality of y, such that the conditions remain satisfied, until x becomes feasible. 140

141 Proposition Primal-dual Schema An approximation guarantee of αβ is achieved using this schema. Proof: ( c i m α i=1 a ij y i ) ( j=1 a ij x j βb i ) n j=1 c j x j α n j=1 m i=1 a ij y i x j = α m i=1 n j=1 a ij x j n y i αβ m i=1 b i y i n j=1 c j x j (primal and dual condition) m αβ b i y i αβ OPT f αβ OPT i=1 141

142 Primal-dual Schema Applied to Set Cover Set Cover(Recall) Standard Primal Program The Dual Program minimize s S c(s) x s maximize e U y e subject to e:e S x s 1, e U x s {0,1}, S S subject to e:e S y e c S, S S y e 0, e U 142

143 Relaxed Complementary Primal-dual Schema Applied to Set Cover Slackness Conditions Definition: Set S is called tight if e:e S y e = c(s) Set α = 1, β = f(to obtain approximation factor f) Primal Conditions Pick only tight sets in the cover S S: x S 0 y e = c(s) e:e S Dual Conditions Each e, y e 0, can be covered at most f times - trivially satisfied e: y e 0 S:e S x S f 143

144 Algorithm (Set Cover factor Primal-dual Schema Applied to Set Cover f) Initialization: x = 0, y = 0. Until all elements are covered, do Pick an uncovered element e, and raise y e until some set goes tight. Pick all tight sets in the cover and update x. Declare all the elements occuring in these sets as covered. Output the set cover x. 144

145 Theorem Primal-dual Schema Applied to Set Cover The algorithm can achieve an approximation factor of f. Proof Clearly, there will be no uncovered and no overpacked sets in the end. Thus, primal and dual solutions will be feasible. Since they satisfy the relaxed complementary slackness conditions with α = 1, β = f, the approximation factor is f. 145

146 Conclusion Combinatorial algorithms are greedy and local. LP-based algorithms are global. 146

147 Conclusion Combinatorial algorithms Greedy algorithms Factor: Hn Layering algorithms Factor: f LP-based algorithms Rounding Factor: f, Hn Primal Dual Schema Factor: f 147

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