1. Approximation Algorithms Chapter 14: Rounding Applied to Set Cover

2. Overview n Set Cover –Approximation by simple rounding f-approx. algorithm (f: the frequency of the most frequent element). –Approximation by randomized rounding O(log n)-approx. algorithm (n: # elements to be covered). n Weighted Vertex Cover –2-approx. algorithm Method based on half-integral solutions of the linear programming Each variable takes only 0, 1/2, or 1.

3. Set Cover n Input –Elements U={a 1,…,a n }, –Subsets of U:S={S 1,…,S m }. –Cost function c: S→Q +. n Output –Subsets of S that cover all elements in U s.t. the sum of costs of chosen subsets in S is minimized. S1S1 S2S2 S3S3 S4S4 S5S5 222 1 4 Cost of a subset Cost:2+2+2=6. Cost:1+4=5. Cost:2+1+4=7. This is not a solution since an element is not covered. a1a1 a2a2 a3a3 a4a4 a5a5 a6a6

4. Set Cover by linear inequalities (1/4) n Objective function –Minimize the sum of costs of subsets chosen: n Constraints –For covers Each element must appear in at least one chosen subset. –For choosing subsets Each subset is either chosen or not chosen. S1S1 S2S2 S3S3 S4S4 S5S5 222 1 4 Cost of a subset Cost: 2+2+2=6. Cost: 1+4=5. Cost: 2+1+4=7. a1a1 a2a2 a3a3 a4a4 a5a5 a6a6 This is not a solution since an element is not covered.

5. Set Cover by linear inequalities (2/4) n Objective function –Minimize the sum of costs of subsets chosen. n Constraints –For covers –For choosing subsets S1S1 S2S2 S3S3 S4S4 S5S5 222 1 4 a1a1 a2a2 a3a3 a4a4 a5a5 a6a6 This is not a solution since an element is not covered. Cost of a subset Cost: 2+2+2=6. Cost: 1+4=5. Cost: 2+1+4=7.

6. Set Cover by linear inequalities (3/4) n Objective function n Constraints –For covers –For choosing subsets S1S1 S2S2 S3S3 S4S4 S5S5 222 1 4 a1a1 a2a2 a3a3 a4a4 a5a5 a6a6 Cost of a subset Cost: 2+2+2=6. Cost: 1+4=5. Cost: 2+1+4=7. This is not a solution since an element is not covered.

7. Set Cover by linear inequalities (4/4) n Objective function n Constraints –For covers –For choosing subsets S1S1 S2S2 S3S3 S4S4 S5S5 222 1 4 a1a1 a2a2 a3a3 a4a4 a5a5 a6a6 Cost of a subset Cost: 2+2+2=6. Cost: 1+4=5. Cost: 2+1+4=7. This is not a solution since an element is not covered.

8. LP-relaxation n Constraints –Each subset is either chosen or not chosen. –It takes a value bet. 0 and 1. From the nature of Set Cover, the upper bound can be eliminated. –It takes a positive value. S1S1 S2S2 S3S3 S4S4 S5S5 222 1 4 a1a1 a2a2 a3a3 a4a4 a5a5 a6a6 Cost of a subset Cost: 2+2+2=6. Cost: 1+4=5. Cost: 2+1+4=7. This is not a solution since an element is not covered.

9. Rounding n To change a natural number into an integer. x1x1 x2x2 x3x3 x4x4 0 1 Solution found by LP x1x1 x2x2 x3x3 x4x4 0 1 x1x1 x2x2 x3x3 x4x4 0 1 Rounded solution with a threshold Probabilistically rounded solution Threshold

10. Overview n Set Cover –Approximation by simple rounding f-approx. algorithm (f: the frequency of the most frequent element). –Approximation by randomized rounding O(log n)-approx. algorithm (n: # elements to be covered). n Weighted Vertex Cover –Method based on half-integral solutions of the linear programming Each variable takes only 0, 1/2, or 1. 2-approx. algorithm

11. Algorithm 14.1 n A simple rounding algorithm A 1 –f: the frequency of the most frequent element. –1. Find an optimal solution to the LP-relaxation. –2. Pick all sets S for which x S ≧ 1/f. x S becomes 1 if x S ≧ 1/f. S1S1 S2S2 S3S3 S4S4 S5S5 22 2 1 3 a1a1 a2a2 a3a3 a4a4 a5a5 f =2. Solution by LP-relax. Solution by LP-relax. 0.5 1.0 0.5 1.0 Rounded solution

12. Theorem 14.2 n A 1 (Algorithm 14.1) is a f -approximation algorithm for Set Cover. –We need to consider the following two properties: A 1 outputs a sound solution, which covers all elements. How much is the cost of the solution with A 1 ?

13. Proof of Theorem 14.2 (1/2) n A 1 outputs a sound solution, which covers all elements. –For any element a, there exists a set S s.t. a is in S and x S ≧ 1/f. From the constraints for covers. –Therefore, every element is chosen. At most f x S の値 From the constraints of covers, the sum of the areas of is at least 1. 1/f S s.t. a is in S = f (1/f )=1. f Area of

14. Proof of Theorem 14.2 (2/2) n How much is the cost of A 1 (COST) ? –Let OPT LP (OPT f in the text) be the cost of a solution by the LP-relaxation. –Let x S be a solution by the LP-relax., and y S rounded one. y S ≦ f x S holds since –x S ≧ 1/f, f x S ≧ 1=y S if y S =1. –x S ≧ 0, f x S ≧ 0=y S if y S =0. Therefore, COST ≦ f OPT LP ≦ f OPT. 1/f f xSf xS xS xS 1 yS yS f xSf xS xS xS 1 yS yS

15. Example 14.3 n A set consists of three connected elements in V i. n A cost of each set is 1. n f = 4. n The optimal cost: 2. n In the bottom figure, the cost is 8. V1V1 V2V2 V3V3 x S =1/4. x S =1.

16. Overview n Set Cover –Approximation by simple rounding f-approx. algorithm (f: the frequency of the most frequent element). –Approximation by randomized rounding O(log n)-approx. algorithm (n: # elements to be covered). n Weighted Vertex Cover –Method based on half-integral solutions of the linear programming Each variable takes only 0, 1/2, or 1. 2-approx. algorithm

17. Randomized rounding C=φ % C is a collection of picked sets. n while (C doesn’t satisfy condition A) –Find C by a manner explained later. This C satisfies condition A with prob. more than 1/2. n end-while –[Condition A] C is a solution of set cover. The cost of C is at most OPT LP ・ 4clog n. –c is some constant. –The expectation T of executing loops in while- statement is at most 2.

18. How to find C (1/2) n Compute a solution x S of the LP-relaxation. n for i=1 to clog n –Construct a family C i of picked sets by choosing S with prob x S. n end-for n C= ∪ C i. Solution by LP-relaxation 1.0 C1C1 S1S1 S2S2 C2C2 S1S1 S2S2 S4S4 C3C3 S1S1 S2S2 S5S5 C4C4 S1S1 S5S5 C S1S1 S2S2 S4S4 S5S5

19. How to find C (2/2) n Compute a solution x S of the LP-relaxation. n for i=1 to clog n –Construct a family C i of picked sets by choosing S with prob x S. n end-for n C= ∪ C i. –C is not a set cover with prob. less than 1/4. –The cost of C is more than OPT LP 4c log n with prob. less than 1/4. Less than 1/4 More than 1/2

20. Prob. that element a is in C i n Consider the example below. n The prob. P that any set S i containing element a is P=(1-x S1 )(1-x S2 )(1-x S3 ). –x S1 +x S2 +x S3 ≧ 1 from the constraints of covers. n P is maximized where x S1 =x S2 =x S3 =1/3. S1S1 S2S2 S3S3 S5S5 a1a1

21. Maximum prob. a is not chosen Suppose an element is in each of k sets. Let Fix d, and replace P k as Then, the partial derivative of log g becomes To simplify the problem, instead of maximize PiPi 0 log g +0 － Max This shows P i =P k makes log g maximized. This property holds for any i, then P i =d/k. Under the constraint that d ≧ 1, g takes the max. ((1-1/k) k ) where d=1. d/kd/k

22. Prob. C is not a set cover n Prob. a is not covered by using C i is at most (1- 1/k) k. n Prob. a is not covered by C is at most (1/e) clogn. –Choose constant c s.t. (1/e) clogn ≦ 1/(4n). –c ≧ 5 ≧ (4/log n)+1 (n ≧ 3).

23. Prob. C is not a set cover n Prob. at least one element is not in C is at most 1/4. Less than 1/4n At least one of a 1, a 2, a 3 is not chosen with prob. less than 1/4. Less than 1/4n a 1 is not chosen.a 2 is not chosen. a 3 is not chosen. n=3

24. The cost of C

25. Markov’s inequality (1/2) n Random variable X takes a non-negative value, and the average of X is μ. ≧0≧0 ≧ε≧ε X P(X=x) x x=ε

26. Markov’s inequality (2/2) n Random variable X takes a non-negative value, and the average of X is μ. ≧0≧0 ≧ε≧ε

27. The value of cost (C) Apply Markov’s inequality to cost (C). Prob. the cost of C becomes more than OPT LP 4clog n is at most 1/4.

28. n Each of the following two events happens with prob. less than 1/4. –C is not a set cover, –The cost of C is more than c=OPT LP 4clog n. n Therefore, the event that C is a set cover and its cost is at most c is at least 1/2. Less than 1/4 More than 1/2

29. Overview n Set Cover –Approximation by simple rounding f-approx. algorithm (f: the frequency of the most frequent element). –Approximation by randomized rounding O(log n)-approx. algorithm (n: # elements to be covered). n Weighted Vertex Cover –Method based on half-integral solutions of the linear programming Each variable takes only 0, 1/2, or 1. 2-approx. algorithm

30. Weighted vertex cover n Weighted vertex cover –Input: graph with weights on vertices G=(V,E). –Output: A ⊆ V. For any (u,v) ∈ E, u ∈ A or v ∈ A. The sum of weights of v ∈ A is minimized. 1 4 1 2 55 1 4 1 2 55 v1v1 v2v2 v3v3 v4v4 v5v5 v6v6 v1v1 v2v2 v3v3 v4v4 v5v5 v6v6

31. Weighted vertex cover n Definition –Input: Graph G=(V,E). –Output: A ⊆ V. For any edge (u,v) ∈ E, u ∈ A or v ∈ A. The sum of weights of v ∈ A is minimized. 1 4 1 2 55 a1a1 a2a2 a3a3 a6a6 a4a4 a5a5 1 4 1 2 55 a1a1 a2a2 a3a3 a6a6 a4a4 a5a5 v1v1 v2v2 v3v3 v4v4 v5v5 v6v6 v1v1 v2v2 v3v3 v4v4 v5v5 v6v6

32. Formulation by linear inequalities n Objective function –Minimize: n Constraints –For covers –For choosing edges 1 4 1 2 55 a1a1 a2a2 a3a3 a6a6 a4a4 a5a5 1 4 1 2 55 a1a1 a2a2 a3a3 a6a6 a4a4 a5a5 v1v1 v2v2 v3v3 v4v4 v5v5 v6v6 v1v1 v2v2 v3v3 v4v4 v5v5 v6v6

33. Formulation by linear inequalities n Objective function –Minimize: n Constraints –For covers –For choosing edges 1 4 1 2 55 a1a1 a2a2 a3a3 a6a6 a4a4 a5a5 1 4 1 2 55 a1a1 a2a2 a3a3 a6a6 a4a4 a5a5 v1v1 v2v2 v3v3 v4v4 v5v5 v6v6 v1v1 v2v2 v3v3 v4v4 v5v5 v6v6

34. LP-relaxation n Objective function –Minimize: n Constraints –For covers –For choosing edges 1 4 1 2 55 a1a1 a2a2 a3a3 a6a6 a4a4 a5a5 1 4 1 2 55 a1a1 a2a2 a3a3 a6a6 a4a4 a5a5 v1v1 v2v2 v3v3 v4v4 v5v5 v6v6 v1v1 v2v2 v3v3 v4v4 v5v5 v6v6

35. LP-relaxation n Objective function –Minimize: n Constraints –For covers –For choosing edges 1 4 1 2 55 a1a1 a2a2 a3a3 a6a6 a4a4 a5a5 1 4 1 2 55 a1a1 a2a2 a3a3 a6a6 a4a4 a5a5 v1v1 v2v2 v3v3 v4v4 v5v5 v6v6 v1v1 v2v2 v3v3 v4v4 v5v5 v6v6

36. Extreme point solution n The optimal solution of Linear Programming. n The solution which cannot be expressed as convex combination of two other feasible solution. –Convex combination: A linear equation s.t. the sum of its coefficients is 1. z is a convex combination of x and y, where z =0.8x+0.2y. Feasible solution Convex combination of feasible solution

37. Half-integral solution n Solution of Linear Programming s.t. each value takes 0, 1/2 or 1.

38. 2-approximation algorithm n Compute an extreme point solution x. n Choose any vertex s.t its corresponding value takes 1/2 or 1. –If x is an extreme point solution, each variable takes 0, 1/2, or 1. (Lemma 14.4)

39. Lemma 14.4 n x: a solution of weighted vertex cover obtained by Linear Programming. n If x is not half-integral, x can be expressed as convex combination of two other feasible solution. –x is not an extreme point solution. –Outline of its proof Construct y and z s.t. x is not half-integral and x=1/2(y+z). –x can be expressed by convex combination of y and z. Show y and z are feasible solutions.

40. Proof of Lemma 14.4 (1/3) n Construct other solutions y and z from x, each of them takes 0, 1/2, or 1. 1.0 x 0.5 1.0 y 0.5 1.0 z 0.5 +ε －ε－ε －ε－ε V + ={v 3 }. v i s.t. x vi > 1/2. V － ={v 4 }. v i s.t. x vi < 1/2. +εin y, － εin z. － εin y, +εin z. holds.

41. Proof of Lemma 14.4 (2/3) n Are y and z feasible solutions? –In any feasible solution, x u +x v ≧ 1 holds. xuxu xvxv 1/21 1 Feasible solution yuyu yvyv 1/21 1 Change from x to y zuzu zvzv 1/21 1 Change from x to z Where εis set to a small value, y and z are feasible solutions.

42. Proof of Lemma 14.4 (3/3) n When x u ＋ x v =1, –x u = x v =1/2. y u = y v =z u = z v =1/2 (no change). –x u =0, x v =1. y u =0, y v =1, z u =0, z v =1 (no change). –x u 1/2. y u ＋ y v = x u +ε ＋ x v -ε =1, z u ＋ z v = x u -ε ＋ x v +ε =1. n Then, y and z are feasible solutions, and any solution can be expressed by a half-integral solution.