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Set Cover via LP-Rounding

Set Cover via LP-Rounding. Yen Fu Chang (張演富) Institute of Bioinformatics, National Chiao Tung University yfchang@cs.pu.edu.tw. Outline. How to formulate the set cover problem into a linear programming LP-relaxation of set cover problem Dual program of set cover problem

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Set Cover via LP-Rounding

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  1. Set Cover via LP-Rounding Yen Fu Chang(張演富) Institute of Bioinformatics, National Chiao Tung University yfchang@cs.pu.edu.tw

  2. Outline • How to formulate the set cover problem into a linear programming • LP-relaxation of set cover problem • Dual program of set cover problem • A simple rounding algorithm • Randomized rounding • Half-integrality of vertex cover Workshop on String Processing and Approximation Algorithms

  3. Set cover problem • Input: given a set U of n elements, a collection of subset of U,S = { S1,…,Sk }, and a cost function c: S→Q+ • Output: a minimum cost subcollection of S that covers all elements of U Workshop on String Processing and Approximation Algorithms

  4. How to formulate the set cover problem into a linear programming • We assign a variable for each set that is allowed 0/1 value. • This variable will be set to 1 iff set Si is picked in the set cover. • The constraint is that for each element we want that at least one of the sets containing it be picked. Workshop on String Processing and Approximation Algorithms

  5. LP-relaxation of set cover problem (1) Workshop on String Processing and Approximation Algorithms

  6. LP-relaxation of set cover problem (cont.) The upper bound on xsi is redundant. (2) Workshop on String Processing and Approximation Algorithms

  7. Example Let U = { 1, 2, 3 } and the specified sets be S1 = { 2, 3 } , S2 = { 1, 3 } , S3 = { 1, 2 }, and each subset has cost 1. minimize subject to • An integral cover must pick two of the subsets for a cost of 2. • On the other hand, picking each set to the extent of 1/2 gives a fractional cover of cost 3/2. Workshop on String Processing and Approximation Algorithms

  8. The idea of LP-rounding • Apparently the optimal solution for LP(2) is less than or equal to that of LP(1), but it can be obtained in polynomial time because LP(2) is a linear program. • However, the optimal values are not necessarily integers in LP(2). • The idea behind the rounding technique is to solve the LP-relaxation of the integer linear programming by a polynomial-time solver, and convert the fractional solution into an integer one. Workshop on String Processing and Approximation Algorithms

  9. Naïve LP-rounding • All nonzero values are rounded up to 1 in the simplest approach. • This approach will increase cost by a factor of Ω(n). Workshop on String Processing and Approximation Algorithms

  10. Proof of the approximation ratio Ω(n) • Consider the following instance of the set cover problem: U={1,2,…n}, S={S1,S2…Sn}, where Si=U \ i, c(Si)=1,i {1,2,…n}. and the following setting for the variable : Workshop on String Processing and Approximation Algorithms

  11. Proof of the approximation ratio Ω(n) (cont.) • The optimal value of objective function is . • Each xs value is rounded up to 1, so the cost becomes n . • However, an optimal solution for this instance is obtained by picking any two sets, which has cost of 2. • Thus, the approximation ratio is Ω(n). Workshop on String Processing and Approximation Algorithms

  12. A simple LP-rounding algorithm • Let f be the frequency of the most frequent element of U. • Theorem: Algorithm 1 achieves an approximation factor of f for the set cover problem. Workshop on String Processing and Approximation Algorithms

  13. Example Let U = { 1, 2, 3 } and the specified sets be S1 = { 2, 3 } , S2 = { 1, 3 } , S3 = { 1, 2 }, and each subset has cost 1. • Picking each set to the extent of 1/2 gives a fractional cover of cost 3/2. • The frequency of most frequent element of U is 2. • Each element is rounded up to 1. Hence, the rounded cost is 3. Workshop on String Processing and Approximation Algorithms

  14. Proof • Let C be the collection of picked sets, and consider an arbitrary element e. • Since e is in at most f sets, one of these sets must be picked, and hence C is a valid set cover. • The rounding process increases by a factor of at most f. • Therefore, the cost of C is at most f times the cost of the fractional cover. Workshop on String Processing and Approximation Algorithms

  15. Randomized LP-rounding • We consider the fractional values obtained by solving LP-relaxation as probabilities of rounding. • The problem with this approach is that we can never be sure if we really end up with a valid cover. • Repeating this process O(logn) times, and picking a set if it is chosen in any of the iterations, we get a set cover with high probability. We provide details in following. Workshop on String Processing and Approximation Algorithms

  16. Randomized LP-rounding (cont.) • Suppose that a occurs in k sets of S, and let the probabilities associated with these sets be p1 ,…,pk. • The probability that a is covered by C is minimized when each of the pi’s is 1/k. • Let C be the cover produced in one round. Workshop on String Processing and Approximation Algorithms

  17. Randomized LP-rounding (cont.) • Hence each element is covered with constant probability by C. • To get a valid set cover, independently pick clogn such subcollections, and compute their union, say C’, where c is a constant such that Workshop on String Processing and Approximation Algorithms

  18. Randomized LP-rounding (cont.) • Summing over all elements , we get • Clearly, . • By applying Markov’s Inequality with t = OPTf · 4clogn, we get Workshop on String Processing and Approximation Algorithms

  19. Randomized LP-rounding (cont.) • The probability of the union of two undesirable events is <=1/2, and hence • Observe that we verify in polynomial time whether C’ satisfies both these conditions. If not, we repeat the entire algorithm. • The expected number of repetitions is at most 2. Workshop on String Processing and Approximation Algorithms

  20. Vertex cover problem • Input: An undirected graph G=(V,E), and a cost function on vertices c : V→Q+. • Output: Find a minimum cost vertex cover, that is, a set V’ V such that every edge has at least one endpoint incident at V’. Workshop on String Processing and Approximation Algorithms

  21. Linear programming of vertex cover • The integer program for the vertex cover problem with arbitrary weights is: • The LP-relaxation of the integer program is: (3) (4) Workshop on String Processing and Approximation Algorithms

  22. Extreme point solution • An extreme point solution of a set of linear inequalities is a feasible solution that can not be expressed as convex combination of 2 other feasible solutions. Workshop on String Processing and Approximation Algorithms

  23. Half-integrality of vertex cover • A half-integral solution to LP(4) is a feasible solution in which each variable is 0, 1, or ½. Workshop on String Processing and Approximation Algorithms

  24. Lemma • Let x be a feasible solution to LP(4) that is not half-integral. Then, x is the convex combination of two feasible solution and is therefore not an extreme point solution for the set inequalities in LP(4). Workshop on String Processing and Approximation Algorithms

  25. Proof • Consider the set of vertices for which solution x does not assign half-integral values. Partition this set as follows. • For ε>0, define the following solutions. Workshop on String Processing and Approximation Algorithms

  26. Proof (cont.) • x is a convex combination of y and z, since • We will show, by choosing ε>0 small enough, that y and z are both feasible solutions for LP(4),thereby establishing the lemma. Workshop on String Processing and Approximation Algorithms

  27. Proof (cont.) • Consider the edge constraints: • xu+xv>1 By choosing εsmall enough, we can ensure that y and z do not violate the constraint for such an edge. • xu+xv=1 three cases for xu and xv: (a) xu = xv=1/2; (b) xu = 0, xv=1; (c) xuV+ ,xv V-; Workshop on String Processing and Approximation Algorithms

  28. Proof (cont.) • In all three cases, for any choice of ε, (a) xu+xv=yu+yv=zu+zv (b) xu+xv=yu+yv=zu+zv (c) yu+yv=xu+ε+xv-ε=xu+xv zu+zv=xu-ε+xv+ε=xu+xv xu+xv=yu+yv=zu+zv The lemma follows. Workshop on String Processing and Approximation Algorithms

  29. Proof (cont.) • This leads to: Theorem Any extreme point solution for the set of inequalities in LP(4) is half-integral. Workshop on String Processing and Approximation Algorithms

  30. Half-integrality of vertex cover • The theorem directly leads to a factor 2 approximation algorithm for weighted vertex cover: find an extreme point solution, and pick all vertices that are set to half or one in this solution. Workshop on String Processing and Approximation Algorithms

  31. Thank you for your attention. Workshop on String Processing and Approximation Algorithms

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