Approximation Algorithms

1. Approximation Algorithms Greedy Strategies

2. Max and Min • min f is equivalent to max –f. • However, a good approximation for min f may not be a good approximation for max –f. • For example, consider a graph G=(V,E). C is a minimum vertex cover of G iff V \ C is a maximum independent set of G. The minimum vertex cover has a polynomial-time 2-approximation, but the maximum independent set has no constant-bounded approximation unless NP=P. • Another example: Minimum Connected Dominating Set and Minimum Spanning Tree with Maximum Number of Leaves I hear, I forget. I learn, I remember. I do, I understand!

3. Greedy for Max and Min • Max --- independent system • Min --- submodular potential function I hear, I forget. I learn, I remember. I do, I understand!

4. Independent System I hear, I forget. I learn, I remember. I do, I understand!

5. Independent System • Consider a set E and a collection C of subsets of E. (E,C) is called an independent system if The elements of C are called independent sets I hear, I forget. I learn, I remember. I do, I understand!

6. Maximization Problem I hear, I forget. I learn, I remember. I do, I understand!

7. Greedy Approximation MAX I hear, I forget. I learn, I remember. I do, I understand!

8. Theorem I hear, I forget. I learn, I remember. I do, I understand!

9. Proof I hear, I forget. I learn, I remember. I do, I understand!

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12. Maximum Weight Hamiltonian Cycle • Given an edge-weighted complete graph, find a Hamiltonian cycle with maximum total weight. I hear, I forget. I learn, I remember. I do, I understand!

13. Independent sets • E = {all edges} • A subset of edges is independent if it is a Hamiltonian cycle or a vertex-disjoint union of paths. • C = a collection of such subsets I hear, I forget. I learn, I remember. I do, I understand!

14. Maximal Independent Sets • Consider a subset F of edges. For any two maximal independent sets I and J of F, |J| < 2|I| I hear, I forget. I learn, I remember. I do, I understand!

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16. Theorem: For the maximum Hamiltonian cycle problem, the greedy algorithm MAX produces a polynomial time approximation with performance ratio at most 2. I hear, I forget. I learn, I remember. I do, I understand!

17. Maximum Weight Directed Hamiltonian Cycle • Given an edge-weighted complete digraph, find a Hamiltonian cycle with maximum total weight. I hear, I forget. I learn, I remember. I do, I understand!

18. Independent sets • E = {all edges} • A subset of edges is independent if it is a directed Hamiltonian cycle or a vertex-disjoint union of directed paths. I hear, I forget. I learn, I remember. I do, I understand!

19. I hear, I forget. I learn, I remember. I do, I understand!

20. Tightness The rest of all edges have a cost ε ε 1 1 1 1+ε I hear, I forget. I learn, I remember. I do, I understand!

21. A Special Case • If c satisfies the following quadrilateral condition: For any 4 vertices u, v, u’, v’ in V, Then the greedy approximation for maximum weight Hamiltonian cycle has the performance ratio 2. I hear, I forget. I learn, I remember. I do, I understand!

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26. Superstring • Given n strings s1, s2, …, sn, find a shortest string s containing all s1, s2, …, sn as substrings. • No si is a substring of another sj. I hear, I forget. I learn, I remember. I do, I understand!

27. An Example • Given S = {abcc, efaab, bccef} • Some possible solutions: • Concatenate all substrings = abccefaabbccef (14 chars) • A shortest superstring is abccefaab (9 chars) I hear, I forget. I learn, I remember. I do, I understand!

28. Relationship to Set Cover? • How to “transform” the shortest superstring (SS) to the Set Cover (SC) problem? • Need to identify U • Need to identify S • Need to define the cost function • The SC instance is an SS instance • Let U = S (a set of n strings). • How to define S ? I hear, I forget. I learn, I remember. I do, I understand!

29. Relationship to SC (cont) | | | | | | Let M be the set that consists of the strings σijk k si sj σijk I hear, I forget. I learn, I remember. I do, I understand!

30. Relationship to SC (cont) Now, define S Define cost of Let C is the set cover of this constructed SC, then the concatenation of all strings in C is a solution of SS. Note that C is a collection of I hear, I forget. I learn, I remember. I do, I understand!

31. Algorithm 1 for SS I hear, I forget. I learn, I remember. I do, I understand!

32. Approximation Ratio • Lemma 1: Let opt be length of the optimal solution of SS and opt’ be the cost of the optimal solution of SC, we have: opt ≤ opt’ ≤ 2opt • Proof: I hear, I forget. I learn, I remember. I do, I understand!

33. Proof of Lemma 1 (cont) I hear, I forget. I learn, I remember. I do, I understand!

34. Approximation Ratio • Theorem1: Algorithm 1 has an approximation ratio within a factor of 2Hn • Proof: We know that the approximation ratio of Set Cover is Hn. From Lemma 1, it follows directly that Algorithm 1 is a 2Hn factor algorithm for SS I hear, I forget. I learn, I remember. I do, I understand!

35. Prefix and Overlap • For two string s1 and s2, we have: • Overlap(s1,s2) = the maximum between the suffix of s1 and the prefix of s2. • pref(s1,s2) = the prefix of s1 that remains after chopping off overlap(s1,s2) • Example: • s1 = abcbcaa and s2= bcaaca, then • overlap(s1,s2) = bcaa • pref(s1,s2) = abc • Note: overlap(s1,s2) ≠ overlap(s2, s1) I hear, I forget. I learn, I remember. I do, I understand!

36. Is there any better approach? • Now, suppose that in the optimal solution, the strings appear from the left to right in the order: s1, s2, …, sn • Define: opt = |pref(s1,s2)| + …+|pref(sn-1,sn)| + |pref(sn,s1)| + |overlap(sn,s1)| Why overlap(sn,s1)? Consider this exampleS={agagag, gagaga}. If we just consider the prefix only, the result would be ag whereas the correct result is agagaga I hear, I forget. I learn, I remember. I do, I understand!

37. Prefix Graph • Define the prefix graph as follows: • Complete weighted directed graph G=(V,E) • V is a set of vertices, labeled from 1 to n (each vertex represents each string si) • For each edge i→j, i ≠ j, assign a weight of |pref(si, sj)| • Example: S={abc, bcd, dab} 1( ) ( )3 2( ) I hear, I forget. I learn, I remember. I do, I understand!

38. Cycle Cover • Cycle Cover: a collection of disjoint cycles covering all vertices (each vertex is in exactly one cycle) • Note that the tour 1 → 2 → … → n → 1 is a cycle cover • Minimum weight cycle cover: sum of weights is minimum over all covers • Thus, we want to find a minimum weight cycle cover I hear, I forget. I learn, I remember. I do, I understand!

39. How to find a min. weight cycle cover • Corresponding to the prefix graph, construct a bipartite graph H=(X,Y;E) such that: • X = {x1, x2, …, xn} and Y = {y1, y2, …, yn} • For each i, j (in 1…n), add edge (xi, yj) of weight |pref(si,sj)| • Each cycle cover of the prefix graph ↔ a perfect matching of the same weight in H. (Perfect matching is a matching which covers all the vertices) • Finding a minimum weight cycle cover = finding a minimum weight perfect matching (which can be found in poly-time) I hear, I forget. I learn, I remember. I do, I understand!

40. s12 s13 s11 How to break the cycle I hear, I forget. I learn, I remember. I do, I understand!

41. A constant factor algorithm Algorithm 2: I hear, I forget. I learn, I remember. I do, I understand!

42. Approximation Ratio • Lemma 2: Let C be the minimum weight cycle cover of S. Let c and c’ be two cycles in C, and let r, r’ be representative strings from these cycles. Then |overlap(r, r’)| < w(c) + w(c’) • Proof: Exercise I hear, I forget. I learn, I remember. I do, I understand!

43. Approximation Ratio (cont) • Theorem 2: Algorithm 2 has an approximation ratio of 4. • Proof: (see next slide) I hear, I forget. I learn, I remember. I do, I understand!

44. Proof I hear, I forget. I learn, I remember. I do, I understand!

45. Modification to 3-Approximation I hear, I forget. I learn, I remember. I do, I understand!

46. 3-Approximation Algorithm • Algorithm 3: I hear, I forget. I learn, I remember. I do, I understand!

47. Superstring via Hamiltonian path • |ov(u,v)| = max{|w| | there exist x and y such that u=xw and v=wy} • Overlap graph G is a complete digraph: V = {s1, s2, …, sn} |ov(u,v)| is edge weight. • Suppose s* is the shortest supper string. Let s1, …, sn be the strings in the order of appearance from left to right. Then si, si+1 must have maximum overlap in s*. Hence s1, …, sn form a directed Hamiltonian path in G. I hear, I forget. I learn, I remember. I do, I understand!

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50. The Algorithm (via Hamiltonian) I hear, I forget. I learn, I remember. I do, I understand!