Slides on Approximation algorithms, part 2: Basic approximation algorithms

Transcription

1 Approximation slides 1 Slides on Approximation algorithms, part 2: Basic approximation algorithms Guy Kortsarz

2 Approximation slides 2 Subjects covered Basics: Approximating TSP, VC and k center Submodular-cover problems Local Ratio Duality and Primal-dual approximation

3 Approximation slides 3 Finding a lower bound; the TSP example The optimum TSP cycle P is an edge plus a spanning tree. Thus, the minimum cost spanning tree T has cost c(t ) c(p ) We use the idea of shortcuts If we are given a (not necessarily simple) a to b path P, replacing the path P by (a, b) can not increase the cost (because of triangle inequality). Claim: Given an Euler cycle (a cycle that crosses all edges once) of cost c there exists a TSP path of cost at most c Pf: Shortcut

4 Approximation slides 4 TSP, Cont. Getting an Euler cycle: double every edge in MST T Gives an Euler cycle (all degrees are even) T * a 4 a 4 b 3 c 4 2 d b 3 3 c 2 2 d e 2 1 f e f c a b a c d c f c e c c a b d f e

5 Approximation slides 5 Steiner tree Input: AgraphG(V,E) a weight function w : E + and a subset S V of terminals Required: A subtree T of G that contains all of S and has minimum cost It is easy to see that we may assume that w is complete: add/replace e =(u, v) bythecostof the shortest u to v path. If an edge (u, v) is used and does not belong to G replace it by the edges of the shortest path. The cost does not increase. Thus, if we restrict ourself to G(S) spanned by S and find a minimum spanning tree on G(S) a ratio 2 approximation results. The reason: The Euler path that result the optimum (by doubling edges) can first be shortcut to contain S vertices only and then shortcut to an Hamiltonian path which in particular is some spanning tree

6 Approximation slides 6 Improving the ratio to 3/2 Idea due to Christophedes Any TSP solution is a Hamilton path; A simple path that contains all vertices. Any TSP tour of even length decomposes into two perfect matchings

7 Approximation slides 7 1. Find MST T (V,E ) The algorithm 2. Let X V be the vertices of odd degree in T 3. Compute the graph G X =(X, X X). Find a minimum cost perfect matching M in G X 4. Find an Euler cycle in T M 5. Shortcut Analysis: First, shortcut the tour to X Let M 1,M 2 be the two perfect matchings of P on X c(p )=c(m 1 )+C(M 2 ) 2c(M) Thus c(t M) 3opt/2 Hence 3/2 ratio

8 Approximation slides 8 Finding lower bound: The Unweighted vertex-cover example Figure 1: VC M The size of any matching lower bounds the minimum VC A maximal matching gives a size 2 M vertex-cover Hence, ratio 2.

9 Approximation slides 9 A lower bound is not always required: certificate of failure Say that for input I we want to find a feasible solution minimizing some integral function µ(i). Let opt be opt = min I µ(i). For an integer x say that we have a procedure P (I,x) that has one of the following inputs: 1. Either it determines that x<optand then returns False 2. Or, it returns a solution S(I) False of cost at most µ(i) ρ x P (I,x) called a certificate of failure procedure (Hochbaum and Shmoys). Claim: The above procedure can be used as an oracle to produce a ρ approximation algorithm.

10 Approximation slides 10 Binary search Assume the costs are integral 1. lb min, ub max /* lb, ub some lower and upper bounds over minimum and maximum possible value of µ(i) */ While max min > 1 do (a) x (lb + up)/2 (b) If P (I,lb) is false, then lb x (c) Else, ub x 4. Return ub Since opt > lb and opt integer, opt ub. So, ρ ratio

11 Approximation slides 11 Running time Claim: If, P (I,x) is polynomial in I and ub lb always O(exp( I )) then polynomial We now use this to give ratio 2 approximation for the k center problem The input a complete graph on the vertices {1,...,n}. A bound k onthenumberof centers. Every i, j have distance d ij.we assume the triangle inequality (otherwise, no approximation possible).

12 Approximation slides 12 A 2 approximation for undirected k center The algorithm. It is a certificate of failure algorithm with (I,x).Due to Hochbaum and Shmoys. 1. S 2. V V 3. While S is not a legal solution, do (a) Add to S and arbitrary vertex i V (b) Delete from V all vertices j so that l ij 2 x 4. If S >kreturn False 5. Else, return S

13 Approximation slides 13 The two properties 2x 2x 2x 2x Figure 2: An illustration of the algorithm in the special case of points in the plane. The distance between any two centers is more than 2x All centers have pairwise distance larger than 2x. Consider our centers that are non-centers in OPT

14 Approximation slides 14 Analysis cont. Let j and p be two centers in our algoritm but not in OPT If q OPT covers both j and p and opt x then by triangle inequality: d j,p d j,q + d p,q 2x. This is a contradiction. If j e.g. computed before p then p is removed So, since S >kthe optimum would have more than k, contradiction. Thus S >kimplies that there could not be a k subset that covers all of S. Inother words, x<opt. This proves that the failure certificate is correct. Remark: It can be that x<optbut the procedure succeeds

15 Approximation slides 15 The two properties. Continued If returns a solution then by construction the solution has radios 2x. Hence ρ =2 We can use binary search, as opt max distance. Thus, 2 approximation Also better than 2 is as hard as solving: The dominating set problem: Input: G(V,E) andk Question: Is there a dominating set of size at most k, namely, a subset U V, U k so that U N(U) =V? This problem is NPC.

16 Approximation slides 16 Why a ratio better than 2 is not possible Give edges of G length 1 and non-edges in V V \ E length 2 This implies the triangle inequality holds There is a dominating set of size k if and only if there is a k centersolutionofsizek Approximating within 2 ɛ implies in the case of a yes instance an optimal solution Figure 3: Edges are given length 1. Non-edges are gives length 2. Not all non-edges are shown

17 Approximation slides 17 Greedy approximation for set-cover 1. X 2. B R B 3. While B R do: (a) Let x be the set in A \ X maximizing N(x, B R ). (b) X X {x}. (c) B R B R \ N(x). 4. Output X

18 Approximation slides 18 Logarithmic gap: X_1 X_2 X_3 Y_1 Y_2 Y_3 The optimum here has 3 sets The greedy algorithm does not notice that: in every iteration half of the elements are covered Ratio Θ(log n) This is the worse case!

19 Approximation slides 19 The largest gap is ln n: Let X be the minimum cover Let B R be the uncovered elements X covers B R Thus, there is a set that covers at least B R / X elements The number of remaining elements: ( B i+1 R Bi r 1 1 ), X B i R B ( 1 1 ) i X, Thus, for i = X ln( B ) a single element remains in B R

20 Approximation slides 20 An analysis of weighted set-cover via charging Consider the optimum cover, and how its elements are covered. 3/11 3/9 3/5 2/2 y The set y belongs to the optimum. The elements that y covers are arranged from left to right by the order the greedy algorithm covers them. In the example, greedy covers 3 elements, then 3 more and 3 more and then the last 2

21 Approximation slides 21 The charging scheme When a set of x is added to X we will charge its cost of 1 to sets in OPT X. Thechargeony OPT X depends on how many mutual neighbors x and y have in B R. Let d = deg BR (x). Let j y be the number of elements in B R that are neighbors of both x and y Charge y by j y /d We can upper bound the charge by j y /deg y (B R ) because by the greedy choice: d deg y (B R ) The previous figure shows an example for such acharge

22 Approximation slides 22 Analysis Replace the charge by a sequence of decreasing fractions as in the figure: 1/11 + 1/10 + 1/9 + 1/8 + 1/7 + 1/6 + 1/5 + 1/4 + 1/3 + 1/2 + 1/1 y For example, we replaced 3/11 by 1/11 + 1/10 + 1/9 > 3/11. This shows every set in OPT is charged by at most H(deg(y)). Thus, the approximation at most H( A )

23 Approximation slides 23 Set Cover with hard capacities: We are given a set cover instance and a load bound L a for a A Every set a of A can not cover more than L a elements of B Example: The usual minimum Set-Cover has size 2. With a load of L = 2 for all, the minimum Set-Cover size is 3

24 Approximation slides 24 A natural Heuristic Take the set in A covering maximum new elements in B R takingintoaccounttheload constraints. This heuristic has unbounded ratio: L = B /2.

25 Approximation slides 25 Best assignment for X is polynomially computable To compute an assignment maximizing the number of elements covered: X B x L x 1 s 1 t G L y 1 y The maximum flow equals the maximum number of elements X can cover obeying load-constraints Denote by f(x) this maximum

26 Approximation slides 26 The idea Add the set maximizing the additional elements covered under the flow computation (this idea is due to Wolsey): X is the current set Consider a set x A Compute the flow f({x} X), for all x X Choose the one maximizing f({x} X) f(x) The fractional argument works. There is always a set covering a 1/ X fraction. The proof is not trivial.

27 Approximation slides 27 A general problem We generalize both set-cover, set-cover with hard capacities and the algorithm Given, a set U (U plays the role of A, namely of the sets). Every subset X U has a cost c(x). Usually, we assume cost function is sum of costs (not always). Define a monotone non-negative function f Feasible sets: All X so that f(x) =f(u). Goal: find a minimum cost X so that f(x) =f(u) For set-cover f(x) = N(X, B R ). For hard capacities set-cover: f(x) =max flow with X vertices as centers.

28 Approximation slides 28 The generic algorithm Let X (x, f) =f(x {x}) f(x). 1. X. 2. While X is not feasible Do: (a) Stopping condition: If f(u) f(x) t, return a ratio t solution (b) Add to X the set x for which is maximum d X (x, f) = (x, S) c(x) 3. Return X

29 Approximation slides 29 The density claim The Density Claim [Johnson-74] Claim: If at any moment the x added into X satisfies: d X (x, f) c(x) f(u) f(x) c(opt) then the ratio of the final X is bounded by: t +ln(max x (,x)). For set-cover t =1. ln A +1ratio.

30 Approximation slides 30 Improvement independance To get the density condition: Improvement independence (Wolsey): For all X, Y it is required that: (f({y} X)) f(x)) f(y X) f(x). y Y \X The density condition follows by averaging. Definition 1 Problems seeking X so that f(x) =f(u), for a monotone nondecreasing f, so that f obeys improvement independence are called submodular-cover problems

31 Approximation slides 31 Improvement independence, some examples For set-cover f(x) = N(X, B R ). (X, x) = N(x, B R ) \ N(X, B R ). Clearly x T \X (X, x) (X, T). In l.h.s of the sum, many B R elements are covered many times For set-cover with loads: Needs to show that the addition of a set T \ X as sources (in the flow) gives no larger advantage than the sum of additions of flow resulting from every T \ X vertex made a source separately. Many ways to prove improvement independence. One way: flow decomposition. Consider Ax b, 0 x i 1, a i,j 0. This generalizes set-cover. The function f(x) defined as i min{ j X a i,j x j,b i }

32 Approximation slides 32 Improvement independence: (X, p) = i min{ j X {p} a i,j x j i j X a i,j x j, b i i j X a i,j x j } If adding T \ X together to X more chance that value of inequality i will be larger than b i (b i i j X a i,j x j } will be the minimum). O(log n +logm) follows

33 Approximation slides 33 A sample of additional problems approximated by these methods 1. Steiner trees with vertex costs: spider decomposition. Density argument. Klein Ravi 2. Sparse 2-spanners: decompose the optimum to stars. Density arguments 3. Finding MST among trees of diameter 5. Reduction to submodular cover 4. Finding min-cost trees among trees of diameter at most D. Density arguments+recursion. 5. Two level facility location. Zhang. Similar to the previous item. 6. Source location: Choose min-cost centers. Every non-center needs a pre-describe number of disjoint paths to centers. Reduction to submodular cover.

34 Approximation slides 34 More examples 1. Directed Steiner Charikar et al. n ɛ ratio. Density arguments+recursion. 2. TSP with time windows. Chekuri and Pal. Density+recursion 3. Min-power vertex k-connectivity. The powerofagraphisthesumoverall vertices of the largest cost edge touching the vertex. Essentially O(log 2 n)ratio using density arguments 4. Non-uniform buy at bulk. Polylog ratio. Density arguments 5. Steiner network vertex disjoint paths version. 0 and 1 edge costs. Essentially an O(log n) ratio approximation by reduction to submodular cover 6. Many more...

35 Approximation slides 35 When does independence fail? Example: G(V,E). f(v ) = #of edges internal to V. The goal: Find a minimum size V so that f(v ) k Start with an arbitrary vertex. Add in every iteration the new vertex that increases the number of internal edges the most. The greedy fails here. Adding many vertices together much better than sum of individual contributions This problem is closely related to finding dense k subgraphs. An approximation of log n is not known Think of a clique attached to a path

36 Approximation slides 36 Other systematic greedy methods: Local ratio First way to look at local ratio: reduction in opt Consider a general covering problem with universe U and a non-decreasing function f(s) for every S U. The goal is to find min-size (cost) S so that f(s) =f(u). Let I be the instance of a covering problem. A local-ratio algorithm maintains a partial (infeasible) solution F and augments F with new elements in iterations. Initially F. At any given iteration the algorithm adds some subset S U\F into F.

37 Approximation slides 37 Local ratio (cont d) Let I = I (S, F) be the residual resulting instance after the addition of S into F. ThusI is the problem of augmenting F S into a feasible solution. If the following holds: 1. cost(s) ρ α i 2. opt(i) opt(i ) α i The step is called a local-ratio step of value ρ Therefore, the local ratio is: ρ = what the algorithm pays reduction in opt.

38 Approximation slides 38 The local ratio theorem Theorem: An algorithm that finds a cover using only local ratio steps all of which have value bounded by ρ is a ρ approximation algorithm. Pf: Clearly, i α i opt Algorithm pays ρ i α i ρ opt. A local point of view. Due to R. Bar-Yehuda

39 Approximation slides 39 The unweighted vertex-cover example a b Figure 4: Add a, b into the solution. The optimum value decreases by at least 1. We pay 2. Local ratio ρ =2.

40 Approximation slides 40 Eliminating length 3 cycles in VC Figure 5: Iteratively add the vertices of a triangle into the solution (remove covered edges). The optimum value decreases by at least 2. We pay 3. Local ratio ρ =3/2. 1. There is no 3/2 ratio known for VC. Some think 2 ɛ, ɛ constant, NP-hard. 2. As hard as in a graph with no triangles

41 Approximation slides 41 Directed multicuts Input: Directed graph and {s i,t i } Goal: Delete as few edges as possible so that for every i no s i to t i path remains. (See later for elaborated discussion on directed multicuts). Say that we try to get a ρ ratio. We may get dist(s i,t i ) ρ at local-ratio ρ. As long as an s i to t i path of length at most ρ exists add all the edges of the path into our solution. We added at most ρ. The optimum added at least 1. Local ratio at most ρ Hence, may assume that the distance is at least the ratio we shoot for

42 Approximation slides 42 A more complex example: basic problem of augmentation connectivity by 1 Given: A connected graph G(V,E) and extra edges E A V V, E E A =. Find: AsetF E A of edges, such that G(V,E F )is2 edge connected Minimize: F Reduction into a tree:

43 Approximation slides 43 A C B A C B Connected components are contracted to vertices

44 Approximation slides 44 The TAP problem Given: A graph G(V,E), a spanning tree T (V,E ) Find: A subset F E \ E so that in G(V,E F ) every edge is contained in at least one cycle. Minimize: The F. This problem is perhaps the most basic augmentation question. Also equivalent to covering of a laminar family and the augmentation of connectivity from k to k +1forodd k (via the Cactus model).

45 Approximation slides 45 Every solution needs a link on every leaf There is a good matching over the leaves, and a bad one. A gap of 2.

46 Approximation slides 46 The DFS example: (all links are backward edges) In this case, the problem is polynomial: For every leaf take its highest link. Contract the cycles. Recurse.

47 Approximation slides 47 The Leaf Lower Bound Let opt be the size of the optimum. Then: opt L /2. Large gap: a path. L =2,opt = n 1

48 Approximation slides 48 Leaf-closed subtrees T A subtree is leaf-closed if all links touching leaves remain within T.

49 Approximation slides 49 Minimally Leaf-closed trees [NI] v u w z

50 Approximation slides 50 Covering leaf-closed subtrees Thus a subtree is minimally leaf-closed iff: All links touching leaves remain within T. T is minimal with respect to containment Covering a leaf-closed tree: If a leaf has a side link (link not leading to ancestors) add the link. Else add the highest reaching (closest to the root) link of the leaf.

51 Approximation slides 51 A local-ratio 2-approximation 1. Compute a minimally leaf-closed tree T. 2. Use the above links to cover T 3. Contract T, recurse on T \ T

52 Approximation slides 52 The (local) ratio 2 argument v u w z opt(t ) L(T ) /2. Alg(T )= L(T ). Induction: opt(t \ T ) Alg(T \ T )/2

53 Approximation slides 53 An algorithm for weighted vertex-cover 1. X, E R E, for every v, c (v) c(v), G G 2. While X is not feasible do (a) Add all v so that c (v) =0intoX (b) Remove all edges touching X from G (c) Pick an edge e =(u, v) and say that c (u) c (v) (d) Set c (u) 0andc (v) c (v) c (u) 3. Return X

54 Approximation slides 54 Analysis Say that you pick an edge e =(u, v) and reduce both costs of u and of v by c (u) Clearly, as the optimum contains at least one of u, v the optimum decreases by at least c (u) The total reduction in the weights of the vertices is 2 c (u) An immediate ratio 2 follows.

55 Approximation slides 55 Steiner forest The Steiner forest: connect at minimum cost all given pairs s i,t i (so add edges E and s i,t i need to be in the same connected component in (V,E )). The idea: Reduce weights of edges. Edges of cost 0 are added into the solution. We maintain connected components X i induced by the so far added edges (edges of weight 0). The components are shrunk into one single node. A connected component X i is active if there is some s j X i and t j X i At the beginning every s i and t i is a trivial connected component. Only edges between different components are considered for addition. Hence edges in the same CC are discarded.

56 Approximation slides 56 A try to get good ratio At every iteration reduce the weight of edges between two active components by 2ɛ If one active one non-active component reduce by ɛ Choose the maximum possible ɛ so that the cost of an edge becomes 0 (not negative). Shrink zero cost components and recurse

57 Approximation slides 57 This algorithm fails b 1 a 1 v 3 u 1 1 z 1 d c

58 Approximation slides 58 The run of the algorithm Say that we want to connect the pairs u, v and u, z. {u}, {v} and {z} are active. Reduce the costs by 1. (v, a), (v, b), (v, c) and (v, d) are added. (z,d) also added. Now, P = a, b, c, d, v, z form one active component. u forms another. The edge between u and P has now cost 1. This edge is added. We can add as many star edges such as v, a. They can make the cost unbounded

59 Approximation slides 59 Paying only for minimal solutions However, the solution we found is not minimal All edges but (z,d), (d, v), (u, v) maybe removed. Total cost: 5 Total reduction: in the final 3 edges remaining, total reduction exactly 5 The solution after minimality is actually optimalinthiscase

60 Approximation slides 60 The ratio is 2 Reduction in OPT: No edge added between any two non-active components Let active=black. Denote this set by B. Number of black components b. Non-active=white: The set is W.Numberof white nodes w. As in OPT every active touched by an edge the decrease in OPT at least ɛb: If an edge between two active nodes belongs to OPT assign one ɛ to each of the 2ɛ reduction Else, assign the single ɛ to the active one Total assigned: ɛb.

61 Approximation slides 61 Reduction in cost in any minimal solution The crucial property: because of minimality, all leaf components must be active The reduction in cost: sum of degrees of vertices in B in the forest of active sets. ɛ v B deg(v) =ɛ v V deg(v) ɛ v W deg(v). As deg(v) 2forv W we get: deg(v) ɛ(2n 2w 2) = ɛ(2b 2). v B Hence, less than 2 local ratio

62 Approximation slides 62 What happened here? Some remarks The forest on which we do the above computation contains ALL active nodes at this time. Namely, it looks on the final minimal solution and checks how much cost it invested in connecting the current active components Because of the recursion, the delete is reverse delete namely edges are removed in inverse order of their addition Thereasontogiveweight2ɛ to edges between 2 active components, etc, is because then we can measure the reduction in a minimal solutions by sum of degrees We have not much knowledge on the final minimal solution, hence we need to argue that any minimal solution is a ratio 2 approximation for appropriate 2ɛ and ɛ weights The game: define weights that are good for any minimal solution

63 Approximation slides 63 More remarks We used two crucial properties: 1. If all edges coming out of the active component get lower cost, the cost of the optimum is reduced 2. We know what to do with edges of cost 0 Example: Say that we are given a weighted graph and want to find the min-cost k-vertex connected subgraph. Reduce the weights of all edges by ɛ. Clearly the optimum cost is reduced by µ ɛ or more, where µ kn/2 is the number of edges in the optimum But we can not continue this: edges of cost 0 can not be contracted. If we contract, the resulting solution will not be vertex k-connected.

64 Approximation slides 64 The decomposition view of LR Consider a minimization covering problem. Let I be the instance. 1. Let the cost function be w. 2. Decompose w = w 1 + w 2, w 1 0,w 2 0 not identically 0 3. Two instances I 1 =(I,w 1 )andi 2 =(I,w 2 ) are formed. The idea: Approximate I 1 and I 2 separately Thus, the problem is separated into two different problems with same feasible solutions but different weights

65 Approximation slides 65 The local ratio theorem If you derive an α approximation for I 1 and for I 2 then the union of the solutions is an α approximation. opt(i,w) = = x OPT(I) x OPT(I) w(x) (1) w 1 (x)+ opt(i 1 )+opt(i 2 ). x OPT(I) w 2 (x) Gives a way to lower bound (OPT): Clearly, (OPT) =opt(i,w) opt(i,w 2 ). Thus: because of (1) (OPT) opt(i 1 )

66 Approximation slides 66 Weighted VC revisited The input is G(V,E) andw. Let v be the least cost vertex. Let (u, v) bean arbitrary edge of v. Define w 1 to be w 1 (u) =w 1 (v) =w(v), w 1 (z) =0,z u, v and w 2 = w w 1. Vertices of cost 0 are added into the solution. Their edges are discarded (covered). Hence the above LR algorithm has ratio 2 if we show that the cost added is at most 2 opt(w 1,G) Clearly opt(w 1,G) w(e) And the added cost at most 2 opt(w 1,G)

67 Approximation slides 67 The the local ratio theorem via minimal solution For minimization problems: Try to find w 1 so that Every minimal feasible solution is a ρ approximation for w 1. Reduce all costs by w 1 (some cost of an edge must become 0) Handle the zero cost edges Gives an immediate ρ ratio solution.

68 Approximation slides 68 Steiner forest revisited For the sake of the example, use another cost function (not as before). w 1 gives the same weight to all edges of E. Namely, the weight for every e is w 1 (e )=w(e) =ɛ (the epsilon would be the smallest cost edge). The optimum for uniform cost, at least b 1 Every Minimal solution hascostatmost the sum of degrees of active components in the final forest. We saw before that since all leaves are black, at most 2b cost for every minimal.

69 Approximation slides 69 Vertex feedback set A useful cost function: give every vertex v ɛ deg(v) (again, the largest possible ɛ so the number do not exceed the minimum weight). We need to show: for any minimal feasible solution A V, deg(v) 2 deg(v). v A Complex proof, but true. v OPT

70 Approximation slides 70 Primal-dual and using weak duality: dual fitting Let op and opt f be the minimum integral and fractional solution for some minimization problem. Let y be a feasible solution for the dual LP and c y its value. Then: c y opt f opt. If we find a feasible solution of cost at most ρ c y a ρ ratio applies. Main point: y does not need to be a dual-optimum solution

71 Approximation slides 71 Dual-fitting for unweighted set-cover The dual to unweighted SC: Maximize b B y b subject to b N(a) y b 1 y b 0 for every a A

72 Approximation slides 72 Intuition An integer solution is a packing: b1 b2 b3 b4 The heads of the stars have no mutual neigbhors Every head of star requires a different covering a A.

73 Approximation slides 73 The fractional ratio The following dual-fitting ratio claim is usually attributed to Lovats, 1975 (but was known way before and is folklore). Was the first Primal-Dual approximation. Let d be the optimal fractional dual Let opt be the minimum size set-cover. Let A be the largest degree in A. Let A be the set returned by the greedy algorithm. Let opt f be the optimal value of the fractional set-cover Claim: A d H( A). Implies: H( A ) approximation ratio A opt A opt f = A d H( A).

74 Approximation slides 74 Proof Consider a star found in greedy versus a star in OPT a a* a A. If the degree of a is now d, the degree of any a OPT \ A is at most d

75 Approximation slides 75 Finding a dual (in)feasible Define a dual: For the i star, let d i be the number of its leaves. Give value y b =1/d i for every b that belongs to the i star in greedy b1 b2 b3 b4 b5 b6 b7 b8 b9 a1 a2 a3 a4 a5 a6 b1 b2 b4 b6 b8 b9 b3 b7 b5 a1 a6 a3 a4 Thus, y b1 = y b2 = y b4 =1/3. And, y b6 = y b8 = y b9 =1/3. And, y b3 = y b7 =1/2, and y b5 =1.

76 Approximation slides 76 Adjustment for the dual Let y denote the above vector. Remark: The vector y cant be dual-feasible. Beacuse the fractional value is exactly A. Andwouldimplygreedyisoptimal. But: y/h( A ) is dual feasible. If we show this, then: y i /H( A ) d = opt f A. And we are done i

77 Approximation slides 77 Proof, Cont. Denote the head of star number i by a i. Look at any a A. whena i is added to A. When a i is added, let the degree of deg(a i,b R )=d i In the same moment, deg(a, B R ) d i. Say that a i covers j elements of N(a) All the j neighbors of a covered get dual value 1/deg(a i ).

78 Approximation slides 78 Analysis continued We change the values (only increasing them). First, replace them by 1/d = 1/deg(a). This may only increase their sum. Further replace their values by 1/d, 1/(d 1), 1/(d 2),...,1/(d j +1). This only increases the sum. Clearly, the next value assigned to a neighbor of a wouldbeatmost1/(d j) because after a i is added, deg(a, B R ) d j. Thus,bytheabove,H(deg(a)) upper bounds the sum for a So, y i / A is dual feasible

79 Approximation slides 79 Example a The greedy algorithm covers 3 neighbors of a, then another 3 and another 3 and then the 2 last. When the first three are covered, the worse case is that the three get 1/11 value. Because the covering element has degree at least 11. This is changed to 1/11, 1/10, 1/9. The next three: The worse case is that they get 1/8, 1/8, 1/8. Changed to 1/8, 1/7, 1/6 Then 1/5, 1/5, 1/5 changed to 1/5, 1/4, 1/3 Finally, 1/2, 1/2 changed to 1/2, 1. Thus the sum is H(11).

80 Approximation slides 80 Growing the variables one by one The following is due to Bar-Yehuda and S. Even for approximating Set-cover: Grow a single dual variable y b untill a constraint (or some constraints) become tight Add the tight variable(s) to the solution Iterate

81 Approximation slides 81 Analysis We encounter again (as in local ratio) the problem of cardinality versus sum of degrees. The lower bound is b y b. However, y b contributes up to deg(b) primal variables: y N(a) y b c(e) And a is in the solution only if the inequality is tight So: a A x a = a A b B deg(b) y b b N(a) y b Hence, B integrality gap and ratio.

82 Approximation slides 82 A general method for getting a dual-feasible: raising the dual variables simultaniously The idea in its full generality is due to Goemans and Williamsson. Was first used by Agrawal, Klein and Ravi without stating its primal-dual. The dual variables are at first set to 0 and thus form a feasible solution. To grow a dual variable is to increase the dual varible by a small enough ɛ so that no dual constraint is violated. To grow a subset of dual variables is to increase all the dual variables in the subset by some ɛ so that all dual constrains are still satisfied

83 Approximation slides 83 The metha-algorithm The algorithm is as follows: 1. S 2. While S is not feasible do: (a) Select a subset Q of dual variables (b) Grow all the dual variables in Q by the maximum possible ɛ Q. (c) Add every x i whose inequality is tight into S 3. Make S into a minimal feasible subset S S by deleting the elements in the reverse order of their addition 4. Return S

84 Approximation slides 84 The Steiner forest example The goal is to connect {s i,t i } for every i. Let δ(s) ={e =(u, v) u S, v S}. Let x(δ(s)) = e δ(s) x e. S is active if for some i, s i S and t i S or vise-versa. The primal LP: minimize x(δ(s)) 1 x e c e e Separating S x e 0

85 Approximation slides 85 The dual There is a variable y S for every active S maximize e δ(s) y S c(e) S e y S x S 0 Intuition: Every active set must be touched by at least one edge. In the case c(e) =1andy S are integers the S with y S > 0 are independent, namely, can notbecoveredbythesameedge.

86 Approximation slides 86 The Primal-dual algorithm 1. The subset of active sets the oracle returns in this case is the set of connected components induced by the already added edges 2. We grow simultaneously the y S for connected components S 3. An edge e whose inequality becomes tight is added into the solution 4. Hence new connected components are formed 5. At the end, we make the feasible solution minimal by reverse delete

87 Approximation slides 87 Preliminaries The algorithm maintains d(e) = S e δ(s) y S. The constraint is d(e) c(e) Let f(s) =1ifS is separating an 0 otherwise. Let V iolated(a) be all edges that touch a connected component in (V,A). Namely, e =(u, v), u S and v S. The maximum reduction in costs is: ɛ e = Initialization procedure: 1. A c(e) d(e) f(s 1 )+f(s 2 ). 2. Implicitly set y S 0 for every separating S 3. C {{v}} /* Connected components are single vertices /* 4. d(e) 0 for every e E

88 Approximation slides Initialize The algorithm 2. While A is not feasible do: (a) Compute V iolated(a) (b) Let e be an edge between S 1,S 2. Compute: 4. Return(A) ɛ e = c(e) d(e) f(s 1 )+f(s 2 ). (c) Let e be the edge minimizing ɛ e.say that e goes from S to S (d) A A {e } C C S S \ (S S ) (e) For every e between some S 1 and S 2 set d(e) =d(e)+(f(s 1 )+f(s 2 )) ɛ e 3. Go over the added edges in the inverse order of their addition, and remove from A every edge whose deletion keeps a feasible solution

89 Approximation slides 89 Why not Michael Jackson? Why do we do reverse delete in the PD algorithm? Why not random order of deletion (like the moves Jackson makes when he dances, Im bad! Im bad!)? Reverse delete emulates recursion. The algorithm can be described as: Increase dual variables Add into the solution edges whose constraint became tight. Recompute active sets Recurse Remove redundant edges

90 Approximation slides 90 The approximation ratio Consider the edges remaining after the reverse delete is finished, touching the components in the current stage This forms a forest F We have increase the cost by A active ɛ deg T (A) witht the tree A belongs to, with ɛ the minimum attained in Line 2c. However, the dual lower bound grows by ɛ b with b number of active We already saw that sum of degrees at most 2b hence, ratio 2 Works for a rather general collection of functions coined by GW as proper functions