Dual fitting approximation for Set Cover, and Primal Dual approximation for Set Cover

Transcription

1 duality 1 Dual fitting approximation for Set Cover, and Primal Dual approximation for Set Cover Guy Kortsarz

2 duality 2 The set cover problem with uniform costs Input: A universe U and a collection of subsets S of U. Objective: Select the minimum size subcollection S S so that S S S = U. There is always a feasible solution as long as S is feasible.

3 duality 3 The greedy algorithm for uniform weights Set-Cover Say that we have a partial solution A of sets. We say that an element e is uncovered if if does not belongs to S A S. It is quite natural to take the set that contains the largest number of uncovered elements, and iterate. Given a choice of some set S, let new(t) be the number of elements that T covers and do not belong to S S S Thus, we pick at every round the T with maximum new(t). Also for simplicity of notation let U(S ) be the union of the sets in S.

4 duality 4 Algorithm Greedy 1. Sol, U R R, S, new(s) S 2. While U(S ) U do (a) Select a set T so that new(t) is maximum (b) S S T (c) For every S set new(s) new(s) T 3. Return S

5 duality 5 The LP for Uniform Cost Set Cover Each element e needs to be covered and so each set S has a variable, x S. Minimize s x S Subject to: For every element e S: e S x S 1 x s 0 for every S. Note that the IP version is equivalent to Set Cover. The dual to Set Cover gives values to elements, such that for every set the sum of its element is at most the cost of the set, namely 1 in this unweighted case (check using the definition of duality!!)

6 duality 6 The dual LP Each element e has a variable x e. Maximize e y e Subject to e e S y e 1. y e 0

7 duality 7 Intuition: why is this a lower bound? See the integral case For integer values of the dual, its clear the sum is at least 3 This follows as we need one vertex from every star.

8 duality 8 What solution can you get from the greedy We define a dual solution using Greedy. Say that the next T covers i new elements. Give each one of these elements value 1/i. Note that the sum of the respective y i is 1. Note that e T x e may be much larger than 1 because of previous elements. Let gr be the number of sets in the greedy solution. At every step both the sum of duals and the size of the solution grow by 1. Thus the sum of the dual values at the end is exactly gr.

9 duality 9 Example Say the largest set a covers 8 elements. Consider b of 6 elements. Only vertices in N(a) N(b) will get a value. If N(A) N(B) = 3 only 3 vertices of N(b) gets value. And its 1/8 each.

10 duality 10 The dual solution generated by the greedy algorithm This solution for the dual we get this way can not be feasible for the dual LP. Indeed, the sum of the y i is exactly g. If the fractional values are feasible for the dual, their dual value is at most g opt d opt f opt. Which means that we solved Set Cover exactly and P = NP. We use a non feasible dual solution

11 duality 11 By how much do we violate the dual constrains? If for every set s, e S y e D for sum D then the solution we have is infeasible. But if you divide it by D, its feasible. Namely, take the values given by the greedy algorithm and divide them by D. Since dual feasible, weak duality gives that the ratio is at most D. We learn from that (in general) that we can relax the dual constrains, prove a bound, divide by the bound and get a feasible dual. Fix some set s and lets see how the greedy algorithm covers it.

12 duality 12 Changing y e for the sake of the analysis We want to show that e s y e D for some D. We will change the value of the duals in the analysis for s. We never change the dual solution. The changes in the dual, when applied to S are part of the analysis (not algorithm). A legal operation is making e S y e Larger Since the sum only grows, it means that we may need a larger D. We change the values, because it makes the analysis simple.

13 duality 13 Changing the dual variables of elements in s Say that the next set chosen by Greedy is t. And t covers i new elements. By our method we set the dual value of every items in s t to 1/i. For analysis sake, we make the sum for S larger. First change. Change to 1/i, 1/(i 1),...,1/(i j) in case N(s) N(t) = j

14 duality 14 Next time Largest set is of size k. If s was chosen by greedy, for t N(s) N(t) get 1/k. Then change to 1/k,1/(k 1),1/(k 2) and so on. In the next slide an example by figure. Four stages.

15 duality 15 The neighbors of s were covered only after 4 greedy iterations The initial dual values and the changes. ALL 1/14 A ALL 1/9 B ALL 1/5 C ALL 1/2 D S 1/14 1/13 1/12 1/11 1/10 1/9 1/8 1/7 1/6 1/5 1/4 1/3 1/2 1

16 duality 16 Bounding the sum of elements of s Do it until the end. Let d be the number of elements in S The values will be at most 1/d+1/(d 1)+1/(d 2)+...+1/2+1//1 = H(d). Let be the number of elements (or degree) of the largest size set in the input. Thus the sum for any S can not be more than H( ). If we divide each y e by H( ) we get that the sum for any S is at most H( S )/H( ) 1, namely, we get a feasible solution for the dual. This means that Greedy has H( ) ratio.

17 duality 17 Why? If we divide by H( ) we get a feasible solution for the dual, and the optimum set S is a feasible solution for the primal. By weak duality i x i/h( ) S. This implies the inequality: i x i/h( ) S Since i x i equals the value g (the size of the set output by greedy) of the greedy solution, has cost at most gr H( ) S. Meaning the algorithm an H( ) ratio.

18 duality 18 Is our analysis tight? In several places it does not seem tight. In the next slide we show an input for Set Cover that will confuse greedy to get a solution of size about log 2 n, when there is a solution of size 3. Recently it was established by Irit Dinur proved that unless P NP H( ) is the best possible approximation in the sense that an approximation of value H( ) implies that P = NP Thus, the ratio we get is the best possible. There is no better than 2 ratio for VC Is 2 the best. Probably yes. NOT KNOWN.

19 duality 19 Example: greedy gives Ω(log n) sets when 3 is optimum Instead of A,B,C since N(A) = N(B) = N(C) = n/3, S 1 = n/2 is taken. In the same way we pick S 2, S 3 and so on. Θ(logn) value solution when OPT = 3 N(A) > N(B) > N(C) > N(S1) N(S2) N(S3)

20 duality 20 What is a primal dual approximation? There is basically two methods. One is finding an optimal solution for the LP. Also computing the dual. Then the algorithm strong duality in the process. Or you raise the dual until an inequality is met. See later. This has the advantage that we do not need to solve the LP exactly. This does not give a good ratio for Set Cover. If an element belongs to at most f sets, the ratio is f. But for VC f = 2 hence its a Primal-Dual approximation for VC

21 duality 21 Primal dual approximation for weighted Set Cover by strong duality Minimize s c s x s Subject to: For all e B, s e s x s 1 x s 0 for every s A. The dual: Each element e has a variable y e. Maximize e y e Subject to e e s y e c s. y e 0

22 duality 22 The most frequent element Let f be the largest times an element appears in a set. We choose S = {s x s 1/j}. Note that the inequality s e s x s 1 is met and so the above is a feasible solution for the primal. s OPT c s = j:midx s >0 c s. We now use strong duality. If x s > 0 c s = b b s y b. Thus s OPT c s = x s >0 b s y b.

23 duality 23 Changing sums we get: s OPT c s = b B y b s b s,x s >0 1 f b B y b = f opt f Note that s b s,x s >01 simply counts the number of times y b appears and thus the equality. Recall that no element appears in more than f sets. The last equality follows from strong duality.

24 duality 24 A primal dual algorithm that does not solve the LP We recall the dual: Maximize e y e Subject to e e s y e c s. y e 0 The algorithm 1. Sol 2. While Sol is not a set cover do: Raise all y e at the same rate, until a collection of dual constraint become tight. (a) Take all tight sets into Sol /* The motivation is complementary slackens */ (b) Remove all tight sets and all the covered elements from the input 3. Return Sol

25 duality 25 The cost of the sets chosen A set s is chosen if e e s = c s. Let S be the sets whose constrains became tight. Then the cost is s S c s = s S e s y e. By changing order of summation we get the cost is at most y e e s, s S f y e since y e appears in at most f sets. This gives f ratio and a ratio of 2 for Vertex Cover.

26 duality 26 There are smarter primal dual algorithms These algorithms bound the average and not the maximum frequency. Albeit, unless P = NP, Set Cover admits no f +1 ǫ approximation for any constant ǫ > 0. This lower bound (is one of the best papers I ever read by) Dinur et al. We cant expect a f ǫ optimal lower bound because this will give an inapproximability of f = 2 for VC. And this a long standing open problem. Easily among the 10 most important open questions in approximation algorithms.