max min z i i=1 x j k s.t. j=1 x j j:i T j

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1 AM 221: Advaned Optimization Spring 2016 Prof. Yaron Singer Leture 22 April 18th 1 Overview In this leture, we will study the pipage rounding tehnique whih is a deterministi rounding proedure that an be used to round the solution of a ontinuous relaxation of a ombinatorial problem. Pipage rounding is an interesting framework in its own right, and reates powerful approximation algorithms. More importantly, pipage rounding provides a gateway to the theory of submodular optimization. In this leture we will desribe pipage rounding for the Max-Cover problem. In the next leture we will generalize this framework for submodular optimization. 2 Relaxations of Max-Cover Let us look again at the Max-Cover problem. Reall that in this problem we are given sets T 1,..., T n over universe of m elements, and the goal is to selet S argmax S: S k i S T i. Integer programming formulation of Max-Cover. Max-Cover an be expressed as the following IP: As we have already seen a few times, max s.t. m z i n k z i x i, z j {0, 1} An equivalent way to write the above program is by expressing the variables assoiated with the elements overed z i in terms of the variables assoiated with the sets that over them x i : m max min 1, s.t. n k {0, 1} 1

2 LP relaxation. The formulation on the right-hand side an be relaxed as max x P L(x) with: m L(x) = min 1, and P = {x [0, 1] n : n = k}. Note that the objetive funtion is onave. Speifially, it is pieewise linear, so the relaxation an be reformulated as an LP and solved effiiently. Randomized relaxation. As we saw in the previous leture, when disussing the dual Min-Cover problem, a frational solution x [0, 1] n an also be interpreted as a random solution where eah set j [n] is inluded independently with probability. Let us ompute the expeted value of this random solution. For eah element i [m], the probability that it is overed by at least one set is: 1 n (1 ) so the expeted value assoiated with frational solution x is: m n F (x) = 1 (1 ) This leads to another relaxation of the Max-Cover problem: max x P F (x). Relationship between the relaxations. Both relaxations (the one with F and the one with L) are over the same polytope P. They are relaxations in the sense that when x is taken to be an integral solution (i.e. x {0, 1} n ), their values oinide and are equal to the Max-Cover value funtion. Note however that the relaxation with L an be effiiently solved (L is pieewise-linear) but the one with F annot. The following lemma shows that the two relaxed objetive funtions are within a onstant multipliative approximation ratio. Lemma 1. x [0, 1] n, F (x) (1 1/e) L(x). Proof. Sine both L and F sum over all elements in the universe a i where i [m], it suffies to show the bound on a given element. For a given element a i, w.l.o.g. suppose a i is overed by sets T 1,..., T for some n. Remember the Arithmeti-Mean-Geometri-Mean (AM-GM) inequality from the previous leture: Arithmeti-Mean-Geometri-Mean(AM-GM) inequality: If x 1,..., x are positive, then: ( (1 ) 1 x ) j In our ase the AM-GM inequality implies: ( 1 (1 ) )

3 We will now distinguish two ases: If 1: 1 ( (1 ) 1 1 ) ( ) 1 1 ( e = 1 1 ) e min 1, If < 1: Then define g(t) = 1 ( 1 t ). It is easy to verify that g is onave, hene: g(t) (1 t)g(0) + tg(1) = t (1 (1 1/) ) t(1 1/e) (1) We then get: ( 1 (1 ) 1 1 x ) j (1 1/e) where the seond inequality used (1) with t =. n = (1 1/e) min 1, 3 The Pipage Rounding Framework Let us now introdue the general theorem of the pipage rounding proedure, that we will then apply to the Max-Cover relaxations introdued above. Theorem 2. Let L : [0, 1] n R and F : [0, 1] n R be two funtions suh that: max x P L(x) and max x P F (x) are two relaxations of the same integer program max x Q I(x); F (x) αl(x) for any x P and some α > 0; max x P L(x) an be solved in poly-time; for all frational x P, integral x Q omputable in poly-time and F ( x) F (x). Then there exists a poly-time α-approximation algorithm for max x Q I(x). Proof. Let x L be an optimal solution to max x P L(x) whih is omputable in poly-time. Then: F ( x L) F (x L) αl(x L) αopt so x L Q is a feasible solution to the integral problem and α-approximation to max x Q I(x). Appliation to Max-Cover. If we want to apply the theorem to the relaxations L and F of the Max-Cover problem, the first three assumptions are learly satisfied using the funtions L and F we defined above with α = 1 1/e. What is left is to show that the last assumption is satisfied. That is, we need to give a magial rounding proedure that an take any vetor x P = {x [0, 1] n : i [n] x i = k} and turn it into a vetor x Q = {x {0, 1 n } : i [n] x i = k} without dereasing the value of F. 3

4 4 The Pipage Rounding Rounding To implement the pipage rounding framework we will use the following simple meta-rounding proedure. The rounding follows n rounds. In eah round we pik two oordinates that are frational; assume we an easily find two points in P, eah with one less frational omponent, and all these points x + α x d x and x β x d x (these are obnoxious names for points, but they will serve us well soon). Assume also, that we know that one of these points has larger value for F than the value of the original point x, i.e. F (x+α x d x ) F (x) or F (x β x d x ) F (x) (or both). Then, the algorithm will update x to be x + α x d x if F (x + α x d x ) F (x) and x β x d x otherwise. Repeating this n times guarantees that we have a rounded solution. We formally desribe this proedure below. Algorithm 1 Pipage Rounding algorithm 1: Input. frational x [0, 1] n, polytope P and funtion F 2: while x is frational do 3: x + α x d x a point in P that has one less frational oordinate than x 4: x β x d x anoter point in P that has one less frational oordinate than x 5: if F (x + α x d x ) F (x) then 6: x x + α x d x 7: else 8: x x β x d x 9: end if 10: end while 11: Return. x The Pipage rounding proedure generates a series of frational points: x (0), x (1),..., x (l) where eah point x (i+1) has at least one frational oordinate less than x (i). The algorithm terminates after n steps sine in eah step the number of frational oordinates dereases by one. Thus, the point x (l) is integral, and l n. What s left is to show that in every iteration of the while loop when we have a point x we an find two points x + α x d x and x β x d x that have one frational oordinate less than x, and that for these points we will always have that F (x) max{f (x + α x d x ), F (x β x d x )}. To do so, we will define the notion of onvexity in a diretion, whih is analogous to diretional derivates: Definition. F : [0, 1] n R is onvex in diretion d x if g(t) = F (x + t d x ) is onvex (in t). The following lemma now unravels some of the mystery behind the obnoxious point naming. Lemma 3. Assume that F : [0, 1] n R is onvex in diretion d x, then for any α x, β x [0, 1]: F (x) max{f (x + α x d x ), F (x β x d x )}. 4

5 Proof. We know that g(t) = F (x + td x ) is onvex in t so: F (x) = = g(0) ( βx = g α x + α ) x β x α x + β x α x + β x β x g(α x ) + α x g( β x ) α x + β x α x + β x β x = F (x + α x d x ) + α x + β ( x βx + α x α x + β x α x + β x = max{f (x β x d x ), F (x + α x d x )}. α x α x + β x F (x β x d x ) ) max{f (x β x d x ), F (x + α x d x )} 4.1 Applying pipage rounding in Max-Cover In our ase, as long as x P is frational, then there must be at least two oordinates that are frational sine x must respet i [n] x i = k. At a given iteration of the pipage rounding proedure, let p and q be frational oordinates that are seleted by the algorithm. For the point x Let us define the diretion d x = e p e q (where e j is the jth basis vetor), α b x = min{1 x p, x q } and β x = min{1 x q, x p }. From the above, what remains is to show that the funtion F is indeed onvex in the diretions e p e q for any p, q [n]. Lemma 4. F (x) = m ( 1 n (1 ) ) is onvex in every diretion e p e q, p, q [n]. Proof. It is suffiient to prove that eah term in the sum is onvex in a given diretion e p e q. Eah term in the sum represents the ontribution on a single element in the universe. Consider the element a i for some i [m]. Define: and F (i) (x) = 1 (1 ) g (i) (t) = F (i) (x + t(e p e q )) We need to prove that g (i) (t) is onvex, for every i [m]. To show this, note that there are four ases to onsider: Neither set T p nor T q over a i : in this ase we have g (i) (t) = 1 (1 ) sine the funtion is independent of t, it is onstant and trivially onvex; 5

6 The set T q overs a i but T p does not: in this ase g (i) (t) = 1 (1 ) (1 (x q t)) j q The funtion is affine in t and therefore onvex; The set T p overs a i but T q does not: in this ase, as above the funtion is affine; Both sets T p and T q over a i : in this ase g (i) (t) = 1 (1 ) (1 (x q t))(1 (x p + t)) = 1 + j p,q j p,q (1 ) ( t 2 + (t + 1)(x p + x q ) 1 ) and the above funtion is learly onvex. So, in all ases we have that the funtion is oves. Sine we have a sum of onvex funtions with positive oeffiients the funtion is onvex. Putting everything together we an onlude with out main theorem: Theorem 5. For the Max-Cover problem the pipage rounding proedure produes a 1 1/e approximation algorithm. 5 Disussion and Further Reading The pipage rounding method we showed atually gives an approximation ration better than 1 1/e: the approximation ratio is 1 (1 1/) where is the frequeny of the instane (reall the definition of frequeny from the previous leture: it is the largest number of set that an over an element in the instane to the problem; in the Max-Cover ase the frequeny is n). For more information about the pipage rounding method see [1]. Referenes [1] Alexander A. Ageev and Maxim Sviridenko. Pipage rounding: A new method of onstruting algorithms with proven performane guarantee. J. Comb. Optim., 8(3),