Lecture 7: Approximation via Randomized Rounding


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1 Lecture 7: Approximation via Randomized Rounding Often LPs return a fractional solution where the solution x, which is supposed to be in {0, } n, is in [0, ] n instead. There is a generic way of obtaining an {0, } n vector from a x [0, ] n one: let the ith variable be with probability x i, 0 with probability ( x i ). The resulting random vector x int has the desirable property that for any i, E[x int i] = x i. In particular, if x minimized a certain linear function c x, then by linearity of expectation, the expected cost of the random {0, } n vector is also equal to c x. Of course, often the resulting x int won t satisfy the constraints, and therein lies the nontriviality. We will need some facts in probability theory, mostly in the concentration of random variables around their means. I will collect these facts down in the appendix; the proofs of which can be found in texts on randomized algorithms or the web. The Set Cover Problem. Recall the LP relaxation for the set cover problem. min subject to m c(s j )x j () j= x j i U (2) x j [0, ] j =...m (3) Let x be a solution to the LP. Consider the experiment where we randomly select set S j with probability x j, independently of each other. Let X j be the indicator random variable which is if we picked S j and 0 otherwise. The random variable m j= c(s j)x j indicates the cost of the sets picked, and thus the expected cost of our solution is precisely m j= c(s j)e[x j ] which is the LP solution. Are all elements covered? Fix an element i. The following claim shows that each element is covered with at least a constant probability. Claim. The probability that an element i is covered is at least ( /e). Proof. Element i is covered iff a set containing i is picked. The probability that i is not covered, thus, is at most ( x j ) due to independence. But this is at most where the inequality follows from (2). e xj = e P x j e
2 So our randomized algorithm returns a collection of sets whose expected cost is the LP cost, and covers each element with probability at least ( /e). Suppose we repeated this 2 ln n times and returned the union of the sets picked in each round. The expected cost of our solution is going to be 2 ln n times the LP solution. What s the probability that an element i is not covered in any of the rounds? Since the rounds are independent, this probability is at most ( e )2 ln n = /n 2. By the union bound (Fact 2), the probability that there exists some element i which is not covered is at most n = /n. Thus, we get the following theorem. n 2 Theorem. The algorithm described above returns a set cover with probability at least ( n ) whose expected cost is 2 ln n times the LP solution. Randomized approximation algorithms often look like the theorem above. They succeed with high probability and return a solution whose expected cost is at most (or at least) a factor of the optimum solution. By taking a slight hit at the approximation factor, the guarantee on the expected cost can often be converted to a guarantee on the cost with high probability. For instance, Markov s inequality (Fact 3) tells us that the probability that the cost of the sets picked exceeds 4 ln n times the optimum is at most /2. Thus, the algorithm returns a set cover of cost 4 ln n times the optimum with probability at least 2 n /3. If we repeat this algorithm t times, then with probability 3 (really, really high probability if t is some polynomial in n) we will encounter such a set cover. t Suppose we look at the set multicover problem where each element i needs to be covered at least b i times for some positive integer b i? The above analysis needs to be modified a bit; in particular, the probability that an element i is not covered in any iteration goes down from /e to /e b i, although, it now needs to be covered b i times. This is true, and I ll leave this calculation checking as an exercise. I now show a slightly different (but similar in spirit) randomized algorithm. It runs in a single round by sampling sets with higher probabilities. Let ˆx j = min(, 6 ln n x j ). Sample each set S j independently with probability ˆx j. That s it. The expected cost now is j c(s j)ˆx j 6 ln n lp. To argue that we cover each element sufficient number of times, we use the Chernoff bound (Fact 4). Let X j be the indicator random variable of the event whether S j is picked or not. For an element i, define Z i = X j. If Z i b i, then i is covered, otherwise not. The following claim, using a union bound, gives that with probability ( /n), the sets picked form a set cover. Claim 2. Pr[Z i < b i ] n 2. Proof. By reordering suppose i S j for j k. We may assume ˆx j < for all such j; since all others (say l of them) are picked with probability, we can move to the residual problem where the element i needs to be covered b i l times. Now, E[Z i ] = k j= ˆx j = 6 ln n b i. Thus, Fact 4 gives us (with δ = 6 ln n 5/6) Pr[Z i < b i ] exp( 6 ln n ) n 2 Facility Location Problem Let s recall the LP relaxation for the facility location problem and it s dual. Let (x, y) and (α, β) be a pair of optimal primaldual solutions. Recall the following claim from last class 2
3 min f i y i + c(i, j)x ij (4) i F i F,j C x ij =, j C (5) i F y i x ij, i F, j C (6) x, y 0 max α j (7) j C β ij f i, i F (8) j C α j β ij c(i, j), i F, j C (9) β 0 Claim 3. If x ij > 0, then c(i, j) α j. Before we describe the algorithm, we are going to make a structural assumption about (x, y). I will leave it as an exercise to prove that the assumption is without loss of generality. Assumption: If x ij > 0, then y i = x ij. Algorithm. (In class, we looked at the clustering algorithm which ordered clients in increasing order of α j, and gave us a (2+2/e) approximation. Below we give a refined argument which attains a ( + 2/e)approximation.) The algorithm proceeds as follows. Let Cj := i c ijx ij. Order the clients in increasing order of (α j + Cj ). Let U denote the set of unassigned clients initialized to C. Let j be the first client in U in the order. Form the cluster N 2 (j) consisting of j, all facilities {i : x ij > 0}, and all clients {j U : x ij > 0 for some i N 2 (j)}. j is called the cluster center of N 2 (j). Remove all clients of N 2 (j) from U and repeat till U is empty. At the end of this step all clients are assigned some cluster N 2 (j), for some client j. Note that in any cluster N 2 (j), i N 2 (j) y i = i N 2 (j) x ij =. In each such cluster, open exactly one facility with probability y i  this is valid since i y i =. Let X be the set of facilities opened. The clients are assigned their nearest facility. Claim 4. The expected facility opening cost is precisely i F f iy i. Proof. The probability a facility i is in X is precisely y i. Let us now calculate the expected connection cost for a client j. Note that a cluster center j pays expected connection cost i N 2 (j) c(i, j)x ij = Cj. Let s take a noncenter client j N 2(j ). Let Γ(j) := {i : x ij > 0}, and let {j,..., j r } be the centers such that N 2 (j l ) Γ(j) is nonempty. For l =,..., r, let p l := i N 2 (j l ) x ij be the probability a facility in N 2 (j l ) Γ(j) is opened. Note that l p l =, and that these r events are independent. Let Cj (l) be the expected cost of connecting to a client in N 2(j l ) Γ(j) conditioned on the event that such a facility is open. Thus, C j (l) := i N 2 (j l ) c(i, j) xij p l Let s order j,..., j l so that Cj () C j (l). Lastly, let s define C j ( ) to be the expected cost of connecting j conditioned on the event that no client in Γ(j) is opened. We ll bound this shortly. Now we are ready to bound the expected connection cost of client j. 3
4 Claim 5. The expected connection cost of client j is at most p C j () + ( p )p 2 C j (2) + + ( p )( p 2 ) ( p r )p r C j (r) + C j ( ) r ( p l ) Proof. We are conditioning on mutually exclusive events: event l is when no facility in N 2 (j l ) Γ(j) has been opened for l < l, and a facility in N 2 (j l ) Γ(j) has been opened; and the last event when no facility in Γ(j) has been opened. Claim 6. C j ( ) C j + 2α j. Proof. If no facility in Γ(j) is opened, then connect j to the facility i opened in N 2 (j ): recall j N 2 (j ) and thus there exists i N 2 (j ) with x ij > 0. The connection cost to i is c(i, j) c(i, j) + c(i, j ) + c(i, j ) 2α j + c(i, j ), where we have used Claim 3 and the fact that α j α j. Taking expectations, we get Cj ( ) 2α j + c(i, j )x i j. i N 2 (j) l= Putting it all together, we get the total connection cost of j is at most r [ pl ( p l ) ( p )Cj (l) ] + (Cj + 2α j ) l= r ( p l ) (0) Since r l= p l =, we have the product r l= ( p l) /e. As a first approximation, note that ( p l ) and r l= p lcj (l) = C j. This gives us total expected facility + connection cost of at most f i y i + Cj + (Cj + 2α j ) opt( + 3/e) e i F j C giving a approximation. One can analyze better. Now we state the following inequality: Lemma. Let A A 2 A r, and let r p i =. Then, ( r ) ( p A + p 2 ( p )A p r ( p )( p 2 ) ( p r )A r p i A i j C l= ) r ( p i ) Proof. Define q i := p i( p )...( p i ) Q r ( p. We need to prove that r i) q ia i r p ia i. Note that r q i = = r p i. Furthermore, there exists a < k < r such that q i p i for all i k and q i p i for all i > k; it s the largest i for which i l= ( p l) exceeds r ( p i). Let δ i = q i p i. Thus, δ i 0 for i k, δ i 0 for i > k and i δ i = 0. ( r r k ) ( r k ) (q i p i )A i = δ i A i A k δ i + A k+ δ i = δ i (A k A k+ ) 0 i=(k+) 4
5 Given the above fact, we can get a stronger approximation factor. Let q := r l= ( p l). Then, the expected connection cost is at most ( q)c j + qc j ( ) C j + 2α j /e Putting it together gives a ( + 2/e)approximation. The Group Steiner Tree Problem (on a tree) In this problem, the input is a rooted tree T on n vertices. There are k disjoint groups g,..., g k, where each group is a subset of leaves. Each edge of the tree has a cost, and the goal is to find the cheapest rooted subtree which contains a vertex from each group. LP Relaxation. min e E(T ) c e x e x 0 () subject to x(δ(s)) S V : g i S for some i, r / S (2) Algorithm. Let the root be r. Given a nonroot vertex v in the tree, let e v denote the edge connecting v to its parent. We ll abuse notation and let x v denote x ev. We call an edge f an ancestor of edge e if the path from r to the end points of e contains f. f is the parent of e if its an ancestor and shares an end point with e. Preprocessing x. We process the solution x as follows. Firstly, remove all vertices v with x v /2g. For any group g i, we still have v g i x v /2. Secondly, increase x e for every edge to the nearest power of (/2). Note that the new LP cost at most doubles. Thirdly, for an edge e with parent f, if x e = x f, then merge the two edges into a single edge with the same x e value. This doesn t change the feasibility of the solution. Note that the height of the tree now is at most 2 log g: any path from root to a vertex v has strictly decreasing x e in powers of /2 ranging from at most to at least /2g. Finally, decrease all the x ev s such that for any group g i, v g i x v =. Since these are leaf edges and groups are disjoint, this can be done without changing the feasibility of the solution. Independently, sample all edges e incident on the root with probability x e. For every other edge, e with parent f, sample e independently with probability xe x f. The resulting graph can have many components  throw away all except the component containing the root. The above step is repeated (6 ln n log g) times independently and the final solution is the union of all edges picked; g is the maximum size of a group. Theorem 2. The above algorithm has an expected cost of O(log n log g) opt and is a feasible group Steiner tree with high probability. Proof. In each iteration the probability that an edge e is sampled is exactly x e. This is because, let the path from root to e be e p, e p,..., e, e. The probability e survives this iteration is at most x e x e x e x e2 xe p x ep x ep = x e 5
6 By linearity of expectation, the expected cost of the edges in any iteration is opt, and the first claim in the theorem follows. Fix a group g i and a iteration. We claim that the probability that the subtree sampled contains a vertex from g is at least 8 log g group g i is not covered in 6 ln n log g iterations is at most. This will prove the second claim of the theorem since the probability ( ) 6 ln n log g 8 log g. Union bound n 2 over all groups (there are n many at most) ends the argument. For a vertex u g i, let E u be the event that u is contained in the subtree. We are trying to lower bound the event Pr[ u g i E u ]. Let s calculate Pr[E u ]. u is contained in the tree if and only if all the edges on the path from u to r are sampled. This probability is x u. Furthermore, the constraint in the LP gives us u g i x u =. So, we have µ := u g i Pr[E u ] =. (Thus, if all the E u s were independent in a group g i, then in fact the probability that none of the vertices are contained in the tree would have been at most /e.) Let u v if the events E u and E v are not independent. Define := u v Pr[E u E v ]. Claim 7. 2 log g Proof. For u and v in the tree, let w be the least common ancestor. Note that conditioned on the event E w, the event E u and E v are independent. Thus the probability Pr[E u E v ] = Pr[E u E v E w ] Pr[E w ] = Pr[E u E w ] Pr[E v E w ] Pr[E w ] = x ux v Now for any interior node w, let T w be the subtree rooted at w. Note that u g i T w x u. This is because of the feasibility of the LP constraint with S containing all vertices in T w and all vertices in g i \ T w : x(δ(s)) = + x u = + x u u g i \T w u g i T w Let A u denote the set of ancestors of u in the tree. Since the height of the tree is at most 2 log g, A u 2 log g. Thus, = u v x u x v u g i w A u x u x v x u v g i T w u g i w A u x v x u A u 2 log g v g i T w u g i Now by Janson s inequality (Fact 5), we get that Pr[ ) ( E u ] exp ( µ2 exp ) µ + 4 log g 8 log g u g i since e x x + x2 2 x/2 when 0 x. So we get that in any iteration and any group g i, the probability we pick a vertex from the group is at least 8 log g. This completes the proof. 6
7 Some Probabilistic Facts. Fact (Linearity of Expectation). Let X,..., X n be random variables and let X := n X i. Then, E[X] = n E[X i]. Fact 2 (The Union Bound). Given n events E,..., E n, the probability that one of them occurs is at most the sum of their probabilities. Pr[E E 2 E n ] n Pr[E i ] Fact 3 (Markov s Inequality). Let X be a nonnegative random variable. Then Pr[X te[x] ] t Fact 4 (Chernoff Bound). X,, X n be n independent {0, } random variables with Pr[X i = ] = p i. Let X := n X i and let µ = n p i = E[X]. Then ( e δ ) µ Pr[X > ( + δ)µ] ( + δ) (+δ) ( e δ ) µ Pr[X < ( δ)µ] ( δ) ( δ) Useful versions of the above: For any δ > 0, For t > 0, For t > 4µ, Pr[X > ( + δ)µ] exp( µδ 2 /3) Pr[X < ( δ)µ] exp( µδ 2 /2) Pr[ X µ > t] 2 exp( 2t 2 /n) Pr[X > t] 2 t Fact 5 (Janson s Inequality). Let X U obtained by sampling element i U with probability p i independently. A,..., A t be subsets of U and let E i be the event that A i X. Then, if µ = t Pr[E i] and = i,j:a i A j Pr[E i E j ], then Pr[ t ) E i ] exp ( µ2. µ + 7
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