1 Unweighted Set Cover

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1 Comp 60: Advanced Algorithms Tufts University, Spring 018 Prof. Lenore Cowen Scribe: Yuelin Liu Lecture 7: Approximation Algorithms: Set Cover and Max Cut 1 Unweighted Set Cover 1.1 Formulations There are equivalent ways of representing the set cover problem: the ordinary set formulation, and the hitting set formulation. The decision version of the set cover problem is NP-complete. Here, however, we specifically consider the optimization/search version of the set cover problem, which is NP-hard Ordinary Set Formulation Consider the following: X = {x 1, x,, x m } P = {P 1, P,, P n } where x j represents a skill, P i represents a person, and x j P i if person P i has the skill x j. As an employer, our goal is to find the minimum number of people to cover all the skills, namely to find R P such that R covers all the x j, and R is minimized: 1.1. Hitting Set Formulation Consider the following: R = argmin R ( R ), s.t. R = X P = {p 1, p,, p n } X = {X 1, X,, X m } 1

2 where p i represents a person, X j represents a skill, and p i X j if person p i has the skill X j. As an employer, our goal is to find the minimum number of people to cover all the skills, namely to find H P such that H hits all skillsets and H is minimized: H = argmin H ( H ), s.t. H X j 0, X j X Weighted Set Cover If we have a set of costs associated with hiring each person in the candidate pool, and our goal is to choose the subset of people to hire that will minimize the total cost of hiring while covering all the skills, then we have a weighted set cover problem..1 Representation of the Problem We first extend the hitting set formulation to represent the weighted set cover problem. Then, we present the weighted set cover as an integer programming (IP) problem. Lastly, we relax this problem to a linear programming (LP) problem that solves for the fractional solution, where we assume that we can hire part-time employees. Solving the LP will get us the optimal fractional solution, only. That is not quite the integer solution we originally set out to get; however, LP problems can be solved in polynomial time, and we will later show how it can guide us towards a good solution to the IP problem..1.1 Hitting Set Formulation (Weighted) Consider the following: P = {p 1, p,, p n } W = {w 1, w,, w n } X = {X 1, X,, X m }

3 where p i represents a person, X j represents a skill, and p i X j if person p i has the skill X j. Additionally, w i represents the cost of hiring candidate p i. As an employer, our goal is to cover all the skills while minimizing the total cost of hiring, namely to find H P such that H hits all skillsets and w i is minimized: H = argmin H ( p i H w i ), s.t. H X j 0, X j X.1. Weighted Set Cover as an IP Problem Let: v i = Our goal is to minimize n i=1 w i v i { 1 if p i H 0 otherwise Definition.1..1: A feasible solution to the integer programming formulation of the weighted set cover problem satisfies X j X, p i X j v j Relaxed Weighted Set Cover as an LP Problem If we relax the constraint to allow fractional part-time hire, we obtain an LP problem as follows: Let: 0 v i 1 Our goal is to get a feasible solution while minimizing n i=1 w i v i. By inspection, we can tell that this is an LP problem: the function to minimize is a linear function of variables, and the inequality constraints also form linear functions. Thus, we can use the fact that LP problems can be solved in polynomial time. Notice in the original setting where we can t hire fractional part-timers; however, we can still use the fractional solution to guide us to a good solution to the IP problem, as we describe in the next section. The optimal fractional solution also serves as a lower bound to the optimal integer solution. 3

4 . An Approximation Algorithm to the Optimal Solution for Weighted Set Cover..1 Algorithm 1. Solve the LP. Choose each v i with probability equal to its value under the optimal fractional solution. Perform this step a total of log m independent rounds, where m is the number of skills. 3. Return the union of all the v i chosen in any round... Proving the Validity of the Algorithm To show that, after log m independent rounds, with high probability, every skill is covered: Let ˆv i be the value assigned to v i in the optimal LP solution. Suppose X j contains p 1,, p k we know k i=1 ˆv i 1 by the definition of feasible solution. Let P r[e] denote the probability of an event e, we observe that: P r[x j is covered] = 1 P r[none of the p i X j is covered] = 1 P r[p 1 isn t chosen p k isn t chosen] = 1 (1 ˆv 1 ) (1 ˆv k ) k = 1 (1 ˆv i ) i=1 1 (1 1 k )k (by the definition of feasible solution) 1 1 e > 1 4

5 We have shown above that, in each round, the each skill has more than half the chance to be covered. Then, we know that: P r[x j uncovered after log m rounds] (1 1 log m ) = ( 1 log m ) Taking all skills into account, we get: = 1 m P r[some skill uncovered after log m rounds] X j X P r[x j uncovered after log m rounds] X j X 1 m = m m = 1 m Therefore, after log m rounds, all skills are covered with high probability...3 Proving the cost of the solution is at most O(log m) OP T Let OP T be the optimal integer solution to the IP formulation of the problem. Recall that the optimal fraction solution to the LP problem serves as a lower bound to the integer solution, OP T. Therefore, we know that the expected cost we pay for each round is at most OP T. After log m rounds, the expected total cost of hiring is at most log m OP T, and we have shown that this algorithm solves for a log m-approximation to the optimal solution. 3 Graph Spanner: An Application to Set Cover When we have a large weighted graph, we might want a sparser subgraph that well approximates the distances in the original graph. Here, we introduce the concept of a graph spanner, which is a sparse subset of a graph with fewer edges, whose shortest path distances approximate the 5

6 shortest paths of the original graph. In particular, given any graph, we will show how to construct a graph on the same vertex set but only a subset of the original edge set such that the new graph has O(n 1.5 log n) edges, and shortest paths in the new sparser graph are at worst 3 times the length of shortest paths in the original graph. 3.1 Formulation Given an undirected graph G = (V, E), our goal is to construct a spanning graph G = (V, E ), such that u, v V, d G (u, v) 3 d G (u, v). The idea is to choose a set of landmarks L, such that: 1. L is not too big. Every vertex has at least one landmark in its local neighborhood; that is, every vertex has at least one landmark among its closest n neighbors in G. 3. Construction 1. For each vertex i, pick S i = {i s n closest neighbors}, and grow a single source shortest path (SSSP) tree of that neighborhood. Namely, for each vertex j S i, we store all edges participating in the SSSP from i to j. Note this set of edges must form a tree by triangle inequality.. Find a small number of landmarks, L, s.t. S i, x L in S i 3. For each landmark, grow a SSSP tree to all other vertices in the graph, and keep all edges. Note this set of edges also must form a tree by triangle inequality. For each landmark, edges kept is O(n). Notice that this is a hitting set problem, where the local neighborhoods are 1 the sets that must be hit by the landmarks. If we assign a weight of n to each vertex, we can solve the LP problem by taking each vertex with probability that equaling its weight. 6

7 Then, applying our log n-approximation algorithm to set cover, we get a set of landmarks L of size: 3.3 Proofs L = O( 1 n n log n) = O( n log n) Claim 3.3.1: There are O(n 1.5 ) edges spanning all local neighborhoods. Proof: By construction, each vertex keeps its n nearest neighbors in G in its local neighborhood. The SSSP tree of each local neighborhood must contain O( n) edges. In a graph, we have n such local neighborhoods; therefore, there are O(n n) = O(n 1.5 ) edges in total spanning the neighborhoods. Claim 3.3.: There are O(n 1.5 log n) edges connecting all landmarks to all the other vertices. Proof: Since we have O( n log n) landmarks, and each of them contains O(n) edges to all other vertices in the graph, we have O(n 1.5 log n) edges connecting all landmarks to all vertices. Claim 3.3.3: In the constructed sparser graph G, the shortest distance between any vertices is no more than 3 times that in the original graph G. Proof: Consider two arbitrary vertices u and v. Denoting the shortest distance between u and v as d(u, v), we have the following cases: 1. Vertex v is in the local neighborhood of u. In G, we have kept the shortest path from u to v from the original graph G by construction; therefore, d G (u, v) = d G (u, v). Vertex v is not in the local neighborhood of u. Since v is not u s n closest neighbors, there exists some landmark l v 7

8 that is closer to v than it is to u, namely: d(l v, v) d G (u, v). By triangle inequality, we know that d(l v, u) d(l v, v) + d G (u, v). Subsitute to get d(l v, u) d G (u, v). Since d G (u, v) = d(l v, u) + d(l v, v), subsitute with inequality to get: d G (u, v) d G (u, v) + d G (u, v) = 3 d G (u, v). We then have proven that, in the sparser graph G with O(n 1.5 log n) edges, the shortest distance between any vertices is no more than 3 times that in G. 4 Max Cut Max cut is an NP-hard problem. We have introduced a deterministic greedy approach in a previous lecture. Here, we first describe a probabilistic randomized 1 -approximation algorithm, then derandomize it to show a deterministic 1 -approximation algorithm. Lastly, we briefly introduce a solution to max cut involving semidefinite programming, which achieves the current best-known approximation. 4.1 Formulation Given a graph G = (V, E), find a bipartition of the vertex set V such that the number of edges crossing the partition is maximized. 4. Probabilistic Randomized Algorithm Define a variable ɛ ij = 1 X ix j, where X i, X j { 1, 1}, representing the respective position of vertex i and j across the cut. Notice that: { 1 if i and j are in different sides of the cut ɛ ij = 0 otherwise With this quadratic expression, ɛ ij counts 1 everytime edge (i, j) crosses the partition. In this probabilistic randomized algorithm, we flip a fair coin to place vertices in either side of the bipartition. Then, let S be the number of 8

9 edges that cross the cut: [ ] E[S] =E ɛ ij = (i,j) E = (i,j) E = (i,j) E = (i,j) E = E (i,j) E E[ɛ ij ] (by linearity of expectation) [ ] 1 Xi X j E ( [ 1 E ] 1 ) E[X ix j ] [ 1 E ] (because X i, X j are independent, E[X i, X j ] = 0) Although we have proved the expectation of the method of flipping a fair coin to partition the vertices is good, it does not yet tell us that this randomized algorithm gives a good partition with high probability. In order to get a bound on the probability of us getting a good partition, we would have to bound the probability that we get a cut that is more than ɛ away from its expectation using Chebychev s inequality; however, even then we still have that ɛ loss off the expectation. So instead, we are going to pursue a different strategy and derandomize this algorithm. 4.3 Deterministic Algorithm E In proving the expected outcome,, from the probabilistic randomized algorithm, we have in fact shown a non-random fact: since the expectation is computed as an average, across all possible outcomes, there must exist one that is at least as good as the expectation. In other words, in any graph, there must always exist at least one way to partition the vertices so that at least half the edges cross the cut. 9

10 4.3.1 Simple Derandomization Let V = n. Suppose we write down all possible outcomes in terms of sequences of coin toss, there would be n of them in total. We can then pick the sequence that gives us the maximum number of edges across the partition With Pairwise Independence Definition : A collection of random variables X 1,, X n are pairwise independent if and only if for any variables X i and X j, we have E[X i X j ] = E[X i ]E[X j ]. Consider a case demonstrated by the following table, where X i and X j represent random coin tosses, and X k is the XOR of the outcome. Any variables from the following table are pairwise independent from each other; however, given any columns, we can predict the third. We then say X i, X j, and X k exhibit pairwise independence, but not total independence. X i X j X k = XOR(X i, X j ) With V = n, assume N = k, where N is the next biggest power of to n. Let w = w 1, w,, w log N be a random log N-bit sequence of 0 s and 1 s. Construction: 1. Let X i = ( 1) bin(i) w, where bin(i) is the binary representation of number i, and a b counts the number of 1-bit s in XOR(a, b).. Try all log N = N settings for w, and at least one of them results in a partition where at least half of the edges cross the cut. 3. Output w where most edges cross the cut. X i { 1, 1} represents the side of the partition i is assigned to. For example, given the random w = , X 7 = ( 1) bin(7) w. Since bin(7) = , bin(7) w = = 5. Therefore X i = ( 1) 5 = 1. 10

11 Claim : At least one sample point in this sample space results in a cut that meets our expectation for the randomized algorithm. Proof: We first need to prove that the X i s as we have defined them are pairwise independent. We can easily see this by comparing X i and X j on a bit in the truly random seed where their binary representations differ. Then, we have to show that our previous analysis of the randomized algorithm only depends on pairwise independence. That is true because we already know E[X i X j ] = 0 since X i and X j are pairwise independent. We know that P r[x i = 1] = P r[x j = 1] = P r[x i = 1] = P r[x j = 1] = 1. Now, i j log N, P r[x i X j ] = P r[x i = 1 X j = 1] + P r[x i = 1 X j = 1]. By pairwise independence, we know this expression is equivalent to P r[x i = 1]P r[x j = 1] + P r[x i = 1]P r[x j = 1] = 1. Since across N E different settings, every pair of vertices has half the chance to be on different sides of a partition, we expect half of the edges to cross the partition. Then, we know there must exist one in all N settings that has at least half of the edges cross the cut. If we try all settings, we have a deterministic algorithm. 4.4 Semidefinite Programming Semidefinite programming (SDP) is a generalization of linear programming (LP). Instead of having constraints with linear equations, we have constraints given by a positive semidefinite matrix. SDP is polynomial-time solvable using ellipsoid method or interior point methods. In solving max cut with SDP, the idea is to place n-dimensional vectors, which represent the vertices in the graph, on the unit sphere, in such a way that all pairs of vertices that share an edge are simultaneously placed as far apart as possible. We model this by the SDP and solve it optimally. To map this to a cut, we pick a random hyperplane and place the vectors into the partition based on which side of the random hyperplane they fall on. We can show an expected size of times the size of the optimal cut. Then, we would need to use a complicatd derandomization procedure to actually come up with a cut that meets the expectation, as before. 11