Approximating some network design problems with node costs

Transcription

1 Approximating some network design problems with node costs Guy Kortsarz Zeev Nutov July 3, 2009 Abstract We study several multi-criteria undirected network design problems with node costs and lengths. All these problems are related to the Multicommodity Buy at Bulk (MBB) problem in which we are given a graph G = (V,E), demands {d st : s,t V }, and a family {c v : v V } of subadditive cost functions. For every s,t V we seek to send d st flow units from s to t on a single path, so that v c v(f v ) is minimized, where f v the total amount of flow through v. In the Multicommodity Cost-Distance (MCD) problem we are also given lengths {l(v) : v V }, and seek a subgraph H of G that minimizes c(h)+ s,t V d st l H (s,t), where l H (s,t) is the minimum l-length of an st-path in H. The approximation for these two problems is equivalent up to a factor arbitrarily close to 2. We give an O(log 3 n)-approximation algorithm for both problems for the case of demands polynomial in n. The previously best known approximation ratio for these problems was O(log 4 n) [Chekuri et al., FOCS 2006] and [Chekuri et al., SODA 2007]. We also consider the Maximum Covering Tree (MaxCT) problem which is closely related to MBB: given a graph G = (V,E), costs {c(v) : v V }, profits {p(v) : v V }, and a bound C, find a subtree T of G with c(t) C and p(t) maximum. The best known approximation algorithm for MaxCT [Moss and Rabani, STOC 2001] computes a tree T with c(t) 2C and p(t) = Ω(opt/log n). We provide the first nontrivial lower bound on approximation by proving that the problem admits no better than Ω(1/(log log n)) approximation assuming NP Quasi(P). This holds true even if the solution is allowed to violate the budget by a universal constant ρ. Our result disproves a conjecture of [Moss and Rabani, STOC 2001]. Another related to MBB problem is the Shallow Light Steiner Tree (SLST) problem, in which we are given a graph G = (V,E), costs {c(v) : v V }, lengths {l(v) : v V }, a set U V of terminals, and a bound L. The goal is to find a subtree T of G containing U with diam l (T) L and c(t) minimum. We give an algorithm that computes a tree Rutgers University, Camden. Partially supported by NSF support grant award number guyk@camden.rutgers.edu The Open University of Israel. nutov@openu.ac.il 1

2 T with c(t) = O(log 2 n) opt and diam l (T) = O(log n) L. Previously, a polylogarithmic bicriteria approximation was known only for the case of edge costs and edge lengths. 1 Introduction Network design problems require finding a minimum cost (sub-)network that satisfies prescribed properties, often connectivity requirements. The most fundamental problems are the ones with 0, 1 connectivity requirements. Classic examples are: Shortest Path, Min-Cost Spanning Tree, Min-Cost Steiner Tree/Forest, Traveling Salesperson, and others. Examples of problems with high connectivity requirements are: Min-Cost k-flow, Min-Cost k-edge/node-connected Spanning Subgraph, Steiner Network, and others. All these problems also have practical importance in applications. Two main types of costs are considered in the literature: the edge costs and the node costs. We consider the latter, which is usually more general than the edge costs variants; indeed, for most undirected network design problems there is a very simple reduction that transforms edge costs to node costs, but the inverse is, in general, not true. The study of network design problems with node costs is already well motivated and established from both theoretical as well as practical considerations [11, 8, 13, 4, 5]. For example, in telecommunication networks, expensive equipment such as routers and switches are located at the nodes of the underlying network, and thus it is natural to model some of these problems by assigning costs on the nodes rather than to the edges. For some previous work on undirected network-design problems with node costs see the work of Klein and Ravi [11], Guha et al. [8], Moss and Rabani [13], and Chekuri et al. [4, 5]. We mostly focus on resolving some open problems posed in these papers. 1.1 Problems considered Given a length function l on edges/nodes of a graph H, let l H (s, t) denote the l-distance between s, t in H, that is, the minimum l-length of an st-path in H (including the lengths of the endpoints). Let diam l (H) = max s,t V(H) l H (s, t) be the l-diameter of H, that is the maximum l-distance between two nodes in H. We consider the following two related problems on undirected graphs. Multicommodity Buy at Bulk (MBB) Instance: A graph G = (V, E), a family {c v : v V } of sub-additive monotone non-decreasing cost functions, a set D of pairs from V, and positive demands {d st : {s, t} D}. Objective: Find a set {P st : {s, t} D} of st-paths so that v V c v (f v ) is minimized, where f v = {d st : {s, t} D, v P st }. 2

3 Multicommodity Cost-Distance (MCD) Instance: A graph G = (V, E), costs {c(v) : v V }, lengths {l(v) : v V }, a set D of pairs from V, and positive integral demands {d st : {s, t} D}. Objective: Find a subgraph H of G that minimizes w(h, D) = c(h) + {s,t} D d st l H (s, t) (1) As linear functions are subadditive, MCD is a special case of MBB. The following statement shows that up to a factor arbitrarily close to 2, MCD and MBB are equivalent w.r.t. approximation. Proposition 1.1 ([2]) If there exists a ρ-approximation algorithm for MCD then there exists a (2ρ + ε)-approximation algorithm for MBB for any ε > 0. We consider two other fundamental problems closely related to MBB (see an explanation below): Maximum Covering Tree (MaxCT) Instance: A graph G = (V, E), costs {c(v) : v E}, profits {p(v) : v V }, and a bounds C. Objective: Find a subtree T of G with c(t) C and p(t) maximum. Shallow-Light Steiner Tree (SLST) Instance: A graph G = (V, E), costs {c(v) : v V }, lengths {l(v) : v V }, a set U V of terminals, and a bound L. Objective: Find a subtree T of G containing U with diam l (T) L and c(t) minimum. Each one of the problems MBB and SLST has an edge version, where the costs/lengths are given on the edges. As was mentioned, the edge version admits an easy approximation ratio preserving reduction to the node version. 1.2 The unifying theme of the problems considered The following problem was defined in [9]: k-buy at Bulk Steiner Tree (k-bbst) Instance: A graph G = (V, E), costs {c(v) : v V }, lengths {l(v) : v V }, a set U V of terminals, a root r U, a diameter bound L, a cost bound c and an integer k. Objective: Find a subtree H of G containing r and at least k 1 additional terminals of U, of diameter L and cost c This problem is related to MBB as follows (this holds for both edge and node costs). Let T be an optimal k-bbst solution for an instance at hand. Thus c(t) = c and T has diameter L. 3

4 Theorem 1.2 [9] There exist two universal constants c 1, c 2 and a polynomial time algorithm that finds a a tree H containing at lest k/8 terminal with cost at most c 1 log 3 n c and diameter at most c 2 log n L Such an algorithm is called a (O(log 3 ), O(logn)) bicriteria approximation algorithm for k- BBST. Theorem 1.3 [5] The algorithm of Theorem 1.2 can be used to design a polynomial time combinatorial O(log 4 n)-ratio approximation algorithm for MBB The importance of Theorem 1.3 is that from a more practical point of view, the combinatorial algorithm is much faster than the algorithm of [5], which repeatedly solves large linear programs. It thus seems important to try and get a better bicriteria approximation for k-bbst than the one in Theorem 1.2. Improved bicriteria algorithm for k-bbst would imply a better combinatorial algorithm for MBB. Thus, if the cost returned is O(ρ 1 ) c and the diameter is O(ρ 2 L) the approximation ratio derived is O(log n) ρ 1 + O(log 3 n) ρ 2. Since we are facing difficulties in improving the results for k-bbst we study MaxCT and SLST which are closely related to, but are are relaxed variants of k-bbst. The hope that it may shed light on k-bbst. Also, MaxCT and SLST are interesting in their own right. Comparing MaxCT and k-bbst: If the costs are uniform, setting cost bound k is the same as seeking a tree spanning a subset of size k of the terminals, and maximizing the profit. However, the big difference between k-bbst and MaxCT is that there are no length constrains in MaxCT and that the primal-dual techniques of [13] do not seem suitable to handle lengths constrains. Comparing SLST and k-bbst: The SLST problem has lengths. And also, the goal is indeed to produce (given a cost and diameter bounds) a tree that is both shallow and light. What seems to make these two problems quite different is that SLST requires covering all terminals (and not only k as in k-bbst). The difference seems quite significant, and thus k-bbst seems significantly harder to handle than SLST. In summary, one may hope that techniques for MaxCT that (in the case of uniform costs) are able to find a tree with k terminals and low cost, but may not be suited to handle distances constrains, could somehow be combined with techniques that do produce a tree that is both shallow and light, but work only for the case k = U. 1.3 Related work We survey some results on relevant network design problems with node costs. Klein and Ravi [12] showed that the Node-Weighted Steiner Tree problem is Set-Cover hard, thus it admits no o(log n) approximation unless P=NP [15]. They also obtained a matching approximation ratio using a greedy merging algorithm. Guha et al. [8] showed O(log n) integrality gap of a natural LP-relaxation for the problem. The MBB problem is motivated by economies of scale that arise in a number of applications, especially in telecommunication. The problem 4

5 is studied as the fixed charge network flow problem in operations research. The first approximation algorithm for the problem is by Salman et al. [16]. For the multi-commodity version MBB the first non-trivial result is due to Charikar and Karagiazova [3] who obtained an O(log D exp(o( log n log log n)))-approximation, where D is the sum of the demands. In [4] an O(log 4 n)-approximation algorithm is given for the edge costs case, and further generalized to the node costs case in [5]. See [1] for an Ω(log 1/2 ε n)-hardness result. The MaxCT problem was introduced in [8] motivated by efficient recovery from power outage. In [8] a pseudo approximation algorithm is presented that returns a subtree T with c(t) 2C and p(t) = Ω(P/ log 2 n), where P is the maximum profit under budget cost C. This was improved in [13] to produce a tree T with c(t) 2C and p(t) = Ω(P/ log n). For a related minimization problem when one seeks to find a minimum cost tree T with p(t) P [13] gives an O(ln n)-approximation algorithm. 1.4 Our results The previously best known ratio for MCD/MBB was O(log 4 n) both for edge costs [4] and node costs [5], and this was also so for polynomial demands. We improve this result by using, among other things, a better LP-relaxation for the problem. Theorem 1.4 MCD/MBB with polynomial demands admits an O(log 3 n)-approximation algorithm. Our next result is for the MaxCT problem. In [13] it is conjectured that MaxCT admits an O(1) approximation algorithm (which would have been quite helpful for dealing with k-bbst). We disprove this conjecture. Theorem 1.5 MaxCT admits no constant approximation algorithm unless NP DTIME(n O(log n) ) even if the algorithm is allowed to use a budget of ρ B for any universal constant ρ. MaxCT admits no o(log log n) approximation algorithm unless NP DTIME(n polylog(n) ) even if the algorithm is allowed to use ρ B budget for any universal constant ρ. Our last result is for the SLST problem. For SLST with edge costs and edge lengths, the algorithm of [12] computes a tree T with c(t) = O(logn) opt and diam l (T) = O(log n) L. We consider the more general case of node costs and node lengths. Theorem 1.6 SLST with node costs and lengths admits an approximation algorithm that computes a tree T with c(t) = O(log 2 n) opt and diam l (T) = O(log n) L. Theorems 1.4 and 1.5 are proved in Sections 2 and 3. 5

6 2 Improved algorithm for MBB (Proof of Theorem 1.4) In this section we prove Theorem 1.4. We give an O(log 2 n log N)-approximation algorithm for MCD with running time polynomial in N, where N is the sum of the demands plus n. If N is polynomial in n, the running time is polynomial in n, and the approximation ratio is O(log 3 n). We may assume (by duplicating nodes) that all demands are 1. Then our problem is: Instance: A graph G = (V, E), costs {c(v) : v V }, lengths {l(v) : v V }, and a set D of node pairs. Objective: Find a subgraph H of G minimizing w(h, D) = c(h) + l H (s, t). {s,t} D For the latter problem, we give an O(log 2 n log D )-approximation algorithm. 2.1 Approximate greedy algorithm and junction trees We use a result about the performance of a Greedy Algorithm for the following type of problems: Covering Problem Instance: A groundset Π and functions ν, w on 2 Π with ν(π) = 0. Objective: Find P Π with ν(p) = ν(π) and with w(p) minimized. Let ρ > 1 and let opt be the optimal solution value for the Covering Problem. The ρ-greedy Algorithm starts with P = and iteratively adds subsets of Π P to P one after the other using the following rule. As long as ν(p) > ν(π) it adds to P a set R Π P so that σ P (R) = w(r) ν(p) ν(p + R) ρ opt ν(p) ν(π). (2) The following known statement follows by a standard set-cover analysis, c.f., [11]. Theorem 2.1 If ν is decreasing and w is increasing and subadditive, then the ρ-greedy Algorithm computes a solution P with w(p) ρ [ln(ν( ) ν(π)) + 1] opt. In our setting, Π is the family of all st-paths, {s, t} D. For a set R Π of paths connecting a set R of pairs in D, let ν(r) = D R be the number of pairs in D not connected by paths in R, and let w(r) = c(r) + {s,t} R l(p st ), where c(r) denotes the cost of the union of the paths in R, and P st is the shortest st-path in R. Note that ν(π) = 0 and ν( ) = D. We will show how to find such R satisfying (2) with ρ = O(log 2 n). W.l.o.g., we may consider the case P =. (Otherwise, we consider the residual instance obtained by excluding from D all pairs connected by P and setting P = ; it is easy to see that if R satisfies (2) for the residual instance, then this is also so for the original instance.) Assuming P =, (2) can be rewritten as: σ(r) = c(r) R + {s,t} R l(p st ) ρ opt R D 6. (3)

7 The quantity σ(r) in (3) is the density of R; it is a sum of cost-part c(r)/ R and the remaining length-part. The following key statement from [4] shows that with O(log n) loss in the length part of the density, we may restrict ourselves to very specific R, as given in the following definition; in [4] it is stated for edge-costs, but the generalization to node-costs is immediate. Definition 2.1 A tree T with a designated node r is a junction tree for a subset R D of node pairs in T if the unique paths in T between the pairs in R all contain r. Lemma 2.2 ([4], The Junction Tree Lemma) Let H be an optimal solution to an MCD instance with {0, 1} demands. Let C = c(h ) and let L = {s,t} D l H (s, t). Then there exists a junction tree T for a subset R Q of pairs, so that diam l (T) = O(log n) L/ D and c(t)/ R = O(C/ D ). If we could find a pair T, R as in Lemma 2.2 in polynomial time, then we would obtain an O(log D log n)-approximation algorithm, by Theorem 2.1. In [4] it is shown how to find such a pair that satisfies (3) with ρ = O(log 3 n). We will show how to find such a pair with ρ = O(log 2 n). Theorem 2.3 There exists a polynomial time algorithm that given an instance of MCD with {0, 1} demands computes a set R of paths connecting a subset R D of pairs satisfying (3) with ρ = O(log 2 n). Motivated by Lemma 2.2, the following LP was used in [4, 5]. Guess the common node r of the paths in R of the junction tree T. Let U be the union of pairs in D. Relax the integrality constraints by allowing fractional nodes and paths. For v V, x v is the fraction of v taken into the solution. For u U, y u is the total amount of flow v delivers to r. In the LP, we require y s = y t for every {s, t} D, so y s = y t amount of flow is delivered from s to t via r. For u U let Π u be the set of all ur-paths in Π, and thus Π = u U Π u. For P Π, f P is the amount of flow through P. Dividing all variables by R (note that this does not affect the objective value), gives the following LP: 2.2 The LP used (LP1) min s.t. v V c(v) x v + P Π l(p) f P u U y u = 1 {P Π u v P } f P x v v V, u U P Π u f P y u u U y s y t = 0 {s, t} D x v, f P, y u 0 v V, P Π, u U Let A log n L/ D be the bound on the lengths of the paths in R guaranteed by Lemma 2.2. We use almost the same LP as (LP1), except that we seek to minimize the cost only, and 7

8 restrict ourselves to paths of length at most A log n L/ D, which reflects better the statement in Lemma 2.2. For Π Π let Π = {P Π : l(p) A log n L/ D }. Again recall that y u is the flow delivered from u to r. The LP we use is: (LP2) min s.t. v V c(v) x v u U y u = 1 {P Π u v P } f P x v v V, u U P Π u f P y u u U y s y t = 0 {s, t} D x v, f P, y u 0 v V, P Π, u U Although the number of variables in (LP2) might be exponential, any basic feasible solution to (LP2) has O(N 2 ) non-zero variables. Lemma 2.4 LP(2) can be solved in time polynomial in n Proof: In [7] we give a proof via duality to a very similar LP. Here we show that another method is applicable (besides the duality arguments that still hold true). We transform our instance into a new one with lengths being integral and bounded by N 4, so that the loss in the approximation ratio is a constant, and hence is negligible in our context. By trying all nodes, we guess a node v that has the largest length among the nodes that carry a positive flow in some optimal solution H. Let M = l(v). Zero the length of all nodes u with l(u) < M/N 4. We claim that the decrease in the length part {s,t} Q l H (s, t) of w(h, Q) due to this modification is negligible. Note that the total demand that goes through one node is at most N 2. Thus if u is a node so that l(u) M/N 4, then the contribution of u to the length part is at most l(u) N 2 M/N 2, and the contribution of all such nodes is at most M/N. Since at least one unit of demand goes through v, {s,t} Q l H (s, t) M. Hence the loss is at most a fraction 1/N of the length part. Now, the minimum (non-zero) length of a node that contributes to the length part is at least M/N 4. Divide the length of every node by M/N 4 and round down the result to the nearest integer. It is easy to see that the rounding down only incurs a loss of factor 2 in the sum of the distances. Thus the total loss incurred is at most 2+1/N, as claimed. We show that (LP2) can be solved by writing an equivalent LP with polynomial number of variables and constraints. Recall that we assume that the lengths are integral and the largest node length is at most N 4. This implies that the maximum length of a path is N 5. Define a variable s(v, u, P, j) for every v V, u U, P Π u, and j N 5, so that s(v, u, P, j) = 1 if, and only if, v at distance j from u in the path P. Define f(v, u, j) = P Π u s(v, u, P, j) f P. Also f(u, j) = v f(v, u, j). Thus f(u, j) is the total amount of flow from u to r at distance j from u. The above variables are used to define an alternative LP. The flow conservation inequality is defined as follows. Every path in Π u in which v is at distance j has a node z prior to v of 8

9 distance j l(v) from u. Thus the conservation of flow equality is: The flow from u to r is f(v, u, j) = zv E f(z, u, j l(v)) v V, u U, j N 5. 1 j N 5,vr E f(u, j l(r)). It is not hard to verify, that the obtained LP is equivalent to (LP2). As its size is polynomial in N, it can be solved in time polynomial in N. Given a solution for the modified LP, the solution for (LP2) is derived by standard flow decomposition By Lemma 2.2 there exists a solution to (LP2) of value O(C/ D ). Indeed, let T, R, R be as in Lemma 2.2; in particular, c(t)/ R = O(C/ D ). For u T let P u be the unique ur-path in T. Define a feasible solution for (LP2) as follows: x v = 1/ R for every v T, y u = f Pu = 1/ R for every u that belongs to some pair in R, and x u, y u, f P are zero otherwise. It easy to see that this solution is feasible for (LP2), and its value (cost) is c(t)/ R = O(C/ D ). 2.3 Proof of Theorem 1.4 We now proceed similarly to [4, 5]. We may assume that max{1/y u : u U} is polynomial in n, see [5]. Partition U into O(logn) sets U j = {u U : 1/2 j+1 y u 1/2 j }. There is some U j that delivers Ω(1/ ln n) flow units to r. Focus on that U j. Clearly, U j = Θ(2 j )/ logn. Setting x v = min{θ(2 j ) x v, 1} for all v V and f P = min{θ(2 j ) f P, 1} for all P Π, gives a feasible solution for the following LP that requires from every node in U j to deliver a flow unit to r. (LP3) min s.t. v V c(v) x v + P Π l(p) f P {P Π u v P } f P x v v V, u U j P Π u f P 1 u U j x v, f P 0 v V, P Π We bound the value of the above solution x, f for (LP3). Since we have v V c(v)x v = O(C/ D ), c(v)x v = O(2j ) C/ D. v V We later see that, since U j = Θ(2 j / log n), an extra log n factor is invoked in the costdensity part of our solution; if, e.g., U j = 2 j would hold, this log n factor would have been saved. Our main point is that the length-part of the density does not depend on the size of U j. We show this as follows. All paths used in (LP2) are of length O(logn L/ D ). First, assure that P Π u f P is not too large. For any u U j the fractional values of {f P : P Π u} only affect u, namely, if u u then Π u Π u =. Therefore, if P Π u f P >> 1, we may 9

10 assure that the sum is at most 3/2 as follows. If a single path carries at least 1/2 a unit of flow then (scaling values by only 2) this path can be used as the solution for u. Else, any minimal collection of paths delivering at least one unit of flow, delivers at most 3/2 units of flow to r. Hence the contribution of a single node u to the fractional length-part is O(log n L/ D ) P Π u f P = O(log n L/ D ). Over all terminals, the contribution is O( U j log n L/ D ). Now, use the main theorem of [5]: Theorem 2.5 ([5]) There exists a polynomial time algorithm that finds an integral solution to (LP3) of value O(log n) times the optimal fractional value of (LP3). Hence we can find in polynomial time a tree T containing r and U j with c(t) = O(log n 2 j C/ D ) and u U j l T (u, r) = O( U j log 2 n L/ D ). Note that if the tree contains i terminals then it contains i/2 pairs. This is due to the constraint y s = y t. Since the tree spans Θ(2 j / log n) pairs, its cost-part density is O(log 2 n) C/ D. Clearly, the length-part density is O(log 2 n) L/ D. This finishes the proof of Theorem 2.3, and thus also the proof of Theorem 1.4 is complete. 3 A lower bound for MaxCT (Proof of Theorem 1.5) Here we prove the following statement that implies Theorem 1.5. Theorem 3.1 MaxCT admits no better than c-approximation algorithm, even if the algorithm is allowed to use budget ρ B for any universal constant ρ, unless NP DTIME(n O(ln c exp(9c)) ). Clearly, this implies that unless P = NP, MaxCT admits no constant approximation algorithm even if the solution may violate the budget by a constant. Also, unless NP DTIME(n polylog n ) the problem admits no µ log log n-ratio approximation for some universal constant µ, even if the algorithm uses budget ρ B for some universal constant ρ. Remark: The size of the instance produced is s = n O(lnc exp(9c)) and thus c = Θ(log log s). Therefore, it is not possible to get a stronger hardness than log log n unless we get a better gap in terms of c. 3.1 The gap of Max-Coverage without budget violating We first show the result assuming the solution does not violate the budget. The extra details for the case the solution violates the budget by ρ are quite simple and gives in the next subsection. For a graph H = (V, E) and X V let Γ(X) = Γ H (X) = {v V X : uv E for some u X} denote the set of neighbors of X in H. We say that A V covers B V if B Γ H (A ). 10

11 We now define the unweighted max-coverage problem. The problem is usually presented by an input that is a collection S of sets and a number opt and the goal is to find subcollection S of S, {S } = opt sets so that S i S S i is maximize. We present the problem via bipartite graphs. Max-Coverage Instance: A bipartite graph H = (A + B, E) whose optimum cover is of cost opt Objective: Find A of cost A = opt and maximize Γ H (A ) A is called the collection of sets and B the collection of em elements. We say that b B belongs to a A is (a, b) E. If b Gamma H (A ) we say that A covers b. In the set-cover problem the goal is to cover all elements with minimum size A (namely find a minimum size A so that Γ H (A ) = B) In [6] it is proved: Theorem 3.2 For any universal constant ǫ so that for every NP C language there is polynomial time a reduction from an instance I to an instance of Max-Coverage so that 1. For a yes instance there exists a subset A A, of size opt so that γ H (A ) = B 2. For every no instance subset of A of size opt can cover more than (1 1/e) B elements In addition there is an O(n loglog n ) time reduction from I to an instance of set cover so that: 1. For a yes instance there exists a subset A of size opt that covers all of B 2. For a no instance any subset of A that covers B has size at least (1 ǫ) ln n opt for any universal constant ǫ Corollary Unless P = NP Max-Coverage admits no polynomial time 1 1/e+ǫ approximation ratio 2. Unless NP DTIME(n O(log log n ), the set-cover problem can not be approximated within (1 ǫ) ln n ratio for any universal constant ǫ approximation algorithm The following claim is essentially identical to Proposition 5.2 in [6] and is given for completeness. Claim 3.4 Unless NP DTIME(n log log n ) for any universal constant 1 < a < lnn there exists an ǫ (that depends on a) so that no algorithm that produces a solution of size a opt and covers more than 1 e a + ǫ fraction of the remaining elements for Proof: If such an algorithm exists, apply it in iterations. Each iterations adds a opt to the solution size and a fraction of at least (1 e a +ǫ) of the elements are covered. Looking at the 11

12 decline of uncovered elements, their number is e a ǫ times the previous number of uncovered elements. Thus, l that satisfies (e a ǫ) l = 1/n is an upper bound on the number of iterations needed to reduce the number of uncovered elements to 0. As ln n l = a ln(1 ǫe a ), it follows that for some constant δ that depends on a, l (1 δ) ln n/a. Since at every iteration a opt sets are added to the solution, the total number of sets in the solution is at most (1 δ)opt contradicting Corollary The reduction Define a sequence of graphs G 1, G 2,... by induction (see Fig. 1). To obtain G 1, take H, add a root r, and connect r to every node in A. Let A 1 = A and B 1 = B. To obtain G i from G i 1, i 2, take G 1 and B copies of G i 1, each corresponding to a node in B 1, and for every copy identify its root with the node corresponding to it in B 1. As the construction resembles a tree, we borrow some terms from the terminology of trees. A copy of H has level i if its A sets have distance 2i 1 to the root r. The copies of H at level i are ordered arbitrarily. A typical copy of H at level i is denoted by H ij = (A ij, B ij, E ij ) with i the level of the copy and j the index of the copy. This means that the A ij sets are at distance 2i 1 from the root and the index j is the order statistic of the copy inside level i. Definition 3.1 The depth of the reduction for a chosen constant c: For any constant c, let Let A i = j A ij and B i = j B ij. h = h(c) = 4e 9c ln(2c) (4) An H ij is an ancestor of a terminal y if y belongs to the subgraph rooted by some v B ij ; such v is called the elements ancestor of y in level i and is denoted ans i (y). Note that ans i (y) is unique. The sets A ij, B ij are the ancestors sets of y in level i and are denoted by A i y, Bi y. We also say that y is a descendant of ans i (y) and A i y and By. i The terminals of G h are j B h,j, and each of them has profit 1; other nodes have profit 0. The cost of every node in A ij is 1/ B i 1 (so the nodes in A 1 = A 11 have cost 1), and the cost of any other node is 0. The cost bound is C = h opt. Fact 3.5 The size (and the construction time) of the construction is n O(h), where n = max{ A, B }. 3.3 Analysis While increasing the level by 1, the number Max-Coverage instances grows up by B but the node costs go down by B. Hence the total cost of every level i is A, and the total cost of 12

13 (a) r (b) r A 1 G 1 A 1 E B 1 1 H E B 1 1 G 2 A 2 E 2 B 2 Figure 1: (a) The graph G 1. (b) The graph G 2 ; if instead of copies of G 1 we attach to nodes in B 1 roots of the copies of G i 1, then we obtain G i. G is h A. We may assume that any solution T to the obtained instance of MaxCT contains r. Otherwise, we may add the shortest path from r to T; the cost added is negligible in our context. Hence a subgraph T of G is a perfect solution if T contains r, T is connected, and c(t) C = h opt. Lemma 3.6 (The YES-instance) The instance G admits a perfect feasible solution T (hence T contains all terminals). Proof: Consider the graph T induced in G by r and all the copies of A B so that A = opt and A set-covers B. This graph is clearly connected and contains all terminals as each A copy covers its copy of B. The cost of all copies of A at any level i is opt. Summing over all levels gives total cost c(t) = h opt = C, as claimed. We now deal with a no instance. Fix a feasible solution T for MaxCT. Our intent is to show that any solution for a NO-instance can cover at most 1/c fraction of the terminals with budget C. Intuitively, T has an average cost of opt to spend on every level i. Averaging over all (A ij, B ij, E ij ) copies, on average the tree T has to select at most opt sets of A ij (the cost of each each set is 1/ B i 1 and there are B i 1 copies of the max coverage instance on level i). Thus if the number of sets chosen at every A ij is opt the total cost would be opt h. Definition 3.2 Level i in G is cheap (w.r.t. T) if c(t A i ) < 2opt and is expensive otherwise. A copy H ij in level i is cheap if T A ij 8 c opt and is expensive otherwise. A terminal y is cheap if for at least h/4 of the cheap levels, (A y i, B y i, Ei i ) is cheap. A cheap level i for terminal y, indicates that T A y i 8 c opt (namely, the number of A y i selected is at most 4c times the expected average 2opt). But a single cheap level for a terminals does not make the terminals cheap. The terminal is cheap only if for at least h/4 of the cheap levels, its (A y i, B y i, E i i) copy is cheap. Because the total budget bound is C we get: 13

14 Fact 3.7 At most half the levels are expensive Claim 3.8 At most 1/2c of the terminals are expensive. Proof: Let s h/2 be the number of cheap levels. Let i be some cheap level. Pick a terminal y at random. Consider the unique ancestor Hy i of y in level i. Note that every Hy i has the same number of terminal ancestors. This follows by symmetry. Plus Hy i, Hi y do not share leaves. Thus by symmetry Hy i is a random copy of the (A ij, B ij, E ij ) at level i. By the definition of a cheap level, the average number of sets chosen over all A ij copies is at most 2opt. As Hy i is a random copy (namely, the sample space is uniform over all set-coverage copies in level i) the expected size of T A i y (which equals the above average) is bounded by 2opt. Using the Markov inequality (c.f., [14]). The event A i y T 8copt holds with probability at least 1 1/(4c). By the linearity of the expectation, this implies that the expected number of cheap levels i for y, namely, levels for which A i y T > 8 c opt, is at most s/(4c) with s the number of cheap levels. Thus, by the Markov inequality, the probability that at least s/2 of the Hy i in cheap levels i are expensive for y is at most 1/(2c). Therefore, with probability at least 1 1/(2c) at least s/2 h/4 of the Hy i (on cheap levels) are cheap for y. The last inequality follows as s is the number of cheap levels and there are at least h/2 cheap levels. Thus the probability of the event that a random y is cheap is at least 1 1/(2c). By symmetry (namely because this sample space is uniform) this probability is exactly the number of cheap y over the number of terminals. Thus the fraction of expensive terminals among all terminals is at most 1/(2c). Intuitively, we ignore the affect of expensive level (namely, assume that every element was covered in such levels). We also assume that expensive terminals belong to T. But we bound the number of cheap terminals in T. The number of expensive terminals is most 1/(2c) fraction of the terminals. Now, check what is the fraction of cheap terminals that can belong to T. Lemma 3.9 (The NO-instance) If Max-Coverage is a no instance T contains at most 1/2c of the cheap terminals. Proof: Consider any cheap level i and a cheap H ij at this level. By Corollary 3.4, for every cheap H ij, as A = A ij T 8 c opt, the set A only covers a subset B = Γ(A ) so that B ( 1 1 ) B. e 1+8c We ignore the ǫ in Corollary 3.4 and add 1 to the power of e. It is easy to see that the effect of ǫ is negligible compared to the +1 in the exponent of e. This gives 1/e 1+8c fraction of the elements descendants of H ij that do not belong to T. This is because every node b B ij has the same number of terminal descendants. Thus the fraction of b B ij lost (namely, that do 14

15 not belong to T) equals the fraction of terminals lost. In summary, at every cheap level i, for every cheap H ij a fraction of at least 1/e 1+8c of its descendants terminals do not belong to T ; indeed, ans i y T for those terminals. We now use Claim 3.8. Mark all H ij that belong to expensive levels and also all expensive H ij. By Claim 3.8, the path from every y G h T to r contains at least h/4 unmarked H ij ancestors. From this we deduce that the fraction of terminals in T is at most: (1 1/e 9c ) h/4. Recall that h = 4 2 9c ln(2c) by Definition 3.1 and so we get that: (1 1/2 9c ) h/4 < 1/(2c). The fraction of terminals that belong to T is at most the number of expensive terminals plus the number of terminals of T namely, at most 1/(2c) + 1/(2c) 1/c. The proof of Lemma 3.9 is now complete. Theorem 1.5 directly follows from Lemma 3.6 and Lemma The details needed to get a bicriteria lower bound We need to define: h = 4 2 9cρ ln(2c). There is no change in the analysis of a yes instance. A layer is called cheap if the cost used in this layer is at most 2cρ opt. For a no instance, A ij in a cheap layer is called cheap if T A ij 2ρ opt and expensive if T A ij 2 c ρ opt. The budget is still h opt. Note that at most 1/2c of the (A j,k, B jk, E jk ) copies at level j may be heavy (ρ is a multiplier both in the definition of a cheap level and in the definition of a heavy copy. The two terms cancel). However, the added ρ to the number of sets chosen in every copy. Thus means that the number of cheap terminals is at most (1 1/e 9cρ ) h/4. But the new definition of h cancels ρ and the claim follows. Thus ρ only influences the degree of the polylog in the quasi-polynomial time solution. Thus the rest of the details remain unchanged. 4 Algorithm for SLST As was mentioned, for SLST with edge costs/lengths the algorithm of [12] computes a tree T with c(t) = O(log n) opt and diam l (T) = O(log n) L. This is done as follows. Let C = opt. Maintain a disjoint partition of the terminals into pairwise disjoint clusters; each cluster has a unique center. Initialize every terminal as a cluster of size 1. Then iterate as follows. Let S be the set of cluster centers. Throughout, only terminals will be centers. For s, t S let c(s, t) be the minimum cost of an st-path among st-paths of length at most L. Although computing c(s, t) is an NP-hard problem, for any ε > 0 we can approximate it using the FPTAS of [10], which computes a path P st with l(p st ) L and c(p st ) (1 + ε)c(s, t). Now, construct an auxiliary complete graph on S with the costs of every edge st being c(p st ). As all terminals 15

16 belong to the solution, by the so called Pairing Lemma (see [12]) there exists a matching on the centers of cost at most (1 + ε)c, with at most one cluster center unmatched. Compute this perfect matching. Replace every edge st in the matching by the corresponding path P st in G and merge the corresponding two clusters together, making its center to be one of s, t. At every iteration, the cost invested is at most (1 + ε)c, the radius of each cluster is increased by at most L, and the number of clusters is roughly halved. The later implies that the number of iterations is O(log n), and the (O(log n), O(log n)) bicriteria approximation follows. In our case of node-costs we use the decomposition of the optimum tree T into disjoint spiders. W.l.o.g., via standard reductions (see [11]), we assume that the terminals are the leaves of T. Definition 4.1 A tree T is a spider if it has at most one node of degree 3. A spider decomposition D of a tree T rooted at r is a collection of node-disjoint spiders, so that each of them is a rooted subtree of T, and the sets of leaves of the spiders in D partition the set of leaves of T. Lemma 4.1 ([11]) Any tree T rooted at r admits a spider decomposition so that every spider has at least two leaves, or in the decomposition there is exactly one spider with one leaf and root r. The problem of finding the cheapest spider decomposition of a graph is at least as hard as the set-cover problem. Instead, we approximate the cost of the best spider decomposition and at the same time control the length invoked. Lemma 4.2 For any ε > 0 there exists a polynomial algorithm that computes a collection of rooted trees of total cost O(log n) opt containing all terminals, so that each tree has l-radius at most (1 + ε)l, and so that every tree, except of maybe one, contains at least two terminals. Proof: While there are at least two terminals not belonging to any tree, iteratively find a tree F with at least 2 terminals of radius (1 + ε)l whose cost-density, (which is its cost over the number of terminals in it) is at most (1+ε) times the one of the best spider decomposition of any optimum solution. We stress that the density in question is only with respect to non-covered terminals (only terminals not covered by previous spiders are considered). Then we remove the covered terminals, and iterate. Finding a tree of low density is done as follows. Guess the root v and the number q of terminals. For every terminal t approximate using [10], the min-cost of a v to t path of length at most L. The candidate tree for v and q is the union of the q paths from v to its best q terminals. Compute the density of the candidate tree for v, q. Then among the trees computed return one with the minimum density. The paths of the tree may intersect but since we are comparing against a spider, whose paths are node disjoint (except for the root), the resulting tree has cost density no larger than 1 + ε times the density of the best spider. An analysis similar to the classical set-cover analysis shows that the total cost of the resulting decomposition is O(log n) times the cost of the minimum cost spider decomposition which is O(log n) opt. 16

17 The other parts of the algorithm are similar to the algorithm of [12] for the edge cost case. Maintain a disjoint partition of the terminals into pairwise disjoint clusters; each cluster has a unique center. Initialize every terminal as a cluster of size 1. Then iterate as follows. Let S be the set of cluster centers. Given ε > 0, compute a family of trees as in Lemma 4.2, considering only the set S of centers as terminals, and for every spider, merge the clusters of the centers it contains; the center of the new cluster is set to be one of these centers. At every iteration, the cost invested is at most (1+ε)C O(logn), the radius of each cluster is increased by at most L, and the number of clusters is roughly halved. The later implies that the number of iterations is O(log n), and the (O(log 2 n), O(logn)) bicriteria approximation follows. References [1] M. Andrews. Hardness of buy-at-bulk network design. In Proc. FOCS, pages , [2] M. Andrews and L. Zhang. Approximation algorithms for access network design. Algorithmica, 34(2): , [3] M. Charikar and A. Karagiozova. On non-uniform multicommodity buy-at-bulk network design. In Proc. STOC, pages , [4] C. Chekuri, M. T. Hajiaghayi, G. Kortsarz, and M. R. Salavatipour. Approximation algorithms for non-uniform buy-at-bulk network design. In Proc FOCS, pages , [5] C. Chekuri, M. T. Hajiaghayi, G. Kortsarz, and M. R. Salavatipour. Approximation algorithms for node-weighted buy-at-bulk network design. In SODA, pages , [6] U. Feige. A threshold of for approximating set cover. J. ACM, 45: , [7] M. Feldman, G. Kortsarz, and Z. Nutov. Improved approximating algorithms for directed steiner forest. In SODA, pages , [8] S. Guha, A. Moss, J. S. Naor, and B. Schieber. Efficient recovery from power outage. In Proc. STOC, pages , [9] M. T. Hajiaghayi, G. Kortsarz, and M. R. Salavatipour. Approximating buy-at-bulk k- Steiner tree. In Proc APPROX, pages , [10] R. Hassin. Approximation schemes for the restricted shortest path problem. Math. Oper. Res., 17(1):36 42,

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