Survey of Some Connectivity Approximation Problems via Survey of Techniques Guy Kortsarz Rutgers University, Camden, NJ.

The talk is based on the comprehensive survey G. Kortsarz and Z. Nutov, Approximating min-cost connectivity problems, Survey Chapter in handbook on approximation, Chapter 58, 30 pages.

Steiner Network Problem Steiner Network: Instance: A complete graph with edge (or node) costs, and connectivity requirements r(u,v) for every pair. Objective: Min-cost subgraph with r(u,v) edge (vertex) disjoint uv - paths for all u,v in V. k-edge-Connected Subgraph: r(u,v) =k for all u,v. k-vertex-Connected Subgraph: r(u,v) =k for all u,v.

Example k=2 vertex 2-connected graph a b c

Previous Work on Steiner Network VERTEX CASE: Labelcover hard. [K, Krauthgamer, Lee, SICOMP] k ε approximation not possible for some universal ε>0 [Chakraborty, Chuzhoy, Khanna,STOC 2008] Undirected and directed problems are equivalent for k>n/2 [Lando & Nutov, APPROX 2008] O(log n)-approximation for metric costs. [Cheriyan & Vetta, STOC 2005] O(k^3log n) (k maximum demand). [Chuzhoy & Khanna, STOC 2009] EDGE CASE: Edge-Connectivity: sequence of papers, until reaching a 2-approximation [Jain, FOCS 98]

Transitivity in Edge Connectivity If (a,b)=k and (b,c)=k then (a,c)=k Proof: a c b b K-1

First Technique: Directed Out-Connectivity The following problem has a polynomial solution: Input: A directed graph G(V,E) a root r and connectivity requirement k Required: Min cost subgraph so that there will be k edge disjoint paths from r to any other vertex Polynomial time algorithm: for the edge case by matroids intersection (Edmonds). Also true for k vertex disjoint paths from r [Frank, Tardo’s] (submodular flow)

Algorithm for k-ECSG If we have k connectivity from a vertex v to all the rest, by transitivity the graph is k-edge-connected Apply the Edmonds algorithm twice: replace every edge with two directed edges Once k-in-connectivity to v Second k-out-connectivity from v Ratio 2 guaranteed

Work on Node k-Vertex Connected Subgraph [Cheriyan, Vempala, Veta, STOC 2002] O(log k)-approximation for undirected graphs with n>6k 2 [K & Nutov STOC 04] n/(n-k) O(log 2 k) for any k, directed/undirected graphs. The ratio is O(log 2 k), unless k = n - o(n). [Fackharoenphol and Laekhanukit, STOC 2008] O(log 2 k)-approximation also for k = n - o(n). O(log k) log (n/(n-k)) [Nutov, SODA 2009]. O(log n) unless k=n-o(n) Many excellent papers about particular cases: –metric costs: (2+k/n) [K & Nutov] –1,∞-costs: (1+1/k) [Cheriyan & Thurimella] –small requirements: [ADNP,DN,KN...]

Technique 2: The Cycle Theorem of Mader Let G(V,E) be a k-vertex connected graph, minimal for edge deletion and let C be a cycle in G Then there is a vertex in C of degree exactly k Strange Claim?

Corolloraly Say that  (G) is at least k-1 Let F be any edge minimal augmentation of G to a k- vertex-connected subgraph Then F is a forest

Proof Consider a cycle in F As all degrees are at least k-1 before F, with F all degrees are at least k+1 which contradicts Mader’s theorem.

Application in Minimum Power Networks In a power setting p(v)= max{ c(e) | e  E(v)} Reasons: transmission range a b c d f g h The power of G is  v p(v)

The Min-Power Vertex k- Connectivity Problem We are given a graph G(V,E) edge costs and an integer k Design a min-power subgraph G(V, E) so that every u,v  V admits at least k vertex-disjoint paths from u to v May seem unrelated to min cost vertex k-connectivity

Previous Work for Min-Power Vertex k - Connectivity Min-Power 2 Vertex-connectivity, heurisitic study [Ramanathan, Rosales-Hain, 2000] 11/3 approximation for k =2 [K, Mirrokni, Nutov, Tsano, 2006] Cone-Based Topology Control for Unit-Disk Graphs [M. Bahramgiri, M. Hajiaghayi and V. Mirrokni, 2002] O(k)- approximation Algorithm and a Distributed Algorithm for Geometric Graphs [M. Hajiaghayi, N. Immorlica, V. Mirrokni, 2003]

Comparing Power And Cost: Spanning Tree Case The case k = 1 is the spanning tree case Hence the min-cost version is the minimum spanning tree problem Min-power spanning tree: even this simple case is NP-hard [Clementi, Penna, Silvestri, 2000] Best known approximation ratio: 5/3 [E. Althaus, G. Calinescu, S.Prasad, N. Tchervensky, A. Zelikovsky, 2004]

The Case k = 1: Spanning Tree The minimum cost spanning tree is a ratio 2 approximation for min-power. Due to: L. M. Kerousis, E. Kranakis, D. Krizank and A. Pelc, 2003

Spanning Tree (cont’) c(T)  p(T): Assign the parent edge e v to v Clearly, p(v)  c(e v ) Taking the sum, the claim follows p(G)  2c(G) (on any graph): Assign to v its power edge e v Every edge is assigned at most twice The cost is at least The power is at exactly

k vertex-conn: Power, Cost Equivalent For Approx’(!) K, Mirrokni, Nutov, Tsano show that the vertex k - connectivity problem is essentially equivalent with respect to approximation for cost and power (somewhat surprising). In all other problem variants, almost, the two problems behave quite differently. Based on a paper by [M. Hajiaghayi, K, V. Mirrokni and Z. Nutov, IPCO 2005].

Reduction to a Forest Solution Say that we know how to approximate by ratio  the following problem: The Min-Power Edge-Cover problem: Input: G(V, E), c(e), degree requirements r(v) for every v  V Required: A subgraph G(V, E) of minimum power so that deg G(v)  r(v) Remark: polynomial problem for cost version

Reduction to Forest (cont’) Clearly, the min power for getting  (G’)  k-1, bounds the optimum power for k-connectivity, from below Say that we have a  approximation for the above problem Hence at cost at most  opt we may start with minimum degree k -1

Reduction to Forest (cont’) Let H be any feasible solution for the Edge-Multicover problem with r(v)  k-1 for all v Recall: let F any minimal augmentation of H into a k vertex-connected subgraph. Then F is a forest

Comparing the Cost and the Power Theorem: If MCKK admits an  approximation then MPKK admits  + 2  approximation. Similarly:  approximation for min-power k- connectivity gives  +  approximation for min-cost k – connectivity. Proof: Start with a β approximation H for the min- power vertex r(v) = k-1 cover problem Apply the best min-cost approximation to turn H to a minimum cost vertex k - connected subgraph H + F, F minimal

Comparing the Cost and the Power (cont’) Since F is minimal, by Mader’s theorem F is a forest Let F* be the optimum augmentation. Then the following inequalities hold: 1) c(F)   c(F*) (this holds because  approximation) 2) p(F)  2c(F) (always true) 3) c(F*)  p(F*) (F* is a forest); 4) p(F)  2c(F)  2  c(F*)  2  p(F*) QED

Very hard technical difficulty: Any edge adds power to both sides. Because of that: take k-1 best edges, ratio k-1 Admits an O(log n) ratio (Mirrokni et al). Proof omited the (quite hard) By The [Nutov 2009] result on min-cost edge k- connectivity O(log n) ratio (almost). SO DOES THE POWER VARIANT We conjecture  (log n) hardness. Approximating the Min-Power  (H)  k-1 Problem

A Result of Khuller and Ragavachari There exists a 2+2(k-1)/n ratio for minimum cost vertex k-connected subgraph in the metric case At most 4 always and tends to 2 for k=o(n) K, Nutov: 2+(k-1)/n ratio At most 3 and tends to 2 for k=o(n) Combines the two techniques shown

The Algorithm Let J k (v 0 ) be cheapest star for any v and its k cheapest edges. Let leaves be {v 1,…..,v k } Averaging gives that best star has cost at most J k (v 0 )  2OPT/n v0v0 v2v2 vkvk v1v1

The Algorithm Continued Let R={v 0,….,v k-1 } Note, that v k is absent from R As in [KR] add a new node s that does not belong to V Similar to [KR] define a new graph G s from G with 0 cost edges for sv i for any vertex v i

The Algorithm continued Compute a k - outconnected graph from s in G s. Let H s be this graph. By [KR] the cost of H s is at most 2opt (remark: our R is different then the one in [KR]) In [KR] it is shown that if we add all edges between the R vertices to H s, the resulting graph is k-connected. Unlike [KR] we add a MINIMAL feasible solution out of E(R) to H s

The Approximation Ratio k out-connectivity from s implies  (H S )  k-1 Thus F is a forest with k nodes. We bound the cost of edges in the forest F. For every v i,v j  v 0 we upper bound c(v i v j )  c(v 0 v i )+c(v j v 0 ) We call these costs the new costs

Upper Bounding c(F) For v i,v j  v 0 we get  vivj  F c(v i v j )   vivj  F c(v i v 0 )+c(v 0 v j )   vivj  F c(v 0 w k-1 )+c(v j, v 0 ) There are k-1 edges in F but we did not take the edges of v 0 which means that c(v 0 w k-1 ) is counted at most k-2 times.

Proof Continued Note that according to the new costs we got a star rooted at v k-1 The node v 0 is (in the worst case) also connected to v k-1 directly. This adds c(v 0 v k-1 ) to the cost of F. Thus c(F)  (k-2)·c(v 0 v k-1 )+c(J k-1 (v 0 ))

Proof Continued c(F)  (k-2) c(v 0 v k-1 ) +  1  i  k-1 c(v 0 v j ) We know that c(J k (v 0 ))  2opt/n Thus c(v 0 v k-1 )+c(v 0 v k )  2opt/n Thus c(v 0 v k-1 )  opt/n c(F)  (k-2) c(v 0 v k-1 ) + c(J k (v 0 )) –c(v 0 v k )  (k-3)· c(v 0 v k-1 ) + c(J k (v 0 )) c(F)  (k-3)opt/n+2opt/n=(k-1)opt/n Thus the final ratio is 2+(k-1)opt/n

Laminar Families We present the Jain result with a simplified proof due to Ravi et. al. The LP: R(S) maximum demand of a separated vertex v  S, u  S d(S)=number of edges going out of S LP= min  w e x e Subject to x(  (S))  R(S)-d(S) x e  0

Jain: one of the x e at least 1/2 For the sake of contradiction assume the contrary May assume tight inequalities in a BFS give laminar family (folklore?). Let L be laminar family and E’ non-zero edges. Thus |E’|=|L|

Charging Total charging equals |E’|=|L| 1-2x e xexe xexe

All Possible Edges All edge types. S S1 C1C1 C2C2 C3C3 C1C1

How Many Tokens S Owns? Let E(S) be edges internal to S. The sets C discussed now are children of S. S owns a vertex in S if does not belong to any child e is assigned to the smallest S so that e  E(S) Define the tokens in S: t(S)=E(S) −  E(C)+x(  (S))−   (C)

Contribution to Both Sides of Every Edge t(S)=E(S) −  E(C)+x(  (S))−   (C) An edge with no endpoint in S or an edge that enters a child of and exits S. Contribution 0. An edge with both end points in S that does not enter a child of S can not exist.

More cases t(S)=E(S)−  E(C)+x(  (S))−   (C) An edge that enters S but not a child of S contributes x e An edge that enters a child of S but not S contributes 1-x e An edge between two children of S contributes 1-2x e.

t(S) IS NOT ZERO It can not be that all edges exit S and enter a child of S. Namely, it can not be that all contributions are 0. Indeed in this case S is the sum of its children In all other cases the contribution is positive.

t(S)  1 Consider: t(S)=E(S)−  E(C)+x(  (S))−   (C) The children C belong to the laminar family, hence they are tight namely their  (C) is integral. Thus t(S)  1.

We Charged Already |L| Because one per S Thus we found t(S) associated with S only, that is at least 1 Clearly the parts associated are disjoint This implies that we found already a fraction of |L|. We are going to show that some fraction remains, contradiction.

The Contradiction Look at the maximum S. Some edges must be leaving it because its violated. The 1-2x e of these edges is positive. Uncharged. This means t(S)  |E’|>|L|, contradiction.

Thank you for attention. Questions?