Survey of Some Connectivity Approximation Problems via Survey of Techniques

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

2. The talk is based on the comprehensive survey • G. Kortsarz and Z. Nutov, Approximating min-cost connectivity problems, • Survey Chapter in handbook on approximation, 2006. Chapter 58, 30 pages.

3. Steiner Network Problem Steiner Network: Instance: A complete graph with edge (or node) costs, and connectivity requirementsr(u,v)for every pair. Objective: Min-cost subgraph withr(u,v)edge (vertex) disjointuv - pathsfor allu,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.

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

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

6. Transitivity in Edge Connectivity • If(a,b)=kand(b,c)=kthen (a,c)=k Proof: b a K-1 c b

7. First Technique: Directed Out-Connectivity • The following problem has a polynomial solution: • Input: AdirectedgraphG(V,E)a rootrand connectivity requirementk • Required: Min cost subgraph so that there will bek edge disjoint paths fromrto any other vertex • Polynomial time algorithm: for the edge case by matroids intersection (Edmonds). • Also true forkvertex disjoint paths fromr [Frank, Tardo’s] (submodular flow)

8. Algorithm for k-ECSG • If we have kconnectivity from a vertexvto all the rest, by transitivity the graph isk-edge-connected • Apply the Edmonds algorithm twice: replace every edge with two directed edges • Oncek-in-connectivitytov • Secondk-out-connectivity fromv • Ratio 2guaranteed

9. Work on Node k-Vertex Connected Subgraph • [Cheriyan, Vempala, Veta, STOC 2002] O(log k)-approximation for undirected graphs with n>6k2 • [K & Nutov STOC 04] n/(n-k) O(log2 k)for anyk, directed/undirected graphs. The ratio isO(log2 k),unlessk = n - o(n). • [Fackharoenphol and Laekhanukit, STOC 2008] O(log2 k)-approximation alsofork = 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...]

10. 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?

11. 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 Fis a forest

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

13. Application in Minimum Power Networks • In a power settingp(v)= max{ c(e) | eE(v)} • Reasons: transmission range. b 7 a f 5 4 2 h 8 5 8 6 9 g c d 3 The power of G is v p(v)

14. 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

15. 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]

16. 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]

17. 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

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

19. 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].

20. 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 requirementsr(v) for everyvV • Required: A subgraphG(V, E)of minimum power so thatdegG(v) r(v) • Remark: polynomial problem for cost version

21. 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

22. Reduction to Forest (cont’) • Let H be any feasible solution for the Edge-Multicover problem with r(v)  k-1for allv • Recall:let Fany minimal augmentation ofHinto akvertex-connected subgraph. ThenFis a forest

23. 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 aβapproximation Hfor the min-power vertexr(v) = k-1cover problem • Apply the best min-cost approximation to turnHto a minimumcost vertexk- connected subgraphH + F, Fminimal

24. Comparing the Cost and the Power (cont’) • Since Fis minimal, by Mader’s theoremFis a forest • LetF*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

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

26. A Result of Khuller and Ragavachari • There exists a2+2(k-1)/nratio for minimum cost vertexk-connected subgraph in the metric case • At most4always and tends to2for k=o(n) • K, Nutov:2+(k-1)/nratio • At most3and tends to2fork=o(n) • Combines the two techniques shown

27. The Algorithm • LetJk(v0)be cheapest star for anyv and its k cheapest edges. Let leaves be{v1,…..,vk} • Averaging gives that best star has cost at mostJk(v0)  2OPT/n v0 vk v1 v2

28. The Algorithm Continued • LetR={v0,….,vk-1} • Note, thatvkis absent fromR • As in [KR] add a new nodes that does not belong toV • Similar to [KR] define a new graphGs fromG with 0 cost edges for svifor any vertex vi

29. The Algorithm continued • Compute ak - outconnected graph fromsinGs. Let Hsbe this graph. • By [KR] the cost ofHsis at most2opt (remark: our R is different thenthe one in[KR]) • In [KR] it is shown that if we add all edges between theR vertices toHs,the resulting graph is k-connected. • Unlike [KR] we add aMINIMALfeasible solution out of E(R) toHs

30. The Approximation Ratio • k out-connectivity from s implies (HS) k-1 • ThusFis a forest withk nodes. • We bound the cost of edges in the forest F. For every vi,vj  v0 we upper bound c(vi vj) c(v0 vi)+c(vj v0) • We call these costs the new costs

31. Upper Bounding c(F) • Forvi,vjv0we get • vivjF c(vivj)  vivjFc(viv0)+c(v0vj)  vivjF c(v0wk-1)+c(vj,v0) • There are k-1 edges in F but we did not take the edges of v0 which means that c(v0wk-1) is counted at most k-2 times.

32. Proof Continued • Note that according to the new costs we got astarrooted atvk-1 • The nodev0is (in the worst case) also connected tovk-1directly. This addsc(v0vk-1)to the cost of F. • Thusc(F) (k-2)·c(v0vk-1)+c(Jk-1(v0))

33. Proof Continued • c(F)  (k-2) c(v0vk-1) + 1 ik-1 c(v0vj ) • We know thatc(Jk(v0))  2opt/n • Thusc(v0vk-1)+c(v0vk)  2opt/n • Thusc(v0vk-1)  opt/n • c(F)  (k-2)c(v0vk-1) +c(Jk(v0)) –c(v0vk)  (k-3)· c(v0vk-1) + c(Jk(v0)) • c(F)  (k-3)opt/n+2opt/n=(k-1)opt/n • Thus the final ratio is 2+(k-1)opt/n

34. 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 vertexvS,uS • d(S)=number of edges going out ofS • LP=min  wexe Subject to x((S))R(S)-d(S) xe0

35. Jain: one of the xeat 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|

36. Charging Total charging equals|E’|=|L| 1-2xe xe xe

37. All Possible Edges • All edge types. C1 S C1 C3 C2 S1

38. How Many Tokens S Owns? • Let E(S)be edges internal toS. The setsCdiscussed now are children ofS. • Sowns a vertex inS if does not belong to any child • eis assigned to the smallestSso thateE(S) • Define the tokens inS: t(S)=E(S) −E(C)+x((S))− (C)

39. Contribution to Both Sides of Every Edge t(S)=E(S) −E(C)+x((S))− (C) • An edge with no endpoint inSor an edge that enters a child of and exitsS. Contribution0. • An edge with both end points inSthat does not enter a child ofScan not exist.

40. More cases • t(S)=E(S)−E(C)+x((S))− (C) • An edge that enters S but not a child of S contributesxe • An edge that enters a child ofS but notScontributes1-xe • An edge between two children of S contributes1-2xe.

41. t(S)IS NOT ZERO • It can not be that all edges exitSand enter a child ofS. Namely, it can not be that all contributions are0. • Indeed in this caseSis the sum of its children • In all other cases the contribution is positive.

42. t(S)1 • Consider: t(S)=E(S)−E(C)+x((S))− (C) The childrenCbelong to the laminar family, hence they are tight namely their(C)isintegral. • Thus t(S)1.

43. We Charged Already |L| Because one per S • Thus we foundt(S)associated withSonly, that is at least1 • 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.

44. The Contradiction • Look at the maximumS. • Some edges must be leaving it because its violated. • The1-2xeof these edges is positive. Uncharged. • This meanst(S)|E’|>|L|,contradiction.

45. Thank you for attention.Questions?