1. Comparing Min-Cost and Min-Power Connectivity Problems Guy Kortsarz Rutgers University, Camden, NJ

2. Motivation-Wireless Networks Nodes in the network correspond to transmitters More power  larger transmission range transmitting to distance r requires r  power, 2  r  4 Transmission range = disk centered at the node Battery operated  power conservation critical Type of problems: Find min-power range assignment so that the resulting communication network satisfies prescribed properties.

3. Directed Networks Define costs c(e) that takes already into account the dependence on the distance. The cost c(e), e = (u,v) would be r  with r the distance and the appropriate . In general, power to send from u to v not the same as v to u Thus power of v in directed graphs: p E' (v)=Max {e  E' leaves v} {c(e)} For example: If no edge leaves v, p(v)=0 p E' ( G)=∑ v p E' (v)

4. Symmetric Networks Networks where the cost to send from u to v or vise-versa is the same Thus graph undirected and: p E' (v)=Max {e  E' touching v} {c(e)} Many classical problems can and have been studied with respect to the (more difficult) min-power model

5. b a c d g f e a b d g f e c Range assignmentCommunication network

6. c(G) = n p(G) = n + 1 c(G) = n p(G) = 1 EXAMPLE UNIT COSTS

7. Why Not Complete Network? 34 6 a b c a is not directly connected to c. Total power is 25

8. Example p(a) = 7, p(b) = 7, p(c) = 9, etc. 7 5 8 9 8 5 4 2 3 6 a b c d f g h

9. Requirements Resilience to node-failures (node- connectivity problems) In the most general case: requirement r (u, v) for every u, v  V r (u, v) = 7 means 7 vertex disjoint paths from u to v are required Edge-disjointness not very relevant

10. The Steiner Network Problem Vertex Version Input: G ( V, E ), costs c(e) for every edge e  E requirements r(u,v) for every u,v  V Required: A subgraph G′ ( V, E′ ) of G so that G′ has r(u,v) vertex disjoint uv-paths for all u,v  V Usual Goal: Mnimize the cost, Alternative Goal: Minimize the power

11. Example r(u, v) = 2 a b c

12. Previous Work on Steiner Network The edge + sum version admits 2 approximation. [Jain, 1998]. The algorithm of Jain: Every BFS has an entry of value at least ½. Hence, iterative rounding The min-cost Steiner network problem vertex version admits no ratio approximation unless NP  DTIME(n polylog n ), [Kortsarz, Krauthgamer and Lee, 2002] The result is based on 1R2P with projection property

13. The Vertex k - Connectivity Problem We are given an integer k The goal is to make the graph resilient to at most k-1 station crashes Design a min-power (min-cost) subgraph G(V, E) so that every u,v  V admits at least k vertex-disjoint paths from u to v

14. 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 (see easy 4 ratio later) [Kortsarz, Mirrokni, Nutov, Tsano, 2006] O(k) approximation [M. Bahramgiri, M. Hajiaghayi and V. Mirrokni, 2002], [M. Hajiaghayi, N. Immorlica, V. Mirrokni 2003]

15. Recent Result Kortsarz, Mirrokni, Nutov, Tsano show that the vertex k-connectivity problem is ″almost″ 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, G. Kortsarz, V. Mirrokni and Z. Nutov, IPCO 2005]

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 network: 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 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

19. Relating the Min-Power and Min-Cost k - Connectivity Problems An Edge e  G is critical for k vertex- connectivity if G-e is not k vertex- connected Theorem (Mader): In a cycle with every edge is critical there exists at least one vertex of degree k

20. Reduction to a Forest Solution Say that we know how to approximate by ratio  the following problem: The Min-Power Edge-Multicover 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

21. Reduction to Forest (cont’) Clearly, the power of a min-power Edge-Multicover solution for r(v) = k-1 for every v is a lower bound on the optimum min-power k-connected graph 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-1 for all v Claim: Let G = H + F with F any minimal augmentation of H into a k vertex-connected subgraph. Then F is a forest

23. Reduction to Forest (cont’) Proof: Say that F has a cycle. Consider a cycle C in F All the edges of C are critical in H + F By Mader’s theorem there must be a vertex v in the cycle with degree k But  H(C) = k - 1, thus  (H+F)(C)  k+1, contradiction

24. 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 [M. Hajiaghayi, G. Kortsarz, V. Mirrokni and Z. Nutov, 2005] 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

25. 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

26. Best Results Known for Min-Cost Vertex k - Connectivity Simple k-ratio approximation [G. Kortsarz, Z Nutov, 2000] Undirected graphs, k  (n/6) 1/2, O(log n) approximation [J. Cheriyan, A.Vetta and S.Vempala, 2002] For any k (directed graphs as well): O(n/(n - k))  log 2 k [G. Kortsarz and Z. Nutov, 2004] For k = n - o(n), k 1/2 [G. Kortsarz and Z. Nutov, 2004]

27. Approximating the Min-Power Edge - Multicover Problem and Related Variants Example: some versions may be difficult. Say that we are given a budget k and all requirements are at least k - 1. All edge costs are 1. Required: a subgraph of power at most k that meets the maximum requirement possible.

28. Approximating the Min-Power Edge- Multiover Problem (cont’) The problem resulting is the densest k-subgraph problem Best known ratio: n 1/3 -  for  about 1/60 [U. Feige, G. Kortsarz and D. Peleg, 1996]

29. Very hard technical difficulty: Any edge adds power to both sides. Because of that: take k best edges, ratio k Usefull first reduction: ab c d 3 6 8 a’b’ C’ d’ a’’b’’ c’’d’’ 3 3 6 6 8 Approximating Edge-Multicover 5 5 5 8

30. An Overview Hence assume input B(X,Y,E) bipartite. Only Y have demands. However: both X and Y have costs Assume opt is known Main idea: Find F so that: p F(V)  3  opt r F(B)  (1 - 1/e)  r(B) / 2 Clearly, this implies O(log n) ratio as r(B)=O(n 2 )

31. Reduction to a Special Variant of the Max-Coverage Problem Let R = r(Y) The edge e = (x,y) is dangerous if cost(e)  2opt  r(y)/R; A dangerous edge requires more than twice “its share” of the cost Dangerous edges can be “ignored”; They cover at most half the demand. Thus

32. The Cost Incurred by Non- Dangerous Edges Since no dangerous edges used the cost is at most Hence, focus on non-dangerous edges because even if every y  Y is touched by its heaviest (non-dangerous) edge the total cost on the Y side is O(opt). Only try to minimize the cost invoked at X This is reducible to a generalization of set-coverage

33. The Max-Coverage Problem With Group Budget Constrains Select at most one of the following sets: 1 27 1125 17 C=5 C=2C=7C=1 25 1 2 5

34. Approximating Set-Coverage with Group Budget Constrains We reduced to a problem similar to the max- coverage algorithm However, we have group constrains: sets are split into groups. At most one set can be selected of every group Can be approximated within (1-1/e) By pipage rounding [Ageev,Sviridenko 2000] Invest opt, cover (1-1/e)/2 of the demand O(log n) ratio approximation

35. Remarks Only Max-SNP hardness is known for min- power edge-coverage For general r ij only 4  r max upper bound is known, [KMNT] The edge case admits n 1/2 approximation [HKMN] Directed variants: even k edge-disjoint path from x to y 1R2p Hard [KMNT]

36. Open Problems The case r(u,v)  {0, 1}. We recently broke the obvious ratio 4 (any solution is a forest so use ratio 2 for min- cost to get 2  2=4). Our ratio is 11/3. What is the best ratio? Does min-cost (min-power) vertex k-connectivity admit  (log n) lower bound? This problem related to deep concepts in graphs known as critical graphs Does the min-power edge-multicover problem admit an  (log n) lower bound? Can we give polylog for k vertex-connectivity directed graphs?