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

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.

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)

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

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

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

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

Example p(a) = 7, p(b) = 7, p(c) = 9, etc a b c d f g h

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

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

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

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

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

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]

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]

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]

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

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

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

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

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

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

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

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

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]

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.

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]

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 a’b’ C’ d’ a’’b’’ c’’d’’ Approximating Edge-Multicover

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 )

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

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

The Max-Coverage Problem With Group Budget Constrains Select at most one of the following sets: C=5 C=2C=7C=

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

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]

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?