Survey of Some Connectivity Approximation Problems via

Survey of Techniques

Guy KortsarzRutgers 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, 2006. Chapter 58, 30 pages.

Steiner Network ProblemSteiner 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 NetworkVERTEX CASE:

Labelcover hard. [K, Krauthgamer, Lee, SICOMP]kε 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 kRequired: 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-ECSGIf we have k connectivity from a vertex v to all the rest, by transitivity the graph is k-edge-connectedApply the Edmonds algorithm twice: replace every edge with two directed edgesOnce 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>6k2

[K & Nutov STOC 04] n/(n-k) O(log2 k) for any k, directed/undirected graphs. The ratio is O(log2 k), unless k = n - o(n).

[Fackharoenphol and Laekhanukit, STOC 2008] O(log2 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) | eE(v)}Reasons: transmission range.

7

5 8

9

8

54 2

3

6

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 vMay 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 caseHence the min-cost version is the

minimum spanning tree problemMin-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 TreeThe 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 ev to vClearly, p(v) c(ev)Taking the sum, the claim follows

p(G) 2c(G) (on any graph):Assign to v its power edge ev

Every edge is assigned at most twice

The cost is at least

The power is at exactly

v

vec2

)(

v

vec )(

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 degG(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 problemApply 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 forestLet 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) 2c(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-1Admits 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 VARIANTWe 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 caseAt most 4 always and tends to 2 for k=o(n)K, Nutov: 2+(k-1)/n ratioAt most 3 and tends to 2 for k=o(n)Combines the two techniques shown

The Algorithm

Let Jk(v0) be cheapest star for any v and its k cheapest edges. Let leaves be

{v1,…..,vk}Averaging gives that best star has cost at most Jk(v0) 2OPT/n

v0

v2vkv1

The Algorithm Continued

Let R={v0,….,vk-1}

Note, that vk 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 Gs from G with 0 cost edges for svi for any vertex vi

The Algorithm continued

Compute a k - outconnected graph from s in Gs. Let Hs be this graph.By [KR] the cost of Hs 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 Hs ,the resulting graph is k-connected.Unlike [KR] we add a MINIMAL feasible solution out of E(R) to Hs

The Approximation Ratio k out-connectivity from s

implies (HS) k-1

Thus F is a forest with k 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

Upper Bounding c(F) For vi,vjv0 we get vivjF c(vivj)

vivjFc(viv0)+c(v0vj) vivjF 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.

Proof Continued Note that according to the new costs we got a star rooted at vk-1

The node v0 is (in the worst case) also connected to vk-1 directly. This adds c(v0vk-1) to the cost of F.

Thus c(F) (k-2)·c(v0vk-1)+c(Jk-1(v0))

Proof Continued c(F) (k-2) c(v0vk-1) + 1 ik-1 c(v0vj )We know that c(Jk(v0)) 2opt/nThus c(v0vk-1)+c(v0vk) 2opt/nThus c(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/nThus the final ratio is 2+(k-1)opt/n

Laminar FamiliesWe present the Jain result with a simplified proof due to Ravi et. al.The LP: R(S) maximum demand of a separated vertex vS, uSd(S)=number of edges going out of SLP= min wexe

Subject to x((S))R(S)-d(S) xe0

Jain: one of the xe 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|

1-2xe

xe xe

All Possible Edges All edge types.

S

S1

C1

C2

C3

C1

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 eE(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 casest(S)=E(S)−E(C)+x((S))− (C)An edge that enters S but not a child of S contributes xe

An edge that enters a child of S but not S contributes 1-xe

An edge between two children of S contributes 1-2xe .

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 childrenIn all other cases the contribution is positive.

t(S)1Consider:

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 disjointThis 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 1-2xe of these edges is positive. Uncharged.This means t(S)|E’|>|L|, contradiction.

Thank you for attention.

Questions?