survey of some connectivity approximation problems via survey of techniques

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?