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Algorithms & LPs for k-EdgeConnected Spanning Subgraphs. David Pritchard Zinal Workshop, Jan 20 09. k-Edge Connected Graph. k edge-disjoint paths between every u, v at least k edges leave S, for all S V even if (k-1) edges fail, G is still connected. | (S)| k. S. - PowerPoint PPT Presentation

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Hypergraphic LP Relaxations for the Steiner Tree Problem

Algorithms & LPs fork-EdgeConnectedSpanning Subgraphs David Pritchard

Zinal Workshop, Jan 20 09

1k-Edge Connected Graphk edge-disjoint paths between every u, vat least k edges leave S, for all S Veven if (k-1) edges fail, G is still connectedS|(S)| k2k-ECSS & k-ECSM Optimization Problemsk-edge connected spanning subgraph problem(k-ECSS): given an initial graph (maybe with edge costs), find k-edge connected subgraph including all vertices, w/ |E| (or cost) minimalk-ecs multisubgraph problem (k-ECSM): can buy asmany copiesas you likeof any edge3-edge-connected multisubgraph of G, |E|=9G3Approximation Algorithmsk-ECSS and k-ECSM are NP-hard for k 2For k-ECS/M or any NP-hard minimization problem, one notion of a good heuristic is a polytime -approximation algorithm: always produces a feasible solution (e.g., k-edge connected subgraph) whose output cost is at most times optimalAlso call the approximation ratio/factorApproximabilityAny problem has eitherno constant-factor approximation algorithm constant 1: -approx alg exists, but no (-)-approx exists for any >0 [=1: in P] constant 1: (+)-approx alg exists for any >0, but no -approx [=1: apx scheme]

QuestionWhat is the approximability of the k-edge-connected spanning subgraph andk-edge-connected spanning multisubgraph problems?It will depend on kExact approximability not known but well determine rough dependence on kAlso look at important unit-cost special caseHistory of k-ECSS2-ECSS is NP-hard, has a 2-apx alg [FJ 80s]& is APX-hard (has no approximation scheme if P NP) even for unit costs [F 97]k-ECSS has a 2-apx alg [KV 92]unit cost k-ECSS: has a (1+2/k)-apx [CT96] & tiny >0, for all k>2, no (1+/k)-apx [G+05]Gets easier to approximate as k increases!How do k-ECSS, k-ECSM depend on k?Main CourseWe prove k-ECSS with general costs does not have approximation ratio tending to 1 as k tends to infinityWe conjecture that k-ECSM with general costs does have approximation ratio tending to 1 as k tends to infinity8Part 1: Approximation Hardness9An Initial ObservationFor the k-ESCM (multisubgraph) problem, we may assume edge costs are metric, i.e. cost(uv) cost(uw) + cost(wv)since replacing uv with uw, wv maintains k-ECuSvw10Whats Hard About Hardness?A 2-VCSS is a 2-ECSS is a 2-ECSM.For metric costs, can split-off conversely, e.g.

All of these are APX-hard [via {1,2}-TSP]2-ECSM2-ECSS2-VCSSvertex-connected11Whats Hard About Hardness?1+ hardness for 2-VCSS implies 1+ hardness for k-VCSS, for all k 2

But this approach fails for k-ECSS, k-ECSMG, a hardinstance for2-VCSS Instance for 3-VCSSwith same hardnessGzero-cost edges to V(G)12Hardness of k-ECSS (slide 1/2) >0, k2, no 1+-apx if P NP Reduce APX-hard TreeCoverByPaths to k-ECSSInput: a tree T, collection X of paths in TA subcollection Y of X is a cover if the union of {E(p) | p in Y} equals E(T)Goal: min-size subcollection of X that is a coversize-2cover13Replace each edge e of T by k-1 zero-cost parallel edges; replace each path p in X by a unit-cost edge connecting endpoints of p

min |X| to cover T = k-ECSS optimum.

0 (k-1)0 (k-1)0 (k-1)0 (k-1)0 (k-1)0 (k-1)1111Hardness of k-ECSS (slide 2/2) >0, k2, no 1+-apx if P NP 14Part 2: Linear Programs15From Hardness to ApproximabilityConjecture [P.]For some constant C, there is a(1+C/k)-approximation algorithm for k-ECSM.For k 2 it is known to hold with C=1. Next: definition of LP-relative, motivationan LP-relative16LP RelaxationIntroduce variables xe 0 for all edges e of G.LP relaxation of k-ECSM problem:min xecost(e) s.t. x((S)) k for all S VSe in (S) xe k0.41.21.417LP-Relativea new name for an old conceptThe LP is a linear relaxation of k-ECSM so LP-OPT OPT (optimal k-ECSM cost)Definition: an LP-relative -approximation algorithm has output cost ALG LP-OPT(Stronger than -approx ALG OPT)+ALGOPTLP-OPT(integrality gap)18MotivationSimilar results for related problems are trueLP-relative (1+C/k)-approx for k-ECSM would imply a (1+C/k)-approx for subset k-ECSMSubset k-ECSM: given a graph with some vertices required, find a graph with edge-connectivity k between all required vertices (k=1: Steiner tree)To show this, use parsimonious property [GB93] of the LP: there is an optimal fractional solution that doesnt use non-required vertices

A Combinatorial Approach? constant C so that for every A, B > 0,every (A + B + C)-edge-connected graph contains a disjoint A-ECSM and B-ECSMwould imply the conjecture. Cousins:k, each k-strongly connected digraph has 2 disjoint strongly connected subdigraphsk, every k-edge connected hypergraph contains 2 disjoint connected hypergraphsOPEN [B-J Y 01]FALSE [B-J T 03]More About the LPLP relaxation of k-ECSM equals (up to scaling)the Held-Karp TSP relaxation, andthe undirected Steiner tree LP relaxation.LP and generalizations are well-studiedBasic (extreme) solutions have few nonzeroesExtreme solution properties are often key to algorithmic results (e.g., [GGTW 05])How unstructured can extreme points get?Extremely Extreme Extreme Point

Edge values of the form Fibi/Fib|V|/2 and 1 - Fibi/Fib|V|/2(exponentially small in |V|) Maximum degree |V|/222Structural LP Property [CFN]The LP has 2|V|-1 constraints, |V|2 vars, but a basic solution which is optimal (all LPs);we can find it in polytime (reformulation);every basic solution has 2|V|-3 nonzero coordinates (laminar family of constraints).

231+O(1/k) Algorithm forUnit-Cost k-ECSM [GGTW]Solve LP to get a basic optimal solution x*Round every value in x* up to the next highest integer and return the corresponding multigraph

(k=4)24AnalysisOptimal k-ECSM has degree k or more at each vertex, hence at least k|V|/2 edgesThe fractional LP solution x* has value (fractional edge count) k|V|/2There are at most 2|V|-3 nonzero coordinatesRounding up increases cost by at most 2|V|-3ALG/OPT (k|V|/2 + 2|V|-3)/(k|V|/2) < 1 + 4/k25Open QuestionsResolve the conjecture!Whats the minimum f(A, B) so that any f(A, B)-edge-connected (multi)graph contains disjoint A- and B-edge-connected subgraphs?Is there a constant k such that every k-strongly connected digraph has 2 disjoint strongly connected subdigraphs?

Thanks for Attending!

27Small Extreme Examples

n=6, denom=2n=7, =4n=8, denom=3n=9, =5n=9, denom=4n=10, denom==528Previously Known Constructions[BP]: minimum nonzero value of x* can be ~1/|V|

[C]: max degreecan be ~|V|1/2

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