Algorithms & LPs for k-Edge Connected Spanning Subgraphs
David Pritchard Zinal Workshop, Jan 20 09
Slide 2
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 |(S)| k
Slide 3
k-ECSS & k-ECSM Optimization Problems k-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) minimal k-ecs multisubgraph problem
(k-ECSM): can buy as many copies as you like of any edge
3-edge-connected multisubgraph of G, |E|=9 G
Slide 4
Approximation Algorithms k-ECSS and k-ECSM are NP-hard for k 2
For 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 optimal Also call the
approximation ratio/factor
Slide 5
Approximability Any problem has either no 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]
Slide 6
Question What is the approximability of the k-edge-connected
spanning subgraph and k-edge-connected spanning multisubgraph
problems? It will depend on k Exact approximability not known but
well determine rough dependence on k Also look at important
unit-cost special case
Slide 7
History of k-ECSS 2-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?
Slide 8
Main Course We prove k-ECSS with general costs does not have
approximation ratio tending to 1 as k tends to infinity We
conjecture that k-ECSM with general costs does have approximation
ratio tending to 1 as k tends to infinity
Slide 9
Part 1: Approximation Hardness
Slide 10
An Initial Observation For 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-EC u S v w
Slide 11
Whats 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-VCSS vertex-connected
Slide 12
Whats 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-ECSM G, a hard instance for 2-VCSS Instance for 3-VCSS
with same hardness G zero-cost edges to V(G)
Slide 13
Hardness of k-ECSS (slide 1/2) >0, k2, no 1+-apx if P NP
Reduce APX-hard TreeCoverByPaths to k-ECSS Input: a tree T,
collection X of paths in T A 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 cover size-2 cover
Slide 14
Replace 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) 1 111 Hardness of
k-ECSS (slide 2/2) >0, k2, no 1+-apx if P NP
Slide 15
Part 2: Linear Programs
Slide 16
From Hardness to Approximability Conjecture [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, motivation an LP-relative
Slide 17
LP Relaxation Introduce variables x e 0 for all edges e of G.
LP relaxation of k-ECSM problem: min x e cost(e) s.t. x((S)) k for
all S V S e in (S) x e k 0.4 1.2 1.4
Slide 18
LP-Relative a new name for an old concept The 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)
Slide 19
Motivation Similar results for related problems are true
LP-relative (1+C/k)-approx for k-ECSM would imply a (1+C/k)-approx
for subset k-ECSM Subset 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
Slide 20
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-ECSM would imply the conjecture. Cousins: k, each
k-strongly connected digraph has 2 disjoint strongly connected
subdigraphs k, every k-edge connected hypergraph contains 2
disjoint connected hypergraphs OPEN [B-J Y 01] FALSE [B-J T
03]
Slide 21
More About the LP LP relaxation of k-ECSM equals (up to
scaling) the Held-Karp TSP relaxation, and the undirected Steiner
tree LP relaxation. LP and generalizations are well-studied Basic
(extreme) solutions have few nonzeroes Extreme solution properties
are often key to algorithmic results (e.g., [GGTW 05]) How
unstructured can extreme points get?
Slide 22
Extremely Extreme Extreme Point Edge values of the form Fib i
/Fib |V|/2 and 1 - Fib i /Fib |V|/2 (exponentially small in |V|)
Maximum degree |V|/2
Slide 23
Structural 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).
Slide 24
1+O(1/k) Algorithm for Unit-Cost k-ECSM [GGTW] 1. Solve LP to
get a basic optimal solution x* 2. Round every value in x* up to
the next highest integer and return the corresponding multigraph
(k=4)
Slide 25
Analysis Optimal k-ECSM has degree k or more at each vertex,
hence at least k|V|/2 edges The fractional LP solution x* has value
(fractional edge count) k|V|/2 There are at most 2|V|-3 nonzero
coordinates Rounding up increases cost by at most 2|V|-3 ALG/OPT
(k|V|/2 + 2|V|-3)/(k|V|/2) < 1 + 4/k
Slide 26
Open Questions Resolve 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?
