approximation and hardness results for packing cycles n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Approximation and Hardness Results for Packing Cycles PowerPoint Presentation
Download Presentation
Approximation and Hardness Results for Packing Cycles

Loading in 2 Seconds...

play fullscreen
1 / 23

Approximation and Hardness Results for Packing Cycles - PowerPoint PPT Presentation


  • 82 Views
  • Uploaded on

Approximation and Hardness Results for Packing Cycles. Mohammad R. Salavatipour Department of Computing Science University of Alberta Joint work with M. Krivelevich (Tel Aviv U.) Z. Nutov (Open U.) J. Verstraete (U. Waterloo) R. Yuster (U. Haifa). Packing problems.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Approximation and Hardness Results for Packing Cycles' - shubha


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
approximation and hardness results for packing cycles

Approximation and Hardness Results for Packing Cycles

Mohammad R. Salavatipour

Department of Computing Science

University of Alberta

Joint work with

M. Krivelevich (Tel Aviv U.)

Z. Nutov (Open U.)

J. Verstraete (U. Waterloo)

R. Yuster (U. Haifa)

packing problems
Packing problems

Example: Edge-disjoint path (EDP) problem

Given a graph and a set of source-sink

10

Pairs

Goal: Find a maximum number of edge-disjoint si,ti-paths.

Classical and very well studied NP-hard problem.

S2

T3

S1

T1

S3

T2

edp known results
EDP known results

The problem has a large integrality gap

even for planar graphs.

10

Fractional solution:

Integral solution: 1

s1

s2

s3

sr

t1

t2

t3

tr

edp known results1
EDP known results:
  • Directed graphs:

Upper bounds: approx

[GKRSY’03,KS04]

Lower bounds: -hardness, unless

[GKRSY’03] and (for DAGS) unless

[MW’00]

  • Undirected graphs:

Upper bounds: has integrality gap [CKS’06]

Lower bounds: -hardness unless

[ACKZ’05]

packing disjoint cycles
Packing Disjoint Cycles
  • What is the maximum number of edge-disjoint cycles (EDC) in a given graph ?
  • What if the input graph is directed/undirected?

Dual problems:

  • For disjoint paths: multi-cut which has (for undirected) [GVY’96] and (for directed) [G’03] approx.
  • For disjoint cycles: Feedback Arc/Vertex Set,

-approx [S’95]

results for edc
Results for EDC:

EDC is APX-hard on undirected graphs and has

approx [CPR’03]

Theorem 1: For undirected graphs, a simple greedy gives -approx.

Theorem 2: For directed graphs there is an -approx.

Theorem 3: For directed graphs the problem

is -hard, unless

We can get -hardness unless

edc on undirected graphs

u

w

w

u

EDC on undirected graphs

Theorem 1: For undirected graphs, a simple greedy gives -approx to the optimal fractional soluion.

Algorithm: Repeat the following until G is empty:

  • Repeatedly, delete degree ≤ 1 vertices.
  • Repeatedly, short-cutevery degree 2 vertex
  • Find and remove shortest cycle.
  • If go to step 1; else step3.

v

hardness of directed edc
Hardness of directed EDC

Theorem 3: For directed graphs the EDC problem

is -hard, unless

First we show the following:

Theorem 4: Directed EDC has an integrality gap of

A natural attempt to prove integrality gap: use the grid construction for EDP

integrality gap for directed edc
Integrality gap for directed EDC

Denote the grid graph with r pairs by Dr

Direct edges from top-to-bottom and left-to-right.

Add a link from the sinks back to sources.

The fractional solution still has size

10

s1

s2

s3

sr

t1

t2

t3

tr

integrality gap for directed edc cont d
Integrality gap for directed EDC (cont’d)

But there is also a large integral solution:

We call these paths “non-canonical” or “cheating” paths.

We have to make it costly to use cheating paths.

10

integrality gap for directed edc cont d1
Integrality gap for directed EDC (cont’d)

10

Take two copies of the grid construction for EDP.

Make graph Hk as follows

Observation: Hkis acyclic.

s1

Hk

s2

s3

sk

t1

t2

t3

tk

z2

z3

zk

z1

We call the triple si,ti,zi a block.

integrality gap for directed edc cont d2
Integrality gap for directed EDC (cont’d)

10

  • Consider Hk and the 2k pairs si,ti and si,zi as an instance of EDP.
  • Fact: Any optimal integral solution to EDP on Hk either
  • has one fully routed block or
  • two partially routed block

s1

Hk

s2

s3

sk

t1

t2

t3

tk

z2

z3

zk

z1

integrality gap for directed edc cont d3
Integrality gap for directed EDC (cont’d)

The idea is to start with several copies of the modified grid graph, say

Also, take several copies of the graph

10

s1

s2

s3

sr

t1

t2

t3

tr

We take a k-uniform r’-regular girth g hypergraph where

integrality gap for directed edc cont d4
Integrality gap for directed EDC (cont’d)

Every copy corresponds to a vertex of

Every copy corresponds to an edge of

Consider an arbitrary edge and let be its corresponding copy of

10

Let’s call the blocks of , where block

consist of triple

with source and two sinks

integrality gap for directed edc cont d5
Integrality gap for directed EDC (cont’d)

10

So we have k blocks

Consider copies of

corresponding to vertices of

say

s1

sr

t1

tr

Replace one intersection block in each with a block of

integrality gap for directed edc cont d7
Integrality gap for directed EDC (cont’d)

The k blocks of will be used to replace one intersection in the k copies , one

from each

Since is r’-regular, every belongs to r’ edges, so each of its intersection will be replaced with blocks of copies of from r’ different edges.

Call this final graph

integrality gap for directed edc cont d8
Integrality gap for directed EDC (cont’d)

10

s1

s1

sr

sr

t1

t1

tr

Considering half integral solution and following canonical paths, every has r cycles with value ½, so fractional solution of has size

integrality gap for directed edc cont d9
Integrality gap for directed EDC (cont’d)

10

s1

s1

sr

sr

t1

t1

tr

The integral solution may have some canonical cycles and some cheating cycles (which will be long).

integrality gap for directed edc cont d10
Integrality gap for directed EDC (cont’d)
  • Canonical cycles: Since each can allow at most two blocks be routed (partially or fully) and since each canonical cycle goes through r blocks there are at most short cycles with

10

s1

sr

t1

tr

  • Long cycles: Each has edges; each long cycles uses g edges → long cycles
  • Thus size of integral solution:
slide21

Integrality gap for directed EDC (cont’d)

So the gap is:

Choosing the parameters:

We give explicit constructions for graph for which

and

Let r be constant and k=g;

Then and gap =

writing the in terms of n:

slide22

Hardness of directed EDC

To turn this into a hardness proof we using the following result of [MW’00]:

Theorem: Given a DAG G and source-sink pairs as an instance of EDP with

it is quasi-NP-hard to decide:

  • All pairs can be routed
  • At most a fraction can be routed.

The construction of will be based on two copies of the instance of EDP on DAGS.

slide23

Conclusion

The upper and lower bounds for directed EDC and undirected EDP are similar:

Questions:

  • What is the correct

upper/lower bound for

directed EDC?

  • What about undir

EDC?