Approximation and Hardness Results for Packing Cycles

Mohammad R. SalavatipourDepartment 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)

2

Packing problems

Example: Edge-disjoint path (EDP) problemGiven a graph and a set of source-sink

10T1

S2

S1

Pairs

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

Classical and very well studied NP-hard problem.

T2S3

T3

3

EDP known results

The problem has a large integrality gap

even for planar graphs.

10

s1

t1

s2

t2

s3

t3

sr

tr

Fractional solution:

Integral solution: 1

4

• 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]

EDP known results:

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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]

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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

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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:1. Repeatedly, delete degree ≤ 1 vertices.2. Repeatedly, short-cut every degree 2 vertex

3. Find and remove shortest cycle.4. If go to step 1; else step3.

v

u wu w

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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

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Integrality gap for directed EDC

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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

s1

t1

s2

t2

s3

t3

sr

tr

10

Integrality gap for directed EDC (cont’d)

10

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.

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Integrality gap for directed EDC (cont’d)

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Take two copies of the grid construction for EDP.

Make graph Hk as follows

Observation: Hk is acyclic.

s1

t1

s2

t2

s3

t3

sk

tkz1 z2 z3 zk

Hk

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

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Integrality gap for directed EDC (cont’d)

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

10

s1

t1

s2

t2

s3

t3

sk

tkz1 z2 z3 zk

Hk

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Integrality gap for directed EDC (cont’d)

10

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

Also, take several copies of the graph

s1

t1

s2

t2

s3

t3

sr

tr

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

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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

Let’s call the blocks of , where block consist of triplewith source and two sinks

10

15

Integrality gap for directed EDC (cont’d)

10

So we have k blocks

Consider copies ofcorresponding to vertices of say

s1

t1

sr

tr

Replace one intersection block in each with a block of

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10

s1

t1

sr

tr

s1

t1

sr

Integrality gap for directed EDC (cont’d)

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Integrality gap for directed EDC (cont’d)

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

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Integrality gap for directed EDC (cont’d)

10

s1

t1

sr

tr

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

s1

t1

sr

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Integrality gap for directed EDC (cont’d)

10

s1

t1

sr

tr

s1

t1

sr

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

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Integrality gap for directed EDC (cont’d)

1.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

t1

sr

tr

2.Long cycles: Each has edges; each long cycles uses g edges → long cycles

Thus size of integral solution:

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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:

Integrality gap for directed EDC (cont’d)

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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:

1. All pairs can be routed

2. At most a fraction can be routed.

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

Hardness of directed EDC

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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?

Conclusion

Upper bound

Lower bound

Undir EDP

Direct EDC

Undir EDC