Simple and Improved Parameterized Algorithms for Multiterminal Cuts

Mingyu Xiao

The Chinese University of Hong Kong

Hong Kong SAR, CHINA

CSR 2008 Presentation, Moscow, Russia, June 2008

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Outline• Problems

— Definitions of Multiterminal Cuts

• History — Previous results and our results

• Methodology — Parameterized algorithm

• Important structural results — Farthest minimum isolating cut and others

• Edge Multiterminal Cut — An simple algorithm

• Vertex Multiterminal Cut —Two algorithms

3

Multiterminal Cut (MTC)

Edge (Vertex) Multiterminal Cut:Given an unweighted graph G=(V,E) and a subset of l terminals, the Edge (respectively, Vertex) Multiterminal Cut problem is to find a set of k edges (respectively, non-terminal vertices), whose removal from G separates each terminal from all the others.

T V

1t

G

2t

3t

4t

Related Problems

Multi-Way Cut: to separate the graph into at least l components.Multicut: to separate l pairs of vertices.

4

History

Results (Approximation ratio) Authors

(2-2/l) for EMTC Dahlhaus et al. (STOC 92)

2 for VMTC Garg et al. (ICALP 94)

2 for directed version Naor & Zosin (FOCS 97)

(1.5-1/l) for EMTC Calinescu et al. (STOC 98)

1.34 for EMTC Karger et al. (STOC 99)

Approximation algorithms:

l=2: the classical minimum (s,t) cut problem.l>2: MTC is NP-hard. (Dahlhaus et al. 1992)

NP-hardness of MTC:

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HistoryExact algorithms:

Dahlhaus et al. (STOC 92) for EMTC in planar

Hartvigsen. (D.A.M. 98) for EMTC in planar

Marx (TCS 06) for VMTC

Chen&Wu (Algorithmica 03) for EMTC in a special case

Yeh (J. ALG 01) for EMTC in planar

AuthorsResults2((4 ) )l lO l n

2(4 )l lO n2( ( ) )l lO l n l

3 (1)(4 )k OO n

3( log )O l n ln n

Chen et al. (Algorithmica, to appear) for VMTC 3(4 )kO kn

Our results in this paper:

and for VMTC

and for Vertex {3,4,5,6}-TC

for EMTC (To be exact, )(2 ( , ))kO lT n m2(( !) ( , ))O k T n m( ( , ))lO k T n m

21(2 ( , ))

ll kO lT n m

(3.349 ( , ))kO T n m (3.857 ( , ))kO T n m(2.059 ( , )),kO T n m (2.772 ( , )),kO T n m

where T(n,m) is the running time for finding a max flow in an unweighted graph.

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Parameterized Algorithm

What is parameterized algorithm?

• Exact algorithm.

• The exponential part of the running time is only related to one or more parameters, but not the input size.

(2 ( , ))kO lT n m 2(( !) ( , ))O k T n m ( ( , ))lO k T n m

k is the parameter k and l are the parameters

Some (parameterized) problems are unlike to have any parameterized algorithms, such as the k-clique problem with parameter k. Those kinds of problems are called W[1]-hard in Parameterized Complexity.

Readers are referred to “Parameterized Complexity” by Downey and Fellow for more details about parameterized algorithms.

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Techniques

Farthest minimum isolating cut

All of our algorithms are based on a simple technique:Branching at an edge (a vertex) in a farthest minimum isolating cut: including it in the solution or excluding it from the solution.

A minimum isolating cut for terminal ti is a minimum cut that separates ti away from all other terminals T-i.

A Minimum isolating cut Ci for terminal ti separates the graph into two components: one that contains ti is called the residual of Ci and denoted by Ri; the other one contains T-i.

The farthest minimum isolating cut for terminal ti is the unique minimum isolating cut that makes the residual of the maximum cardinality.

it iT

Minimum isolating cutsFarthest minimum isolating cut

iR

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A Structural Property

Lemma: Let Ci be the (farthest) minimum isolating cut for terminal ti in G,and G’ be the graph after merging Ri into a new terminal ti. Then any minimum multiterminal cut in G’ is a minimum multiterminal cut in G.

This lemma holds for both edge and vertex version.

Ci

it

iT

G

Ri

itiT

G’

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Rule 1: For each terminal ti, let Ci and Ri be its farthest minimumisolating cut and the corresponding residual, then we can contract Ri

in the graph to form terminal ti.

Edge Multiterminal Cut

Data reduction rules:

Rule 2: We can remove all the edges that connect two terminals fromthe graph and put them into the solution.

Rule 3: If , then is a multiterminal cut with size atmost k, where satisfying'iC

'i i iC

' 1max li i iC C

3t

2t

1t

4t

3t

2t

1t

4t A solution

1 1| |l li i lC k

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Proof: Let S be a minimum multiterminal cut and the minimal isolating cut for ti. We have

Edge Multiterminal Cut

Lemma: is a 2-approximation solution. 1li iC

iS S

S

1 1| | | | 2 | | .

l l

i ii iC S S

3t

2t

1t

4tRule 4: Let Ci be a minimum isolating cut for terminal ti. If , then there is no multiterminal cut with size .k

12

l

iiC k

1S

We are ready to design our algorithm now.

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Edge Multiterminal Cut

Main steps of our recursive algorithm:

Step 1: applying the 4 reduction rules to reduce the input size.Step 2: Let , branching at an edge e in B by including it in the solution or excluding it from the solution.

1li iB C

G

G-e G*e

…… …… ……

G*e is the graph obtained by shrinking e in G.

Does this simple algorithm work efficiently?

How to analyze the running time?

12Note: The notation system is different. Here l denotes the solution size.

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We will use a control value to build up a recurrence relation.

Edge Multiterminal CutAnalysis of our algorithm

( ) ( 1) ( 1) 1.C p C p C p

12 | |

l

iip k C

It is easy to see that in Step 1 (applying reduction rules), p will not increase. We can further prove that in each branch of Step 2, p decreases by at least 1. Then we get

It is easy to see that satisfies it.( ) (2 )pC p O

If , we will find a solution when applying Rule 3. Else we have

1 1| |l li i lC k

21 .

llp k

G

G-e G*e

…… …… ……

is the size of the tree.( )C p

p

1p 1p

Lemma: Edge Multiterminal Cut can be solved in time. 21(2 ( , ))

ll kO lT n m

Corollary: Edge 3-Terminal Cut can be solved in time. (1.415 ( , ))kO T n m

Previous result3 (1)(4 )k OO n

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Rule 1: For each terminal ti, let Ci and Ri be its farthest minimumisolating cut and the corresponding residual, then we can contract Ri

in the graph to form terminal ti.

Vertex Multiterminal Cut

Do data reduction rules still hold?

Rule 2: We can remove all the edges that connect two terminals fromthe graph and put them into the solution.

Rule 3: If , then is a multiterminal cut with size atmost k, where satisfying'iC

'i i iC

' 1max li i iC C1 1| |l l

i i lC k

Rule 2’: There is no solution if one terminal is adjacent to anotherterminal. We can remove all the vertices that are common neighborsof two terminals and put them into the solution.

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Vertex Multiterminal Cut

1 1| | | | | | .

l l

i ii iC S l S

1 1

| | | | 2 | | .l l

i ii iC S S

S

3t

2t

1t

4t

Rule 4: Let Ci be a minimum isolating cut for terminal ti. If , then there is no multiterminal cut with size .k

12

l

iiC k

S

3t

2t

4t

1t

Edge version Vertex version

EMTC: every edge in S will appear in exactly two isolating cuts.VMTC: a vertex in S will appear in up to l isolating cuts.

1

l

iiC lk

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Vertex Multiterminal Cut

The algorithm is almost the same as the algorithm for EMTC.

Step 1: applying the 4 reduction rules to reduce the input size.Step 2: Let , branching at a vertex v in B by including it in the solution or excluding it from the solution.

1li iB C

Let be the control value.

( ) ( ( 1)) ( 1) 1.C p C p l C p

1| |

l

iip lk C

In Step 1, p will not increase. In Step 2, when v is included into the solution, p will decrease by l-1; when v is excluded from the solution, p will decrease by 1. We get recurrence relation

If , we will find a solution when applying Rule 3. Else we have1 1| |l li i lC k

11 1| | ( 1) .l

i i lp lk C l k k

Analysis

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Vertex Multiterminal Cut

( ) ( ( 1)) ( 1) 1,C p C p l C p

We can verify that when l=3,4,5,6, and respectively satisfy (1) and (2).

11 1| | ( 1) .l

i i lp lk C l k k

Now we get two relations:

(1)

(2)

Lemma: Vertex {3,4,5,6}-Terminal Cut can be solved in and time respectively.

( ) (2.059 ), (2.772 ), (3.349 )k k kC p O O O(3.857 )kO

(3.349 ( , ))kO T n m (3.857 ( , ))kO T n m(2.059 ( , )),kO T n m

(2.772 ( , )),kO T n m

Furthermore, we can prove that ( ) ( ).kC p O l

Lemma: Vertex Multiterminal Cut can be solved in time. ( ( , ))kO l T n m

The exponential part is related to l and k.

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The jth layer farthest minimum isolating cut

The first layer farthest minimum isolating cut for t i is just the farthest minimum isolating cut for ti. The jth layer farthest minimum isolating cut is the farthest minimum for ti’, where ti’ is formed by merge and together.

An Alternative Algorithm for VMTC

(1)iC

( 1)jiR

( )jiC

( 1)jiC

Obviously,

itiT

(1)iC

(2)iC

(3)iC

(1) (2) (3)| | | | | | ......i i iC C C

Let b be the smallest number such that does not exist or , and ( 1)biC

( 1)| |biC k

( )1 .b j

j iB C

Claim: If there is a solution (a multiterminal cut with size ≤k), then at least one vertex in B is contained in a solution.

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Since we get

Recursive algorithm:

Recursive step: Branching on B by including each vertex in B into the solution.

An Alternative Algorithm for VMTC

(1) (2) ( )| | | | ...... | | ,bi i iC C C k

where C(k) is the size of the search tree when our algorithm finds a solution of size≤k.

( )1| | | | ( 1) / 2.b j

j iB C k k Analysis:

Then

( 1)( ) ( 1) 1,

2

k kC k C k

To compute B, we need at most b<k farthest minimum isolating cut computations.Lemma: Vertex Multiterminal Cut can be solved in time.

2( !)

2( ( , ))k

kO T n m

The exponential part is only related to k.

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We present a simple reduction from Multicut to Multierminal Cut:For each instance of Multicut, we can reduce it to at most instances of Muliterminal Cut with at most terminals.

Multicut

• Objective: to separate l pairs {si, ti} of vertices (terminals).

• Measure: the cardinality of the deletion set (solution size not greater then k).

G

2l(2 )ll

By using our results on MTC, we can also improve previously known results on Mulicut.