Course Presentation:

A Polynomial Kernel for Multicut in Trees, STACS 2009

Authors: Nicolas Bousquet, Jean Daligault,

Stephan Thomasse, Anders Yeo

Speaker: Jia-Hao Fan

Outline

Introduction:Definitions:Reduction rules:Analysis:

Problem:

Input: A undirected tree T = (V, E) and a set of requests P, an integer k. A request is a pair of vertices (u, v)

Output: k edges, whose removal separates each pair of vertices in P

Example

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Given T = (V, E), A set of Request P = {(v1, v3), (v4, v5), (v6, v8), (v5, v10)}, and k = 3

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Caterpillar A tree is a caterpillar if its internal tree is a path

General Tree

Example of Caterpillar

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Outline

Introduction:Definitions:Reduction rules:Analysis:

Some definitions

Bad leaf A group of leaves is the set of leaves

connected to the same internal node A group request is a request (x,y) where x and

y belong to the same group Bad leaf which is an endpoint of a group

request

Inner node An internal node with no leaves attached to it.

Bad leaves and inner node

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Given T = (V, E), A set of Request P = {(v1, v3), (v4, v5), (v6, v8), (v5, v10)}, and k = 3Group leaves:{v4,v5,v6},Group request: (v4,v5)Bad leaves: v4,v5

Inner node: v7

Some definitions

Disjoint request: two request are disjoint if their edge sets are disjoint

Common factor: the common factor of two requests is the intersection of their paths.

Disjoint Request and common factor

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Given T = (V, E), A set of Request P = {(v1, v3), (v4, v5), (v6, v8), (v5, v10)}, and k = 3

Disjoint Request (v1,v3), (v6, v8)Common factor of (v6,v8), (v5,v10) are {(v3,v7), (v7,v8)}

Some definitions for caterpillars

The internal tree T’ of caterpillar T,The set I1 of leaves of T’,

The set I2 of degree two nodes of T’,

The set L1 of leaves rooted at I1,

The set L’2 of bad leaves rooted at I2,

The set L2 of other leaves rooted at I2.

Example of Caterpillar

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I1 = {v2, v8},I2 = {v3, v7},L1 = {v1, v9 , v10},L’2 = {v4, v5},L2 = {v6 , v11}

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Some definitions

R-neighbors: we say two nodes x, y are R-neighbor if there exists a request (x, y).

Quasi-R-neighbors: A leaf x and an internal node y are quasi-R-neighbors if there exists a request (x, y), or a request (x, z) where z is a leaf rooted at y.

Example of R-neighbors, and quasi-R-neighbors

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Given T = (V, E), A set of Request P = {(v1, v3), (v4, v5), (v6, v8), (v5, v10)}(v5, v10) are R-neighbors(v5, v8) are quasi-R-neighbors

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Some definitions

Internal path of a request is the intersection between the path of request and the internal tree

R1 dominates a request R2 if the internal path of R1 contains the internal path of R2.

Example of internal path

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Given T = (V, E), A set of Request P = {(v1, v3), (v4, v5), (v6, v8), (v5, v10) , (v2, v5) , (v4, v11)}The internal path of request, (v5, v10) is:(v5, v10) dominates (v4, v11)

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Some definitions

Wingspan W of a leaf x is the path between the closest quasi-R-neighbor on the right of x and the closest quasi-R-neighbor on the left of x (if no such neighbor exists, we take the father f(x) of x by convention)

The size of Wingspan is the number of L2 leaves pending from it.

The wingspan

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Given T = (V, E), A set of Request P = { (v6, v8), (v6, v11) , (v6, v2)}

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The wingspan of v6 is (v2, v7)

Outline

Introduction:Definitions:Reduction rules:Analysis:

Reduction Rules

Unit Request: a request of length 1Disjoint Request: more than k+1 disjoint

requestsUnique Direction:

Leaf: Inner node:

Unique Direction: leaf

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Unique Direction: inner node

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Reduction Rules

Inclusion

Inclusion

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Consider two requests(v5, v10), (v3, v8),

Reduction Rules

Common factor: Let R be a request. If there are k+1 requests R1,

R2, …, Rk+1 different form R but intersecting R such that the common factor of any pair of them is the subset of R, then remove R

Common Factor

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Given a requestR=(v1, v10), R1 = (v4, v11), R2 = (v5, v9), and k = 1

Dominating Wingspans

Dominating wingspans: If x is an L2-leaf with a wingspan dominating at

least k+1 endpoint-disjoint requests, then contract e(x).

The wingspan

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Supposed that the wingspan of v6 is (v2, v8)Two request (v4, v1),(v11,v9)If k = 1, then …

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Time complexity for Reduction Rules

Unit Request: Detecting: O(|R|*|E|) Processing: O(1)

Disjoint Request: Detecting: Polynomial time Processing: O(1)

Unique Direction: Detecting: O(|V|*|R|) Processing: O(1)

Time complexity for Reduction Rules

Inclusion: Detecting: O(|R|^2 * |E|^2) Processing: O(1)

Common factor: Detecting: O(|R|^2 * |E|^2) Processing: O(1)

Dominating Wingspan: Detecting: O(|R|^2 * |E|^2) Processing: O(1)

Time complexity for Reduction Rules

Each operation requires polynomial timeEach operation decrease k, |R|, or |E| by 1Overall time complexity is polynomial

Outline

Introduction:Definitions:Reduction rules:Analysis:

Two lemmas

Let G be an graph having m edges, of maximal degree d. Then G has a matching of size: Floor(m/(2d - 1)).

First Lemma

The maximal degree is d

Two Lemmas

Let H be an graph having n vertices, of maximal degree d. Then H has an independent set of size: Floor(n/(d + 1)).

First Lemma

The maximal degree is d

Number of bad leaves

Create an new graph G’For each bad leaf, v in T, set an vertex v’

in G’.For each group request of (u, v) in T, set

an edge (u’, v’) in G’

Number of bad leaves: example

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v4v5 v6

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v9 v10

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Given T = (V, E), A set of Request P = {(v1, v3), (v4, v5), (v6, v8), (v9, v10), (v5,v6)}, and k = 3

Group leaves:{v4,v5,v6}, {v9, v10} Group request: (v4,v5), (v5, v6), (v9, v10)Bad leaves: v4,v5, v6, v9, v10,

Number of bad leaves: example

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v’5v’6

Group of leaves:{v4,v5,v6}, {v9, v10} Group request: (v4,v5), (v5, v6), (v9, v10)Bad leaves: v4,v5, v6, v9, v10,

v’9 v’10

Number of bad leaves

For bad leaf in G’, the maximal degree is k+1 by Common factor rule, the minimum degree is 1

If there are at least 2(k+1)(2k+1) bad leaves, then there are at least (k+1)(2k+1) edges in G’

There is a matching of size (k+1) by First lemma, floor(m/(2d-1)).

The size of bad leaves is at most 2(k+1)(2k+1)

The size of wingspan

Consider a maximal wingspan W, and the leaves W’ pending from it.

Create a new graph G’For each v in W or W’, set an vertex v’ in

G’For each request (u, v) where u, v in W or

W’, set an edge (u’, v’) in G’

Maximal wingspan

Maximal wingspan, a wingspan, every vertex v in W’ has at least an request (u, v) where u is in W or W’

The size of wingspan: Example

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Supposed that the wingspan of v6 is (v2, v8)Requests, (v4, v1), (v11,v9), (v6, v9), (v6, v2), (v4, v11)

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W ={v2, v3, v7, v8}W’={v1, v4, v5, v6, v11, v9, v10}

The size of wingspan: Example

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Supposed that the wingspan of v6 is (v2, v8)Requests, (v4, v1), (v11,v9), (v6, v9), (v6, v2), (v4, v11)

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W ={v2, v3, v7, v8}W’={v1, v4, v5, v6, v11, v9, v10}

The size of wingspan

For bad leaf in G’, the maximal degree is 2(k+1), the minimum degree is 1

If there are at least 2(k+1)(4k+3) L2-leaves pending from W, then there are at least (k+1)(4k+3) edges in G’

There is a matching of size (k+1) by First lemma, floor(m/(2d-1)).

The size of a L2-leaves pending from W is at most 2(k+1)(4k+3)

Number of L2-leaves

There are less than 2(k+1)(4k+3) L2-leaves not pending from W have wingspan intersection with W for each direction.

Example

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Number of L2-leaves

Create a new graph G’For each L2 leave, v, set an vertex v’ in G’

For any two wingspan of u, v intersect with each other, set an edge (u’, v’) in G’

The maximal degree of G’ is 6(k+1)(4k+3)If the number of vertices is at least (6(k+1)

(4k+3)+1)(k+1), then it has at least (k+1) disjoint wingspans.

Number of I2 nodes

Each inner node of requests in two directions by Unique direction rule.

The union of a set of inner nodes is called as a request path.

The length of each request path is at most k.

Number of I2 nodes

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Requests, (v2, v3), (v3,v8)

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Number of I2 nodes

Each end of a request path is a vertex of I1, L1, L2, or L’2.

The number of vertices of I1, L1, L2, or L’2 is at most O(k3)

The number of inner nodes of with a neighbor in I1, L1, L2, or L’2 is at most O(k4)

The number of inner nodes is at most O(k5)

Questions

Thanks