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Refined Search Tree Technique for Dominating Set on Planar Graphs

Jochen Alber, Hongbing Fan, Michael R. Fellows, Henning Fernau, Rolf Niedermeier, Fran Rosamond, and Ulrike Stege

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Outline

Background Techniques to cope with NP-hardness

Paper Technique Analysis

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Background

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Introduction

Techniques used to cope with NP-hardness: approximate solutions randomization quantum mechanics, bio-molecular chemistry …

… and Fixed Parameter Tractability (FPT)

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What is FPT?

Tractability of a problem having a large input size up to a small fixed parameter

i.e. “exactly” solving a limited version of an NP-hard problem

The problem has to have an algorithm of running time

O(f(k).nO(1)) [k<<n]

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Classification

NP = O(cn) FPT = O(f(k).nO(1)) P = O(nO(1))

P [subset of] FPT [subset of] NP

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Foundations of FPT

3 basic problems of FPT theory –

Vertex Cover Independent Set Dominating Set

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Determine a vertex cover of at most k vertices that cover all edges

Complete enumeration: O(nk) possibilities

k-Vertex Cover

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FPT Solution techniques

Kernelization (or data-reduction) Remove redundancy Reduce the problem to a simpler version

Ex: In k-vertex cover, simply include any vertex

having degree > k

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… FPT solution techniques

Bounded-Search Tree O(2kn) possibilities – linear in n!

Start with 2 sets: Set of included vertices I = Ø Set of excluded vertices E = G

Build a binary tree of height k using “edges” At each step add 1 vertex from edges in E to I Bound: Remove that edge and neighboring edges

for a smaller tree

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k-Vertex Cover Search Tree

Tree height ≤ k Number of nodes O(2k) Hence complete search requires 2k * poly steps

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k-Independent Set

A set of k-disconnected vertices

Complete enumeration: O(nk) possibilities

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k-Independent Set Search Tree

Bound: Choose any vertex from [a vertex + its neighbors], delete them all and continue

Complete search requires O((d+1)k.n) Just like every vertex has 2 branches in k-vertex

cover tree - vertex itself IN or its neighbor IN

Here there are d+1 branches

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k-Dominating Set

Either a vertex or its connected neighbor is in the dominating set

(So vertices cover other vertices)

Searching in O((d+1)k.n) time

like k-Independent set

(i.e. bounding) not possible!

Only possible for planar graphs

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The Paper

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Previous research

An O(c√kn) algorithm was proposed before large c (= 36√34) used advance techniques like tree-width thus - theoretically efficient but not easy to

implement

This paper gives an O(ckn) algorithm (c = 8) uses a combination of kernelization and bounded-

search

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Annotated Dominating Set

A planar black and white graph i.e. G = (B [du] W, E)

Find a set of k vertices that dominate all black vertices

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Definitions

Open Neighborhood: N(u) = {v Є V | {u, v} Є E}

Closed Neighborhood: N[u] = N(u) U {u}

Pendant vertex - a vertex of degree one

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The basics

Branching according to a low-degree black vertex

Use of reduction rules (re-kernelization)

Whenever a new vertex u is IN, k-1 more vertices are required …

… and whiten the neighbors of u (Why?)

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The Idea

A dominating set problem has 3 types of vertices:

Dominated (black) Dominating (black or white) -DS Untouched (black or white)

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The Idea

At any point, add the vertex adjacent to an untouched black vertex to DS + reduce k The added vertex could be black or white

Once that is added, whiten its neighbors (as they are also dominated) If any neighbor is black, then it is already

dominated anyway and k shouldn’t reduce If it is white, we don’t care anyways!

Going on like this will guarantee an exact solution having k vertices

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7 Reduction Rules - R1

Delete edges between white vertices After this white vertices connected only to black

vertices

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R2

Delete a pendant white vertex

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R3 - Branching on a low-degree black vertex

Delete a pendant black vertex w, put its neighbor u (black or white) in D, whiten neighbors of u and lower k to k − 1

This is the only rule that builds D

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R4

Delete a white vertex u of degree 2 having two black neighbors u1 and u2 connected by an edge {u1, u2}

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R5

Delete a white vertex u of degree 2 with black neighbors u1, u3, if there is a black vertex u2 with edges {u1, u2} and {u2, u3}

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R6

Delete a white vertex u of degree 2 with black neighbors u1, u3, if there is a white vertex u2 and edges {u1, u2} and {u2, u3}

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R7

Delete a white vertex u of degree 3, with black neighbors u1, u2, u3 for which the edges {u1, u2} and {u2, u3} are present

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A Few Observations

Rules are sound Reduction by each rule is linear time

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Definitions

G is reduced if any of the above rules cannot be applied anymore to it

G is nearly reduced if (R1), (R2), (R4)–(R7) cannot be applied anymore to it

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Plane Embedding

Planar Graph embedding Plane Graph

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Main Resultand

Intuition of the Proof

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A New Branching Theorem

If G is nearly reduced planar black and white graph with b black and w white vertices, then max degree of black vertices is 7

Main technique: Euler’s formula for planar graphs extended for

plane black and white graphs

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Embed the Graph

H = G[B] is plane sub-graph of black vertices of the nearly reduced plane graph G

F is the set of faces (i.e. “face-graph”) of H fH = |F| = number of faces of H

cH = number of connected components of H

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Extended Euler Formula

For graph G: Let:

v = b + w e = ebb + ebw

v − e + f = 2 (- original Euler formula)

For graph H: b − ebb + fH = 1 + cH (- extended Euler

formula)

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Let z = (3(b + w) − 6) − e = number of edges for which G fails to be a

triangulation of the plane

Triangulation Error

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Main Result

Lemma 4:

If 3w − 4b − z + fH − cH < 7 is satisfied - EQN (A)

then max degree of black vertices is 7

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Proof Outline

Take a nearly reduced black and white plane graph G

Determine H and its “face graph” Build equations inductively from face graph to

entire graph Determine cH in terms of cF of each face F

cF = total connected components on boundary of F - 1 Determine z in terms of zF of each face F of G

zF = total edges to be added to F for triangulation

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Proof Outline

Substitute them in EQN (A) to get: 3 ∑FЄҒ(wF +cF/3 −zF/3 +1/3)−4b −2cH< 6 - EQN (B)

(wF = number of white vertices embedded in F)

Determine tF in terms of ebb- EQN (C) tF = total edges needed to triangulate F of H (i.e.

to triangulate only black vertices on F’s boundary as they may be disconnected)

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Proof Outline

Consider that for any face of G wF + cF ≤ zF + 1 - EQN (D)

wF ≤ tF - EQN (E)

Use EQN (C, D, E) to solve the first term of EQN (B)

Thus prove the inequality

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New Search Tree algorithm

Apply Reduction Rules Construct a k-DS from the reduced graph

Here, running time - O(8kn)

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Degree of each black vertex = 7 Degree of each white vertex = 4

Optimality of Rules

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Future Work

Adding more involved reduction rules

Investigating if and how more technical approaches such as tree decompositions could be used in the algorithm

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Questions?

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Ponder this (Question 1)

We know that there is one NP-hard problem called minimum Dominating Set. Suppose we solve it by brute force and get an answer where the minimum number of vertices that dominate all others are n.

Now, for the same graph, if we are solving the k-Dominating Set problem, if k < n then what will happen?

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Food for thought (Question 2)

Finding min DS is NP-hard Take a graph and run a breath-first-search on it

Running time = O(# of Vertices + # of Edges) The interior nodes of the resulting tree form a

“connected” DS Pick such smallest tree – it will form min “connected”

DS running time O(# of Vertices)

Total Running time = O(V2 + EV) Hence P = NP! What is the flaw in this argument?

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Thanks for your patience!