Zero Forcing and Power Domination in Graphs

ZERO FORCING AND POWER DOMINATION IN GRAPHS

SUDEEP STEPHEN Master of Science (Mathematics)(University of Madras) School of Mathematical and Physical Sciences The University of Newcastle, Australia

A thesis submitted in total fulﬁlment of the requirements for the degree of Doctor of Philosophy October 2017

Statements Statement of Originality

The thesis contains no material which has been accepted for the award of any other degree or diploma in any university or other tertiary institution and, to the best of my knowledge and belief, contains no material previously published or written by another person, except where due reference has been made in the text. I give consent to the ﬁnal version of my thesis being made available worldwide when deposited in the University’s Digital Repository, subject to the provisions of the Copyright Act 1968.

Statement of Authorship

I hereby certify that the work embodied in this thesis contains published papers of which I am a joint author. I have included as part of the thesis a written statement, endorsed by my supervisor, attesting to my contribution to the joint publication.

Signed: SUDEEP STEPHEN

The candidate has six papers, of which he is a joint author, embodied in the thesis. As principal supervisor I can attest that the candidate’s contribution, in all cases were at least 80% of the ﬁnal article.

Signed: Dr JOE RYAN iv

List of Publications

Publications Arising from this Thesis (1) D. Ferrero, T. Kalinowski and S. Stephen. Zero forcing in iterated line digraphs. Submitted to Discrete Applied Mathematics, 2017 [37]. (2) R. Davila, T. Kalinowski and S. Stephen. Proof of a conjecture of Davila and Kenter regarding a lower bound for the forcing number in terms of girth and minimum degree. Submitted to Discrete Applied Mathematics, 2016 [23]. (3) D. Ferrero, C. Grigorious, T. Kalinowski, J. Ryan and S. Stephen. Minimum rank and zero forcing number for butterﬂy networks. Submitted to Discrete Applied Mathematics, 2016 [38]. (4) C. Grigorious, T. Kalinowski, J. Ryan and S. Stephen. On the power domination number of de Bruijn and Kautz digraphs. To appear in Lecture Notes in Computer Science, 2017 [45]. (5) S. Stephen, B. Rajan, C. Grigorious and A. William. Resolving-power dominating sets. Applied Mathematics and Computation, 256, 778–785, 2015 [85]. (6) S. Stephen, B. Rajan, J. Ryan, C. Grigorious and A. William. Power domination in certain chemical structures. Journal of Discrete Algorithms, 33, 10–18, 2015 [86].

Other Publications Produced During my Candidature (1) C. Grigorious, T. Kalinowski, J. Ryan and S. Stephen. Metric dimension of Circulant graph C(n, ±{1, 2, 3, 4}). The Australasian Journal of Combinatorics, 69 (3), 417– 441, 2017 [46]. (2) C. Grigorious, S. Stephen, B. Rajan and M. Miller. On the Partition Dimension of Circulant Graphs. The Computer Journal, 60 (2), 180–184, 2016 [48]. (3) S. Klavˇzar, P. Manuel, M. J. Nadjaﬁ-Arani, R. S. Rajan, C. Grigorious and S. Stephen. Average Distance in Interconnection Networks via Reduction Theorems for Vertex-Weighted Graphs. The Computer Journal, 59(12), 1900–1910, 2016 [68]. (4) C. Grigorious, P. Manuel, M. Miller, B. Rajan and S. Stephen. On the metric dimension of circulant and Harary graphs. Applied Mathematics and Computation, 248, 47–54, 2014 [47]. (5) P. Manuel, B. Rajan, C. Grigorious and S. Stephen. On the Strong Metric Dimen- sion of Tetrahedral Diamond Lattice. Mathematics in Computer Science, 9, 201–208, 2015 [76]. (6) C. Grigorious, S. Stephen, B. Rajan, M. Miller and A. William. On the Partition Dimension of a Class of Circulant Graphs. Information Processing Letters, 114, 353– 356, 2014 [49]. To my parents, Stephen P.M and Mary Stephen And in memory of my most beloved supervisor, Mirka Miller.

Acknowledgements

First and foremost, I would like to express my sincere gratitude to my su- pervisors, (L) Prof. Mirka Miller, Dr. Joe Ryan, Dr. Thomas Kalinowski and Dr. Paul Manuel, who have supported me with their encouragement, guidance and patience. My sincere gratitude to Mirka and Joe, without them, I would never have been in Newcastle. Thank you for trusting my research capabilities and above all for the emotional and practical support whenever I needed them. Thank you for teaching how to build collaborations and introducing me to all your collaborators. I can never forget the good times we had spent together. Your motivations and encouragements helped me to be a better person and a researcher. It was indeed a tough phase for me and Joe after Mirka lost her battle with cancer. Thank you Joe for being there always and guiding me through the completion of my candidature. I am happy we did this together for Mirka. I specially like to acknowledge Thomas Kalinowski for his insightful sug- gestions and comments on my research and talks at every stage of my Ph.D. I am always inspired by his work. To be very honest, I have always looked up at him as my role model. I also thank Dr. Paul Manuel for contributing 25% towards my scholarship. My special thanks to Dr. Bharati Rajan, my mentor, for introducing me to graph theory and research. I owe what I am today to you. I am thankful to Dr. Cristina Dalfo Simo, my external supervisor for her inputs and support during my Ph.D candidature. I am grateful to Prof. Indra Rajasingh and Dr. Sundara Rajan for their encouragement and support during my initial stages of research in Chennai. Discussions and chats with Prof. Brian Alspach were always interesting and helpful. Thank you. I am also thankful to Renee Mam and the entire Maths department of St. Joseph’s College, Bangalore for inspiring me to do research in mathematics. I like to thank Prof. Zdenek Ryacek and Prof. Dafik for hosting me during my stay in The University of West Bohemia, Czech Republic and University of Jember, Indonesia respectively. I also like to thank my collaborators Randy Davila, Prof. Daniela Ferrero and Prof. Sandi Klavzar for accepting my request to work with me and sharing their knowledge. It was indeed wonderful 3 years in Newcastle. I would like to acknowledge Dr. Benny Sudakov for pointing out the argument for Theorem 5.2.3 on his visit to the University of Newcastle during the 40 ACCMCC. Thanks to the University of Newcastle for providing scholarships (UNIPRS and UNRSC 50-25) to pursue my dream. I thank the International Leader- ship Experience and Development (ILEAD) Plus Program for the generous support for participating in various conferences. I acknowledge the Faculty of Sciences for the Conference scholarship. My sincere thanks to the school of mathematical and physical sciences. All the staff, both academic and administrative staff created a healthy and comfortable environment for me to pursue my research. Many thanks to all of them for their supports and efforts. Special thanks to Vicki, Anna Rosa, Irene, Rosemary and Felicity for their administrative help whenever I needed them. To my friend and colleague Cyriac Grigorious, thank you so much for being there with me for the past 8 years. You have influenced my life more than you can imagine. Thank you for the support and friendship during my stay here. To my office mates, Bong, Dushyant, Oudone and Wulan, thank you for bearing me. Thanks to the Malayali community in Newcastle and Sydney for making me welcome among them and making me feel at home away from home. I am indebted to my parents, Stephen P. M and Mary Stephen, my wife, Jinu Joseph, my brother, Suraj Stephen, my sister-in-law, Rakhi Ashok and my nephew, Aswin Suraj, for their love and support during these years. A special word of acknowledgement to Mr. Simon, Saly amme, Ashwin and Ada for considering me a part of their family. Thank you for the unconditional love you have shown towards me. Saly amme, thank you so much for being there like a mother encouraging, supporting and guiding me through various frustrations and disappointments. Last but not least, I thank God Almighty for blessing me with this opportunity to pursue my research and for giving me sound body and mind to complete this successfully.

Sudeep Stephen Contents

List of Figures xi

Nomenclature xv

Chapter 1. Introduction1 1.1. Contributions of the thesis3 1.2. Structure of the thesis3

Chapter 2. Background Theory5 2.1. Graph theory5 2.1.1. Undirected graphs5 2.1.2. Graph operations8 2.1.3. Directed graphs 10 2.2. Zero forcing in graphs 11 2.2.1. Bounds on the zero forcing number 17 2.2.2. Zero forcing in directed graphs 21 2.2.3. Variants of zero forcing problem 23 2.3. Power domination in graphs 24 2.3.1. Bounds on the power domination number 26 2.3.2. Power domination in directed graphs 31 2.3.3. Variants of power domination problem 32 2.4. Zero forcing and power domination 33

Chapter 3. Zero forcing and power domination in iterated line digraphs 35 3.1. An alternate interpretation 35 3.2. Zero forcing and power domination in de Bruijn digraphs 38 3.3. Zero forcing and power domination number in Kautz digraphs 41 3.4. Zero forcing in iterated line digraphs 45 3.5. Power domination in iterated line digraphs 48 3.6. Applications 52

Chapter 4. Zero forcing and power domination in butterﬂy networks 55 4.1. An alternate interpretation 55 4.2. Zero forcing in butterﬂy network 57

ix x CONTENTS

4.2.1. The upper bound for Z(BF(r)) 58 4.2.2. The lower bound for Z(BF(r)) 62 4.3. Power domination in butterﬂy network 71

Chapter 5. Zero forcing number in terms of girth and minimum degree 77 5.1. Proof of the conjecture 77 5.2. Better bounds for large girth and minimum degree 82

Chapter 6. Power domination and resolving-power domination in graphs 85 6.1. Power domination-subgraph relation 85 6.1.1. Silicate networks 88 6.1.2. Rhenium trioxide lattice 90 6.2. Resolving-power dominating set 94 6.2.1. Complexity results of resolving-power domination problem 97 6.2.2. Resolving-power domination in trees 101

Chapter 7. Conclusion 105

Bibliography 111 List of Figures

2.1 A graph G with zero forcing set S = {4, 6} 11 2.2 Exhibition of the colour change rule 12

2.3 Graphs G1 and G2 speciﬁed in Theorem 2.2.21 19 2.4 A directed graph with empty set as the zero forcing set 23 2.5 A graph G with power dominating set S = {5, 6} 26 2.6 Exhibition of the power dominating steps 27

3.1 A directed graph G~ . 36 3.2 Illustration of the construction of the power dominating set S for d = 5 and ` = 7. For

the two columns (a1, . . . , a5) = (1, 3, 4, 4, 2) and (a1, . . . , a5) = (3, 1, 0, 2, 4) we show

the elements of S (black squares), and we indicate for the sets X(a1, . . . , a6) (enclosed by rectangles) the elements of S having them as their out-neighbourhood. 40

4.1 The butterﬂy network BF(4). 58 4.2 The set S(4) indicated by squares. 59 4.3 Labelling of vertices and structure of adjacency matrix in BF(3). 65

4.4 The second level of the recursion for Ar. 67 4.5 The construction of K˜ for r = 4 and i = 1. Here K+ = {2, 6}, K0− = {9, 12}, ˜ − ˜ + ˜ − 00+ K1 = {14, 18, 46, 50}, K1 = {21, 24, 53, 56}, K2 = {76, 77, 80}, K2 = ∅. 69 4.6 The critical sets in diﬀerent layers of BF(4) . For convenience, we identify the binary r Pr i−1 vector x = (x1, . . . , xr) ∈ {0, 1} with the number i=1 xi2 . 72

6.1 WK(3, 3) containing three copies of WK(3, 2). 87 6.2 Illustration of Step 1 of the algorithm on S(2) 89 6.3 Step 2 and Step 3 of the algorithm 89 6.4 Subgraph X of Theorem 6.1.8 90 6.5 Unit cell of Rhenium trioxide 91 6.6 Square bullets depict (i) Metric basis (ii) Minimum power dominating set (iii) Minimum metric-locating-dominating set and (iv) Minimum resolving-power-dominating set of G. 95

xi xii LIST OF FIGURES

6.7 Variable gadget for each xi 98

6.8 Clause gadget for each Cj 98

6.9 Reduction ofx ¯1 ∨ x2 ∨ x¯3 99 6.10Variable gadget for bipartite graphs 101

7.1 Critical sets in 3 × 3 grid 105 Abstract

This thesis investigates the dynamic colouring of vertices in a graph G. Dy- namic colouring of the vertices in a graph has seen a rise in application and relations to well studied graph theoretic parameters in recent years. By dynamic colouring, we mean a colouring of the vertices in a graph which may propagate (change the colour of) vertices that were not initially coloured in the graph. Of these dynamic colourings, and of relation to this thesis, we highlight that zero forcing sets and power dominating sets, and the associated graph invariants known as the zero forcing number and power domination number respectively, are of particular interest. Both zero forcing number and power domination number have been shown to relate to well studied graph invariants such as the domination number, the total domination number, the connected domination number, the path cover number, the chromatic number, the independence number, and the minimum rank. Moreover, it has been established that both these problems lie within the class of NP- complete decision problems. Thus, it is desirable to ﬁnd easily computable bounds for the zero forcing number and power domination number.

In this thesis, we give an alternate interpretation of zero forcing and power domination problems as set covering problems. These interpretations were used in solving both these problems for de Bruijn and Kautz digraphs.

Moreover the same technique was used to extend the results to obtain these graph invariants for iterated line digraphs. The zero forcing problem is solved for butterﬂy networks and bounds were obtained in the case of power domination number.

Another important result in this thesis is settling the conjecture of Davila and Kenter connecting zero forcing number, girth and minimum degree. A general lower bound technique for power domination number by looking at subgraphs satisfying certain conditions is described and used eﬀectively in

ﬁnding the power domination number of some speciﬁc graphs. In the last section of the thesis, we introduce a new variant of power domination problem called resolving-power domination problem. We show that the problem is

NP-complete. Nomenclature

Table 0.1. General Notation

N The natural numbers Z The integers Zd The cyclic group of order d R The real numbers [j] {1, 2, . . . , j}

G An undirected graph mrF (G) The minimum rank of G over ﬁeld F MF (G) The maximum nullity of G over ﬁeld F P(G) The path number of G Z(G) The zero forcing number of G

γp(G) The power domination number of G γ(G) The domination number of G dim(G) The metric dimension of G

ηp(G) The resolving-power domination number of G G~ A directed graph mrF (G~ ) The minimum rank of G~ over ﬁeld F MF (G~ ) The maximum nullity of G~ over ﬁeld F P(G~ ) The path number of G~ Z(G~ ) The zero forcing number of G~

γp(G~ ) The power domination number of G~

CHAPTER 1

Introduction

Many real world situations can conveniently be described by means of a diagram consisting of a set of points (nodes) together with lines (edges) joining certain pairs of these points. Points joined by a line are referred to as neighbours. For example, the points might be communication centres, with lines representing communication links between the centres. Note that we are interested mainly in whether or not two given points are joined by a line; the manner in which they are joined is immaterial. A mathematical abstraction of situations of this type gives rise to the concept of a graph. Graph theory is the study of graphs. The wide variety of applications has increased the popularity of graph theory in many ﬁelds like natural sciences, engineering, genetics and even in soft sciences like psychology and sociology.

This thesis deals with the study of two minimisation problems - zero forcing problem and power domination problem. In the zero forcing problem, we are given a graph G and the goal is to ﬁnd a minimum-sized set of nodes S that covers all of the nodes, where a node v is covered if

(1) v is an element of the set S, or

(2) v has a neighbour u such that u and all of its neighbours except v are covered.

Rule (2) is called the colour change rule and is applied iteratively. The cardinality of any such minimum sized set is called the zero forcing number.

The power domination problem can be thought as a variant of zero forcing problem. In this case, we are given a graph G and the goal is to ﬁnd a minimum-sized set of nodes S that covers all of the nodes, where a node v is covered if

(1) v is in the closed neighbourhood of the set S, or

(2) v has a neighbour u such that u and all of its neighbours except v are covered.

The cardinality of any such minimum sized set is called the power domination number.

1 2 1. INTRODUCTION

Although the power domination and the zero forcing problem are closely related, they were introduced independently and have been studied by diﬀerent groups of researchers using diﬀerent approaches until Benson et al. [11] connected the concepts and combined techniques and results from both to obtain new results on several graph products.

Both the zero forcing problem and the power domination problem are modelled as graph theory problems from a real world scenario.

The zero forcing process was introduced in [16] and used in [15] as a criterion for quantum controllability of a system. Independently, the authors in [3] have introduced it to bound the minimum rank, or equivalently, the maximum nullity of a graph G. Given an n-vertex graph

G, let M(G) denote the maximum nullity over all symmetric real-valued matrices A whose zero-nonzero pattern of the off-diagonal entries is described by the graph G. This means that for i 6= j, the entry Aij is non-zero if and only if ij is an edge in G, whereas the diagonal entries are chosen freely. The minimum rank of G is n − M(G). This parameter has been extensively studied in the last fifteen years, largely due to its connection to inverse eigenvalue problems for graphs, singular graphs, biclique partitions and other problems. Among several tools introduced to study the minimum rank, the zero forcing number has the advantage that its definition is purely combinatorial.

The motivation for the power domination problem comes from modelling electrical power systems. Electric power companies need to continually monitor their system state as deﬁned by a set of state variables (for example, the voltage magnitude at loads and the machine phase angle at generators). One method of monitoring these variables is to place phase measurement units (PMUs) at selected locations in the system. Because of the high cost of a PMU, it is desirable to minimize their number while maintaining the ability to monitor (observe) the entire system. A system is said to be observed if all of the state variables of the system can be determined from a set of measurements (e.g., voltages and currents). This was modelled as a graph theory problem by Haynes et al. in [55].

Both zero forcing problem and power domination problem have been extended to digraphs in [9] and [1] respectively. 1.2. STRUCTURE OF THE THESIS 3

1.1. Contributions of the thesis

The main objective of this research is to exploit the similarity between zero forcing and power domination problem in order to obtain an eﬃcient technique of lower bound for both these parameters and use them eﬀectively in obtaining the zero forcing number and power domination number in several classes of graphs. To this end the following results were obtained:

• Zero forcing number and power domination number for de Bruijn and Kautz digraphs.

• Bounds on the zero forcing number and power domination number of iterated line

digraphs.

• Zero forcing number and minimum rank of butterﬂy networks.

• Proof of the conjecture connecting zero forcing number, girth and minimum degree.

• A lower bound technique for power domination number in relation to subgraphs of

the graph satisfying certain condition.

• A new variant of power domination called resolving-power domination.

1.2. Structure of the thesis

This thesis is structured as follows.

Chapter 2 on page 5: Background Theory- This chapter contains relevant and useful concepts from graph theory and an extended literature review of zero forcing and power domination problems needed for the thesis.

Chapter 3 on page 35: Zero forcing and power domination in iterated line digraphs-

In this chapter, we obtain the zero forcing and power domination numbers of de Bruijn and

Kautz digraphs by interpreting them as set cover problems. The later sections of the chapter extends the results obtained for de Bruijn and Kautz digraphs to iterated line digraphs.

Chapter 4 on page 55: Zero forcing and power domination in butterﬂy networks-

In this chapter, we obtain the zero forcing number of butterﬂy networks. A lower bound and upper bound for power domination number for butterﬂy networks is provided in the second section.

Chapter 5 on page 77: Zero forcing number in terms of girth and minimum degree

- In this chapter we resolve the conjecture connecting zero forcing number, girth and minimum degree posed by Davila and Kenter in [24]. 4 1. INTRODUCTION

Chapter 6 on page 85: Power domination and resolving-power domination in graphs- In the first section of this chapter, a lower bound technique for power domination is introduced and is used effectively in finding the power domination numbers of certain chemical structures. In the second section, a new variant of power domination called resolving-power domination is introduced. We show that the problem is NP-complete and obtain bounds for trees.

Chapter 7 on page 105: Conclusion- In this chapter, we summarise the results and provide a list of open problems that arise from this thesis.

In this thesis, all original results are indicated by the symbol ♦ and end of proofs are marked by symbol . CHAPTER 2

Background Theory

This chapter presents some basic deﬁnitions, concepts and facts from graph theory that we would use in this thesis. For terms not deﬁned here, readers may refer to [14, 50, 70]. We also give a literature review of zero forcing and power domination problems.

2.1. Graph theory

In mathematics, graph theory is the study of graphs, which are mathematical structures used to model pairwise relations between objects. A graph consists of a ﬁnite set of vertices

(points, nodes), a finite set of edges (lines, linkages) and a rule that defines which edge joins which pair of vertices. It is customary to represent a graph by means of a diagram and refer to the diagram as the graph. Graphs in graph theory portray topological quantities, patterns and relationships. Structures that can be represented as graphs exist in nature and many problems of practical interest can be represented by graphs. For example, the vertices could be different cities in a country and the edges could be the roads joining them.

2.1.1. Undirected graphs. An undirected graph (or simply graph) G is a ﬁnite nonempty set of objects called vertices together with a set of unordered pairs of distinct vertices of G, called edges. The vertex set and the edge set of G are denoted by V (G) and E(G) respectively. The order of a graph G is the number of vertices in G, denoted by n = |V (G)|; the number of edges in G is called the size of G and denoted by m = |E(G)|. If e = (u, v) is an edge of G, we say that u and v are adjacent and that each vertex is incident with e. We may also say that two vertices u and v are neighbours, if they are adjacent. The open neighbourhood of a vertex v, denoted N(v), is given by N(v) = {u ∈ V :(u, v) ∈ E}. The closed neighbourhood of a vertex v, denoted N[v], is given by N[v] = N(v) ∪ {v}. The idea of neighbourhood can be generalized to include set of vertices as follows. For a set S, let N(S) = ∪v∈SN(v) − S and N[S] = N(S) ∪ S denote the open and close neighbourhood of S respectively.

5 6 2. BACKGROUND THEORY

An edge with identical ends is called a loop and an edge with distinct ends a proper edge.

A multi-edge is a collection of two or more edges having identical endpoints. A graph is simple if it has no loops or multi-edges. A simple graph on n vertices, in which each pair of distinct vertices is joined by an edge is called a complete graph, denoted by Kn. A graph is ﬁnite if both its vertex set and edge set are ﬁnite. We call a graph with just one vertex trivial and all other graphs nontrivial.A null graph is a graph whose vertex and edge sets are empty.

The degree of a vertex v in G, denoted by degG(v), is the number of edges incident with v. The maximum degree of G, denoted by ∆(G), is the maximum degree over all vertices of G, while the minimum degree of G, denoted by δ(G), is deﬁned as the minimum degree over all vertices of G. A graph G is called r-regular if degG(v) = r, for each v ∈ V (G). A vertex v is called a pendant vertex if degG(v) = 1 and an isolated vertex if degG(v) = 0. An edge e in a graph G is called a pendant edge if it is incident with a pendant vertex.

A walk W of length ` in a graph G is an alternating sequence of vertices and edges W = v0, v0v1, v1, v1v2, v2, . . . , v`−2, v`−2v`−1, v`−1 containing ` vertices and ` − 1 edges. A walk in which no edges are repeated is called a trail and a walk of length ` in which no vertices are repeated is called a path P`. A walk is closed if it has positive length and its origin and terminus are the same. A closed path is a cycle. A cycle of length ` is called a `-cycle, denoted C`; a `-cycle is odd or even according as ` is odd or even. A 3-cycle is often called a triangle. The length of the smallest cycle in a graph G is called the girth of G. If the graph G contains no triangles, alternatively, if G has girth g ≥ 4, then G is called triangle-free graph.

A trail in G is called an Eulerian trail if it includes every edge of G.A tour of G is a closed walk of G which includes every edge of G at least once. An Eulerian tour of G is a tour which includes each edge of G exactly once. A graph G is said to be Eulerian if it has an Eulerian tour. A graph is Eulerian if and only if it has no vertex of odd degree.

A Hamiltonian path in a graph G is a path which contains every vertex of G.A Hamiltonian cycle in a graph G is a cycle which contains every vertex of G. A graph G is called Hamiltonian if it has a Hamiltonian cycle.

The distance distG(u, v) ( or simply dist(u, v) if the graph G is clear in the context) between two vertices u and v in a graph G is the length of the shortest path between u and v. The 2.1. GRAPH THEORY 7 eccentricity of a vertex v, denoted by e(v), in a graph G, is the longest distance from a vertex v to any other vertex in G, that is, e(v) = max{distG(v, u): u ∈ V (G)}. The diameter, denoted by λ (or D) of a graph G is the maximum eccentricity of all the vertices in G, that is,

λ(G) = max{e(v): v ∈ V }. The radius of G, written as rad(G), is the minimum eccentricity of any vertex in G, that is, rad(G) = min{e(v): v ∈ V }. A vertex whose eccentricity is equal to the radius is called a central vertex or simply a center.

A graph S is a subgraph of G if V (S) ⊆ V (G) and E(S) ⊆ E(G). If V (S) = V (G) then

S is a spanning subgraph of G. If V (S) is a proper subset of V (G) or E(S) is a proper subset of E(G) then S is called a proper subgraph of G. If U is a nonempty subset of the vertex set

V (G) of a graph G, then the subgraph hUi of G induced by U is the graph having vertex set U and whose edge set consists of those edges of G incident with two elements of U. A subgraph

H of G is called vertex-induced or simply induced if H = hUi for some subset U of V (G). By deleting from G all loops and, for every pair of adjacent vertices, all but one edge joining them, we obtain a simple spanning subgraph of G, called the underlying simple graph of G.

A vertex u is said to be connected to a vertex v in a graph G if there exists a u−v path in G.

A graph G is connected if every two of its vertices are connected. A graph that is not connected is called disconnected. Connection is an equivalence relation on the vertex set V . Thus there is a partition of V into nonempty subsets V1,V2 ...Vω such that two vertices u and v are connected if and only if both u and v belong to the same set Vi. The subgraphs hV1i, hV2i ... hVωi are called the components of G. If G has exactly one component, G is connected; otherwise G is disconnected. The number of components of G is denoted by ω(G). A cut-vertex is a vertex whose removal increases the number of components. A cut-edge is an edge whose removal increases the number of components.

A graph G1 is isomorphic to a graph G2 if there exists a one-to-one mapping φ, called an isomorphism from V (G1) onto V (G2) such that φ preserves adjacency and non-adjacency; that is, uv ∈ E(G1) if and only if φ(u)φ(v) ∈ E(G2). As structural properties of graphs are determined by their adjacency relations, we can prove that each G1 and G2 are not isomorphic by finding some structural properties of one that are not true of the other. If they have different vertex degrees, different sizes of the largest clique or smallest cycle, and so on, then they cannot 8 2. BACKGROUND THEORY be isomorphic, because these properties are preserved by isomorphism. On the other hand, no known list of common structural properties implies that G1 is isomorphic to G2. An automorphism of G is a permutation of V (G), that is, an isomorphism from G into G.

A graph is vertex transitive if for every pair u, v ∈ V (G) there is an automorphism that maps u to v. A graph G is edge transitive if for every pair e, f ∈ E(G) there is an automorphism that maps e to f. For example, every cycle of size n is not only vertex transitive but also edge transitive.

An acyclic graph is one that contains no cycles. An acyclic graph is also called a forest.A tree is a connected acyclic graph, denoted T . Every tree T has at least two vertices of degree one; indeed, two end-vertices of any longest path in a tree are of degree one. A tree in which one vertex (called the root), is distinguished is called a rooted-tree. In a rooted-tree any vertex of degree one, unless it is the root, is called a leaf and the other vertices are called internal vertices.A complete binary tree is deﬁned as a tree in which there is exactly one vertex of degree two, and each of the remaining vertices is of degree one or three. Since the vertex of degree two is distinct from all other vertices, this vertex serves as a root. Thus every complete binary tree is a rooted tree.

A bipartite graph is one whose vertex set can be partitioned into two subsets X and Y , so that each edge has one end in X and the other end in Y ; such a partition (X,Y ) is called a bipartition of the graph. A graph is bipartite if and only if it contains no odd cycle. A complete bipartite graph is a bipartite graph G(X ∪ Y,E) in which each vertex of X is joined by an edge to each vertex of Y . If |X| = s and |Y | = t, a complete bipartite graph is denoted by Ks,t.

2.1.2. Graph operations. Let G = (V,E) be a graph. The operation of adding the vertex u, u∈ / V to a graph G, yields a new graph G ∪ {u} with vertex set V ∪ {u} and edge set

E. Note that the new vertex u has no neighbours.

The operation of deleting the vertex v from a graph G, not only removes the vertex v but also removes every edge incident to v. The resulting graph is denoted G − v. The operation of adding an edge e (or uv) to a graph G = (V,E) joining the vertices u and v yields a new graph with vertex set V and edge set E ∪ {e} (or E ∪ {uv}), which is denoted G ∪ {e} (or G ∪ {uv}). 2.1. GRAPH THEORY 9

The operation of deleting an edge e (or uv) from a graph G = (V,E) removes only that edge.

The resulting graph is denoted as G − e (or G − uv).

A subdivision of an edge e = uv in a graph is the graph obtained from G by replacing e by the path uwv where w is a new vertex of degree 2. The resulting graph is denoted as Ge. The operation of edge contraction produces a graph with edge set E \{e} but with a vertex set obtained by identifying the vertices deﬁning e in G, thus creating a new single vertex where the latter inherits all the adjacencies of the pair of replaced vertices, without introducing loops or multiple edges. The resulting graph is denoted as G/e.

The graph union of two graphs G and H is the graph G ∪ H whose vertex-set and edge-set are the disjoint unions, respectively, of the vertex-sets and the edge-sets of G and H. The complement G of a graph G also has V (G) as its vertex set, and two vertices are adjacent in

G if and only if they are not adjacent in G. For disjoint graphs G and H, G + H is the graph obtained from G and H by joining every vertex of G with every vertex of H.

The Cartesian product of two graphs G and H, denoted G × H is deﬁned as the graph with vertex set V (G) × V (H) and edge set {((u, v), (x, y)) : u = x and vy ∈ E(H) or ux ∈ E(G) and v = y}. The rth hypercube, Qr is deﬁned inductively by Q1 = K2 and Qr+1 = Qr × K2.

The Cartesian product of Ps × Pt, Ps × Cs and Cs × Ct are often called s × t grid, cylinder and torus respectively.

The direct product of two graphs G and H, denoted G · H is deﬁned as the graph with vertex set V (G) × V (H) and edge set {((u, v), (x, y)) : ux ∈ E(G) and vy ∈ E(H)}.

The strong product of two graphs G and H, denoted G ⊗ H, is the graph with vertex set

V (G) × V (H) and edge set {((u, v), (x, y)) : u = x and vy ∈ E(H) or ux ∈ E(G) and v = y or ux ∈ E(G) and vy ∈ E(H)} .

The lexicographic product of two graphs G and H, denoted G ∗ H, is the graph with vertex set V (G) × V (H) and edge set {((u, v), (x, y)) : ux ∈ E(G) or u = x and vy ∈ E(H)} .

The corona of G with H, denoted G ◦ H, is the graph of order |G||H| + |G| obtained by taking one copy of G and |G| copies of H, and joining all the vertices in the ith copy of H to the ith vertex of G. 10 2. BACKGROUND THEORY

2.1.3. Directed graphs. An edge between two vertices creates a connection in two opposite senses at once. Assigning a direction makes one of these senses forward and the other backward. Viewing direction as an edge attribute is partly motivated by its impact on computer implementations of graph algorithms. A directed edge (or arc) is an edge e, one of whose endpoints is designed as the tail, and whose other endpoint is designed as the head. A directed edge is said to be directed from its tail to its head. A digraph (or directed graph) is a graph each of whose edges is directed, denoted by G~ = (V,A) with vertex set V = V (G~ ) and arc set

A = A(G~ ). An arc in the form (u, u) is called a loop.A simple digraph is a directed graph that does not allow loops and multiple edges, and a loop digraph is a directed graph that allow loops but not multiple edges. For each vertex v the open out-neighbourhood of v is N out(v) = {u ∈ G~ V (G~ ): vu ∈ A(G~ )} and the open in-neighbourhood of v is N in(v) = {w ∈ V (G~ ): wv ∈ A(G~ )}. G~ The closed out-neighbourhood and the closed in-neighbourhood of v are respectively deﬁned as

N out[v] = N out(v) ∪ {v} and N in[v] = N in(v) ∪ {v}. Analogously, for each vertex v, Aout(v) = G~ G~ G~ G~ G~ {vw ∈ A(G~ ): w ∈ N out(v)} and Ain(v) = {uv ∈ A(G~ ): u ∈ N in(v)}. For a vertex v, the out- G~ G~ G~ degree is degout(v) = |N out(v)| = |Aout(v)| and the in-degree is degin(v) = |N in(v)| = |Ain(v)|. G~ G~ G~ G~ G~ G~ ~ ~ out in A digraph G is r-regular if for every vertex v in G, degG~ (v) = degG~ (v) = r. The maximum ~ out ~ out ~ out-degree of G is denoted as ∆ (G) and it is deﬁned as max{degG~ (v): v ∈ V (G)}. Anal- ~ in ~ in ~ ogously, the maximum in-degree of G is ∆ (G) = max{degG~ (v): v ∈ V (G)}. The minimum ~ out ~ out ~ out-degree of G is denoted as δ (G) and it is deﬁned as min{degG~ (v): v ∈ V (G)}. Analo- ~ in ~ in ~ ~ gously, the minimum in-degree of G is δ (G) = min{degG~ (v): v ∈ V (G)}. When the graph G can be deduced from the context we will omit the sub-indices.

Let u, v be two vertices in a digraph G~ = (V,A). A walk of length n from u to v is a sequence of vertices u = x1, . . . , xn = v such that xixi+1 ∈ A(G~ ) for every i = 1, . . . n − 1. A trail is a walk that does not repeat arcs, and a path is a trail that does not repeat vertices. A walk is a circuit if u = v and a circuit that does not repeat vertices is a cycle. The distance between two diﬀerent vertices u and v in a digraph G~ = (V,A) is the length of a shortest path ~ from u to v and it is denoted as distG~ (u, v) or simply dist(u, v) if the graph G is clear in the context. The maximum distance between any two diﬀerent vertices in G~ is the diameter of G~ and it is denoted as D(G~ ). 2.2. ZERO FORCING IN GRAPHS 11

For a digraph G~ = (V,A), the line digraph of G~ is the digraph L(G~ ) where V (L(G~ )) = A(G~ ) and there is an arc from a vertex x to a vertex y in L(G~ ) if and only if x = uv and y = vw for some vertices v, w in V (G~ ). Iterated line graphs are naturally deﬁned by the following rules:

L0(G~ ) = G~ and for every integer i > 0, Li(G~ ) = L(Li−1(G~ )). Two famous examples of line digraphs are de Bruijn and Kautz digraphs, discussed in detail in Chapter3.

2.2. Zero forcing in graphs

The notion of a zero forcing set, as well as the associated zero forcing number, of a simple undirected graph was introduced by the “AIM Minimum Rank Special Graphs Work Group” in [3] to bound the minimum rank of associated matrices for numerous families of graphs.

For a red/blue colouring of the vertex set of a simple undirected graph G = (V,E) consider the following colour-change rule: a red vertex u is converted to blue if it is the only red neighbour of some blue vertex v. We call such a blue vertex v a forcing vertex or in other words, we say v forces u, and write v → u. Given a two-colouring of G, the derived set is the set of blue vertices obtained by applying the colour-change rule until no more changes are possible. A zero forcing set for G is a subset of vertices S ⊆ V such that if initially the vertices in S are coloured blue and the remaining vertices are coloured red, then the derived set is the complete vertex set V .

The minimum cardinality of a zero forcing set for the undirected graph G is called the zero forcing number of G, denoted by Z(G).

5 6

4 3

1 2

Figure 2.1. A graph G with zero forcing set S = {4, 6}

In Figure 2.1, the set S = {4, 6} is a zero forcing set because at the ﬁrst time step, vertex

6 forces vertex 3, vertex 3 forces vertex 2 in the second time step, vertex 2 forces vertex 1 at third time step and lastly, vertex 4 forces vertex 5 in the fourth time step. See Figure 2.2. 12 2. BACKGROUND THEORY

5 6 5 6

4 3 4 3

1 2 1 2 (a) First time step (b) Second time step 5 6 5 6

4 3 4 3

1 2 1 2 (c) Third time step (d) Fourth time step

Figure 2.2. Exhibition of the colour change rule

The term “zero forcing” is based on an algebraic property of these sets. Consider a vector with the entries corresponding to the vertices of a graph. Further, assume the entries corresponding to a set of vertices in a zero forcing set for the graph are equal to zero. The zero forcing property of the set guarantees that such a vector is in the kernel of the adjacency matrix of the graph only if the vector is the zero vector. The term “zero forcing” refers to the fact that the remaining entries of the vector are forced to be zero for the vector to be in the kernel of the adjacency matrix [3, 12].

Let F be a ﬁeld, and denote by Sn(F ) the set of symmetric n × n matrices over F . For a simple undirected graph G(V,E) with vertex set V = {1, . . . , n}, let S(F,G) be the set of matrices in Sn(F ) whose non-zero oﬀ-diagonal entries correspond to edges of G, i.e.,

S(F,G) = {A ∈ Sn(F ): i 6= j =⇒ (ij ∈ E(G) ⇐⇒ aij 6= 0)}. 2.2. ZERO FORCING IN GRAPHS 13

The minimum F -rank of an undirected graph G is deﬁned as the minimum rank over all matrices

A in S(F,G):

mrF (G) = min {rank(A): A ∈ S(F,G)} .

The maximum F -corank of an undirected graph G is deﬁned as the maximum corank over all matrices A in S(F,G):

MF (G) = max {corank(A): A ∈ S(F,G)} .

If the index F is omitted then it is understood that F = R. The minimum rank problem and maximum corank problem for an undirected graph G is to determine mr(G) and M(G) respectively (and more generally, mrF (G) and MF (G)), and has been studied intensively for more than ten years. See [35, 36] for surveys of known results and an extensive bibliography.

It is a crucial observation that for a zero forcing set S and a matrix A ∈ S(F,G), the rows of A that correspond to the vertices in V \S must be linearly independent, so rank(A) > n−|S|, F and consequently mr (G) > n − Z(G), or equivalently,

F Theorem 2.2.1 ([3]). For any graph G and ﬁeld F , Z(G) > M (G).

Based on this insight, the authors of [3] determined MF (G) and established equality in

Theorem 2.2.1 for many interesting graph classes. In [63], this equality is proved for block- clique graphs and unit interval graphs. See Table 2.1 for zero forcing number of various graph classes obtained. The American Institute for Mathematics maintains the minimum rank graph catalogue [59] in order to collect the information about minimum rank problem for various graph classes.

In general, characterising equality for Theorem 2.2.1 is diﬃcult. Addressing this issue, the authors in [3] posed the following question which is still open.

Question 1 ([3]). What is the class of graphs G for which Z(G) = MF (G) for some ﬁeld F ?

Prior to the development of zero forcing number, another graph parameter was studied in conjunction with maximum nullity. A path cover of an undirected graph G is a set of vertex disjoint induced paths that cover all the vertices of G. The path cover number of an undirected 14 2. BACKGROUND THEORY graph G, P(G), is the minimum size of a path cover. An undirected graph G is a graph of two parallel paths if there exist two independent induced paths of G that cover all the vertices of G and such that G can be drawn in the plane in such a way that the two paths are parallel and the edges (drawn as segments, not curves) between the two paths do not cross. A simple path is not considered to be such a graph. A graph that consists of two connected components, each of which is a path, is considered to be such a graph.

The relation between zero forcing number and path cover was brought out in [7] where the concept of forcing chain, a sequence of forcing vertices obtained through iterative application of the colour-change rule, was introduced. The vertices in a forcing chain induce a path in G because the forcing process in a forcing chain occur chronologically in the order of the chain.

This implies that if S ⊂ V (G) is a zero forcing set of G, then S induces a path cover for G.

Theorem 2.2.2 ([7]). For any graph G, Z(G) ≥ P(G).

The most important family of graphs that satisfy Z(G) = P(G) are trees [3] and block-cycle graphs [87]. However, the diﬀerence between the two parameters could be arbitrarily large, for example,

P(Kn) = dn/2e < n − 1 = Z(Kn).

Since its introduction the zero forcing number has been studied for its own sake as an interesting graph invariant [7,8, 32, 73]. Similarly in this thesis, we treat zero forcing problem as a graph theory problem in its own right and not in relation with minimum rank problem. The zero forcing problem is closely related to a number of topics of current interest. Physicists have independently studied the zero forcing parameter, referring to it as the graph infection number, in conjunction with the control of quantum systems [82]. Current applied and theoretical research on social networks has focussed on diﬀusion processes (spread of inﬂuence) and their characteristics that led to the introduction of models such as threshold model [66], very closely related to zero forcing model.

It is very clear that the zero forcing set of an undirected graph G should have at least δ vertices otherwise the forcing process cannot begin. 2.2. ZERO FORCING IN GRAPHS 15

Graph G order Z(G) Reference r r−1 Qr (hypercube) 2 2 [3] Ks × Pt st s [3] Ps × Pt st min{s, t} [3] Cs × Pt st min{s, 2t} [3] Ks × Kt st st − s − t + 2 [3] Cs × Kt, s ≥ 4 st 2t [3] Ks × Kt, t ≥ 2 st + t st − 1 [3] Cs × Ct, s = t and s is odd st 2s − 1 [11] Cs × Ct, s 6= t or s = t and s is even st 2s [11] Ps ⊗ Pt st s + t − 1 [3] P2s+1 · Kt 2st + t (2s + 1)t − 4s [63] Ks · Kt st st − 4 [63] Cn, n ≥ 5 n n − 3 [3] n(n−1) n2−3n+4 L(Kn) 2 2 [3] L(T ), T a tree and ` = # pendant vertices of T |T | − 1 ` − 1 [3] Tree T |T | P(T )[3] Block-cycle graph G |G| P(G)[87] Unicyclic graph G |G| P(G)[87] Block clique graphs G |G| M(G)[63] Unit interval graphs G |G| M(G)[63] Table 2.1. Zero forcing number of various graph classes

Theorem 2.2.3 ([3]). For any graph G of order n ≥ 2,

Z(G) ≥ δ.

Graphs that satisfy equality in Theorem 2.2.3 are cycles, paths and complete graphs amongst others.

Theorem 2.2.4. Let G be a connected graph of order n ≥ 2. Then

(1) [3, 33] Z(G) = 1 if and only if G = Pn. (2) [79] Z(G) = 2 if and only if G is a graph of two parallel paths.

(3) [3, 33] Z(G) = n − 1 if and only if G = Kn.

Table 2.1 gives the summary of the results that has been obtained for various graph classes.

In [7], it was proved that there exists no graph of order greater than one that has a unique minimum zero forcing set. 16 2. BACKGROUND THEORY

Theorem 2.2.5 ([7]). For any connected graph G of order more than one, no vertex is in every optimal zero forcing set of G.

The interesting consequence of Theorem 2.2.5 is the following result.

Theorem 2.2.6 ([7]). If G is a non trivial graph, then G does not have a unique minimal zero forcing set.

To validate the above theorem, in Figure 2.2, both {3, 5} and {4, 6} are zero forcing sets.

Theorem 2.2.7 ([3]). For any graphs G and H,

Z(G × H) ≤ min{Z(G)|H|, Z(H)|G|}.

Theorem 2.2.8 ([3]). For any graphs G and H,

Z(G ◦ H) ≤ Z(G)|H| + Z(H)|G| − Z(G) Z(H).

Recently, the authors in [44] characterised all connected graphs on n vertices with zero forcing number at least n − 2. A graph G is complement reducible (cograph) if every induced subgraph of G with at least two vertices is either disconnected or is the complement of a disconnected graph.

Theorem 2.2.9 ([44]). A graph G with Z(G) = n − 2 is a cograph.

Theorem 2.2.10 ([44]). Every connected graph G with Z(G) ≥ n − 2 can be constructed as follows. Pick two, not necessarily distinct, graphs from the list below and connect all vertices from one to the other.

(1) Kn;

(2) The complement of Kn; (3) Disjoint union of two complete graphs;

(4) A connected graph H on n vertices with zero forcing number n − 2.

In [17], a relationship between zero forcing number of graph G and number of edges of G is found. 2.2. ZERO FORCING IN GRAPHS 17

Theorem 2.2.11 ([17]). For a graph G with n vertices, if Z(G) ≤ k then,

k + 1 |E(G)| ≤ kn − . 2

The bound is tight in the case of path Pn as Z(Pn) = 1 with E(Pn) = n − 1. The bound is also tight if G is an outerplanar graph. An interesting consequence that follows from the above theorem is the following result.

m Theorem 2.2.12 ([17]). For a graph G with n vertices and m edges, Z(G) > n , in particular Z(G) is greater than half of the average degree of the graph.

This bound works best for graphs with highly irregular degree sequences.

2.2.1. Bounds on the zero forcing number. Theorem 2.2.1, Theorem 2.2.2 and The- orem 2.2.3 are the best general lower bounds we have on zero forcing number in literature.

However, there have been many attempts to improve the lower bounds by imposing certain conditions on the graph G.

Theorem 2.2.13 ([3]). If G is a strongly regular graph, then

j|V (G)|k Z(G) ≥ . 2

Theorem 2.2.14 ([13]). If G is bipartite and k-regular, then

Z(G) ≥ 2(k − 1).

Theorem 2.2.15 ([24]). Let G be a triangle-free graph with minimum degree δ ≥ 3. Then

Z(G) ≥ δ + 1.

Theorem 2.2.16 ([24]). Let G be a graph with girth g and minimum degree δ ≥ 2. Then,

Z(G) ≥ 2δ − 2. 18 2. BACKGROUND THEORY

Theorem 2.2.17 ([24]). Let G be a graph with δ ≥ 3 and a cut vertex v such that G \ v has a component with girth at least 5. Then,

Z(G) ≥ 3δ − 6.

Theorem 2.2.18 ([24]). Let G be a graph with girth g ≥ 5, minimum degree δ ≥ 3 and a cut edge e. Then,

Z(G) ≥ 4δ − 9.

Davila and Kenter in [24] conjectured that for graphs G with girth g ≥ 3 and minimum degree δ ≥ 2,

(2.2.1) Z(G) ≥ (g − 3)(δ − 2) + δ.

Gentner, Penso, Rautenbach, and Souzab [42], Gentner and Rautenbach [41] and Davila and Henning [22] have shown that inequality (2.2.1) is true for small girth g ≤ 10, while Davila and Kenter [24] have proven that inequality (2.2.1) is true for girth g ≥ 7 and suﬃciently large minimum degree.

One of the contributions of this thesis is proving that inequality (2.2.1) is true for all graphs with girth g ≥ 11 and minimum degree δ ≥ 2. We shall discuss this in detail in Chapter4.

It is a well known fact that Z(Kn) = n − 1 and thus we have

Observation 1. Let G be a graph with non-empty components and order n. Then,

Z(G) ≤ n − 2,

whenever G 6= Kn.

Theorem 2.2.19 ([5]). For any graph G with maximum degree ∆ ≥ 1,

n∆ Z(G) ≤ . ∆ + 1

Theorem 2.2.20 ([5]). If G is connected with maximum degree ∆ ≥ 2, then

(∆ − 2)n + 2 Z(G) ≤ . ∆ − 1 2.2. ZERO FORCING IN GRAPHS 19

(a) Graph G1 (b) Graph G2

Figure 2.3. Graphs G1 and G2 speciﬁed in Theorem 2.2.21

It was shown that the only extremal graph satisfying the inequality in Theorem 2.2.19 is the complete graph K∆+1 of order ∆ + 1 [42] and that the only extremal graphs satisfying inequality in Theorem 2.2.20 are Cn, Kn and K∆,∆ [42, 73]

2 The authors in [41] have shown that the additive term in the inequality of ∆−1 in Theo- rem 2.2.20 could be done away with.

Theorem 2.2.21 ([41]). If G is a connected graph of order n and maximum degree ∆ ≥ 3, then (∆ − 2)n Z(G) ≤ ∆ − 1 if and only if G/∈ {K∆+1,K∆,∆,K∆−1,∆,G1,G2} where G1 and G2 are the two speciﬁc graphs illustrated in Figure 2.3.

The authors in [41] have been able to improve the upper bound stated in Theorem 2.2.20 by some lower order term for sub cubic graphs of girth at least 5.

Theorem 2.2.22 ([41]). If G is a connected graph of order n, maximum degree ∆ ≥ 3, and girth g ≥ 5, then n n Z(G) ≤ − + 2. 2 24 log2(n) + 6

Some upper bounds are obtained for special classes of graphs. In [34], zero forcing in line graphs were studied in which the authors introduced an edge equivalent notion of zero forcing, edge zero forcing. They made an observation that each edge zero forcing set of G corresponds to a zero forcing set of line graph of G, L(G). Using this observation, they proved the following theorem. 20 2. BACKGROUND THEORY

Theorem 2.2.23 ([34]). For a connected graph G,

Z(G) ≤ 2 Z(L(G)).

A connected graph G is called a cycle-tree if it consist of q vertex disjoint cycles that are connected by q − 1 edges. Thus, a cycle-tree with n vertices will have m = n + q − 1 edges and each edge between two cycles is a cut-edge. In [5], they obtained the following results.

Theorem 2.2.24 ([5]). Let G be a cycle-tree with q ≥ 1 cycles. Then,

Z(G) ≤ 2q.

Theorem 2.2.25 ([5]). Let G be a Hamiltonian graph with t ≥ 1 chords and n ≥ 4. Then,

Z(G) ≤ t + 1.

In general, the study of the zero forcing number is challenging for many reasons. First, it is diﬃcult to compute exactly, as it is NP-hard [20]. Further, many of the known bounds leave a wide gap for graphs in general. For example, given the minimum and maximum degree of a graph, δ and ∆ respectively, the zero forcing number on a graph with n vertices can be as low

n∆ as δ (Theorem 2.2.3) and as high as ∆+1 (Theorem 2.2.19). It is thus important to devise a good lower bound technique that would enable us to compute zero forcing number of graphs.

We have made our attempts towards this end in Chapter5.

The next couple of results examine what happens to the zero forcing number under certain graph operations.

Theorem 2.2.26 ([32]). If G − e is the graph obtained from G by deleting an edge e = uv, then

Z(G) − 1 ≤ Z(G − e) ≤ Z(G) + 1.

Theorem 2.2.27 ([32]). If G/e is the graph obtained from G by contracting an edge e = uv, then

Z(G) − 1 ≤ Z(G/e) ≤ Z(G) + 1. 2.2. ZERO FORCING IN GRAPHS 21

Theorem 2.2.28 ([32]). If Ge is the graph obtained from G by subdividing an edge e = uv, then

Z(G) ≤ Z(Ge) ≤ Z(G) + 1.

It is evident that if the subdivision is carried out on an edge incident to degree 1 or 2 vertex, then there will be no change in the zero forcing number.

Theorem 2.2.29 ([32]). If G − v is the graph obtained from G by deleting a vertex v, then

Z(G − v) − 1 ≤ Z(G) ≤ Z(G − v) + 1.

2.2.2. Zero forcing in directed graphs. This notion was extended to digraphs in [9] with the same motivation i.e., to bound the minimum rank of a digraph. Recall that a simple digraph is a pair G~ = (V,A) where A ⊆ (V × V ) \{(i, i): i ∈ V }, and a loop digraph is a pair G~ = (V,A) where A ⊆ V × V . In fact, the deﬁnition of zero forcing in directed graphs is identical to deﬁnition in undirected graphs except for the colour-change rule. For a red/blue colouring of the vertex set of a digraph G~ consider the following colour-change rule:

(1) For a simple digraph G~ , a red vertex w is converted to blue if it is the only red

out-neighbour of some blue vertex u.

(2) For a loop digraph G~ , a red vertex w is converted to blue if it is the only red out-

neighbour of some vertex u.

We say u forces w and denote this by u → w. A vertex set S ⊆ V is called zero forcing if, starting with the vertices in S blue and the vertices in the complement V \ S red, all the vertices can be converted to blue by repeatedly applying the colour-change rule. The minimum cardinality of a zero forcing set for the digraph G is called the zero forcing number of G~ , denoted by Z(G~ ).

Remark 2.2.30. We note that Z(G) and Z(G~ ) is used to denote the zero forcing number of an undirected and a directed graph respectively.

Note that there is a diﬀerence in the forcing process in simple and loop directed graphs.

While in simple digraphs, only a blue vertex can force a colour change, in the loop digraphs, a red or a blue vertex can force a colour change. 22 2. BACKGROUND THEORY

The motivation of deﬁning this way comes from bringing about a relation between zero forcing in directed graphs and minimum rank in directed graphs.

For a digraph G~ = (V,A) of order n, the qualitative class of G~ is the set of matrices   n×n ~ {X ∈ R : i 6= j =⇒ (Xi,j 6= 0 ⇐⇒ (i, j) ∈ A)} if G does not have any loops, Q(G~ ) =  n×n ~ {X ∈ R : Xi,j 6= 0 ⇐⇒ (i, j) ∈ A} if G has at least one loop.

Note that in the family of matrices described by a simple digraph, the diagonal entries of the matrix are free, whereas in the family of matrices described by a loop digraph, the the absence or presence of loops in the digraph describes the zero-nonzero pattern of the diagonal entries of the matrix. The asymmetric minimum rank problem for a digraph asks us to determine the minimum rank among all real matrices whose zero-nonzero pattern of entries is described by a given digraph. The maximum nullity of G~ is M(G~ ) = max{corank(X): X ∈ Q(G~ )}, and the minimum rank of G is mr(G~ ) = min{rank(X): X ∈ Q(G~ )}; clearly M(G~ ) + mr(G~ ) = |V (G~ )|.

The concept of zero forcing models the process to force zeros in a null vector of a matrix

X ∈ Q(G~ ), implying M(G~ ) ≤ Z(G~ )[58].

Just as it is possible for the maximum nullity of a digraph to be zero, it is possible for the empty set to be a zero forcing set for a digraph (note that both of these are impossible for an undirected graph). For example, in Figure 2.4, the digraph G~ = (V,A) with V = {1, 2} and A = {(1, 2), (2, 1)} can be a simple digraph as well as a loop digraph, and its zero forcing number depends on this: As a loop digraph its zero forcing number is 0, and as a simple digraph its zero forcing number is 1. We have      0 m12    ~ {  : m12, m21 6= 0} if we interpret G as a loop digraph,     m21 0 Q(G~ ) =     m11 m12    ~ {  : m12, m21 6= 0} if we interpret G as a simple digraph.     m21 m22

Clearly, the maximum nullity is 0 for the loop digraph and 1 for the simple digraph and the bound M(G~ ) ≤ Z(G~ ) remains valid in all cases. 2.2. ZERO FORCING IN GRAPHS 23

1 2

Figure 2.4. A directed graph with empty set as the zero forcing set

Not much has been studied on the zero forcing set in the directed case. The only known results are listed below. The analogous results of Theorem 2.2.1 and Theorem 2.2.2 holds for directed version of the problem as well. The minimum rank problem or maximum nullity problem has been studied over ﬁelds F other than the real numbers.

Theorem 2.2.31 ([9]). For any digraph G~ ,

Z(G~ ) ≥ MF (G~ ).

Theorem 2.2.32 ([9]). For any digraph G~ ,

Z(G~ ) ≥ P(G~ ).

It is also shown in [9] that for any directed tree T , Z(T ) = P(T ).

2.2.3. Variants of zero forcing problem. In this section, we give an overview of the many variants of zero forcing that have been studied but that are beyond the scope of this thesis.

Propagation time [60]. Propagation time of a graph is introduced as the number of steps it takes for a zero forcing set to turn the entire graph blue. Propagation time of a zero forcing set was implicit in [16] and explicit in [82]. Chilakamarri et al. [21] determined the propagation time, which they call the iteration index, for a number of families of graphs including Cartesian products and various grid graphs. Control of an entire network by sequential operations on a subset of particles is valuable [82] and the number of steps needed to obtain this control

(propagation time) is a signiﬁcant part of the process was the main motivation towards the study of this topic.

Positive semideﬁnite zero forcing set [7]. The positive semideﬁnite colour change rule is: Let B be the set consisting of all the blue vertices. Let W1,...,Wk be the sets of vertices of the k components of G \ B (note that it is possible that k = 1). Let w ∈ Wi. If u ∈ B 24 2. BACKGROUND THEORY

and w is the only red neighbour of u in G[Wi ∪ B], then change the colour of w to blue. The positive semidefinite zero forcing number of an undirected graph G, denoted by Z+(G), is the minimum of over all positive semidefinite zero forcing sets in G (using the positive semidefinite colour change rule).

Theorem 2.2.33 ([7]). Since any zero forcing set is a positive deﬁnite zero forcing set

Z+(G) ≤ Z(G).

k-forcing set [5]. Given a simple undirected graph G and a positive integer k, the k-forcing number of G, denoted Fk(G), is the minimum number of vertices that need to be initially coloured blue in a red/blue colouring of the vertex set of the graph G so that all vertices eventually become blue during the discrete dynamical process described by the following rule.

Starting from an initial set of blue coloured vertices and stopping when all vertices are coloured blue: if a blue coloured vertex has at most k red neighbours, then each of its red neighbours becomes blue. The minimum cardinality of this set is called k-forcing number of G. When k = 1, the problem reduces to zero forcing problem.

2.3. Power domination in graphs

Power domination was introduced by Baldwin et al. in [6], then described as a graph theoretical problem by Haynes et al. in [55]. The problem is motivated by the requirement for constant monitoring of power systems. Electric power companies need to continually monitor their systems state as deﬁned by a set of state variables (for example, the voltage magnitude at loads and the machine phase angle at generators). One method of monitoring these variables is to place phase measurement units (PMUs) at selected locations in the system. Because of the high cost of a PMU, it is desirable to minimize their number while maintaining the ability to monitor (observe) the entire system. A system is said to be observed if all of the state variables of the system can be determined from a set of measurements (e.g., voltages and currents). 2.3. POWER DOMINATION IN GRAPHS 25

Let G = (V,E) be a graph representing an electric power system, where a vertex represents an electrical node (a substation bus where transmission lines, loads, and generators are connected) and an edge represents a transmission line joining two electrical nodes. The problem is to locate the minimum number of PMUs.

The next question is then to ask: How does a PMU measure the state variables? A PMU measures the state variable (voltage and phase angle) for the vertex at which it is placed and its incident edges and their end vertices. (These vertices and edges are said to be observed.)

The other observation rules are as follows:

(1) Any vertex that is incident to an observed edge is observed.

(2) Any edge joining two observed vertices is observed.

(3) If a vertex is incident to a total of k > 1 edges and if k −1 of these edges are observed,

then all k of these edges are observed.

It was noticed in [29, 31] that the power domination problem can be studied considering only vertices.

Let G be a connected graph and S a subset of its vertices. Then we denote the set observed by S with X(S) and deﬁne it recursively as follows:

(1) (domination)

X(S) ← S ∪ N(S)

(2) (propagation)

As long as there exists v ∈ X(S) such that

N(v) ∩ (V (G) − X(S)) = {w}

set X(S) ← X(S) ∪ {w}

A set S is called a power dominating set of G if X(S) = V (G). The power domination number

γp(G) is the minimum cardinality of a power dominating set of G. A power dominating set of

G with the minimum cardinality is called a γp(G)-set. The ﬁrst step in the power domination process can be thought of as domination process.

A dominating set of a graph G(V,E) is a set S of vertices of G such that S ∪ N(S) = V (G) and the minimum cardinality of such a set is called domination number denoted γ(G). The second step of the power domination process is called the propagation process and is similar to 26 2. BACKGROUND THEORY the zero forcing process described in Section 2.2 on page 11. Thus, power domination is closely related to domination problem and zero forcing problem.

5 6

4 3

1 2

0 7

Figure 2.5. A graph G with power dominating set S = {5, 6}

In Figure 2.5, the set S = {5, 6} is clearly a power dominating set as exhibited in Figure 2.6.

At ﬁrst time step (Figure 2.6a), the neighbours of 5 and 6 are observed. In Figure 2.6b, vertex

4 observes vertex 1 and in Figure 2.6c, vertex 3 observes vertex 2. In Figure 2.5, blue vertices denote observed vertices and red vertices are vertices that are not observed.

The problem of deciding if a graph G has a power dominating set of cardinality k has been shown to be NP-complete even for bipartite graphs, chordal graphs [55] or even split graphs [72].

The power domination problem has eﬃcient polynomial time algorithms for the classes of trees [55], graphs with bounded tree-width [51], block graphs [89], block cactus graphs [61], interval graphs [72], and circular-arc graphs [71].

2.3.1. Bounds on the power domination number.

Theorem 2.3.1 ([55]). For any graph G,

1 ≤ γp(G) ≤ γ(G).

Also γp(G) = 1 for G ∈ {Kn,Cn,Pn,K2,k}.

It is evident that any dominating set is trivially a power dominating set and thus we have the trivial upper bound yet the authors proved that there is no forbidden subgraph characterization of the graphs reaching this bound.

The next result allows us to restrict vertices that belong to any power dominating set. 2.3. POWER DOMINATION IN GRAPHS 27

5 6 5 6

4 3 4 3

1 2 1 2

0 7 0 7

(a) First time step (b) Second time step 5 6

4 3

1 2

0 7

Figure 2.6. Exhibition of the power dominating steps

Theorem 2.3.2 ([55]). Let G be a graph with ∆(G) ≥ 3. Then there is a minimum power dominating set S in which each vertex in S has degree at least 3.

In [55], the authors studied extensively the power dominating set in trees. For a tree T , deﬁne the spider number of T , denoted sp(T ), to be the minimum number of subsets V (T ) can be partioned so that each subset induces a graph homeomorphic to K1,k for some k ∈ N.

Theorem 2.3.3 ([55]). For any tree T , sp(T ) = γp(G).

The following sharp upper bound for the power domination number of a graph was initially shown for trees in [55] and was later generalized to all graphs in [91].

Theorem 2.3.4 ([91]). Let T be the family of graphs obtained from connected graphs H by adding two new vertices v0 and v00 to each vertex v of H and new edge vv0 and vv00, while v0v00 28 2. BACKGROUND THEORY

n may be added or not. If G is a connected graph of order n ≥ 3, then γp(G) ≤ 3 with equality if and only if G ∈ T ∪ {K3,3}.

Zhao and Kang addressed the question whether one can bound the power domination number of a graph with some condition on the diameter, that is the maximum distance between two vertices of the graph. In [90], they gave some general bounds for the power domination number of planar graphs with diameter 2 or 3. They showed that in outerplanar graphs, if the diameter is at most 2, then the graph admits a power dominating set of size one, while if the diameter is 4 or more, the power domination number can be arbitrarily large.

It was shown in [91] by Zhao, Kang and Chang that the bound could be improved for regular graphs.

Theorem 2.3.5 ([91]). If G is a connected claw-free cubic graph on n vertices, then

n γ (G) ≤ . p 4

Zhao et al. in the same paper also characterized the graphs for which the bound is tight which can be described as follows: take an even cycle and replace every second edge by a K4 minus an edge, using the degree 2 vertices of K4 − e as end vertices of the edge. In [27], the result obtained in Theorem 2.3.5 was improved by removing the claw free condition.

Theorem 2.3.6 ([27]). If G is a connected cubic graph on n vertices, then either G = K3,3

n or γp(G) ≤ 4 .

The above result in Theorem 2.3.6 motivated the authors to pose the following conjecture which is still open to be resolved.

Conjecture 1 ([27]). For r ≥ 3, if G 6= Kr,r is a connected r-regular graph on n vertices, then n γ (G) ≤ . p r + 1

Another area where the power domination of graphs was extensively studied was on graph products. We shall highlight some of the known results in literature. 2.3. POWER DOMINATION IN GRAPHS 29

Theorem 2.3.7 ([31]). The power domination number of s × t grid Ps × Pt for s ≥ t ≥ 1 is   s+1 d 4 e if t ≡ 4 (mod 8), γp(Ps × Pt) =  s d 4 e otherwise.

This result is quite surprising in some sense because domination problem on grid graphs is still an open problem. Motivated by this result, Ferrero et al. [39] extended the results of grid networks to hexagonal honeycomb grid graph and triangular grid graph.

The power domination numbers for cylinders Ps × Ct for integers s ≥ 2, t ≥ 3, tori Cs × Ct for integers s, t ≥ 3 were studied in [10] and tight bounds were found.

Theorem 2.3.8 ([10]). The power domination number of s × t cylinder Ps × Ct for s ≥ 2, t ≥ 3 is   t+1 s+1 min d 4 e, d 2 e if s ≡ 4 (mod 8), γp(Ps × Ct) ≤  t s+1 min d 4 e, d 2 e otherwise.

Theorem 2.3.9 ([10]). The power domination number of s × t torus Cs × Ct for s ≤ t is   s d 2 e if s ≡ 0 (mod 4), γp(Cs × Ct) ≤  s+1 d 2 e otherwise.

Recently, Benson et al. [11] showed that the upper bound obtained for torus in [10] is tight.

Theorem 2.3.10 ([11]). For t ≥ s ≥ 3,

l s m γ (C × C ) = . p s t 2

In [29], this study on the products of paths is continued with other graph products. For the direct product (which has two connected components), the bound obtained is as follows:

Theorem 2.3.11 ([29]). The power domination number of the direct product Ps · Pt for t ≥ s ≥ 1 is   s 2d 4 e if s is even, γp(Ps · Pt) ≤  t 2d 4 e if s is odd and t even, 30 2. BACKGROUND THEORY

If both s and t are odd,

l t m lt − 2m lt + sm lt + s − 2m γ (P · P ) ≤ max + , + . p s t 4 4 6 6

Theorem 2.3.12 ([88]). The power domination number of the direct product Cs · Kt for t ≥ 3, s ≥ 4 is   2k if s = 4k,   γp(Cs · Kt) ≤ 2k + 1 if s = 4k + 1,    2k + 2 if s = 4k + 2 or n = 4k + 3.

Theorem 2.3.13 ([88]). The power domination number of the direct product Ps ·Ct, s even, is   s 2d 3 e if t is even, γp(Ps · Ct) =  s d 3 e if t is odd. The situation for the strong product of two paths is a little simpler, though not completely solved either.

Theorem 2.3.14 ([29]). The power domination number of the direct product Ps ⊗ Pt for t ≥ s ≥ 2 is l t m ls + 1m γ (P ⊗ P ) = min , p s t 4 2 unless 3s − t − 6 ≡ 4 (mod 8) in which case

l t m lt + s − 2m l t m lt + s − 2m max , ≤ γ (P ⊗ P ) ≤ max , + 1 . 3 4 p s t 3 4

Dean et al. [25] gave a lower bound and upper bound for the power domination number in hypercubes.

Theorem 2.3.15 ([25]). For the r-dimensional hypercube Qr,

2r−1 ≤ γ (Q ) ≤ 2r−dlog re−1. r p r

The exact values of the power domination for hypercube network Qr has been known only for values of r ≤ 7 as shown in [78]. This is an indication of how diﬃcult it is ﬁnding the exact values of power domination number of graphs. 2.3. POWER DOMINATION IN GRAPHS 31

Power domination problem has been studied eﬀectively for Sierpi´nskigraphs [28], generalized Petersen graphs [10], undirected de Bruijn and Kautz graphs [69] and Mycielskian of graphs [88].

The next couple of results examine what happens to the power domination number under certain graph operations. The results below are obtained independently by Benson et al. [11] and Dorbec et al. [30].

Theorem 2.3.16 ([11, 30]). For every graph G and for every vertex v,

γp(G) − 1 ≤ γp(G − v).

There is no upper bound for γp(G−v) in terms of γp(G). In fact, the authors in [11] proved the following proposition to justify the claim.

Proposition 1 ([11]). For every integer r ≥ −1, there is a graph Gr with a vertex v such that

γp(Gr − v) = γp(Gr) + r.

Theorem 2.3.17 ([11, 30]). If G − e is the graph obtained from G by deleting an edge e = uv, then

γp(G) − 1 ≤ γp(G − e) ≤ γp(G) + 1.

Theorem 2.3.18 ([11, 30]). If G/e is the graph obtained from G by contracting an edge e = uv, then

γp(G) − 1 ≤ γp(G/e) ≤ γp(G) + 1.

Theorem 2.3.19 ([11, 30]). If Ge is the graph obtained from G by subdividing an edge e = uv, then

γp(G) ≤ γp(Ge) ≤ γp(G) + 1.

2.3.2. Power domination in directed graphs. The directed version of the power domination problem was initiated in [1]. Following the deﬁnition in [37], we deﬁne a set S ⊆ V is a power dominating set of G~ if and only if N out[S] is a zero forcing set of G~ .A minimum 32 2. BACKGROUND THEORY power dominating set is a power dominating set of minimum cardinality. The power dominating number of G is the cardinality of a minimum power dominating set and is denoted by γp(G~ ).

1− In [1], the authors show hardness of approximation threshold of 2log n for directed acyclic graphs. They also show that the power domination problem for directed graphs can be solved optimally in linear time if the underlying undirected graph has bounded tree-width.

Theorem 2.3.20 ([1]). Given a directed graph G~ and a tree decomposition of width k of its underlying undirected graph, power dominating set problem for directed graphs can be optimally solved in O((ck)2.n) time for a global constant c.

Remark 2.3.21. We note that γp(G) and γp(G~ ) is used to denote the power domination number of an undirected and a directed graph respectively.

2.3.3. Variants of power domination problem. In this section, we give an overview of the variants of power domination problem that have been studied and are beyond the scope of this thesis.

k-power domination. This notion of k-power domination was introduced by Chang et al. [18] which generalizes the concept of power domination. Let G be a connected graph and

i S a subset of its vertices. Then we denote the set monitored by S at step i as PG,k(S) and deﬁne it recursively as follows:

0 (1) PG,k(S) = N[S] i+1 S i i (2) PG,k(S) = {N[v]: v ∈ PG,k(S)} such that |N[v] \ PG,k(S)| ≤ k

i i+1 i0 i0+1 Note that PG,k(S) ⊆ PG,k(S) ⊆ V (G) for any i. If PG,k(S) = PG,k (S) for some i0, then

j i0 ∞ i0 PG,k(S) = PG,k(S) for any j ≥ i0. Deﬁne PG,k(S) = PG,k(S). ∞ A set S such that PG,k(S) = V (G) is a k-power dominating set of G. The least cardinality of such a set is called the k-power domination number of G, written γp,k(G). A γp,k(G)-set is a minimum k-power dominating set of G.

It is to be noted that when k = 0, the k-power domination problem reduces to the domination problem and when k = 1, the problem reduces to the power domination problem. This is a beautiful generalisation which brings both domination problem and power domination problem under a single roof. Results on k-power domination problem may be found in [27, 28, 30]. 2.4. ZERO FORCING AND POWER DOMINATION 33

Propagation radius. Propagation radius was introduced in [28] as a measure of the eﬃciency of power dominating sets and is deﬁned for k-power dominating set as follows:

i radp,k(G) = 1 + min{i : PG(S) = V (G),S is a k-power dominating set of G}.

2.4. Zero forcing and power domination

Both zero forcing problem and power domination problem are very closely related problems.

It is not until recently that a connection between these problems was established in undirected graphs.

Theorem 2.4.1 ([25]). The set S is a power dominating set if and only if N[S] is a zero forcing set.

Theorem 2.4.1 implies that every zero forcing set is a power dominating set. That is,

γp(G) ≤ Z(G). It is also known that every dominating set is also a power dominating set (Theorem 2.3.1). But there exists no relationship between zero forcing number and domination number.

Observation 2. Let G be an undirected graph. Then, the zero forcing number Z(G) is neither bounded above nor bounded below by domination number γ(G).

If G = P6, path on 6 vertices, γ(G) = 2 and Z(G) = 1 whereas if G = Kn, complete graph on n vertices, γ(G) = 1 and Z(G) = n − 1.

The next theorem due to Benson et al. [11] was deduced from Theorem 2.4.1.

Theorem 2.4.2 ([11]). Let G be a connected non trivial graph. Then

l Z(G) m ≤ γ (G) ∆(G) p and this bound is is tight.

The graph satisfying Theorem 2.4.2 is the complete graph Kn as Z(Kn) = ∆(Kn) = n − 1 and γp(G) = 1. The next result is immediate from the fact that M(G) ≤ Z(G)[3]. Although weaker than Theorem 2.4.2, it can sometimes be applied using a well known matrix such as the adjacency or Laplacian matrix of the graph, even if M(G) and Z(G) are not known. 34 2. BACKGROUND THEORY

Theorem 2.4.3 ([11]). For a graph G that has an edge and any matrix A ∈ S(R,G),

lcorank Am ≤ γ (G). ∆(G) p

As an application to Theorem 2.4.2, the authors solved the power domination of the torus described in Theorem 2.3.10. They also exhibited that many of the proofs of the values of the power domination number for families can be simpliﬁed by application of the relationship between power domination and zero forcing.

In this thesis, we also establish a relation between power domination and zero forcing and exploit it to solve these problems for de Bruijn and Kautz digraphs in Chapter3. CHAPTER 3

Zero forcing and power domination in iterated line

digraphs

In this chapter we make the following contributions.

• In Section 3.1, we interpret zero forcing and power domination in directed graphs as

set covering problems.

• In Sections 3.2 and 3.3, we solve the zero forcing and power domination problem in

de Bruijn and Kautz digraphs.

• In Sections 3.4 and 3.5, we extend the results of de Bruijn and Kautz digraphs to

obtain bounds for zero forcing number and power domination number in iterated line

digraphs respectively.

3.1. An alternate interpretation

We give an interpretation of the zero forcing problem and the power domination problem as a set cover problem. We call a vertex set W strongly critical if there is no vertex in G~ which has exactly one out neighbour in W . We call a vertex set W weakly critical if there is no vertex outside W which has exactly one out-neighbour in W . If W is strongly (weakly) critical, but no proper subset of W is strongly (weakly) critical, then we call W minimal strongly (weakly) critical.

♦ Remark 3.1.1. Every strongly critical set is a weakly critical set but the converse need not be true. As a consequence, in any digraph, the maximum number of pairwise disjoint strongly critical sets is less than or equal to the maximum number of pairwise disjoint weakly critical sets.

The contents of this chapter reproduce original results from [37] and [45].

35 36 3. ZERO FORCING AND POWER DOMINATION IN ITERATED LINE DIGRAPHS

3 4 1 2

Figure 3.1. A directed graph G~ .

In Figure 3.1, the set of vertices {1, 2} is both a strongly and weakly critical set. But the set {3, 4} is a weakly critical set but not a strongly critical set.

It is interesting to note that in Figure 3.1, vertex 5 and vertex 6 have to belong in any zero forcing and power dominating set because their in-degree is zero and cannot be forced by any other vertex but itself.

~ ~ in ♦ Remark 3.1.2. Let G be a digraph and let v be a vertex in G such that degG~ (v) = 0, then v must belong to any zero forcing and power dominating set.

Generalising this observation, we now characterise zero forcing sets of digraphs in terms of strongly and weakly critical sets. The singleton {v} is strongly critical if and only if the in-degree of v is 0. Recall that for a loop digraph, a red or a blue vertex can force a colour change.

♦ Lemma 3.1.3. A vertex set S is a zero forcing set in a loop digraph G~ if and only if S ∩ W 6= ∅ for every strongly critical set W ⊆ V , and therefore

Z(G~ ) = min {|S| : S ∩ W 6= ∅ for every strongly critical set W ⊆ V } .

Proof. Let S be a zero forcing set with S ∩ W = ∅ for some strongly critical set W ⊆ V . This implies that even if all vertices of V \ W are in S, no vertices of W can be forced by any vertex in V \ W , a contradiction. By pigeon hole principle, we conclude that at least one vertex from every strongly critical set must belong to any minimum zero forcing set. To prove the other direction, we need to show that if a set S intersects all minimal strongly critical sets, then

S is a zero forcing set. Assume the counter example. Since S is not a zero forcing set, not at 3.1. AN ALTERNATE INTERPRETATION 37 all vertices of G~ are turned blue after the last forcing step possible. This implies that all blue vertices in G~ after the last forcing step are adjacent to at least two red vertices or no red vertex.

Thus, the collection of all red vertices will form a strongly critical set, a contradiction.

For simple digraphs, we recall that only a blue vertex can force a colour change and following the same argument of Lemma 3.1.3, we have

♦ Lemma 3.1.4. A vertex set S is a zero forcing set in a simple digraph G~ if and only if S ∩ W 6= ∅ for every weakly critical set W ⊆ V , and therefore

Z(G~ ) = min {|S| : S ∩ W 6= ∅ for every weakly critical set W ⊆ V } .

♦ Corollary 3.1.5. In any digraph G~ = (V,A), its zero forcing number Z(G~ ) is at least the maximum number of disjoint strongly critical sets in G~ . Moreover, if G~ does not have any loops, then Z(G~ ) is at least the maximum number of disjoint weakly critical sets in G~ .

Proof. Let S be a minimum zero forcing set of G~ and let {W1,...,Wr} be a set of pairwise disjoint strongly critical sets. By Lemma 3.1.3, |S ∩ Wi| ≥ 1 for every i = 1, . . . , r, and since the sets W1,...,Wr are pairwise disjoint, it must be |S| ≥ r. If G~ does not have any loops we apply the same argument with a collection of pairwise disjoint weakly critical sets.

A set S ⊆ V is a power dominating set of G~ if and only if N out[S] is a zero forcing set of

G~ . Thus, one can characterise power dominating sets as follows.

♦ Lemma 3.1.6. A vertex set S is a power dominating set in a simple (loop) digraph G~ if and only if N out[S] ∩ W 6= ∅ for every weakly (strongly) critical set W ⊆ V , and therefore G~

n o γ (G~ ) = min |S| :(S ∪ N out(S)) ∩ W 6= ∅ for every weakly (strongly) critical set W ⊆ V . p G~

In the next section, we apply this new interpretation in solving the zero forcing and power domination of de Bruijn and Kautz digraphs. Due to their attractive connectivity features these digraphs have been widely studied as a topology for interconnection networks [62], and some generalizations of these digraphs were proposed [64]. Recently, Dong et al. (2015) [26] investigated the domination number of generalized de Bruijn and Kautz digraphs. Kuo et al. (2015) [69] gave an upper bound for power domination in undirected de Bruijn and Kautz graphs. This 38 3. ZERO FORCING AND POWER DOMINATION IN ITERATED LINE DIGRAPHS motivated us to study the directed versions, i.e., the zero forcing number and power domination number of de Bruijn and Kautz digraphs.

3.2. Zero forcing and power domination in de Bruijn digraphs

For an integer d ≥ 2, let Zd = {0, 1, . . . , d − 1} denote the cyclic group of order d. The de Bruijn digraph, denoted B(d, `), with parameters d ≥ 2 and ` ≥ 2 is deﬁned to be the graph

G~ = (V,A) with vertex set V and arcs set A where

` V = Zd = {(a1, . . . , a`): ai ∈ Zd for i = 1, . . . , `} ,

A = {((a1, a2, . . . , a`), (a2, . . . , a`, b)) : (a1, a2, . . . , a`) ∈ V, b ∈ Zd} .

Our main result in this section is the following theorem.

♦ Theorem 3.2.1. Let G~ be a de Bruijn digraph with parameters d, ` ≥ 2. Then the zero forcing number and power domination number of G~ are (d−1)d`−1 and (d−1)d`−2, respectively.

Let us deﬁne the sets

X(a1, . . . , a`−1) = {(a1, . . . , a`−1, α): α ∈ Zd}

`−1 out which partition the vertex set V into d sets of size d. Furthermore, N (v) = X(a1, . . . , a`−1) for every vertex v of the form (α, a1, a2, . . . , a`−1).

♦ Lemma 3.2.2. Let G~ be a de Bruijn digraph with parameters d and `. Then Z(G~ ) ≥ (d − 1)d`−1.

Proof. Every 2-element subset of each of the sets X(a1, . . . , a`−1) is strongly critical, and therefore, any zero forcing set S needs to intersect X(a1, . . . , a`−1) in at least d − 1 elements, and the result follows.

♦ Lemma 3.2.3. Let G~ be a de Bruijn digraph with parameters d and `. Then Z(G~ ) ≤ (d − 1)d`−1.

Proof. Consider the vertex set S = {(a1, . . . , a`−1, a`) ∈ V : a1 6= a`}. To show that S is a zero forcing set, it is suﬃcient to verify that each vertex v = (a1, . . . , a`−1, a`) is either in S 3.2. ZERO FORCING AND POWER DOMINATION IN DE BRUIJN DIGRAPHS 39

or is the unique out-neighbour in V \ S for some vertex w. If a1 6= a`, then v ∈ S. If a1 = a`, then for any vertex of the form w = (β, a1, . . . , a`−1), v is the only neighbour of w in V \ S.

Lemmas 3.2.2 and 3.2.3 imply the ﬁrst statement of Theorem 3.2.1. In order to prove the second part of this theorem we recall that S ⊆ V is a power dominating set if and only if S ∪ N out(S) is a zero forcing set. In particular, it is necessary that |(S ∪ N out(S)) ∩

`−1 X(a1, . . . , a`−1)| > d − 1 for every (a1, . . . , a`−1) ∈ Zd .

♦ Lemma 3.2.4. Let G~ be a de Bruijn graph with parameters d and `. Then every power dominating set has size at least (d − 1)d`−2.

Proof. Let S be a power-dominating set, suppose |S| < (d − 1)d`−2 and set Z = S ∪ N out(S). We have

(Z \ S) ∩ X(a1, . . . , a`−1) 6= ∅ =⇒ X(a1, . . . , a`−1) ⊆ Z.

For k = 0, 1, . . . , d, we set αk = |{(a1, . . . , a`−1): |S ∩ X(a1, . . . , a`−1)| = k}|, and get

|S| = α1 + 2α2 + ··· + (d − 1)αd−1 + dαd.

Now let I0 = {(a1, . . . , a`−1): X(a1, . . . , a`−1) ⊆ Z}. Then

|I0| 6 |S| + αd = α1 + 2α2 + ··· + (d − 1)αd−1 + (d + 1)αd.

For (a1, . . . , a`−1) ∈/ I0 we must have |Z ∩ X(a1, . . . , a`−1)| = d − 1, and this implies that

`−1 |S ∩ X(a1, . . . , a`−1)| = d − 1. We conclude |I0| + αd−1 > d . Therefore

`−1 α1 + 2α2 + ··· + (d − 2)αd−2 + dαd−1 + (d + 1)αd > d , and together with |S| < (d − 1)d`−2 this yields

`−1 `−2 `−2 αd−1 + αd > d − (d − 1)d = d .

`−2 But then |S| > (d − 1)(αd−1 + αd) > (d − 1)d , which is the required contradiction. 40 3. ZERO FORCING AND POWER DOMINATION IN ITERATED LINE DIGRAPHS

We deﬁne a set S ⊆ V by   {(0, 1), (0, 2),..., (0, d − 1)} if ` = 2,   (3.2.1) S = {(a1, a2, a3) ∈ V : a2 = a1, a3 6= a1} if ` = 3,    {(a1, . . . , a`) ∈ V : a`−1 = a1 + a`−2, a` 6= a1 + a2 + a`−2} if ` > 4.

Note that |S| = (d − 1)d`−2. The construction of the set S deﬁned in (3.2.1) can be visualized

2 `−2 by arranging the vertices of G~ in a d ×d -array where the rows are indexed by pairs (a`−1, a`) and the columns are indexed by (` − 2)-tuples (a1, . . . , a`−2). Then column (a1, . . . , a`−2) is the the union of the d sets X(a1, . . . , a`−2, a`−1) over a`−1 ∈ Zd, and the set S contains d − 1 elements from each column. More precisely, the intersection of S with column (a1, . . . , a`−2) is

X(a1, . . . , a`−2, a1 + a`−2) \{(a1, . . . , a`−2, a1 + a`−2, a1 + a2 + a`−2)}.

In Figure 3.2 this is illustrated for two columns with d = 5 and ` = 7.

(1, 3, 4, 4, 2) (3, 1, 0, 2, 4)

X(1, 3, 4, 4, 2, 0) X(3, 1, 0, 2, 4, 0)

a6 = 0 = N out(3, 1, 3, 4, 2, 2, 0) = N out(2, 3, 1, 0, 2, 4, 0)

X(1, 3, 4, 4, 2, 1) X(3, 1, 0, 2, 4, 1)

a6 = 1 = N out(3, 1, 3, 4, 2, 2, 1) = N out(2, 3, 1, 0, 2, 4, 1)

X(1, 3, 4, 4, 2, 2)

a6 = 2 = N out(3, 1, 3, 4, 2, 2, 2)

X(3, 1, 0, 2, 4, 3)

a6 = 3 = N out(2, 3, 1, 0, 2, 4, 3)

X(1, 3, 4, 4, 2, 4) X(3, 1, 0, 2, 4, 4)

a6 = 4 = N out(3, 1, 3, 4, 2, 2, 4) = N out(2, 3, 1, 0, 2, 4, 4)

Figure 3.2. Illustration of the construction of the power dominating set S for d = 5 and ` = 7. For the two columns (a1, . . . , a5) = (1, 3, 4, 4, 2) and (a1, . . . , a5) = (3, 1, 0, 2, 4) we show the elements of S (black squares), and we indicate for the sets X(a1, . . . , a6) (enclosed by rectangles) the elements of S having them as their out-neighbourhood. 3.3. ZERO FORCING AND POWER DOMINATION NUMBER IN KAUTZ DIGRAPHS 41

♦ Lemma 3.2.5. The set S deﬁned in (3.2.1) is a power dominating set for G~ = B(d, `).

out Proof. For Z = S ∪ N (S) it is suﬃcient to show that |Z ∩ X(a1, . . . , a`−1)| > d − 1 for every (a1, . . . , a`−1). We provide the full argument for ` > 4 (the cases ` = 2 and ` = 3 are easy to check).

Case 1.: If a`−1 = a1 + a`−2, then by (3.2.1),

S ∩ X(a1, . . . , a`−1) = {(a1, . . . , a`): a` ∈ Zd \{a1 + a2 + a`−2}},

hence |Z ∩ X(a1, . . . , a`−1)| ≥ |S ∩ X(a1, . . . , a`−1)| = d − 1.

Case 2.: If a`−1 6= a1 + a`−2, then X(a1, . . . , a`−1) ⊆ Z because

out X(a1, . . . , a`−1) = N ((a`−2 − a`−3, a1, a2, . . . , a`−1))

and (a`−2 − a`−3, a1, a2, . . . , a`−1) ∈ S.

The second part of Theorem 3.2.1 follows from Lemmas 3.2.4 and 3.2.5.

3.3. Zero forcing and power domination number in Kautz digraphs

The Kautz digraph, denoted K(d, `), with parameters d ≥ 2 and ` ≥ 2 is deﬁned to be the graph G~ = (V,A) with vertex set V and arcs set A where

V = {(a1, . . . , a`): ai ∈ Zd+1, ai 6= ai+1}

A = {((a1, a2, . . . , a`), (a2, . . . , a`, b)) : (a1, a2, . . . , a`) ∈ V, b ∈ Zd+1 \{a`}} .

Our main result in this section is the following theorem.

♦ Theorem 3.3.1. Let G~ be a Kautz digraph with parameters d ≥ 2 and ` ≥ 3. Then, the zero forcing number and power domination number of G~ are (d − 1)(d + 1)d`−2 and (d −

1)(d + 1)d`−3, respectively.

Let us deﬁne the sets

X(a1, . . . , a`−1) = {(a1, . . . , a`−1, a`): a` ∈ Zd+1 \{a`−1}} 42 3. ZERO FORCING AND POWER DOMINATION IN ITERATED LINE DIGRAPHS

`−1 for (a1, . . . , a`−1) ∈ Zd+1 with ai 6= ai+1 for all i. These sets partition the vertex set V into `−2 out (d + 1)d sets of size d. Furthermore, N (v) = X(a1, . . . , a`−1) for every vertex v of the form (a0, a1, a2, . . . , a`−1).

~ ~ ♦ Lemma 3.3.2. Let G be a Kautz digraph with parameters d, ` > 2. Then Z(G) ≥ (d − 1)(d + 1)d`−2.

~ ~ ♦ Lemma 3.3.3. Let G be a Kautz digraph with parameters d, ` > 2. Then Z(G) ≤ (d − 1)(d + 1)d`−2.

Proof. Consider the vertex set   {(a1, a2) ∈ V : a2 6= a1 + 1} if ` = 2, S =  {(a1, . . . , a`) ∈ V : a` 6= a`−2} if ` > 3.

We have |S| = (d − 1)(d + 1)d`−2, and to show that S is a zero forcing set, it is suﬃcient to verify that each vertex v = (a1, . . . , a`−1, a`) is either in S or is the unique out-neighbour in V \ S for some vertex w.

Case 1 : ` = 2. If a2 6= a1 + 1 then v ∈ s. If a2 = a1 + 1 then for any vertex of the form

w = (β, a1), v is the only neighbour of w in V \ S.

Case 2 : ` ≥ 3. If a` 6= a`−2, then v ∈ S. If a` = a`−2, then for any vertex of the form

w = (β, a1, . . . , a`−1), v is the only neighbour of w in V \ S.

Lemmas 3.3.2 and 3.3.3 imply the ﬁrst statement of Theorem 3.3.1.

♦ Lemma 3.3.4. Let G~ be a Kautz digraph with parameters d ≥ 2 and ` ≥ 3. Then, every power dominating set has size at least (d − 1)(d + 1)d`−3.

Proof. Let S be a power-dominating set, suppose |S| < (d − 1)(d + 1)d`−3 and set Z = S ∪ N out(S). We have

(Z \ S) ∩ X(a1, . . . , a`−1) 6= ∅ =⇒ X(a1, . . . , a`−1) ⊆ Z. 3.3. ZERO FORCING AND POWER DOMINATION NUMBER IN KAUTZ DIGRAPHS 43

For k = 0, 1, . . . , d, we set αk = |{(a1, . . . , a`−1): |S ∩ X(a1, . . . , a`−1)| = k}|, and get

|S| = α1 + 2α2 + ··· + (d − 1)αd−1 + dαd.

Now let I0 = {(a1, . . . , a`−1): X(a1, . . . , a`−1) ⊆ Z}. Clearly,

|I0| 6 |S| + αd = α1 + 2α2 + ··· + (d − 1)αd−1 + (d + 1)αd.

For (a1, . . . , a`−1) ∈/ I0 we must have |Z ∩ X(a1, . . . , a`−1)| = d − 1 because Z intersects every weakly critical set. This implies that |S ∩ X(a1, . . . , a`−1)| = d − 1, and we conclude

`−2 |I0| + αd−1 > (d + 1)d . Therefore

`−2 α1 + 2α2 + ··· + (d − 2)αd−2 + dαd−1 + (d + 1)αd > (d + 1)d , and together with |S| < (d − 1)(d + 1)d`−3 this yields

`−2 `−3 `−3 αd−1 + αd > (d + 1)d − (d − 1)(d + 1)d = (d + 1)d .

`−3 But then |S| > (d−1)(αd−1 +αd) > (d−1)(d+ 1)d , which is the required contradiction.

We deﬁne a set S ⊆ V by

(3.3.1)   {(0, 1), (0, 2),..., (0, d)} if ` = 2,    {(a1, a2, a3) ∈ V : a2 = a1 + 1, a3 6= a1 + 2} if ` = 3, S =  {(a1, a2, a3, a4) ∈ V : a3 = a1, a4 6= a2} if ` = 4,    {(a1, . . . , a`) ∈ V : ((a`−2, a`−1) = (a1, a2) ∧ an 6= a3) ∨ (a`−1 = a1 ∧ a` 6= a2)} if ` > 5.

  d if ` = 2, ♦ Lemma 3.3.5. |S| =  `−3 (d − 1)(d + 1)d if ` > 3.

Proof. For ` ≤ 4 this is easy to check. For ` ≥ 5 we proceed by the following argument.

We consider the partition S = S1 ∪ S2 where

S1 = {(a1, . . . , a`) ∈ S : a`−3 = a1},S2 = {(a1, . . . , a`) ∈ S : a`−3 6= a1}. 44 3. ZERO FORCING AND POWER DOMINATION IN ITERATED LINE DIGRAPHS

Let sk be the number of words a1 . . . ak over the alphabet Zd+1 which satisfy ak = a1 and k−2 ai 6= ai+1 for all i ∈ {1, . . . , k − 1}. Then s2 = 0 and sk = (d + 1)d − sk−1 for k ≥ 3. It

k−1 k `−3 follows by induction on k that sk = d − (−1) d. Every vector (a1, . . . , a`−3) ∈ Zd+1 with ai 6= ai+1 and a`−3 = a1 can be extended to an element of S1 by choosing a`−2 ∈ Zd+1 \{a1}, a`−1 = a1 and a` ∈ Zd+1 \{a1, a2}, hence

`−4 `−3 |S1| = s`−3d(d − 1) = d − (−1) d d(d − 1).

If a`−3 6= a1 then we can choose (a`−2, a`−1) = (a1, a2) and a` ∈ Zd+1 \{a2, a3}, or a`−2 ∈

Zd+1 \{a1, a`−3}, a`−1 = a1 and a` = Zd+1 \{a1, a2}, hence

`−4 2 |S2| = (d + 1)d − s`−3 (d − 1) + (d − 1)

= (d + 1)d`−4 − d`−4 + (−1)`−3d d(d − 1)

= d`−3 + (−1)`−3d d(d − 1).

Finally,

`−4 `−3 `−3 `−3 `−3 |S| = |S1| + |S2| = d(d − 1) d − (−1) d + d + (−1) d = (d + 1)(d − 1)d .

♦ Lemma 3.3.6. The set S deﬁned in (3.3.1) is a power dominating set for G~ = K(d, `).

out Proof. For Z = S ∪ N (S) it is suﬃcient to show that |Z ∩ X(a1, . . . , a`−1)| > d − 1 for every (a1, . . . , a`−1). We provide the full argument for ` > 5 (the cases ` = 2, ` = 3 and ` = 4 are easy to check).

Case 1.: If a`−2 = a1 and a`−1 = a2 then

|S ∩ X(a1, . . . , a`−1)| = |{(a1, . . . , a`): a` ∈ Zd+1 \{a2, a3}}| = d − 1,

and the claim follows from Z ⊇ S.

Case 2.: If a`−2 = a1 and a`−1 6= a2, then X(a1, . . . , a`−1) ⊆ Z because

out X(a1, . . . , a`−1) = N ((a`−3, a1, a2, . . . , a`−1))

and (a`−3, a1, a2, . . . , a`−1) ∈ S. 3.4. ZERO FORCING IN ITERATED LINE DIGRAPHS 45

Case 3.: If a`−2 6= a1 and a`−1 = a2, then X(a1, . . . , a`−1) ⊆ Z because

out X(a1, . . . , a`−1) = N ((a`−2, a1, a2, . . . , a`−1))

and (a`−2, a1, a2, . . . , a`−1) ∈ S.

Case 4.: If a`−2 6= a1 and a`−1 = a1 then

|S ∩ X(a1, . . . , a`−1)| = |{(a1, . . . , a`): a` ∈ Zd+1 \{a1, a2}}| = d − 1,

and the claim follows from Z ⊇ S.

Case 5.: If a`−2 6= a1 and a`−1 6∈ {a1, a2}, then X(a1, . . . , a`−1) ⊆ Z because

out X(a1, . . . , a`−1) = N ((a`−2, a1, a2, . . . , a`−1))

and (a`−2, a1, a2, . . . , a`−1) ∈ S.

The second part of Theorem 3.3.1 follows from Lemmas 3.3.4, 3.3.5 and 3.3.6.

3.4. Zero forcing in iterated line digraphs

In this section, we extend the results of zero forcing number obtained for de Bruijn and

Kautz digraphs in Sections 3.2 and 3.3 to iterated line digraphs. Recall the deﬁnition of iterated line digraphs in Section 2.1.3 on page 10. For a digraph G~ = (V,A), the line digraph of G~ is the digraph L(G~ ) where V (L(G~ )) = A(G~ ) and there is an arc from a vertex x to a vertex y in

L(G~ ) if and only if x = uv and y = vw for some vertices v, w in V (G~ ).

~ ~ out ♦ Lemma 3.4.1. Let G be a digraph. For every vertex uv in L(G) with degG~ (uv) ≥ 2, every subset T ⊆ N out(uv) with |T | ≥ 2 is a strongly critical set in L(G~ ).

Proof. Let xy be any vertex of L(G~ ). If y = v then T ⊆ N out(xy), hence |N out(xy)∩T | =

out |T | ≥ 2, and if y 6= v, then N (xy) ∩ T = ∅.

♦ Lemma 3.4.2. Let G~ = (V,A) be a digraph and let S be a zero forcing set of L(G~ ). For ~ out out every vertex uv in L(G), |S ∩ N (uv)| ≥ degL(G~ )(uv) − 1.

out out out Proof. If |S ∩N (uv)| < degL(G~ )(uv)−1 then T = N (uv)\S is a strongly critical set by Lemma 3.4.1. Then S ∩ T 6= ∅ by Lemma 3.1.3, and this is the required contradiction. 46 3. ZERO FORCING AND POWER DOMINATION IN ITERATED LINE DIGRAPHS

The following simple observation is needed in the proof of the next theorem.

♦ Observation 3.4.3. Let G~ = (V,A) be a digraph. Then every vertex of G~ is obtained in at most one directed cycle C such that every vertex of C has in-degree 1.

♦ Theorem 3.4.4. Let G~ = (V,A) be a digraph with minimum in-degree δin(G~ ) ≥ 1 and minimum out-degree δout(G~ ) ≥ 2. Then

Z(L(G~ )) = |A(G~ )| − |V (G~ )|.

Proof. Let us assume V = {v1, . . . , vn} and let S ⊆ V (L(G~ )) be a zero-forcing set. By assumption, there are vertices u1, . . . , un ∈ V such that uivi ∈ A for all i ∈ {1, . . . , n}. By

out out Lemma 3.4.2, |S ∩ N (uivi)| ≥ degG~ (uivi) − 1 for all i, and using the fact that the sets out N (uivi) form a partition of V (L(G~ ))

n n X out X out ~ ~ |S| = |S ∩ N (uivi)| ≥ (degG~ (uivi) − 1) = |A(G)| − |V (G)|. i=1 i=1

This implies Z(G~ ) ≥ |A(G~ )| − |V (G~ )|. In order to prove the reverse inequality, we describe how to obtain a zero-forcing set S of the required size. From the assumption δout(G~ ) ≥ 2 and

out Observation 3.4.3 it follows that for every i ∈ {1, . . . , n} there exists a vertex wi ∈ N (vi) such that the arc viwi is not contained in a cycle consisting only of vertices of in-degree 1. We set n [ out S = {viw : w ∈ N (vi) \{wi}}. i=1 ~ ~ Then S is a zero-forcing set with |S| = |A(G)| − |V (G)|.

♦ Corollary 3.4.5. Let G~ = (V,A) be a digraph with minimum in-degree δin(G~ ) ≥ 1, minimum out-degree δout(G~ ) ≥ 2 and maximum out-degree ∆out(G~ ). Then

(δout(G~ ) − 1)|V | ≤ Z(L(G~ )) ≤ (∆out(G~ ) − 1)|V |.

Proof. From Theorem 3.4.4,

Z(L(G~ )) = |A| − |V |. 3.4. ZERO FORCING IN ITERATED LINE DIGRAPHS 47

P out Since |A| = v∈V deg (v),

δout(G~ )|V | ≤ |A| ≤ ∆out(G~ )|V |.

As a consequence,

(δout(G~ ) − 1)|V | ≤ |A| − |V | = Z(L(G~ )).

~ out ~ Analogously, Z(L(G)) = |A| − |V | ≤ (∆ (G) − 1)|V |.

The following result extends Theorem 3.4.4 and Corollary 3.4.5 to iterated line digraphs.

♦ Theorem 3.4.6. Let G~ be a digraph with minimum in-degree δin(G~ ) ≥ 1 and minimum out-degree δout(G~ ) ≥ 2. For every integer k ≥ 1,

Z(Lk(G~ )) = |V (Lk(G~ ))| − |V (Lk−1(G~ ))|.

♦ Corollary 3.4.8. Let G~ be a digraph with minimum in-degree δin(G~ ) ≥ 1, minimum out-degree δout(G~ ) ≥ 2 and maximum out-degree ∆out(G~ ). For every integer k ≥ 1,

(δout(G~ ) − 1)(δout(G~ ))k−1|V (G~ )| ≤ Z(Lk(G~ )) ≤ (∆out(G~ ) − 1)|(∆out(G~ ))k−1|V (G~ )|.

The following results follow immediately from Theorem 3.4.6 or from Corollary 3.4.8. 48 3. ZERO FORCING AND POWER DOMINATION IN ITERATED LINE DIGRAPHS

♦ Corollary 3.4.9. If G~ is a d-regular digraph with d ≥ 2, for every integer k ≥ 1,

Z(Lk(G~ )) = d Z(Lk−1(G~ )) = (d − 1)dk−1|V (G~ )|.

The following result by Gimbert and Wu in [43], together with our results on zero forcing of line digraphs allow us to show that for any digraph G~ and any positive integer k, M(Lk(G~ )) =

Z(Lk(G~ )) where M(Lk(G~ )) denote the maximum nullity of Lk(G~ ).

Lemma 3.4.10 ([43]). A d-regular digraph of order p is a line digraph if and only if the rank

p of its adjacency matrix is equal to d .

♦ Theorem 3.4.11. If G~ is a d-regular digraph with d ≥ 2, for every integer k ≥ 1,

M(Lk(G~ )) = Z(Lk(G~ )) = (d − 1)dk−1|V (G~ )|.

k k ~ d |V (G~ )| Proof. Let X is the adjacency matrix of L (G). By Lemma 3.4.10, X has rank d = dk−1|V (G~ )|. Then, the equality M(X) + rank(X) = |V (Lk(G~ ))| = dk|V (G~ )| implies that

M(X) = dk|V (G~ )| − dk−1|V (G~ )| = (d − 1)dk−1|V (G~ )|.

Since M(X) ≤ M(Lk(G~ )) ≤ Z(Lk(G~ )), and by Corollary 3.4.9, Z(Lk(G~ )) = (d − 1)dk−1|V (G~ )|, we conclude

(d − 1)dk−1|V (G~ )| = M(X) ≤ M(Lk(G~ )) ≤ Z(Lk(G~ )) = (d − 1)dk−1|V (G~ )| and the result follows immediately.

3.5. Power domination in iterated line digraphs

Let G~ = (V,A) be a digraph and S ⊆ V . By deﬁnition, S is a power dominating set of G~ if

out out and only if N [S] is a zero forcing set of G~ . This implies that Z(G~ ) ≤ γp(G~ )(∆ (G~ )+1). We will improve this upper bound for Z(G~ ) in terms of γp(G~ ) in the case that G~ is a line digraph.

♦ Theorem 3.5.1. If G~ is a digraph with minimum in-degree δin(G~ ) ≥ 1, minimum out-

out out out degree δ (G~ ) ≥ 2, and maximum out-degree ∆ (G~ ), then Z(L(G~ )) ≤ ∆ (G~ )γp(L(G~ )). 3.5. POWER DOMINATION IN ITERATED LINE DIGRAPHS 49

Proof. Let S = {u1v1, . . . , urvr} be a minimum power dominating set of L(G~ ). We

out prove Z(L(G~ )) ≤ ∆ (G~ )γp(L(G~ )) by constructing a zero forcing set of L(G~ ) with cardinality ∆out(G~ )|S|.

Let S1 = {uivi ∈ S : vi 6= vj for all j with 1 ≤ j < i} and S2 = S \ S1. Since S is a power ~ out r out out dominating set of L(G), N [S] = ∪i=1N [uivi] = S2 ∪ (∪uv∈S1 N [uv]) is a zero forcing set of L(G~ ).

out out Since δ (G~ ) ≥ 2, for every uivi ∈ S1 we can select an arbitrary vertex viwi ∈ N (uivi).

out Let P = ∪uivi∈S1 (N [uivi] \{viwi}).

out out out Then P can force ∪uv∈S1 N [uv], so S2 ∪ P can force S2 ∪ (∪uv∈S1 N [uv]) = N [S].

out Since N [S] is a zero-forcing set of L(G~ ), then S2 ∪ P is also a zero-forcing set of L(G~ ) and

Z(G~ ) ≤ |S2 ∪ P |. Now,

X out out out |S2 ∪ P | ≤ |S2| + |N [uivi] \{wiui})| ≤ |S2| + |S1|∆ (G~ ) ≤ |S|∆ (G~ ).

uivi∈S1

~ ~ out ~ Therefore, Z(L(G)) ≤ γp(L(G))∆ (G).

2 ♦ Corollary 3.5.2. If G~ is a d-regular digraph with d ≥ 2, then Z(L(G~ )) ≤ γp(L (G~ )).

Proof. By deﬁnition of line digraph, G~ being d-regular implies that L(G~ ) is also d-regular. Since d ≥ 2, we apply 3.5.1 to L(G~ ) and obtain

2 2 Z(L (G~ )) ≤ dγp(L (G~ )).

By Corollary 3.4.9,

Z(L2(G~ )) = d Z(L(G~ )), and this implies

2 Z(L(G~ )) ≤ γp(L (G~ )).

out Next we will show that for a d-regular digraph with d ≥ 2, Z(L(G~ )) = ∆ (G~ )γp(L(G~ )). To do so, we introduce the following terminology.

A 1-factor of a digraph G~ is a spanning subdigraph H~ (i.e. V (H~ ) = V (G~ ) and A(H~ ) ⊆

A(G~ )) that is the union of disjoint directed cycles. Equivalently, a 1-factor is a spanning 50 3. ZERO FORCING AND POWER DOMINATION IN ITERATED LINE DIGRAPHS subgraph whose matrix is a permutation matrix. Moreover, H~ is a good 1-factor if for each cycle in H~ , there exists at least one vertex with in-degree at least 2 in G~ .

♦ Observation 3.5.3. The following are examples of digraphs that have a 1-factor.

• Any Hamiltonian digraph G~ with δin(G~ ) ≥ 2.

• Any digraphs G~ with min{δout(G~ ), δin(G~ )} ≥ 2 that admits a 1-factor [52].

• Any d-regular digraph with d ≥ 2 [81].

• Digraphs described by unitary matrices [80].

2 ♦ Theorem 3.5.4. If a digraph G~ admits a good 1-factor, then Z(L(G~ )) ≥ γp(L (G~ )).

Proof. By hypothesis, G~ admits a good 1-factor, say H~ . Then, H~ induces a permutation f on the set of vertices of G~ where f(v) = u if and only if (u, v) is an arc in H~ . In addition, since H~ is a good 1-factor, each orbit in the permutation f contains a vertex of in-degree at least 2. Let us denote V (L(G~ )) = {uv :(u, v) ∈ A} and V (L2(G~ )) = {uvw :(v, v) ∈ A(G~ ), (v, w) ∈ A(G~ )}. Let

S := {f(u)uv ∈ V (L2(G~ )) : u 6= f(v)}. Then, |S| = |V (L(G~ ))| − |V (G~ )| and by Theorem 3.4.4 it is suﬃcient to prove that S is a power dominating set of L2(G~ ).

Given xyz in V (L2(G~ )), it is suﬃcient to show that either xyz ∈ N out[S] or xyz is obtained by a sequence of forcing steps starting from N out[S].

Case 1.: If x 6= f(y) then f(x)xy ∈ S, hence xyz ∈ N out(f(x)xy) ⊆ N out[S].

Case 2.: If x = f(y) and y 6= f(z) then xyz ∈ S ⊆ N out[S].

Case 3.: If x = f(y) and y = f(z) then xyz = f 2(z)f(z)z.

In the last case, for notational convenience, denote the elements of the orbit of z by zi =

i f (z), i.e., z = z0 = z`, y = z1 = z`+1, etc., where ` is the length of the orbit of z. By

in in assumption, deg (zk) ≥ 2 for some k ≥ 2. This implies that there exists a ∈ N (zk) such that

out out a 6= zk+1. Consequently, f(a)azk ∈ S and azkzk−1 ∈ N (f(a)azk) ⊆ N [S].

out • If deg (zk−1) = 1 then azkzk−1 forces its only out-neighbour zkzk−1zk−2.

out • If deg (zk−1) ≥ 2 every out-neighbour of azkzk−1 diﬀerent from zkzk−1zk−2 is in

out the form zkzk−1b for some vertex b ∈ N (zk−1) \{zk−2}. Since f is bijective and

zk−2 6= b, we have zk−1 6= f(b), and therefore, zkzk−1b ∈ S. So every out-neighbour 3.5. POWER DOMINATION IN ITERATED LINE DIGRAPHS 51

of azkzk−1 diﬀerent from zkzk−1zk−2 is in S and this means that azkzk−1 forces

zkzk−1zk−2.

Repeating this argument, we obtain that all the vertices zizi−1zi−2 for i = k, k − 1,... are forced, and in particular xyz = z2z1z0.

As proven by Hasunuma and Shibata in [54], if a digraph has a 1-factor, so does its line digraph. Then, combining Theorem 3.5.1 and Theorem 3.5.4 we obtain the following result for iterated line digraphs.

♦ Corollary 3.5.5. Let G~ be a digraph with minimum in-degree δin(G~ ) ≥ 2, minimum out-degree δout(G~ ) ≥ 2, and maximum out-degree ∆out(G~ ). For every positive integer k,

k+1 ~ k k+1 Z(L (G)) Z(L (G~ )) ≥ γp(L (G~ )) ≥ . ∆out(G~ )

♦ Corollary 3.5.6. If G~ is a d-regular digraph with d ≥ 2. For any integer k ≥ 1,

k k+1 k−1 Z(L (G~ )) = γp(L (G~ )) = (d − 1)d |V (G~ )|.

Proof. Since G~ is d-regular, G~ is also Eulerian and as a consequence, L(G~ ) is Hamiltonian. In addition, in a digraph with at least one vertex of in-degree 2, a Hamiltonian cycle induces a good 1-factor in G~ . Therefore, L(G~ ) admits a good 1-factor. Since L(G~ ) is d-regular, d ≥ 2, by

Corollary 3.5.5 we have

Z(Lk+1(G~ )) Z(Lk(G~ )) ≥ γ (Lk+1(G~ )) ≥ p d

By Corollary 3.4.9

Z(Lk+1(G~ )) (d − 1)dk|V (L(G~ ))| = = (d − 1)dk−1|V (L(G~ ))| = Z(Lk(G~ )) d d and the result follows immediately.

If G~ is a digraph with minimum in-degree δin(G~ ) ≥ 1 and minimum out-degree δout(G~ ) ≥ 1, we deﬁne µ(G~ ) as the minimum number of cycles with only vertices of in-degree 1 in any 1-factor of G~ . 52 3. ZERO FORCING AND POWER DOMINATION IN ITERATED LINE DIGRAPHS

Note that G~ admits a good 1-factor if and only if µ(G~ ) = 0. Therefore, the following result generalizes Theorem 3.5.4.

♦ Theorem 3.5.7. If G~ = (V,A) is a digraph with minimum in-degree δin(G~ ) ≥ 1 and

out 2 minimum out-degree δ (G~ ) ≥ 1, then Z(L(G~ )) ≥ γp(L (G~ )) + µ(G~ ).

To prove Theorem 3.5.7 one can construct a power dominating set of L2(G~ ) following the following steps:

1. Start with the set S as in the proof of Theorem 3.5.4.

2. For each cycle consisting exclusively of vertices of in-degree 1, select a vertex v adjacent

to one of the vertices in the cycle. Add v to S

Proving that the set S constructed is a power dominating set of L2(G~ ) is analogous to the proof of Theorem 3.5.4.

3.6. Applications

Next we present some families of iterated line digraphs and use Corollary 3.4.9 to obtain their zero forcing number, Theorem 3.4.11 to obtain their maximum nullity and their minimum rank, and Corollary 3.5.6 to determine their power domination number.

• Generalized de Bruijn digraphs

For each pair of integers d ≥ 2 and n ≥ 1 the generalized de Bruijn graph GB(d, n)

is deﬁned as follows:

V (GB(d, n)) = {0, 1, . . . , n − 1},

A(GB(d, n)) = {(x, y): y ≡ dx + i (mod n), 0 ≤ i ≤ d − 1} .

They contain the de Bruijn digraphs as a sub-family since B(d, `) = GB(d, d`). The

generalized de Bruijn digraphs, also known as Reedy-Pradhan-Kuhl digraphs, have the

property that GB(d, dn) = L(GB(d, n)), so for every integer k ≥ 1, GB(d, dkn) =

Lk(GB(d, n)).

♦ Corollary 3.6.1. For any integers d ≥ 2, n ≥ 1 and k ≥ 1, i)Z( GB(d, dkn)) = (d − 1)dk−1n. 3.6. APPLICATIONS 53

ii)M( GB(d, dkn)) = (d − 1)dk−1n.

iii) mr(GB(d, dkn)) = dk−1n.

k k−2 iv) γp(GB(d, d n)) = (d − 1)d n.

• Generalized Kautz digraphs

For each pair of integers d ≥ 2 and n ≥ 1, the generalized Kautz digraph GK(d, n)

can be deﬁned as follows:

V (GK(d, n)) = {0, 1, . . . , n − 1},

A(GK(d, n)) = {(x, y); y ≡ −dx − i (mod n), 1 ≤ i ≤ d} .

Notice that they contain the Kautz digraphs as a sub-family since K(d, `) = GK(d, d`+

d`−1). The generalized Kautz digraphs, also known as Imase-Itoh digraphs, have the

property that GK(d, dn) = L(GK(d, n)) so, for every integer k ≥ 1, GK(d, dkn) =

Lk(GK(d, n)).

♦ Corollary 3.6.2. For any integers d ≥ 2, n ≥ 1 and k ≥ 1, i)Z( GK(d, dkn)) = (d − 1)dk−1n.

ii)M( GK(d, dkn)) = (d − 1)dk−1n.

iii) mr(GK(d, dmn)) = dk−1n.

k k−2 iv) γp(Z(GK(d, d n))) = (d − 1)d n.

• Directed wrapped butterﬂy

Given two integers d, ` ≥ 2, the wrapped butterﬂy WB(d, `) has for vertices the

ordered pairs (x, i) where x is an `-tuple of integers in Zd and i is an integer, 0 ≤ i ≤

` − 1. A vertex (x1 . . . x`, l) is adjacent to d vertices in the form (x1 . . . xi−1αxi+1x`, i)

` where α ∈ Zd. It is clear that |V (WB(d, `))| = `d . The wrapped butterﬂies WB(d, `) can be obtained as a line digraph, but we need some deﬁnitions prior to explain this

in detail.

If G~ and H~ are two digraphs, the direct product of G~ and H~ is the digraph

G~ · H~ , whose vertex set corresponds to V (G~ ) × V (H~ ); a vertex (g, h) is adjacent to a

vertex (g0, h0) if (g, g0) is an arc in G~ and (h, h0) is an arc in H~ . Then, WB(d, `) = 54 3. ZERO FORCING AND POWER DOMINATION IN ITERATED LINE DIGRAPHS

`−1 L (K~d · C~`) where K~d denotes the complete digraph of order d and C~` denotes the

directed cycle of order `. Also |K~d · C~`| = d`.

♦ Corollary 3.6.3. For any integers d ≥ 2 and ` ≥ 2, i)Z( WB(d, `)) = (d − 1)d`−1`.

ii)M( WB(d, `)) = (d − 1)d`−1`.

iii) mr(WB(d, `)) = d`−1`.

`−2 iv) γp(WB(d, `)) = (d − 1)d `. CHAPTER 4

Zero forcing and power domination in butterﬂy networks

In this chapter, we make the following contributions:

• In Section 4.1, we introduce an alternate interpretation of zero forcing and power

domination problems in undirected graphs as set covering problems.

• In Section 4.2, we solve the zero forcing and minimum rank problem for butterﬂy

networks.

• In Section 4.3, we obtain bounds for power domination in butterﬂy networks.

4.1. An alternate interpretation

The interpretation is a bit different from the one defined for directed graphs in Section 3.1 on page 35. This is due to the fact that the definitions of zero forcing problem in directed graphs and undirected graphs are different. Recall that in the directed version, either a red or a blue vertex could force a colour change but in undirected version, only a blue vertex can force the colour change. To account for the colour change in directed graphs, we needed the concept of strongly critical set. But it is not needed here in the case of undirected graphs.

We call a vertex set W critical if there is no vertex outside W which has exactly one neighbour in W . If W is critical, but no proper subset of W is critical, then we call W minimal critical.

Whenever W is critical and S is a zero forcing set then S ∩ W 6= ∅. More precisely, we have the following characterization of zero forcing sets in undirected graphs similar to the one we had in directed graphs in Lemma 3.1.3. The proofs are similar and therefore we omit the proof.

♦ Lemma 4.1.1. A vertex set S is a zero forcing set of G if and only if S ∩ W 6= ∅ for every minimal critical set W in G.

The contents of this chapter reproduce original results from [38].

55 56 4. ZERO FORCING AND POWER DOMINATION IN BUTTERFLY NETWORKS

Recall Theorem 2.4.1 that the set S is a power dominating set if and only if N[S] is a zero forcing set and thus we have the following characterisation of power dominating sets in undirected graphs.

♦ Lemma 4.1.2. A vertex set S is a power dominating set of G if and only if N[S] ∩ W 6= ∅ for every minimal critical set W in G.

The importance of critical sets in obtaining bounds was exhibited in Chapter3. We believe that a better knowledge of critical sets would help us improve bounds for zero forcing number and power domination number in graphs.

Let C denote the collection of minimal critical sets, and let Ci denote the collection of minimal critical sets that contain vertex i. We have

( n ) X X n Z(G) = min xi : xi ≥ 1 ∀ W ∈ C, x ∈ {0, 1} i=1 i∈W

We can deﬁne the dual quantity

( ) 0 X X C Z (G) = max yW : yW ≤ 1 ∀ i ∈ V, y ∈ {0, 1}

W ∈Ci W ∈Ci We have that Z0(G) is a lower bound for the zero forcing number i.e.

Z(G) ≥ Z0(G).

This lower bound is certainly not tight in general: For a star G = K1,t with t > 2 leaves we have Z(G) = t − 1 while Z0(G) = dt/2e.

In the next section, we solve the zero forcing number of butterfly network by solving the minimum rank of butterfly network. The motivation to looking into the butterfly network was due to the open problem posed in [3]: “What is the class of graphs G for which Z(G) =

M F (G) for some ﬁeld F ”? The authors of [3] found many interesting graph classes which satisﬁes the equality. In [63], this equality is proved for proved for block-clique graphs and unit interval graphs. Recently, the zero forcing number of Cartesian product of cycles was established by constructing a matrix with the required corank [11]. The American Institute for

Mathematics maintains the minimum rank graph catalog [59] in order to collect the information 4.2. ZERO FORCING IN BUTTERFLY NETWORK 57 about minimum rank problem for various graph classes. Though we have not been able to solve the open problem, we have identiﬁed another graph which satisfy this equality.

4.2. Zero forcing in butterﬂy network

Butterfly networks was introduced as a variant of hypercube network, which is considered quite powerful from a computational point of view, to overcome some disadvantages to its use as an architecture for parallel computation. One of the most obvious disadvantages is that the node degree of the hypercube grows with its size. In order to circumvent the difficulties associated with node degrees in hypercubes, several variations of the hypercube have been devised that have similar computational properties but bounded degree (usually 3 or 4 ) such as cube connected cycles, butterfly networks and Benes networks [70].

For a positive integer r, the butterﬂy network BF(r) = V (r),E(r) has vertex set V (r) =

(r) (r) (r) (r) (r) (r) (r) V0 ∪ V1 ∪ · · · ∪ Vr and edge set E = E1 ∪ E2 ∪ · · · ∪ Er , where

(r) r Vi = {(x, i): x ∈ {0, 1} } for i = 0, 1, . . . , r,

(r) r Ei = {{(x, i − 1), (y, i)} : x ∈ {0, 1} , y ∈ {x, x + ei}} for i = 1, 2, . . . , r.

Here addition is modulo 2, and ei is the binary vector of length r with a one in position i and zeros in all other components. For convenience, we identify the binary vector x = (x1, . . . , xr) ∈

r Pr i−1 {0, 1} with the number i=1 xi2 . Using this identiﬁcation the butterﬂy network BF(4) is shown in Figure 4.1.

Our main result in this section is the following theorem.

♦ Theorem 4.2.1. The minimum rank of the butterfly network BF(r) over any field F equals 2 mrF (BF(r)) = [(3r + 1)2r − (−1)r] , 9 and this is equal to the rank of the adjacency matrix of BF(r). Furthermore, for the butterfly network we have equality in Theorem 2.2.1, i.e.,

1 Z (BF(r)) = (r + 1)2r − mrF (BF(r)) = [(3r + 7)2r + 2(−1)r] . 9 58 4. ZERO FORCING AND POWER DOMINATION IN BUTTERFLY NETWORKS

0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Figure 4.1. The butterﬂy network BF(4).

1 4.2.1. The upper bound for Z(BF(r)). Let (Jn) denote the Jacobsthal sequence which is deﬁned by J0 = 0, J1 = 1 and Jn = Jn−1 + 2Jn−2 for n > 2. We will need the following relation which follows immediately from the deﬁnition:

n (4.2.1) Jn+2 = 2 + Jn for every integer n > 0.

(r) (r) (r) (r) For every r, we deﬁne a set S = S0 ∪ S1 ∪ · · · ∪ Sr by

(r) i+1 i+1 Si = (x, i) : 2 ` 6 x 6 2 ` + Ji+1 − 1 for some ` for i = 0, 1, . . . , r − 1

(r) Sr = {(x, i) : 0 6 x 6 Jr+1 − 1} .

For r = 4 this is illustrated in Figure 4.2. Solving the recurrence relation for the numbers Jn, we ﬁnd a closed form expression for the size of the set S(r).

r X 1 ♦ Lemma 4.2.2. We have S(r) = J + 2r−iJ = [(3r + 7)2r + 2(−1)r]. r+1 i 9 i=1

1OEIS:A001045 4.2. ZERO FORCING IN BUTTERFLY NETWORK 59

0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Figure 4.2. The set S(4) indicated by squares.

Next we want to verify that S(r) is a zero forcing set for BF(r). For this purpose we set

(r) X0 = S and deﬁne a sequence X1,X2,...,X2r of vertex sets by

(4.2.2)

Xk = Xk−1 ∪ {(x, r − k): {(x, r − k)} = N(v) \ Xk−1 for some v = (y, r − k + 1) ∈ Xk−1}

(4.2.3)

Xr+k = Xr+k−1 ∪ {(x, k): {(x, k)} = N(v) \ Xr+k−1 for some v = (y, k − 1) ∈ Xr+k−1}

for k = 1, . . . , r. After applying the color-change rule to the coloring with Xk−1 blue and

(r) V \ Xk−1, all the vertices in Xk are blue and therefore it is suﬃcient to prove that X2r = V .

♦ Lemma 4.2.3. For k ∈ {0, 1, . . . , r}, Xk = Xk(0) ∪ Xk(1) ∪ · · · ∪ Xk(r) with

(r) Xk(i) = Si for i ∈ {0, 1, . . . , r − k} ∪ {r}

i i Xk(i) = (x, i) : 2 k 6 x 6 2 k + Ji+1 − 1 for some ` for i ∈ {r − k + 1, . . . , r − 1}.

Proof. We proceed by induction on k. For k = 0, there is nothing to do since X0 =

(r) (r) (r) S = S0 ∪ · · · ∪ Sr . Let k > 1 and set i = r − k. By (4.2.2), we have Xk(j) = Xk−1(j) for 60 4. ZERO FORCING AND POWER DOMINATION IN BUTTERFLY NETWORKS all j 6= i. By induction, this implies

(r) Xk(j) = Sj for j ∈ {0, 1, . . . , r − k − 1} ∪ {r}

j j Xk(j) = (x, i) : 2 k 6 x 6 2 k + Ji+1 − 1 for some ` for j ∈ {r − k + 1, . . . , r − 1}, and it remains to be shown that

i i (4.2.4) Xk(i) = (x, i) : 2 ` 6 x 6 2 ` + Ji+1 − 1 for some ` .

i+1 i+1 Let (x, i) be an arbitrary element of the RHS of (4.2.4). If 2 ` 6 x 6 2 ` + Ji+1 − 1 for some integer ` then (x, i) ∈ X0 ⊆ Xk. Otherwise

i+1 i i+1 i 2 ` + 2 6 x 6 2 ` + 2 + Ji+1 − 1

i for some integer `. By induction, the vertex (y, i + 1) with y = x − 2 is in Xk−1 because

i+1 i+1 i+1 2 ` 6 y 6 2 ` + Ji+1 − 1 6 2 ` + Ji+2 − 1.

Let ` = 2`0 + ε with ε ∈ {0, 1}. The neighbourhood of (y, i + 1) is   {(y, i), (x, i)} if k = 1, N((y, i + 1)) =  ε i+1 {(y, i), (x, i), (y, i + 2), (y + (−1) 2 , i + 2)} if k > 1.

Now (y, i) ∈ X0 ⊆ Xk−1, and for k = 1 that’s all we need. Using (4.2.1), we have

i+2 0 i+1 i+1 i+1 i+2 0 i+1 i+2 0 2 ` = 2 (`−ε) 6 2 ` 6 y 6 2 `+Ji+1 −1 = 2 ` +2 ε+Ji+1 −1 6 2 ` +Ji+3 −1,

and therefore (y, i + 2) ∈ Xk−1. Similarly,

i+2 0 i+1 ε i+1 ε i+1 i+2 0 i+1 i+2 0 2 ` = 2 ` + (−1) 2 6 y + (−1) 2 6 2 ` + 2 + Ji+1 − 1 6 2 ` + Ji+3 − 1,

ε i−1 and therefore (y + (−1) 2 , i + 2) ∈ Xk−1. Consequently, {(x, i)} = N((y, i + 1)) \ Xk−1, and this implies

i i Xk(i) ⊇ (x, i) : 2 ` 6 x 6 2 ` + Ji+1 − 1 for some ` . 4.2. ZERO FORCING IN BUTTERFLY NETWORK 61

To prove the converse, consider (x, i) with

i i 2 ` + Ji+1 6 x 6 2 (` + 1) − 1. and ` = 2`0 + ε as before. We have

(r) ε i N((x, i)) ∩ Si+1 = {(x, i + 1), (x + (−1) 2 , i + 1)}.

If (x, i + 1) ∈ Xk−1 then

i+1 • (x + 2 , i + 2) ∈ N((x, i + 1)) \ Xk−1 if ` ≡ 0 or 1 (mod 4), and

• (x, i + 2) ∈ N((x, i + 1)) \ Xk−1 if ` ≡ 2 or 3 (mod 4).

ε i Similarly, if (x + (−1) 2 , i + 1) ∈ Xk−1 then ` is odd and

i+1 • (x + 2 , i + 2) ∈ N((x, i + 1)) \ Xk−1 if ` ≡ 1 (mod 4), and

• (x, i + 2) ∈ N((x, i + 1)) \ Xk−1 if ` ≡ 3 (mod 4).

In all cases it follows that (x, i) 6∈ Xk, and this concludes the proof.

♦ Lemma 4.2.4. For k ∈ {0, 1, . . . , r}, Xr+k = Xr+k(0) ∪ Xr+k(1) ∪ · · · ∪ Xr+k(r) with   (r) Vi for i ∈ {0, . . . , k}, Xr+k(i) =  Xr(i) for i ∈ {k + 1, k + 2, . . . , r}.

Proof. We proceed by induction on k. For k = 0, there is nothing to do since Xr =

Xr(1) ∪ Xr(2) ∪ ... ∪ Xr(r) ∪ V0(r), which is true by Lemma 4.2.3. Let k > 1 and set i = k.

By (4.2.3), we have Xr+k(j) = Xr+k−1(j) for all j 6= i. By induction, this implies

(r) Xr+k(j) = Vj for j ∈ {0, 1, . . . , k − 1}

Xr+k(j) = Xr(j) for j ∈ {k + 1, . . . , r − 1}, and it remains to be shown that

(r) Xr+k(k) = Vk . 62 4. ZERO FORCING AND POWER DOMINATION IN BUTTERFLY NETWORKS

(r) i+1 i+1 i+1 i Let (x, i) be an arbitrary element of Vk . If 2 ` 6 x 6 2 ` + Ji+1 − 1 and 2 ` + 2 6 x 6 i+1 i 2 ` + 2 + Ji+1 − 1 for some integer `, then (x, i) ∈ Xr−k ⊆ Xr+k. Otherwise

i+1 i i+1 i 2 ` + 2 + Ji+1 − 1 6 x 6 2 ` + 2

i−1 for some integer `. By induction, the vertex (y, i − 1) with y = x − 2 is in Xr+k−1 because (r) y ∈ Vk−1. More precisely,

i+1 i i−1 i+1 i i−1 2 ` + 2 + Ji+1 − 1 − 2 6 y 6 2 ` + 2 − 2 .

Let `0 = 2` − ε with ε ∈ {0, 1}. The neighbourhood of (y, i − 1) is   {(y, i), (x, i)} if k = 1, N((y, i − 1)) =  ε i−2 {(y, i), (x, i), (y, i − 2), (y + (−1) 2 , i − 2)} if k > 1.

ε i−2 Now (y, i − 2), (y + (−1) 2 , i − 2) ∈ Xr+k−1. Using (4.2.1), we have

i 0 i i−1 i 0 i i−1 2 ` + 2 + Ji+1 − 1 − 2 6 y 6 2 ` + 2 − 2 .

and therefore (y, i) ∈ Xr+k−1. Consequently, {(x, i)} = N((y, i − 1)) \ Xr+k−1, and this implies

(r) Xr+k(i) = Vk .

Combining Lemmas 4.2.3 and 4.2.4, we have proved that S(r) is indeed a zero forcing set for BF(r).

(r) ♦ Lemma 4.2.5. For every r > 1, S is a zero forcing set for the butterﬂy network

BF(r).

From Lemmas 4.2.2 and 4.2.5 we obtain an upper bound for the zero forcing number of the butterﬂy network. 1 ♦ Proposition 4.2.6. For every r 1, Z (BF(r)) [(3r + 7)2r + 2(−1)r]. > 6 9 4.2.2. The lower bound for Z(BF(r)). By Theorem 2.2.1, the corank of the adjacency matrix of a graph G provides a lower bound for the zero forcing number of G, and consequently we can conclude the proof of Theorem 4.2.1 by establishing the following result. 4.2. ZERO FORCING IN BUTTERFLY NETWORK 63

♦ Proposition 4.2.7. Let F be a ﬁeld, and let Ar denote the adjacency matrix of BF(r) over F . Then

1 2 rank(A ) (r + 1)2r − [(3r + 7)2r + 2(−1)r] = [(3r + 1)2r − (−1)r] . r 6 9 9

We will prove this by verifying that the rows corresponding to vertices in S(r) are linear combinations of the rows corresponding to vertices in the complement of S(r). For this purpose it turns out to be convenient to number the vertices recursively as indicated in Figure 4.3a.

Formally this vertex numbering is given by a bijection f : {0, 1, 2,...}2 → {1, 2, 3,...} deﬁned

ρx−1 ρ(x) as follows. For a positive integer x, let ρ(x) be the unique integer such that 2 6 i < 2 . In addition, let ρ(0) = −1. Then   i i2 + i + 1 if i > ρ(x), (4.2.5) f(x, i) =  ρ(x)−1 ρ(x)−1 ρ(x)2 + f x − 2 , i if i < ρ(x). 64 4. ZERO FORCING AND POWER DOMINATION IN BUTTERFLY NETWORKS

With respect to the vertex numbering given by (4.2.5) the adjacency matrices for BF(1) and

BF(2) are

  0 0 1 1 0 0 0 0 0 0 0 0      0 0 1 1 0 0 0 0 0 0 0 0           1 1 0 0 0 0 0 0 1 0 1 0         1 1 0 0 0 0 0 0 0 1 0 1            0 0 1 1  0 0 0 0 0 0 1 1 0 0 0 0          0 0 1 1  0 0 0 0 0 0 1 1 0 0 0 0      A1 =   ,A2 =   , 1 1 0 0          0 0 0 0 1 1 0 0 1 0 1 0    1 1 0 0      0 0 0 0 1 1 0 0 0 1 0 1           0 0 1 0 0 0 1 0 0 0 0 0       0 0 0 1 0 0 0 1 0 0 0 0       0 0 1 0 0 0 1 0 0 0 0 0      0 0 0 1 0 0 0 1 0 0 0 0 and in general, Ar has the structure illustrated in Figure 4.3b where I is the identity matrix of size 2r−1 × 2r−1.

Using the vertex numbering given by (4.2.5) the upper bound construction for a zero forcing set S(r) ∈ V (BF(r)) can be written recursively as S(1) = {1, 3} and

(r) (r−1) n r−1 (r−1) r−1o r r (4.2.6) S = S ∪ i + r2 : i ∈ S , i 6 (r − 1)2 ∪ {r2 + 1, . . . , r2 + Jr+1}

(r) for r > 2. In order to prove that S is a minimum zero forcing set for BF(r) it is suﬃcient (r) to show that every row i ∈ S of Ar can be written as a linear combination of the rows in

(r) r (r) S = {1,..., (r + 1)2 }\ S . We proceed by induction on r. Let Ar(i) denote the i-th row of

(1) Ar. The induction base is provided by checking the cases r = 1 and r = 2. For S = {1, 3}, 4.2. ZERO FORCING IN BUTTERFLY NETWORK 65

0 Ar−1 0

I I

0 0 Ar−1

I I I I 0 0 I I

(b) The structure of the adjacency ma- (a) The butterﬂy network BF(3). trix of the butterﬂy graph.

Figure 4.3. Labelling of vertices and structure of adjacency matrix in BF(3).

we have

(4.2.7) A1(1) = A1(2),

(4.2.8) A1(3) = A1(4), and for S(2) = {1, 5, 3, 9, 10, 11} we have

(4.2.9) A2(1) = A2(2),

(4.2.10) A2(5) = A2(6),

(4.2.11) A2(3) = A2(4) + A2(7) − A2(8),

(4.2.12) A2(9) = A2(2) + A2(6) − A2(12),

(4.2.13) A2(10) = A2(12),

(4.2.14) A2(11) = A2(2) + A2(6) − A2(12)..

The next two lemmas follow directly from the recursive structure illustrated in Figure 4.3b. 66 4. ZERO FORCING AND POWER DOMINATION IN BUTTERFLY NETWORKS

r−1 ♦ Lemma 4.2.8. If i 6 (r − 1)2 and

X X + − (r−1) Ar−1(i) = Ar−1(j) − Ar−1(j) for some K ,K ⊆ S , j∈K+ j∈K− then

X X Ar(i) = Ar(j) − Ar(j) and j∈K+ j∈K−

r−1 X X Ar i + r2 = Ar(j) − Ar(j) j∈K0+ j∈K0−

+ − (r) 0ε r−1 ε (r) where K ,K ⊆ S and K = {j + r2 : j ∈ K } ⊆ S for ε ∈ {+, −}.

r−1 r−1 ♦ Lemma 4.2.9. If (r − 1)2 + 1 6 i 6 (r − 1)2 + Jr and

X X + − (r−1) Ar−1(i) = Ar−1(j) − Ar−1(j) for some K ,K ⊆ S , j∈K+ j∈K− then X X Ar(i) = Ar(j) − Ar(j) j∈K0+ j∈K0− where

(r) K0+ = K+ ∪ j + r2r−1 : j ∈ K− ∪ i + r2r−1 ⊆ S ,

(r) K0− = K− ∪ j + r2r−1 : j ∈ K+ ⊆ S .

Lemmas 4.2.8 and 4.2.9 take care of the ﬁrst two components in the recursion for S(r)

r r−1 in (4.2.6). It remains to check the rows r2 + i for i ∈ {1,...,Jr+1}. For Jr−1 + 1 6 i 6 2 r r r−1 r−1 the required linear dependence is Ar (r2 + i) = Ar r2 + i + 2 , because i + 2 > Jr+1

r r−1 (r) r−1 r−1 r and therefore r2 + i + 2 ∈ S . For i > 2 we have i 6 2 + Jr−1 and Ar(r2 + i) = r r−1 Ar(r2 +i−2 ), and consequently it is suﬃcient to consider i ∈ {1,...,Jr−1}. The induction step for these cases will be from BF(r − 2) to BF(r), so we have to take the recursion for the adjacency matrix one step further which is illustrated in Figure 4.4. The basic idea is as

r−2 (r−2) follows. Let i ∈ {1,...,Jr−1}. Then (r − 2)2 + i ∈ S , and by induction there are sets 4.2. ZERO FORCING IN BUTTERFLY NETWORK 67 (

r A r−2 0 r−2

− (r − 2)2

1)2 I I 0 (r − 1)2r−2 r −

2 0 0 Ar−2 (2r − 3)2r−2 (3 I I r (r − 1)2r−1 − I I 1)2 I I

r I I

− r−1

2 r2

Ar−2 0 (3r − 2)2r−2 I I 0 (3r − 1)2r−2 0 0 Ar−2 (4r − 3)2r−2 (3

r I I r−1

4)2 + (2r − 1)2 I I I I

r I I − r

2 r2 I I 0 0 0 (2r + 1)2r−1 I I (r + 1)2r

Figure 4.4. The second level of the recursion for Ar.

(r−2) K+,K−S such that

r−2 X X (4.2.15) Ar−2 (r − 2)2 + i = Ar−2(j) − Ar−2(j), j∈K+ j∈K− or equivalently

X X (4.2.16) Ar−2(j) − Ar−2(j), j∈K+ j∈K0−

0− − r−2 where K = K ∪ (r − 2)2 + i . This is a linear dependence of the rows of Ar−2 with

r−2 r−2 coeﬃcients in {1, −1} and involving exactly one of the rows (r−2)2 +1,..., (r−2)2 +Jr−1, 68 4. ZERO FORCING AND POWER DOMINATION IN BUTTERFLY NETWORKS namely (r − 2)2r−2 + i. Putting K = K+ ∪ K0− we have

r−2 r−2 r−2 (4.2.17) K ∩ (r − 2)2 + 1,..., (r − 2)2 + Jr−1 = (r − 2)2 + i .

We now translate the |K| rows in this linear dependence by (r − 1)2r−2 and (3r − 1)2r−2 as indicated in Figure 4.4. The combination of the 2|K| translated rows is a {0, 1, −1}-vector x which has all its nonzero entries in columns with indices in {(r −1)2r−1 +1, . . . , r2r−1}∪{(2r −

r−1 r r−1 r−1 1)2 + 1, . . . , r2 }, and has xk = 1 for k ∈ (r − 1)2 + i, (2r − 1)2 + i which are the

r r r one-entries of the row Ar(r2 + i). Finally we use some of the rows r2 + Jr+1 + 1,. . . , (r + 1)2 with the appropriate sign to eliminate the other nonzero entries of x.

˜ ˜ + ˜ − r ˜ + ˜ + ˜ + More precisely, we deﬁne K = K ∪ K ⊆ {1,..., (r + 1)2 } with K = K1 ∪ K2 and ˜ − ˜ − ˜ − K = K1 ∪ K2 where

˜ + r−2 0− r−2 0− (4.2.18) K1 = j + (r − 1)2 : j ∈ K ∪ j + (3r − 1)2 : j ∈ K

˜ − r−2 + r−2 + (4.2.19) K1 = j + (r − 1)2 : j ∈ K ∪ j + (3r − 1)2 : j ∈ K

˜ + r−2 + r−2 K2 = j + (3r + 4)2 : j ∈ K with j > (r − 2)2

(4.2.20) ∪ j + (3r + 5)2r−2 : j ∈ K+ with j > (r − 2)2r−2 .

˜ − r−2 − r−2 K2 = j + (3r + 4)2 : j ∈ K with j > (r − 2)2

(4.2.21) ∪ j + (3r + 5)2r−2 : j ∈ K0− with j > (r − 2)2r−2 .

The construction of K˜ is illustrated for r = 4 and i = 1 in Figure 4.5. The next two lemmas show that K˜ has the required properties.

(r−2) ♦ Lemma 4.2.10. Let r > 3, i ∈ {1,...,Jr−1}, suppose K ⊆ S satisﬁes (4.2.15), and (r) deﬁne K˜ by (4.2.18) to (4.2.21). Then K˜ ⊆ S .

Proof. Note that by construction

˜ ˜ + ˜ − r−2 r−1 r−2 r−1 (4.2.22) K1 = K1 ∪ K1 ⊆ (r − 1)2 + 1, (r − 1)2 ∪ (3r − 1)2 + 1, (2r − 1)2 . 4.2. ZERO FORCING IN BUTTERFLY NETWORK 69

Figure 4.5. The construction of K˜ for r = 4 and i = 1. Here K+ = 0− ˜ − ˜ + ˜ − {2, 6}, K = {9, 12}, K1 = {14, 18, 46, 50}, K1 = {21, 24, 53, 56}, K2 = 00+ {76, 77, 80}, K2 = ∅.

r−1 (r) Suppose there is an element j ∈ K such that k = j + (r − 1)2 ∈ K˜1 ∩ S . Using (4.2.6), we obtain

(r) (r−1) n r−1 (r−1) r−1o r r k ∈ S = S ∪ p + r2 : p ∈ S , p 6 (r − 1)2 ∪ {r2 + 1, . . . , r2 + Jr+1}

(r−1) (r−2) n r−2 (r−2) r−2o =⇒ k ∈ S = S ∪ p + (r − 1)2 : p ∈ S , p 6 (r − 2)2

r−2 r−1 ∪ (r − 1)2 + 1,..., (r − 1)2 + Jr

r−2 (r−2) r−2 =⇒ k = p + (r − 1)2 for some p ∈ S with p 6 (r − 2)2 ,

(r−2) which contradicts the assumption that j ∈ K ⊆ S ∪ {(r − 2)2r−2 + i}. Similarly, for

r−1 (r) k = j + (3r − 1)2 ∈ K˜1 ∩ S we obtain

(r) (r−1) n r−1 (r−1) r−1o r r k ∈ S = S ∪ p + r2 : p ∈ S , p 6 (r − 1)2 ∪ {r2 + 1, . . . , r2 + Jr+1}

r−1 (r−1) r−1 =⇒ k = p + r2 for some p ∈ S with p 6 (r − 1)2

r−2 (r−2) r−2 =⇒ k = q + (r − 1)2 for some q ∈ S with q 6 (r − 2)2 , where we use k > (3r − 1)2r−2 for the last implication. Again we obtain a contradiction to the

(r−2) r−2 ˜ ˜ + ˜ − assumption that j ∈ K ⊆ S ∪ {(r − 2)2 + i}. Finally, the elements of K2 = K2 ∪ K2 70 4. ZERO FORCING AND POWER DOMINATION IN BUTTERFLY NETWORKS

(r) are in S since for j ∈ K+ ∪ K− we have

r−2 r−2 r−2 r r−1 r j > (r−2)2 =⇒ j > (r−2)2 +Jr−1 =⇒ j+(3r+4)2 > r2 +2 +Jr−1 = 2 +Jr+1, and for j ∈ K0,

r−2 r−2 r r−1 r−2 r j > (r − 2)2 =⇒ j + (3r + 5)2 > r2 + 2 + 2 > r2 + Jr+1, and this concludes the proof of the lemma.

+ − (r−2) ♦ Lemma 4.2.11. Let r > 3, i ∈ {1,...,Jr−1}, suppose K ,K ⊆ S satisfy (4.2.15), and deﬁne K˜ + and K˜ − by (4.2.18) to (4.2.21). Then

r X X (4.2.23) Ar (r2 + i) = Ar(j) − Ar(j). j∈K˜ + j∈K˜ −

Proof. Setting

X X r X X x = Ar(j) − Ar(j), y = Ar(r2 + i) − Ar(j) + Ar(j) ˜ + ˜ − ˜ + ˜ − j∈K1 j∈K1 j∈K2 j∈K2 equation (4.2.23) is equivalent to x = y. From (4.2.22) and (4.2.15) it follows that

supp(x) ⊆ (r − 1)2r−1 + 1, r2r−1 ∪ (2r − 1)2r−1 + 1, r2r , and by construction, for every j ∈ {1,..., 2r−2}, x (r − 1)2r−1 + j = x (r − 1)2r−1 + 2r−2 + j = x (2r − 1)2r−1 + j = x (2r − 1)2r−1 + 2r−2 + j .

Denoting this value byx ˜(j), we have   r−2 0− 1 if (r − 2)2 + j ∈ K ,   (4.2.24)x ˜(j) = −1 if (r − 2)2r−2 + j ∈ K+,    0 otherwise.

From (4.2.20) and (4.2.21) it follows that

˜ + ˜ − r−1 r+1 r K2 ∪ K2 ∪ (2r + 1)2 + i ⊆ (2r + 1)2 + 1, (r + 1)2 , 4.3. POWER DOMINATION IN BUTTERFLY NETWORK 71 and therefore

supp(y) ⊆ (r − 1)2r−1 + 1, r2r−1 ∪ (2r − 1)2r−1 + 1, r2r ,

r r r−1 After replacing Ar (r2 + i) by Ar r2 + 2 + i (which we can do since the two rows are equal), the rows contributing to y come in pairs j, j + 2r−2 where both rows in each pair have the same sign in y. Therefore y (r − 1)2r−1 + j = y (r − 1)2r−1 + 2r−2 + j = y (2r − 1)2r−1 + j = y (2r − 1)2r−1 + 2r−2 + j .

Finally, for j ∈ {1,..., 2r−2} we have y (r − 1)2r−1 + j = 1 if and only if (2r + 1)2r−1 + j = j0 + (3r + 4)2r−1 for some j0 ∈ K0−, or equivalently j0 = (r − 2)2r−2 + j ∈ K0−. Similarly, we have y (r − 1)2r−1 + j = −1 if and only if j0 = (r − 2)2r−2 + j ∈ K+, and comparing this with (4.2.24) we conclude x = y, as required.

Proof of Proposition 4.2.6. The statement follows by induction with base (4.2.7)–

(4.2.14), using Lemmas 4.2.8, 4.2.9, 4.2.10 and 4.2.11 for the induction step.

Finally, Theorem 4.2.1 is a consequence of Propositions 4.2.6 and 4.2.7.

4.3. Power domination in butterﬂy network

It was shown in [11] that Z(G)/∆ provides a lower bound for the size of a power dominating set in G where ∆ is the maximum degree of G. This implies that the power domination number

1 γ (BF(r)) ≥ [(3r + 7)2r + 2(−1)r] . p 36

We shall use the knowledge of critical sets to improve the lower bound for butterﬂy networks.

We shall do this by partitioning the vertex set into disjoint minimal critical sets. Recall that a set S is power dominating if and only if N[S] ∩ W 6= ∅ for every minimal critical set W .

r−1 For every vector x = (x1, . . . , xr−1) ∈ {0, 1} , let

(r) C0(x) = {((0, x1, . . . , xr−1), 0), ((1, x1, . . . , xr−1), 0)} ⊆ V0 ,

(r) Cr(x) = {((x1, . . . , xr−1, 0), r), ((x1, . . . , xr−1, 1), r)} ⊆ Vr . 72 4. ZERO FORCING AND POWER DOMINATION IN BUTTERFLY NETWORKS

0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Figure 4.6. The critical sets in diﬀerent layers of BF(4) . For convenience, r we identify the binary vector x = (x1, . . . , xr) ∈ {0, 1} with the number Pr i−1 i=1 xi2 .

r−2 For every vector x = (x1, . . . , xi−1, xi+2, xi+3, . . . , xr) ∈ {0, 1} , and every index i ∈ {1, . . . , r− 1}, let

Ci(x) = {((x1, . . . , xi−1, 0, 0, xi+2, . . . , xr), i) , ((x1, . . . , xi−1, 0, 1, xi+2, . . . , xr), i) ,

(r) ((x1, . . . , xi−1, 1, 0, xi+2, . . . , xr), i) , ((x1, . . . , xi−1, 1, 1, xi+2, . . . , xr), i)} ⊆ Vi .

♦ Lemma 4.3.1. The levels of BF(r) can be partitioned into minimal critical sets as follows:

(r) [ Vi = Ci(x) i ∈ {0, r}, x∈{0,1}r−1

(r) [ Vi = Ci(x) i ∈ {1, . . . , r − 1}. x∈{0,1}r−2

r−1 Proof. For every x = (x1, . . . , xr−1) ∈ {0, 1} , we have

N(C0(x)) = {((0, x1, . . . , xr−1), 1), ((1, x1, . . . , xr−1), 1)}

= N(((0, x1, . . . , xr−1), 0)) = N(((1, x1, . . . , xr−1), 0)) 4.3. POWER DOMINATION IN BUTTERFLY NETWORK 73

N(Cr(x)) = {((x1, . . . , xr−1, 0), r − 1), ((x1, . . . , xr−1, 1), r − 1)}

= N(((x1, . . . , xr−1, 0), r)) = N(((x1, . . . , xr−1, 1), r))

r−1 We note that |C0(x)| = |Cr(x)| = 2 . It is clear that the sets C0(x) and Cr(x) form critical sets as there are only two vertices in the neighbourhood of these sets and each of these vertex in the neighbourhood is adjacent to both vertices from the set.

r−2 For every vector x = (x1, . . . , xi−1, xi+2, xi+3, . . . , xr) ∈ {0, 1} , and every index i ∈ {1, . . . , r − 1},

N(Ci(x)) = N(((x1, . . . , xi−1, 0, 0, xi+2, . . . , xr), i)) ∪ N(((x1, . . . , xi−1, 0, 1, xi+2, . . . , xr), i))

∪ N(((x1, . . . , xi−1, 1, 0, xi+2, . . . , xr), i)) ∪ N(((x1, . . . , xi−1, 1, 1, xi+2, . . . , xr), i))

To show it is a critical set, we need to prove every vertex is either adjacent to at least two vertices from the set or none.

N ((x1, . . . , xi−1, 0, 0, xi+2, . . . , xr), i) = {((x1, . . . , xi−1, 0, 0, xi+2, . . . , xr), i − 1) ,

((x1, . . . , xi−1, 1, 0, xi+2, . . . , xr), i − 1) , ((x1, . . . , xi−1, 0, 0, xi+2, . . . , xr), i + 1) ,

((x1, . . . , xi−1, 0, 1, xi+2, . . . , xr), i + 1)}

N ((x1, . . . , xi−1, 1, 0, xi+2, . . . , xr), i) = {((x1, . . . , xi−1, 0, 0, xi+2, . . . , xr), i − 1) ,

((x1, . . . , xi−1, 1, 0, xi+2, . . . , xr), i − 1) , ((x1, . . . , xi−1, 1, 0, xi+2, . . . , xr), i + 1) ,

((x1, . . . , xi−1, 1, 1, xi+2, . . . , xr), i + 1)} 74 4. ZERO FORCING AND POWER DOMINATION IN BUTTERFLY NETWORKS

N ((x1, . . . , xi−1, 0, 1, xi+2, . . . , xr), i) = {((x1, . . . , xi−1, 0, 1, xi+2, . . . , xr), i − 1) ,

((x1, . . . , xi−1, 1, 1, xi+2, . . . , xr), i − 1) , ((x1, . . . , xi−1, 0, 0, xi+2, . . . , xr), i + 1) ,

((x1, . . . , xi−1, 0, 1, xi+2, . . . , xr), i + 1)}

N ((x1, . . . , xi−1, 1, 1, xi+2, . . . , xr), i) = {((x1, . . . , xi−1, 0, 1, xi+2, . . . , xr), i − 1) ,

((x1, . . . , xi−1, 1, 1, xi+2, . . . , xr), i − 1) , ((x1, . . . , xi−1, 1, 0, xi+2, . . . , xr), i + 1) ,

((x1, . . . , xi−1, 1, 1, xi+2, . . . , xr), i + 1)}

It is clear that there are only eight vertices in the neighbourhood of these sets and each of these eight vertex is adjacent to exactly two vertices from the set.

(r) For a vertex v ∈ Vi , i ∈ {1, 2, . . . , r − 1}, deg(v) = 4. The vertex v is adjacent to two vertices in level i + 1 and two vertices in level i − 1. It is evident that the two vertices that v is adjacent to in level i + 1 has to be in the same critical set. The same argument follows for the two vertices in level i − 1 and thus we have the following observation.

Sr ♦ Proposition 4.3.2. Let C = i=0 Cj where Cj is the collection of disjoint minimal (r) critical sets in level j as deﬁned in Lemma 4.3.1. For a vertex v ∈ Vi , i ∈ {1, 2, . . . , r − 1},

|N(v) ∩ C| = 3.

In particular, |N(v) ∩ Cj| = 1 for j ∈ {i − 1, i, i + 1}.

We note that the power domination numbers of BF(1) and BF(2) are 1 and 2 respectively.

We now discuss the power domination of BF(r) such that r ≥ 3.

♦ Lemma 4.3.3. Let BF(r) denote butterﬂy network with r ≥ 3. Then   r 2 if r ∈ {3, 4, 5}, γp(BF(r)) ≥  r r−5 r−2 2 + 3 2 if r ≥ 6.

Proof. We recall that for any graph G with ∆ ≥ 3, there exists a minimum power dominating set S in G such that each vertex in S has degree at least 3. Let the vertex set of BF(r) 4.3. POWER DOMINATION IN BUTTERFLY NETWORK 75

(r) (r) (r) (r) Sr be represented as V = V0 ∪ V1 ∪ · · · ∪ Vr and let S = i=0 Si be the minimum power (r) dominating set of BF(r) where Si = S ∩ Vi . We note that S0 and Sr are empty sets. We r−1 r−1 now claim that |S1| ≥ 2 . Assume the contrary, |S1| < 2 , then by Proposition 4.3.2, there exists at least one critical set in C0 which is not contained in the neighbourhood of S1 and thus

r−1 S is not a power dominating set, a contradiction. In a similar way, |Sr−1| ≥ 2 . We note that

r−2 r−2 r−1 r−1 |C2| = 2 and |Cr−2| = 2 and since |S1| ≥ 2 and |Sr−1| ≥ 2 , there is a possibility that N(S1) and N(Sr−1) hits all minimal critical sets in C2 and Cr−2 respectively. Thus, for 3 ≤ r ≤ 5,

r γp(BF(r)) ≥ 2 .

It is to be noted that N[S1] and N[Sr−1] can possibly hit all critical sets in C0, C1, C2, Cr,

r−2 Cr−1 and Cr−2. Thus, for every three levels `, 3 ≤ ` ≤ r − 3, at least 2 vertices have to be chosen in S by Proposition 4.3.2. Thus,

r − 5 γ (BF(r)) ≥ 2r + 2r−2. p 3

r ♦ Lemma 4.3.4. For r ∈ {3, 4, 5}, γp(BF(r)) ≤ 2 .

Proof. For i ∈ {2, r − 1}, deﬁne Si = {(x, i):(xi, xi+1) ∈ {(0, 0), (1, 1)}}, and then put S S = i∈{2,r−1} Si. Then N[S] is a zero forcing set.

r + 1 ♦ Lemma 4.3.5. For r ≥ 6, γ (BF(r)) ≤ 2r−1. p 3

Proof. Let I = {i : 1 ≤ i ≤ r−1, i ≡ 1 (mod 3)}∪{r−1}, and note that |I| = d(r+1)/3e. S For i ∈ I, set Si = {(x, i):(xi, xi+1) ∈ {(0, 0), (1, 1)}}, and then put S = i∈I Si. Then S (r) N(S) = i∈[0,r]\I Vi , and N[S] is a zero-forcing set.

CHAPTER 5

Zero forcing number in terms of girth and minimum

degree

In this chapter, we make the following contributions:

• In Section 5.1, we prove a conjecture which was originally posed by Davila and Kenter

in [24].

• In Section 5.2, we provide better bounds for zero forcing number in graphs with large

girth and minimum degree.

5.1. Proof of the conjecture

In this section, we prove a conjecture which was originally posed by Davila and Kenter in [24].

Conjecture 2. ([24]) If G is a graph with girth g ≥ 3 and minimum degree δ ≥ 2, then

(5.1.1) Z(G) ≥ δ + (δ − 2)(g − 3).

There was many attempts to resolve conjecture2.

• Genter, Penso, Rautenbach, and Souzab [42] showed that inequality (5.1.1) is true for

graphs with girth g = {3, 4}.

• Genter and Rautenbach [41] showed that inequality (5.1.1) is true for graphs with

girth g = {5, 6}.

• Davila and Henning [22] showed that inequality (5.1.1) is true for graphs with girth

g = {7, 8, 9, 10}.

• Davila and Kenter [24] showed that inequality (5.1.1) is true for graphs with girth

g ≥ 7 and suﬃciently large minimum degree δ.

The contents of this chapter reproduce original results from [23].

77 78 5. ZERO FORCING NUMBER IN TERMS OF GIRTH AND MINIMUM DEGREE

We remark that Conjecture2 has been resolved whenever g ≤ 10. Next we provide our main result which resolves Conjecture2 in the aﬃrmative.

♦ Theorem 5.1.1. Let G be a graph with girth g ≥ 11 and minimum degree δ ≥ 2. Then inequality (5.1.1) is true.

Before proceeding to the proof of Theorem 5.1.1, we require some tools that will aid us in our proof.

The maximum number of edges in a simple graph of order n and girth at least `+1 is denoted by ex(n; {C3,C4,...,C`}), often referred to as extremal function. The following theorem will be used in the proof of our main result.

Theorem 5.1.2 ([2]). Let ` ≥ 4 and ` + 1 ≤ n ≤ 2` be integers. Then   n if ` + 1 ≤ n ≤ b3`/2c,   ex (n; {C3,C4,...,C`}) = n + 1 if b3`/2c + 1 ≤ n ≤ 2` − 1,    n + 2 if n = 2`.

This statement will be used in form of the following corollaries.   k − 1 for k ∈ {11, 12, 13}, Corollary 5.1.3. For k ≥ 11, ex(k−2, {C3,C4,...,Ck−6}) =  k − 2 for k ≥ 14.

Proof. Using Theorem 5.1.2 with n = k − 2 and ` = k − 6, it can easily veriﬁed that

b(3k − 18)/2c + 1 ≤ k − 2 ≤ 2k − 13 for k ∈ {11, 12, 13}

k − 5 ≤ k − 2 ≤ b(3k − 18)/2c for k ≥ 14.

Corollary 5.1.4. For k ≥ 11, ex(k − 2, {C3,C4,...,Ck−4}) = k − 2.

Proof. Using Theorem 5.1.2 with n = k − 2 and ` = k − 4, it can easily veriﬁed that

k − 3 ≤ k − 2 ≤ b(3k − 12)/2c for k ≥ 11. 5.1. PROOF OF THE CONJECTURE 79

We are now ready to proceed with the proof of our main theorem.

Proof of Theorem 5.1.1. Let G be a graph with minimum degree δ ≥ 2 and girth g ≥ 11. Suppose S ⊆ V is a zero forcing set with cardinality |S| ≤ δ + (δ − 2)(g − 3) − 1. Let x1, . . . , xt be a chronological list of forcing vertices resulting in all of V becoming blue starting with S as an initial set of blue vertices. Let S¯ = V \ S be the set of initially red vertices. Since

G is a graph with minimum degree δ ≥ 2, and girth g ≥ 5, we have n ≥ g(δ − 1). Hence, we obtain the chain of inequalities

|S¯| = n − |S| ≥ g(δ − 1) − (δ − 2)(g − 3) − δ + 1 = g + 2δ − 5 ≥ g − 1.

We next set X = {x1, . . . , xg−2}, and slightly modifying the notation of [22], we set S1 =

S ∩ N(x1) and

 i−1  [ Si = S ∩ N(xi) \ N[xj] for i = 2, . . . , g − 2, j=1

g−2 ∗ [ SX = Si, i=1

∗ SX = X ∩ (S \ SX ).

∗ Then SX and SX are disjoint subsets of S, and consequently

∗ (5.1.2) |S| ≥ |SX | + |SX |.

Since x1 ∈ S ∩ X, and x1 6∈ Si for all i ∈ [g − 2], we have x1 ∈ SX and thus |SX | ≥ 1. Let H = (X,E0), be the graph with vertex set X and edge set

0 E = {{xi, xj} : {xi, xj} ∈ E(G) or v ∈ N(xi) ∩ N(xj) for some v ∈ V \ X} .

Let m0 = |E0| be the size of H. 80 5. ZERO FORCING NUMBER IN TERMS OF GIRTH AND MINIMUM DEGREE

♦ Remark 5.1.5. Variants of the following two lemmas (Lemma 5.1.6 and Lemma 5.1.7) have been proved in [22]. Since our deﬁnition of the graph H is slightly diﬀerent from the one in [22] (and in order to keep this thesis self-contained), we present complete proofs.

∗ 0 Lemma 5.1.6. |SX | ≥ (δ − 1)(g − 2) − m .

Proof. Since xi is a forcing vertex in step i, we have   i−1 [ N(xi) ∩ S ∪ N[xj] = degG(xi) − 1.

j=1

0 Note that for every edge {xj, xi} ∈ E with j < i, we have either

•{ xj, xi} ∈ E and N(xi) ∩ N(xj) = ∅, or

•{ xj, xi} 6∈ E and |N(xi) ∩ N(xj)| = 1.

This implies

i−1 [ 0 N(xi) ∩ N[xj] ≤ |{xj : j < i and {xi, xj} ∈ E }| ,

j=1 and consequently   i−1 [ 0 |Si| = S ∩ N(xi) \ N[xj] ≥ degG(xi) − 1 − |{xj : j < i and {xi, xj} ∈ E }|.

j=1

Using the fact that the sets Si are pairwise disjoint, we obtain

g−2 g−2 ∗ X X 0 0 |SX | = |Si| ≥ (degG(xi) − 1 − |{xj : j < i and {xi, xj} ∈ E }|) ≥ (δ − 1)(g − 2) − m , i=1 i=1

Pg−2 0 0 where we used i=1 |{xj : j < i and {xi, xj} ∈ E }| = m .

0 0 Lemma 5.1.7. If x ∈ X \ SX , then distG(x, x ) = 1 for some x ∈ X.

¯ ∗ ¯ Proof. If x ∈ X \ SX , then x ∈ S or x ∈ SX . If x ∈ S, then at some point in the forcing 0 0 ∗ process a vertex x ∈ X forces x, which implies distG(x, x ) = 1. If x ∈ SX , then x belongs to some Si, i ∈ [g − 2], which implies x ∈ N(xi), i.e., distG(x, xi) = 1. 5.1. PROOF OF THE CONJECTURE 81

Next observe that Lemma 5.1.6, inequality (5.1.2), and our assumption on the cardinality of S, together provide the inequality

∗ 0 (δ − 2)(g − 3) + δ − 1 ≥ |S| ≥ |SX | + |SX | ≥ |SX | + (δ − 1)(g − 2) − m which implies

0 (5.1.3) m ≥ g − 3 + |SX |.

Note that the girth of H is at least g/2, since every edge in H corresponds to an edge or a path of length 2 in G, and thus

0 m ≤ ex(g − 2, {C3,...,Cb(g−1)/2c}).

Since g−2 ≤ 2b(g−1)/2c, it follows by Theorem 5.1.2, that m0 ≤ g. Thus, m0 ∈ {g−2, g−1, g}.

That is, we have three separate cases to consider. We handle these cases next.

0 Case 1.: If m = g − 2 then |SX | = 1, hence |X \ SX | = g − 3 and by Lemma 5.1.7 at least g − 3 edges of H correspond to edges of G. Consequently, a cycle of length k in

H leads to a cycle of length at most k + 1 in G (as there is at most one edge in the

cycle that corresponds to a path of length two in G). Therefore, H does not contain

0 any cycle, hence m 6 g − 3, which is the required contradiction. 0 Case 2.: If m = g − 1 then |SX | 6 2, hence |X \ SX | ≥ g − 4 and by Lemma 5.1.7 at least g − 4 edges of H correspond to edges of G. Consequently, a cycle of length k in

H leads to a cycle of length at most k + 3 in G, and therefore the girth of H is at least

0 g − 3. By Corollary 5.1.4 this implies m 6 g − 2, which is the required contradiction. 0 Case 3.: If m = g then |SX | 6 3, hence |X \ SX | ≥ g − 5 and by Lemma 5.1.7 at least g − 5 edges of H correspond to edges of G. Consequently, a cycle of length k in H

leads to a cycle of length at most k + 5 in G, and therefore the girth of H is at least

0 g − 5. By Corollary 5.1.3 this implies m 6 g − 1 which is the required contradiction.

This completes the proof of the theorem, and the conjecture presented in [24] is resolved in the aﬃrmative. 82 5. ZERO FORCING NUMBER IN TERMS OF GIRTH AND MINIMUM DEGREE

5.2. Better bounds for large girth and minimum degree

Let f(g, δ) denote the minimum zero forcing number over all graphs of girth g and minimum degree δ. Theorem 5.1.1 provides a lower bound for f, and from [24] we know that this bound is tight in the following cases:

• f(g, 2) = 2 for all g ≥ 3 (the g-cycle),

• f(3, δ) = δ for all δ ≥ 1 (the complete graph Kδ+1),

• f(4, δ) = 2δ − 2 for all δ ≥ 2 (the complete bipartite graph Kδ,δ),

• f(4, 3) = 4 (the 3-cube),

• f(5, 3) = 5 (the Petersen graph),

• f(6, 3) = 6 (the Heawood graph).

Consequently, the smallest open cases are the following.

♦ Question 5.2.1. We know 7 ≤ f(7, 3) ≤ 8 and 8 ≤ f(8, 3) ≤ 10. Can we close these gaps?

♦ Question 5.2.2. We know f(5, 4) ≥ 8. What is the best upper bound we can come up with?

Using essentially the same argument as in the proof of Theorem 5.1.1, one can prove that the bound (5.1.1) is not sharp in general (for instance f(g, δ) ≥ δ + (g − 3)(δ − 2) + 1 for g ≥ 14,

δ ≥ 3). This motivated us to look for better lower bounds especially for graphs with large girth.

In the following we derive some lower bounds that are better than (5.1.1) for large values of δ and g. Let n(g, δ) be the minimum number of vertices in a graph with girth at least g and minimum degree δ. In [4] it is proved that for δ ≥ 2,   2 k−1 1 + δ 1 + (δ − 1) + (δ − 1) + ··· + (δ − 1) if g = 2k + 1, (5.2.1) n(g, δ) ≥  2 k−1 2 1 + (δ − 1) + (δ − 1) + ··· + (δ − 1) if g = 2k.

♦ Theorem 5.2.3. For any positive integer α,

αδ (α + 1)f(g, δ) ≥ n g, . α + 1 5.2. BETTER BOUNDS FOR LARGE GIRTH AND MINIMUM DEGREE 83

Proof. Suppose G = (V,E) has minimum degree δ and girth g, and S ⊆ V is a zero forcing set with (α + 1)|S| < n (g, αδ/(α + 1)). This implies (α + 1)|S| < n(g, δ) ≤ |V |, and therefore at least r = α|S| forcing steps are needed. Let S0 be the set of vertices that are colored after r steps. Then |S0| = (α + 1)|S| and at least r vertices in S0 have at least δ neighbors in

S0. Therefore, the average degree of the graph induced by S0 is at least rδ/|S0| = αδ/(α + 1).

Since the girth of this induced subgraph is at least g, we have

αδ (α + 1)|S| = |S0| ≥ n g, , α + 1 and this is the required contradiction.

(δ − 2)k ♦ Corollary 5.2.4. If g = 2k + 1 and δ ≥ 4, then f(g, δ) ≥ . δ

Proof. Applying Theorem 5.2.3 with α = δ − 1 and using (5.2.1), we obtain

1 1 (δ − 2)k − 1 (δ − 2)k f(g, δ) ≥ n(g, δ − 1) ≥ 1 + (δ − 1) ≥ . δ δ δ − 3 δ

2 ♦ Corollary 5.2.5. If g = 2k and δ ≥ 4, then f(g, δ) ≥ (δ − 2)k−1 − 1. δ

Proof. Applying Theorem 5.2.3 with α = δ − 1 and using (5.2.1), we obtain

1 2 (δ − 2)k − 1 2 f(g, δ) ≥ n(g, δ − 1) ≥ ≥ (δ − 2)k−1 − 1 . δ δ δ − 3 δ 5 h i ♦ Corollary 5.2.6. If g = 2k + 1, then f(g, 3) ≥ (3/2)k − 1 . 6

Proof. Applying Theorem 5.2.3 with α = 5 and using (5.2.1), we obtain

1 1 5 (3/2)k − 1 5 f(g, 3) ≥ n(g, 5/2) ≥ 1 + ≥ (3/2)k − 1 . 6 6 2 1/2 6

2 h i ♦ Corollary 5.2.7. If g = 2k, then f(g, 3) ≥ (3/2)k − 1 . 3

Proof. Applying Theorem 5.2.3 with α = 5 and using (5.2.1), we obtain

1 1 (3/2)k − 1 2 f(g, 3) ≥ n(g, 5/2) ≥ ≥ (3/2)k − 1 . 6 3 1/2 3 84 5. ZERO FORCING NUMBER IN TERMS OF GIRTH AND MINIMUM DEGREE

♦ Remark 5.2.8. A stronger result of Theorem 5.2.3 is established in [65]. Also an alterna- tive proof of Conjecture2 which does not depend on the previous results has recently appeared in [40]. CHAPTER 6

Power domination and resolving-power domination in

graphs

In this chapter, we make the following contributions:

• In Section 6.1, we establish a lower bound technique for power domination number.

• In Section 6.2, we introduce a new variant of power domination problem called resolving-

power domination problem.

As stated in Chapter 2.3 on page 24, power domination problem is a minimisation problem and thus finding good lower bounds are of interests. An effort in that direction is made in Section 6.1 in which a technique for finding lower bound is presented. Unlike critical sets mentioned in

Section 4.1 which applies to all graphs in general, this lower bound technique depends heavily on the structure of the graph. Thus, in some graphs, it is possible that we yield good lower bounds, while in other graphs, it is possible that we do not.

6.1. Power domination-subgraph relation

We begin this section with a fundamental result and illustrate its application by ﬁnding the power domination number of some graphs modelling chemical structures. Recall that in a graph G with a power dominating set S, a vertex u is said to be observed if either u ∈ N[S] or u is forced by some vertex v during the propagation process of power domination.

♦ Theorem 6.1.1. [Power Domination-Subgraph Relation]

Let H1,H2,...,Hk be pairwise disjoint subgraphs of G satisfying the following conditions S (1) V (Hi) = V1(Hi) V2(Hi) where V1(Hi) = {x ∈ V (Hi): x ∈ N(y) for some y ∈

V (G) \ V (Hi)} and V2(Hi) = {x ∈ V (Hi): x∈ / N(y) for any y ∈ V (G) \ V (Hi)}.

(2) V2(Hi) 6= ∅ and |N(x) ∩ V2(Hi)| ≥ 2 for each x ∈ V1(Hi).

The contents of this chapter reproduce original results from [85] and [86].

85 86 6. POWER DOMINATION AND RESOLVING-POWER DOMINATION IN GRAPHS

If V1(Hi) is observed and `i is the minimum number of vertices required to observe V (Hi), then

k X γp(G) ≥ `i. i=1

Proof. We need to show that from each subgraph Hi in G, at least `i vertices belong to any power dominating set S. We prove by the method of contradiction. Let us assume that S is a power dominating set of G with S ∩ V (Hi) = ∅ for some i. Now two cases arise:

(1) N(S) ∩ V (Hi) 6= ∅

(2) N(S) ∩ V (Hi) = ∅

In either of these cases, vertices in V2(Hi) are not observed as every vertex in V1(Hi) is adjacent to at least two vertices in V2(Hi). Further no other vertices outside V (Hi) can observe V2(Hi) as |N(S) ∩ V (Hi)| is at most |V1(Hi)|, this implies that at least `i vertices must belong to S, where `i is the minimum number of vertices required to observe V (Hi) with V1(Hi) observed.

Thus, S is not a power dominating set contradicting the assumption that S ∩ V (Hi) = ∅. As k P the argument holds true for all subgraphs Hi, i ∈ {1, 2, . . . , k}, we have γp(G) ≥ `i. i=1

The rest of the section deals with the illustrations of Theorem 6.1.1 deducing already proved results.

Let us recall that a vertex in a tree adjacent to a leaf is called a support vertex and a vertex adjacent to two or more leaves is called a strong support vertex.

Theorem 6.1.2 ([55]). If v is a strong support vertex in a graph G, then v is in every minimum power dominating set.

This readily follows from Theorem 6.1.1 by considering the strong support vertex along with its adjacent leaves as the subgraph together with the fact that for any graph G with

∆ ≥ 3, there exists a minimum power dominating set S in G such that each vertex in S has degree at least 3 [55].

Definition 6.1.3. An rth complete binary tree BT (r) is a graph whose node set is {0, 1, 2,..., 2r−

j 2} and edge set is {(i, j): b 2 c = i}. A vertex v of a tree is said to be at level j if its distance from the root is j − 1. There are r levels in BT (r). 6.1. POWER DOMINATION-SUBGRAPH RELATION 87

h−2 Theorem 6.1.4 ([55]). Let G be a complete binary tree of height h. Then γp(G) ≥ 2 .

Theorem 6.1.4 can be deduced from Theorem 6.1.1 by taking each Hi as K1,2, induced by the vertices of complete binary tree in level h − 1 and h.

Definition 6.1.5. The WK-recursive networks can be constructed recursively by grouping basic modules. Any d-node complete graph can serve as the basic module. We use WK(d, t) to denote a WK-recursive network of level t whose basic modules are some d-node complete graph, where d > 1 and t ≥ 1. We deﬁne WK(d, t) formally as follows:

V (WK(d, t)) = {(atat−1 . . . a1): ai ∈ Zd for i ∈ [t]}

Vertex (atat−1 . . . a1) is adjacent to

• to all vertices (atat−1 . . . a2b), where b ∈ Zd, b 6= a1 and j−1 • to a vertex (atat−1 . . . aj+1aj−1(aj) ) if there exists j such that 2 ≤ j ≤ t, aj−1 =

j−1 aj−2 = ... = a1 and aj 6= aj−1 where (aj) denotes j − 1 consecutive aj’s.

Figure 6.1. WK(3, 3) containing three copies of WK(3, 2).

We note that WK(d, t) is a recursive structure. It consists of d copies of WK(d, t − 1) or d2 copies of WK(d, t − 2) and so on. Thus, WK(d, t) contains dt−i copies of WK(d, i). See

Figure 6.1 for WK(3, 3). By Theorem 6.1.1, at least one vertex from each copy of WK(3, 2) should belong to any power dominating set. Thus generalising the above observation, one can 88 6. POWER DOMINATION AND RESOLVING-POWER DOMINATION IN GRAPHS show that at least d − 2 vertices from each copy of WK(d, 2) in WK(d, t) should belong to any power dominating set [84, Theorem 2.3].

Theorem 6.1.6 ([28],[84]). Let G be WK-recursive network WK(d, t) with d, t ≥ 3, then

t−2 γp(G) ≥ (d − 2) × d .

In the remaining subsections of this section, we outline the application of Theorem 6.1.1 to solve the power domination problem of certain graphs modelled from chemical structures.

6.1.1. Silicate networks. Silicate networks were proposed as ﬁxed interconnection parallel architectures in [74]. The authors motivated its construction from the molecular structure of a chemical compound SiO4. The authors describe the construction of a silicate network from a honeycomb network. The honeycomb network HC(1) is a hexagon. The honeycomb network

HC(2) is obtained by adding six hexagons to the boundary edges of HC(1). Inductively, honeycomb network HC(r) is obtained from HC(r − 1) by adding a layer of 6(r − 1) hexagons around the boundary edges of HC(r − 1). The parameter r of HC(r) is called the dimension of

HC(r). The number of vertices and edges of HC(r) are 6r2 and 9r2 −3r respectively. Consider a honeycomb network HC(r) of dimension r. Place silicon ions on all the vertices of HC(r).

Subdivide each edge of HC(r) once. Place oxygen ions on the new vertices. Introduce 6r new pendant edges one each at the 2-degree silicon ions of HC(r) and place oxygen ions at the pendent vertices. With every silicon ion associate the three adjacent oxygen ions and form a tetrahedron. The resulting network is a silicate network of dimension r, denoted SL(r). The number of nodes in SL(r) is 15r2 + 3r and the number of edges of SL(r) is 36r2 [74].

6.1.1.1. Algorithm to generate a power dominating set in silicate networks. We use the following terminology in this section. It is clear that the graph HC(r) contains HC(k) for every k, 1 ≤ k ≤ r. The graph induced by V (HC(k)) \ V (HC(k − 1)), 1 ≤ k ≤ n, is called a bounding cycle of HC(k) with V (HC(0)) = ∅. We now propose an algorithm to generate a power dominating set in silicate networks.

Input: SL(r).

Step 1: Consider each K4 in the silicate network SL(r) as a vertex. Join two vertices

0 by an edge if the corresponding K4s share a common vertex. The resulting graph obtained is HC(r). 6.1. POWER DOMINATION-SUBGRAPH RELATION 89

Figure 6.2. Illustration of Step 1 of the algorithm on S(2)

Step 2: From each bounding cycle of HC(i), 1 ≤ i ≤ r, in HC(r), choose alternate edges.

Step 3: For every edge chosen in Step 2, take the common vertex in the corresponding silicate network SL(r) and put it in S.

Figure 6.3. Step 2 and Step 3 of the algorithm

Output: A power dominating set S of SL(r) with cardinality 3r2.

Proof of Correctness: To show |D| = 3r2, we note that the length of the bounding cycle of HC(k), 1 ≤ k ≤ r is 6k2 − 6(k − 1)2 = 12k − 6. From each bounding cycle, alternate edges

Pr 2 are chosen. Thus, |D| = k=1(6k − 3) = 3r . To prove that each vertex in SL(r) is observed, it is enough to show that the common vertex considered in Step 3 observes the two copies of

K4. This is obvious as the common vertex is adjacent to all the vertices in the two copies of

2 K4. Hence, the set S is power dominating set with cardinality 3r . This algorithm leads to the following result.

2 ♦ Theorem 6.1.7. Let G be SL(r). Then γp(G) ≤ 3r . 90 6. POWER DOMINATION AND RESOLVING-POWER DOMINATION IN GRAPHS

2 We call a one point union of two copies of K4 in SL(r) as a twin K4. There are 3r edge disjoint copies of twin K4’s in SL(r).

2 ♦ Theorem 6.1.8. Let G be SL(r). Then γp(G) ≥ 3r .

Proof. In order to prove the above theorem, it is enough to show that at least one vertex from each copy of a twin K4 belongs to any power dominating set S. Consider a copy of twin

K4, X, as shown in Figure 6.4. Applying Theorem 6.1.1 on X with V1(X) = {x1, x2, x3, x4} and V2(X) = {y1, y2, y3}, we have |S ∩ V (X)| ≥ 1.

x1 x4

y1 y2 y3

x2 x3

Figure 6.4. Subgraph X of Theorem 6.1.8

2 Since this is true for all 3r edge disjoint copies of twin K4 in SL(r), we have γp(G) ≥

2 3r .

In view of Theorem 6.1.7 and Theorem 6.1.8 we can now state the following theorem.

2 ♦ Theorem 6.1.9. Let G be SL(r). Then γp(G) = 3r .

6.1.2. Rhenium trioxide lattice. Rhenium trioxide is an inorganic compound with formula ReO3 [77]. It consists of rhenium atoms and oxygen atoms. It is a red solid with a metallic lustre. Rhenium trioxide forms a crystal with a primitive cubic unit cell. The unit cell of ReO3 is shown in Figure 6.5. The vertices marked in solid circles are oxygen atoms and the vertices marked in hollow circles are rhenium atoms.

We now model rhenium trioxide as a graph.

♦ Definition 6.1.10. A p×q×r ReO3 lattice denoted by RO(p, q, r) is a three-dimensional array of p rows of unit cells along the X-axis, q columns of unit cells along the Y-axis and r pages of unit cells along the Z-axis. 6.1. POWER DOMINATION-SUBGRAPH RELATION 91

Figure 6.5. Unit cell of Rhenium trioxide

The structure of ReO3 is similar to the famous mathematical architecture, 3-dimensional grid (3d-grid) of the Euclidean space. A 3d-grid M(p, q, r) is deﬁned as the cartesian product

Pp × Pq × Pr of paths. The rhenium trioxide of dimension p × q × r, denoted as RO(p, q, r), is constructed as follows: (1) Draw the corresponding 3d-grid, M(p, q, r). These vertices correspond to rhenium atoms. (2) Subdivide each edge of M(p, q, r) to account for all oxygen atoms. We shall now label vertices of RO(p, q, r). The rhenium atoms will receive the same

0 0 0 label as the vertices of the 3d-grid. The oxygen atom between two rhenium atoms (x , y , z )

0 00 0 00 0 00 00 00 00 x +x y +y z +z and (x , y , z ) will receive the label (x, y, z) where x = 2 , y = 2 and z = 2 . 0 0 0 The number of vertices and edges in RO(p, q, r) is respectively. Two vertices (u , v , w ) and

00 00 00 0 00 0 00 0 00 1 (u , v , w ) are adjacent if |(u − u ) + (v − v ) + (w − w )| = 2 .

♦ Theorem 6.1.11. Each unit cell in RO(p, q, r) contributes at least three vertices to any power dominating set S.

Proof. This is an application of Theorem 6.1.1 by considering each unit cell as a subgraph. We need to show at least three vertices are required to observe the vertices of the unit cell. Let us assume that there exists a unit cell U in RO(p, q, r) such that U has two vertices say v1, v2 in S. Let V (U) = X ∪Y where X and Y are sets of rhenium and oxygen atoms, respectively, in

U. We recall that for any graph G with ∆ ≥ 3, there exists a minimum power dominating set S 92 6. POWER DOMINATION AND RESOLVING-POWER DOMINATION IN GRAPHS

in G such that each vertex in S has degree at least 3. Thus, v1, v2 ∈ X and by Theorem 6.1.1, assume N(D − {v1, v2}) ∩ V (U) = X. Then, the following cases arise.

Case 1.: If v1 and v2 lie in the same face of U, then clearly all the oxygen atoms of U in

the face opposite to the one containing v1 and v2 are not observed.

Case 2.: If v1 and v2 do not lie on the same face of U then there are two oxygen atoms in each face of U that are not observed.

By Theorem 6.1.11, at least three vertices from a unit cell belong to any power dominating set. In what follows, we look at the distribution of these vertices.

♦ Corollary 6.1.12. At least three rhenium atoms of a unit cell belong to any power dominating set.

♦ Corollary 6.1.13. The three vertices of a unit cell U of RO(p, q, r) belonging to a power dominating set cannot lie in the same face. (Follows immediately from Case 1 of the proof.)

♦ Corollary 6.1.14. Every face in each unit cell has at least one vertex in any power dominating set.

We shall now discuss the lower bounds obtained for power domination number in RO(p, 1, 1).

♦ Theorem 6.1.15. Let G = RO(p, 1, 1) with p ≥ 1, then   3(p+1)  2 , if p is odd; γp(G) ≥  3p  2 + 1, if p is even.

Proof. RO(p, 1, 1) is composed of p copies of RO(1, 1, 1). By Theorem 6.1.11, every unit cell contributes at least three vertices to any power dominating set S. Also, by Corollary 6.1.13, at most two vertices in each unit cell lie in a face. Thus, each intersecting face can have a

3(p+1) maximum of two vertices. Thus, γp(G) ≥ 2 when p is odd since there are (p + 1)/2 vertex disjoint copies of RO(1, 1, 1). When p is even, RO(p, 1, 1) can be decomposed into p/2 vertex disjoint copies of RO(1, 1, 1) and a face of RO(1, 1, 1). By Corollary 6.1.14, at least one vertex

3p from each face must belong to any power dominating set. Thus, γp(G) ≥ 2 + 1. 6.1. POWER DOMINATION-SUBGRAPH RELATION 93

It is not diﬃcult now to see how we extend the result of Theorem 6.1.15 for RO(p, 1, 1) to

RO(p, q, 1). We basically count for the number of disjoint copies of RO(p, 1, 1) in RO(p, q, 1).

♦ Theorem 6.1.16. Let G = RO(p, q, 1) with p, q ≥ 1, then   3pq+2q+3p+2 , p even, q odd;  4   3pq+2q+2p  4 , p even, q even; γp(G) ≥  3pq+3q+2p+2 , p odd, q even;  4   3pq+3q+3p+3  4 , p odd, q odd.

Proof. We have the following cases.

q+1 Case 1.: When p is even and q is odd, there are 2 disjoint copies of RO(p, 1, 1) in 3p q+1 3pq+2q+3p+2 RO(p, q, 1). Thus, γp(G) ≥ ( 2 + 1) × ( 2 ) = 4 . q Case 2.: When p and q are even, there are 2 disjoint copies of RO(p, 1, 1) in RO(p, q, 1). p Moreover, in addition 2 vertices have to be included in the power dominating set S 3p so that each face of the unit cell satisﬁes Corollary 6.1.14. Thus, γp(G) ≥ ( 2 + 1) × q p 3pq+2q+2p ( 2 ) + ( 2 ) = 4 . q+1 Case 3.: When p and q are odd, there are 2 disjoint copies of RO(p, 1, 1) in RO(p, q, 1). 3(p+1) q+1 3pq+3q+3p+3 Thus, γp(G) ≥ ( 2 ) × ( 2 ) = 4 . p+1 Case 4.: When p is odd and q is even, there are 2 disjoint copies of RO(1, q, 1) in 3q p+1 3pq+2p+3q+2 RO(p, q, 1). Thus, γp(G) ≥ ( 2 + 1) × ( 2 ) = 4 .

Repeating the same arguments for RO(p, q, r) on can obtain the following result.

♦ Theorem 6.1.17. Let G = RO(p, q, r) with p, q, r ≥ 1, then   r+1  γp(RO(p, q, 1)) × 2 , when r is odd; γp(RO(p, q, r)) ≥  r  γp(RO(p, q, 1)) × 2 + k, when r is even.

p+1 q+1 where k = b 2 c × b 2 c.

♦ Remark 6.1.18. We have not been able to show that these bounds are tight. We leave it as an open problem. 94 6. POWER DOMINATION AND RESOLVING-POWER DOMINATION IN GRAPHS

6.2. Resolving-power dominating set

In this section, we introduce a new variant of power domination problem called resolving- power domination problem. As mentioned in Section 2.3, the power domination problem arose in the context of monitoring electric power networks. Electric power networks are complex and are open to possible failures. For studies involving safeguards applications for graphical models of facilities or for multiprocessor networks, various types of protection sets have been studied where the objective is to precisely locate an intruder such as a thief, saboteur, or ﬁre, or a faulty processor. Locating sets (for which a protection device can determine the distance to a faulty vertex) were introduced by Slater in [83] and subsequently by Harary and Melter [53] where they were called metric bases. This motivated us to merge the concepts of resolving set and power dominating set to form resolving-power dominating set wherein the vertices in the set not only observe the vertices but also locates the position of each vertex.

Recall that a dominating set of a graph G(V,E) is a set S of vertices such that every vertex

(node) in V \ S has at least one neighbor in S. The problem of ﬁnding a dominating set of minimum cardinality is an important problem that has been extensively studied. The minimum cardinality of a dominating set of G is its domination number, denoted by γ(G)[56]. We recall, in a connected graph G, the distance dist(u, v) between two vertices u, v ∈ V is the length of a shortest path between them. Let W = {w1, w2 . . . wk} be an ordered set of vertices of G and let v be a vertex of G. The representation r(v|W ) of v with respect to W is the k-tuple

(dist(v, w1), dist(v, w2),..., dist(v, wk)). If distinct vertices of G have distinct representations with respect to W , then W is called a locating/resolving set for G [19, 83]. A resolving set of minimum cardinality is called a metric basis for G and this cardinality is the metric dimension of G, denoted by dim(G)[53]. A set S ⊆ V is called a metric-locating-dominating set of a connected graph G if the set S is both a locating as well as a dominating set. The minimum cardinality of such a set is called metric-locating-domination number denoted by η(G)[57]. Our focus is on a variation called the resolving-power dominating set problem.

A set S ⊆ V is called a resolving-power dominating set of a connected graph G if the set

S is both a resolving as well as a power dominating set. The minimum cardinality of such a 6.2. RESOLVING-POWER DOMINATING SET 95

set is called resolving-power domination number denoted by ηp(G). We shall illustrate with an example. In the example in Figure 6.6, dim(G) = 2, γp(G) = 2, η(G) = 4 and ηp(G) = 3.

(i) (ii) (iii ) (iv)

Figure 6.6. Square bullets depict (i) Metric basis (ii) Minimum power dominating set (iii) Minimum metric-locating-dominating set and (iv) Minimum resolving-power-dominating set of G.

The following results are relatively straightforward.

♦ Theorem 6.2.1. max{γp(G), dim(G)} ≤ ηp(G) ≤ γp(G) + dim(G)

Proof. Suppose ηp(G) < max{γp(G), dim(G)}. Let S ⊆ V be a resolving-power dominating set of cardinality ηp(G). But this is a contradiction to the deﬁnition of γp(G) and dim(G). Consider a set S which comprises of all vertices in metric basis B and all vertices in the power dominating set D. It is easy to see that S is a resolving-power dominating set. Thus,

ηp(G) ≤ γp(G) + dim(G)

♦ Theorem 6.2.2. ηp(G) = 1 if and only if G is a path.

Proof. For any graph G, ηp(G) ≥ 1. For the reverse inequality, it is easy to see that any pendant vertex of G will observe and locate all vertices of G.

Suppose now ηp(G) = 1 and G is not a path. Then G is either a cycle or has a vertex v such that deg(v) ≥ 3. In either case, dim(G) ≥ 2 (A graph G = (V,E) has dim(G) = 1 if and only if

G is a path [67]) and thus by Theorem 6.2.1, ηp(G) ≥ 2, a contradiction to the assumption.

♦ Theorem 6.2.3. ηp(G) = n − 1 if and only if G is the complete graph Kn.

Proof. It is shown in [19] that dim(Kn) = n − 1 and by Theorem 6.2.1, ηp(G) ≥ n − 1.

Also, choosing any set of n − 1 vertices of Kn would be a resolving-power dominating set, thus

ηp(Kn) ≤ n − 1, which proves the theorem in one direction. 96 6. POWER DOMINATION AND RESOLVING-POWER DOMINATION IN GRAPHS

To prove the other direction i.e. if ηp(G) = n−1 then G is the complete graph Kn, we assume the counter example. Suppose ηp(G) = n − 1 and G 6= Kn. It follows that dim(G) ≤ n − 2 [19]. Now choose n−2 vertices of G in D such that D is a resolving set. Let x, y ∈ V (G)\D. Clearly

D is a power dominating set as at least one of the vertices x or y is adjacent to some vertex in

D and thus ηp(G) ≤ n − 2, a contradiction.

♦ Corollary 6.2.4. 1 ≤ ηp(G) ≤ n − 1 and the bounds are tight.

♦ Theorem 6.2.5. Let G be a complete bipartite graph Km,n. Then ηp(G) = m + n − 2

Proof. Let (X,Y ) be a bipartition of G with |X| = m and |Y | = n. It is known that dim(G) = m + n − 2. Hence by Theorem 6.2.1, ηp(G) ≥ m + n − 2. The reverse inequality is obvious.

The following theorem is an extension of the result in [67].

♦ Theorem 6.2.6. If a graph G has ηp(G) = 2 then G cannot have K5 nor K3,3 as a subgraph.

Proof. Since the resolving-power dominating set is both a resolving and power dominating set, it is evident that the set satisﬁes all properties of a resolving set. It is shown by

Khuller et al. in [67] that if G has a resolving set of size two, then G cannot have K5 nor K3,3 as a subgraph.

It was shown in [67] that if a vertex has r leaves, then at least r − 1 of them should belong to any resolving set. We extend this result to resolving-power dominating set.

♦ Remark 6.2.7. If a vertex has r leaves, then at least r − 1 of them should belong to any resolving-power dominating set.

♦ Remark 6.2.8. ηp(G) ≤ η(G) where ηp(G) and η(G) are the resolving-power domination number and metric-locating-domination number respectively.

The problem of deciding if a graph G has a resolving-power dominating set of cardinality k is shown to be NP-complete in the next section. 6.2. RESOLVING-POWER DOMINATING SET 97

6.2.1. Complexity results of resolving-power domination problem. The basic idea of this proof is due to Khuller et al. [67] and Manuel et al. [75].

Question : Given a graph G(V,E) and an integer k does there exist a set C of k vertices such that G admits a resolving-power dominating set.

♦ Theorem 6.2.9. The resolving-power domination problem is NP-complete.

Proof. It is clear that the resolving-power domination problem is in NP. We need to ﬁnd a function f that maps instances of 3-SAT into instances of the resolving-power domination problem such that:

• Given an arbitrary instance I of 3-SAT, f(I) will be an instance of the resolving-power

domination problem, i.e., f(I) will be a graph G together with an integer k, written

as f(I) = (G, k).

• f(I) can be computed in polynomial time |I|.

• f(I) will be a positive instance if and only if I is a positive instance.

We now deﬁne a function f by construction that, given I, produces G and k. Suppose the variables of I are x1, x2, . . . , xn and the m clauses in the Boolean formula:

(u11 ∨ u12 ∨ u13) ∧ (u21 ∨ u22 ∨ u23) ∧ ... ∧ (um1 ∨ um2 ∨ um3)

The variables ujl are literals and could be any one of xi. The labels j, l refer to the particular position of the literal in the Boolean formula. In what follows 1 ≤ i ≤ n, 1 ≤ j ≤ m and

1 ≤ l ≤ 3.

For each variable xi, we construct a gadget as in Figure 6.7. The nodes Ti and Fi are the ‘true’ and ‘false’ ends of the gadget. The gadget is attached to the rest of the graph only through these nodes. For each clause Cj = (uj1 ∨ uj2 ∨ uj3), we construct a gadget as in Figure 6.8.

1 1 3 If ujl = xi in clause Cj, we add the edges (Ti, cj ), (Fi, cj ) and (Fi, cj ). If ujl =x ¯i in 1 1 3 clause Cj, we add the edges (Ti, cj ), (Fi, cj ) and (Ti, cj ). To complete the construction, for all 1 1 3 k such that neither xk norx ¯k occur in Cj, add the following edges (Tk, cj ), (Fk, cj ), (Tk, cj ) 3 and (Fk, cj ). Thus, the graph G is constructed from the formula F with n variables and m clauses has 6n + 4m nodes. 98 6. POWER DOMINATION AND RESOLVING-POWER DOMINATION IN GRAPHS

1 1 ai bi

2 2 ai bi

Ti Fi

Figure 6.7. Variable gadget for each xi

2 cj

1 3 cj cj

4 cj

Figure 6.8. Clause gadget for each Cj

We begin with a few results required for the remaining study.

1 2 1 2 ♦ Lemma 6.2.10. Let xi be an arbitrary variable. At least one of the vertices {ai , ai , bi , bi } must belong to any resolving-power dominating set D.

Proof. Suppose that some resolving-power dominating set D does not contain any vertex

1 2 1 2 1 2 of {ai , ai , bi , bi }. Then ai and ai are equidistant from every vertex of the graph constructed.

Also, the variable gadget is connected to the rest of the graph only through the vertices Ti and

1 2 1 2 Fi. So even if Ti and Fi are observed, the vertices ai , ai , bi and bi will not be observed by the propagation step, contradicting D is a resolving-power dominating set.

Using essentially the same argument as in Lemma 6.2.10, one can show that at least one

2 4 vertices {cj , cj } of an arbitrary clause Cj belong to any resolving-power dominating set D.

2 4 ♦ Lemma 6.2.11. Let Cj be an arbitrary clause. At least one of the vertices {cj , cj } should belong to any resolving-power dominating set D.

Proof. Suppose that some resolving-power dominating set D does not contain any vertex

2 4 2 4 of {cj , cj }. Then cj and cj are equidistant from every vertex of the graph constructed. The 6.2. RESOLVING-POWER DOMINATING SET 99

T1 F1 T2 F2 T3 F3

1 3 cj cj

Figure 6.9. Reduction ofx ¯1 ∨ x2 ∨ x¯3

1 3 clause gadget is connected to the rest of the graph only through the vertices cj and cj . So even 1 3 2 4 if cj and cj are observed, the vertices cj and cj will not be observed by the propagation step, contradicting D is a resolving-power dominating set.

From Lemma 6.2.10 and Lemma 6.2.11, we have

♦ Lemma 6.2.12. The resolving-power domination number of G is at least m + n.

♦ Lemma 6.2.13. If the clauses are satisﬁable, then there is a resolving-power dominating set of cardinality k.

Proof. Suppose we have a satisfying assignment. We deﬁne a set S as follows. For each

4 clause Cj, we put cj in S and for every variable xi,

1 • If xi is true, put ai in the set S. 1 • If xi is false, put bi in the set S.

We now show that the set S is a resolving-power dominating set i.e., we need to show it is both a resolving set as well as a power dominating set. The set S is clearly a power dominating

1 1 4 set as each variable gadget has either ai or bi in S and each clause gadget has cj in S. These vertices observes all vertices of the graph. To show S is a resolving set, we only need to show

1 3 that the vertices {cj , cj } have distinct representation. For any other pair of nodes, it is easy to ﬁnd a vertex in S which distinguishes them.

1 3 For any clause Cj, we show that cj and cj have diﬀerent representations if the vertices are chosen according to the satisfying assignment. Suppose Cj is satisﬁed by the variable 100 6. POWER DOMINATION AND RESOLVING-POWER DOMINATION IN GRAPHS

xi, a variable occurring as a positive literal in Cj and has the value true in the assignment.

1 1 Corresponding to xi being true, we placed a vertex on ai . From this vertex, cj will be at 3 distance 2 and cj will be at distance 3. Suppose Cj is satisﬁed by the variable xi, a variable occurring as a negative literal in Cj and has the value true in the assignment. Corresponding

1 1 3 to xi being false, we placed a vertex on bi . From this vertex, cj will be at distance 2 and cj will be at distance 3. Thus all nodes are observed and have distinct representations and therefore we have a resolving-power dominating set of size m + n.

♦ Lemma 6.2.14. If there is a resolving-power dominating set of cardinality k, then the clauses are satisﬁable

Proof. Suppose we have a resolving-power dominating set S of cardinality m + n. By Lemmas 6.2.10 and 6.2.11, we know that in any resolving-power dominating set, at least one vertex must be placed within each variable and each clause gadget. We now set an assignment as follows:

1 2 • If ai or ai belongs to the set S, then set xi to be true. 1 2 • If bi or bi belongs to the set S, then set xi to be false.

We now show that this is a satisfying assignment. Consider an arbitrary clause Cj. We will show that at least one of its literals is true. The main idea is in tracing which vertex

1 3 distinguishes cj and cj and showing that the corresponding variable assignment satisﬁes Cj. 4 1 3 For each clause Ck, without loss of generality, put ck in S. cj and cj are at unit distance from 4 4 cj and at distance 3 from ck, k 6= j.

For any variable xp which does not occur in Cj, the vertex in resolving-power dominating

1 3 set in the variable gadget xp is at distance 2 from each of cj and cj . Therefore the only vertex 1 3 that could distinguish cj and cj must be on the variable xq which occurs in Cj. Thus, a vertex distinguishes two nodes only if one of the following two statements hold.

1 (1) xq occurs as a positive literal in Cj and a vertex in the set is placed on either aq or 2 aq; in this case xq is set to true. 1 (2) xq occurs as a negative literal in Cj and a vertex in the set is placed on either bq or 2 bq; in this case xq is set to false.

In either case, the setting of xq is such that it satisﬁes Cj. 6.2. RESOLVING-POWER DOMINATING SET 101

Lemma 6.2.13 and Lemma 6.2.14 together complete the reduction from 3-SAT to resolving- power domination problem. This completes the proof of Theorem 6.2.9.

Repeating the argument in Theorem 6.2.9 for the variable gadget as in Figure 6.10, the resultant graph obtained is a bipartite graph and thus we have the following result.

1 ai 1 2 bi bi 2 ai

3 ai

3 4 bi bi 4 ai

Ti Fi

Figure 6.10. Variable gadget for bipartite graphs

♦ Theorem 6.2.15. Resolving-power domination problem is NP-complete even when re- stricted to bipartite graphs.

6.2.2. Resolving-power domination in trees. We borrow the following deﬁnitions from [67].

Definition 6.2.16. Let T = (V,E) be a tree, and u a speciﬁed vertex in T . Partition the edges of T by the equivalence relation ≡v, deﬁned as follows: two edges e ≡v f if and only if there is a path in T including e and f that does not have v as an internal vertex. The subgraphs induced by the edges of the equivalence classes of E are called the channels of T relative to v.

Definition 6.2.17. For each node v ∈ V of a tree T , the legs of v are the channels of T which are paths. Denote the number of legs at v by `v.

It was shown in [67] that for any tree T , such that T 6= Pn,

X (6.2.1) dim(T ) = (`v − 1) v∈V 102 6. POWER DOMINATION AND RESOLVING-POWER DOMINATION IN GRAPHS

Definition 6.2.18. A vertex of a tree T with degree 3 or greater is called a branching vertex of T .

P ♦ Theorem 6.2.19. Let T be a tree such that T 6= Pn. Then ηp(T ) ≥ (`v − 1) v∈V

Proof. We claim that for every vertex v, at least `v − 1 of its legs contribute to any resolving-power dominating set S. In case there exists a vertex v of G such that only `v − 2 of its legs contribute to some resolving-power dominating set S, then there are two legs of v, say

1 2 1 2 `v and `v such that v1 ∈ `v and v2 ∈ `v are neither observed nor resolved by any vertex in S. P Thus, ηp(T ) ≥ (`v − 1). v∈V P ♦ Theorem 6.2.20. Let T be a tree. Then ηp(T ) ≤ `v v∈V

Proof. Let S be the set of all leaf vertices of the legs of v, for all v ∈ V . Then S is clearly a resolving set (Equation (6.2.1)). It is shown in [3], that we can always ﬁnd a minimum zero forcing set in T such that degree of each vertex is one (Follows from the fact that Z(T ) = P(T )).

This implies that S is a zero forcing set and hence a power dominating set as well.

From Theorem 6.2.19 and Theorem 6.2.20, we have the bounds for the resolving-power domination number for trees.

♦ Theorem 6.2.21. Let T be a tree such that T 6= Pn. Then

X X (`v − 1) ≤ ηp(T ) ≤ `v v∈V v∈V

There is a striking diﬀerence between metric dimension and resolving-power domination number in the case of trees. We shall now identify classes of trees such that dim(T ) = ηp(T ). In Theorem 6.2.20 we deﬁned S as the set of all leaf vertices of the legs of v, for all v ∈ V .

We now remove from S one leaf vertex from each v ∈ V . We show that S is a resolving-power P dominating set of T given the conditions of Theorem 6.2.22. Clearly, |S| = (`v − 1). v∈V P ♦ Theorem 6.2.22. Let T be a tree. Then ηp(T ) = (`v − 1) if and only if v∈V

(1) there is no path u0u1 . . . uk−1uk in T such that deg(u0) ≥ 3, deg(ui) = 2, 1 ≤ i ≤ k−1,

deg(uk) ≥ 3 and 6.2. RESOLVING-POWER DOMINATING SET 103

(2) every branching vertex has at least two legs.

P Proof. (⇒) Assume that S is a resolving-power dominating set with cardinality (`v−1) v∈V as deﬁned above. Suppose that T contains a path u0u1 . . . uk−1uk such that deg(u0) ≥ 3, deg(ui) = 2, 1 ≤ i ≤ k − 1, deg(uk) ≥ 3. The following cases arise:

Case 1.: If the vertices u0 and uk have at least one leg then clearly by the choice of

vertices in S, both u0 and uk have two unobserved vertices as neighbours and hence the

two degree vertices between the branching vertices u0 and uk can never be observed.

Case 2.: At least one of u0 and uk have no leg. We note that by the choice of vertices in S, at ﬁrst time step, all branching vertices with at least two legs are observed. Without

loss of generality, let us assume that u0 has no leg. Then, even if u0 is adjacent to

observed branching vertices, u0 cannot be observed as each of these observed branching

vertices have one unobserved leg and cannot force u0. Similarly, uk cannot be observed

if uk has no leg. If uk has at least one leg, the two degree vertices between u0 and uk cannot be observed by Case 1.

In both these cases, not all vertices of T are observed contradicting that S is a resolving-power dominating set.

Again assume S be a resolving-power dominating such that T has at least one branching vertex with one leg incident to it. In this case, similar to the proof of Case 2, this branching vertex can never be observed, a contradiction to the assumption.

(⇐) Conversely, let us take a tree T such that T does not contain a path u0u1 . . . u`−1u` in

T such that deg(u0) ≥ 3, deg(ui) = 2, 1 ≤ i ≤ `−1, deg(u`) ≥ 3 and every branching vertex has at least two legs. By Remark 6.2.7, at least `v − 1 leaf vertices from each vertex v ∈ T should belong to resolving-power dominating set. It is easy to see that this set is a resolving-power dominating set. This is because each branching vertex is observed and since there is no path u0u1 . . . uk−1uk in T such that deg(u0) ≥ 3, deg(ui) = 2, 1 ≤ i ≤ k − 1, deg(uk) ≥ 3, all the neighbours of a branching vertex are observed except for the vertices from its leg from which no vertex was chosen in S. These vertices are observed as a result of propagation.

♦ Theorem 6.2.23. Let BT (r) be a complete binary tree of height r ≥ 4. Then, ηp(BT (r)) = 2r−1 + 2r−4 104 6. POWER DOMINATION AND RESOLVING-POWER DOMINATION IN GRAPHS

Proof. In a binary tree, there are 2r−2 strong support vertices (The count is basically the number of vertices in level r − 1). These strong support vertices have two leaves each.

By Remark 6.2.7, one leaf vertex from each support vertex should belong to any resolving- power dominating set S. Clearly S is a resolving set by Equation (6.2.1), but it is not a power dominating set. The set with one pendant vertex from each strong support vertex can observe itself and the strong support vertex only. Now our aim is to add minimum vertices into the set

S such that it becomes a power dominating set as well. By Theorem 6.1.1 on page 85, at least one vertex from each copy of BT (4) (induced by vertices of level r − 3, r − 2, r − 1 and r) should belong to the power dominating set. There are 2r−4 copies of BT (4) in BT (r) (The count is

r−2 r−4 the number of vertices in level r − 3). Thus, ηp(BT (r)) ≥ 2 + 2 . To prove the other direction, we construct a resolving-power dominating set of the required cardinality. Construct S such that it consists of one pendant vertex from each support vertex and all vertices in level r − 3. Note that |S| = 2r−2 + 2r−4. Clearly S is a resolving set. We next show S is a power dominating set. By the choice of the vertices in S, the pendant vertices chosen in S will dominate all vertices in level r − 1. The vertices of level r − 3 chosen in S will dominate all vertices of level r − 2 and r − 4. Thus, each strong support vertex will observe its unobserved leaf vertex in level r. Thus, all vertices in level `, r − 4 ≤ ` ≤ r are observed.

A vertex in level r − 5 will be observed by its adjacent vertex in level r − 4. This is due to the fact that each child has only one parent and all vertices in levels `, r − 4 ≤ ` ≤ r are observed. Repeating the same arguments for vertices in level `, 1 ≤ ` ≤ r − 6, we see that all vertices are observed and thus the set S is a resolving-power dominating set and we have

r−2 r−4 ηp(BT (r)) ≤ 2 + 2 . CHAPTER 7

Conclusion

As mentioned in Chapter2, zero forcing and power domination problems were introduced in different contexts and were studied independently by different researchers. In this thesis, we exploited the similarity between the two problems where one can be considered as a variant of the other by interpreting them as set cover problems. The characterisation of zero forcing sets and power dominating sets, with the help of critical sets, provided an effective tool in computing the lower bound of their respective graph invariants, zero forcing number and power domination number.

Having said that, the lower bound obtained in general for graphs by counting disjoint critical sets need not be sharp at all. For example consider a 3 × 3 grid, the number of disjoint critical sets is 2 whereas the zero forcing number is 3. See Figure 7.1.

We believe having a better understanding of critical sets will help us improve the bounds for zero forcing number and power domination number of graphs in general. Below are some questions for which we do not have the solutions but even partial results will deﬁnitely be an advancement in developing a better understanding of critical sets. Let C denote the collection of minimal critical sets, and let Ci denote the collection of minimal critical sets that contain

(b) The third critical set not vertex dis- (a) Two vertex disjoint critical sets joint

Figure 7.1. Critical sets in 3 × 3 grid

105 106 7. CONCLUSION vertex i. We have

( n ) X X n Z(G) = min xi : xi ≥ 1 ∀ W ∈ C, x ∈ {0, 1} i=1 i∈W

We can deﬁne the dual quantity

( ) 0 X X C Z (G) = max yW : yW ≤ 1 ∀ i ∈ V, y ∈ {0, 1}

W ∈Ci W ∈Ci We have that Z0(G) is a lower bound for the zero forcing number i.e.

Z(G) ≥ Z0(G).

This lower bound is certainly not tight in general: For a star G = K1,t with t > 2 leaves we have Z(G) = t − 1 while Z0(G) = dt/2e.

♦ Open Problem 7.0.1. Classify graphs for which Z0(G) = Z(G)?

♦ Open Problem 7.0.2. Is the size of C polynomially bounded?

♦ Open Problem 7.0.3. Is there an eﬃcient way of generating minimal critical sets?

In Chapter 3 on page 35, the zero forcing number and power domination number of iterated line digraphs are obtained using essentially the notion of critical sets.

One of the main results obtained in this chapter is:

♦ Theorem 7.0.4. Let G~ = (V,A) be a digraph with minimum in-degree δin(G~ ) ≥ 1 and minimum out-degree δout(G~ ) ≥ 2. Then

Z(L(G~ )) = |A(G~ )| − |V (G~ )|.

The condition δout(G~ ) ≥ 2 is necessary for the above theorem to be true and therefore we pose this question.

♦ Open Problem 7.0.5. Let G~ = (V,A) be a digraph with minimum in-degree δin(G~ ) ≥ 1 and minimum out-degree δout(G~ ) = 1. What can be said about Z(L(G~ ))? 7. CONCLUSION 107

We have shown in Chapter 3.5 that for a d-regular digraph with d ≥ 2,

2 Z(L(G~ )) = γp(L (G~ ))

We recall the result of Theorem 3.5.7 as follows:

♦ Theorem 7.0.6. If G~ = (V,A) is a digraph with minimum in-degree δin(G~ ) ≥ 1 and

out 2 minimum out-degree δ (G~ ) ≥ 1, then Z(L(G~ )) ≥ γp(L (G~ )) + µ(G~ ), where µ(G~ ) is the minimum number of cycles with only vertices of in-degree 1 in any 1-factor of G~ .

Note that for a d-regular digraph, µ(G~ ) = 0.

♦ Open Problem 7.0.7. What can we say about µ(G~ )? Maybe under some mild additional assumptions, we might have µ(G~ ) = 1. If so, it might be interesting to classify such graphs?

In Chapter 4 on page 55, we solved the zero forcing number and minimum rank problem for butterﬂy graphs.

♦ Theorem 7.0.8. The minimum rank of the butterfly network BF(r) over any field F is given by 2 mrF (BF(r)) = [(3r + 1)2r − (−1)r] , 9 and this is equal to the rank of the adjacency matrix of BF(r). Furthermore, for the butterfly network we have

1 Z (BF(r)) = (r + 1)2r − mrF (BF(r)) = [(3r + 7)2r + 2(−1)r] . 9

To date the best known bounds for the power domination number of butterﬂy networks is given by

♦ Theorem 7.0.9. Let BF(r) be a butterﬂy network with dimension r ≥ 5. Then,

r − 5 r + 1 2r + 2r−2 ≤ γ (BF(r)) ≤ 2r−1. 3 p 3

It is very interesting that for r > 5, we were unable to close the gap between the lower bound and upper bound. We believe that the true value lies in between these bounds. 108 7. CONCLUSION

An r-dimensional Benes network has 2r + 1 levels, each level with 2r nodes. The level 0 to level r nodes in the network form an r-dimensional butterﬂy. The middle level of the Benes network is shared by these butterﬂies [70]. An r-dimensional Benes is denoted by B(r). The

Benes network is often referred to as back to back butterﬂy network.

Our computations with Sage1 shows that the zero forcing number of Benes network is closely related to the zero forcing number of butterﬂy network. We have been able to verify for small dimensions but we were unable to come up with a conclusive proof. Thus we state the following conjecture.

♦ Conjecture 7.0.10. The zero forcing number of Benes network B(r) is given by

Z(B(r)) = 2(Z(BF(r) − Jr−2)

where (Jr) denotes the Jacobsthal sequence which is deﬁned by J0 = 0, J1 = 1 and Jr =

Jr−1 + 2Jr−2 for r > 2.

In Chapter 5 on page 77, we solved the conjecture of Davila and Kenter [24] and thus improved the lower bound of Z(G) ≥ δ for graphs with large girth and minimum degree.

♦ Theorem 7.0.11. Let G be a graph with girth g ≥ 3 and minimum degree δ ≥ 2. Then

Z(G) ≥ δ + (δ − 2)(g − 3).

We used the knowledge of Moore bounds to give a better estimate of Z(G) for graphs with large girth.

Let f(g, δ) denote the minimum zero forcing number over all graphs of girth g and minimum degree δ. Theorem 7.0.11 provides a lower bound for f, and as discussed in Section 5.2, f(7, 3) is the smallest case where the lower bound obtained is not sharp. This leads us to the following questions.

♦ Open Problem 7.0.12. What are upper bounds for f(g, δ)?

♦ Open Problem 7.0.13. What can be said about the asymptotic behaviour of f(g, δ)?

1 Sage: Open Source Mathematical Software. Available at http://www.sagemath.org 7. CONCLUSION 109

♦ Open Problem 7.0.14. In a slightly diﬀerent direction, we can ask for better lower bounds on the zero forcing number under additional assumptions on the structure of the graph.

For instance, it is known that Z(G) ≥ Z(G − e) − 1 for every edge e of G [32]. If G contains a bridge e whose removal leads to two connected components G1 and G2, then this implies

Z(G) ≥ Z(G1) + Z(G2) − 1, and if G has girth g and minimum degree δ then

Z(G) ≥ 2 [(δ − 1) + (g − 3)(δ − 3)] − 1 which is a stronger bound than Theorem 7.0.11 whenever δ ≥ 4.

In Chapter 6 on page 85, we introduced a new variant of power domination problem called resolving power domination problem. We showed that the problem is NP complete.

It is shown that the resolving power domination number ηp(G) satisﬁes

♦ Theorem 7.0.15. max{γp(G), dim(G)} ≤ ηp(G) ≤ γp(G) + dim(G)

It would be interesting to identify graphs in which either of these bounds is tight.

♦ Open Problem 7.0.16. Characterise graphs such that ηp(G) = max{γp(G), dim(G)}?

♦ Open Problem 7.0.17. Characterise graphs such that ηp(G) = γp(G) + dim(G)?

In Theorem 6.2.23, we showed that for binary trees, BT (r), r ≥ 4,

r−2 r−4 ηp(BT (r)) = 2 + 2

This leads us to the next open problem.

♦ Open Problem 7.0.18. Devise an algorithm to ﬁnd the resolving-power domination for general trees.

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