enumeration of the number of spanning trees in some ... - Toubkal

toubkal.imist.ma

enumeration of the number of spanning trees in some ... - Toubkal

1To my daughter Hanin.


ACKNOWLEDGEMENTSThe work presented in this thesis was performed at the Laboratory of MIA "Mathematics,Computer Science and Applications" - UFR of Mathematics, Computer Science andApplications, Department of Mathematics and Computer Science at the Faculty ofSciences Rabat - University of Mohammed V-Agdal.First of all, thanks to ALLAH who awarded me with patienace, courage and selfconsitancy in order to successfully carry out this research work. I would like to warmlythank my worthy of respect supervisor, Mr. Mohamed El MARRAKI, Professor ofHigher Education in the Faculty of Sciences Rabat, for his support, his availability,patience, close collaboration and kind assistance that allowed me to complete this thesis.I express my profound gratitude to Mr. Driss ABOUTAJDINE, Professor of HigherEducation at the Faculty of Sciences Rabat, for the honor that he gave me to evaluatethis work, on one hand, and to chair the thesis committee, on the other hand. Therefore,he finds the expression of my deep gratitude here.I wish to thank Mr. El Mamoun SOUIDI, Professor of Higher Education at theFaculty of Sciences Rabat, and Mr. Abdelmalek AZIZI, professor of Higher Educationat the Faculty of Sciences - University of Mohammed I - Oujda, for having accepted theburden to be the penal secretaries of thesis committee and for their valuable commentsthat helped me to improve this manuscript.I am also quite grateful to Professor Hussain BEN-AZZA, University of Mollay Ismal,ENSAM in Meknes, for giving me an honor to accept the evaluation burden of my thesisbeing thesis examiner and for his kind help on spanning trees and for patiently answeringmy several questions. Very simply, thanks a lot Sir.I extend my thanks to Mr. Mohamed EL KAMILI, Professor in the Faculty ofScience - University of Sidi Mohammed Ben Abdellah - Fes by agreeing his consent toparticipate in the examination committee as its member and also for the discussions wehad together.Finally, my warmest thanks go to my family for always having faith in me; who gaveme non conditional support and encouragement throughout my studies. Also, thanks to3


4my wife who has been very supportive during the period of my research. At last but notleast, I remain unable to conclude without mentioning the kind assistance and thankingall the members of our research group "Laboratory of Mathematics, Computer Scienceand Applications" at the Faculty of sciences Rabat. My thanks of course to all my dears, Icame into contact during these four years and who helped me directly or indirectly duringthe course of this thesis.


ContentsIntroduction 15I Preliminaries 191 Definitions and Properties 211.1 Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.1.1 Vertex Degrees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.1.2 Planar Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301.2 Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301.2.1 Preliminaries and notations on maps . . . . . . . . . . . . . . . . . 301.2.2 Faces and Euler’s formula . . . . . . . . . . . . . . . . . . . . . . . 321.2.3 Euler’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331.3 Trees and Forests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341.3.1 Properties of Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . 351.3.2 Spanning Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381.4 Distance in trees and graphs . . . . . . . . . . . . . . . . . . . . . . . . . . 382 Spanning trees 412.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.2 Basic concepts and research background . . . . . . . . . . . . . . . . . . . 412.3 Spanning Trees and Enumeration . . . . . . . . . . . . . . . . . . . . . . . 452.3.1 Enumeration of Trees . . . . . . . . . . . . . . . . . . . . . . . . . . 452.3.2 Spanning trees in graphs . . . . . . . . . . . . . . . . . . . . . . . . 452.4 Matrices associated to a graph . . . . . . . . . . . . . . . . . . . . . . . . . 462.4.1 Adjacency Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462.4.2 Degree Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472.4.3 Incidence Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . 482.4.4 Laplacian Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492.4.5 Notation: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515


6 CONTENTSII Theoretical Part 533 How to count the number of spanning trees in graphs 553.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.2 Problematic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.3 Counting the number of spanning trees in graphs combinatorically . . . . . 563.4 Counting the number of spanning trees in graphs algebraically . . . . . . . 583.4.1 The Matrix-Tree Theorem . . . . . . . . . . . . . . . . . . . . . . . 583.4.2 Other algebraic methods for counting τ(G) . . . . . . . . . . . . . . 633.5 More known results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 664 New methods to compute the number of spanning trees of Planar Maps 674.1 Introdution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.2 Complexity of maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 684.3.1 Counting the number of spanning trees in a map of type C = C 1 • C 2 684.3.2 Counting the number of spanning trees in a map C . . . . . . . . . 704.3.3 Counting the number of spanning trees in a map of type C= C 1 : C 2 764.3.4 Counting the number of spanning trees in a map of type C= C 1 | C 2 784.3.5 Counting the number of spanning trees in a map of type C= C 1 ‡ C 2 80III Use of Derived Theoretical Results 835 The Number of Spanning Trees of Certain Families of Planar Maps 855.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 855.2 The case of one cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 855.3 The case of two cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 865.4 The case of n cycles (n ≥ 3) . . . . . . . . . . . . . . . . . . . . . . . . . . 875.5 Particular cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 885.5.1 The case of k i = k and h i = h (h ≥ 2k + 1) . . . . . . . . . . . . 885.5.2 The case of k i = 1 and h i = h . . . . . . . . . . . . . . . . . . . . . 895.6 Other values of k i = 1 and h i = h . . . . . . . . . . . . . . . . . . . . . . . 905.7 Other uses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 935.7.1 Formula for the Number of Spanning Trees in the n-Home chains . 935.7.2 Formula for the Number of Spanning Trees in The n-Barrel chains . 955.7.3 Formula for the Number of Spanning Trees in the n-Light chains . . 965.8 The number of spanning trees in some particular planar maps . . . . . . . 975.8.1 Formulae for the number of spanning trees in particular planar map 985.8.2 Formulae for the number of spanning trees in the n-Kite chains . . 995.8.3 Formulae for the number of spanning trees in the n-Envelope chains 1025.8.4 Formulae for the number of spanning trees in the n-Diphenylenechains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104


CONTENTS 76 Counting the number of spanning trees in the star flower planar map 1076.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1076.2 The star flower planar map . . . . . . . . . . . . . . . . . . . . . . . . . . . 1086.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1086.4 An explicit formula for the number of spanning trees in S n,k . . . . . . . . 1107 Maximal Planar Maps 1157.1 Introdution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1157.2 Calculating the Wiener index in the maximal planar maps . . . . . . . . . 1187.2.1 Introdution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1187.2.2 Calculation of the Wiener index in the planar maps . . . . . . . . . 1187.3 Formulae for the Number of Spanning Trees in a Maximal Planar Map . . 1237.3.1 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1247.3.2 An explicit formula for the number of spanning trees in E n . . . . . 127Conclusion 133Bibliography 135


8 CONTENTS


List of Figures1.1 An example of graph G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.2 Some examples of graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.3 Example of graph with adjacent vertices . . . . . . . . . . . . . . . . . . . 231.4 An example of vertex degrees . . . . . . . . . . . . . . . . . . . . . . . . . 241.5 An example of vertex degrees . . . . . . . . . . . . . . . . . . . . . . . . . 241.6 An example of complete graph K 5 . . . . . . . . . . . . . . . . . . . . . . . 251.7 From left to right, the graphs K 4 , K 2,2 , P 4 , C 4 . . . . . . . . . . . . . . . . . 261.8 A graph G and subgraphs of G. . . . . . . . . . . . . . . . . . . . . . . . . 261.9 Some r-regular graphs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.10 Complete Graphs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.11 Null Graphs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.12 Cycle Graphs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.13 Two graphs G and H are not the same, but they are isomorphic. . . . . . . 291.14 Two graphs G and H are not isomorphic. . . . . . . . . . . . . . . . . . . . 291.15 The graph K 4 drawn as a plane graph without edge crossing. . . . . . . . . 301.16 One graph gives two planar maps . . . . . . . . . . . . . . . . . . . . . . . 311.17 (a) A representation of a graph; its set of vertices is {1, 2, 3, 4}, and(multi)set of edges is {{1, 2}, {2, 3}, {2, 4}, {2, 4}, {3, 3}, {3, 4}}. (b) Twoembeddings of this graph in the sphere, which are not homeomorphic sincethe second has a triangular face, unlike the first. . . . . . . . . . . . . . . . 311.18 The degree of the faces of this planar map are written inside the faces . . . 321.19 An example of map C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331.20 Path graphs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341.21 Simple trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351.22 Path and Star trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351.23 A graph G with its 3 spanning trees . . . . . . . . . . . . . . . . . . . . . . 381.24 An example of a graph G whose diameter is 2 . . . . . . . . . . . . . . . . 392.1 A graph(left) and all spanning trees (right) of this graph. . . . . . . . . . . 422.2 A graph G gives rise to five spanning trees . . . . . . . . . . . . . . . . . . 432.3 A graph G and three of its spanning trees . . . . . . . . . . . . . . . . . . 432.4 A spanning tree of graph G . . . . . . . . . . . . . . . . . . . . . . . . . . 442.5 A spanning tree of graph G . . . . . . . . . . . . . . . . . . . . . . . . . . 449


12 LIST OF FIGURES7.11 The maps F n and G n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1297.12 The crystal planar map C n . . . . . . . . . . . . . . . . . . . . . . . . . . . 1317.13 The maps E n , E n − e and E n .e . . . . . . . . . . . . . . . . . . . . . . . . . 132


List of Tables5.1 Some values of τ(F n ), τ(G n ), τ(T n ), τ(H n ) and τ(E n ) . . . . . . . . . . . . 935.2 Some values of τ(H n ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 945.3 Some values of τ(B n ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 965.4 Some values of τ(L n ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 975.5 Some values of τ(C n ) and τ(Q n ) . . . . . . . . . . . . . . . . . . . . . . . . 995.6 Some values of τ(K n ) and τ(Q n ) . . . . . . . . . . . . . . . . . . . . . . . . 1025.7 Some values of τ(E n ) and τ(Q n ) . . . . . . . . . . . . . . . . . . . . . . . . 1045.8 Some values of τ(D n ) and τ(Q n ) . . . . . . . . . . . . . . . . . . . . . . . . 1066.1 Some values of τ(S n,1 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1136.2 Some values of τ(S n,2 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1147.1 Some values of τ(E n ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1297.2 Some values of τ(F n ) and τ(G n ) . . . . . . . . . . . . . . . . . . . . . . . . 1317.3 Some values of τ(C n ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13213


14 LIST OF TABLES


16 LIST OF TABLESand in random walks in graphs [114]. The research of the number of spanning trees in agraph has a long history. The cornerstone of the research, the Matrix Tree Theorem, datedback to 1847, which is attributed to Kirchhoff. The Matrix Tree Theorem is a famousand classic result on the study of τ(G). It is known that Kirchhoff Matrix Tree Theorem[81, 85], can be applied to any map C to determine τ(C) by taking the determinant ofLaplacian matrix L of C, i.e., all cofactors of L are equal, and their common value is τ(C),but this requires evaluation of a determinant of a corresponding characteristic matrix.However, for a few special families of graphs there exist simple formulae which makeit much easier to calculate and determine the number of corresponding spanning treesespecially when these numbers are very large. One of the first such results is due toCayley [25] who showed that complete graph on n vertices, K n , has n n−2 spanning trees[65], that is he showed τ(K n ) = n n−2 for n ≥ 2. Another combinatorial method ofcounting the number of spanning trees in a graph G is the Feussner’s recursive formula[49], for counting τ(G) in a graph G, is quite intuitive. For an undirected simple graph G,let e be any edge of G. All spanning trees in G can be separated into two parts: one partcontains all spanning trees without e as a tree edge; another contains all spanning treeswith e as a tree edge. The first part has the same number of spanning trees as graph G 1 ,where G 1 is the graph G with e deleted; the second part has the same number of spanningtrees as graph G 2 , where G 2 is the graph created from G by shrinking the two vertices of einto one vertex. Both G 1 and G 2 have fewer edges than G, so the number of spanning treesin G can be counted recursively in this way. It is then clear that τ(G) = τ(G 1 ) + τ(G 2 ).In this thesis, we have generalized this formula by replacing the edge e by a simple pathp that contains k edges, see [89]. Let G be a graph with multiple edges and self-loops.When we count τ(G), we first neglect all self-loops in G because they have no contributionto any spanning tree. If G itself is a tree then τ(G) = 1, and if G is disconnected, thenτ(G) = 0.Our research theme in this thesis focuses on the counting of the number of spanningtrees in connected planar maps, a subject in combinatorial graph theory; and to find newmethods to calculate the number of spanning trees in any map (network). Most researchabout the number of spanning trees is devoted to determining exact formulae for thenumber of spanning trees in many kinds of special graphs, see [6, 14, 62, 63, 65, 66, 127].In this thesis, we start by stating the general methods for counting the number ofspanning trees in graphs; we then provide our new results.Spanning trees are relevant to various aspects of graphs (networks). Generally, thenumber of spanning trees in a network can be obtained by computing a related determinantof the Laplacian matrix L of the network. However, for a large map (network),evaluating the relevant determinant is computationally intractable. In this thesis, we givenew methods to facilitate the calculation of the number of spanning trees in planar mapsand prove novel simplified results. Then, we apply these methods on certain planar mapsto derive several explicit simple formulae for calculating the number of spanning treesin some special families of planar maps which are called the n-Fan chains, the n- Gridchains, the n-tent chains, the n-Hexagonal chains, the n-Eight chains, the n-Home chains,


LIST OF TABLES 17the n-Barrel chains, the n-Light chains, the n-Kite chains, the n-Envelope chains and then-Diphenylene, ... etc.Outline of the thesisThis thesis is organized as follows. In the first chapter, some general graph theoryis described. In the second chapter, we introduce the background and research historyof our problem and definitions and general properties of basic objects studied (spanningtrees, complexity, matrices associated to a graph, ... etc). Some basic results are alsointroduced. The third chapter introduces the background and research of the calculationof the number of spanning trees in graphs and some methods for counting spanning treesare introduced. In this chapter, we firstly state the general methods for counting thenumber of spanning trees in graphs, and then our new results are discussed. In the fourthchapter, we provide new non-trivial methods for calculating the number of spanningtrees in planar maps in general then in particular in some special connected planarmaps and prove new simplified results, we then apply these methods on some specialsplanar maps to give explicit simple formulae to calculate the number of spanning treesin certain families of planar maps in chapter five. In the sixth chapter, we consider thestar flower planar map, and derive a simple explicit formulae for counting the numberof spanning tree in it. in this chapter we use one of the methods which mentioned inthe fourth chapter to find the number of spanning tree in the star flower planar map.Finally in the last chapter, we are going to focus on the maximal planar maps. Thischapter will be divided into two sections; we devote the first section for calculating theWeiner index (the sum of distances between all pairs of vertices) in the case of planarmaps in general then in particular in the maximal planar maps and in the other sectionshall study how to count the number of spanning trees in this type of this maps, as wellas enumeration of spanning trees in the maximal planar maps. For that, we develop amethod, by using the Matrix Tree Theorem as the base and manipulating the matrices,to prove that the number of spanning trees in a maximal planar map which satisfies arecurrence relation. This recurrence relation can be determined exactly, i.e., we reach ata formula to calculate the number of spanning trees in the maximal planar maps. In theend, some possible future works are proposed.New resultsThe new results obtained in the framework of this thesis are concentrated in Chapters 4,5, 6 and 7.The results of Chapter 4 have been partially published in [89]. This paper is thecornerstone of the research subject in this thesis, as it includes new methods to calculatethe number of spanning trees in connected planar maps.The results of Chapter 5 are published in the articles [85, 87, 88, 89]. In these articles,we have applied our new methods that we have reached in the article [89] on certain


18 LIST OF TABLESspecial panar maps. We were able to obtain explicit simple formulae for calculating thenumber of spanning trees in some special planar maps.The results of Chapter 6 are published in article [86]. In this paper, we have beeninterested in calculating the number of spanning trees in the star flower planar map andhave derived the explicit formula to calculate the number of spanning trees in the starflower planar maps.The results of Chapter 7 were divided into two parts. The first part emphasizes uponthe determination of Weiner index in the case of planar maps, in general, and in maximalplanar maps particularly, as can be seen in [90]. While, the second part focuses on thederivation of an explicit formula to calculate the number of spanning trees in a maximalplanar map by employing the Laplacian matrix of planar maps (Matrix Tree Theorem)which has already been published in [84].


Part IPreliminaries19


Chapter 1Definitions and PropertiesThis chapter includes a brief introduction to graph theory. Some of the basics concerningthis theory are presented in [19] and [119]. In case of rare terminologies, the reader isreferred to consult [69] and [117].1.1 GraphsIn this section, we are going to present some useful definitions related to our work asfollows. A graph will usually be denoted by a capital letter G.Definition 1.1.1 An undirected graph G is a triplet (V G ,E G ,δ) where V G is the set ofvertices of the graph G, E G is the set of edges of the graph G (E G ⊆ V 2 , meaning thatan edge e ∈ E G is a 2-element subset {v i ; v j } of V G ) and δ is the application:δ : E G → P(V G )e i ↦→ δ(e i ) = {v j , v k }with v j and v k are end vertices of the edge e i (not necessarily distinct). When u and vare the endpoints of an edge, they are neighbors. We notice that the set {v j , v k } as amultiset (if v j = v k , the same vertex appears twice in δ(e i )). A loop is an edge e i ∈ E Gwith v j = v k (an edge whose endpoints are equal), if δ(e i ) = δ(e j ) with i ≠ j then theedges e i and e j are called multiple edges (edges having the same pair of endpoints).Example 1.1.2 In the graph shown in Figure 1.1, We have: E G = {e 1 , e 2 , e 3 , e 4 }, V G ={v 1 , v 2 , v 3 }, δ(e 1 ) = {v 1 , v 1 } (multiset) δ(e 2 ) = {v 1 , v 2 } and δ(e 3 ) = δ(e 4 ) = {v 2 , v 3 }, theedges e 3 and e 4 are multiple edges, the edge e 1 is a loop, then the graph G admits a loopand two multiple edges.In a graph G, a path is a sequence of vertices and edges p = v 0 , e 1 , v 1 , e 2 , ..., v n−1 , e n , v nsuch that δ(e i ) = {v i−1 , v i }. We say that this path attached both ends v 0 and v n . Acycle is a path such that v 0 = v n . A graph G is finite if its vertex set and edge set are finite.21


22 CHAPTER 1. DEFINITIONS AND PROPERTIESFigure 1.1: An example of graph GWe adopt the convention that every graph mentioned in this thesis is finite, unlessexplicitly constructed otherwise.Remark 1.1.3 For brevity we denote the edge {u, v} ∈ E G simply by uv. In particularwhen uv ∈ E G we say that the vertices u and v are adjacent. Then the degree of a vertexv, written deg(v), is the number of vertices of G which are adjacent to v. When there isno ambiguity, we sometimes write V (G) and E(G) instead of V G and E G , respectively.In a graph, as already explained, two or more edges joining the same pair of verticesare multiple edges. An edge joining a vertex to itself is a loop. A graph with no multipleedges or loops is a simple graph. For example, graph (a) below (see Figure 1.2) hasmultiple edges and graph (b) has a loop; therefore, it is not a simple graph. Graph (c)has no multiple edges or loops, and is therefore a simple graph.Figure 1.2: Some examples of graphDefinition 1.1.4The vertices v and u of a graph are adjacent vertices if they are joined by an edge e. Thevertices v and u are incident with the edge e, and the edge e is incident with the verticesv and u.Example 1.1.5 In the graph below (see Figure 1.3), the vertices u and x are adjacent,vertex w is incident with edges 2, 3, 4 and 5, and edge 6 is incident with the vertex x.


1.1. Graphs 23Figure 1.3: Example of graph with adjacent verticesDefinition 1.1.6 A graph is finite if its vertex set and edge set are finite. We adopt theconvention that every graph mentioned in this thesis is finite, unless explicitly constructedotherwise.Definition 1.1.7 We say that a graph G is connected if any two of its vertices may beconnected by a path (if each pair of vertices in G belongs to a path); otherwise, G isdisconnected [69, 84].Remark 1.1.8 If G has a u, v-path (Tow vertices u and v connected by a path in G),then u is connected to v in G.The connection relation on V (G) consists of the ordered pairs (u, v) such that u isconnected to v. "Connected" is an adjective we apply only to graphs and to pairs ofvertices (we never say "v is disconnected" when v is a vertex).For a graph G, and a set S of vertices and edges in G, we shall denote G − S as thegraph obtained by removing the elements of S from G. Removing a vertex from a graph,is defined as removing the vertex, and all edges incident to it.Definition 1.1.9 A graph G is said to be k-connected, if the removal of any (k − 1)vertices will not disconnect G, but there exist a set S of k vertices such that G − S isdisconnected.The graphs that we consider are in most cases connected but may contain multipleedges.1.1.1 Vertex DegreesIn many applications of graph theory we need a term for the number of edges meeting at avertex. For example, we may wish to specify the number of roads meeting at a particularintersection, the number of wires meeting at a given terminal of an electrical network, orthe number of chemical bonds joining a given atom to its neighbors. Such situations areillustrated below (see Figure 1.4):


26 CHAPTER 1. DEFINITIONS AND PROPERTIES3. A path P n is a graph with vertices {v 1 , ..., v n } and edges {v 1 v 2 , v 2 v 3 , ..., v n−1 v n }. Itcan also be called a path from v 1 to v n .4. For n ≥ 3, C n is the graph P n with one additional edge: v n v 1 . Consequently it iscalled a closed path or a cycle.Figure 1.7: From left to right, the graphs K 4 , K 2,2 , P 4 , C 4 .In mathematics, we often study complicated objects by looking at simpler objects of thesame type contained in them - subsets of sets, subgroups of groups, and so on. In graphtheory we make the following definition.Definition 1.1.16 A subgraph of a graph G is a graph, such that, all of whose verticesare vertices of G and all of whose edges are edges of G, i.e., a graph H is a subgraph ofanother graph G if V H ⊆ V G and E H ⊆ E G and the assignment of endpoints to edges inH is the same as in G. We denote this relation by H ⊆ G and say that G contains H.For example, P 3 ⊆ C 3 and K 2,2 ⊆ K 2,4 .Remark 1.1.17 Note that G is a subgraph of itself.Example 1.1.18 The following graphs (see Figure 1.8) are subgraphs of the graph G onthe left, with vertices {u, v, w, x} and edges {1, 2, 3, 4, 5}.Figure 1.8: A graph G and subgraphs of G.A graph in which all the vertex degrees are the same is given a special name.Definition 1.1.19 A graph G is regular if its vertices all have the same degree.


1.1. Graphs 27Definition 1.1.20 A graph G is said to be r-regular, or regular of degree r, or simplyregular, if every vertex in G has degree r, that is, every vertex has r edges incident to it.For example K 3 is 2-regular and K n is n − 1-regular. A regular 3-regular graph is calledcubic. A cubic graph has an even number of vertices.Example 1.1.21 In the following graphs, we illustrate some r-regular graphs, for variousvalues of r:Figure 1.9: Some r-regular graphs.A useful consequence of the proposition (Degree-Sum Formula) 1.1.12 is the followingresult.Theorem 1.1.22 Let G be an r-regular graph with n vertices. Then G has nr/2 edges.Proof: Let G be a graph with n vertices, each of degree r; then the sum of the degreesof all the vertices is nr. By the Proposition (Degree-Sum Formula) 1.1.12, the number ofedges is one-half of this sum, which is nr/2.□Examples of Regular GraphsWe now consider some important classes of regular graphs.1. Complete Graphs, a complete graph is a graph in which each vertex is joined to eachof the others by exactly one edge. The complete graph with n vertices is denotedby K n . The graph K n is regular of degree n − 1, and therefore has n(n − 1)/2 edges,by Theorem 1.1.22.


28 CHAPTER 1. DEFINITIONS AND PROPERTIESFigure 1.10: Complete Graphs.2. Null Graphs, A null graph is a graph with no edges. The null graph with n verticesis denoted by N n . The graph N n is regular of degree 0.Figure 1.11: Null Graphs.3. Cycle Graphs, A cycle graph is a graph consisting of a single cycle of vertices andedges. The cycle graph with n vertices is denoted by C n . The graph C n is regularof degree 2, and has n edges. For n ≥ 3, C n can be drawn in the form of a regularpolygon.Figure 1.12: Cycle Graphs.Definition 1.1.23 If V = {v 1 , ..., v n } then the degree sequence of G is deg(v 1 ), ..., deg(v n )arranged in decreasing order. For example the degree sequence of P 5 is 2, 2, 2, 1, 1.Definition 1.1.24 An isomorphism from a simple graph G to a simple graph H is abijection f : V (G) → V (H) such that uv ∈ E(G) if and only if f(u)f(v) ∈ E(H). Wesay "G is isomorphic to H", written G ∼ = H, if there is an isomorphism from G to H.Definition 1.1.25 Two graphs are isomorphic to each other, written G ∼ = H, if there isa bijection f : V G → V H such that a, b ∈ V G are adjacent if and only if f(a), f(b) ∈ V Hare adjacent. For example K 3∼ = C3 and both are called triangles. Also K 2∼ = K1,1∼ = P2 .


1.1. Graphs 29Two graphs G and H are isomorphic if H can be obtained by relabelling the verticesof G that is, if there is a one-one correspondence between the vertices of G and those ofH, such that the number of edges joining each pair of vertices in G is equal to the numberof edges joining the corresponding pair of vertices in H. Such a one-one correspondenceis an isomorphism.Example 1.1.26 For example, the graphs G and H represented by the diagramsFigure 1.13: Two graphs G and H are not the same, but they are isomorphic.are not the same, but they are isomorphic, since we can relabel the vertices in the graphG to get the graph H, using the following one-one correspondence:G ↔ H, u ↔ 4, v ↔ 3, w ↔ 2 and x ↔ 1Note that edges in G correspond to edges in H, for example: the two edges joiningu and v in G correspond to the two edges joining 4 and 3 in H; the edge uw in Gcorresponds to the edge 42 in H; the loop ww in G corresponds to the loop 22 in H.Remark 1.1.27 Note that if G ∼ = H then |V G | = |V H |, |E G | = |E H |, and their degreesequences must be identical. However none of these is a sufficient condition for isomorphism.Example 1.1.28 For example, the graphs G and H represented by the diagrams below(see Figure 1.14); which have |V G | = |V H | and |E G | = |E H |, but they are not isomorphic.Figure 1.14: Two graphs G and H are not isomorphic.


30 CHAPTER 1. DEFINITIONS AND PROPERTIES1.1.2 Planar GraphsDefinition 1.1.29 A graph is planar if it can be drawn in the plane such that no edgesare crossing each other (its edges do not cross). This particular drawing of the graphis called a plane graph. For example K 4 is planar graph but K 3,3 is not planar graph.Although the complete graph with four vertices K 4 is usually pictured with crossing edgesas in Figure. 1.15(a), it can also be drawn with noncrossing edges as in Figure. 1.15(b);hence K 4 is planar.Figure 1.15: The graph K 4 drawn as a plane graph without edge crossing.Note that if G is disconnected, then G is planar if and only if each component is planar,hence we may assume well that G is connected throughout this thesis.Proposition 1.1.30 If H ⊆ G and H is not planar then neither is G. In particular K m,nis not planar if m, n ≥ 3.Definition 1.1.31 A planar graph partitions the plane into subsets called regions (faces).For example the plane graph of K 4 has four regions, one of which is exterior to the graph.1.2 MapsThe aim of this section is to provide a short and accessible presentation of planar maps.For a more detailed introduction, see the introductory chapter in [69] and the thesis of ElMarraki [43].1.2.1 Preliminaries and notations on mapsWe begin with some vocabulary on maps. A map is a proper embedding of a connectedgraph into the two-dimensional sphere, considered as continuous deformations. A map isrooted if one of its edges is distinguished as the root-edge and attributed an orientation.Unless otherwise specified, all maps under consideration in this thesis are rooted. Theface at the right of the root-edge is called the root-face and the other faces are said tobe internal. Similarly, the vertices incident to the root-face are said to be external andthe others are said to be internal. Graphically, the root-face is usually represented as theinfinite face when the map is projected on the plane.


1.2. Maps 31Definition 1.2.1 (Map) A map C is a graph G drawn on a surface X or embedded intoit (that is, a compact 2-dimensional orientable variety) in such a way that:• the vertices of graph are represented as distinct points of the surface.• the edges are represented as curves on the surface that intersect only at the vertices.• if we cut the surface along the graph thus drawn, what remains (that is, the set X\G)is a disjoint union of connected components, called faces, each homeomorphic to anopen disk (for more information on the faces of a map see [43] and [69]).A planar map is a map drawn on the plane. Through this thesis, all maps are planar andconnected.Figure 1.16: One graph gives two planar mapsA planar map (hereafter shortly called a map) is an isotopy class of planar embeddingsof a connected planar graph. Notice that the graphs embedded are unlabelled. To stateit simply, a planar map is a connected unlabelled graph drawn in the plane without edgecrossingsand up to continuous deformation. Planar maps are often called plane graphs inthe literature [43, 69]. As illustrated in Figure 1.17 (a)-(b), a planar graph can have nonisotopicplanar embeddings, so that it gives rise to several maps. Due to the topologicalembedding, a map has more structure than a graph. In particular, a map has faces, eachface corresponding to a connected component of the plane splits by the embedding.Figure 1.17: (a) A representation of a graph; its set of vertices is {1, 2, 3, 4}, and (multi)setof edges is {{1, 2}, {2, 3}, {2, 4}, {2, 4}, {3, 3}, {3, 4}}. (b) Two embeddings of this graphin the sphere, which are not homeomorphic since the second has a triangular face, unlikethe first.


32 CHAPTER 1. DEFINITIONS AND PROPERTIESThe unique unbounded face is called the outer (or infinite) face. Vertices, edges, andfaces are called the 0-cells, 1- cells, and 2-cells of the map, respectively. The numbers |V |,|E|, and |F | of vertices, edges, and faces (including the outer (external) face) of a mapare related by the well known Euler’s relation:|V | − |E| + |F | = 2.1.2.2 Faces and Euler’s formulaLet C be a planar map. If we omit the line segments of C from the plane surface on whichC is drawn, the remainder splits into a number of connected open regions; the closure ofsuch a region is called a face.An edge is incident to a face if it belongs to the boundary of this face. If both "banks"of the edge belong to the same face, then such an edge is called an isthmus; we say thatan isthmus is incident to the corresponding face twice.Definition 1.2.2 The degree of a face f denoted by deg(f) is the number of edgesincident to this face (isthmus being counted twice).The notion of face degree is illustrated in Figure.1.18). If we go around the boundary ofa face slightly inside the face, then the number of times we pass along an edge is exactlythe degree of the face.Figure 1.18: The degree of the faces of this planar map are written inside the facesThe following proposition is similar to the previous proposition 1.1.12.Proposition 1.2.3 The sum of the degrees of all faces of a map C is equal to twice thenumber of its edges, i.e.,∑f∈F Cdeg(f) = 2|E C |,where the sum is taken over the set F C of the faces of the map.


1.2. Maps 33Theorem 1.2.4 (Euler characteristic) Let us associate to a map C the numberχ(C) = |V C | + |F C | − |E C |which is called its Euler characteristic. Then χ(C) does not depend on the map C itselfbut only on its genus g and is equal to 2 − 2g.1.2.3 Euler’s formulaEuler’s formula for planar maps is:|V C | + |F C | − |E C | = 2 − 2g,where F C is the set of faces of the planar map C and g is the genus of a map C, in theplanar case g = 0. A map of genus zero is called planar map. This formula is valid forplanar maps (graphs embedded in the plane or in the surface without edge-crossings).For the genus zero case this theorem was already observed by Descartes, and was provedby Euler in 1752 [125].The following theorem gives another famous result due to Euler.Theorem 1.2.5 (Euler’s formula) Let G be a connected planar map with n vertices, medges and f faces (regions). Then n − m + f = 2.Proof: Let’s apply the induction on m . For m = 0 we have n = 1 and f = 1, so thatthe statement holds. Now let m ≠ 0. If G contains a cycle, we discard one of the edgescontained in this cycle and get a graph G ′ with n ′ = n, m ′ = m − 1 and f ′ = f − 1. Byinduction hypothesis, n ′ −m ′ +f ′ = 2 and hence n−m+f = 2. If G is acyclic, then G is atree so that m = n−1, by Theorem 1.2.8; as f = 1, we still obtain n−m+f = 2. □Example 1.2.6 In the map C shown in Figure. 1.19, we have: the number of the verticesis 11, the number of the faces is 13, the number of the edges is 22, then 11 + 13 − 22 = 2.Figure 1.19: An example of map C


34 CHAPTER 1. DEFINITIONS AND PROPERTIESCorollary 1.2.71. If G is planar with |V | ≥ 3 then |E| ≤ 3|V | − 6. In particular K 5 is not planar andneither is K n for all n ≥ 6.2. If G is planar then there is a vertex of degree 5 or less.3. If G is planar and contains no triangles then |E| ≤ 2|V | − 4.Remark 1.2.8 In this thesis, we are only interested in planar graphs. For further detailand more explanation on graphs we refer the reader to [1, 10, 13, 19, 20, 26, 39, 56, 117,118, 119, 121].1.3 Trees and ForestsIn this section, we focus our attention on one particularly important and useful type ofgraph - a tree. Although trees are relatively simple structures, they form the basis ofmany of the practical techniques used to model and to design large-scale systems.The concept of a tree is one of the most important and commonly used ideas in graphtheory, especially in the applications of the subject. It arose in connection with the workof Gustav Kirchhoff on electrical networks in the 1840, and later with Arthur Cayley’swork on the enumeration of molecules in the 1870. More recently, trees have proved to beof value in such areas as computer science, decision making, linguistics, and the design ofgas pipeline systems.The word "tree" suggests branching out from a root and never completing a cycle. Trees asgraphs have many applications, especially in data storage, searching, and communication.Remark 1.3.1 A path graph P n is a tree consisting of a single path through all itsvertices. The graph P n has n − 1 edges, and is obtained from the cycle graph C n byremoving any of its edges (see Figure 1.20).Figure 1.20: Path graphs.Definition 1.3.2 (Tree) A tree is a connected simple graph that contains no cycle, i.e.,without cycle. For example P 4 .Definition 1.3.3 (A plan tree) A plan tree is a tree designed in the plane or is a mapwith only one face, the outer face (see Figure 1.21).


1.3. Trees and Forests 35Figure 1.21: Simple treesRemark 1.3.4 Tree graphs form an important class of planar graphs.In general a graph which contains no cycles is called acyclic. A tree is a connected acyclicgraph. A tree is a path if and only if its maximum degree is 2. A star is a tree consistingof one vertex adjacent to all the others K 1,n−1 , see Figure 1.22.Figure 1.22: Path and Star treesAn acyclic graph, one not containing any cycles, is called a forest. A connected forestis called a tree (Thus, a forest is a graph whose components are trees). The vertices ofdegree 1 in a tree are its leaves. Every nontrivial tree has at least two leaves-take, forexample, the ends of a longest path. This little fact often comes in handy, especially ininduction proofs about trees: if we remove a leaf from a tree, what remains is still a tree.Example 1.3.5 A tree is a connected forest, and every component of a forest is a tree.1.3.1 Properties of TreesTrees have many equivalent characterizations, any of which could be taken as thedefinition. Such characterizations are useful because we need only verify that a graphsatisfies anyone of them to prove that it is a tree, after which we can use all the otherproperties.We first prove that deleting a leaf from a tree yields a smaller tree.Proposition 1.3.6 A tree contains a vertex of degree 1, which is called a leaf.Lemma 1.3.7 Every tree with at least two vertices has at least two leaves. Deleting aleaf from an n-vertex tree produces a tree with n − 1 vertices.


36 CHAPTER 1. DEFINITIONS AND PROPERTIESProof: A connected graph with at least two vertices has an edge. In an acyclic graph,an endpoint of a maximal nontrivial path has no neighbor other than its neighbor on thepath. Hence the endpoints of such a path are leaves. Let v be a leaf of a tree G, andlet G ′ = G − v. A vertex of degree 1 belongs to no path connecting two other vertices.Therefore, for u, w ∈ V (G ′ ), every u, w-path in G is also in G ′ . Hence G ′ is connected.Since deleting a vertex cannot create a cycle, G ′ also is acyclic. Thus G ′ is a tree withn−1 vertices.□Remark 1.3.8 The previous Lemma implies that every tree with more than one vertexarises from a smaller tree by adding a vertex of degree 1 (all our graphs are finite).This rescues some proofs from the induction trap: growing an n + 1-vertex tree from anarbitrary n-vertex tree by adding a new neighbor at an arbitrary old vertex generates alltrees with n + 1 vertices. The word "arbitrary" means that the discussion considers allways of making the choice.The proof of equivalence of characterizations of trees uses induction, prior results,a counting argument, extremality, and contradiction.Theorem 1.3.9 For an n-vertex graph G (with n ≥ 1), the following are equivalent (andcharacterize the trees with n vertices).(A) G is connected and has no cycles (G is acyclic).(B) G is connected and has n − 1 edges (|V | = |E| + 1).(C) G has n − 1 edges and no cycles.(D) For u, v ∈ V (G), G has exactly one u, v-path.Proof: We first demonstrate the equivalence of A, B, and C by proving that any twoof connected, acyclic, n − 1 edges together imply the third. A ⇒ B, C. We use inductionon n. For n = 1, an acyclic 1-vertex graph has no edge. For n > 1, we suppose that theimplication holds for graphs with fewer than n vertices. Given an acyclic connected graphG, the previous Lemma provides a leaf v and states that G ′ = G − v also is acyclic andconnected (see figure above). Applying the induction hypothesis to G ′ yields E G ′ = n − 2.Since only one edge is incident to v, we have E G = n − 1.B ⇒ A, C. Delete edges from cycles of G one by one until the resulting graph G ′ is acyclic.Since no edge of a cycle is a cut-edge (the previous Theorem), G ′ is connected. Now the


1.3. Trees and Forests 37preceding paragraph implies that E G ′ = n − 1. Since we are given E G = n − 1, no edgeswere deleted. Thus G ′ = G, and G is acyclic.C ⇒ A, B. Let G l , ..., G k be the components of G. Since every vertex appears in onecomponent, ∑ n(G i ) = n. Since G has no cycles, each component satisfies property A.iThus E Gi = n(G i ) − 1. Summing over i yields E G = ∑ [n(G i ) − 1] = n − k. We are giveniE G = n − 1, so k = 1, and G is connected.A ⇒ D. Since G is connected, each pair of vertices is connected by a path. If some pairis connected by more than one, we choose a shortest (total length) pair P , Q of distinctpaths with the same endpoints. By this extremal choice, no internal vertex of P or Q canbelong to the other path (see figure below).This implies that P ∪ Q is a cycle, which contradicts the hypothesis A.D ⇒ A. If there is a u, v-path for every u, v ∈ V G , then G is connected. If G has a cycleC, then G has two u, v-paths for u, v ∈ V C ; hence G is acyclic (this also forbids loops). □Definition 1.3.10 A cut-edge or cut-vertex of a graph is an edge or vertex whose deletionincreases the number of components. We write G − e or G − M for the subgraph of Gobtained by deleting an edge e or set of edges M. We write G − v or G − S for thesubgraph obtained by deleting a vertex v or set of vertices S.Next, we characterize cut-edges in terms of cycles.Theorem 1.3.11 An edge is a cut-edge if and only if it belongs to no cycle.Corollary 1.3.12a) Every edge of a tree is a cut-edge.b) Adding one edge to a tree forms exactly one cycle.c) Every connected graph contains a spanning tree.Proof: (a) A tree has no cycles, so Theorem 1.3.11 implies that every edge is acut-edge. (b) A tree has a unique path linking each pair of vertices (Theorem 1.3.9 D),so joining two vertices by an edge creates exactly one cycle. (c) As in the proof of B⇒ A, C in Theorem 1.3.9, iteratively deleting edges from cycles in a connected graphyields a connected acyclic subgraph.□


38 CHAPTER 1. DEFINITIONS AND PROPERTIESTheorem 1.3.13 The following assertions are equivalent for a graph T :(i) T is a tree;(ii) any two vertices of T are linked by a unique path in T ;(iii) T is minimally connected, i.e. T is connected but T − e is disconnected for everyedge e ∈ T ;(iv) T is maximally acyclic, i.e. T contains no cycle but T + xy does, for any twonon-adjacent vertices x, y ∈ T .1.3.2 Spanning TreesAn important concept that we need later is that of a spanning tree in a graph.Definition 1.3.14 Let G be a connected graph. Then a spanning tree in G is a subgraphof G that includes every vertex and is also a tree. For example, the following diagramshows a graph G and its three spanning trees.Figure 1.23: A graph G with its 3 spanning trees1.4 Distance in trees and graphsWhen using graphs to model communication networks, we want vertices to be close togetherto avoid communication delays. We measure distance using lengths of paths.Many applications of graphs involve getting from one vertex to another. For example,we may wish to find the shortest route between one town and another. Other examplesinclude the routing of a telephone call between one subscriber and another, the flow ofcurrent between two terminals of an electrical network, and the tracing of a maze. Wenow make this idea precise by defining a walk in a graph. It is sometimes useful to beable to refer to a walk under more restrictive conditions in which we require all the edges,or all the vertices, to be different.


1.4. Distance in trees and graphs 39Definition 1.4.1 (Walk) A walk is a sequence of edges v 1 v 2 , v 2 v 3 , ..., v n−1 v n which arenot necessarily distinct (unless the walk is a path). In this case we say that the walk isfrom v 1 to v n of length n − 1.Definition 1.4.2 (Distance) The distance between two distinct vertices v i and v j ofa graph G, denoted by d(v i , v j ) is equal to the length of the shortest path (number ofedges in) that connects v i and v j (the least length between v i and v j ). Conventionally,d(v i , v i ) = 0. If G has a u, v-path, then the distance from u to v, written d G (u, v) or simplyd(u, v), is the least length of a u, v-path. If G has no such path, then d(u, v) = ∞ [87].The distance between two vertices v and u, written d(v, u), is the length of theshortest walk from v to u, if it exists, otherwise let d(v, u) = ∞. Note that the shortestwalk is necessarily a path. Furthermore, in a weighted graph, d(v, u) is understood to bethe minimum total weights of all possible walks from v to u.Definition 1.4.3 (Weight) The weight, denoted by p(v i , v j ) is the number of edges thatconnects v i with v j .We use the word "diameter" due to its use in geometry, where it is the greatest distancebetween two elements of a set.Definition 1.4.4 (Diameter) The diameter of a graph G is defined as the maximum ofthe set of all shortest walks joining any two vertices, i.e.,Diam(G) = max{d(v, u)|v, u ∈ V }.For example diam(G) = 1 if and only if G is complete but the diameter in a plane tree isthe number of edges of the longest path. Note also that diam(G) = ∞ if and only if G isdisconnected.Example 1.4.5 The diameter of the graph below is 2 (see Figure. 1.24). This is becausefor any vertex not directly connected to another, there is a path of length two connectingthe two. By looking at the graph, it can be seen that this is true.Figure 1.24: An example of a graph G whose diameter is 2


40 CHAPTER 1. DEFINITIONS AND PROPERTIES


Chapter 2Spanning trees2.1 IntroductionSpanning trees have always been of great interest in various areas of computer science.The same is true for the idea of shortest paths in a graph. The number of spanning treesof a graph (network) G, also known as the complexity, denoted by τ(G), is an important,well-studied quantity in graph theory, and appears in a number of applications. Mostnotable application fields are network reliability. In order to motivate the research inthis thesis, we begin with some basic definitions and properties about spanning tree. Animportant concept which is needed in this chapter is the spanning tree in a graph. Forfurther detail and more explanation on properties of trees and graphs we refer the readerto [117]. In this chapter, we are going to begin by introducing the basic notions anddefinitions needed for spanning trees and to study many different properties of them.2.2 Basic concepts and research backgroundThe central concept of the research presented here the spanning tree. A spanning treein a graph G is a spanning subgraph of G, which is a tree. We define a spanning treeformally as follows.Definition 2.2.1 For a graph G, a spanning tree in G is a tree which has the same vertexset as G (tree that passing through all the vertices of the graph), i.e., a spanning tree Tof a graph G = (V, E) is a graph T = (V, E ′ ) such that T is a tree and E ′ ⊆ E. (seeFigure. 2.1).Remark 2.2.2 A spanning tree of a graph G is a connected subgraph without cyclesthat includes every vertex of G, assuming G is connected. Otherwise, G has spanningforests, which are maximal subgraphs without cycles (so a forest is a subgraph withoutcycles). Here we shall discuss spanning trees since we are only concerned with connectedgraphs. Figure 2.1 gives an example of spanning trees of a graph. The subgraphs on theright is spanning trees of the graph on the left.41


42 CHAPTER 2. SPANNING TREESFigure 2.1: A graph(left) and all spanning trees (right) of this graph.The number of spanning trees of a finite graph or multigraph G, also known as thecomplexity τ(G), is certainly one of the most important graph-theoretical parameters,and also one of the oldest. Its applications range from the theory of networks, wherethe number of spanning trees is used as a measure for network reliability [34] totheoretical chemistry, in connection with the enumeration of certain chemical isomers[22]. Kirchhof’s celebrated matrix tree theorem [65] that relates the properties of anelectrical network to the number of spanning trees in the underlying graph.The number of spanning trees in a graph (network) is an important, well studied quantity[36]. As well as being of combinatorial interest, several application uses, mentionedin the following, are adduced in [62].1. Kirchhoff’s laws, well know as Matrix Tree Theorem, provide an effective methodfor designing electrical circuits, which are enormously useful in the analysis andsynthesis of networks.2. Suppose we are given a network of communication lines, which can break. Theprobability of a single line breaking is 1−p. It is necessary to estimate the reliabilityof such a network. If the reliability is the probability of connectedness of the network,P , thenm∑P = A k p k (1 − p) m−k ,k=n−1where n is the number of vertices, m the number of edges, of the graph; A k is thenumber of connected subgraphs with n vertices and k edges. It is clear that, if thereliability of each line is small, thenP ≈ A n−1 p n−1 (1 − p) m−n+1 ,where A n−1 is the number of spanning trees of the graph.


2.2. Basic concepts and research background 43Thus, with low reliability of each of the line, the network’s reliability is determined,basically, by the number of spanning trees of the network.3. In building a maser, one must investigate the possible particle transitions. For this,one constructs a graph in which the vertices correspond to energy levels and edges topossible particle transitions. Then for the analysis of the maser’s energetics, it turnsout to be very useful to know the number of spanning trees in the correspondinggraph.The research of the number of spanning trees in a graph has a long history. Thecornerstone of the research, the Matrix Tree Theorem, dated back to 1847, is attributedto Kirchhoff. Most research about the number of spanning trees is devoted to determiningexact formulae for the number of spanning trees in many kinds of special graphs, see[6, 14, 62, 63, 65, 66, 127]. We now state the general methods for counting the numberof spanning trees in graphs in the following.Now, the problem is: given a map (graph embedded into surfaces), how manyspanning trees does it have? As example, a graph G with all its spanning trees isdisplayed in Figure 2.2.Figure 2.2: A graph G gives rise to five spanning treesLet G be a connected graph. Then a spanning tree in G is a subgraph of G thatincludes every vertex and is also a tree. For example, the following diagram shows agraph G and three of its spanning trees; (see Figure. 2.3).Figure 2.3: A graph G and three of its spanning trees


44 CHAPTER 2. SPANNING TREESThe number of spanning trees in a graph can be very large; for example, the abovegraph G has twenty-one spanning trees and the Petersen graph has 2000 labelled spanningtrees. Given a connected graph, we can construct a spanning tree by using either of thefollowing two methods. We illustrate these by applying them to the graph G above.Building-up method:Select edges of the graph one at a time, in such a way that no cycles are created; repeatthis procedure until all vertices are included.Example 2.2.3 In the above graph G in Figure 2.3, we select the edges vz, wx, xy, yz;then no cycles are created. We obtain the following spanning tree; (see Figure. 2.4).Figure 2.4: A spanning tree of graph GCutting-down method:Choose any cycle and remove any one of its edges; repeat this procedure until no cyclesremain.Example 2.2.4 From the above graph G in Figure 2.3, we remove the edges vy (destroyingthe cycle vwyv), yz (destroying the cycle vwyzv), xy (destroying the cycle wxyw).We obtain the following spanning tree; (see Figure. 2.5).Figure 2.5: A spanning tree of graph G


2.3. Spanning Trees and Enumeration 45Remark 2.2.5 If two spanning trees in a graph have different edge (arc) sets, we considerthem as different trees, even though, they may be isomorphic.2.3 Spanning Trees and EnumerationThere are 2 (n 2) simple graphs with vertex set [n] = {1, ..., n}, since each pair may or maynot form an edge. How many of these are trees? In this section, we solve this countingproblem, count spanning trees in arbitrary graphs, and discuss several applications.2.3.1 Enumeration of TreesWith one or two vertices, only one tree can be formed. With three vertices there is stillonly one isomorphism class, but the adjacency matrix is determined by which vertex isthe center. Thus there are three trees with vertex set [3]. With vertex set [4], there arefour stars and 12 paths, yielding 16 trees. With vertex set [5], a careful study yields 125trees.Now we may see a pattern. With vertex set [n], there are n n−2 trees; this is Cayley’sFormula. Prüfer, Kirchhoff, Pólya, Renyi, and others found proofs of that whileJ.W. Moon [92] wrote a book about enumerating classes of trees. We present a bijectiveproof, establishing a one-to-one correspondence between the set of trees with vertex set[n] and a set of known size.Remark 2.3.1 In the mathematical discipline of graph theory, a graph labeling is theassignment of labels, traditionally represented by integers, to the edges or vertices, orboth of a graph.Theorem 2.3.2 (Cayley’s Formula [1889]). For a set S ⊆ N of size n, there are n n−2trees with vertex set S.Corollary 2.3.3 The number of labeled trees with n ≥ 2 vertices is the number ofspanning trees of a labeled K n , which is n n−2 .2.3.2 Spanning trees in graphsWe can interpret Cayley’s Formula in another way. Since the complete graph with vertexset [n] has all edges that can be used in forming trees with vertex set [n], the numberof trees with a specified vertex set of size n equals the number of spanning trees in acomplete graph on n vertices.We now consider the more general problem of computing the number of spanningtrees in any graph G. In general, G will not have as much symmetry as a completegraph, so it is not reasonable to expect as simple a formula as for K n , but we can hopefor an algorithm that provides a simple way to compute the answer for a given graph G.


46 CHAPTER 2. SPANNING TREESExample 2.3.4 Below is the kite. To count the spanning trees, observe that four arepaths around the outside cycle in the drawing. The remaining spanning trees use thediagonal edge. Since we must include an edge to each vertex of degree 2, we obtain fourmore spanning trees. The total is eight.Figure 2.6: A graph G and its all spanning treesIn Example 2.3.4, we counted separately the trees that did or did not contain the diagonaledge. This suggests a recursive procedure to count spanning trees. It is clear that thespanning trees of G not containing e are simply the spanning trees of G − e, but how dowe count the trees that contain e? The answer uses an elementary operation on graphs.Remark 2.3.5 We denote by τ(G) the number of spanning trees of a graph G, sometimescalled the complexity of G. If G is not connected then τ(G) = 0, so we can assume thatG is connected from now on. If G ′ is obtained from G by removing all the loops of G,then τ(G ′ ) = τ(G), since a loop can never occur in a spanning tree, i.e., loops don’t affectspanning trees, so we delete them before the computation. Thus, we may assume that Gcontains no loops as well. Multiple edges, however, do remain a possibility.2.4 Matrices associated to a graphHow do we specify a graph? We can list the vertices and edges (with end-points), butthere are other useful representations. Saying that a graph is loopless means that multipleedges are allowed but loops are not. We need to define some matrices as follows.2.4.1 Adjacency MatrixThe adjacency matrix of a graph G is the square matrix A = A(G) indexed by V × V ,which has its entries as: A vv = 0 for all v ∈ V , and if v ≠ u in V then A vu is the numberof edges of G which have vertices v and u at their ends. In other words, the adjacencymatrix A of a graph G, is a (0, 1)-matrix, where 1’s represent adjacent vertices. A has arow for each vertex, and a column for each vertex. If a vertex v is connected to a vertexu, the entry in row v, column u is 1, and so is the entry in row u, column v. For instance,a graph G and its adjacency matrix; see Figure 2.7


2.4. Matrices associated to a graph 47Figure 2.7: A graph G with its adjacency matrix.Remark 2.4.1 An adjacency matrix is determined by a vertex ordering. Every adjacencymatrix is symmetric (a ij = a ji for all i, j). An adjacency matrix of a simple graph G hasentries 0 or 1, with 0s on the diagonal. The degree of v is the sum of the entries in therow for v in either A or M.Example 2.4.2 Consider the labelled graph G in Figure 2.8.Figure 2.8: A graph with 6 vertices.which has the adjacency matrix A2.4.2 Degree Matrix⎡⎤0 1 0 0 1 01 0 1 0 1 1A =0 1 0 1 0 1⎢0 0 1 0 0 1⎥⎣1 1 0 0 0 1⎦0 1 1 1 1 0The degree of a vertex v ∈ V of G denoted by deg G (v); is the number of edges of G whichare incident with v. The degree matrix of G is the diagonal V -by-V matrix D = D(G)such that D vv = deg G (v) for all v ∈ V , and D vu = 0 if v ≠ u; more precisely, the degreematrix D of a graph, is a diagonal matrix with the vertex degrees in the diagonal.


48 CHAPTER 2. SPANNING TREESFor example, a graph G that shown in Figure 2.7 has degree matrix as follows:⎛⎞2 0 0 0D = ⎜ 0 3 0 0⎟⎝ 0 0 4 0 ⎠0 0 0 1Example 2.4.3 The degree matrix D of the graph G shown in Figure 2.8 is as follows:2.4.3 Incidence Matrix⎡⎤2 0 0 0 0 00 4 0 0 0 0D =0 0 3 0 0 0⎢0 0 0 2 0 0⎥⎣0 0 0 0 3 0⎦0 0 0 0 0 4Let G be a finite undirected graph without loops. The (vertex-edge) incidence matrix ofG is the 0-1 matrix M, with rows indexed by the vertices and columns indexed by theedges, where M ve = 1 when vertex v is an endpoint of edge e. For example the incidencematrix of P 4 can be given by:Figure 2.9: P 4 and its incidence matrix.The incidence matrix M is the n-by-m matrix in which entry m ij is 1 if v i is an endpointof e j and otherwise is 0. If vertex v is an endpoint of edge e, then v and e are incident.Example 2.4.4 For the loopless graph G below (see Figure 2.10), we exhibit the incidencematrix that result from the vertex ordering w, x, y, z and the edge ordering a, b,c, d, e. The degree of y is 4, by viewing the graph or by summing the row for y in eithermatrix.


2.4. Matrices associated to a graph 49Figure 2.10: The incidence matrix of a graph GDefinition 2.4.5 The incidence matrix M for a graph with n vertices and m edges is an × m matrix that indicates which edges are incident on which vertices. We assume thatboth the edges and vertices are given an ordering. Using the ordering of the vertices, weimpose a direction on each edge such that the edge points from the lower-ordered vertexto the higher ordered vertex. The entries of the incidence matrix are defined as follows:⎧⎨ 1 if edge e j points out from v ia i,j := −1 if edge e j points to v i⎩0 otherwise.Example 2.4.6 The underlying graph of the digraph below, is the graph of Example2.4.4; note the similarity and difference in their matrices.Figure 2.11: The incidence matrix M of a directed graph G2.4.4 Laplacian MatrixIn the mathematical field of graph theory the Laplacian matrix or the matrix Laplace,sometimes called admittance matrix or Kirchhoff matrix, is a matrix representationof a graph. Together with Kirchhoff’s theorem it can be used by Kirchhoff [65, 81]to calculate the number of spanning trees for a given graph. The Laplacian matrixcan be used to find many other properties of the graph, see e.g., spectral graphtheory. Cheeger’s inequality from Riemannian Geometry has a discrete analogueinvolving the Laplacian Matrix, which is perhaps the most important theorem inSpectral Graph theory and one of the most useful facts in algorithmic applications.It approximates the sparsest cut of a graph through the second eigenvalue of its Laplacian.


50 CHAPTER 2. SPANNING TREESIn the following work, we consider only the graphs that do not have loops.Definition 2.4.7 The Laplacian matrix L of an undirected graph G is defined as thedifference between its degree matrix D and its adjacency matrix A.L = D − A.More formally, an undirected graph G = (V G , E G ) of n vertices, weighted by the weightfunction at any edge (v i , v j ) associated weight p(v i , v j ).The Laplacian matrix of G verifies :⎧⎨ deg(v i ) = ∑ nk=1 p(v i, v k ) if i = jL i,j := −p(v i , v j )if i ≠ j and (v i , v j ) ∈ E G⎩0 otherwise.That is, the diagonal elements have values equal to the degree of the corresponding vertices,and the off-diagonal elements are −p(v i , v j ) (the number of edges that connects v iwith v j ) if an edge connects the two vertices, and 0 otherwise.Example 2.4.8 The Laplacian matrix of a labeled graph G shown in Figure 2.7 is givenby:⎛⎞ ⎛⎞ ⎛⎞2 0 0 0 0 1 1 0 2 −1 −1 0L = ⎜ 0 3 0 0⎟⎝ 0 0 4 0 ⎠ − ⎜ 1 0 2 0⎟⎝ 1 2 0 1 ⎠ = ⎜ −1 3 −2 0⎟⎝ −1 −2 4 −1 ⎠0 0 0 1 0 0 1 0 0 0 −1 1Example 2.4.9 The graph G pictured in Figure 2.8 gives:⎡⎤2 −1 0 0 −1 0−1 4 −1 0 −1 −1L =0 −1 3 −1 0 −1⎢ 0 0 −1 2 0 −1⎥⎣−1 −1 0 0 3 −1⎦0 −1 −1 −1 −1 4Remark 2.4.10 Let G be a finite undirected graph without loops. The Laplace matrix ofG is the matrix L indexed by the vertex set of G, with zero row sums, where L vu = −A vufor v ≠ u. If D is the diagonal matrix, indexed by the vertex set of G such that D vv isthe degree (valency) of v, then L = D − A. The matrix Q = D + A is called the signlessLaplace matrix of G. An important property of the Laplace matrix L and the signlessLaplace matrix Q is that they are positive semidefinite. Indeed, one has Q = MM T andL = NN T , if M is the incidence matrix of G and N the directed incidence matrix of thedirected graph obtained by orienting the edges of G in an arbitrary way.


2.4. Matrices associated to a graph 51Depth-First Search AlgorithmGoal: To determine whether or not a given graph G is connected, in which case it producesa spanning tree of G.1. We shall label the vertices of G as 1, 2, . . . based upon the order of traversal asfollow.2. Start with an arbitrary vertex 1.3. Move to any vertex adjacent to the current selection which have not been labeled.If no such vertex exists backtrack to a previously visited vertex.4. Repeat until no more move is possible.5. If all vertices of G are labeled then G is connected and this traversal generates aspanning tree of G.Kirchoff’s AlgorithmGoal: To count the number of spanning trees of a connected labeled graph G.1. Assume that V G = {v 1 , v 2 , ..., v n }.2. Let L be an n × n matrix given by⎧⎨ deg(v i ) if i = j or else(L) ij = −1 if ij ∈ E G⎩0 if ij /∈ E G .3. Compute any cofactor of L. This is the number of spanning trees of G.2.4.5 Notation:1. One more piece of notation is required. If M is a matrix and i is a row-index for Mand j is a column-index for M, let M(i|j) denotes the submatrix of M obtained bydeleting row i and column j from M.We denote by det L(C)[v i ; v j ] = det L(C)(v 1 , ..., ̂v i , ..., ̂v j , ..., v n ).where L(C)[v i ; v j ] is the submatrix, obtained by deleting the v i -th, v j -th rows andthe v i -th, v j -th columns from the Laplacian matrix of C.2. When choosing two vertices from a set of size n, we can pick one and then the otherbut don’t care about the order, so the number of ways is n(n − 1)/2. (The notationfor the number of ways to choose k elements from n elements is ( nk)or Cnk , read "nchoose k". These numbers are called binomial coefficients.


52 CHAPTER 2. SPANNING TREES


Part IITheoretical Part53


Chapter 3How to count the number of spanningtrees in graphsIn this chapter, we shall focus on the general methods for counting the number of spanningtrees in graphs while discussing the spanning trees, as well as the enumeration of spanningtrees.3.1 IntroductionSpanning trees are relevant to various aspects of graphs (networks). Generally, the numberof different spanning trees in a graph G turns out to be a very useful number leading tointeresting results and applications. Most of the classical theories of interest concerningspanning trees is the number of spanning trees of a graph G. We denote the number ofspanning trees in a graph G by τ(G) and then, we shall investigate different methods forcalculating τ(G). The research of the number of spanning trees in a graph has a longhistory. The cornerstone of the research, the Matrix Tree Theorem, dated back to 1847,is attributed to Kirchhoff. Most research about the number of spanning trees is devotedto determining exact formulae for the number of spanning trees in many kinds of specialgraphs, see [6, 14, 62, 63, 65, 66, 127]. We first state the problem of counting the numberof spanning trees in graphs then display the general methods for counting the number ofspanning trees in graphs.3.2 ProblematicHow to count the number of spanning trees in graphs?Statement of problem: given a graph, how many spanning trees does it have?As an example, a graph G with all its spanning trees is displayed in Figure 3.155


56CHAPTER 3.HOW TO COUNT THE NUMBER OF SPANNING TREES INGRAPHSFigure 3.1: A map C gives rise to eight spanning treesNext, we introduce general methods for counting the number of spanning trees ingraphs, combinatorically and algebraically.3.3 Counting the number of spanning trees in graphscombinatoricallyThe basic combinatorial idea, Feussner’s recursive formula [49], for counting τ(G) in agraph G is quite intuitive. For an undirected simple graph G, let e be any edge of G.All spanning trees in G can be separated into two parts: one part contains all spanningtrees without e as a tree edge; another contains all spanning trees with e as a tree edge.The first part has the same number of spanning trees as graph G 1 , where G 1 is the graphG with e deleted; the second part has the same number of spanning trees as graph G 2 ,where G 2 is the graph created from G by shrinking the two vertices of e into one vertex.Both G 1 and G 2 have fewer edges than G, so the number of spanning trees in G can becounted recursively in this way. It is then clear that τ(G) = τ(G 1 ) + τ(G 2 ). An exampleis shown in Figure 3.2.Figure 3.2: Graph G and its reduced graphs G 1 and G 2Now, let G be a graph with multiple edges and self-loops. When we count τ(G), we firstneglect all self-loops in G because they have no contribution to any spanning tree. Forany graph G (directed or undirected, simple or multiple), we denote τ(G) as the numberof spanning trees in G, sometimes called the complexity of G. If G itself is a tree thenτ(G) = 1, and if G is disconnected, then τ(G) = 0.


3.3. Counting the number of spanning trees in graphs combinatorically 57Another combinatorial method of counting the number of spanning trees in a graph Gis the Prüfer’s code or Prüfer’s sequence approach, which is first described by Prüfer[100] for proving Cayles’s theorem [25]: τ(K n ) = n n−2 , for the complete graph K n . Asan example, the 16 spanning trees of K 4 are shown in Figure 3.3. The key idea of thePrüfer’s code approach is to build a bijection between the set of all trees in a graph anda set of some sequences, then count the spanning trees by counting the sequences in thesequences set.Figure 3.3: The 16 different spanning trees in K 4 .Theorem 3.3.1 (Cayley’s Theorem) The number of spanning trees of the complete graphK n is n (n−2) .The proof of Cayley’s Theorem presented here is the work of Jim Pitman, who used clevercounting in two ways to prove the theorem 3.3.1. Cayley’s theorem can also be provedusing the Matrix Tree Theorem [117] presented below, but here we give an alternate proofto show a different way of proving the theorem. Before presenting the proof, we need tostate a few definitions. A rooted forest on the vertex set {1, ..., n} is a forest togetherwith a choice of a root in each component tree. Additionally, a forest F is said to containanother forest F ′ if F contains F ′ as a directed graph. Then clearly if F properly containsF ′ , then F has fewer components than F ′ . And finally, let F n,k be the set of all rootedforests that consist of k rooted trees. Then we call a sequence F 1 , ..., F k of forests a refiningsequence if F i ∈ F n,i and F i contains F i+1 , for all i. Now we can begin the proof.


58CHAPTER 3.HOW TO COUNT THE NUMBER OF SPANNING TREES INGRAPHSProof: First we notice that F n,1 is the set of all rooted trees (since they only consist ofone component). Note that |F n,1 | = nT n , since in every tree there are n choices for theroot. We now regard F n,k ∈ F n,k as a directed graph with all of the edges directed awayfrom the roots. Now let F k be a fixed forest in F n,k and denote by N(F k ) the number ofrooted trees containing F k , and by N ∗ (F k ) the number of refining sequences ending in F k .We count N ∗ (F k ) in two ways, first by starting at a tree and secondly by starting atF k . Suppose F 1 ∈ F n,1 contains F k . Since we may delete the k - 1 edges of F 1 \F k in anypossible order to get a refining sequence from F 1 to F k , we findN ∗ (F k ) = N(F k )(k − 1)!.Let us now start at the other end. To produce from F k an F k−1 we have to add a directededge, from any vertex v, to any of the k - 1 roots of the trees that do not contain v. Thuswe have n(k − 1) choices. Similarly, for F k−1 we may produce a directed edge from anyvertex u to any of the k - 2 roots of the trees not containing u. For this we have n(k − 2)choices. Continuing this way, we arrive atand thus we have thatN ∗ (F k ) = n k−1 (k − 1)!,N(F k ) = n k−1 for any F k ∈ F n,k .For k = n, F n consists of just n isolated vertices. Hence N(F n counts the number of allrooted trees, thus |F n,1 | = n n−1 , and thus T n = (1/n)(|F n,1 |) = n n−2 .□The method of Prüfer’s code has also been used to investigate the spanning treeformulae for bipartite graphs [42], k-partite graphs [96], extended graphs [61] andcomplete multipartite graphs [41, 72].3.4 Counting the number of spanning trees in graphsalgebraicallyAlgebraic graph theory is a branch of mathematics that studies graphs by using algebraicproperties of associated matrices. More in particular, spectral graph theory studies therelation between graph properties and the spectrum of the adjacency matrix or Laplacematrix. There are some known algebraic methods for counting the number of spanningtrees.3.4.1 The Matrix-Tree TheoremAnother way to compute the number of spanning trees in a graph is using linear algebra.By a theorem of Gustav Robert Kirchoff in 1847, the number of spanning trees can becomputed by finding the cofactor of a special matrix called the Laplacian matrix L. Oneof the earliest uses of the matrix L proper was the Matrix-Tree Theorem, proposed by


3.4. Counting the number of spanning trees in graphs algebraically 59Kirchhoff (in fact, L is sometimes called the Kirchhoff matrix). The classic result knownas the Matrix Tree Theorem [65, 81] states that, the Laplacian matrix L which is definedas L = D − A, has all its co-factors equal to the number of spanning trees of a graph G.Denoting the number of spanning trees of G by τ(G). We may now proceed the MatrixTree Theorem by stating:Theorem 3.4.1 [84] (The Matrix-Tree Theorem) Let A be an adjacency matrix of agraph G, and D be a diagonal matrix with the diagonal entry in row i equal to the degreeof vertex i, and let L = D − A. If L ∗ (G) is a matrix obtained by deleting the i-th rowand j-th column of the Laplacian matrix L(G), thenτ(G) = (−1) i+j det L ∗ (G).Example 3.4.2 The Kirchhoff matrix of the graph G in the following Figure 3.4 is:Figure 3.4: A simple example graph⎛L =⎜⎝4 -1 -1 0 -1 -1-1 3 -1 -1 0 0-1 -1 3 -1 0 00 -1 -1 4 -1 -1-1 0 0 -1 3 -1-1 0 0 -1 -1 3All its co-factors are equal to 128. So, there are 128 spanning trees in the graph.One could prove this theorem by induction on the number of edges, by showing that theright-hand side of the formula satisfies the same deletion - contraction recursion as doesτ(G). The initial conditions forming the base case of the induction are easily checked.However, there is a more informative proof which also yields various generalizations ofthis Matrix-Tree Theorem.We begin by expressing the Laplacian matrix of a graph in a different form. Considera graph G = (V, E), and orient each edge of G arbitrarily by putting an arrow onit pointing towards one of its two ends. The signed incidence matrix of G (with respectto this orientation) is the V - by - E indexed matrix M with entries:⎧⎨ 1 if e points in to v but not out,M ve := −1 if e points out of v but not in,⎩0 otherwise.⎞⎟⎠


60CHAPTER 3.HOW TO COUNT THE NUMBER OF SPANNING TREES INGRAPHSFigure 3.5: A graph G and its graph orientedFigure 3.5 (b) shows the graph of Figure 3.5 (a) with the edges oriented arbitrarily andlabelled with the letters from a to h. Relative to this orientation the signed incidencematrix of the graph is:⎡⎤−1 1 −1 0 0 0 0 01 0 0 −1 −1 0 0 0M =⎢ 0 0 1 0 1 0 1 0⎥⎣ 0 0 0 0 0 −1 −1 1 ⎦0 −1 0 1 0 1 0 −1(The rows are indexed by 1 ... 5 and the columns by a ... h.) For any matrix Mwe denote the conjugate transpose of M by M T .Lemma 3.4.3 Let G = (V, E) be a graph, orient G arbitrarily, and let M be the correspondingsigned incidence matrix. ThenL(G) = MM T .The proof of Matrix Tree Theorem can be shown to follow from the basic combinatorialidea, the multilinearity of the determinant and induction on the number of edges in thegraph.Proof: First, we show that if G ′ is an orientation of G and M is the incidence matrixof G ′ then L = MM T . Label the now directed edges by e 1 , ..., e m . By definition of theincidence matrix, since every entry in the n by n matrix MM T is the dot product of rowsof M, diagonal entries in the product count vertex degrees and off-diagonal entries count−1 for every edge of G between two vertices.


3.4. Counting the number of spanning trees in graphs algebraically 61Now we want to show that if B is an (n − 1) × (n − 1) submatrix of M, thendet B = 0 if the corresponding n − 1 edges contain a cycle, and det B = ±1 if they forma spanning tree of G. If the edges corresponding to the columns contain a cycle C, thenthe columns sum to the zero vector when weighted with +1 or −1 determined by whetherthe directed edge is followed forward or backward when following the cycle. This columndependency implies det B = 0.For the other case, we use induction on n. For n = 1, by convention a 0 × 0 matrixhas determinant 1. Now suppose n > 1, and let T be the spanning tree whose edgesare the columns of B. Since T has at least two leaves, B contains a row correspondingto a leaf x of T . This row has only one nonzero entry in B. When computing thedeterminant by expanding along that row, the only submatrix B ′ given nonzero weightin the expansion corresponds to the spanning subtree of G − x obtained by deleting xand its incident edge from T . Since B ′ is an (n − 2) × (n − 2) submatrix of the incidencematrix for an orientation of G − x, the induction hypothesis implies that the determinantof B ′ is ±1, and multiplying it by ±1 gives the same result for B.Finally, we need to compute det L ∗ . Let M ∗ be the matrix obtained by deletingrow t of M, so L ∗ = M ∗ (M ∗ ) T . We may assume m ≥ n − 1, else both sides havedeterminant 0 and there are no spanning subtrees. The Binet-Cauchy formula expressesthe determinant of a product of matrices, not necessarily square, in terms of the determinantsof submatrices of the factors. In particular, if m ≥ p, A is a p × m matrix, andB is an m × p matrix, then det AB = ∑ S det A S det B S , where the summation runs overall S ⊆ [m] consisting of p indices, A S is the submatrix of A having the columns indexedby S, and B S is the submatrix of B having the rows indexed by S. When we apply theBinet-Cauchy formula to L ∗ = M ∗ (M ∗ ) T , the submatrix A S is an (n − 1) × (n − 1)submatrix of M as discussed before, and B S = A T S . Hence the summation counts1 = (±1) 2 for each set of n − 1 edges corresponding to a spanning tree and 0 for eachother set of n − 1 edges.□Example 3.4.4 Consider the graph G that shown in the following Figure 3.6, we shallcompute the number of spanning trees of this graph by the matrix-tree theorem.Figure 3.6: A graph G and its adjacency matrix, degree matrix and Laplacian matrixThe degree (diagonal) matrix D for this graph is the 8 × 8 matrix and the adjacencymatrix A is the 8 × 8 matrix as follows:


62CHAPTER 3.HOW TO COUNT THE NUMBER OF SPANNING TREES INGRAPHS⎛D =⎜⎝2 0 0 0 0 0 0 00 2 0 0 0 0 0 00 0 3 0 0 0 0 00 0 0 3 0 0 0 00 0 0 0 3 0 0 00 0 0 0 0 3 0 00 0 0 0 0 0 2 00 0 0 0 0 0 0 2⎞⎛, A =⎟ ⎜⎠ ⎝Then the matrix L = D − A is the 8 × 8 matrix⎛L =⎜⎝2 −1 0 −1 0 0 0 0−1 2 −1 0 0 0 0 00 −1 3 −1 0 −1 0 0−1 0 −1 3 −1 0 0 00 0 0 −1 3 −1 0 −10 0 −1 0 −1 3 −1 00 0 0 0 0 −1 2 −10 0 0 0 −1 0 −1 20 1 0 1 0 0 0 01 0 1 0 0 0 0 00 1 0 1 0 1 0 01 0 1 0 1 0 0 00 0 0 1 0 1 0 10 0 1 0 1 0 1 00 0 0 0 0 1 0 10 0 0 0 1 0 1 0Let now s = 3. By deleting the s-th row and s-th column of L we have the 7 × 7 matrix⎛L ∗ =⎜⎝2 −1 −1 0 0 0 0−1 2 0 0 0 0 0−1 0 3 −1 0 0 00 0 −1 3 −1 0 −10 0 0 −1 3 −1 00 0 0 0 −1 2 −10 0 0 −1 0 −1 2The determinant of this matrix is 56. Thus the Matrix Tree Theorem states that thenumber of spanning trees for this graph is 56. These computations were performed byusing Maple.Example 3.4.5 In the labelled graph G shown in Figure 2.8, by using the matrix-treetheorem , we find a cofactor by removing the last row and column, and then taking thedeterminant.2 −1 0 0 −1−1 4 −1 0 −1τ(G) =0 −1 3 −1 0= 55.0 0 −1 2 0∣−1 −1 0 0 3 ∣⎞⎟⎠⎞⎟⎠⎞⎟⎠


3.4. Counting the number of spanning trees in graphs algebraically 633.4.2 Other algebraic methods for counting τ(G)Some other methods for counting τ(G) are as follows. Let A be an n × n matrix. Theeigenvalues of A are defined as the n roots of the characteristic polynomial det(λI − A).The spectrum of A consists of all eigenvalues of A. All eigenvalues of a real symmetricmatrix are real. For further detail and more explanation on properties and significance ofthe theory of spectra, we refer the reader to [5, 11, 21, 35, 57, 52, 53, 92, 80, 81, 83, 91, 93].The Laplacian matrix of a graph and its eigenvalues can be used in several areas ofmathematical research and have a physical interpretation in various physical andchemical theories. The related matrix: the adjacency matrix of a graph and its eigenvalueswere extensively investigated in the past than the Laplacian matrix. From theLaplace spectrum of a graph, one can determine the number of spanning trees (whichwill be nonzero only if the graph is connected). Furthermore, by basic knowledge oflinear algebra, all eigenvalues of the Kirchhoff matrix L of a graph with n verticesare non-negative, and 0 is one of its eigenvalues because all its row vectors add up tothe 0 vector. So we can let λ 1 ≥ λ 2 ≥ ... ≥ λ n (= 0) denote all eigenvalues of L.Kel’mans and Chelnokov [64] have shown that the number of distinct spanning treesof a graph G is equal to the product of nonzero eigenvalues of the combinatorial Laplacian.For a graph G, the combinatorial Laplacian has non-negative eigenvalues exceptone, λ 1 ≥ λ 2 ≥ ... ≥ λ n (= 0). The number of spanning trees, can be related to theeigenvalues of L as follows: (A proof can be found in [11]. For completeness, we brieflydescribe the proof here). We now can state the following theorem and its proof.Theorem 3.4.6 (Kirchhoff’s Matrix Tree Theorem) For a given undirected connectedgraph G with n vertices, let λ 1 , . . . , λ n−1 be the non-zero eigenvalues of L(G). Then,the number of distinct spanning trees τ(G) is equal to:τ(G) = 1 n−1∏λ i .nEquivalently, τ(G) is equal to the absolute value of any cofactor of the Laplacian matrixof G.Proof: Suppose we consider the characteristic polynomial p(x) of the combinatorialLaplacian L.p(x) = det(L − xI).The coefficient of the linear term is exactlyn−1−i=1∏λ i .i=1On the other hand, the coefficient of the linear term of p(x) is -1 times the sum of thedeterminant of n principal submatrix of L obtained by deleting the i-th row and i-th


64CHAPTER 3.HOW TO COUNT THE NUMBER OF SPANNING TREES INGRAPHScolumn. By the matrix-tree theorem, the product n−1 ∏of spanning trees of G.i=1λ i is exactly n times the number□Remark 3.4.7 [21] Let G be an undirected (multi)graph with at least one vertex, andLaplace matrix L with eigenvalues 0 = λ 1 ≥ λ 2 ≥ ... ≥ λ n . Let L ∗ vu be the (v, u)-cofactorof L. Then the number of spanning trees of G equalsτ(G) = L ∗ vu = det(L + 1 n 2 J) = 1 n λ 2...λ n for any v, u ∈ V (G).(The (i, j)-cofactor of a matrix M is by definition (−1) i+j det M(i, j), where M(i, j) isthe matrix obtained from M by deleting row i and column j. Note that L ∗ vu does notdepend on an ordering of the vertices of G)For example, the graph G of valency k on 2 vertices has Laplace matrix L as follows:( ) k −kL =,−k kso that λ 1 = 0, λ 2 = 2k, and τ(G) = 1 2 .2k = k. If we consider the complete graph K n,then λ 2 = ... = λ n = n, and therefore K n has τ(G) = n n−2 spanning trees. This formulawas proposed by Cayley [24]. Remark 4.3.14 is implicit in Kirchhoff [65] and it is knownas the Matrix-Tree Theorem [21].We consider two examples, shown in Figure 3.7 and Figure 3.8.Example 3.4.8 Consider the graph in Figure 3.7Figure 3.7: Example graph, with Laplacian matrix and eigenvalues. Numbers near eachvertex indicate the chosen ordering. The total number of spanning trees can be seen tobe 8 by inspection, which matches with Kirchhoff’s theorem.


3.4. Counting the number of spanning trees in graphs algebraically 65Example 3.4.9 Consider the graph in Figure 3.8Figure 3.8: The 2nd example graph, with Laplacian matrix and eigenvalues. Numbersnear each vertex indicate the chosen ordering. The total number of spanning trees can beseen to be 3 by inspection, which matches with Kirchhoff’s theorem.From Theorem 3.4.6, it is easy to derive the result of Sachs [102], which states that if Gis regular of degree r, thenτ(G) = 1 n−1∏(r − µ i ),ni=1where µ 1 ≤ µ 2 ≤ ... ≤ µ n = r are the eigenvalues of the adjacency matrix A. For example,the Petersen graph in Figure 3.9 is a regular graph. Its adjacency matrix has eigenvalues3, 1, 1, 1, 1, 1, -2, -2, -2, -2, and its Kirchhoff matrix has eigenvalues 0, 2, 2, 2, 2, 2, 5, 5,5, 5. It is concluded that the Petersen graph has 2000 spanning trees [113].Figure 3.9: The Petersen graph.Finally, we mention Temperley’s result [111],τ(G) = 1 n2det(L + J),where J is the n × n matrix all of whose elements are unity.


66CHAPTER 3.HOW TO COUNT THE NUMBER OF SPANNING TREES INGRAPHSRemark 3.4.10 As we have seen, calculating the number of spanning trees of planargraphs by using the determinant of Laplacian matrix or using the Laplace spectrum istedious and impractical. Hereafter (see chapters 4, 5, 6 and 7), we are going to providenew methods to facilitate the calculation of the number of spanning trees in planar maps(graphs embedded into surfaces) to find the number of spanning trees exactly in somespecial planar maps.3.5 More known resultsThe Feussner’s theorem, Matrix Tree Theorem, and Kel’mans and Chelnokov’s result provideefficient algorithms for calculating τ(G) for any graph G, but they are infeasible forlarge graphs because of the computational difficulty of large matrices. For some specialclasses of graphs, it is possible to give explicit, simple formulae for the numbers of spanningtrees. The best known is Cayley’s tree formula [25] which states that, if K n is thecomplete graph with n vertices then τ(K n ) = n n−2 . Closed formulae for the numbers ofspanning trees in some special graphs, such as, complete prism, fan, point wheel, ladderand Moebius ladder in [14], cocktail party graph, and k-dimensional lattice in [36], completemultipartite graph in [72], multi-star related graph in [95, 122], multi-complete/starrelated graph in [30], circulant graphs Cn 1,2 , Cn 1,3 , Cn1,4 in [123, 124], generalized de Bruijngraphs in [73], and iterated line digraphs in [128], can be found. The range of the numbersof spanning trees in a regular graph [4, 82] and a graph with a given degree sequence [68]has been explored. The asymptotic property of the numbers of spanning trees in somecirculant graphs has been investigated in [74, 75, 126]. The numbers of spanning trees inAztec diamonds has been investigated in [67, 110], and the problem has been generalizedin [27].As the literature regarding the numbers of spanning trees in graphs is extensive. It isnot possible to mention it all here, however, some of it and our results may be seen in thefollowing chapters.


Chapter 4New methods to compute the numberof spanning trees of Planar MapsThe number of spanning trees of a graph G is certainly one of the most important graphtheoreticalparameters, and also one of the oldest. It is an important, well-studied quantityin graph theory, that appears in a number of applications. Its applications range fromthe theory of networks, where the number of spanning trees is used as a measure fornetwork reliability. In this chapter, we start by stating the general methods for countingthe number of spanning trees in maps; we then discuss our new results.4.1 IntrodutionAs mentioned previously, calculating the number of spanning trees of a planar map by thedeterminant of Laplacian matrix is tedious and impractical. Hereafter, we give some newmethods to facilitate the calculation of the number of spanning trees of planar maps. Inother words, this chapter presents some simple results on the calculation of the numberof spanning trees in planar maps. In the following, we assume our maps are loopless butmultiple edges are allowed.4.2 Complexity of mapsIn this work, we denote the number of spanning trees of a map C by τ(C), sometimescalled the complexity of C. The complexity of a map C is the total number of spanningtrees of C (trees that passing through all the vertices of the map); see Figure 4.1.67


68CHAPTER 4.NEW METHODS TO COMPUTE THE NUMBER OF SPANNINGTREES OF PLANAR MAPSRemark 4.2.1Figure 4.1: A map C and its complexity which has five spanning trees1. The number of spanning trees of a tree is equal to 1 (a graph that is a tree hasexactly one spanning tree; the full graph itself).2. Each map has two faces, possesses one cycle. Let C be a planar map with two faces(a cycle) with n vertices, then the complexity of this map C is equal to the lengthof its cycle (the length of a cycle is the number of edges that form this cycle); seeFigure. 5.1.Figure 4.2: Example of a map C which has two faces (Cycle)3. If C is not connected then τ(C) = 0 (i .e., a disconnected map has no spanningtrees.), so we can assume that C is connected from now on. If C ′ is obtained fromC by removing all the loops of C, then τ(C ′ ) = τ(C), since a loop can never occurin a spanning tree, i.e., loops don’t affect spanning trees, so we delete them beforethe computation. Thus, we may assume that C contains no loops as well. Multipleedges, however, do remain a possibility.4.3 Main Results4.3.1 Counting the number of spanning trees in a map of typeC = C 1 • C 2Definition 4.3.1 We denote by C = C 1 • C 2 the map obtained by connecting C 1 andC 2 with a single vertex v 1 , i.e., C 1 and C 2 are connected maps which intersect exactly in


4.3. Main Results 69one vertex (any path connecting a vertex of C 1 to a vertex of C 2 must pass through thisvertex); see Figure 4.3.Figure 4.3: A map C = C 1 • C 2Property 4.3.2 Let C be a map of type C = C 1 • C 2- C 1 and C 2 have a common vertex v 1 and a common face (the external face).- V C =V C1 +V C2 -1, E C =E C1 +E C2 and F C =F C1 +F C2 -1- A path from a vertex of C 1 to a vertex of C 2 must pass through v 1 .- If we remove the vertex v 1 of the map C, the resulting map is not connected.Theorem 4.3.3 If we have a map C such that C = C 1 • C 2 , thenτ(C) = τ(C 1 • C 2 ) = τ(C 1 ) × τ(C 2 ).Proof: Each path that connects a vertex of C 1 to a vertex of C 2 must pass through v 1 .The laplacian matrix associated with a previous map C = C 1 • C 2 is as follows:After deleting the row and the column of the vertex v 1 , we obtain the following matrix:⎛⎞M n1 ,n 10⎝⎠0 M n2 ,n 2


70CHAPTER 4.NEW METHODS TO COMPUTE THE NUMBER OF SPANNINGTREES OF PLANAR MAPSIn calculating the determinant, we obtain:τ(C) = τ(C 1 ) × τ(C 2 ).□Theorem 4.3.4 (Generalization of Theorem 4.3.3) Let C be one of the following maps(see Figure 4.4), thenn∏τ(C) = τ(C i ).i=1Figure 4.4: Star map and chain map4.3.2 Counting the number of spanning trees in a map CDefinition 4.3.5 Let C be a map that contains an edge e = (v 1 , v 2 ), we denote by C − ethe map obtained by deleting the edge e = (v 1 , v 2 ) and the map remains connected. Wedenote by C.e or C.v 1 v 2 the map obtained by deleting the edge e and pasting the vertexv 1 with v 2 (contracting the two vertices v 1 and v 2 ).Example 4.3.6 Here is an example of maps C, C − e and C.e (see Figure. 4.5).Figure 4.5: An example of maps C, C − e and C.eLemma 4.3.7 If C is a connected planar map, then for any e ∈ E(C), there is a spanningtree containing e.


4.3. Main Results 71Lemma 4.3.8 Let C be a map and let e = v i v j be an edge of C (v i ≠ v j ) then: thenumber of spanning trees of map C passing through e (containing an edge e) is equal to:τ(C.e) = det L(C)[v i ; v j ].Proof: Let C a map, e = v i v j an edge of C(e is not a loop), τ(C.e) is the number ofspanning trees of map C which contains the edge e is given by:Remark 4.3.9 In case where C = T is a tree, τ(T.v i v j ) is the distance between twovertices v i and v j [87].Lemma 4.3.10 The number of spanning trees in a map C not containing an edge e isequal to τ(C − e).One way to compute the number of spanning trees in a map (graph embedded into surfaces),is to recursively degenerate the map by removing and contracting an edge. Inthis way, the number of spanning trees of a map C is reduced to a sum of the number ofspanning trees in small map which can be computed by hand. A Contraction-Deletiondegeneration could look like this□Figure 4.6: A Contraction-Deletion methodTheorem 4.3.11 (Contraction-Deletion) [89, 117] Let C be a map and τ(C) denote thenumber of spanning trees of C. If e = v i v i+1 ∈ E(C) (e is not a loop), thenτ(C) = τ(C − e) + τ(C.e).


72CHAPTER 4.NEW METHODS TO COMPUTE THE NUMBER OF SPANNINGTREES OF PLANAR MAPSProof: Let T be the set of spanning trees of C,τ(C) = |T |, T = T 1 ∪ T 2 , T 1 ∩ T 2 = ∅, withT 1 = the set of spanning trees of C which contains the edge eT 2 = the set of spanning trees of C which does not contain the edge e|T | = |T 1 | + |T 2 |,|T 1 | = τ(C.e) and |T 2 | = τ(C − e) then τ(C) = τ(C.e) + τ(C − e).We cannot apply the recurrence of Theorem 4.3.11 when e is a loop. For example, a mapconsisting of one vertex and one loop has one spanning tree, but deleting and contractingthe loop would count it twice. Since loops do not affect the number of spanning trees, wecan delete loops as they arise.Remark 4.3.12 The quantity τ(C) can be computed recursively, using the formulaτ(C) = τ(C − e) + τ(C.e), in which C − e is the map obtained from C by deleting theedge e and C.e is the map obtained from C by contracting the edge e (and removing anyloops produced). [Proof of this formula is a simple case analysis: τ(C − e) counts thespanning trees of C that do not contain the edge e, and τ(C.e) counts the spanning treesof C that do contain the edge e.] Furthermore, if C 1 and C 2 are connected maps whichintersect in exactly one vertex (and no edges) then τ(C 1 • C 2 ) = τ(C 1 ) × τ(C 1 ) (Theorem4.3.3), as is also easily seen. See Figure 4.7 for an example of computation using thismethod. (For each map in the figure, an edge which is deleted-contracted is marked withan asterisk.)□Figure 4.7: Computing of τ(C) by deletion-contraction.


4.3. Main Results 73This recursion shows that the function τ is recursively enumerable, but the resultingalgorithm in general requires on the order of 2 |E(C)| arithmetic operations, and so it isnot suitable for large computations. It is a fortune that we have several computationallysimplified and totally generalized formulae for τ(C) which we are going to derive them inthis chapter and in upcoming chapters.Example 4.3.13 Here is an example of a map C with Recursive calculation of τ(C)Figure 4.8: Recursive calculation of τ(C)We consider finite undirected simple maps and multimaps with no loops; the term multimapis used when multiple edges are allowed in a map. For a map C, we denote by V (C)and E(C) the vertex set and edge set of C, respectively. The multiplicity of a vertex-pair(v, u) of a map C, denoted by l C (vu), is the number of edges joining the vertices v and uin C.Remark 4.3.14 If C is a connected loopless map with no cycle of length at least 3, thenτ(C) is the product of the edge multiplicities. A disconnected map has no spanning trees.Counting trees recursively requires initial conditions for maps (graphs) in which all edgesare loops. Such a map has one spanning tree if it has only one vertex, and it has nospanning trees if it has more than one vertex. If a computer completes the computationby deleting or contracting every edge in a loopless map C, then it may compute as manyas 2 E(C) terms. Even with savings from Remark 4.3.14, the amount of computation growsexponentially with the size of the map; this is impractical, therefore, we shall providenew non-trivial methods for calculating the number of spanning trees in maps in general,then in particular in some special maps and prove new simplified results in this chapter.Now, we consider the case where the edge e is replaced by a simple path p = v 1 ,v 2 , ..., v k , v k+1 of length k which connects v 1 with v k+1 , i.e., If there are k edges betweentwo vertices v 1 , v 2 of a map C.


74CHAPTER 4.NEW METHODS TO COMPUTE THE NUMBER OF SPANNINGTREES OF PLANAR MAPSDefinition 4.3.15 Let C be a map that contains a simple path p formed by the verticesv 1 , v 2 , ..., v k , v k+1 . We denote by C − p the map obtained by deleting the simple pathp and the map remains connected. We denote by C.p or C.v 1 v k+1 the map obtained bydeleting the simple path p and pasting two vertices v 1 and v k+1 .Example 4.3.16 Here is an example of maps C, C − p and C.p; see Figure. 4.9.Figure 4.9: An example of maps C, C − p and C.pLemma 4.3.17 (Generalization of Lemma 4.3.7) If C is a connected planar map, thenfor any simple path p from C, there is a spanning tree containing p.Lemma 4.3.18 (Generalization of Lemma 4.3.8) Let C be a map and let v 1 , v k+1 be twovertices of map C connected by a simple path p, then: the number of spanning trees of Cthat contains the path p isτ(C.p) = det L(C)[v 1 ; v 2 ; ...; v k ; v k+1 ].Proof: It has the same proof as Lemma 4.3.8.Lemma 4.3.19 (Generalization of Lemma 4.3.10) The number of spanning trees in amap C not containing a simple path p is equal to τ(C − p).If there are k edges between two vertices v 1 , v 2 (a simple path p) of a map C. Let C 1 bethe map derived from C by deleting all edges between v 1 and v 2 , and C 2 the map derivedfrom C by shrinking v 1 and v 2 into one vertex. It is then clear that τ(C) = kτ(C 1 ) + τ(C 2 )Theorem 4.3.20 (Generalization of Theorem 4.3.11) [89, 87] Let C be a map, and letp = v 1 , v 2 , ..., v k , v k+1 be a simple path in the map C, thenwhere k is the length of the path p.τ(C) = kτ(C − p) + τ(C.p),□


4.3. Main Results 75Proof: It is the same proof as that of Theorem 4.3.11 with |T 1 | = τ(C.p) and |T 2 | =kτ(C − p) since there are k ways to cut the path (v 1 , v 2 , ..., v k , v k+1 ), then τ(C) =τ(C.p) + kτ(C − p).Theorem 4.3.21 Let C be a map, then|E C |1 ∑τ(C) =τ(C.e i ), where e i ∈ E C .|V C | − 1i=1Proof: Let C a map, T a spanning tree of C and e i an edge of T (there are |V C | − 1edges in T ).If in C, we look for all spanning trees that must pass through e i we find τ(C.e i ) trees witha single copy of the tree T .If we sum on the e i with i = 1, ..., |E C |, we are going to have |V C | − 1 times the tree T(because the other remaining edges |F C | − 1 do not belong to T ) then□|E C |∑τ(C.e i ) = (|V C | − 1)τ(C),i=1hence the result.□Corollary 4.3.22 Let C be a map, then|E C |1 ∑τ(C) =τ(C − e i ), such that C − e i is connected.|F C | − 1i=1Proof:From Theorem 4.3.11, we have:|E C ||E∑C ||E∑C |∑τ(C) = τ(C.e i ) + τ(C − e i )i=1i=1i=1From the previous Theorem 4.3.21, we have:|E C |∑|E C |τ(C) − (|V C | − 1)τ(C) = τ(C − e i )By Euler’s formula, we obtain the result.i=1□


76CHAPTER 4.NEW METHODS TO COMPUTE THE NUMBER OF SPANNINGTREES OF PLANAR MAPS4.3.3 Counting the number of spanning trees in a map of typeC= C 1 : C 2Now, we are interested in the maps of type C= C 1 : C 2 , C is a map such that if a pathconnects a vertex v k of C 1 and a vertex v l of C 2 must pass through v 1 or v 2 ; C 1 and C 2have two common vertices v 1 , v 2 (see Figure 4.10).Figure 4.10: A map C= C 1 : C 2Example 4.3.23 Here is an example of this type of maps (see Figure 4.11):Figure 4.11: An example of map C= C 1 : C 2Property 4.3.24 Let C be a map of type C= C 1 : C 2- C 1 and C 2 have two common vertices v 1 , v 2 and a common face (the external face).- V C =V C1 +V C2 -2, E C =E C1 +E C2 and F C =F C1 +F C2 .- A path that connects a vertex of C 1 and a vertex of C 2 must pass through v 1 or v 2 .- If we remove the two vertices v 1 and v 2 of the map C, the resulting map is not connected.Example 4.3.25 Here is an example of maps C= C 1 : C 2 , C 1 , C 2 , C 1 .v 1 v 2 and C 2 .v 1 v 2 (seeFigure. 4.12).


4.3. Main Results 77Figure 4.12: An example of maps C= C 1 : C 2 , C 1 , C 2 , C 1 .v 1 v 2 and C 2 .v 1 v 2Theorem 4.3.26 [87, 89] Let C= C 1 : C 2 be a map, v 1 and v 2 two vertices of the map Cwhich is formed by two maps C 1 and C 2 (see Figure 4.10), thenτ(C) = τ(C 1 ) × τ(C 2 .v 1 v 2 ) + τ(C 1 .v 1 v 2 ) × τ(C 2 ).Proof: Let T be the set of spanning trees of C, τ(C) = |T |.Let T 1 be the set of spanning trees of C which has a path from the vertex v 1 to the vertexv 2 in C 1 and let T 2 be the set of spanning trees of C which has a path from the vertex v 1to the vertex v 2 in C 2 .We have T 1 ∩ T 2 = ∅ and T 1 ∪ T 2 = T , then τ(C) = |T | = |T 1 | + |T 2 ||T 1 | is the number of spanning trees of C which has a path from v 1 to v 2 in C 1 and hasnot any path from v 1 to v 2 in C 2 .On the other hand, all spanning trees of C 1 have a path from v 1 to v 2 in C 1 , all spanningtrees of C 2 .v 1 v 2 does not have any path from v 1 to v 2 then |T 1 | = τ(C 1 ) × τ(C 2 .v 1 v 2 ) thesame for |T 2 | = τ(C 2 )×τ(C 1 .v 1 v 2 ) hence the result.□Example 4.3.27 In Example 4.3.25, we apply Theorem 4.3.26, we then obtain:τ(C) = τ(C 1 ) × τ(C 2 .v 1 v 2 ) + τ(C 1 .v 1 v 2 ) × τ(C 2 )= 3 × 4 + 5 × 4= 32.Remark 4.3.28 Let C be one of the following maps (see Figure 4.13), thenτ(C) = deg v = deg uor,τ(C)= The number of edges of the map = The number of faces of the map.


78CHAPTER 4.NEW METHODS TO COMPUTE THE NUMBER OF SPANNINGTREES OF PLANAR MAPSFigure 4.13: Case of Particular maps4.3.4 Counting the number of spanning trees in a map of typeC= C 1 | C 2Now, we are interested in the maps of type C= C 1 | C 2 , C which is a map such that v 1 andv 2 two vertices of map C connected by an edge e (see Figure 4.14).Figure 4.14: A map C= C 1 | C 2Property 4.3.29 Let C be a map of type C = C 1 | C 2- C 1 and C 2 have two common vertices v 1 , v 2 , a common edge e and a common face (theexternal face).- V C =V C1 +V C2 -2, E C =E C1 +E C2 -1 and F C =F C1 +F C2 -1.Example 4.3.30 Here is an example of maps C, C 1 , C 2 , C 1 − e, C 2 − e, C 1 .e and C 2 .e (seeFigure. 4.15).Figure 4.15: An example of maps C, C 1 , C 2 , C 1 − e, C 2 − e, C 1 .e and C 2 .e


4.3. Main Results 79Theorem 4.3.31 [87, 89] Let C be a map of type C= C 1 | C 2 (v 1 and v 2 two vertices ofthe map C connected by an edge e) (see Figure 4.14), thenτ(C) = τ(C 1 ) × τ(C 2 ) − τ(C 1 − e) × τ(C 2 − e).Proof: We transform the map C as follows (see Figure 4.16):τ(C 1 | C 2 )= τ((C 1 − e) : C 2 )From Theorem 4.3.26, we haveFigure 4.16: The map C after the tranformationτ(C) = τ((C 1 − e).v 1 v 2 ) × τ(C 2 ) + τ(C 1 − e) × τ(C 2 .v 1 v 2 ).From Theorem 4.3.11, we have τ(C 2 .e) = τ(C 2 ) − τ(C 2 − e) and since τ((C 1 − e).v 1 v 2 ) =τ(C 1 .e), C 2 .v 1 v 2 = C 2 .e thenτ(C) = τ(C 1 .e) × τ(C 2 ) + τ(C 1 − e) × (τ(C 2 ) − τ(C 2 − e))We take τ(C 2 ) as a factor, then= τ(C 1 .e) × τ(C 2 ) + τ(C 1 − e) × τ(C 2 ) − τ(C 1 − e) × τ(C 2 − e).τ(C) = [τ(C 1 .e) + τ(C 1 − e)]τ(C 2 ) − τ(C 1 − e)τ(C 2 − e)We use Theorem 4.3.11, we obtain : τ(C) = τ(C 1 )τ(C 2 )−τ(C 1 −e)τ(C 2 −e).□Example 4.3.32 Here is an example of a map C of type C= C 1 | C 2 with calculation ofτ(C) (see Figure 4.17):Figure 4.17: An example of a map C of type C = C 1 | C 2we apply Theorem 4.3.31 in C 1 to calculate τ(C 1 ), we then obtain τ(C 1 )= 8, (see Figure4.18):


80CHAPTER 4.NEW METHODS TO COMPUTE THE NUMBER OF SPANNINGTREES OF PLANAR MAPSFigure 4.18: A map C of type C= C 1 | C 2 with calculation of τ(C)The same goes for C 2 , we obtain τ(C 2 )= 8. Now, we calculate τ(C), we again applyTheorem 4.3.31, we then obtain τ(C) = 55, (see Figure 4.19):Figure 4.19: A map C of type C= C 1 | C 2 with calculation of τ(C)4.3.5 Counting the number of spanning trees in a map of typeC= C 1 ‡ C 2We take the previous map shown in the figure 4.14, by replacing the edge e by ‡ a simplepath p = v 1 , v 2 , ..., v k , v k+1 (see Figure 4.20).Figure 4.20: A map C = C 1 ‡ C 2


4.3. Main Results 81Property 4.3.33 Let C be a map of type C = C 1 ‡ C 2 , where ‡ a simple path thatcontains k + 1 vertices and k edges, then- C 1 and C 2 have k + 1 common vertices v 1 , v 2 ,..., v k , v k+1 , k common edges (simple pathp) and a common face (the external face).- V C =V C1 +V C2 -(k + 1), E C =E C1 +E C2 -k and F C =F C1 +F C2 -1.Theorem 4.3.34 (Generalization of Theorem 4.3.31) Let C be a map, v 1 and v k+1 twovertices of the map C connected by a simple path p = v 1 , v 2 , ..., v k , v k+1 that contains kedges; see Figure 4.20, thenτ(C) = τ(C 1 ) × τ(C 2 ) − k 2 τ(C 1 − p) × τ(C 2 − p).Proof:From Theorem 4.3.20, and we use Theorems 4.3.3 and 4.3.26, we have:τ(C) = τ(C.p) + kτ(C − p)= τ(C 1 .p)τ(C 2 .p) + k[τ(C 1 .p)τ(C 2 − p) + τ(C 1 − p)τ(C 2 .p)]= [τ(C 1 .p) + kτ(C 1 − p)]τ(C 2 .p) + kτ(C 1 .p)τ(C 2 − p)= τ(C 1 )[τ(C 2 ) − kτ(C 2 − p)] + kτ(C 1 .p)τ(C 2 − p)= τ(C 1 )τ(C 2 ) − kτ(C 1 )τ(C 2 − p) + kτ(C 1 .p)τ(C 2 − p)= τ(C 1 )τ(C 2 ) − k[τ(C 1 ) − τ(C 1 .p)]τ(C 2 − p)= τ(C 1 )τ(C 2 ) − k 2 τ(C 1 − p)τ(C 2 − p), hence the result.Example 4.3.35 Here is an example of a map C of type C= C 1 ‡ C 2 with calculation ofτ(C) (see Figure 4.21):□Figure 4.21: An example of a map C of type C = C 1 ‡ C 2we apply Theorem 4.3.34 in C 1 to calculate τ(C 1 ), we then obtain τ(C 1 ) = 16, (see Figure4.22):


82CHAPTER 4.NEW METHODS TO COMPUTE THE NUMBER OF SPANNINGTREES OF PLANAR MAPSFigure 4.22: A map C of type C= C 1 ‡ C 2 with calculation of τ(C)and, we apply Theorem 4.3.31 in C 2 to calculate τ(C 2 ), (see Figure 4.21); we thenobtain τ(C 2 )= 15. Now, we calculate τ(C), we again apply Theorem 4.3.34, we thenobtain τ(C) = 176, (see Figure 4.23):Figure 4.23: A map C of type C= C 1 ‡ C 2 with calculation of τ(C)In the next chapter, we are going to apply our methods which we have seen in this chapteron some special planar maps (maps complicated with n vertices) to give the number ofspanning trees in them and derive several simple formulae which facilitate the calculationof the number of spanning trees in some special families of planar maps.


Part IIIUse of Derived Theoretical Results83


Chapter 5The Number of Spanning Trees ofCertain Families of Planar MapsIn this chapter, we are going to derive several easily computable formulae for certainfamilies of planar maps, and similar formulae for related generating series.5.1 IntroductionThe number of spanning trees of a map C is the total number of distinct spanning subgraphsof C that are trees. Even though the classic result, Matrix Tree Theorem expressesthe number of spanning trees τ(C) of a map C as a function of the determinant of amatrix that can be easily constructed from C’s incidence matrix, in practice, this methodof counting spanning trees by calculating determinants is infeasible for large planar maps(graphs embedded in the plane without edge-crossings). For some special classes of planarmaps, it is possible to give explicit, simple formulae for the number of spanning trees. Inthis chapter, we apply our methods which we have seen in Chapter 4 to give the numberof spanning trees in some special planar maps and derive several simple formulae forthe number of spanning trees of special families of planar maps. They are known as then-Fan chains, the n-Grid chains, the n-tent chains, the n-Hexagonal chains, the n-Eightchains, the n-Barrel chains the n-Light chains, the n-Home chains, the n-Kite chains, then-Envelope chains and the n-Diphenylene chains ... etc.5.2 The case of one cycleEach map having two faces possesses one cycle. Let C be a planar map with two faces(i.e., a cycle) with n vertices, then the complexity of C is equal to the length of its cycle(the length of a cycle is the number of edges that form this cycle); see Figure. 5.1.85


CHAPTER 5.86THE NUMBER OF SPANNING TREES OF CERTAIN FAMILIES OFPLANAR MAPSFigure 5.1: Example of A map C which has two faces (Cycle)5.3 The case of two cyclesTheorem 5.3.1 Let C be the following map; see Figure 5.2 (It contains two cycles C 1with h 1 edges and C 2 with h 2 edges, and k is the length of the simple path p), thenτ(C) = τ(C 1 ) × τ(C 2 ) − k 2 .Figure 5.2: Case of two cyclesProof:From Theorem 4.3.20, we haveτ(C) = τ(C.p) + kτ(C − p) = (h 1 − k)(h 2 − k) + k[(h 1 − k) + (h 2 − k)]= h 1 h 2 − k 2 = τ(C 1 ) × τ(C 2 ) − k 2 .□Particular case:In the previous Theorem 5.3.1, if k = 1, then τ(C) = τ(C 1 ) × τ(C 2 ) − 1.Example 5.3.2 Consider the following simple map (see Figure. 5.3); using the previousTheorem 5.3.1, it is clear that the number of spanning trees of this map is 23.


5.4. The case of n cycles (n ≥ 3) 87Figure 5.3: A simple planar map has 23 spanning trees5.4 The case of n cycles (n ≥ 3)Let C n be a sequence of maps defined by:• C 1 is a cycle with h 1 edges and it contains the path v 1 , v 2 , ..., v k , v k+1 of length k 1 .• C n is formed by n cycles of length h 1 , h 2 , ..., h n , the i − th cycle and the (i + 1) − thcycle have common path of length k i (see Figure 5.4).Figure 5.4: Case of n cycles


CHAPTER 5.88THE NUMBER OF SPANNING TREES OF CERTAIN FAMILIES OFPLANAR MAPSTheorem 5.4.1 For the sequence maps C n in Figure 5.4, we have:⎧⎨ τ(C n ) = h n τ(C n−1 ) − τ(C n−2 )kn−1, 2 with n ≥ 3(1) τ(C 1 ) = h 1 ,⎩τ(C 2 ) = h 1 h 2 − k12where k n is the number of common edges between the i − th and the (i + 1) − th cyclesof C n and h i is the number of edges of i − th cycle (h i ≥ 3, for i = 1, ..., n).Proof: τ(C 1 ) = h 1 , τ(C 2 ) = h 1 h 2 − k1, 2 in the sequence of maps C n , we cut the last cycleof length h n (see Figure 5.4) and we use Theorem 4.3.34, then we obtain:τ(C n ) = τ(C n−1 )h n − τ(C n−2 )k 2 n−1 = h n τ(C n−1 ) − τ(C n−2 )k 2 n−1,hence we obtain the system (1).□5.5 Particular casesIn this section, we apply the previous Theorem 5.4.1 to obtain some particular cases ofk i and h i .5.5.1 The case of k i = k and h i = h (h ≥ 2k + 1)Theorem 5.5.1 In the previous Theorem 5.4.1, if we take k i = k for i = 1, ..., n − 1 andh i = h (h ≥ 2k + 1) for i = 1, ..., n, n ≥ 1, we obtain the sequence of maps C n in Figure5.5, thenFigure 5.5: Case of n cycles whose lengths are the same with h i = h and k i = kτ(C n ) =(1√h2 − 4k 2( h + √ h 2 − 4k 22) n+1 − ( h − √ h 2 − 4k 2)), n+1 n ≥ 1.2


5.5. Particular cases 89Proof: τ(C 1 ) = h, τ(C 2 ) = h 2 − k 2 , in the sequence of maps C n , we cut the last cycle oflength h (see Fig. 5.5) and we use Theorem 4.3.34, then we obtain: τ(C n ) = τ(C n−1 )h −τ(C n−2 )k 2 = hτ(C n−1 ) − τ(C n−2 )k 2 , hence we obtain the system:⎧⎨⎩τ(C n ) = hτ(C n−1 ) − τ(C n−2 )k 2 , with n ≥ 3τ(C 1 ) = h,τ(C 2 ) = h 2 − k 2The characteristic equation is r 2 − hr + k 2 = 0, ∆ = h 2 − 4k 2 , if h ≥ 2k ⇒ ∆ ≥ 0,therefore the solutions of this equation are: r 1 = h−√ h 2 −4k 2and r2 2 = h+√ h 2 −4k 2, hence:2τ(C n ) = α( h−√ h 2 −4k 2) n + β( h+√ h 2 −4k 2) n , α, β ∈ R, n ≥ 1 Using the initial conditions2 2τ(C 1 ) = h and τ(C 2 ) = h 2 − k 2 , we obtain:α = −h2 +h √ √ h 2 −4k 2 +2k 2h 2 −4k 2 (h− √ and β = h2 +h √ √ h 2 −4k 2 −2k 2h 2 −4k 2 ) h 2 −4k 2 (h+ √ , hence the result.h 2 −4k 2 )□5.5.2 The case of k i = 1 and h i = hTheorem 5.5.2 In the previous Theorem 5.5.1, if we take k i = 1 for i = 1, ..., n − 1 andh i = h (h ≥ 3) for i = 1, ..., n, n ≥ 1, we obtain the sequence of maps C n in Figure 5.6,Figure 5.6: Case of n cycles whose lengths are the same with h i = h and k i = 1thenτ(C n ) =(1√h2 − 4( h + √ h 2 − 42) n+1 − ( h − √ h 2 − 4)), n+1 n ≥ 1.2Proof: By replacing k i = 1 and h i = h for i = 1, ..., n in the system (1) of Theorem5.4.1, we get :⎧⎨ τ(C n ) = hτ(C n−1 ) − τ(C n−2 )τ(C 1 ) = h⎩τ(C 2 ) = h 2 − 1


CHAPTER 5.90THE NUMBER OF SPANNING TREES OF CERTAIN FAMILIES OFPLANAR MAPSThe characteristic equation is r 2 − hr + 1 = 0, ∆ = h 2 − 4, if h ≥ 2 ⇒ ∆ ≥ 0 thereforethe solutions are: r 1 = h−√ h 2 −4and r2 2 = h+√ h 2 −4, hence:2τ(C n ) = α( h − √ h 2 − 42) n + β( h + √ h 2 − 4) n , α, β ∈ R, and n ≥ 12Using the initial condition τ(C 1 ) = h, τ(C 2 ) = h 2 − 1, we obtain:α = (h+√ h 2 −4)(−h 2 +2+h √ h 2 −4)4 √ h 2 −4, β = (h−√ h 2 −4)(h 2 −2+h √ h 2 −4)4 √ , hence the result.h 2 −4□Remark 5.5.3 In Theorem 5.5.1, if we replace k = 1, we obtain the sequence of mapsC n shown in Figure 5.6, thenτ(C n ) =(1√h2 − 4( h + √ h 2 − 425.6 Other values of k i = 1 and h i = h) n+1 − ( h − √ h 2 − 4)), n+1 n ≥ 1.2In this section, we fix the value of k i by 1 and change the value of h i ; subsequently, wederive a simple formula for the number of spanning trees of special families of planarmaps (classical results), called the n-Fan chains, the n-Grid chains, the n-Tent chains,the n-Hexagonal chains and the n-Eight chains, ... etc.• The n-Fan chains (k = 1 and h = 3)If we take h = 3 in the sequence of maps C n in Figure 5.6, we obtain the sequencen-Fan chains F n , where n is the number of triangles (|V Fn | = n + 2) [39]; (seeFigure 5.7).Figure 5.7: The n-Fan chains F n


5.6. Other values of k i = 1 and h i = h 91Corollary 5.6.1 (The n-Fan chains) The complexity of the n-Fan chains F n is given bythe following formula:τ(F n ) = 1 √5(( 3 + √ 52) n+1 − ( 3 − √ 5)), n+1 n ≥ 1.2• The n-Grid chains (k = 1 and h = 4)If we take h = 4 in the sequence of maps C n in Figure 5.6, we obtain the sequencen-Grid chains G n , where n is the number of squares (|V Gn | = 2n + 2) [125, 106];(see Figure 5.8).Figure 5.8: The n-Grid chains G nCorollary 5.6.2 (The n-Grid chains) The complexity of the n-Grid chains G n (|V Gn | =2n + 2), is given by the following formula:τ(G n ) = 1 ((22 √ + √ 3) n+1 − (2 − √ )3) n+1 , n ≥ 1.3• The n-Tent chains (k = 1 and h = 5)If we take h = 5 in the sequence of maps C n in Figure 5.6, we obtain the sequence n-Tentchains T n (see Figure. 5.9).Figure 5.9: The n-Tent chains T n


CHAPTER 5.92THE NUMBER OF SPANNING TREES OF CERTAIN FAMILIES OFPLANAR MAPSCorollary 5.6.3 (The n-Tent chains) The complexity of the n-Tent chains T n is givenby the following formula:τ(T n ) = √ 1 (21( 5 + √ 212• The n-Hexagonal chains (k = 1 and h = 6)) n+1 − ( 5 − √ 21)), n+1 n ≥ 1.2If we take h = 6 in the sequence of maps C nn-Hexagonal chains H n (see Figure 5.10).in Figure 5.6, we obtain the sequenceFigure 5.10: The n-Hexagonal chains H nCorollary 5.6.4 (The n-Hexagonal chains) The complexity of the n-Hexagonal chainsH n is given by the following formula:τ(H n ) = 1 ((34 √ + 2 √ 2) n+1 − (3 − 2 √ )2) n+1 , n ≥ 1.2• The n-Eight chains (k = 1 and h = 8)If we take h = 8 in the sequence of maps C n in Figure 5.6, we obtain the sequence n-Eightchains E n (see Figure 5.11).Figure 5.11: The n-Eight chains E nCorollary 5.6.5 (The n-Eight chains) The complexity of the n-Eight chains E n is givenby the following formula:τ(E n ) = 1 ((42 √ + √ 15) n+1 − (4 − √ )15) n+1 , n ≥ 1.15


5.7. Other uses 93Numerical results The Table 5.1 illustrates some of the values of the number of spanningtrees in the n-Fan chains F n , the n-Grid chains G n , the n-Tent chains T n , then-Hexagonal chains H n and in the n-Eight chains E n , by using the formulae given in theprevious Corollaries:n 1 2 3 4 5 6 7 8 9 10τ(F n ) 3 8 21 55 144 377 987 2584 6765 17711τ(G n ) 4 15 56 209 780 2911 10864 40545 151316 564719τ(T n ) 5 24 115 551 2640 12649 60605 290376 1391275 6665999τ(H n ) 6 35 204 1189 6930 40391 235416 1372105 7997214 46611179τ(E n ) 8 63 496 3905 30744 242047 1905632 15003009 118118440 929944511Table 5.1: Some values of τ(F n ), τ(G n ), τ(T n ), τ(H n ) and τ(E n )5.7 Other usesIn this section, we firstly consider some other cases of maps which are not cycles. Wethen show that the number of spanning trees in some special planar maps which calledthe n-Home chains, the n-Barrel chains and the n-Light chains always satisfy recurrencerelations and describe how to derive these relations, i.e., we find three simple formulae forcalculating the number of spanning trees in the n-Home chains, the n-Barrel chains andthe n-Light chains.5.7.1 Formula for the Number of Spanning Trees in the n-HomechainsIn this section, we derive a simple formula for the number of spanning trees of a specialfamily of maps called the n-Home chains.Now, by applying Theorem 4.3.31, we obtain the following results:Theorem 5.7.1 (The n-Home chains) The complexity of the n-Home chains H n (seeFigure. 5.12) is given by the following formula:τ(H n ) = ( 1 2 − 11√ 85170 ) (( 11 − √ 852) n − 1 9 (11 + √ 85)), n+2 n ≥ 1.2Proof: τ(H 1 ) = 11, τ(H 2 ) = 112, in the sequence of maps Home chains H n , wecut the last Home (see Figure. 5.12) and we use Theorem 4.3.31, then we obtain:


CHAPTER 5.94THE NUMBER OF SPANNING TREES OF CERTAIN FAMILIES OFPLANAR MAPSFigure 5.12: The n-Home chains H nτ(H n ) = 11τ(H n−1 ) − 3 × 3τ(H n−2 ), hence we obtain the system:⎧⎨ τ(H n ) = 11τ(H n−1 ) − 9τ(H n−2 )τ(H 1 ) = 11⎩τ(H 2 ) = 112The characteristic quadratic equation is r 2 − 11r + 9 = 0, so the solutions of this equationare: r 1 = 11−√ 85and r2 2 = 11+√ 85, hence:2τ(H n ) = α( 11 − √ 852) n + β( 11 + √ 85) n , α, β ∈ R, n ≥ 1.2Using the initial condition τ(H 1 ) = 11, τ(H 2 ) = 112, we obtain: α = 1 − 11√ 85,2 170β = 1 + 11√ 85, hence the result. □2 170Numerical results The Table 5.2 illustrates some of the values of the number of spanningtrees in the n-Home chains H n by using the formula given in Theorem 5.7.1:n τ(H n )1 112 1123 11334 114555 1158086 11707937 118364518 1196638249 120977400510 12230539639Table 5.2: Some values of τ(H n )


5.7. Other uses 955.7.2 Formula for the Number of Spanning Trees in The n-BarrelchainsIn this section, we derive a simple formula for the number of spanning trees of a specialfamily of planar maps which is called the n-Barrel chains.Now, by applying Theorem 4.3.31, we obtain the following results:Theorem 5.7.2 (The n-Barrel chains) The number of spanning trees of the n-Barrelchains τ(B n ) (see Figure. 5.13) is given by the following formula:τ(B n ) = ( 33 + 7√ (33) ( 7 + √ 33) n − 1 66 2 4 (7 + √ 33)), n+2 n ≥ 1.2Figure 5.13: The n-Barrel chains B nProof: τ(B 1 ) = 7, τ(B 2 ) = 45, in the sequence of maps Barrel chains B n , we cut thelast Barrel (see Figure. 5.13) and we use Theorem 4.3.31, then we obtain: τ(B n ) =7 × τ(B n−1 ) − 2 × 2 × τ(B n−2 ), hence we obtain the following system:⎧⎨⎩τ(B n ) = 7τ(B n−1 ) − 4τ(B n−2 )τ(B 1 ) = 7τ(B 2 ) = 45The characteristic quadratic equation is r 2 − 7r + 4 = 0, so the solutions of this equationare: r 1 = 7−√ 33and r2 2 = 7+√ 33, hence: τ(B2 n ) = α( 7−√ 33) n + β( 7+√ 33) n , α, β ∈ R, n ≥2 21. Using the initial condition τ(B 1 ) = 7, τ(B 2 ) = 45, we obtain: α = 1 − 7√ 33and2 66β = 1 + 17√ 33, hence the result.2 66□


CHAPTER 5.96THE NUMBER OF SPANNING TREES OF CERTAIN FAMILIES OFPLANAR MAPSNumerical result The Table 5.3 illustrates some of the values of the number of spanningtrees in the n-Barrel chains τ(B n ) by using the formula given in Theorem 5.7.2:n 1 2 3 4 5 6 7 8 9 10τ(B n ) 7 45 287 1829 11655 74269 473263 3015765 19217303 122458061Table 5.3: Some values of τ(B n )5.7.3 Formula for the Number of Spanning Trees in the n-LightchainsNow, we derive a simple formula for enumerating the number of spanning trees in aspecial family of planar maps which is called the n-Light chains.Now, by applying Theorem 4.3.31, the following results are obtained:Theorem 5.7.3 (The n-Light chains) The the number of spanning trees of the n-Lightchains τ(L n ) (see Figure. 5.14) is given by the following formula:τ(L n ) = 1 )(38 3n 2n+2 − 1 , n ≥ 1.Figure 5.14: The n-Light chains L n


5.8. The number of spanning trees in some particular planar maps 97Proof: τ(L 1 ) = 30, τ(L 2 ) = 819, in the sequence of maps Light chains L n , we cutthe last Light (see Figure. 5.14) and we use Theorem 4.3.31, then we obtain: τ(L n ) =30 × τ(L n−1 ) − 9 × 9 × τ(L n−2 ), hence, we obtain the following system:⎧⎨⎩τ(L n ) = 30τ(L n−1 ) − 81τ(L n−2 )τ(L 1 ) = 30τ(L 2 ) = 819The characteristic quadratic equation is r 2 −30r+81 = 0; so, the solutions of this equationare: r 1 = 3 and r 2 = 27, hence: τ(L n ) = α(3) n + β(27) n , α, β ∈ R, n ≥ 1. Using theinitial condition τ(L 1 ) = 30, τ(L 2 ) = 819, we obtain: α = −1,β = 9, hence the result. 8 8□Numerical results The Table 5.4 illustrates some of the values of the number of spanningtrees in the n-Light chains τ(L n ) by using the formula given in Theorem 5.7.3:n τ(L n )1 302 8193 221404 5978615 161424806 4358476597 117678888908 3177330063219 857879118954010 231627362174199Table 5.4: Some values of τ(L n )5.8 The number of spanning trees in some particularplanar mapsIn this section, we derive simple formulae for the number of spanning trees in some specialplanar maps. In the following cases of particular planar maps, each case need anotherplanar map for help us in the proof, for this reason we shall obtain in each case twoformulae for calculation the number of spanning trees in it.


CHAPTER 5.98THE NUMBER OF SPANNING TREES OF CERTAIN FAMILIES OFPLANAR MAPS5.8.1 Formulae for the number of spanning trees in particularplanar mapLet C n and Q n be the sequences of planar maps as following (see Figure 5.15):Figure 5.15: Particular case of maps C n and Q nTheorem 5.8.1 For the sequence maps C n and Q n in Figure 5.15, we have:τ(C n ) = √ 1 ((9 + 4 √ 5)( 7 + 3√ 5) n−1 − (9 − 4 √ 5)( 7 − 3√ 5)), n−1 n ≥ 15 22τ(Q n ) = 1 √5(( 7 + 3√ 52) n − ( 7 − 3√ 5)), n n ≥ 1.2Proof: We put C n = τ(C n ) and Q n = τ(Q n ). C 1 = 8, Q 1 = 3, in the sequence of mapsC n ; we then cut the last triangle (see Figure 5.15), and use the Theorem 4.3.31 (the samegoes for the sequence of maps Q n ), then we obtain:τ(Q n ) = 3τ(C n−1 ) − τ(Q n−1 ), τ(C n ) = 3τ(Q n ) − τ(C n−1 )therefore, we have the following system:{Qn = 3C n−1 − Q n−1C n = 3Q n − C n−1By replacing Q n by its value in the second equation we get:{Cn = 8C n−1 − 3Q n−1Q n = 3C n−1 − Q n−1 , with C 1 = 8 and Q 1 = 3(Cn)= MQ n(Cn−1Q n−1), where M =( 8 -33 -1)


5.8. The number of spanning trees in some particular planar maps 99(Cn)= MQ n(Cn−1Q n−1)= ... = M n−1 (C1Q 1), we compute M n−1 :det (M − λI 2 ) = λ 2 −7λ + 1 = 0, λ 1 = 7−3√ 5and λ2 2 = 7+3√ 5, λ2 1 ≠ λ 2 then there is amatrix P invertible such that M = P DP −1 where( )λ1 0D =0 λ 2and, P is the invertible transformation matrix formed by eigenvectors( 1 1P =3+ √ 523− √ 52M n−1 = P D n−1 P −1 where D n−1 =(M n−1 = √ −15hence the result.)(; P −1 = √ 15(( 7−3√ 53− √ 52-1−3− √ 512)) n−1 02, from which:0 ( 7+3√ 5) n−1 2)( 7−3√ 5) n−1 ( 3−√ 5) + ( 7+3√ 5) n−1 ( −3−√ 5) −( 7−3√ 5) n−1 + ( 7+3√ 5) n−12 2 2 2 2 2( 7−3√ 5) n−1 − ( 7+3√ 5) n−1 ( 3−√ 5)( 7+3√ 5) n−1 − ( 3+√ 5)( 7−3√ 5) n−1 2 2 2 2 2 2)□Numerical results The Table 5.5 illustrates some of the values of the number of spanningtrees in the sequence of planar maps τ(C n ) and τ(Q n ) by using the formula given inTheorem 5.8.1:n 1 2 3 4 5 6 7 8 9 10τ(C n ) 8 55 377 2584 17711 121393 832040 5702887 39088169 267914296τ(Q n ) 3 21 144 987 6765 46368 317811 2178309 14930352 102334155Table 5.5: Some values of τ(C n ) and τ(Q n )5.8.2 Formulae for the number of spanning trees in the n-KitechainsWe derive simple formulae for the number of spanning trees of a special family of planarmaps which is called the n-Kite chains.


CHAPTER 5.100THE NUMBER OF SPANNING TREES OF CERTAIN FAMILIES OFPLANAR MAPSTheorem 5.8.2 (The n-Kite chains) The complexity of the n-Kite chains K n and Q n(see Figure. 5.16) is given by the following formulae:τ(K n ) = 3 × ((43 6n−14 √ + 30 √ 2)(3 + 2 √ 2) n−1 − (43 − 30 √ 2)(3 − 2 √ )2) n−1 , n ≥ 12andτ(Q n ) =21 × 6n−14 √ 2((3 + 2 √ 2) n − (3 − 2 √ 2) n ), n ≥ 1.Figure 5.16: The n-Kite chains K nProof: We put K n = τ(K n ) and Q n = τ(Q n ). K 1 = 45, Q 1 = 21, in the sequence ofmaps K n ; we cut the last Kite (see Figure 5.18), and use Theorem 4.3.31 (the same goesfor the sequence of maps Q n ), the obtained is: τ(Q n ) = 21τ(K n−1 ) − 9τ(Q n−1 ), τ(K n ) =45τ(K n−1 ) − 21τ(Q n−1 ) therefore, we have the following system:{Kn = 45K n−1 − 21Q n−1Q n = 21K n−1 − 9Q n−1 with K 1 = 45 and Q 1 = 21(Kn)= MQ n(Kn)= MQ n(Kn−1Q n−1), where M =( 45 -2121 -9(Kn−1Q n−1)= ... = M n−1 (K1Q 1),)


5.8. The number of spanning trees in some particular planar maps 101we compute M n−1 :det (M − λI 2 ) = λ 2 −36λ + 36 = 0, λ 1 = 18 − 12 √ 2 and λ 2 = 18 + 12 √ 2, λ 1 ≠ λ 2 thenthere is a matrix P invertible such that M = P DP −1 where( )λ1 0D =0 λ 2and, P is the invertible transformation matrix formed by eigenvectors( )1 1P =M n−1 = P D n−1 P −1 whereD n−1 =P −1 = −78 √ 29+4 √ 27from which we obtain M n−1 , hence the result.(9−4 √ 279−4 √ 27-1−9−4 √ 217( √ )(18 − 12 2)n−100 (18 + 12 √ 2) n−1Example 5.8.3 Here is an example of the planar map 2-Kite chains and its Laplacianmatrix; (see Figure. 5.17). We can calculate the number of spanning tree in this exampleby using the matrix tree theorem (see Chapter 3) or using the previous Theorem 5.8.2;in both ways we obtain τ(K 2 ) = 1584.)□Figure 5.17: A planar map 2-Kite chains and its Laplacian matrix


CHAPTER 5.102THE NUMBER OF SPANNING TREES OF CERTAIN FAMILIES OFPLANAR MAPSNumerical results The Table 5.6 illustrates some of the values of the number of spanningtrees in the n-Kite chains K n and Q n by using the formula given in Theorem 5.8.2:n τ(K n ) τ(Q n )1 45 212 1584 7563 55404 264604 1937520 9253445 67756176 323598246 2369471616 11316412807 82861755840 395741324168 2897722232064 13839296808969 101334977144064 4839679974528010 3543741176832000 1692463322317824Table 5.6: Some values of τ(K n ) and τ(Q n )5.8.3 Formulae for the number of spanning trees in the n-Envelope chainsIn this section, we derive simple formulae for the number of spanning trees of a specialfamily of planar maps which is called the n-Envelope chains.Theorem 5.8.4 (The n-Envelope chains) The complexity of the n-Envelope chains E nand Q n (see Figure. 5.18) is given by the following formulae:1τ(E n ) =((4841+2284 √ √ 434)(45+2 √ 434) n−1 −(4841−228 √ 434)(45−2 √ )434) n−1 ,434n ≥ 1, andτ(Q n ) = 55 ((454 √ + 2 √ 434) n − (45 − 2 √ )434) n , n ≥ 1.434Proof: We put E n = τ(E n ) and Q n = τ(Q n ). E 1 = 114, Q 1 = 55, in the sequenceof maps E n , we cut the last Envelope (see Figure 5.18), and we use Theorem 4.3.31(the same goes for the sequence of maps Q n ), then we obtain: τ(Q n ) = 55τ(E n−1 ) −24τ(Q n−1 ), τ(E n ) = 114τ(E n−1 ) − 55τ(Q n−1 ) therefore, we have the following system:{En = 114E n−1 − 55Q n−1Q n = 55E n−1 − 24Q n−1 with E 1 = 114 and Q 1 = 55


5.8. The number of spanning trees in some particular planar maps 103(EnFigure 5.18: The n-Envelope chains E n)= MQ n( )En= MQ n(En−1), where M =Q n−1( )En−1Q n−1( 114 -5555 -24= ... = M n−1 (E1Q 1),we compute M n−1 :det (M − λI 2 ) = λ 2 −90λ + 289 = 0, λ 1 = 45 − 2 √ 434 and λ 2 = 45 + 2 √ 434, λ 1 ≠ λ 2 thenthere is a matrix P invertible such that M = P DP −1 where( )λ1 0D =0 λ 2and, P is the invertible transformation matrix formed by eigenvectors()1 1P =M n−1 = P D n−1 P −1 whereD n−1 =P −1 =69+2 √ 43455−554 √ 434from which we obtain M n−1 , hence the result.(69−2 √ 4345569−2 √ 434-155−69−2 √ 434155( √ )(45 − 2 434)n−100 (45 + 2 √ 434) n−1))□


CHAPTER 5.104THE NUMBER OF SPANNING TREES OF CERTAIN FAMILIES OFPLANAR MAPSNumerical results The Table 5.7 illustrates some of the values of the number of spanningtrees in the n-Envelope chains E n and Q n by using the formula given in Theorem5.8.4:n τ(E n ) τ(Q n )1 114 552 9971 49503 864444 4296054 74918341 372339005 6492826374 32268951556 562702973111 2796599668507 48766840757904 242368243167058 4226394508982281 21004924580838009 366281888829371034 18203987900001425510 31743941981547513851 15776546789615064750Table 5.7: Some values of τ(E n ) and τ(Q n )5.8.4 Formulae for the number of spanning trees in the n-Diphenylene chainsWe derive two simple formulae for the number of spanning trees of a special family ofplanar maps which is called the n-Diphenylene chains.Theorem 5.8.5 (The n-Diphenylene chains) [87] The complexity of the n-Diphenylenechains D n and Q n (see Figure. 5.19) is given by the following formulae:τ(D n ) = 12 √ 30 (126 + 23√ 30)((11 + 2 √ 30) n−1 − (11 − 2 √ )30) n+2 ,andτ(Q n ) = 3 ((112 √ + 2 √ 30) n − (11 − 2 √ )30) n , n ≥ 1.30Proof: We put d n = τ(D n ) and q n = τ(Q n ). d 1 = 23, q 1 = 6, in the sequence of mapsD n , we cut the last Diphenylene (see Figure 5.19), and we use Theorem 4.3.31 (the samegoes for the sequence of maps Q n ), then we obtain: τ(Q n ) = 6τ(D n−1 )−τ(Q n−1 ), τ(D n ) =23τ(D n−1 ) − 4τ(Q n−1 ) therefore, we have the following system:{dn = 23d n−1 − 4q n−1q n = 6d n−1 − q n−1 with d 1 = 23 and q 1 = 6


5.8. The number of spanning trees in some particular planar maps 105(dnFigure 5.19: The n-Diphenylene chains D n)= Mq n( )dn= Mq n(dn−1), where M =q n−1( )dn−1q n−1( 23 -46 -1= ... = M n−1 (d1q 1),we compute M n−1 :det (M − λI 2 ) = λ 2 −22λ + 1 = 0, λ 1 = 11 − 2 √ 30 and λ 2 = 11 + 2 √ 30, λ 1 ≠ λ 2 thenthere is a matrix P invertible such that M = P DP −1 where( )λ1 0D =0 λ 2and, P is the invertible transformation matrix formed by eigenvectors( )(1 1P =, P −1 = √ −1 6− √ )30-12306+ √ 302M n−1 = P D n−1 P −1 where,D n−1 =6− √ 302from which we obtain M n−1 , hence the result.−6− √ 3012( √ )(11 − 2 30)n−100 (11 + 2 √ 30) n−1Numerical results The Table 5.8 illustrates some of the values of the number of spanningtrees in the n-Diphenylene chains D n and Q n by using the formula given in Theorem5.8.5:)□


CHAPTER 5.106THE NUMBER OF SPANNING TREES OF CERTAIN FAMILIES OFPLANAR MAPSn τ(D n ) τ(Q n )1 23 62 505 1323 11087 28984 243409 636245 5343911 13968306 117322633 306666367 2575754015 6732691628 56549265697 147812549289 1241508091319 32451433925410 27256628743321 7124534208660Table 5.8: Some values of τ(D n ) and τ(Q n )


Chapter 6Counting the number of spanning treesin the star flower planar mapIn this chapter, we shall consider the star flower planar maps (graphs embedded in theplane without edge-crossings). We look at this map classes, and find a formula for countingthe number of spanning trees in the star flower planar map. Next, we show that thenumber of spanning trees in this type of planar maps always satisfy recurrence relationsand describe how to derive this relation, i.e., we find the simple explicit formulae forcalculating the number of spanning trees in the star flower planar map.6.1 IntroductionAs mentioned in Chapter 3, the number of spanning trees in a graph (network) is animportant quantity to estimate the reliability of the graph (network). A famous andclassic result on the study of the number of spanning trees in graph τ(G) is the followingtheorem, known as the Matrix tree Theorem [81, 106]. The Laplacian matrix (also calledKirchhoff matrix) of a graph G is defined as L(G) = D(G) − A(G), where D(G) andA(G) are the degree matrix and the adjacency matrix of G, respectively. In graph theory,Laplacian matrix is a matrix representing a graph. It has been used by Kirchhoff tocalculate the number of spanning trees of a graph.The Matrix Tree Theorem, which is given by Kirchhoff defines the number of spanningtrees in graph G as the determinant of its Laplacian matrix. This theorem simplifies thecomputation of the number of spanning trees but in large graphs, it is not practical. Inthe following we shall describe a general method to count the number of spanning treesin the star flower planar map. Our approach is to use our main results [89] that we havederived in Chapter 4 to calculate the number of spanning trees in the star flower planarmap.107


CHAPTER 6.108COUNTING THE NUMBER OF SPANNING TREES IN THE STARFLOWER PLANAR MAP6.2 The star flower planar mapLet C n be a cycle with n vertices. The Star flower planar map is a simple graph G formedfrom a cycle C n by adding a vertex adjacent to every edge of C n and we connect thisvertex with two end vertices of each edge of C n , i.e., we replace each edge of C n by atriangulation. If there are k edges between every two vertices of each edge of the cycleC n , then we obtain the star flower planar map in the general case. In this chapter, wedenote the star flower planar map by S n,k where n is the number of triangles of the starflower planar map, k is the number of edges between each two vertices of each edge of thecycle C n ; and derive the explicit formula for τ(S n,k ) the number of spanning trees in S n,kto be τ(S n,k ) = 2kn(k + 2) n−1 , n ≥ 2.6.3 Main ResultsThe idea of this work is that how we can calculate the number of spanning trees in thestar flower planar map (graph) shown in the Figure. 6.1, when we do not want to usethe determinant of Laplacian matrix of planar graphs (Matrix Tree Theorem).Figure 6.1: A star flower planar graph (map)This work has been extended to develop new techniques for the calculation of the numberof spanning trees in a star flower planar map.Let C n be a cycle with n vertices, the number of spanning trees of this cycle isequal to the length of its cycle (the length of a cycle is the number of edges that form thiscycle). The star flower planar map is a simple graph G formed from a cycle C n by addinga vertex adjacent to every edge of C n and we connect this vertex with two end verticesof each edge of C n , i.e., we replace each edge of C n by a triangulation (see Figure. 6.2).


6.3. Main Results 109Figure 6.2: A cycle C n and the star flower planar mapIf there are k edges (simple path in the general case) between every two vertices of eachedge of the cycle C n , then we obtain the star flower planar map shown in Figure. 6.3.In this work, we denote the star flower planar map by S n,k where n is the number oftriangles of the star flower planar map, k is the number of edges (the length of simplepath) between each two vertices of each edge of the cycle C n ; the star flower planar maphas n(k + 1) vertices (n vertices of degree 4 and n + (k − 1)n of degree 2), n(k + 2) edgesand n + 2 faces (n faces of degree k + 2, one face of degree nk and the other face of degree2n)(see Figure. 6.3).Figure 6.3: The star flower planar map S n,kExample 6.3.1 Here is an example of a star flower planar map S 2,k with n = 2 (2triangles) and k edges, and a star flower planar map S 3,k with n = 3 (3 triangles) and kedges (see Figure. 6.4).


CHAPTER 6.110COUNTING THE NUMBER OF SPANNING TREES IN THE STARFLOWER PLANAR MAPFigure 6.4: The star flower planar maps S 2,k and S 3,k6.4 An explicit formula for the number of spanningtrees in S n,kNow, we derive the explicit formula for τ(S n,k ) the number of spanning trees in the starflower planar map S n,k .Theorem 6.4.1 The number of spanning trees of the star flower planar map S n,k (seeFigure. 6.3) is given by the following formula:τ(S n,k ) = 2kn(k + 2) n−1 , n ≥ 2.Proof: We cut a triangle as shown in Figure. 6.3, then we obtain the star flower planarmap S n−1,k after cutting (see Figure. 6.5):Figure 6.5: The star flower planar map S n,k after cutting


6.4. An explicit formula for the number of spanning trees in S n,k 111and we apply Theorem 4.3.26, then we obtain:τ(S n,k ) = (k + 2) n−1 ∗ 2k + τ(S n−1,k ) ∗ (k + 2)τ(S n,k ) = 2k(k + 2) n−1 + (k + 2)τ(S n−1,k )(k + 2)τ(S n−1,k ) = 2k(k + 2) n−1 + (k + 2) 2 τ(S n−2,k )(k + 2) 2 τ(S n−2,k ) = 2k(k + 2) n−1 + (k + 2) 3 τ(S n−3,k )(k + 2) 3 τ(S n−3,k ) = 2k(k + 2) n−1 + (k + 2) 4 τ(S n−4,k ). . . . . . . . .(k + 2) n−4 τ(S 4,k ) = 2k(k + 2) n−1 + (k + 2) n−3 τ(S 3,k )(k + 2) n−3 τ(S 3,k ) = 2k(k + 2) n−1 + (k + 2) n−2 τ(S 2,k )(k + 2) n−2 τ(S 2,k ) = 2k(k + 2) n−1 + (k + 2) n−1 2kBy the sum of all the previous equations, we obtain:τ(S n,k ) = 2k(k + 2) n−1 (n − 1) + (k + 2) n −1 2kWe take 2k(k+2) n−1 as a factor, hence the result.□Particular cases:1) In the previous Theorem 6.4.1, if we take k = 1 (see the star flower planarmap S n,k shown in figure 6.3), then we obtain the star flower planar map S n,1 , which has2n vertices (n vertices of degree 4 and n of degree 2), 3n edges and n + 2 faces (n facesof degree 3, one face of degree n and the other face of degree 2n)(see Figure. 6.6):Figure 6.6: The star flower planar map S n,1Example 6.4.2 Here is an example of the star flower planar map S 2,1 with n = 2 (2triangles) and the star flower planar map S 3,1 with n = 3 (3 triangles) (see Figure. 6.7).


CHAPTER 6.112COUNTING THE NUMBER OF SPANNING TREES IN THE STARFLOWER PLANAR MAPFigure 6.7: The star flower planar maps S 2,1 and S 3,1Corollary 6.4.3 The number of spanning trees of the star flower planar map S n,1 (seeFigure. 6.6) is given by the following formula:τ(S n,1 ) = 2n(3) n−1 , n ≥ 2.Proof: We cut a triangle as shown in Figure. 6.6, we then obtain the star flower planarmap S n−1,1 after cutting (see Figure. 6.8):Figure 6.8: The star flower planar map S n,1 after cuttingand we apply Theorem 4.3.26, then we obtain:τ(S n,1 ) = (3) n−1 ∗ 2 + τ(S n−1,1 ) ∗ (3)τ(S n,1 ) = 2(3) n−1 + (3)τ(S n−1,1 )(3)τ(S n−1,1 ) = 2(3) n−1 + (3) 2 τ(S n−2,1 )(3) 2 τ(S n−2,1 ) = 2(3) n−1 + (3) 3 τ(S n−3,1 )(3) 3 τ(S n−3,1 ) = 2(3) n−1 + (3) 4 τ(S n−4,1 ). . . . . . . . .(3) n−4 τ(S 4,1 ) = 2(3) n−1 + (3) n−3 τ(S 3,1 )(3) n−3 τ(S 3,1 ) = 2(3) n−1 + (3) n−2 τ(S 2,1 )(3) n−2 τ(S 2,1 ) = 2(3) n−1 + (3) n−1 ∗ 2


6.4. An explicit formula for the number of spanning trees in S n,k 113By the sum of all the previous equations, we obtain:τ(S n,1 ) = 2(3) n−1 (n − 1) + 2(3) n −1 then, we take 2(3) n−1 as a factor, hence the result. □Example 6.4.4 In the star flower planar maps S 2,1 and S 3,1 shown in Figure. 6.7; wehave n = 2 in S 2,1 and n = 3 in S 3,1 then, we apply Corollary 6.4.3 to find the number ofspanning trees of these star flower planar maps S 2,1 and S 3,1 , we obtain:τ(S 2,1 ) = 2 ∗ 2(3) 2−1 = 12 and τ(S 3,1 ) = 2 ∗ 3(3) 3−1 = 54.Numerical results The Table 6.1 illustrates some of the values of the number ofspanning trees in the star flower planar map S n,1 by using the formula given in Corollary6.4.3:n 2 3 4 5 6 7 8 9 10τ(S n,1 ) 12 54 216 810 2916 10206 34992 118098 393660Table 6.1: Some values of τ(S n,1 )2) In the previous Theorem 6.4.1, if we take k = 2 (see the star flower planar mapS n,k shown in figure 6.3), then we obtain the star flower planar map S n,2 (see Figure. 6.9).Figure 6.9: The star flower planar map S n,2Corollary 6.4.5 The number of spanning trees of the star flower planar map S n,2 (seeFigure. 6.9) is given by the following formula:τ(S n,2 ) = n(2) 2n , n ≥ 2.


CHAPTER 6.114COUNTING THE NUMBER OF SPANNING TREES IN THE STARFLOWER PLANAR MAPProof: See the proof of Corollary 6.4.3. □Example 6.4.6 In the star flower planar map shown in Figure. 6.1; we have n = 8 andk = 2, we apply Corollary 6.4.5 to find the number of spanning trees of this star flowerplanar map S 8,2 (the idea of this work), thenτ(S 8,2 ) = 8(2) 2∗8 = 524288.Numerical results The Table 6.2 illustrates some of the values of the number of spanningtrees in the star flower planar map S n,2 by using the formula given in Corollary 6.4.5:n 2 3 4 5 6 7 8 9 10τ(S n,2 ) 32 192 1024 5120 24576 114688 524288 2359296 10485760Table 6.2: Some values of τ(S n,2 )Another result is due to Sedlacek [103] who derived a formula for the wheel on n + 1vertices, W n+1 , which is formed from a cycle C n on n vertices by adding a vertex adjacentto every vertex of C n . In addition, other formulae are related to our work can also befound in [84, 103, 105].


Chapter 7Maximal Planar MapsIn this chapter, we shall focus on the maximal planar maps. This chapter will be dividedinto two sections; we devote the first section for calculating the Weiner index in the caseof planar maps in general then in particular in the maximal planar maps and in the othersection will study how to count the number of spanning trees in this type of this maps,as well as enumeration of spanning trees in the maximal planar maps, i.e., we derive theexplicit formulae for the number of spanning trees of the maximal planar map.7.1 IntrodutionA maximal planar map is a planar map to which no new edge can be added withoutviolating the planarity of C. The number of the edges in any maximal planar map with nvertices is 3n − 6. Furthermore the (facial) faces (including the infinite outerface) are alltriangles, i.e., cycles of length three. A simple planar map is called maximal planar if itis planar but adding any edge (on the given vertex set) would destroy that property. Allfaces (even the outer ones) are then bounded by three edges, explaining the alternativeterms triangular and triangulated for these graphs. If a triangular graph has v verticeswith v > 2, then it has precisely 3v − 6 edges and 2v − 4 faces. In this chapter, we areinterested in the maximal planar map with two complete vertices to isomorphism, thereis only one maximal planar map which has n vertices, two complete vertices of degreen − 1, two vertices of degree 3 and n − 4 vertices of degree 4, 2(n − 2) faces of degree 3(all faces having degree 3) and 3(n − 2) edges, (see Figure 7.5).Isomorphism It follows from the definition that a graph is completely determinedwhen we know its vertices and edges, and that two graphs are the same if they have thesame vertices and edges. Once we know the vertices and edges, we can draw the graphand, in principle, any picture we draw is as good as any other; the actual way in whichthe vertices and edges are drawn is irrelevant, although, some pictures are easier to usethan others!115


116 CHAPTER 7. MAXIMAL PLANAR MAPSFor example, recall the utilities graph, in which three houses A, B and C are joinedto the three utilities gas (g), water (w) and electricity (e). This graph is specifiedcompletely by the following sets:vertices: {A, B, C, g, w, e},edges: {Ag, Aw, Ae, Bg, Bw, Be, Cg, Cw, Ce},and can be drawn in many ways, such as the following:Figure 7.1: The utilities graph can be drawn in many ways.Each of these diagrams has six vertices and nine edges, and conveys the same information.Each house is joined to each utility, but no two houses are joined, and no twoutilities are joined. It follows that these two dissimilar diagrams represent the same graph.On the other hand, two diagrams may look similar, but represent different graphs. Forexample, the diagrams below look similar, but they are not the same graph: for example,AB is an edge of the second graph, but not the first.Figure 7.2: Two diagrams look similar, but they are not the same graph.We express this similarity by saying that the graphs represented by these two diagramsare isomorphic. This means that the two graphs have esse1ltially the same structure: wecan relabel the vertices in the first graph to get the second graph - in this case, we simplyinterchange the labels w and B.This leads to the following definition.Definition 7.1.1 Two graphs G and H are isomorphic if H can be obtained by relabellingthe vertices of G that is, if there is a one-one correspondence between the vertices of Gand those of H, such that the number of edges joining each pair of vertices in G is equalto the number of edges joining the corresponding pair of vertices in H. Such a one-onecorrespondence is an isomorphism.


7.1. Introdution 117Example 7.1.2 For example, the graphs G and H represented by the diagramsFigure 7.3: Two graphs G and H are not the same, but they are isomorphic.are not the same, but they are isomorphic, since we can relabel the vertices in the graphG to get the graph H, using the following one-one correspondence:G ↔ H, u ↔ 4, v ↔ 3, w ↔ 2 and x ↔ 1Note that edges in G correspond to edges in H, for example: the two edges joining u andv in G correspond to the two edges joining 4 and 3 in H; the edge uw in G correspondsto the edge 42 in H; the loop ww in G corresponds to the loop 22 in H.Remark 7.1.3 To check whether two graphs are the same, we must check whether all thevertex labels correspond. However, to check whether two graphs are isomorphic, we mustinvestigate whether we can relabel the vertices of one graph to give those of the other.In order to do this, we first check that the graphs have the same numbers of vertices andedges, and then look for special features in the two graphs, such as a loop, multiple edges,or the number of edges meeting at a vertex. For example, the following two graphs bothhave five vertices and six edges, but are not isomorphic, as the first has two vertices wherejust two edges meet, whereas the second has only one.Figure 7.4: Two graphs both have five vertices and six edges, but are not isomorphic.Example 7.1.4 The graphs G and H, represented in the following Figure 7.5 are notthe same, but they are isomorphic.The graphs G and H represent the maximal planar map with 5 vertices. Hereafter, wedenote the maximal planar map with n vertices by E n .


118 CHAPTER 7. MAXIMAL PLANAR MAPSFigure 7.5: The graphs G and H are not the same, however, they are isomorphic.7.2 Calculating the Wiener index in the maximal planarmapsIn this section, we shall focus on the Wiener index in case of planar maps and give aformula which calculates the Wiener index in the maximal planar map, then we give aninequality, which minimizes and maximizes any planar map by the maximal planar mapwith n vertices and the path of n vertices; like the inequality in the trees.7.2.1 IntrodutionIn a communication network, large diameter may be acceptable if most pairs can communicatevia short paths. This leads us to study the average distance instead of themaximum. Since the average is the sum divided by ( n2)(the number of vertex pairs), itis equivalent to study D(G) = ∑ {v i ,v j }⊆V (G) d(v i, v j ). The sum D(G) has been called theWiener index of G (also written W (G)). It is used by Wiener to study the boiling pointof paraffin. Molecules can be modeled by graphs with vertices for atoms and edges foratomic bonds. Many chemical properties of molecules are related to the Wiener indexof the corresponding graphs. We study the extreme values of W (G). In the next, wefocus on the Wiener index in planar maps, the particular case of trees has been echoedby several people [39, 69, 125]. We study the extreme values of W (G). Refer to [2] and[117] for the significance of the Wiener index.7.2.2 Calculation of the Wiener index in the planar mapsHerein, we give some basic definitions and properties about the Wiener index. An importantconcept that we need in this section is that of the Wiener index in the case of planarmaps.Definition 7.2.1 The distance between two distinct vertices v i and v j of a map C, denotedby d(v i , v j ) is equal to the length of (number of edges in) the shortest path thatconnects v i and v j . Conventionally, d(v i , v i ) = 0.


7.2. Calculating the Wiener index in the maximal planar maps 119Definition 7.2.2 We define a complete vertex in a planar map C by the vertex v 0 suchthat d(u, v 0 ) = 1 for each u ∈ V (C). In a complete graph, all the vertices are complete.The Wiener index of a connected graph is the sum of distances between all pairs of∑vertices [39, 87, 69, 125], the Wiener index of a connected graph G is defined as W (G) ={v i ,v j }⊆V (G) d(v i, v j ). The Wiener index of a vertex u in G denoted by W (u, G), is the sumof distances of vertex u to each vertex of vertices of G, i.e., W (u, G) = ∑ v∈V (G)d(u, v).Definition 7.2.3 In the same way as graphs, we define the Wiener index for planar mapsas follows:∑W (C) = d(v i , v j ) and W (u, C) = ∑d(u, v){v i ,v j }⊆V (G)Remark 7.2.4 We notice that:W (C) = 1 2= 1 2∑∑u∈V (C) v∈V (C)∑u∈V (C)W (u, C)d(u, v)v∈V (C)Example 7.2.5 Let C 5 be a planar map with |V (C 5 )| = 5 (see Figure 7.6), we have:deg(v 1 ) = 2, deg(v 2 ) = 4, deg(v 4 ) = 4, W (v 1 , C 5 ) = 6, W (v 2 , C 5 ) = 4, W (v 3 , C 5 ) = 6,W (v 4 , C 5 ) = 4, W (v 5 , C 5 ) = 6, W (C 5 ) = 13, v 2 and v 4 are complete vertices.Figure 7.6: Example of a map C 5Let C be a planar map and e be an edge in C (e ∈ E(C)), we denote by C − e the mapobtained after deleting the edge e from the map C and the resulting map is connected.Lemma 7.2.6 Let C be a planar map and let e 1 , e 2 be two edges in C that connect thevertices v 1 and v 2 (multiple edges in C), thenW (C − e i ) = W (C), i = 1, 2where C − e i is the map obtained by deleting the edge e i from the map C.


120 CHAPTER 7. MAXIMAL PLANAR MAPSRemark 7.2.7 We notice that deleting one edge of multiple edges does not affect theWiener index; through this section, we consider only the simple planar maps, i.e., mapswithout loops and multiple edges.Lemma 7.2.8 Let C n be a simple planar map with n vertices and v be a complete vertexin the map C, thendeg(v) = n − 1 and W (v, C n ) = n − 1.Lemma 7.2.9 Let C n be a simple planar map with n vertices (n ≥ 2) and let v be avertex not complete of C n , thenW (v, C n ) ≥ n.Remark 7.2.101. Let C n be a simple planar map and v be a vertex of C n , then W (v, C n ) ≥ n − 1.2. Let C n be a simple planar map with n vertices, e be an edge of C n and let C n − e bethe map obtained by deleting the edge e such that the map C n −e remains connected,then W (C n − e) ≥ W (C n ).Theorem 7.2.11 Let C n be a simple planar map with n vertices, thenW (C n ) ≥n(n − 1).2Proof: Let C n be a planar map with n vertices and let v be a vertex of C n . By Remark7.2.10, we have: W (v, C n ) ≥ n − 1;W (C n ) = 1 2≥ 1 2≥∑v∈V (C n )∑v∈V (C n )n(n − 1).2W (v, C n )(n − 1) = 1 2 (n − 1) ∑v∈V (C n )Definition 7.2.12 (A maximal planar map [39]) Let E n be a family of planar maps thatcontains:• n vertices, two complete vertices of degree n − 1, two vertices of degree 3 and n − 4vertices of degree 4,• 2(n − 2) faces of degree 3 (all faces having degree 3),• 3(n − 2) edges,1□


7.2. Calculating the Wiener index in the maximal planar maps 121then the family of this maps is called the maximal planar maps if to which no new edgecan be added without violating the planarity of this maps.The maps E 3 and E 4 are presented in the example 7.2.13. For E 5 , E 6 , and E n (see Figure.7.7)Figure 7.7: The maps E 5 , E 6 and E nExample 7.2.13 In the planar maps E 3 and E 4 , all the vertices are completes. In otherwords, we say that the map is complete (see Figure 7.8).Figure 7.8: The maps E 3 and E 4Remark 7.2.14 For all u, v ∈ V (E n ), we have d(u, v) ≤ 2.Proposition 7.2.15 Let E n be the maximal planar map and v be a vertex of E n , thenwe have:1. W (v, E n ) = 2n − deg(v) − 2, 2. W (E n ) = (n − 2) 2 + 2.


122 CHAPTER 7. MAXIMAL PLANAR MAPSProof:1. W (v, E n ) = ∑u∈V (E n)= ∑u∈V (En)d(u,v)=1d(u, v) (we use Remark 7.2.14)d(u, v) += deg(v) + 2 ∑u∈V (En)d(u,v)=2∑u∈V (En)d(u,v)=21d(u, v)= deg(v) + 2(n − deg(v) − 1)= 2n − deg(v) − 22. W (E n ) = 1 2= 1 2∑W (v, E n )v∈V (E n )∑v∈V (E n )= (n − 1)(2n − deg(v) − 2) (from 1)∑1 − 1 2v∈V (E n)∑deg(v)v∈V (E n)= n(n − 1) − 1 2 × 2|E(E n)|= (n − 2) 2 + 2□Lemma 7.2.16 Let C n be a simple planar map with n vertices (n ≥ 2) and let v be avertex of C n , thenW (v, C n ) ≥ 2n − deg(v) − 2.Proof:W (v, C n ) = ∑u∈V (C n )= ∑u∈V (Cn)d(u,v)=1d(u, v)d(u, v) +∑u∈V (Cn)d(u,v)≥2d(u, v)≥ deg(v) + 2(n − deg(v) − 1)≥ 2n − deg(v) − 2□Let T n be a tree with n vertices, then the Wiener index W (T n ) = ∑ u,vd(u, v) is minimizedby star tree with n vertices and maximized by path with n vertices, both uniquely, i.e.,


7.3. Formulae for the Number of Spanning Trees in a Maximal Planar Map 123(n − 1) 2 ≤ W (T n ) ≤ n(n2 −1)6[117]. The goal of this work is to give an inequality similar inthe case of planar maps. Let C n be a planar map with n vertices, then the Wiener indexof W (C n ) is minimized by the maximal planar map W (E n ) with n vertices and maximizedby the path W (P n ) with n vertices. Now we can state the following theorem.Theorem 7.2.17 Let C n be a simple planar map with n vertices, thenW (E n ) ≤ W (C n ) ≤ W (P n )Proof: By Remark 7.2.10, In each deleted edge of C, we expand the Wiener index. Theconnected planar map obtained after deleting all possible edges is a spanning tree of C n ,On the other hand in the trees, the Wiener index is maximized by the path P n with nvertices, hence W (C n ) ≤ W (P n ).The remaining premises is to show that: W (E n ) ≤ W (C n )• For n = 2, 3 or 4 :W (E n ) = 1 2 n(n − 1) and as W (C n) ≥ 1 2 n(n − 1) = W (E n), hence the result.• For n ≥ 5Since W (v, C n ) ≥ 2n − deg(v) − 2, we have :W (C n ) = 1 2≥ 1 2∑v∈V (E n )∑v∈V (E n )≥ n(n − 1) − 1 2W (v, E n )(2n − deg(v) − 2)∑v∈V (E n )deg(v)≥ n(n − 1) − |E(E n )| = W (E n )Remark 7.2.18 For a planar map C with n vertices, W (C n ) ≤ W (P n ) ( P n is the pathwith n vertices). The lower bound of W (C n ) is not yet known [54, 117]. But in case ofplanar maps, we have given in this section the upper bound that is the Wiener index inthe maximal planar map W (E n ).7.3 Formulae for the Number of Spanning Trees in aMaximal Planar MapIn this section, we derive the explicit formula for the number of spanning trees of themaximal planar map and deduce a formula for the number of spanning trees of the crystalplanar map.□


124 CHAPTER 7. MAXIMAL PLANAR MAPS7.3.1 Main ResultsIt is known that Kirchhoff Matrix Tree Theorem [81, 85], can be applied to any map Cto determine τ(C) by taking a determinant of Laplacian matrix of C, but this requiresevaluating a determinant of a corresponding characteristic matrix. However, for a fewspecial families of maps, there exist simple formulae which make it much easier to calculateand determine the number of corresponding spanning trees especially when thesenumbers are very large, as we have already seen in previous chapters. In this section, weare going to describe a way to calculate the number of spanning trees by an extensionof Kirchhoff’s formula, also known as the matrix tree theorem. In this work, we use thedeterminant of Laplacian matrix of planar maps (Matrix Tree Theorem) to derive theexplicit formula for calculating the number of spanning trees in the maximal planar mapand deduce a formula to calculate the number of spanning trees in the crystal planar map.Before presenting the main results, we need the following lemmas:Lemma 7.3.1 Let E n be a maximal planar map with n vertices, thenn/τ(E n ), n ≥ 3.Proof: We first form the Kirchhoff characteristic matrix L n (n × n) corresponding tothe labeling of E n shown as follows (in Figure 7.9):Figure 7.9: The principal matrix of E nNext, we focus our attention on the principal submatrix L n−1 which obtained by cancelingits last row and column corresponding to vertex v n . So, the number of spanning trees ofa maximal planar map E n equals τ(E n ) = det(L n−1 ).


7.3. Formulae for the Number of Spanning Trees in a Maximal Planar Map 125n − 1 −1 −1 −1 −1 −1 . . . −1 −1−1 n − 1 −1 −1 −1 −1 . . . −1 −1−1 −1 3 −1 0 0 . . . 0 0−1 −1 −1 4 −1 0 . . . 0 0τ(E n ) =−1 −1 0 −1 4 −1 . . . 0 0−1 −1 0 0 −1 4 . . . 0 0. . . . . . . .. . .−1 −1 0 0 0 0 . . . 4 −1∣ −1 −1 0 0 0 0 . . . −1 4 ∣we denote r i by the i-th row and c i by the i-th column of the determinant. In previousdeterminant, we replace c 1 by c 1 + c 2 + ... + c n−1 , i.e., we add to the first column the sumof other (transformation is symbolized as follows: c 1 ← ∑ n−1i=1 c i, ; this does not changethe determinant, then we obtain:1 −1 −1 −1 −1 −1 . . . −1 −11 n − 1 −1 −1 −1 −1 . . . −1 −10 −1 3 −1 0 0 . . . 0 00 −1 −1 4 −1 0 . . . 0 0τ(E n ) =0 −1 0 −1 4 −1 . . . 0 00 −1 0 0 −1 4 . . . 0 0. . . . . ... . . .0 −1 0 0 0 0 . . . 4 −1∣1 −1 0 0 0 0 . . . −1 4 ∣Next, we replace c j by c 1 + c j for j = 2, ..., n − 1, i.e., c j ← c 1 + c j , we obtain:1 0 0 0 0 0 . . . 0 01 n 0 0 0 0 . . . 0 00 −1 3 −1 0 0 . . . 0 00 −1 −1 4 −1 0 . . . 0 0τ(E n ) =0 −1 0 −1 4 −1 . . . 0 00 −1 0 0 −1 4 . . . 0 0. . . . . . . .. . .0 −1 0 0 0 0 . . . 4 −1∣1 0 1 1 1 1 . . . 0 5 ∣Expanding L n−1 along the first row we obtain the determinant of order (n − 2) × (n − 2)


126 CHAPTER 7. MAXIMAL PLANAR MAPSas follows:n 0 0 0 0 . . . 0 0−1 3 −1 0 0 . . . 0 0−1 −1 4 −1 0 . . . 0 0−1 0 −1 4 −1 . . . 0 0τ(E n ) =−1 0 0 −1 4 . . . 0 0. . . . . . .. . .−1 0 0 0 0 . . . 4 −1∣ 0 1 1 1 1 . . . 0 5 ∣and expanding the determinant obtained along the first row we obtain the determinantof order (n − 3) × (n − 3) as follows:3 −1 0 0 . . . 0 0−1 4 −1 0 . . . 0 00 −1 4 −1 . . . 0 0τ(E n ) = n0 0 −1 4 . . . 0 0. . . . . .. . .0 0 0 0 . . . 4 −1∣ 1 1 1 1 . . . 0 5 ∣hence the result.□Lemma 7.3.2 Let E n be a maximal planar map with n vertices, thenτ(E n )n= 4τ(E n−1)n − 1− τ(E n−2)n − 2 , n ≥ 5.Proof: By Lemma 7.3.1, we have the determinant of order (n − 3) × (n − 3):τ(E n )n3 −1 0 0 . . . 0 0−1 4 −1 0 . . . 0 00 −1 4 −1 . . . 0 0=0 0 −1 4 . . . 0 0. . . . . .. . .0 0 0 0 . . . 4 −1∣ 1 1 1 1 . . . 0 5 ∣In the previous determinant, we denote by c i by the i-th column and r i by the i-th row ofthe determinant, c i ← c i −c i+1 for i = 1, ..., n−4, we obtain the determinant as follows:


7.3. Formulae for the Number of Spanning Trees in a Maximal Planar Map 127τ(E n )n4 −1 0 0 . . . 0 0−5 5 −1 0 . . . 0 01 −5 5 −1 . . . 0 0=0 1 −5 5 . . . 0 0. . . . . . . . .0 0 0 0 . . . 5 −1∣ 0 0 0 0 . . . −5 5 ∣In the end, r i ← ∑ ij=1 r j for i = 2, ..., n − 3, we obtain the determinant as follows:τ(E n )n4 −1 0 0 . . . 0 0−1 4 −1 0 . . . 0 00 −1 4 −1 . . . 0 0=0 0 −1 4 . . . 0 0. . . . . .. . .0 0 0 0 . . . 4 −1∣ 0 0 0 0 . . . −1 4 ∣then τ(En)n= 4∆ n−1 − ∆ n−2 , where ∆ i = det(D i ) (∆ i = 4∆ i−1 − ∆ i−2 because ∆ iis tri-diagonal matrix), and D i is defined as following:⎡⎤4 −1−1 4 −1−1 4 −1D i =−1 4 −1. .. . .. . ..⎢⎥⎣−1 4 −1⎦−1 4i×iWith the same technique, we obtain:result.τ(E n−1 )n−1= ∆ n−1 and τ(E n−2)n−2= ∆ n−2 , hence the□7.3.2 An explicit formula for the number of spanning trees in E nNow, we derive the explicit formula for the number of spanning trees in the maximalplanar map τ(E n ).


128 CHAPTER 7. MAXIMAL PLANAR MAPSTheorem 7.3.3 (A maximal planar map) The complexity of the maximal planar mapE n (see Figure. 7.7) is given by the following formula:τ(E n ) =n ((22 √ + √ 3) n−2 − (2 − √ )3) n−2 , n ≥ 3.3Proof: τ(E 3 ) = 3, τ(E 4 ) = 16, τ(E 5 ) = 75, τ(E n)nhence we obtain the system:⎧⎨⎩τ(E n )n= 4τ(E n−1)n−1τ(E 3 ) = 3τ(E 4 ) = 16= 4τ(E n−1)n−1− τ(E n−2)n−2 , with− τ(E n−2)n−2(by Lemma 7.3.2),Then we get a sequence such that u n = 4u n−1 − u n−2 , n ≥ 3, therefore the characteristicquadratic equation is r 2 − 4r + 1 = 0, so the solutions of this equation are: r 1 = 2 − √ 3and r 2 = 2 + √ 3, hence:τ(E n ) = α(2 − √ 3) n + β(2 + √ 3) n , α, β ∈ R, n ≥ 1.Using the initial conditions τ(E 3)3= 1 and τ(E 4)4= 4, we obtain:α = −2− 7 6√3, β = −2+76√3, hence the result. □Example 7.3.4 The following figure (see Figure. 7.10), illustrates a maximal planar mapwith 4 vertices E 4 and its all spanning trees.Figure 7.10: A maximal planar map E 4 gives rise to 16 spanning trees


7.3. Formulae for the Number of Spanning Trees in a Maximal Planar Map 129Numerical results The Table 7.1 illustrates some of the values of the number of spanningtrees in the maximal planar map E n by using the formula given in Theorem 7.3.3:n 3 4 5 6 7 8 9 10τ(E n ) 3 16 75 336 1463 6240 26199 108640Table 7.1: Some values of τ(E n )Let F n and G n be the maps as follows (see Figure 7.11):Figure 7.11: The maps F n and G nTheorem 7.3.5 (A Fan planar map) The complexity of the Fan planar maps F n and G n(see Figure. 7.11) is given by the following formulae:τ(F n ) = √ 1 ((2 + √ 3) n−1 − (2 − √ 3)), n−1 n ≥ 2,3τ(G n ) = 1 2 (√ 3 − 1)((2 + √ 3) n−1 − (2 − √ )3) n−2 , n ≥ 2.


130 CHAPTER 7. MAXIMAL PLANAR MAPSProof: We put f n = τ(F n ) and g n = τ(G n ). f 2 = 2, g 2 = 1, in the map F n , we cut thefirst cycle (see Figure. 7.11), and we use Theorem 4.3.31 (the same goes for the mapG n ), then we obtain: τ(G n ) = 3τ(F n−1 ) − τ(G n−1 ), τ(F n ) = 2τ(G n ) − τ(F n−1 ) therefore,we have the following system:{fn = 2g n − f n−1g n = 3f n−1 − g n−1 with f 2 = 2 and g 2 = 1,we replace by the value of g n in the first equation, we get:{fn = 5f n−1 − 2g n−1g n = 3f n−1 − g n−1 with f 2 = 2 and g 2 = 1(fn)= Mg n)= Mg n(fn(fn−1), where M =g n−1)g n−1(fn−1( 5 -23 -1= ... = M n−2 (f2g 2),then, we compute M n−2 :det (M − λI 2 ) = λ 2 −4λ + 1 = 0, λ 1 = 2 − √ 3 and λ 2 = 2 + √ 3, λ 1 ≠ λ 2 then there isP invertible such that M = P DP −1 where( )λ1 0D =,0 λ 2P is the transformation matrix formed by eigenvectors( 1 1P =M n−2 = P D n−2 P −1 whereD n−2 =3+ √ 32(P −1 = √ −133− √ 33− √ 322-1−3− √ 312),)( √ )(2 − 3)n−200 (2 + √ 3) n−2 ,from which we obtain M n−2 , then(fng n)= M n−2 (f2g 2),,),hence the result.□


7.3. Formulae for the Number of Spanning Trees in a Maximal Planar Map 131Numerical results The Table 7.2 illustrates some of the values of the number of spanningtrees in the sequence of planar maps, the Fan planar maps τ(F n ) and τ(G n ), byusing the formula given in Theorem 7.3.5:n 2 3 4 5 6 7 8 9 10τ(F n ) 2 8 30 112 418 1560 5822 21728 81090τ(G n ) 1 5 19 71 265 989 3691 13775 51409Table 7.2: Some values of τ(F n ) and τ(G n )Let E n be a maximal planar map shown in Figure. 7.7. If we delete the edge e = v 1 v 2 ofE n , we obtain the crystal planar map C n with n vertices (see Figure. 7.12).Figure 7.12: The crystal planar map C nTheorem 7.3.6 (A crystal planar map) The complexity of the crystal planar map C n(see Figure. 7.12) is given by the following formula:τ(C n ) = 12 √ 3 (n − 2) ((2 + √ 3) n−2 − (2 − √ 3) n−2 ), n ≥ 3.Proof: By Theorem 4.3.11, we have: τ(E n ) = τ(E n −e)+τ(E n .e), since τ(E n −e) = τ(C n )and τ(E n .e) = τ(F n ) (see Figure. 7.13):then τ(C n ) = τ(E n ) − τ(F n ). We replace by the value of τ(E n ) from Theorem 7.3.3 andthe value of τ(F n−1 ) from Theorem 7.3.5, thus our result follows.□


132 CHAPTER 7. MAXIMAL PLANAR MAPSFigure 7.13: The maps E n , E n − e and E n .eNumerical results The Table 7.3 illustrates some of the values of the number of spanningtrees in the crystal planar maps τ(C n ) by using the formula given in Theorem 7.3.6:n 3 4 5 6 7 8 9 10τ(C n ) 1 8 45 224 1045 4680 20377 86912Table 7.3: Some values of τ(C n )


ConclusionThis thesis presents our contribution in the research area of calculating the number ofspanning trees in connected planar maps (graphs embedded into surfaces without edgecrossings).In this thesis, we have surveyed the several methods as the Matrix Tree Theorm, thedeletion - contraction theorem, Cayley’s Theorem and other methods that can calculatethe number of spanning trees in planar maps. However, we noticed that, the calculation ofthe number of spanning trees of planar maps by using the determinant of Laplacian matrixor the Laplace spectrum is tedious and impractical, therefore, we gave new methods inthe Chapter 4 which facilitate the calculation of the number of spanning trees in planarmaps. Then, we applied these methods on certain planar maps and as a result, weobtained explicit formulae to find the number of spanning trees exactly in some specialplanar maps.The first three chapters are devoted on definitions and general properties of basicobjects studied (graphs, maps, trees, spanning trees, complexity, matrices associated to agraph, etc), and some methods for counting spanning trees are introduced.In Chapter 4, we gave some new methods to calculate the number of spanning treesin planar maps in general then in particular in some special planar maps.The fifth Chapter is devoted on derivative explicit formulae which facilitate the calculationof the number of spanning trees in some families of planar maps which are called then-Fan chains, the n- Grid chains, the n-tent chains, the n-Hexagonal chains, the n-Eightchains, the n-Home chains, the n-Barrel chains, the n-Light chains, the n-Kite chains,the n-Envelope chains and the n-Diphenylene, ... etc. These results obtained came as aresult of the use of our methods that we have derived in Chapter 4.In chapter 6, we focused on a particular type of planar maps which is called the starflower planar map, and we derived an explicit formula which calculates the number ofspanning trees in this type of planar maps.Finally, in Chapter 7, we were interested in the maximal planar map. We calculatedthe Wiener index in general in the case of planar maps, in particular, in the maximalplanar maps and gave an inequality, which minimizes and maximizes any planar map bythe maximal planar map with n vertices and the path with n vertices. We then derivedan explicit formula to calculate the number of spanning trees in the maximal planar map,then we have deduced the formula for the number of spanning trees in the crystal planarmap.133


134 CHAPTER 7. MAXIMAL PLANAR MAPSIn future, we plan to work on the planar maps of type C= C 1 ‡ C 2 with ‡ is notat all a simple path, and shall try to generalize the theorem deletion - contraction byworking on deletion more than one edge, i.e., the theorem deletion - contraction workedon one edge but we shall work on n edges and associate it with our results in this thesis.Keywords graphs, maps, trees, spanning trees, complexity, Laplacian matrix, Matrix-Tree Theorem, the n-Fan chains, the n- Grid chains, star flower planar map, maximalplanar map, Wiener index.


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Auteur :Titre :Directeurs de thèse :Abdulhafidh MODABISHÉnumération de nombre d’arbres couvrants dans certainescartes planaires spéciales.Pr. Mohamed EL MARRAKIRésuméLe nombre d’arbres couvrants dans une carte planaire - graphe plongé dans une surfacesans croisement d’arêtes - (réseau) est un important bien étudié la quantité et invariantdu graphe (réseau); de plus c’est aussi une mesure importante de la fiabilité d’un réseauqui joue un rôle central dans la théorie classique de Kirchhoff des réseaux électriques.Dans un graphe (réseau) qui contient plusieurs cycles, il faut supprimer les redondancesdans ce réseau, i.e., on obtient un arbre couvrant. Un arbre couvrant dans une carte Cest un arbre qui a le même ensemble de sommets en tant que C (arbre qui passe par tousles sommets de la carte C).Notre thème de recherche dans cette thèse se concentre sur le calcul du nombred’arbres couvrants dans les cartes planaires connexes, un sujet dans la théorie des graphescombinatoire; ainsi que, pour trouver de nouvelles méthodes pour calculer le nombred’arbres couvrants dans une carte planaire (réseau).Arbres couvrants sont pertinents pour les différents aspects de graphes (réseaux). Engénéral, le nombre d’arbres couvrants dans un réseau peut être obtenu par le calcul ledéterminant de la matrice laplacien liée ou le calcul du spectre de Laplace du réseau.Cependant, pour une grande carte (réseau), l’évaluation du déterminant pertinent est uncalcul difficile. Dans ce travail, nous fournissons de nouvelles méthodes pour faciliter lecalcul du nombre d’arbres couvrants dans les cartes planaires et de prouver de nouveauxrésultats simplifiée et généralisée. Enfin, nous appliquons ces méthodes sur certainescartes planaires à fin de dériver plusieurs formules explicites pour calculer le nombred’arbres couvrants dans certaines familles particulières des cartes planaires.Mots-clés : graphes, cartes, arbres, arbres couvrants, complexité, laplacien matrice,théorème de Kirchhoff, les chaînes de n-Fan, les chaînes de n-Grille, fleur d’étoile carteplanaire, maximale carte planaire, indice de Wiener.


ENUMERATION OF THE NUMBER OF SPANNING TREESIN SOME SPECIAL PLANAR MAPSAbstractThe number of spanning trees in a planar map - graph embedded into the surfaceswithout edge-crossings - (network) is an important well-studied quantity and invariantof the graph (network); moreover it is also an important measure of reliability of anetwork which plays a central role in Kirchhoff’s classical theory of electrical networks.In a graph (network), that contains several cycles, we must remove the redundanciesin this network, i.e., we obtain a spanning tree. A spanning tree in a map C is a treewhich has the same vertex set as C (tree that passing through all the vertices of the map C).Our research theme in this thesis focuses on the counting of the number of spanningtrees in connected planar maps, a subject in combinatorial graph theory; as well as, tofind new methods to calculate the number of spanning trees in any map (network).Spanning trees are relevant to various aspects of graphs. Generally, the number ofspanning trees in a network can be obtained by computing a related determinant ofthe Laplacian matrix or computing the Laplace spectrum of the network. However,for a large map, evaluating the relevant determinant is computationally intractable.In this work, we provide new methods to facilitate the calculation of the number ofspanning trees in planar maps and to prove new simplified and generalized results.Finally, we apply these methods on certain planar maps to derive several explicit formulaefor calculating the number of spanning trees in some special families of planar maps.Keywords: graphs, maps, trees, spanning trees, complexity, Laplacian matrix,Matrix-Tree Theorem, the n-Fan chains, the n- Grid chains, star flower planar map,maximal planar map, Wiener index.

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