MMNA GRAPHS ON EIGHT

VERTICES OR FEWER

A Thesis

Presented

to the Faculty of

California State University, Chico

In Partial Fulfillment of the Requirements for the Degree

Master of Science

in

Mathematics Education

by

Hugo E. Ayala

Summer 2014

TABLE OF CONTENTS

PAGE

List of Tables………………...... iv

List of Figures..……………...... v

Abstract.……………………...... xxi

CHAPTER

I. Introduction.………………………………………………………………………….. 1

Statement of Problem.…………………………………………………. 1 Map of Thesis..……………………………………………………………... 4

II. Definitions and Preliminary Constructions.…………………………….. 5

Definition of Terms.………………………………………….…..……... 5 Non-planar (7, 12) graphs..………………………………………….. 13 Triangulations on Seven Vertices.………………………………… 35 Petersen Graphs and their Complements.……………………... 57

III. Proof of Theorem A.……………………………………………………………….. 72

Lemmas and Prior Results.…………………………………………... 72 Theorem 1.………………………………………………………………….. 75 Theorem 2.………………………………………………………………….. 84

References Cited.………………………………………………………………………………………….. 187

iii LIST OF TABLES

TABLE PAGE

2.3.1 The Five Triangulations…………………………………………………………… 52

3.3.1 The Candidate MMNA Graphs…………………………………………………... 86

iv LIST OF FIGURES

FIGURE PAGE

1.1.1 The Utility Problem.…………………………………………………………………. 1

1.1.2 The K3,3 Graph……………………..…………………………………………………… 2

1.1.3 The K5 Graph………...…………………………………………………………………. 2

1.1.4 The J1 Graph, the K6 Graph, and the K3,3,1 Graph………………………… 3

1.1.5 The P7 Graph, the K4,4 – e Graph, and the P8 Graph…………..…………. 4

2.1.1 Graph G………..……………………..…………………………………………………… 9

2.1.2 Graph G’..……..……………………..……………………………………………………. 9

2.1.3 Graph H...……..……………………..……………………………………………………. 10

2.1.4 Graph H with Vertex B Split…...………………………………………………….. 10

2.1.5 Graph J.....……..……………………..……………………………………………………. 11

2.1.6 Graph J with a (1,2) Split of Vertex B…………………………………………. 11

2.1.7 Graph J with a Second (1,2) Split of Vertex B…………………...…………. 12

2.1.8 Graph J with a Third (1,2) Split of Vertex B.……………………….………. 12

2.1.9 A Degree Four Vertex, B.……..……………………………………………………. 12

2.1.10 A Degree Four Vertex, B, Split into (0,4).……………………………………. 13

2.2.1 The K3,3 Graph……….…………………………………………………………………. 14

2.2.2 The Only Non-Planar (6,10) Graph with δ(G) > 2.………………………. 14

2.2.3 First Derived Non-Planar (6,11) Graph……..……………………….………. 15

v 2.2.4 Second Derived Non-Planar (6,11) Graph.…………………………………. 16

2.2.5 The K5 Graph……………………..………………..……………………………………. 16

2.2.6 Non-Planar (6,11) Graph Derived from K5…………………………………. 17

2.2.7 Comparing Two Non-Planar Graphs. ………………………………………… 18

2.2.8 Third Derived Non-Planar (6,11) Graph…...……………………….………. 18

2.2.9 The Only Non-Planar (7,10) Graph with δ(G) > 2……………………….. 19

2.2.10 First Non-Planar (7,11) Graph……………..……………………………………. 20

2.2.11 Second Non-Planar (7,11) Graph. ..……………………………………………. 21

2.2.12 Third Non-Planar (7,11) Graph. ..………………………………………………. 21

2.2.13 Fourth Non-Planar (7,11) Graph. ..……………………………….……………. 22

2.2.14 Fifth Non-Planar (7,11) Graph. ..…….…………………………………………. 22

2.2.15 First (5, 42, 33, 2) Graph...……..……………………………………………………. 23

2.2.16 Second (5, 42, 33, 2) Graph...…….…………………………………………………. 24

2.2.17 Third (5, 42, 33, 2) Graph...…...….…………………………………………………. 24

2.2.18 The (5, 4, 35) Graph………………………………….……………………….………. 25

2.2.19 First (43, 34) Graph……….……..……………………………………………………. 25

2.2.20 Second (43, 34) Graph…...……..……………………………………………………. 26

2.2.21 Third (43, 34) Graph……….…...……………………………………………………. 26

2.2.22 First (44, 32, 2) Graph…………………………..……………………………………. 27

2.2.23 Second (44, 32, 2) Graph…………..…………..……………………………………. 28

2.2.24 Third (44, 32, 2) Graph……..…………………..……………………………………. 28

2.2.25 Fourth (44, 32, 2) Graph…………...…………..……………………………………. 29

vi 2.2.26 Fourth (43, 34) Graph…………………………..……………………………………. 29

2.2.27 Fifth (43, 34) Graph……………………….……..……………………………………. 30

2.2.28 First (45, 22) Graph.…………………….……..……………………………...………. 31

2.2.29 Second (45, 22) Graph.…………………..……..……………………………………. 31

2.2.30 Third (45, 22) Graph..…………………….……..……………………………………. 32

2.2.31 Fourth (5, 42, 33, 2) Graph…………….……..……………………………………. 33

2.2.32 The (52, 34, 2) Graph.…………………….……..……………………………………. 34

2.2.33 Fifth (44, 32, 2) Graph..………………….……..……………………………………. 34

2.3.1 A (5, 15) Graph…….……………………….……..……………………………………. 37

2.3.2 A (52, 25) Graph…...……………………….……..……………………………………. 38

2.3.3 A (52, 4, 32, 22) Graph..………………….……..……………………………………. 38

2.3.4 The (52, 45) Triangulation.…………….……..……………………………………. 39

2.3.5 A (6, 16) Graph…….……………………….……..……………………………………. 39

2.3.6 A (6, 36) Graph…….……………………….……..……………………………………. 40

2.3.7 A (6, 44, 32) Graph….…………………………………………………………………. 40

2.3.8 The (6, 46) Graph is Not a Triangulation.…………………………………… 41

2.3.9 A (5, 15) Graph……..…………………….……..……………………………...………. 41

2.3.10 A (5, 4, 24, 1) Graph…..…………………..……..……………………………………. 42

2.3.11 A (5, 43, 33) Graph…..…………………….……..……………………………………. 42

2.3.12 A (52, 43, 32) Graph……………………….……..……………………………………. 43

2.3.13 The (53, 43, 3) Triangulation..…….….……..……………………………………. 43

2.3.14 A (6, 16) Graph……….....………………….……..……………………………………. 44

vii 2.3.15 A (6, 36) Graph.…………………………….……..……………………………………. 44

2.3.16 A (6, 5, 42, 33) Graph.…………………….……..……………………………………. 45

2.3.17 The (6, 5, 44, 3) Graph is Not a Triangulation.……………………………. 45

2.3.18 A (6, 36) Graph……………….…………….……..……………………………………. 46

2.3.19 The (62, 43, 32) Triangulation….…….……..……………………………………. 46

2.3.20 A (6, 36) Graph……..…………………….……..……………………………...………. 47

2.3.21 A (6, 5, 42, 33) Graph...…………………..……..……………………………………. 47

2.3.22 The (6, 52, 42, 32) Triangulation…….……..……………………………………. 48

2.3.23 A (5, 35, 2) Graph………………………….……..……………………………………. 48

2.3.24 A (54, 32, 2) Graph……………....…….….……..……………………………………. 49

2.3.25 The (54, 4, 32) Graph is Not a Triangulation….……………………………. 49

2.3.26 A (6, 36) Graph..…………...……………….……..……………………………………. 50

2.3.27 The (6, 53, 33) Triangulation.……….……..………………..……………………. 50

2.3.28 A (6, 16) Graph…………………………………………...……………………………. 51

2.3.29 A (62, 25) Graph.…………….…………….……..……………………………………. 51

2.3.30 The (63, 34) Graph is Not a Triangulation….……….………………………. 52

2.3.31 (52, 45) Graph and its Complement..……..……………………………………. 53

2.3.32 The “Pentagon – Segment” Graph….……..……………………………………. 53

2.3.33 Graph and its Complement………………………....……………………………. 53

2.3.34 The “Thin – Y” Graph.……..…………….……..……………………………………. 54

2.3.35 (62, 43, 32) Graph and its Complement.....……………………………………. 54

2.3.36 The “House” Graph…………………….……..……………………………...………. 55

viii 2.3.37 (6, 52, 42, 32) Graph and its Complement..…..………………………………. 55

2.3.38 The “Fat – Y” Graph………………..…….……..……………………………………. 55

2.3.39 (6, 53, 33) Graph and its Complement…..……………………………………. 56

2.3.40 The “Hat” Graph………………....…….….……..……………………………………. 56

2.3.41 A Listing of Complements of the Five Triangulations on Seven Vertices………………………………………………………………………………. 56

2.4.1 The K6 Graph, the K3,3,1 Graph, and the P7 Graph………...………………. 57

2.4.2 The K4,4 – e Graph, the P8 Graph, and the P9 Graph.….…………………. 57

2.4.3 The Petersen Graph.…………………………………...……………………………. 58

2.4.4 The Petersen Graph P8 and its Complement.…..…………………………. 59

2.4.5 The Petersen Graph K4,4 – e and its Complement...….…………………. 59

2.4.6 The Petersen Graph P7 and its Complement...…….………………………. 59

2.4.7 The Complement of the P7 Graph…..……..……………………………………. 60

2.4.8 The P7 Graph Plus a Degree Zero Vertex, X………………...………………. 60

2.4.9 The P7 Complement Plus a Vertex X of Full Degree….…………………. 60

2.4.10 The Petersen Graph P7…………………………………………………..…………. 61

2.4.11 The First Graph on Eight Vertices Derived from the P7 Graph, and its Complement..…..……………………………………………...………. 61

2.4.12 The Second Graph on Eight Vertices Derived from the P7 Graph, and its Complement..…..……………………………………………...………. 62

2.4.13 The Third Graph on Eight Vertices Derived from the P7 Graph, and its Complement..…..……………………………………………...………. 62

2.4.14 The Fourth Graph on Eight Vertices Derived from the P7 Graph, and its Complement..…..……………………………………………...………. 63

ix 2.4.15 The Fifth Graph on Eight Vertices Derived from the P7 Graph, and its Complement..…..……………………………………………...………. 63

2.4.16 The Sixth Graph on Eight Vertices Derived from the P7 Graph, and its Complement..…..……………………………………………...………. 64

2.4.17 The Petersen Graph K3,3,1 and its Complement…....….…………………. 65

2.4.18 The K3,3,1 Complement Plus a Vertex X of Full Degree...………………. 65

2.4.19 The First Graph on Eight Vertices Derived from the K3,3,1 Graph, and its Complement..…..……………………………………………...………. 66

2.4.20 The Second Graph on Eight Vertices Derived from the K3,3,1 Graph, and its Complement..…..……………………………………………...………. 66

2.4.21 The Third Graph on Eight Vertices Derived from the K3,3,1 Graph, and its Complement..…..……………………………………………...………. 67

2.4.22 The K6 Graph……………………..………………..…………………………………… 68

2.4.23 A Graph on Seven Vertices Derived from the K6 Graph…….………… 68

2.4.24 A Graph on Eight Vertices Derived from the K6 Graph..……...………. 69

2.4.25 The Complement Derived from a Graph of Eight Vertices that has a K6 Minor..…..……..…………………………….…………………………. 69

2.4.26 Complements Constructed from the Two Petersen Graphs of Eight Vertices..…..………………..…………………………….…………………………. 70

2.4.27 Complements Created from the P7 Graph on Eight Vertices……..…. 70

2.4.28 Complements Created from the K3,3,1 Graph on Eight Vertices….…. 71

2.4.29 The Complement Created from the K6 Graph on Eight Vertices..…. 71

3.1.1 The K6 Graph……………………………………………………...….…………………. 72

3.1.2 The Graph K6 – (A, B), on the Left, and G – C on the Right..…………. 73

3.1.3 The Graph K6/(A, B), on the Left, and G – C on the Right....…………. 74

x 6 3.2.1 The Graph G = (5 , 4), on the Left, G – V1 on the Right…...... …………. 76

3.2.2 The Three Graphs on the Left are of Degree Sequence (54, 43), and the Two Graphs on the Right are of Degree Sequence (55, 4, 3). 77

4 3 3.2.3 The First (5 , 4 ) Graph, on the Left, G – W1 on the Right..…….……. 77

4 3 3.2.4 The Second (5 , 4 ) Graph, on the Left, G – W1 on the Right..………. 78

4 3 3.2.5 The Third (5 , 4 ) Graph, on the Left, G – W1 on the Right…….……. 78

3.2.6 The K6 Graph (left), the K6 Expansion Graph on Seven Vertices (center), and the Complement of the K6 Expansion Graph on Seven Vertices (right)…………………………………………………………. 79

3.2.7 The First (55, 4, 3) Graph and its Complement...….……..………………. 79

3.2.8 The Second (55, 4, 3) Graph and its Complement...….…………………. 80

3.2.9 The Non-Planar (6,10) Graph of Minimum Degree at Least Two…. 80

3.2.10 The Sub Case 2 Graph G on the Left, and G – W1 on the Right…...…. 81

3.2.11 The Sub Case 3 Graph G on the Left, and G – V1 on the Right….....…. 82

5 3.2.12 The (5, 4 , 3) Graph on the Left, and G – W3 on the Right...... ………. 82

7 3.2.13 The (4 ) Graph on the Left, and G – W3 on the Right……...... ………. 83

5 2 3.2.14 The (4 , 3 ) Graph on the Left, and G – W3 on the Right...... ………. 84

3.3.1 The Non-Planar (7,10) Graph with δ(G) > 2………...….…………………. 88

3.3.2 The (7,10) Graph with Vertex A Added, has Vertex V2 Removed and is Found Planar in Two Different Constructions…………..…. 88

3.3.3 The (7,10) Graph with Vertex A Added, has Vertex W2 Removed and is Found to be Planar in Two Different Constructions.....…. 89

3.3.4 When G – a is One of the (7,12) Graphs Shown, Deleting the Starred Vertex from G Leaves a Planar Graph……………………...... …. 91

3.3.5 G – V3 Yields a Planar Graph………………………...... ….……..………………. 91

xi 3.3.6 If G1 and G2 are Both K4 then G is Planar…………...... ….…………………. 92

3.3.7 If (V1, W1) ∈E(G), then G is Apex by Deleting the Starred Vertex (left). If (V1, W1) ∉E(G), G is Planar (right)……………………………. 93

3.3.1.1 The First Non-Planar (7,12) Graph, a (5, 42, 33, 2) Graph, and its Complement…………………………………………………………...…...… 97

3.3.1.2 The (5, 42, 33, 2) Graph Plus Vertex A, Creates a Graph on Eight Vertices.……………………………………………….……………………...…...… 97

3.3.1.3 The First (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar………..…………………………………...…...… 98

3.3.1.4 The First (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found Once Again to be Planar………..………………………… 99

3.3.1.5 These Six Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A)….…………...…...… 100

3.3.1.6 These Four Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).…………...…...… 101

3.3.1.7 The Second Non-Planar (7,12) Graph and its Complement.....…...… 102

3.3.1.8 The Second (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar……………………………...…...… 103

3.3.1.9 The Second (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found Once Again to be Planar…………...…...… 103

3.3.1.10 The Second (7,12) Graph with Vertex A Added, has Vertex W1 Removed and is Found to be Planar……………………………...…...… 104

3.3.1.11 These Three Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A)….…………...…...… 105

3.3.1.12 These Three Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).…………...…...… 106

3.3.1.13 The Third Non-Planar (7,12) Graph and its Complement…….…...… 107

xii 3.3.1.14 The Third (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar in Three Different Constructions..… 107

3.3.1.15 These Three Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A)….…………...…...… 108

3.3.1.16 These Three Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).…………...…...… 109

3.3.1.17 The Fourth Non-Planar (7,12) Graph and its Complement…..…...… 110

3.3.1.18 The Fourth (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar in Two Different Constructions..…… 110

3.3.1.19 The Fourth (7,12) Graph with Vertex A Added, has Vertex W3 Removed and is Found to be Planar in Two Different Constructions.…………...…...…………………………………………...……. 111

3.3.1.20 The Graphs Represent Complement of G’ When We Consider that U, W1 ∉N(A)…………………………………………………………………….. 112

3.3.1.21 The Graphs Represent Complement of G’ When We Consider that V3, W1 ∉N(A)...………………………………………………………………….. 113

3.3.1.22 The Graphs Represent Complement of G’ When We Consider that V2, W1 ∉N(A)...………………………………………………………………….. 113

3.3.1.23 These Three Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A)….…………...…...… 114

3.3.1.24 These Four Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).…………...…...… 115

3.3.1.25 The Fifth Non-Planar (7,12) Graph and its Complement……...…...… 116

3.3.1.26 The Fourth (7,12) Graph with Vertex A Added, has Vertex V1 Removed and is Found to be Planar in Two Different Constructions..…… 116

3.3.1.27 The Graphs Represent Complement of G’ When We Consider that the Given Pair of Vertices are Not in N(A)....………………………… 117

3.3.1.28 The Graphs Represent the Complement of G’ When We Consider that V3, W1 ∉N(A)...…………………………………………………………………… 118 xiii 3.3.1.29 The Graphs Represent the Complement of G’ When We Consider that U, W1 ∉N(A)....………………………………………………………………….. 118

3.3.1.30 These Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A)….……………….....…...… 119

3.3.1.31 The Graphs Represent the Complement of G’ When We Consider that U, V1 ∉N(A)...... …………………………………………………………………... 119

3.3.1.32 These Three Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A).……………...…...… 120

3.3.1.33 These Five Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).…………...…...… 121

3.3.1.34 The Sixth Non-Planar (7,12) Graph and its Complement……..…...… 122

3.3.1.35 These Two Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A)….…………...…...… 123

3.3.1.36 The Graphs Represent the Complement of G’ When We Consider that W1, W2 ∉N(A).…………………………………………………………………... 123

3.3.1.37 These Two Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).…………...…...… 124

3.3.1.38 The Seventh Non-Planar (7,12) Graph and its Complement…...... … 125

3.3.1.39 The Seventh (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar in Two Different Constructions…………………………………………………………………… 125

3.3.1.40 These Graphs Represent the Complement of G’ When we Assume that the Given Pair of Vertices are Not in N(A).……………...…...… 126

3.3.1.41 The Graphs Represent the Complement of G’ When We Assume that U, V2 ∉N(A)…..…………………………………………………………………... 127

3.3.1.42 The Graphs Represent the Complement of G’ When We Consider that V2, W3 ∉N(A).…………………………..………………………………………... 127

3.3.1.43 The Graphs Represent the Complement of G’ When We Consider that U, V1 ∉N(A).…………...………………..………………………………………... 128 xiv 3.3.1.44 The Graph Represents the Complement of G’ When We Assume that U, W3 ∉N(A).……..…...………………..………………………………………... 128

3.3.1.45 The Graphs Represent the Complement of G’ When We Consider that V1, W3 ∉N(A).…………………………..………………………………………... 129

3.3.1.46 These Five Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).…………...…...… 130

3.3.1.47 The Eighth Non-Planar (7,12) Graph and its Complement…….....… 131

3.3.1.48 The Eighth (7,12) Graph with Vertex A Added, has Vertex W1 Removed and is Found to be Planar in Two Different Constructions…………………………………………………………………… 131

3.3.1.49 The Eighth (7,12) Graph with Vertex A Added, has Vertex V1 Removed and is Found to be Planar in Two Different Constructions…………………………………………………………………… 132

3.3.1.50 The Graph Represents the Complement of G’ When We Consider that V3, W2 ∉N(A).…………………………..………………………………………... 133

3.3.1.51 These Two Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).…………...…...… 133

3.3.1.52 The Ninth Non-Planar (7,12) Graph and its Complement………....… 134

3.3.1.53 The Ninth (7,12) Graph with Vertex A Added, has Vertex W1 Removed and is Found to be Planar in Two Different Constructions…………………………………………………………………… 135

3.3.1.54 The Ninth (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar…...……………………………… 136

3.3.1.55 These Three Graphs Represent the Complement of G’ When we Assume that the Given Pair of Vertices are Not in N(A).……...… 136

3.3.1.56 These Three Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).…………...…...… 137

3.3.1.57 The Tenth Non-Planar (7,12) Graph and its Complement...……....… 138

xv 3.3.1.58 The Tenth (7,12) Graph with Vertex A Added, has Vertex V1 Removed and is Found to be Planar in Two Different Constructions…………………………………………………………………… 138

3.3.1.59 These Graphs Represent the Complement of G’ When we Assume that the Given Pair of Vertices are Not in N(A).………………...... … 139

3.3.1.60 The Graphs Represent the Complement of G’ When We Consider that V3, W2 ∉N(A)…..………………….……………………………………... 140

3.3.1.61 These Three Graphs Represent the Complement of G’ When we Assume that the Given Pair of Vertices are Not in N(A)..…...... … 140

3.3.1.62 These Two Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).…………...…...… 141

3.3.1.63 The Eleventh Non-Planar (7,12) Graph and its Complement...... … 142

3.3.1.64 The Eleventh (7,12) Graph with Vertex A Added, has Vertex V1 Removed and is Found to be Planar…...……………………………… 143

3.3.1.65 The Eleventh (7,12) Graph with Vertex A Added, has Vertex W3 Removed and is Found to be Planar…...……………………………… 143

3.3.1.66 These Graphs Represent the Complement of G’ When we Assume that the Given Pair of Vertices are Not in N(A).………………...... … 144

3.3.1.67 The Graphs Represent the Complement of G’ When We Consider that V3, W2 ∉N(A)…..………………….……………………………………... 144

3.3.1.68 These Three Graphs Represent the Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A)..…...... … 145

3.3.1.69 These Four Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).…………...…...… 146

3.3.1.70 The Twelfth Non-Planar (7,12) Graph and its Complement...... … 146

3.3.1.71 The Twelfth (7,12) Graph with Vertex A Added, has Vertex W3 Removed and is Found to be Planar…...……………………………… 147

xvi 3.3.1.72 The Twelfth (7,12) Graph with Vertex A Added, has Vertex V1 Removed and is Found to be Planar…...……………………………… 148

3.3.1.73 These Graphs Represent the Complement of G’ When we Assume that the Given Pair of Vertices are Not in N(A).………………...... … 148

3.3.1.74 The Graphs Represent the Complement of G’ When We Consider that V3, W1 ∉N(A)…..………………….……………………………………... 149

3.3.1.75 The Graphs Represent the Complement of G’ When We Consider that U, W1 ∉N(A)…..………………….…………………………..…………... 149

3.3.1.76 The Graph Represents the Complement of G’ When We Assume that V1, W2 ∉N(A)…..………………….…………..……………..…………... 150

3.3.1.77 The Graph Represents the Complement of G’ When We Assume that V1, V3 ∉N(A)…..………………….…………………………..…………... 150

3.3.1.78 The Graphs Represent the Complement of G’ When We Consider that U, V1 ∉N(A)…..………………….……….…………………..…………... 151

3.3.1.79 The Graph Represents the Complement of G’ When We Assume that U, V3 ∉N(A)…..………………….………………………..…..…………... 152

3.3.1.80 The Graph Represents the Complement of G’ When We Assume that U, W2 ∉N(A)…..………………….…………………………..…………... 152

3.3.1.81 The Graph Represents the Complement of G’ When We Consider that V3, W2 ∉N(A)…..…...…………….…………………………..…………... 153

3.3.1.82 These Five Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).…………...…...… 154

3.3.1.83 The Thirteenth Non-Planar (7,12) Graph and its Complement....… 155

3.3.1.84 The Thirteenth (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar in Two Different Constructions…………………………………………………………………… 155

3.3.1.85 The Thirteenth (7,12) Graph with Vertex A Added, has Vertex W1 Removed and is Found Once Again to be Planar…...….………… 156

xvii 3.3.1.86 The Graph Represents the Complement of G’ When We Assume that U, W2 ∉N(A)…..…...…………….………………….………..…………... 157

3.3.1.87 The Graph Represents the Complement of G’ When We Assume that U, V3 ∉N(A)…..….....…………….………………….………..…………... 157

3.3.1.88 The Graphs Represent the Complement of G’ When We Consider that U, V1 ∉N(A)…..….....…………….………………….………..…………... 158

3.3.1.89 The Graph Represents the Complement of G’ When We Assume that V1, V3 ∉N(A)…..….....…………...………………….………..…………... 158

3.3.1.90 The Graph Represents the Complement of G’ When We Assume that V1, W2 ∉N(A)…..….....…………...…..…………….………..………….. 159

3.3.1.91 The Graphs Represent the Complement of G’ When We Consider that V3, W2 ∉N(A)…..….....………….………………….………..…………... 159

3.3.1.92 These Four Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).…………...…..... 160

3.3.1.93 The Fourteenth Non-Planar (7,12) Graph and its Complement.... 161

3.3.1.94 The Fourteenth (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar in Two Different Constructions…………………………………………………………………… 161

3.3.1.95 These Three Graphs Represent the Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A).…...... 162

3.3.1.96 These Three Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).…………...…..... 163

3.3.1.97 The Fifteenth Non-Planar (7,12) Graph and its Complement...... 164

3.3.1.98 The Fifteenth (7,12) Graph with Vertex A Added, has Vertex V3 Removed and is Found to be Planar…………………………………... 164

3.3.1.99 These Six Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A).…………………... 165

3.3.1.100 These Four Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).…………...……. 166 xviii 3.3.1.101 The Sixteenth Non-Planar (7,12) Graph and its Complement...... 167

3.3.1.102 The Sixteenth (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar Three Different Constructions………………………………………………...…...….………… 167

3.3.1.103 The Graph Represents the Complement of G’ When We Assume that V1, V2 ∉N(A)…..…...…………….………………….………..…………... 168

3.3.1.104 These Two Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).…………...……. 169

3.3.1.105 The Seventeenth Non-Planar (7,12) Graph and its Complement.. 170

3.3.1.106 The Seventeenth (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar Two Different Constructions………………………………………………...…...….………… 170

3.3.1.107 The Seventeenth (7,12) Graph with Vertex A Added, has Vertex W3 Removed and is Found Once Again to be Planar…...….………… 171

3.3.1.108 These Three Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A).…………………... 172

3.3.1.109 These Three Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).………………... 172

3.3.1.110 The Eighteenth Non-Planar (7,12) Graph and its Complement…. 173

3.3.1.111 The Eighteenth (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar Three Different Constructions………………………………………………...…...….………… 174

3.3.1.112 These Three Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A).…………………... 175

3.3.1.113 These Three Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).………………... 175

3.3.1.114 The Nineteenth Non-Planar (7,12) Graph and its Complement…. 176

xix 3.3.1.115 The Nineteenth (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar Two Different Constructions………………………………………………...…...….………… 177

3.3.1.116 The Nineteenth (7,12) Graph with Vertex A Added, has Vertex W3 Removed and is Found to be Planar Two Different Constructions………………………………………………...…...….………… 177

3.3.1.117 The Graph Represents the Complement of G’ When We Assume that V1, W1 ∉N(A)…..…...…………….……………….………..…………..... 178

3.3.1.118 These Two Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A).………………... 179

3.3.2.1 The J1 Graph………………………………………………………..………………... 180

3.3.2.2 J1 – V1 and its Complement………………….…...…………..………………... 180

3.3.2.3 The Fifteenth Non-Planar (7,12) Graph of Degree at Least Two from Chapter II, and its Complement……………………………………….…. 181

3.3.2.4 J1 – A1 and its Complement………………….…...…………..……………….... 181

3.3.2.5 The Fourteenth Non-Planar (7,12) Graph of Degree at Least Two from Chapter II, and its Complement……………………………………….…. 182

3.3.2.6 The Graph G = J1 – V1 (left) and G – A2 (right)…....…..……………….... 182

3.3.2.7 The Graph G = J1 – A1 (left) and G – A2 (right).…...…..……………….... 183

3.3.2.8 The Graph G = J1 – (A1, V1) (left) and G – V5 (right)..…...... ……….... 183

3.3.2.9 The Graph G = J1 – (V1, V2) (left) and G – V4 (right).…...…..……….... 184

3.3.2.10 The Graph G = J1/(A1, V1) (left) and G – A2 (right)...…...…..……….... 184

3.3.2.11 The Graph G = J1/(V1, V2) (left) and G – V6 (right)....…...…..……….... 185

xx ABSTRACT

MMNA GRAPHS ON EIGHT

VERTICES OR FEWER

by

Hugo E. Ayala

Master of Science in Mathematics Education

California State University, Chico

Summer 2014

We prove that the only minor minimal non-apex (MMNA) graphs on eight vertices or fewer are the Jorgensen J1 graph and five Petersen family graphs, P8, K4,4

– e, K3,3,1, P7, and K6. We determine the non-planar (7,12) graphs of minimum degree at least two. We argue that there are five planar triangulations on seven vertices.

We make use of the Petersen graphs of eight and seven vertices: P8, K4,4 – e, K3,3,1, and P7, and determine all graphs on eight vertices with one of these Petersen graphs as a minor. With these in hand we give a complete classification of MMNA graphs on seven or fewer vertices. For graphs on eight vertices, we illustrate our proof by contradiction argument with a detailed analysis of (8,17) and (8,18) graphs constructed by adding a vertex to a non-planar (7, 12) graph.

xxi CHAPETR I

INTRODUCTION

1.1. Statement of Problem

Let us begin with the utility problem that asks for three utilities, electricity, water, and gas for example, to be connected to three homes without any of the utility lines crossing. It turns out this is impossible. In Figure 1.1.1 we cannot connect house 3 with the water utility without lines intersecting.

E W G

1 2 3

Figure 1.1.1. The Utility Problem.

In graph theory terms, the problem is modeled by the bipartite graph K3,3, and we say K3,3 is non-planar (In this intro, we will give an overview, leaving precise definitions for chapter II).

1 2

W1 W2 W3

V V V 1 2 3 Figure 1.1.2. The K3,3 Graph.

A fundamental result in graph theory, Kuratowski’s theorem is a forbidden graph characterization of planar graphs.

Theorem (Kuratowski): A finite graph is planar if and only if it does not contain

K3,3 or K5. In other words every non-planar graph contains either the K3,3 graph or the K5 graph.

V2 V3

V1

V4

V 5 Figure 1.1.3. The K5 Graph.

A very powerful generalization of Kuratowski’s theorem is the Robertson-Seymour

Graph Minor theorem. It states every property of graphs (e.g. non-planar) is characterized by a finite set of forbidden graphs. In particular, for the apex property described in the following paragraph, there are a finite number of forbidden graphs called the MMNA graphs.

3

If you look at the K3,3 and K5 graphs above, you will notice that both are non-planar, meaning that there is no way to draw them without having edges crossing each other. But you may also notice that both the K3,3 and K5 graphs are very close to being planar. If you were to remove any vertex and its edges from either graph, you would make that graph planar. A graph like this is called Apex.

If you delete a vertex or an edge from a graph your resulting graph is called a minor. You can also create a minor by contracting an edge. We will go over contracting an edge in detail in chapter II. Any combination of these three moves also results in a minor.

If we begin with a graph G, which is NA (non apex), and every minor of G is apex then we can say that G is a minor minimal non apex (MMNA) graph. This means that G has the characteristic of being NA, but none of its minors have this characteristic. Now we can state our main theorem.

Theorem A: There are six MMNA graphs on eight vertices or fewer. They are J1, K6,

K3,3,1, P7, K4,4 – e, and P8.

A1 D C C D V1 V5 V3 B E V4 B E A F V2 V6 A F G A 2 Figure 1.1.4. (From left to right) The J1 Graph, the K6 Graph, and the K3,3,1 Graph.

4

C D C D C D

E B E E B G H G B A H F

A F A F G

Figure 1.1.5. (From left to right) The P7 Graph, the K4,4 – e Graph, and the P8 Graph

1.2. Map of Thesis

Our goal is to prove theorem A. In this introductory chapter we give a brief overview of the problem and state our theorem. In chapter II, we give definitions and some preliminary constructions. The third chapter is where we prove theorem A in two parts. Part one deals with graphs on seven or fewer vertices and part two with those on eight vertices.

CHAPTER II

DEFINITIONS AND PRELIMINARY

CONSTRUCTIONS

There are four sections in this chapter. In section one we define terms. In section two we prove that there are exactly nineteen non-planar (7,12) graphs of minimum degree at least two. In section three we prove that there are exactly five triangulations on seven vertices. Finally, in section four we will look at Petersen graphs with at most eight vertices. We construct every graph with eight vertices that contains a minor that is a Petersen graph of seven or eight vertices. Also, we will construct a single graph on eight vertices with a K6 minor. In each case, we also construct the complement.

2.1. Definition of Terms

Definition 2.1.1. Graph – A graph G = (V,E) is a set of vertices V and a set of edges E.

The edges are pairs of vertices {A,B}, where A is not equal to B. A given pair of vertices gives rise to at most one edge. We will be using the notation, (A, B), to refer to an edge between vertices A and B so that (A, B) and (B, A) denote the same edge.

We write V = V(G) for the set of vertices and E = E(G) for the set of edges. The

5 6 number of vertices |V| is the order of the graph and |E| is its size. We often refer to a graph as a (|V|,|E|) graph.

Definition 2.1.2. Subgraph – A graph G’ whose vertices and edges form subsets of those of a given graph G.

Definition 2.1.3. Planar Graph – A graph is planar if it can be drawn in the plane without edges crossing.

Theorem 2.1.1. Euler Formula of a Planar Graph – If a finite, connected, planar graph is drawn in the plane without any edge crossings, and V is the number of vertices, E the number of edges, and F the number of faces then V – E + F = 2.

Note: This is called Euler’s formula in honor of the similar formula for polyhedron.

Definition 2.1.4. Complete graph Kn – A complete graph is a graph in which each pair of vertices is connected by an edge. The complete graph is denoted Kn, where n represents the number of vertices in the graph.

Definition 2.1.5. Triangulation – A triangulation is an embedding of a planar graph such that every face is a triangle.

7 Definition 2.1.6. Graph Complement – The complement of a graph G, denoted as Gc; has the same vertex set but whose edge set consists of the edges not present in G.

Definition 2.1.7. Apex Graph – A graph G that is planar or can be made planar by the removal of a single vertex and its edges. The deleted vertex is called an apex of the graph.

Definition 2.1.8. Graph Isomorphism – Let V(G) be the vertex set of a graph and

E(G) its edge set. Then a graph isomorphism from graph G to graph H is a bijection: f: V(G) → V(H) such that (A, B) ∈ E(G) iff (f(A), f(B)) ∈ E(H). If there is a graph isomorphism from G to H, then G is said to be isomorphic to H, written G ≅ H.

Definition 2.1.9. Graph Automorphism – A graph automorphism is an isomorphism of a graph G with itself, i.e., a mapping from V(G) to V(G) that is an isomorphism.

Definition 2.1.10. Isomorphic Vertices – Vertices A and B in V(G) are isomorphic or in the same isomorphism class if there is a graph automorphism f with f(A) = B. An isomorphism class is a collection of vertices isomorphic to each other.

Definition 2.1.11. The Neighborhood of A, N(A) – Let G be a graph and let A be a vertex in G. The neighborhood of A is defined as the set of all v in G that are adjacent to A via an edge, N(A) = {v ∈ V(G)| (v, A) ∈ E(G)}.

8 Definition 2.1.12. Degree of a Vertex – The degree of a vertex v of a graph G is the number of edges incident to v. The degree of a vertex v is denoted as deg(v). The maximum degree of a graph G is denoted by Δ(G), and the minimum degree of a graph G is denoted by δ(G).

Definition 2.1.13. Degree Sequence – A listing of the degrees of vertices in a graph.

We’ll use notation (AB), where A = the degree of a vertex, while B = the number of vertices with that degree.

Example 2.1.1. Consider a graph on seven vertices of degree sequence: (6, 53, 42, 3)

The first entry, 6, represents a single degree six vertex.

The second entry, 53, represents three degree five vertices.

The third entry, 42, represents two degree four vertices.

The fourth entry, 3, represents a single degree three vertex.

This is a total of seven vertices. We usually write the degrees in descending order from left to right.

Definition 2.1.14. Edge Contraction – In a graph G, contraction of an edge e with endpoints A, B is the replacement of A and B with a single vertex V such that the edges incident to V are the edges other than e that were incident with A or B. The resulting graph G’ has one less edge than G. We denote a contraction of the edge (A,

B) in G as G/(A, B).

9 Example 2.1.2. Consider the following graph G, with vertices A, B, C, D, and F. Edge e in graph G has endpoints of A and B.

C D

e A B F

Figure 2.1.1. Graph G.

Let us contract edge e with endpoints A and B and let us construct the resulting graph G’. Contracting edge e results in replacing vertices A and B with vertex V.

C D

V F

Figure 2.1.2. Graph G’.

Definition 2.1.15. Minor – If G is a graph; we say that H is a minor of G if H can be obtained from G by a non-empty sequence of the following three moves: deleting or contracting an edge of G or deleting a vertex of G.

Definition 2.1.16. Minor Minimal (MM) – We say that a graph G is minor minimal with respect to a certain property if G has that particular property, but no minor of G has that property.

10 Definition 2.1.17. Splitting a Vertex – Let G be a graph and v a vertex of G. Splitting v means forming a new graph, G’ by replacing v with two new vertices, v1 and v2, that are connected by a single edge. Vertices adjacent to v in G are adjacent to v1 or v2 in G’.

Note: This is the inverse of an edge contraction.

Example 2.1.3. Consider graph H with three vertices A, B, and C and two edges:

A C B

Figure 2.1.3. Graph H.

Now let us split vertex B in Figure 2.1.3, which is a degree two vertex.

A C B1 B2

Figure 2.1.4. Graph H with Vertex B Split.

We have now created two new vertices, B1 and B2, which are connected by a single edge, to replace the original vertex B. Since B was a degree two vertex where one edge connected B to A and the second edge connected B to C, when we split B these connections are preserved and as you see in Figure 2.1.4 there is an edge connecting

B1 to A and a second edge connecting B2 to C. We could have switched the positions of B1 and B2 and had B2 connect to A and B1 connect to C. Another option is to connect A and C to B1 and have B2 be a degree 1 vertex with B1 as its only neighbor.

11 Example 2.1.4. The degree three vertex B.

Consider the following where B is a degree three vertex.

A C B

D

Figure 2.1.5. Graph J.

Now let us split vertex B. Vertex B will be split into two vertices, B1 and B2, which are connected by a single edge, but how do we align the remaining vertices A, C, and D?

Since B is a degree three vertex we have options on how we can split B. Those options are to split B into (0,3) or (1,2). But even within each split, (0,3) or (1,2), we find further options on how to align vertices A, C, and D. Consider splitting B into a

(1,2) configuration. Splitting B into (1,2) offers a few possibilities since B1 is connected to “1” of the vertices A, C, or D with an edge, and B2 is connected to the remaining “2” vertices of {A, C, D} with two edges. Please note that splitting the vertex into (1,2) is the same as splitting the vertex into (2,1). An example of splitting

B into (1,2) is having B1 connected to A and B2 adjacent to C and D.

A C B1 B2

D

Figure 2.1.6. Graph J with a (1,2) Split of Vertex B.

Another option for splitting B into (1,2), is having “1” be the connection between B2 and C and “2” the connections between B1 and both A and D.

12

A C B1 B2 D

Figure 2.1.7. Graph J with a Second (1,2) Split of Vertex B.

Or another possibility is “1” connection, B2 to D and “2” edges, B1 to A and C.

A D B1 B2 C

Figure 2.1.8. Graph J with a Third (1,2) Split of Vertex B.

As you can see, a simple (1,2) split of vertex B yields a few options on how to align the remaining vertices A, C, and D with the two new vertices, B1 and B2.

Example 2.1.5. The degree four vertex B, which we split into (0,4).

Consider the following where B is a degree four vertex.

C B A D

E

Figure 2.1.9. A Degree Four Vertex, B.

Now let us split B into a (0,4) configuration. Splitting B into (0,4) would construct a graph, where “0” connects B1 to no other vertex besides B2 and the “4” refers to B2 connecting with all four vertices A, C, D, and E.

13 A

C B1 B2

D

E

Figure 2.1.10. A Degree Four Vertex, B, Split into (0,4).

This is a perfectly acceptable split of the vertex B, as is any (0,Y) split where Y equals the degree of the vertex being split. However, we will usually require graphs to have minimum degree two. We can ensure that by only splitting a vertex into (X,Y) where

X + Y = the degree of the vertex and both X, Y ≥ 1.

Regardless of the degree of the vertex, two new vertices that are connected by a single edge replace it. Also, it is necessary to specify what you are going to be splitting your vertex into and how you will align the adjacent vertices to the vertex you are splitting. As in the examples above, it is often convenient to do this simply by drawing the resulting graph.

2.2. Non-planar (7, 12) graphs

In this section we prove:

Theorem 2.2. There are exactly nineteen distinct non-planar (7,12) graphs with

δ(G) > 2.

The proof requires four lemmas.

14 Lemma 2.2.1. There is exactly one non-planar (6,10) graph with δ(G) > 2.

Proof. By Kuratowski, a non-planar graph has a K5 or K3,3 minor. This means, a non- planar (6,10) has a non-planar minor formed by deleting or contracting an edge, or deleting a vertex. Since K5 is a (5,10) graph, the only possibility is to add an edge to the (6,9) graph K3,3.

V3 W3

W2 V2

V1 W1 Figure 2.2.1. The K3,3 Graph.

For any pair of vertices A, B of K3,3, there is an automorphism with f(A) = B, so all vertices are isomorphic. Since all vertices in a K3,3 are isomorphic there is exactly one non-planar (6,10) graph with δ(G) > 2. 

V3 W3

W2 V2

V1 W1 Figure 2.2.2. The Only Non- Planar (6,10) Graph with δ(G) > 2.

15 Lemma 2.2.2. There are exactly three non-planar (6,11) graphs with δ(G) > 2.

Proof. To construct a non-planar (6,11) graph we either add an edge to a non-planar

(6,10) graph or split a vertex in a non-planar (5,10) graph. An edge could be added to a (6,10) graph with δ(G) = 1, however no such graph exists (Mattman 5). The only

(6,10) graph available to us is the one in Figure 2.2.2. This graph has three isomorphism classes: {V1, V2}, {V3}, and {W1, W2, W3}. Since not all vertices are isomorphic we have options for where we add this edge. Suppose we take a vertex from {V1, V2}, WLOG V1, and connect it to the only vertex it is not connected to, V3.

This is our first non-planar (6,11) graph.

V3 W3

W2 V2

V1 W1 Figure 2.2.3. First Derived Non- Planar (6,11) Graph.

If we connect two vertices from {W1, W2, W3} with an edge, WLOG let us connect W1 and W3, we construct a second non-planar (6,11) graph.

16

V3 W3

W2 V2

V1 W1 Figure 2.2.4. Second Derived Non- Planar (6,11) Graph.

The other option is to split a vertex of K5, the non-planar (5,10) graph. All five vertices lie in a single isomorphism class: {V1, V2, V3, V4, V5}.

V2 V3

V1

V4

V 5 Figure 2.2.5. The K5 Graph.

When we split a vertex in a graph we replace an existing vertex with two vertices that are connected by a single edge. The two new vertices will be labeled U1 and Vi (I

= 1, 2, 3, 4, or 5), where Vi refers to the vertex that is being split. Since the vertices are isomorphic suppose we split V1. The vertex V1 is a degree four vertex, so there are only two options in how to split V1 and still produce a graph of minimum degree

17 two. We will either split the vertex into (2,2) or (1,3). Let us begin by splitting V1 into (2,2) where the first “2” refers to the edges between U and, WLOG V2 and V3, and the second “2” refers to the edges between V1 and both V4 and V5.

V2 V3

U1

V1

V4

V 5 Figure 2.2.6. Non-Planar (6,11) Graph Derived from K5.

Figure 2.2.6 shows a non-planar (6,11) graph; however it is one that we have already derived. Compare the complements of this (6,11) graph with our second

(6,11) graph from Figure 2.2.4 and you will find that they are the same. Since the complements are the same, the original graphs that created the complements are the same.

18

V2 V3 V2 V3 U1 U U V2 V3 V1 V1

V4 V4 V5 V4 V1

V5 V5

W3 V3 W3 V3 V3 W3 W1

W V2 2 W2 V2

V V1 2 W2 V1 W1 V1 W1 Figure 2.2.7. Comparing Two Non-Planar Graphs. (Complements are in orange).

Let us go back to the K5 graph and split V1 into (1,3). Let “1” connect U to, WLOG V2, and “3” connect V1 to V3, V4, and V5. This split creates another unique non-planar

(6,11) graph.

V2 V3 U1

V1

V4

V 5 Figure 2.2.8. Third Derived Non- Planar (6,11) Graph.

19

By adding an edge to the non-planar (6,10) graph or splitting a vertex of K5 we find three distinct non-planar (6,11) graphs. 

Lemma 2.2.3. There is exactly one non-planar (7,10) graph with δ(G) > 2.

Proof. To construct a non-planar (7,10) graph with δ(G) > 2 we must either add an edge to a non-planar (7,9) graph with δ(G) > 1 or split a vertex of a non-planar (6,9) graph with δ(G) > 2. However, there is no such (7,9) graph (Mattman 4). So split a vertex of the non-planar (6,9) graph, K3,3. Recall that the vertices in a K3,3 are isomorphic, so suppose we split V1. The degree three vertex V1 can only be split into

(1,2) and still produce a graph of minimum degree two. Let “1” connect U to, WLOG

W1, and “2” connect V1 to W2 and W3. Then this is the only non-planar (7,10) graph with δ(G) > 2. 

V3 W3

W2 V2

V1 U W1 Figure 2.2.9. The Only Non-Planar (7,10) Graph with δ(G) > 2.

Lemma 2.2.4. There are exactly five non-planar (7,11) graphs with δ(G) > 2.

20 Proof. Each of the five non-planar (7,11) graphs can be attained by adding an edge to a non-planar (7,10) graph, splitting a vertex of a non-planar (6,10) graph, or both methods. Although there may be two or more ways to construct a (7,11) graph, here

I will describe a single method for each of the five graphs. The non-planar (6,10) graph from Figure 2.2.2 has isomorphism classes: {V1, V2}, {V3}, and {W1, W2, W3}.

We split vertices to form a non-planar (7,11) graph. We replace vertex Vi or Wi (I =

1, 2, or 3) with U and Vi or Wi. Suppose we split a vertex from {V1, V2}, WLOG V1.

Since deg(V1) = 4, V1 can only be split into (2,2) or (1,3) and still produce a graph with δ(G) > 2. Let us begin by splitting V1 into (1,3). Let “1” connect U to a vertex from {W1, W2, W3}, WLOG W1, and let “3” connect V1 to V2, W2, and W3. This creates our first non-planar (7,11) graph.

V3 W3

W2 V2

U V1 W1 Figure 2.2.10. First Non-Planar (7,11) Graph.

There is another (1,3) split of V1. Let “1” connect U to V2 and let the “3” connect V1 to all three vertices in {W1, W2, W3}. This produces a second non-planar (7,11) graph.

21

V3 W3

W2 V2

U

V1 W1 Figure 2.2.11. Second Non-Planar (7,11) Graph.

Now let us split V1 into (2,2). Let the first “2” connect U to a vertex from {W1, W2,

W3}, WLOG W1, and to V2. Let the second “2” connect V1 to W2 and W3. This is our third non-planar (7, 11) graph.

V3 W3

W2 V2

V1 U W1 Figure 2.2.12. Third Non-Planar (7,11) Graph.

The remaining two (7,11) graphs come from adding an edge to the non-planar

(7,10) graph. The isomorphism classes of the (7,10) graph of Figure 2.2.9 are: {V2,

V3, W2, W3}, {V1, W1}, and {U}. Let us connect the two elements in the isomorphism class {V1, W1} with an edge to create our fourth non-planar (7,11) graph.

22

V3 W3

W2 V2

U

V1 W1 Figure 2.2.13. Fourth Non-Planar (7,11) Graph.

Connecting two vertices from the isomorphism class {V2, V3, W2, W3}, WLOG V2 and

V3, finds our fifth non-planar (7,11) graph.

V3 W3

W2 V2

V1 U W1 Figure 2.2.14. Fifth Non-Planar (7,11) Graph.

The only way to construct a non-planar (7,11) graph of δ(G) > 2 is to either split a vertex of a non-planar (6,10) graph of δ(G) > 2, or to add an edge to a non-planar

(7,10) graph of δ(G) > 1. There is exactly one such (6,10) graph and exactly one such

(7,10) graph (Mattman 4). By systematically going through all possible splits and

23 edge additions, we found exactly five (7,11) graphs, even though only one procedure per construction was given above. 

Theorem 2.2. There are exactly nineteen distinct non-planar (7,12) graphs with

δ(G) > 2.

Proof. The only way to construct non-planar (7,12) graphs with δ(G) > 2 is to either add an edge to a non-planar (7,11) graph with δ(G) > 1 or to split a vertex of a non- planar (6,11) graph with δ(G) > 2. We found the (6,11) and (7,11) graphs in lemmas

2.2.2 and 2.2.4. Let us first split a vertex of the non-planar (6,11) graph from Figure

2.2.3. The isomorphism classes of vertices are: {V1}, {V2, V3}, and {W1, W2, W3}. We replace the existing vertex Vi or Wi (I = 1, 2, or 3) with two vertices labeled U and either Vi or Wi. First split a vertex from {W1, W2, W3}, WLOG W1. To produce a graph with δ(G) > 2, vertex W1 must be split into (1,2). For example the “1” could connect

U to V1 and the “2” connect W1 to {V2, V3}. This is our first (7,12) graph:

V3 W3

W2 V2

V1 U W1 Figure 2.2.15. First (5, 42, 33, 2) Graph.

24 For each example of a non-planar (7,12) graph, we will show it in a Figure, give it a number, and give its degree sequence as above. There is another way to split W1 into (1,2): U is connected to a vertex from {V2, V3}, WLOG V2, and W1 is adjacent to

V1 and V3. This is another (7,12) graph:

V3 W3

W2 V2

U

V1 W1 Figure 2.2.16. Second (5, 42, 33, 2) Graph.

Suppose we take a vertex from the next isomorphism class {V2, V3}, WLOG V2, and split it. The degree four vertex V2 can be split into (2,2) or (1,3). First split it into

(1,3) where “1” connects U to V1 and “3” connects V2 to {W1, W2, W3}. This split creates our third (7,12) graph:

V3 W3

W2 V2

U

V1 W1 Figure 2.2.17. Third (5, 42, 33,2) Graph.

25

We can split V2 into (1,3) a second way. Let “1” be the edge that connects U to a vertex from {W1, W2, W3}, WLOG W1, and let “3” connect V2 to V1, W2, and W3. But this recreates the graph from Figure 2.2.16 on our list. Next split V2 into (2,2), where the first “2” connects U to V1 and an element of {W1, W2, W3}, WLOG W1, and the second “2” connects V2 to W2 and W3. This is our fourth (7,12) graph:

V3 W3

W2 V2

U

V1 W1 Figure 2.2.18. The (5, 4, 35) Graph.

There remains one isomorphism class: {V1}. The degree five vertex V1 can be split into (1,4) or (2,3). First split V1 into (2,3). Let “2” connect U to a vertex each from

{W1, W2, W3} and {V2, V3}, WLOG W1 and V2, and “3” connect V1 to V3, W2, and W3.

This creates a fifth (7,12) graph:

V3 W3

W2 V2

V1 U W1 Figure 2.2.19. First (43, 34) Graph.

26

But there is another (2,3) split of V1. Let “2” connect U to both elements of {V2, V3} and let “3” connect V1 to all the elements of {W1, W2, W3}. This gives our sixth (7,12) graph:

V3 W3

W2 V2

U

V1 W1 Figure 2.2.20. Second (43, 34) Graph.

There is yet another (2,3) split of V1. Let “2” connect U to two vertices of {W1, W2,

W3}, WLOG W1 and W2, and let “3” connect V1 to V2, V3, and W3. This yields our seventh (7,12) graph:

V3 W3

W2 V2

V1 U W1 Figure 2.2.21. Third (43, 34) Graph.

Next let us split V1 into (1,4). Let “1” connect U to a vertex from {W1, W2, W3}, WLOG

W1, and “4” connect V1 to W2, W3, and the two vertices {V2, V3}. This reconstructs the graph from Figure 2.2.15 on our list. There is another (1,4) split of V1. Have “1”

27 connect U to a vertex from {V2, V3}, WLOG V2, and “4” will connect V1 to V3 and all the elements of {W1, W2, W3}. But this creates the graph from Figure 2.2.17 on our list. This exhausts all the possible splittings of vertices in the first (6,11) graph.

The second (6,11) graph from Figure 2.2.4 has the isomorphism classes:

{V1, V2, W1, W3}, and {V3, W2}. We split vertex Vi or Wi (I = 1, 2, or 3) into two new vertices labeled U and either Vi or Wi. Let’s split a vertex from the isomorphism class

{V3, W2}, WLOG V3. The degree three vertex V3 must be split into (1,2). Let us have

“1” connect U to the vertex in {W2} and “2” connect V3 to the two vertices in {W1,

W3}. This creates our eighth (7,12) graph:

V3 W3

U

W2 V2

V1 W1 Figure 2.2.22. First (44, 32, 2) Graph.

But that is not the only (1,2) split of V3. Let “1” be the edge that connects U to a vertex from {W1, W3}, WLOG W3, and let “2” connect V3 to W1 and W2. This produces our ninth (7,12) graph:

28

U V3 W3

W2 V2

V1 W1 Figure 2.2.23. Second (44, 32, 2) Graph.

Next split a vertex from the isomorphism class {V1, V2, W1, W3}, WLOG let us split V1.

The degree four vertex V1 can be split into either (2,2) or (1,3). First split V1 into

(1,3) where “1” connects U to W2 and “3” connects V1 to V2 and {W1, W3}. This reconstructs the graph from Figure 2.2.23. There is another (1,3) split of V1. Have

“1” connect U to V2 and “3” connect V1 to W2 and the two vertices in {W1, W3}. This creates a tenth (7,12) graph:

V3 W3

W2 V2

U

V1 W1 Figure 2.2.24. Third (44, 32, 2) graph.

29

There is still another (1,3) split of V1. Let “1” connect U to an element of {W1, W3},

WLOG W1, and let “3” connect V1 to V2, W2, and W3. This produces an eleventh (7,12) graph:

V3 W3

W2 V2

U V1 W1 Figure 2.2.25. Fourth (44, 32, 2) Graph.

Now let us split V1 into (2,2). Let the first “2” connect U to V2 and a vertex from {W1,

W3}, WLOG W1, and the second “2” connect V1 to W2 and W3. This constructs our twelfth (7,12) graph:

V3 W3

W2 V2

V1 U W1 Figure 2.2.26. Fourth (43, 34) Graph.

But there is a second (2,2) split of V1. Let the first “2” connect U to W2 and V2, and the second “2” connect V1 to {W1, W3}. This yields our thirteenth (7,12) graph:

30

V3 W3

W2 V2

U

V1 W1 Figure 2.2.27. Fifth (43, 34) Graph.

We now look at the (6,11) graph of Figure 2.2.8 whose isomorphism classes are: {U1}, {V1, V2}, and {V3, V4, V5}. We will replace vertex U1 or Vi (I = 1, 2, 3,

4, or 5) with two vertices labeled U2 and either U1 or Vi (I = 1, 2, 3, 4, or 5). Let’s split a vertex from {V3, V4, V5}, WLOG V3. The degree four V3 can be split into (1,3) or

(2,2). First split V3 into (2,2) where the first “2” connects U2 to {V1, V2}, and the second “2” connects V3 to V4 and V5. This (2,2) split recreates the (7,12) graph from

Figure 2.2.24. There is a second (2,2) split of V3. Let the first “2” connect U2 to a vertex each from {V3, V4, V5} and {V1, V2}, WLOG V5 and V1. Let the second “2” connect V3 to V2 and V4. But this recreates the (7,12) graph from Figure 2.2.25. Now let us take V3 and split it into (1,3) where “1” connects U2 to an element from {V1,

V2}, WLOG V2 and “3” connects V3 to V1, V4, and V5. This creates the fourteenth (7,12) graph:

31

V2 U2 V3 U1

V1

V4

V 5 Figure 2.2.28. First (45, 22) Graph.

There is a second (1,3) split of V3. Let “1” connect U2 to an element from {V3, V4, V5},

WLOG V4 and let “3” connect V3 to V5 and both vertices from {V1, V2}. This yields a fifteenth (7,12) graph:

V2 V3 U1

V1 U2

V4

V5 Figure 2.2.29. Second (45, 22) Graph.

Suppose that we split a vertex from {V1, V2}, WLOG V1. The degree four V1 can split into (1,3) or (2,2). First split V1 into (2,2) where the first “2” connects U2 to two vertices from {V3, V4, V5}, WLOG V4 and V5, and the second “2” connects V1 to U1 and

V3. This reconstructs the graph from Figure 2.2.23. But we can split V1 into (1,3) where “1” connects U2 to an element of {V3, V4, V5}, WLOG V5, and “3” connects V1 to

32

U1, V2, and V3. This recreates the graph from Figure 2.2.28. There is another (1,3) split of V1. Let “1” connect U2 to U1 and let “3” connect V1 to {V3, V4, V5}. This yields our sixteenth (7, 12) graph:

V2 U 1 V3 U2

V1

V4

V 5 Figure 2.2.30. Third (45, 22) Graph.

The final isomorphism class of the third (6,11) graph is {U1} a degree two vertex that can only be split in (1,1). If we let the first “1” connect U2 to V1 and the second

“1” connect U1 to V2 we recreate the graph from Figure 2.2.30.

Remember that we are trying to find the nineteen non-planar (7,12) graphs by either adding an edge to a (7,11) graph, or splitting a vertex of a (6,11) graph. We have split each of the three (6,11) graphs in all possible ways and found sixteen distinct non-planar (7,12) graphs with δ(G) > 2. By lemma 2.2.4, there are five non-planar (7,11) graphs with δ(G) > 2. By adding edges to these five non- planar (7,11) graphs, we will find the remaining three non-planar (7,12) graphs.

Please note that there are four non-planar (7,11) graphs with δ(G) > 1 that we did

33 not construct (Mattman 5). I have checked that adding edges to these four (7,11) graphs yields (7,12) graphs that already appear in the list of nineteen.

When we add edges to the (7,11) graphs we often reconstruct (7,12) graphs previously found. I will not take us through that redundancy; instead I will explain how to find the remaining three (7,12) graphs. First let me begin by stating that both the third and fourth (7,11) graphs from Figures 2.2.12 and 2.2.13, respectively, yield no new non-planar (7,12) graphs. I tried all possible ways of adding an edge and found only (7,12) graphs already on our list. Let us now consider the first (7,11) graph from Figure 2.2.10. Connecting W1 and V1; we find our seventeenth (7,12) graph:

V3 W3

W2 V2

U V1 W1 Figure 2.2.31. Fourth (5, 42, 33, 2) Graph.

Now let us consider the second (7,11) graph from Figure 2.2.11. Connecting V1 and

V2 we construct our eighteenth (7,12) graph:

34

V3 W3

W2 V2

U V1 W1 Figure 2.2.32. The (52, 34, 2) Graph.

Finally in the fifth (7,11) graph from Figure 2.2.14 if we connect W1 and V1 we find our nineteenth (7,12) graph:

V3 W3

W2 V2

U V1 W1 Figure 2.2.33. Fifth (44, 32, 2) Graph.

There are only two methods to construct non-planar (7,12) graphs with δ(G) > 2 and that is to either split a vertex of a (6,11) graph or to add an edge to a (7,11) graph.

We found all the non-planar (6,11) graphs of degree at least two and systematically split all the vertices in all the different cases to find distinct non-planar (7,12) graphs with δ(G) > 2. We found all the non-planar (7,11) graphs of degree at least one and systematically added an edge between vertices in all the different cases to

35 find distinct non-planar (7,12) graphs of degree at least two. By exhausting these two methods we find there are nineteen distinct non-planar (7,12) graphs with δ(G)

> 2. 

2.3. Triangulations on Seven Vertices

The following shows that there are exactly five triangulations on seven vertices. These triangulations will help us determine if a graph on eight vertices is an apex graph or not. A triangulation on seven vertices presents a planar graph with the most edges. An apex graph can be made planar by the removal of a single vertex.

So, if we begin with an apex graph on eight vertices, remove a single vertex, the planar graph that results will be a graph on seven vertices that is equal to or a subgraph of one of the triangulations that we will discover in this section.

In section 2.3 we prove theorem 2.3, there are exactly five triangulations on seven vertices. For this we must prove the following two lemmas. Lemma 2.3.1 proves that every triangulation on seven vertices has exactly fifteen edges and ten faces. Lemma 2.3.2 proves that every vertex v in a triangulation on seven vertices has deg(v) > 3. To prove theorem 2.3 we find all the degree sequences with fifteen edges such that deg(v) > 3. We then test each degree sequence to see if it gives ten faces.

Lemma 2.3.1. Every triangulation on seven vertices has exactly fifteen edges and ten faces.

36 Proof. Consider Euler’s formula V – E + F = 2 for a planar graph where (V), (E), and

(F) represent the number of vertices, edges, and faces. Since each face of a

2E triangulation has three edges and each edge is used in two faces 3F = 2E or F = . 3

Substituting into Euler’s formula:

V – E + F = 2

2E 7 − E + = 2 3

21 – 3E + 2E = 6

-E = -15

E = 15

2E Every triangulation on seven vertices has exactly fifteen edges. Then F = shows 3 every triangulation on seven vertices has exactly ten faces. 

Lemma 2.3.2. Every vertex v in a triangulation on seven vertices has deg(v) > 3.

Proof. Triangulations are made up of triangular faces. A degree zero vertex means the face in which it appears is not a triangle. A degree one vertex in a triangular face means the other two vertices of the face are the same and joined by a loop, which is not allowed in a graph. The only triangulation with a degree two vertex is the one on three vertices with only two faces. Since no vertex can be of degree zero, one, or two, the minimum degree a vertex in a triangulation on seven vertices is three. 

37 Theorem 2.3. There are exactly five triangulations on seven vertices.

Proof. In a triangulation on seven vertices, the highest possible degree is six, and by lemma 2.3.2, the least a degree can be is three. We now look at different cases based on the number of degree three vertices.

Case 1: Let us assume there are no degree three vertices. There are only two degree sequences with seven vertices, fifteen edges, and 4 < deg(v) < 6: (52, 45) and (6, 46).

To construct a (52, 45) graph, first consider a single degree five vertex.

B C

A D

F E

Figure 2.3.1. A (5, 15) Graph.

Let us assume that our seventh vertex, G, is our second degree five vertex. This will allow us to efficiently create this (52, 45) graph. Note that there are a few different approaches to constructing this graph and those that follow but I will make assumptions during the constructions in order to simplify my explanation. However, they do mean this argument is more a sketch than a complete proof. In particular, a graph with adjacent degree five vertices will not be planar, but we omit the proof.

38 Since deg(A) = 5, the second degree five vertex G must be connected to vertices B, C, D, E, and F.

B C

G A D

F E

Figure 2.3.2. A (52, 25) Graph.

We have our two degree five vertices and now turn to the five degree four vertices.

Which must be B, C, D, E, and F. In order for B to have degree four and to avoid edge crossings, B must be connected to C and F.

B C

A G D

E F

Figure 2.3.3. A (52, 4, 32, 22) Graph.

Similarly, the only way to bring C, D, E, and F up to degree four is by adding edges as in Figure 2.3.4. This is a triangulation on seven vertices that is a (52, 45) graph. There

39 are fifteen edges and ten faces, including ΔDGE on the outside of the graph. Although we made some assumptions along the way, we checked that this is the only (52, 45) triangulation.

7

8 B C

3 4 1 2 A G D 5 6 10 E

F 9

Figure 2.3.4. The (52, 45) Triangulation.

Next we construct a (6, 46) graph. First consider a single degree six vertex.

C D

A B E

G F

Figure 2.3.5. A (6, 16) Graph.

Vertices B, C, D, E, F, and G must all have degree four. Connect each vertex to the two adjacent vertices. This is not the only option, however it efficiently brings each vertex to degree three without any edge intersections.

40

C D

A B E

G F

Figure 2.3.6. A (6, 36) Graph.

To bring B to degree four it must be connected to D, E or F. Let us connect B to D and similarly connect G to E.

C D

A B E

G F

Figure 2.3.7. A (6, 44, 32) Graph.

To form a (6, 46) graph, we must connect C to F. However there is no way to do this and not intersect an existing edge. Also BDEG is a quadrilateral and cannot be a face of a triangulation.

41

C D

A E B

G F

Figure 2.3.8. The (6, 46) Graph is Not a Triangulation.

We have verified that a (6, 46) graph cannot be a triangulation, but we omit further details here. The (52, 45) graph of Figure 2.3.4 is the only triangulation on seven vertices, with no degree three vertex.

Case 2: Assume we have exactly one degree three vertex. The degree sequences with seven vertices, fifteen edges, and exactly one degree three vertex are, (53, 43, 3) or (6, 5, 44, 3).

To construct a (53, 43, 3) graph, start with a degree five vertex.

C B

A D

F E

Figure 2.3.9. A (5, 15) Graph.

42 Assume our seventh vertex is the degree four G and is connected to B, D, E, and F.

C B

G A D

F E

Figure 2.3.10. A (5, 4, 24, 1) Graph.

The most efficient way to raise each vertex’s degree is to connect to nearby vertices.

Since deg(A) = 5 it requires no more edges, nor will our degree four vertex G.

B C

G A D

F E

Figure 2.3.11. A (5, 43, 33) Graph.

We are missing two degree five vertices. Connect D to B so deg(D) = 5.

43

B C

G A D

F E

Figure 2.3.12. A (52, 43, 32) Graph.

Now we are missing one degree five vertex. Connecting B to F makes deg(B) = 5.

With this final edge construction we have a (53, 43, 3) graph. In fact, this is the only

(53, 43, 3) triangulation although we omit further details.

7 B C

10 3 4 G 1 2 A D 5 6

F E 8

9

Figure 2.3.13. The (53, 43, 3) Triangulation.

To construct a (6, 5, 44, 3) graph, start with a degree six vertex.

44 C D

A B E

G F

Figure 2.3.14. A (6, 16) Graph.

Again, construct edges between adjacent vertices.

C D

A B E

G F

Figure 2.3.15. A (6, 36) Graph.

Suppose B is our single degree five vertex. Then B is connected to two of the three D,

E, and F, say D and F.

45

C D

A B E

G F

Figure 2.3.16. A (6, 5, 42, 33) Graph.

Finally, to construct a (6, 5, 44, 3) graph we must connect two of the three degree three vertices. However no matter which you try (C to E, E to G, or G to C) that edge intersects an already existing edge.

C D

A B E

G F

Figure 2.3.17. The (6, 5, 44, 3) Graph is Not a Triangulation.

Although we omit further details here, we checked that there are no (6, 5,

44, 3) triangulations. The only triangulation with a single degree three vertex is the

(53, 43, 3) graph from Figure 2.3.13.

46 Case 3: Assume we have exactly two degree three vertices. There are three degree sequences that have seven vertices, fifteen edges, exactly two degree three vertices, and no degree greater than six vertex: (62, 43, 32), (6, 52, 42, 32), or (54, 4, 32).

To construct a (62, 43, 32) graph, start with a degree six vertex and construct edges between adjacent vertices.

C D

A B E

G F

Figure 2.3.18. A (6, 36) Graph.

Assume vertex B is our second degree six vertex.

C 7 D

2 10 1 A 3 B E 6 4 5 F G 8

9

Figure 2.3.19. The (62, 43, 32) Triangulation.

47 Constructing the edges connecting B to D, B to E, and B to F creates a (62, 43, 32) graph that satisfies all our criteria. We have verified that this is the only (62, 43, 32) triangulation.

To construct a (6, 52, 42, 32) graph, start with a degree six vertex and construct edges between adjacent vertices.

C D

A B E

G F

Figure 2.3.20. A (6, 36) Graph.

Assume that B is a degree five vertex and construct an edge from B to E and from B to F.

C D

A B E

G F

Figure 2.3.21. A (6, 5, 42, 33) Graph.

Now let us assume the second degree five vertex is E. We must connect E to C.

48

C 7 D

10 2 1 A 3 B E 6 4 5

G 8 F 9

Figure 2.3.22. The (6, 52, 42, 32) Triangulation.

We have verified that this is the only (6, 52, 42, 32) triangulation.

To construct a (54, 4, 32) graph, start with a degree five vertex A, and construct a seventh vertex G adjacent to B and F.

B C

G A D

F E

Figure 2.3.23. A (5, 35, 2) Graph.

Assume that B, D, and F are all degree five vertices. Add edges from B to D, D to F, and F to B.

49

B C

G A D

F E

Figure 2.3.24. A (54, 32, 2) Graph.

To complete our construction of the (54, 4, 32) graph we must construct an edge connecting C or E to G. However there is no way to construct an edge between C and

G or E and G. In fact there is no (54, 4, 32) triangulation.

B C

G A D

F E

Figure 2.3.25. The (54, 4, 32) Graph is Not a Triangulation.

Although (54, 4, 32) graphs are not triangulations, we were able to construct triangulations for the other two degree sequences (62, 43, 32) and (6, 52,

42, 32).

50 Case 4: Assume we have exactly three degree three vertices. There is only one degree sequence with seven vertices, fifteen edges, exactly three degree three vertices and no degree greater than six vertex, (6, 53, 33).

To construct a (6, 53, 33) graph, begin with a degree six vertex and construct edges between adjacent vertices.

C D

A B E

G F

Figure 2.3.26. A (6, 36) Graph.

Assume that the degree five vertices are B, D, and F and construct edges that connect these three vertices with each other. With these edges we finish our (6, 53,

33) graph. In fact, this is the only (6, 53, 33) triangulation.

C 7 D

2 10 1 3 A E B 6 4 8 5

G 9 F

Figure 2.3.27. The (6, 53, 33) Triangulation.

51 Case 5: Assume we have more than three degree three vertices. There is only one degree sequence with seven vertices, fifteen edges, at least four degree three vertices and no degree greater than six vertex, (63, 34).

To construct a (63, 34) graph, start with a degree six vertex.

C D

A B E

G F

Figure 2.3.28. A (6, 16) Graph.

Assume a second degree six vertex is B and construct edges connecting B to vertices

C, D, E, F, and G.

C D

A B E

G F

Figure 2.3.29. A (62, 25) Graph.

52 Now assume a third degree six vertex is E and construct edges connecting E to vertices C, D, F, and G.

C D

A B E

G F

Figure 2.3.30. The (63, 34) Graph is Not a Triangulation.

Constructing the third degree six vertex requires edges that intersect already existing edges. Although we omit further details, we checked that no graph with fifteen edges, seven vertices, at least four degree three vertices, and no vertex of degree greater than six can be a triangulation.

We have exhausted all cases in trying to find triangulations for graphs that must have seven vertices, ten faces, fifteen edges, and vertices of at least degree three and at most degree six. We found exactly five that satisfy the criteria, they have the degree sequence shown in Table 2.3.1. 

Table 2.3.1. The Five Triangulations a. (52, 45) b. (53, 43, 3) c. (62, 43, 32) d. (6, 52, 42, 32) e. (6, 53, 33)

53 We now construct the complements of each triangulation found above.

B B B B C C C G E

A A G A D G D G D D C E E E A F F F F

Figure 2.3.31. (52, 45) Graph and its Complement.

As you can see above, we fill in the (52, 45) graph with edges so that all vertices are all of maximum degree. These new edges, which are in orange, create the complement to the (52, 45) graph, which we will call “Pentagon – Segment”.

B G E

D

C A F

Figure 2.3.32. The “Pentagon – Segment” Graph.

B B C D B C B C

E C F G A D G A D G A D G

F E F E F E A

Figure 2.3.33. (53, 43, 3) Graph and its Complement.

54

Using the same procedure we construct the complement of the (53, 43, 3) triangulation, called “Thin – Y”.

B D

E C F

G

A

Figure 2.3.34. The “Thin – Y” Graph.

E C D C D C D A G C A A A B E B E B E B

D F G F G F G F

Figure 2.3.35. (62, 43, 32) Graph and its Complement.

The (62, 43, 32) triangulation has two vertices that are of maximum degree six, A and

B. In the complement A and B are of degree zero. Due to the graph created by vertices C, D, E, F, and G, we will refer to this as the “House” graph.

55 E

A G C B

D F

Figure 2.3.36. The “House” Graph.

D C D C D C D F

A A E A E E G C B B B E A

G F G F G F B Figure 2.3.37. (6, 52, 42, 32) Graph and its Complement.

The (6, 52, 42, 32) triangulation has a single maximum degree vertex, vertex A, and in the complement A has degree zero. The complement will be referred to as the “Fat –

Y” graph.

D F

G C E A

B

Figure 2.3.38. The “Fat – Y” Graph.

56

C 7 D C D C D A 10 2 F C 1 A 3 A A B E B E B E 6 4 5 B E D G G 8 F G F G F 9

Figure 2.3.39. (6, 53, 33) Graph and its Complement.

The (6, 53, 33) triangulation will also have a degree zero vertex A in the complement and is called the “Hat” graph.

A F C

B E D G

Figure 2.3.40. The “Hat” Graph.

The complements of the five triangulations are as follows: a is called “Pentagon –

Segment”, b is called “Thin – Y”, c is called “House”, d is called “Fat – Y”, and e is called “Hat”.

a.) b.) c.) d.) e.)

B B E D F A G E D F C A D E F G C C G C B B E C E D G A G F A D F B A

Figure 2.3.41. A Listing of Complements of the Five Triangulations on Seven Vertices.

57 2.4. Petersen Graphs and their Complements

In this section we construct all possible complements of the Petersen graphs on eight, seven, and six vertices. But first let us look at all seven Petersen graphs.

D C D C D C

E B E B E B G

A F

A A F F G Figure 2.4.1. (From Left to right) The K6 Graph, the K3,3,1 Graph, and the P7 Graph.

D C C D C D

B E H G E B E B I A F H G A F A H F G

Figure 2.4.2. (From Left to right) The K4,4 – e Graph, the P8 Graph, and the P9 Graph.

58

H A F G

I D J C B E

Figure 2.4.3. The Petersen Graph.

As we show below in section 3.1, the Petersen graphs are MMNA and therefore useful in proving theorem 2. We compile a list of graphs on eight vertices with

Petersen graph minors. If graph G has one of these graphs on eight vertices as a subgraph, then G has a Petersen graph minor. Although G is NA, having a Petersen graph minor implies it is not MM (unless G is itself a Petersen graph).

Here is how we construct graphs on eight vertices that have P8, K4,4 – e,

K3,3,1, P7, or K6 as a minor and construct their complements. Both the P8 and K4,4 – e

Petersen graphs are already graphs on eight vertices, so constructing their complements is done directly. However, in order to construct a graph on eight vertices that either has a K3,3,1 or P7 graph as a minor, we begin by taking K3,3,1 or P7 and splitting a vertex or adding a vertex. This leads to multiple graphs (and complements) on eight vertices that have either a K3,3,1 or P7 graph as a minor. For

K6 we require only one complement on eight vertices formed by splitting a vertex and then adding a vertex.

59

Start with the Petersen graphs on eight vertices, P8 and K4,4 – e. The following shows the P8 graph, the P8 graph together with all edges in K8, and finally just the complement of P8.

C D C D C D H

D B E B E B E C B E

A F A F A F F A H H H G G G G Figure 2.4.4. The Petersen Graph P8 and its Complement.

The following shows the complement of K4,4 – e.

C D C D C D

B E E E C F H G G H H G H G A E B B B D

A F A A F F Figure 2.4.5. The Petersen Graph K4,4 – e and its Complement.

There are two Petersen graphs on seven vertices: P7 and K3,3,1.

C D C D C D

C D E B G B G B G B G E E E F A

A F A A F F Figure 2.4.6. The Petersen Graph P7 and its Complement.

60

C D B G E F A

Figure 2.4.7. The Complement of the P7 Graph.

We make P7 an eight vertex graph by adding an eight vertex, X.

C D

E B G X

A F

Figure 2.4.8. The P7 Graph Plus a Degree Zero Vertex, X.

The complement of P7 with vertex X is a graph on eight vertices that has a P7 complement as a subgraph.

C E B X D G

F A

Figure 2.4.9. The P7 Complement Plus a Vertex X of Full Degree.

61

Another way to form an eight vertex graph is by splitting a vertex of P7.

The isomorphism classes of P7 are: {A, B, C}, {D, E, F}, and {G}.

C D

E B G

A F

Figure 2.4.10. The Petersen Graph P7.

We will split a vertex replacing it with vertices labeled U and A, B, C, D, E, F, or G according to the vertex split. First split a vertex from {A, B, C}, WLOG A. Since deg(A)

= 5, it can be split into (1,4) or (2,3). A (0,5) split is the same as adding a degree one vertex, and is included in the previous case. Consider a (1,4) where “1” connects U to a vertex from {B, C}, WLOG B, and “4” connects A to C, D, E, and F:

C D C B D E B G E U G A U F

A F Figure 2.4.11. The First Graph on Eight Vertices Derived from the P7 Graph, and its Complement.

62 But that is not the only way to split A into (1,4). Let “1” connect U to a vertex from

{D, E, F}, WLOG F, and let “4” connect A to B, C, D, and E:

C D U C E D B G E B G F A

A U F Figure 2.4.12. The Second Graph on Eight Vertices Derived from the P7 Graph, and its Complement.

Now consider splitting A into (2,3). Let “2” connect U to the other two vertices from {A, B, C}, B and C, and let “3” connect A to the three vertices from {D, E,

F}:

C D D C B E U G B G E

A U F

A F

Figure 2.4.13. The Third Graph on Eight Vertices Derived from the P7 Graph, and its Complement.

This is a subgraph of the complement constructed from the K4,4 – e graph, so we will not consider it further. There is another way to split A into (2,3). Let “2” connect U to two vertices from {D, E, F}, WLOG E and F, and let “3” connect A to B, C, and D:

63

C D A G

E B C B G F E

D U

A U F

Figure 2.4.14. The Fourth Graph on Eight Vertices Derived from the P7 Graph, and its Complement.

There is a third way to split A into (2,3). Let “2” connect U to a vertex from each {A,

B, C} and {D, E, F}, WLOG B and F and consider “3” to be the connection between A and C, D, and E:

C D U C

E B G G D E

F A B

A U F

Figure 2.4.15. The Fifth Graph on Eight Vertices Derived from the P7 Graph, and its Complement.

A degree four vertex from {D, E, F}, WLOG F, can be split into (1,3) or

(2,2). Consider splitting F into (1,3) where “1” connects U to a vertex from {A, B, C},

WLOG A, and “3” connects F to B, C, and G. This replicates Figure 2.4.12, and is not a

64 new graph. Let us split F into (1,3) again now with “1” connecting U to G, and “3” connecting F to A, B, and C:

C D U C E A B G D E B U G F A F

Figure 2.4.16. The Sixth Graph on Eight Vertices Derived from the P7 Graph, and its Complement.

Suppose we split F into (2,2). Let the first “2” connect U to two vertices from {A, B,

C}, WLOG A and B, and the second “2” connect F to C and G. This gives a complement that is a sub graph of the P8 complement so we will not use it.

Suppose we split the only degree three vertex in P7, G into (1,2). Let “1” connect U to a vertex from {D, E, F}, WLOG F, and let “2” connect G to D and E. This split reconstructs the graph of Figure 2.4.16.

We have exhausted all the possible ways to split vertices in P7 and have come up with five graphs on eight vertices and their complements. These five complements are neither subgraphs of the P8 complement nor subgraphs of the K4,4

– e complement.

The second Petersen graph of seven vertices is K3,3,1.

65

C D C D C D C D F E B E B E B E B G A A F A F A F

G G G Figure 2.4.17. The Petersen Graph K3,3,1 and its Complement.

As with the P7 graph, let us add an eighth vertex, X, of degree zero to the K3,3,1 graph and construct the complement:

C D F X E

B A G

Figure 2.4.18. The K3,3,1 Complement Plus a Vertex X of Full Degree.

Now just as for P7, consider splitting a K3,3,1 vertex to create a graph on eight vertices. There are two isomorphism classes: {A, B, C, D, E, F} and {G}. When we split a vertex we replace it with two new vertices labeled U and A, B, C, D, E, F, or G depending on the vertex we are splitting. First split a vertex from {A, B, C, D, E, F},

WLOG A. Consider splitting A into (1,3) where “1” connects U to a vertex from {A, B,

C, D, E, F}, WLOG B, and “3” connects A to C, F, and G:

66

C D B A

B E F C E D U A F U

G G

Figure 2.4.19. The First Graph on Eight Vertices Derived from the K3,3,1 Graph, and its Complement.

But that is not the only way to split A into (1,3). Let “1” be the edge that connects U to G, and let “3” connect A to B, C, and F:

C D

D B U A B E F G C E A F U

G Figure 2.4.20. The Second Graph on Eight Vertices Derived from the K3,3,1 Graph, and its Complement.

Now consider splitting A into (2,2). Let the first “2” connect U to the other two vertices from {A, B, C}, B and C, and let the second “2” connect A to F and G. This results in a complement that is a subgraph of the P8 complement, so we will use the

P8 complement instead.

67 Suppose we split the single degree six vertex, G. It can be split into (1,5),

(2,4) or (3,3). Splitting G into (1,5) reconstructs the graph from Figure 2.4.20.

Splitting G into (2,4) creates a complement that is a subgraph of the complement constructed from the P8 graph. Now consider a (3,3) split on G, where the first “3” connects U to three vertices from {A, B, C, D, E, F}, WLOG A, B, and D and the second

“3” connects G to the remaining three vertices in that set:

C D U E

A B E F C D

A F B G

U G Figure 2.4.21. The Third Graph on Eight Vertices Derived from the K3,3,1 Graph, and its Complement.

We have exhausted all the possible ways to split vertices in K3,3,1 and have come up with three additional graphs on eight vertices.

The remaining Petersen graph is K6.

68

C D

B E

A F

Figure 2.4.22. K6 Graph.

Since K6 only has six vertices we need to add two more vertices in order to construct a graph of eight vertices. We can split a vertex or add a degree zero vertex to construct a graph on seven vertices and then repeat the construction to complete a graph on eight vertices. There are a few different combinations that include one or both of these constructions. Rather than make a complete list we describe one graph with K6 minor that was useful in proving theorem 2.

We split A into (1,4) where “1” connects U to F and “4” connects A to B, C,

D, and E.

C D

B E

A U F

Figure 2.4.23. A Graph on Seven Vertices Derived from the K6 Graph.

69 Now add an eighth vertex X of degree zero:

C D

B E

A U F

X

Figure 2.4.24. A Graph on Eight Vertices Derived from the K6 Graph.

Here is the complement of the graph of eight vertices from Figure 2.4.24:

B

A X C U E F

D

Figure 2.4.25. The Complement Derived from a Graph of Eight Vertices that has a K6 Minor.

There are other complements that one can derive from K6 with eight vertices, but this is the only one we need.

70 As a recap, we found one complement each from the two Petersen graphs on eight vertices P8 and K4,4 – e, six complements from the Petersen graph P7 on eight vertices, four complements from the Petersen graph K3,3,1 on eight vertices, and one complement from a K6 graph on eight vertices. They are illustrated below:

a.) P8 complement b.) K4,4 – e complement

H C F G H A E D C B E B D

F A G Figure 2.4.26. Complements Constructed from the Two Petersen Graphs of Eight Vertices.

a.) b.) c.)

C B U A G D C E D U G A F B E B E C F G F D U A d.) e.) f.)

U U C C E C B X A D G D E D G E B

B G F A F A F

Figure 2.4.27. Complements Created from the P7 Graph on Eight Vertices.

71 a.) b.) c.) d.)

B A D U E C D B X U A F E F G F C E D A F C E C D B A G U B G G

Figure 2.4.28. Complements Created from the K3,3,1 Graph on Eight Vertices.

B

A X C U E F

D

Figure 2.4.29. The Complement Created from the K6 Graph on Eight Vertices.

CHAPTER III

PROOF OF THEOREM A

In this chapter we prove our main theorem. We begin in section 3.1 with some lemmas. In section 3.2 we prove theorem 1: there are no MMNA graphs on seven or fewer vertices besides the Petersen graphs. In section 3.3 we prove theorem 2: there are no MMNA graphs on eight vertices besides J1 and the Petersen graphs. This includes a proof that J1 is MMNA in section 3.3.2. Together, theorem 1 and theorem 2 prove our main theorem, theorem A: There are six MMNA graphs one eight vertices or fewer, they are J1 and the Petersen graphs K6, K3,3,1, P7, K4,4 – e, and

P8.

3.1. Lemmas and Prior Results

Lemma 3.1.1. The Petersen graphs are MMNA.

Proof. Let us begin with the K6 graph.

C D

B E

A F Figure 3.1.1. The K6 Graph.

72 73

We first show that K6 is NA by removing a vertex and observing that the resultant graph K5 is non-planar.

Now let us show that K6 is MM by showing that the simple minors of K6 are apex. To find the simple minors remove a vertex, remove an edge, or contract an edge of K6.

Again removing a vertex gives K5, which is apex.

Let us continue to find minors of K6, but now let us remove an edge.

Recall that the isomorphism class of vertices in K6 is {A, B, C, D, E, F}. WLOG remove the edge (A, B). Let us call the K6 – (A, B) graph, G. Now remove any vertex besides A or B, for example C, and you will find that the resultant graph is planar.

D C D

E B E B

A F A F

Figure 3.1.2. The Graph K6 – (A, B), on the Left, and G – C on the Right.

Next we contract an edge of K6. WLOG let us contract the edge (A, B).

Then removing any vertex, for example C, will find that the resultant graph is planar.

74

C D D

E E

A/B F A/B F

Figure 3.1.3. The Graph K6/(A, B), on the Left, and G – C on the Right.

Since all K6 – v, K6 – (M, N), and K6/(M, N) are apex then, all minors of K6 are apex graphs. Since K6 is NA, if all minors of K6 are apex graphs then K6 is MMNA.

Similarly the remaining six Petersen graphs are also MMNA. 

Lemma 3.1.2. If G is MMNA and has n vertices, then it has no vertex of full degree n

– 1 unless G is either K6 or K3,3,1.

Proof. Let G have a vertex v of full degree. If G is NA then G – v is non-planar. Sachs

[S] showed G is then intrinsically linked. This means, by Robertson & Seymour [RS]

G has a Petersen graph minor. Since G and the Petersen graphs are MMNA (lemma

3.1.1), G must be a Petersen graph. The graphs K6 and K3,3,1 are the only Petersen graphs with a vertex of full degree. 

Lemma 3.1.3. If G is MMNA and has n vertices, with n > 7, then it has fewer than 4n

– 9 edges.

75 Proof. As per [CMOPRW], a graph on n vertices and 4n – 9 or more edges contains a

K6 minor. By lemma 3.1.1 K6 is MMNA. For G to be MMNA, it can have no proper K6 minor, so G must have fewer than 4n – 9 edges. 

Lemma 3.1.4. The minimal degree of a MMNA graph is at least three.

Proof. If G is NA with a vertex, v, of degree zero, one, or two, then deleting v or contracting one of its edges gives an NA minor. So, G is not MM. 

3.2. Theorem 1

In this section we prove theorem 1, the only MMNA graphs on seven vertices or fewer are the Petersen graphs. This requires two lemmas. Lemma 3.2.1 proves that the only MMNA graph on six vertices or fewer is the Petersen graph, K6.

Lemma 3.2.2 proves that the only MMNA graphs on seven vertices are the Petersen graphs, K3,3,1 and P7.

Lemma 3.2.1. The only MMNA graph on six vertices or fewer is the Petersen graph,

K6.

Proof. By lemma 3.1.1 the Petersen graph K6 is MMNA. Any other graph on six vertices or fewer is a minor of K6 and therefore apex. 

Lemma 3.2.2. The only MMNA graphs on seven vertices are the Petersen graphs,

K3,3,1 and P7.

76

Proof. By lemma 3.1.1 the Petersen graphs, K3,3,1 and P7, are MMNA. Suppose G is

MMNA and not K3,3,1 or P7. By lemma 3.1.2 G has no vertex of full degree, six. Then the maximum degree of a vertex in G is five. If all seven vertices are of degree five

7(5) then there are a total of ⎛ ⎞ or seventeen and a half edges. This is not possible ⎝⎜ 2 ⎠⎟ but shows seventeen edges are the maximum in a graph on seven vertices of maximum degree five or less. The degree sequence (56, 4) meets the criteria and

6 there is one such graph. However, removing V1 shows the (5 , 4) graph is apex.

W3 V3 W3 U V3 U W2 V2 W2 V2

V W W 1 1 1 Figure 3.2.1. The Graph G = (56, 4), on the Left, G – V1 on the Right.

Suppose G has sixteen edges. By lemma 3.1.2, G has a maximum degree of five or less and by lemma 3.1.4 it has a minimum degree of at least three. There are two degree sequences that meet the criteria, (54, 43) and (55, 4, 3). There are three graphs with the degree sequence (54, 43) and two with (55, 4, 3).

77

V3 V3 W V3 W V V3 W3 W3 3 3 3 W3

W W2 W2 V W2 V W2 2 V2 V2 2 2 V2

W V W1 V W1 V1 W1 V1 1 1 1 V1 W1 A A A A A Figure 3.2.2. The Three Graphs on the Left are of Degree Sequence (54, 43), and the Two Graphs on the Right are of Degree Sequence (55, 4, 3).

Case 1: If you look at the first graph of degree sequence (54, 43), and remove vertex

4 3 W1 you will find that the resultant graph is planar, so our first (5 , 4 ) graph is not

MMNA.

A W V3 W3 3 V3 W W2 2 V2

V V2 V1 W1 1 A Figure 3.2.3. The First (54, 43) Graph, on the Left, G – W1 on the Right.

Case 2: If you look at the second graph of degree sequence (54, 43), and remove

4 3 vertex W1 you will find that the resultant graph is planar, so our second (5 , 4 ) graph is not MMNA.

78

V3 W3 A W3

W2 V2 V3

W2 W V1 1 V1 V2 A Figure 3.2.4. The Second (54, 43) Graph, on the Left, G – W1 on the Right.

Case 3: If you look at the third graph of degree sequence (54, 43), and remove vertex

4 3 V1 you will find that the resultant graph is planar, so our third (5 , 4 ) graph is not

MMNA.

V3 W3 V3 W3

W2 W 2 V2 V2

V1 W1 A W1 A Figure 3.2.5. The Third (54, 43) Graph, on the Left, G – V1 on the Right.

Case 4: If you look at the first graph of degree sequence (55, 4, 3), you will notice

5 that it is an expansion of the K6 graph (i.e., K6 is a minor of the (5 , 4, 3) graph). Let us look at K6 and let us split one of its vertices, say V1. Since deg(V1) = 5 we can split it into (1,4) or (2,3). Let us split V1 into (2,3). Then let us take the complement of the

K6 expansion graph on seven vertices.

79

V3 W V3 W3 V3 W3 3 A

W V2 W2 V2 W2 V2 2

W V W V1 A W1 V1 1 1 1 Figure 3.2.6. The K6 Graph (left), the K6 Expansion Graph on Seven Vertices (center), and the Complement of the K6 Expansion Graph on Seven Vertices (right).

Let us look at the first graph of degree sequence (55, 4, 3) and its complement. You will notice that it contains a K6 minor, and is therefore not MMNA.

A V V3 W3 1

V3 W3 W2 V2 V2

W1 W2 V1 W1 A

Figure 3.2.7. The First (55, 4, 3) Graph and its Complement.

Case 5: If you look at the second graph of degree sequence (55, 4, 3), and remove

5 vertex W1 you will find that the resultant graph is planar, so our second (5 , 4, 3) graph is not MMNA.

80

V3 W3 V3 W3

W V2 2 V2 W2

V1 W1 V1 A A

Figure 3.2.8. The Second (55, 4, 3) Graph, on the Left, G – W1 on the Right.

Suppose G has fifteen edges. If the maximum degree is four then there are

7(4) at most ⎛ ⎞ = fourteen edges, so G must have a degree five vertex. If A is a vertex ⎝⎜ 2 ⎠⎟ of degree five in G then G – A is the (6,10) graph (see lemma 2.2.1).

V3 W3

W2 V2

V1 W1 Figure 3.2.9. The Non-Planar (6,10) Graph of Minimum Degree at Least Two.

Adding a degree five vertex A to the (6,10) graph will result in one of three (7,15) graphs. The isomorphism classes of the (6,10) graph are {V1, V2}, {V3}, and {W1, W2,

W3}. If A is adjacent to {W1, W2, W3} then one of two graphs is constructed. One sub case is that both vertices in the isomorphism class {V1, V2} are also adjacent to A.

81

The other sub case is that only one of {V1, V2}, WLOG V1, and V3 are also adjacent to

A.

Sub case 1: V1, V2, W1, W2, W3 ∈N(A).

Sub case 2: V1, V3, W1, W2, W3 ∈N(A).

The third case is if only two vertices in {W1, W2, W3}, WLOG W1 and W2, are adjacent to A. Then all the vertices {V1, V2} and {V3} must also be adjacent to A.

Sub case 3: V1, V2, V3, W1, W2 ∈N(A).

Sub case 1: Let V1, V2, W1, W2, W3 ∈N(A). This gives the P7 graph, which is MMNA.

Sub case 2: Let V1, V3, W1, W2, W3 ∈N(A). Let us construct this graph and call it G.

Then G – W1 is planar, so G is not MMNA.

V3 W3 A W3

V3 W2 V2 V2 W2

V1 W1 V1 A Figure 3.2.10. The Sub Case 2 Graph G on the Left, and G – W1 on the Right.

Sub case 3: Let V1, V2, V3, W1, W2 ∈N(A). Let us construct this graph and call it G.

Then G – V1 is planar, so G is not MMNA.

82

V3 W3 V3 W3

W2 W2 V2 V2

V1 W1 A W1

A Figure 3.2.11. The Sub Case 3 Graph G on the Left, and G – V1 on the Right. Note: K3,3,1 is not constructed here as it has a degree six vertex, see lemma 3.1.2.

Suppose G has fourteen edges. There are two cases to consider. First let us assume that there is a degree five vertex in G. Then G – v is K3,3 so G’s degree sequence is, (5, 45, 3). Removing a degree four neighbor w of the degree three vertex means G – w is a (6,10) graph with a degree two vertex. By lemma 2.2.1 G – w is planar and G is NA.

V3 W3 V3

W2 V2 W2 V2

A

V1 W1 V1 W1

A Figure 3.2.12. The (5, 45, 3) Graph on the Left, and G – W3 on the Right.

83 Our second case is if all vertices have degree four. If we call the (47) graph G then, G

– v is a non-planar (6,10) graph, the graph of lemma 2.2.1 and Figure 3.2.9. Then G is the graph on the left on Figure 3.2.13, so G – W3 is planar and G is apex.

V3 W3 V3

W2 V2 W2 V2 A

V1 W1 V1 W1

A Figure 3.2.13. The (47) Graph on the Left, and G – W3 on the Right.

Suppose G has thirteen edges. If the maximum degree in G is three there

7(3) are at most ⎛ ⎞ edges, so G must have a vertex of degree at least four. Since G is ⎝⎜ 2 ⎠⎟

NA every G – v has at least nine edges, this implies that the maximum degree of G is four. By lemma 3.1.4 we assume that the minimum degree is at least three, so the only possible degree sequence is (45, 32). Then there is a degree four vertex v adjacent to a degree three vertex. Then G – v is a non-planar (6,9) graph with a degree two vertex. Since K3,3 is the only non-planar (6,9), this is a contradiction and

G is not NA.

84

V3 W3 V3

W2 V2 W2 V2

A

V1 W1 V1 W1

A Figure 3.2.14. The (45, 32) Graph on the Left, and G – W3 on the Right.

Suppose G has twelve edges. If the maximum degree in G is three there

7(3) are at most ⎛ ⎞ edges, so G must have a vertex of degree at least four, which we ⎝⎜ 2 ⎠⎟ will call v. Then G – v has eight edges and is planar by Kuratowski, so G is apex. 

Theorem 1. The only MMNA graphs on seven vertices or fewer are the Petersen graphs.

Lemma 3.2.1 and lemma 3.2.2. prove theorem 1.

3.3. Theorem 2

In this section we prove theorem 2, the only MMNA graphs on eight vertices are J1 and the Petersen graphs, P8 and K4,4 – e, through a series of lemmas.

We begin by stating those lemmas. We next prove the theorem and conclude this introduction to Section 3.3 with proofs of lemmas 3.3.1 through 3.3.5. In Section

3.3.1, we prove lemma 3.3.1.1 (stated below) and in Section 3.3.2, we prove theorem

3.3.2: J1 is MMNA.

85 Here are the lemmas we will use to prove theorem 2.

Lemma 3.3.1. If G is MMNA and has eight vertices, then G has at least fifteen edges.

Lemma 3.3.2. The (8,15) MMNA graphs are P8 and K4,4 – e.

Lemma 3.3.3. There are no (8,16) MMNA graphs nor any (8,17) with Δ(G) > 5.

Lemma 3.3.4. There are no (8,22) or (8,21) graphs with Δ(G) < 5.

Lemma 3.3.5. There are no (8,20), (8,19), (8,18), or (8,17) graphs with Δ(G) < 4.

Lemma 3.3.6. There is no G that is MMNA and (8,18) with Δ(G) = 5 or (8,19) with

Δ(G) = 6.

Lemma 3.3.7. There is no G that is MMNA and (8,19) with Δ(G) = 5 or (8,20) with

Δ(G) = 6.

Lemma 3.3.8. There is no G that is MMNA and (8,20) with Δ(G) = 5 or (8,21) with

Δ(G) = 6, besides the J1 graph.

Lemma 3.3.9. There is no G that is MMNA and (8,22) with Δ(G) = 6.

Lemma 3.3.1.1. There is no G that is MMNA and (8,17) with Δ(G) = 5 or (8,18) with

Δ(G) = 6.

These lemmas, together with lemma 3.1.1, which shows that both the K4,4,

- e and P8 graphs are MMNA, and theorem 3.3.2 (proved in Section 3.3.2 below), that shows J1 is MMNA, will allow us to complete the proof of theorem 2.

Theorem 2. The only MMNA graphs on eight vertices are J1 and the Petersen graphs, P8 and K4,4 – e.

86 Proof. Let G’ be an MMNA graph on eight vertices. Let us look at some of the lemmas and prior results in order to determine the number of edges in such a graph. By lemma 3.3.3, G’ has at least seventeen edges. By lemma 3.1.3, if a graph is MMNA and has n vertices then it has less than 4n – 9 edges. Since we know that n = 8 the number of edges is less than 23. These two lemmas allow us to assume that G’ has between seventeen and twenty-two edges. Let A be the vertex of maximum degree in G’. By the lemma 3.1.2, if a graph is MMNA and has n vertices where n > 7, then it has no vertex of full degree n – 1. This allows us to conclude that no vertex, including

A, can be a degree seven vertex. By lemma 3.1.4, the minimal degree of a MMNA graph is at least three. By lemma 3.3.3, if G’ has seventeen edges, then the graph has maximum degree five. This means that there are no (8,17) graphs with a degree six vertex. Using lemmas 3.3.4 and 3.3.5, Table 3.3.1 gives the maximum degree value(s) for vertex A in each of the candidate MMNA graphs on eight vertices.

Table 3.3.1. The Candidate MMNA Graphs Graph Maximum degree value(s) for A (8,22) 6 (8,21) 6 (8,20) 6 or 5 (8,19) 6 or 5 (8,18) 6 or 5 (8,17) 5

Our strategy is to go through the number of edges and size of G’ starting from seventeen through twenty-two. In each case we assume G’ is a MMNA graph of the given size and deduce a contradiction, except when G’ is the J1 graph. In Section

87 3.3.1, we will go through (8,17) and (8,18) graphs arising from (7,12) graphs, from

Chapter II, in some detail as we prove lemma 3.3.1.1. The remaining graphs are discussed at http://www.csuchico.edu/~tmattman/HAAppendix.pdf, where we prove lemmas 3.3.6 through 3.3.9. This shows that there are no MMNA graphs on eight vertices except for K4,4 – e, P8, and J1. Lemma 3.1.1 and theorem 3.3.2 complete the proof by showing those three graphs are, in fact, MMNA. 

We now conclude this introduction of Section 3.3 by proving lemmas

3.3.1 through 3.3.5.

Lemma 3.3.1. If G is MMNA and has eight vertices, then G has at least fifteen edges.

Proof. Let v be a vertex in G. By lemma 3.1.4, δ(G) > 3. If we assume deg(v) > 3, then

G – v has at most three fewer edges than G. However, a graph on seven vertices of minimum degree at least two has at least ten edges [Ma]. So, G must have at least thirteen edges.

Suppose G has thirteen edges. If the maximum degree in G is three then

8(3) all vertices are of degree three and there are ⎛ ⎞ twelve edges. G must have a ⎝⎜ 2 ⎠⎟ degree four vertex. But then G – v has fewer than ten edges and G is apex.

Suppose G has fourteen edges. Again G must have a degree four vertex. If deg(v) > 5 then G – v has fewer than ten edges and G is apex. So, the maximum degree of a vertex in G is four. Let deg(v) = 4 then G – v is the non-planar (7,10) graph of lemma 2.2.3.

88

V3 W3

W2 V2

V1 U W1

Figure 3.3.1. The Non-Planar (7,10) Graph with δ(G) > 2.

Take the (7,10) graph from Figure 3.3.1 and add a seventh vertex A of degree four.

We must determine N(A). If we consider G – V2 we will find that it is planar unless both W2 and W3 are also adjacent to A.

a.) G – V2 b.) G – V2

V3 V3 W3

W3 W2 A W2 A

W V U W V1 U 1 1 1

Figure 3.3.2. The (7,10) Graph with Vertex A Added, has Vertex V2 Removed and is Found Planar in Two Different Constructions.

But now consider G – W2. The resultant graph is planar and G is apex.

89 a.) G – W2 b.) G – W2

W3 V3 W3 A V3 V2 A V2

V U W1 V U W 1 1 1

Figure 3.3.3. The (7,10) Graph with Vertex A Added, has Vertex W2 Removed and is Found to be Planar in Two Different Constructions.

No graph on eight vertices and either thirteen or fourteen edges is

MMNA. So, if G is MMNA then it has at least fifteen edges. 

Lemma 3.3.2. The (8,15) MMNA graphs are P8 and K4,4 – e.

Proof. Let G be an (8,15) MMNA graph. If there were a vertex of degree six or more, then G – v has at most nine edges and cannot be planar (for example, see [Ma]: there is no (7,9) graph with δ(G) > 2). So G has degrees ranging from three to five. Suppose there's a degree five vertex A. Then G – A is a non-planar (7,10) graph with δ(G) > 2.

By lemma 2.2.3, there is only one such graph, shown in Figure 3.3.1. Since U has degree three in G, it is a neighbor of A. Then G – V2 is planar unless W2, W3 ∈N(A) and, similarly, G – W2 implies V2, V3 ∈N(A). So, N(A) = {U, V2, V3, W2, W3} and G = P8.

Thus, we can assume G has no degree five vertex, meaning its degree sequence is (46, 32). Suppose the degree three vertices a and b are adjacent. Then, G

– a is a non-planar (7,12) graph with δ(G) > 2 and degree sequence (44, 32, 2). There are five such graphs (See theorem 2.2). Given G – a, it's easy to reconstruct G since

90 N(a) is the three vertices of degree less than four. In Figure 3.3.4, we star a vertex of

G – a that shows G is apex.

W W3 W3 V W3 3 V 3 V3 b 3 V3 V3 W3* b V2 V V2 V2 V2 W2 W2 2 W2 W2 W2 b b

W V1* W1 b W V1* 1 V1* W1 V1* 1 V1 W1 Figure 3.3.4. When G – a is One of the (7,12) Graphs Shown, Deleting the Starred Vertex from G Leaves a Planar Graph.

So, (a, b) ∉E(G). Next suppose c ∈N(a) ∩ N(b). Then G – c is a non-planar

(7,11) graph with degree sequence (43, 32, 22). There are no such graphs [Ma]. So,

N(a) = {V1, V2, V3} and N(b) = {W1, W2, W3} are disjoint. Our goal is to argue G is K4,4

– e. By symmetry, for this, it's enough to show that there is no edge between the

Vi's.

For a contradiction, suppose (V1, V2) ∈E(G). We can argue by symmetry that there is also an edge between the Wi's. Indeed, let G1 be the induced subgraph on {a, V1, V2, V3} and G2 that on {b, W1, W2, W3}. An induced subgraph is a subset of the vertices of G together with any edges whose endpoints are both in the subset.

There are at most seven edges leaving G1 toward G2 (three from V3 and two each from V1 and V2) so at most seven leaving G2 to meet them. This shows that there must be an edge between the Wi's, say, W1 and W2.

If V3 is not adjacent to the two other Vi's, that is, N(V3) = {a, W1, W2, W3}, then G – V3 must be planar.

91

W3 b

V W2 2 a

V1 W1

Figure 3.3.5. G – V3 Yields a Planar Graph.

So, WLOG (V1, V3) ∈E(G). By symmetry and WLOG, (W1, W3) ∈E(G), too. If G1 is K4, by symmetry G2 is too and G is planar.

V2 W2 V W a 1 1 b

V3 W3

Figure 3.3.6. If G1 and G2 are Both K4 then G is Planar.

So we can assume (V2, V3) ∉E(G) (and similarly, (W2, W3) ∉E(G)). Then there are two more cases depending on whether or not (V1, W1) ∈E(G). Figure 3.3.7 shows that G is apex or planar, respectively.

92 W V2 W2* V2 2 W V1 1 b b V1 a W1 a

V3 W3 V3 W3

Figure 3.3.7. If (V1, W1) ∈E(G), then G is Apex by Deleting the Starred Vertex (left). If (V1, W1) ∉E(G), G is Planar (right).

Thus, assuming (V1, V2) ∈E(G) contradicts our assumption that G is MMNA. This completes the proof. 

Lemma 3.3.3. There are no (8,16) MMNA graphs nor any (8,17) with Δ(G) > 5.

Proof. Suppose G is a MMNA (8,16) graph. By lemma 3.1.2, G has no vertex of full degree, and Δ(G) < 6. Let A denote a vertex of maximal degree.

If deg(A) = 6, then G – A is the non-planar (7,10) graph of lemma 2.2.3.

Since deg(U) = 2, U ∈N(A). Also G – V1 implies W2, W3 ∈N(A) while G – W1 gives V2,

V3 ∈N(A). But then G has a P8 subgraph and is not minor minimal.

If deg(A) = 5, then G – A is one of the five non-planar (7,11) graphs of lemma 2.2.4. Suppose G – A is the graph of Figure 2.2.10. Again, deg(U) = 2 means U

∈N(A) and G – V1 shows W2, W3 ∈N(A). Also, G – W3 forces V2 ∈N(A). Then, no matter which is the fifth neighbor of A, G will contain a P8 subgraph and is not minor minimal.

Suppose G – A is the graph of Figure 2.2.11. Again, U ∈N(A) and G – V1 shows W1, W2, W3 ∈N(A). Then G has the Petersen family graph K4,4 – e as a

93 subgraph and is not minor minimal.

If G – A is the graph of Figure 2.2.12, G – V2 shows that W2, W3 ∈N(A) and at least one of U and W1 is also adjacent to A . Suppose first that U ∈N(A). Then G – U is planar unless either V2, V3 ∈N(A), which means G has a P8 subgraph, or else W1

∈N(A). So we can assume W1 ∈N(A). Then G – W3 shows V2 and either U or V1

∈N(A). In the first case, N(A) = {U, V2, W1, W2, W3} and contracting (V1, W3) shows that G has the Petersen graph P7 as a minor. If it's V1 that's adjacent to A, then N(A) =

{V1, V2, W1, W2, W3} and G has a P8 subgraph.

Next, suppose G – A is Figure 2.2.13. Then U ∈N(A) while G – W1 and G –

V1 show that V2, V3, W2, W3 round out the neighborhood of A. This means G has a

P8 subgraph and is not minor minimal. Similarly, if G – A is Figure 2.2.14, we can argue N(A) = {U, V2, V3, W2, W3} which gives a P8 subgraph.

So, it must be that deg(A) = 4 and G is 4-regular (All vertices have degree four). Then G – A is a non-planar (7,12) graph with degree sequence (43, 34). By theorem 2.2, there are five such graphs. If G – A is graph 5 or 7 of theorem 2.2, then

G is the graph of Figure 2.4.21 and has the Petersen family graph K3,3,1 as a minor.

Thus, G is not minor minimal. Graph 6 of the theorem means G is K4,4 so has the

Petersen family graph K4,4 – e as a minor. The other two graphs G – A result in an apex G. If G – A is graph 12, then deleting U shows G is apex. For graph 13, vertex W1 makes G an apex graph. This completes the argument when G has 16 edges.

If G has 17 edges, since there can be no vertex of full degree, Δ(G) < 6. We argue here that Δ(G) = 6 is not possible. Indeed, if deg(A) = 6, then G – A is one of the

94 five (7,11) graphs of lemma 2.2.4. Earlier in this proof, we investigated adding a degree five vertex to these graphs. In each case, in constructing a non-apex graph, we arrive at one that is not minor minimal. Adding A of degree six instead will only add a further edge to the resulting graphs that were already not minor minimal. This shows there are no (8,17) MMNA graphs with Δ(G) = 6. 

Lemma 3.3.4. There are no (8,22) or (8,21) graphs with Δ(G) < 5.

Proof. Assume G is a graph on eight vertices where all eight vertices are of degree

8(5) five. If all eight vertices are of degree five then there are a total of ⎛ ⎞ twenty ⎝⎜ 2 ⎠⎟ edges. Twenty edges are the maximum amount of edges one can construct in a graph on eight vertices of maximum degree five or less. 

Lemma 3.3.5. There are no (8,20), (8,19), (8,18), or (8,17) graphs with Δ(G) < 4.

Proof. Assume that G is a graph on eight vertices where all eight vertices are of

8(4) degree four. If all eight vertices are of degree four then there are a total of ⎛ ⎞ ⎝⎜ 2 ⎠⎟ sixteen edges. Sixteen edges are the maximum amount of edges one can construct in a graph on eight vertices of maximum degree four or less. 

3.3.1. (8,17) and (8,18) MMNA Graphs

In section 3.3.1 we prove lemma 3.3.1.1, there is no G that is MMNA and

(8,17) with Δ(G) = 5 or (8,18) with Δ(G) = 6.

95

Lemma 3.3.1.1. There is no G that is MMNA and (8,17) with Δ(G) = 5 or (8,18) with

Δ(G) = 6.

Proof. Consider a MMNA (8,17) graph G’. By lemma 3.1.4 δ(G’) > 3 and by lemmas

3.3.3 and 3.3.5; Δ(G’) = 5. If we remove vertex A from this graph, we are removing a degree five vertex. The result is a non-planar (7,12) graph. Is this the only way to get a (7,12) graph from a graph on eight vertices? The answer is no. We can take a

(8,18) MMNA graph and remove vertex A, where A is of degree six, to find a non- planar (7,12) graph as a result. By theorem 2.2 there are exactly nineteen non- planar (7,12) graphs of minimum degree at least two. We can regard these nineteen

(7,12) graphs as a result of either taking a MMNA (8,18) and removing a degree six vertex, or taking a (8,17) MMNA graph and removing a degree five vertex.

Now let us use these nineteen non-planar (7,12) graphs of minimum degree at least two in order to work backwards in a sense. Let us add vertex A to each non-planar (7,12) graph in order to construct candidate MMNA (8,17) and

(8,18) graphs. We will assume that A is of degree five in order to construct MMNA

(8,17) graphs, and we will assume that A is of degree six in order to construct

MMNA (8,18) graphs. Let G refer to a non-planar (7, 12) graph and let v refer to a vertex in G. Since A is of degree five or of degree six, we will have to determine which vertices in G are adjacent to A. The neighborhood of A, N(A), will be defined as the set of all v in G that are adjacent to A. Let us call the resulting graph on eight vertices G’. After N(A) has been determined we can construct the complement of

96 each G’. We construct the complement of G’ to observe two things. First we check if the complement of G’ has any of the five triangulation complements, that were discovered in chapter II, as a subgraph. If a triangulation complement from Chapter

II is found to be a subgraph of the complement of G’ then there is a v such that G’ - v is planar, meaning that G’ is an apex graph. This contradicts G’ being NA. Second, we check if the complement of G’ is a subgraph of any of the Petersen graph complements that were discussed in Chapter II. If a G’ complement is found to be a subgraph of a Petersen graph complement then, a Petersen graph is a minor of G’. If a Petersen graph is found to be a minor of G’ then, G’ is not minor minimal. Since we are assuming G’ is MMNA then we are in search of a contradiction in order to prove that the only MMNA graphs on eight or fewer vertices are J1 and the Petersen graphs. In particular, If G’ is a (8,17) or (8,18) graph, it cannot be J1 (J1 has twenty- one edges) nor a Petersen graph (both P8 and K4,4 – e have fifteen edges each). So we need only find a contradiction. Let us look at all the nineteen cases individually.

Case 1: Let G be the first non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (5, 42, 33, 2) graph of Figure

3.3.1.1, which also shows its complement.

97

V3 W3 V1

W1

V3 V2 W2 W3 W2 V2

U

V1 U W1 Figure 3.3.1.1. The First Non-Planar (7,12) Graph, a (5, 42, 33, 2) Graph, and its Complement (complement is in orange).

We think of the non-planar (7,12) graphs as a result of either removing vertex A of degree six from a (8,18) MMNA graph with δ(G) > 3 and Δ(G) = 6 or from removing vertex A of degree five from a (8,17) MMNA graph with δ(G) > 3 and Δ(G) = 5. Let us take the (7,12) graph from Figure 3.3.1.1, and work backwards to describe vertex A in order to create a graph on eight vertices, G’.

V3 W3

W2 V2 A

V1 U W1 Figure 3.3.1.2. The (5, 42, 33, 2) Graph Plus Vertex A, Creates a Graph on Eight Vertices.

Next we will determine N(A) assuming A is a vertex of degree five or six. We first argue that W2, W3 ∈N(A) by showing this is necessary for G’ – V2 to be non-planar.

Consider G’ – V2. If A is of degree six, assume that V2 is an element of N(A) as

98 otherwise N(A) = V(G’) – V2 and W2, W3 ∈N(A), which is what we want to show.

Since V2 would be adjacent to A, when V2 is removed so would the edge that connects V2 to A. Since we are assuming that A is a degree six vertex then it is still adjacent to five other vertices in G. If A is of degree five, we can assume that V2 is not an element of N(A) otherwise A has only degree four in G’ – V2, which is even easier to make planar. When we remove V2 from G’ vertex A remains a degree five vertex and is still adjacent to five vertices in G. Now let us take G’ – V2 and let us arrange the remaining vertices so that five are adjacent to A in order to create a planar graph. If G’ – V2 is found to be a planar graph then, that is a contradiction to the assumption that G’ is MMNA because it means that G’ is an apex graph.

W V3 3

W2 A

V1 U W1

Figure 3.3.1.3. The First (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar.

We can construct a planar graph if V2 is removed from G’ and A is adjacent to U, V1,

V3, W1, and W2. We have found a contradiction to our assumption that G’ is MMNA because it shows that G’ is an apex graph. In order to avoid having G’ as an apex

99 graph we must have the vertex W3 as an element of N(A). Let us look at G’ – V2 once again. Now that we know W3 ∈N(A), we still can construct a planar graph after we take G’ – V2 and A is adjacent to U, V1, V3, W1, and W3. This too is a contradiction to our assumption that G’ is MMNA, unless the vertex W2 is also in N(A). In order to avoid contradictions to the assumption that G’ is MMNA, we conclude that W2, W3

∈N(A).

V3 W3

W2 W1 A

U V1

Figure 3.3.1.4. The First (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found Once Again to be Planar.

Besides W2, W3 ∈N(A), vertex U must also be an element of N(A). By lemma 3.1.4, the minimal degree of a MMNA graph is at least three. This means that all vertices in

G’ must be of degree at least three. Since U is of degree two in the (7,12) graph, it must be adjacent to A in order to be a degree three vertex in G’. As of now we have determined three elements in N(A), U, W2, and W3.

Let us assume that A is of degree five. If A is of degree five, then there are two vertices in G that are not elements of N(A). Since we’ve determined that U, W2,

100

W3 ∈N(A) one of the following combinations of vertices, is not in N(A), {V1, V2}, {V1,

V3}, {V1, W1}, {V2, V3}, {V2, W1}, or {V3, W1}. However, we find a contradiction with every one of the combinations. When we construct the complement of G’, for each pair of vertices not in N(A), we find that the complement is a subgraph of the P8 graph complement. If the complement of G’ is a subgraph of the P8 graph complement, then G’ has the P8 graph as a subgraph. This is a contradiction to the assumption that G’ is a MMNA graph since it implies that G’ is not MM. Let us construct the complement in each case. When a pair of vertices is not an element of

N(A) then A is not adjacent to those two vertices in the graph G’, but when taking the complement of G’ we find that A is adjacent to only those two vertices.

a.) V1, V2 ∉N(A) b.) V1, V3 ∉N(A c.) V1, W1 ∉N(A)

V1 V1 V1 A A A W1 W1 W1 V3 V3 V3 V2 W2 W3 V2 W2 W3 V2 W2 W3

U U U d.) V2, V3 ∉N(A) e.) V2, W1 ∉N(A) f.) V3, W1 ∉N(A)

V1 V1 V1 A A W A W1 1 W1 V V 3 W3 3 W3 V2 W2 V3 V2 W2 V2 W2 W3

U U U

Figure 3.3.1.5. These Six Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Each complement is a subgraph of the P8 complement.

101 When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Assume that A is of degree six. Since U, W2, W3 ∈N(A) then, one of the following vertices is not an element of N(A), V1, V2, V3, or W1. However when we look at the complement of G’ for every case, we find that each complement is a subgraph of the P8 graph complement. So, G’ is not MM.

a.) V1 ∉N(A) b.) V2 ∉N(A) c.) V3 ∉N(A) d.) W1 ∉N(A)

V1 V1 V1 V1 A A A A W1 W1 W1 W1 V3 V3 V3 V3 V2 W2 W3 V2 W2 W3 V2 W2 W3 V2 W2 W3

U U U U

Figure 3.3.1.6. These Four Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A). Each complement is a subgraph of the P8 complement.

When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the first (7,12) graph. However, a MMNA G’ may arise from a different non-planar (7,12) graph of minimum degree at least two. Let us look at case two.

102 Case 2: Let G be the second non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (5, 42, 33, 2) graph of

Figure 3.3.1.7, which also shows its complement.

V3 W3 V1

U V3

W3 W2 V2 W2 V2

W U 1

V1 W1 Figure 3.3.1.7. The Second Non-Planar (7,12) Graph and its Complement.

Let us construct vertex A, where A is either of degree five or degree six, in order to create a graph on eight vertices which we will refer to as G’. Now let us determine

N(A) for G’. Consider G’ – V2. If A is of degree six, assume that V2 ∈N(A). Remember that if V2 ∈N(A) then, V2 is adjacent to A. If we remove V2 from G’ then the edge connecting A to V2 is also removed. Since A is a degree six vertex then it is still adjacent to five other vertices in G. However, if A is of degree five, assume that V2

∉N(A). When we remove V2 from G’ vertex A remains a degree five vertex and is still adjacent to five vertices in G. Now let us take G’ – V2 and let us arrange the remaining vertices so that five are adjacent to A in order to create a planar graph.

103

V3 W3

A W2 U

V W 1 1 Figure 3.3.1.8. The Second (7,12) graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar.

We can construct a planar graph if V2 is removed from G’ and A is adjacent to U, V1,

V3, W1, and W2. This is a contradiction to our assumption that G’ is MMNA, that shows vertex W3 ∈N(A). Now consider G’ – V2 once again with W3 ∈N(A). We can still construct a planar graph after we take G’ – V2 and A is adjacent to U, V1, V3, W1, and W3. This is still a contradiction to our assumption that G’ is MMNA, and shows the vertex W2 is also in N(A).

W3 V3

A U W2

V1 W1

Figure 3.3.1.9. The Second (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found Once Again to be Planar.

104

We next argue V1 ∈N(A). Consider G – W1 and W2, W3 ∈N(A). If A is of degree six, assume that W1 ∈N(A). If A is of degree five, assume that W1 ∉N(A). We still can construct a planar graph after we take G’ – W1 and A is adjacent to U, V2, V3, W2, and

W3. This is another contradiction to our assumption that G’ is MMNA, that shows V1

∈N(A). In order to avoid all contradictions and to continue to assume that G’ is

MMNA, we conclude that all three vertices V1, W2, and W3 are elements of N(A).

V3 W3 A

U V2 W2

V 1 Figure 3.3.1.10. The Second (7,12) Graph with Vertex A Added, has Vertex W1 Removed and is Found to be Planar.

Besides V1, W2, W3 ∈N(A), vertex U must also be an element of N(A), since U is of degree two in G and must be of degree three in G’.

Let us assume that A is of degree five. Since U, V1, W2, W3 ∈N(A) one of the following combinations of vertices is not in N(A), {V2, V3}, {V2, W1}, or {V3, W1}.

However, we find a contradiction with every one of the combinations. When we

105 construct the complement of G’ for each pair of vertices not in N(A), we find that the complement is a subgraph of the P8 graph complement. So, G’ is not MM.

a.) V2, V3 ∉N(A) b.) V2, W1 ∉N(A) c.) V3, W1 ∉N(A)

V1 V1 V1

U V3 U V3 U V3

W2 W2 W2 W3 A V2 W3 A V2 W3 A V2

W W 1 1 W1

Figure 3.3.1.11. These Three Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Each complement is a subgraph of the P8 complement.

When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Assume that A is of degree six. Since U, V1, W2, W3 ∈N(A) we can say that one of the following vertices is not an element of N(A), V2, V3, or W1. However when we look at the complement of G’ for every case we find that each complement is a subgraph of the P8 graph complement. So, G’ is not MM.

106 a.) V2 ∉N(A) b.) V3 ∉N(A) c.) W1 ∉N(A)

V1 V1 V1

U V3 U V3 U V3

W W2 W2 2 W3 W3 W3 A V2 A V2 A V2

W W W1 1 1

Figure 3.3.1.12. These Three Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A). Each complement is a subgraph of the P8 complement.

When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the second (7,12) graph. However, a MMNA G’ may arise from a different non-planar (7,12) graph of minimum degree at least two. Let us look at case three.

Case 3: Let G be the third non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (5, 42, 33, 2) graph of

Figure 3.3.1.13, which also shows its complement.

107

V3 W3

W2 V2 U V3 W1 W2 V2 W3 V1

U

V1 W1 Figure 3.3.1.13. The Third Non-Planar (7,12) Graph and its Complement.

Let us construct vertex A, where A is either of degree five or degree six, in order to create a graph on eight vertices which we will refer to as G’. Consider G’ – V2. If A is of degree six, assume that V2 ∈N(A). If A is of degree five, assume that V2 ∉N(A).

Then let us arrange the remaining vertices so that five are adjacent to A in order to create a planar graph. We can construct G’ – V2 three different ways and still find a construction that is planar.

a.) G’ – V2 b.) G’ – V2 c.) G’ – V2

V W3 V3 3 V3 W3 A

W3 A W2 A W W 2 2 U U U W V W V1 1 V1 W1 1 1

Figure 3.3.1.14. The Third (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar in Three Different Constructions.

In all three constructions in Figure 3.3.1.14 we find a contradiction to G’ being a

MMNA graph. In construction (a) from Figure 3.3.1.14 we find that G’ – V2 is planar

108 so W3 ∈N(A). In construction (b) from Figure 3.3.1.14 we find that G’ – V2 is planar so W1 ∈N(A). In construction (c) from Figure 3.3.1.14 we find that G’ – V2 is planar so W2 ∈N(A). In order to avoid these contradictions and continue to assume that G’ is MMNA, we conclude that W1, W2, W3 ∈N(A). Also, let us not forget that U is of degree two in G, so it must also be in N(A).

Let us assume that A is of degree five. Since U, W1, W2, W3 ∈N(A) one of the following combinations of vertices is not in N(A), {V1, V2}, {V1, V3}, or {V2, V3}.

However, we find a contradiction with every one of the combinations. When we construct the complement of G’ for each pair of vertices not in N(A), we find that the complement is a subgraph of the K4,4 – e graph complement. So, G’ is not MM.

a.) V1, V2 ∉N(A) b.) V1, V3 ∉N(A) c.) V2, V3 ∉N(A)

W2 V2 W2 V2 W2 V2 U V3 U V3 U V3 W1 W1 W1 A A A W V1 W V1 W V1 3 3 3

Figure 3.3.1.15. These Three Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Each complement is a subgraph of the K4,4 – e complement.

When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Assume that A is of degree six. Since U, W1, W2, W3 ∈N(A) then, one of the following vertices is not an element of N(A), V1, V2, or V3. However when we look at

109 the complement of G’ for every case we find that each complement is a subgraph of the K4,4 – e graph complement. This is a contradiction as it means G’ is not MM.

a.) V1 ∉N(A) b.) V2 ∉N(A) c.) V3 ∉N(A)

W2 V2 W2 V2 W2 V2 U V3 U V3 U V3 W1 W1 W1 A A A W V1 W V1 W V1 3 3 3

Figure 3.3.1.16. These Three Graphs Represent Complements of G’ When we Assume that the Given vertex is Not an Element of N(A). Each complement is a subgraph of the K4,4 – e complement.

When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the third (7,12) graph. However, a MMNA G’ may arise from a different non-planar (7,12) graph of minimum degree at least two. Let us look at case four.

Case 4: Let G be the fourth non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (5, 4, 35) graph of Figure

3.3.1.17, which also shows its complement.

110

V3 W3 U V3

W3 W2 V1 W2 V2

U V W1 2

V1 W1 Figure 3.3.1.17. The Fourth Non-Planar (7,12) Graph and its Complement.

Let us construct vertex A, where A is either of degree five or degree six, in order to create a graph on eight vertices which we will refer to as G’. Consider G’ – V2. If A is of degree six, assume that V2 ∈N(A). If A is of degree five, assume that V2 ∉N(A). We can construct G’ – V2 two different ways and still find a construction that is planar.

a.) G’ – V2 b.) G’ – V2

W W3 3 V3 V3

A W2

W2

W1 U V1 U A V W 1 1

Figure 3.3.1.18. The Fourth (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar in Two Different Constructions.

In both constructions in Figure 3.3.1.18 we find a contradiction to G’ being a MMNA graph. In construction (a) from Figure 3.3.1.18 we find that G’ – V2 is planar so W3

∈N(A). In construction (b) from Figure 3.3.1.18 we find that G’ – V2 is planar so W2

111

∈N(A). We next argue V1 ∈N(A). Consider G’ – W3, where A is either of degree five or degree six. Then let us arrange the remaining vertices so that five are adjacent to A in order to create a planar graph. This is another contradiction to our assumption that G’ is MMNA, that shows vertex V1 ∈N(A). In order to avoid contradictions to or assumption we conclude that V1, W2, W3 ∈N(A).

V3 A

W2 V2 U

V1 W1

Figure 3.3.1.19. The Fourth (7,12) Graph with Vertex A Added, has Vertex W3 Removed and is Found to be Planar.

Assume that that A is of degree five. Suppose that W1 ∉N(A). We want to show that if W1 ∉N(A) then, W1 along with another vertex in G, are the two vertices that are not elements in N(A). Since we know that V1, W2, W3 ∈N(A) our only options are to check {U, W1}, {V3, W1}, and {V2, W1} to see if any of these pairs of vertices do not create a contradiction to our assumption that G’ is MMNA. If we test all three options, {U, W1}, {V3, W1}, and {V2, W1}, and find that every option creates a contradiction to the assumption that G’ is MMNA then, W1 must be an element of

112

N(A). So, let us consider U, W1 ∉N(A). Now let us construct the complement of G’ with this consideration. We will find that the complement of G’ has a “Pentagon –

Segment” subgraph. We found one of the triangulation complements as a subgraph in the complement of G’. If one of the triangulation complements is found to be a subgraph of the complement of G’ then, G’ is a subgraph of one of the triangulations.

This is a contradiction to the assumption that G’ is MMNA, because it implies that G’ is an apex graph.

U V3 U V3

A W3 W A W 2 V1 3 W2 V1

W V W V 1 2 1 2 Figure 3.3.1.20. The Graphs Represent Complement of G’ When We Consider that U, W1 ∉N(A). The complement of G’ has a triangulation complement, known as “Pentagon – Segment” as a subgraph that is highlighted in black on the graph to the right.

Now let us consider V3, W1 ∉N(A). Let us construct the complement of G’ with this consideration. We will find that the complement of G’ has a “Pentagon – Segment” subgraph. This is a contradiction to the assumption that G’ is MMNA, because this implies that G’ is an apex graph.

113

U V3 U V3

W W3 W2 3 W2 V V1 A 1 A W1 W1 V2 V2

Figure 3.3.1.21. The Graphs Represent the Complement of G’ When We Consider that V3, W1 ∉N(A). The complement of G’ has a triangulation complement, known as “Pentagon – Segment” as a subgraph that is highlighted in black on the graph to the right.

Consider V2, W1 ∉N(A). Let us construct the complement of G’ with this consideration. We will find that the complement of G’ is a minor of one of the P7 complements on eight vertices. This is a contradiction to the assumption that G’ is

MMNA, because this implies that G’ is not MM.

A W1

W3 V2 W2

U V3 V1

Figure 3.3.1.22. The Graph Represents the Complement of G’ When We Assume that V2, W1 ∉N(A). The complement is a subgraph of a P7 complement on eight vertices.

Since all three cases where we consider W1 ∉N(A) construct a complement that contradicts G’ being MMNA, we must conclude that W1 ∈N(A). Since V1, W1, W2, W3

114

∈N(A) one of the following combinations of vertices is not in N(A), {U, V2}, {U, V3}, or

{V2, V3}. However, we find a contradiction with every one of the combinations. When we construct the complement of G’ for each pair of vertices not in N(A), we find that the complement is a subgraph of the P8 graph complement. This is a contradiction as it means G’ is not MM.

a.) U, V2 ∉N(A) b.) U, V3 ∉N(A) c.) V2, V3 ∉N(A)

V1 V1 V1

W1 V2 W1 V2 W1 V2 W W2 W W2 W3 2 V W3 A V3 3 A V3 A 3

U U U

Figure 3.3.1.23. These Three Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Each complement is a subgraph of the P8 complement.

When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Assume that A is of degree six. Since V1, W2, W3 ∈N(A) then, one of the following vertices is not an element of N(A), U, V2, V3, or W1. However when we look at the complement of G’ for every case we find that each complement is a subgraph of either the P8 graph complement or a P7 graph complement on eight vertices. This is a contradiction as it means G’ is not MM.

115 a.) U ∉N(A) b.) V2 ∉N(A) c.) V3 ∉N(A) d.) W1 ∉N(A)

V1 V1 V1 A W1 W1 V2 W1 V2 W1 V2 W3 V2 W2 W2 W2 W2 W3 A V3 W3 A V3 W3 A V3 V U V3 1 U U U

Figure 3.3.1.24. These Four Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A). The graphs (a)-(c) are complements of G’ that are subgraphs of the P8 complement, while graph (d) is a complement of G’ that is a subgraph of a P7 complement on eight vertices.

When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the fourth (7,12) graph. However, a MMNA G’ may arise from a different non-planar (7,12) graph of minimum degree at least two. Let us look at case five.

Case 5: Let G be the fifth non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (43, 34) graph of Figure

3.3.1.25, which also shows its complement.

116

V3 W3

U V3

W V W3 2 2 W2 V2

W1 V1

V1 U W1 Figure 3.3.1.25. The Fifth Non-Planar (7,12) Graph and its Complement.

Let us construct vertex A, where A is either of degree five or degree six, in order to create a graph on eight vertices which we will refer to as G’. Consider G’ – V1. If A is of degree six, assume that V1 ∈N(A). If A is of degree five, assume that V1 ∉N(A). We can construct G’ – V1 two different ways and still find a construction that is planar.

a.) G’ – V1 b.) G’ – V1

V W3 V3 3 W3 A W 2 W2 V2 V2

U U A W1 W1

Figure 3.3.1.26. The Fifth (7,12) Graph with Vertex A Added, has Vertex V1 Removed and is Found to be Planar in Two Different Constructions.

In both constructions in Figure 3.3.1.26 we find a contradiction to G’ being a MMNA graph. In construction (a) from Figure 3.3.1.26 we find that G’ – V1 is planar so W3

∈N(A). In construction (b) from Figure 3.3.1.26 we find that G’ – V1 is planar so W2

117 ∈N(A). In order to avoid these contradictions and continue to assume that G’ is

MMNA, we must have W2, W3 ∈N(A).

Assume that A is of degree five. Suppose that W1 ∉N(A). Consider V1, W1

∉N(A) and consider V2, W1 ∉N(A). Now let us construct a complement of G’ with each consideration. We will find that the complement of G’, in both cases, is a subgraph of the P8 graph complement. So, G’ is not MM.

a.) V1, W1 ∉N(A) b.) V2, W1 ∉N(A)

V3 V3 U U V2 V2

W3 W2 A W3 W2 A V1 V1

W W 1 1

Figure 3.3.1.27. The Graphs Represent the Complement of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Both complements are subgraphs of the P8 complement.

Now let us consider V3, W1 ∉N(A). Let us construct the complement of G’ with this consideration. We will find that the complement of G’ has a “Pentagon – Segment” subgraph. This is a contradiction to the assumption that G’ is MMNA, because this implies that G’ is an apex graph.

118

U V3 U V3

W3 W2 V2 W3 W2 V2 A A W1 W1 V V 1 1 Figure 3.3.1.28. The Graphs Represent the Complement of G’ When We Consider that V3, W1 ∉N(A). The complement of G’ has a triangulation complement, known as “Pentagon – Segment” as a subgraph that is highlighted in black on the graph to the right.

Now let us consider U, W1 ∉N(A). Let us construct the complement of G’ with this consideration. We will find that the once again the complement of G’ has a

“Pentagon – Segment” subgraph. This is a contradiction to the assumption that G’ is

MMNA, because this implies that G’ is an apex graph.

U V3 U V3 A A W3 W2 V2 W3 W2 V2

W1 V W V 1 1 1 Figure 3.3.1.29. The Graphs Represent the Complement of G’ When We Consider that U, W1 ∉N(A). The complement of G’ has a triangulation complement, known as “Pentagon – Segment” as a subgraph that is highlighted in black on the graph to the right.

Since all four cases where we consider W1 ∉N(A) construct a complement that contradicts G’ being MMNA, we must conclude that W1 ∈N(A). Now Suppose that U

∉N(A). Consider U, V2 ∉N(A) and U, V3 ∉N(A). Now let us construct a complement of

G’ with each consideration. We will find that the complement of G’, in both cases, is a subgraph of the P8 graph complement. So, G’ is not MM.

119 a.) U, V2 ∉N(A) b.) U, V3 ∉N(A)

V1 V1 V V W1 2 W1 2 V V W3 W2 A 3 W3 W2 A 3

U U

Figure 3.3.1.30. The Graphs Represent the Complement of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Both complements are subgraphs of the P8 complement.

Now let us consider U, V1 ∉N(A). Let us construct the complement of G’ with this consideration. We will find that the complement of G’ has a “Pentagon – Segment” subgraph. This is a contradiction to the assumption that G’ is MMNA, because this implies that G’ is an apex graph.

V3 V3 U U A A W V W3 2 2 W3 W2 V2

W1 V1 W1 V 1 Figure 3.3.1.31. The Graphs Represent the Complement of G’ When We Consider that U, V1 ∉N(A). The complement of G’ has a triangulation complement, known as “Pentagon – Segment” as a subgraph that is highlighted in black on the graph to the right.

Since all three cases where we consider U ∉N(A) construct a complement that contradicts G’ being MMNA, we must conclude that U ∈N(A). Since U, W1, W2, W3

∈N(A) one of the following combinations of vertices is not in N(A), {V1, V2}, {V1, V3}, or {V2, V3}. However, we find a contradiction with every one of the combinations.

120 When we construct the complement of G’ for each pair of vertices not in N(A), we find that the complement is a subgraph of a K3,3,1 graph complement on eight vertices. So, G’ is not MM.

a.) V1, V2 ∉N(A) b.) V1, V3 ∉N(A) c.) V2, V3 ∉N(A)

U V3 U V3 U V3

A A A W3 V2 W3 V2 W3 V2 W2 W2 W2

W V W1 V1 W1 V1 1 1

Figure 3.3.1.32. These Three Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Each complement is a subgraph of a K3,3,1 complement on eight vertices.

When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Assume that A is of degree six. Since W2, W3 ∈N(A) then, one of the following vertices is not an element of N(A), U, V1, V2, V3, or W1. However when we look at the complement of G’, for every case, we find that each complement is a subgraph of either the P8 graph complement or a K3,3,1 graph complement on eight vertices. This is a contradiction as it means G’ is not MM.

121 a.) U ∉N(A) b.) V1 ∉N(A) c.) V2 ∉N(A)

V1 U V3 U V3 V W1 2 W2 W2 V W A V3 W3 V2 W3 2 W3 2 A A

W1 V1 W1 V1 U d.) V3 ∉N(A) e.) W1 ∉N(A)

U V3 V3 U V2 W2 W3 V2 W A A W3 2 V1

W1 V W 1 1

Figure 3.3.1.33. These Five Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A). The two graphs (a) and (e) are subgraphs of the P8 complement, while graphs (b)-(d) are subgraphs of a K3,3,1 complement on eight vertices.

When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the fifth (7,12) graph. However, a MMNA G’ may arise from a different non-planar (7,12) graph of minimum degree at least two. Let us look at case six.

Case 6: Let G be the sixth non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (43, 34) graph of Figure

3.3.1.34, which also shows its complement.

122

V3 W3

W3 V3

U W1 V1 W V2 2 W 2 V2

U

V1 W1 Figure 3.3.1.34. The Sixth Non-Planar (7,12) Graph and its Complement.

Let us look at the isomorphism classes of G, {V1, V2, V3} and {U, W1, W2,

W3}. Let us assume that A is a degree five vertex. If all the elements in the isomorphism class {U, W1, W2, W3} are elements of N(A) then, at least one element from the class {V1, V2, V3} is an element of N(A). So, WLOG let V3 ∈N(A). Also, if all the elements in the isomorphism class {V1, V2, V3} are elements of N(A) then, at least two elements from the class {U, W1, W2, W3} are elements of N(A). So, WLOG let U,

W3 ∈N(A). If we consider that U, V3, W3 ∈N(A) and that A is of degree five, we find ourselves with three sub cases for N(A).

Sub case 1: U, V3, W1, W2, W3 ∈N(A) and V1, V2 ∉N(A)

Sub case 2: U, V1, V3, W2, W3 ∈N(A) and V2, W1 ∉N(A)

Sub case 3: U, V1, V2, V3, W3 ∈N(A) and W1, W2 ∉N(A)

In both sub case 1 and sub case 2, if we construct the complement of G’, we find that it is a subgraph of the K4,4 – e graph complement. So, G’ is not MM.

123 a.) Sub case 1: V1, V2 ∉N(A) b.) Sub case 2: V2, W1 ∉N(A)

W3 V3 W3 V3

W1 A W1 A U V1 U V1

W2 W2 V V2 2

Figure 3.3.1.35. These Two Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Each complement is a subgraph of the K4,4 – e complement.

In sub case 3, if we construct the complement of G’, we find that the complement of

G’ has a “Pentagon – Segment” subgraph. This is a contradiction to the assumption that G’ is MMNA, because this implies that G’ is an apex graph.

A A W V3 3 W3 V3

W V1 U 1 U W1 V1 W 2 V2 W2 V 2 Figure 3.3.1.36. The Graphs Represent the Complement of G’ When We Consider that W1, W2 ∉N(A). The complement of G’ has a triangulation complement, known as “Pentagon – Segment” as a subgraph that is highlighted in black on the graph to the right.

When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Let us once again look at the isomorphism classes of G, {V1, V2, V3} and {U,

W1, W2, W3}. Now let us assume that A is a degree six vertex. If all the elements in the isomorphism class {U, W1, W2, W3} are elements of N(A) then, at least two

124 elements from the class {V1, V2, V3} are elements of N(A). So, WLOG let V2, V3 ∈N(A).

Also, if all the elements in the isomorphism class {V1, V2, V3} are elements of N(A) then, at least three elements from the class {U, W1, W2, W3} are elements of N(A). So,

WLOG let U, W2, W3 ∈N(A). If we consider that U, V2, V3, W2, W3 ∈N(A) and that A is of degree six, one of the following vertices is not an element of N(A), V1 or W1.

However when we look at the complement of G’ for both cases we find that each complement is a subgraph of the K4,4 - e graph complement. So, G’ is not MM.

a.) V1 ∉N(A) b.) W1 ∉N(A)

W3 V3 W3 V3

W1 A W1 A U V1 U V1

W2 W V2 2 V 2

Figure 3.3.1.37. These Two Graphs Represent the Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A). The two graphs are subgraphs of the K4,4 - e complement.

When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the sixth (7,12) graph. However, a MMNA G’ may arise from a different non-planar (7,12) graph of minimum degree at least two. Let us look at case seven.

125 Case 7: Let G be the seventh non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (43, 34) graph of Figure

3.3.1.38, which also shows its complement.

V3 W3

U W3

V 3 V2 W 2 W W2 V2 1

V1

V1 U W1 Figure 3.3.1.38. The Seventh Non-Planar (7,12) Graph and its Complement.

Let us construct vertex A, where A is either of degree five or degree six, in order to create a graph on eight vertices which we will refer to as G’. Consider G’ – V2. If A is of degree six, assume that V2 ∈N(A). If A is of degree five, assume that V2 ∉N(A). We can construct G’ – V2 two different ways and still find a construction that is planar.

a.) G’ – V2 b.) G’ – V2

W3 V3 W3 V3

A W A 2 W1 W1 W2 U

U V1 V1

Figure 3.3.1.39. The Seventh (7,12) Graph with Vertex A added, has Vertex V2 Removed and is Found to be Planar in Two Different Constructions.

126 In both constructions in Figure 3.3.1.39 we find a contradiction to G’ being a MMNA graph. In construction (a) from Figure 3.3.1.39 we find that G’ – V2 is planar so W2

∈N(A). In construction (b) from Figure 3.3.1.39 we find that G’ – V2 is planar so W1

∈N(A). In order to avoid these contradictions and continue to assume that G’ is

MMNA, we must have W1, W2 ∈N(A).

Let us look at the isomorphism classes of G, {U}, {V1}, {V2, V3}, {W1, W2}, and {W3}. Assume that that A is of degree five. Suppose that V2 ∉N(A). Consider V1,

V2 ∉N(A) and V2, V3 ∉N(A). Now let us construct a complement of G’ with each consideration. We will find that the complement of G’, in both cases, is a subgraph of a K3,3,1 graph complement on eight vertices. So, G’ is not MM.

a.) V1, V2 ∉N(A) b.) V2, V3 ∉N(A)

U U W3 W3

W2 W2 V3 V3 V2 W1 V2 W1

A V A V1 1

Figure 3.3.1.40. The Graphs Represent the Complement of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Both complements are subgraphs of a K3,3,1 complement graph on eight vertices.

Consider U, V2 ∉N(A). Let us construct the complement of G’ with this consideration.

We will find that the complement of G’ is a subgraph of one of the P7 graph complements on eight vertices. So, G’ is not MM.

127

U W3

W2 V3 A W1

V2 V1

Figure 3.3.1.41. The Graph Represents the Complement of G’ When We Assume that U, V2 ∉N(A). The complement is a subgraph of a P7 complement on eight vertices.

Now let us consider V2, W3 ∉N(A). Let us construct the complement of G’ with this consideration. We will find that the complement of G’ has a “House” subgraph. This is a contradiction to the assumption that G’ is MMNA, because this implies that G’ is an apex graph.

U W3 W1 U W3 W1

V3 V3

W2 V1 W2 V1 V2 A V2 A

Figure 3.3.1.42. The Graphs Represent the Complement of G’ When We Consider that V2, W3 ∉N(A). The complement of G’ has a triangulation complement, known as “House” as a subgraph that is highlighted in black on the graph to the right.

Since all four cases where we consider V2 ∉N(A) construct a complement that contradicts G’ being MMNA, we must conclude that V2 ∈N(A). Since V2 and V3 are in the same isomorphism class then, V3 ∈N(A). Now that we know that V2, V3, W1, W2

∈N(A) one of the following combinations of vertices is not in N(A), {U, V1}, {U, W3},

128 or {V1, W3}. However, when we construct the complement of G’ for each pair of vertices not in N(A), we find a contradiction to the assumption that G’ is MMNA.

When we consider U, V1 ∉N(A), we find that the complement of G’ has a “Pentagon-

Segment” subgraph. This is a contradiction to the assumption that G’ is MMNA, because this implies that G’ is an apex graph.

V 3 U V3 W3 U W3

W2 W2 V2 V2 W1 W1

A V1 A V 1 Figure 3.3.1.43. The Graphs Represent the Complement of G’ When We Consider that U, V1 ∉N(A). The complement of G’ has a triangulation complement, known as “Pentagon-Segment” as a subgraph that is highlighted in black on the graph to the right.

When we consider U, W3 ∉N(A), we find that the complement of G’ is a subgraph of the P8 graph complement. This is a contradiction as it means G’ is not MM.

A U W3

V3 W2 V 2 W1

V1

Figure 3.3.1.44. The Graph Represents the Complement of G’ When We Assume that U, W3 ∉N(A). The complement is a subgraph of the P8 complement.

129

When we consider V1, W3 ∉N(A), we find that the complement of G’ has a “Pentagon-

Segment” subgraph. This is a contradiction to the assumption that G’ is MMNA, because this implies that G’ is an apex graph.

U W 3 U W3 W V 2 W2 3 W V3 V 1 W1 2 A V2 A V 1 V1

Figure 3.3.1.45. The Graphs Represent the Complement of G’ When We Consider that V1, W3 ∉N(A). The complement of G’ has a triangulation complement, known as “Pentagon-Segment” as a subgraph that is highlighted in black on the graph to the right.

When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Assume that A is of degree six. Since W1, W2 ∈N(A) then, one of the following vertices is not an element of N(A), U, V1, V2, V3, or W3. However, when we look at the complement of G’ for every case we find a contradiction to the assumption that G’ is MMNA. We find that each complement of G’ is a subgraph for either the P8 graph complement or a K3,3,1 graph complement on eight vertices. So, G’ is not MM.

130 a.) U ∉N(A) b.) V1 ∉N(A) c.) V2 ∉N(A)

A U W3 U W3 U W3 W W2 V 2 V3 V 3 3 W2 V W V2 W1 V 2 1 2 W1 A V1 A V1 V1 d.) V3 ∉N(A) e.) W3 ∉N(A)

A U W3 U W3 V W2 3 V V W 3 W2 2 1 V 2 W1 A V 1 V1

Figure 3.3.1.46. These Five Graphs Represent the Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A). The two graphs (a) and (e) are subgraphs of the P8 complement, while graphs (b)-(d) are subgraphs of a K3,3,1 complement on eight vertices.

When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the seventh (7,12) graph. However, a MMNA G’ may arise from a different non-planar (7,12) graph of minimum degree at least two. Let us look at case eight.

Case 8: Let G be the eighth non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (44, 32, 2) graph of Figure

3.3.1.47, which also shows its complement.

131

V3 W3

V3 W2 U

V2 W3 V1 W1 W2 V2

U

V1 W1 Figure 3.3.1.47. The Eighth Non-Planar (7,12) Graph and its Complement.

Let us construct vertex A, where A is either of degree five or degree six, in order to create a graph on eight vertices which we will refer to as G’. Consider G’ – W1. If A is of degree six, assume that W1 ∈N(A). If A is of degree five, assume that W1 ∉N(A).

We can construct G’ – W1 two ways and still find a construction that is planar.

a.) G’ – W1 b.) G’ – W1

W3 W3 V3 A A V 3 U U

V2 W2 V2 W2

V1 V1

Figure 3.3.1.48. The Eighth (7,12) Graph with Vertex A Added, has Vertex W1 Removed and is Found to be Planar in Two Different Constructions.

In both constructions in Figure 3.3.1.48 we find a contradiction to G’ being a MMNA graph. In construction (a) from Figure 3.3.1.48 we find that G’ – W1 is planar so V2

∈N(A). In construction (b) from Figure 3.3.1.48 we find that G’ – W1 is planar so V1

132 ∈N(A). In order to avoid these contradictions and continue to assume that G’ is

MMNA, we must have V1, V2 ∈N(A). We next argue W1 ∈N(A) and W3 ∈N(A).

Consider G’ – V1. We can construct G’ – V1 two different ways and still find a construction that is planar.

a.) G’ – V1 b.) G’ – V1

V3 V3 W W3 3 A U U W 2 V V 2 W2 2

A

W1 W1

Figure 3.3.1.49. The Eighth (7,12) Graph with Vertex A Added, has Vertex V1 Removed and is Found to be Planar in Two Different Constructions.

In both constructions in Figure 3.3.1.49 we find a contradiction to G’ being a MMNA graph. In construction (a) from Figure 3.3.1.49 we find that G’ – V1 is planar so W1

∈N(A). In construction (b) from Figure 3.3.1.49 we find that G’ – V1 is planar so W3

∈N(A). Also, let us not forget that U is of degree two in G, so it must also be in N(A) in order to be a vertex of degree three in G’.

Assume that that A is of degree five. Since U, V1, V2, W1, W3 ∈N(A) we can say that V3, W2 ∉N(A). However, when we construct the complement of G’ we find that the complement is a subgraph of the P8 graph complement. So, G’ is not MM.

133 A V3 W2

V2 W3 V1 W1

U

Figure 3.3.1.50. The Graph Represents the Complement of G’ When we Assume that V3, W2 ∉N(A). The complement is a subgraph of the P8 complement.

When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Assume that A is of degree six. Since U, V1, V2, W1, W3 ∈N(A) then, one of the following vertices is not an element of N(A), V3 or W2. However, when we look at the complement of G’ for every case we find a contradiction to the assumption that

G’ is MMNA. We find that each complement of G’ is a subgraph of the P8 graph complement. This is a contradiction as it means G’ is not MM.

a.) V3 ∉N(A) b.) W2 ∉N(A) A A V3 W2 V3 W2

V2 W3 V2 W3 V1 W1 V1 W1

U U

Figure 3.3.1.51. These Two Graphs Represent the Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A). The two graphs are a subgraph of the P8 complement.

134 When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the eighth (7,12) graph. However, a MMNA G’ may arise from a different non-planar (7,12) graph of minimum degree at least two. Let us look at case nine.

Case 9: Let G be the ninth non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (44, 32, 2) graph of Figure

3.3.1.52, which also shows its complement.

U V W3 3 W1

W2 U

W3 V1 V2

W2 V2

V3

V1 W1 Figure 3.3.1.52. The Ninth Non-Planar (7,12) Graph and its Complement.

Let us construct vertex A, where A is either of degree five or degree six, in order to create a graph on eight vertices which we will refer to as G’. Consider G’ – W1. If A is of degree six, assume that W1 ∈N(A). If A is of degree five, assume that W1 ∉N(A).

We can construct G’ – W1 two ways and still find a construction that is planar.

135 a.) G’ – W1 b.) G’ – W1

A W U 3 W3 U V3 V3 A V2 W2 V2 W2

V V1 1

Figure 3.3.1.53. The Ninth (7,12) Graph with Vertex A Added, has Vertex W1 Removed and is Found to be Planar in Two Different Constructions.

In both constructions in Figure 3.3.1.53 we find a contradiction to G’ being a MMNA graph. In construction (a) from Figure 3.3.1.53 we find that G’ – W1 is planar so V1

∈N(A). In construction (b) from Figure 3.3.1.53 we find that G’ – W1 is planar so V2

∈N(A). In order to avoid these contradictions and continue to assume that G’ is

MMNA, we must have V1, V2 ∈N(A). We next argue W1 ∈N(A). Consider G’ – V2, where A is either of degree five or degree six. Then let us arrange the remaining vertices so that five are adjacent to A in order to create a planar graph. This is another contradiction to our assumption that G’ is MMNA, that shows W1 ∈N(A).

Also, let us not forget that U is of degree two in G, so it must also be in N(A) so, U, V1,

V2, W1 ∈N(A).

136

U V3 W3

W2 A

V1 W1

Figure 3.3.1.54. The Ninth (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar.

Assume that that A is of degree five. Since U, V1, V2, W1 ∈N(A) one of the following combinations of vertices is not in N(A), {V3, W2}, {V3, W3}, or {W2, W3}.

However, we find a contradiction with every one of the combinations. When we construct the complement of G’ for each pair of vertices not in N(A) we find that the complement is a subgraph of the P8 graph complement. So, G’ is not MM.

a.) V3, W2 ∉N(A) b.) V3, W3 ∉N(A) c.) W2, W3 ∉N(A)

W1 W1 W1 W2 U W2 U W2 U

W 3 A V1 V W3 W3 V 2 A V1 V2 A 1 V2

V3 V V3 3

Figure 3.3.1.55. These Three Graphs Represent the Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Each complement is a subgraph of the P8 complement.

137 When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Assume that A is of degree six. Since U, V1, V2, W1 ∈N(A) then, one of the following vertices is not an element of N(A), V3, W2, or W3. However when we look at the complement of G’ for every case we find that each complement is a subgraph of the P8 graph complement. This is a contradiction as it means G’ is not MM.

a.) V3 ∉N(A) b.) W2 ∉N(A) c.) W3 ∉N(A)

W1 W1 W1 W2 U W2 U W2 U

W W3 W3 V 3 A V1 V A V1 V2 A 1 V2 2

V V3 V3 3

Figure 3.3.1.56. These Three Graphs Represent the Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A). Each complement is a subgraph of the P8 complement.

When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the ninth (7,12) graph. However, a MMNA G’ may arise from a different non-planar (7,12) graph of minimum degree at least two. Let us look at case ten.

138 Case 10: Let G be the tenth non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (44, 32, 2) graph of Figure

3.3.1.57, which also shows its complement.

V3 W3

V W2 2 U V3 W1 W2 V2 W3 V1 U

V1 W1 Figure 3.3.1.57. The Tenth Non-Planar (7,12) Graph and its Complement.

Let us construct vertex A, where A is either of degree five or degree six, in order to create a graph on eight vertices which we will refer to as G’. Consider G’ – V1. If A is of degree six, assume that V1 ∈N(A). If A is of degree five, assume that V1 ∉N(A). We can construct G’ – V1 two different ways and still find a construction that is planar.

a.) G’ – V1 b.) G’ – V1

V3 W3 V3 W3 A

W2 V2 U V2

W2 A U W1 W1

Figure 3.3.1.58. The Tenth (7,12) Graph with Vertex A Added, has Vertex V1 Removed and is Found to be Planar in Two Different Constructions.

139 In both constructions in Figure 3.3.1.58 we find a contradiction to G’ being a MMNA graph. In construction (a) from Figure 3.3.1.58 we find that G’ – V1 is planar so W3

∈N(A). In construction (b) from Figure 3.3.1.58 we find that G’ – V1 is planar so W1

∈N(A). Also, let us not forget that U is of degree two in G, so it must also be in N(A) so, U, W1, W3 ∈N(A).

Assume that that A is of degree five. Suppose that W2 ∉N(A). Consider V1,

W2 ∉N(A) and V2, W2 ∉N(A). Now let us construct a complement of G’ with each consideration. We will find that the complement of G’, in both cases, is a subgraph of a P7 graph complement on eight vertices. So, G’ is not MM.

a.) V1, W2 ∉N(A) b.) V2, W2 ∉N(A)

U V3 U V3

W1 W1 V1 V1 V2 V2 W3 W3

W A 2 W2 A

Figure 3.3.1.59. The Graphs Represent the Complement of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Both complements are a subgraph of a P7 graph complement on eight vertices.

Now let us consider V3, W2 ∉N(A). Let us construct the complement of G’ with this consideration. We will find that the complement of G’ has a “Pentagon – Segment” subgraph. This is a contradiction to the assumption that G’ is MMNA, because this implies that G’ is an apex graph.

140

V2 V2 U V3 U V3

W1 W1

W3 V1 W3 V1 W A W A 2 2 Figure 3.3.1.60. The Graphs Represent the Complement of G’ When We Consider that V3, W2 ∉N(A). The complement of G’ has a triangulation complement, known as “Pentagon – Segment” as a subgraph that is highlighted in black on the graph to the right.

Since all three cases where we consider W2 ∉N(A) construct a complement that contradicts G’ being MMNA, we must conclude that W2 ∈N(A). Since U, W1, W2, W3

∈N(A) one of the following combinations of vertices is not in N(A), {V1, V2}, {V1, V3}, or {V2, V3}. However, we find a contradiction with every one of the combinations.

When we construct the complement of G’ for each pair of vertices not in N(A) we find that the complement is a subgraph of the K4,4 - e graph complement. This is a contradiction as it means G’ is not MM.

a.) V1, V2 ∉N(A) b.) V1, V3 ∉N(A) c.) V2, V3 ∉N(A)

V W2 2 W2 V2 W2 V2 U V3 U V3 U V3 A A A W1 W1 W1 W 3 V1 W3 V W3 V 1 1

Figure 3.3.1.61. These Three Graphs Represent the Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Each complement is a subgraph of the K4,4 - e complement.

141 When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Assume that A is of degree six. Since U, W1, W3 ∈N(A) then, one of the following vertices is not an element of N(A), V1, V2, V3, or W2. However when we look at the complements of G’, each complement is a subgraph of either the K4,4 – e graph complement or a P7 graph complement on eight vertices. So, G’ is not MM.

a.) V1 ∉N(A) b.) V2 ∉N(A)

V W2 V2 W2 2 U V3 U V3 A A W1 W1

W3 W3 V V1 1

c.) V3 ∉N(A) d.) W2 ∉N(A)

V W2 V2 U 3 U V3 W1 A V1 W1 V2 W3 W3 V 1 W2 A

Figure 3.3.1.62. These Four Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A). The three graphs (a)-(c) are subgraphs of the K4,4 – e complement, while graph (d) is a subgraph of a P7 complement on eight vertices.

When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

142 We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the tenth (7,12) graph. However, a MMNA G’ may arise from a different non-planar (7,12) graph of minimum degree at least two. Let us look at case eleven.

Case 11: Let G be the eleventh non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (44, 32, 2) graph of

Figure 3.3.1.63, which also shows its complement.

W V3 3 U V2

W2 W3 V3

W2 V2 W1 V1

U V1 W1 Figure 3.3.1.63. The Eleventh Non-Planar (7,12) Graph and its Complement.

Let us construct vertex A, where A is either of degree five or degree six, in order to create a graph on eight vertices which we will refer to as G’. Consider G’ – V1. If A is of degree six, assume that V1 ∈N(A). If A is of degree five, assume that V1 ∉N(A).

Then let us arrange the remaining vertices so that five are adjacent to A in order to create a planar graph.

143

V3 W3

W2 V2

U A W 1 Figure 3.3.1.64. The Eleventh (7,12) Graph with Vertex A Added, has Vertex V1 Removed and is Found to be Planar.

We can construct a planar graph if V1 is removed from G’ and A is connected to U, V2,

V3, W1, and W2. This is a contradiction to our assumption that G’ is MMNA, that shows vertex W3 ∈N(A). We next argue V2 ∈N(A). Consider G’ – W3. We still can construct a planar graph after we take G’ – W3 and A is adjacent to U, V1, V3, W1, and

W2. This is still a contradiction to our assumption that G’ is MMNA, that shows the vertex V2 is also in N(A). Also, let us not forget that U is of degree two in G, so it must also be in N(A) in order to be a degree three vertex in G’.

V3

V W2 2

V1 U A W 1 Figure 3.3.1.65. The Eleventh (7,12) Graph with Vertex A Added, has Vertex W3 Removed and is Found to be Planar.

144

Assume that that A is of degree five. Suppose that W2 ∉N(A). Consider V1,

W2 ∉N(A) and W1, W2 ∉N(A). Now let us construct a complement of G’ with each consideration. We will find that the complement of G’, in both cases, is a subgraph of the P8 graph complement. So, G’ is not MM.

a.) V1, W2 ∉N(A) b.) W1, W2 ∉N(A)

W3 W3 W2 U W2 U

A V A V2 W1 2 V3 W1 V3

V V1 1 Figure 3.3.1.66. The Graphs Represent the Complement of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Both complements are a subgraph of the P8 complement graph.

Now let us Consider V3, W2 ∉N(A). Let us construct the complement of G’ with this consideration. We will find that the complement of G’ has a “Pentagon – Segment” subgraph. This is a contradiction to the assumption that G’ is MMNA, because this implies that G’ is an apex graph.

U U W3 W3 V2 V2 A A W2 W2 V3 V3

W1 V1 W1 V 1 Figure 3.3.1.67. The Graphs Represent the Complement of G’ When We Consider that V3, W2 ∉N(A). The complement of G’ has a triangulation complement, known as “Pentagon – Segment” as a subgraph that is highlighted in black on the graph to the right.

145

Since all three cases where we consider W2 ∉N(A) construct a complement that contradicts G’ being MMNA, we must conclude that W2 ∈N(A). Since U, V2, W2, W3

∈N(A) one of the following combinations of vertices is not in N(A), {V1, V3}, {V1, W1}, or {V3, W1}. However, we find a contradiction with every one of the combinations.

When we construct the complement of G’ for each pair of vertices not in N(A) we find that the complement is a subgraph of the P8 graph complement. This is a contradiction as it means G’ is not MM.

a.) V1, V3 ∉N(A) b.) V1, W1 ∉N(A) c.) V3, W1 ∉N(A)

V2 V2 V2 U V3 U V3 U V3

W2 A W2 A W A W3 V1 W3 V1 2 W3 V1 W W 1 1 W 1

Figure 3.3.1.68. These Three Graphs Represent the Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A Each complement is a subgraph of the P8 complement.

When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Assume that A is of degree six. Since U, V2, W3 ∈N(A) then, one of the following vertices is not an element of N(A), V1, V3, W1, or W2. However when we look at the complement of G’ for every case we find that each complement is a subgraph of the P8 graph complement. So, G’ is not MM.

146 a.) V1 ∉N(A) b.) V3 ∉N(A) c.) W1 ∉N(A) d.) W2 ∉N(A)

V V V W3 2 2 2 W U U V3 U V3 U V3 2

W2 A W A W2 A A V W 2 W W1 2 V3 3 V1 W3 V1 3 V1 W W 1 W1 1 V1

Figure 3.3.1.69. These Four Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A). Each complement is a subgraph of the P8 complement.

When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the eleventh (7,12) graph. However, a MMNA G’ may arise from a different non-planar (7,12) graph of minimum degree at least two. Let us look at case twelve.

Case 12: Let G be the twelfth non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (43, 34) graph of Figure

3.3.1.70, which also shows its complement.

V W3 3 U

V3 W2 W3

W2 V2 V2 W1 V1

V1 U W1 Figure 3.3.1.70. The Twelfth Non-Planar (7,12) Graph and its Complement.

147 Let us construct vertex A, where A is either of degree five or degree six, in order to create a graph on eight vertices which we will refer to as G’. Consider G’ – W3. If A is of degree six, assume that W3 ∈N(A). If A is of degree five, assume that W3 ∉N(A).

Then let us arrange the remaining vertices so that five are adjacent to A in order to create a planar graph.

V3

W2 V2

V1 U A W1

Figure 3.3.1.71. The Twelfth (7,12) Graph with Vertex A Added, has Vertex W3 Removed and is Found to be Planar.

We can construct a planar graph if W3 is removed from G’. This is a contradiction to our assumption that G’ is MMNA, that shows vertex V2 ∈N(A). We next argue W3

∈N(A). Consider G’ – V1. If A is of degree six, assume that V1 ∈N(A). If A is of degree five, assume that V1 ∉N(A). We still can construct a planar graph after we take G’ –

V1. This is still a contradiction to our assumption that G’ is MMNA, that shows the vertex W3 is also in N(A).

148

V3 W3

W2 V2

U A

W 1

Figure 3.3.1.72. The Twelfth (7,12) Graph with Vertex A Added, has Vertex V1 Removed and is Found to be Planar.

Assume that that A is of degree five. Suppose that W1 ∉N(A). Consider V1,

W1 ∉N(A) and W1, W2 ∉N(A). Now let us construct a complement of G’ with each consideration. We will find that the complement of G’, in both cases, is a subgraph of the P8 graph complement. So, G’ is not MM.

a.) V1, W1 ∉N(A) b.) W1, W2 ∉N(A)

W3 W3 W2 U W2 U V V V V A 2 3 A 2 3 W1 W1 V V 1 1

Figure 3.3.1.73. These Graphs Represent the Complement of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Both complements are a subgraph of the P8 graph complement.

149

Now let us consider V3, W1 ∉N(A). Let us construct the complement of G’ with this consideration. We will find that the complement of G’ has a “House” subgraph. This is a contradiction to the assumption that G’ is MMNA, because this implies that G’ is an apex graph.

U U W3 W3 A A V3 V3 W2 W2

V2 V2 W1 W1 V V1 1 Figure 3.3.1.74. The Graphs Represent the Complement of G’ When We Consider that V3, W1 ∉N(A). The complement of G’ has a triangulation complement, known as “House” as a subgraph that is highlighted in black on the graph to the right.

Now let us consider U, W1 ∉N(A). Let us construct the complement of G’ with this consideration. We will find that the complement of G’ has a “House” subgraph. This is a contradiction to the assumption that G’ is MMNA, because this implies that G’ is an apex graph.

W3 U W3 U

V3 V3 W2 A W2 A

V2 V2 W1 W1 V V1 1 Figure 3.3.1.75. The Graphs Represent the Complement of G’ When We Consider that U, W1 ∉N(A). The complement of G’ has a triangulation complement, known as “House” as a subgraph that is highlighted in black on the graph to the right.

150

Since all four cases where we consider W1 ∉N(A) construct a complement that contradicts G’ being MMNA, we must conclude that W1 ∈N(A). Suppose that V1

∉N(A). Consider V1, W2 ∉N(A). Now let us construct the complement of G’ with this consideration. We will find that the complement of G’ is a subgraph of the P8 graph complement. So, G’ is not MM.

W3 W2 U V V A 2 3 W1

V1

Figure 3.3.1.76. The Graph Represents the Complement of G’ When We Assume that V1, W2 ∉N(A). The complement is a subgraph of the P8 complement.

Consider V1, V3 ∉N(A). Let us construct the complement of G’ with this consideration. We will find that the complement of G’ is a subgraph of one of the P7 graph complements on eight vertices. So, G’ is not MM.

U V3

A W2 V2 W3

W1 V 1 Figure 3.3.1.77. The Graph Represents the Complement of G’ When We Assume that V1, V3 ∉N(A). The complement is a subgraph of a P7 complement on eight vertices.

151

Now let us consider U, V1 ∉N(A). Let us construct the complement of G’ with this consideration. We will find that the complement of G’ has a “House” subgraph. This is a contradiction to the assumption that G’ is MMNA, because this implies that G’ is an apex graph.

W3 U W3 U

V3 V3 W2 A W2 A

V2 V2 W1 W1 V1 V1

Figure 3.3.1.78. The Graphs Represent the Complement of G’ When We Consider that U, V1 ∉N(A). The complement of G’ has a triangulation complement, known as “House” as a subgraph that is highlighted in black on the graph to the right.

Since all three cases where we consider V1 ∉N(A) construct a complement that contradicts G’ being MMNA, we must conclude that V1 ∈N(A). Since V1, V2, W1, W3

∈N(A) one of the following combinations of vertices is not in N(A), {U, V3}, {U, W2}, or {V3, W2}. However, when we construct the complement of G’ for each pair of vertices not in N(A), we find a contradiction to the assumption that G’ is MMNA.

When we consider U, V3 ∉N(A), we find that the complement of G’ is a subgraph of the P8 graph complement. This is a contradiction as it means G’ is not MM.

152 A U V3

W2 V1 W3 V2

W1

Figure 3.3.1.79. The Graph Represents the Complement of G’ When We Assume that U, V3 ∉N(A). The complement is a subgraph of the P8 complement.

When we consider U, W2 ∉N(A), we will find that the complement of G’ is a subgraph of one of the P7 graph complements on eight vertices. So, G’ is not MM.

U V3 W A 3 V2 V1

W2 W1

Figure 3.3.1.80. The Graph Represents the Complement of G’ When We Assume that U, W2 ∉N(A). The complement is a subgraph of a P7 complement on eight vertices.

Finally let us consider V3, W2 ∉N(A). Let us construct the complement of G’ with this consideration. We will find that the complement of G’ has a “House” subgraph. This

153 is a contradiction to the assumption that G’ is MMNA, because this implies that G’ is an apex graph.

U W3 W3 U

V3 V3 W2 W2 A A V2 V2 W1 W1 V1 V1

Figure 3.3.1.81. The Graphs Represent the Complement of G’ When We Consider that V3, W2 ∉N(A). The complement of G’ has a triangulation complement, known as “House” as a subgraph that is highlighted in black on the graph to the right.

When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Assume that A is of degree six. Since V2, W3 ∈N(A) then, one of the following vertices is not an element of N(A), U, V1, V3, W1, or W2. However when we look at the complement of G’ for every case we find that each complement is a subgraph of the P8 graph complement. This is a contradiction as it means G’ is not

MM.

154 a.) U ∉N(A) b.) V1 ∉N(A) c.) V3 ∉N(A)

A W3 A U V U V 3 W2 U 3 V2 V3 W V W V1 2 W V 1 A 2 W V 3 2 W1 3 2

W1 V W1 1 d.) W1 ∉N(A) e.) W2 ∉N(A)

W3 W3 W2 U W2 U V V V2 V3 A 2 3 A W1 W1 V V 1 1

Figure 3.3.1.82. These Five Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A). All the graphs represent a subgraph of the P8 graph complement.

When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the twelfth (7,12) graph. However, a MMNA G’ may arise from a different non-planar (7,12) graph of minimum degree at least two. Let us look at case thirteen.

155 Case 13: Let G be the thirteenth non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (43, 34) graph of

Figure 3.3.1.83, which also shows its complement.

V3 W3

U V3

W 3 W1 V2

W2 V2 W2 V1 U

V1 W1 Figure 3.3.1.83. The Thirteenth Non-Planar (7,12) Graph and its Complement.

Let us construct vertex A, where A is either of degree five or degree six, in order to create a graph on eight vertices which we will refer to as G’. Consider G’ – V2. If A is of degree six, assume that V2 ∈N(A). If A is of degree five, assume that V2 ∉N(A). We can construct G’ – V2 two different ways and still find a construction that is planar.

a.) G’ – V2 b.) G’ – V2

W3 V3 V3 W3

W2 A W2

U A U

V W1 W 1 V1 1

Figure 3.3.1.84. The Thirteenth (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar in Two Different Constructions.

156 In both constructions in Figure 3.3.1.84 we find a contradiction to G’ being a MMNA graph. In construction (a) from Figure 3.3.1.84 we find that G’ – V2 is planar so W3

∈N(A). In construction (b) from Figure 3.3.1.84 we find that G’ – V2 is planar so W1

∈N(A). We next argue V2 ∈N(A). Consider G’ – W1. We still can construct a planar graph after we take G’ – W1. This is still a contradiction to our assumption that G’ is

MMNA, that shows vertex V2 is also in N(A). In order to avoid these contradictions and continue to assume that G’ is MMNA, we must have V2, W1, W3 ∈N(A).

W3 A

V3

V2 W2

U

V1

Figure 3.3.1.85. The Thirteenth (7,12) Graph with Vertex A Added, has Vertex W1 Removed and is Found Once Again to be Planar.

Assume that that A is of degree five. Suppose that U ∉N(A). Consider U,

W2 ∉N(A). We will find that the complement of G’ is a subgraph of one of the P7 graph complements on eight vertices. So, G’ is not MM.

157 U V3 W1 V2 A

W3 V1 W2

Figure 3.3.1.86. The Graph Represents the Complement of G’ When We Assume that U, W2 ∉N(A). The complement is a subgraph of a P7 complement on eight vertices.

Now consider U, V3 ∉N(A). When we construct the complement of G’ we find that it is a subgraph of the P8 graph complement. So, G’ is not MM.

A V3 U

W3 W1 V1 V2

W2

Figure 3.3.1.87. The Graph Represents the Complement of G’ When We Assume that U, V3 ∉N(A). The complement is a subgraph of the P8 complement.

Finally let us consider U, V1 ∉N(A). Let us construct the complement of G’ with this consideration. We will find that the complement of G’ has a “House” subgraph. This

158 is a contradiction to the assumption that G’ is MMNA, because this implies that G’ is an apex graph.

U V3 U V3 V2 W3 V2 W3 W1 A W1 A V1 V W2 W 1 2 Figure 3.3.1.88. The Graphs Represent the Complement of G’ When We Consider that U, V1 ∉N(A). The complement of G’ has a triangulation complement, known as “House” as a subgraph that is highlighted in black on the graph to the right.

Since all three cases where we consider U ∉N(A) construct a complement that contradicts G’ being MMNA, we must conclude that U ∈N(A). Since U, V2, W1, W3

∈N(A) one of the following combinations of vertices is not in N(A), {V1, V3}, { V1, W2}, or {V3, W2}. However, when we construct the complement of G’ for each pair of vertices not in N(A), we find a contradiction to the assumption that G’ is MMNA.

When we consider V1, V3 ∉N(A), we find that the complement of G’ is a subgraph of one of the P7 graph complements on eight vertices. So, G’ is not MM.

U V3

W1 W3 V2 A W 2 V 1 Figure 3.3.1.89. The Graph Represents the Complement of G’ When We Assume that V1, V3 ∉N(A). The complement is a subgraph of a P7 complement on eight vertices.

159

When we consider V1, W2 ∉N(A), we find that the complement of G’ is a subgraph of the P8 graph complement. So, G’ is not MM.

A V1 W2

W1 V3 W3 V2

U

Figure 3.3.1.90. The Graph Represents the Complement of G’ When We Assume that V1, W2 ∉N(A). The complement is a subgraph of the P8 complement.

Now, consider V3, W2 ∉N(A). Let us construct the complement of G’ with this consideration. We will find that the complement of G’ has a “House” subgraph. This is a contradiction to the assumption that G’ is MMNA, because this implies that G’ is an apex graph.

U U V3 V3

W3 W A 1 V2 W3 W A 1 V2

W2 V 1 W2 V 1 Figure 3.3.1.91. The Graphs Represent the Complement of G’ When We Consider that V3, W2 ∉N(A). The complement of G’ has a triangulation complement, known as “House” as a subgraph that is highlighted in black on the graph to the right.

160 When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Assume that A is of degree six. Since V2, W1, W3 ∈N(A) then, one of the following vertices is not an element of N(A), U, V1, V3, or W2. However when we look at the complement of G’ for every case we find that each complement is a subgraph of the P8 graph complement. This is a contradiction as it means G’ is not MM.

a.) U ∉N(A) b.) V1 ∉N(A) c.) V3 ∉N(A) d.) W2 ∉N(A)

A A A A V1 V1 V3 U W2 V3 U W2

W W3 W1 W W3 W1 W 1 V3 W 1 V3 3 V2 V1 V2 3 V2 V1 V2 U W U W2 2

Figure 3.3.1.92. These Four Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A). Each complement is a subgraph of the P8 complement.

When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the thirteenth (7,12) graph. However, a MMNA G’ may arise from a different non-planar (7,12) graph of minimum degree at least two. Let us look at case fourteen.

161 Case 14: Let G be the fourteenth non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (45, 22) graph of

Figure 3.3.1.93, which also shows its complement.

V2 U2 V3 U U1 1 V3

V1 V 5 V4 V2

V4

U2 V1

V5 Figure 3.3.1.93. The Fourteenth Non-Planar (7,12) Graph and its Complement.

Let us construct vertex A, where A is either of degree five or degree six, in order to create a graph on eight vertices which we will refer to as G’. Consider G’ – V2. If A is of degree six, assume that V2 ∈N(A). If A is of degree five, assume that V2 ∉N(A). We can construct G’ – V2 two different ways and still find a construction that is planar. a.) G’ – V2 b.) G’ – V2

A V3 V3

U2 U U1 2 V1 U1 V1 A V4 V4

V V5 5

Figure 3.3.1.94. The Fourteenth (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar in Two Different Constructions.

162 In both constructions in Figure 3.3.1.94 we find a contradiction to G’ being a MMNA graph. In construction (a) from Figure 3.3.1.94 we find that G’ – V2 is planar so V4

∈N(A). In construction (b) from Figure 3.3.1.94 we find that G’ – V2 is planar so V5

∈N(A). Also, let us not forget that U1 and U2 are both of degree two in G, in order for each vertex to be of degree three in G’ they both must be in N(A). So, the elements in

N(A) so far are, U1, U2, V4, and V5.

Let us assume that A is of degree five. Since U1, U2, V4, V5 ∈N(A) one of the following combinations of vertices is not in N(A), {V1, V2}, {V1, V3}, or {V2, V3}.

However, we find a contradiction with every one of the combinations. When we construct the complement of G’ for each pair of vertices not in N(A), we find that the complement is a subgraph of a P7 graph complement on eight vertices. This is a contradiction as it means G’ is not MM.

a.) V1, V2 ∉N(A) b.) V1, V3 ∉N(A) c.) V2, V3 ∉N(A)

U1 V3 U1 V3 U1 V3

V4 V4 V4 V V2 V V2 V5 2 V5 5 A A A

V V U2 V1 U2 1 U2 1

Figure 3.3.1.95. These Three Graphs Represent the Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Each complement is a subgraph of a P7 complement on eight vertices.

When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

163

Assume that A is of degree six. Since U1, U2, V4, V5 ∈N(A) then, one of the following vertices is not an element of N(A), V1, V2, or V3. However when we look at the complement of G’ for every case we find that each complement is a subgraph of a

P7 graph complement on eight vertices. So, G’ is not MM.

a.) V1 ∉N(A) b.) V2 ∉N(A) c.) V3 ∉N(A)

U1 V3 U1 V3 U1 V3

V4 V4 V4 V V2 V V2 V V2 5 A 5 A 5 A

V V V U2 1 U2 1 U2 1

Figure 3.3.1.96. These Three Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A). Each complement is a subgraph of a P7 graph complement on eight vertices.

When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the fourteenth (7,12) graph. However, a MMNA G’ may arise from a different non-planar (7,12) graph of minimum degree at least two. Let us look at case fifteen.

Case 15: Let G be the fifteenth non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (45, 22) graph of

Figure 3.3.1.97, which also shows its complement.

164

V2 V3 U1 V5 U U2 1 V1 U2

V4 V1 V2 V3 V4

V5 Figure 3.3.1.97. The Fifteenth Non-Planar (7,12) Graph and its Complement.

Let us construct vertex A, where A is either of degree five or degree six, in order to create a graph on eight vertices which we will refer to as G’. Consider G’ – V3. If A is of degree six, assume that V3 ∈N(A). If A is of degree five, assume that V3 ∉N(A).

Then let us arrange the remaining vertices so that five are adjacent to A in order to create a planar graph.

V2 U1

U2 V1 A

V5 V 4 Figure 3.3.1.98. The Fifteenth (7,12) Graph with Vertex A Added, has Vertex V3 Removed and is Found to be Planar.

We can construct a planar graph if V3 is removed from G’. This is a contradiction to our assumption that G’ is MMNA, that shows vertex V5 ∈N(A). Also, let us not forget

165 that U1 and U2 are both of degree two in G, so they both must be in N(A). As of now the known elements in N(A) are, U1, U2, and V5.

Let us assume that A is of degree five. Since U1, U2, V5 ∈N(A) one of the following combinations of vertices is not in N(A), {V1, V2}, {V1, V3}, {V1, V4}, {V2, V3},

{V2, V4}, or {V3, V4}. However, we find a contradiction with every one of the combinations. When we construct the complement of G’ for each pair of vertices not in N(A), we find that the complement is a subgraph of the P8 graph complement.

This is a contradiction as it means G’ is not MM.

a.) V1, V2 ∉N(A) b.) V1, V3 ∉N(A) c.) V1, V4 ∉N(A)

V5 V5 V5 U1 U2 U1 U2 U1 U2

V V3 V2 V3 V V3 2 2 V1 V4 V1 V4 V1 V4 A A A

d.) V2, V3 ∉N(A) e.) V2, V4 ∉N(A) f.) {V3, V4} ∉N(A)

V5 V5 V5 U2 U1 U2 U1 U2 U1

V V V3 V V2 3 2 V V2 3 V1 V4 V1 4 V1 V4

A A A

Figure 3.3.1.99. These Six Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Each complement is a subgraph of the P8 complement.

166 When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Assume that A is of degree six. Since U1, U2, V5 ∈N(A) then, one of the following vertices is not an element of N(A), V1, V2, V3, or V4. However when we look at the complement of G’ for every case we find that each complement is a subgraph of the P8 graph complement. This is a contradiction as it means G’ is not MM.

a.) V1 ∉N(A) b.) V2 ∉N(A) c.) V3 ∉N(A) d.) V4 ∉N(A)

V5 V5 V5 V5 U U U U2 U1 U2 1 U2 1 U2 1

V V3 V2 V3 V2 V3 V2 V3 2 V V4 V4 V1 V4 V1 4 V1 V1 A A A A

Figure 3.3.1.100. These Six Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not of N(A). Each complement is a subgraph of the P8 graph complement.

When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the fifteenth (7,12) graph. However, a MMNA G’ may arise from a different non-planar (7,12) graph of minimum degree at least two. Let us look at case sixteen.

167 Case 16: Let G be the sixteenth non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (45, 22) graph of

Figure 3.3.1.101, which also shows its complement.

V2 U1 V3 V5 U2 U1

V4 V1

V1 U2 V3 V4

V2 V5 Figure 3.3.1.101. The Sixteenth Non-Planar (7,12) Graph and its Complement.

Let us construct vertex A, where A is either of degree five or degree six, in order to create a graph on eight vertices which we will refer to as G’. Consider G’ – V2. If A is of degree six, assume that V2 ∈N(A). If A is of degree five, assume that V2 ∉N(A). We can construct G’ – V2 three different ways and still find a construction that is planar.

a.) G’ – V2 b.) G’ – V2 c.) G’ – V2

V3 A V3 V1 V V3 A U2 U1 4 U V1 1 U2 A U2 V4 U1 V4 V1 V V5 V 5 5

Figure 3.3.1.102. The Sixteenth (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar Three Different Constructions.

168 In all three constructions in Figure 3.3.1.102 we find a contradiction to G’ being a

MMNA graph. In construction (a) from Figure 3.3.1.102 we find that G’ – V2 is planar so V3 ∈N(A). In construction (b) from Figure 3.3.1.102 we find that G’ – V2 is planar so V4 ∈N(A). In construction (c) from Figure 3.3.1.102 we find that G’ – V2 is planar so V5 ∈N(A). Also, let us not forget that U1 and U2 are both of degree two in G, so they both must be in N(A). So, as of now the known elements in N(A) are, U1, U2, V3,

V4, and V5.

Let us assume that A is of degree five. Since U1, U2, V3, V4, V5 ∈N(A) we can say that the two vertices V1 and V2 are not elements of N(A). However, we find a contradiction. When we construct the complement of G’, we find that the complement is a subgraph of a P7 graph complement on eight vertices. This is a contradiction as it means G’ is not MM.

U1 V5

V4 A V3 V1 U2

V2

Figure 3.3.1.103. The Graph Represents the Complement of G’ When We Assume that V1, V2 ∉N(A). The complement is a subgraph of a P7 graph complement.

169 When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Assume that A is of degree six. Since U1, U2, V3, V4, V5 ∈N(A) then, one of the following vertices is not an element of N(A), V1 or V2. However, when we look at the complement of G’ for every case we find a contradiction to the assumption that

G’ is MMNA. We find that each complement of G’ is a subgraph of a P7 graph complement on eight vertices. This is a contradiction as it means G’ is not MM.

a.) V1 ∉N(A) b.) V2 ∉N(A)

U1 V5 U1 V5

V4 V4 A A V3 V3 U V1 U2 V1 2

V V2 2

Figure 3.3.1.104. These Two Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A). The two graphs are subgraphs of a P7 graph complement on eight vertices.

When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the sixteenth (7,12) graph. However, a MMNA G’ may arise from a

170 different non-planar (7,12) graph of minimum degree at least two. Let us look at case seventeen.

Case 17: Let G be the seventeenth non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (5, 42, 33, 2) graph of Figure 3.3.1.105, which also shows its complement.

V3 W3 U V2

W2 V3 W W2 V2 3

W1 V1

U V1 W1 Figure 3.3.1.105. The Seventeenth Non- Planar (7,12) Graph and its Complement.

Let us construct vertex A, where A is either of degree five or degree six, in order to create a graph on eight vertices which we will refer to as G’. Consider G’ – V2. If A is of degree six, assume that V2 ∈N(A). If A is of degree five, assume that V2 ∉N(A). We can construct G’ – V2 two different ways and still find a construction that is planar.

a.) G’ – V2 b.) G’ – V2

W3 W3 V3 V3

W W2 A 2 W1 A

U V1 U V1 W1 Figure 3.3.1.106. The Seventeenth (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar Two Different Constructions.

171 In both constructions in Figure 3.3.1.106 we find a contradiction to G’ being a MMNA graph. In construction (a) from Figure 3.3.1.106 we find that G’ – V2 is planar so W3

∈N(A). In construction (b) from Figure 3.3.1.106 we find that G’ – V2 is planar so W2

∈N(A). We next argue V2 ∈N(A). Consider G’ – W3. We still can construct a planar graph after we take G’ – W3. This is a contradiction to our assumption that G’ is

MMNA that shows vertex V2 is also in N(A). Also, let us not forget that U is of degree two in G, so it must be in N(A). In order to avoid contradictions and to have vertices of at least degree three in G’ we must have U, V2, W2, W3 ∈N(A).

V3

W2 V2

V1

A U W1

Figure 3.3.1.107. The Seventeenth (7,12) Graph with Vertex A Added, has Vertex W3 Removed and is Found Once Again to be Planar.

Let us assume that A is of degree five. Since U, V2, W2, W3 ∈N(A) one of the following combinations of vertices is not in N(A), {V1, V3}, {V1, W1}, or {V3, W1}.

However, we find a contradiction with every one of the combinations. When we construct the complement of G’ for each pair of vertices not in N(A), we find that the

172 complement is a subgraph of a P7 graph complement on eight vertices. This is a contradiction as it means G’ is not MM.

a.) V1, V3 ∉N(A) b.) V1, W1 ∉N(A) c.) V3, W1 ∉N(A) U U U V2 V2 V2

W2 W2 W2 V3 V3 V3 W3 W3 W3

W A V1 A V1 A V1 1 W1 W1 Figure 3.3.1.108. These Three Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Each complement is a subgraph of a P7 graph complement on eight vertices.

When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Assume that A is of degree six. Since U, V2, W2, W3 ∈N(A) then, one of the following vertices is not an element of N(A), V1, V3, or W1. However when we look at the complement of G’ for every case we find that each complement is a subgraph of a

P7 graph complement on eight vertices. So, G’ is not MM.

a.) V1 ∉N(A) b.) V3 ∉N(A) c.) W1 ∉N(A) U U U V2 V2 V2

W W2 W2 2 V V3 V3 3 W W3 W3 3

V V1 W A 1 A V1 W1 A 1 W1 Figure 3.3.1.109. These Three Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A). Each complement is a subgraph of a P7 graph complement on eight vertices.

173 When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the seventeenth (7,12) graph. However, a MMNA G’ may arise from a different non-planar (7,12) graph of minimum degree at least two. Let us look at case eighteen.

Case 18: Let G be the eighteenth non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (52, 34, 2) graph of

Figure 3.3.1.110, which also shows its complement.

V3 W3

W2 V2

U V3 W1 W2 V2 W3 V1

U V1 W1 Figure 3.3.1.110. T The Eighteenth Non-Planar (7,12) Graph and its Complement.

Let us construct vertex A, where A is either of degree five or degree six, in order to create a graph on eight vertices which we will refer to as G’. Consider G’ – V2. If A is of degree six, assume that V2 ∈N(A). If A is of degree five, assume that V2 ∉N(A). We can construct G’ – V2 three different ways and still find a construction that is planar.

174 a.) G’ – V2 b.) G’ – V2 c.) G’ – V2

W3 V3 V3 W3 W3 V3

W A A 2 W2 W2 A U W 1 U V1 U V1 W1 V W1 1

Figure 3.3.1.111. The Eighteenth (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar Three Different Constructions.

In all three constructions in Figure 3.3.1.111 we find a contradiction to G’ being a

MMNA graph. In construction (a) from Figure 3.3.1.111 we find that G’ – V2 is planar so W1 ∈N(A). In construction (b) from Figure 3.3.1.111 we find that G’ – V2 is planar so W2 ∈N(A). In construction (c) from Figure 3.3.1.111 we find that G’ – V2 is planar so W3 ∈N(A). Also, let us not forget that U is of degree two in G, so it must be in N(A) in order for U to be a vertex of degree three in G’. The known elements of N(A) are,

U, W1, W2, and W3.

Let us assume that A is of degree five. Since U, W1, W2, W3 ∈N(A) one of the following combinations of vertices is not in N(A), {V1, V2}, {V1, V3}, or {V2, V3}.

However, we find a contradiction with every one of the combinations. When we construct the complement of G’ for each pair of vertices not in N(A), we find that the complement is a subgraph of the K4,4 – e graph complement. So, G’ is not MM.

175 a.) V1, V2 ∉N(A) b.) V1, V3 ∉N(A) c.) V2, V3 ∉N(A)

W2 V2 W2 V2 W2 V2 U V3 A U V3 A U V3 A W1 W1 W1 W W W 3 V1 3 V1 3 V1

Figure 3.3.1.112. These Three Graphs Represent Complements of G’ When we Assume that the Given Pair of Vertices are Not in N(A). Each complement is a subgraph of the K4,4 – e graph complement.

When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Assume that A is of degree six. Since U, W1, W2, W3 ∈N(A) then, one of the following vertices is not an element of N(A), V1, V2, or V3. However when we look at the complement of G’ for every case we find that each complement is a subgraph of the K4,4 – e graph complement. This is a contradiction as it means G’ is not MM.

a.) V1 ∉N(A) b.) V2 ∉N(A) c.) V3 ∉N(A)

W2 V2 W2 V2 W2 V2 U V3 A U V3 A U V3 A W1 W1 W1 W3 V1 W3 V1 W3 V1

Figure 3.3.1.113. These Three Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A). Each complement is a subgraph of the K4,4 – e graph complement.

When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

176 We’ve shown no MMNA (8,17) or MMNA (8,18) graph comes from adding a vertex A to the eighteenth (7,12) graph. However, a MMNA G’ may arise from a different non-planar (7,12) graph of minimum degree at least two. Let us look at case nineteen.

Case 19: Let G be the nineteenth non-planar (7,12) graph of minimum degree of at least two from our list of nineteen from chapter II. Then G is the (44, 32, 2) graph of

Figure 3.3.1.114, which also shows its complement.

V3 W3 U V3

W3

W2 V2 W2 V2 V1

W1 U V1 W1 Figure 3.3.1.114. The Nineteenth Non-Planar (7,12) Graph and its Complement.

Let us construct vertex A, where A is either of degree five or degree six, in order to create a graph on eight vertices which we will refer to as G’. Consider G’ – V2. If A is of degree six, assume that V2 ∈N(A). If A is of degree five, assume that V2 ∉N(A). We can construct G’ – V2 two different ways and still find a construction that is planar.

177 a.) G’ – V2 b.) G’ – V2

W3 W3 V3 V3

A W2 A W2 W1 V U 1 U V 1 W1

Figure 3.3.1.115. The Nineteenth (7,12) Graph with Vertex A Added, has Vertex V2 Removed and is Found to be Planar Two Different Constructions.

In both constructions in Figure 3.3.1.115 we find a contradiction to G’ being a MMNA graph. In construction (a) from Figure 3.3.1.115 we find that G’ – V2 is planar so W2

∈N(A). In construction (b) from Figure 3.3.1.115 we find that G’ – V2 is planar so W3

∈N(A). We next argue V2 ∈N(A) and V3 ∈N(A). Consider G’ – W3. We can construct G’

– W3 two different ways and still find a construction that is planar.

a.) G’ – W3 b.) G’ – W3

V3 V3

W V2 V2 2 W2 A

V1 A U U W1 V W 1 1

Figure 3.3.1.116. The Nineteenth (7,12) Graph with Vertex A Added, has Vertex W3 Removed and is Found to be Planar Two Different Constructions.

In both constructions in Figure 3.3.1.116 we find a contradiction to G’ being a MMNA graph. In construction (a) from Figure 3.3.1.116 we find that G’ – W3 is planar and

178 shows V2 ∈N(A). In construction (b) from Figure 3.3.1.116 we find that G’ – W3 is planar and shows V3 ∈N(A). Also, let us not forget that U is of degree two in G, so it must be in N(A). So, the known elements of N(A) are, U, V2, V3, W2, and W3.

Let us assume that A is of degree five. Since U, V2, V3, W2, W3 ∈N(A) we can see that, V1, W1 ∉N(A). However, we find a contradiction. When we construct the complement of G’, we find that the complement is a subgraph of a P7 graph complement on eight vertices. Showing G’ is not MM.

U V3

W3

W2 V2 V1

W1 A Figure 3.3.1.117. The Graph Represents the Complement of G’ When We Assume that V1, W1 ∉N(A). The complement is a subgraph of a P7 graph complement.

When we assume A is a degree five vertex, there is no way to choose N(A) without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,17)

MMNA graph from this non-planar (7,12) graph.

Assume that A is of degree six. Since U, V2, V3, W2, W3 ∈N(A) then, one of the following vertices is not an element of N(A), V1 or W1. However, when we look at the complement of G’ for every case we find a contradiction to the assumption that

179

G’ is MMNA. We find that each complement of G’ is a subgraph of a P7 graph complement on eight vertices. This is a contradiction as it means G’ is not MM.

a.) V1 ∉N(A) b.) W1 ∉N(A)

U U V3 V3

W3 W3

W V2 W2 V2 2 V V1 1

W W1 A 1 A

Figure 3.3.1.118. These Two Graphs Represent Complements of G’ When we Assume that the Given Vertex is Not an Element of N(A). The two graphs are subgraphs of a P7 graph complement on eight vertices.

When we assume A is a degree six vertex, no N(A) is possible without contradicting the assumption that G’ is a MMNA graph. We cannot derive a (8,18) MMNA graph from this non-planar (7,12) graph.

We have argued that we cannot construct a (8,17) MMNA graph of maximum degree five nor a (8,18) graph of maximum degree six by adding a vertex

A to one of our nineteen non-planar (7,12) graphs. It follows there are no such

(8,17) or (8,18) MMNA graphs. 

3.3.2. The J1 Graph

In this section we prove that J1 is MMNA. We first show that it is NA by removing each of the eight vertices in turn and finding that the result is always a non-planar graph on seven vertices. We then show that J1 is MM by checking that all

180 simple minors of J1 are apex. The J1 graph is a non-planar (8,21) graph of maximum degree six and minimum degree five.

A1

V1 V5 V3 V4

V2 V6

A2

Figure 3.3.2.1. The J1 Graph.

Theorem 3.3.2. J1 is MMNA.

Proof. We first show that J1 is NA by removing each vertex in turn to find that the resultant graph is non-planar. Since J1 has two isomorphism classes, {V1, V2, V3, V4,

V5, V6}, and {A1, A2}, we have two cases.

Case 1: Remove a vertex from the isomorphism class {V1, V2, V3, V4, V5, V6}, WLOG

V1, to construct J1 – V1 and its complement.

A1 A1 A1 V5 V2

V5 V5 V5 V3 V3 V3 V V V V4 V4 V4 3 6 4

A1 V2 V6 V2 V6 V2 V6 A2 A A A 2 2 2 Figure 3.3.2.2. J1 – V1 and its Complement (complement is in orange).

181

The complement of J1 – V1 is a subgraph of the complement of the fifteenth non- planar (7,12) graph in chapter II, Figure 2.2.29.

U1 V V2 V5 2 V V3 U 3 U1 1

V1 U V1 U2 2 V3 U2 V4 V V4 4

V1 V2 V V5 5 Figure 3.3.2.3. The Fifteenth Non-Planar (7,12) Graph of Degree at Least Two from Chapter II, and its Complement.

Since the J1 – V1 complement is a subgraph of the complement of the graph from

Figure 2.2.29 then, the fifteenth non-planar graph is a subgraph of J1 – V1 graph. So the J1 – V1 is non-planar.

Case 2: Remove a vertex from the isomorphism class {A1, A2}, WLOG A1, to construct

J1 – A1 and its complement.

V V V5 V1 V6 V1 V5 V1 5 1 V V V3 3 V3 3 V V V4 4 V4 4 A2

V V6 V2 V6 V2 V6 2 V2 V5

A A2 A2 2 Figure 3.3.2.4. J1 – A1 and its Complement (complement is in orange).

The complement of J1 – A1 is a subgraph of the complement of the fourteenth non- planar (7,12) graph in chapter II, Figure 2.2.28.

182 V 2 U2 V2 U2 U1 V3 V3 U1 U1 V3

V1 V2

V5 V4

V4 V1 V1 U2 V4 V 5 V5 Figure 3.3.2.5. The Fourteenth Non-Planar (7,12) Graph of Degree at Least Two from Chapter II, and its Complement.

Since the J1 – A1 complement is a subgraph of the complement of the graph from

Figure 2.2.28 then, the fourteenth non-planar graph is a subgraph of J1 – A1. So the

J1 – A1 graph is non-planar.

Since for each isomorphism class of J1, J1 – v is non-planar then, J1 is NA.

Now let us show that J1 is MM by showing its simple minors are apex. We find these minors by removing a vertex or edge, or contracting an edge.

Let us begin by removing a vertex. Since there are two isomorphism classes of vertices there are two cases to check.

Case 1: Remove a vertex from the isomorphism class {V1, V2, V3, V4, V5, V6}, say V1.

Let us call the J1 – V1 graph, G. Then G – A2 is planar and G is apex.

A1 V3 V4 V5 V3 A1 V V4 5

V2 V6 V2 V6 A 2 Figure 3.3.2.6. The Graph G = J1 – V1 (left) and G – A2 (right).

183

Case 2: Remove a vertex from the isomorphism class {A1, A2}, say A1. Let us call the

J1 – A1 graph, G. Then G – A2 is planar and G is apex.

V1 V5 V1 V5 V3 V3 V4 V4

V2 V6 V2 V6 A 2 Figure 3.3.2.7. The Graph G = J1 – A1 (left) and G – A2 (right).

So, any minor of J1 constructed by removing a vertex results in an apex graph.

We next find minors of J1 by removing an edge. Since the isomorphism classes of vertices in J1 are {V1, V2, V3, V4, V5, V6}, and {A1, A2}, the isomorphism classes for the edges of J1 are {(A1, V1)} and {(V1, V2)}. There are two cases to check.

Case 1: Remove an edge from the isomorphism class {(A1, V1)}, so let us remove the edge (A1, V1). Let us call the J1 – (A1, V1) graph, G. Then G – V5 is planar and G is apex.

A1 V3 V4 V1 V5 V3 A 2 V1 A1 V4

V2 V6 V6 V A 2 2 Figure 3.3.2.8. The graph G = J1 – (A1, V1) (left) and G – V5 (right).

184

Case 2: Remove an edge from the isomorphism class {(V1, V2)}, so let us remove the edge (V1, V2). Let us call the J1 – (V1, V2) graph, G. Then G – V4 is planar and G is apex.

A1 A1

V1 V5 V1 V5 V3 V3 V4 A2

V2 V6 V2 V6 A 2 Figure 3.3.2.9. The Graph G = J1 – (V1, V2) (left) and G – V4 (right).

So, any minor of J1 constructed by removing an edge will result in an apex graph.

Let us continue to find minors of J1, but now let us contract an edge. As in the previous case, there are two isomorphism classes of edges {(A1, V1)} and {(V1,

V2)}.

Case 1: Contract the edge from the isomorphism class {(A1, V1)}. Let us call the

J1/(A1, V1) graph, G. Then G – A2 is planar and G is apex.

A1/V1 A1/V1

V5

V3 V4 V3 V5 V4 V2 V6 V2 V6 A 2 Figure 3.3.2.10. The Graph G = J1/(A1, V1) (left) and G – A2 (right).

185

Case 2: Contract the edge from the isomorphism class {(V1, V2)}. Let us call the

J1/(V1, V2) graph, G. Then G – V6 is planar and G is apex.

A 1 V1/V2 V5

V1/V2 A1 V5 V3 V3 V4 V4

V6

A2 A2

Figure 3.3.2.11. The Graph G = J1/(V1, V2) (left) and G – V6 (right).

So, any minor of J1 constructed by contracting an edge will result in an apex graph.

Since all simple minors J1 – v, J1 – (A, B), and J1/(A, B) are apex graphs then, all minors of J1 are apex graphs. Recall that all J1 – v graphs are non-planar so that J1 is NA. Since all minors are apex, J1 is MMNA. 

REFERENCES CITED

REFERENCES CITED

Adams, Colin C. The Knot Book (An Elementary Introduction to the Mathematical Theory of Knots).

[BBFFHL] P. Blain, G. Bowlin, T. Fleming, J. Foisy, J.Hendricks, and J. LaCombe, ‘Some Results on Intrinsically Knotted Graphs,’ J. Knot Theory Ramifications, 16 (2007), 749-760.

[CMOPRW] J. Campbell, T.W. Mattman, R. Ottman, J. Pyzer, M. Rodrigues, and S. Williams, ‘Intrinsic knotting and linking of almost complete graphs,’ Kobe J. Math, 25 (2008), 39-58. math.GT/0701422.

Diestel, Reinhard Graph Theory. Fourth edition. Graduate Texts in Mathematics, 173. Springer, Heidelberg, 2010. xviii+437 pp. ISBN: 978-3-642-14278-9.

Flapan, Erica. When Topology Meets Chemistry. published by MAA.

Gardner, M. The Sixth Book of Mathematical Games from Scientific American. Chicago, IL: University of Chicago Press, pp. 92-94, 1984.

Kuratowski, Kazimierz (1930), "Sur le problème des courbes gauches en topologie", Fund. Math. (in French) 15: 271–283.

[Ma] T.W. Mattman. Graphs of 20 edges are 2-apex, hence unknotted. Alg. Geom. Top., 11 (2011) 691-718. arxiv.org/0910.1575.

[OT] M. Ozawa and Y. Tsutsumi, ‘Primitive Spatial Graphs and Graph Minors,’ Rev. Mat. Comlut. 20 (2007), 391-406.

[RS] Fellows, M. R. "The Robertson-Seymour Theorems: A Survey of Applications." Comtemp. Math. 89, 1-18, 1987.

[S] H. Sachs, ‘On Spatial Representation of Finite Graphs’, in: A. Hajnal, L. Lovasz, V.T. Sos (Eds.), Colloq. Math. Soc. Janos Bolyai, North-Holland, Amsterdam, (1984), 649-662.

187