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Weaver, Robert Wooddell

SOME PROBLEMS IN STRUCTURAL GRAPH THEORY

The Ohio State University Ph.D. 1986

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SOME PROBLEMS IN STRUCTURAL GRAPH THEORY

DISSERTATION

Presented in Partial Fulfillment of the Requirements for the Degree Doctor of Philosophy in the Graduate School of the Ohio State University

By

Robert Wooddell Weaver, B. S., M. Sci.

* * * * *

The Ohio State University 1986

Dissertation Committee: Approved by

N. Robertson T. Dowling {\SU-X Adviser P. Huneke Department of Mathematics D. Ray-Chaudhuri To the memory of Herbert John Ryser ACKNOWLEDGEMENTS

I express my sincerest appreciation to Dr. Neil Robertson, to whom I owe whatever knowledge of structural graph theory I possess. Thanks for his time and patience during the many years of study. Thanks also go to Dr. Thomas Zaslavsky and

Joseph Fiedler, for helpful conversations that led to the development of chapters two and five. A special thanks to Dr. Paul Seymour, for his work that led to chapter four and for some exciting courses of study. To my dissertation committee, I offer gratitude for spending several hours of their summer reading this dissertation. Finally I would like to express my appreciation for Richard Vitray, fellow student of graph theory, co-conspirator, and friend, who made the time at OSU a little more pleasant and successful. VITA

August 19t 1958 ...... Bom — San Diego, California 1980 ...... B. S., California Institute of Technology, Pasadena, CA 1982 ...... M. Sci., The Ohio State University, Columbus, OH

PUBLICATIONS

Seymour, P. D., and Weaver, R. W., A generalization of chordal graphs, J. Graph

Theory, Vol. 8 , No. 2, Summer 1984.

FIELDS OF STUDY Major Field: Structural Graph Theory Studies in: Matroid Theory with Paul Seymour TABLE OF CONTENTS

DEDICATION...... ii ACKNOWLEDGEMENTS...... Hi VITA...... iv

LIST OF FIGURES...... vii LIST OF PLATES...... ix LIST OF NUMBERED ITEMS...... x CHAPTER PAGE I. INTRODUCTION...... 1 H. HEXAGONS AND NOWHERE ZERO FIVE FLOWS . 13 Si. The main theorem ...... 13 §2. Ornaments and the isthmus tree...... 17 53. Sets that prune trees ...... 2 2 54. Establishing the main theorem ...... 31 55. Applications ...... 38

ID. GIRTH 7 GRAPHS AND NOWHERE ZERO FIVE FLOWS. 46 51. The minimal counterexample ...... 46 52. The case v=24 ...... 60 53. The case v«26...... 62 54. The case v=28 ...... 90 55. A corollary ...... 108

v IV. A GENERALIZATION OF CHORDAL GRAPHS. . . . I l l 51.Histoiy of the problem...... I l l 52. Contracted strangulations ...... 116 53. Small cutsets...... 118 54. The main proof...... 124

V. A FLOW POLYNOMIAL IDENTITY FOR VERTEX 3-SEPARATED GRAPHS...... 129 51. The identity ...... 129 APPENDIX

PASCAL PROGRAMS USED IN CHAPTER H I 137 Results of the v«28 automated search ...... 188

LIST OF REFERENCES...... 211

vi LIST OF FIGURES

1 Some elementaiy definitions ...... 2 2 Some special graphs ...... 3 3 How contractions are obtained...... 4 4 A typical 3-separation ...... 5 5 An isthmus ...... 6 6 A Z^flow on a 4-clique ...... 8 7 Converting from cage, the McGee graph ...... 60 23 D2 with vertex labeling ...... 62 24 T2 with vertex labeling ...... '...... 63 25 T3 with vertex labeling ...... 64 26 Two indecomposable girth 5 graphs with 12 vertices 65 27 Sj with ab, cd, a l, b3, c 2 , d4,67 subdivided ...... 6 8 28 Beth ...... 70 29 C asper ...... 71 30 Si with ab, cd, al, c2,34,56,78 subdivided ...... 72 31 Si with ab, cd,al,c2,45,67,81 subdivided ...... 73 32 Si with abt cd, al, c 2 ,18,34; 56 subdivided ...... 74 33 Si with ab, cd, al, c 6 ,12,34,56 subdivided ...... 75 34 Si with ab, cd, al, c 6 ,18,34,67 subdivided ...... 76 35 Si with ab, cd, a l, 12,23,45,67 subdivided ...... 77 36 Si with ab, cd, al, 12,34,45,67 subdivided ...... 78 37 Si with ab,cd,al, 12,34,56,78 subdivided ...... 79 38 Hamiltonian display of Alfred...... 81 39 Another display of Alfred ...... 82 40 Si with ab, c2, d 4 ,12,45,67,78 subdivided ...... 83 41 S2 with 23,34,45,67,78,89,91 subdivided ...... 85 42 S2 with al, 12,23,45,56,67,89 subdivided ...... 8 6 43 S2 with al, b2, c3,12,45,67,78 subdivided ...... 87 44 S2 with a l, c3, b 5 ,12,45,67,89 subdivided ...... 89 45 Heawood graph, with Hamiltonian path labeling ...... 92 46 The indecomposable cubic girth 5 graphs with 14 vertices... 93 47 The graph D j 1 ...... 97 48 Two labeled copies of Tg ...... 101 49 The graph Tg*Tg ...... ; ...... 102 50 G] with edge labeling ...... 103 51 G2 with edge labeling ...... 104 52 G3 with edge labeling ...... 105 53 G4 with vertex labeling ...... 106 54 A graph that is not strangulated ...... 114

viii LIST OF PLATES

Plate 1 ...... 52 Plate 2, the graph europa ...... 192 Plate 3, the graph am althea ...... 193 Plate 4, the graph callisto ...... 194 Plate 5, the graph hesda ...... 195 Plate 6 , the graph h e ra ...... 196 Plate 7, the graph dem eter ...... 197 Plate 8 , the graph p a n ...... 198 Plate 9, the graph adrastea...... 199 Plate 10, the graph enceladus ...... 200 Plate 11, the graph poseidon ...... 201 Plate 12, the graph hades ...... 202 Plate 13, the graph tethys ...... 203 Plate 14, the graph dione ...... 204 Plate 15, the graph m im as...... 205 Plate 16, the graph rh e a ...... 206 Plate 17, the graph hyperion ...... 207 Plate 18, the graph ganymede ...... 208 Plate 19, the graph io ...... 209 Plate 20, the graph tita n ...... 210

ix LIST OF NUMBERED ITEMS

Eq 1.1.1 Union constraint on separations ...... S Eql.1.2 Intersection constraint on separations ...... 5 Eql.1.3 Nontriviality constraint on separations ...... 5 Fact 1.1.4 Isthmi and circuits ...... 6 Fact 1.1.5 Isthmi and subgraphs ...... 6 Fact 1.1.6 Isthmi and contraction m inors ...... 6 Eq 1.1.7 Kirchhoffs law ...... 7 Eq 1.1.8 Flows and isthmi ...... 9 Eq 1.1.9 The cyclomatic number ...... 10 Eq 1.1.10 A formula for the flow polynomial ...... 10 Th II. 1.1 The main theorem ...... 13 Prop II. 1.2 Inductive step for small circuits ...... 14 EqII.1.3 Verifying the flow property ...... 14 Prop IL 1.4 Inductive step for hexagons with diameters IS LemII.2.1 Uniqueness of ornaments ...... 17 LemII.2.2 Isthmi and ornaments ...... 17 Lem II.2.3 Isth(G/c) s Isth(G ) ...... 18 Lem E.2.4 Isth(G) is a contraction minor of G ...... 19 CorII.2.5 Isth(G) is a forest ...... 19 LemH.2.6 Isthmus forest and connected components 20 Lem II.2.7 Edge Addition ...... 20 Propn.2.8 Tree Pruning ...... 21 LemII.2.9 Monovalent vertices of Isth(GVE(H)) ...... 21 PropIL3.1 Pruning trees ...... 22 Lem II.3.2 Monovalent vertices and pruning ...... 24 Lemn.3.3 Pruning forests with 4 monovalent vertices 25 LemII.3.4 Pruning forests with 5 monovalent vertices 26

x Degree sequences and forests ...... 26

Pruning forests with 6 monovalent vertices ...... 28 5-flow conjecture for the torus or klein bottle ...... 39 Euler characteristic of a surface ...... 39 Two basic counts ...... 39 Two additional counts ...... 39 Euler characteristic again ...... 39 Euler characteristic with face sizes and valencies .. 40 Face valency count for minimal counterexample .. 40 5-flow conjecture for the surface with 3 cross caps . 40 Face count on the minimal counterexample, again . 40 5-flow conjecture for nearly S graphs, x(S)£0 .... 41 Flows and clique extensions ...... 43 An observation on flows in contractions ...... 44 Face count on the minimal counterexample ...... 47 An odd fact about girth ...... 47 Robertson's theorem ...... 53 Pentagons of S j ...... 66 Constraints on attachment vertices ...... 67 Restricted count on pentagons of S j ...... 80 Pentagons of S j ...... 84 Graphs from the Hcawood graph arc Hamiltonian . 91 Pentagons of C 7 ...... 93 Mechanical search of C 7 ...... 94 Pentagons of S 3 ...... 96 Pentagons of D j 1 ...... 97 A selection criterion for Dj* ...... 97 Mechanical search of D 3( ...... 98 Pentagons of T 5 T 5 ...... 102 A selection criterion of T g T g ...... 102 Pentagons of G j ...... 104 A selection criterion for G j ...... 104

xi IneqUI.4.12 A selection criterion for O 2 ...... 105 Ineq IH.4.13 A selection criterion for O 3 ...... 106 Ineq m.4.14 Pentagons of O 4 ...... 107 Th in.4.15 5-flow conjecture for surfaces of characteristic £ -2. 107 CorIII.5.1 The minimal counterexample has £ 30 vertices 109 ThIV.1.1 Dirac's theorem ...... I l l ThIV.1.2 Another theorem of D irac ...... 112 ThIV.1.3 A conjecture of Robertson ...... 113 ThIV.1.4 A theorem of Tu tte ...... 113 PropIV.1.5 Clique sums of strangulated graphs ...... 113 LemIV.2.1 Contractions of strangulations ...... 116 LemIV.3.1 Bridge building ...... 118 Prop IV.3.2 Cutvcrtices on C in (G\e)\W(C) ...... 118 PropIV.3.3 Attachments of the maximal bridge ...... 119 LemIV.3.4 Vertex 3-separations of strangulations ...... 120 Th IV.3.5 The sj, tj, S 2 , t2 theorem ...... 121 LemIV.3.6 Vertex 4-separations of strangulations ...... 121 PropIV.3.7 Pairs of paths from 4-separations ...... 121 ThIV.4.1 A theorem on chordal graphs ...... 124 ThIV.4,2 Planar strangulations ...... 124 PropIV.4.3 Minimal counterexamples ...... 125 PropIVA4 5-connectivity of the minimal counterexample 125 Prop IV.4.5 A contraction of the minimal counterexample 126 PropIV.4.6 Implications ...... 126 PropIV.4.7 Another implication ...... 126 PropIV.4.8 A property of the minimal counterexample ...... 127 PropIV.4.9 The final implication ...... 127 ThV.1.1 The flow polynomial theorem ...... 129 EqV.1.2 The flow polynomial equality ...... 129 EqV.1.3 The sets Sq, S 4 ...... 130 Eq V.1.4 The definition of 1 ...... 130 EqV.1.5 The dependence of <}> on

xii Eq V.1.6 The definition of a and p ...... 131 EqV.1.7 A consequence of the bijection i ...... 131 EqV.1.8 An count of Sq ...... 132 Eq V. 1.9 A fact about automoiphisms ...... 132 EqV.1.10 Another bijection result ...... 132 Eq V.1.11 A count of S j ...... 132 EqV.1.12 The definition of Fy 5 ...... 133 EqV.1.13 A dissection of some flow s ...... 133 EqV.1.14 Another count ...... 133 EqV.1.15 The resulting count ...... 134 EqV.1.16 The definition of K ...... 134 EqV.1.17 The definition of r | ...... 134 EqV.1.18 Another dissection...... 135 EqV.1.19 An observation about F(OjUT;2n) ...... 135 EqV.1.20 A count of S 4 ...... 135 Cor V. 1.21 Vertex 3-separations of minimal graphs ...... 136 CHAPTERI

INTRODUCTION

A graph is defined by a set V(G) whose elements arc called vertices, asetE(G) whose elements are called edges, and a relation of incidence which associates with each edge one or two vertices called its endvertices. If v c V(G), then the valency of v is the number of edges that have it as an endvertex (with loops counted twice incident). The endvertices of an edge are said to be adjacent. The neighbors of a vertex v are the vertices adjacent to v. [See Figure 1.] If an edge is incident with only one vertex, it is called a loop. [See Figure 1.] If two edges share both endvertices they are said to be parallel. [See Figure 1.] Any edge that is parallel to another edge is said to be m ultiple. A graph without loops or multiple edges is said to be sim ple. Because there is a natural identification of an edge of a simple graph with its endvertices, when convenient we shall represent edges as a set of two vertices, or by juxtaposing the two vertices when this is unambiguous. Two edges that share an endvertex are said to be coincident. [See Figure

1.] If two edges are coincident at a vertex of valency two, they are said to be series edges.

[See Figure 1.] All graphs considered in this dissertation are fin ite, that is, ]V(G)|, |E(G)|

< For some purposes it is convenient to consider the null graph, having no vertices or edges, but that will not occur here.

1 2

^ the neighbors of v ( ) a i00p

these edges are coincident at v s

^ two parallel edges two series edges

Figure 1. Some elementary definitions.

Let G and H be graphs. A homomorphism from G to is H a pair of maps

9 =(V(H) and 9 2 :E(G)->E(H) such that for any edge e of E(G) with endvertices x and y, the endvertices of 9 2 (e) are 9 j(x) and 9 }(y). A graph isomorphism is a bijective homomorphism. An automorphism 9 of a graph G is an isomorphism 9 G-»G. Let S(X) denote the symmetric group on the set X. One can verify that an automoiphism of G can be viewed as an element of the direct product S(V(G))*S(E(G)). The automorphism group of a graph G is the set of automoiphisms of

G given a group structure as a subgroup of S(V(G))S(E(G)).

Some families of graphs deserve special notation. Let n be a positive integer. Let x and y be vertices of a graph G. A path from x toy o f length is a nsubgraph of G whose vertex set may be named x = v q , v j , ...»vn = y and edge set e j ,..., such that ej is incident with vj. j and vj (l£i£n). [See Figure 2.] A path in a simple graph is sometimes indicated by giving the vertex sequence vq, v j , ..., vn. An n-star is a graph whose vertex set may be named v q , v j , ..., vn and edge set e j , ..., e,} such that ej is incident with vq and Vj (l£i£n), An n-circuit is a graph whose vertex set may be named V} ...... vn and edge set e j , ..., Cjj such that ej is incident with Vj and v$+i (l£i

a chord of the hexagon

of length 6 a 5_staf a 6-circuit (hexagon)

Figure 2. Some special graphs.

A subgraph of a graph G is a graph H with V(H) ? V(G), E(H) c E(G), with incidence inherited from the larger graph, i. e. if c e E(H) and e has endvertices x and y in G then it has endvertices x and y in H. One can then perform set-theoretic operations on graphs: if H and K arc subgraphs of a graph G then HnK is the subgraph of G with vertex set V(Hr\K) = V(H)nV(K) and edge set E(HnK)=E(H)nE(K). The graph HuK is the subgraph of G with vertex set V(HuK)=V(H)uV(K) and edge set E(HuK)=E(H)uE(K). If X is a subset of the vertex set of G, then G[X], called the subgraph induced isby X the subgraph of O with vertex set V(G[X])=X and edge set E(G[X])={e t E(G): e has both endvertices in X}, and G\X is defined to be G[V(G)\X]. If Y c E(G) then G:Y is the subgraph of G with vertex set V(G:Y)=V(G) and edge set Y, and G\Y is defined to be G:(E(G)\Y).

Let G be a graph and c be an edge of G with endvertices >; and y. Then the contractionGate, o f written G/e, is defined to be the graph with vertex set V(G/e)=(V(G)\{x,y})u{z} for a new vertex z f V(G), with edge set E(G/e)=E(G)\{e}, where incidence is as follows: if f * E(G/e) has neither x nor y as an endvertex, then it has the same endvertices as in G, while if f has x or y as an endvertex then that endvertex is replaced by z. (Edges parallel to e become loops at z.) [See Figure 3.J The graph H is said to be a contraction minor of G provided there exists a series of edges e j, ej* ..., en such that G=Gg, Gj+j=Gi/ej+i (0Si

G

Figure 3. How contractions are obtaim

A graph is said to be connected provided there is a path between any pair of vertices. A component of a graph G is a maximal connected subgraph. For a 5 non-negative integer n, we define an n-separation of 0 to be a pair of subgraphs H and K of O with disjoint edge sets such that the following holds:

(1.1) G=HuK, (1.2) |V(HnK)|« n,

(1.3 ) H and K each have a vertex not belonging to the other.

G Figure 4. A typical 3-separation.

If G has no n-separation for all n < m, we say that G is m-connected. It follows that a graph is connected it is 1-connected. Note that any n-clique is infinitely connected. An edge e of G is called an isthmus of G provided V(G) can be partitioned into two sets X, Y where e is the only edge with one end in X and the other in Y. 6

all other edges remain within the two ellipses.

XY

Figure 5. An isthmus.

For X, Y s V(G), we denote by (X,Y)q the set of edges of G with one end in X and the other in Y. Thus e is an isthmus o there exist X,YcV(G) with XnY=* 0 , XuY»V(G), and (X,Y)q«={c}. It follows that:

(1.4) an edge e is an isthmus <=> e is in no circuit of G, and (1.5) if e is an isthmus of G and an edge of HcG then c is an isthmus of H.

If e is in a circuit C of G and fre is an edge of G then e is in a circuit (either C or C/0 of Gif, hence no isthmi are created by contraction. Suppose {e} a (X,Y)q and erf. Without loss of generality we may assume that both ends of f are in Y. Let Y' be the subset of V(G/0 complimentary to X, i. e. the set Y with the ends of f deleted and the new vertex created by contraction added. Then (X,Y) 0 *=(X,Y')G/f* so no isthmi are lost by contraction (except under contraction by an isthmus). Thus

(1.6) In a contraction minor, no isthmi are gained or lost except by contraction by an isthmus. A graph in which every edge is an isthmus is called a fo re st, A connected forest is called a tree. Note that n-stars are trees. A directed graph D is something like a graph, in that it is defined by a set V(D) of vertices, a set E(D) of elements called dafts and two incidence relations h and t of darts with vertices. For a dart e, the vertex h(e) is called the head of e, and t(e) is called the tail of e. The dart e is said to be directed from the tail of e into the head of e. For a vertex v c V(D), the invalency of v is the number of darts directed into v, and the outvalcncy of v is the number of darts directed out of v. Each directed graph D has an underlying undirected graph G with vertex set V(D), edge set E(D), and the endvertices of the edge e arc the head and tail of the dart e. Given a graph G, an orientation o f isG a directed graph D with vertex set V(G), and dart set E(G), with incidence such that G is the underlying undirected graph of D. Note that there are 2 l* ^ )| orientations of a loopless graph to a directed graph.

An A-flow on D, or simply & flow, is a map q>:E(D)->A, for some abelian group A such that Kirchhoffs current law is satisfied, i.e.

(1.7) Z

Figure 6 . A Z^fiow on a 4-clique.

Note for any abelian group that eveiy directed graph has the zero flow, where each edge is given the value zero. Chapters II and III are concerned with flows that never take on the value zero, called nowhere zero flows. It can be shown that if a directed graph supports such a nowhere zero A-flow for some abelian group A of order k, then it supports a nowhere zero A’-flow for any abelian group A' of order k. Moreover, a z^-flow is equivalent to an integer flow using only the values -k+ 1, ..., k-1. Thus it makes sense to define a k-flow as a z-flow using the values -k+1,..., k-1. If a nowhere zero k-flow q> exists for a directed graph D, and D* is obtained from D by reversing the orientation of one dart e, then by defining

(1.8 ) 0 = 1 ( I 9(e) - 1 9(e)) =

Let 0 j, C2 be two subgraphs of a graph G such that G j r ^ is an n-clique. Then

G juG 2 is said to be a clique-sum of the graphs G j and G 2 . A simple graph is said to be chordal provided the only circuits induced by their vertex sets are triangles, i. e. any circuit of size greater than 3 in G has a chord in G. Complete graphs are chordal, as are clique sums of chordal graphs. A theorem of Dirac [1, Theorem 1.1] states that all chordal graphs are built up by taking clique sums of cliques. A subgraph H of a 3-connected graph G is said to be peripheral in providedG either V(H)=V(G) and there is at most one edge of G not in H, or else that H is induced and G\V(H) is connected. Note that in a chordal graph (of more than 4 vertices) no n-circuit with t&4 is peripheral. Chapter IV is concerned with characterizing all graphs with this property, that no n-circuit with i£4 is peripheral.

The flow polynomial F(G;X) of a graph G is a polynomial in the indeterminate X such that F(G;k) is the number of nowhere zero k-flows supported by some orientation of

G. (Recall this number is independent of the orientation.) Chapter V gives an expression for the flow polynomial of graphs that have a separation of size at most three in terms of smaller related graphs. Besides its usefulness for computational purposes, this leads to an identification of the structure of minimal graphs not supporting a nowhere zero k-flow. An explicit formulation of the flow polynomial, due to Tutte [11], is as follows: let Pq (G ) be the number of connected components of a graph G, and define the cyclomatic number 10

Pl(G) to be

(1.9) Pi(G) = |E(G)| - |V(G)| + p 0 (G). Then

(1.10) F(G;X) = (-1)IE(G>I X (-l)'S'>.p l(G:S>. SsE(G)

While computationally complex, this equation yields insight into the flow polynomial of a graph. It is fairly easy to show that if e is an isthmus of H then Pl(H\e)epi(H), and if e is not an isthmus of H then pi(H\e)«pi(H) -1. Thus in (1.10) if G is without isthmus the only term of degree £ pi(G) occurs when S=E(G), and so the flow polynomial is not identically zero. Thus for any isthmusless G there exists some k for which F(G;k)*0. In 1954 Tutte [ 11] advanced the following two related conjectures: Any isthmusless graph supports a nowhere zero 5-flow, known as the 5-flow conjecture, and that any isthmusless graph not containing a subdivided Petersen graph supports a nowhere zero 4-flow, known as the 4-flow conjecture. (A subdivided Petersen graph does not support a nowhere zero 4-flow.) Chapter n and III resolve the 5-flow conjecture for a variety of classes of graphs that embed into certain surfaces. Both conjectures are still open, and in particular the 4-flow conjecture is an extension of the 4-color theorem. A nice exposition of the significance of these open problems is given in Welsh's paper [13]. There are several questions raised in this thesis that are not completely answered.

Most immediate is a precise set of sufficient conditions that a graph support a nowhere zero 5-flow, i. e. a resolution to Tutte's 5-flow conjecture. The main result of Chapter n, essentially that hexagons are a reducible configuration, is certainly a step in the right direction, but does not seem to extend. It is probable that some new idea will be necessary to complete the problem. It is possible that the 4-flow conjecture will be more easily answered, possibly using the 4-color theorem as a basis. It is my belief that the result of 11

Chapter V will be of some assistance in this area. The efforts of Chapter m are essentially to generate most of the girth 7 cubic graphs on 28 or fewer vertices to verify Tutte's 5-flow conjecture in those cases: a natural extension is to compile a complete list of all girth 7 cubic graphs on 28 or fewer vertices to check other conjectures; this will be completed in the near future. Of course, the examination requires a mechanized search; it is plausible that this may be avoided in Theorem ID.4.15 and Theorem m.5.1 using a proof technique as was used to handle the Heawood graph. [See Theorem m.4.1.] If so, this would obviate the use of a computer and provide a humanly verifiable proof. One natural extension of Chapter HI would be to continue searching cubic girth 7 graphs for a larger number of vertices; however, to achieve Tutte's conjecture for the next surface, the sphere with five cross-caps, would require examination of graphs with up to 34 vertices. Since even at 28 vertices days of computations are required, I expect that the computation would prove exceedingly tedious and not very enlightening. Extensions of Chapter IV are possible, but the basic concept does not extend uniquely. One such extension is Reinhard Diestal's [2], Another would be to use the

Tutte concept of peripheral extended to graphs of connectivity less than three; this leads to a different extension of the classification. Since both are reasonable extensions, it would seem best to leave the matter where it is uniquely defined, i. e. at the level of

3-connectivity.

Extensions of the formula of Chapter V do not seem possible; the flow polynomial of 4-scparatcd graphs does not seem to depend in any natural way on the flow polynomials of the subgraphs. This is not surprising; if the flow polynomial of a 4-scparatcd graph could be explicitly defined, since a result of Jaeger [4] yields that any 4-connected graph must support a nowhere zero 4-flow, together this would yield an "easy" proof of the

4-color theorem. However, it may be that an attack on Tutte's 4-flow conjecture may be 12 possible using the ideas of Chapter V as a springboard. In any case, further examination of what can be said about local structures determining a zero of the flow polynomial seems called for. CHAPTER n

HEXAGONS AND NOWHERE ZERO FIVE FLOWS

§1. The main theorem.

In 1984, Steinberg [10] proved a special case of Tutte’s 5-flow conjecture dealing with graphs that embed into the (real) projective plane. He used a theorem of

Fleischner [3, Satz 1], the "splitting lemma" to show that a minimal counterexample to

Tutte's 5-flow conjecture must have girth at least 6 , and so no counterexample exists on the projective plane. This chapter will establish a series of lemmas in the spirit of

Fleischner’s lemma to examine the structure of isthmi of a 2-connected graph after a small circuit has been deleted. This will provide an inductive step to push the girth of a minimal counterexample up to at least six or seven. Of importance will be examination of a related graph, Isth(G) whose edges correspond to the isthmi of G.

A class of graphs Q is said to be closed under subgraph inclusion provided if G e G and H = H' g G then H c G. The main result is the following:

Theorem 1.1: Let G be a class of graphs closed under subgraph inclusion. For

G e G let H q be the set of hexagons P of G such that if d j , d £ , d 3 are the three chords connecting opposite vertices in P then at least two of Gu{dj}, G u ^ } , 13 14

G ufdj} are in G. Then either every isthmusless graph in G supports a nowhere zero 5-flow, or there exists a minimal graph G in G that does not support a nowhere zero

5-flow, is two-connected, is of girth £ 6 , and has Hq=0.

As stated, the proof is inductive and requires several lemmas. To begin with, we have the following two propositions used to enlarge nowhere zero 5-flows on a subgraph of G to G:

Proposition 1.2: Let P be a circuit of a graph G. Suppose there exists M«P such that |E(M)|£3 and (G\E(P))uM supports a nowhere zero 5-flow. Then G supports a nowhere zero 5-flow.

pf: Orient G so that Pis a directed circuit. Orient the rest of G arbitrarily. Let Z5 be a nowhere zero z^-flow. Since |E(M)|£3,

|{cpo(m): mcE(M)}|£3 and so there is a nonzero element x of Z 5 not used in M.

Extend q\) to a flow on G by setting unspecified edges to zero. Define

if e «E(P), and 9 i(e)==v. Then

(1.3) £ i(e) -

=-x + £ %(c) = -x + £

“ 90(c2>’x + ^ 9l(e) = £

Consequently,

9 1 is nowhere zero on E(G\E(P)). Since 9i(c)=9q(c)'x f°r e £ E(P) and 9 q does not take on the value x on the edges of P, 9 j is nowhere zero on P. Thus 9 j is a nowhere zero Zyflow on G. I

Proposition 1.4: Let P be a hexagon of a graph G. Let M be a graph with V(M)=V(P) consisting of two opposite edges of P together with an edge connecting the remaining two vertices (a diameter). If (G\E(P))uM supports a nowhere zero 5-flow then G supports a nowhere zero 5-flow.

pf: Let {vj,...,vg} be V(M)=V(P) with the vertices numbered in order in P and without loss of generality let the two opposite edges of P be {vj, vj} and { V 4 , V 5 } ; then the diameter of P in M is { V 3 , vg}. Orient G so that P is a directed circuit, and arbitrarily orient the rest of G. Orient { V 3 , vg) from V 3 to vg, Let 9 g be a nowhere zero ^-flow on (G\E(P))uM. Suppose 9 o({vj, V2 » = x, 9 o((v 4 . V5 »=y, and

G\E(P), and 9 j as indicated below in figure 7 on P. It is a simple calculation to verify

9 j is a nowhere zero flow on G. I 16

..-Ox Z+W w

V5?”’* oV. yn X y+z+w»\ X+W

V OV 4 * ■•O'*. z+w w

Figure 7. Converting from q>Q to (pj.

Propositions 1.2 and 1.4 provide the inductive step to establish the theorem. It is of course necessary that the hypotheses of the theorem together with the inductive hypothesis imply the hypotheses of the proposition. With this in mind, it is necessary to examine the structure of isthmi in G\E(P) to ensure the existence of suitable M. 17

Chapter 1182. Ornaments and the isthmus tree.

An ornament of a graph O is a maximal connected subgraph not containing an isthmus of O.

Lemma 2.1: Let H be a connected isthmusless subgraph of O. Then H is contained in a unique ornament

pf: Since H contains no isthmus, from 1.1.5, H contains no isthmus of G. That H is contained in some ornament follows from the finiteness of G. Suppose O j, O 2 are ornaments of G containing H. Then no edge of Oj or O 2 is an isthmus of G, so no edge of 0 |U 0 2 is an isthmus of G. Since O j, O 2 arc connected and have the vertices of H in common, O juC ^ is connected. Thus is a connected subgraph not containing an isthmus of G that contains the maximal such subgraph O j. Thus O i= O ju 0 2 and similarly 0 2 = 0 iu O j, hence 0 ^=0 2 - 11

Lemma 2.2: An edge of G is an isthmus of G e* it is incident with vertices of exactly two ornaments and is contained in no ornament.

2 f: An isthmus is contained in no ornament, by definition. A vertex graph is connected and without isthmus, so the endvertices of an isthmus are contained in unique ornaments, by lemma 2.1. Let e be an isthmus of G with endvertices x, y. Suppose x, y are elements of the same ornament O. Since O is connected, there is a path P from x to y in O, hence in

G. Now e f E(P) s E(0), so Pu{e} is a circuit of G through the edge e, contradicting e an isthmus of G. Thus e is incident with the vertices of exactly two ornaments, as desired.

Conversely, suppose e is not an isthmus. Then by Lemma 2.1 it is contained in a unique ornament, so it is not incident with two ornaments. ■

We define the graph Isth(G) where the vertices are the ornaments of G, edges are the isthmi of G, and an edge of Isth(G) is incident with a vertex of Isth(G) precisely when an endvertex is contained within the ornament Lemma 2.1 shows this definition to be meaningful. Lemma 2.2 guarantees that an edge is incident with two vertices in Isth(G), and so Isth(G) is a well-defined graph.

Lemma 2.3: If e is not an isthmus of G, then Isth(G/e)=Isth(G). p£: Let C | be the ornament of G containing e. Let x, y be the endvertices of e in G and z be the new vertex of G/e. Let C 2 , ..., Cn be the remaining ornaments of G. Then C 2 ,

..., Cn are connected, isthmusless subgraphs of G and disjoint from x, y, so C 2 , ..., Cn are connected, isthmusless graphs of G/e. Let H be an ornament of G/e containing Cj, (2 5 i £ n). Suppose H properly contains Cj. Since H is connected,

(V(Cj),V(H)\V(Cj))(j/c is nonempty. Let f be any edge in this set. Since CjU{f} is the union of two connected graphs at a common vertex (the endvertex of f in V(Cj)) it is a connected subgraph of G properly containing the maximal connected isthmusless subgraph Cj of G, hence f is an isthmus of G. Then f is an isthmus of G/e; so f f E(H), a contradiction. Thus C 2 Cn are ornaments of G/e. Since Cj is connected and isthmusless, C j/c is connected and isthmusless. Every edge of G that is not an isthmus is an edge of some Cj, thus every edge of G/e that is not an isthmus is an edge of Cj/e or C 2 ,

..., Cn. Thus Cj/e is maximal connected without isthmus, hence an ornament of G/e. 19

Let 9 i be the bijecdon between C j, C j. •••. Cn and Cj/e, € 3 ,..., Cn. The isthmi of G are the isthmi of G/e, so the identity map 9 2 is a bijection between edges of Isth(G) and

Isth(G/e). Let f be an edge of Isth(G) with endvertices Cj, Cj. If neither i nor j is 1, then the endvertices of f in G are in ornaments that are ornaments of G/e, and hence the endvertices o ff in Isth(G/e) are (p j(Cj),q>j(Cj). Suppose i«l. Let v be the endvertex of f in C j. If v=x or v=y then in G/e, f is incident with z in G/e, hence the endvertices of f in Isth(G/e) are 9j(C j), cpj(Cj). If v*x and v*y then veV(Cj/e) and again the endvertices of f in Isth(G/e) are 9j(C j), (pj(Cj). Thus (cpj, 9 2 ) provides the necessary isomorphism. I

Corollary 2.4: The graph Isth(G) is isomorphic to a contraction minor of G whose only edges are isthmi. pf: We proceed by induction on the number of edges that are not isthmi of G: if every edge is an isthmus then no ornament contains an edge, so every ornament is a vertex graph, and so there is the identity bijection between G and Isth(G). So suppose e is any edge that is not an isthmus. Then by induction Isth(G/e) is isomorphic to a contraction minor of G/e whose only edges are isthmi, and from the previous lemma, Isth(G/e) = Isth(O). Hence Isth(G) is isomorphic to a contraction minor of G whose only edges are isthmi. I

Corollary 2.S: Isth(G) is a forest pf: Immediate from lemma 2.4. I 20

Lemma 2.6: The isthmus graphs of the connected components of G are the connected components of Isth(G). pf: Let C be a component of G, and let O be an ornament of C. Then O is a connected, isthmusless subgraph of G, hence contained in an ornament O' of G. Since O' is a connected subgraph of G containing O and C is a maximal connected subgraph of G containing O, O'cC. But then O' is a connected, isthmusless subgraph of C containing O, hence 0 = 0 . Thus the ornaments of C are the ornaments of G contained in C, and Isth(C)=Isth(G)[V(Isth(C))]. Since C is connected and by corollary 2.4 Isth(C) is a contraction minor of C, Isth(C) is connected. Since (V(C), V(G)\V(C)) q is edgcless,

(V(Isth(C)), V(Isth(G)\V(Isth(C))))o is edgeless. Thus Isth(C) is a component of Isth(G). Since every ornament of G is contained in some component of G, every component of Isth(G) is Isth(C) for a component C of G. I

In light of lemma 2.6, for most applications we consider connected graphs.

Lemma 2.7: (Edge addition) Let {f j...... fjc> be a set of edges disjoint from E(G) with endvertices assigned in V(G). Let (fj%..., fk'} be a set of edges disjoint from E(Isth(G)) and define the endvertices of fj' as the ornaments of G containing the endvertices of fj. Then an edge is an isthmus of G'=Gu{f j, ..., fjJ <=>it is an isthmus o fT M sth tO u tfj' fk'}. pf: From corollary 2.4 there exist edges e j, ..., en of G, none of which are isthmi, such that G=Gq, Gj+j=Gj/cj+j (0£i

G u{f j, ..., fn}=Ho, Hj+ 1=Hj/cj+j (0£i

Proposition 2.8: (Tree Pruning) Suppose {f ..., f^'} is a set of edges such that Isth(G )u{fj',..., fjj’} is without isthmus. Let { f j, ..., fjJ be any set of edges such that the endvertices of fj are venices in the ornaments of G that correspond to endvertices of f ;1 (1 £ i £k). Then G u f f j,..., fjJ is without isthmus. pf: Immediate from Lemma 2.7. I

Lemma 2.9: Let G be a graph without isthmus, and H«G. Set G’=G\E(H). Then any monovalent vertex of Isth(G') is an ornament of G' containing a vertex of H. pf: Let O be a monovalent vertex of Isth(G'), and c be the unique edge incident with O.

Let X»V(0), Y=V(G)\V(0). Then e is one edge of G with one end in X and the other in

Y. Suppose e' c (X,Y) q ». Then e' is incident with a vertex contained in V(O) but is not itself contained in the ornament O, hence e' is an isthmus incident with O in Isth(G').

Thus e'*=e. So G' has no other edge with one end in X and the other in Y. However, e is not an isthmus of G, so there is some edge f of H with f e (X,Y) q . The endvertex of f in

X is a vertex of H in O. I 22

Chapter n S3. Sets that prune trees.

Wc wish to investigate the structure of minimal graphs M which added to graphs, such as G\E(P), produce a graph (G\E(P))uM without isthmus. In light of the tree pruning proposition, it suffices to consider graphs which when added to a forest F produce a graph without isthmus. In fact, it suffices to consider the case when F is a tree. Lemmas

3.2 - 3.7 establish the structure of such graphs. While proposition 3.1 is not essential to the sequel, it is an interesting observation. Let K(X) denote the |X|-clique with vertex set X.

Proposition 3.1: Let F be a forest and X be its set of monovalent vertices. Then there exists a spanning subgraph M of K(X) of no more than (|X|+l)/2 edges such that FUM has no isthmus.

Remade: Any edge incident with a monovalent vertex x of any graph will be an isthmus of that graph (with partition {xJ'fV-tx}). Thus if FUM has no isthmus, M must be incident with every vertex of X. Consequently 2|E(M)| £ |X| and hence any such M as above must have at least |X|/2 edges. Proposition 3.1 is thus the sharpest possible result

pf: We proceed by induction on [Xj, and among all forests F with |X| minimal, on |E(F)|.

Note that if |X|=0 then F has no edges, and so no isthmi, and if |X|=2 then adding the single edge of K(X) produces a graph without isthmus. We may assume |X| £ 3 in what follows. 23

Case 1; F is disconnected.

Let xj,X 2 be monovalent vertices of different trees of F. Letej be the edge of K(X) with endvertices xj, xj. Then Fu{cj} is a forest with |X|-2 monovalent vertices, so by induction there exists NT« K(X\{xj, xj}) of at most (|X|-l)/2 edges so that (F uteiJJuN f is without isthmus. Set MaM'u{ej}. Then M c K(X) is of the proper size and FUM has no isthmus. Case 2: F has a divalent vertex v.

Let e^, C2 be the edges adjacent to the vertex v. Then F/ej is a forest with the same monovalent vertices and one fewer edge. So by induction there exists M c K(X) of the proper size such that (F/e j)uM has no isthmus. Now FUM may be obtained from

(F/e j)uM by subdividing the edge Since ej *s not an isthmus, we do not create any new isthmi by the edge subdivision. Thus FUM is without isthmus. Case 3: F is a |X|-star.

If |X|=3 then one easily verifies the addition of any two edges of K(X) provides the required M. So assume |X|*3. Let xj,X 2 be any two monovalent vertices, e j,e 2 the edges they are incident with. Then F\{xj, X 2 } is a forest with |X|-2 monovalent vertices, and so by induction there exists M* c K(X\{x j, xj}) of at most (|X|-l)/2 edges such that

(R{xj, X2 » u M ' is without isthmus. Define M as M* together with the edge e with endvertices xj, X 2 . Then FUM = [(F\{xj, X 2 })uM,]u{ej, C2 , e} is the union of an isthmusless graph with a triangle, hence has no isthmus.

Case 4: F is a tree that is without divalent vertices and is not a vertex star.

Delete all monovalent vertices from F: what is left is a tree but is more than a single vertex. Let vj, v j be two monovalent vertices of FVX and xj, X 2 « X; c j, e2 « E(F) with cj having endvertices xj, vj (i=l, 2 ), Let 6 3 be the edge of K(X) with endvertices x j, xj, and let C be the unique circuit of FU fej). Then ej, e 2 , e j t E(C). LetF=R {xj, X 2 >; 24 since vj, V2 are of valency at least three, vj and V 2 are not monovalent in F , so F has

|X|-2 monovalent vertices. By induction there exists M' c K(X\{xj, xj}) of at most (|X|-l)/2 edges such that FulvT is without isthmi. Define M as M* together with the edge

0 3 . Then FUM = [(F\{ej, e 2 » u M ’]uC is the union of an isthmusless graph together with a circuit, hence has no isthmus. I

What proposition 3.1 asserts is the existence of graphs M that barely span K(X) yet remove all isthmi from F. In fact, many of the maximal matchings (for |X| even) and maximal matchings plus an edge (for |X| odd) will suffice: if F is a subdivided vertex star, any such M will work. Thus for example, as in the proof of lemma 3.1 if |X|=3 then F is necessarily a subdivided star and any pair of edges of K(X) will work. (See lemma 3.2.) While it is not true that for any F with |X|«4 that any maximal matching will work, at least two of the three maximal matchings must work. What affects the structure of sets that prune trees are the location of the edges of the tree not incident with a monovalent vertex, denoted internal edges.

Lemma 3.2: Let F be a tree. Let M s K(V(F)) be any set of edges such that every monovalent vertex of F is incident with a member of M. Then any edge of F incident with a monovalent vertex of F is not an isthmus in FUM. pf: Let c be incident with the monovalent vertex x in F. By hypothesis, M contains an edge f with endvertices x, y. Since F is a tree, F contains a unique path P from x to y, and since x is monovalent in F, c t E(P). Thus Pu{f}«PuM is a circuit containing the edge e, and so e is not an isthmus. I Lemma 3.3: Let F be a forest and X be its set of monovalent vertices. If |X|=4 then for two of the three matchings M of K(X), FUM has no isthmus and for die third, if e is any other edge in K(X) then FU(Mu{e}) has no isthmus.

pf: We may assume F is without divalent vertices, as edges are to be added only to monovalent verdces and an edge incident with a divalent vertex is an isthmus precisely when the other edge incident with this divalent vertex is an isthmus. We may also ignore isolated vertices, as such subgraphs are without isthmi. We consider several cases:

Case 1: F is disconnected. Then F is two paths (edges). Adding the matching gives a graph all of whose valencies are two, which has no isthmus. Case 2: F is a tree. Then F is isomorphic to one of

1

2 2 a 2b

Figure 8 . The two trees with 4 monovalent vertices.

In 2a, by lemma 3.2 any matching will suffice. In 2b, the only matching that fails is

{1,2}, {3,4} in which the internal edge remains an isthmus. Thus two of the three matchings suffice. If any edge is added, it must cross the two endgraphs of the internal edge, and so put the internal edge in a circuit I 26

Lemma 3.3 is a minor reformulation of Fleishner's "splitting lemma". A similar, but

more tedious result holds for |X|=5.

Lemma 3.4: Let F be a tree and X be the set of monovalent vertices of F. Suppose |X|=5. Then:

(i) there exist two independent edges ej, e 2 such that for any spanning set M of three edges avoiding ej, 6 2 , FuM is without isthmus.

(ii) Suppose H c K(X) where H consists of a quadrilateral of K(X) together with an edge incident with the fifth vertex of X. Then H contains a subset M of three edges so that FUM is without isthmus.

pf: Again we may assume F is without divalent or isolated vertices. Since F is a tree, F has one more vertex, than edge and the sum of the valencies is twice the number of edges, so if we let d(v) be the valency of the vertex v, then

(3.5) -2 = Z(d(v)- 2 ) or |X|-2o Z (d(v)-2). veV(F) v«V(F), d(v)*l

There are no vertices of degree two, so the degree sequence of the non-monovalent vertices is one of a) 5; b) 4,3; or c) 3,3,3. Thus F is isomorphic to one of 2a, 2b, or 2c given in Figure 9. 27

3

2 a 2b 2 c

Figure 9. The three trees with five monovalent vertices.

By lemma 3.2, any spanning set M on three edges will suffice in 2a to provide FuM without isthmus. In 2b, if we may avoid the edge with endvertices 4 and 5 then the edges incident with 4 and 5 will cross the endgraphs of the internal edge and no edge will be an

isthmus. In 2c, if we choose M as three spanning edges avoiding the edges {1,2} and {4,5} then one of the two edges incident with 1,2 must be incident with 4 or 5 (if the two edges incident with 1,2 are both incident with 3 the third is {4,5}) and FuM has no isthmus. This satisfies part (i) of the lemma. Next consider part (ii). If F is isomorphic to 2a then any spanning subset of H

will suffice. If, in the case where F is isomorphic to 2b, we are obliged to use the edge

{4,5} then without loss we may assume 1,2,3,4 are the vertices of the quadrilateral in that order, and M={{1,2},{3,4},{4,5}} will suffice. Otherwise, any spanning set of three edges avoiding {4,5} will suffice. So suppose F is isomorphic to 2c. Under the automorphisms of 2 c, we distinguish three cases:

Case (i) The edge not in the quadrilateral is {1,2}. Without loss of generality, the quadrilateral is 2,3,4,5 or 2,4,3,5 in that order, and in either case we may set M={{ 1,2},{2,5},{3,4}}.

Case (ii) The edge not in the quadrilateral is {1,3}. 28

Then without loss of generality the quadrilateral is one of 1,2,4,5; 1,4,2,5; 2,3,4,5; 2,4,3,5. In each case M consisting of {1,3}; one of {2,4} or {2,5}; and any edge incident with the remaining vertex will provide a subgraph of H whose addition to M is without isthmus. Case (iii) The edge not in the quadrilateral is {1,4}. The edge {1,4} puts all internal edges into circuits; any spanning set M containing

{1,4} will yield FuM without isthmus, i

We shall also require a lemma describing the structure of sets that prune trees for

|X|=6 . There are several possible phrasings of an avoidance criterion to guarantee that the resulting graph FUM has no isthmus. The following is just a convenient formulation for the main theorem.

Lemma 3.6. Let F be a forest with X as its set of monovalent vertices. If |X |= 6 then there exist three edges Cj, C 2 > 6 3 of K(X) that are either a triangle (in which case F is the disjoint union of two subdivided 3-stars) or a matching such that if M is any matching of K(X) avoiding all of ej, C 2 , 6 3 then FuM is without isthmus.

pf: Again, we may assume that F is without divalent or isolated vertices. Case 1. F is disconnected.

Then F is one of (a) two 3-stars, (b) three edge components, (c) an edge component and a 4-star, (d) an edge and an H graph with vertices denoted 1,2,3,4 as in figure 8 2b. In case (a), let the triangle to be avoided be the subgraph of K(X) whose vertices are the monovalent vertices of one of the stars: then if M is a matching avoiding the triangle each edge crosses the stars and so FuM is without isthmus. In case (b), FuM for 29 arbitrary M is the union of two matchings, hence 2-regular and without isthmus. In case

(c) any matching will work. Finally, in case (d) set ej to be the edge of K(X) in parallel with the edge component of F, and 03 to be the edges with endvertices 1,2 and 3,4 respectively. Then any matching M avoiding e j, C 3 when added to F will produce a graph without isthmus.

Case 2. F is a tree.

As in lemma 3.4, equation 3 .5 is valid and so the degree sequence of the non-monovalent vertices is one of (a) 3,3,3,3; (b) 3,3,4; (c) 4,4; (d) 5,3; or (c) 6 .

Consequently, F is isomorphic to one of the seven graphs in Figure 10:

Figure 10. The seven trees with six monovalent vertices.

It will be shown that any matching of K(X) avoiding the edges indicated below will provide a graph without isthmus.

Figure 11. Edges that must be avoided. 30

In a j, we set {ej, ^ e3 }={{ 1,2}, {3,4}, {5,6}}. Let M be any matching of K(X) avoiding these three edges. Then there must be two edges incident with the vertices 1 and 2; these edges cannot be incident with both of 3,4 or 5,6 since then the remaining edge of the matching would be 03 or respectively, so one edge must be adjacent with one of 3,4 and one of 5,6; repeating the argument with each of the other pairs 3,4 and 5,6 yields that

FuM is isomorphic to a subdivision of a 4-clique hence has no isthmus. In 561 t c l» e2 ) sa{{l*^}*{ 5 *6 }}, and letM be any matching of K(X) that avoids {ej, ej}. Now M has three edges and at most two edges are incident with 3,4, so one edge must cross from the set {1,2} to the set {5,6}. But then all internal edges are in the circuit created by this crossing edge, so by lemma 3.2, FuM has no isthmus. The argument for bj is identical, with {e^, ej} and M taken as above. In b j the situation is simpler let C|={ 1,2} and M be any matching avoiding e j. Then M has two edges incident with the vertices 1,2 and at most one of these are incident with the vertex 3, so that there must be an edge crossing from the vertex set {1,2} to the vertex set {4,5,6}. This crossing edge creates a circuit that contains both internal edges, so by lemma 3.2, FUM has no isthmus. The argument for c is identical to that for t >2 with e [ as above. In d, there is only one internal edge. Since the endgraphs of the internal edge have odd size, in any matching there must be at least one edge that crosses the endgraphs, and so puts the internal edge in a circuit.

Thus for any matching M, FUM is without isthmus. In e, since there is no internal edge, lemma 3.2 immediately gives us that FuM is without isthmus. ■ 31

Chapter I I S4. Establishing the main theorem.

We now prove the main result

Proof to the main theorem 1.1: Let G be a family of graphs as specified in theorem 1.1. Suppose the theorem is false for G. Then there must be an edge minimal G « G that has no isthmus and does not support a nowhere zero 5-flow. Without loss of generality, we may assume that G is connected. Moreover, for this graph either (1) G is not 2-connected

(2) G has girth < 6 , or

(3) G has girth six and Hq * 0.

Suppose that there exists a 1-separation G = HuK, Let c be an edge of H. Since e is not an isthmus of G, by 1.1.4 there is a circuit CcG with ecE(C), Since |V(HnK)|=l, |V(CnK)|£l andCsH. Thus e is not an isthmus of H. Similarly K has no isthmus. It can be easily seen that if both H and K support nowhere zero 5-flows, then G does also. Thus one of them does not support a nowhere zero 5-flow. But then

G is not edge minimal, a contradiction. It follows that G is 2 -connected. Let P be a circuit of minimum size in G. By the girth constraint,

|V(P)| eS 6 . Ifgirth(G)- 6 , choose P « Hq.

Let T - Isth(G\E(P)), and let X be the set of monovalent vertices of T. By lemma

2.9, |Xj £ | V(P)| £6 , and each ornament of X contains a vertex of P. We consider several cases depending on the size of X. 32

Case 1. |X| = 0.

Since X=0, G\E(P) has no isthmus. Set M=0; then (G\E(P))uM is a proper

isthmusless subgraph of G and so by induction supports a nowhere zero 5 -flow. By proposition 1.2, G also supports a nowhere zero 5-flow, a contradiction.

Case 2 . |X|-1.

This is impossible, since no tree has exactly one monovalent vertex. Case 3. |X|=2.

Let x, y be vertices of P in the two ornaments of X. Since |V(P)| £ 6 , there exists a shortest path in P from x to y of length at most 3. Let M be the set of edges of this path with incidence inherited from the path, and let M* be the same set of edges with

incidence defined as follows: the cndvertices of an edge e in M' arc the ornaments of

G\E(P) containing the endvertices of e. (The set M 1 may contain loops.) Then TOM' has even valency, so it follows that eveiy edge is in a circuit and hence TOM' is without isthmus (by 1.1.4). Then (G\E(P))uM has no isthmus by the tree pruning proposition

2.8 and is a proper subgraph of G. By the induction hypothesis (G\E(P))uM supports a

nowhere zero 5-flow. By proposition 1.2, G supports a nowhere zero 5-flow, a contradiction.

Case 4. |X|=3.

Let X consist of the ornaments Op O 2 , 0 3 . By lemma 2.9, there exist vertices

xl» x2> x3 of P with xj t V(Oj) (i=l,2,3). Let Pj be the path in P from xj to X 2

including X 3 . Let cj be the first edge of Pj with one end in Oj and the other not in Oj

(i=l,2,3). Since P j has edges incident with each ornament, such e p 0 3 ,6 3 must exist

(but need not be distinct). Let e;' be a new edge whose endvertices are the ornaments containing the endvertices of ej (i=1,2,3). By lemma 3.2, T u {cj\ e 2 ’, 6 3 '} is without

isthmus. By the tree pruning lemma 2.9, (G\E(P))w{ep e 2 > 6 3 } is without isthmus. 33

By the inductive hypothesis, (GVE(P))u{e j; C 2 , 6 3 } supports a nowhere zero 5-flow, and by proposition 1.2, G supports a nowhere zero 5-flow, a contradiction. Case 5. |X|=4.

Let X be {O j,..., O 4 }. By lemma 2.9 there exist vertices x j ,..., X 4 of P with

xj € V(Oj) (l£i£4). Suppose we may choose x j ,..., X 4 consecutive in P. The edges

{Oi, O2 }, {O2 , 0 3 }, {O3 , 0 4 } contain a matching of K(X); by lemma 3.3, either

Tu{{ Oj, 0 2}, {0 3 , 0 4 }} or 1 \ j { { Oj, O2 }, {0 2 , 0 3 }, {03 , 0 4}} is without isthmus (say the former). By the tree pruning lemma 2.9,

G'=(G\E(P))u{{ X|, X2>» {X3, X4}} is without isthmus. By the inductive hypothesis,

G' supports a nowhere zero 5-flow, and so by proposition 1.2, G supports a nowhere zero 5-flow, a contradiction.

Thus we may assume that x j ,..., X 4 may not be chosen consecutively; it follows

that |V(P)|b 6 and so P £ Hq. Suppose three consecutive vertices of P arc in different

monovalent ornaments. We may assume that the vertices of P are x, x j, X 2 > X3 , y, X4 in

that order. By lemma 3.3, either Tu{{Oj, O 2 MO 3 , 0 4 }} or Tu{{Oj, O 4 ),

{O 2,03}} is without isthmus, say the former. Then Gj=(G\E(P))u {{xj, X2>,

{ X 3 , X 4 } } is without isthmus. Suppose G 2 =(GVE(P))u{{x1, X 2 M X 3 , y}, (y, X 4 } }

has an isthmus e. If e » { x 3 , y} or e*{y, X 4 ) let C be a circuit of G j through { X 3 , X 4 } . Otherwise e is in Gj and is not an isthmus of Gj so is in a circuit C of Gj. If

{X3, X4} e E(C), then replace { X 3 , X 4 } with { X 3 , y} and {y, X 4 } in C to form C,

Then C contains a circuit through e, so e is not an isthmus, a contradiction. Thus G 2 is

without isthmus and by the induction hypothesis supports a nowhere zero 5-flow. By proposition 1.2, G supports a nowhere zero 5-flow, a contradiction.

Suppose no three consecutive vertices are in different monovalent ornaments of

GYE(P). Then we may take {xj, X2>, {X3, X4} as edges of P and {Oj, O 2 }, {O3 , 0 4 } 34

is a matching of K(X). Since P t Hq, one of the diameters {xj, X 3 } , {xj, X 4 } can be

added to G to form a graph in G, say (xj, X 3 } . Then by lemma 3 . 3 ,

TU{{Oi, O2 }, {O3 ,0 4}} orTUftO!, 0 2 >, {0 3 , 0 4 }, {Oj, O3 }} is without

isthmus. But then by the tree pruning proposition 2.9, (Q\E(P))u{{X|, X 2 }, { X 3 , X 4 } }

or (G\E(P))u{{xi, X 2 }, { X 3 , X 4 } , {xj, X 3 } } is without isthmus. By the inductive

hypothesis one of (G\E(P))u{{xj, X 2 }, {X3 , X 4 } } or

(G\E(P))u{{xi, X2 }, { X 3 , X 4 } , {xj, X 3 } } supports a nowhere zero 5-flow, and so by proposition 1.2 or 1.4, we have thatG supports a nowhere zero 5-flow, a contradiction.

Case 6 . |X| - 5.

First suppose | V(P)| = 5. Let P' be the graph with edge set E(P')=E(P) and vertex

set V(F)»X where the edge e of F has as endvertices the two ornaments in X containing the endvertices of e in P.

Suppose T is disconnected. Since T has five monovalent vertices, T consists of a path component and a subdivided 3-star. Now P' has two independent edges incident with the path component; let M consist of these two edges together with any other edge of P' incident with the fifth vertex of P\ Then TUM 1 consists of a subdivided circuit with a chord, which is without isthmus.

Suppose T is connected. By Lemma 3.4, there are two independent edges e j, e 2 of K(X) such that any spanning set M 1 of three edges avoiding cj, C 2 is such that TUM is without isthmus. Now P\{ej, C 2 } spans K(X) and hence there is a spanning set M' of independent edges of Pt{ej, C 2 > (since ej, C2 are independent and the valency of each vertex in P* is two) providing such an M'.

Let M be the edges of G corresponding to M\ Then by the tree pruning lemma

2.9 we have that (G\E(P))uM is without isthmus. By the inductive hypothesis (GVE(P))uM supports a nowhere zero 5-flow, and by proposition 1.2 we have that G 35 supports a nowhere zero 5-flow, a contradiction. Next suppose that |V(P)|=6 and hence that P c Hq. Let x j,..., xg be the five representative vertices of P from the monovalent vertices Oj ..... O5 of T in order in P, x j c V(Oj) (l£i£5), and x be the remaining vertex. One of the diameters {Xj, X 4 } , {X2,

X 5 } may be added to G to form a graph O' still in G, say {Xj,X 4 } . The set of edges {{Oj, O2}, {O2,03}, {O3,04}, {Oj, O4}, {O4,05}} is a quadrilateral plus an edge incident with the fifth vertex, hence by lemma3.4 there is an M' c E(P)u{{Oj, O4}} such that TOM' has no isthmus. Let M be set of the edges ftom E(P)u{{x j , X 4 } } corresponding to M'. Then (G\E(P))uM c O' t G has no isthmus. Then (G\E(P))uM is without isthmus and has fewer edges than O, so by induction (GVE(P))uM supports a nowhere zero 5-flow. By proposition 1.2 or 1.4, we have that G supports a nowhere zero 5-flow, a contradiction.

Case 7. |X| =» 6.

By lemma2.9 and the choice of P, |V(P)|» 6 and P t Hq. Without loss of generality we may let X be {Oj, O2 Og} where V(P)»{xi,..., xg) and txj Oj (l£i£6). Let P' be the subgraph of K(X) with edge set {{Oj, O2}, {O 2,03}...... {O5, Og}, {Og, Oj}}; so there is a natural correspondence between P and F. By lemma 3.6, there exists a matching or triangle e j, C3 of K(X) such that any matching M of K(X) avoiding these three edges has the property that TOM (and hence GwM) has no isthmus. If |E(F)n{ei, e2,63)1=0,1, or 3 then P"\{Cj, 62, C3} contains such a matching of K(X), So suppose E(P')n{ej, 62, C3} = {ej, e2): there are two cases where (a) e j, e2 are opposite in P', or else (b) they are coincident in P'. (If ej and e2 are neither coincident nor opposite, then the third edge 63 must not be coincident with cj or C2 and so would be in P', contradicting C3 f E(P').) 36

Suppose we are in case (a). Then the edges do not come from a triangle, and so

63 must be parallel to e j and tq. Let d be one of the diameters connecting the

endvertices of C j, 0 3 such that Gu{d} c G, and let d' be the edge of K(X) whose endvertices are the ornaments containing the endvertices of d. Then d’*^ . Define M'

to be the edge d' together with the two edges of F whose endvertices are disjoint from

the endvertices of d'; then M' is a matching of K(X) avoiding {c^, cj, 6 3 }, and so by lemma 3.6, we have that TuM' has no isthmus. Let M be d and the corresponding edges of P. By the tree pruning lemma 2.8, (G\E(P))uM is without isthmus, and by the minimality of G supports a nowhere zero 5-flow. Then proposition 1.4 implies G

supports a nowhere zero 5 -flow, a contradiction.

Suppose we are in case (b). From lemma 3.6 the three edges are from a triangle,

and T is the union of two disjoint subdivided 3-stars. Let c, f be the two edges of P'

adjacent to c j, cp, so that four edges of P' are e, ej, ej, f in that order. Since the vertex set of ornaments on one component of T lies entirely on one side of the cutset e, f of F,

Tu(P’\{c,f}) is a disconnected isthmusless graph. Consequently, by lemma 2 .6 we

have that G\{e,f} is a disconnected isthmusless graph. Since it is a proper subgraph of

G, it has a nowhere zero Z 5 »flow, and in particular one with the orientation taken so that

the edges of P (with e, f replaced) form a directed circuit If all four nonzero values of

Z5 occur on P\(e, f}, apply one of the automorphisms of Z 5 to every edge in one of

the two components to create a nowhere zero z^-flow

9'(g)=9(g) on edges g not in P, and 9 '(g)=9 (g)-x on the edges g of P. This preserves the flow property at each vertex, since the edges of P are a directed circuit and so x is subtracted exactly once from the sum of the flows directed into each vertex of P and subtracted exactly once from the sum of the flow directed out of each vertex of P. The new flow 9' is also nowhere zero, since 9 is nowhere zero off P, the value x did not appear in F\{c,f}, and the edges e, f are given the value -x. Thus O supports a nowhere zero Z^-flow, a contradiction. I 38

Chapter I I 85. Applications.

Corollary 5.1. Any graph without isthmus that embeds in the torus or Klein bottle supports a nowhere-zero 5-flow. pf: Let G be the family of graphs which embed in the toms or klein bottle. Since an embedding of a graph provides an embedding of all its subgraphs, G satisfies the hypotheses of the theorem. For O c G, fix an embedding and note that as we can add any chord across a face extending the embedding, Hq contains the set of hexagonal faces of that embedding. By applying the theorem we conclude that either every isthmusless graph in G has a nowhere zero 5-flow or else the smallest isthmusless graph

G in G without a nowhere zero 5-flow is of girth at least six with H q empty, and hence has no hexagonal faces.

First we observe that G has no divalent vertices. Suppose e, f arc the two edges incident with a divalent vertex. Then G/e embeds into the same surface as G and has fewer edges, so that G/e has a nowhere zero 5-flow. One obtains the graph G from

G/e by subdividing the edge f; given a flow 9 on G/e, to lift it to a flow on G direct e in the same direction as f and assign it the same value as f. One readily verifies that the new mapping is a nowhere zero flow on G, a contradiction.

Now for a bit of topological graph theory. Let G be a connected graph and S be a surface of maximal Euler characteristic with G embedding into S. Then G satisfies Euler's equation in its embedding into S, 39

(5.2) I V| - |E| + |F| = x(S)»

where F is the set of faces of O and % is the Euler characteristic of S. Let O be the graph given above. Let f; be the number of faces of size i and dj be the number of vertices of valency i. Then

(5.3) Z fj =* [F| and Z d i = |V|. i=l i=l

Each edge is in exactly two faces of the embedding, and each edge is incident with two

vertices (counting multiplicities). Thus

(5.4) Z i f i = 2|E| and Z id i = 2|E|. i=l i=l

Multiplying (5.2) by 6 and rearranging

(5.5) (6|V|-2|E|) + (6 |F| - 4|E|) = 6 X(S),

or

(5.6) Z (6-i) fj + Z(6-2i)di = 6x(S). i=l i=l

Let us apply (5.6) to the problem at hand. There are no isolated, monovalent, or

divalent vertices. Since G has girth at least 6 and no hexagonal faces, fi=...=f£= 0 . Hence for the embedding of our minimal counterexample, 40

6 0 0 6 (5.7) 1 (6 -!)^ + I (6-2i) dj = 0. i=7 i=3

But as some fj*0, the left side of (5.7) is negative, a contradiction. ■

Corollary 5.8. Any graph without isthmus that embeds in the unorientable surface of characteristic *1 (three cross caps) supports a nowhere zero 5-flow.

Ef: Let G be the family of graphs which embed into any surface of characteristic at most

- 1. As in corollary 5.1, G satisfies the hypotheses of the theorem. For G 6 G, fix an embedding and note as before H q contains the set of hexagonal faces of the embedding.

Let G be a minimal counterexample if one exists. By the same minimality argument of corollary 5.1, G does not have a divalent vertex. Thus (5.6) holds with x(S)£-l. Multiplying by -1,

( 5.9) fj + 2fg + 3fg + 4f jo + 5f j j + 6 fj 2 +... £6 .

From (5.3) and (5.9) we have 2| E | £ 6 + 6 | F |, or | E | £ 3 + 3| F |. Also, since the values of fj must be nonnegative integers, we have that | F | £ 6 , and that equality holds only when all faces are 7-circuits. Thus | E | £ 21.

A simple cubic graph of girth n and of smallest size is called an n-cage. The

3-cage is a 4-clique, the 4-cage is the graph whose six vertices may be denoted xj, X 2 » x3» yi» ?2» y3 BHd whose 9 edges adjoin each pair of xj, yj (l£ij£3) exactly once. The

5-cage is the Petersen graph. It is well-known that the 6 -cage is the Heawood graph, a bipartite graph obtained from the point-line incidences of the projective plane of order 41

two, which has 14 vertices and 21 edges. It is the unique such on 21 edges. However the Heawood graph is bipartite and so cannot have circuits of size seven, hence no faces of size seven. Thus no counterexample exists. I

A graph is nearly planar (nearly S ) provided deletion of one vertex allows embedding of the remaining graph in the plane (surface S).

Corollary 5.10. Let S be any surface with X(S)£0. Then any nearly S graph without isthmus has a nowhere zero 5-flow.

pf: Let G be the family of nearly S graphs, where x(S)£0. Let G < G; then there exists a vertex v such that G\{v} embeds into S. If H c G then H\{v} c G\{v} aiso embeds into S. Thus G is closed under subgraph inclusion and so satisfies the hypotheses of the main theorem. Fix an embedding of G\{v}. Note that H q contains the set of hexagonal faces of G\{v}, since adding a chord across a face still permits embedding into the same surface, so that for an arbitrary diameter d of P, Gu{d} e G. Also, H q contains the set of hexagons of G consisting of at least four vertices from a single face of G\{v} together with v and any other vertex completing the circuit: if a hexagon P of

G has at least four vertices from one face and also includes the special vertex, then by the pigeonhole principle at least one pair from the face must be opposite in P. The diameter dj whose endvertices are this pair is a chord across a face, and so (Gu{dj })\{v} still embeds into S. The diameter d 2 with one endvertex v has (Gu{d 2 ))\{v}=G\{v} and so embeds into S. Thus P t H q as desired. If the corollary is false, then the minimal counterexample G has no isthmus, no divalent vertex, supports no nowhere zero 5-flow, 42

is of girth at least six, and has Hq=0. Let Oj be the graph obtained from O by deleting the special vertex v and sequentially contracting one of every pair of series edges until no divalent vertices remain. (Thus Gj is topologically equivalent to G\{v}, and Gj

embeds into S.) Applying equation (5.6) to O j, the right-hand side is nonnegative, and

so there is a face F with at most six (non-divalent) vertices. Note that in (5.6) if there are

no faces of Gj with five vertices, the left hand side of (5.6) is nonpositive and so must be zero term by term: hence all faces are hexagons. The face F corresponds (possibly after subdivision of edges) to a circuit P of G\{v).

Case 1. There exists a face F of size at most 5.

Since G has girth 6 , there is at least one divalent vertex adjacent to v in P. If there is exactly one, then P is a hexagonal face of G\{v}, so P c Hq contradicting H q = 0. So P must have at least two divalent vertices adjacent to v. Choose two, say x and y, at minimal distance in P. Then there are at most 2 vertices between x and y (since P has at most five non-divalent vertices). Thus the shortest path from x to y in P together with the two edges with endvertices {x, v} and {y, v} is a circuit in G of size at most 5, which contradicts the girth of G being at least six.

Case 2. All faces of G j are hexagons.

If | V(P)|»6 , so that there were no divalent vertices adjacent to v on F, then P is a hexagonal face of G\{v), contradicting Hq = 0. Suppose some P has at least two divalent vertices adjacent to v. Then again choose x and y two vertices adjacent to v at shortest distance in P. If there are at most two vertices between x and y in P, then as in case 1, this would create a circuit of size at most 5, contradicting the girth constraint If there are exactly three vertices between x and y in P let Pj be the circuit consisting of these three vertices, x, y, and v. Then Pj consists of five vertices from a single face and the special vertex, so Pj c Hq, =><=, hence all hexagonal faces have exactly one 43 divalent vertex adjacent to v. Let x be this attachment, and x j, X 2 »..., xg be the remaining vertices of F, in order. Let F be the face not F using the edge (x j, xj}. By assumption, F has a vertex adjacent to v, say y. The shortest path in F from y to xj is a path of length at most 3. If x«y then the shortest path in F from y to xj together with the edge {x, xj} forms a circuit of length at most 4, a contradiction. If x*y, this path, together with the edges {xj, x}, {x, v}, and {v, y} make up a circuit P j, and hence the path must be of length at least three. Thus Pj is a hexagon of Q. But then Pj consists of four vertices from a face of the embedding and includes the vertex v, so Pj t H q , contradicting H q»0. Thus no counterexamples exist. I

Corollary 5.11. Suppose H is an n-clique subgraph of an isthmusless graph O where n*2 and G\V(H) embeds into a surface S with nonnegative characteristic. Then G supports a nowhere zero 5 -flow. pf: We proceed by induction on n » | V(H) |. The case n « 1 is corollary 5.10. So we may assume that n k 3. Fix an orientation of G. Number the vertices of H as 1,2,

..., n. Let G | be the directed graph obtained from G by contracting all edges of H.

Clearly Gj has no isthmus. By corollary 5.10, G] supports a nowhere zero z^-flow

9 j (c)= 0 on the darts of H. Now 9 ] is not necessarily a flow. At every vertex v of

G\V(H) the edges incident at v in Gj and G are identical, so has the sum of the values of the darts directed into v equal to the sum of the darts directed out of v. At the vertices of H there may be an imbalance. Define 04 to be the sum of the values of the darts directed into vertex i minus the sum of the values of the darts directed out of i

(l£i£n). Since 9 q is a flow, 44

(5.12) + 0 2 +... + ctn = 0 (in 2 5 ).

Alter the orientation of G (and negate the values of

with endvertices 1 and i, set 9 2 (f) = 9 j(f) - 0 4, and 9 2 (e) - 9 ^ ) on all other darts.

Again, at all vertices of G\V(H) the sum of the values of the darts directed into v is equal to the sum of the darts directed out of v. At vertex 1, an additional - 0 2 -0 3 -...-04, is directed out, but since Oi=- 0 2 -...-an, there is no imbalance at vertex 1. At a vertex i * 1, an additional -04 is directed into vertex i, so there is no imbalance at i. Thus 9 2 is a flow. We now define a sequence 9 3 ,... of flows so that has at least one more nonzero edge than 9 ^ Choose ej t E(H) with 9 i(ej)*= 0 . Let the ej be directed from x to y. Let z be any other vertex of H. Let C - {ej, ej, 6 3 } be the circuit of H induced by x, y, z. We may assume, by reversing the orientation and negating the value of 9 ^ on those edges, that C is a directed circuit. Now |{ 9 j(e^), 9 ^ ) , 9 ^ 3 ) }| £ 3, so there exists a nonzero element w not in this set Define 9 j+j by 9 i+i(cj)=9 i(cj)-w for j=l, 2,3 and 9 j+] (e)«9 j(e) elsewhere. Then it is an easy verification that 9 f+j is a z^-flow and that while none of the darts cj are made zero, the dart c j is now assigned the value -w * 0. This process must terminate after at most n(n-l)/2 steps with a nowhere zero 5-flow. I

Remark: The above is part of a general phenomenon — if G is any graph that supports a nowhere zero k-flow (k£4), then one may “expand" a vertex v by a complete graph on vl* ••• vn replying the endvertex v in an edge with an arbitrary choice of an endvertex vj and still have a graph that supports a nowhere zero k-flow. It would seem that proving a similar result for n = 2 would be very hard; in particular the result itself is 45

HOI true for n = 2 since an arbitrary graph can be contracted to a graph with one vertex and many loops, which supports a nowhere zero k-flow for any k£ 2 , and then the contraction process reversed by expanding vertices by edges. But of course not all isthmusless graphs support nowhere zero 4-flows. CHAPTER ffi

GIRTH 7 GRAPHS AND NOWHERE ZERO FIVE FLOWS

§1. The minimal counterexample.

When theorem H. 1 .1 is applied to the set G of all graphs, we conclude that either

the Tuttc conjecture is true or else a minimal counterexample is of girth at least 6 and has

no hexagons, i.e. is of girth at least 7. The 7-cage, (also known as the McGee graph

[5]) has 24 vertices. Thus in any class of graphs closed under subgraph inclusion, any

minimal counterexample must have at least 24 vertices. Moreover, if a girth 7 graph with minimum valency at least 3 contains a vertex of valency 4 it must contain at least 29 vertices; this is seen by noting that all the vertices at distance at most 3 from the 4-vaIcnt vertex must be distinct This provides lower bounds on the number of vertices. Depending upon the structure of the class G, the theorem may also provide an upper bound. Let G be the set of all graphs that embed into the sphere with two handles attached or the sphere with four cross caps attached. Certainly G is closed under subgraph inclusion. Choose a counterexample G as given by the theorem, and fix an embedding of that graph. Note that we can add any chord across a face and preserve the embedding, so certainly all hexagonal faces are contained in Hq . As before, (11.5,6) holds for this embedding: setting x(S)=-2 and multiplying through by -1 yields 46 47

(1.1) f7 + 2 f8 + ... + 6 f1 2 ^ 12.

It follows that there are at most 7-12/2=42 edges, so the minimal counterexample has at most 42-2/3=28 vertices. If we can show that every graph of girth 7 with no more than 28 vertices has a nowhere zero 5-flow, then no minimal counterexample to Tutte’s 5-flow conjecture can exist in the set of graphs that embed into the sphere with two handles adjoined or four cross caps adjoined. Deleting the vertices of a 7-circuit from the 7-cage and replacing each of the resulting 7 pairs of series edges at the divalent vertices with a single edge results in the Petersen graph, the 5-cage. This odd fact is generalized in the following theorem:

Theorem 1.2: Let G be a graph of girth 7, and P be any 7-circuit of 0. LetG*bethe graph obtained from G by deleting the vertices of P and contracting any series edge until no divalent vertices remain. Then G* has girth at least 5.

pft Let Q* be a circuit of G* of length less than 5. Let Q be the corresponding circuit of

G. Define an attachment o/P to be a vertex not in P adjacent to a vertex of P, i. c. a vertex of attachment of G\V(P), Since G has no circuit of length less than 7, Q must have at least three attachments of P. Suppose two attachments of P are consecutive in Q.

Since there is a path in P of length at most three joining any pair of vertices of P, this path, together with the two attachments in Q would determine a circuit of length at most six in G, a contradiction. So no edge of Q* corresponds to a path of Q with more than one attachment of P. This implies that Q* is a quadrilateral, and that at least three of the edges of Q* are subdivided with attachments of P. Suppose V(Q*)={qj, q2, q2, q ^ and V(Q)j{qj, x, q2, y, q 3 , z, q ^ in that order. Set V(P)={vj,v 2 , v 7) in that 48 order. Without loss of generality, we may assume that y is adjacent to v j. Since x, qj, y is a path of length two in G, the path joining x and y in P must be of length at least 3; considering similarly y, < 13, z we may assume that x is adjacent to V 4 and z is adjacent to

V 5 . If Q is a 7-circuit then V 4 , x, q j, q 4 , z, V 5 is a hexagon of G, a contradiction. If Q is an 8 -circuit then there must be another attachment w of P between q 4 and in Q; since x is adjacent to V 4 we must have w adjacent to v j and V5 , vg, V7 , w, q4 , z is a hexagon of G. Thus no circuit of G* of length less than five can exist, and G* has girth at least 5. I

This leads to a natural method of construction and examination of small graphs of girth 7. Recall all girth 7 graphs with minimum valency at least 3 on less than 29 vertices are cubic. The 7-cage has 24 vertices and is unique. Since cubic graphs have an even number of vertices, this leaves two possibilities for cubic graphs of girth 7 with at most 28 vertices: either 26 or 28 vertices.

Suppose G, without isolated, monovalent, or divalent vertices, is a girth 7 graph of 26 vertices. Let P be any 7-circuit of G. Now G is cubic, and after deletion of P and replacing each of the 7 pairs of series edges with a single edge, the resultant graph G* is cubic of girth at least 5 and has exactly 12 vertices. Suppose any edge c is in no pentagon. Then there is a binary branching tree through e ° {0,1} [see Figure 12] that is a subgraph of G*,where all these vertices must be distinct since e is in no pentagon. Thus G* has at least 14 vertices, a contradiction. Thus, every edge is in some pentagon. Figure 12. A binary branching tree.

To examine such graphs, we use a classification by Robertson [ 6 ]: suppose G is a cubic graph of girth S where every edge is in some pentagon of G. Let E^fG) be the set of edges of G in exactly k pentagons (l£k£4). A pentagon P of G is said to be singular provided |E j(G)nE(P)|«2. The edge of a singular pentagon P coincident with the two edges in only one pentagon is called a pivot edge of G. We say that pentagons P, Q of G are related when there exist pentagons Qq, Q j Qn of G such that Qq=P, Qn=Q and E(Qj.j)nE(Qj) is neither empty nor a pivot edge of G (l£i£n). This forms an equivalence class on the pentagons of G. The constituents of G are the graphs obtained by unions over equivalence classes of related pentagons. By definition, constituents are isthmusless and generated by pentagons. We say G is decomposable if G has a singular pentagon and indecomposable otherwise. A part L of G is a graph, every edge of which is in a pentagon, for which there exists a mapping f:L-»G such that fL = H is a constituent of G and f is 2-to-l on and one-to-one off the pivot edges of H whose singular pentagons are in G. Then L is said to represent H under f:L-»G and we write

L-»G. Let W be the set of graphs defined by the diagrams below: 50

abab a b a b a b a b

B A B A BABABABA C| C2 Cj C, C5 C6 C7 C8 C9 C1Q C„

Figure 13. An infinite family.

These graphs are drawn as though embedded in a cylinder or Moebius band, the dotted line AB to the left in the figure being indentified with the dotted line AB

immediately above Cj. When i is even, the darkened path forms two (i/2)-circuits in Cj,

when i is odd, the darkened line forms a single i-circuit. The darkened edges indicate edges contained in an odd number of pentagons.

Figure 14. Some sporadic examples.

These three graphs, labeled Sj, S 2 , and S 3 on 12,12 and 14 vertices respectively, do not fit into the family of Cj above, but are sporadic examples of indecomposable graphs. In Sj, the darkened edges are in four pentagons. The darkened edges in S 2 and S 3 are in an odd number of pentagons. In the plate that appears on page 52, two infinite sequences of graphs are 51 suggested in the families D j, D 3 , ... and D 2 , D4 , ... where the internal linkage is extended arbitrarily. Also given is a finite sequence T j ,..., Ty of graphs. Note that Dj has 10+2i vertices, and thatTj through T 7 have 12,14,14,16,22,10, and 16 vertices respectively. Each Dj has two pivots, as do Tj through T 5 . The graphs Tg and T 7 each have three pivots. These graphs act as representative parts for the constituents of any decomposable O. si

i \

T 4

‘T3 53

T5

• n

M

VT6 l y

Figure 15. Three of the T parts

Again, darkened edges represent edges in an odd number of pentagons.

Robertson's classification is based upon the following theorem:

Theorem 1.3: If H is a constituent of G, a cubic girth 5 graph generated by pentagons, then there exists a unique L in the set W above with L-»H.

We are trying to determine all cubic girth 5 graphs G* with 12 vertices and the property that every edge is in some pentagon. Thus Robertson's theorem applies. If the graph is 54

decomposable, it must have a decomposable part H where if H has t pivots then the vertex contribution to the graph is |V(H)| - L Since G* is decomposable, it must be that

either it decomposes into one part, so |V(H)| - 1 ■ 12, or else it decomposes into two or

more parts, so |V(H)| - 15 6 . The graphs with |V(H)| - 1 = 12 are D2 , T2 , and T 3 . No

graph in W has |V(H)| - 1 S 6 . If the graph G* is indecomposable it must be one of S j or s* Suppose G, without monovalent or divalent vertices, is a girth 7 graph of 28 vertices. Let P be any 7-circuit of G. Now G is cubic, and after deletion of P and replacing each of the 7 pairs of series edges with a single edge, the resultant graph G* is cubic of girth at least 5 and has exactly 14 vertices. We distinguish three possibilities:

( 1) Every edge e e E(G*) is in some pentagon of G*.

In this case, Robertson's theorem applies. If the graph is indecomposable, it

either is one of the sporadic examples, hence S 3 , or comes from the infinite family,

hence C 7 . If the graph is decomposable, it must have a decomposable part H where if H

has t pivots then the vertex contribution to the graph is |V(H)| - 1 Since G* is

decomposable, it must be that either it decomposes into one part, so |V(H)| - 1 « 14, or

else it decomposes into two or more parts, so |V(H)| - 1 S 7. In the first possibility, the part must be D 3 or T 4 . In the second possibility, the part must be Tg joined to itself.

(2) No edge is in a pentagon (G* has girth 6 ).

Then all edges leaving (aj, bj, b 2 > on the left of the bipartition suggested by figure 12 cross to (xj, X 2 » yj, y2 >» and one can easily verify this yields the Heawood graph. (3) There exists an edge in no pentagon, but some edges are in pentagons. Using the notation of Figure 12, the proof breaks into two subcases:

(3i) The vertex 0 is in exactly one pentagon. 55

Without loss of generality, we may assume that {a2, bj} is an edge and all

unspecified edges incident with a vertex in {a^ a2. b j, b 2 > cross from (a ^ a2, bj, b2} to (xj, x2, yj, y2). Hence there exist exactly six crossing edges. Thus there must be a

pentagon through 1, say (x2, y j }. Both edges leaving cannot be incident with

{x2, y j} (or we would have a triangle) so without loss of generality, one of the edges incident with aj is (a j, xj). Now the other edge leaving aj may not be incident with

x2, so we have the following two possibilities:

Figure 16. Two possibilities.

Consider (3i.a). If a 2 is adjacent to x2, then b 2 would be adjacent to xj and y2> and b j would be adjacent to y 2 creating the quadrilateral b2, b, bj, y2. So a 2 is adjacent to y2.

Now b] is not adjacent to y 2 or we have the triangle a2, bj, y2. If b 2 is adjacent with xj, then we are in case (3i.b) (under the automorphism (a, b)(aj, b 2 )(a2, bj)). So b 2 is adjacent to x 2 and bj is adjacent to Xj. This yields the graph Gj: Figure 17. The graph Oj.

Consider (3i.b). Now &2 >s not adjacent to xj or y 2 , or we have a quadrilateral through aj and ^ So without loss of generality, &2 is adjacent to xj. If b 2 is adjacent to both yj and y j then we have the quadrilateral b 2 , yj, y. y2 so *>2 must be adjacent to xj. If b | is adjacent to yj we have the quadrilateral ^ x 2 * y 1* 50 W must be adjacent to y2 > Thus b 2 *s remaining adjacency is to yj. This yields the graph G 2 :

Figure 18. The graph G 2 .

(3ii) The vertex 0 is in exactly 2 pentagons.

As before, we may assume that {a 2 , b j) is one edge. If the other pentagon through 0 also includes a 2 , say {a2 i b 2 >, then a 2 , bj, b, b 2 would be a quadrilateral. 57

So the other edge creating a pentagon through 0 is (aj, b 2 >. Similarly, without loss of

* generality we may assume that the graph contains the edges {xj* y i) and {x j, y 2 >. By symmetry, one of the crossing edges is {aj, x j). From this, we distinguish two types of crossing pairs:

creating pentagons creating hexagons

Figure 19. Two types of crossing pairs.

There are thus three possibilities: both pairs of edges incident with (aj, b 2 > and (xj, y2 > create hexagons, one pair creates pentagons and one creates hexagons, or both create pentagons. The first possibility yields the graph G 2 again. The remaining two give two different graphs, G 3 and G 4 . These two complete the classification when G* has 14 vertices. Figure 20. The graph Q 3 .

Figure 21. The graph G 4 .

Thus, if v=26, then G *« {D2l, T2l, T3l, Sj, S2}; or if v=28, then G* « {S3, C7, D3l,

T4 *» Tg'Tg, G j, G2, G3, G4 , Heawood}, where Mt" denotes the constituent twisted upon itself in some fashion with the corresponding pivots identified, and denotes the union of the constituents with the pivots identified in some fashion. It is possible that these operations may give rise to several or no decomposable graphs. This list, while somewhat lengthy, is finite. This means that it is possible to classify all graphs that are 59

of girth seven, without isolated, monovalent or divalent vertices, with at most 28 vertices

by considering each of these smaller topological subgraphs in turn, finding (up to isomorphism) all sets of seven edges that meet every hexagon of the subgraph and meet

every pentagon twice, and then finding all ways to reattach the deleted 7-circuit to these seven edges to create a girth 7 graph.

/ 60

Chapter in 92. The case v=24.

It is well known [5] that there is, up to isomoiphism, exactly one girth 7 graph on 24 vcrdcies. A drawing is given below. Notice that this graph is Hamiltonian;

Figure 22. The 7-cage, the McGee graph.

thus there exists a set of three perfect matchings whose union is the edge set Such matching decompositions of cubic graphs are called Tait colorings. Tait colorings lead in a natural way to nowhere zero z^flows; index the color classes with the non-zero values of Z4. Direct the graph arbitrarily; since each element of Z4 is of characteristic 2 the orientation is irrelevant Since each vertex is incident with exactly one edge of each color class, the sum of the flow values at each vertex is zero. 61

Thus for cubic graphs, Tait colorings coincide with nowhere zero 4-flows. In particular, any Hamiltonian cubic graph supports a nowhere zero 4-flow. Thus the

7-cage supports a nowhere zero 4-flow. 62

Chapter in 83. The case v=26.

"Any problem that can be cut into a structure with finitely many cases is a combinatorial problem, in the worst sense.”

- Neil Robertson, June 1981

Recall from the discussion of § 1 that there are five graphs to consider three decomposable graphs D 2 l, T2 l, T j1 and two indecomposable graphs Sj, and S 2 . (Sec §1, plate 1, and Figure 14.) Let us consider the decomposable examples first

Suppose G* is isomoiphic to D 2 l.

2 3 4

a

b 9 10

Figure 23. D2 with veitex labeling.

We need to identify the two unlabeled vertices with the vertices labeled a and b in such a fashion that no new pentagons are introduced. Suppose the upper unlabeled vertex is identified with a. Then the pentagon a, 5,9,10,4 is created. Suppose the upper unlabeled vertex is identified with b. Then the pentagon b, 1,2,3,4 is created. 63

Thus any possible identification of the unlabeled vertices will make the graph indecomposable, i. e. one of Sj or Sj. Thus this case need not be further considered. Next suppose that G* is isomorphic to

Figure 24. T j with vertex labeling.

Again, we need to identify the two unlabeled vertices with the vertices labeled a, b in such a fashion that no new pentagons are created. If the upper unlabeled vertex is a, then a, 1, 2 ,3, is a quadrilateral which is not allowed; if the upper unlabeled vertex is b then a, 1,2,3, b is a new pentagon. Thus this assignment leads to an indecomposable graph, hence must be one of S j or S 2 , considered below. 64

Finally, suppose G* is isomorphic to T j1.

1 2 3 Q

b

8 9 1 0

Figure 25. T3 with vertex labeling.

Again, we need to identify the two unlabeled vertices with the labeled vertices a, b in such a fashion that no new pentagons are created. As before, the path a, 1,2,3 is of length 4, so the upper vertex can not be identified with a (or a quadrilateral would be created) nor with b (or a pentagon would be created). Thus this case does not arise.

This leaves only the two indecomposable girth five cubic graphs on 12 vertices. S s 1 2

Figure 26. Two indecomposable girth 5 graphs with 12 vertices.

First, consider the graph S,. Let the octagon 1,..., 8 be denoted O. The edges ab and cd are each in four pentagons: any path of length 2 of O together with the path of length 3 through a central edge, ab or cd. Edges of O are in two pentagons each determined by the two paths of length two in O that contain them. Hexagons consist of any path of length 4 in O together with the edges leading in to the appropriate vertex not in O. Refer to the edges al, a5, b3, b7 as being on "the ab side", and the edges c2, c 6 , d4, d8 as being on "the cd side". The automorphism group of Sj is easy to describe: the automorphism group of O is D jg, the dihedral group of sixteen elements. The subgraph O is stabilized by the automorphism group of S j. Each automorphism of O extends uniquely to an automorphism of S j. 66

Define the incidence variable Xg to be 1 if the edge e is to be subdivided to

create an attachment to the 7-circuit P, and 0 otherwise. Let P be the set of all pentagons inSj. Then

(3.1) 1 6 - 2 |P |X (Ixg). I ( I xe ) = 2’7 + 2 xab + !>;„, Q c P eiE(Q) c

so that the proof breaks naturally into two parts: either both ab and cd are subdivided to form attachments of P in G, or else only one, say ab is to be an attachment

EfllLL Xrt-Xgi-1. There are five remaining attachments to determine. Suppose none of the remaining attachments are on O. If four attachments are on one side, say the ab side, then there will be only one on the cd side. Thus there will be a pair of non-adjacent edges incident with c, d, without attachment, so there will be a pentagon with only one attachment So it may be assumed that there are three attachments on the ab side and two attachments on the cd side. If the two attachments on the cd side are coincident, then there will be a hexagon consisting of the other two edges on the cd side plus half of O, without attachment So there must be one edge incident with c and one edge incident with d, neither of which is an attachment; then there will be a pentagon through these edges with only one attachment, which is impossible.

Suppose there is exactly one attachment on O. Suppose there are three attachments on one side, say the ab side; then there is only one attachment on the cd side, say c2. The pentagons through 4,5,6 and 6,7,8 would each need an attachment on O, but there is only one attachment available. So there are two attachments on each side. First consider the case that the two attachments on one side are coincident, say al and a5. Then the two hexagons using b3 and b7 would each need attachments, which is impossible. So both the attachments on either side have no common incident vertices. 67

Without loss of generality choose al and b3 on the ab side. By symmetry, the two attachments on the cd side must be c2, d4; or c 6 , d4. Since the pentagon 5 , 6 ,7, a, b does not yet have a second attachment, the attachment on O must be 56 or 67. If c 6 and d4 are the attachments on the cd side then the pentagon 1,2, c, d, 8 would need an attachment on O, which is impossible. If c2 and d4 are the attachments on the cd side the pentagon 6 ,7 , 8 , d, c also needs an attachment on O, so the attachment must be 67. This configuration satisfies the pentagon and hexagon conditions, but does not give rise to a girth 7 graph, as P cannot be put back in to produce a girth 7 graph. This may be seen as follows: let v j ,..., vy be the seven vertices of the deleted polygon P, with x j x -7 the seven vertices adjacent to P, xj adjacent to Vj. Define two subdivision vertices xj and xj to be opposite provided i«j+3 or i=j-3 (mod 7). Define two subdivision vertices xj and xj to be consecutive provided i=j+1 or i«*j-l (mod 7).

Lemma 3.2: (i) If two subdivided edges share a common vertex v, then their

subdivision vertices xj, xj must be opposite.

(ii) If the subdivided edges are coincident with an edge, say uv, then their

subdivision vertices Xj, xj must not be consecutive.

pf: In (i), if xj and xj are not opposite, then the path vj, xj, v, xj, vj together with the shortest path from Vj to vj in P will constitute a hexagon. Similarly, in (ii), we may assume Xj is adjacent to v and xj is adjacent to u. If i and j are consecutive, say i=j+l, the circuit Xj, vj, vj, xj, u, v is a hexagon. I 68

Figure 27. Sj with ab, cd, al, b3, c2, d4,67 subdivided.

Lemma 3.2 is exploited heavily to determine the manner in which P may be reattached to the subdivision vertices. Returning to the current configuration, namely the one where the edges ab, cd, al, b3, c2, d4, and 67 are subdivided, we may assume by

symmetry that the edge ab is subdivided with xj. Since b3 and al are to be subdivided,

they must be assigned vertices X 4 and X5 opposite xj by Lemma 3.2 (i). Using

symmetry, assume b3 is subdivided by X 4 and al is subdivided by X 5 Now 67 shares

an edge with ab and b3, so can not be subdivided with X 2 » X3 , or x-j, so is subdivided with X5 . Also 2c shares an edge with 3b, 67 so cannot be subdivided with X 3 or x-j, so is subdivided with X 2 - Then cd shares a vertex with 2c so can not be subdivided with X 3 or X7 , a contradiction.

Next consider the case where there are two attachments on O. If there are three attachments on one side, say the ab side, then there are no attachments on the cd side.

Each of the four paths 2,3,4; 4 ,5 , 6 ; 6 ,7 , 8 ; 8 ,1 ,2 is then in a hexagon with only one attachment identified so far, so would require an attachment on O. But there are only 69

two attachments on O, so this is impossible. Thus the attachments must be balanced,

say two on the ab side and one on the cd side. By symmetry, we may assume the single

attachment on the cd side is 2c. Then since 6 ,7 , 8 , d, c and 4 ,5 , 6 , c, d are two pentagons with only one attachment identified, the two attachments on O must occur in the paths 4,5 , 6 and 6 ,7 , 8 . But then the hexagon 8 ,1 ,2 ,3 ,4 , d has no attachment, a contradiction.

Now consider the case where there are three attachments on O. Two graphs will arise out of this configuration: 15 25

Figure 28. Beth.

The Hamiltonian graph above is denoted Beth. This graph is characterized by the number of 7-circuits through each vertex: there are eight vertices in nine 7-circuits: these have been numbered 1 through 8 . (Notice that 1 and 5 are different from the other six vertices in that they are adjacent only to vertices in nine 7-circuits. The vertices 21 through 26 are in seven 7-circuits each. All other vertices are in eight 7-circuits. 71

0

7 6

Figure 29. Casper.

This graph, with two disjoint circuits xq, xj, ..., x j 2 and xq', x^, ..., x jj' of length

13, is denoted Casper. The vertex xj is adjacent to xj\ where j = 5 i (mod 13) for i = 0 ,

..., 12. Casper is also Hamiltonian; the circuit is, up to isomorphism, unique and given

b y x q , x l t X2, X 3 , X 4 , x7', xg', x j2, xt j, x3', X2*, xj', xg, x7, X 9 ', x10', !*, x j q *

X 9 , xg*, xg', X 4 ', xg, xg, X j 2 ' , xq*. Casper may be distinguished in that every vertex

is in exactly seven 7-circuits. (The automorphism group is vertex, but not edge transitive.)

First assume that both of the attachments on the ab or cd side are on the same side, say the ab side. The four pentagons through cd intersect only in edges incident with c, d thus these four pentagons would each need an attachment cm O, contradicting the fact that there are only three attachments on O. So there is one attachment on each side. Without loss of generality, we may assume that the attachment on the ab side is al, and 72 the attachment on the cd side is one of c2, c 6 . Then because of the two pentagons through aS, there is one attachment on the path 3,4 ,5 and one on the path 5 , 6 ,7; because of the hexagon 7 , 8 ,1 ,2 ,3 , b there is also an attachment on the path 7 , 8 ,1 ,2,

3.

Suppose that the attachment on the cd side is c2. Then the two pentagons through c6 each require an attachment on O, so we have that either 45 is not subdivided or else both 45 and 67 are subdivided. Suppose the former. Then 34,56, and 78 must be subdivided.

c

d

Figure 30. Sj with ab, cd, al, 2c, 34,56,78 subdivided.

Without loss of generality we may assume that ab is subdivided with xj and al is subdivided with X 4 . The vertex subdividing 56 is not consecutive with xj or X 4 and so must be X 5 . Then the vertex subdividing 78 will be consecutive with one of xj, X 4 , or xg, contradicting lemma 3.2 (ii).

So we may assume that 45 is subdivided. Because of the hexagon 8 ,1,2,3,4, d 73

we must have 67 subdivided, and then there are three possibilities for the remaining attachment on 0 : 81,12, or 23.

Figure 31. Si with ab, cd, al, c2,45,67,81 subdivided.

Case 1. The attachments are ab, cd, al, c2,45,67,81.

Without loss of generality, we may assume that the vertex subdividing al is xj, the vertex subdividing ab is X 4 , and the vertex subdividing 81 is X 5 Now the vertex subdividing 45 is at distance 2 from a, so may not be X 2 , * 3 , or X7 by Lemma 3.2 (i). Thus the vertex subdividing 45 is xg. By Lemma 3.2 (i), the vertices subdividing cd and c2 are opposite; the only remaining opposite pair is 3,7. But since the remaining vertex, X 2 » would then subdivide 67, X 3 and X 2 would be in a hexagon X 2 , 6 , c, X 3 , V 3 ,

V2 « Thus this case is impossible. Case 2. The attachments are ab, cd, al, c2,45,67,12.

Since the only fact used in the previous case was that 81 and al share a common vertex, replacing 81 and 12 mutatis mutandis in the previous case yields the same 74

impossibility.

Case 3. Attachments are ab,cd, 81,02,45,67,23.

Applying the automorphism (ac)(bd)(12)(38)(47)(56) yields case 1.

Suppose alternatively, the attachment on the cd side is c 6 . The two pentagons

through c 2 require an attachment, so there must be an attachment on the path 8 , 1 ,2 and

on 2 ,3,4. Combined with the previous information, this yields that 34 must be

subdivided, and that the two remaining attachments are one of 81,12 and one of 56,67. This will give rise to the graphs Beth and Casper. Let us consider these one at a time.

Figure 32. Sj with ab, cd, al, c 6 ,18,34,36 subdivided.

Case 1. The attachments are ab, cd, al, c 6 ,18,34,36.

Without loss of generality we may assume the vertices subdividing al, ab, and 18 are xj, X4 , and X5 respectively. Since the vertex subdividing 36 can not be consecutive with xj or X 4 by lemma 3.2 (ii), it must be xg, The vertex subdividing 34 can not be consecutive with xg or X 4 , so must be xj. Since the vertex subdividing cd can not be 75 consecutive with x2, the edges cd and c 6 are subdivided with x 7 and X3 respectively.

This yields Beth under the isomorphism 1,2,..., 26 «->b, X 4 , 3,7, v7, vg, vj, x7, v4, a, 2, x 2 , 6 , 8 , v5, xg, xlt v2, c, d, v3, x 3 , 5 ,4 ,1, x5.

Figure 33. Sj with ab, cd, al, c 6 ,12,34,56 subdivided.

Case 2. The attachments are ab, cd, al, c 6 ,12,34,56.

Without loss of generality we may assume the vertices subdividing al, ab, 12 are xj, x 4 and xg respectively. Since the vertex subdividing 56 can not be consecutive with xj or X 4 by lemma 3.2 (ii), it must be xg. The vertex subdividing 34 can not be consecutive with xg or x4, so must be x2. Since the vertex subdividing cd can not be consecutive with x2, the edges cd and c 6 are subdivided with x 7 and x 3 respectively. This yields Beth under the isomorphism 1,2 ...... 26 <4 x4, a, b, v4, x7, v7, d, c, xj,

5,3,7, v5, v3, vj, vg, 4 , 8 ,2, x 3 , 1, x5, xg, 6 , x2, v2.

Case 3. The attachments are ab, cd, al, c 6 ,12,34,67. 76

Applying the automorphism (ac)(bd)(16)(25)(34)(78) yields case 1.

Figure 34. Sj with ab, cd, al, c 6 , 18,34,67 subdivided.

Case 4. The attachments are ab, cd, al, c 6 , 18,34,67.

Without loss of generality we may assume the vertices subdividing al, ab, and 18 are xj, X4 , and xg respectively. The edge c 6 is coincident with 67 and cd, so their attachments must be pairwise opposite: the only remaining sequences of length three of sequentially opposite vertices are 6 ,3,7 or 2,3, 6 . Since neither neither attachment of cd or 67 can be consecutive with X 5 by lemma 3.2 (ii), it must be that c 6 is subdivided with xg, and 34 subdivided with xy. Since the vertex subdividing 67 can not be consecutive with X 4 , it must be X2 and so cd is subdivided with X 3 . This yields the graph Casper under the isomorphism 0 ,..., 12, O',... 12' <-> c, xg, vg, vy, vj, x j, a,

X4 , b, 7 , 8 , d, x3, 2,3, xy, 4 ,5 , 6 , x2, v2, v3, v4, vg, x g ,1. 77

The next possibility under consideration is when there are four attachments on O.

Without loss of generality we may assume the one attachment on the ab or cd side is a l. The four pentagons through cd all require attachments on O, so there must be an attachment in each of the four paths 2,3,4; 4 ,5 , 6 ; 6 ,7 , 8 ; and 8 ,1,2 . Exploiting symmetry, we may assume that 12 is an attachment Since the two pentagons through a5 require O attachments, two of the attachments must be on the paths 3,4,5 and 5 , 6 , 7. If 45 is subdivided, then 67 is as well and we have a choice of 23 or 34. If 45 is not subdivided then 56 and 34 are subdivided and we have a choice of 67 or 78. This leaves four cases:

Figure 35. S} with ab, cd, al, 12,23,45,67 subdivided.

Case 1. The subdivided edges are ab, cd, al, 12,23,45,67.

Without loss of generality the edge ab is subdivided by X 4 , al by xj, and 12 by

X3 The vertex subdividing 23 must be opposite X 5 and cannot be xj, so must be X 2 .

The vertex subdividing cd may not be consecutive with Xj or X 5 so must be xy. But then the vertices subdividing 67 and 45 must be X 3 , xg, and so the edge subdivided by 78 xg and the edges subdivided by xy are coincident with an edge, contrary to lemma 3.2 (ii).

Figure 36. Sj with ab, cd, al, 12,34,45,67 subdivided.

Case 2. The subdivided edges are ab, cd, al, 12,34,45,67.

Again, without loss of generality we may assume that the vertices subdividing ab, al, 12 are X4 , xj, xg respectively. Since c is adjacent to 2, the vertex subdividing cd cannot be xg, hence must be one of X 2 » X 3 , xy. But then since the only remaining opposite pairs of vertices are xy, xg; X 3 , xg; and X 3 , xy; and these must subdivide the pair 34,45, so the vertex subdividing cd is consecutive with one of the vertices subdividing 34 or 45, contrary to Lemma 3.2 (ii).

Case 3. The subdivided edges are ab, cd, a l, 12,34,56,67.

Replacing 34,45 with 56,67 in case 2, mutatis mutandis, eliminates this case. 79

Figure 37. S j with ab, cd, al, 12,34,56,78 subdivided.

Case 4. The subdivided edges are ab, cd, al, 12,34,56,78. As before, we may without loss of generality assume that ab, al, 12 are subdivided by X4, X |, X5 respectively. The remaining vertices are X 2 , X 3 , xg, x 7 and cd must be subdivided by one of this set Moreover, since all other attachments are at distance 2 from cd, this means the remaining attachments must not be consecutive with the vertex subdividing cd by lemma 3.2 (ii). But each index is consecutive with one of the other three, which is a contradiction. There remains only the possibility where all five remaining attachments are on O.

This means that the vertex subdividing ab will be at distance two from five other subdivision vertices. This would imply that the vertex subdividing ab will be consecutive with a subdivision vertex at distance two, which is contrary to lemma 3.2

(ii). Thus no more configurations arise assuming that both ab and cd are subdivided. 80

Partll; *ab=^* There are six remaining attachments to determine. However, there is a useful piece of information that did not exist before: since cd is not subdivided, (3.1) holds with equality, and so every pentagon has exactly two attachments. A helpful observation is that considering only the four pentagons through cd the same double count as in (3.1) yields

(3.3) I ^ + 2 2 xe« 8. e c E(O) e on cd side

Thus the only possible configurations must have either two attachments on the ab side

and four attachments on the cd side, one attachment on the ab side and three attachments

on the cd side, or no attachments on the ab side and two attachments on the cd side. Suppose we have two attachments on the ab side, thus four on the cd side and none on O. If the two on the ab side are coincident with one vertex, say al and aS, then

b, 3 ,4 ,5 , 6 ,7 will be a hexagon without attachment If the two are not coincident, say

al and b3, then a, 5 , 6 ,7, b will be a pentagon with only one attachment Thus this configuration can not occur.

Suppose there is only one attachment on the ab side, say a l, and thus three

attachments on the cd side and two attachments on O. Since b, 3,4,5, a and a, 5 , 6 ,7,

b both need attachments on O, both attachments on O occur on the path 3,4, S, 6 ,7.

The hexagon 7 , 8 ,1 ,2 ,3 , b has then no attachment, a contradiction. Thus this configuration can not occur. So we may assume there are no attachments on the ab side, two attachments on the cd side, and four attachments on O. This will lead to the graph given on the following page, denoted Alfred. The display explicitly gives the automorphism group of Alfred: the automorphism group stabilizes the cube, and any automorphism of the cube extends to an automorphism of Alfred (The automorphism group of the cube is of order 48.)

Alfred can be distinguished from Beth by the number of vertices in nine 7-circuits (on which the graph is transitive). This display shows Alfred to be Hamiltonian.

Figure 38. Hamiltonian display of Alfred 82 A B

G

W

K

F E

Figure 39. Another display of Alfred.

There are two possibilities for the attachments on the cd side: either the two

attachments are on edges incident with the same vertices, say c 2 and c 6 , or else they are

not, say c2 and d4. If the attachments are c2 and c 6 , then without loss of generality the

vertex subdividing ab is xj, so since xj may not be consecutive with the four

attachments on O, we must have the vertices subdividing c2 and c 6 are x

these vertices must also be opposite, which is not true. Thus we may assume the subdivided edges on the cd side are c2 and d4. Again, since every pentagon through ab has only one attachment not on O, there must be exactly one attachment in each of the 83

four paths 1,2,3; 3,4,5; 5,6 ,7; and 7 , 8 ,1. The pentagon 2,3,4, d, c already has

two attachments, so'12 and 45 must be subdivided. The pentagon 6 ,7 , 8 , d, c needs two O attachments, so the remaining subdivided edges are 67 and 78.

Figure 40. Sj with ab, c2, d 4 ,12,45,67,78 subdivided.

Up to isomorphism, this has a unique assignment of subdivided vertex indices to attachments: exploiting symmetry, we may assume that the vertices subdividing ab, 2 c,

4d, arc xj, X2 and x-j respectively. Then the vertex subdividing 78 must have index 3,

4, or 5; and the vertex subdividing 67 must have index 4,5, or 6 ; since this pair must be opposite, we have 78 and 67 are subdivided by X 3 , xg respectively, hence 45 and 12 are subdivided by X 4 and respectively. This yields Alfred, under the isomorphism A, B,

•••» Z w V4 , Vj, c, d, Xj, b, Vji xg, xj, 2,4,5,1, vj, vg» 7, vj, 3, X 4 * a, x^» X2 » X7,6,8. 84

This completes the analysis of graphs whose underlying girth 5 graph is Sj. Next is S 2> Let N be the nonagon 1, 2, •••>9. There are nine pentagons, consisting of a path of length 4 in N together with the vertex from (a, b, c} incident with the ends of the

path. The hexagons consist of two pairs of edges incident with a, b, or c and the two

edges of N connecting them. Edges of N arc in three pentagons each, edges incident with a, b, or c are in two pentagons.

Once again, the automorphism group is easy to describe: the subgraph N is

stabilized by the automoiphism group of S 2 . Each automorphism of N uniquely extends to an automoiphism of Sg A 40 degree rotation of N cycles (a, b, c), and a flip fixes one of a, b, c and exchanges the other two vertices.

As before, let P be the set of pentagons of S 2 . Looking once again at the double sum,

(3.4) 18=2| P | £ £ (Xxc) = X ( I x c )» 2*7 + £ xe Q« P e t E(Q) e e E(S2) E(Q) > c ecE(N)

so that at least four edges of N arc attachments. Suppose some outer triad (i. e. a triple of edges incident with one of a, b, or c) has no attachment Then restricting (3.4) to the

three pentagons through that triad (since then edges of N are in only one pentagon) we

have that there must be at least six attachments on N. Thus there are three possibilities: either no outer triad has an attachment; there exists one triad with exacdy one attachment,

and no other triad has an attachment; or all triads have exactly one attachment. Suppose no triad has an attachment. Then seven out of the nine edges of N are attachments. Without loss of generality, one edge without attachment is 12 and the other is one of 23,34,45, or 56. Since the pentagon 1,2 ,3 ,4 , a must have two attachments, 23 and 34 must be subdivided. Since 1,2, b, 5,4, a needs an attachment, 85

45 must be subdivided. Thus 12 and 56 arc the only edges of N that arc not subdivided.

Figure 41. S2 with 23,34,45,67,78,89,91 subdivided.

Exploiting symmetry we may assume that the vertices subdividing 89,91,78 arc xj, X 4 ,

X5 respectively. Since the vertex subdividing 67 must be opposite X 5 and is not Xj, it

must be x7. The remaining indices arc 3 , 6 , and 7; since the vertex subdividing 34 must be opposite the vertices subdividing 23 and 45, it must be given index 3 and the other

two edges 6 and 7. One may verify that in both cases, the resulting graph is isomorphic

to Beth; in the first case an isomorphism is given by 1,2 ,..., 26 <-» b, 8 ,2 ,5 , V3 , V2 ,

v4, X 3 , xj, x5, xg, 1, 6 , x7, Vj, x2, v5, X 4 , 3 , 4 , 9, c, 7, a, vg, v 7 and the second given by 8 , x lt b, x5, x3, v 3 , 3,4, vlt 9,5, 2 , v5 , 7, v2, v4, c, x7, xg, a, v7, vg, x4,

1,6, x7. Next suppose that al is subdivided, but no other triad has a subdivided edge.

Since a, 4 ,5 , 6 ,7 has no attachment off N, two o f45,56,67 arc subdivided. Suppose

45 is not subdivided. Then 78 is subdivided, or there will be a hexagon using a, b without attachment Also 56 and 67 are subdivided and considering the pentagon 2,3,

4,5, b we have that 23 and 34 are subdivided. This accounts for six attachments, but the pentagon 8 ,9 ,1 , 2, b has no attachment identified, a contradiction. Next consider when 56 is not subdivided. Then 34,45,67, and 78 must be subdivided so that 3,4,5,

6 , c and 5 , 6 ,7 , 8 , b will have two attachments. Since the hexagons between b and c require attachments, the edges 23 and 89 must be subdivided. This accounts for all the attachments, but the pentagon 8 ,9 , 1, 2 , b has only one attachment identified so far, a contradiction. So finally suppose all of 45,56, and 67 are subdivided. Suppose 23 is not subdivided. Since 2,3,4,5, b needs two attachments, 34 must be subdivided, and since 9,1,2,3, c needs two attachments, 91 and 12 must be subdivided. But then 2,3, c, 9 , 8 , b will be a hexagon without attachment, which is impossible. So 23, and by symmetiy, 89 must be subdivided. Since 9,1,2,3, c needs two attachments, one of 12 and 91 must be subdivided, by symmetiy, say 12.

t

Figure 42. S2 with al, 12,23,45,56,67,89 subdivided. 87

Exploiting symmetiy, we may assume that the vertices subdividing 12, al, 23 are

xj, X4 , and X5 respectively. Since 89 is at distance two from xj and X 4 , we may conclude that 89 is subdivided by xg, But then we cannot assign vertices to 45,56,67 so that the vertex subdividing 56 will be opposite both vertices subdividing 45 and 67.

So finally we may consider the case where all outer triads have exactly one attachment Equality holds in (3.4) so that each pentagon will have exactly two attachments. There are three possible cases: Case l : The subdivided edges not in N are incident with consecutive vertices in N.

Without loss of generality we may assume that al, b2, and c3 are subdivided.

Considering the pentagons a, 4 ,5 , 6 ,7; b, 5 , 6 ,7 , 8 ; and c, 6 ,7 , 8 ,9 we have that two

each of {45,56,67), (56,67,78}, and {67,78,89} are subdivided. Suppose 67 is

not subdivided. Then 45,56,78,89 must all be subdivided, but this means a, 1,2,3,4

has only one attachment So 67 must be subdivided. Suppose 45 is subdivided. Then

56 is not subdivided, so 78 is subdivided and 89 is not subdivided. Since 8 , 9 , 1,2, b and 1, 2 ,3,4, a both need another attachment, 12 must be an attachment

1

Figure 43. S2 with al, b2, c 3 ,12,45,67,78 subdivided. 88

We may assume by symmetry that the vertices subdividing 12, al, and b 2 are xj,

X4 , and X5 respectively. Since the vertices subdividing 45,67, and 78 are at distance two from X4 , we must have c3 is subdivided by X 3 . Since the vertices subdividing 45 and 78 would be at distance two from X 5 , we must have 67 subdivided by xg. But then

X7 is at distance at most two from xg, contradicting Lemma 3.2 (ii).

Case 2: Exactly one pair of the subdivided edges not in N arc incident with consecutive vertices in N.

Without loss of generality, we may assume the edges al, b2, c 6 are subdivided.

Considering the three pentagons a, 4 ,5 , 6 ,7; b, 5 , 6 ,7 , 8 and 9 ,1,2,3, c we have that two each of {45,56,67}, {56,67,78}, and {91,12,23} are subdivided. Since the first two and last one are disjoint sets, and the first two intersect only in 56 and 67, we have that 56 and 67 must be subdivided. But then all three edges incident with 6 would be subdivided, and it is impossible for three subdivision vertices to be mutually opposite, so no graphs arise out of this configuration.

Case 3: None of the subdivided edges not in N are incident with consecutive vertices in N.

Without loss of generality, we may assume that al, c3, and b5 are subdivided.

Considering the three pentagons 4 ,5 , 6 ,7, a; 8 ,9 ,1 ,2 , b; and 6 , 7 , 8 ,9, c we have that two each of {45,56,67}, {67,78,89}, {89,91,12} are subdivided. Since these sets intersect in two elements, 67 and 89 are subdivided, and 78 is not subdivided. The pentagons 1,2,3,4, a and 2 ,3,4,5, b each need one more attachment, so 12 and 45 are subdivided. Figure 44. S2 with al, c3, b 5 ,12,45,67,89 subdivided.

By symmetry, we may assume that 12 and al are subdivided by X 4 and xj respectively. The pair b5 and 45 must be su.bdivided by opposite vertices, say X 2 and X5 or X2 and xg. Since 45 is coincident with a l, 45 is not subdivided by X 2 . So b5 is subdivided by X 2 * But then every remaining subdivided edge is coincident with an edge coincident with an edge subdivided by X 2 or X4 ; thus no assignment of X 3 is possible by lemma 3.2 (ii). So no new graphs arise.

This shows that there are, up to isomorphism, precisely three cubic girth seven graphs on 26 vertices. Since all of these graphs are Hamiltonian, they have Tait colorings, and so nowhere zero four flows. Thus the minimum counterexample has more than 26 vertices. ■ 90

Chapter III S4. The v=28 case.

Recall from § 1 that there are ten constructions to consider: the decomposable constructs D 3 l, T ^, and Tg-Tg; the indecomposable graphs 8 3 , C7 , the girth five graphs containing an edge in no pentagon Gj, G 2 . 0 3 , and 0 4 , and the Heawood graph. These graphs will be considered individually. It will turn out that T 4 l will not exist, in the sense that the dccompostion can not occur without creating new pentagons and collapsing to one of the indecomposable graphs. With the exception of the

Heawood graph, the same approach as in S3 could be used: ( 1) symmetiy exploiting casework on the choice of subdivided edges so that each pentagon has at least two subdivided edges and each hexagon has at least one subdivided edge; and then ( 2 ) giving arguments to establish which collection of subdivided edges will extend to attachments of the deleted 7-circuit P. The structure of the argument would be identical, and extremely tedious. Moreover, some of the graphs G j,..., G 4 have relatively many subsets that satisfy (1). To avoid the tedium, the arguments will not be given; instead "good" computer algorithms will be provided so that the reader may verify the results listed in appendix B including all cubic girth 7 graphs on 28 vertices with topological subgraphs of the first nine types. For the Heawood graph, a different sort of problem arises: since there are no pentagons, much of the force of arguments of type (1) disappear. Computer algorithms would still work, but would reveal the existence of thousands (many isomorphic, but still a large number in human terms) of graphs whose deletion leaves the Heawood graph. However, a helpful theorem comes to the rescue. 91

Theorem 4.1: Let G be a cubic, girth 7 graph. Suppose there exists a 7-circuit P of G so that after replacing pairs of series edges by a single edge in G\V(P) the resulting graph is isomorphic to the Heawood graph. Then O is Hamiltonian. pf: Suppose that there are not two consecutive attachments that are on edges of the Heawood graph at distance exactly 3, i. e. there is a path of length 2 between them. In the Heawood graph for every pair of edges there is a path of length 4 containing them, so every pair of consecutive vertices in P occurs on edges of the Heawood graph at distance 1 or 2. Let v j ,..., V 7 be the seven vertices of P, in that order, and let x j ,...,

X7 be their respective attachments. From before, we know that each edge of the

Heawood graph is subdivided by at most one of the xj. Let e; be the edge subdivided by

Xj (lfiiS7). Now Cj and ej+j can not be coincident with a single vertex v by lemma 3.2

(i). So every e; and ej^j arc at distance 2, and there is exactly one other edge, say f, with endvertices say x arid y, where ej and f are coincident at x and ej+i and f are coincident at y. Now xj, vj, Vj+j, xj+j, y, x may not be a hexagon of 0, so f must be subdivided, so f must be some ej. As xj is opposite to both xj and xj+j by lemma 3.2

(i), it must be that j=i+4 (mod 7). Thus ej, C 5 , ej, eg, 6 3 , 6 7 , C4 are the edges of a circuit of the Heawood graph. But the Heawood graph is bipartite, and contains no circuit of length 7.

Without loss of generality we may assume that cj and C2 are two edges of the Heawood graph at distance exactly three. Since the Heawood graph is edge transitive, and transitive on the edge neighborhood at distance 3 from a fixed edge, we have the following "butterfly" layout below: 92

Figure 45. Heawood graph, with Hamiltonian path labeling.

The path xj, vj, vy v2, x 2 together with the path indicated in the Heawood graph

by the labeling (including subdivided edge vertices, where appropriate) provides a Hamiltonian circuit in G. I

Consider first the two indecomposable girth 5 graphs. 93

Figure 46. The indecomposable cubic giith 5 graphs with 14 vertices.

The graph on the left is C 7 , similar in structure to the Petersen graph, and the graph on the right is S 3 . Consider first the possible attachment of a 7-circuit P to C 7 to create a girth 7 graph.

There are seven pentagons in C 7 . They arise out of a pair of edges of the (inner) star coincident with a single central vertex, together with the edges from their other ends out to the (outer) 7-circuit Q, and the edge of Q between them. Thus every edge in Q is in but one pentagon, and all other edges are in two pentagons. Hexagons consist of paths of length three of Q, together with the path of length three through the star connecting their endpoints. Let P be the set of pentagons of C 7 . As before, defining incidence variables for the edges we have

(4.2) 14 = 2 1 P | I Zxc « 7+ £ x e 5 1 4 , PtP eiE(P) e*E(Q) so that equality holds and none of the edges of Q may be subdivided. Note also that no pentagon may have more than two subdivided edges, since equality does hold. Thus an algorithm may be formulated as follows: 94

Algorithm 4.3

1. Number the edges not in Q arbitrarily; for convenience, number the star edges before the edges connecting the star to Q.

2. Create the following data structure:

A stack, to consist of a maximum of seven entries of edges.

An array, consisting of boolean variables indexed by the edges.

3. Choose the smallest star edge: make it the first entry in the stack, and mark the edge in the array as examined. 4. Perform the following loop:

If the stack is full, verify that this selection satisfies the hexagon attachment condition (i. e. that for any edge of the star plus the two coincident edges extending out

to Q there is at least one attachment) and if so, write the contents. Then pop the top entry off the stack and repeat step 4.

If the stack is not full, check for the first unmarked edge past the top of the stack. If there is no such, proceed to step 5. If there is one, mark it as examined and push it on top of the stack. Then verify that this is an acceptable initial configuration: check the seven paths of length four to ensure that none has more than two edges in the stack. If one does have two edges in the stack, pop the top entry back off. Repeat step 4. 5. Backup:

Unmark all edges after the top of the stack, then pop the top entry back off. If the stack is now empty, we are done. Otherwise, repeat step 4.

This algorithm has the effect of examining lexicographically all subsets of the edges of Cy\E(Q) that have the property that no pentagon of Cy contains more than two of the edges. If we get a subset of size 7 with this property, then it follows that all 95

pentagons have exactly two subdivided edges, and the pentagon girth constraint is

satisfied. The check for hexagon constraints before printing verifies that both constraints will be satisfied. Thus any output from this stage will provide a collection of edges that,

if the 7-circuit can be assigned, will provide a girth 7 graph. Next, simply check which

of the 6 !/2 circular arrangements will extend: for each subset, store the distance between

any pair of subdivided edges. Then for each of the permutations, check that lemma 3.2 is not violated; if not, this will yield a girth 7 graph.

This algorithm is not especially useful for large graphs; in particular examining the

subsets requires exponential time. But at this level it works in reasonable time. The algorithms were implemented in an interpreted Pascal on an apple Macintosh: the first pass required about eight minutes of computation, and for each resulting subset, checking the assignment of P took about five minutes (three of which were used by the operator, typing in the 21 pairs of distances). The result of the search is the two graphs Ganymede and Io, listed in appendix B.

The graph Ganymede is non-Hamiltonian, but does possess two disjoint circuits of length 14, so by two-coloring the circuits and giving the remaining matching a third color, the graph has a Tait coloring and so a nowhere zero 4-flow, hence can not be a

counterexample. The graph Io is Hamiltonian. Thus C-j gives rise to no counterexamples.

Next consider the graph S 3 . Number the vertices in the (outer) 12-circuit 1,2,

... ,9, a, b, c clockwise from the top. One may readily verify that the edges 12, lc, 67,

78 are in only one pentagon, the vertical chords 26 and 8 c are in three pentagons, and

that all other edges are in two pentagons. Let P be the set of pentagons of S 3 , There are eight pentagons, and we have as before 96

(4.4) 1 6 - 2 j P | S xc = 14 + X26+X8c-xl2-x67‘x78*xl c 5 16- P t P c« P

Thus both vertical chords 26 and 8 c must be subdivided, and none of the edges in only one pentagon may be subdivided. For the remaining 15 edges, the previously discussed algorithm will work, with the change that we start the stack with the two vertical chords, and if the stack ever has only one entry, we may stop. This search will verify that there are no extensions of S 3 to a cubic girth 7 graph.

Next consider the decomposable graphs T ^, and T 5 T 5 . Easiest to deal with is numbering T 4 in similar fashion to T 3 (see figure 15) we again have a path of length 4 across the top of the graph, and so as in the case of T 3 l, the identification of vertices will give rise to a quadrilateral or pentagon, and so need not be considered.

Consider D j1: referring to Plate 1, note that in D 3 there is a path between the lower of the two vertices to be identified across the top of the diagram using four of the internal trivalent vertices: thus the identification, if it is to avoid creation of new pentagons (and so reduce to one of the indecomposable graphs above) must not identify that pair. This leads to the graph drawn below: Fig 47. The graph D j1.

Edges have been thickened to represent the number of pentagons containing that edge. The two extra thick edges are in three pentagons each. The thinnest edges are in only one pentagon. Let the two edges in three pentagons be numbered 1 and 2, and the edges in two pentagons numbered 3 through 12 in any order. If we let P be the set of pentagons of Dj*, then as in (4.2),

21 (4.5) 14b»2|P|SZ £ xc = 14 + xi+X 2 -Z x j . PeP c«P i=I3

Algorithm 4.3 can be slightly modified to deal with this (and the further graphs).

Unfortunately, equality need not exist, so a decision criterion as simple as in 4.3 will not suffice. Let { i\ it } (l£i£t) be the indices from an initial subset of a set of seven edges that satisfies the pentagon and hexagon constraint Let yj be 1 or 0 depending on whether i is in this set Then

(4.6) 0. 98

(This follows because if edges 1 or 2 are to be subdivided, they are chosen first) This leads to the following, easily automated algorithm:

Algorithm 4.7

First Pass: Finding Feasible Subsets

1. Create the following data structure: A stack, to consist of a maximum of seven entries of edges.

An array, consisting of boolean variables indexed by the edges. 3. Make the first entry in the stack 1, and mark it as examined 4. Perform the following loop:

If the stack is full, determine if this selection meets each of the seven pentagons of

D3 l at least twice and each of the seven hexagons at least once: if so, write the selection to disk. Then pop the top entry off the stack and repeat step 4.

If the stack is not full, check for the first unmarked edge past the top of the stack.

If there is no such, proceed to step 5. If there is one, mark it as examined and push it on top of the stack. Then verify that this is an acceptable initial configuration: check equation 4.6. If the right hand side of the inequality is negative, pop the top entry back off. Repeat step 4.

5. Backup:

Unmark all edges after the top of the stack, then pop the top entry back off. If the stack is now empty, we are done. Otherwise, repeat step 4. Second Pass: Removing Duplicate Entries

1. Create an array large enough to hold all entries, and read from disk the results of the first pass into the array, with rows corresponding to different selections and columns running from 1 to 7 to hold the index of the chosen edges. 99

2 . Find the first row where the first column entry is non-zero. If all the entries in the first column are zero, we are done.

3. Copy this row to disk, and store a copy of this selection in memory. Zero the first column of this row.

4. Apply the single nontrivial automorphism of D 3 * to the copy of the stored selection,

sort it, and find the row of the array corresponding to this selection. Zero the first

column of this row. Repeat step 2. Third Pass: Extending to Girth 7 Graphs

1. For each edge e store in memoiy all edges coincident with e and all edges coincident with an edge coincident with e. 2. Read from disk an entry from the results of the second pass. If all entries have been processed, we are done. 3. Compute the distance matrix for the new vertices a j, ..., ay created by subdivision

of the selected edges: if the two edges corresponding to 14 and aj are coincident, their

distance is 1; if the two edges corresponding to aj and aj have an edge e that they are mutually coincident with, and that edge is itself not subdivided, then a; and aj are at

distance two. In all other cases aj and aj are at distance at least 3. 4. Applying the cyclic subgroup of the dihedral group to the deleted 7-circuit if

necessary, we may assume that Run through all 6 ! bijections of {aj ay} with { x j,..., X 7 } such that a p x j to see which satisfy lemma 3.2.

Any time mathematics is reduced to a computer algorithm, the possibility for error occurs. Naturally one tries to make the algorithms and programs as simple as possible, but the need to verify results is always present This algorithm has several nice redundant features to allow checking of accuracy. In the first pass, one can slowly step 100

through the algorithm for a few dozen cases, making sure that the right flow through the

program occurs and the proper decisions are made at each step. (This algorithm uses integer arithmetic and relies mostly on logic to obtain results.) Two possible sources of error are typographical (the program is fed the wrong data about pentagons and hexagons) and through oversight (some pentagons or hexagons are not checked). Gveiy attempt to eliminate typographical error has been made. If some pentagons or hexagons are not checked, the error is in the direction of safety, since it only means that additional cases are checked that do not need to be checked. In the second pass, the cases are being reduced by the automoiphism group of the graph. One possible source of error is that the program is given the permutation group with some typographical error. The result would be that a feasible entry from the first section would be sent under the false

automoiphism to an entry not in the list: as the program was run the flow of the program

was observed to see that each entry after the automoiphism had been applied

corresponded to an entry in the list This also serves to detect errors of symmetiy from the first pass. The third stage is perhaps most sensitive to erron typographical error due

to incorrect entry of the edge neighborhoods would cause the distance matrix

computation to be false and disallow possible girth 7 graphs. A separate program was

written to visually verify that the edge neighborhood data was correctly entered. The

portion of the algorithm dealing with enumeration of the permutation group was ported

from other applications, and is probably without flaw. Moreover, algorithm 4.3 was

applied to results from a previous computation to verify that the two corresponded. So while one can never be absolutely sure of computer algorithms, this algorithm is relatively simple and straightforward, and has ample redundancy to detect error.

The result of the investigation is that there are four ways to extend D j 1 by a

7-circuit to obtain a girth 7 graph. Complete results are listed in appendix B. 101

Next we consider a graph constructed by identification of divalent vertices of two copies ofTg.

0 b o' b'

Figure 48. Two labeled copies of Tg.

We must investigate how the primed and unprimed vertices can be identified to produce a decomposable girth 5 cubic graph on 12 vertices. Without loss of generality, we may assume that a and a' are identified, and b and b' are identified. If c is identified with c' or d', then there is either a quadrilateral or pentagon through the vertex b=b’; hence (c, d} must be identified with {c\ f }. If c is identified with e', again there is a pentagon through b; thus c is identified with e 1 and d is identified with f ; similarly e must be identified with d' and f with c'. This leads to the following graph: Figure 49. The graph Tg-Tg,

The pentagons of this graph are the pentagons of the original Tg's: there are thus exactly six: the edges of the central triad of each Tg are in two pentagons, as are the pivot edges. The remaining twelve edges are in only one pentagon. Let the edges of the triads be numbered 1,..., 6 and the pivots 7 , 8 ,9 with the remaining edges numbered arbitrarily. Then as in (4.2),

9 (4.8) 12 = 2 | P | £ l £ xc = 7 + £ xj . P « P c e P M

In the same fashion as before, we can define yj for the initial subsets of sets of seven edges that satisfy the pentagon and hexagon constraints. Then analogously to

(4.6) for a subset of size t,

(4.9) yi+ y 2 + ... + yg ^ min(5, t). 103

We need then only slightly modify algorithm (4.10): the decision criterion of the

first pass, step 4 when the stack is full is replaced by checking the pentagon and hexagon constraints in this graph: the pentagons are already explicitly given, and the nine

hexagons arise from a path of length 4 through one of the Tg’s to opposite vertices in the intersection, the path either through the central triad or around the graph, together with

the pair of edges of the other triad incident with those opposite vertices. When the stack

is not full, the decision criterion of (4.6) is replaced by that of (4.9). In the second pass,

the only change is to replace the automoiphism group of D j 1 by the automoiphism group ofTgTg: a group of order twelve instead of two. In the third pass, the edge

neighborhoods from Tg*Tg should be used instead of that of D 3 *. The end result of this search is that there are two ways to extend Tg'Tg to a girth 7 graph.

Next come the four graphs that contain some edges in pentagons and some edges in no pentagons. Let us consider the first graph, G j.

2 20

Figure 50. G i with edge labeling.

Edges 1 and 2 are in three pentagons each, edges 3 through 12 are in two pentagons each, edges 13 through 20 are in one pentagon, and edge 21 is in no pentagons. There are a total of 6 pentagons in the graph. Then as in (4.2), 104

20 (4.10) 12 = 2 |P| £ £ 2 Xg = 14 + Xj + 2x - Z Xj - 2 *x 2 i. P i P e i P i=13

This leads to the analogous inequality as (4.6),

(4.11) 2 + y 1 + y2 - (yi3 + y 14 +... + y20) - 2*y2i * 0.

The modifications to algorithm 4.7 are: in the first pass stack full check to replace the pentagon and hexagon structure with that of this graph, and when the stack is not full to replace with the constraint given by (4.11). In the second pass, one needs only replace the automorphism group with that of G j (which is isomoiphic to z 2 xz2) and in the third pass give the edge neighborhood structure for that of Gj. The end result of this search is that there are 11 extensions of Gj to girth 5 graphs. Next, consider the graph G2.

3

2 4

Figure 51. G2 with edge labeling. 105

The graph O j contains only four pentagons: 1,2,7,13,5; 1,2,8,14,6; and the two symmetrically opposite pairs. It does have sixteen hexagons. Edges 1 through 4 are in two pentagons each, and edges 5 through 16 are in one each. This leads to the

inequality as in (4.6),

(4.12) yi+y2+y3+y4'(yi7+yi8+yi9+y2o+y2i)sl-

The automorphism group is elementary abelian of order 8 (each automorphism fixes the edge 21; but there exist involutions that exchange 1,3 and 2,4; exchange 1,2 and 3,4 and an involution that fixes 1,2,3,4 but exchanges 5 , 6 ). Adjusting algorithm 4.7 to reflect the different hexagon and pentagon structure, inequality (4.12) and the different permutation group and edge neighborhoods will yield an algorithm to examine girth seven cubic graphs arising from Gj. There are 39 such extensions.

Let us next consider the graph G 3 .

3

2 4

Figure 52. G3 with edge labeling. 106

The graph G 3 contains six pentagons: edges 1 through 10 are in two pentagons

each, edges 11 through 20 are in one pentagon, and edge 21 is in no pentagon. This yields the decision criterion for a subset of size t,

(4.13) yi + ... + yi0-y2i^min(5,t).

There are eight hexagons in G 3 , and the automoiphism group is of order 4. (There is an involution exchanging 1,2 but fixing 3,4; and an involution exchanging 3,4 but fixing

1,2.) Changing algorithm 4.7 to reflect the different hexagon and pentagon structure,

inequality (4.13) and the different permutation group and edge neighborhoods will yield

an algorithm to examine girth seven cubic graphs arising from G 3 . There are 15 such extensions.

Finally, we consider the graph G 4 . Recall that G 4 arises from the configuration where both crossing edges create pentagons, and so G 4 has more pentagons than any of the graphs G j,..., G 4 . For this reason, G 4 will not occur as a topological subgraph of a girth 7 cubic graph on 28 vertices. G K

C E

A 6

D F

J N

Figure 53. G4 with vertex labeling. 107

The graph O 4 contains eight pentagons: two are A, C, G, H, D and C, G, K, L, I; six others follow from left-right and up-down symmetry. Every edge is in two of these pentagons excepting AB which is in no pentagon. Consequently, we have the inequality,

(4.14) 16 = 2 |P |S I I xe = 2 £*e-2*AB* M. PeP e

Theorem 4.15: Every isthmusless graph that embeds into the sphere with two handles or four cross caps supports a nowhere zero 5-flow. I 108

Chapter in $5. A corollary.

During the long history of the 4-color theorem, there were two main attacks on the

problem: one school would develop configurations that, if they existed in a planar graph, would be 4-colorable under any conditions on their periphery and hence could not exist

in a minimal counterexample. The other school would work up from the bottom, showing that minimal counterexamples would require a large number of vertices, and so the hope was that at some point the two schools would meet: the minimal counterexample would be so large that a reducible configuration would occur and so no minimal counterexample would exist

There is some parallelism between the 4 -color theorem and the two flow conjectures of Tutte. Of course, the two problems are dual problems in the sense of matroid duality, and Tuttc's 4-flow conjecture is an extension of the 4-color theorem.

Theorem II. 1.1 may be interpreted to state that in a sufficiently general class of graphs, a hexagon is a reducible configuration. The equivalent theorem in the 4-color problem was that circuits of size at most 4 are reducible, and this launched a study of planar graphs of girth 5. The defect in the current Appel and Haken proof of the 4-color theorem is that the two schools finally did meet, but outside the region of computation easily accessible to humans unaided by computing devices. It may be that the same will occur with the

5-flow conjecture. 109

In any case, I offer the following corollary to the computations of this chapter a lower bound in the spirit of the second school.

Corollary 5.1. If Tutte’s 5-flow conjecture is not true, the minimal counterexample is cubic of girth at least 7 and has at least 30 vertices. pf: Let G be the class of all finite graphs; certainly G is closed under subgraphs. As before, for O c G we have that H q contains all hexagons of the the graph. Applying

theorem 1.1.1 to G, we And that a minimum counterexample is 2-connected (in fact, internally 4-connected: see V.1.21) and G* has girth at least 7.

Suppose G* is not cubic, and let v be a vertex incident with edges cj, C 2 , 6 3 ,... en. Define a new graph Gj replacing the vertex v with a 4-circuit H with vertex set vj, v2» v3’ v4 *n ^ at order, and setting edge Cj incident with vj (l£i£3) and edges C 4 through ep incident with V 4 . We will show that there is a graph isomorphic to a subdivision of a splitting of the vertex v that does not support a nowhere zero 5 -flow, contradicting the minimality of G*. Any edge of H is in a circuit of G j, so none of its edges are isthmi, and G*=Gj/E(H), so by 1.1.6, Gj is without isthmus. By n.2.9, the isthmus graph of Gj\E(H) has at most 4 monovalent vertices, and by U.3.2 or n.3.3 and the tree pruning lemma n.2.8 we may add one of the pairs of edges {{vj, vj}, { V 3 ,

V 4 } } or {{V j, V4}, {vj, V 3 } } , say the former, to Gj\E(H) to produce a graph G 2 without isthmus. Orient G 2 so that ej is directed into vj, 62 is directed out of V2 , 63 is directed into V 3 , (vj, V 2 } is directed from v j to V 2 and {V3 , V 4 } is directed from V 3 to

V 4 , Suppose G 2 supports a nowhere zero 5-flow

supporting a flow is a topological property, we can replace the two divalent vertices vj

and V2 and edges cj, cj, {vj, V2 > with a single edge whose endvertices are the

endvertices of ej and e 2 not v, and replace the vertex V 3 and edges 63, {V3, V4} with a

single edge whose endvertices are the endvertex of e 3 not v and V 4 . The resulting graph

O3 has the same number of vertices but one fewer edge than G*, but is 2-connected and

does not support a nowhere zero 5-flow, contradicting the minimality of G*. Thus G*

is cubic.

We know from the computations of this chapter that no cubic girth 7 graph on 28 or fewer vertices does not support a nowhere zero 5-flow. Thus since cubic graphs must have an even number of vertices, the minimal counterexample must have at least 30 vertices. I CHAPTER IV

A GENERALIZATION OF CHORDAL GRAPHS

§1. History of the problem.

During the summer of 1981, Dr. Neil Robertson offered a course dealing with

selected topics in graph theory. One of the points covered was an analysis of several

proofs of Dirac's theorem: Define the k-whecl as W j^G 1UG2 where G j is an

n-star, G 2 is an n-circuit, and G j r ^ is the edgeless graph with vertex set V(G 2 >. Then Dirac's theorem states:

Theorem 1.1. Let G be a 3-connected simple graph. Then one of the following is true:

i. G has two vertex disjoint circuits,

ii. G-Wfc, for some k£3,

iii. there exists x, y, z t V(G) with G\{x, y, z} without an edge,

iv. G is isomorphic to the 5-cIiquc.

He was able to show that if cases i, iii, or iv did not occur, then there existed a peripheral n-circuit with r&4 and then completed the proof. He then asked which 3-connected graphs have only triangles as peripheral circuits? He remarked that:

111 112

i. All planar triangulations, the class of graphs that arc 3-connected and which can be

embedded into the plane such that every face is a triangle, are in this class since if G is a planar graph then peripheral circuits must bound a face;

ii. Any complete graph is in this class, since the only induced circuit is a triangle;

iii. Clique sums (where the cliques arc of size at least three) preserve this property, since any induced circuit must be on one side or the other of the separation.

This is a generalization of another theorem of Dirac [1, theoreml], classifying

chordal graphs:

Theorem 1.2. G is chordal if and only if G can be obtained by repeated clique-sums, starting from complete graphs.

During investigation of the problem, the properties of being 3-connected, having only triangles as peripheral circuits, and not separating at a clique were frequently called upon; for convenience a set S was defined as that of all graphs having these properties. The strategy of proof was to examine minimal members of S. At one point, the minimal member was found to be (partially) triangulated, and during the presentation referred to as S-triangulated. Dr. Robertson immediately began calling all simple graphs having the properties of 3-connection, and having only triangles as peripheral circuits as s-triangulated, hence strangulated. This terminology has now become the standard nomenclature.

Since chordal graphs have no induced circuits larger than triangles, one may phrase Robertson's conjecture in a manner similar to Dirac's, as follows: 113

Theorem 1.3 A 3-connected graph O is strangulated if and only if G can be obtained by repeated clique-sums, starting from complete graphs and 4-connected planar triangulations.

Lest one believe that the property of being peripheral in a graph is difficult to

satisfy, consider the following result of Tutte [ 12], which forms an important part of the background for theorem 1.3, although it is not required for the proof:

Theorem 1.4 If O is 3-connected, then every edge of G is in at least two peripheral circuits.

In light of theorem 1.4, since locally at every edge the graph tends either to be complete (if the edge is in more than two peripheral triangles) or a planar triangulation (if the edge

is in exactly two peripheral triangles) the characterization of theorem 1.3 is not

surprising. The "if 1 part of theorem 1.3 follows from proposition 1.5 below and

Robertson's observations. The "only if' part requires examination of contraction minors of strangulated graphs and the occurrence of small cutsets to push the connectivity high enough to force a chordal graph.

Proposition 1.5 If G is 3-connected, and G is the clique-sum of Gj and G 2 , and G j,

G2 are both strangulated, then so is G.

pf: Let Vj be the vertex set of Gj, so that Gj = G[Vj] (1=1,2) and G K V jr^ )] is complete. Let C be any peripheral circuit of G. We must show that C has length 3. Suppose that C meets both VjW^ and Then two non-consecutive vertices of C 114 are in V jnV 2 , since G j, G 2 is a separation of G, and they are adjacent, since

G[(VjnV 2 )] is complete. Thus C has a chord. But C is peripheral, and so G consists of C together with one chord, which is impossible since G is 3-connected. Thus we may assume, without loss of generality, that V(C) c V j, and hence C is a circuit of Gj.

If C is peripheral in G j, then it has length 3 since G] is strangulated, as required. We suppose therefore (for a contradiction) that C is not peripheral inG j. Hence C has at least two bridges in G j; B, B' say. Now G[(V jr ^ ) ] is complete, and so we may assume without loss of generality that no edge of B is incident with a vertex of

(V inV 2 >\V(C). It follows that B is a bridge of C in G, and B* is a subset of another bridge, and so C is not peripheral in G, a contradiction. I

At first sight it may seem more natural to omit the "3-connected" condition in the definition of "strangulated", and to ask for an appropriate generalization of theorem 1.3. This has not been done because the form of the characterization would need to be changed in an essential way. The graph of Figure 54 for example has no peripheral circuit of length > 3 and cannot be expressed as a clique-sum of two smaller graphs with the same property, but it is neither chordal nor a planar triangulation.

Figure 54. A graph that is not strangulated. 115

Another type of example may be produced as follows. Let H be any graph, and for each edge e of H, add a new vertex vc to H with valency 2, adjacent to the ends of e. Every peripheral circuit in the new graph has length 3, as is easily seen. However, a generalization is possible. Reinhard Diestal [2] considers the class of graphs with the property that the vertex set of every induced circuit of size greater than 3 is a minimal separator of the graph. He shows that this definition extends the concept of strangulation, in the sense that every strangulated graph has this property. Note that neither Figure 54 nor the examples in the paragraph following the figure have this property, since while the induced circuits do disconnect the graph, they are not minimal. His conclusion involves analysis of the simpticial decomposition of a graph and gives a result that graphs with his property are constructed by taking clique sums of chordal graphs with 4-connected planar triangulations with the restriction that any clique sum involving a planar triangulation must be at a triangle. 116

Chapter IV 52. Contracted Strangulations

This section consists of a proof of the following inductive step:

Lemma 2.1 Let e be an edge of a strangulated graph O such that G/e is 3-connected. Then G/e is strangulated.

pf: Let C be a circuit of G/e of length £ 4, and suppose for a contradiction that C is peripheral. Now G/e is 3-connected, and so if V(C) = V(G/e) then C has at least two

chords, which is impossible. Thus V(C) * V(G/e), and so C has no chords.

Let H « (G/e)\V(C). Then H is nonnull, and is connected since C has only one

bridge. Every vertex of C is adjacent to some vertex of H, since G/e is 3-connected.

Either E(C) or E(C)u{e} is the edge-set of a circuit of G. We denote this circuit by D. Now D has length £ 4 and so it is not peripheral in G. There are three cases: Case 1: Neither end of e is in D.

Evidently G\V(D) is connected, because (G\V(D))/e = (G\V(C))/e = H and H is connected. Moreover D has no chords in G since C has no chords in G/e. Hence D is peripheral in G, a contradiction. Case 2: exactly one end of e is in D.

Let the ends of e be u, v with v < V(D). Let the neighbors of v in D be x, y. Now

G\(V(D)u{u}) = H and so G\(V(D)u{u}) is connected. Moreover, D has no chords. Thus u is not adjacent to any vertex of G not in D, since D is not peripheral. It is not adjacent to any vertices of D except v and possibly x, y, since C has no chords. But G is 3-connected, and so u is adjacent to x, y. Let D' be the circuit induced by (V(D)\{v»u{u}. By the same argument applied to D', we find that v is adjacent only to x, u, y. But then G[{u, v, x, y}]uG\{u,v) is a 2-separation of G contradicting the fact that G is 3-connected. Case 3: Both ends of e are in D.

It follows that e t E(D). Let the ends of e be u, v and let x, y be vertices of D such that x * v is adjacent to u and y *u is adjacent to v. In G/e every vertex of C is adjacent to a vertex of H, and so at least one of u and v is adjacent to a vertex of H. Now G\V(D) b H and so G\V(D) is connected; but D is not peripheral and so has at least one chord. However C has no chords, and so every chord of D joins x to v or u to y.

Thus we may assume that x is adjacent to v. We may also assume that u is adjacent to a vertex of H; for if u is not adjacent to y this follows from 3-connectivity, and if u is adjacent to y it follows by the symmetry between u and v. Let D' be the circuit of G induced by V(D)\{u}. Then D 1 has length £ 4; but G\V(D') is connected, and D* has no chords and so is peripheral, a contradiction, a 118

Chapter IV 83. Small Cutsets.

We shall need the following:

Lemma 3.1. (Bridge building) Let O be a graph, let X s V(G), and let e be an edge with neither endvertex in X. Suppose that there is a circuit of G using e disjoint from X. Then at least one of the following is true:

(i) there is a circuit C of G using e disjoint from X with no chords, such that every component of G\V(C) contains a vertex of X.

(ii) there is a separation G], G 2 of G with |V(GinG 2 )|£2 , such that

X c V(G2 >\V(G j) and e has both ends in V(G 2 >.

pf: For any circuit C using e and avoiding X, let W(C) denote the union of the vcrtcx-scts of those components of G\V(C) which contain a vertex of X. Thus

X c W(C)« V(G)\V(C) for any such circuit C. Choose C with W(C) maximal. We shall show that either C satisfies (i), or (ii) is true. Suppose then that (ii) is false. Let y j, y2 be the ends of e, and let Y = {y j, y 2 >. Let Z - {v c V(C)\Y: v is adjacent to some vertex in W(C)}. For any v t V(C), let Pj(v), P 2 (v) be the paths of C from v to yj, y2 respectively which do not use e. We first establish the following proposition:

Proposition 3.2: For any z e Z, W(C)u{z) separates V(Pj(z)) and V(P 2 (z)) in GVe.

pf: Suppose not, and let P be a path of G from V(Pj(z)) to V(P 2 (z)) disjoint from W(C)u{z> and not using e, and choose P so that it has only two vertices in common 119

with C. Let the ends of P be pj, P 2 labeled so that either = y j, or P 2 = y2 > or y j,

Pi* z» P2* y i distinct and occur on C in that order. Let C' be the circuit consisting

of P together with P ^p i), P 2 (P2 )» and e. Clearly W(C)u{z) c W(C), and this is impossible by the maximality of W(C). ■

We require a second proposition. Let the vertices of Z be zj (12i

thatyi.zj, ...^ . 1, y2 occur on C in that order. (It is possible that n = 1.) Put *0“yi»*n“y2-

Proposition 3.3: The vertices of C are z q , z j , ..., zn.

pf: For suppose that v e V(C)\(YuZ). Choose an integer r with 0 £ r £ n maximal so

that Zf is a vertex of Pi(v). Then r < n, because v f Y. By the falsity of (ii), {Zp z ^ j } does not separate {v} from W(C), and so there is a path P from v to W(C) avoiding {Zp

Zr+i }. Now P has no vertex in common with PjCzj.), because v is a vertex of P 2 (zr)

and by lemma 3.2, VfQufZj.) separates the vertex sets of P^Zj) and P 2 (Zj-). Similarly,

P has no vertex in common with PjCXr+l)* Let v' be the last vertex of P in V(C). Then

there is a path from v 1 to W(C) using no vertex of C except v', and so v ' t Z.

However, Z c V(P j (z,.))uV (?2(zr+ i)) and P avoids this set, a contradiction. I

Now the completion of the proof of lemma 3.1. Suppose for a contradiction that

C does not satisfy (i), and let B be a connected component of G\V(C) not containing any

edge incident with a vertex of X. Let U be the set of vertices of G incident with edges in B, so that UuX * 0 . If |UnV(C)| £ 1, we find that (ii) is true, a contradiction. Thus

IUnV(C)| £ 2. Choose i, j with 0 £ i j £ n such that zj, zj c U, with i minimum and j 120

maximum. By propositions 3.2 and 3.3, j =» i + 1, and so B is not a chord, and hence

U i V(C). Choose u i U\V(C). Then { z j, Zj} separates {u} from X, and so (ii) is true, a contradiction. I

We use lemma 3.1 to study the structure of small cutsets in strangulated graphs.

Lemma 3.4. If G*G jL<32 is a separation of a strangulated graph O with

|V(G^nG 2 )| - 3, then G is a clique sum of G | and G 2 .

pf: Let V(GjnG 2 ) - (vj, V2 , vgjsay. Suppose for a contradiction that v j, V 2 say are not adjacent F o ri« 1,2, let Hj be the graph obtained from Gj by adding a new edge ej

joining vj, V2. There is a path in Gj from Vj to V2 avoiding {V3} since G is

3-connected, and so there is a circuit of Hj using ej and avoiding {V3}. By lemma 3.1 applied to Hj with X -fvj} there is such a chordless circuit (Cj say) with GW(Cj)

connected (again, since G is 3-connected). Let C be the circuit of G consisting of C j\C|

together with Thcn C has length £ 4 but is peripheral in G, a contradiction. I

A similar result is true for cutsets of size 4, but the necessaiy tool for establishing

this result is the following lemma, which was stated in [7] and proved in [ 8 ]. If

X « V(G), A(X) denotes the set of vertices of V(G)\X which are adjacent to vertices in

X. 121

Theorem 3.5 Let 8jt tj, 82 , t2 be distinct vertices of a graph O. Then just one of the

following is true:

(i) there are vertex disjoint paths joining sj to tj, and S 2 to t 2 respectively, (ii) for some k > 0 there are pairwise disjoint sets

A j,..., Afc c V(G)\{slt tj, S2 » t2 > such that (a) for i * j, (AAfJnAj « 0 ,

(b) for 1 £ i £ k, |AA;| £ 3

(c) if O' is the graph obtained by (for each i) deleting Aj and adding new

edges joining every pair of distinct vertices in AAj, and also for j- 1 ,2 adding an edge ej joining sj to tj, then O' may be drawn in the plane with

no pairs of edges crossing except ej, e 2 which cross once.

We use theorem 3.5 to prove the following:

Lemma 3.6. Let G-G jU O j be a 4-separation of the 4-connected strangulated graph G.

Then G jnG 2 is isomorphic to one of a 4-clique or a 4-circuit; and the second case occurs only if G is planar.

pf: LetV (G inG 2 ) “ {si,ti,S2 ,t2 ). Let V(Gj) be Vj (i=l,2). We first prove the

following:

Proposition 3.7. If Pj, P 2 arc vertex disjoint paths of Gj, each with at least two

vertices, and all four of their terminal vertices are in V(GjoG 2 ), then forj=l, 2 the terminal vertices of Pj are adjacent 122

fif: Let the terminal vertices of Pj be, say, Sj and tj (j = 1,2), and suppose that Sj and

say, are not adjacent Put Xj = V(P 2 ) and X 2 = {S2 , t2li and for i = 1,2 add a new edge ej to Gj joining sj and tj, forming Hj. By lemma 3.1 there is a circuit Cj of Hi

using cj and avoiding X j, with only one bridge; and there is a circuit C 2 of H 2 using e 2

and avoiding X 2 , such that every bridge has an edge incident with S 2 or t 2 « Let C be the

circuit of G consisting of CjNej together with C 2Ve2 . Then C has length £ 4, but has only one bridge, a contradiction. I

Continuing the proof of lemma 3.6, either some pair of vertices in V f G jr ^ ) (say s j, tj) are not adjacent, or else the theorem is true. There is a path of G within Gj joining s j, tj and avoiding S 2 , t2 , since G is 4-connected, and hence by proposition 3.7,

S2 , t2 are not adjacent (for otherwise we would have two disjoint paths, and sj, t^ j would need to be adjacent).

| Suppose that some other pair (s j, $2 say) of sj, tj, $2 , t2 are not adjacent Then by proposition 3.7 there do not exist vertex disjoint paths of G j from sj to s j and fro n

t j to t2 respectively. Hence there do not exist two vertex disjoint paths from {sj, t 2 > to

{S2 , tj}, and so by Monger's theorem there is a single vertex vj say which meets all paths of Gj from (sj, t 2 > to {52, tj>. If v' c V(Gj)\ {vj, sj, tj, S2 , t2 > then there ant four paths from v* to s j, t j, S 2 , t2 respectively, vertex disjoint except for v'; but then vj does not meet all paths of Gj from {s j, t 2 > to {S2 , tj}, a contradiction. Hence

V(Gj) = {vj, sj, tj, S2 , t2 >; but V(Gj) \ V(G 2 > * 0 , and so Vi * sj, tj, S 2 , t2 and

|V(G|)| = 5. • Similarly |V(G2 )| = 5, and so |V(G)|« 6, and Sj has valency at most 3, contradicting G being 4-connected We have therefore proved that any two of sj, tj, S 2 . t2 are adjacent except sjt], sjtj. Hence G jn G 2 is a 4-circuit 123

In G, there do not exist vertex disjoint paths linking s^ to tj and $21° *2 respectively. Moreover, for X « V(G j )\{8j , tj, sj, ^ ^ then |AX| £4 , since G is 4-connected. Prom theorem 3.5 we deduce that Gj is planar and may be drawn in the plane so that the boundary of the infinite region is the circuit sj, $ 2 * tj, t 2 *

Similarly, G2 is planar, and may be drawn in the same way. Thus G is planar. ■ 124

Chanter IV 54. The Main Proof.

Now we prove theorem 1.4. Only two further outside facts are required

Theorem 4.1 If O is a chordal graph and v is a vertex of G, then one of the following is true: (i) v is adjacent to every other vertex of O.

(ii) G is expressible as a clique-sum G=G 1OG2 with v t V(Gj)\V(G 2 ).

This follows immediately from theorem 1.2.

Proposition 4.2 If G is strangulated and planar then G is a planar triangulation.

pf: Take a planar drawing of G. Since G is 3-connected, the boundary of every region is a peripheral circuit; and therefore has length 3, since G is strangulated Thus G is a planar triangulation.

Proof of Theorem 1.3: The "if1 part was proved in section 1. Now we prove the "only if 1 part, which is more difficult If possible, let G* be a strangulated graph which is not obtainable by means of clique sums starting from complete graphs and 4-connected planar triangulations.

Choose G* with this property with |V(G*)| minimum. 125

Proposition 4.3 G* is not expressible as a clique sum.

pf: Suppose that it is expressible as a clique sum, say G*=G JUG 2 . Now

|V(GinG2)ls 3, since G* is 3-connected, and so |V(G j)|, |V(G 2 )| £ 4. It is easy to see that Gj, G 2 are 3-connected We claim that they are both strangulated For let C be a peripheral circuit of G j say. If C has £ 3 vertices in G jn G 2 and has length £ 4, then C has a chord and at least one other bridge, which is impossible. We assume then without loss of generality that C has S 2 vertices in G j jG 2 * Now |V(GjnG 2 )| ^ 3, and so there is a vertex v say of GjnG 2 which is not in C. Let B be the (unique) connected component of G j\C. Every edge of G j incident with v is in B. Now for every vertex u 6 V(G*)\V(C), there is a path in G* from u to v avoiding V(C), since G j, G 2 are

3-connected. Hence every edge of G* not in C is in the same component of G*\V(C) and so C has length 3, since G* is strangulated.

Hence G j and G 2 are 3-connected and strangulated. By the minimality of

|V(G*)|, Gj and G 2 satisfy the theorem and hence can be obtained by repeated clique sums in the required way. But then so can G*t a contradiction. I

It follows from proposition 4.3 that

Eroposilign 4.4. G* is 5-connected. pf: Now G* is 4-connected because of proposition and lemma 3.4. It is not planar, by proposition 4.2; and so it is 5-connected, by proposition 4.3 and lemma 3.6. 126

Proposition 4 .5. For every edge e of G*, G*/e is 4-connected and strangulated.

pf: For G*/e is 4-connected because G* is 5-connected, and it is strangulated by Lemma 2.1.

Proposition 4.6. For every edge e of G*, either G*/e is a 4-connected planar triangulation or it is a chordal graph.

pf: For G*/c is obtainable by clique sums, starting from complete graphs and

4-connected planar triangulations, by proposition 4.5 and the minimality of |V(G*)|.

Each clique sum utilized here corresponds to an expression G*/e=GjuG 2 with

|V(G jnG 2 )| ^ 4, because G*/e is 4-connected. But no 4-connected planar triangulation has a 4-clique as a subgraph, and so none is involved in a clique sum when constructing G*/e. Hence either G*/e is a 4-connected planar triangulation, or G*/e may be obtained by clique sums starting from complete graphs alone, that is, it is chordal.

Proposition 4.7. For every edge e of G*, G*/e is chordal.

pf: Suppose that for some edge e, G*/e is not chordal. Then by proposition 4.6, G*/e is a 4-connected planar triangulation. Let the ends of e be V}, V 2 , and let Aj be the set of vertices of V(G*)\{vj, V 2 > adjacent to vj (i = 1,2). Let G be G*\{vj, V 2 }. Then G is

3-connected and planar, and A j*jA 2 is a subset of the vertices on the boundary C of some region R say. There are two cases:

Case 1: There are two vertices u j, U 2 of C, dividing C into two paths Pj, P 2 , such that Aj s V(Pj) (i = 1,2). 127

Then G* is planar, contrary to proposition 4.2.

Case 2: There are no such vertices. Now |Aj|, |Ajl S 4, since G* is 5-connected, and it

follows easily that there exist s j, S 2 , tj, t2 of C, in that order, such that sj, tj e A j, S 2 ,

t2 e Aj. Let X be V(C)\{s j, tj}. Since G* is 5-connected, there is a path from S} to tj in G avoiding X; and hence by theorem 3.1 (by adding a new edge to G joining S}, tj)

there is a path P from s j to tj in G avoiding X such that every component of G\V(P)

contains a vertex of X and no two non-consecudve vertices of P are adjacent But then, by combining P and the edges joining s j and tj to vj, we obtain a circuit of length et 4 that is peripheral, a contradiction.

Proposition 4.8. For every edge e of G* and every vertex v of G*, either c is incident with v or v is adjacent to an end of e.

pf: Let u be the vertex of G*/e produced by the identification of the ends of e under contraction. Then by proposition 4.3, u t V f G jr ^ ) for every decomposition

G*/e=GjUG 2 such that G ^ r ^ >s complete. Hence by Theorem 4.1 and proposition 4.7, u is adjacent to every other vertex of G*/c, I

Proposition 4.9. G* is chordal.

pf: Let C be a circuit of G*. If C has length £ 5, then it has a chord by proposition 4.8.

If it has length 4, then G*\V(C) is connected, since G* is 5-connected; but C is not peripheral and again it has a chord. I 128

But proposition 4.9 contradicts our choice of G*; and hence the choice of G* is impossible. This completes the proof of theorem 1.4. CHAPTERV

A FLOW POLYNOMIAL IDENTITY FOR VERTEX 3-SEPARATED GRAPHS

$1. The Identity.

Let F(G;n) be the set of nowhere zero (Z 2 )n-flows on G without regard to an orientation of G; since every element of (Z 2 )n has characteristic 2 this definition is meaningful. Note that | F(G;n) | « F(G;2n), where F(G;X) is the flow polynomial of G. This section is devoted to a proof of the following theorem:

Theorem 5.1: Suppose G»G i u G 2, V(G1nG 2H x»y.2)» and E(GjnG2)=0. Let el* c2» c3 ^ disjoint from G with endvertices x,y; y,z; and x,z; respectively. Let T be a 3-star with monovalent vertices x,y, z. Then

(5.2) F(G;X) = F(G i ;X)F(G 2;X) + [F(G1u{ei};X)F(G2u{e1};X) +

F(GiU{e2};X)F(G2U{e2>;X) + F(Giu{c3};X)F(G2U{e3);X)]/(X-l) + F(G j UT:X)F(G2OT;X)/(X-1 )(X-2).

129 130

Of: Define the following sets:

(5.3) S q o { ( 9 i , : 9i * F(Gi;n), i = 1,2 }

Sj = {(91*92): 9 i£ F(GjU(ej};n), i = 1 ,2 ; 9[(Cj) = ^ (e j)} , j = 1,2,3

s4 ° ((9i»92): 9 ie F(GM ei,e 2 };n), i - 1,2 ;

Let S = S q S j w S2 w S3 ^ S4 . Define 1 : S -* F(G;n) by

(5.4) 1(9 1 ,9 2 ) “ 9 where 9 (e) = 9 1 (e) if c c E(Gj), and 9 2 (c) if e c E(G2 >.

Certainly 1 is well defined, since the edge sets of G] and G 2 are disjoint and their union

is all of E(G). It is necessary to observe that 1(9 1 *9 2 ) is a nowhere zero (Z 2 )n flow on

G. Define 9i* 4*2 fore £ E(G)u{ei* c2 > c3 ) &y ^i(e)“9i(c) if 9j(c) is defined and

on G u {ei, e 2 * 0 3 }. In each of Sj, M),...*4 we have ( $ 1 + t^Hcj) *» 0, j=l,2,3 as they are either specified to be equal or both are undefined (recall we are in characteristic

2 ). Thus 9 ■ $ 1 ♦ <1*2 Ig *s a fi°w on G* Since the edge sets of G 1 and G 2 are disjoint,

the values of <}> arc all values of 9 1 or 9 2 , hence nowhere zero. Thus 1(9 1 *9 2 ) *s a nowhere zero flow on G.

We assert 1 is injective. Suppose 1(9 1 *9 2 ) 8 1 (9 3 *9 4 ). Considering edges as a multiset of 1- or 2 -subsets of vertices, we have the three equations

(5.5) 9i(ei) + 4»i(e3)= X 9i(e). c e E(Gi), x € e

9i(ei) + 9 i (C2> 8 X 9i(c)» e eE(Gi),y «e 131

and

4 >l(e2 ) + $i(e3) “ Z

Moreover, by (5.3) cither i(e 3 ) * 0 and <> ^(ej)=1(e2 ) or else only one of $ j(e j),

l(e2 ), ^ 1(0 3 ) is nonzero. Thus $ 1 (and hence (> 2 ) is determined by its respective

values on the edges of O. Since 9 j, 9 3 and 9 2 . 9 4 agree on the edges of Gj, G j respectively, 9 j « 9 3 and 9 2 “ 9 4 *

We assert 1 is suijective. Let $ 1 F(G,n), and let 9 j, 9 2 be defined by setting

9 j = on the edges of Gj (i=l, 2 ) and on ej, C 2 » 6 3 by defining 9 ; as follows: let

(5.6) a = E <{>(e), P = E (c). ecE(Gj),xce etE(Gj),yte

These values represent the flow from Gj to G 2 through x and y. If a ** P *» 0, set

9 j(cj) = 0 (i=l,2; j=l,2,3). If a = 0, P * 0, set 9 |(e j)=9 ^(0 3 ) =0, 9i(c2)°P* If a * 0, p = 0, set 9 j(ej) = 9 j(C2 ) = 0 ,9 ^ 3 )» a. If a = p * 0, set 9 j(e j) «9 ^(0 3 ) ° 0,

9 j(e2 ) = a. If ot, P * 0, a*p, set 9 ^(6 3 )« 0 ,9j(Cj) = a, 9 ^ ) » a + p. One may readily verify that in each case 9 j, 9 2 are flows on G u{Cj, C 2 >0 3 } and that 9 j is nonzero on the edges of Gj. Consequently by abusing the notation to restrict 9 j to its support, which includes the edges of Gj (isl,2), we get that ( 9 1 ,9 2 ) « S. Thus 1 is injective and suijective, hence a bijection.

As 1 establishes a bijection between F(G,n) and S, we have

(5.7) F(0;2n) = |F(G.n)| - |S| = |S0| + JSj| +■ |S2| + |S3| + |S4|. 132

We consider each term of the right hand side of (5.7) in turn. Now using the usual technique of summing over the first component in the Cartesian product,

(5.8) |S0| - X 1(92« F(G2;n)}l - I F(G2;2n) - F(Gi;2n)F(G2;2n). 9 i«f(Gj;n) 9 i t F(G2 ;n)

Let Fg = { 9 e F(G2 vj cj;n): 9(cl) ** 8 ). Note F(G 2 u cj;n) = U Fg where the

union is taken over all non-zero elements 5 of (Z2)n. Let yt 8 be two nonzero elements of (z2)n, and x c ^ut((Z 2)n) & GL(n,2) be any of the nonsingular linear

transformations exchanging y and 8 . If 9 1 F(G2 uej;n), set 9 X: E(G2 )u{ei}-*(Z2)n

to be 9 x(e) = 1 9 (e). Since 9 is nowhere zero, and x is nonsingular, 9 X is nowhere zero. Since x is linear, for each veitex v,

(5.9) X

Thus x induces a permutation of F(G 2 u ei;n). If 9 e Fg, then 9 x(e) = x 8 = y so x exchanges Fg and Fy. Consequently |Fg| |f^| and so

(5.10) F(G2 u e i ; 2n)« (2 n -l)|Fg| for any 5 * 0 . This yields

(5.11) |S j|= X |{9 2 €5r(G2 VJCl I n) :

= £ 1(92 * f 6 : ® " *Kcl)}l 9 1 iF(G 1 u e 1 ;n)

= [ F(02 u C j; 2n)F(G 2 u e j ; 2 n) ] / ( 2 n - 1).

In similar fashion one replaces c j with e 2 or 6 3 throughout to get an equivalent formula for the cardinality of S 2 and S 3 . Let

(5.12) Fy5 ={

We note

(5.13) F(G2 u { e 1>c2} ;n) = U F 5 5 * U

5 c ((Z2)n), 5/0 S,y e ((z2)n), 5*y, Y»6*0

Let y. 8 and C, ^ be two pairs of elements of ((z2)n)^ with y * and 8 * Then if

r£2, GL(n,2) contains a (nonsingular) linear transformation t taking y to 8 and £ to §.

Then as above x induces a permutation of F(G 2 u {ej, e2} ; n) that takes Fyy to F 5 5

and takes F ^ to Fgj:, so x induces a permutation of F(G 2 u (cj, c2} ; n) and thus

|Fy£| = |Fgj:| whenever and 8 /Jj. Hence

(5.14) |F(G2 u (Cl, C2 ) ; n)| = (2n - DtFggl + (2n - l)(2n - 2)|F ^|

for any pair of nonzero, nonequal elements 8 , £ of (z2)n. Thus 134

(5.15) lF(G2 U{c1(e2 };n)| = |U F ^ | + (2n-l)( 2 ^ 2 ) ^ .

W * 2)n), C*>

Let

(5.16) k : UF^£ -> F(G2 Ue3 ;n)

where for

Thus K is injective. If <(> £ F(G 2 u e 3 ;n) define (c) for e « E(G2) and <(>(0 3 ) to be the common value of

(5.17) 1 1 : UFY5 -*F(G 2 uT;n) 5,y « « z2)n), 6^ y.8"0

where if 9 e F^g then Tl=

T) define <|>(fi) = Y, W 3 ) = 8 , (f2) = 7 + 6 . Since y*8 this map is nowhere zero. The value of ^ at the trivalent vertex of T is y + 6 + (y + 8 ) = 0. The value of

is a nowhere zero (Z2)n flow on G 2 uT. Again, T|

$ e F(G2 uT;n) then by defining 9 where 9 |G2 =<)>|G2, (f3 ) = 5 it

is easily seen that 9 c Fyg. [Note that the value of at the trivalent vertex of T is

(fl) +

9 at y is equal to the value of $ at y.] This yields q's suijectivity, hence bijecdvity and

(5.18) |U F ^| = F(G2a r ; 2 n). y,8 *(z2)n, 7^5

By exchanging Gj for G 2 we And

(5.19) |{ 9 ! c F(G1u [ e 1,e2 };n): 9i(ei)*Pi(c2»l» F ^ U T # 11),

which is precisely the number of choices for 9 j in S 4 . We compute

(5.20) IS4 J » 21(92 * *r(G 2 VJ(e lic2 }in)! 92(c l)=

» 2 (F^gl where y = 9 i(ei), 8 = 9 !(c2)

9 j £ F(G2 U{ej, e2 );n), 9i(ei)?t9i(c2)

= 2 F(G2 UT;2n)/(X-1) (X-2)

9 j £ F(G2 u ( e lf e2 };n), 9 1 ^ 1)^9 1 ^ 2 )

- F(G 1 uT;2n)F(G 2 uT;2n)/(X-1) (X-2).

This means that the left and right hand sides of equation (5.2) agree whenever the necessary t exist, i.e. for n£2. Thus equation (5.2) is valid for infinitely many X. But for a fixed graph G, equation (5.2) is a statement about the values of two polynomials. 136

If two polynomials agree at infinitely many values, they are equal; thus the left and right hand sides of (5.2) are equal and the proposition is true. ■

Corollary 5.21. Let O be a vertex minimal 3-connected graph that does not support a nowhere zero k-flow, and G=HuK be a 3-separation of G. Then one of H or K is a

3-star.

* pf: Since G «HuK is a 3-separation of G, we can write HnK as the edgeless graph with vertex set {x, y, z}. Let e j, e 2 , ej, T be as specified in theorem 5.1. Applying the theorem, we see that F(G; k) = 0 implies that every term on the right hand side of equation (5.2) with X>k is zero (since the sum has nonnegative terms). In particular, F(HUT; k) F(KuT; k) - 0, ( i» 1,2). Suppose e is an edge of H in no circuit of HUT.

Now e is in a circuit of G, say C. If C«H, then e would be in the circuit C of HUT, a contradiction. Thus C intersects both H\V(K) and K\V(H), hence HnK in at least two vertices. Let Cj be the shortest path containing e in C with endvertices say x and y in

HnK; then CjsHsHUT. Let C 2 be the path in T from x to y. Then e is an edge of the circuit C juC 2 » hence no edge of H is an isthmus of HUT. Since G is 3-connected, there is a path in H between any pair of vertices of HnK avoiding the third, so every edge of T is in a circuit of HUT. Thus HUT is without isthmus. Similarly, KUT is without isthmus. Suppose both of H and K have two vertices other than x, y, z. Then

HUT and KUT are isthmusless and at least one does not support a nowhere zero k-flow, contradicting the minimality of G. So one of H or K has at most one vertex, say H, and by 3-connectivity H is a 3-star, as desired. 1 Appendix A

PASCAL programs used in the graphical search

As discussed in Chapter m , the case v=28 required a mechanical search to determine the different isomorphism classes of all cubic girth 7 graphs on 28 vertices that have a cubic girth 5 subgraph of 14 vertices as a topological subgraph. (The case where a Heawood graph was a topological subgraph was handled directly.) In addition, some programs were used to reduce the many copies of the girth 7 graphs to find the isomorphically distinct copies, and investigate the structure of these graphs.

The overview of the computer algorithms is as follows:

From the analysis of Chapter m , there were seven girth 5 topological subgraphs to be considered as possible extensions to girth 7 graphs, namely C 7 , S3 , Tg*Tg, Dgl, Q j,

G2 , and G 3 . For each graph, two decision functions were created: one, that served as a pruning criterion for the branch-and-bound routine, denoted check_pentagons, and a second that served as final acceptance or rejection of each collection of edges developed by the search routine as meeting all pentagons at least twice and all hexagons at least once, denoted checkjfokay. Then these functions were inserted inside a generation program. Output from the program was a textfile consisting of all subsets of seven edges that meet all hexagons of the graph and meet all pentagons at least twice. (There is one exception to this. Because of the automoiphism group of C 7 , two cases of attachments were considered. Either the seven edges of the attachments consisted of all seven edges of the internal 7-circuit, or else at least one of the edges incident with the 137 138

outermost 7-circuit Q must be an attachment, say edge 1. Thus the program considered

only attachments using the edge 1, and the earlier case where the attachments were

8 ,..., 14 was considered separately.)

The output was then fed into the program removeJsomorphlc_copies,

together with a table of automorphisms of the graph. The list was reduced under the

automorphisms of the graph to eliminate unnecessary computation. Output from this

program was a textfile consisting of a representative sample of subsets of seven edges that meet every hexagon and every pentagon at least twice.

This output was fed into automated_attachments_check, together with information that gave the edge neighborhood to distance 2 from each edge, so that the program would be able to evaluate the distance between two attachments in order to verify the conditions of lemma in .3 .6 . (The program check_edge_data was used as a visual display of the edge neighborhood data, to ensure that no errors were introduced at this stage.) The program then looks at all 6 ! bijections between the seven edges of the attachments and x j , ..., xy with xj chosen to be the first attachment, to see which bijections satisfy lemma m .3 .6 . Output from this program is a textfile listing for each topological subgraph the ways to attach the 7-circuit to obtain a girth 7 graph. There is still a redundancy in this output: since the circuit v j, vj,...', v-j has automorphism group of order 14 but so far only the choice of x j has been fixed, there will be two copies of each possible attachment, corresponding to clockwise and counterclockwise orientations of attachments to x j ,..., xy. This allowed a final inspection of the results to ensure that no peculiarities have arisen. Duplicate entries were manually eliminated at this point. The collection of lists now consists of a complete

(although repetitive) set of all cubic girth 7 graphs arising from the seven basic topological subgraphs. 139

To determine which graphs were isomorphic and to create files that gave the vertex neighborhood information, the lists together with information about the edge-vertex neighborhoods of each vertex were fed into the program create_vertex_neighborhoods. This program runs sequentially through all the

vertices building the induced subgraph of vertices of distance at most three, counting the number of edges connecting two vertices at distance exactly three. This number is the

number of heptagons through the root vertex. This number is of course a graphical invariant associated with the vertex. The program dumps to the line printer a table of vertices and the number of heptagons through each vertex. Each vertex is in at least four and no more than eight heptagons. This led to an initial sorting of the graphs based on the 5-tuple ( 3 4 , 8 5 , a^, ay, 8 3 ) where aj is the number of vertices that are in j of the heptagons. From this information, each graph was redrawn and the vertices colored according to the number of heptagons containing it From this coloring, it was easy to determine by hand that in fact all graphs with the same 5-tuple were isomorphic, and so the invariant completely determined the isomorphism class of the graph. This leaves a total of 19 different isomorphism classes.

The textfiles of vertex neighborhood information from each of the 19 classes were then fed sequentially into the program hamiltonlanjcheck to determine if the graphs were in fact Hamiltonian. (As was observed in chapter m section 2, cubic Hamiltonian graphs support a nowhere zero 4-flow, and so satisfy the conditions of the 5-flow conjecture.) In all graphs except Io, a Hamiltonian circuit was detected. However, since Io has two vertex disjoint circuits of length 14, it too supports a nowhere zero 4-flow. Thus no counterexamples to the 5-flow conjecture have 28 vertices or less. 140

The programs follow. First, the programs that generate the original list of subsets of edges that are then manipulated to get the attachments of the girth 7 graph: program C7_generator; {This generates the attachments for the subgraph C7} const num_edges ■ 14; type edges -1 ..num.edges; {Only 14 edges can be attachments} edge_cholce - array[edges] of 0..1; {1 if an attachment} var d ep th : 1 ..7; {number of edges chosen} ch o ice: array[1..7]of edges;{the attachments, in order} selection: edge_choice; {Incidence army of chosen edges} check ed: array[1..7,edges] of boolean; {at each level of the tree,the checked branches} alLdone: boolean; outputjile: text;

procedure write_selectlons; {dump an answer to disk} var j : integer; begin for ] 1 to 7 do write(output_file, cholceQ]: 2 ,''); writeln(output_flle); end;

function check.pentagons: boolean; var bad: boolean; begin bad > true; if selection^ ] + selectton[2] + selection[8] + selection[9] < 3 then if selectlon[2] + selection{3] + selection^ 0] + selection}! 1] < 3 then If selection[3] + selection[4] + selection[12] + selection[13] < 3 then if selectlon[4] + selection[5] + selectlon[8] + selection[14] < 3 then if selectlon[5] + selectlon[6] + selectlon[9] + selection[10] < 3 then If selection{6] + selection[7] + selection^ 1] + selection[12] < 3 then If selection[1] + selectlon[7] + selectlon[13] + seledion[14] < 3 then bad:»false; check.pentagons:« bad; end; {of function check pentagons} procedure up;{end of branches to try, up to previous} begin If depth ■ 2 then alljdonetrue else begin selection[choice[depth]]:« 0; {unmark this last edge} d ep th :« depth -1 ; {back up} selection[choice[depth]] > 0; {unmark that edge} end; end; {up} procedure down; var j : edges; begin If depth < 7 then begin for | > 1 to numjedges do checked[depth +1, j]:« checked[depth, j];{no need to} depth := depth +1; {check eliminated edges} end {depth<7} else {depth-7, so print} begin write.selections; selectlon[choice[depth]]:* 0; end; {if} end; {down} procedure pick; var try: edges; got_one: boolean; begin try :■ 1; {this is the edge we are going to try to use} gotjone := false; repeat while checked[depth, try] and (try < numjedges) do begin try :»try + 1; If not checked[depth, try] then {unchecked edge} begin 142

selectlon[try] > 1; {so select it, and test it} checked[depth, try] :■ check_pentagons; selection[try] :■ 0;{then deselect, for choice later} end; end; {while} if checked[depth, try] then {try*14 and all checked} begin up; {so back up to previous level} try := 1; {and try again} end else {we found something to use} begin choice[depth] > try; gotjone:«true; end; until alljdone or got_one; If goLone then begin {go to next level of tree} selection[choice[depth]]:«1; checked[depth, choice[depth]] := true; down; end; end; {pick}

begin {main program} Showtext; {specific to macintosh; initializes display} d e p t h 1; rewrite(output_file, 'C7 attachments'); repeat pick until alljdone; end. This program is the generator for the possible subsets of seven edges of the graph

C7 that will meet all pentagons twice and hexagons once. Recall from the analysis of chapter 3, section 4, that the seven edges in the 7-circuit Q (whose edges are characterized by being in only one pentagon each) can not be subdivided, and that when all seven attachments have been chosen, each attachment will be in exactly two pentagons. Consequently, no pentagon may contain more than two attachments; the function check_pentagons looks at each of the six pentagons of C j and verifies that none 143

contain 3; it returns true if any pentagon contains more than two attachments and false otherwise. This program does not have a function checkJfokay; it was the first of the generator programs written, and it was decided that it would be more rapid to check the hexagons and to just verify the pentagon constraint at this time. However, note that since each edge is in exactly two pentagons and by selection no pentagon contains more than two attachments, any set of seven edges that have been selected must hit every pentagon exactly twice, and so does satisfy the pentagon criterion. The data structures for the program are relatively straightforward. The variable depth is the size of the subset currently being examined: if depth is seven, then the current subset is a candidate for being the attachments from a girth 7 graph. The array selection directly codes whether or not an edge is an attachment. The array choice is an

array that lists the index numbers of the subset (of which only 1 through depth are significant) currently under examination. The information in checked is the heart of the algorithm: it stores information about which branches have been examined and which need to be examined. Only rows 1 through depth are significant An entry is false if it has not been examined, and true otherwise. As an extension i of depth j is examined (with the function check_pentagons) the entry checked[i j] is marked true. If all possible extensions of the subset (of size depth) are not suitable, then every entry of the row depth of checked will be marked true, and so the program moves upward, via the procedure up, and picks a new element for the depth-1 position of choice. If an entry is acceptable and we do not have a subset of size seven (dcpth=7), the program moves downward in checked: the depth is incremented, and the new row is marked with the same true/false values as the previous row. (This ensures that previously examined edges are not reexamined.) If we do have a subset of size seven then the subset satisfies 144

the pentagon constraint and is dumped to disk. The variable a lljbne is a flag for the

stopping criterion: as discussed earlier we may assume that edge 1 is an attachment, so that if the program tries to deselect the edge 1, all_done is set to true. Output Jile contains the path name to direct results to disk. Hopefully, this explanation of the data structure and the commentary embedded in the program will suffice to make the program comprehensible.

Next is the program S3j>enerator:

program S3_g0nerator; co n st num_edges« 17; typo edges *> 1..num_edges; edge.choice * array[edges] of 0..1; var depth :1„7; choice: array[1..7] of edges; selection: edge_cholce; checked: arrayfl..7( edges] of boolean; alljdone: boolean; output_fils: text;

procedure write.selectlons; var j : Integer; begin for ] :■ 1 to 7 do write(output_flle, cholceQ]: 2 ,''); writeln(output_fll 0); end;

function check_pentagons: boolean; var bad: boolean; begin bad := true; If selectlon[1] + selectlon[3] + selectlon[4] + selection[12] + selectlon[18] <= 2 then 145

If selection!*!] + selection[6] + selection[7] + selectlon[9] + selection^ 9] <= 2 then If selection[1] + seiection{9) + selection!) 0) + selection^ 1] + solectionI12] < - 2 then If selection[6] + selection^] + selection[10] + selection^ 5] + selection[17] < - 2 then If selectlon[4] + selection[5] + selection^ 1] + selection[14] + selection[17] <» 2 then if selection^] + selection[3] + selection[5] + selection[13] + selection[21] <— 2 then If selectlon[2] + selection[8] + selection[7] + selection[16] + selection[20] <= 2 then If selectlon[2] + selection[13] + selection[14] + selection[15] + selection[16] <- 2 then bad :- false; checK_pentagons:« bad; end; {of function check pentagons) procedure up; begin If depth - 3 then a1l_done true else begin selection[choice[depth]] :- 0; d e p th :* depth -1 ; selection[choice[depth]] := 0; end; end; {up} procedure down; var j : edges; begin If depth < 7 then begin for j :«1 to numjedges do checked[depth +1, j] checked[depth, j); depth:- depth + 1; end {depth<7} e lse {depth-7, so print) begin write_selections; selection[choice[depth]]0; end; {if} end; {down}

procedure pick; var try: edges; gotjone: boolean; begin try :*» 1; gotjone > false; repeat while checked[depth, try] and (try < num jedges) do begin try :*tiy + 1; If not checked[depth, try] then begin selectlonltry]:»1; checked[depth, try]:« check_pentagons; setection[try]:« 0; end; end; {while} If checked[deptht try] then begin up; try :«* 1; end else begin choice[depth] := try; gotjone := true; end; until alljdone or got_one; If got_one then begin selection[choice[depth]] :■ 1; checkedfdepth, cholce[depth]] := true; down; end; end; {pick} begin {main program} Showtext; rewrite (output_file, 'S3 attachments'); depth :o 3; 147

selection[1]:»1; selection[2] := 1; check 0d[depth, 1] > true; checked[depth, 2] := true; choice[1] > 1; cholce[2] > 2; repeat pick until alljdone; end.

There are but a few changes between this program and the previous program. Firstly, as discussed in chapter three, we need to consider all but the four edges in only one pentagon, hence 17 rather than 14 edges. Moreover, the first two edges (each in three pentagons) must be in any subset, so that the stopping criterion is that if the edge at depth 2 is deselected, then all necessary subsets have been examined. Thus the constant num_edges has been changed, and the stopping procedure has been altered accordingly. The main program block has the additional statements to initialize using the two forced attachments. The function check_pentagons has been altered to list the eight pentagons of S3. However, the rest of the structures are identical.

Next, the decomposable graphs D j 1 and Tg-Tg. program D3_generator; const num_edges ■ 21; type edges ■ 1..num_edges; edge_cholce = array [edges] of 0..1; var d e p th : 1..7; choice: array[1 ..7] of edges; selection: edge_choice; checked: array[1 ..7, edges] of boolean; all_done: boolean; ou tp u tJile: text; 148

procedure write.selections; var j : integer; begin for j := 1 to 7 do write{output_file, choice[fl : 2, *'); writeln{outputJile); end;

function check_if_okay: boolean; var good: boolean; begin good:«false; if selection[1] + selectlon[3] + selection[5] + selection^ 1] + selection^ 9] > 1 then if selection[1] + se1ectlon[3] + selectlon[7] + selectlon[18] + selectlon[9] > 1 then If selection[1] + selection^] + selectlon[16] + select!on[12] + selectlon[13] > 1 th en If selectton[2] + selectlon[4] + selection[6] + selection^ 1] + selectlon[20] > 1 then If selection[2] + select!on[4] + selectlon[8] + selection[17] + se!ectlon[10] > 1 then if selectlon[2] + selectlon[10] + se!eclion[15] + selection[12] + selection[14] > 1 then If selection[5] + selection[6] + selectlon[8] + selectton[21] + selectlon[7] + selectlon[5] > 1 then (all the pentagons are okay} If selection[1] + selectlon[13] + select!on[15] + selectlon[10] + selectlon[20] + se!ection[19] > 0 then If se!ection[2] + selectlon[14] + selection[16] + selection[9] + selection[19] + selectton[20] > 0 then If selectlon[3] + selectlon[7] + selectlon[18] + selectlon[16] + selection[12] + selection[13] > 0 then If selectlon[4] + selectlon[8] + selection[17] + selectlon[15] + selection[12] + se!ectlon[14] > 0 then If selection[5J + selection[7] + selectlon[18] + selectlon[9] + selection[19] + selection^ 1] > 0 then If select!on[6] + select!on[8] + selection[17] + selection[10] + selectlon[20] + selection^ 1] > 0 then If selection[12] + selectlon[15] + selectlon[17] + selection^ 1] + select!on[18] + selection[16] > 0 then {all the hexagons are okay} good := true; 149

checkjfjokay:« good; end; {of function chock for print}

function ch9ck_pontagons: boolean; begin check_pentagons :« (selection[1] + selection[2] - selection[13] * setectlon[14] - selection^ 5] - selection[16] - selection[17] - selection[18] - selection[19] - selection[20] - selection[21] < 0); end; procedure up; begin If depth = 1 then all_done:»true else begin selection[choice[depth]] := 0; depth :«* depth - 1; selection[cho!ce[depth]] > 0; end; end; {up} procedure down; var j : edges; begin If depth < 7 then begin for j :■ 1 to num_edges do checked[depth +1, j] := checked{depth, J}; depthdepth +1; end {depth<7} else {depth—7, so print} begin If checkJf_okay then {it really is a subset that works} write_selections; selection[choice[depth]]:« 0; end; {if} end; {down} procedure pick; var try; edges; got_one; boolean; begin t r y 1; gotjone :■ false; repeat while checked[depth, try] and (try < num jedges) do begin try try +1; If not checked[depth, try] then begin selection[try] :« 1; checkedfdepth, try]check_pentagons; selectlon[try] > 0; end; end; {while} If checked[deptht try] then begin up; try > 1 ; en d e lse begin cholce[depth]:» try; got_one:«true; end; until alljdone or got_one; If gotjone then begin select!on[choice[depth]] > 1; checked[depth, cholce[depth]] :•* true; down; end; end; {pick} begin {main program} Showtext; d e p th :»1; r0write(output_fil9( ’big disk;D3 attachments'); repeat pick until alljdone; end. 151

This program uses a slightly different tactic, in that there are two functions,

4 checkjyentagons and checkJf_okay, that determine whether a subset can possibly be extended and if it actually is a set of seven edges that satisfy the pentagon and hexagon constraints. The function check_pcntagons verifies that 0 £ y j + y 2 - (y 1 3 +... y ji) as was pointed out in chapter III to be necessary. The function checkjfokay merely runs through all the pentagons to verify that each is met at least twice by the selected seven edges and that each hexagon is met at least once. The function checkjpentagons is employed when an extension is tried out, and the more complicated function check_if_okay is employed just prior to printing. Note that the stopping criterion is much weaker: now, the only sure stopping criterion is when all subsets of edges have been examined, i. e., when depth - 1 and the program tries to go up a level in checked.

Notice how similar this is to the program that generates subsets of attachments for % T 6: program T6xT6jgenerator; const num_edges » 21; type edges - 1..num_edges; edge_choice * array[edges] of 0..1; var depth :1..7; ch o ice: array[1..7] of edges; selection; edge_choice; ch eck ed : array[1 ..7,edges] of boolean; alljdone: boolean; o u tp u tjile : text; procedure write_S9lectlons; var j : integer; begin for j :»1 to 7 do write(output_file, cho!ce[j]: 2 ,''); 152

writeln(output„file); end;

function check_if_okay: boolean; var good: boolean; begin good:«false; If selection^] + selection[2] + selection[11] + selectlon[7] + selection[10] > 1 then If selectlon[1] + selection[3] + selectlon[14] + selection[9] + selection[15] > 1 then If selection[2] + selection[3] + selectlon[13] + selection[8] + selection[12] > 1 then If selectlon[4] + selection^] + selection[18] + selectlon[8] + selection^ 9] > 1 then If selection[4] + selection[6] + selection!! 7] + selection[7] + selection[16] > 1 then If selection[5] + selection[6] + selection[20] + se!ection[9] + selection[2l] > 1 then If selection[7] + selection[11] + seIectlon[12] + selection[8] + selectlon[19] + selectlon[16] >0 then If setection{8] + selection!! 3] + selection[14] + selection^] + selection[21] + selection[18] > 0 then If selection[7] + selectlon[10] + selection[15] + selection[9] + selectlon[20] + selectlon[17] > 0 then If selection[1] + selection[2] + selectlon[12] + selectlon[18] + selQction[21] + selection[15] > 0 then If selection!5] + selection^] + selection!! 7] + selection!!!] + selection!! 2] + selection!! 8] > 0 then If selection^] + selection^] + selection!! 4] + selectlon[20] + selection!! 7] + selection!!!] > 0 then If selection!!] + selection^] + selection!!3] + selection!!9] + selection!! 6] + selection!! 0] > 0 then If selection[4] + selection^] + selection[20] + selection!! 4] + selection!! 3] + selection!! 9] > 0 then If selection[4] + select!on[5] + selection[2l] + selection!! 5] + selection!! 0] + selection!! 6] > 0 then good:«true; checkjfjokay > good; end; (of function check for printing] function check_pentagons: boolean; var incount: integer; begin incount := selection^ ] + selection[2] + setection[3] + selection[4] + selection[5] + selection[6] + selection[7] + selection[8] + selection[9]; checkjaentagons:« (incount < depth) and (incount < 5); end; procedure up; begin If depth « 1 then alljdone > true else begin selection[cholce[depth]]0; depth :« depth -1; selection[cholce[depth]] :« 0; end; end; {up} procedure down; var j : edges; begin If depth < 7 then begin for] := 1 to num_edges do checked[depth +1, j] > checked[depth, j]; d ep th := depth + 1; end (depth<7) else {depth=7, so print) begin If check_if_okay then writejselections; selection[cholce[depth]]:» 0; end; (if) end; {down} procedure pick; var try: edges; got_one: boolean; begin try :** 1; got_one := false; repeat 154

while chacked[depth, try] and (try < num_edges) do begin try :« try + 1; if not checked[depth, try] then begin select!on[try]1; checked[depth, try]:* check.pentagons; selectlon[try]:» 0; end; end; {while} if checked[depth, try] then begin up; try :■ 1; end else begin cholce[depth]:« try; gotjone := true; end; until alljdone or got_one; If got_one then begin selection[choice[depth]]1; checked[depth, cholcejdepth]]true; down; end; end; {pick}

begin {main program} d e p th :«1; rewrite(output_file, 'big disk:T6xT6 attachments*); repeat pick until alljdone; end. Careful examination of the two programs will show that the only differences are in the functions checkjjcntagons and check_if_okay, reflecting the fact that the algorithms are identical, only the structure of the graphs are different The same holds true for the remaining three programs generating subsets for Gj, G j, and G 3 . The two 155

functions are straightforward: the function check_pentagons is precisely the pruning algorithm listed in the discussion of chapter three, and the function checkjfokay simply looks at each pentagon and hexagon to make sure that all pentagons contain at

least two attachments and hexagons contain at least one attachment The remaining functions are: ForG j:

function checKJLok&y: boolean; var good: boolean; begin g o o d :« false; If select!on[1] + selection[3] + selection[12] + selection^ 4] + selectlon[8] > 1 then If selection[1] + selection[4] + selection[7] + selection^ 1] + selectlon[13] > 1 then If selectlon[1] + selection^ +■ selection^] + selection[17] + selectlon[18] > 1 then If selectlon[2] + selectlon[6] + select!on[9] + selectlon[11] + selectlon[15] > 1 then If selectlon[2] + select)on[5] + selectlon[10] + selection[12] + selectlon[16] > 1 then If selection[2] + se!ection[9] + selection! 10] + selection^ 9] + selectlon[20] >1 then If select!on[11] + selectlon[13] + selectlon[15] + select!on[17] + selectlon[19] + selectlon[21] > 0 then If select!on[4] + selectlon[8] + selection[15] + selectlon[18] + selection[19] + selection[21 ] > 0 then If selectlon[5] + selection[9] + selection[14] + selectlon[18] + selection[19] + selection[21] > 0 then if selectlon[12] + selectlon[14] + selection[16] + select!on[18] + selection[20] + selectlon[21] > 0 then If selectlon[3] + selection[7] + selection[16] + selection[17] + selectlon[20] + selectlon[21] > o then If select!on[6] + selection^ 0] + selectlon[13] + selectlon[17] + selectlon[20] + selectlon[21] > 0 then If selectlon[4] + select!on[5] + selectlon[8] + selection[9] + selection[14] + selectlon[15] > 0 then If selection[3] + selectlon[6] + selection[7] + selection[10] + selectlon[13] + se!ection[16] > 0 then 156

good :«true; checkJL okay := good; end; {of function check for printing}

function check_pentagons: boolean; begin If 2 + selection[1] + selection[2] - selection[13] - selection[14] - selectlon[15] - se!ection[16] - selection[17] - select!on[18] - selection[19] - selection[20] - 2 * selection[21 j >» 0 then check_pentagons:=false else check oentagons := true; end;

For O 2 :

function checkJfjokay: boolean; var good: boolean; begin good false; If selection^ ] + selection[2] + selection[8J + setection[14] + selection[6] > 1 then If select!on[1] + selection^] + selection[7] + selection[13] + selection[5] > 1 then If selection[3] + selection[4] + selection[12] + selection[16] + selection[10] > 1 then If selection[3] + selection[4] + selectlon[11] + select!on[15] + selection[9] > 1 then If selection[1] + selection[5] + selectlon[17] + selection^ 1] + selection[4] + selectlon[21] > 0 then If selection^ ] + selection[6] + selection^ 9] + selection[12] + selectlon[4] + selection[21] > 0 then If se(ection[2] + selectlon[7] + selection[18] + selection[10] + selection[3] <1- se!ectlon[21] > 0 then if selection[2] + selection[8] + selectlon[20] + selectton[9] + selection[3J + selection[21] > 0 then If selection[5] + seIection[6] + selection[14] + selection [8] + selectlon[7] + selection[13] > 0 then If selection[5] + selection[6] + selection[19] + selection[16] + selection[18] + selection[13] > 0 then If selection[7] + selectlon[8] + selection[20] + selectlon[15] + selectlon[17] + selectlon[13] > 0 then 157

If selection{7] + selection[8] + sel©ction[14] + s©lection[19] + selection[16] + selection[18] > 0 then if selection^] + seIection[6] + selection^4] + selectlon[2(J] + selectlon[15] + setection[17] > 0 then If selection[5] + selection[6] + selection[19] + setect!on[12] + selection^ 1] + selection! 17] > 0 then if selection[7] + selection[8] + selection[20] + selection[9] + selection[10] +■ selection[18] > 0 then if selection[9] + selection^ 0] + selection[18] + setection[13] + selection^ 7] + selectlon[15] > 0 then if selection^] + selection[10] + selection[16] + selectlon[19] + selection^ 4] + selection[20] > 0 then If setection[9] + selection[10] + selection[16] + selection[12] + selectlon[11 ] + selection[15] > 0 then if selection^ 1] + selection[12] + selection[16] + selection^ 8] + selectlon[13] + selectlon[17] > 0 then If selectlon[11] + selection[12] + selection[19] + selection[14] + selectlon[20] +■ selection[15] > 0 then good :«true; checkJf_okay > good; end; {of function check for printing}

function check.pentagons: boolean; begin checkjjentagons := (selection[1] + selection[2] + selection[3] + seiection[4] - selection[17] - selection[18] - selection[19] - selection[20] - selection[21] < 1); end;

For G3 :

function checkjf.okay: boolean; var good: boolean; begin good :« false; if selectlon[1] + selection[2] + selectlon[11] + selectlon[5] + selection[13] > 1 then If selection[1] + selectlon[2] + selection[12] + selection[6] + select!on[14] > 1 then If select!on[3] + se!ectlon[4] + selection[9] + selectlon[15J + selection[7] > 1 then if selection^] + selectlon[4] + selection[10] + selection[16] + selection[8] > 1 then if se!ection[5] + selectlon[17] + selection[9] + selection[10] + selection[18] 158

> 1 then if selection[6] + selection[20] + selection[7] + se!ection[8] + selection[19] > 1 then If selection[1] + selection^ 3] + se!ectlon[l 7] + selection[9] + selection[4] + selection[21] > 0 then if selection[1] + selection[14] + selection[19] + selection[8] + selection[3] + seIectlon[21] > 0 then if selection[2] + selection[11] + selection[18] + selectlon[10] + selection[4] + selection[21] > 0 then if selection[2] + seIect!on[12] + selection[20] + selection[7] + seledion[3] + selection[21 ] > 0 then if selection[13] + se!ectlon[14] + selection[6] + selection[20] + selection[15] + selection! 17] > 0 then If selectlon[13] + selectlon[14] + selection^ 9] + selection[16] + selection[18] + select ion [5] > 0 then If selectlon[11] + selectlon[12] + selection[6] + selectlon[19] + selectlon[16] + selection[18] > 0 then if selection^ 1] + selection^ 2] + selectlon[20] + selection^ 5] + selection[17] + selection[5] > 0 then If selectlon[6] + selection[12] + selection^ 1] + selection[5] + selection[13] + selectlon[14] > 0 then If selection[7] + selection[8) + se!ection[1 6] + selection^ 0] + setection[9] + selection[15] > 0 then good :«true; checkJLokay :■ good; end; {of function check for printing}

function check_pentagons: boolean; var incount: integer; begin incount :« selection[1] + seIectlon[2] + seiection[3] + selection[4] + selection[5] + selection[6] + selection[7] + selection[8] + selectIon{9] + seiection[10] - selectlon[21]; check_pentagons{Incount < depth) and (incount < 5); end;

Output from these programs was a text file that was fed into the following program: program removeJsomorphlcjcoptes; co n st blgnum = 2000; perms *=11; {should be of size |perm group|-1, change with each group} type edges »0..21; var edgellst: arrayl1..7,1..bignum] of edges; permutation :array[1 ..perms, 1 ..21] of edges; thls_array, kill_array: array[1..7] of edges; index, i, printout, entry, perm jndex: Integer; mln, m lnjndex, j, start_point: integer; all_done; boolean; permfile, infile, outfile: text; perm_name, oid_name, new_name; string; begin {main program block) perm_name:« OldRleName('Where is the perm group?1); {Specific to the macintosh, allows interactive selection of file name} Reset(permflle, permjiame); writeln('The permutations are:'); for index > 1 to perms do begin for i 1 to 21 do begin read(permfi!e, permutation[index, I]); write(permutation[index, I]: 3); end; readln(permfile); writeln; end; close(permflle); oid_name:» OldFileName('Wheres the list ?'); new_name NewFileName('Wheres it going?'); Reset(inflle, old_name); Rewrite(outfile, new_name); index :» 0; while not eof(infile) d o {read in the list} begin index:*index +1; for i :«1 to 7 do read(infile, edgelistfi, index]); readln(infile); end; writeln('Total o fin d e x ,' entries.'); {count on input} printout := 0; start_polnt := 1; 160 a lljd o n e :■ false; while not alljdone do begin entry:» start_polnt; while (edgelist[1, entry]■ 0) and not (entry « index) do entry:» entry +1; {find first nonzapped entry) If edgelist[1, entry]** 0 then all_done := true; If not all.done then begin for I :** 1 to 7 do begin write(outfile, edgellst[i, entry]: 3); thls_array[i]:» edgelist[i, entry]; end; printout:« printout +1; writeln(outfile); start_polntentry; edgellst[1, entry] > 0 ; for permjndex :■ 1 to perms do begin for I :■ 1 to 7 do kill_arraypj:« permutatlonfpermjndex, thls_array[i]]; {kill entiy from list) for I 1 to 6 do begin m ln:« kllljarrayp]; mlnjndex ;■ i; for] >1 + 1 to 7 do If mln > kllLarrayU] then begin mln := kllLarrayU]; m lnjndex := j; end; kllLarray{mlnJndex]:« kill_arrayp]; kilLarray[l]:« mln; end; entry:« starLpoInt; while entry <« index do begin If edge!ist[1, entry] - kill_array[1] then If edgeltst[2, entry] *= k]|l_array[2] then If edgellst[3, entry] - kill_array[3] then If edgellst[4, entry] = kilLarray[4] then If edgelist[5, entry]« kill_array[5] then 161

If edgelist{6, entry]« kill_array[6] then If edgelist[7, entry]« kill_array[7] then begin edgelist[1, entry] > 0 ; entryindex; end; entry :■ entry+ 1; end; {while entry <« index} end; {for permjndex] end; {if not alljdone) end; {while not all_done} writeln(Totat ofprintout,1 entries.'); end. While not the best programming style, this portion of the algorithm is sufficiently straightforward so that it was written in a single block. The data structures are as

follows: type edges runs from 0 to 2 1 ; values from 1 to 21 indicate the edge of attachment, while the value zero indicates that this entry has already been processed.

The array edgelist stores as a rectangular array the full list of all attachments from the previous program. If the first entry of a column of edgelist is zero, that column has already been processed; if the first entry is nonzero, that column is still active. The array permutation contains in functional form the set of all nonidentity edge isomorphisms.

The arrays thisjirray contains the current column under examination, and Idlljzrray

contains the current column under a permutation that needs to be deleted from edgelist

The program is rather simple. The algorithm can be summarized as follows: (1) Read in the permutation group. Display the group.

(2) Read in the list of attachments.

(3) Find the first set of attachments that haven't had their first entry set to zero

(and hence had a copy dumped to disk). Dump this column to disk and copy it into this_array. Mark the first entry of the column zero to indicate it has been examined. (4) Apply all the nonidentity automorphisms to eliminate the isomorphic copies of the column. This is accomplished by applying sequentially elements of permutation to 162

this_array, sorting into increasing order, and then searching for the appropriate column in edgelist If found, zero the first entry.

(5) Repeat step 3 until all entries have been examined.

The only remaining concerns are tactical: since the search first eliminates entries from the start of the list, there is no need to look for copies of the examined attachment

set before the index of the eliminated column; the variable start_point contains the first unexamined entry to begin the search. Since the number of entries of edgelist varies

from graph to graph, the variable index contains the largest significant entry of edgelist This also serves as a stopping criterion: when an entry is found and set to zero (in step

5), the point of examination (contents of the variable entry) is set to index, so that the

increment pushes the point of examination past the number of significant entries, so the program can continue on with the next permutation, or if that was the last permutation, on with the next unexamined column.

Reducing under the permutation group of the underlying topological subgraph was helpful: it cut the necessary time for computation of the next step by a factor of 2 to 8 ; the next step, checking for assignments of attachments to positions in the deleted 7-circuit took about two minutes per set of attachments examined, so even though additional computation time was spent on the reduction, overall it reduced the amount of computation time considerably. (It also served as a check on the accuracy of the output from the first pass: see the discussion of chapter three.)

This program was not applied to the attachment list for C7 and S3, as these lists were fairly shot, and hand computation was deemed better to facilitate understanding of the procedures involved. The permutation groups for the remaining graphs follow: ForD j1: 2 1 4 3 6 5 8 7 10 9 11 12 14 13 16 15 18 17 20 19 21

ForT6 T6: 2 3 1 5 6 4 8 9 7 12 13 14 15 10 11 18 19 20 21 16 17 3 1 2 6 4 5 9 7 8 14 15 10 11 12 13 20 21 16 17 18 19 2 1 3 6 5 4 7 9 8 11 10 15 14 13 12 17 16 21 20 19 18 1 3 2 5 4 6 9 8 7 15 14 13 12 11 10 21 20 19 18 17 16 3 2 1 4 6 5 8 7 9 13 12 11 10 15 14 19 18 17 16 21 20 6 5 4 3 2 1 9 8 7 20 21 18 19 16 17 14 15 12 13 10 11 5 4 6 2 1 3 8 7 9 18 19 16 17 20 21 12 13 10 11 14 15 4 6 5 1 3 2 7 9 8 16 17 20 21 18 19 10 11 14 15 12 13 5 6 4 1 2 3 9 7 8 21 20 17 16 19 18 15 14 11 10 13 12 6 4 5 2 3 1 7 8 9 17 16 19 18 21 20 11 10 13 12 15 14 4 5 6 3 1 2 8 9 7 19 18 21 20 17 16 13 12 15 14 11 10

ForG j: 1 2 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 21 2 1 5 6 3 4 9 10 7 8 11 12 15 16 13 14 19 20 17 18 21 2 1 6 5 4 3 10 9 8 7 12 11 16 15 14 13 20 19 18 17 21

For Gj: 4 3 2 1 11 12 9 10 7 8 5 6 15 16 13 14 17 20 19 18 21 1 2 3 4 6 5 8 7 10 9 12 11 14 13 16 15 19 20 17 18 21 2 1 4 3 7 8 5 6 12 11 10 9 13 14 16 15 18 17 20 19 21 2 1 4 3 8 7 6 5 11 12 9 10 14 13 15 16 20 19 18 17 21 3 4 1 2 9 10 11 12 6 5 8 7 15 16 14 13 20 17 IB 19 21 4 3 2 1 12 11 10 9 8 7 6 5 16 15 14 13 19 18 17 20 21 3 4 1 2 10 9 12 11 5 6 7 8 16 15 13 14 18 19 20 17 21

ForG j: 2 1 3 4 5 6 8 7 10 9 13 14 11 12 16 15 18 17 20 19 21 1 2 4 3 6 5 10 9 8 7 12 11 14 13 16 15 19 20 17 18 21 2 1 4 3 6 5 9 10 7 8 14 13 12 11 15 16 20 19 18 17 21

The results of the previous pass were then fed into the program automated_attachments_check to determine how (if at all) the subdivision vertices of the edges of attachment can be identified with x j ,..., x j to create a cubic girth 7 graph.

* program automated_attachments_check; {given the graph and vertices of attachment,} {find sequencing in 7-gon} const n * 7 ; 6 b 2 1 ; type perm » array[1..n] of 1..n; edges « 1..e; var distance_matrix: array[1..n, 1..n] of integer; neighbors: array[edges, 1 ..12] of edges; used: array[edges] of boolean; assignment: perm; e d g e : array[1 ..7] of edges; i, j, count, completed: integer; infile, outfile: text;

procedure nextperm (var next: perm; prev: perm); var i, j, min : integer; u sed : array[1..n] of boolean; begin i := n +1; repeat l:«l-1 until (prevp - 1] < prev[ij) or (I« 2); {find place to make switch} for j 1 to i - 2 do next£j] := prev[jj; {initial seg same} mlnn +1; forj :°ito ndo If (prev[j] > prev[l -1]) and (prev[j] < min) then m in:» prev[fl; If min s n 4 1 then {at end of sequence. Cycle to first} for I :»1 to n do next[i] > i e lse begin for j :«1 ton do used[j]:»false; nextp -1 ]:» min; f o r j 1 toi- 1 do used[next[j]]true; i :«i -1; 165

repeat i :*» i +1; j :** 0; repeat j :«j + 1 until not used[fl; next[i] :«j; usedfl] :« true; until I ** n; end; end; {procedure nextperm} procedure read_edge_nelghborhoods; var I, j : Integer; edgename; string; edgefile: text; begin edg enam e:» Old File Name {'Where are the edges?'); Reset(edgefile, edgename); for I :■ 1 to e do begin for j :«1 to 12 do read(edgefile, neighbors^, j]); readln(edgefile); end; close(edgefile); end; procedure write Jhlsjasslgnment; var I : integer; begin for I :«1 to 7 do write (outfile,' x', 1:1 , v , edge[assignment[i]]: 2); writetn(outflle); end; {write assignment) procedure get_dlstance_matrix; var I, j : integer; function distance_between (indexl, Index2 ; integer): integer; var 166

i, ]: integer; begin distance_between :«* 3; for I :«= 1 to 4 do If netghbors[edge[index1]( i]« edge[index2] then distanceJbetween :■ 1; If (neighbors[edge[index1], 5]» edge[index2]) or (neighbors[edge[index1], 6 ]- edge[index2]) then If not used[neighbors[edge[index1]t 1]] then distance__between > 2; If (neighbors[edge[index1]t 7]» edge[index2]) or (neighbors[edge[index1], 8]» edge[index2]) then If not used[neighbors[edge[index1], 2]] then distance_between > 2; If (neighbors{edge[index1], 9] - edge{index2]) or (ne!ghbors[edge[index1], 10] - edge[index2]) then If not used[nefghbors[edge[index1], 3]] then distance_between :« 2; If (neighbors[edge[index1], 11]« edge[index2]) or (ne1ghbors[edge[index1]r 12] - edge[index2]) then if not used[nelghbors[edge[fndex1], 4]] then dlstance.between:« 2; end; {of procedure to compute distance}

begin {get distance matrix main block} for i :■ 1 to 6 do for j :«i +1 to 7 do begin distancejnatrix[i, j]distance_between(i, ]); distance_matrix[j, i] :■ distance_matrix[i, j]; end; for I :■ 1 to 7 do d!stance_matrix[i, i] > 0; end; {of getting the distance matrix} function checkjthis_perm: boolean; var ok: boolean; begin o k f a l s e ; if distance_matrix[assignment[1], assignment^]] > 2 then if distance_matrix[assignment[2], assignment^]] > 2 then If distance_matrix[assignment[3]t assignment^]] > 2 then If d!stance_matrix[assignment[4], assignment[5]] > 2 then If dlstance_matrix[asstgnment{5], assignment^]] > 2 then if distancejnatrix[assignment[6], assignment^]] > 2 then if d!stance_matrix[assignment[7]1 assign ment[lj] > 2 then if dlstance_matrix[asslgnment[1], assignment^]] > 1 then If distance_matrix[assignment[2], assignment^]] > 1 then If distance_matrix[asslgnment[3], assignment^]] > 1 then If distanc8_matrix[asslgnment[4], assignment^]] > 1 then If distance_matiix[assignment[5], assignment [7]] > 1 then If distance_matrix[assignment[6], assignment[1]] > 1 then if distance_matf1x[asslgnment[7], assignment^]] > 1 then ok > true; check_thls_perm:» ok; end; {check this perm}

procedure set_up_files; var nam e: string; begin name > OldFlleNameCEdge selection list:'); reset(infile, name); name :« NewFlleName('Direct output to:'); rewrite(outfile, name); end;

procedure get_next_subset; var I : integer; begin for I :«1 to n do read(infile, edge[i]); readln(inflle); for i :■ 1 to e do used[i]:«false; for I :■ 1 to n do used[edge[i]] > true; end; {procedure to get next subset} begin {MAIN} showtext; read_edge_neighborhoods; set_up_files; completed > 0; while not eof(infile) do begin 168

get_next_subset; get_dlstance_matrix; for i :» 1 to 7 do assignment^]:*i; repeat If check_thls_perm then write_this_asslgnment; nextperm(assignment, assignment); until assignments ] - 2; completedcompleted +1; end; end.

The data structures this time are only slightly more complicated than in previous programs. The array distancejnatrix contains the distance between pairs of attachments

or 3, whichever is smaller. (Two edges are at distance 1 provided they are coincident, at

distance 2 provided there is an edge that is not subdivided coincident with both, and at distance 3 or larger if the two previous cases are not met.) The array used contains a direct reference to whether or not a particular edge is used as an attachment. The airay neighbors contains the edge neighborhood at distance up to two from the fixed edge: entries neighbors[i j] for l£j£4 contain the four edges coincident with edge i; entries ncighbors[ij] for j ** 4 + 2t or 4 + 2t -1 are the two edges at distance two from i

coincident with neighbois[i,t]. The array assignment of type perm contains the

attempted identification of attachments with x j ,..., x j. The array edge contains the

current subset of seven edges under examination.

The first procedure, nextperm, is a general purpose procedure to lexicographically

run through the symmetric group on n letters. The basic idea is that if a is a permutation

and i is the largest index such that o(i)

CT1(i)=min{o(j): a(j)>o(i), j>i), and filling out the rest of o’ in increasing order then o' is the next permutation lexicographically. It has its own array used (this procedure was 169

from a previous program) that contains a direct reference of the o(j)'s used in the initial

segment to determine how to fill out the remainder of the o*(j) (j>i). The second procedure, read_cdgc_ncighborhoods, sets up the array neighbors.

The procedure writeJhis_assignment is also simply I/O, and controls the dumping of the answers to disk. There are two other I/O related procedures, sct_up_filcs, which

initializes the input (attachment subsets) and output files, and get_next_subset which reads the next subset for examination into the array edge. The procedure get_distancc_matrix sets up the distance matrix between pairs of attachment edges (with paradigm as discussed above) with function distance_between that gives as result the

value 1,2,3 of the smaller of the distance between pairs of attachment edges or 3. The

function check_this_perm verifies the conditions of Lemma m.2.3; if all the distances are sufficiently large then the identification of subdivided vertices with x j ,..., x j as reflected by the perm assignment yields a cubic girth 7 graph. The main block consists of setting everything up, initializing assignment to the identity permutation, then running sequentially until assignment 1] is no longer 1.

The array neighbors is essential to the creation of the distance matrices, and is very sensitive to errors. The program check_edge_data was written to verify that no errors were passed on to automatcd_attachments_check. program check_edge_data; var edgefile: text; neighbors; array[1..21,1 ..12] of integer; i, j : integer; k : string;

procedure read_edge_neighborhoods; var i, j : Integer; edgename: string; edgefile: text; begin efdgename :> OldFileNamefWhere are the edges?'); Res0t(edgefll9, edgename); for I 1 to 21 do begin fo rj > 1 to 12 do read(edgeflle, neighbors^, j]); readln(edgefile); end; close(edgefile); end; begin {main} read_edge_nelghborhoods; ShowDravving; pensize(16,16); {command to control how "pen" writes} Penpat(white); for i > 1 to 21 do begin MoveTo(10,10); {move the pen to the selected pixel} drawstring(StringOf(neighbors[it 5]: 3)); {write the edge number} moveto(10,30); {move to the next point on the screen) drawstring(StringOf(nelghbors[1,6]: 3)); {write the next number, etc} MoveTo(30,50); drawstring(stringof(neIghbors[1,1]: 3)); moveto(30,70); drawstring(stringof(neighbors[i, 2]: 3)); moveto(10,90); drawstring(stringof(nelghbors[I,7]: 3)); moveto(10,110); drawstring(stringof(neIghbors{i, 8]: 3)); MoveTo(150,10); drawstring(StringOf(neIghbors[1,9]: 3)); moveto{150,30); drawstring(StringOf(neIghbofs[i( 10]: 3)); MoveTo(90,50); drawstring(stringof(neIghbors[i, 3]: 3)); moveto(90,70); drawstring(stringof(neighbors[i, 4]: 3)); moveto(150,90); drawstring(stringof(nelghbors{1,11] :3)); 171

moveto(150,110); drawstring(stringof(neighbors[i, 12]: 3)); moveto(60,60); drawstring(stringof(i :3)); while not button do {nothing} f moveto(0,0);{move above where the first number occurred) llne(20t 0); {and draw over it with white.erasing) moveto(0,14); {then the same with the other positions) line(20,0); moveto(14,34); line(20,0); moveto(14p 54); line(20.0); moveto(0,74); line(20» 0); moveto{0,94); line(20,0); moveto(134,0); line(20.0); m oveto(134,14); line{20,0); moveto(74,34); llne(20, 0); moveto(74,54); iine(20.0); moveto(134,74); !ine(20,0); moveto(134,94); line(20,0); moveto(44,44); line{20,0); while button do

l end; end.

What this program does is to run through all edges, displaying the edge neighborhood at distance two from the fixed edge, then waits until the mouse button is pressed to give the operator time to compare the numbers on the screen with the drawn 172

graph. The program is hopefully self explanatoiy. The data that was fed to the program is as follows:

For D-a1: 3 13 9 19 5 7 12 15 16 18 11 20 4 14 10 20 6 8 12 16 15 17 11 19 1 13 5 7 9 19 12 15 6 11 18 21 2 14 6 8 10 20 12 16 5 11 17 21 3 7 6 11 1 13 18 21 4 8 19 20 4 8 5 11 2 14 17 21 3 7 19 20 3 5 18 21 1 13 6 11 9 16 8 17 4 6 17 21 2 14 5 11 10 15 7 18 1 19 16 18 3 13 11 20 12 14 7 21 2 20 15 17 4 14 11 19 12 13 8 21 S 6 19 20 3 7 4 8 1 9 2 10 13 15 14 16 1 3 10 17 2 4 9 18 1 3 12 15 9 19 5 7 14 16 10 17 2 4 12 16 10 20 6 8 13 15 9 18 10 17 12 13 *O 20 8 21 14 16 1 3 9 18 12 14 1 19 7 21 13 15 2 4 8 21 10 15 4 6 7 18 2 20 12 13 7 21 9 16 3 5 8 17 1 19 12 14 1 9 11 20 3 13 16 18 5 6 2 10 2 10 11 19 4 14 15 17 5 6 1 9 7 18 8 17 3 5 9 16 4 6 10 15

ForT 6 -T6: 2 3 10 15 11 12 13 14 7 16 9 21 1 3 11 12 10 15 13 14 7 17 8 18 1 2 13 14 10 15 11 12 8 19 9 20 5 6 16 19 18 21 17 20 7 10 8 13 4 6 18 21 16 19 17 20 8 12 9 15 4 5 17 20 16 19 18 21 7 11 9 14 10 16 11 17 1 15 4 19 2 12 9 14 12 18 13 19 2 11 5 21 3 14 4 16 14 20 15 21 3 13 6 17 1 10 5 18 1 15 7 16 2 3 9 21 11 17 4 19 2 12 7 17 1 3 8 18 10 16 6 20 2 11 8 18 1 3 7 17 13 19 5 21 3 14 8 19 1 2 9 20 12 18 4 16 3 13 9 20 1 2 8 19 15 21 6 17 1 10 9 21 2 3 7 16 14 20 5 18 4 19 7 10 5 6 8 13 11 17 1 15 6 20 7 11 4 5 9 14 10 16 2 12 5 21 8 12 4 6 9 15 13 19 2 11 4 16 8 13 5 6 7 10 12 18 3 14 6 17 9 14 4 5 7 11 15 21 3 13 5 18 9 15 4 6 8 12 14 20 1 10

ForGi: 3 7 4 8 12 16 13 17 11 15 14 18 5 9 6 10 12 14 19 15 11 13 16 20 1 7 12 16 4 8 13 17 5 14 10 20 1 8 11 15 3 7 14 18 6 13 9 19 2 9 12 14 6 10 15 19 3 16 8 18 2 10 11 13 5 9 16 20 4 15 7 17 1 3 13 17 4 8 12 16 6 11 18 21 1 4 14 18 3 7 11 15 5 12 17 21 173

2 5 15 19 6 10 12 14 4 11 20 21 2 6 16 20 5 9 11 13 3 12 19 21 4 15 6 13 1 8 9 19 2 10 7 17 3 16 5 14 1 7 10 20 2 9 8 18 6 11 7 17 2 10 4 15 1 3 18 21 5 12 8 18 2 9 3 16 1 4 17 21 4 11 9 19 1 8 6 13 2 5 20 21 3 12 10 20 1 7 5 14 2 6 19 21 7 13 18 21 1 3 6 11 8 14 19 20 8 14 17 21 1 4 5 12 7 13 19 20 9 15 20 21 2 5 4 11 10 16 17 18 10 16 19 21 2 6 3 12 9 15 17 18 17 16 19 20 7 13 8 14 9 15 10 16

F o r G ?: 2 21 5 6 7 8 3 4 13 17 14 19 1 21 7 8 5 6 3 4 13 18 14 20 4 21 9 10 11 12 1 2 15 20 16 18 3 21 11 12 9 10 1 2 15 17 16 19 1 6 13 17 2 21 14 19 7 18 11 15 1 5 14 19 2 21 13 17 8 20 12 16 2 8 13 18 1 21 14 20 5 17 10 16 2 7 14 20 1 21 13 18 6 19 9 15 3 10 15 20 4 21 16 18 11 17 8 14 3 9 16 18 4 21 15 20 12 19 7 13 4 12 15 17 3 21 16 19 9 20 5 13 4 11 16 19 3 21 15 17 10 18 6 14 5 17 7 18 1 6 11 15 2 8 10 16 6 19 8 20 1 5 12 16 2 7 9 15 9 20 11 17 3 10 8 14 4 12 5 13 10 18 12 19 3 9 7 13 4 11 6 14 5 13 11 15 1 6 7 18 4 12 9 20 7 13 10 16 2 8 5 17 3 9 12 19 6 14 12 16 1 5 8 20 4 11 10 18 8 14 9 15 2 7 6 19 3 10 11 17 1 2 3 4 5 6 7 8 9 10 11 12

For G 3 : 2 21 13 14 11 12 3 4 5 17 6 19 1 21 11 12 13 14 3 4 5 18 6 20 4 21 7 8 9 10 1 2 15 20 19 16 3 21 9 10 7 8 1 2 15 17 18 16 11 18 13 17 2 12 10 16 1 14 9 15 12 20 14 19 2 11 7 15 1 13 8 16 3 8 15 20 4 21 16 19 9 17 6 12 3 7 16 19 4 21 15 20 10 18 6 14 4 10 15 17 3 21 16 18 7 20 5 13 4 9 16 18 3 21 15 17 8 19 5 11 2 12 5 IB 1 21 6 20 13 17 10 16 2 11 6 20 1 21 5 18 14 19 7 15 5 17 1 14 11 IB 9 15 2 21 6 19 1 13 6 19 2 21 5 17 12 20 8 16 7 20 9 17 3 8 6 12 4 10 5 13 8 19 10 18 3 7 6 14 4 9 5 11 5 13 9 15 11 18 1 14 4 10 7 20 5 11 16 10 13 17 2 12 8 19 4 9 6 14 8 16 12 20 1 13 3 7 18 10 6 12 7 15 14 19 2 11 3 8 9 17 1 2 3 4 13 14 11 12 7 8 9 10 174

After a complete list of assignments has been formed (and hence a complete set of graphs formed), while one might stop at this point and examine flow polynomials of the graphs, it seems sensible to try to sort out the distinct isomorphic copies. The program create_vertex_neighborhoods serves two purposes. One is to take the attachment data and create a text file consisting of vertex neighborhoods (instead of edge neighborhoods). The other is to give for each vertex the number of 7-circuits containing it (and verify that the graphs are of girth 7). The theory for the former task is trivial: read in from a file the edges around each vertex and the other cndvcrticcs (the monovalent vertices of the vertex star), to be stored in the array n b h d jn. fo For a vertex v, i runs from 1 to 3 denoting the three edges incident with the vertex v. Here, nbhd_info[v,i,l] is the number of the edge incident with the vertex v, and nbhd_info[v,i, 2 ] is the number of the vertex that is the endvertex of nbhd_info[v,i,l] not v. We then run through the vertices v from 1 to 14 and three edges e incident with v: if e is not an attachment then v remains adjacent to the endvertex not v of e. If e is an attachment, say of xj, then v is set adjacent to i+14, and i+14 set adjacent to v. The vertex i+14 is then made adjacent to two vertices, the endvertices of the attachment edge. The remaining adjacency of i+14 is to vertex vj, given number i+21. Finally, the vertices vj given number 22 through 28 are made adjacent to one another (cyclically) and down to the vertex Xj.

The theory of the second task is only slightly more sophisticated. Since these graphs are of girth 7, all vertices at distance at most 3 from a fixed vertex must be distinct, and hence one can count the number of 7-circuits through a fixed vertex by counting the number of edges incident between pairs of vertices at distance three from that fixed vertex. (Moreover, the pair of vertices must come from different "branches" from the fixed vertex.) The program builds a tree from a fixed vertex in the procedure buildjrcc (a subprocedure of procedure analyzeJt) by denoting the root vertex as vertex 175

0, vertices 1 through 3 the three vertices adjacent to vertex 0, vertices 4 and 5 as the two vertices not vertex 0 adjacent to vertex 1, vertices 6 and 7 (resp. 8 and 9) as the two vertices not vertex 0 adjacent to vertex 2 (resp. 3), and vertices 10 and 11 (resp 12,13; 14,15; 16,17; 18,19; 20,21) as the two vertices not vertex 1 (resp. 1; 2; 2; 3; 3) adjacent to vertex 4 (resp. 5; 6; 7; 8; 9). Because of the repetition of the defining relation, the procedure branch was written, that finds the two vertices {left and right) that are descendents of a fixed vertex {dad) not the predecessor vertex {gramps). Thus to count the total number of 7-circuits through vertex 0, one counts the number of edges crossing the sets of vertices {10,11,12,13}, {14,15,16,17}, and {18,19,20,21}.

(This program, in a slightly different format with the actual trees displayed, was used to find the two circuits of size 13 that characterize the graph Casper; the tree overlay edge structure was different for edges that crossed the two circuits from edges in the two circuits.) The information can be dumped to the line printer, providing some insight into an isomorphism invariant structure of the graph; in actual practise, this information is enough to determine the precise automoiphism group of these graphs. The program: program create_vertex_nelghborhoods; {takes data of edge nbhds & vertices for underlying graphs} {then takes each overlay to create vertex neighborhoods} {of girth 7 graphs} const number_of_vert!ces ■ 28; num berjofjedges ■ 21; typo vertices ■ 1..number_of_vertIces; edges « 1..number_of_edges; tree_yertex » 0..21; var Inpufjile, outputjile, LPT: text; vertex_neighborhoods; arrayfvertices, 1..3] of vertices; nbhdjnfo ; array[vertices, 1..3,1..2] of 1 ..21; subdivided: array{edges) of vertices; i : vertices; output_file_name: string; filename: string; tree : array[tree_vertex] of vertices; filenumber: integer; detail, LPT_on: boolean; an tijree: array[verticesj of tree_vertex; m arks: array[vertices] of boolean; procedure init_llne_printer; begin rewrite(LPT, 'Printer:'); writelnCPrinter initialized'); LPT_on :■ true; writeln; end; procedure analyzeJt; {provide analysis of graph just created) var v : vertices; i : Integer; seven_structure : array[4..8] of Integer;

procedure buildjree (root: vertices); procedure branch (var left, right: vertices; gramps, d ad : vertices); begin left := vertex_neighborhoods[dad, 1]; right > vertex_neighborhoods[dad, 2); If left■ gramps then left:« vertex_neighborhoods[dad, 3] (my three sons) else If right « gramps then right > vertex_neighborhoods[dad, 3]; end; (decides which two of the three edges are new) begin (tree building procedure) tree[0] := root; tree[1]:» vertex_neighborhoods[tree[OJ( 1); tree[2] :« vertex.neighborhoodsftreeioj, 2); tree[3] > vertex_nelghborhoods[tree[0], 3); branch(tree[4], tree [5], tree[0], tree[1]); branch(tree[6], tree[7], tree[0], tree[2]); branch(tree[8], tree[9], tree[0], tree[3)); branch(tree[10), tree[11], tree[1]t tree[4]); branch (tree[12j, tree[13], tree[1 j, tree[5J); branch (t ree[ 14], treafisj, tree[2], tree[6]); branch(tree[16], tree[17], tree[2], tree(7]); branch(tree[18], tree[19], tree[3], tree[8]); branch(tree[20], tree[21 j, tree[3], tree[9]); end; {of constructing the tree)

procedure dumpjree; var v, n : tree_vertex; procedure dump_vert (v: tree_vertex); begin If LPT_on then write(LPT, v : 2,' istree[v]: 2,' ') else write(v: 2 ,' is \ tree[v]: 2 ,' '); end; (procedure to print entry)

begin {dump tree procedure) for v :« 0 to 2 do begin for n :« 0 to 5 do dump_vert(n + v * 6); write In; If LPT on then writeln(LPT,''); end; {forv} for v := 18 to 21 do dump_vert(v); writeln; If LPT_on then write!n(LPT,''); end; {procedure dumpjree)

procedure bombout (bad.vertex; vertices); {error detected; display the tree to find the problem.) begin writeln{Tree vertex \ bad_vertex: 1,' is in a hexagon. The tree Is;'); dum pjree; end; {the graph isn't of girth 7}

procedure mark_verts; 178

var iv: vertices; it: tree_vertex; i, count: integer;

begin for iv := 1 to number_of_vertices do markspv]:«false; for i t := 0 to 21 do If marks[tree[it]] then bombout(it) {because we've seen this vertex before} else marks[tree[it]] :« true; {mark this vertex as seen} for i t :« 0 to 9 do marks[tree[it]] > false; for it:** 10 to 21 do anti_tree[tree[it}]:« it; {an inverse array to tree} count > 0; for i t 10 to 21 do for i:«1 to 3 do If marks[vertex_neighborhoods[tree[it], IJJ then {its an overlay edge} begin count:« count + 1; If detail then {we want to see the fine structure of the overlays} If LPTjon then writeln(LPT, 'overlay vertex \ it: 2 ,' adjacent to anti_tree[vertex_nelghborhoods[treept], i]]) else writeln('overlay v e rte x it: 2,' [', tree[it]: 2,1 adjacent to anti_tree[vertex_ne!ghborhoods[tree[it], i]}: 3 ,' [\ tree[antLtree{vertex_neighborhoods[tree[itJ, ij}}: 2,']'); end; {for if} count := count div 2; {since each edge has two ends, halve the total} If LPT_on then writeln(LPT, 'Vertex', v : 2 ,' In', count: 1, * 7-circuits.') else writeln('Vertexv: 2 ,' in', count: 1,' 7-clrcuits.'); seven_structure[count]:« seven_structure[count] +1; end; {checking girth and finding overlay}

begin {main segm ent for analysis routine} detail := false; {dont give overlay Info, just count} for i := 4 to 8 do seven_structure[i] := 0; {this will hold number of vertices in l-many 7-circuits} for v ^ 1 to number_of_vertices do begin buildJree(v); mark_verts; {this also increases seven_structure} end; {for} write{Type:'); for i :■ 4 to 8 do write(seven_structure[i]: 3); write In; end; {procedure analyze setup) procedure set_upjile (comment: string); var nam e: string; begin nameOldRleName(comment); reset(inputjile, name); end; {of procedure to set up files} procedure read_vertex_lnfo; {calls set_up_file} var I ; vertices; j : 1..3; k : 1..2; begin set_up„fileCWhere is the edge-vertex data?’); for i :«1 to 14 do begin forj:«1 to3 do for k :s 1 to 2 do read(lnput_file, nbhdJnfo[i, J, k]); readln(inputjile); {one entry per line} end; {for} close(input_file); end; {procedure reading vertex info} procedure get_nbhd (left.vertex: vertices); var right_vertex, attachment: vertices; edge: edges; i :1..3; begin for I :« 1 to 3 d o begin right_vertex :o nbhdJnfo{left_vertex, 1,2]; edge := nbhdJnfo[left_vertex, i, 1J; attachment:» subdlvided[edge]; If attachm ent■ 28 then {the edge isn't subdivided} vertex_neighborhoods[left_vertex, i}:« right_vertex else begin vertex_ne!ghborhoods[left_VQrtex, |] :s attachment; vertex_nelghboitioods[attachment, 1} := leftj/Grtex; vertex_nelghborhoods[attachment, 2 ]:» right.vertex; vertex__nelghborhoods[attachment, 3] := attachment + {vertex in 7-circuit} end; {else} end; {for} end; {procedure get neighborhood} procedure dump_this_to_disk; var i : vertices; j :1..3; begin rewrite(output_file, filename); for i :« 1 to 28 do for j 1 to 3 do write(outputJile, vertex_neighborhoods[i, j]); writeln(output_file); close(outputjile); end; {procedure dumping computed graph to disk} procedure get_nexLinput; var i : vertices; edge: edges; begin for i > 1 to 21 d o . subdivlded[i]28; {28 means not subdivided. } {Other is attachment.} f o r i 15to 21 do begin get(input_file); {format is * x#«" then the number} {of the attachment} get{tnput_fiie); 181

get(lnputjile); get(lnputjile); read(input_fil8, edge); subdivided[edge] :« I; end; {fori} readln(input_file); {enough of this file} filenumber > filenumber +1; If filenumber < 10 then filename :■ concat(output_file_name, 'O’, StringOf(filenumber: 1)) else filename:« concat{output_file_name1 StringOf{filenumber: 1)); write In (’Exam iningfilenam e) ;{filename contains the} If LPT_on then {name of the vertex nbhd as it will be} write!n(LPT, 'Examiningfilename); {stored to disk} end; {getting the next overlay}

begin {MAIN} showtext; lnitjlne_printer; read_vertexJnfo; set_up_file(,Where are the overlays?'); output_file_name:« NewRleNameCFamlly name?'); filenumber:■ 0; while not eof(lnputjile) do begin get_next_input; fori > 1 to 14do geLnbhd(l); {note 15..21 already done} for I :** 22 to 27 do begin vertex_nelghborhoods{i, 1] := I -7; V9rt0x_neighborhoods[i, 2] := i -1; vertex_neighborhoods[i, 3 ]:« I +1; end; {for} vertex_ne!ghborhoods[22,2}:« 28;{fix first of 7-gon} vertex_nelghborhoods[26,1} > 21; vertex_neighborhoods[28,2]:« 27; vertex_neighborhoodsi28,3] := 22; {this is why we need to do 28 differently} { dumpJhis_to_disk; ) analyzejt; end; {while} close(inputjile); end. {main program} 182

Finally, once the different graphs had been sorted into equivalence classes, yet another program was used to see if the graph contained a Hamiltonian circuit. Existence of a Hamiltonian circuit in a cubic graph implies the existence of a nowhere zero 4-flow, hence a nowhere zero 5-flow. No real subtlety is intended in the next program: most of these graphs have lota of Hamiltonian circuits, and so no subtlety is required. (The one exception is the graph Io, in which no Hamiltonian circuit was detected.)

program hamlltonianjcheck; c o n st num_verts ■ 28; valence - 3; typo vertex* 1..num_verts; valency * 1..valence; vertex_cholce * array[vertex] of 1..num_verts; var nbhd: array[vertex, valency] of vertex; d e p th : vertex; selection: array[vertex] of vertex; checked: array[vertex, valency] of boolean; hamlltonlan_drcuit, end_search: boolean; used: array[vertex] of boolean;

procedure readjrom_dlsk; var nam e: string; dev: text; v : vertex; begin nam e:* OldFIIeNamefName of graph'); reset(dev, name); for v > 1 to num_verts do read(dev, nbhdfv, 1], nbhd[v, 2], nbhd[v, 3]); close(dev); end;

procedure save_positlon; {The search was done interactively; it was thus possible to] {Interrupt normal program flow and Invoke this routine then} {later invoke restore position to continue examination} var i : vertex; j : valency; save_file: file of integer; begin rewrite(save_file, tem p hamiltonian'); write(save_file, depth); for i :»1 to num_verts do write(save_file, selection[i]); for i :■ 1 to num_verts do for j :* 1 to valence do If checkedp, j] then write(save_filet 1) else write(savejF!le, 0); end; {of procedure save the position} procedure restore_positlon; var i : vertex; j : valency; flag: integer; save_file: file of integer; begin reset(save_file, tem p hamiltonian'); read(save_file, depth); for i 1 to num_verts do read(save_file, selectionfi]); for i :■ 1 to num_verts do for j :»1 to valence do begin read(save_flle, flag); checkedp, j] :« (flag ®1); end; for i:® 1 to depth do used[selection[i]]:«tree; end; {of procedure restore the position} procedure write_path; var i : vertex; begin for I :** 1 to num verts do write(selection[i]: 3); end; procedure up; begin If depth « 2 then endjsearchtrue else begin If depth < 8 then writef"); {an indication of completion of search} used[selection[depth -1]] := false; depth :« depth -1; end; end; {up} procedure down; var neighbor, j : vertex; I : valency; begin If depth < num_verts then begin depth :« depth +1; for I:«1 to valence do checked[depth, I]:» used[nbhd[selection[depth -1], I]]; end {depthcmax} else {depth»max, so hamiltonian path} begin for I > 1 to valence do If nbhd[selectIon[depth], I]■1 then hamiltonian_cifcult:»true; If hamlltonian.clrcuit then begin write_path; writeln(' Is a hamiltonian circuit.*); end; used[selection[depth]1false; end; {if} end; {down} procedure pick; var try: valency; 185

gotjone: boolean; begin try >> 1; got_one := false; repeat while checked[depth, try] and (try < valence) do try :*try + 1; If checked[depth, try] then begin up; try :b 1; end else begin selection[depth] > nbhd[se lectio n[depth -1], try]; used[selection[depth]] :» true; gotjone :** true; end; until end_search or got_one; If got_one then begin checked[depth, try] :■ true; down; end; end; {pick} begin {main program) read_from_disk; selectlon[1] := 1; used[1] :=tme; d e p th :«1; down; repeat pick until hamiltonlanjclrcult or end_search; sysbeep(20); {user alert} If end_search then writelnfNo hamiltonian circuit detected.'); end.

The data structures are similar to those in the original search programs. The array nbhd contains the vertex neighborhoods of vertices; rows corresponding to the 186

neighborhoods and columns corresponding to different vertices. The variable depth contains the length of the path built so fan if depth = num_verts then the path is

Hamiltonian, and if depth = num_verts and the first and last vertices are adjacent, we have a Hamiltonian circuit. The array selection is the current path we are extending. The array checked codes the search algorithm: if checkedfij] is true, then either the jth adjacency is a vertex that has been used earlier in the path or else the jth adjacency has already been tried out as an extension of the path at level i. If all of checked[i j] is true for a fixed i, then the ith extension was a blind alley, and the corresponding entry of checked[i-l, *] should be marked as true; otherwise, try extending the path with the unchecked edge. The array used directly codes the edges used in the path so far, so that it is not necessary to reexamine the path each time a downward move is made.

Procedure rcad_from_disk reads the vertex neighborhoods to fill the array nbhd.

Procedures save_position and restorejx>sition were written so that the computations could be interrupted without loss of calculations; the program was nin in an interactive environment where the flow could be halted or run one line at a time; thus (for example, on the computation of Io) if something else needed to be done, the program could be halted, then stepped line by line to a convenient stopping place, and the values of all variables saved; at a later time the program could be rerun, the procedure restore jxisidon invoked, and computation proceeded from the halted line. The procedures down and up extend the path and clip off the end, marking it as examined, respectively. Procedure pick makes one extension of the path, with as many invocations of the procedure up as necessary. The main block simply does the necessary initializations and then keeps picking until a circuit is found or the search is complete. 187

In the one case of the graph Io, it took approximately eight hours to complete the entire search and discover that no Hamiltonian circuit was detected In the remaining 18 cases, Hamiltonian circuits were detected after about 10 minutes. Appendix B

Results of the Automated Search

After completion of all the searches, a total of 19 cubic girth 7 graphs on 28 vertices were detected that arise from graphs other than the Heawood graph. As mentioned before, no graph arose from S3, but from 2 to 37 graphs arose from the other girth 5 subgraphs. Interestingly enough, they are completely determined by their

7-circuit type vector, that is the 5-tuple ( 8 4 , a5, ag, ay, ag) where aj is the number of vertices in exactly i of the heptagons of the girth 7 graph. The results of the search are summarized in the following tables:

Subgraph Attachments & Assignments

C 7 - i n n e r x l - 1 x 2 - 4 x 3 - 7 x 4 - > 3 x 5 - 6 x 6" 2 x 7 - 5 C 7 - o u t e r x l - 8 x 2 - 1 0 X 3 - 1 2 x 4 - 1 4 x 5 - 9 x 6 » l l X 7 - 1 3

D 3 - 0 1 x l - 1 x 2 « 2 x 3 « 7 x 4 - 1 5 x S - l l x 6 » 1 6 x 7 - 8 D 3 - - 0 2 x l - 1 x 2 - 4 x 3 - 7 x 4 - ■20 X 5 - 1 2 x 6 - 5 X 7 - 1 0 D 3 - - 0 3 x l ~ 1 x 2 - 4 x 3 - 7 x 4 " 1 9 x S - 1 2 x 6- 6 x 7 « * 1 0 D 3 - - 0 4 x l - 1 x 2 - 6 X 3 - 1 2 x 4 - 1 9 x 5 “ 4 x 6- 7 X 7 - 1 0

G l - - 0 1 x l - 1 x 2 - 2 x 3 - 1 8 x 4 “ 3 X 5 - 1 1 x 6 - 5 x 7 - 2 0 G l - - 0 2 x l - 1 x 2 - 2 x 3 - 1 7 x4«16 X5-11 X6-14 X 7 - 1 9 G l - - 0 3 x l - 1 x 2 - 2 x 3 - 1 8 x 4 - 1 6 x 5 - l l x 6 - 1 4 X 7 - 2 0 G l - - 0 4 x l - 1 x 2 - 5 X 3 - 1 1 x 4 « 3 x5- 9 x6“18 x 7 - 1 0 G l - - 0 5 x l " 1 x 2 - 9 x 3 - 1 7 x 4 - 1 2 X 5 - 1 1 x 6 - 1 8 X 7 - 1 0 G l - - 0 6 x l - 1 x 2 - 9 X 3 - 1 8 x 4 - 1 1 x 5 « 1 2 x 6 - 1 7 X 7 - 1 0 G l - 0 7 x l - 1 x 2 - 9 x 3 - 1 8 x 4 - 1 6 x 5 - l l x 6 - 1 4 X 7 - 1 0 G l - - 0 8 x l - 7 x 2 - 9 x 3 « 8 x 4 - 1 6 X 5 - 1 1 x 6 - 1 4 X 7 - 1 0 G l - • 0 9 x l * * 7 X 2 - 1 0 x 3 » 8 x 4 “ 9 x 5 - 1 6 x 6 - l l X 7 - 1 4

G 2 - -0 1 x l - 1 x 2 « 9 X 3 - 1 3 x 4 ™ 1 2 x 5 - 2 X 6 - 1 5 X 7 - 1 6 G 2 - 0 2 x l - 1 X 2 - 1 2 X 3 - 1 3 x 4 = 9 x5- 2 X6-16 x7“15 G 2 - • 0 3 x l - 1 x 2 - 1 0 x 3 - 8 x 4 " 5 x 5 » 3 x 6= 7 x 7 = l l 188 189

G 2 - 0 4 x l - 1 X 2 - 1 5 x 3 “ 7 x 4 « 3 x 5 " 5 x 6- 8 X 7 - 1 6 G 2 - 0 5 x l - 1 x 2 « 9 x3«12 x4“ 5 x5** 3 x 6- 8 x 7 - 1 6 G 2 - 0 6 x l “ 1 X 2 - 1 2 x 3 “ 8 x 4 “ 3 x 5 - 5 X 6 - 1 6 X 7 - 1 5 G 2 - 0 7 x l « 1 X 2 - 1 2 x 3 - 8 x 4 - 5 x 5 - 3 X 6 - 1 9 X 7 - 1 5 G 2 - 0 8 x l - 1 x 2 - 1 5 X 3 - 1 9 x 4 « 3 x 5 " 5 x 6“ 8 X 7 - 1 6 G 2 - 0 9 x l - 1 X 2 - 1 5 X 3 - 1 6 x 4 « 5 x S - 3 x 6 - 1 4 X 7 - 1 8 G 2 - 1 0 x l “ 1 x 2 « l l x 3 “ 7 x 4 » 3 X 5 - 1 7 x 6“ 8 x 7 = 1 6 G 2 - 1 1 x l - 1 x 2 - 1 5 x 3 - 7 x 4 - 3 X 5 - 1 7 x 6- 8 x 7 » » 1 6 G 2 - 1 2 x l - 1 x 2 - l l X 3 - 1 4 x 4 - 3 x 5 « 7 x 6 - 1 5 X 7 - 1 6 G 2 - 1 3 x l - 1 X 2 - 1 1 x 3 - 7 x 4 - 3 X 5 - 1 4 x 6 - 1 7 X 7 - 1 6 G 2 - 1 4 x l - 1 X 2 - 1 5 x 3 - 7 x 4 » 3 X 5 - 1 4 X 6 - 1 7 X 7 - 1 6 G 2 - 1 5 x l - 1 x 2 - 9 x 3 - 7 x 4 - 4 x 5 « 5 x 6- 8 X 7 - 1 2 G 2 - 1 6 x l - 1 x 2 - 1 5 x 3 - 7 x 4 - 4 x 5 - 5 x 6- 8 X 7 - 1 6 G 2 - 1 7 x l - 1 x 2 - 9 x 3 - 1 2 x 4 - 5 x 5 - 8 x 6- 4 X 7 - 1 8 G 2 - 1 8 x l - 1 x 2 « 9 x 3 - 1 2 x 4 - 8 x 5 « 5 x 6« 4 X 7 - 1 8 G 2 - 1 9 x l - 1 X 2 - 1 2 x 3 - 9 x 4 “ 5 x 5 - 8 x 6- 4 X 7 - 1 8 G 2 - 2 0 x l - 1 X 2 - 1 0 X 3 - 1 4 x 4 « 4 x 5 - 5 x 6- 8 x 7 - l l G 2 - 2 1 x l - 1 X 2 - 1 1 x 3 - 8 x 4 - 1 0 x 5 - 5 x 6 - 4 X 7 - 1 8 G 2 - 2 2 x l - 1 X 2 - 1 1 X 3 - 1 0 x 4 - 5 x 5 - 8 X 6 - 4 X 7 - 1 8 G 2 - 2 3 x l - 1 X 2 - 1 1 x 3 - 1 0 x 4 - 8 x 5 « 5 x 6 - 4 X 7 - 1 8 G 2 - 2 4 x l - 1 x 2 - 1 0 x 3 * 8 x 4 " 4 x 5 - 5 x 6 - 1 6 X 7 - 1 5 G 2 - 2 4 x l - X X 2 - 1 0 x 3 - 8 x 4 - 5 x 5 - 4 x 6 - 1 8 X 7 - 1 5 G 2 - 2 6 x l - 1 X 2 - 1 1 x 3 - 8 X 4 - 1 6 x 5 - 5 x 6- 4 X 7 - 1 8 G 2 - 2 7 x l - 1 X 2 - 1 5 X 3 - 1 6 x 4 “ 5 x 5 - 8 x 6 - 4 X 7 - 1 8 G 2 - 2 8 x l - 1 X 2 - 1 5 X 3 - 1 6 x 4 - 6 x 5 - 5 x 6 " 4 X 7 - 1 8 G 2 - 2 9 x l “ 1 X 2 - 1 5 X 3 - 1 8 x 4 « 4 x 5 - 5 x 6- 8 X 7 - 1 6 G 2 - 3 0 x l - 1 x 2 - 9 x3- 7 x4“ 4 xS-14 x 6 " l 7 X 7 - 1 6 G 2 - 3 1 x l « 1 x 2 - 9 X 3 - 1 2 x 4 - 5 x 5 - 8 x 6 - 1 0 X 7 - 1 1 G 2 - 3 2 x l “ 1 x 2 - 9 X 3 - 1 2 x 4 “ 5 x S - 1 0 x 6 - 8 X 7 - 1 1 G 2 - 3 3 x l - 1 x 2 « 9 x 3 - 1 2 x 4 - 8 x 5 - 5 x 6 - 1 0 X 7 - 1 1 G 2 - 3 4 x l - 1 X 2 - 1 0 X 3 - 1 1 x 4 » 8 x 5 - 5 x 6- 9 X 7 - 1 2 G 2 - 3 5 x l - 1 x 2 - 1 0 X 3 - 1 1 x 4 - 8 x 5 - 5 x 6 “ 1 6 X 7 - 1 5 G 2 - 3 6 x l - 1 x 2 - 9 x 3 - 1 2 X 4 - 1 3 x 5 * 1 4 x 6 - 1 0 X 7 - 1 1 G 2 - 3 7 x l - 1 X 2 - 1 0 x 3 - l l X 4 - 1 4 X 5 - 1 3 x 6 - 9 x 7 - 1 2

G 3 - 0 1 x l - 1 x 2 - 8 x 3 - 5 x 4 - 6 x 5 - 3 X 6 - 1 1 x 7 “ 9 G 3 - 0 2 x l “ 1 x 2 - 9 x 3 - 8 X 4 - 5 x 5 - 6 x 6- 3 X 7 - 1 8 G 3 - 0 3 x l - 1 x 2 - 8 x 3 - 5 x 4 - 6 x 5 « 3 x 6* l 8 x 7 - 1 5 G 3 - 0 4 x l - 1 X 2 - 1 0 x 3 - 6 x 4 - 3 x 5 - 5 x 6 - 1 9 x 7 - 1 5 G 3 - 0 5 x l - 1 x 2 - 8 x 3 - 9 X 4 - 1 1 x 5 - 3 x 6 - 6 X 7 - 1 8 G 3 - 0 6 x l “ 1 x 2 - 8 x 3 - l l x 4 - 9 x 5 - 6 x 6 - 3 X 7 - 1 8 G 3 - 0 7 x l - 1 x 2 - 9 x 3 " 6 x 4 - 3 x S - 1 3 x 6 - 1 0 X 7 - 2 0 G 3 - 0 8 x l - 1 x 2 « 7 x 3 - 1 0 x 4 - 6 x 5 - 5 x 6“ 8 x 7 - 9 G 3 - 0 9 x l " 1 x 2 - 8 x 3 - 9 x 4 « 6 x 5 « 5 x 6- 7 X 7 - 1 0 G 3 - 1 0 x l - 1 x 2 - 7 x 3 « 1 0 X 4 - 1 4 x S - l l x 6« 8 x 7 - 9 G 3 - 1 1 x l - 1 x 2 « 8 x 3 - 9 X 4 - 1 1 X 5 - 1 4 x 6= 7 x 7 - 1 0 G 3 - 1 2 x l “ 1 x 2 - 8 x 3 - 9 X 4 - 1 4 x 5 - l l x 6« 7 x 7 “ 1 0 G 3 - 1 3 x l “ 1 x 2 - 8 X 3 - 1 1 x 4 " 9 X 5 - 1 4 x 6« 7 X 7 - 1 0 G 3 - 1 4 x l * » 3 x 2 « 5 x 3 " 6 X 4 - 1 0 x 5 « 7 x 6 = 1 3 x 7 = 1 2 G 3 - 1 5 x l « * 3 x 2 = 6 x 3 - 5 x 4 “ 7 X 5 - 1 0 x 6 “ l 2 x 7 “ 1 3 190

T6-01 xl«* 1 x2“ 4 x3“ 9 x4» 2 x5* 5 x6- 7 x7- 8 T6-02 xl« 1 x2« 8 x3“ 9 x4- 4 x5- 2 x6-21 x7*17

There are 2 graphs arising out of C j, 4 graphs arising out of D 3 l, 9 graphs arising out of Gj, 37 graphs arising out of G 2 , 15 graphs arising out of G 3 , and 2 graphs arising out of T 5 T 5 . The breakdown into type is as follows:

SubgraDh 1 £ £ 2 a. i_ name C7-inner 0 0 0 14 14 30 ganymede C7-outer 0 0 28 0 0 24 lo

D3-1 0 6 16 6 0 24 demeter D3-2 0 6 13 5 4 25 hera D3-3 1 1 16 10 0 25 poseidon D3-4 0 6 13 5 4 25 hera

Gl-01 1 11 10 6 0 23 tethya Gl-02 0 6 6 12 4 26 callisto Gl-03 12 5 7 0 4 21 hyperion Gl-04 0 6 13 5 4 25 hera Gl-05 0 6 16 6 0 24 demeter Gl-06 0 6 6 12 4 26 callisto Gl-07 0 10 11 4 3 24 adraatea Gl-08 12 5 7 0 4 21 hyperion Gl-09 1 11 10 6 0 23 tethys

G2-01 0 6 16 6 0 24 demeter G2-02 0 6 6 12 4 26 callisto G2-03 0 11 13 4 0 23 enceladus G2-04 3 14 8 0 3 22 rhea G2-05 0 8 16 0 4 24 pan G2-06 0 10 11 4 3 24 adrastea G2-07 1 12 9 5 1 23 dione G2-08 3 14 8 0 3 22 rhea G2-09 0 8 16 0 4 24 pan G2-10 0 6 13 5 4 25 hera G2-11 12 5 7 0 4 21 hyperion G2-12 0 2 8 13 5 27 europa G2-13 1 1 16 10 0 25 poseidon G2-14 2 8 8 8 2 24 mimas G2-15 0 3 17 6 2 25 amalthea G2-16 1 12 9 5 1 23 dione G2-17 0 3 17 6 2 25 amalthea G2-18 0 3 17 6 2 25 amalthea 191

G2-19 0 3 17 6 2 25 amalthea G2-20 0 10 11 4 3 24 adrastea G2-21 12 5 7 0 4 21 hyperion G2-22 1 12 9 5 1 23 dione G2-23 3 14 8 0 3 22 rhea G2-24 2 8 8 8 2 24 mimas G2-25 1 12 9 5 1 23 dione G2-26 0 6 13 5 4 25 hera G2-27 1 12 9 5 1 23 dione G2-20 0 11 13 4 0 23 enceladus G2-29 0 11 13 4 0 23 enceladus G2-30 0 2 8 13 5 27 europa G2-31 6 10 4 8 0 22 titan G2-32 1 11 10 6 0 23 tethys G2-33 2 8 8 8 2 24 mimas G2-34 0 6 12 7 3 25 hestia G2-35 12 5 7 0 4 21 hyperion G2-36 0 6 12 7 3 25 hestia G2-37 0 6 12 7 3 25 hestia

G3-01 1 1 16 10 0 25 poseidon G3-02 1 4 10 13 0 25 hades G3-03 0 10 11 4 3 24 adrastea G3-04 0 10 11 4 3 24 adrastea G3-05 1 4 10 13 0 25 hades G3-06 0 10 11 4 3 24 adrastea G3-07 0 10 11 4 3 24 adrastea G3-08 0 2 8 13 5 27 europa G3-09 0 2 8 13 5 27 europa G3-10 0 2 8 13 5 27 europa G3-11 2 8 8 8 2 24 mimas G3-12 0 6 6 12 4 26 callisto G3-13 1 1 16 10 0 25 poseidon G3-14 0 10 11 4 3 24 adrastea G3-15 0 6 13 5 4 25 hera

T6-1 1 4 10 13 0 25 hades T6-2 0 8 16 0 4 24 pan Explicit drawings of the 19 graphs follow on the next 19 plates. The vertex colorings

indicate the number of heptagons through each vertex. 192

Plate 2

The graph europa automorphism group has order 1 contains 27 heptagons Plate 3 193

11 # 5 0 6 0 7

® 8

24

The £raph amalthea automorphism group has order 2 contains 25 heptagons Hamiltonian circuit: (numbering from G2-15) 1, 15, 3, 19, 7, 6, 9, 14, 13,18, 12, 10. 11, 16, 23,24,17,5, 8, 12, 16, 10, 20, 27, 26, 25, 18, 13 194

Plate 4

©

The graph callisto automorphism group has order 4: left-right and exchanging 5,4 with 28,27 contains 26 heptagons

Hamiltonian circuit: (numbering from G2-02) 1, 15, 3, 7, 17, 6 , 9, 20, 27, 28, 22, 23, 24, 25, 26, 19, 2 ,4 ,5 , 8 , 16, 12,10, 21, 11, 18, 14, 13. 195 Plate 5

The graph hestia automorphism group of order 1 contains 25 heptagons 4

The graph hera automorphism group of order 2 contains 25 heptagons Hamiltonian cycle:(numbering from G2-10) 1, 15, 3, 7, 6, 17, 2, 20, 27,26, 19, 10, 12, 8, 5,4, 11, 14, 9, 21, 28, 22, 23, 24, 25, 18, 13 197 Plate 7 (D 5

©6

The graph dcmctcr automorphism group has order 12 (free rotation on central triad, and one flip of outer H graphs) contains 24 heptagons.

Hamiltonian circuit: (numbering from G2-01) 1, 15, 3, 7, 17, 6, 9, 21, 8, 5 ,4 ,2 ,1 9 , 26, 27, 28, 22, 23, 24, 25, 18, 12,10,20, 11, 16,14, 13 198

Plate 8

Hie graph pan automorphism group of order 16: dihedral of order 8 with flip of vertices interior to 12-circuit contains 24 heptagons

Hamiltonian circuit: (G2-0S labeling) 1, 15, 3, 18, 7, 6, 2, 20, 27, 28,22, 23, 16, 14 9, 21, 8 ,5 ,4 , 11, 10,12, 17, 24, 25, 26, 19, 13 Plate 9 199

The paph adrastea automorphism group of order 2 contains 24 heptagons

Hamiltonian circuit: (G2-06 numbering) 1, 15, 3, 18, 7, 6, 2, 17,24, 23, 22, 28, 21, 10, 12, 16, 8 ,5 ,4 , 11, 14, 9, 20, 27, 26, 25, 18, 13 200

Plate 10

The graph enceladus automorphism group of order 1 contains 23 heptagons 201

Platell # 4 4D 5

18 9 26 28

The graph poseidon automorphism group of order 2 contains 25 heptagons

Hamiltonian circuit: (G2-13 labeling) 1, 15, 3, 7, 6, 17, 24, 23,22, 28, 27, 20, 10, 16, 12,13, 18, 25,26, 19, 5, 8,21,9, 14, 11,4, 2 Plate 12 202 • 4 (D 5

9 7

The graph hades automorphism group has order 2 contains 25 heptagons Hamiltonian circuit (T6-01 labeling) 1, 18,3, 7, 20, 6, 2,5, 17, 24, 25, 26. 27, 28, 21, 8, 12, 19,11, 14,10,4, 9, 13, 16, 23, 22, 15 14 18

The graph tethys automorphism group has order 2 contains 23 heptagons Hamiltonian circuit: (G2-32 labeling) 1, 15, 3,18, 7, 6 , 9, 19, 14, 13, 12, 21, 10, 11, 16, 23, 22, 28, 27, 26, 25, 24, 17, 8 ,5 ,4 , 20, 2 204

Plate 14

The graph dione automorphism group of size 1 contains 23 heptagons 205

Plate 15

The graph mi mas automorphism group of order 2 (axis of symmetry the two thickened edges) contains 24 heptagons 206

Plate 16

The graph rhea automorphism group of order 2 contains 2 2 heptagons

Hamilton circuit: (G2-04 numbering) 1, 15, 3, 19,7, 6 , 17, 24,23, 22, 28, 27, 26, 25, 18, 13, 12, 10, 16, 11, 14, 9, 21, 8 ,5 ,4 , 20, 2 207 Plate 17

JT 2 ^ C 15 C / 0

1 7^ 17

4 - 4 24 5 T I P

19

\ Q 25 \ / 8 d /

/ V ' X v 1 8 \ \ / h \ 13

26 \ \ 27 7% 1 / 21

14 10

The graph hypcrion automorphism group of order 2 contains 2 1 heptagons

Hamilton circuit: (G 2-11 labeling) 1, 15, 3, 7. 6 , 17, 24, 25,18, 13, 12, 10, 19, 26, 27, 28,22, 23, 16, 11, 14,9, 21, 8 ,5 ,4 , 20,2 Plate 18

The graph ganymedc

• 4 210 Plate 20 (Q) 5 0 6

0 7 22

The graph titan contains 22 heptagons automoiphism group elem. abelian of order 4

Hamiltonian circuit: (G2-31 labeling) 1,15, 3,18,7, 6. 2,19,26,25, 24, 17, 12, 21, 10,11,4,5, 8, 9, 20,27,28,22, 23, 16, 14, 13 REFERENCES

[1] Dirac, O. A., On rigid circuit graphs, Abh. Math. Sem. Univ. Hamburg 25 (1961), 71-76. [2] Diestal, R., A generalization of planar triangulations, J. Graph Theory, to appear.

[3] Fleischner, H., Eine gemeinsame Basis ftir die Theorie der Eulerschen Graphen und der Satz von Petersen, Monatsh. Math. 81 (1976) 267-278. [4] Jaeger, F., Flows and generalized coloring theorems in graphs, J. Comb. Theoiy B 26, (1979), 205-216.

[5] McGee, W. F„ A minimal cubic graph of girth seven, Canad. Math. Bull. 3 (1960), 149-152.

[6] Robertson, N., Pentagon generated bivalent graphs with girth 5, Can. J. Math. Vol. X X ni (1971), 36-47.

[7] Seymour, P. D., Disjoint paths in graphs, Discrete Math. 29 (1980), 293-309.

[8] Seymour, P. D., Adjacency in binary matroids, European J. Comb., to appear.

[9] Seymour, P. D., and Weaver, R. W., A generalization of chordal graphs, J. Graph Theoiy, Vol. 8, No. 2, Summer 1984,241-251.

[10] Steinberg, R., Tutte's 5-flow conjecture for the projective plane, J. Graph Theory, Vol. 8, No. 2, Summer 1984,277-285.

[11] Tutte, W. T., A contribution to the theory of chromatic polynomials, Canad. J. Math 6 (1954), 80-91. [12] Tutte, W. T„ How to draw a graph, Proc. London Math. Soc. (3), 13 (1963), 743-768. i [13] Welsh, D. J., Colourings, flows, and projective geometry, Niew Archief Voor Wiskunde (3), XXVIII (1980), 159-176