Prologue

“I wonder why problems about map-colourings are so fascinating? I know several people who have made more or less serious attempts to prove the Four-Colour Theorem, and I suppose many more have made collections of maps in the hope of hitting upon a counter-example. I like P.G. Tait's approach myself; he removed the problem from the plane so that it could be discussed in terms of more general figures. He showed that the Four-Colour Theorem is equivalent to the proposition that if N is a connected cubic graph, without an isthmus, in the plane, then the edges of N can be coloured in three colours so that the colours of the three meeting at any vertex are all different. It was at first conjectured that every cubical graph having no isthmus could be ‘three-coloured’ in this way, but this was disproved by reference to the Petersen graph, for which it may readily be verified that no three-colouring exists.

“I have often tried to find other cubic graphs which cannot be three-coloured. I do think that the right way to attack the Four-Colour Theorem is to classify the exceptions to Tait's Conjecture and see if any correspond to graphs in the plane. I did find some, but they were mere trivial modifications of the Petersen graph, obtained by detaching the three edges meeting at some vertex from one another so that the vertex becomes three vertices, and joining these three by additional edges and vertices so as to obtain another cubic graph. (Figure 0.1 is an example of such a trivial modification.)

Prologue

In the court of King Arthur there dwelt 150 knights and 150 ladies-in-waiting. The king decided to marry them off, but the trouble was that some pairs hated each other so much that they would not even get married, let alone speak! King Arthur tried several times to pair them off but each time he ran into conflicts. So he summoned Merlin the Wizard and ordered him to find a pairing in which every pair was willing to marry. Now Merlin had supernatural powers and he saw immediately that none of the 150! possible pairings was feasible, and this he told the king. But Merlin was not only a great wizard, but a suspicious character as well, and King Arthur did not quite trust him. “Find a pairing or I shall sentence you to be imprisoned in a cave forever!” said Arthur. Fortunately for Merlin, he could use his supernatural powers to find the reason why such a pairing could not exist. He asked a certain 56 ladies to stand on one side of the king and 95 knights on the other side, and asked: “Is any one of you ladies, willing to marry any of these knights?”, and when all said “No!”, Merlin said: “O King, how can you command me to find a husband for each of these 56 ladies among the remaining 55 knights?” So the king, whose courtly education did include the pigeonhole principle, saw that in this case Merlin had spoken the truth and he graciously dismissed him.

Prologue

“The Four Colour Problem has been solved by K. Appel, W. Haken and J. Koch. But what about the other mathematicians who have been working on the problem? I imagine one of them outgribing in despair, crying ‘What shall I do now?’ To which the proper answer is ‘Be of good cheer. You can continue in the same general line of research. You can study the Hajós and Hadwiger Conjectures. You can attack the problem of 5-flows and you can try to classify the tangential 2-blocks.’”

In this optimistic vein, Tutte [wT 78], rallied possibly disheartened ‘Mapmen’. We have already given some space to Hajos and to Hadwiger in Section 2.5. Now we turn our attention to the last two problems mentioned by Tutte above, namely the problem of 5-flows and the classification of tangential 2-blocks.

Some real progress has been made, not only on the questions themselves in [bD76], [bD 81], [fJ 76] and [pS 81b], but also on the intimate relationship between them - [dW 79], [dW 80], [pS 81a] .

In fact P.D. Seymour [pS 81b] has shown that every bridgeless graph has a nowhere-zero 6-flow.

Flows

Let G be a finite pseudograph. Orient G by putting arrows on each edge e ∈ EG, so that one end of e is distinguished as the tail t(e) of e and the other as the head h(e) of e. Hence t(e) = h(e) if and only if e is a loop.

Prologue

Anon.

The Petersen graph has exactly 120 automorphisms. All of its vertices are the same in the sense that any vertex can be mapped into any other by an automorphism (in fact by exactly 12 automorphisms). As we saw in Chapter 6 it has much more symmetry than this. All of its 120 paths of length 3 are the same. In fact there is exactly one automorphism which maps any one such path into any other. The Petersen graph is a Tutte graph since the constraint for the s–transitivity of a graph G given by s ≤ [½(γ(G) + 2)] is satisfied as an equality when G = P. In this sense P is as symmetric as it can be.

The Petersen graph has the property that any two pairs of vertices which are the same distance apart are also the same in the sense above. We say P is distance transitive. More precisely if u, v, x, y ∈ VP and d(u, v) = d(x, y) then there is an automorphism of P which maps u to x and v to y. Very surprisingly there are only twelve finite connected cubic distance transitive graphs.

The automorphism group of P acts primitively on its vertices. Roughly speaking this means that the automorphism group acts transitively on the vertex set and there is no k-subset of vertices (2 ≤ k < |VP|) which always stays together under the action of the automorphism group.

Prologue

(From “The lion and Albert” by Marriott Edgar (1932))

Cubic graphs are a class of graphs which have attracted considerable interest, not least because of the intimate relationship between planar cubic graphs and the Four Colour Theorem. One particular field of interest has always been the symmetries of cubic graphs. Another has been the existence of cubic graphs with specified constraints on their diameter or girth. These two fields of interest coincide in the study of cages i.e. the smallest possible cubic graphs with given constraints on their girth. In a certain sense some cages turn out to be the most symmetric graphs of all. Of course, P is a cage.

Cages

A regular graph of degree 1 has no cycles. A regular graph of degree 2 has arbitrary girth. So we define a graph which is regular of degree r and of girth g to be an (r, g)-graph only for r ≥ 3. Clearly g is also greater than or equal to 3. By the arguments of Section 1.5 the Petersen graph is a (3, 5)-graph.

The Petersen graph has fascinated many graph theorists over the years because of its appearance as a counterexample in many places. Because of its ubiquity, it seemed a natural graph to use as a central theme for a book. As a result of using this graph as our centre piece, much of this book deals with the properties of cubic 3-connected graphs and the ideas that generalize from them.

Incidentally, a biography of Julius Petersen can be found in J.Lützen, G.Sabidussi, B. Toft, Julius Petersen 1839–1910, A Biography, Preprint, Odense University, 1990.

The book has grown out of lecture courses that we have given in various places over a number of years. In all cases the audience were final honours year students, graduate students or research colleagues. Hence we would expect the current volume to be useful for senior students and research workers. Because of this we have included a reasonably large number of Exercises and an extensive set of references.

In citing references we have used the form [A-B 34] for the 1934 paper of the authors Able and Baker and [aB 34] for the 1934 (or perhaps 1834) paper of the single author A. Board. When two authors have the same initials an extra letter has been inserted to distinguish between them so [aB 89] becomes [aaB 89] if there is another author with initials A, B. References for each chapter are at the end of that chapter.

Figures, lemmas, theorems, corollaries, conjectures, etc., are numbered consecutively throughout each section of each chapter.

Prologue

(In The Elusive Pimpernel by the Baroness Orczy)

Like the Scarlet Pimpernel the Petersen graph turns up all over the place and often unexpectedly. This chapter is a by no means all-inclusive list of some of these venues. There are exactly 19 connected cubic graphs on 10 vertices. The number of elements in the set, C(n), of connected cubic graphs on n vertices grows rapidly with n; for example |C(20)| = 510489, |C(30)| = 845480228069. The Petersen graph is the only graph in C(10) with 120 automorphisms; the only graph in C(10) with girth 5; the only graph in C(10) with diameter 2; the only bridgeless graph in C(10) with chromatic index 4 and finally it is the only bridgeless non-hamiltonian graph in C(10). These many ways in which P is unique within C(10), are also reflected in the unique role that P plays within the theory of graphs. We now show some other sides to Petersen's character, and hope our discussions will not only support our central theme but also expose the reader to some other interesting areas of graph theory. This chapter makes no claim to being exhaustive. Its only claim is to enforce the well known caveat: graph theorists should always consider P and its generalizations before making conjectures.

The ubiquitous nature of the Petersen graph is further pursued in [C–W 85], [C–H–W92].

Prologue

It appears that there was a club and the president decided that it would be nice to hold a dinner for all the members. In order not to give any one member prominence, the president felt that they should be seated at a round table.

But at this stage he ran into some problems. It seems that the club was not all that amicable a little group. In fact each member only had a few friends within the club and positively detested all the rest. So the president thought it necessary to make sure that each member had a friend sitting on either side of him at the dinner.

Unfortunately, try as he might, he could not come up with such an arrangement. In desperation he turned to a mathematician. Not long afterwards, the mathematician came back with the following reply.

‘It's absolutely impossible! However, if one member of the club can be persuaded not to turn up, then everyone can be seated next to a friend.’

‘Which member must I ask to stay away?’ the president queried.

‘It doesn't matter’, replied the mathematician. ‘Anyone will do.’

‘By the way, if you had fewer members in the club you wouldn't be faced with this strange combination of properties.’

So the president, on some pretext, excused himself from the dinner and was easily able to seat the members of the club so they all had a friend on either side.

How many club members were there? Who likes whom and who dislikes whom? Show that the solution is unique (to within the obvious symmetries).