This section is devoted to the problem of drawing graphs in the plane. First, we need the notion of isomorphism. Two graphsG= (V, E) andG= (V, E) are called isomorphic if there is a bijection α:V →V such that we have {a, b} ∈E if and only if {α(a), α(b)} ∈E for all a, b in V. Let E be a set of line segments in three-dimensional Euclidean space and V the set of end points of the line segments inE. Identifying each line segment with the two-element set of its end points, we can consider (V, E) as a graph. Such a graph is calledgeometric if any two line segments inE are disjoint or have one of their end points in common.
Lemma 1.5.1 Every graph is isomorphic to a geometric graph.
Proof Let G= (V, E) be a graph on nvertices. Choose a set V of npoints inR3such that no four points lie in a common plane (Why is that possible?) and mapV bijectively toV. Let E contain, for each edge ein E, the line segment connecting the images of the vertices on e. It is easy to see that (V, E) is a geometric graph isomorphic toG.
As we have only a plane piece of paper to draw graphs, Lemma1.5.1does not help us a lot. We call a geometric graph plane if its line segments all
22 1 Basic Graph Theory
lie in one plane. Any graph isomorphic to a plane graph is called planar.7 Thus, the planar graphs are exactly those graphs which can be drawn in the plane without additional points of intersection between the edges; see the comments after Fig.1.2. We will see that most graphs are not planar; more precisely, we will show that planar graphs can only contain comparatively few edges (compared to the number of vertices).
LetG= (V, E) be a planar graph. If we omit the line segments ofGfrom the plane surface on which Gis drawn, the remainder splits into a number of connected open regions; the closure of such a region is called aface. The following theorem gives another famous result due to Euler [Eul52/53].
Theorem 1.5.2 (Euler’s formula) Let G be a connected planar graph with nvertices,m edges andf faces.Then n−m+f= 2.
Proof We use induction onm. Form= 0 we haven= 1 andf= 1, so that the statement holds. Now letm= 0. IfGcontains a cycle, we discard one of the edges contained in this cycle and get a graphGwithn=n,m=m−1 and f=f−1. By induction hypothesis,n−m+f= 2 and hencen−m+f= 2.
IfGis acyclic, thenGis a tree so thatm=n−1, by Theorem1.2.8; asf= 1,
we again obtainn−m+f= 2.
Originally, Euler’s formula was applied to the vertices, edges and faces of a convex polyhedron; it is used, for example, to determine the five regular polyhedra (or Platonic solids, namely the tetrahedron, octahedron, cube, icosahedron and dodecahedron); see, for instance, [Cox73]. We will now use Theorem 1.5.2 to derive bounds on the number of edges of planar graphs.
We need two more definitions. An edgeeof a connected graphGis called a bridge ifG\eis not connected. Thegirth of a graph containing cycles is the length of a shortest cycle.
Theorem 1.5.3 Let G be a connected planar graph on n vertices. If G is acyclic, then G has precisely n−1 edges. If G has girth at least g, then G can have at most g(ng−−22) edges.
Proof The first claim holds by Theorem 1.2.8. Thus let G be a connected planar graph having n vertices, m edges and girth at least g. Then n≥3.
We use induction on n; the case n= 3 is trivial. Suppose first that G con-tains a bridge e. Discarde so thatG is divided into two connected induced subgraphsG1 andG2on disjoint vertex sets. Letni andmi be the numbers of vertices and edges ofGi, respectively, for i= 1,2. Then n=n1+n2 and m=m1+m2+ 1. As e is a bridge, at least one of G1 and G2 contains a
7In the definition of planar graphs, one often allows not only line segments, but curves as well. However, this does not change the definition of planarity as given above, see [Wag36].
For multigraphs, it is necessary to allow curves.
cycle. If bothG1 and G2 contain cycles, they both have girth at leastg, so that by induction
m=m1+m2+ 1≤g((n1−2) + (n2−2))
g−2 + 1<g(n−2) g−2 . If, say,G2 is acyclic, we have m2=n2−1 and
m=m1+m2+ 1≤g(n1−2)
g−2 +n2<g(n−2) g−2 .
Finally suppose thatG does not contain a bridge. Then each edge of G is contained in exactly two faces. If we denote the number of faces whose border is a cycle consisting ofiedges byfi, we get
2m=
i
ifi≥
i
gfi=gf,
as each cycle contains at leastgedges. By Theorem1.5.2, this implies m+ 2 =n+f≤n+2m
g and hencem≤g(n−2)
g−2 .
In particular, we obtain the following immediate consequence of Theo-rem1.5.3, sinceGis either acyclic or has girth at least 3.
Corollary 1.5.4 Let G be a connected planar graph with n vertices, where n≥3. ThenGcontains at most 3n−6 edges.
Example 1.5.5 By Corollary 1.5.4, the complete graph K5 is not planar, as a planar graph on five vertices can have at most nine edges. The complete bipartite graphK3,3 has girth 4; this graph is not planar by Theorem1.5.3, as it has more than eight edges.
Exercise 1.5.6 Show that the graphs which arise by omitting one edge e from either K5 or K3,3 are planar. Give plane realizations for K5\e and K3,3\ewhich use straight line segments only.
For the sake of completeness, we will state one of the most famous re-sults in graph theory, namely the characterization of planar graphs due to Kuratowski [Kur30]. We refer the reader to [Har69, Tho81] or [Die10] for the elementary but rather lengthy proof. Again we need some definitions.
A subdivision of a graph G is a graph H which can be derived from Gby applying the following operation any number of times: replace an edgee=ab by a path (a, x1, . . . , xk, b), wherex1, . . . , xk are an arbitrary number ofnew vertices; that is, vertices which were not in a previous subdivision. For con-venience, G is also considered to be a subdivision of itself. Two graphs H
24 1 Basic Graph Theory
Fig. 1.11 K3,3, a subdivision and a contraction
andHare calledhomeomorphicif they are isomorphic to subdivisions of the same graphG. Figure1.11shows a subdivision ofK3,3.
Exercise 1.5.7 Let (V, E) and (V, E) be homeomorphic graphs. Show that
|E| − |V|=|E| − |V|.
Result 1.5.8(Kuratowski’s theorem) A graphGis planar if and only if it does not contain a subgraph which is homeomorphic to K5 or K3,3.
In view of Example1.5.5, a graph having a subgraph homeomorphic toK5 orK3,3cannot be planar. For the converse we refer to the sources given above.
There is yet another interesting characterization of planarity. If we identify two adjacent verticesuandv in a graphG, we get anelementary contraction ofG; more precisely, we omit uand v and replace them by a new vertexw which is adjacent to all vertices which were adjacent to uor v before;8 the resulting graph is usually denoted by G/e, where e=uv. Figure 1.11 also shows a contraction ofK3,3. A graphGis calledcontractible to a graphH if H arises fromGby a sequence of elementary contractions. For the proof of the following theorem see [Wag37, Aig84], or [HarTu65].
Result 1.5.9(Wagner’s theorem) A graphGis planar if and only if it does not contain a subgraph which is contractible toK5 orK3,3.
Exercise 1.5.10 Show that thePetersen graph(see Fig.1.12, cf. [Pet98]) is not planar. Give three different proofs using1.5.3,1.5.8, and1.5.9.
Exercise 1.5.11 Show that the Petersen graph is isomorphic to the com-plement of the triangular graphT5.
The isomorphisms of a graphGto itself are calledautomorphisms; clearly, they form a group, the automorphism group ofG. In this book we will not study automorphisms of graphs, except for some comments on Cayley graphs
8Note that we introduce only one edgewx, even ifxwas adjacent tobothuandv, which is the appropriate operation in our context. However, there are occasions where it is actually necessary to introduce two parallel edgeswxinstead, so that a contracted graph will in general become a multigraph.
Fig. 1.12 The Petersen graph
in Chap. 9; we refer the reader to [Yap86, Har69], or [CamLi91]. However, we give an exercise concerning this topic.
Exercise 1.5.12 Show that the automorphism group of the Petersen graph contains a subgroup isomorphic to the symmetric groupS5. Hint: Use Exer-cise1.5.11.
Exercise 1.5.13 What is the minimal number of edges which have to be removed from Kn to get a planar graph? For each n, construct a planar graph having as many edges as possible.
The final exercise in this section shows that planar graphs have to contain many vertices of small degree.
Exercise 1.5.14 Let G be a planar graph on n vertices and denote the number of vertices of degree at most dbynd. Prove
nd≥n(d−5) + 12 d+ 1
and apply this formula to the cases d= 5 and d= 6. (Hint: Use Corol-lary1.5.4.) Can this formula be strengthened?
Much more on planarity (including algorithms) can be found in the mono-graph by [NisCh88].
1.6 Digraphs
For many applications—especially for problems concerning traffic and transportation—it is useful to give a direction to the edges of a graph, for example to signify a one-way street in a city map. Formally, adirected graph
26 1 Basic Graph Theory
Fig. 1.13 (Directed) multigraphs
or, for short, adigraph is a pairG= (V, E) consisting of a finite setV and a setE of ordered pairs (a, b), wherea=bare elements ofV. The elements of V are again calledvertices, those ofEedges; the termarcis also used instead ofedge to distinguish between the directed and the undirected case. Instead ofe= (a, b), we again writee=ab;ais called the start vertex ortail, and b theend vertex orheadofe. We say thataandbareincident withe, and call two edges of the form ab andba antiparallel. To draw a directed graph, we proceed as in the undirected case, but indicate the direction of an edge by an arrow. Directed multigraphs can be defined analogously to multigraphs;
we leave the precise formulation of the definition to the reader.
There are some operations connecting graphs and digraphs. LetG= (V, E) be a directed multigraph. Replacing each edge of the form (a, b) by an undi-rected edge{a, b}, we obtain theunderlying multigraph|G|. Replacing parallel edges in|G| by a single edge, we get the underlying graph (G). Conversely, letG= (V, E) be a multigraph. Any directed multigraphH with|H|=Gis called anorientationofG. Replacing each edgeabinE by two arcs (a, b) and (b, a), we get theassociated directed multigraph →G; we also call →Gthe com-plete orientationofG. The complete orientation ofKn is called thecomplete digraphonnvertices. Figure 1.13illustrates these definitions.
We can now transfer the notions introduced for graphs to digraphs. There are some cases where two possibilities arise; we only look at these cases ex-plicitly and leave the rest to the reader. We first consider trails. Thus let G= (V, E) be a digraph. A sequence of edges (e1, . . . , en) is called atrail if the corresponding sequence of edges in|G|is a trail. We define walks, paths, closed trails and cycles accordingly. Thus, if (v0, . . . , vn) is the corresponding sequence of vertices,vi−1vi orvivi−1must be an edge ofG. In the first case, we have aforward edge, in the second abackward edge. If a trail consists of forward edges only, it is called a directed trail; analogous definitions can be given for walks, closed trails, etc. In contrast to the undirected case, there may exist directed cycles of length 2, namely cycles of the form (ab, ba).
A directed Euler tour in a directed multigraph is a directed closed trail containing each edge exactly once. We want to transfer Euler’s theorem to the directed case; this requires some more definitions. Theindegree din(v) of a vertexv is the number of edges with headv, and theoutdegree dout(v) of
v is the number of edges with tail v. A directed multigraph is called pseu-dosymmetric ifdin(v) =dout(v) holds for every vertex v. Finally, a directed multigraphG is calledconnected if|G| is connected. We can now state the directed analogue of Euler’s theorem. As the proof is quite similar to that of Theorem1.3.1, we shall leave it to the reader and merely give one hint: the part (b)implies (c) needs a somewhat different argument.
Theorem 1.6.1 LetGbe a connected directed multigraph.Then the following statements are equivalent:
(a) Ghas a directed Euler tour.
(b) Gis pseudosymmetric.
(c) The edge set ofGcan be partitioned into directed cycles.
For digraphs there is another obvious notion of connectivity besides simply requiring that the underlying graph be connected. We say that a vertexbof a digraph G is accessible from a vertex a if there is a directed walk with start vertexaand end vertex b. As before, we allow walks to have length 0 so that each vertex is accessible from itself. A digraph G is called strongly connected if each vertex is accessible from every other vertex. A vertexafrom which every other vertex is accessible is called aroot ofG. Thus a digraph is strongly connected if and only if each vertex is a root.
Note that a connected digraph is not necessarily strongly connected. For example, a tree can never be strongly connected; here, of course, a digraph Gis called atreeif|G|is a tree. If Ghas a rootr, we callGadirected tree, anarborescence or a branching with root r. Clearly, given any vertex r, an undirected tree has exactly one orientation as a directed tree with rootr.
We now consider the question which connected multigraphs can be oriented in such a way that the resulting graph is strongly connected. Such multigraphs are calledorientable. Thus we ask which connected systems of streets can be made into a system of one-way streets such that people can still move from each point to every other point. The answer is given by the following theorem [Rob39].
Theorem 1.6.2(Robbins’ theorem) A connected multigraph is orientable if and only if it does not contain any bridge.
We will obtain Theorem1.6.2 by proving a stronger result which allows us to orient the edges one by one, in an arbitrary order. We need some more terminology. Amixed multigraph has edges which are either directed or undirected. (We leave the formal definition to the reader.) Adirected trail in a mixed multigraph is a trail in which each oriented edge is a forward edge, but the trail might also contain undirected edges. A mixed multigraph is calledstrongly connected if each vertex is accessible from every other vertex by a directed trail. The theorem of Robbins is an immediate consequence of the following result due to Boesch and Tindell [BoeTi80].
28 1 Basic Graph Theory
Theorem 1.6.3 LetGbe a mixed multigraph andean undirected edge ofG.
Suppose that Gis strongly connected. Thene can be oriented in such a way that the resulting mixed multigraph is still strongly connected if and only ife is not a bridge.
Proof Obviously, the condition thateis not a bridge is necessary. Thus sup-pose thate is an undirected edge ofG for which neither of the two possible orientations of e gives a strongly connected mixed multigraph. We have to show thate is a bridge of|G|. Let uand w be the vertices incident withe, and denote the mixed multigraph we get by omittingefrom GbyH. Then there is no directed trail inH fromutow: otherwise, we could orientefrom wtouand get a strongly connected mixed multigraph. Similarly, there is no directed trail inH from wtou.
LetS be the set of vertices which are accessible fromuinH by a directed trail. Thenuis, for any vertexv∈S, accessible fromvinH for the following reason:uis accessible inGfromvby a directed trailW; supposeW contains the edge e, then wwould be accessible in H from u, which contradicts our observations above. Now put T=V \S; as wis in T, this set is not empty.
Then every vertex t∈T is accessible from w in H, because t is accessible fromwinG, and again: if the trail fromwtotin Gneeded the edgee, then t would be accessible fromuin H, and thus twould not be inT.
We now prove that eis the only edge of |G| having a vertex inS and a vertex inT, which shows thateis a bridge. By definition ofS, there cannot be an edge (s, t) or an edge {s, t} with s∈S and t∈T in G. Finally, if G contained an edge (t, s), then u would be accessible in H from w, as t is
accessible fromwanduis accessible froms.
Mixed multigraphs are an obvious model for systems of streets. However, we will restrict ourselves to multigraphs or directed multigraphs for the rest of this book. One-way streets can be modelled by just using directed multi-graphs, and ordinary two-way streets may then be represented by pairs of antiparallel edges.
We conclude this section with a couple of exercises.
Exercise 1.6.4 Let G be a multigraph. Prove that G does not contain a bridge if and only if each edge of Gis contained in at least one cycle. (We will see another characterization of these multigraphs in Chap. 7: any two vertices are connected by two edge-disjoint paths.)
Exercise 1.6.5 LetGbe a connected graph all of whose vertices have even degree. Show thatGhas a strongly connected, pseudosymmetric orientation.
Some relevant papers concerning (strongly connected) orientations of graphs are [ChvTh78, ChuGT85], and [RobXu88].
Exercise 1.6.6 Prove that any directed closed walk contains a directed cy-cle. Also show that any directed walkW with start vertex aand end vertex b, where a=b, contains a directed path fromatob.
Hint: The desired path may be constructed fromW by removing directed cycles.
We will be frequently concerned with digraphs throughout this book. They constitute an extremely important part of Graph Theory, both from a the-oretical and an applied point of view. Thus it is a bit surprising that there are almost no books concentrating on digraphs. However, we can recommend one book, namely the truly comprehensive monograph by Bang-Jensen and Gutin [BanGu09].