Graph Theory (II) ABC

Transcription

1 J.A. Bondy U.S.R. Murty Graph Theory (II) ABC

2 J.A. Bondy, PhD Université Claude-Bernard Lyon 1 Domaine de Gerland 50 Avenue Tony Garnier Lyon Cedex 07 France U.S.R. Murty, PhD Mathematics Faculty University of Waterloo 200 University Avenue West Waterloo, Ontario, Canada N2L 3G1 Editorial Board S. Axler Mathematics Department San Francisco State University San Francisco, CA USA K.A. Ribet Mathematics Department University of California, Berkeley Berkeley, CA USA Graduate Texts in Mathematics series ISSN: ISBN: e-isbn: DOI: / Library of Congress Control Number: Mathematics Subject Classification (2000): 05C; 68R10 c J.A. Bondy & U.S.R. Murty 2008 Apart from any fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act 1988, this publication may only be reproduced, stored or transmitted, in any form or by any means, with the prior permission in writing of the publishers, or in the case of reprographic reproduction in accordance with the terms of licenses issued by the Copyright Licensing Agency. Enquiries concerning reproduction outside those terms should be sent to the publishers. The use of registered name, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant laws and regulations and therefore free for general use. The publisher makes no representation, express or implied, with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any errors or omissions that may be made. Printed on acid-free paper springer.com

3 Contents 1 Graphs Subgraphs Connected Graphs Trees Nonseparable Graphs Tree-Search Algorithms Flows in Networks Complexity of Algorithms Connectivity Planar Graphs The Four-Colour Problem Stable Sets and Cliques The Probabilistic Method Vertex Colourings Colourings of Maps Matchings Edge Colourings...451

4 XII Contents 18 Hamilton Cycles Coverings and Packings in Directed Graphs Electrical Networks Integer Flows and Coverings Unsolved Problems References General Mathematical Notation Graph Parameters Operations and Relations Families of Graphs Structures Other Notation Index...637

5 10 Planar Graphs Contents 10.1 Plane and Planar Graphs The Jordan Curve Theorem Subdivisions Duality Faces Duals Deletion Contraction Duality Vector Spaces and Duality Euler s Formula Bridges Bridges of Cycles Unique Plane Embeddings Kuratowski s Theorem Minors Wagner s Theorem Recognizing Planar Graphs Surface Embeddings of Graphs Orientable and Nonorientable Surfaces The Euler Characteristic The Orientable Embedding Conjecture Related Reading Graph Minors Linkages Brambles Matroids and Duality Matroid Minors Plane and Planar Graphs A graph is said to be embeddable in the plane, orplanar, if it can be drawn in the plane so that its edges intersect only at their ends. Such a drawing is called

6 Planar Graphs a planar embedding of the graph. A planar embedding G of a planar graph G can be regarded as a graph isomorphic to G; the vertex set of G is the set of points representing the vertices of G, the edge set of G is the set of lines representing the edges of G, and a vertex of G is incident with all the edges of G that contain it. For this reason, we often refer to a planar embedding G of a planar graph G as a plane graph, and we refer to its points as vertices and its lines as edges. However, when discussing both a planar graph G and a plane embedding G of G, in order to distinguish the two graphs, we call the vertices of G points and its edges lines; thus, by the point v of G we mean the point of G that represents the vertex v of G, and by the line e of G we mean the line of G that represents the edge e of G. Figure 10.1b depicts a planar embedding of the planar graph K 5 \ e, shown in Figure 10.1a. (a) (b) Fig (a) The planar graph K 5 \ e, and (b) a planar embedding of K 5 \ e The Jordan Curve Theorem It is evident from the above definition that the study of planar graphs necessarily involves the topology of the plane. We do not attempt here to be strictly rigorous in topological matters, however, and are content to adopt a naive point of view toward them. This is done so as not to obscure the combinatorial aspects of the theory, which is our main interest. An elegant and rigorous treatment of the topological aspects can be found in the book by Mohar and Thomassen (2001). The results of topology that are especially relevant in the study of planar graphs are those which deal with simple curves. By a curve, we mean a continuous image of a closed unit line segment. Analogously, a closed curve is a continuous image of a circle. A curve or closed curve is simple if it does not intersect itself (in other words, if the mapping is one-to-one). Properties of such curves come into play in the study of planar graphs because cycles in plane graphs are simple closed curves. A subset of the plane is arcwise-connected if any two of its points can be connected by a curve lying entirely within the subset. The basic result of topology that we need is the Jordan Curve Theorem.

7 10.1 Plane and Planar Graphs 245 Theorem 10.1 The Jordan Curve Theorem Any simple closed curve C in the plane partitions the rest of the plane into two disjoint arcwise-connected open sets. Although this theorem is intuitively obvious, giving a formal proof of it is quite tricky. The two open sets into which a simple closed curve C partitions the plane are called the interior and the exterior of C. We denote them by int(c) and ext(c), and their closures by Int(C)andExt(C), respectively (thus Int(C) Ext(C) =C). The Jordan Curve Theorem implies that every arc joining a point of int(c) toa point of ext(c) meets C in at least one point (see Figure 10.2). C int(c) ext(c) Fig The Jordan Curve Theorem Figure 10.1b shows that the graph K 5 \e is planar. The graph K 5, on the other hand, is not planar. Let us see how the Jordan Curve Theorem can be used to demonstrate this fact. Theorem 10.2 K 5 is nonplanar. Proof By contradiction. Let G be a planar embedding of K 5, with vertices v 1,v 2,v 3,v 4,v 5. Because G is complete, any two of its vertices are joined by an edge. Now the cycle C := v 1 v 2 v 3 v 1 is a simple closed curve in the plane, and the vertex v 4 must lie either in int(c) orinext(c). Without loss of generality, we may suppose that v 4 int(c). Then the edges v 1 v 4,v 2 v 4,v 3 v 4 all lie entirely in int(c), too (apart from their ends v 1,v 2,v 3 ) (see Figure 10.3). Consider the cycles C 1 := v 2 v 3 v 4 v 2, C 2 := v 3 v 1 v 4 v 3,andC 3 := v 1 v 2 v 4 v 1. Observe that v i ext(c i ), i =1, 2, 3. Because v i v 5 E(G) andg is a plane graph, it follows from the Jordan Curve Theorem that v 5 ext(c i ), i =1, 2, 3, too. Thus v 5 ext(c). But now the edge v 4 v 5 crosses C, again by the Jordan Curve Theorem. This contradicts the planarity of the embedding G. A similar argument can be used to establish that K 3,3 is nonplanar, too (Exercise b).

8 Planar Graphs C v 3 int(c 2) v 1 v 4 int(c 1) int(c 3) ext(c) v 2 Fig Proof of the nonplanarity of K 5 Subdivisions Any graph derived from a graph G by a sequence of edge subdivisions is called a subdivision of G or a G-subdivision. Subdivisions of K 5 and K 3,3 are shown in Figure (a) (b) Fig (a) A subdivision of K 5, (b) a subdivision of K 3,3 The proof of the following proposition is straightforward (Exercise ). Proposition 10.3 A graph G is planar if and only if every subdivision of G is planar. Because K 5 and K 3,3 are nonplanar, Proposition 10.3 implies that no planar graph can contain a subdivision of either K 5 or K 3,3. A fundamental theorem due to Kuratowski (1930) states that, conversely, every nonplanar graph necessarily contains a copy of a subdivision of one of these two graphs. A proof of Kuratowski s Theorem is given in Section As mentioned in Chapter 1 and illustrated in Chapter 3, one may consider embeddings of graphs on surfaces other than the plane. We show in Section 10.6

9 10.1 Plane and Planar Graphs 247 that, for every surface S, there exist graphs which are not embeddable on S. Every graph can, however, be embedded in the 3-dimensional euclidean space R 3 (Exercise ). Planar graphs and graphs embeddable on the sphere are one and the same. To see this, we make use of a mapping known as stereographic projection. Consider a sphere S resting on a plane P, and denote by z the point that is diametrically opposite the point of contact of S and P. The mapping π : S \{z} P, defined by π(s) =p if and only if the points z, s, andp are collinear, is called a stereographic projection from z; it is illustrated in Figure s z p Fig Stereographic projection Theorem 10.4 A graph G is embeddable on the plane if and only if it is embeddable on the sphere. Proof Suppose that G has an embedding G on the sphere. Choose a point z of the sphere not in G. Then the image of G under stereographic projection from z is an embedding of G on the plane. The converse is proved similarly. On many occasions, it is advantageous to consider embeddings of planar graphs on the sphere; one instance is provided by the proof of Proposition 10.5 in the next section. Exercises Show that: a) every proper subgraph of K 3,3 is planar, b) K 3,3 is nonplanar Show that a graph is planar if and only if every subdivision of the graph is planar.

10 Planar Graphs a) Show that the Petersen graph contains a subdivision of K 3,3. b) Deduce that the Petersen graph is nonplanar a) Let G be a planar graph, and let e be a link of G. Show that G/e is planar. b) Is the converse true? Let G be a simple nontrivial graph in which each vertex, except possibly one, has degree at least three. Show, by induction on n, that G contains a subdivision of K Find a planar embedding of the graph in Figure 10.6 in which each edge is a straight line segment. (Wagner (1936) proved that every simple planar graph admits such an embedding.) Fig Find a straight-line planar embedding of this graph (Exercise ) A k-book is a topological subspace of R 3 consisting of k unit squares, called its pages, that have one side in common, called its spine, but are pairwise disjoint otherwise. Show that any graph G is embeddable in R 3 by showing that it is embeddable in a k-book, for some k Consider a drawing G of a (not necessarily planar) graph G in the plane. Two edges of G cross if they meet at a point other than a vertex of G. Each such point is called a crossing of the two edges. The crossing number of G, denoted by cr(g), is the least number of crossings in a drawing of G in the plane. Show that: a) cr(g) = 0 if and only if G is planar, b) cr(k 5 ) = cr(k 3,3 )=1, c) cr(p 10 ) = 2, where P 10 is the Petersen graph, d) cr(k 6 )=3.

11 10.2 Duality Show that cr(k n )/ ( n 4) is a monotonically increasing function of n AgraphG is crossing-minimal if cr(g \ e) < cr(g) for all e E. Show that every nonplanar edge-transitive graph is crossing-minimal A thrackle is a graph embedded in the plane in such a way that any two edges intersect exactly once (possibly at an end). Such an embedding is called a thrackle embedding. Show that: a) every tree has a thrackle embedding, b) the 4-cycle has no thrackle embedding, c) the triangle and every cycle of length five or more has a thrackle embedding Show that every simple graph can be embedded in R 3 in such a way that: a) each vertex lies on the curve {(t, t 2,t 3 ):t R}, b) each edge is a straight line segment. (C. Thomassen) 10.2 Duality Faces A plane graph G partitions the rest of the plane into a number of arcwise-connected open sets. These sets are called the faces of G. Figure 10.7 shows a plane graph with five faces, f 1,f 2,f 3,f 4,andf 5. Each plane graph has exactly one unbounded face, called the outer face. In the plane graph of Figure 10.7, the outer face is f 1. We denote by F (G) andf(g), respectively, the set of faces and the number of faces of a plane graph G. The notion of a face applies also to embeddings of graphs on other surfaces. f 1 v 1 v 2 e 1 e 6 f 2 e 9 e 2 v 7 e 8 v 6 v 3 f 5 f 3 e 10 e 5 e 7 f 4 e 3 v 5 e 4 v 4 Fig A plane graph with five faces

12 Planar Graphs The boundary of a face f is the boundary of the open set f in the usual topological sense. A face is said to be incident with the vertices and edges in its boundary, and two faces are adjacent if their boundaries have an edge in common. In Figure 10.7, the face f 1 is incident with the vertices v 1,v 2,v 3,v 4,v 5 and the edges e 1,e 2,e 3,e 4,e 5 ; it is adjacent to the faces f 3,f 4,f 5. We denote the boundary of a face f by (f). The rationale for this notation becomes apparent shortly, when we discuss duality. The boundary of a face may be regarded as a subgraph. Moreover, when there is no scope for confusion, we use the notation (f) to denote the edge set of this subgraph. Proposition 10.5 Let G be a planar graph, and let f be a face in some planar embedding of G. ThenG admits a planar embedding whose outer face has the same boundary as f. Proof Consider an embedding G of G on the sphere; such an embedding exists by virtue of Theorem Denote by f the face of G corresponding to f. Letz be a point in the interior of f, and let π( G) be the image of G under stereographic projection from z. Clearly π( G) is a planar embedding of G with the desired property. By the Jordan Curve Theorem, a planar embedding of a cycle has exactly two faces. In the ensuing discussion of plane graphs, we assume, without proof, a number of other intuitively obvious statements concerning their faces. We assume, for example, that a planar embedding of a tree has just one face, and that each face boundary in a connected plane graph is itself connected. Some of these facts rely on another basic result of plane topology, known as the Jordan Schönfliess Theorem. Theorem 10.6 The Jordan Schönfliess Theorem Any homeomorphism of a simple closed curve in the plane onto another simple closed curve can be extended to a homeomorphism of the plane. One implication of this theorem is that any point p on a simple closed curve C can be connected to any point not on C by means of a simple curve which meets C only in p. We refer the reader to Mohar and Thomassen (2001) for further details. A cut edge in a plane graph has just one incident face, but we may think of the edge as being incident twice with the same face (once from each side); all other edges are incident with two distinct faces. We say that an edge separates the faces incident with it. The degree, d(f), of a face f is the number of edges in its boundary (f), cut edges being counted twice. In Figure 10.7, the edge e 9 separates the faces f 2 and f 3 and the edge e 8 separates the face f 5 from itself; the degrees of f 3 and f 5 are 6 and 5, respectively. Suppose that G is a connected plane graph. To subdivide afacef of G is to add a new edge e joining two vertices on its boundary in such a way that, apart from its endpoints, e lies entirely in the interior of f. This operation results in a plane graph G + e with exactly one more face than G; all faces of G except f are

13 10.2 Duality 251 f e f 2 f 1 Fig Subdivision of a face f by an edge e also faces of G + e, and the face f is replaced by two new faces, f 1 and f 2,which meet in the edge e, as illustrated in Figure In a connected plane graph the boundary of a face can be regarded as a closed walk in which each cut edge of the graph that lies in the boundary is traversed twice. This is clearly so for plane trees, and can be established in general by induction on the number of faces (Exercise ) In the plane graph of Figure 10.7, for instance, (f 3 )=v 1 e 1 v 2 e 2 v 3 e 10 v 5 e 7 v 6 e 6 v 1 e 9 v 1 and (f 5 )=v 1 e 6 v 6 e 8 v 7 e 8 v 6 e 7 v 5 e 5 v 1 Moreover, in the case of nonseparable graphs, these boundary walks are simply cycles, as was shown by Whitney (1932c). Theorem 10.7 In a nonseparable plane graph other than K 1 or K 2, each face is bounded by a cycle. Proof Let G be a nonseparable plane graph. Consider an ear decomposition G 0,G 1,...,G k of G, where G 0 is a cycle, G k = G, and, for 0 i k 2, G i+1 := G i P i is a nonseparable plane subgraph of G, where P i is an ear of G i. Since G 0 is a cycle, the two faces of G 0 are clearly bounded by cycles. Assume, inductively, that all faces of G i are bounded by cycles, where i 0. Because G i+1 is a plane graph, the ear P i of G i is contained in some face f of G i. (More precisely, G i+1 is obtained from G i by subdividing the face f by an edge joining the ends of P i and then subdividing that edge by inserting the internal vertices of P i.) Each face of G i other than f is a face of G i+1 as well, and so, by the induction hypothesis, is bounded by a cycle. On the other hand, the face f of G i is divided by P i into two faces of G i+1, and it is easy to see that these, too, are bounded by cycles. One consequence of Theorem 10.7 is that all planar graphs without cut edges have cycle double covers (Exercise ). Another is the following. Corollary 10.8 In a loopless 3-connected plane graph, the neighbours of any vertex lie on a common cycle.

14 Planar Graphs e 2 e 6 f 2 e 9 e 1 f 5 e 8 f 3 e 7 e 10 e 4 f 4 e 3 e 5 f 1 Fig The dual of the plane graph of Figure 10.7 Proof Let G be a loopless 3-connected plane graph and let v be a vertex of G. Then G v is nonseparable, so each face of G v is bounded by a cycle, by Theorem If f is the face of G v in which the vertex v was situated, the neighbours of v lie on its bounding cycle (f). Duals GivenaplanegraphG, one can define a second graph G as follows. Corresponding to each face f of G there is a vertex f of G, and corresponding to each edge e of G there is an edge e of G. Two vertices f and g are joined by the edge e in G if and only if their corresponding faces f and g are separated by the edge e in G. Observe that if e is a cut edge of G, then f = g, soe is a loop of G ;conversely, if e is a loop of G, the edge e is a cut edge of G. The graph G is called the dual of G. The dual of the plane graph of Figure 10.7 is drawn in Figure In the dual G of a plane graph G, the edges corresponding to those which lie in the boundary of a face f of G are just the edges incident with the corresponding vertex f. When G has no cut edges, G has no loops, and this set is precisely the trivial edge cut (f ); that is, (f )={e : e (f)} It is for this reason that the notation (f) was chosen. It is easy to see that the dual G of a plane graph G is itself a planar graph; in fact, there is a natural embedding of G in the plane. We place each vertex f in the corresponding face f of G, and then draw each edge e in such a way that it crosses the corresponding edge e of G exactly once (and crosses no other edge of G). This procedure is illustrated in Figure 10.10, where the dual is indicated by heavy lines. It is intuitively clear that we can always draw the dual as a plane graph in this way, but we do not prove this fact. We refer to such a drawing of the dual as a plane dual of the plane graph G.

15 10.2 Duality 253 Fig The plane graph of Figure 10.7 and its plane dual Proposition 10.9 The dual of any plane graph is connected. Proof Let G be a plane graph and G aplanedualofg. Consider any two vertices of G. There is a curve in the plane connecting them which avoids all vertices of G. The sequence of faces and edges of G traversed by this curve corresponds in G to a walk connecting the two vertices. Although defined abstractly, it is often convenient to regard the dual G of a plane graph G as being itself a plane graph, embedded as described above. One may then consider the dual G of G. When G is connected, it is not difficult to prove that G = G (Exercise ); a glance at Figure indicates why this is so. It should be noted that isomorphic plane graphs may well have nonisomorphic duals. For example, although the plane graphs in Figure are isomorphic, their duals are not: the plane graph shown in Figure 10.11a has two faces of degree three, whereas that of Figure 10.11b has only one such face. Thus the notion of a dual graph is meaningful only for plane graphs, and not for planar graphs in general. We show, however (in Theorem 10.28) that every simple 3-connected planar graph has a unique planar embedding (in the sense that its face boundaries are uniquely determined) and hence has a unique dual. The following relations are direct consequences of the definition of the dual G. v(g )=f(g), e(g )=e(g), and d G (f )=d G (f) for all f F (G) (10.1) The next theorem may be regarded as a dual version of Theorem 1.1. Theorem If G is a plane graph, d(f) =2m f F

16 Planar Graphs (a) (b) Fig Isomorphic plane graphs with nonisomorphic duals Proof Let G be the dual of G. By (10.1) and Theorem 1.1, d(f) = d(f )=2e(G )=2e(G) =2m f F (G) f V (G ) A simple connected plane graph in which all faces have degree three is called a plane triangulation or, for short, a triangulation. The tetrahedron, the octahedron, and the icosahedron (depicted in Figure 1.14) are all triangulations. As a consequence of (10.1) we have: Proposition A simple connected plane graph is a triangulation if and only if its dual is cubic. It is easy to show that every simple plane graph on three or more vertices is a spanning subgraph of a triangulation (Exercise ). On the other hand, as we show in Section 10.3, no simple spanning supergraph of a triangulation is planar. For this reason, triangulations are also known as maximal planar graphs. They play an important role in the theory of planar graphs. Deletion Contraction Duality Let G be a planar graph and G a plane embedding of G. For any edge e of G, a plane embedding of G \ e can be obtained by simply deleting the line e from G. Thus, the deletion of an edge from a planar graph results in a planar graph. Although less obvious, the contraction of an edge of a planar graph also results in a planar graph (Exercise b). Indeed, given any edge e of a planar graph G and a planar embedding G of G, the line e of G can be contracted to a single point (and the lines incident to its ends redrawn) so that the resulting plane graph is a planar embedding of G/e. The following two propositions show that the operations of contracting and deleting edges in plane graphs are related in a natural way under duality. Proposition Let G be a connected plane graph, and let e be an edge of G that is not a cut edge. Then (G \ e) = G /e

17 10.2 Duality 255 Proof Because e is not a cut edge, the two faces of G incident with e are distinct; denote them by f 1 and f 2. Deleting e from G results in the amalgamation of f 1 and f 2 into a single face f (see Figure 10.12). Any face of G that is adjacent to f 1 or f 2 is adjacent in G \ e to f; all other faces and adjacencies between them are unaffected by the deletion of e. Correspondingly, in the dual, the two vertices f1 and f2 of G which correspond to the faces f 1 and f 2 of G are now replaced by a single vertex of (G \ e), which we may denote by f, and all other vertices of G are vertices of (G \ e). Furthermore, any vertex of G that is adjacent to f1 or f2 is adjacent in (G \ e) to f, and adjacencies between vertices of (G \ e) other than v are the same as in G. The assertion follows from these observations. e f 1 e f f 2 (a) (b) Fig (a) G and G,(b)G \ e and G /e Dually, we have: Proposition Let G be a connected plane graph, and let e be a link of G. Then (G/e) = G \ e Proof Because G is connected, G = G (Exercise ). Also, because e is notaloopofg, the edge e is not a cut edge of G,soG \ e is connected. By Proposition 10.12, (G \ e ) = G /e = G/e The proposition follows on taking duals. We now apply Propositions and to show that nonseparable plane graphs have nonseparable duals. This fact turns out to be very useful. Theorem The dual of a nonseparable plane graph is nonseparable. Proof By induction on the number of edges. Let G be a nonseparable plane graph. The theorem is clearly true if G has at most one edge, so we may assume

18 Planar Graphs that G has at least two edges, hence no loops or cut edges. Let e be an edge of G. Then either G \ e or G/eis nonseparable (Exercise 5.3.2). If G \ e is nonseparable, so is (G \ e) = G /e, by the induction hypothesis and Proposition Applying Exercise 5.2.2b, we deduce that G is nonseparable. The case where G/e is nonseparable can be established by an analogous argument. The dual of any plane graph is connected, and it follows from Theorem that the dual of a loopless 2-connected plane graph is 2-connected. Furthermore, one can show that the dual of a simple 3-connected plane graph is 3-connected (Exercise ). The notion of plane duality can be extended to directed graphs. Let D be a plane digraph, with underlying plane graph G. Consider a plane dual G of G. Each arc a of D separates two faces of G. Asa is traversed from its tail to its head, one of these faces lies to the left of a and one to its right. We denote these two faces by l a and r a, respectively; note that if a is a cut edge, l a = r a.for each arc a of D, we now orient the edge of G that crosses it as an arc a by designating the end lying in l a as the tail of a and the end lying in r a as its head. The resulting plane digraph D is the directed plane dual of D. An example is shown in Figure Fig An orientation of the triangular prism, and its directed plane dual Vector Spaces and Duality We have seen that the cycle and bond spaces are orthogonal complements (Exercise 2.6.4a). In the case of plane graphs, this relationship of orthogonality can also be expressed in terms of duality, as we now explain. As usual in this context, we identify cycles, trees, and cotrees with their edge sets. We observed earlier that all duals are connected (Proposition 10.9). A similar argument, based on the fact that the interior of a cycle in a plane graph is arcwiseconnected, establishes the following proposition.

19 10.2 Duality 257 Proposition Let G be a plane graph, G a plane dual of G, C acycleof G, andx the set of vertices of G that lie in int(c). ThenG [X ] is connected. For a subset S of E(G), we denote by S the subset {e : e S} of E(G ). Theorem Let G be a connected plane graph, and let G be a plane dual of G. a) If C is a cycle of G, then C is a bond of G. b) If B is a bond of G, then B is a cycle of G. Proof a) Let C be a cycle of G, and let X denote the set of vertices of G that lie in int(c). Then C is the edge cut (X )ing. By Proposition 10.15, the subgraph of G induced by X is connected. Likewise, the subgraph of G induced by V (G ) \ X is connected. It follows from Theorem 2.15 that C is a bond of G. We leave part (b) (the converse of (a)) as an exercise (Exercise ). As a straightforward consequence of Theorem 10.16, we have: Corollary For any plane graph G, the cycle space of G is isomorphic to the bond space of G. The relationship between cycles and bonds expressed in Theorem may be refined by taking orientations into account and considering directed duals, as defined above. Let D be a plane digraph and D its directed plane dual. For a subset S of A(D), denote by S the subset {a : a S} of A(D ). Theorem Let D be a connected plane digraph and let D be a plane directed dual of D. a) Let C be a cycle of D, with a prescribed sense of traversal. Then C is a bond (X )of D. Moreover the set of forward arcs of C corresponds to the outcut + (X ) and the set of reverse arcs of C to the incut (X ). b) Let B := (X) be a bond of D. ThenB is a cycle of D. Moreover the outcut + (X) corresponds to the set of forward arcs of B and the incut (X) corresponds to the set of reverse arcs of B (with respect to a certain sense of traversal of B ). The proof of Theorem is left to the reader (Exercise ). Exercises a) Show that a graph is planar if and only if each of its blocks is planar. b) Deduce that every minimal nonplanar graph is both simple and nonseparable.

20 Planar Graphs Prove that the boundary of a face of a connected plane graph can be regarded as a closed walk in which each cut edge of the graph lying in the boundary is traversed twice Determine the duals of the five platonic graphs (Figure 1.14) Let G be a plane graph. Show that G = G if and only if G is connected Show that every simple connected plane graph on n vertices, where n 3, is a spanning subgraph of a triangulation Let G be a triangulation on at least four vertices. Show that G is a simple nonseparable cubic plane graph Show that the dual of a simple 3-connected plane graph is both simple and 3-connected Show that every planar graph without cut edges has a cycle double cover Show that if B is a bond of a plane graph G, then B is a cycle of its plane dual G a) Show that the dual of an even plane graph is bipartite (and conversely). b) A Hamilton bond of a connected graph G is a bond B such that both components of G \ B are trees. Let G be a plane graph which contains a Hamilton cycle, and let C be (the edge set of) such a cycle. Show that C is a Hamilton bond of G Outerplanar Graph AgraphG is outerplanar if it has a planar embedding G in which all vertices lie on the boundary of its outer face. An outerplanar graph equipped with such an embedding is called an outerplane graph. Show that: a) if G is an outerplane graph, then the subgraph of G induced by the vertices corresponding to the interior faces of G is a tree, b) every simple 2-connected outerplanar graph other than K 2 has a vertex of degree two Let T be a spanning tree of a connected plane graph G. Show that (E\T ) is a spanning tree of G Prove Theorem A Halin graph is a graph H := T C, where T is a plane tree on at least four vertices in which no vertex has degree two, and C is a cycle connecting the leaves of T in the cyclic order determined by the embedding of T. Show that:

21 10.3 Euler s Formula 259 a) every Halin graph is minimally 3-connected, b) every Halin graph has a Hamilton cycle The medial graph of a plane graph G is the 4-regular graph M(G) with vertex set E(G) in which two vertices are joined by k edges if, in G, they are adjacent edges which are incident to k common faces (k =0, 1, 2). (The medial graph has a natural planar embedding.) Let G be a nonseparable plane graph. Show that: a) M(G) is a 4-regular planar graph, b) M(G) = M(G ) Euler s Formula There is a simple formula relating the numbers of vertices, edges, and faces in a connected plane graph. It was first established for polyhedral graphs by Euler (1752), and is known as Euler s Formula. Theorem Euler s Formula For a connected plane graph G, v(g) e(g)+f(g) = 2 (10.2) Proof By induction on f(g), the number of faces of G. Iff(G) = 1, each edge of G is a cut edge and so G, being connected, is a tree. In this case e(g) =v(g) 1, by Theorem 4.3, and the assertion holds. Suppose that it is true for all connected plane graphs with fewer than f faces, where f 2, and let G be a connected plane graph with f faces. Choose an edge e of G that is not a cut edge. Then G \ e is a connected plane graph with f 1 faces, because the two faces of G separated by e coalesce to form one face of G \ e. By the induction hypothesis, Using the relations v(g \ e) e(g \ e)+f(g \ e) =2 v(g \ e) =v(g), e(g \ e) =e(g) 1, and f(g \ e) =f(g) 1 we obtain The theorem follows by induction. v(g) e(g)+f(g) =2 Corollary All planar embeddings of a connected planar graph have the same number of faces.

22 Planar Graphs Proof Let G be a planar embedding of a planar graph G. By Euler s Formula (10.2), we have f( G) =e( G) v( G)+2=e(G) v(g)+2 Thus the number of faces of G depends only on the graph G, and not on its embedding. Corollary Let G be a simple planar graph on at least three vertices. Then m 3n 6. Furthermore, m =3n 6 if and only if every planar embedding of G is a triangulation. Proof It clearly suffices to prove the corollary for connected graphs. Let G be a simple connected planar graph with n 3. Consider any planar embedding G of G. Because G is simple and connected, on at least three vertices, d(f) 3 for all f F ( G). Therefore, by Theorem and Euler s Formula (10.2) 2m = d(f) 3f( G) =3(m n + 2) (10.3) f F ( G) or, equivalently, m 3n 6 (10.4) Equality holds in (10.4) if and only if it holds in (10.3), that is, if and only if d(f) =3foreachf F ( G). Corollary Every simple planar graph has a vertex of degree at most five. Proof This is trivial for n<3. If n 3, then by Theorem 1.1 and Corollary 10.21, δn d(v) =2m 6n 12 v V It follows that δ 5. We have already seen that K 5 and K 3,3 are nonplanar (Theorem 10.2 and Exercise b). Here, we derive these two basic facts from Euler s Formula (10.2). Corollary K 5 is nonplanar. Proof If K 5 were planar, Corollary would give 10 = e(k 5 ) 3v(K 5 ) 6=9 Thus K 5 must be nonplanar. Corollary K 3,3 is nonplanar.

23 10.3 Euler s Formula 261 Proof Suppose that K 3,3 is planar and let G be a planar embedding of K 3,3. Because K 3,3 has no cycle of length less than four, every face of G has degree at least four. Therefore, by Theorem 10.10, we have 4f(G) f F d(f) =2e(G) =18 implying that f(g) 4. Euler s Formula (10.2) now implies that 2=v(G) e(g)+f(g) 6 9+4=1 which is absurd. Exercises Show that the crossing number satisfies the inequality cr(g) m 3n+6, provided that n a) Let G be a connected planar graph with girth k, where k 3. Show that m k(n 2)/(k 2). b) Deduce that the Petersen graph is nonplanar Deduce Euler s Formula (10.2) from Exercise a) Show that the complement of a simple planar graph on at least eleven vertices is nonplanar. b) Find a simple planar graph on eight vertices whose complement is planar A plane graph is face-regular if all of its faces have the same degree. a) Characterize the plane graphs which are both regular and face-regular. b) Show that exactly five of these graphs are simple and 3-connected. (They are the platonic graphs.) The thickness θ(g) of a graph G is the minimum number of planar graphs whose union is G. (Thus θ(g) = 1 if and only if G is planar.) a) Let G be a simple graph. Show that θ(g) m/(3n 6). b) Deduce that θ(k n ) (n +1)/6 + 1 and show, using Exercise b, that equality holds for all n 8. (Beineke and Harary (1965) proved that equality holds for all n 9, 10; Battle et al. (1962) showed that θ(k 9 ) = 3.)

24 Planar Graphs c) Express the Turán graph T 6,12 (defined in Exercise ) as the union of two graphs, each isomorphic to the icosahedron. d) Deduce from (b) and (c) that θ(k 12 )= a) Let G be a simple bipartite graph. Show that θ(g) m/(2n 4). b) Deduce that θ(k m,n ) mn/(2m +2n 4). (Beineke et al. (1964) showed that equality holds if mn is even. It is conjectured that equality holds in all cases.) A plane graph is self-dual if it is isomorphic to its dual. a) Show that: i) if G is self-dual, then e(g) =2v(G) 2, ii) the four plane graphs shown in Figure are self-dual. b) Find four infinite families of self-dual plane graphs of which those four graphs are members. (Smith and Tutte (1950) proved that every self-dual plane graph belongs to one of four infinite families.) Fig Self-dual plane graphs a) Let S be a set of n points in the plane, where n 3 and the distance between any two points of S is at least one. Show that no more than 3n 6 pairs of points of S can be at distance exactly one. (P. Erdős) b) By considering the triangular lattice (shown in Figure 1.27) find, for each positive integer k, asets of 3k 2 +3k + 1 points in the plane such that the distance between any two points of S is at least one, and such that 9k 2 +3k pairs of points of S are at distance exactly one The Sylvester Gallai Theorem a) Let L be a finite set of lines in the plane, no two of which are parallel and not all of which are concurrent. Using Euler s Formula (10.2), show that some point is the point of intersection of precisely two lines of L. b) Deduce from (a) the Sylvester Gallai Theorem: if S is a finite set of points in the plane, not all of which are collinear, there is a line that contains precisely two points of S. (E. Melchior)

25 10.4 Bridges Bridges In the study of planar graphs, certain subgraphs, called bridges, play an important role. We now define these subgraphs and discuss their properties. Let H be a proper subgraph of a connected graph G. The set E(G) \ E(H) may be partitioned into classes as follows. For each component F of G V (H), there is a class consisting of the edges of F together with the edges linking F to H. Each remaining edge e (that is, one which has both ends in V (H)) defines a singleton class {e}. The subgraphs of G induced by these classes are the bridges of H in G. It follows immediately from this definition that bridges of H can intersect only in vertices of H, and that any two vertices of a bridge of H are connected by a path in the bridge that is internally disjoint from H. ForabridgeB of H, the elements of V (B) V (H) are called its vertices of attachment to H; the remaining vertices of B are its internal vertices. A bridge is trivial if it has no internal vertices (that is, if it is of the second type). In a connected graph, every bridge has at least one vertex of attachment; moreover, in a nonseparable graph, every bridge has at least two vertices of attachment. A bridge with k vertices of attachment is called a k-bridge. Two bridges with the same vertices of attachment are equivalent bridges. Figure shows a variety of bridges of a cycle in a graph; edges of different bridges are distinguished by different kinds of lines. Bridges B 1 and B 2 are equivalent 3-bridges; B 3 and B 6 are trivial bridges. B 1 B 2 B 5 B 6 B 3 B 4 Fig Bridges of a cycle

26 Planar Graphs Bridges of Cycles We are concerned here with bridges of cycles, and all bridges are understood to be bridges of a given cycle C. Thus, to avoid repetition, we abbreviate bridge of C to bridge in the coming discussion. The vertices of attachment of a k-bridge B with k 2 effect a partition of C into k edge-disjoint paths, called the segments of B. Two bridges avoid each other if all the vertices of attachment of one bridge lie in a single segment of the other bridge; otherwise, they overlap. In Figure 10.15, B 2 and B 3 avoid each other, whereas B 1 and B 2 overlap, as do B 3 and B 4. Two bridges B and B are skew if there are distinct vertices of attachment u, v of B, andu,v of B, which occur in the cyclic order u, u,v,v on C. In Figure 10.15, B 3 and B 4 are skew, whereas B 1 and B 2 are not. Theorem Overlapping bridges are either skew or else equivalent 3-bridges. Proof Suppose that bridges B and B overlap. Clearly, each must have at least two vertices of attachment. If either B or B is a 2-bridge, it is easily verified that they must be skew. We may therefore assume that both B and B have at least three vertices of attachment. If B and B are not equivalent bridges, then B has a vertex u of attachment between two consecutive vertices of attachment u and v of B. Because B and B overlap, some vertex of attachment v of B does not lie in the segment of B connecting u and v. It follows that B and B are skew. If B and B are equivalent k-bridges, then k 3. If k 4, B and B are skew; if k = 3, they are equivalent 3-bridges. B 4 B 3 B 1 B 2 Fig Bridgesofacycleinaplanegraph We now consider bridges of cycles in plane graphs. Suppose that G is a plane graph and that C is a cycle in G. Because C is a simple closed curve in the plane, each bridge of C in G is contained in one of the two regions Int(C) orext(c). A

27 10.4 Bridges 265 bridge contained in Int(C) is called an inner bridge, a bridge contained in Ext(C) an outer bridge. In Figure 10.16, B 1 and B 2 are inner bridges, and B 3 and B 4 are outer bridges. Theorem Inner (outer) bridges avoid one another. Proof Let B and B be inner bridges of a cycle C in a plane graph G. Suppose that they overlap. By Theorem 10.25, they are either skew or equivalent 3-bridges. In both cases, we obtain contradictions. Case 1: B and B are skew. By definition, there exist distinct vertices u, v in B and u,v in B, appearing in the cyclic order u, u,v,v on C. LetuP v be a path in B and u P v a path in B, both internally disjoint from C. Consider the subgraph H := C P P of G (see Figure 10.17a). Because G is plane, so is H. LetK be the plane graph obtained from H by adding a vertex in ext(c) and joining it to u, u,v,v (see Figure 10.17b). Then K is a subdivision of K 5. But this is impossible, K 5 being nonplanar. u P u P v v (a) (b) Fig Proof of Theorem 10.26, Case 1: (a) the subgraph H, (b) the subdivision K of K 5 Case 2: B and B are equivalent 3-bridges. Denote by S := {v 1,v 2,v 3 } their common set of vertices of attachment. By Exercise 9.2.3, there exists a (v, S)-fan F in B, for some internal vertex v of B; likewise, there exists a (v,s)-fan F in B, for some internal vertex v of B. Consider the subgraph H := F F of G. Because G is plane, so is H. LetK be the plane graph obtained from H by adding a vertex in ext(c) and joining it to the three vertices of S. Then K is a subdivision of K 3,3. But this is impossible, because K 3,3 is nonplanar (see Figure 10.18). We conclude that inner bridges avoid one another. Similarly, outer bridges avoid one another. It is convenient to visualize the above theorem in terms of the bridge-overlap graph. Let G be a graph and let C be a cycle of G. Thebridge-overlap graph of C is the graph whose vertex set is the set of all bridges of C in G, two bridges being

28 Planar Graphs v 3 v v 1 v v 2 (a) (b) Fig Proof of Theorem 10.26, Case 2: (a) the subgraph H, (b) the subdivision K of K 3,3 adjacent if they overlap. Theorem simply states that the bridge-overlap graph of any cycle of a plane graph is bipartite. Thus, a necessary condition for a graph to be planar is that the bridge-overlap graph of each of its cycles be bipartite. This condition also suffices to guarantee planarity (Exercise ). Unique Plane Embeddings Just as there is no unique way of representing graphs by diagrams, there is no unique way of embedding planar graphs in the plane. Apart from the positions of points representing vertices and the shapes of lines representing the edges, two different planar embeddings of the same planar graph may differ in the incidence relationships between their edge and face sets; they may even have different facedegree sequences, as in Figure We say that two planar embeddings of a planar graph G are equivalent if their face boundaries (regarded as sets of edges) are identical. A planar graph for which any two planar embeddings are equivalent is said have an unique embedding in the plane. Using the theory of bridges developed above, we show that every simple 3-connected planar graph is uniquely embeddable in the plane; note that the graph of Figure is not 3-connected. The notion of a nonseparating cycle plays a crucial role in this proof. A cycle is nonseparating if it has no chords and at most one nontrivial bridge. Thus, in a loopless graph G which is not itself a cycle, a cycle C is nonseparating if and only if it is an induced subgraph of G and G V (C) is connected. In the case of simple 3-connected plane graphs, Tutte (1963) proved that facial and nonseparating cycles are one and the same. Theorem A cycle in a simple 3-connected plane graph is a facial cycle if and only if it is nonseparating. Proof Let G be a simple 3-connected plane graph and let C be a cycle of G. Suppose, first, that C is not a facial cycle of G. Then C has at least one inner

29 10.4 Bridges 267 bridge and at least one outer bridge. Because G is simple and connected, these bridges are not loops. Thus either they are both nontrivial or at least one of them is a chord. It follows that C is not a nonseparating cycle. Now suppose that C is a facial cycle of G. By Proposition 10.5, we may assume that C bounds the outer face of G, so all its bridges are inner bridges. By Theorem 10.26, these bridges avoid one another. If C had a chord xy, the set {x, y} would be a vertex cut separating the internal vertices of the two xy-segments of C. Likewise, if C had two nontrivial bridges, the vertices of attachment of one of these bridges would all lie on a single xy-segment of the other bridge, and {x, y} would be a vertex cut of G separating the internal vertices of the two bridges. In either case, the 3-connectedness of G would be contradicted. Thus C is nonseparating. A direct consequence of Theorem is the following fundamental theorem, due to Whitney (1933). Theorem Every simple 3-connected planar graph has a unique planar embedding. Proof Let G be a simple 3-connected planar graph. By Theorem 10.27, the facial cycles in any planar embedding of G are precisely its nonseparating cycles. Because the latter are defined solely in terms of the abstract structure of the graph, they are the same for every planar embedding of G. The following corollary is immediate. Corollary Every simple 3-connected planar graph has a unique dual graph. Exercises Let G 1 and G 2 be planar graphs whose intersection is isomorphic to K 2. Show that G 1 G 2 is planar Let H be a subgraph of a graph G. Consider the binary relation on E(G) \ E(H), where e 1 e 2 if there exists a walk W in G such that: the first and the last edges of W are e 1 and e 2, respectively, W is internally disjoint from H (that is, no internal vertex of W is a vertex of H). Show that: a) the relation is an equivalence relation on E(G) \ E(H), b) the subgraphs of G\ E(H) induced by the equivalence classes under this equivalence relation are the bridges of H in G.

30 Planar Graphs A3-polytope is the convex hull of a set of points in R 3 which do not lie on a common plane. Show that the polyhedral graph of such a polytope is simple, planar, and 3-connected. (Steinitz (1922) proved that, conversely, every simple 3-connected planar graph is the polyhedral graph of some 3-polytope.) Show that any 3-connected cubic plane graph on n vertices, where n 6, may be obtained from one on n 2 vertices by subdividing two edges in the boundary of a face and joining the resulting new vertices by an edge subdividing the face A rooting of a plane graph G is a triple (v, e, f), where v is a vertex, called the root vertex, e is an edge of G incident with v, called the root edge, andf is a face incident with e, called the root face. a) Show that the only automorphism of a simple 3-connected plane graph which fixes a given rooting is the identity automorphism. b) Let G be a simple 3-connected planar graph. Deduce from (a) that: i) aut(g) divides 4m, ii) aut(g) =4m if and only if G is one of the five platonic graphs. (F. Harary and W.T. Tutte; L. Weinberg) 10.5 Kuratowski s Theorem Planarity being such a fundamental property, the problem of deciding whether a given graph is planar is clearly of great importance. A major step towards this goal is provided by the following characterization of planar graphs, due to Kuratowski (1930). Theorem Kuratowski s Theorem A graph is planar if and only if it contains no subdivision of either K 5 or K 3,3. A subdivision of K 5 or K 3,3 is consequently called a Kuratowski subdivision. We first present a proof of Kuratowski s Theorem due to Thomassen (1981), and then explain how it gives rise to a polynomial-time decision algorithm for planarity. Before proving the theorem, we reformulate it in terms of minors. Minors A minor of a graph G is any graph obtainable from G by means of a sequence of vertex and edge deletions and edge contractions. Alternatively, consider a partition (V 0,V 1,...,V k )ofv such that G[V i ] is connected, 1 i k, and let H be the graph obtained from G by deleting V 0 and shrinking each induced subgraph G[V i ], 1 i k, to a single vertex. Then any spanning subgraph F of H is a minor of G. For instance, K 5 is a minor of the Petersen graph because it can be obtained by contracting the five spoke edges of the latter graph (see Figure 10.19).

31 10.5 Kuratowski s Theorem 269 Fig Contracting the Petersen graph to K 5 If F is a minor of G, we write F G. ByanF -minor of G, where F is an arbitrary graph, we mean a minor of G which is isomorphic to F. It is important to point out that any graph which contains an F -subdivision also has an F -minor: to obtain F as a minor, one simply deletes the vertices and edges not in the subdivision, and then contracts each subdivided edge to a single edge. For example, because the Petersen graph contains a K 3,3 -subdivision (Exercise ), it also has a K 3,3 -minor. Conversely, provided that F is a graph of maximum degree three or less, any graph which has an F -minor also contains an F -subdivision (Exercise a). Wagner s Theorem As observed in Section 10.2, the deletion or contraction of an edge in a planar graph results again in a planar graph. Thus we have: Proposition Minors of planar graphs are planar. A minor which is isomorphic to K 5 or K 3,3 is called a Kuratowski minor. Because K 5 and K 3,3 are nonplanar, Proposition implies that any graph which has a Kuratowski minor is nonplanar. Wagner (1937) proved that the converse is true. Theorem Wagner s Theorem A graph is planar if and only if it has no Kuratowski minor. We remarked above that a graph which contains an F -subdivision also has an F -minor. Thus Kuratowski s Theorem implies Wagner s Theorem. On the other hand, because K 3,3 has maximum degree three, any graph which has a K 3,3 -minor contains a K 3,3 -subdivision (Exercise a). Furthermore, any graph which has a K 5 -minor necessarily contains a Kuratowski subdivision (Exercise b). Thus Wagner s Theorem implies Kuratowski s Theorem, and the two are therefore equivalent. It turns out to be slightly more convenient to prove Wagner s variant of Kuratowski s Theorem. Before doing so, we need to establish two simple lemmas.