Lecture Notes on GRAPH THEORY Tero Harju

Transcription

1 Lecture Notes on GRAPH THEORY Tero Harju Department of Mathematics University of Turku FIN Turku, Finland

2 Contents 1 Introduction Graphs and their plane figures Subgraphs Paths and cycles Connectivity of Graphs Bipartite graphs and trees Connectivity Tours and Matchings Eulerian graphs Hamiltonian graphs Matchings Colourings Edge colourings Ramsey Theory Vertex colourings Graphs on Surfaces Planar graphs Colouring planar graphs Genus of a graph Directed Graphs Digraphs Network Flows Index 96

3 1 Introduction Graph theory can be said to have its beginning in 1736 when EULER considered the (general case of the) Königsberg bridge problem: Is there a walking route that crosses each of the seven bridges of Königsberg exactly once? (Solutio Problematis ad geometriam situs pertinentis, Commentarii Academiae Scientiarum Imperialis Petropolitanae 8 (1736), pp ) It took 200 years before the first book on graph theory was written. This was done by KÖNIG in ( Theorie der endlichen und unendlichen Graphen, Teubner, Leipzig, Translation in English, 1990.) Since then graph theory has developed into an extensive and popular branch of mathematics, which has been applied to many problems in mathematics, computer science, and other scientific and not-so-scientific areas. For the history of early graph theory, see N.L. BIGGS, R.J. LLOYD AND R.J. WILSON, Graph Theory , Clarendon Press, There seem to be no standard notations or even definitions for graph theoretical objects. This is natural, because the names one uses for these objects reflect the applications. So, for instance, if we consider a communications network (say, for ) as a graph, then the computers, which take part in this network, are called nodes rather than vertices or points. On the other hand, other names are used for molecular structures in chemistry, flow charts in programming, human relations in social sciences, and so on. These lectures study finite graphs and majority of the topics is included in J.A. BONDY AND U.S.R. MURTY, Graph Theory with Applications, Macmillan, R. DIESTEL, Graph Theory, Springer-Verlag, F. HARARY, Graph Theory, Addison-Wesley, D.B. WEST, Introduction to Graph Theory, Prentice Hall, R.J. WILSON, Introduction to Graph Theory, Longman, (3rd ed.) In these lectures we study combinatorial aspects of graphs. For more algebraic topics and methods, see N. BIGGS, Algebraic Graph Theory, Cambridge University Press, (2nd ed.) and for computational aspects, see S. EVEN, Graph Algorithms, Computer Science Press, 1979.

4 In these lecture notes we mention several open problems that have gained respect among the researchers. Indeed, graph theory has the advantage that it contains easily formulated open problems that can be stated early in the theory. Finding a solution to any one of these problems is on another layer of difficulty. Sections with a star ( ) in their heading are optional. Notations and notions For a finite set, denotes its size (cardinality, the number of its elements). Let ½ Ò ½ ¾ Ò and in general, Ò ½ Ò for integers Ò. For a real number Ü, the floor and the ceiling of Ü are the integers Ü ÑÜ ¾ Ü and Ü ÑÒ ¾ Ü A family ½ ¾ of subsets of a set is a partition of, if ¾½ For two sets and, is their Cartesian product. For two sets and, and for all different and Ü Ýµ Ü ¾ Ý ¾ Ò µ Ò µ is their symmetric difference. Here Ò Ü Ü ¾ Ü ¾. Two numbers Ò ¾ Æ (often Ò and for sets and ) have the same parity, if both are even, or both are odd, that is, if Ò ÑÓ ¾µ. Otherwise, they have opposite parity. 3 Graph theory has abundant examples of NP-complete problems. Intuitively, a problem is in P 1 if there is an efficient (practical) algorithm to find a solution to it. On the other hand, a problem is in NP 2, if it is first efficient to guess a solution and then efficient to check that this solution is correct. It is conjectured (and not known) that P NP. This is one of the great problems in modern mathematics and theoretical computer science. If the guessing in NP-problems can be replaced by an efficient systematic search for a solution, then PNP. For any one NP-complete problem, if it is in P, then necessarily PNP. ½ Solvable by an algorithm in polynomially many steps on the size of the problem instances. ¾ Solvable nondeterministically in polynomially many steps on the size of the problem instances.

5 1.1 Graphs and their plane figures Graphs and their plane figures Let Î be a finite set, and denote by Î µ Ù Ú Ù Ú ¾ Î Ù Ú the subsets of Î of two distinct elements. DEFINITION. A pair Î µ with Î µ is called a graph (on Î ). The elements of Î are the vertices, and those of the edges of the graph. The vertex set of a graph is denoted by Î and its edge set by. Therefore Î µ. In literature, graphs are also called simple graphs; vertices are called nodes or points; edges are called lines or links. The list of alternatives is long (but still finite). A pair Ù Ú is usually written simply as ÙÚ. Notice that then ÙÚ ÚÙ. In order to simplify notations, we also write Ú ¾ instead of Ú ¾ Î. DEFINITION. For a graph, we denote Î and The number of the vertices is called the order of, and is the size of. For an edge ÙÚ ¾, the vertices Ù and Ú are its ends. Vertices Ù and Ú are adjacent or neighbours, if ÙÚ ¾. Two edges ½ ÙÚ and ¾ ÙÛ having a common end, are adjacent with each other. A graph can be represented as a plane figure by drawing a line (or a curve) between the points Ù and Ú (representing vertices) if ÙÚ is an edge of. The figure on the right is a drawing of the graph with Î Ú ½ Ú ¾ Ú Ú Ú Ú and Ú ½ Ú ¾ Ú ½ Ú Ú ¾ Ú Ú ¾ Ú Ú Ú. Ú ½ Ú ¾ Ú Ú Ú Ú Often we shall omit the identities (names Ú) of the vertices in our figures, in which case the vertices are drawn as anonymous circles. Graphs can be generalized by allowing loops ÚÚ and parallel (or multiple) edges between vertices to obtain a multigraph Î µ, where ½ ¾ Ñ is a set (of symbols), and Î µ ÚÚ Ú ¾ Î is a function that attaches an unordered pair of vertices to each ¾ : µ ÙÚ. Note that we can have ½ µ ¾ µ. This is drawn in the figure of by placing two (parallel) edges that connect the common ends. On the right there is (a drawing of) a multigraph with vertices Î and edges ½ µ, ¾ µ, µ, and µ.

6 1.1 Graphs and their plane figures 5 Later we concentrate on (simple) graphs. DEFINITION. We also study directed graphs or digraphs Î µ, where the edges have a direction, that is, the edges are ordered: Î Î. In this case, ÙÚ ÚÙ. The directed graphs have representations, where the edges are drawn as arrows. A digraph can contain edges ÙÚ and ÚÙ of opposite directions. Graphs and digraphs can also be coloured, labelled, and weighted: DEFINITION. A function «Î à is a vertex colouring of by a set à of colours. A function «Ã is an edge colouring of. Usually, à ½ for some ½. If Ã Ê (often à Æ), then «is a weight function or a distance function. Isomorphism of graphs DEFINITION. Two graphs and À are isomorphic, denoted by À, if there exists a bijection «Î Î À such that ÙÚ ¾ µ «Ùµ«Úµ ¾ À for all Ù Ú ¾. Hence and À are isomorphic if the vertices of À are renamings of those of. Two isomorphic graphs enjoy the same graph theoretical properties, and they are often identified. In particular, all isomorphic graphs have the same plane figures (excepting the identities of the vertices). This shows in the figures, where we tend to replace the vertices by small circles, and talk of the graph although there are, in fact, infinitely many of such graphs. Example 1.1. The following graphs are isomorphic. Indeed, the required isomorphism is given by Ú ½ ½, Ú ¾, Ú, Ú ¾, Ú. Ú ¾ Ú Ú Ú ½ Ú ½ ¾ Isomorphism Problem. Does there exist an efficient algorithm to check whether any two given graphs are isomorphic or not? The following table lists the number ¾ Ò ¾ µ of graphs on a given set of Ò vertices, and the number of nonisomorphic graphs on Ò vertices. It tells that at least for computational purposes an efficient algorithm for checking whether two graphs are isomorphic or not would be greatly appreciated.

7 1.1 Graphs and their plane figures 6 Ò graphs ½ ¾ ½¼¾ ¾ ¾ ¼ ½¾ ¾ ¾ ½¼ ½¼ nonisomorphic ½ ¾ ½½ ½ ½¼ ½¾ ¾ Other representations Plane figures catch graphs for our eyes, but if a problem on graphs is to be programmed, then these figures are (to say the least) unsuitable. Matrices of integers are ideal for computers, since every respectable programming language has array structures for these, and computers are good in crunching numbers. Let Î Ú ½ Ú Ò be ordered. The adjacency matrix of is the Ò Ò-matrix Å with entries Å ½ or Å ¼ according to whether Ú Ú ¾ or not. For instance, the graphs of Example 1.1 has an adjacency matrix on the right. Notice that the adjacency matrix is always symmetric (with respect to its diagonal consisting of zeros). ¼ ¼ ½ ½ ¼ ½ ½ ¼ ¼ ½ ½ ½ ¼ ¼ ½ ¼ ¼ ½ ½ ¼ ¼ ½ ½ ¼ ¼ ¼ ½ A graph has usually many different adjacency matrices, one for each ordering of its set Î of vertices. The following result is obvious from the definitions. Theorem 1.1. Two graphs and À are isomorphic if and only if they have a common adjacency matrix. Moreover, two isomorphic graphs have exactly the same set of adjacency matrices. Graphs can also be represented by sets. For this, let ½ ¾ Ò be a family of subsets of a set, and define the intersection graph as the graph with vertices ½ Ò, and edges for all and ( ) with. Theorem 1.2. Every graph is an intersection graph of some family of subsets. Proof. Let be a graph, and define, for all Ú ¾, a set Ú Ú Ù ÚÙ ¾ Then Ù Ú if and only if ÙÚ ¾. Let µ be the smallest size of a base set such that can be represented as an intersection graph of a family of subsets of, that is, µ ÑÒ for some ¾ How small can µ be compared to the order (or the size ) of the graph? It was shown by KOU, STOCKMEYER AND WONG (1976) that it is algorithmically difficult to determine the number µ the problem is NP-complete.

8 1.2 Subgraphs 7 Example 1.2. As yet another example, let Æ be a finite set of natural numbers, and let µ be the graph defined on Î such that Ö ¾ µ if and only if Ö and (for Ö ) have a common divisor ½. As an exercise, we state: All graphs can be represented in the form for some set of natural numbers. 1.2 Subgraphs Ideally, in a problem the local properties of a graph determine a solution. In such a situation we deal with (small) parts of the graph (subgraphs), and a solution can be found to the problem by combining the information determined by the parts. For instance, as we shall see later on, the existence of an Euler tour is very local, it depends only on the number of the neighbours of the vertices. Degrees of vertices DEFINITION. Let Ú ¾ be a vertex a graph. The neighbourhood of Ú is the set Æ Úµ Ù ¾ ÚÙ ¾ The degree of Ú is the number of its neighbours: Úµ Æ Úµ If Úµ ¼, then Ú is said to be isolated in, and if Úµ ½, then Ú is a leaf of the graph. The minimum degree and the maximum degree of are defined as Æ µ ÑÒ Úµ Ú ¾ and µ ÑÜ Úµ Ú ¾ The following lemma, due to EULER (1736), tells that if several people shake hands, then the number of hands shaken is even. Lemma 1.1 (Handshaking lemma). For each graph, Ú¾ Úµ ¾ Moreover, the number of vertices of odd degree is even. Proof. Every edge ¾ has two ends. The second claim follows immediately from the first one. Lemma 1.1 holds equally well for multigraphs, when Úµ is defined as the number of edges that have Ú as an end, and when a loop ÚÚ is counted twice. Note that the degrees of a graph do not determine. Indeed, there are graphs Î µ and À Î À µ on the same set of vertices that are not isomorphic, but for which Úµ À Úµ for all Ú ¾ Î.

9 1.2 Subgraphs 8 DEFINITION. Let be a graph. A ¾-switch Ù Ú Ü Ýµ of, for ÙÚ ÜÝ ¾ and ÙÜ ÚÝ ¾, replaces the edges ÙÚ and ÜÝ by ÙÜ and ÚÝ. Ú Ù Ý Ü Ú Ù Ý Ü Before proving Berge s switching theorem we need the following tool. Lemma 1.2. Let be a graph of order Ò with a degree sequence ½ ¾ Ò, where Ú µ. There is a graph ¼ which is obtained from by a sequence of ¾-switches such that Æ ¼ Ú ½ µ Ú ¾ Ú ½ ½. Proof. Denote µ ½ µ. Suppose that there exists a vertex Ú with ¾ ½ such that Ú ½ Ú ¾. Since Ú ½ µ, there exists a Ú with ¾ such that Ú ½ Ú ¾. Here, since. Since Ú ½ Ú ¾, there exists a Ú Ø (¾ Ø Ò) Ú ½ Ú Ú such that Ú Ú Ø ¾, but Ú Ú Ø ¾. We can now perform a ¾-switch with respect to the vertices Ú ½ Ú Ú Ú Ø. This Ú Ø gives a new graph À, where Ú ½ Ú ¾ À and Ú ½ Ú ¾ À, and the other neighbours of Ú ½ remain to be its neighbours. When we repeat this process for all indices with Ú ½ Ú ¾ for ¾ ½, we obtain a graph ¼ as in the claim. Theorem 1.3 (BERGE (1973)). Two graphs and À on a common vertex set Î satisfy Úµ À Úµ for all Ú ¾ Î if and only if À can be obtained from by a sequence of ¾-switches. Proof. If a graph À is obtained from by a ¾-switch, then clearly À has the same degrees as. In the other direction, we use induction on the order. Let and À have the same degrees, and let µ. By Lemma 1.2, there are sequences of ¾-switches that transform to ¼ and À to À ¼ such that Æ ¼ Ú ½ µ Ú ¾ Ú ½ Æ À ¼ Ú ½ µ. Now the graphs ¼ Ú ½ and À ¼ Ú ½ have the same degrees. By induction hypothesis, ¼, and thus also, can be transformed to À ¼ by a sequence of ¾-switches. Finally, we observe that À ¼ can be transformed to À by the inverse sequence of ¾-switches, and this proves the claim. DEFINITION. Let ½ ¾ Ò be a descending sequence of nonnegative integers, that is, ½ ¾ Ò. Such a sequence is said to be graphical, if there exists a graph Î µ with Î Ú ½ Ú ¾ Ú Ò such that Ú µ for all. Using the next result recursively one can decide whether a sequence of integers is graphical or not.

10 1.2 Subgraphs 9 Theorem 1.4 (HAVEL (1955), HAKIMI (1962)). A sequence ½ ¾ Ò (with ½ ½ and Ò ¾) is graphical if and only if ¾ ½ ½ ½ ½ ½ ½ ¾ ½ Ò (1.1) is graphical (when put into nonincreasing order). Proof. ( ) Consider of order Ò ½ with vertices (and degrees) Ú ¾ µ ¾ ½ Ú ½ ½µ ½ ½ ½ Ú ½ ¾µ ½ ¾ Ú Ò µ Ò as in (1.1). Add a new vertex Ú ½ and the edges Ú ½ Ú for all ¾ ¾ ½ ½. Then in this new graph À, À Ú ½ µ ½, and À Ú µ for all. (µ) Assume Ú µ. By Lemma 1.2 and Theorem 1.3, we can suppose that Æ Ú ½ µ Ú ¾ Ú ½ ½. But now the degree sequence of Ú ½ is in (1.1). Example 1.3. Consider the sequence ¾ ½. By Theorem 1.4, is graphical µ ¾ ½ ½ is graphical ¾ ½ ½ ¼ is graphical ¼ ¼ ¼ is graphical Ú ¾ Ú Ú The last sequence corresponds to a discrete graph Ã, and hence also our original sequence is graphical. Indeed, the graph on the right has this degree sequence. Ú ½ Ú Ú Special graphs DEFINITION. A graph Î µ is trivial, if it has only one vertex, i.e., ½; otherwise is nontrivial. The graph à Πis the complete graph on Î, if every two vertices are adjacent: Î µ. All complete graphs of order Ò are isomorphic with each other, and they will be denoted by à Ò. à The complement of is the graph on Î, where ¾ Î µ ¾. The complements à Πof the complete graphs are called discrete graphs. In a discrete graph. Clearly, all discrete graphs of order Ò are isomorphic with each other. A graph is said to be regular, if every vertex of has the same degree. If this degree is equal to Ö, then is Ö-regular or regular of degree Ö.

11 1.2 Subgraphs 10 Note that a discrete graph is 0-regular, and a complete graph Ã Ò is Ò ½µ-regular. In particular, ÃÒ Ò Ò ½µ¾, and therefore Ò Ò ½µ¾ for all graphs that have order Ò. Example 1.4. The graph on the right is the Petersen graph that we will meet several times (drawn differently). It is a -regular graph of order ½¼. Example 1.5. Let ½ be an integer, and consider the set of all binary strings of length. For instance, ¼¼¼ ¼¼½ ¼½¼ ½¼¼ ¼½½ ½¼½ ½½¼ ½½½. Let É be the graph, called the -cube, with Î É, where ÙÚ ¾ É if and only if the strings Ù and Ú differ in exactly one place. The order of É is É ¾, the number of binary strings of length. Also, É is -regular, and so, by the handshaking lemma, É ¾ ½. On the right we have the -cube, or simply the cube Example 1.6. Let Ò be any even number. We show by induction that there exists a - regular graph with Ò. Notice that all -regular graphs have even order by the handshaking lemma. If Ò, then à is -regular. Let be a -regular graph of order ¾Ñ ¾, and suppose that ÙÚ ÙÛ ¾. Let Î À Î Ü Ý, and À Ò ÙÚ ÙÛµ ÙÜ ÜÚ ÙÝ ÝÛ ÜÝ Then À is -regular of order ¾Ñ. Ü Û Ù Ý Ú Subgraphs DEFINITION. A graph À is a subgraph of a graph, denoted by À, if Î À Î and À. A subgraph À spans (and À is a spanning subgraph of ), if every vertex of is in À, i.e., Î À Î. Also, a subgraph À is an induced subgraph, if À Î À µ. In this case, À is induced by its set Î À of vertices. In an induced subgraph À, the set À of edges consists of all ¾ such that ¾ Î À µ. To each nonempty subset Î, there corresponds a unique induced subgraph µµ

12 1.3 Paths and cycles 11 To each subset of edges there corresponds a unique spanning subgraph of, Î µ subgraph spanning induced For a set of edges, let Ò be the subgraph of obtained by removing (only) the edges ¾ from. In particular, is obtained from by removing ¾. Similarly, we write, if each ¾ (for Î µµ is added to. For a subset Î of vertices, we let be the subgraph induced by Î Ò, that is, Î Ò and, e.g., Ú is obtained from by removing the vertex Ú together with the edges that have Ú as their end. Many problems concerning (induced) subgraphs are algorithmically difficult. For instance, to find a maximal complete subgraph (a subgraph Ã Ñ of maximum order) of a graph is unlikely to be even in NP. Reconstruction Problem. The famous open problem, Kelly-Ulam problem or the Reconstruction Conjecture, states that a graph of order at least is determined up to isomorphism by its vertex deleted subgraphs Ú (Ú ¾ ): if there exists a bijection «Î Î À such that Ú À «Úµ for all Ú, then À. 1.3 Paths and cycles The most fundamental notions in graph theory are practically oriented. Indeed, many graph theoretical questions ask for optimal solutions to problems such as: find a shortest path (in a complex network) from a given point to another. This kind of problems can be difficult, or at least nontrivial, because there are usually choices what branch to choose when leaving an intermediate point.

13 1.3 Paths and cycles 12 Walks DEFINITION. Let Ù Ù ½ ¾ be edges of for ¾ ½. Here and ½ are compatible in the sense that is adjacent to ½ for all ¾ ½ ½. The sequence is a walk of length from Ù ½ to Ù ½. We write, more informally, Ï ½ ¾ Ï Ù ½ Ù ¾ Ù Ù ½ or Ï Ù ½ Ù ½ Write Ù Ú to say that there is a walk of some length from Ù to Ú. Here we understand that Ï Ù Ú is always a specific walk, Ï ½ ¾, although we sometimes do not care to mention the edges it uses. The length of a walk Ï is denoted by Ï. DEFINITION. Let Ï ½ ¾ ( Ù Ù ½ ) be a walk. Ï is closed, if Ù ½ Ù ½. Ï is a path, if Ù Ù for all. Ï is a cycle, if it is closed, and Ù Ù for except that Ù ½ Ù ½. Ï is a trivial path, if its length is 0. A trivial path has no edges. For a walk Ï Ù Ù ½ Ù ½ Ú, also Ï ½ Ú Ù ½ Ù ½ Ù is a walk in, called the inverse walk of Ï. A vertex Ù is an end of a path È, if È starts or ends in Ù. The join of two walks Ï ½ Ù Ú and Ï ¾ Ú Û is the walk Ï ½ Ï ¾ Ù Û. (Here the end Ú must be common to the walks.) Paths È and É are disjoint, if they have no vertices in common, and they are independent, if they can share only their ends. Clearly, the inverse walk È two paths need not be a path. ½ of a path È is a path (the inverse path of È ). The join of A (sub)graph, which is a path (cycle) of length ½ (, resp.) having vertices is denoted by È (, resp.). If is even (odd), we say that the path or cycle is even (odd). Clearly, all paths of length are isomorphic. The same holds for cycles of fixed length. È Lemma 1.3. Each walk Ï Ù Ú with Ù Ú contains a path È Ù Ú, that is, there is a path È Ù Ú that is obtained from Ï by removing edges and vertices.

14 1.3 Paths and cycles 13 Proof. Let Ï Ù Ù ½ Ù ½ Ú. Let be indices such that Ù Ù. If no such and exist, then Ï, itself, is a path. Otherwise, in Ï Ï ½ Ï ¾ Ï Ù Ù Ù Ú the portion Í ½ Ï ½ Ï Ù Ù Ù Ú is a shorter walk. By repeating this argument, we obtain a sequence Í ½ Í ¾ Í Ñ of walks Ù Ú with Ï Í ½ Í Ñ. When the procedure stops, we have a path as required. (Notice that in the above it may very well be that Ï ½ or Ï is a trivial walk.) DEFINITION. If there exists a walk (and hence a path) from Ù to Ú in, let Ù Úµ ÑÒ Ù Ú be the distance between Ù and Ú. If there are no walks Ù Ú, let Ù Úµ ½ by convention. A graph is connected, if Ù Úµ ½ for all Ù Ú ¾ ; otherwise, it is disconnected. The maximal connected subgraphs of are its connected components. Denote µ the number of connected components of If µ ½, then is, of course, connected. The maximality condition means that a subgraph À is a connected component if and only if À is connected and there are no edges leaving À, i.e., for every vertex Ú ¾ À, the subgraph Î À Ú is disconnected. Apparently, every connected component is an induced subgraph, and Æ Úµ Ù Ú Ùµ ½ is the connected component of that contains Ú ¾. In particular, the connected components form a partition of. Shortest paths DEFINITION. Let «be an edge weighted graph, that is, «is a graph together with a weight function «Ê on its edges. For À, let «Àµ ¾ À «µ be the (total) weight of À. In particular, if È ½ ¾ is a path, then its weight is «È µ È ½ «µ. The minimum weighted distance between two vertices is «Ù Úµ ÑÒ«È µ È Ù Ú In extremal problems we seek for optimal subgraphs À satisfying specific conditions. In practice we encounter situations where might represent a distribution or transportation network (say, for mail), where the weights on edges are distances, travel expenses, or rates of flow in the network; a system of channels in (tele)communication or computer architecture, where the weights present the rate of unreliability or frequency of action of the connections; a model of chemical bonds, where the weights measure molecular attraction.

15 1.3 Paths and cycles 14 In these examples we look for a subgraph with the smallest weight, and which connects two given vertices, or all vertices (if we want to travel around). On the other hand, if the graph represents a network of pipelines, the weights are volumes or capacities, and then one wants to find a subgraph with the maximum weight. We consider the minimum problem. For this, let be a graph with an integer weight function «Æ. In this case, call «ÙÚµ the length of ÙÚ. The shortest path problem: Given a connected graph with a weight function «Æ, «find Ù Úµ for given Ù Ú ¾. Assume that is a connected graph. Dijkstra s algorithm solves the problem for every pair Ù Ú, where Ù is a fixed starting point and Ú ¾. Let us make the convention that «ÙÚµ ½, if ÙÚ ¾. Dijkstra s algorithm: (i) Set Ù ¼ Ù, Ø Ù ¼ µ ¼ and Ø Úµ ½ for all Ú Ù ¼. (ii) For ¾ ¼ ½ : for each Ú ¾ Ù ½ Ù, replace Ø Úµ by ÑÒØ Úµ Ø Ù µ «Ù Úµ Let Ù ½ ¾ Ù ½ Ù be any vertex with the least value Ø Ù ½ µ. «(iii) Conclusion: Ù Úµ Ø Úµ. Example 1.7. Consider the following weighted graph. Apply Dijkstra s algorithm to the vertex Ú ¼. Ù ¼ Ú ¼, Ø Ù ¼ µ ¼, others are ½. Ø Ú ½ µ ÑÒ½ ¾ ¾, Ø Ú ¾ µ ÑÒ½, others are ½. Thus Ù ½ Ú ½. Ø Ú ¾ µ ÑÒ Ø Ù ½ µ «Ù ½ Ú ¾ µ ÑÒ, Ø Ú µ ¾ ½, Ø Ú µ ¾, Ø Ú µ ¾ ¾. Thus choose Ù ¾ Ú. Ø Ú ¾ µ ÑÒ ½, Ø Ú µ ÑÒ ¾, Ø Ú µ ÑÒ ½. Thus set Ù Ú ¾. Ø Ú µ ÑÒ ½, Ø Ú µ ÑÒ ½. Thus choose Ù Ú. Ø Ú µ ÑÒ ½. The algorithm stops. Ú ¼ ¾ Ú ½ Ú ¾ ¾ ½ ½ Ú Ú ¾ ¾ ¾ ½ Ú We have obtained: Ø Ú ½ µ ¾ Ø Ú ¾ µ Ø Ú µ Ø Ú µ Ø Ú µ These are the minimal weights from Ú ¼ to each Ú.

16 1.3 Paths and cycles 15 The steps of the algorithm can also be rewritten as a table: Ú ½ Ú ¾ Ú ½ Ú ½ Ú ½ The correctness of Dijkstra s algorithm can verified be as follows. Let Ú ¾ Î be any vertex, and let È Ù ¼ Ù Ú be a shortest path from Ù ¼ to Ú, where Ù is any vertex Ù Ú on such a path, possibly Ù Ù ¼. Then, clearly, the first part of the path, Ù ¼ Ù, is a shortest path from Ù ¼ to Ù, and the latter part Ù Ú is a shortest path from Ù to Ú. Therefore, the length of the path È equals the sum of the weights of Ù ¼ Ù and Ù Ú. Dijkstra s algorithm makes use of this observation iteratively.

17 2 Connectivity of Graphs 2.1 Bipartite graphs and trees In problems such as the shortest path problem we look for minimum solutions that satisfy the given requirements. The solutions in these cases are usually subgraphs without cycles. Such connected graphs will be called trees, and they are used, e.g., in search algorithms for databases. For concrete applications in this respect, see T.H. CORMEN, C.E. LEISERSON AND R.L. RIVEST, Introduction to Algorithms, MIT Press, Certain structures with operations are representable as trees. These trees are sometimes called construction trees, decomposition trees, factorization trees or grammatical trees. Grammatical trees occur especially in linguistics, where syntactic structures of sentences are analyzed. On the right there is a tree of operations for the arithmetic formula Ü Ý Þµ Ý. Ü Ý Þ Ý Bipartite graphs DEFINITION. A graph is called bipartite, if Î has a partition to two subsets and such that each edge ÙÚ ¾ connects a vertex of and a vertex of. In this case, µ is a bipartition of, and is µ-bipartite. A bipartite graph (as in the above) is a complete Ñ µbipartite graph, if Ñ,, and ÙÚ ¾ for all Ù ¾ and Ú ¾. All complete Ñ µ-bipartite graphs are isomorphic. Let Ã Ñ denote such a graph. A subset Î is stable, if is a discrete graph. The following result is clear from the definitions. à ¾ Theorem 2.1. A graph is bipartite if and only if Î has a partition to two stable subsets. Example 2.1. The -cube É of Example 1.5 is bipartite for all. Indeed, consider Ù Ù has an even number of ½ ¼ s and Ù Ù has an odd number of ½ ¼ s Clearly, these sets partition, and they are stable in É.

18 2.1 Bipartite graphs and trees 17 Theorem 2.2. A graph is bipartite if and only if it has no odd cycles. Proof. (µ) Let be µ-bipartite. For a cycle Ú ½ Ú ½ Ú ½ of length, Ú ½ ¾ implies Ú ¾ ¾, Ú ¾,..., Ú ¾ ¾, Ú ¾ ½ ¾. Consequently, ½ ¾Ñ ½ is odd, and is even. ( ) Suppose that all cycles in are even. First, we observe that it suffices to show the claim for connected graphs. Indeed, if is disconnected, then each cycle of is contained in one of the connected components, ½ Ô, of. If is µ-bipartite, then ½ ¾ Ô ½ ¾ Ô µ is a bipartition of. Assume thus that is connected. Let Ú ¾ be a chosen vertex, and define Ü Ú Üµ is even Ý Ú Ýµ is odd Since is connected, Î. Also, by the definition of distance,. Let Ù Û ¾ be both in or both in, and let È Ú Ù and É Ú Û be (among the) shortest paths from Ú to Ù and Û. Assume that Ü is the last common vertex of È and É: È È ½ È ¾, É É ½ É ¾, where È ¾ Ü Ù and É ¾ Ü Û are independent. Since È and É are shortest paths, È ½ and É ½ are shortest paths Ú Ü. Consequently, È ½ É ½. So È ¾ and É ¾ have the same parity. Therefore É ½ È ¾ ¾ Û Ù is an even path. It follows that Ù and Û are not adjacent in, since otherwise É ½ È ¾ ¾ ÙÛµ would be an odd cycle. Therefore and are discrete induced subgraphs, and is bipartite as claimed. Ú È ½ É ½ Ü È ¾ É ¾ Ù Û ÙÛ Checking whether a graph is bipartite is easy. Indeed, this can be done by using two opposite colours, say ½ and ¾. Start from any vertex Ú ½, and colour it by ½. Then colour the neighbours of Ú ½ by ¾, and proceed by colouring all neighbours of an already coloured vertex by an opposite colour. If the whole graph can be coloured, then is µ-bipartite, where consists of those vertices with colour ½, and of those vertices with colour ¾; otherwise, at some point one of the vertices gets both colours, and in this case, is not bipartite. ¾ ½ ½ ½ ½ ¾ ¾ ¾ ½ ¾ Theorem 2.3 (ERDÖS (1965)). Each graph has a bipartite subgraph À such that À ½ ¾. Proof. Let Î be a partition such that the number of edges between and is as large as possible. Denote ÙÚ Ù ¾ Ú ¾

19 2.1 Bipartite graphs and trees 18 and let À. Obviously À is a spanning subgraph, and it is bipartite. By the maximum condition, À Úµ ½ ¾ Úµ since, otherwise, Ú is on the wrong side. (That is, if Ú ¾, then the pair ¼ ¼ Ú does better that the pair.) Now À ½ À Úµ ½ ½ Úµ ½ ¾ ¾ ¾ ¾ Ú¾À Ú¾ Ò Ú, Bridges DEFINITION. An edge ¾ is a bridge of the graph, if has more connected components than, that is, if µ µ. In particular, and most importantly, an edge in a connected is a bridge if and only if is disconnected. On the right the two horizontal lines are bridges. The rest are not. Theorem 2.4. An edge ¾ is a bridge if and only if is not in any cycle of. Proof. First of all, note that ÙÚ is a bridge if and only if Ù and Ú belong to different connected components of. (µ) If there is a cycle in containing, then there is a cycle È Ù Ú Ù, where È Ú Ù is a path in, and so is not a bridge. ( ) Assume that ÙÚ is not a bridge. Hence Ù and Ú are in the same connected component of. If È Ú Ù is a path in, then È Ù Ú Ù is a cycle in that contains. Lemma 2.1. Let be a bridge in a connected graph. (i) Then µ ¾. (ii) Let À be a connected component of. If ¾ À is a bridge of À, then is a bridge of. Proof. For (i), let ÙÚ. Since is a bridge, the ends Ù and Ú are not connected in. Let Û ¾. Since is connected, there exists a path È Û Ú in. This is a path of, unless È Û Ù Ú contains ÙÚ, in which case the part Û Ù is a path in. For (ii), if ¾ À belongs to a cycle of, then does not contain (since is in no cycle), and therefore is inside À, and is not a bridge of À.

20 2.1 Bipartite graphs and trees 19 Trees DEFINITION. A graph is called acyclic, if it has no cycles. An acyclic graph is also called a forest. A tree is a connected acyclic graph. By Theorem 2.4 and the definition of a tree, we have Corollary 2.1. A connected graph is a tree if and only if all its edges are bridges. Example 2.2. The following enumeration result for trees has many different proofs, the first of which was given by CAYLEY in 1889: There are Ò Ò ¾ trees on a vertex set Î of Ò elements. We omit the proof. On the other hand, there are only a few trees up to isomorphism: Ò ½ ¾ trees ½ ½ ½ ¾ ½½ ¾ Ò ½¼ ½½ ½¾ ½ ½ ½ ½ trees ½¼ ¾ ½ ½ ¼½ ½ ½ ½ ¾¼ The nonisomorphic trees of order are: Theorem 2.5. The following are equivalent for a graph Ì. (i) Ì is a tree. (ii) Any two vertices are connected in Ì by a unique path. (iii) Ì is acyclic and Ì Ì ½. Proof. Let Ì Ò. If Ò ½, then the claim is trivial. Suppose thus that Ò ¾. (i)µ(ii) Let Ì be a tree. Assume the claim does not hold, and let È É Ù Ú be two different paths between the same vertices Ù and Ú. Suppose that È É. Since È É, there exists an edge which belongs to È but not to É. Each edge of Ì is a bridge, and therefore Ù and Ú belong to different connected components of Ì. Hence must also belong to É; a contradiction. (ii)µ(iii) We prove the claim by induction on Ò. Clearly, the claim holds for Ò ¾, and suppose it holds for graphs of order less than Ò. Let Ì be any graph of order Ò satisfying (ii). In particular, Ì is connected, and it is clearly acyclic. Let È Ù Ú be a maximal path in Ì, that is, there are no edges, for which È or È is a path. Such paths exist, because Ì is finite. It follows that Ì Úµ ½, since, by maximality,

21 2.1 Bipartite graphs and trees 20 if ÚÛ ¾ Ì, then Û belongs to È ; otherwise È ÚÛµ would be a longer path. In this case, È Ù Û Ú, where ÚÛ is the unique edge having an end Ú. The subgraph Ì Ú is connected, and therefore it satisfies the condition (ii). By induction hypothesis, Ì Ú Ò ¾, and so Ì Ì Ú ½ Ò ½, and the claim follows. (iii)µ(i) Assume (iii) holds for Ì. We need to show that Ì is connected. Indeed, let the connected components of Ì be Ì Î µ, for ¾ ½. Since Ì is acyclic, so are the connected graphs Ì, and hence they are trees, for which we have proved that Î ½. È È Now, Ì Î ½, and Ì ½. Therefore, Ò ½ Ì ½ Î ½µ ½ Î Ò which gives that ½, that is, Ì is connected. Example 2.3. Consider a cup tournament of Ò teams. If during a round there are teams left in the tournament, then these are divided into pairs, and from each pair only the winner continues. If is odd, then one of the teams goes to the next round without having to play. How many plays are needed to determine the winner? So if there are ½ teams, after the first round teams continue, and after the second round teams continue, then ¾. So ½ plays are needed in this example. The answer to our problem is Ò ½, since the cup tournament is a tree, where a play corresponds to an edge of the tree. Spanning trees Theorem 2.6. Each connected graph has a spanning tree, that is, a spanning graph that is a tree. Proof. Let À be a minimal connected spanning subgraph, that is, a connected spanning subgraph of such that À is disconnected for all ¾ À. Such a subgraph is obtained from by removing nonbridges: To start with, let À ¼. For ¼, let À ½ À, where is a not a bridge of À. Since is not a bridge, À ½ is a connected spanning subgraph of À and thus of. À À, when only bridges are left. By Corollary 2.1, À is a tree. Corollary 2.2. For each connected graph, is a tree if and only if ½. ½. Moreover, a connected graph Proof. Let Ì be a spanning tree of. Then Ì Ì ½ ½. The second claim is also clear.

22 2.1 Bipartite graphs and trees 21 Corollary 2.3. Each tree Ì with Ì ¾ has at least two leaves. Proof. Let be the number of leaves of Ì. By Corollary 2.2 and the handshaking lemma, ¾ Ì ¾ ¾ Ì Ì Úµ Ú¾Ì Ì Úµ½ ¾ Ì µ ¾ Ì Ì Úµ from which it follows that ¾, as required. Example 2.4. In Shannon s switching game a positive player È and a negative player Æ play on a graph with two special vertices: a source and a sink Ö. È and Æ alternate turns so that È designates an edge by, and Æ by. Each edge can be designated at most once. It is È s purpose to designate a path Ö (that is, to designate all edges in one such path), and Æ tries to block all paths Ö (that is, to designate at least one edge in each such path). We say that a game Öµ is positive, if È has a winning strategy no matter who begins the game, negative, if Æ has a winning strategy no matter who begins the game, neutral, if the winner depends on who begins the game. Ö The game on the right is neutral. LEHMAN proved in 1964 that Shannon s switching game Öµ is positive if and only if there exists À such that À contains and Ö and À has two spanning trees with no edges in common. In the other direction the claim can be proved along the following lines. Assume that there exists a subgraph À containing and Ö and that has two spanning trees with no edges in common. Then È plays as follows. If Æ marks by an edge from one of the two trees, then È marks by an edge in the other tree such that this edge reconnects the broken tree. In this way, È always has two spanning trees for the subgraph À with only edges marked by in common. In converse the claim is considerably more difficult to prove. There remains the problem to characterize those Shannon s switching games Öµ that are neutral (negative, respectively). The connector problem To build a network connecting Ò nodes (towns, computers, chips in a computer) it is desirable to decrease the cost of construction of the links to the minimum. This is the connector problem. In graph theoretical terms we wish to find an optimal spanning subgraph of a weighted

23 2.1 Bipartite graphs and trees 22 graph. Such an optimal subgraph is clearly a spanning tree, for, otherwise a deletion of any nonbridge will reduce the total weight of the subgraph. Let then «be a graph together with a weight function «Ê (positive reals) on the edges. Kruskal s algorithm (also known as the greedy algorithm) provides a solution to the connector problem. Kruskal s algorithm: For a connected and weighted graph «of order Ò: (i) Let ½ be an edge of smallest weight, and set ½ ½. (ii) For each ¾ Ò ½ in this order, choose an edge ¾ ½ of smallest possible weight such that does not produce a cycle when added to ½, and let ½. The final outcome is Ì Î Ò ½ µ. By the construction, Ì Î Ò ½ µ is a spanning tree of, because it contains no cycles, it is connected and has Ò ½ edges. We now show that Ì has the minimum total weight among the spanning trees of. Suppose Ì ½ is any spanning tree of. Let be the first edge produced by the algorithm that is not in Ì ½. If we add to Ì ½, then a cycle containing is created. Also, must contain an edge that is not in Ì. When we replace by in Ì ½, we still have a spanning tree, say Ì ¾. However, by the construction, «µ «µ, and therefore «Ì ¾ µ «Ì ½ µ. Note that Ì ¾ has more edges in common with Ì than Ì ½. Repeating the above procedure, we can transform Ì ½ to Ì by replacing edges, one by one, such that the total weight does not increase. We deduce that «Ì µ «Ì ½ µ. The outcome of Kruskal s algorithm need not be unique. Indeed, there may exist several optimal spanning trees (with the same weight, of course) for a graph. Example 2.5. When applied to the weighted graph on the right, the algorithm produces the sequence: ½ Ú ¾ Ú, ¾ Ú Ú, Ú Ú, Ú ¾ Ú and Ú ½ Ú ¾. The total weight of the spanning tree is thus 9. Also, the selection ½ Ú ¾ Ú, ¾ Ú Ú, Ú Ú, Ú Ú, Ú ½ Ú ¾ gives another optimal solution (of weight 9). Ú ½ Ú ¾ ¾ Ú ½ ½ ¾ ¾ ¾ Ú ½ Ú ¾ Ú Problem. Consider trees Ì with weight functions «Ì Æ. Each tree Ì of order Ò has exactly ¾ Ò paths. (Why is this so?) Does there exist a weighted tree Ì «of order Ò such that Ò the (total) weights of its paths are ½ ¾ ¾?

24 2.1 Bipartite graphs and trees 23 In such a weighted tree Ì «different paths have different weights, and each ¾ ½ Ò ¾ is a weight of one path. Also, «must be injective. No solutions are known for any Ò. ¾ ½ TAYLOR (1977) proved: if Ì of order Ò exists, then necessarily Ò ¾ or Ò ¾ ¾ for some ½. Example 2.6. A computer network can be presented as a graph, where the vertices are the node computers, and the edges indicate the direct links. Each computer Ú has an address Úµ, a bit string (of zeros and ones). The length of an address is the number of its bits. A message that is sent to Ú is preceded by the address Úµ. The Hamming distance Úµ Ùµµ of two addresses of the same length is the number of places, where Úµ and Ùµ differ. For example, ¼¼¼½¼ ¼½½¼¼µ and ½¼¼¼¼ ¼¼¼¼¼µ ½. It would be a good way to address the vertices so that the Hamming distance of two vertices is the same as their distance in. In particular, if two vertices were adjacent, their addresses should differ by one symbol. This would make it easier for a node computer to forward a message. A graph is said to be addressable, if it has an addressing such that Ù Úµ Ùµ Úµµ. ¼¼¼ ¼½¼ ½¼¼ ½½¼ ½½½ We prove that every tree Ì is addressable. Moreover, the addresses of the vertices of Ì can be chosen to be of length Ì ½. The proof goes by induction. If Ì ¾, then the claim is obvious. In the case Ì ¾, the addresses of the vertices are simply 0 and 1. Let then Î Ì Ú ½ Ú ½, and assume that Ì Ú ½ µ ½ (a leaf) and Ú ½ Ú ¾ ¾ Ì. By the induction hypothesis, we can address the tree Ì Ú ½ by addresses of length ½. We change this addressing: let be the address of Ú in Ì Ú ½, and change it to ¼. Set the address of Ú ½ to ½ ¾. It is now easy to see that we have obtained an addressing for Ì as required. The triangle à is not addressable. In order to gain more generality, we modify the addressing for general graphs by introducing a special symbol in addition to 0 and 1. A star address will be a sequence of these three symbols. The Hamming distance remains as it was, that is, Ù Úµ is the number of places, where Ù and Ú have a different symbol 0 or 1. The special symbol does not affect Ù Úµ. So, ½¼ ¼½ ¼ ½¼½µ ½ and ½ ¼¼ µ ¼. We still want to have Ù Úµ Ù Úµ.

25 2.2 Connectivity 24 We star address this graph as follows: Ú ½ µ ¼¼¼¼, Ú ¾ µ ½¼ ¼, Ú µ ½ ¼½, Ú µ ½½. These addresses have length 4. Can you design a star addressing with addresses of length 3? Ú ½ Ú ¾ Ú Ú WINKLER proved in 1983 a rather unexpected result: The minimum star address length of a graph is at most ½. For the proof of this, see VAN LINT AND WILSON, A Course in Combinatorics. 2.2 Connectivity Spanning trees are often optimal solutions to problems, where cost is the criterion. We may also wish to construct graphs that are as simple as possible, but where two vertices are always connected by at least two independent paths. These problems occur especially in different aspects of fault tolerance and reliability of networks, where one has to make sure that a breakdown of one connection does not affect the functionality of the network. Similarly, in a reliable network we require that a break-down of a node (computer) should not result in the inactivity of the whole network. Separating sets DEFINITION. A vertex Ú ¾ is a cut vertex, if Úµ µ. A subset Î is a separating set, if is disconnected. We also say that separates vertices Ù and Ú, if Ù and Ú belong to different connected components of. If is connected, then Ú is a cut vertex if and only if Ú is disconnected, that is, Ú is a separating set. We remark also that if Î separates Ù and Ú, then every path È Ù Ú visits a vertex of. Lemma 2.2. If a connected graph has no separating sets, then it is a complete graph. Proof. If ¾, then the claim is clear. For, assume that is not complete, and let ÙÚ ¾. Now Î Ò Ù Ú is a separating set. The claim follows from this. DEFINITION. The (vertex) connectivity number µ of is defined as µ ÑÒ disconnected or trivial Î A graph is -connected, if µ.

26 2.2 Connectivity 25 In other words, µ ¼, if is disconnected, µ ½, if is a complete graph, and otherwise µ equals the minimum size of a separating set of. Clearly, if is connected, then it is 1-connected. DEFINITION. An edge cut of consists of edges so that is disconnected. Let ¼ µ ÑÒ disconnected For trivial graphs, let ¼ µ ¼. A graph is -edge connected, if ¼ µ. A minimal edge cut is a bond ( Ò is not an edge cut for any ¾ ). Example 2.7. Again, if is disconnected, then ¼ µ ¼. On the right, µ ¾ and ¼ µ ¾. Notice that the minimum degree is Æ µ. Lemma 2.3. Let be connected. If ÙÚ is a bridge, then either à ¾ or one of Ù or Ú is a cut vertex. Proof. Assume that à ¾ and thus that, since is connected. Let Ù Æ Ùµ and Ú Æ Úµ be the connected components of containing Ù and Ú. Now, either Ù ¾ (and Ù is a cut vertex) or Ú ¾ (and Ú is a cut vertex). Lemma 2.4. If be a bond of a connected graph, then µ ¾. Proof. Since is disconnected, and is minimal, the subgraph Òµ is connected for each ¾. Hence is a bridge in Ò µ. By Lemma 2.1, has exactly two connected components. Theorem 2.7 (WHITNEY (1932)). For any graph, µ ¼ µ Æ µ Proof. Assume is nontrivial. Clearly, ¼ µ Æ µ, since if we remove all edges with an end Ú, we disconnect. If ¼ µ ¼, then is disconnected, and in this case also µ ¼. If ¼ µ ½, then is connected and contains a bridge. By Lemma 2.3, either à ¾ or has a cut vertex. In both of these cases, also µ ½. Assume then that ¼ µ ¾. Let be an edge cut of with ¼ µ, and let ÙÚ ¾. Then is a bond, and has two connected components.

27 2.2 Connectivity 26 Consider the connected subgraph À Ò µ µ where is a bridge.. Now for each ¾ Ò choose an end different from Ù and Ú. (The choices for different edges need not be different.) Note that since, either end of is different from Ù or Ú. Let Ë be the collection of these choices. Thus Ë ½ ¼ µ ½, and Ë does not contain edges from Ò. If Ë is disconnected, then Ë is a separating set and so µ Ë ¼ µ ½ and we are done. On the other hand, if Ë is connected, then either Ë Ã ¾ ( ), or either Ù or Ú (or both) is a cut vertex of Ë (since À Ë Ë, and therefore Ë À is an induced subgraph of À). In both of these cases, there is a vertex of Ë, whose removal results in a trivial or a disconnected graph. In conclusion, µ Ë ½ ¼ µ, and the claim follows. Menger s theorem Theorem 2.8 (MENGER (1927)). Let Ù Ú ¾ be nonadjacent vertices of a connected graph. Then the minimum number of vertices separating Ù and Ú is equal to the maximum number of independent paths from Ù to Ú. Proof. If a subset Ë Î separates Ù and Ú, then every path Ù Ú of visits Ë. Hence Ë is at least the number of independent paths from Ù to Ú. Conversely, we use induction on Ñ to show that if Ë Û ½ Û ¾ Û is a minimum set (that is, a subset of the smallest size) that separates Ù and Ú, then has at least (and thus exactly) independent paths Ù Ú. The case for ½ is clear, and this takes care of the small values of Ñ, required for the induction. (1) Assume first that Ù and Ú have a common neighbour Û ¾ Æ Ùµ Æ Úµ. Then necessarily Û ¾ Ë. In the smaller graph Û the set Ë Ò Û is a minimum set that separates Ù and Ú, and so the induction hypothesis yields that there are ½ independent paths Ù Ú in Û. Together with the path Ù Û Ú, there are independent paths Ù Ú in as required. Æ (2) Assume then that Æ Ùµ Æ Úµ, and denote by À Ù Æ Ë Ùµ and À Ú Ë Úµ the connected components of Ë for Ù and Ú. (2.1) Suppose next that Ë Æ Ùµ and Ë Æ Úµ.. À..

28 2.2 Connectivity 27 Let Ú be a new vertex, and define Ù to be the graph on À Ù ËÚ having the edges of À Ù Ë together with ÚÛ for all ¾ ½. The graph Ù is connected and it is smaller than. Indeed, in order for Ë to be a minimum separating set, all Û ¾ Ë have to be adjacent to some vertex in À Ú. This shows that Ù, and, moreover, the assumption (2.1) rules out the case À Ú Ú, and therefore À Ú ¾ and so Ù in the present case. If Ë ¼ is any subset that separates Ù and Ú in Ù, then Ë ¼ will separate Ù from all Û ¾ Ë ÒË ¼ in. This means that Ë ¼ separates Ù and Ú in. Since is the size of a minimum separating set, Ë ¼. We noted that Ù is smaller than, and thus by the induction hypothesis, there are independent paths Ù Ú in Ù. This is possible only if there exist paths Ù Û, one for each ¾ ½, that have only the end Ù in common. By the present assumption, also Ù is nonadjacent to some vertex of Ë. A symmetric argument applies to the graph Ú (with a new vertex Ù), which is defined similarly to Ù. This yields that there are paths Û Ú that have only the end Ú in common. When we combine these with the above paths Ù Û, we obtain independent paths Ù Û Ú in. (2.2) There remains the case, where for all separating sets Ë of elements, either Ë Æ Ùµ or Ë Æ Úµ. (Note that then, by (2), Ë Æ Úµ or Ë Æ Ùµ.) Let È É be a shortest path Ù Ú in, where ÙÜ, ÜÝ, and É Ý Ú. Notice that, by the assumption (2), È, and so Ý Ú. In the smaller graph, let Ë ¼ be a minimum set that separates Ù and Ú. If Ë ¼, then, by the induction hypothesis, there are independent paths Ù Ú in. But these are paths of, and the claim is clear in this case. If, on the other hand, Ë ¼, then Ù and Ú are still connected in Ë ¼. Indeed, every path Ù Ú in Ë ¼ necessarily travels along the edge ÜÝ, and so Ü Ý ¾ Ë ¼. Let Ë Ü Ë ¼ Ü and Ë Ý Ë ¼ Ý These sets separate Ù and Ú in (by the above fact), and they have size. By our current assumption, the vertices of Ë Ý are adjacent to Ú, since the path È is shortest and so ÙÝ ¾ (meaning that Ù is not adjacent to all of Ë Ý ). The assumption (2) yields that Ù is adjacent to all of Ë Ü, since ÙÜ ¾. But now both Ù and Ú are adjacent to the vertices of Ë ¼, which contradicts the assumption (2). Ù Û Û ¾ Û ½ Ú Theorem 2.9 (MENGER (1927)). A graph is -connected if and only if every two vertices are connected by at least independent paths. Proof. If any two vertices are connected by independent paths, then it is clear that µ. In converse, suppose that µ, but that has vertices Ù and Ú connected by at most ½ independent paths. By Theorem 2.8, it must be that ÙÚ ¾. Consider the graph. Now Ù and Ú are connected by at most ¾ independent paths in, and by

29 2.2 Connectivity 28 Theorem 2.8, Ù and Ú can be separated in by a set Ë with Ë ¾. Since (because µ ), there exists a Û ¾ that is not in Ë Ù Ú. The vertex Û is separated in by Ë from Ù or from Ú; otherwise there would be a path Ù Ú in µ Ë. Say, this vertex is Ù. The set Ë Ú has ½ elements, and it separates Ù from Û in, which contradicts the assumption that µ. This proves the claim. We state without a proof the corresponding separation property for edge connectivity. DEFINITION. Let be a graph. A ÙÚ-disconnecting set is a set such that every path Ù Ú contains an edge from. Theorem Let Ù Ú ¾ with Ù Ú in a graph. Then the maximum number of edgedisjoint paths Ù Ú equals the minimum number of edges in a ÙÚ-disconnecting set. Corollary 2.4. A graph is -edge connected if and only if every two vertices are connected by at least edge disjoint paths. Example 2.8. Recall the definition of the cube É from Example 1.5. We show that É µ. First of all, É µ Æ É µ. In converse, we show the claim by induction. Extract from É the disjoint subgraphs: ¼ induced by ¼Ù Ù ¾ ½ and ½ induced by ½Ù Ù ¾ ½. These are (isomorphic to) É ½, and É is obtained from the union of ¼ and ½ by adding the ¾ ½ edges ¼Ù ½Ùµ for all Ù ¾ ½. Let Ë be a vertex cut of É. Then Ë. If both ¼ Ë and ½ Ë were connected, also É Ë would be connected, since one pair ¼Ù ½Ùµ necessarily remains in É Ë. So we can assume that ¼ Ë is disconnected. (The case for ½ Ë is symmetric.) By the induction hypothesis, ¼ µ ½, and hence Ë contains at least ½ vertices of ¼ (and so Ë ½). If there were no vertices from ½ in Ë, then, of course, ½ Ë is connected, and the edges ¼Ù ½Ùµ of É would guarantee that É Ë is connected; a contradiction. Hence Ë. Example 2.9. We have ¼ É µ for the -cube. Indeed, by Whitney s theorem, µ ¼ µ Æ µ. Since É µ Æ É µ, also ¼ É µ. Algorithmic Problem. The connectivity problems tend to be algorithmically difficult. In the disjoint paths problem we are given a set Ù Ú µ of pairs of vertices for ½ ¾, and it is asked whether there exist paths È Ù Ú that have no vertices in common. This problem was shown to be NP-complete by KNUTH in (However, for fixed, the problem has a fast algorithm due to ROBERTSON and SEYMOUR (1986).)

30 2.2 Connectivity 29 Dirac s fans DEFINITION. Let Ú ¾ and Ë Î such that Ú ¾ Ë in a graph. A set of paths from Ú to a vertex in Ë is called a Ú Ëµ-fan, if they have only Ú in common. Theorem 2.11 (DIRAC (1960)). A graph is -connected if and only if and for every Ú ¾ and Ë Î with Ë and Ú ¾ Ë, there exists a Ú Ëµ-fan of paths. Ú Ë Proof. Exercise. Theorem 2.12 (DIRAC (1960)). Let be a -connected graph for ¾. Then for any vertices, there exists a cycle of containing them. Proof. First of all, since µ ¾, has no cut vertices, and thus no bridges. It follows that every edge, and thus every vertex of belongs to a cycle. Let Ë Î be such that Ë, and let be a cycle of that contains the maximum number of vertices of Ë. Let the vertices of Ë Î be Ú ½ Ú Ö listed in order around so that each pair Ú Ú ½ µ (with indices modulo Ö) defines a path along (except in the special case where Ö ½). Such a path is referred to as a segment of. If contains all vertices of Ë, then we are done; otherwise, suppose Ú ¾ Ë is not on. It follows from Theorem 2.11 that there is a Ú Î µ-fan of at least ÑÒ Î paths. Therefore there are two paths È Ú Ù and É Ú Û in such a fan that end in the same segment Ú Ú ½ µ of. Then the path Ï Ù Û (or Û Ù) along contains all vertices of Ë Î. But now È Ï É ½ is a cycle of that contains Ú and all Ú for ¾ ½ Ö. This contradicts the choice of, and proves the claim.

31 3 Tours and Matchings 3.1 Eulerian graphs The first proper problem in graph theory was the Königsberg bridge problem. In general, this problem concerns about travels around a graph such that one tries to avoid using the same edge twice. In practice these eulerian problems occur, for instance, in optimizing distribution networks such as delivering mail, where in order to save time each street should be travelled only once. The same problem occurs in mechanical graph plotting, where one avoids lifting the pen off the paper while drawing the lines. Euler tours DEFINITION. A walk Ï ½ ¾ Ò is a trail, if for all. An Euler trail of a graph is a trail that visits every edge once. A connected graph is eulerian, if it has a closed trail containing every edge of. Such a trail is called an Euler tour. Notice that if Ï ½ ¾ Ò is an Euler tour (and so ½ ¾ Ò ), also ½ Ò ½ ½ is an Euler tour for all ¾ ½ Ò. A complete proof of the following Euler s Theorem was first given by HIERHOLZER in Theorem 3.1 (EULER (1736), HIERHOLZER (1873)). A connected graph is eulerian if and only if every vertex has an even degree. Proof. (µ) Suppose Ï Ù Ù is an Euler tour. Let Ú Ùµ be a vertex that occurs times in Ï. Every time an edge arrives at Ú, another edge departs from Ú, and therefore Úµ ¾. Also, Ùµ is even, since Ï starts and ends at Ù. ( ) Assume is a nontrivial connected graph such that Úµ is even for all Ú ¾. Let Ï ½ ¾ Ò Ú ¼ Ú Ò with Ú ½ Ú be a longest trail in. It follows that all Ú Ò Û ¾ are among the edges of Ï, for, otherwise, Ï could be prolonged to Ï. In particular, Ú ¼ Ú Ò, that is, Ï is a closed trail. (Indeed, if it were Ú Ò Ú ¼ and Ú Ò occurs times in Ï, then Ú Ò µ ¾ ½µ ½ and that would be odd.) If Ï is not an Euler tour, then, since is connected, there exists an edge Ú Ù ¾ for some, which is not in Ï. However, now ½ Ò ½ is a trail in, and it is longer than Ï. This contradiction to the choice of Ï proves the claim.

32 3.1 Eulerian graphs 31 Example 3.1. The -cube É is eulerian for even integers, because É is -regular. Theorem 3.2. A connected graph has an Euler trail if and only if it has at most two vertices of odd degree. Proof. If has an Euler trail Ù Ú, then, as in the proof of Theorem 3.1, each vertex Û ¾ Ù Ú has an even degree. Assume then that is connected and has at most two vertices of odd degree. If has no vertices of odd degree then, by Theorem 3.1, has an Euler trail. Otherwise, by the handshaking lemma, every graph has an even number of vertices with odd degree, and therefore has exactly two such vertices, say Ù and Ú. Let À be a graph obtained from by adding a vertex Û, and the edges ÙÛ and ÚÛ. In À every vertex has an even degree, and hence it has an Euler tour, say Ù Ú Û Ù. Here the beginning part Ù Ú is an Euler trail of. The Chinese postman The following problem is due to GUAN MEIGU (1962). Consider a village, where a postman wishes to plan his route to save the legs, but still every street has to be walked through. This problem is akin to Euler s problem and to the shortest path problem. Let be a graph with a weight function «Ê. The Chinese postman problem is to find a minimum weighted tour in (starting from a given vertex, the post office). If is eulerian, then any Euler tour will do as a solution, because such a tour traverses each edge exactly once and this is the best one can do. In this case the weight of the optimal tour is the total weight of the graph, and there is a good algorithm for finding such a tour: Fleury s algorithm: Let Ú ¼ ¾ be a chosen vertex, and let Ï ¼ be the trivial path on Ú ¼. Repeat the following procedure for ½ ¾ as long as possible: suppose a trail Ï ½ ¾ has been constructed, where Ú ½ Ú. Choose an edge ½ ( for ¾ ½ ) so that (i) ½ has an end Ú, and (ii) ½ is not a bridge of ½, unless there is no alternative. Notice that, as is natural, the weights «µ play no role in the eulerian case. Theorem 3.3. If is eulerian, then any trail of constructed by Fleury s algorithm is an Euler tour of. Proof. Exercise.

33 3.2 Hamiltonian graphs 32 If is not eulerian, the poor postman has to walk at least one street twice. This happens, e.g., if one of the streets is a dead end, and in general if there is a street corner of an odd number of streets. We can attack this case by reducing it to the eulerian case as follows. An edge ÙÚ will be duplicated, if it is added to parallel to an existing edge ¼ ÙÚ with the same weight, «¼ µ «µ. ¾ ½ ¾ ¾ ¾ ½ ¾ ¾ ¾ ½ ¾ Above we have duplicated two edges. The rightmost multigraph is eulerian. There is a good algorithm by EDMONDS AND JOHNSON (1973) for the construction of an optimal eulerian supergraph by duplications. Unfortunately, this algorithm is somewhat complicated, and we shall skip it. 3.2 Hamiltonian graphs In the connector problem we reduced the cost of a spanning graph to its minimum. There are different problems, where the cost is measured by an active user of the graph. For instance, in the travelling salesman problem a person is supposed to visit each town in his district, and this he should do in such a way that saves time and money. Obviously, he should plan the travel so as to visit each town once, and so that the overall flight time is as short as possible. In terms of graphs, he is looking for a minimum weighted Hamilton cycle of a graph, the vertices of which are the towns and the weights on the edges are the flight times. Unlike for the shortest path and the connector problems no efficient reliable algorithm is known for the travelling salesman problem. Indeed, it is widely believed that no practical algorithm exists for this problem. Hamilton cycles DEFINITION. A path È of a graph is a Hamilton path, if È visits every vertex of once. Similarly, a cycle is a Hamilton cycle, if it visits each vertex once. A graph is hamiltonian, if it has a Hamilton cycle. Note that if Ù ½ Ù ¾ Ù Ò is a Hamilton cycle, then so is Ù Ù Ò Ù ½ Ù ½ for each ¾ ½ Ò, and thus we can choose where to start the cycle. Example 3.2. It is obvious that each Ã Ò is hamiltonian whenever Ò. Also, as is easily seen, à ÒÑ is hamiltonian if and only if Ò Ñ ¾. Indeed, let à ÒÑ have a bipartition

34 3.2 Hamiltonian graphs 33 µ, where Ò and Ñ. Now, each cycle in à ÒÑ has even length as the graph is bipartite, and thus the cycle visits the sets equally many times, since and are stable subsets. But then necessarily. Unlike for eulerian graphs (Theorem 3.1) no good characterization is known for hamiltonian graphs. Indeed, the problem to determine if is hamiltonian is NP-complete. There are, however, some interesting general conditions. Lemma 3.1. If is hamiltonian, then for every nonempty subset Ë Î, ˵ Ë Proof. Let Ë Î, Ù ¾ Ë, and let Ù Ù be a Hamilton cycle of. Assume Ë has k connected components,, ¾ ½. The case ½ is trivial, and hence suppose that ½. Let Ù be the last vertex of that belongs to, and let Ú be the vertex that follows Ù in. Now Ú ¾ Ë for each by the choice of Ù, and Ú Ú Ø for all Ø, because is a cycle and Ù Ú ¾ for all. Thus Ë as required. Example 3.3. Consider the graph on the right. In, ˵ ¾ Ë for the set Ë of black vertices. Therefore does not satisfy the condition of Lemma 3.1, and hence it is not hamiltonian. Interestingly this graph is µ-bipartite of even order with. It is also -regular. Example 3.4. Consider the Petersen graph on the right, which appears in many places in graph theory as a counter example for various conditions. This graph is not hamiltonian, but it does satisfy the condition ˵ Ë for all Ë. Therefore the conclusion of Lemma 3.1 is not sufficient to ensure that a graph is hamiltonian. The following theorem, due to ORE, generalizes an earlier result by DIRAC (1952). Theorem 3.4 (ORE (1962)). Let be a graph of order, and let Ù Ú ¾ be such that Ùµ Úµ Then is hamiltonian if and only if ÙÚ is hamiltonian. Proof. Denote Ò. Let Ù Ú ¾ be such that Ùµ Úµ Ò. If ÙÚ ¾, then there is nothing to prove. Assume thus that ÙÚ ¾. (µ) This is trivial since if has a Hamilton cycle, then is also a Hamilton cycle of ÙÚ. ( ) Denote ÙÚ and suppose that has a Hamilton cycle. If does not use the edge, then it is a Hamilton cycle of. Suppose thus that is on. We may then assume

35 3.2 Hamiltonian graphs 34 that Ù Ú Ù. Now Ù Ú ½ Ú ¾ Ú Ò Ú is a Hamilton path of. There exists an with ½ Ò such that ÙÚ ¾ and Ú ½ Ú ¾ For, otherwise, Úµ Ò Ùµ would contradict the assumption. Ú ½ Ú ¾ Æ Æ Ú ½ Ú Æ Æ ÚÒ But now Ù Ú ½ Ú ½ Ú Ò Ú Ò ½ Ú ½ Ú Ú ½ Ù is a Hamilton cycle in. Closure DEFINITION. For a graph, define inductively a sequence ¼ ½ of graphs such that ¼ and ½ ÙÚ where Ù and Ú are any vertices such that ÙÚ ¾ and Ùµ Úµ. This procedure stops when no new edges can be added to for some, that is, in, for all Ù Ú ¾ either ÙÚ ¾ or Ùµ Úµ. The result of this procedure is the closure of, and it is denoted by Ð µ ( ). In each step of the construction of Ð µ there are usually alternatives which edge ÙÚ is to be added to the graph, and therefore the above procedure is not deterministic. However, the final result Ð µ is independent of the choices. Lemma 3.2. The closure Ð µ is uniquely defined for all graphs of order. Proof. Denote Ò. Suppose there are two ways to close, say À ½ Ö and À ¼ ½ where the edges are added in the given orders. Let À ½ and À ¼ ½. For the initial values, we have À ¼ À ¼. Let ¼ ÙÚ be the first edge such that for all. Then À ½ Ùµ À ½ Úµ Ò, since ¾ À, but ¾ À ½. By the choice of, we have À ½ À ¼, and thus also À ¼ Ùµ À ¼ Úµ Ò, which means that ÙÚ must be in À ¼ ; a contradiction. Therefore À À ¼. Symmetrically, we deduce that À ¼ À, and hence À ¼ À. Theorem 3.5. Let be a graph of order. (i) is hamiltonian if and only if its closure Ð µ is hamiltonian. (ii) If Ð µ is a complete graph, then is hamiltonian. Proof. First, Ð µ and spans Ð µ, and thus if is hamiltonian, so is Ð µ. In the other direction, let ¼ ½ Ð µ be a construction sequence of the closure of. If Ð µ is hamiltonian, then so are ½ ½ and ¼ by Theorem 3.4. The Claim (ii) follows from (i), since each complete graph is hamiltonian.

36 3.2 Hamiltonian graphs 35 Theorem 3.6. Let be a graph of order. Suppose that for all nonadjacent vertices Ù and Ú, Ùµ Úµ. Then is hamiltonian. In particular, if Æ µ ½ ¾, then is hamiltonian. Proof. Since Ùµ Úµ for all nonadjacent vertices, we have Ð µ Ã Ò for Ò, and thus is hamiltonian. The second claim is immediate, since now Ùµ Úµ for all Ù Ú ¾ whether adjacent or not. Chvátal s condition The hamiltonian problem of graphs has attracted much attention, at least partly because the problem has practical significance. (Indeed, the first example where DNA computing was applied, was the hamiltonian problem.) There are some general improvements of the previous results of this chapter, and quite many improvements in various special cases, where the graphs are somehow restricted. We become satisfied by two general results. Theorem 3.7 (CHVÁTAL (1972)). Let be a graph with Î Ú ½ Ú ¾ Ú Ò, for Ò, ordered so that ½ ¾ Ò, for Ú µ. If for every Ò¾, then is hamiltonian. µ Ò Ò (3.1) Proof. First of all, we may suppose that is closed, Ð µ, because is hamiltonian if and only if Ð µ is hamiltonian, and adding edges to does not decrease any of its degrees, that is, if satisfies (3.1), so does for every. We show that, in this case, à Ò, and thus is hamiltonian. Assume on the contrary that à Ò, and let ÙÚ ¾ with Ùµ Úµ be such that Ùµ Úµ is as large as possible. Because is closed, we must have Ùµ Úµ Ò, and therefore Ùµ Ò¾. Let Û ÚÛ ¾ Û Ú. By our choice, Ûµ for all Û ¾, and, moreover, Ò ½µ Úµ Ùµ Consequently, there are at least vertices Û with Ûµ, and so Ùµ. Similarly, for each vertex from Û ÙÛ ¾ Û Ù, Ûµ Úµ Ò Ùµ Ò, and Ò ½µ Ùµ Ò ½µ Also Ùµ Ò, and thus there are at least Ò vertices Û with Ûµ Ò. Consequently, Ò Ò. This contradicts the obtained bound and the condition (3.1). Note that the condition (3.1) is easily checkable for any given graph.

37 3.3 Matchings Matchings In matching problems we are given an availability relation between the elements of a set. The problem is then to find a pairing of the elements so that each element is paired (matched) uniquely with an available companion. A special case of the matching problem is the marriage problem, which is stated as follows. Given a set of boys and a set of girls, under what condition can each boy marry a girl who cares to marry him? This problem has many variations. One of them is the job assignment problem, where we are given Ò applicants and Ñ jobs, and we should assign each applicant to a job he is qualified. The problem is that an applicant may be qualified for several jobs, and a job may be suited for several applicants. Maximum matchings DEFINITION. For a graph, a subset Å is a matching of, if Å contains no adjacent edges. The two ends of an edge ¾ Å are matched under Å. A matching Å is a maximum matching, if for no matching Å ¼, Å Å ¼. The two vertical edges on the right constitute a matching Å that is not a maximum matching, although you cannot add any edges to Å to form a larger matching. This matching is not maximum because the graph has a matching of three edges. DEFINITION. A matching Å saturates Ú ¾, if Ú is an end of an edge in Å. Also, Å saturates Î, if it saturates every Ú ¾. If Å saturates Î, then Å is a perfect matching. It is clear that every perfect matching is maximum. On the right the horizontal edges form a perfect matching. DEFINITION. Let Å be a matching of. An odd path È ½ ¾ ¾ ½ is Å-augmented, if È alternates between Ò Å and Å (that is, ¾ ½ ¾ Å and ¾ ¾ Å), and the ends of È are not saturated. Lemma 3.3. If is connected with µ ¾, then is a path or a cycle. Proof. Exercise.

38 3.3 Matchings 37 We start with a result that states a necessary and sufficient condition for a matching to be maximal. One can use the first part of the proof to construct a maximum matching in an iterative manner starting from any matching Å and from any Å-augmented path. Theorem 3.8 (BERGE (1957)). A matching Å of is a maximum matching if and only if there are no Å-augmented paths in. Proof. (µ) Let a matching Å have an Å-augmented path È ½ ¾ ¾ ½ in. Here ¾ ¾ ¾ Å, ½ ¾ ½ ¾ Å. Define Æ by Æ Å Ò ¾ ¾ ½ µ ¾ ½ ¾ ¼ Now, Æ is a matching of, and Æ Å ½. Therefore Å is not a maximum matching. ( ) Assume Æ is a maximum matching, but Å is not. Hence Æ Å. Consider the subgraph À Å Æ for the symmetric difference Å Æ. We have À Úµ ¾ for each Ú ¾ À, because Ú is an end of at most one edge in Å and Æ. By Lemma 3.3, each connected component of À is either a path or a cycle. Since no Ú ¾ can be an end of two edges from Æ or from Å, each connected component (path or a cycle) alternates between Æ and Å. Now, since Æ Å, there is a connected component of À, which has more edges from Æ than from Å. This cannot be a cycle, because an alternating cycle is even, and it thus contains equally many edges from Æ and Å. Hence Ù Ú is a path, which starts and ends with an edge from Æ. Because is a connected component of À, the ends Ù and Ú are not saturated by Å, and, consequently, is an Å-augmented path. This proves the theorem. Example 3.5. Consider the -cube É for ½. Each maximum matching of É has ¾ ½ edges. Indeed, the matching Å ¼Ù ½Ùµ Ù ¾ ½, has ¾ ½ edges, and it is clearly perfect. Hall s theorem For a subset Ë Î of a graph, denote Æ Ëµ Ú ÙÚ ¾ for some Ù ¾ Ë If is µ-bipartite, and Ë, then Æ Ëµ. The following result, known as the Theorem 3.9 (HALL (1935)). Let be a µ-bipartite graph. Then contains a matching Å saturating if and only if Ë Æ Ëµ for all Ë (3.2)

39 3.3 Matchings 38 Proof. (µ) Let Å be a matching that saturates. If Ë Æ Ëµ for some Ë, then not all Ü ¾ Ë can be matched with different Ý ¾ Æ Ëµ. ( ) Let satisfy Hall s condition (3.2). We prove the claim by induction on. If ½, then the claim is clear. Let then ¾, and assume (3.2) implies the existence of a matching that saturates every proper subset of. If Æ Ëµ Ë ½ for every nonempty Ë with Ë, then choose an edge ÙÚ ¾ with Ù ¾, and consider the induced subgraph À Ù Ú. For all Ë Ò Ù, Æ À ˵ Æ Ëµ ½ Ë and hence, by the induction hypothesis, À contains a matching Å saturating Ò Ù. Now Å ÙÚ is a matching saturating in, as was required. Suppose then that there exists a nonempty subset Ê with Ê such that Æ Êµ Ê. The induced subgraph À ½ Ê Æ Êµ satisfies (3.2) (since does), and hence, by the induction hypothesis, À ½ contains a matching Å ½ that saturates Ê (with the other ends in Æ Êµ). Also, the induced subgraph À ¾ Î Ò, for Ê Æ Êµ, satisfies (3.2). Indeed, if there were a subset Ë Ò Ê such that Æ À¾ ˵ Ë, then we would have Æ Ë Êµ Æ À¾ ˵ Æ À½ ʵ Ë Æ Êµ Ë Ê Ë Ê (since Ë Ê ), which contradicts (3.2) for. By the induction hypothesis, À ¾ has a matching Å ¾ that saturates Ò Ê (with the other ends in Ò Æ Êµ). Combining the matchings for À ½ and À ¾, we get a matching Å ½ Å ¾ saturating in. Second proof. This proof of the direction µ uses Menger s theorem. Let À be the graph obtained from by adding two new vertices Ü Ý such that Ü is adjacent to each Ú ¾ and Ý is adjacent to each Ú ¾. There exists a matching saturating if (and only if) the number of independent paths Ü Ý is equal to. For this, by Menger s theorem, it suffices to show that every set Ë that separates Ü and Ý in À has at least vertices. Let Ë, where and. Now, vertices in Ò are not adjacent to vertices of Ò, and hence we have Æ Òµ, and thus that Ò Æ Òµ using the condition (3.2). Ü Ò Ò Ý We conclude that Ë. Corollary 3.1 (FROBENIUS (1917)). If is a -regular bipartite graph with ¼, then has a perfect matching. Proof. Let be -regular µ-bipartite graph. By regularity,, and hence. Let Ë. Denote by ½ the set of the edges with an end in Ë, and by ¾ the set of the edges with an end in Æ Ëµ. Clearly, ½ ¾. Therefore, Æ Ëµ ¾ ½ Ë, and so Æ Ëµ Ë. By Theorem 3.9, has a matching that saturates. Since, this matching is necessarily perfect.

40 3.3 Matchings 39 Applications of Hall s theorem DEFINITION. Let Ë Ë ½ Ë ¾ Ë Ñ be a family of finite nonempty subsets of a set Ë. (Ë need not be distinct.) A transversal (or a system of distinct representatives) of Ë is a subset Ì Ë of Ñ distinct elements one from each Ë. As an example, let Ë ½, and let Ë ½ Ë ¾ ½ ¾, Ë ¾ and Ë ½. For Ë Ë ½ Ë ¾ Ë Ë, the set Ì ½ ¾ is a transversal. If we add the set Ë ¾ to Ë, then it is impossible to find a transversal for this new family. The connection of transversals to the Marriage Theorem is as follows. Let Ë and ½ Ñ. Form an µ-bipartite graph such that there is an edge µ if and only if ¾ Ë. The possible transversals Ì of Ë are then obtained from the matchings Å saturating in by taking the ends in of the edges of Å. Corollary 3.2. Let Ë be a family of finite nonempty sets. Then Ë has a transversal if and only if the union of any of the subsets Ë of Ë contains at least elements. Example 3.6. An Ñ Ò latin rectangle is an Ñ Ò integer matrix Å with entries Å ¾ ½ Ò such that the entries in the same row and in the same column are different. Moreover, if Ñ Ò, then Å is a latin square. Note that in a Ñ Ò latin rectangle Å, we always have that Ñ Ò. We show the following: Let Å be an Ñ Ò latin rectangle (with Ñ Ò). Then M can be extended to a latin square by the addition of Ò Ñ new rows. The claim follows when we show that Å can be extended to an Ñ ½µ Ò latin rectangle. Let ½ Ò be the set of those elements that do not occur in the -th column of Å. Clearly, Ò Ñ for each, and hence È ¾Á Á Ò Ñµ for all subsets Á ½ Ò. Now ¾Á Á, since otherwise at least one element from the union would be in more than Ò Ñ of the sets with ¾ Á. However, each row has all the Ò elements, and therefore each is missing from exactly Ò Ñ columns. By Marriage Theorem, the family ½ ¾ Ò has a transversal, and this transversal can be added as a new row to Å. This proves the claim. Tutte s theorem The next theorem is a classic characterization of perfect matchings. DEFINITION. A connected component of a graph is said to be odd (even), if it has an odd (even) number of vertices. Denote by Ó µ the number of odd connected components in. Denote by Ñ µ be the number of edges in a maximum matching of a graph. Theorem 3.10 (Tutte-Berge Formula). Each maximum matching of a graph has elements. Ñ µ ÑÒ ËÎ Ë Ó Ëµ ¾ (3.3)

41 3.3 Matchings 40 Note that the condition in (ii) includes the case, where Ë. Proof. We prove the result for connected graphs. The result then follows for disconnected graphs by adding the formulas for the connected components. We observe first that holds in (3.3), since, for all Ë Î, Ñ µ Ë Ñ Ëµ Ë Î Ò Ë Ó Ëµ ¾ Ë Ó Ëµ ¾ Indeed, each odd component of Ë must have at least one unsaturated vertex. The proof proceeds by induction on. If ½, then the claim is trivial. Suppose that ¾. Assume first that there exists a vertex Ú ¾ such that Ú is saturated by all maximum matchings. Then Ñ Úµ Ñ µ ½. For a subset Ë ¼ Ú, denote Ë Ë ¼ Ú. By the induction hypothesis, for all Ë ¼ Ú, Ñ µ ½ ½ ¾ ½µ Ë ¼ Ó Ë ¼ Úµµ ½ ¾ Ë Ó Ëµµµ ½ The claim follows from this. Suppose then that for each vertex Ú, there is a maximum matching that does not saturate Ú. We show that each maximum matching leaves exactly one vertex unsaturated. Suppose to the contrary, and let Å be a maximum matching having two different unsaturated vertices Ù and Ú, and choose Å so that the distance Ù Úµ is as small as possible. Now Ù Úµ ¾, since otherwise ÙÚ ¾ could be added to Å, contradicting the maximality of Å. Let Û be an intermediate vertex on a shortest path Ù Ú. By assumption, there exists a maximum matching Æ that does not saturate Û. We can choose Æ such that the intersection Å Æ is maximal. Since Ù Ûµ Ù Úµ and Û Úµ Ù Úµ, Æ saturates both Ù and Ú. The (maximum) matchings Æ and Å leave equally many vertices unsaturated, and hence there exists another vertex Ü Û saturated by Å but which is unsaturated by Æ. Let ÜÝ ¾ be the edge in Å incident with Ü. If Ý is also unsaturated by Æ, then Æ is a matching, contradicting maximality of Æ. It also follows that Ý Û. Therefore there exists an edge ¼ ÝÞ in Æ, where Þ Ü. But now Æ ¼ Æ Ò ¼ is a maximum matching that does not saturate Û. However, Æ Å Æ ¼ Å contradicts the choice of Æ. Therefore, every maximum matching leaves exactly one vertex unsaturated, i.e., Ñ µ ½µ¾. In this case, for Ë, the right hand side of (3.3) gets value ½µ¾, and hence, by the beginning of the proof, this must be the minimum of the right hand side. For perfect matchings we have the following corollary, since for a perfect matching we have Ñ µ ½¾µ. Theorem 3.11 (TUTTE (1947)). Let be a nontrivial graph. The following are equivalent. (i) has a perfect matching. (ii) For every proper subset Ë Î, Ó Ëµ Ë.

42 3.3 Matchings 41 Tutte s theorem does not provide a good algorithm for constructing a perfect matching, because the theorem requires too many cases. Its applications are mainly in the proofs of other results that are related to matchings. There is a good algorithm due to EDMONDS (1965), which uses blossom shrinkings, but this algorithm is somewhat involved. Example 3.7. The simplest connected graph that has no perfect matching is the path È. Here removing the middle vertex creates two odd components. The next 3-regular graph (known as the Sylvester graph) does not have a perfect matching, because removing the black vertex results in a graph with three odd connected components. This graph is the smallest regular graph with an odd degree that has no perfect matching. Using Theorem 3.11 we can give a short proof of PETERSEN s result for 3-regular graphs (1891). Theorem 3.12 (PETERSEN (1891)). If is a bridgeless -regular graph, then it has a perfect matching. Proof. Let Ë be a proper subset of Î, and let, ¾ ½ Ø, be the odd connected components of Ë. Denote by Ñ the number of edges with one end in and the other in Ë. Since is 3-regular, Úµ and Úµ Ë Ú¾ Ú¾Ë The first of these implies that Ñ Ú¾ Úµ ¾ is odd. Furthermore, Ñ ½, because has no bridges, and therefore Ñ. Hence the number of odd connected components of Ë satisfies Ø ½ Ø ½ Ñ ½ Ú¾Ë and so, by Theorem 3.11, has a perfect matching. Stable Marriages Úµ Ë DEFINITION. Consider a bipartite graph with a bipartition µ of the vertex set. In addition, each vertex Ü ¾ supplies an order of preferences of the vertices of Æ Üµ. We write Ù Ü Ú, if Ü prefers Ú to Ù. (Here Ù Ú ¾, if Ü ¾, and Ù Ú ¾, if Ü ¾.) A matching Å of is said to be stable, if for each unmatched pair ÜÝ ¾ Å (with Ü ¾ and Ý ¾ ), it is not the case that Ü and Ý prefer each other better than their matched companions: ÜÚ ¾ Å and Ý Ü Ú or ÙÝ ¾ Å and Ü Ý Ù

43 3.3 Matchings 42 We omit the proof of the next theorem. Theorem For bipartite graphs, a stable matching exists for all lists of preferences. Example 3.8. That was the good news. There is a catch, of course. A stable matching need not saturate and. For instance, the graph on the right does have a perfect matching (of edges). Suppose the preferences are the following: ¾ ½ ½ ¾ ½ ¾ ¾ ¾ Then there is no stable matchings of four edges. A stable matching of is the following: Å ¾ which leaves ½ and unmatched. (You should check that there is no stable matching containing the edges ½ and ¾.) Theorem Let à ÒÒ be a complete bipartite graph. Then has a perfect and stable matching for all lists of preferences. Proof. Let the bipartition be µ. The algorithm by GALE AND SHAPLEY (1962) works as follows. Procedure. Set Å ¼, and È Üµ for all Ü ¾. Then iterate the following process until all vertices are saturated: Choose a vertex Ü ¾ that is unsaturated in Å ½. Let Ý ¾ be the most preferred vertex for Ü such that Ý ¾ È Üµ. (1) Add Ý to È Üµ. (2) If Ý is not saturated, then set Å Å ½ ÜÝ. (3) If ÞÝ ¾ Å ½ and Þ Ý Ü, then set Å Å ½ Ò Þݵ ÜÝ. First of all, the procedure terminates, since a vertex Ü ¾ takes part in the iteration at most Ò times (once for each Ý ¾ ). The final outcome, say Å Å Ø, is a perfect matching, since the iteration continues until there are no unsaturated vertices Ü ¾. Also, the matching Å Å Ø is stable. Note first that, by (3), if ÜÝ ¾ Å and ÞÝ ¾ Å for some Ü Þ and, then Ü Ý Þ. Assume the that ÜÝ ¾ Å, but Ý Ü Þ for some Þ ¾. Then ÜÝ is added to the matching at some step, ÜÝ ¾ Å, which means that Þ ¾ È Üµ at this step (otherwise Ü would have proposed Þ). Hence Ü took part in the iteration at an earlier step Å, (where Þ was put to the list È Üµ, but ÜÞ was not added). Thus, for some Ù ¾, ÙÞ ¾ Å ½ and Ü Þ Ù, and so in Å the vertex Þ is matched to some Û with Ü Þ Û. Similarly, if Ü Ý Ú for some Ú ¾, then Ý Ú Þ for the vertex Þ ¾ such that ÚÞ ¾ Å.

44 4 Colourings 4.1 Edge colourings Colourings of edges and vertices of a graph are useful, when one is interested in classifying relations between objects. There are two sides of colourings. In the general case, a graph with a colouring «is given, and we study the properties of this pair ««µ. This is the situation, e.g., in transportation networks with bus and train links, where the colour (buss, train) of an edge tells the nature of a link. In the chromatic theory, is first given and then we search for a colouring that the satisfies required properties. One of the important properties of colourings is properness. In a proper colouring adjacent edges or vertices are coloured differently. Edge chromatic number DEFINITION. A -edge colouring «½ of a graph is an assignment of colours to its edges. We write «to indicate that has the edge colouring «. A vertex Ú ¾ and a colour ¾ ½ are incident with each other, if «ÚÙµ for some ÚÙ ¾. If Ú ¾ is not incident with a colour, then is available for Ú. The colouring «is proper, if no two adjacent edges obtain the same colour: «½ µ «¾ µ for adjacent ½ and ¾. The edge chromatic number ¼ µ of is defined as ¼ µ ÑÒ there exists a proper -edge colouring of A -edge colouring «can be thought of as a partition ½ ¾ of, where «µ. Note that it is possible that for some. We adopt a simplified notation «½ ¾ Ø ½ ¾ Ø for the subgraph of consisting of those edges that have a colour ½, ¾,..., or Ø. That is, the edges having other colours are removed. Lemma 4.1. Each colour set in a proper -edge colouring is a matching. Moreover, for each graph, µ ¼ µ. Proof. This is clear.

45 4.1 Edge colourings 44 Example 4.1. The three numbers in Lemma 4.1 can be equal. This happens, for instance, when à ½Ò is a star. But often the inequalities are strict. A star, and a graph with ¼ µ. Optimal colourings We show that for bipartite graphs the lower bound is always optimal: ¼ µ µ. Lemma 4.2. Let be a connected graph that is not an odd cycle. Then there exists a 2-edge colouring (that need not be proper), in which both colours are incident with each vertex Ú with Úµ ¾. Proof. Assume that is nontrivial; otherwise, the claim is trivial. (1) Suppose first that is eulerian. If is an even cycle, then a 2-edge colouring exists as required. Otherwise, since now Úµ is even for all Ú, has a vertex Ú ½ with Ú ½ µ. Let ½ ¾ Ø be an Euler tour of, where Ú Ú ½ (and Ú Ø ½ Ú ½ ). Define ½ if is odd «µ ¾ if is even Hence the ends of the edges for ¾ ¾ Ø ½ are incident with both colours. All vertices are among these ends. The condition Ú ½ µ guarantees this for Ú ½. Hence the claim holds in the eulerian case. (2) Suppose then that is not eulerian. We define a new graph ¼ by adding a vertex Ú ¼ to and connecting Ú ¼ to each Ú ¾ of odd degree. In ¼ every vertex has even degree including Ú ¼ (by the handshaking lemma), and hence ¼ is eulerian. Let ¼ ½ Ø be an eulerian tour of ¼, where Ú Ú ½. By the previous case, there is a required colouring «of ¼ as above. Now, «restricted to is a colouring of as required by the claim, since each vertex Ú with odd degree Ú µ is entered and departed at least once in the tour by an edge of the original graph : ½. DEFINITION. For a -edge colouring «of, let «Úµ Ú is incident with ¾ ½ A -edge colouring is an improvement of «, if Ú ¼ ¾ ½ ¾ ½ ¾ ½

46 4.1 Edge colourings 45 Ú¾ Úµ Ú¾ Also, «is optimal, if it cannot be improved. «Úµ Notice that we always have «Úµ Úµ, and if «is proper, then «Úµ Úµ, and in this case «is optimal. Thus an improvement of a colouring is a change towards a proper colouring. Note also that a graph always has an optimal -edge colouring, but it need not have any proper -edge colourings. The next lemma is obvious. Lemma 4.3. An edge colouring «of is proper if and only if «Úµ Úµ for all vertices Ú ¾. Lemma 4.4. Let «be an optimal -edge colouring of, and let Ú ¾. Suppose that the colour is available for Ú, and the colour is incident with Ú at least twice. Then the connected component À of «that contains Ú, is an odd cycle. Proof. Suppose the connected component À is not an odd cycle. By Lemma 4.2, À has a 2-edge colouring À, in which both and are incident with each vertex Ü with À ܵ ¾. (We have renamed the colours ½ and ¾ to and.) We obtain a recolouring of as follows: µ if ¾ À µ «µ if ¾ À Since À Úµ ¾ (by the assumption on the colour ) and in both colours and are now incident with Ú, Úµ «Úµ ½. Furthermore, by the construction of, we have È È Ùµ «Ùµ for all Ù Ú. Therefore Ù¾ Ùµ Ù¾ «Ùµ, which contradicts the optimality of «. Hence À is an odd cycle. Theorem 4.1 (KÖNIG (1916)). If is bipartite, then ¼ µ µ. Proof. Let «be an optimal -edge colouring of a bipartite, where µ. If there were a Ú ¾ with «Úµ Úµ, then by Lemma 4.4, would contain an odd cycle. But a bipartite graph does not contain such cycles. Therefore, for all vertices Ú, «Úµ Úµ. By Lemma 4.3, «is a proper colouring, and ¼ µ as required. Vizing s theorem In general we can have ¼ µ µ as one of our examples did show. The following important theorem, due to VIZING, shows that the edge chromatic number of a graph misses µ by at most one colour. Theorem 4.2 (VIZING (1964)). For any graph, µ ¼ µ µ ½. Proof. Let µ. We need only to show that ¼ µ ½. Suppose on the contrary that ¼ µ ½, and let «be an optimal ½µ-edge colouring of.

47 4.1 Edge colourings 46 We have (trivially) Ùµ ½ ¼ µ for all Ù ¾, and so Claim 1. For each Ù ¾, there exists an available colour Ùµ for Ù. Moreover, by the counter hypothesis, «is not a proper colouring, and hence there exists a Ú ¾ with «Úµ Úµ, and hence a colour ½ that is incident with Ú at least twice, say «ÚÙ ½ µ ½ «Úܵ (4.1) Claim 2. There is a sequence of vertices Ù ½ Ù ¾ such that «ÚÙ µ and ½ Ù µ Indeed, let Ù ½ be as in (4.1). Assume we have already found the vertices Ù ½ Ù, with ½, such that the claim holds for these. Suppose, contrary to the claim, that Ú is not incident with Ù µ ½. We can recolour the edges ÚÙ by ½ for ¾ ½, and obtain in this way an improvement of «. Here Ú gains a new colour ½. Also, each Ù gains a new colour ½ (and may loose the colour ). Therefore, for each Ù either its number of colours remains the same or it increases by one. This contradicts the optimality of «, and proves Claim 2. Now, let Ø be the smallest index such that for some Ö Ø, Ø ½ Ö. Such an index Ø exists, because Úµ is finite.. Ù ¾ Ù Ö ½ Ù ½ ¾ ½ Ù Ö Ú Ü ½... Ö ½ Ö Ø ½ Ø Ù Ø Let be a recolouring of such that for ½ Ö ½, ÚÙ µ ½, and for all other edges, µ «µ. Claim 3. is an optimal ½µ-edge colouring of. Indeed, Úµ «Úµ and Ùµ «Ùµ for all Ù, since each Ù (½ Ö ½) gains a new colour ½ although it may loose one of its old colours.. Ù ¾ Ù Ö ½ Ù ½ Ö Ù Ö... Ö Ø ½ Ú Ø Ù Ø ¾ ½ Ü Let then the colouring be obtained from by recolouring the edges ÚÙ by ½ for Ö Ø. Now, ÚÙ Ø is recoloured by Ö Ø ½. Claim 4. is an optimal ½µ-edge colouring of. Indeed, the fact Ö Ø ½ ensures that Ö is a new colour incident with Ù Ø, and thus that Ù Ø µ Ù Ø µ. For all other vertices, Ùµ Ùµ follows as for.. Ù ¾ Ù Ö ½ Ù ½ Ö Ù Ö... Ö ½ Ú Ö Ù Ø ¾ ½ Ü

48 4.2 Ramsey Theory 47 By Claim 1, there is a colour ¼ Úµ that is available for Ú. By Lemma 4.4, the connected components À ½ of ¼ Ö and À ¾ of ¼ Ö containing the vertex Ú are cycles, that is, À ½ is a cycle ÚÙ Ö ½ µè ½ Ù Ö Úµ and À ¾ is a cycle ÚÙ Ö ½ µè ¾ Ù Ø Úµ, where both È ½ Ù Ö ½ Ù Ö and È ¾ Ù Ö ½ Ù Ø are paths. However, the edges of È ½ and È ¾ have the same colours with respect to and (either ¼ or Ö ). This is not possible, since È ½ ends in Ù Ö while È ¾ ends in a different vertex Ù Ø. This contradiction proves the theorem. Example 4.2. We show that ¼ µ for the Petersen graph. Indeed, by Vizing theorem, ¼ µ or. Suppose colours suffice. Let Ú ½ Ú Ú ½ be the outer cycle and ¼ Ù ½ Ù Ù ½ the inner cycle of such that Ú Ù ¾ for all. Observe that every vertex is adjacent to all colours ½ ¾. Now uses one colour (say ½) once and the other two twice. This can be done uniquely (up to permutations): Ú ½ ½ Ú ¾ ¾ Ú Ú ¾ Ú Ú ½ Hence Ú ½ ¾ Ù ½, Ú ¾ Ù ¾, Ú ½ Ù, Ú ½ Ù, Ú ½ Ù. However, this means that ½ cannot be a colour of any edge in ¼. Since ¼ needs three colours, the claim follows. Edge Colouring Problem. Vizing s theorem (nor its present proof) does not offer any characterization for the graphs, for which ¼ µ µ ½. In fact, it is one of the famous open problems of graph theory to find such a characterization. The answer is known (only) for some special classes of graphs. By HOLYER (1981), the problem whether ¼ µ is µ or µ ½ is NP-complete. The proof of Vizing s theorem can be used to obtain a proper colouring of with at most µ ½ colours, when the word optimal is forgotten: colour first the edges as well as you can (if nothing better, then arbitrarily in two colours), and use the proof iteratively to improve the colouring until no improvement is possible then the proof says that the result is a proper colouring. 4.2 Ramsey Theory In general, Ramsey theory studies unavoidable patterns in combinatorics. We consider an instance of this theory mainly for edge colourings (that need not be proper). A typical example of a Ramsey property is the following: given 6 persons each pair of whom are either friends or enemies, there are then 3 persons who are mutual friends or mutual enemies. In graph theoretic terms this means that each colouring of the edges of à with 2 colours results in a monochromatic triangle. Turan s theorem for complete graphs We shall first consider the problem of finding a general condition for Ã Ô to appear in a graph. It is clear that every graph contains à ½, and that every nondiscrete graph contains à ¾.

49 4.2 Ramsey Theory 48 DEFINITION. A complete Ô-partite graph consists of Ô discrete and disjoint induced subgraphs ½ ¾ Ô, where ÙÚ ¾ if and only if Ù and Ú belong to different parts, and with. Note that a complete Ô-partite graph is completely determined by its discrete parts, ¾ ½ Ô. Let Ô, and let À À ÒÔ be the complete Ô ½µ-partite graph of order Ò Ø Ô ½µ Ö, where Ö ¾ ½ Ô ½ and Ø ¼, such that there are Ö parts À ½ À Ö of order Ø ½ and Ô ½ Ö parts À Ö ½ À Ô ½ of order Ø (when Ø ¼). (Here Ö is the positive residue of Ò modulo Ô ½µ, and is thus determined by Ò and Ô.) By its definition, Ã Ô À. One can compute that the number À of edges of À is equal to Ì Ò Ôµ Ô ¾ ¾ Ô ½µ Ò¾ Ö ¾ ½ Ö Ô ½ The next result shows that the above bound Ì Ò Ôµ is optimal. (4.2) Theorem 4.3 (TURÁN (1941)). If a graph of order Ò has contains a complete subgraph à Ô. Ì Ò Ôµ edges, then Proof. Let Ò Ô ½µØ Ö for ½ Ö Ô ½ and Ø ¼. We prove the claim by induction on Ø. If Ø ¼, then Ì Ò Ôµ Ò Ò ½µ¾, and there is nothing to prove. Suppose then that Ø ½, and let be a graph of order Ò such that is maximum subject to the condition à Ô. Now contains a complete subgraph Ã Ô ½, since adding any one edge to results in a à Ô, and Ô ½ vertices of this Ã Ô induce a subgraph Ã Ô ½. Each Ú ¾ is adjacent to at most Ô ¾ vertices of ; otherwise Ú Ã Ô. Furthermore, à Ô, and Ò Ô ½. Because Ò Ô ½ Ø ½µ Ô ½µ Ö, we can apply the induction hypothesis to obtain Ì Ò Ô ½ Ôµ. Now Ô ½µ Ô ¾µ Ì Ò Ô ½ Ôµ Ò Ô ½µ Ô ¾µ ¾ which proves the claim. Ì Ò Ôµ When Theorem 4.3 is applied to triangles Ã, we have the following interesting case. Corollary 4.1 ( MANTEL (1907)). If a graph has triangle Ã. ½ ¾ edges, then contains a

50 4.2 Ramsey Theory 49 Ramsey s theorem DEFINITION. Let «be an edge colouring of. A subgraph À is said to be (-) monochromatic, if all edges of À have the same colour. The following theorem is one of the jewels of combinatorics. Theorem 4.4 (RAMSEY (1930)). Let Ô Õ ¾ be any integers. Then there exists a (smallest) integer Ê Ô Õµ such that for all Ò Ê Ô Õµ, any 2-edge colouring of Ã Ò ½ ¾ contains a ½-monochromatic Ã Ô or a ¾-monochromatic à Õ. Before proving this, we give an equivalent statement. Recall that a subset Î is stable, if is a discrete graph. Theorem 4.5. Let Ô Õ ¾ be any integers. Then there exists a (smallest) integer Ê Ô Õµ such that for all Ò Ê Ô Õµ, any graph of order Ò contains a complete subgraph of order Ô or a stable set of order Õ. Be patient, this will follow from Theorem 4.6. The number Ê Ô Õµ is known as the Ramsey number for Ô and Õ. It is clear that Ê Ô ¾µ Ô and Ê ¾ Õµ Õ. Theorems 4.4 and 4.5 follow from the next result which shows (inductively) that an upper bound exists for the Ramsey numbers Ê Ô Õµ. Theorem 4.6 (ERDÖS and SZEKERES (1935)). The Ramsey number Ê Ô Õµ exists for all Ô Õ ¾, and Ê Ô Õµ Ê Ô Õ ½µ Ê Ô ½ Õµ Proof. We use induction on Ô Õ. It is clear that Ê Ô Õµ exists for Ô ¾ or Õ ¾, and it is thus exists for Ô Õ. It is now sufficient to show that if is a graph of order Ê Ô Õ ½µ Ê Ô ½ Õµ, then it has a complete subgraph of order Ô or a stable subset of order Õ. Let Ú ¾, and denote by Î Ò Æ Úµ Úµ the set of vertices that are not adjacent to Ú. Since has Ê Ô Õ ½µ Ê Ô ½ Õµ ½ vertices different from Ú, either Æ Úµ Ê Ô ½ Õµ or Ê Ô Õ ½µ (or both). Assume first that Æ Úµ Ê Ô ½ Õµ. By the definition of Ramsey numbers, Æ Úµ contains a complete subgraph of order Ô ½ or a stable subset Ë of order Õ. In the first case, Ú induces a complete subgraph Ã Ô in, and in the second case the same stable set of order Õ is good for. If Ê Ô Õ ½µ, then contains a complete subgraph of order Ô or a stable subset Ë of order Õ ½. In the first case, the same complete subgraph of order Ô is good for, and in the second case, Ë Ú is a stable subset of of Õ vertices. This proves the claim. A concrete upper bound is given in the following result.

51 4.2 Ramsey Theory 50 Theorem 4.7 ( ERDÖS and SZEKERES (1935)). For all Ô Õ ¾, Ô Õ ¾ Ê Ô Õµ Ô ½ Proof. For Ô ¾ or Õ ¾, the claim is clear. We use induction on Ô Õ for the general statement. Assume that Ô Õ. By Theorem 4.6 and the induction hypothesis, which is what we wanted. Ê Ô Õµ Ê Ô Õ ½µ Ê Ô ½ Õµ Ô Õ Ô Õ Ô ½ Ô ¾ Ô Õ ¾ Ô ½ In the table below we give some known values and estimates for the Ramsey numbers Ê Ô Õµ. As can be read from the table 1, not so much is known about these numbers. ÔÒÕ The first unknown Ê Ô Ôµ (where Ô Õ) is for Ô. It has been verified that Ê µ, but to determine the exact value is an open problem. Generalizations Theorem 4.4 can be generalized as follows. Theorem 4.8. Let Õ ¾ be integers for ¾ ½ with ¾. Then there exists an integer Ê Ê Õ ½ Õ ¾ Õ µ such that for all Ò Ê, any -edge colouring of Ã Ò has an -monochromatic Ã Õ for some. Proof. The proof is by induction on. The case ¾ is treated in Theorem 4.4. For ¾, we show that Ê Õ ½ Õ µ Ê Õ ½ Õ ¾ Ôµ, where Ô Ê Õ ½ Õ µ. Let Ò Ê Õ ½ Õ ¾ Ôµ, and let «ÃÒ ½ be an edge colouring. Let ÃÒ ½ ½ be obtained from «by identifying the colours ½ and : «µ if «µ ½ µ ½ if «µ ½ or By the induction hypothesis, Ã Ò has an -monochromatic Ã Õ for some ½ ¾ (and we are done, since this subgraph is monochromatic in Ã Ò «) or Ã Ò has a ½µ-monochromatic subgraph À à Ô. In the latter case, by Theorem 4.4, À «and thus Ã Ò «has a ½µmonochromatic or a -monochromatic subgraph, and this proves the claim. ½ S.P. RADZISZOWSKI, Small Ramsey numbers, Electronic J. of Combin., 2000 on the Web

52 4.2 Ramsey Theory 51 Since for each graph À, À Ã Ñ for Ñ À, we have Corollary 4.2. Let ¾ and À ½ À ¾ À be arbitrary graphs. Then there exists an integer Ê À ½ À ¾ À µ such that for all complete graphs Ã Ò with Ò Ê À ½ À ¾ À µ and for all -edge colourings «of à Ò, à «Ò contains an -monochromatic subgraph À for some. This generalization is trivial from Theorem 4.8. However, the generalized Ramsey numbers Ê À ½ À ¾ À µ can be much smaller than their counter parts (for complete graphs) in Theorem 4.8. Example 4.3. We leave the following statement as an exercise: If Ì is a tree of order Ñ, then Ê Ì Ã Ò µ Ñ ½µ Ò ½µ ½ that is, any graph of order at least Ê Ì Ã Ò µ contains a subgraph isomorphic to Ì, or the complement of contains a complete subgraph à Ò. Examples of Ramsey numbers Some exact values are known in Corollary 4.2, even in more general cases, for some dear graphs (see RADZISZOWSKI s survey). Below we list some of these results for cases, where the graphs are equal. To this end, let Ê µ Ê µ times µ The best known lower bound of Ê ¾ µ for connected graphs was obtained by BURR AND ERDÖS (1976), ½ Ê ¾ µ connectedµ Here is a list of some special cases: Ê ¾ È Ò µ Ò Ê ¾ Ò µ Ò ¾ ¾Ò ½ Ò¾ ½ ¾Ò ½ Ê ¾ à ½Ò µ ¾Ò Ê ¾ à ¾ µ ½¼ ½ if Ò or Ò if Ò and Ò odd if Ò and Ò even if Ò is even if Ò is odd Ê ¾ à µ ½ The values Ê ¾ à ¾Ò µ are known for Ò ½, and in general, Ê ¾ à ¾Ò µ Ò ¾. The value Ê ¾ à ¾½ µ is either or. Let Ï Ò denote the wheel on Ò vertices. It is a cycle Ò ½, where a vertex Ú with degree Ò ½ is attached. Note that Ï Ã. Then Ê ¾ Ï µ ½ and Ê ¾ Ï µ ½

53 4.3 Vertex colourings 52 For three colours, much less is known. In fact, the only nontrivial result for complete graphs is: Ê Ã µ ½. Also, ½¾ Ê Ã µ ¾, and Ê Ã µ, but no nontrivial upper bound is known for Ê Ã µ. For the square, we know that Ê µ ½½. Needless to say that no exact values are known for Ê Ã Ò µ for and Ò. It follows from Theorem 4.4 that for any complete à Ò, there exists a graph (well, any sufficiently large complete graph) such that any ¾-edge colouring of has a monochromatic (induced) subgraph à Ò. Note, however, that in Corollary 4.2 the monochromatic subgraph À is not required to be induced. The following impressive theorem improves the results we have mentioned in this chapter and it has a difficult proof. Theorem 4.9 (DEUBER, ERDÖS, HAJNAL, PÓSA, and RÖDL (around 1973)). Let À be any graph. Then there exists a graph such that any ¾-edge colouring of has an monochromatic induced subgraph À. Example 4.4. As an application of Ramsey s theorem, we shortly describe Schur s theorem. For this, consider the partition ½ ½¼ ½ ¾ ½½ ½¾ of the set Æ ½ ½ ½. We observe that in no partition class there are three integers such that Ü Ý Þ. However, if you try to partition Æ ½ into three classes, then you are bound to find a class, where Ü Ý Þ has a solution. SCHUR (1916) solved this problem in a general setting. The following gives a short proof using Ramsey s theorem. For each Ò ½, there exists an integer Ë Òµ such that any partition Ë ½ Ë Ò of Æ Ë Òµ has a class Ë containing two integers Ü Ý such that Ü Ý ¾ Ë. Indeed, let Ë Òµ Ê µ, where occurs Ò times, and let à be a complete on Æ Ë Òµ. For a partition Ë ½ Ë Ò of Æ Ë Òµ, define an edge colouring «of à by «µ if ¾ Ë By Theorem 4.8, à «has a monochromatic triangle, that is, there are three vertices ½ Ø Ë Òµ such that Ø Ø ¾ Ë for some. But Ø µ µ Ø proves the claim. There are quite many interesting corollaries to Ramsey s theorem in various parts of mathematics including not only graph theory, but also, e.g., geometry and algebra, see R.L. GRAHAM, B.L. ROTHSCHILD AND J.L. SPENCER, Ramsey Theory, Wiley, (2nd ed.) Vertex colourings The vertices of a graph can also be classified using colourings. These colourings tell that certain vertices have a common property (or that they are similar in some respect), if they share the same colour. In this chapter, we shall concentrate on proper vertex colourings, where adjacent vertices get different colours.

54 4.3 Vertex colourings 53 The chromatic number DEFINITION. A -colouring (or a -vertex colouring) of a graph is a mapping «Î ½. The colouring «is proper, if adjacent vertices obtain a different colour: for all ÙÚ ¾, we have «Ùµ «Úµ. A colour ¾ ½ is said to be available for a vertex Ú, if no neighbour of Ú is coloured by. A graph is -colourable, if there is a proper -colouring for. The (vertex) chromatic number µ of is defines as µ ÑÒ there exists a proper -colouring of If µ, then is -chromatic. Each proper vertex colouring «Î ½ provides a partition Î ½ Î ¾ Î of the vertex set Î, where Î Ú «Úµ. Example 4.5. The graph on the right, which is often called a wheel (of order ), is -chromatic. By the definitions, a graph is ¾-colourable if and only if it is bipartite. Again, the names of the colours are immaterial: Lemma 4.5. Let «be a proper -colouring of, and let be any permutation of the colours. Then the colouring «is a proper -colouring of. Proof. Indeed, if «Î ½ is proper, and if ½ ½ is a bijection, then ÙÚ ¾ implies that «Ùµ «Úµ, and hence also that «Ùµ «Úµ. It follows that «is a proper colouring. Example 4.6. A graph is triangle-free, if it has no subgraphs isomorphic to Ã. We show that there are triangle-free graphs with arbitrarily large chromatic numbers. The following construction is due to GRÖTZEL: Let be any triangle-free graph with Î Ú ½ Ú ¾ Ú Ò. Let Ø be a new graph obtained by adding Ò ½ new vertices Ú and Ù ½ Ù ¾ Ù Ò such that Ø has all the edges of plus the edges Ù Ú and Ù Ü for all Ü ¾ Æ Ú µ and for all ¾ ½ Ò. Claim. Ø is triangle-free and it is ½-chromatic Indeed, let Í Ù ½ Ù Ò. We show first that Ø is triangle-free. Now, Í is stable, and so a triangle contains at most (and thus exactly) one vertex Ù ¾ Í. If Ù Ú Ú induces a triangle, so does Ú Ú Ú by the definition of Ø, but the latter triangle is already in ; a contradiction. For the chromatic number we notice first that Ø µ ½µ. If «is a proper -colouring of, extend it by setting «Ù µ «Ú µ and «Úµ ½.

55 4.3 Vertex colourings 54 Secondly, Ø µ. Assume that «is a proper -colouring of Ø, say with «Úµ. Then «Ù µ. Recolour each Ú by «Ù µ. This gives a proper ½µ-colouring to ; a contradiction. Therefore Ø µ ½. Now using inductively the above construction starting from the triangle-free graph à ¾, we obtain larger triangle -free graphs with high chromatic numbers. Critical graphs DEFINITION. A -chromatic graph is said to be -critical, if Àµ for all À with À. In a critical graph an elimination of any edge and of any vertex will reduce the chromatic number: µ µ and Úµ µ for ¾ and Ú ¾. Each Ã Ò is Ò- critical, since in Ã Ò ÙÚµ the vertices Ù and Ú can gain the same colour. Example 4.7. The graph à ¾ È ¾ is the only 2-critical graph. The 3-critical graphs are exactly the odd cycles ¾Ò ½ for Ò ½, since a 3-chromatic is not bipartite, and thus must have a cycle of odd length. Theorem If is -critical for ¾, then it is connected, and Æ µ ½. Proof. Note that for any graph with the connected components ½ ¾ Ñ, µ ÑÜ µ ¾ ½ Ñ Connectivity claim follows from this observation. Let then be -critical, but Æ µ Úµ ¾ for Ú ¾. Since is critical, there is a proper ½µ-colouring of Ú. Now Ú is adjacent to only Æ µ ½ vertices. But there are colours, and hence there is an available colour for Ú. If we recolour Ú by, then a proper ½µ-colouring is obtained for ; a contradiction. The case (iii) of the next theorem is due to SZEKERES AND WILF (1968). Theorem Let be any graph with µ. (i) has a -critical subgraph À. (ii) has at least vertices of degree ½. (iii) ½ ÑÜ À Æ Àµ. Proof. For (i), we observe that a -critical subgraph À is obtained by removing vertices and edges from as long as the chromatic number remains. For (ii), let À be -critical. By Theorem 4.10, À Úµ ½ for every Ú ¾ À. Of course, also Úµ ½ for every Ú ¾ À. The claim follows, because, clearly, every -critical graph À must have at least vertices. For (iii), let À be -critical. By Theorem 4.10, µ ½ Æ Àµ, which proves this claim.

56 4.3 Vertex colourings 55 Lemma 4.6. Let Ú be a cut vertex of a connected graph, and let, for ¾ ½ Ñ, be the connected components of Ú. Denote Ú. Then µ ÑÜ µ ¾ ½ Ñ. In particular, a critical graph does not have cut vertices. Proof. Suppose each has a proper -colouring «. By Lemma 4.5, we may take «Úµ ½ for all. These -colourings give a -colouring of. Brooks theorem For edge colourings we have Vizing s theorem, but no such strong results are known for vertex colouring. Lemma 4.7. For all graphs, µ µ ½. In fact, there exists a proper colouring «Î ½ µ ½ such that «Úµ Úµ ½ for all vertices Ú ¾. Proof. We use greedy colouring to prove the claim. Let Î Ú ½ Ú Ò be ordered in some way, and define «Î Æ inductively as follows: «Ú ½ µ ½, and «Ú µ ÑÒ «Ú Ø µ for all Ø with Ú Ú Ø ¾ Then «is proper, and «Ú µ Ú µ ½ for all. The claim follows from this. Although, we always have µ µ ½, the chromatic number µ usually takes much lower values as seen in the bipartite case. Moreover, the maximum value µ ½ is obtained only in two special cases as was shown by BROOKS in The next proof of Brook s theorem is by LOVÁSZ (1975) as modified by BRYANT (1996). Lemma 4.8. Let be a ¾-connected graph. Then the following are equivalent: (i) is a complete graph or a cycle. (ii) For all Ù Ú ¾, if ÙÚ ¾, then Ù Ú is a separating set. (iii) For all Ù Ú ¾, if Ù Úµ ¾, then Ù Ú is a separating set. Proof. It is clear that (i) implies (ii), and that (ii) implies (iii). We need only to show that (iii) implies (i). Assume then that (iii) holds. We shall show that either is a complete graph or Úµ ¾ for all Ú ¾, from which the theorem follows. First of all, Úµ ¾ for all Ú, since is ¾-connected. Let Û be a vertex of maximum degree, Ûµ µ. If the neighbourhood Æ Ûµ induces a complete subgraph, then is complete. Indeed, otherwise, since is connected, there exists a vertex Ù ¾ Æ ÛµÛ such that Ù is adjacent to a vertex Ú ¾ Æ Ûµ. But then Úµ Ûµ, and this contradicts the choice of Û. Assume then that there are different vertices Ù Ú ¾ Æ Ûµ such that ÙÚ ¾. This means that Ù Úµ ¾ (the shortest path is Ù Û Ú), and by (iii), Ù Ú is a separating

57 4.3 Vertex colourings 56 set of. Consequently, there is a partition Î Ï Ù Ú Í, where Û ¾ Ï, and all paths from a vertex of Ï to a vertex of Í go through either Ù or Ú. We claim that Ï Û, and thus that µ ¾ as required. Suppose on the contrary that Ï ¾. Since Û is not a cut vertex (since has no cut vertices), there exists an Ü ¾ Ï with Ü Û such that ÜÙ ¾ or ÜÚ ¾, say ÜÙ ¾. Since Ú is not a cut vertex, there exists a Ý ¾ Í such that ÙÝ ¾. Hence Ü Ýµ ¾, and by (iii), Ü Ý is a separating set. Accordingly, Î Ï ½ Ü ÝÍ ½, where all paths from Ï ½ to Í ½ pass through Ü or Ý. Assume that Û ¾ Ï ½, and hence that also Ù Ú ¾ Ï ½. (Since ÙÛ ÚÛ ¾ Î ÜÝ). Û Ü Ù Ú Ý There exists a vertex Þ ¾ Í ½. Note that Í ½ Ï Í. If Þ ¾ Ï (or Þ ¾ Í, respectively), then all paths from Þ to Ù must pass through Ü (or Ý, respectively), and Ü (or Ý, respectively) would be a cut vertex of. This contradiction, proves the claim. Theorem 4.12 ( BROOKS (1941)). Let be connected. Then µ µ ½ if and only if either is an odd cycle or a complete graph. Proof. ( ) Indeed, ¾ ½ µ, ¾ ½ µ ¾, and Ã Ò µ Ò, Ã Ò µ Ò ½. (µ) Assume that µ. We may suppose that is -critical. Indeed, assume the claim holds for -critical graphs. Let µ ½, and let À be a -critical proper subgraph. Since Àµ µ ½ Àµ, we must have Àµ Àµ ½, and thus À is a complete graph or an odd cycle. Now is connected, and therefore there exists an edge ÙÚ ¾ with Ù ¾ À and Ú ¾ À. But then Ùµ À Ùµ, and µ Àµ, since À Ã Ò or À Ò. Let then be any -critical graph for ¾. By Lemma 4.6, it is ¾-connected. If is an even cycle, then ¾ µ. Suppose now that is neither complete nor a cycle (odd or even). We show that µ µ. By Lemma 4.8, there exist Ú ½ Ú ¾ ¾ with Ú ½ Ú ¾ µ ¾, say Ú ½ Û ÛÚ ¾ ¾ with Ú ½ Ú ¾ ¾, such that À Ú ½ Ú ¾ is connected. Order Î À Ú Ú Ú Ò such that Ú Ò Û, and for all, À Ú Ûµ À Ú ½ Ûµ Therefore for each ¾ ½ Ò ½, we find at least one such that Ú Ú ¾ (possibly Ú Û). In particular, for all ½ Ò, Æ Ú µ Ú ½ Ú ½ Ú µ µ (4.3) Then colour Ú ½ Ú ¾ Ú Ò in this order as follows: «Ú ½ µ ½ «Ú ¾ µ and «Ú µ ÑÒÖ Ö «Ú µ for all Ú ¾ Æ Ú µ with

58 4.3 Vertex colourings 57 The colouring «is proper. By (4.3), «Ú µ µ for all ¾ ½ Ò ½. Also, Û Ú Ò has two neighbours, Ú ½ and Ú ¾, of the same colour ½, and since Ú Ò has at most µ neighbours, there is an available colour for Ú Ò, and so «Ú Ò µ µ. This shows that has a proper µ-colouring, and, consequently, µ µ. Example 4.8. Suppose we have Ò objects Î Ú ½ Ú Ò, some of which are not compatible (like chemicals that react with each other, or worse, graph theorists who will fight during a conference). In the storage problem we would like to find a partition of the set Î with as few classes as possible such that no class contains two incompatible elements. In graph theoretical terminology we consider the graph Î µ, where Ú Ú ¾ just in case Ú and Ú are incompatible, and we would like to colour the vertices of properly using as few colours as possible. This problem requires that we find µ. Unfortunately, no good algorithms are known for determining µ, and, indeed, the chromatic number problem is NP-complete. Already the problem if µ is NP-complete. (However, as we have seen, the problem whether µ ¾ has a fast algorithm.) The chromatic polynomial A given graph has many different proper vertex colourings «Î ½ for sufficiently large natural numbers. Indeed, see Lemma 4.5 to be certain on this point. DEFINITION. The chromatic polynomial of is the function Æ Æ, where µ ««Î ½ a proper colouring This notion was introduced by BIRKHOFF (1912), BIRKHOFF AND LEWIS (1946), to attack the famous -Colour Theorem, but its applications have turned out to be elsewhere. If µ, then clearly µ ¼, and, indeed, µ ÑÒ µ ¼ Therefore, if we can find the chromatic polynomial of, then we easily compute the chromatic number µ just by evaluating µ for ½ ¾ until we hit a nonzero value. Theorem 4.13 will give the tools for constructing. Example 4.9. Consider the complete graph à on Ú ½ Ú ¾ Ú Ú. Let à µ. The vertex Ú ½ can be first given any of the colours, after which ½ colours are available for Ú ¾. Then Ú has ¾ and finally Ú has available colours. Therefore there are ½µ ¾µ µ different ways to properly colour à with colours, and so à µ ½µ ¾µ µ On the other hand, in the discrete graph à has no edges, and thus any -colouring is a proper colouring. Therefore à µ

59 4.3 Vertex colourings 58 Remark. The considered method for checking the number of possibilities to colour a next vertex is exceptional, and for more nonregular graphs it should be avoided. DEFINITION. Let be a graph, ÙÚ ¾, and let Ü Ü ÙÚµ be a new contracted vertex. The graph on Î Î Ò Ù Úµ Ü is obtained from by contracting the edge, when ¾ has no end Ù or Ú ÛÜ ÛÙ ¾ or ÛÚ ¾ Hence is obtained by introducing a new vertex Ü, and by replacing all edges ÛÙ and ÛÚ by ÛÜ, and the vertices Ù and Ú are deleted. (Of course, no loops or parallel edges are allowed in the new graph.) Theorem Let be a graph, and let ¾. Then Ù Ú Ü µ µ µ Proof. Let ÙÚ. The proper -colourings «Î ½ of can be divided into two disjoint cases, which together show that µ µ µ: (1) If «Ùµ «Úµ, then «corresponds to a unique proper -colouring of, namely «. Hence the number of such colourings is µ. (2) If «Ùµ «Úµ, then «corresponds to a unique proper -colouring of, namely «, when we set «Üµ «Ùµ for the contracted vertex Ü Ü ÙÚµ. Hence the number of such colourings is µ. Theorem The chromatic polynomial is a polynomial. Proof. The proof is by induction on. Indeed, ÃÒ µ Ò for the discrete graph, and for two polynomials È ½ and È ¾, also È ½ È ¾ is a polynomial. The claim follows from Theorem 4.13, since there and have less edges than. The connected components of a graph can be coloured independently, and so Lemma 4.9. Let the graph have the connected components ½ ¾ Ñ. Then µ ½ µ ¾ µ Ñ µ Theorem Let Ì be a tree of order Ò. Then Ì µ ½µ Ò ½. Proof. We use induction on Ò. For Ò ¾, the claim is obvious. Suppose that Ò, and let ÚÙ ¾ Ì, where Ú is a leaf. By Theorem 4.13, Ì µ Ì µ Ì µ. Here Ì is a tree of order Ò ½, and thus, by the induction hypothesis, Ì µ ½µ Ò ¾. The graph Ì consists of the isolated Ú and a tree of order Ò ½. By Lemma 4.9, and the induction hypothesis, Ì µ ½µ Ò ¾. Therefore Ì µ ½µ Ò ½.

60 4.3 Vertex colourings 59 Example Consider the graph of order from the above. Then we have the following reductions. µ Theorem 4.13 reduces the computation of to the discrete graphs. However, we know the chromatic polynomials for trees (and complete graphs, as an exercise), and so there is no need to prolong the reductions beyond these. In our example, we have obtained and so µ µ ½µ µ µ ½µ ¾ ½µ ¾ ¾µ µ µ µ ½µ ¾ ¾µ ½µ ¾µ ½µ ¾µ ¾ ¾ For instance, for colours, there are proper colourings of the given graph. Chromatic Polynomial Problems. It is difficult to determine of a given graph, since the reduction method provided by Theorem 4.13 is time consuming. Also, there is known no characterization, which would tell from any polynomial È µ whether it is a chromatic polynomial of some graph. For instance, the polynomial ¾ is not a chromatic polynomial of any graph, but it seems to satisfy the general properties (that are known or conjectured) of these polynomials. REED (1968) conjectured that the coefficients of a chromatic polynomial should first increase and then decrease in absolute value. REED (1968) and TUTTE (1974) proved that for each of order Ò: The degree of µ equals Ò. The coefficient of Ò equals ½. The coefficient of Ò ½ equals. The constant term is ¼. The coefficients alternate in sign. ѵ Ñ Ñ ½µ Ò ½ for all positive integers Ñ, when is connected. ܵ ¼ for all real numbers ¼ Ü ½.

61 5 Graphs on Surfaces 5.1 Planar graphs The plane representations of graphs are by no means unique. Indeed, a graph can be drawn in arbitrarily many different ways. Also, the properties of a graph are not necessarily immediate from one representation, but may be apparent from another. There are, however, important families of graphs, the surface graphs, that rely on the (topological or geometrical) properties of the drawings of graphs. We restrict ourselves in this chapter to the most natural of these, the planar graphs. The geometry of the plane will be treated intuitively. A planar graph will be a graph that can be drawn in the plane so that no two edges intersect with each other. Such graphs are used, e.g., in the design of electrical (or similar) circuits, where one tries to (or has to) avoid crossing the wires or laser beams. Planar graphs come into use also in some parts of mathematics, especially in group theory and topology. There are fast algorithms (linear time algorithms) for testing whether a graph is planar or not. However, the algorithms are all rather difficult to implement. Most of them are based on an algorithm designed by AUSLANDER AND PARTER (1961) see Section 6.5 of S. SKIENA, Implementing Discrete Mathematics: Combinatorics and Graph Theory with Mathematica, Addison-Wesley, Definition DEFINITION. A graph is a planar graph, if it has a plane figure È µ, called the plane embedding of, where the lines (or continuous curves) corresponding to the edges do not intersect each other except at their ends. The complete bipartite graph à ¾ is a planar graph. DEFINITION. An edge ÙÚ ¾ is subdivided, when it is replaced by a path Ù Ü Ú of length two by introducing a new vertex Ü. A subdivision À of a graph is obtained from by a sequence of subdivisions.

62 5.1 Planar graphs 61 The following result is clear. Lemma 5.1. A graph is planar if and only if its subdivisions are planar. Geometric properties It is clear that the graph theoretical properties of are inherited by all of its plane embeddings. For instance, the way we draw a graph in the plane does not change its maximum degree or its chromatic number. More importantly, there are as we shall see some nontrivial topological (or geometric) properties that are shared by the plane embeddings. We recall first some elements of the plane geometry. Let be an open set of the plane Ê Ê, that is, every point Ü ¾ has a disk centred at Ü and contained in. Then is a region, if any two points Ü Ý ¾ can be joined by a continuous curve the points of which are all in. The boundary µ of a region consists of those points for which every neighbourhood contains points from and its complement. Let be a planar graph, and È µ one of its plane embeddings. Regard now each edge ÙÚ ¾ as a line from Ù to Ú. The set Ê Êµ Ò is open, and it is divided into a finite number of disjoint regions, called the faces of È µ. DEFINITION. A face of È µ is an interior face, if it is bounded. The (unique) face that is unbounded is called the exterior face of È µ. The edges that surround a face constitute the boundary µ of. The exterior boundary is the boundary of the exterior face. The vertices (edges, resp.) on the exterior boundary are called exterior vertices exterior edges, resp.). Vertices (edges, resp.) that are not on the exterior boundary are interior vertices interior edges, resp.). ¼ ¾ ½ Embeddings È µ satisfy some properties that we accepts at face value. Lemma 5.2. Let È µ be a plane embedding of a planar graph. (i) Two different faces ½ and ¾ are disjoint, and their boundaries can intersect only on edges. (ii) È µ has a unique exterior face. (iii) Each edge belongs to the boundary of at most two faces. (iv) Each cycle of surrounds (that is, its interior contains) at least one internal face of È µ. (v) A bridge of belongs to the boundary of only one face. (vi) An edge that is not a bridge belongs to the boundary of exactly two faces.

63 5.1 Planar graphs 62 If È µ is a plane embedding of a graph, then so is any drawing È ¼ µ which is obtained from È µ by an injective mapping of the plane that preserves continuous curves. This means, in particular, that every planar graph has a plane embedding inside any geometric circle of arbitrarily small radius, or inside any geometric triangle. Euler s formula Lemma 5.3. A plane embedding È µ of a planar graph has no interior faces if and only if is acyclic, that is, if and only if the connected components of are trees. Proof. This is clear from Lemma 5.2. The next general form of Euler s formula was proved by LEGENDRE (1794). Theorem 5.1 (Euler s formula). Let be a connected planar graph, and let È µ be any of its plane embeddings. Then ³ ¾ where ³ is the number of faces of È µ. Proof. We shall prove the claim by induction on the number of faces ³ of a plane embedding È µ. First, notice that ³ ½, since each È µ has an exterior face. If ³ ½, then, by Lemma 5.3, there are no cycles in, and since is connected, it is a tree. In this case, by Theorem 2.5, we have ½, and the claim holds. Suppose then that the claim is true for all plane embeddings with less than ³ faces for ³ ¾. Let È µ be a plane embedding of a connected planar graph such that È µ has ³ faces. Let ¾ be an edge that is not a bridge. The subgraph is planar with a plane embedding È µ È µ obtained by simply erasing the edge. Now È µ has ³ ½ faces, since the two faces of È µ that are separated by are merged into one face of È µ. By the induction hypothesis, ³ ½µ ¾, and hence ½µ ³ ½µ ¾, and the claim follows. In particular, we have the following invariant property of planar graphs. Corollary 5.1. Let be a planar graph. Then every plane embedding of has the same number of faces: ³ ¾ Maximal planar graphs Lemma 5.4. If is a planar graph of order, then. Moreover, if has no triangles, then ¾.

64 5.1 Planar graphs 63 Proof. If is disconnected with connected components, for ¾ ½, and if the claim holds for these smaller (necessarily planar) graphs, then it holds for, since ½ ½ It is thus sufficient to prove the claim for connected planar graphs. Also, the case where ¾ is clear. Suppose thus that. Each face of an embedding È µ contains at least three edges on its boundary µ. Hence ³ ¾, since each edge lies on at most two faces. The first claim follows from Euler s formula. The second claim is proved similarly except that, in this case, each face of È µ contains at least four edges on its boundary (when is connected and ). An upper bound for Æ µ for planar graphs was achieved by HEAWOOD. Theorem 5.2 (HEAWOOD (1890)). If is a planar graph, then Æ µ. Proof. If ¾, then there is nothing to prove. Suppose. By the handshaking lemma and the previous lemma, Æ µ Ú¾ Úµ ¾ ½¾ It follows that Æ µ. DEFINITION. A planar graph is maximal, if is nonplanar for every ¾. Example 5.1. Clearly, if we remove one edge from Ã, the result is a maximal planar graph. However, if an edge is removed from Ã, the result is not maximal! Lemma 5.5. Let be a face of a plane embedding È µ that has at least four edges on its boundary. Then there are two nonadjacent vertices on the boundary of. Proof. Assume that the set of the boundary vertices of induces a complete subgraph Ã. The edges of à are either on the boundary of or they are not inside (since is a face.) Add a new vertex Ü inside, and connect the vertices of à to Ü. The result is a plane embedding of a graph À with Î À Î Ü (that has as its induced subgraph). The induced subgraph ÀÃ Ü is complete, and since À is planar, we have à as required. By the previous lemma, if a face has a boundary of at least four edges, then an edge can be added to the graph (inside the face), and the graph remains to be planar. Hence we have proved Corollary 5.2. If is a maximal planar graph with, then is triangulated, that is, every face of a plane embedding È µ has a boundary of exactly three edges.

65 5.1 Planar graphs 64 Theorem 5.3. For a maximal planar graph of order, Proof. Each face of an embedding È µ is a triangle having three edges on its boundary. Hence ³ ¾, since there are now no bridges. The claim follows from Euler s formula. Kuratowski s theorem Theorem 5.5 will give a simple criterion for planarity of graphs. This theorem (due to KURA- TOWSKI in 1930) is one of the jewels of graph theory. In fact, the theorem was proven earlier by PONTRYAGIN ( ), and also independently by FRINK AND SMITH (1930). For history of the result, see J.W. KENNEDY, L.V. QUINTAS, AND M.M. SYSLO, The theorem on planar graphs. Historia Math. 12 (1985), Theorem 5.4. à and à are not planar graphs. Proof. By Lemma 5.4, a planar graph of order 5 has at most 9 edges, but à has 5 vertices and 10 edges. By the second claim of Lemma 5.4, a triangle-free planar graph of order 6 has at most 8 edges, but à has 6 vertices and 9 edges. The graphs à and à are the smallest nonplanar graphs, and, by Lemma 5.1, if contains a subdivision of à or à as a subgraph, then is not planar. We prove the converse of this result in what follows. Therefore Theorem 5.5 (KURATOWSKI (1930)). A graph is planar if and only if it contains no subdivision of à or à as a subgraph. We prove this result along the lines of THOMASSEN (1981) using -connectivity. Example 5.2. The cube É is planar only for ½ ¾. Indeed, the graph É contains a subdivision of Ã, and thus by Theorem 5.5 it is not planar. On the other hand, each É with has É as a subgraph, and therefore they are nonplanar. The subgraph of É that is a subdivision of à is given below

66 5.1 Planar graphs 65 DEFINITION. A graph is called a Kuratowski graph, if it is a subdivision of à or Ã. Lemma 5.6. Let be the set of the boundary edges of a face in a plane embedding of. Then there exists a plane embedding È µ, where the edges of are exterior edges. Proof. This is a geometric proof. Choose a circle that contains every point of the plane embedding (including all points of the edges) such that the centre of the circle is inside the given face. Then use geometric inversion with respect to this circle. This will map the given face as the exterior face of the image plane embedding. Lemma 5.7. Let be a nonplanar graph without Kuratowski graphs such that is minimal in this respect. Then is -connected. Proof. We show first that is ¾-connected. On the contrary, assume that Ú is a cut vertex of, and let ½ be the connected components of Ú. Since is minimal nonplanar with respect to, the subgraphs Ú have plane embeddings È µ, where Ú is an exterior vertex. We can glue these plane embeddings together at Ú to obtain a plane embedding of, and this will contradict the choice of. Assume then that has a separating set Ë Ù Ú. Let ½ and ¾ be any subgraphs of such that ½ ¾, Ë Î ½ Î ¾, and both ½ and ¾ contain a connected component of Ë. Since is ¾-connected (by the above), there are paths Ù Ú in ½ and ¾. Indeed, both Ù and Ú are adjacent to a vertex of each connected component of Ë. Let À ÙÚ. (Maybe ÙÚ ¾.) If both À ½ and À ¾ are planar, then, by Lemma 5.6, they have plane embeddings, where ÙÚ is an exterior edge. It is now easy to glue À ½ and À ¾ together on the edge ÙÚ to obtain a plane embedding of ÙÚ, and thus of. We conclude that À ½ or À ¾ is nonplanar, say À ½. Now À½, and so, by the minimality of, À ½ contains a Kuratowski graph À. However, there is a path Ù Ú in À ¾, since ¾ À ¾. This path can be regarded as a subdivision of ÙÚ, and thus contains a Kuratowski graph. This contradiction shows that is -connected. Lemma 5.8. Let be a -connected graph of order. Then there exists an edge ¾ such that the contraction is -connected. Proof. On the contrary suppose that for any ¾, the graph has a separating set Ë with Ë ¾. Let ÙÚ, and let Ü Ü ÙÚµ be the contracted vertex. Necessarily Ü ¾ Ë, say Ë Ü Þ (for, otherwise, Ë would separate already). Therefore Ì Ù Ú Þ separates. Assume that and Ë are chosen such that Ì has a connected component with the least possible number of vertices. ½ À ½ ¾ À ¾

67 5.1 Planar graphs 66 There exists a vertex Ý ¾ with ÞÝ ¾. (Otherwise Ù Ú would separate.) The graph Þݵ is not - connected by assumption, and hence, as in the above, there exists a vertex Û such that Ê Þ Ý Û separates. It can be that Û ¾ Ù Ú, but by symmetry we can suppose that Û Ù. Ì Ù Ú Þ Ý Since ÙÚ ¾, Ê has a connected component such that Ù Ú ¾. For each Ý ¼ ¾, there exists a path È Ù Ý ¼ in Þ Û, since is -connected, and hence this È goes through Ý. Therefore Ý ¼ is connected to Ý also in Ì, that is, Ý ¼ ¾, and so. The inclusion is proper, since Ý ¾. Hence, and this contradicts the choice of. By the next lemma, a Kuratowski graph cannot be created by contractions. Lemma 5.9. Let be a graph. If for some ¾ the contraction has a Kuratowski subgraph, then so does. Proof. The proof consists of several cases depending on the Kuratowski graph, and how the subdivision is made. We do not consider the details of these cases. Let À be a Kuratowski graph of, where Ü Ü ÙÚµ is the contracted vertex for ÙÚ. If À ܵ ¾, then the claim is obviously true. Suppose then that À ܵ or. If there exists at most one edge ÜÝ ¾ À such that ÙÝ ¾ (or ÚÝ ¾ ), then one easily sees that contains a Kuratowski graph. There remains only one case, where À is a subdivision of Ã, and both Ù and Ú have neighbours in the subgraph of corresponding to À. In this case, contains a subdivision of Ã. Ú ¾ Ú Ú ¾ Ú Ü Ù Ú Ú ½ Ú Ú ½ Ú Lemma Every -connected graph without Kuratowski subgraphs is planar. Proof. The proof is by induction on. The only -connected graph of order is the planar graph Ã. Therefore we can assume that. By Lemma 5.8, there exists an edge ÙÚ ¾ such that (with a contracted vertex Ü) is -connected. By Lemma 5.9, has no Kuratowski subgraphs, and hence has a plane embedding È µ by the induction hypothesis. Consider the part È µ Ü, and let be the boundary of the face of È µ Ü containing Ü (in È µ). Here is a cycle of (since is -connected). Now since Ù Ú µ Ü, È µ Ü is a plane embedding of Ù Ú, and Æ Ùµ Î Ú and Æ Úµ Î Ù. Assume, by symmetry, that Úµ Ùµ. Let

68 5.2 Colouring planar graphs 67 Æ Úµ Ò Ù Ú ½ Ú ¾ Ú in order along the cycle. Let È Ú Ú be the path along from Ú to Ú. We obtain a plane embedding of Ù by drawing (straight) edges ÚÚ for ½. (1) If Æ Ùµ Ò Ú ¾ È ½ ( ½ is taken modulo ) for some, then, clearly, has a plane embedding (obtained from È µ Ù by putting Ù inside the triangle Ú Ú Ú ½ µ and by drawing the edges with an end Ù inside this triangle). (2) Assume there are Ý Þ ¾ Æ Ùµ Ò Ú such that Ý ¾ È and Þ ¾ È for some and, where Ý Þ ¾ Ú Ú. Now, Ù Ú Ú ½ Ú Þ Ý form a subdivision of Ã. Þ Ù Ý Ú By (1) and (2), we can assume that Æ ÙµÒÚ Æ Úµ. Therefore, Æ ÙµÒÚ Æ ÚµÒÙ by the assumption Úµ Ùµ. Also, by (1), Úµ Ùµ. But now Ù Ú Ú ½ Ú ¾ Ú give a subdivision of Ã. Ù Ú Proof of Theorem 5.5. By Theorem 5.4 and Lemma 5.1, we need to show that each nonplanar graph contains a Kuratowski subgraph. On the contrary, suppose that is a nonplanar graph that has a minimal size such that does not contain a Kuratowski subgraph. Then, by Lemma 5.7, is -connected, and by Lemma 5.10, it is planar. This contradiction proves the claim. Example 5.3. Any graph can be drawn in the plane so that three of its edges never intersect at the same point. The crossing number µ is the minimum number of intersections of its edges in such plane drawings of. Therefore is planar if and only if µ ¼, and, for instance, à µ ½. We show that à µ. For this we need to show that à µ. For the equality, one is invited to design a drawing with exactly crossings. Let à µ be a drawing of à using crossings so that two edges cross at most once. Add a new vertex at each crossing. This results in a planar graph on vertices and ¾ ½ edges. Now, since ¾ ½ µ. 5.2 Colouring planar graphs The most famous problem in the history of graph theory is that of the chromatic number of planar graphs. The problem was known as the -Colour Conjecture for more than 120 years, until it was solved by APPEL AND HAKEN in 1976: if is a planar graph, then µ. The -Colour Conjecture has had a deep influence on the theory of graphs during the last 150 years. The solution of the -Colour Theorem is difficult, and it requires the assistance of a computer.

69 5.2 Colouring planar graphs 68 The -colour theorem We prove HEAWOOD s result (1890) that each planar graph is properly -colourable. Lemma If is a planar graph, then µ. Proof. The proof is by induction on. Clearly, the claim holds for. By Theorem 5.2, a planar graph has a vertex Ú with Úµ. By the induction hypothesis, Úµ. Since Úµ, there is a colour available for Ú in the -colouring of Ú, and so µ. The proof of the following theorem is partly geometric in nature. Theorem 5.6 (HEAWOOD (1890)). If is a planar graph, then µ. Proof. Suppose the claim does not hold, and let be a -critical planar graph. Recall that for -critical graphs À, Æ Àµ ½, and thus there exists a vertex Ú with Úµ Æ µ. By Theorem 5.2, Úµ. Let «be a proper -colouring of Ú. Such a colouring exists, because is -critical. By assumption, µ, and therefore for each ¾ ½, there exists a neighbour Ú ¾ Æ Úµ such that «Ú µ. Suppose these neighbours Ú of Ú occur in the plane in the geometric order of the figure. Ú Ú Ú Ú ½ Ú ¾ Ú È ½ Consider the subgraph made of colours and. The vertices Ú and Ú are in the same connected component of (for, otherwise we interchange the colours and in the connected component containing Ú to obtain a recolouring of, where Ú and Ú have the same colour, and then recolour Ú with the remaining colour ). Let È Ú Ú be a path in, and let ÚÚ ½ µè ½ Ú Úµ. By the geometric assumption, exactly one of Ú ¾, Ú lies inside the region enclosed by the cycle. Now, the path È ¾ must meet at some vertex of, since is planar. This is a contradiction, since the vertices of È ¾ are coloured by ¾ and, but contains no such colours. The final word on the chromatic number of planar graphs was proved by APPEL AND HAKEN in Theorem 5.7 (4-Colour Theorem). If is a planar graph, then µ. By the following theorem, each planar graph can be decomposed into two bipartite graphs. Theorem 5.8. Let Î µ be a -chromatic graph, µ. Then the edges of can be partitioned into two subsets ½ and ¾ such that Î ½ µ and Î ¾ µ are both bipartite. Proof. Let Î «½ µ be the set of vertices coloured by in a proper -colouring «of. The define ½ as the subset of the edges of that are between the sets Î ½ and Î ¾ ; Î ½ and Î ; Î and Î. Let ¾ be the rest of the edges, that is, they are between the sets Î ½ and Î ; Î ¾ and Î ; Î ¾ and Î. It is clear that Î ½ µ and Î ¾ µ are bipartite, since the sets Î are stable.

70 5.2 Colouring planar graphs 69 Map colouring The -Colour Conjecture was originally stated for maps. In the map-colouring problem we are given several countries with common borders, and we wish to colour each country so that no neighbouring countries obtain the same colour. How many colours are needed? A border between two countries is assumed to have a positive length in particular, countries that have only one point in common are not allowed in the map colouring. Formally, we define a map as a connected planar (embedding of a) graph with no bridges. The edges of this graph represent the boundaries between countries. Hence a country is a face of the map, and two neighbouring countries share a common edge (not just a single vertex). We deny bridges, because a bridge in such a map would be a boundary inside a country. The map-colouring problem is restated as follows: How many colours are needed for the faces of a plane embedding so that no adjacent faces obtain the same colour. The illustrated map can be -coloured, and it cannot be coloured using only colours, because every two faces have a common border. Let ½ ¾ Ò be the countries of a map Å, and define a graph with Î Ú ½ Ú ¾ Ú Ò such that Ú Ú ¾ if and only if the countries and are neighbours. It is easy to see that is a planar graph. Using this notion of a dual graph, we can state the map-colouring problem in new form: What is the chromatic number of a planar graph? By the -Colour Theorem it is at most four. Map-colouring can be used in rather generic topological setting, where the maps are defined by curves in the plane. As an example, consider finitely many simple closed curves in the plane. These curves divide the plane into regions. The regions are ¾-colourable. That is, the graph where the vertices correspond to the regions, and the edges correspond to the neighbourhood relation, is bipartite. To see this, colour a region by ½, if the region is inside an odd number of curves, and, otherwise, colour it by ¾. ½ ¾ ¾ ½ ½ ¾ ¾ ½ ¾ ½ ¾ ½ History of the 4-Colour Theorem That four colours suffice planar maps was conjectured around 1850 by FRANCIS GUTHRIE, a student of DE MORGAN at University College of London. During the following 120 years many outstanding mathematicians tried to solve the problem, and some of them even thought that they had been successful.

71 5.2 Colouring planar graphs 70 In 1879 CAYLEY pointed out some difficulties that lie in the conjecture. The same year ALFRED KEMPE published a paper, where he claimed a proof of the 4CC. The basic idea in KEMPE s argument (known later as Kempe chains) was the same as later used by HEAWOOD to prove the -Colour Theorem, (Theorem 5.6). For more than 10 years KEMPE s proof was considered to be valid. For instance, TAIT published two papers on the 4CC in the 1880 s that contained clever ideas, but also some further errors. In 1890 HEAWOOD showed that KEMPE s proof had serious gaps. As we shall see in the next chapter, HEAWOOD discovered the number of colours needed for all maps on other surfaces than the plane. Also, he proved that if the number of edges around each region is divisible by, then the map is -colourable. One can triangulate any planar graph (drawn in the plane), by adding edges to divide the faces into triangles. BIRKHOFF introduced one of the basic notions (reducibility) needed in the proof of the 4CC. In a triangulation, a configuration is a part that is contained inside a cycle. An unavoidable set is a set of configurations such that any triangulation must contain one of the configurations in the set. A configuration is said to be reducible, if it is not contained in a triangulation of a minimal counter example to the 4CC. The search for avoidable sets began in 1904 with work of WEINICKE, and in 1922 FRANKLIN showed that the 4CC holds for maps with at most ¾ regions. This number was increased to ¾ by REYNOLDS (1926), to by WINN (1940), to by ORE AND STEMPLE (1970), to by MAYER (1976). The final notion for the solution was due to HEESCH, who in 1969 introduced discharging. This consists of assigning to a vertex Ú the charge Úµ. From Euler s formula we see that for the sum of the charges, we have Ú Úµµ ½¾ Now, a given set Ë of configurations can be proved to be unavoidable, if for a triangulation, that does not contain a configuration from Ë, one can redistribute the charges so that no Ú comes up with a positive charge. According to HEESCH one might be satisfied with a set of ¼¼ configurations to prove the 4CC. There were difficulties with his approach that were solved in 1976 by APPEL AND HAKEN. They based the proof on reducibility using Kempe chains, and ended up with an unavoidable set with over ½¼¼ configurations and some ¼¼ discharging rules. The proof used ½¾¼¼ hours of computer time. (KOCH assisted with the computer calculations.) A simplified proof by ROBERTSON, SANDERS, SEYMOUR AND THOMAS (1997) uses configurations and ¾ discharging rules. Because of these simplifications also the computer time is much less than in the original proof. The following book contains the ideas of the proof of the -Colour Theorem. T.L. SAATY AND P.C. KAINEN, The Four-Color Problem, Dover, 1986.

72 5.2 Colouring planar graphs 71 List colouring DEFINITION. Let be a graph so that each of its vertices Ú is given a list (set) Úµ of colours. A proper colouring «Î ½ Ñ of is a ( -)list colouring, if each vertex Ú gets a colour from its list, «Úµ ¾ Úµ. The list chromatic number µ is the smallest integer such that has a -list colouring for all lists of size, Úµ. Also, is -choosable, if µ. Example 5.4. The bipartite graph à is not ¾- choosable. Indeed, let the bipartition of à be µ, where Ü ½ Ü ¾ Ü and Ý ½ Ý ¾ Ý. The lists for the vertices shown in the figure show that à µ ¾. ½ ¾ Ý ½ Ü ½ ½ ¾ ½ Ý ¾ Ü ¾ ½ ¾ Ý Ü ¾ Obviously µ µ, since proper colourings are special cases of list colourings, but equality does not hold in general. However, it was proved by VIZING (1976) and ERDÖS, RUBIN AND TAYLOR (1979) that µ µ ½ For planar graphs we do not have a -list colour theorem. Indeed, it was shown by VOIGT (1993) that there exists a planar graph with µ. At the moment, the smallest such a graph was produced by MIRZAKHANI (1996), and it is of order. Theorem 5.9 (THOMASSEN (1994)). Let be a planar graph. Then µ. In fact, THOMASSEN proved a stronger statement: Theorem Let be a planar graph and let be the cycle that is the boundary of the exterior face. Let consist of lists such that Úµ for all Ú ¾, and Úµ for all Ú ¾. Then has a -list colouring «. Proof. We can assume that the planar graph is connected, and that it is given by a neartriangulation; an embedding, where the interior faces are triangles. (If the boundary of a face has more than edges, then we can add an edge inside the face.) This is because adding edges to a graph can only make the list colouring more difficult. Note that the exterior boundary is unchanged by a triangulation of the interior faces. The proof is by induction on under the additional constraint that one of the vertices of has a fixed colour. (Thus we prove a stronger statement than claimed.) For, the claim is obvious. Suppose then that. Let Ü ¾ be a vertex, for which we fix a colour «Üµ ¾ ܵ. Let Ú ¾ be a vertex adjacent to Ü, that is, Ú Ü Ú.

73 5.2 Colouring planar graphs 72 Let Æ Úµ Ü Ú ½ Ú Ý, where Ý ¾, and Ú are ordered such that the faces are triangles as in the figure. It can be that Æ Úµ Ü Ý, in which case ÜÝ ¾. Consider the subgraph À Ú. The exterior boundary of À is the cycle Ü Ú ½ Ú Ý Ü Since Úµ, there are two colours Ö ¾ Úµ that differ from «Üµ. We define new lists for À as follows: ¼ Ú µ Ú µ Ò Ö such that ¼ Ú µ for each ¾ ½, and otherwise ¼ Þµ Þµ. Ý Ú Ú Ú ¾ Now À ½, and by the induction hypothesis (with «Üµ still fixed), À has a ¼ -list colouring «. For the vertex Ú, we choose «Úµ Ö or such that «Úµ «Ýµ. This gives a ¼ -list colouring for. Since ¼ Þµ Þµ for all Þ, we have that «is a -list colouring of. Straight lines and kissing circles We state an interesting result of WAGNER, the proof of which can be deduced from the above proof of Kuratowski s theorem. The result is known as Fáry s Theorem. Theorem 5.11 (WAGNER (1936)). A planar graph has a plane embedding, where the edges are straight lines. This raises a difficult problem: Integer Length Problem. Can all planar graphs be drawn in the plane such that the edges are straight lines of integer lengths? Ú ½ Ü We say that two circles kiss in the plane, if they intersect in one point and their interiors do not intersect. For a set of circles, we draw a graph by putting an edge between two midpoints of kissing circles. The following improvement of the above theorem is due to KOEBE (1936), and it was rediscovered independently by ANDREEV (1970) and THURSTON (1985). Theorem 5.12 (KOEBE (1936)). A graph is planar if and only if it is a kissing graph of circles. Graphs can be represented as plane figures in many different ways. For this, consider a set Ë of curves of the plane (that are continuous between their end points). The string graph of Ë is the graph Ë µ, where ÙÚ ¾ if and only if the curves Ù and Ú intersect. At first it might seem that every graph is a string graph, but this is not the case. It is known that all planar graphs are string graphs (this is a trivial result).

74 5.2 Colouring planar graphs 73 Line Segment Problem. A graph is a line segment graph if it is a string graph for a set Ä of straight line segments in the plane. Is every planar graph a line segment graph for some set Ä of lines? Note that there are also nonplanar graphs that are line segment graphs. Indeed, all complete graphs are such graphs. The above question remains open even in the case when the slopes of the lines are ½, ½, ¼ and ½. A positive answer to this -slope problem for planar graphs would prove the -Colour Theorem. ½ ½ ¼ ½ The Minor Theorem DEFINITION. A graph À is a minor of, denoted by À, if À is isomorphic to a graph obtained from a subgraph of by successively contracting edges. A recent result of ROBERTSON AND SEYMOUR ( ) on graph minors is (one of) the deepest results of graph theory. The proof goes beyond these lectures. Indeed, the proof of Theorem 5.13 is around 500 pages long. a subgraph a contraction Note that every subgraph À is a minor, À. The following properties of the minor relation are easily established: (i), (ii) À and À imply À, (iii) À Ä and Ä imply À. The conditions (i) and (iii) ensure that the relation is a quasi-order, that is, it is reflexive and transitive. It turns out to be a well-quasi-order, that is, every infinite sequence ½ ¾ of graphs has two graphs and with such that. Theorem 5.13 (Minor Theorem). The minor order is a well-quasi-order on graphs. In particular, in any infinite family of graphs, one of the graphs is a (proper) minor of another. Each property È of graphs defines a family of graphs, namely, the family of those graphs that satisfy this property.

75 5.3 Genus of a graph 74 DEFINITION. A family of graphs is said to be minor closed, if every minor À of a graph ¾ is also in. A property È of graphs is said to be inherited by minors, if all minors of a graph satisfy È whenever does. The following families of graphs are minor closed: the family of (1) all graphs, (2) planar graphs (and their generalizations to other surfaces), (3) acyclic graphs. The acyclic graphs include all trees. However, the family of trees is not closed under taking subgraphs, and thus it is not minor closed. More importantly, the subgraph order of trees (Ì ½ Ì ¾ ) is not a well-quasi-order. WAGNER proved a minor version of Kuratowski s theorem: Theorem 5.14 (WAGNER (1937)). A graph is nonplanar if and only if à or Ã. Proof. Exercise. ROBERTSON AND SEYMOUR (1998) proved the Wagner s conjecture: Theorem 5.15 (Minor Theorem 2). Let È be a property of graphs inherited by minors. Then there exists a finite set of graphs such that satisfies È if and only if does not have a minor from. One of the impressive application of Theorem 5.15 concerns embeddings of graphs on surfaces, see the next chapters. By Theorem 5.15, one can test (with a fast algorithm) whether a graph can be embedded onto a surface. Every graph can be drawn in the -dimensional space without crossing edges. An old problem asks if there exists an algorithm that would determine whether a graph can be drawn so that its cycles do not form (nontrivial) knots. This problem is solved by the above results, since the property knotless is inherited by minors: there exists a fast algorithm to do the job. However, this algorithm is not known! Hadwiger s Problem. HADWIGER conjectured in 1943 that for every graph, à µ that is, if µ Ö, then has a complete graph Ã Ö as its minor. The conjecture is trivial for Ö ¾, and it is known to hold for all Ö. The cases for Ö and follow from the -Colour Theorem. 5.3 Genus of a graph A graph is planar, if it can be drawn in the plane without crossing edges. A plane is an important special case of a surface. In this section we study shortly drawing graphs in other surfaces.

76 5.3 Genus of a graph 75 There are quite many interesting surfaces many of which are rather difficult to draw. We shall study the easy surfaces those that are compact and orientable. These are surfaces that have both an inside and an outside, and can be entirely characterized by the number of holes in them. This number is the genus of the surface. There are also non-orientable compact surfaces such as the Klein bottle and the projective plane. Background on surfaces We shall first have a quick look at the general surfaces and their classification without going into the details. Consider the space Ê, which has its (usual) distance function Ü Ýµ ¾ Ê of its points. Two figures (i.e., sets of points) and are topologically equivalent (or homeomorphic) if there exists a bijection such that and its inverse ½ are continuous. In particular, two figures are topologically equivalent if one can be deformed to the other by bending, squeezing, stretching, and shrinking without tearing it apart or gluing any of its parts together. All these deformations should be such that they can be undone. A set of points is a surface, if is connected (there is a continuous line inside between any two given points) and every point Ü ¾ has a neighbourhood that is topologically equivalent to an open planar disk µ Ü Ø Üµ ½. We deal with surfaces of the real space, and in this case a surface is compact, if is closed and bounded. Note that the plane is not compact, since it it not bounded. A subset of a compact surface is a triangle if it is topologically equivalent to a triangle in the plane. A finite set of triangles Ì, ½ ¾ Ñ, is a triangulation of if Ñ Ì ½ and any nonempty intersection Ì Ì with is either a vertex or an edge. The following is due to RADÓ (1925). Theorem Every compact surface has a triangulation. Each triangle of a surface can be oriented by choosing an order for its vertices up to cyclic permutations. Such a permutation induces a direction for the edges of the triangle. A triangulation is said to be oriented if the triangles are assigned orientations such that common edges of two triangles are always oriented in reverse directions. A surface is orientable if it admits an oriented triangulation. Equivalently, orientability can be described as follows. Theorem A compact surface is orientable if and only if it has no subsets that are topologically equivalent to the Möbius band. In the Möbius band (which itself is not a surface according the above definition) one can travel around and return to the starting point with left and right reversed. A connected sum of two compact surfaces is obtained by cutting an open disk off from both surfaces and then gluing the surfaces together along the boundary of the disks. (Such a deformation is not allowed by topological equivalence.) The next result is known as the classification theorem of compact surfaces.

77 5.3 Genus of a graph 76 Theorem 5.18 (DEHN AND HEEGAARD (1907)). Let be a compact surface. Then (i) if is orientable, then it is topologically equivalent to a sphere Ë Ë ¼ or a connected sum of tori: Ë Ò Ë ½ Ë ½ Ë ½ for some Ò ½, where Ë ½ is a torus. (ii) if is nonorientable, then is topologically equivalent to a connected sum of projective planes: È Ò È È È for some Ò ½, where È is a projective plane. It is often difficult to imagine how a figure (say, a graph) can be drawn in a surface. There is a helpful, and difficult to prove, result due to RADÓ (1920), stating that every compact surface (orientable or not) has a description by a plane model, which consists of a polygon in the plane such that each edge of the polygon is labelled by a letter, each letter is a label of exactly two edges of the polygon, and each edge is given an orientation (clockwise or counter clockwise). Given a plane model Å, a compact surface is obtained by gluing together the edges having the same label in the direction that they have. Sphere Torus Klein bottle Projective plane From a plane model one can easily determine if the surface is oriented or not. It is nonoriented if and only if, for some label, the edges labelled by have the same direction when read clockwise. (This corresponds to the Möbius band.) A plane model, and thus a compact surface, can also be represented by a (circular) word by reading the model clockwise, and concatenating the labels with the convention that ½ is chosen if the direction of the edge is counter clockwise. Hence, the sphere is represented by the word ½ ½, the torus by ½ ½, the Klein bottle by ½ and the projective plane by ½. These surfaces, as do the other surfaces, have many other plane models and representing words as well. A word representing a connected sum of two surfaces, represented by words Ï ½ and Ï ¾, is obtained by concatenating these words to Ï ½ Ï ¾. By studying the relations of the representing words, Theorem 5.18 can be proved. Klein bottle