Graph Theory. Thomas Bloom. February 6, 2015

Graph Theory Thomas Bloom February 6, 2015 1

Lecture 1 Introduction A graph (for the purposes of these lectures) is a finite set of vertices, some of which are connected by a single edge. Most importantly, we don t allow multiple edges between two vertices, or an edge from a vertex to itself. Remember: no multiple edges or loops. One certainly can study graphs with these allowed, but we won t do so here. More formally, a graph G is defined by two pieces of data: a finite set V (called the vertices of G) and a set E of unordered pairs of elements of V (called the edges of G) so every e E has the form {x, y} for some x, y V with x y. In this case we will write things like G = (V, E), and V (G) for the set of vertices of G and E(G) for the set of edges of G. Note that such a graph is undirected, so the set of edges defines a symmetric relationship on the vertices V. If {x, y} is an edge of G then we will write x y. Thus the study of graphs is precisely the study of symmetric non-reflexive relations on finite sets but the pictorial interpretation is certainly more intuitive! The order of a graph, denoted by G, is the number of vertices. The size of a graph, denoted by e(g), is the number of edges. Lemma 1.1. For any graph G, 0 e(g) ( ) G. 2 The degree of a vertex x, denoted by d(x), is the number of y such that x y that is, the number of edges adjacent to x. The neighbourhood of x, denoted by N(x), is the set of such y, so that d(x) = N(x). As we are only concerned with these objects in a combinatorial way, the set of vertices can be a set of whatever we please. It is often convenient to use {1,..., n} as the set of vertices of a graph of order n. Example 1.1. The empty graph on n vertices is that which has no edges at all; i.e. G = n and e(g) = 0. 3

4 LECTURE 1. INTRODUCTION (a) The empty graph on 1 vertex (b) The empty graph on 5 vertices Example 1.2. The complete graph on n vertices is the graph with n vertices and every possible edge, so that e(g) = ( n 2) and x y whenever x y. It is denoted by K n. (a) K 3, the complete graph on 3 vertices (b) K 5, the complete graph on 5 vertices If G = (V, E) is a graph we say that H = (V, E ) is a subgraph of G if V V and E E. It is an induced subgraph if H includes every possible edge (i.e. those edges present in G) between its vertices; that is, if x, y V and {x, y} E(G) then {x, y} E(H). As usual in mathematics, if two objects are trivially the same, then we don t distinguish between them this is formalised with the notion of an isomorphism. An isomorphism between two graphs φ : G 1 G 2 is a bijection between V (G 1 ) and V (G 2 ) such that x G1 y if and only if φ(x) G2 φ(y). We treat isomorphic graphs as the same graph. Lemma 1.2. For any graph G d(v) = 2e(G). v V Proof. If one thinks about it, this is obvious each edge contributes 1 to the degree of each of its vertices, so it contributes exactly 2 to the sum over all degrees. It s not hard to turn this into a more formal proof, as below. We can write d(v) = 1 v e, and hence d(v) = v V v V e E(G) e E(G) 1 v e = e E(G) v V 1 v e. The inner sum, however, is always exactly 2, for it counts, for any fixed edge, the number of vertices incident to the edge e.

5 A graph is bipartite if we can partition the vertex set V (G) into two disjoint sets V 1 and V 2 such that every edge of G goes between the two sets, and there are no interior edges that is, if x y then x V 1 and y V 2 or vice versa. A path of length l, denoted by P l, is a graph with vertex set {0,..., l} and edge set E(P l ) = {{i, i + 1} : 0 i < l}. Equivalently, a path can be thought of as a sequence of vertices v 0 v l such that v i v i+1 for 0 i < l (this interpretation is useful when trying to find paths as subgraphs of other graphs). Observe that P l has size l and order l+1. A cycle of size l (and order l), denoted by C l, is a path of length l 1 with an added edge between 0 and l 1 or, again, a sequence of vertices v 1 v l such that v 1 v l and v i v i+1 for 1 i < l. Observe that K 3 = C 3. (a) P 3, the path of length 3 (b) C 4, the cycle of length 4 Lemma 1.3. A bipartite graph contains no odd cycles. Proof. Let G be a bipartite graph with vertex classes V 1 and V 2. Let x 1... x n be a cycle in G. Without loss of generality we may suppose that x 1 V 1. Since x 1 x 2 and G is bipartite we must have x 2 V. In general (use induction to be formal), we observe that x i V 1 if i is odd and x i V 2 if i is even. Furthermore, since it is a cycle we have x 1 x n and hence x n V 2, and so n is even as claimed. Remark 1.1. The converse also holds so if a graph contains no odd cycles then it must be bipartite. For a proof, see the exercise sheet (though this result is not examinable). One last definition: we say that a graph is connected if there exists a path between any two vertices. Intuitively, the point is that no part of the graph is cut-off from another. Recap of terminology make sure you re comfortable with all these terms! order of a graph size of a graph degree of a vertex, d(v)

6 LECTURE 1. INTRODUCTION neighbourhood of a vertex, N(v) bipartite graph complete graph path cycle connected There are lots of directions one can take in the study of graphs; in these lectures we focus on so-called extremal problems: given a graph G, give a fairly simple global criterion for G to contain some specified subgraph (for example, a path, a cycle, or a complete graph), which is a local property. The simplest criterion is having enough edges if we fix the order of the graph, then how many edges do we need to force G to contain the subgraph. To whet your appetites, here is a classical theorem of this type. Theorem 1.1 (Mantel s theorem). If G is a graph of order n and e(g) > n 2 /4 then G contains a triangle. Proof. Suppose that G is a graph of order n which contains no triangles. This means that if x y then there is no z G such that x z and y z, or else xyz forms a triangle in G. Thus N(x) N(y) =, and so N(x) + N(y) = N(x) N(y) n. Thus, for any x y we have d(x) + d(y) n. If we sum this inequality over all edges then we have 1 x e1 1 x e2 and hence d(x) 2 = x G x G = = e 1,e 2 E(G) e E(G) x G {x,y} e(g) 1 x e f E(G) (d(x) + d(y)) d(x) 2 ne(g). x G By the Handshaking Lemma and Cauchy s inequality 1 x f ( 2 ( ) (2e(G)) 2 = d(x)) n d(x) 2. x G x G

7 Thus 4e(G) 2 n 2 e(g), and so e(g) n 2 /4 and the proof is complete. We should check to see if this theorem can be improved but this is where Lemma 1.3 is useful, since a bipartite graph will contain no triangles. To have the most edges possible, it therefore seems sensible to split the graph into two disjoint sets and add every possible edge between those two sets. Indeed, it is easy to check that if n is even then such a graph contains no triangles and contains precisely n 2 /4 edges. Figure 1.4: A graph on 8 vertices with 16 edges with no triangle Both Mantel s theorem and the simple bipartite construction showing that it is sharp can be generalised for, not just triangles (which are complete graphs on 3 vertices) but to complete graphs on any number of vertices which we will see in Lecture 4! Exercises Nothing defined solely in an exercise will be examinable, and the exercises will not count towards your final mark. Nonetheless, I strongly recommend that you attempt all of the exercises for each lecture this will ensure that you properly understand the material. Exercise 1.1. For any x, y G let d(x, y) be the distance between x and y, defined as the length of the shortest path between x and y. For example, if x y then d(x, y) = 1. By considering the parity of d(x, y), prove that if a graph contains no odd cycles then it must be bipartite. Exercise 1.2. A tree is a connected graph with no cycles. By induction on n, show that a tree of order n has size n 1. Deduce using Lemma 1.2 that a tree has at least 2 vertices of degree 1. Exercise 1.3. Prove that every graph of order at least 2 contains at least two vertices with the same degree.

Lecture 2 Matchings Let G be a bipartite graph, with vertex classes V 1 and V 2. A matching from V 1 to V 2 is a set of independent edges (that is, edges which do not touch each other) which cover every vertex in V 1 (that is, every vertex of V 1 is in one of the edges). More formally, if V 1 = {v 1,..., v m } then a matching from V 1 is a set of edges e 1,..., e m E(G) such that v i e i for 1 i m and if i j then e i e j =. In this lecture we will consider the problem of finding necessary and sufficient conditions for such a graph to contain a matching. This problem is popularly called the marriage problem we can think of one vertex set as a set of men, the other as a set of women, and join two vertices if they are amenable to the idea of marriage. Asking if this graph contains a matching is then asking if we can pair everybody up to their mutual satisfaction. Let s first consider what is a necessary condition. Obviously, there must be at least one vertex in V 2 for every vertex in V 1, since a matching gives an injective function from V 1 into V 2. This condition is clearly not sufficient, however trivially, we might have the graph with no edges, and then there is no hope of a matching! The next necessary condition we observe, then, is that if V 1 = m, then the graph must have at least m edges, one for each vertex of V 1 we need to match. This condition is necessary, but it too fails to be sufficient for example, consider the graph where every y V 2 is joined to the same x V 1, and there are no other edges. If we consider what fails in this graph, we are led to a more general necessary condition not only must there be people available to match with everyone in V 1, but the same must be true for any subset of V 1. More formally, if X V 1 then N(X) X, where N(X) = N(x) = {w G : w v for some v X}. v X It is a striking theorem that this simple condition is not only necessary, but also sufficient. 9

10 LECTURE 2. MATCHINGS Theorem 2.1 (Hall s theorem, 1935). A bipartite graph G with vertex sets V 1 and V 2 has a matching from V 1 to V 2 if and only if for all X V 1. N(X) X (2.1) The condition (2.1) is also known as Hall s condition, or Hall s criterion. That Hall s condition is necessary for the presence of a matching is clear, as mentioned above. The hard part of the theorem is that this condition is also sufficient; that is, if (2.1) holds for all X V 1 then there exists a matching. This theorem is important enough that we give two, quite different, proofs. Proof. We use induction on m = V 1. For m = 1 the condition is clearly sufficient (check this!). Suppose that m 2 and that the theorem holds for all m < m. If strict inequality holds in (2.1) for every X V 1, i.e. N(X) X + 1 (2.2) for every X V 1, then we can choose any edge as part of the matching, say {x, y}, and remove it and its two vertices from the graph. We now claim that the remaining graph, G\{x, y}, still satisfies (2.1). For suppose not; then there exists a set W V 1 \{x} such that N(W )\{y} < W, but then N(W ) < W + 1, which contradicts (2.3). By induction there therefore exists a matching from V 1 \{x} to V 2 \{y}, and adding the edge between x and y gives a matching from V 1 to V 2 as required. Finally, suppose that (2.3) does not hold for every X V 1 ; then since G does satisfy (2.1) there must exist some X V 1 such that N(X) = X < m. By induction there exists a matching from X to N(X) fix any such matching. It remains to show that the graph induced by G\(X N(X)) still satisfies (2.1). If not, say for some X V 1 \X we have N(X )\N(X) < X, then N(X X ) N(X) N(X ) N(X) + N(X )\N(X) < X + X = X X, which contradicts the condition (2.1) of the original graph. As before, we now have a matching from X to N(X) and a matching from V 1 \X to V 2 \N(X). Combining these matchings gives a matching from V 1 to V 2 as required. Proof. For the second proof, we instead use induction on the number of edges. The important ingredient is the following claim: if e and e are two adjacent (in V 1 ) edges then either G e or G e satisfies the Hall condition. The

11 theorem now follows easily by induction, as follows. Suppose that there exist no two edges which are adjacent in V 1. Then this means that every vertex in V 1 has degree 1. Moreover, the neighbourhoods of each v V 1 are disjoint, for if w v w v then N({v, v }) = {w}, which contradicts (2.1). If there do exist two edges which are adjacent in V 1 then, by our earlier claim, we can remove one of these and preserve (2.1), and so by induction there exists a matching on this smaller graph, and hence one for the original graph. It remains to prove our initial claim: that if e 1 = {x, y 1 } and e 2 = {x, y 2 } are two distinct edges of G, with x V 1 and y 1, y 2 V 2, either G 1 = G e 1 or G 2 = G e 2 satisfies the Hall condition. Suppose not. Then there exist W 1, W 2 V 1 \{x} such that N(W 1 ) (N(x)\{y 1 }) < W 1 + 1 (2.3) and In particular, N(W 2 ) (N(x)\{y 2 }) < W 2 + 1. N(W 1 ) (N(x)\{y 1 }) W 1 N(W 1 ), and hence N(x)\{y 1 } N(W 1 ), so that N(W 1 ) = W 1, and similarly for W 2, and so N(W 1 ) N(W 2 ) N(x), since y 1 y 2, and hence N(W 1 W 2 ) = N(W 1 W 2 {x}). Finally, this implies that W 1 + W 2 = N(W 1 ) + N(W 2 ) = N(W 1 W 2 {x}) + N(W 1 ) N(W 2 )) N(W 1 W 2 {x}) + N(W 1 W 2 ) W 1 W 2 {x} + W 1 W 2 W 1 + W 2 + 1, which contradicts the assumption that G satisfies (2.1), and the proof is complete. Exercises Nothing defined solely in an exercise will be examinable, and the exercises will not count towards your final mark. Nonetheless, I strongly recommend that you attempt all of the exercises for each lecture this will ensure that you properly understand the material. Exercise 2.1. Let G be a bipartite graph between V 1 and V 2 where V 2 V 1 such that there exist positive integers d 1, d 2 1 such that for all v V 1 we have d(v) = d 1 and for all w V 2 we have d(w) = d 2. Show that there is a matching from V 1 to V 2.

12 LECTURE 2. MATCHINGS Exercise 2.2. Let G be a bipartite graph between V 1 and V 2 which satisfies Hall s condition, and such that d(v) k for all v V 1. Show that G contains at least k! different matchings. (Hint: adapt the first proof of Hall s theorem). Exercise 2.3. An n n Latin square is an n n array such that the numbers 1,..., n appear exactly once in each row and column. Show that there exist at least n! such Latin squares.

Lecture 3 Extremal behaviour of paths and cycles Recall that an extremal problem is one where we fix a graph H and ask for simple conditions on general graphs G to contain H as a subgraph and, moreover, what can we say about the extreme cases? In the last lecture we saw a fairly simple condition for bipartite graphs that guaranteed containing a matching as a subgraph, for example. In this lecture, we will look at the extremal problem for paths and Hamiltonian cycles cycles which contain every vertex of G. That is, can we give a simple sufficient condition for G to contain a path of a specified length or a Hamiltonian cycle? And, moreover, can we show this condition to be sharp? The Hamiltonian cycle extremal problem you may have encountered before as a game someone draws a graph, and the challenge is to start at one vertex and go around the graph, visiting every vertex and ending up at the beginning, without visiting a vertex more than once. In fact, in 1857 William Hamilton marketed this problem for a dodecahedron as a game (google the Icosian game for more). We will first consider the problem for paths we d like to find a simple sufficient condition for G to contain a path of some specified length. Theorem 3.1. Let G be a connected graph of order G 3 such that for any two vertices x and y we have d(x)+d(y) k. Then G contains a path of length k. Proof. Let P = v 1 v l be a path of maximal length in G, and suppose that l < k + 1. By maximality every neighbour of v 1 and v l must already be in {v 1,..., v l }. Consider the sets Γ 1 = {v j : v 1 v j } and Γ 2 = {v j+1 : v j v l }. Then Γ 1 Γ 2 l 1 < k. Clearly, however, Γ 1 d(v 1 ) and Γ 2 d(v l ), and hence Γ 1 + Γ 2 d(v 1 ) + d(v l ) k > Γ 1 Γ 2 13

14 LECTURE 3. EXTREMAL BEHAVIOUR OF PATHS AND CYCLES by hypothesis. It follows that there is some v j Γ 1 Γ 2, so that v j v 1 and v j 1 v l. Since G is connected there exists some w P which is adjacent to some vertex v i P. We may then construct a longer path: either if i j 1, or wv i v i+1 v j 1 v l v l 1 v j v 1 v 2 v i 1 wv i v i+1 v l v j 1 v 1 v j v i 1 otherwise, and hence we have constructed a path using l + 1 vertices, which contradicts maximality and completes the proof. That this simple condition on degrees is enough to guarantee a long path is quite striking; nonetheless, we may hope to find a simpler condition using just the number of edges. It is not too difficult to deduce such a bound from this theorem. Theorem 3.2. If G satisfies then G contains a path of length k. e(g) > k 1 2 Proof. We use induction on n = G. The case n k is obvious, since we always have e(g) ( n 2) (k 1)n/2. Furthermore, if G is not connected then the result follows by induction: let G 1,..., G r be the connected components of G and suppose that G (and hence each G i ) contains no P k. We have e(g) = r e(g i ) i=1 r i=1 k 1 2 G G i = k 1 2 G. We therefore may assume that G is connected. If every vertex has degree d(x) > (k 1)/2 then we are done by Theorem 3.1; otherwise, let x G be a vertex of degree at most (k 1)/2 and suppose that e(g) > (k 1)n/2. Let G be the graph with x and all its incident edges removed; then clearly G = n 1 and e(g ) = e(g) d(x) > k 1 n k 1 2 2 = k 1 (n 1), 2 and hence by induction G contains a path of length k and we are done. We now consider a similar problem for Hamiltonian cycles. In fact, a very similar argument works in this case also. Theorem 3.3. Let G be a connected graph of order G 3 such that for any two vertices x and y we have d(x)+d(y) G. Then G contains a Hamiltonian cycle, i.e. a cycle that includes every vertex exactly once.

15 Proof. Suppose G does not contain a Hamiltonian cycle, and let P = v 1 v l be a path of maximum length. We first claim that every cycle in G has length less than l. For otherwise, if x 1 x r is a cycle, then there exists some y G\{x 1,..., x r } which is adjacent to some x i for 1 i r, because G is connected and does not have a cycle which includes every vertex. We can then make a path using r+1 vertices as yx i x i 1 x 1 x r x i+1, and hence r+1 l as required. Arguing as in the proof of Theorem 3.1 we consider the sets Γ 1 = {v j : v 1 v j } and Γ 2 = {v j+1 : v j v l }. We now observe that these sets are disjoint; for otherwise, we have some v j such that v j v 1 and v j 1 v l, but then v 1 v j 1 v l v l 1 v j is a cycle of length l, which contradicts the claim in the first paragraph. Furthermore, every vertex of v 1 and v l is amongst {v 1,..., v l } by maximality of P. It follows that G d(v 1 ) + d(v l ) = Γ 1 + Γ 2 = Γ 1 Γ 2 l 1 G 1, which is a contradiction which completes the proof. The following easy corollary is a very simple criterion to remember for the presence of a Hamiltonian cycle. Corollary 3.1 (Dirac s theorem). If G 3 and d(x) G /2 for all x G then G contains a Hamiltonian cycle. Exercises Nothing defined solely in an exercise will be examinable, and the exercises will not count towards your final mark. Nonetheless, I strongly recommend that you attempt all of the exercises for each lecture this will ensure that you properly understand the material. Exercise 3.1. Show that Theorem 3.2 is best possible by constructing, for any even n and k 1, a graph of order n and size (k 1)n/2 with no paths of length k. Can you show that your construction is unique? Exercise 3.2. Show that every graph such that d(x) G /2 for all x G is connected. (Note: this result is implicit in the statement of Dirac s theorem!) Exercise 3.3. Solve William Hamilton s Icosian game, i.e. find a Hamiltonian cycle for the graph below. Find a graph for which finding a Hamiltonian cycle is more fun.

16 LECTURE 3. EXTREMAL BEHAVIOUR OF PATHS AND CYCLES

Lecture 4 Extremal behaviour of complete graphs We now consider the extremal question for complete graphs that is, given some fixed k 2 and n 1, what is the maximum number of edges that a graph on n vertices can have and yet contain no K k? The case k = 2 is trivial, so we will consider the case k 3. When trying to construct a graph with many edges and yet no K k, there is an obvious candidate namely, K k 1. A moment s thought shows that we can generalise this to handle more vertices by blowing up the graph. This means we replace each vertex by a set of independent vertices (i.e. a set of vertices with no edges between them), and replace each edge between two vertices by adding an edge between every vertex in the two corresponding vertex classes. For example, a complete bipartite graph is nothing more than a blow-up of K 2. To have as many edges as possible, it is sensible to choose these vertex classes to have roughly the same number of vertices. The blow-up of K r with n vertices and vertex classes that differ by at most 1 in size is the Turán graph, denoted by T r (n). (a) T 2(8) (b) T 3(11) Let us make things a little more precise. For any r 1 an r-partite graph 17

18 LECTURE 4. EXTREMAL BEHAVIOUR OF COMPLETE GRAPHS is one where we can partition V (G) into disjoint sets V 1,..., V r such that if x, y V i then x y. Lemma 4.1. There is a unique r-partite graph of order n with maximal size, which we denote by T r (n). Proof. Let G be an r-partite graph with vertex classes V 1,..., V r of maximal size. Clearly, it must be a complete r-partite graph. We order the vertex classes so that V 1 V r. It suffices to show that V 1 V r 1, for such a graph is unique up to isomorphism. Suppose that V r V 1 + 2. If we move one vertex from V r to V 1 then we add ( V 1 + 1)( V r 1) many edges, and lose V 1 V r many edges, and hence have a total gain of V r V 1 1 1 edges, which contradicts the maximality of G. An equivalent definition, which is more easily applied, is the following: T r (n) is the complete r-partite graph with n vertices, where the sizes of the r vertex classes differ by at most 1 from each other. Thus, if n = rq + k and k < r then (r k) of the classes have q vertices and the other k contain q + 1 vertices. Let t r (n) denote the number of edges of T r (n). For example, let s calculate t 2 (n). The Turán graph T 2 (n) is a complete bipartite graph with classes of size n/2 and n/2 if n is even, and (n 1)/2 and (n + 1)/2 if n is odd. Thus { n 2 /4 if n is even, and t 2 (n) = (n 2 1)/4 otherwise. Similarly, t 3 (n) = { n 2 /3 if n 0 (mod 3) (n 2 1)/3 otherwise. As discussed above, it is obvious that T r (n) is a graph which contains no K r+1. It is a striking theorem that T r (n) is also the largest graph which has this property that is, every graph with more edges must contain a K r+1. Theorem 4.1 (Turán s theorem). If G = n and e(g) > t r (n) then G contains a K r+1. Furthermore, if e(g) = t r (n) and G contains no K r+1 then it is isomorphic to T r (n). Proof. We use induction on n + r. Let G be a graph with e(g) t r (n) which contains no K r+1. Clearly, we can remove edges from G without destroying this property; thus there is a subgraph G with e(g ) = t r (n) such that G contains no K r+1. It suffices to show that G is isomorphic to T r (n) for if e(g) > e(g ) then it contains a T r (n) with an extra edge, which clearly contains a K r+1, which is a contradiction.

19 Since t r (n) > t r 1 (n), by induction G contains a K r, say with vertex set W. Let H = G W, and observe that e(h, W ) (n r)(r 1), since each vertex of H is connected to at most r 1 of the vertices of W, or else we d have a K r+1. It follows that e(h) = e(g) ( ) r 2 e(h, W ) t r (n) ( ) r (n r)(r 1) = t r (n r). 2 since if we remove a K r from T r (n) then we are left with a T r (n r) and have removed exactly ( r 2) + (n r)(r 1) edges. By induction H is isomorphic to T r (n r), and e(h, W ) = (n r)(r 1) so that every vertex of H is joined to precisely r 1 vertices of W, and thus G is T r (n). Exercises Nothing defined solely in an exercise will be examinable, and the exercises will not count towards your final mark. Nonetheless, I strongly recommend that you attempt all of the exercises for each lecture this will ensure that you properly understand the material. Exercise 4.1. Calculate the Turán number t r (n) explicitly when r n. Show also that, for any n, r 1, ( t r (n) 1 1 ) ( ) n. r 2 Exercise 4.2. Show that if G = n and for all v G then G contains a K r. d(v) > r 2 r 1 n Exercise 4.3. Let G = n and suppose that e(g) > t r (n) + 1. Show that G contains at least two copies of K r which do not share any edges.

Lecture 5 Ramsey theory Ramsey theory, in its most broad sense, is concerned with finding structure in large disordered sets that is, if we have enough of something, we are guaranteed to find some kind of structure. We will look at just the tip of the Ramsey iceberg in this lecture, and will be concerned with the following question: if fix some complete graph K n, and colour the edges red and blue, what kind monochromatic subgraphs will be forced? It is not obvious that any non-trivial monochromatic subgraphs must be created the first theorem of Ramsey theory is that, in fact, if n is large enough, we will be forced to contain all kinds of monochromatic subgraphs. Let s be precise. For any graphs G and H the Ramsey number R(G, H) is the smallest n such that every colouring of the edges of K n red and blue will create either an all red copy of G or an all blue copy of H. We may actually create both, but the point is at least one must be created. Complete graphs play a special role here we let R(s, t) = R(K s, K t ), i.e. the smallest n such that every red-blue colouring of the edges of K n contains either a red K s or a blue K t. Obviously R(s, t) = R(t, s) and R(2, s) = s. We also just write R(s) = R(s, s). So far, it s not obvious that R(s, t) is even well-defined perhaps there exist no such n! The following theorem guarantees the existence of such an n and gives an upper bound for it. Theorem 5.1. If s > 2 and t > 2 then R(s, t) R(s 1, t) + R(s, t 1). Proof. Let n = R(s 1, t) + R(s, t 1). Unpacking the definitions, what we need to show is that if we colour the edges of K n red and blue there will either be a red K s or a blue K t. Let v K n be any vertex. The degree of v is n 1, and hence v either has at least R(s 1, t) red neighbours or R(s, t 1) blue neighbours. Suppose the first case holds, and let G be the graph (with the inherited colouring) on those vertices which are joined to v by red edges. Since G has at least R(s 1, t) vertices it either contains a red K s 1 or a blue K t. If it has 21

22 LECTURE 5. RAMSEY THEORY a blue K t then we are done. Otherwise, by adding v to the red K s 1 we have found a red K s, and we are also done. Corollary 5.1. R(3) = 6. Proof. Theorem 5.1 and the trivial equality R(2, 3) = R(3, 2) = 3 imply that R(3) 6. To finish the proof we need to show that R(3) > 5, i.e. there is a colouring of the edges of K 5 with no creating a monochromatic triangle. Here it is: Figure 5.1: A colouring of K 5 with no monochromatic triangles The next non-trivial case is R(3, 4) Theorem 5.1 implies that R(3, 4) 10 but even now this falls short! In fact, R(3, 4) = 9. This then implies that R(4) 18 now this is actually sharp. It is known that R(4) = 18... but then we don t know! It is known that 43 R(5) 49. Finding the exact value is a very difficult problem. Erdős asks us to imagine an alien force, vastly more powerful than us, landing on Earth and demanding the value of R(5) or they will destroy our planet. In that case, he claims, we should marshal all our computers and all our mathematicians and attempt to find the value. But suppose, instead, that they ask for R(6). In that case, he believes, we should attempt to destroy the aliens. Joel Spencer Finally, we recall the Ramsey number for general graphs R(G, H). We observe that its existence is guaranteed by the existence of R(s, t) for if G = s and K n contains a red K s then it will certainly contain a red G. Lemma 5.1. If G = s and H = t then R(G, H) R(s, t). This is often far short of the truth, however. For example, let s consider the Ramsey number for a 4-cycle that is, R(C 4, C 4 ). It s easy to show that R(C 4 ) > 5 (look at the colouring above!). Since C 4 = 4 the bound above implies that R(C 4 ) R(4) = 18, but this is terrible in fact, R(C 4 ) = 6.

23 Exercises Nothing defined solely in an exercise will be examinable, and the exercises will not count towards your final mark. Nonetheless, I strongly recommend that you attempt all of the exercises for each lecture this will ensure that you properly understand the material. Exercise 5.1. Show that R(3, 4) = 9. Exercise 5.2. Show that R(C 4 ) = 6. Exercise 5.3. Let K 1,l be the complete bipartite graph between 1 vertex and l independent vertices. Using Turán s theorem, show that, for any r, l 1, R(K r, K 1,l ) = (r 1)l + 1.