1 Cliques and Independent Sets. Definition 1

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1 1 Cliques and Independent Sets Definition 1 A set of vertices is called independent if no two vertices in the set are adjacent. A set of vertices is called a clique if every two vertices in the set are adjacent. An independent set (resp. a clique) is called maximal, if no other independent set(resp. a clique) contains it. An independent set (resp. a clique) is called maximum, if its cardinality is maximal among all independent sets (resp. cliques) in the graph. α(g) (resp. ω(g)) denotes the maximum size of an independent set (resp. a clique) in a graph G; π(g) denotes the maximum size of a matching in a graph G; 1

2 Find a maximum independent set and a maximum clique Problem: Given n 1 and p (1 p n), construct a graph with n vertices and clique-size p, which contains the maximum number of edges. Dual Problem: Given n 1 and p (1 p n), construct an n-vertex graph with the minimal number of edges for which the size of any independent set is p. Special Case: (p = 2) Find the maximum number of edges in a graph with n vertices and without triangles. 2

3 Theorem 1 (a) Every triangle-free graph with 2k vertices has at most k 2 edges. (b) The only triangle-free graph with 2k vertices and k 2 edges is the complete bipartite graph with the partitions of sizes k each. Proof. Induction on k. PART (a). Base. The result is straightforward for k = 1. Inductive step. Let the statement be correct for all triangle-free graphs with 2k 2 vertices, and let G be a triangle-free graph with 2k vertices. Select an edge (a,b) E(G). The set of all edges of G consists of (i) the edges in G {a,b}; (ii) the edges connecting {a,b} with the rest of G; and (iii) the edge (a,b) itself. By induction, E(G {a,b}) (k 1) 2. Furthermore, there are at most 2k 2 edges of type (ii). (Explain why) a b 2k 2 vertices; no triangles e(g) e(g {a,b})+2k 2+1 (k 1) 2 +2k 1 = k 2 2k +1+2k 1 = k 2. 3

4 PART (b). DenoteT(2k,2)thecompletebipartitegraphon2k verticeswiththe partitions of sizes k each. It is easy to see that for k = 1, the only triangle-free graph with k 2 edges is T(2k,2). Assume, inductively, that the second part of the Theorem holds for all graphs with 2k 2 vertices, and let G be a 2k-vertex triangle-free graph with k 2 edges. Then, as before consider G {a,b} for some edge ab and analyze the inequality e(g) e(g {a,b})+2k 2+1 (k 1) 2 +2k 1 = k 2. The following is obvious: for this inequality to be an equality, G {a,b} must have exactly (k 1) 2 edges and the number of edges connecting {a,b} with the rest must be equal to 2k 2. The first requirement implies, inductively, that G {a,b} = T(2k 2,2), which in turn, implies that a (resp. b) is adjacent to the vertices of one part of G {a,b} only. Those parts must be distinct which proves that G = T(2k,2). 4

5 Comment. The Theorem above can be expanded to graphs with an arbitrary number n of vertices (even or odd): Theorem 2 (a) Every triangle-free graph with n vertices has at most n2 4 edges. (b) The only triangle-free graph with n vertices and n2 4 edges is the complete bipartite graph with the partitions of sizes n 2 and n

6 Definition 2 Givennandp,Turán sgrapht(n,p)obtainedbypartitioningnverticesintopdisjointsetsofalmostthesamesize(within1)andsetting edges to be all pairs comprised of vertices from different partitions. The sizes of the partitions of T(n,p) are obtained by dividing n by p with a remainder: n = p q +r, where 0 r p 1. The r partitions of T(n,p) are of size q+1, and the remaining p r partitions are of size q. It is easy to prove the following Lemma 1 If n p, then T(n,p) is a complete graph on n vertices. For any n 2 and 1 p < n e(t(n+p,p)) = p +n(p 1)+e(T(n,p)). 2 Theorem 3 (Turán[1944]) Given positive integers n and p, the number of edges of any graph with n vertices and without a clique of size p+1 is at most e(t(n,p)). 6

7 Theorem 4 (Turán[1944]) Every graph of order n and size m contains an independent set of size n 2 /(2m+n). Proof. We present a greedy algorithm which constructs an independent set whose size is n 2 /(2m+n). I = ; H = G; while (not done) select a vertex v V(H) of the minimal degree in H; I = I {v}; H = H {v} {all the vertices adjacent to v}; To prove that the algorithm constructs an independent set of size n 2 /(2m+n), we use induction on n. The statement is obviously true for the 1-vertex graph. Suppose, the theorem holds for every graph with < n vertices and let G be a graph with n vertices and m edges. Let d be the degree of the vertex v chosen by the algorithm. Consider the graph H resulting from the deletion of v and all vertices adjacent to v. Clearly, the degree of every deleted vertex is at least d. Therefore, thetotalnumberofedgesdeletedisatleastd(d+1)/2. Thus, m(h) m d(d+1)/2 and n(h) = n d 1. Since the algorithm constructs an independent set in H and adjoins v to it, the theorem will be proved if we verify the following inequality 1+ (n d 1) 2 2(m d(d+1)/2)+n d 1 n2 2m+n 7

8 The left part can be transformed as follows: 1+ (n d 1) 2 2(m d(d+1)/2)+n d 1 = 1+ (n (d+1))2 2m (d+1) 2 +n = 2m+n+n2 2n(d+1) 2m+n (d+1) 2 Let us now denote 2m+n by Q. We must show that Indeed Q+n 2 2n(d+1) Q (d+1) 2 n2 Q. Q 2 +Qn 2 2nQ(d+1) Qn 2 n 2 (d+1) 2 (Q n(d+1))

9 Problem 1 Determine the disconnected n-vertex graphs (n 2) that have the maximum number of edges. Problem 2 Determine the maximum number of edges in an n- vertex graph (without parallel edges) that has an independent set of size α. Problem 3 Let G be a simple graph with n 4 vertices. Prove that if G has more than n 2 /4 edges, then it has a vertex whose deletion leaves a graph with more than (n 1) 2 /4 edges. Problem 4 Prove that every n-vertex triangle-free simple graph with the maximum number of edges is isomorphic to K n/2, n/2. Problem 5 A flat circular city of radius six miles is patrolled by eighteen police cars, which communicate with one another by radio. If the range of a radio is nine miles, show that at any time, there is always at least two cars each of which can communicate with at least five other cars. 9

10 2 Dominating sets Definition 3 For a graph G, a set D V(G) of vertices is called dominating if N G (D) = V(G), that is if every vertex in V(G) is either in D or adjacent to a veretex in D. A dominating set is called minimal if no subsets of D is dominating. A dominating set is called minimum if no smaller set in G is dominating. γ(g) denotes the minumal size of a dominating set in a graph G. Lower and upper bounds for γ(g). Theorem. For a simple graph G, letα(g) denote the maximal size of an independent set, and let diam(g) denote the diameter of G. Then diam(g)+1 γ(g) α(g). 3 10

11 3 Discrete Mathematics revisited. Facts to remember Given set X, the number of subsets of X is given by 2 X = 2 X. The number of all permutations of a set X with n elements is n! = n (n 1) (n 2) nn e n 2πn. The number ( ) n k of k-subsets of a set X with n elements n k = n! k!(n k)!. n 2 n n/2 2 πn = 2n+1/2. πn 11

12 4 Ramsey Theory The Simplest Ramsey Type Theorem. In any collection of six people either three of them mutually know each other or three of them mutually do not know each other. Theorem 5 (Ramsey [1930]) Foreverytwopositiveintegersk andl, thereexistsasmallestinteger R(k,l)suchthateverygraphoforderR(k,l)containseitheraclique on k vertices or an independent set on l vertices. Proof (Erdős and Szekeres [1935]). By induction on k and l. The statement is correct if k = 2 or if l = 2: R(2,l) = l and R(k,2) = k. Letk,l 3andassume, inductively, thattheexistenceofr(k 1,l) and R(k,l 1) has been established. We will prove that R(k,l) R(k 1,l)+R(k,l 1). Let G be a graph with n = n(g) = R(k 1,l)+R(k,l 1) vertices and let v be an arbitrary vertex of G. Denote N(v) and N(v) the set of vertices adjacent to v and the set of vertices that are not adjacent to v, respectively. Then either N(v) R(k 1,l) or N(v) R(k,l 1). 12

13 Case of N(v) R(k 1,l). By induction, N(v) must contain either a clique of size k 1, or an independent set of size l. In the former case, the clique and v yield a clique of size k in G; in the latter case, the same independent set is is an independent set of size l. Case of N(v) R(k,l 1). By induction, N(v) must contain either a clique of size k, or an independent set of size l 1. In the former case, an independent set and v yield an independent set of size k in G, in the latter case, the same set of size l is a clique of size l in G. 13

14 Definition 1: The Ramsey number R(p, q) is the smallest integer n such that in any 2-coloring of the edges of a complete graph on n vertices,k n,byredandblue,thereiseitheraredk p (i.e. acomplete subgraph on p vertices all of whose edges are colored red) or a blue K q. Corollary 1. For all p,q 1, R(p,q) R(p,q 1)+R(p 1,q). If both R(p,q 1) and R(p 1,q) are even, then R(p,q) < R(p,q 1)+R(p 1,q). Corollary 2. For all p,q 1, R(p,q) p+q 2. p 1 R(3,3) = 6; R(3,4) = 9; R(3,5) = 14: R(3,6) = R(5,5) 49; 35 R(4.6) 41; 102 R(6,6)

15 Theorem 6 Erdős and Szekeres [1935] R(p,p) > 2 p/2 for all p 3. Proof (1). Consider the set of all graphs on {1,2,...,n}. There are 2 (n 2) of them. Each clique occurs in 2 ( n 2) ( p 2) of these graphs. Similarly, there are 2 (n 2) ( p 2) graphs for which a particular set occurs as independent set. Together, there are 2 (n 2) ( p 2)+1 graphs for which a particular set is either independent or a clique. Since there are ( ) n p subsets of size p, out of 2 (n ( 2) graphs at most n p) 2 ( n 2) ( p 2)+1 contain either a clique of size p or an independent set of size p. If n < 2 p/2, then ( ) n p 2 ( n 2) ( p 2)+1 < 2 (n 2), which proves the theorem. Proof (2). Consider a random coloring of K n, where each edge is colored independently and with the same probability in either red or blue. For a fixed set D of p vertices, let A D be the event that the induced subgraph of K n on D is monochromatic; denote the probability of A D by Pr(D A ). To compute Pr(D A ), note that the event happens iff the colors of ( ) p 2 1 edges coincide with that of the remaining edge. Thus, Pr[D A ] = 2 ( ( p 2) 1) = 2 1 ( p 2). Sincethereare ( ) n p possiblechoicesfora,theprobabilitythatatleast one of these events occurs is at most If n is selected so that n 2 1 ( p 2). p n 2 1 ( p 2) < 1, p 15 ( )

16 thereisapositiveprobabilitythatnoeventd A occurs, andthereisa 2-coloring of K n without a monochromatic K p, that is R(p,p) > n. To finish the proof, we show that any n 2 p/2 would satisfy (*). Indeed for all p > 3. n 2 1 ( p 2) n p < p p!2 2(p 2)/2 (p 2) 1 p! < 1 16

17 Theorem 7 (Ramsey [1930]) For every two positive integers p and q, (a) there exists a finite Ramsey number R(p,q) = R(K p,k q ). (b) R(p,q) R(p 1,q)+R(p,q 1); and (c) if both R(p 1,q) and R(p,q 1) are even, then R(p,q) < R(p 1,q)+R(p,q 1). Proof. The only part which we didn t prove before is (c). To prove (c), we need to establish that every graph with N = R(p 1,q)+ R(p,q 1) 1 vertices contains either a clique of size p, or an independent set of size q. Let V(G) = N. Since N is odd, there is a vertex v V(G) of even degree. If deg(v) R(p 1,q), then either G contains a clique C of size p 1, or it contains an independent set of of size q 1. In the former case, C and v yield a p-clique C of G; in the latter case, G contains an independent set of size q. On the other hand, if deg(v) < R(p 1,q), then deg(v) < R(p 1,q) 1, since deg(v) is even. Then the degree of v in G is at least R(p,q 1) which implies the result. 17

18 Examples. From theorem 1, R(2, q) = q; R(p, 2) = p; and R(3,3) = 6, we compute R(3,4) R(2,4)+R(3,3) = 4+6 thm 3 = R(3,4) 9. R(3,5) R(2,5)+R(3,4) = 5+9 = 14. R(3,6) R(2,6)+R(3,5) = 6+14 thm 3 = R(3,6) 19. R(3,7) R(2,7)+R(3,6) = 7+19 = 26. R(3,8) R(2,8)+R(3,7) = 8+26 thm 3 = R(3,8) 33. R(p, q)

19 Theorem 8 For every integer n 3, Proof. Induction on n. R(3,n) n Base. n = 3. R(3,3) = 6 = Inductive step. Assume R(3,n 1) (n 1) 2 +3)/2 for some n > 3, and consider R(3,n). By theorem 1, Inductively, Then R(3,n) R(2,n)+R(3,n 1) = n+r(3,n 1). R(3,n 1) = (n 1) R(3,n) R(2,n)+R(3,n 1) n+ (n 1)2 +3 = n To complete the proof, we need to show that the last inequality is strict. This is obvious, if n is odd. It is also true if R(3,n 1) < (n 1) Thus, let n = 2k for some integer k and let R(3,n 1) = ((n 1) 2 +3)/2. Then, R(3,n 1) = (n 1)2 +3 = 4k2 4k +1+3 = 2k 2 2k Since, R(2,n) and R(3,n 1) are both even, theorem 3 applies yielding the result. 19

20 5 Ramsey Theory (generalizations) Definition 4 Given graphs G 1,...,G k, the Ramsey number R(G 1,...,G k ) is the smallest integer n such that for every k- coloring of the edges E(K n ), G contains a copy of G i for some i = 1,...,k all of whose edges are of the same color (monochromatic copy). Theorem 9 (Chvátal[1977]) IfT isap-vertex tree, then R(T,K q ) = (p 1)(q 1)+1. The Chvatal graph for 6-tree and 4-clique Definition 5 Given graphs G and H without common vertices, G + H denotes the graph R with V(R) = V(G) V(H) and E(R) = E(G) E(H). For an integer m > 0 and a graph G, mg denotes G+...G. }{{} m Theorem 10 (Burr-Erdős-Spencer[1975]) For m > 1, R(mK 3,mK 3 ) = 5m. 20

21 I2m-1 K 3m-1 The Burr-Erdos-Spencer graph for R(mK,mK )

22 Problem 6 Show that (a) R(5,5) 70; (b) R(5,6) 126. Problem 7 Show that if the edges of the complete graph are colored red, white, blue, green, brown, and purple, then if there are sufficiently many vertices, then there is a 4-gon all of whose edges are colored the same color. Problem 8 Show that if the edges of the complete graph are colored red, white, blue, and green, then if there are sufficiently many vertices, and there is no red, white, or blue triangle, then there is a complete 12-gon all of whose edges are colored green 22