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 cardi- nality 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 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 Theorem 1 (a) Every triangle-free graph with 2k vertices has at most k2 edges. (b) The only triangle-free graph with 2k vertices and k2 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= k2 2k +1+2k 1= k2. ≤ − − − − 3 PART (b). Denote T (2k, 2) the complete bipartite graph on 2k vertices with the partitions of sizes k each. It is easy to see that for k = 1, the only triangle-free graph with k2 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 bea2k-vertex triangle-free − graph with k2 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= k2. ≤ − { } − ≤ − − 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 Comment. The Theorem above can be expanded to graphs with an arbitrary number n of vertices (even or odd): Theorem 2 2 (a) Every triangle-free graph with n vertices has at most n ⌊ 4 ⌋ edges. 2 (b) The only triangle-free graph with n vertices and n edges ⌊ 4 ⌋ is the complete bipartite graph with the partitions of sizes n ⌊ 2 ⌋ and n+1 . ⌊ 2 ⌋ 5 Definition 2 Given n and p, Tur´an’s graph T (n,p) obtained by partitioning n ver- tices into p disjoint sets of almost the same size (within 1) and setting 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 ver- ≤ tices. For any n 2 and 1 p<n ≥ ≤ p e(T (n + p,p)) = + n(p 1) + e(T (n,p)). 2 − Theorem 3 (Tur´an[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 Theorem 4 (Tur´an[1944]) Every graph of order n and size m contains an independent set of size n2/(2m + n). ≥ Proof. We present a greedy algorithm which constructs an inde- pendent set whose size is n2/(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 n2/(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. Con- sider 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, the total number of edges deleted is at least d(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 (n d 1)2 n2 1+ − − 2(m d(d + 1)/2) + n d 1 ≥ 2m + n − − − 7 The left part can be transformed as follows: (n d 1)2 1+ − − = 2(m d(d + 1)/2) + n d 1 − − − (n (d + 1))2 2m + n + n2 2n(d + 1) 1+ − = − 2m (d + 1)2 + n 2m + n (d + 1)2 − − Let us now denote 2m + n by Q. We must show that Q + n2 2n(d + 1) n2 − . Q (d + 1)2 ≥ Q − Indeed Q2 + Qn2 2nQ(d + 1) Qn2 n2(d + 1)2 − ≥ − (Q n(d + 1))2 0. − ≥ 8 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 n2/4 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 2 Dominating sets Definition 3 For a graph G, a set D V (G) of vertices is called dominating if ⊆ NG(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 dominat- ing. 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, let α(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 3 Discrete Mathematics revisited. Facts to remember Given set X, the number of subsets of X is given by • X X 2 =2| |. | | The number of all permutations of a set X with n elements is • nn n!= n (n 1) (n 2) ... 2 1 √2πn. × − × − × × × ≈ en The number n of k-subsets of a set X with n elements • k n n! = . k k!(n k)! − n+1/2 • v n nu 2 2 2 u = . u n/2 ≈ tπn √πn 11 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]) For every two positive integers k and l, there exists a smallest integer R(k,l) such that every graph of order R(k,l) contains either a clique on k vertices or an independent set on l vertices. Proof (Erd˝os 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. Let k,l 3 and assume, inductively, that the existence of R(k 1,l) ≥ − and R(k,l 1) has been established.