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

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1 1 Discrete Mathematics revisited. Facts to remember Give set X, the umber of subsets of X is give by X = X. The umber of all permutatios of a set X with elemets is! = ( 1) ( )... 1 e π. The umber ( ) k of k-subsets of a set X with elemets k =! k!( k)!. / π = +1/. π 1

2 Ramsey Theory The Simplest Ramsey Type Theorem. I ay collectio of six people either three of them mutually kow each other or three of them mutually do ot kow each other. Theorem 1 (Ramsey [1930]) For every two itegers p ad q (p,q ), there exists a smallest iteger R(p,q) such that every graph of order R(p,q) cotais either a clique o p vertices or a idepedet set o q vertices. Proof (Erdős ad Szekeres [1935]). By iductio o p ad q. The statemet is correct if p = or if q = : Check it! R(,q) = q ad R(p,) = p.

3 Letp,q 3adassume,iductively,thattheexisteceofR(p 1,q) ad R(p,q 1) has bee established. We will prove that R(p,q) R(p 1,q)+R(p,q 1). Let G be a graph with = (G) = R(p 1,q)+R(p,q 1) vertices ad let v be a arbitrary vertex of G. Deote N(v) ad N(v) the set of vertices adjacet to v ad the set of vertices that are ot adjacet to v, respectively. The Check it! either N(v) R(p 1,q) or N(v) R(p,q 1). Case of N(v) R(p 1,q). By iductio, N(v) must cotai either a clique of size p 1, or a idepedet set of size q. I the former case, the clique ad v yield a clique of size p i G; i the latter case, the same idepedet set is is a idepedet set of size q. Case of N(v) R(p,q 1). By iductio, N(v) must cotai either a clique of size p, or a idepedet set of size q 1. I the former case, a idepedet set ad v yield a idepedet set of size p i G, i the latter case, the same set of size q is a clique of size q i G. 3

4 Defiitio 1: The Ramsey umber R(p, q) is the smallest iteger such that i ay -colorig of the edges of a complete graph o vertices,k,byredadblue,thereiseitheraredk p (i.e. acomplete subgraph o 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) ad R(p 1,q) are eve, the R(p,q) < R(p,q 1)+R(p 1,q). Proof. Suppose G has R(p,q 1)+R(p 1,q) 1vertices ad has either a p-clique, or a q-idset. The for every vertex x V(G), N(x) = R(p 1,q) ad N(x) = R(p,q 1). The, V(G) is odd ad the degree of all vertices are odd, which is impossible. Corollary For all p,q 1, R(p,q) p+q. p 1 4

5 R(3,3) = 6; R(3,4) = 9; R(3,5) = 14: R(3,6) = R(5,5) 49; 35 R(4.6) 41; 10 R(6,6)

6 Theorem Erdős ad Szekeres [1935] R(p,p) > p/ for all p 3. Proof (1). Cosider the set of all graphs o {1,,...,}. There are ( ) of them. Each clique occurs i ( ) ( p ) of these graphs. Similarly, there are ( ) ( p ) graphs for which a particular set occurs as idepedet set. Together, there are ( ) ( p )+1 graphs for which a particular set is either idepedet or a clique. Sice there are ( ) p subsets of size p, out of ( ( ) graphs at most p) ( ) ( p )+1 cotai either a clique of size p or a idepedet set of size p. If < p/, the ( ) p ( ) ( p )+1 < ( ), which proves the theorem. Proof (). Cosider a radom colorig of K, where each edge is colored idepedetly ad with the same probability i either red or blue. For a fixed set D of p vertices, let A D be the evet that the iduced subgraph of K o D is moochromatic; deote the probability of A D by Pr(D A ). To compute Pr(D A ), ote that the evet happes iff the colors of ( ) p 1 edges coicide with that of the remaiig edge. Thus, Pr[D A ] = ( ( p ) 1) = 1 ( p ). Sicethereare ( ) p possiblechoicesfora,theprobabilitythatatleast oe of these evets occurs is at most If is selected so that 1 ( p ). p 1 ( p ) < 1, p 6 ( )

7 thereisapositiveprobabilitythatoevetd A occurs, adthereisa -colorig of K without a moochromatic K p, that is R(p,p) >. To fiish the proof, we show that ay p/ would satisfy (*). Ideed for all p > 3. 1 ( p ) p < p p! (p )/ (p ) 1 p! < 1 7

8 Examples. From theorem 1, R(, q) = q; R(p, ) = p; ad R(3,3) = 6, we compute R(3,4) R(,4)+R(3,3) = 4+6 corollary 1 = R(3,4) 9. R(3,5) R(,5)+R(3,4) = 5+9 = 14. R(3,6) R(,6)+R(3,5) = 6+14 corollary 1 = R(3,6) 19. R(3,7) R(,7)+R(3,6) = 7+19 = 6. R(3,8) R(,8)+R(3,7) = 8+6 corollary 1 = R(3,8)

9 Theorem 3 For every iteger 3, Proof. Iductio o. R(3,) +3. Base. = 3. R(3,3) = 6 = Iductive step. Assume R(3, 1) (( 1) +3)/ for some > 3, ad cosider R(3,). By theorem 1, Iductively, The R(3,) R(,)+R(3, 1) = +R(3, 1). R(3, 1) = ( 1) +3. R(3,) R(,)+R(3, 1) + ( 1) +3 = +4 To complete the proof, we eed to show that the last iequality is strict. This is obvious, if is odd. It is also true if R(3, 1) < ( 1) +3. Thus, let = k for some iteger k ad let R(3, 1) = (( 1) +3)/. The, R(3, 1) = ( 1) +3 = 4k 4k +1+3 = k k +. Sice, R(,) ad R(3, 1) are both eve, theorem 3 applies yieldig the result. 9

10 Problem 1 Show that (a) R(5,5) 6; (b) R(5,6) 111. Problem Show that if the edges of the complete graph are colored red, white, blue, gree, brow, ad purple, the if there are sufficietly may vertices, the there is a 4-go all of whose edges are colored the same color. Problem 3 Show that if the edges of the complete graph are colored red, white, blue, ad gree, the if there are sufficietly may vertices, ad there is o red, white, or blue triagle, the there is a complete 1-go all of whose edges are colored gree 10