Combinatorics Dan Hefetz, Richard Mycroft (lecturer)

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1 Combiatorics Da Hefetz, Richard Mycroft (lecturer) These are otes of the lectures give by Richard Mycroft for the course Commuicatio Theory ad Combiatorics (3P16/4P16) at the Uiversity of Birmigham, o from Jauary to March These otes have bee typed by studets, who are solely resposible for the mistakes below (we claim o credit for the material). We will gladly correct ay errors you fid i the text Notatio Some of the otatio we use is fairly stadard, but ot all of it. N the atural umbers 1, 2,... equality by defiitio, e.g.: N {1, 2, 3,...}. [] the caoical set with elemets: [] {1, 2,..., }. 1 Ramsey theory Ramsey theory deals (morally) with existece argumets that some substructure is uavoidable i (large) combiatorial objects. Oe could say Ramsey-type questios seek order i (large) chaos. A rudimetary example: let a ad b be positive umbers, with a+b = c. The max{a, b} c/2. (You may argue this is a fact of averages, ad you would be right.) The ext example is much less artificial. Example I ay group of six people, there are either 3 people who each kow each other or three people who each do ot kow each other Proof. Represet each perso by a poit, coect two poits with a red lie if those people kow each other, ad with a blue lie if they do t. (Let us suppose we are allowed to draw i 3 dimesios, so that lies ca oly itersect at their edpoits.) The there is a lie (of some coulour) betwee each two poits. Fix some perso u. The there are 5 lies leavig u, ad at least 3 of those are of the same colour. Without loss of geerality, we assume they are red. The there are people a, b ad c which are coected to u by red lies. If ay of the lies ab, ac or bc is red, the with u we have formed a red triagle. Otherwise, all these three lies are blue, ad they form a blue triagle, which proves the claim. 1.1 Defiitios A graph G is a pair of sets (V, E), where V is the vertex set ad E is the edge set, ad the elemets of E are uordered pairs {u, v} of vertices (elemets of V ). We usually abbreviate the edge {u, v} as uv (ad this is the same edge as vu). If uv is a edge, we say that v is a eighbour of u. The degree d(u) of a vertex u is d(u) {uv : uv E}. The complete graph K has vertices, each pair of which forms a edge. See Figure 1. Let C be a set of colours. A colourig of a set X with C iw a fuctio f : X C, i.e., a assigmet of a colour i C to each member of X. A edge-colourig of a graph with C is a colourig of the edge set (ad o restrictios o which edge gets some colour i C). Give a colourig of the edges of a graphm, we shall use specialized otios of some cocepts. For 1 de 7 Geerated February 6, 2015, at

2 Figure 1: Some complete graphs. K 1 K 2 K 3 K 4 K 5 istace, blue eighbour of a vertex u is a eighbor of u coected to u by a blue edge, ad the blue degree of u is the umber of blue eighbors of u. (Similarly to other colours.) With this termiology we ca restate our iitial example as follows: i ay edge colourig of K 6 with colours red ad blue, there is a moochromatic copy of K 3 (i.e., a subgraph isomorphic to K 3 with all edges red or all edge blue). 1.2 Ramsey theory for fiite graphs For at t N, the diagoal Ramsey umber R(t) is the smallest iteger such that wheever the edges of K are coloured red ad blue there exists a moochromatic copy of K t. For ay s, t N, the Ramsey umber R(s, t) is the smallest iteger s such that wheever the edges of K are coloured red ad blue there is either a red K s or a blue K t. It is perhaps ot immediately obvious that these umbers exist; but Ramsey s theorem, proved by Frak Plumpto Ramsey i 1930 states that they do. Our earlier example shows that R(3) 6. The followig example shows that R(3) > 5. Some simple facts about the Ramsey umbers (exercises!). 1. R(t) = R(t, t). 2. R(s, t) = R(t, s) for ay s, t N. 3. R(1, t) = 1 for ay t N. 4. R(2, t) = t for ay t N. Theorem 1.1 (Ramsey theorem). R(s, t) exists for ay s, t N. Moreover, if s, t 2 the R(s, t) R(s 1, t) + R(s, t 1). Proof by Erdős ad Szekeres, Proof by iductio o s + t. Note that if s 2 or t 2 the R(s, t) exists by the facts above. So we oly eed do cosider s, t 3. Base case: if s + t 5 the either s 2 or t 2, so R(s, t) exists. Iductive step: fix s, t 3. By the iductive hypothesis, R(i, j) exists wheever i+j < s+t. I particular, R(s 1, t) ad R(s, t 1) both exist. Let = R(s 1, t) + R(s, t 1). We will show that wheever the edges of K are coloured red ad gree there is a red K s or a gree K t. This will prove that R(s, t) exists ad that R(s, t). Suppose that the edges of K are coloured red ad gree ad choose a vertex v arbitrarily. The v has degree 1, ad has either red degree R(s 1, t) or gree degree R(s, t 1). I the former case, the red eighbors of v have either a red copy of K s 1 or a gree copy of K t, which (together with v) demostrate that K has the desired moochromatic complete graph. The latter case is symmetric. 2 de 7 Geerated February 6, 2015, at

3 Corollary 1.2. For ay s, t N we have R(s, t) ( s+t 2) ( s 1. I particular, R(t, t) 2t 2 ) t 1 < 2 2t 2 = 4 t 1. Proof. The proof is by iductio o s + t. First ote that R(s, t) ( s+t 2) s 1. If s 2 or t 2 (see Problem sheet 1). This icludes our base case whe s + t 5. Now suppose s, t 3. The ( ) ( ) ( ) ( ) s 1 + t 2 s + t 3 s + t 1 2 s + t 3 R(s 1, t) = ad R(s, t 1) =. s 1 s 2 s 1 s 1 Ad the ( ) ( ) ( ) s + t 3 s + t 3 s + t 2 R(s, t) R(s 1, t) + R(s, t 1) + =, s 2 s 1 s 1 where the first iequality follows from Theorem 1.1 ad the secod from the iductio hypothesis. For the secod part of the claim, ote that R(t, t) ( 2t 2 t 1 ) 2 2t 2 = 4 t 1. Propositio 1.3. For t N we have R(t, t) 2 2t Proof. We begi with the followig observatio. (1) If is eve ad the edges of K are coloured red ad blue the ay of its vertices has either at least /2 red eighbours or at least /2 blue eighbours (as it has 1 > 2(/2 1) eighbours i total). Cosider a red/blue edge colourig of K 2 2t 1. The it suffices to fid either a red K t or a blue K t. We do this as follows. Choose a vertex x 1 arbitrarily. The x 1 has at least 2 2t 2 red eighbours or at least 2 2t 2 blue eighbours, by (1). I the first case, mark x 1 red ad let N 1 be a set of 2 2k 2 red eighbours of x 1. I the secod case, mark x 1 blue ad let N 1 be a set of 2 2t 2 blue eighbours of x 1. Ad repeat the followig. At step i, choose a vertex x i i N i 1 arbitrarily. The x i has at least 2 2t 1 i red eighbours or at least 2 2t 1 i blue eighbours, by (1). I the first case, mark x i red ad let N i be a set of 2 2k 1 i red eighbours of x i. I the secod case, mark x i blue ad let N i be a set of 2 2t 1 i blue eighbours of x i. As N i has half the size of N i 1, ad we start with 2 2t 1 vertices, we may cotiue for 2t 1 steps to obtai x 1, x 2,..., x 2t 1. These vertices have the property that for ay i < j the edge x i x j has the colour we used to mark x i. Fially, sice there are 2t 1 marked vertices, some t of them have bee marked with the same colour, ad these form a moochromatic K t. I particular, sice we may mark the last vertex of the sequece with either colour, we ca actually obtai a better boud (by a factor of 2). 1.3 Ramsey theory with more tha 2 colours For k N ad s 1, s 2,..., s k N we defie R k (s 1, s 2,..., s k ) to be the smallest atural umber such that wheever the edges of K are coloured with colours c 1, c 2,..., c k there exists a copy of K si i colour c i, for some i. For istace, R 2 (s 1, s 2 ) = R(s 1, s 2 ) ad R1(s 1 ) = s 1 R(s 1 ) = R 2 (s 1, s 1 ). 3 de 7 Geerated February 6, 2015, at

4 Table 1: Kow Ramsey umbers R(s, t) for s, t Theorem 1.4 (Ramsey, 1930). Let k N, ad s 1, s 2,..., s k N. The the Ramsey umber R k (s 1, s 2,..., s k ) exists. Oe possible proof mimicks the proof of existece of R 2 (s, t) by Erdős ad Szekeres. Fid aother (exercise!). [Hit: preted you ca t distiguish betwee two of the colours.] 1.4 Small Ramsey umbers Not much is kow about the Ramsey umbers R(s, t) i geeral. For t N, we kow R(1, t) = R(t, 1) = 1, ad R(2, t) = R(t, 2) = r. Otherwise, the kow Ramsey umbers are i Table 1. There are, however, geeral bouds o the Ramsey umbers (their gap is large!). E.g., 43 R(5, 5) 49. Example We will see that R(3, 4) = 9. To begi, we show that ay colourig of K 9 has the sought moochromatic subgraph. To prove this, cosider some red/gree colourig of the edes of K 9, ad cosider 3 cases. 1) some vertex has at least 4 red eighbors; 2) some vertex has at most 2 red eighbors; 3) every vertex has exactly 3 red eighbors. Which are left as exercises. The first case should be clear; check cases 2) [hit: use R(3, 3)] ad 3) [hit: cout the edges!]. To coclude, oe must show that R(3, 4) > 8 (see figure 2) Figure 2: Colourig of the edges of R(3, 4) without red K 3 or blue K Lower bouds for Ramsey umbers Our aim i this sectio is to give lower bouds o R(t, t) i terms of t. Theorem 1.5 (Erdős, 1947). For, t N, if ( t) 2 1 ( t 2) < 1 the R(t) >. Key observatio: if X is a radom variable whose value must be a oegative iteger ad such that E(X) < 1 the Pr(X = 0) > Proof. Fix such ad t ad colour the edges of K radomly, with colour choices made idepedetly of oe aother: a edge is coloured red with probability 1/2 ad is coloured 4 de 7 Geerated February 6, 2015, at

5 blue with the same probability. Now let X be the umber of moochromatic copies of K t i the colourig. So X is a radom variable whose value must be a oegative iteger. Sice there are ( t) sets of t vertices of K. For ay such set T, the probability that T iduces a red K t is 2 (t 2), ad the same holds for the probability that T iduces a blue Kt. So E(X) = ( t) 2 1 ( 2) t < 1 ad therefore Pr(X = 0) > 0. This, i tur, implies that there exists a colourig of the edges of K with o moochromatic K t. Proof. There are 2 ( 2) possible ways to colour the edges of K with red ad blue. A set T of t vertices moochromatic i 2 ( 2) ( 2)+1 t of them. So the umber of colourigs without moochromatic K t is at least 2 ( 2) ( t ) 2 ( 2) ( t 2)+1 = 2 ( 2) ( 1 ( ) 2 1 (t 2) ) = 2 ( 2) ( 1 t ( ) 2 1 (t 2) ) > 0. t Corollary 1.6. For all t N with t 4 we have R(t, t) > 2 t/2. Proof. Let = 2 t/2. The by 1.5 it suffices to prove that ( t) 2 1 ( 2) t < 1. Ideed, ( ) t t t! t(t 1) t2 /2 2 1 t2 t 2 = 21+t/2 2t 2 < 1, t! t! t! where we use that t t/2 wheever t 4. Defiitio 1.1. For graphs G, H, the Ramsey umber R(G, H) is the smallest such that wheever the edges of K are coloured red ad gree there is either a red copy of G or a gree copy of H Therefore R(K s, K t ) = R(s, t). Also, if G has s vertices ad H has t vertices R(G, H) R(K s, K t ), so R(G, H) exists for ay G, H. 1.6 Ramsey theory for umbers We will briefly see Ramsey-type behaviour for systems of equatios icreasig ad decreasig sequeces arithmetic progressios Systems of equatios Theorem 1.7 (Schur, 1916). For ay r N, there exists N such that wheever [] is coloured with r colours there exists a moochromatic solutio to x + y = z. Example Six umbers suffice for for r = 2 (check!). [Hit: if 1 is coloured red, o cosecutive umbers ca be coloured red if oe tries to avoid a red solutio to the equatio: 1, 2, 3, 4, 5, 6]. We ote that colourig is the same as partitioig the umbers i classes (umbers i the same class receive the same colour). 5 de 7 Geerated February 6, 2015, at

6 Proof. Let R r (3, 3,..., 3) ad c: : [] [r] be a arbitrary colourig of []. Defie a colourig c of the edges of K with r colours by c (ij) c( i j ), where c(m) is the colour of the iteger m. By defiitio of, this colourig has a moochromatic copy of K 3, say with vertices i < j < k. The c (ij) = c (ik) = c (jk), so c(j i) = c(k i) = c(k j). Let x = j i, y = k j ad z = k i: the x + y = z is a moochromatic solutio we sought. Remark 1. With a bit more work oe ca get a moochromatic solutio with x, y, z distict. We say that a a system of equatios is partitio regular if for all r N there exists N such that wheever [] is r-coloured there is a moochromatic solutio to the system of equatios. So Schur s theorem states that x + y = z is partitio regular. This is ot the case i geeral; x = 2y, for istace, does ot have that property. Rado s theorem characterizes precisely the partitio regular systems of liear equatios. Icreasig ad decreasig sequeces Suppose we have a sequece a 1, a 2,..., a of distict real umbers. Give s ad t, how large must be to guaratee either a icreasig sequece of legth s or a decreasig sequece of legth t? Note that = R(s, t) suffices, as we ca colour the edges of K by colourig ij red if a i < a j ad gree otherwise. By defiitio of, there is either a red K s or a gree K t i this colourig, ad these are mootoic subsequeces. However, if = (s 1)(t 1), o such subsequece eed exist. We will defie a sequece a 1, a 2,..., a of (s 1)(t 1), such that oe of its icreasig subsequeces has legth s ad oe of its decreasig subsequeces has legth t. The sequece is formed as follows: take s 1 decreasig sequeces A 1, A 2,..., A s 1 with t 1 umbers each, such that the umbers i A j are greater tha the umbers i A i, wheever 1 i < j s 1. Form the sequece by listig first the elemets of A 1 (i decreasig order), followed by the elemets of A 2 (i decreasig order), followed by... the elemets of A s 1 (i decreasig order). Lemma 1.8 (Erdős ad Szekeres, 1935). Let, s, t N, with > (s 1)(t 1). Ay sequece of distict umbers cotais either a icreasig subsequece of at least s elemets, or a decreasig subsequece of at least t elemets. Proof. Let, s, t be as above ad let a 1,..., a be a sequece of distict umbers. Place the umbers i piles p 1, p 2,... i the followig maer (top(p j ) deotes the umber at the top of pile p j, or if p j is empty): 1. a 1 goes i p 1 ; 2. for i > 1, put a i at the top of the pile pile p j of least idex such that a i is greater tha ay elemet at p j (that is j mi{k : top(p k ) < a i }). Sice > (s 1)(t 1), there is either a pile with more tha s elemets, or more tha t piles, ad the theorem follows (exercise!). Proof. Observe that for ay i < j, sice either a i < a j or a j < a i we have that either the logest icreasig subsequece startig with a i is loger tha the logest icreasig subsequece startig with a j, or the logest decreasig subsequece startig with a i is loger tha the logest decreasig subsequece startig with a j. Let l icreasig (i) be the legth of logest icreasig subsequece startig at a i, ad de 7 Geerated February 6, 2015, at

7 l decreasig (i) be the legth of logest decreasig subsequece startig at a i. Defie a fuctio f : [] N N by f(i) (l icreasig, l decreasig ). Our iitial observatio implies that f is ijective, as we caot have f(i) = f(j) with i j. Sice > (s 1)(t 1) = [s 1] [t 1], ad f is ijective, there must be some i [] such that f(i) / { [s 1] [t 1] }. 7 de 7 Geerated February 6, 2015, at