Topics in Extremal Combinatorics - Notes

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1 Topics i Extremal Combiatorics - Notes Asaf Shapira Cotets 1 Lecture Ramsey umbers of bouded degree graphs: a improved boud Lecture Lower boud for liear Ramsey umbers Lecture Krivelevich s proof that r(3, k) ck 2 / log 2 k Ajtai-Komlós-Szemerédi s r(3, k) < ck 2 / log k AKS Ituitio AKS formal proof Improved boud for r(4, ) Lecture Ajtai-Komlós-Szemerédi a-la Shearer Erdős-Hajal Cojecture/Theorem Lecture Erdős-Szemerédi Theorem (Ramsey s Theorem for sparse graphs) Beck s Theorem (Size-Ramsey umber of a path) Pósa s rotatio-extesio lemma Hamiltoicity of sparse radom graphs Lecture Depedet radom choice basic lemma Variat of basic method Degeerate bipartite graphs

2 7 Lecture Discrepacy i graphs - Erdős Goldberg Pach Specer Six stadard deviatios suffice Tight examples for six-stadard-deviatios Lidsy s lemma The eigevalue boud Beck-Fiala Theorem Lecture Discrepacy of arithmetic progressios: Roth s 1 4-Theorem Discrepacy of arithmetic progressios: Beck s upper boud Liear ad hereditary discrepacy

3 1 Lecture Ramsey umbers of bouded degree graphs: a improved boud We recall that Ramsey s Theorem states that 2 t/2 r(k t ) 2 2t. I the first few lectures we will see several variats of this theorem, with the emphasize o usig tools/techiques. We proved i the previous course that r(g) c( ) if G is a -vertex graph of maximum degree (this was cojectured by Burr ad Erdős). The proof used the regularity-lemma so it oly gave the very weak boud c( ) = tower( ). We will ow prove a much better boud, due to Graham-Rödl-Ruciński 01, usig a iterestig techique that avoids usig the regularity lemma. However, the moral of the regularity lemma will be importat here. We also recall at this poit statemet of the the regularity lemma ad Gowers s lower boud, which ca be foud i the previous lecture otes. We also recall the so called triagle Coutig Lemma. Lemma 1: If A, B, C are three vertex sets ad the three bipartite graphs coectig them are ɛ-regular with desities a, b, c > 2ɛ the they cotai at least (1 2ɛ)(a ɛ)(b ɛ)(c ɛ) A B C triagles. Proof: Prove this lemma while observig that we oly eed oe side of the desity coditio guarateed by the otio of ɛ-regularity (this simple observatio will be importat soo). Furthermore, it is eough to assume that oly 2 of the pairs are ɛ-regular. I the previous course we actually proved the followig more geeral result. Observe that the umber of clusters as well as the parameters deped oly o ad d ad ot o the size of H. Lemma 2: For every d ad there is ɛ = ɛ(d, ) ad c = c(d, ) so that if V 1,..., V +1 are all ɛ-regular with d(v i, V j ) d ad if V i c the they cotai a copy of every -vertex graph H of maximum degree. Furthermore, we ca take ɛ = (d/2) /2 ad c = 1/ɛ. Observe that the assumptio i the above lemma imply that for every U i V i ad U j V j of sizes at least ɛ V i ad ɛ V j we have d(u i, U j ) d ɛ > d/2. We ow prove a oe-sided versio of the above lemma which makes oly this weaker assumptio. Lemma 2: Give δ,, set ɛ = δ /2 ad c = 2/δ. If V 1,..., V +1 have the property that every pair of subsets U i V i ad U j V j of sizes at least ɛ V i ad ɛ V j have desity at least δ ad V i c the they cotai a copy of every -vertex graph H with maximum degree. Proof: Suppose H s vertices are []. Sice H is ( + 1)-colorable, we ca fix a colorig ito + 1 sets ad let σ(i) deote the color of vertex i. We will greedily embed H ito G by placig each vertex i ito V σ(i). Start ruig the algorithm ad metio what eeds to happe. Fix some 0 j 1 ad suppose we have already foud vertices x 1,..., x j to play the role of 1,..., j. For every i > j let Y j i deote the vertices of V σ(i) which ca play the role of i give the choice of x 1,..., x j. Whe j = 0 we have Yi 0 = V σ(i). 3

4 For every 0 j < i let b j,i deote the umber of eighbours of i amog [j]. We claim that we ca pick the vertices x 1,..., x j so that Y j i δb j,i V σ(i). This is true whe j = 0 (sice b 0,i = 0), so we proceed by iductio o j. Suppose the claim holds for j 1. The defiitio of Y j 1 j guaratees that ay choice of vertex x j Y j 1 j amog [j 1]. We just eed to pick x j so that the sets Y j i. If (i, j) E(H) the choice of x j does ot affect Y j 1 Y j 1 i ote that there ca be at most such i. We eed to make sure that Y j i will be coected to the required vertices i will ot shrik by much compared to. Suppose the that (i, j) E(H) ad j 1 δ Yi. For each such j ɛ V σ(j). i we kow that Y j 1 i δ b(j 1,i) V σ(i) δ V σ(i) ɛ V σ(i) ad i particular Y j 1 The assumptio o the sets V 1,..., V +1 thus meas that Y j 1 j cotais at most ɛ V σ(j) vertices with less tha δ Y j 1 i eighbors i Y j 1 i. Sice there are at most relevat i, there are at most ɛ V σ(j) vertices that violate this coditio for some i. Hece, there are at least Y j 1 j ɛ V σ(j) δ V σ(j) 1 2 δ V σ(j) = 1 2 δ V σ(j) = choices of x j V σ(j) which will make sure that Y j j 1 i δ Yi for all i. Furthermore, sice we have choices for x j we will always be able to pick x j V σ(j) that will be distict from all other vertices chose from V σ(j). We say that G is (σ, δ)-dese if d(x) δ for every set X V (G) of size at least δ V. We say that G is bi-(σ, δ)-dese if d(x, Y ) δ for every pairs of disjoit sets X, Y V (G) of size at least δ V. The followig is a easy corollary of the above lemma. Corollary 4: If G is a bi-(δ /4 2, δ)-dese graph o at least (4 /δ ) vertices the G cotais a copy of every vertex graph of maximum degree. There is a very simple aive recursive algorithm that fids a very sparse subset i a graph that has o large bi-dese subgraph. Hece we wat to be able to deal with the case of very dese graphs. This is doe i the followig lemma. Lemma 5: If G has 4 vertices ad desity at least 1 1/8 the it cotais a copy of every -vertex graph of maximum degree. Proof: The desity of the of the complemet graph G is at most 1/8. Hece, there are at most G /2 vertices with a degree (i G) greater tha G /4. We remove those vertices (ad maybe a few more) ad remai with exactly half the vertices. I the remaiig graph, every vertex has a degree greater tha G /2 G /4. Now we embed the vertices of H oe by oe. Suppose we embedded 1,..., i ad we attempt to embed i + 1. vertex i + 1 has at most eighbors amog 1,..., i (it has at most eighbors i geeral). For each such eighbor, there are at most G /4 vertices which are ot coected to it. Hece there are at most G /4 forbidde vertices. Therefore we ca pick oe of G /2 G /4 = G /4 vertices, which meas we ca pick a vertex for i + 1 which has t bee picked before. 4

5 We ow show how to actually fid a large dese subset i a graph that has o large bi-dese subgraph. The precise statemet is the followig. Lemma 6: The followig holds for every 0 < σ, δ < 1 ad s 2/δ: If G is ((σ/2) s, δ)-dese graph o N vertices, the there is U V (G) of size at least (σ/2) s N so that G[U] is bi-(σ, δ/4)-dese. Proof: If G is bi-(σ, δ/4)-dese, we are doe. Otherwise, there is a pair of vertex sets U 1,1, U 1,2 of size σn such that d(u 1,1, U 1,2 ) < δ/4. By averagig, U 1,1 cotais a subset, W 1 of size (σ/2) s N such that d(w 1, U 1,2 ) < δ/4. Agai, by averagig, at most half of U 1,2 s vertices have desity > δ/2 i regard to W 1. Let V 2 deote those vertices of U 1,2 whose desity (to W 1 ) is δ/2. The, V 2 σ/2n ad if it is bi-(σ, δ/4)-dese, we are doe. If V 2 is ot bi-(σ, δ/4)-dese, the there are two subsets of V 2, deote them by U 2,1, U 2,2, of size (σ V 2 (σ/2) s N such that d(u 2,1, U 2,2 ) δ/4. As before, take W 2 U 2,1 of size (σ/2) s N such that d(w 2, U 2,2 ) δ/4, ad as before defie V 3 U 2,2 with desity δ/2 to W 2. The, V 3 (σ/2) 2 N. We ote that each vertex i V 3 has also desity < δ/2 to W 1. We cotiue this process for at most s = 2/δ iteratios. bi-(σ, δ/4)r-dese, we are doe. If somewhere alog the way V i is Otherwise, we ve built vertex sets W 1,..., W s each oe of size (σ/2) s N such that every vertex i W i has at most (δ/2) W j eighbors i W j for j < i. Therefore, if we take W = s i=1 W i, there are few edges betwee the W i s, the desity betwee them is less tha δ/2. Hece, eve if every W i is a clique, the desity of W is strictly less tha δ, but as W (σ/2) s N, it is a cotradictio to the assumptio that G is ((σ/2) s, δ)-dese. We are ow ready to state ad prove our mai result. Theorem: c( ) O( 2 ) Proof: Take δ = 1/8, σ = ( δ 4 ) ad s = 16 2/δ. Cosider a colorig of K N with N = 4 (2/σ) s (4/δ). We eed to show that that either the black edges or the red edges spa a copy of H, where H is a arbitrary vertex graph of max degree. If the red edges form a (( σ 2 )s, δ)-dese the by Lemma 6 we ca fid a set of vertices of size at least ( σ 2 )s N = (4 /δ ) o which the red edges form a bi-(σ, δ/4)-dese=bi-( ( δ 4 ), δ/4)-dese graph. Corollary 4 the say that we ca fid a red H. Otherwise, there is a subset of size at least ( σ 2 )s N 4 o which the black desity is at least 1 δ = Hece we ca use Lemma 5 to fid a black H. 2 Lecture Lower boud for liear Ramsey umbers We ow wat to prove a 2 Ω( ) lower boud for Ramsey umbers of bouded degree graphs. Let s the try to reverse-egieer the above proof, that is, fid a graph G ad colorig of K N, with 5

6 N = 2 Ω( ), so that the above proof will ot fid a copy of G i the colorig of K N. Explai why a o()-blowup of a radom graph of size 2 Ω( ) is a reasoable approach: The clusters are of size o() so they do ot cotai H (that is, we do t care if they cotai desity as i the secod case of the proof). We wat to make sure there will ot be a ( + 1)-clique as otherwise the first case of the proof will allow to embed H (sice all the pairs betwee the sets will be bi-dese). Therefore, a radom graph of size 2 Ω( ) is a good choice sice it has o -clique. Oe ca of course say that perhaps we do ot eed all pairs of the clusters to be dese (sice perhaps there are o edges betwee some pairs) ad perhaps we decide to break some color classes betwee various clusters. All of these are ruled out by the followig fractioal geeralizatio of Eröds s lower boud for Ramsey s theorem. Lemma 1: For every k 4 there is a Red/Black colorig of [K k ] so that for all fuctios w : [k] [0, 1] with k i=1 w(i) = x > ( ) log k we have B = ( ) x w(i)w(j) < 0.51 ad R = ( ) x w(i)w(j) < (i,j) B (i,j) R Proof: Prove that R/B is maximized whe the vertices with weight i (0, 1) form a B/R clique. Argumet is similar to the weight-shiftig proof of Matel s theorem. Take a colorig with o R/B copy of K 2 log k ad where every set of size y 10 7 log k has [0.499 ( y 2), ( y 2) ] red/black edges 1. Suppose w maximizes R. The at most 2 log k vertices have weight i (0, 1). All the other t x 2 log k 10 7 log k have weight 1 so they cotribute at most ( t 2) ( x 2). The cotributio of edges with oe vertex havig weight i (0, 1) is at most x 2 log k < 0.09 ( x 2) so all together R < 0.51 ( x 2). Same argumet also applies to B. To use the above lemma, we would wat our graph to have the property that o matter how oe breaks it ito pieces, there would be may large pairs with at least oe edge betwee them. The ext lemma does exactly that. Lemma 1: There is a absolute costat c > 1 so that the followig holds for every > 0, t > 100, > 0 ( ) ad k = c. There exists a -vertex graph H of maximum degree so that for every partitio V = V 1... V k i which V i /t we have ( ) V i V j > i<j : e(v i,v j )>0 Proof: Set d = /300. We pick a radom graph G(m, d/m) where m = Expected umber of edges is md/2, so with very high probability we will ot have more tha dm edges. Observe that a graph with at most dm = 1.01d edges cotais at most /100 vertices of degree more tha 300d. 1 Prove that such a graph exists usig Cheroff sayig P [B(, p) > (1 + ɛ)p] < 2 ɛ2 p/3. (1) 6

7 Hece removig the /100 vertices of highest degree gives a graph o vertices with maximum degree at most 300d =. Take ay partitio of the m vertices ito k + 1 sets; oe distiguished set B of size /100 (this is the set we will remove) ad the other k sets V 1,..., V k of size at most /t each. Sice t > 100 the umber of pairs of vertices withi the sets V 1,..., V k is bouded by t(/t) 2 < 2 /100. Sice B = /100 the umber of pairs of vertices with at least oe vertex i B is at most 1.01(/100) 2 /99. Hece there are at least ( ) /99 pairs coectig two vertices from differet sets V i, V j. Hece, if V 1,..., V k violate (1), the ( ) ( ) 1.01 V i V j 2 2 / (2) 2 2 i<j : e(v i,v j )=0 Therefore, the probability that a fixed partitio B, V 1,..., V k will violate (1) with respect to some fixed choice of pairs V i, V j as i (2) is at most (1 d/m) 0.12 = (1 /303) Sice there are at most (k + 1) m such partitios, ad for each such partitio there are at most 2 (k 2) ways to pick the pairs V i, V j i (2), we see that the probability that i some partitio B, V 1,..., V k, the sets V 1,..., V k will violate (1) is at most (k + 1) m 2 k2 (1 /303) 0.12 < c 2 2 k2 e /3030. Hece for some c > 1 ad 2 large eough this is o(1). We see that with positive probability, the graph has at most dm edges ad i every partitio B, V 1,..., B k, the sets V 1,..., V k satisfy (1). Therefore, removig the /100 vertices of highest degree, gives a -vertex graph of maximum degree that satisfies (1). All that is left is to combie the above two lemmas to prove our lower boud. Mai Result: There are c, 0 > 1 so that for every 0 ad large eough 0( ) there is a vertex graph H of maximum degree satisfyig r(h) c. Proof: Take > 0 ad t > 100 to be chose later, ad suppose 0 ( ). Let H be the graph from Lemma 2 (which satisfies (1)). Let k = c be the costat from Lemma 2 ad let R be the R/B colorig of [K k ] give by Lemma 1. Defie a R/B colorig of K N where N = k/t = c /t, (3) as follows. Partitio [N] ito k sets U 1,..., U k of size /t. Iside U i color black. Color (U i, U j ) with color R(i, j). Suppose the graph cotais a black H (same argumet works for red H). Let V i = V (H) U i. The V i U i = /t so V 1,..., V k satisfies the coditio of Lemma 1, hece it must satisfy (1). Set w(i) = V i / U i. The x = k w(i) = i=1 k i=1 V i U i = t k V i = t > ( ) log k (4) i=1 2 I the iequality we use the assumptio that > 0 to get k + 1 = c + 1 c 2 7

8 where we choose t = 10 8 log k i order to satisfy the last iequality. Note that this meas that N = c /10 8 log(c ) > c for some 3 c > 1 ad all large eough > 0. cotradicts (1), sice it gives i<j : e(v i,v j )>0 V i V j = (i,j) B V i V j = N 2 k 2 (i,j) B Lemma 1 the w(i)w(j) < N 2 ( ) ( ) x k 2 (0.51) < 0.51, 2 2 where the secod equality uses (3) which gives N/k = /t, ad last iequality uses (3) ad (4) which together give N/k = /x. 3 Lecture Krivelevich s proof that r(3, k) ck 2 / log 2 k Equivaletly there is a -vertex triagle-free graphs with α(g) = O( log ). If we wat G(, p) to be triagle-free, the we eed p = c/ but the G(, p) will cotai c edges ad thus cotai a idepedet set of liear size. We thus wat to use the alteratio method. We wat the expected umber of triagles (which is p 3 3 /6) to be less tha /2. We ca the remove oe vertex from each triagle ad get half the vertices. This gives p 2/3. The probability of havig a idepedet set of size k is at most ( k) (1 1/ 2/3 ) k2 /2. If k 2/3 log the this is o(1). Let s cosider the followig more subtle alteratio method. Give a set of vertices S, let M(S) deote the umber of edges i S, ad let T (S) deote the size of the largest collectio of edge disjoit triagles 1, 2,... with the property that each triagle i has at least oe edge i S. Suppose G has the property that for every subset of vertices S of size k, we have M(S) > 3 T (S). The, if we tur G ito a triagle free graph G by iteratively removig triagles from it (thus costructig a collectio of edge disjoit triagles), the the above property guaratees that α(g ) < k. Eough to prove that there are 0 < α < 1 < β so that G(, α/ ) satisfies the above property with k = β log. Sice there are at most k 2 β log 2 sets S of size k, it is eough to show that for ay give S, the probability S violates the above coditio is at most 2 2β log 2. Fix S of size k, ad let M deote the umber of edges i S. The E[M] = pk 2 /2 = 1 2 αβ2 log 2. By Cheroff 4 we get P[M 1 2 E[M]] < E[M]. Let T be the largest collectio of edge disjoit triagles with at least oe edge i S. We wat T < E[M]. The probability that some fixed choice of 1 6E[M] pairs of vertices i S will form a collectio of edge disjoit triagles is at most 1 6 E[M] (p 3 ) 1 6 E[M]. Sice there are at most ( S 2 ) 1 6 E[M] ways to pick the collectio of edges withi S, we get that the probability that T 1 6E[M] is at most ( S 2 ) 1 6 E[M] (p 3 ) 1 6 E[M] < (6ep 3 S 2 /E[M]) 1 6 E[M]. 3 If c = 1 + ɛ we ca take c = 1 + ɛ/2 4 P[B(, p) < (1 ɛ)p] < e ɛ2 p/2 8

9 Sice S = k, p = α/ ad E[M] = pk 2 /2, this is bouded from above by (12eα 2 ) 1 6 E[M] so if α = 0.1 the this is at most E[M]. Hece, the probability that S will ot satisfy T 1 3 M is at most E[M] E[M] < E[M]. Recallig that E[M] = 1 2 αβ2 log 2 (ad that we have already chose α), we ca choose β > 1 so that 1 9 E[M] will be larger tha 2β log 2, as eeded. Observe that i G(, 1/ ) with very high probability, all vertices v will satisfy N(v) so the probability of v ot belogig to ay triagle is at most the probability of N(x) beig idepedet which is thus at most (1 1/ ) Θ() < 2 so we expect all vertices to belog to at least oe triagle. This tells us that i this regime we have o chace of efficietly makig G triagle free by removal of vertices ad thus have to aalyze the more subtle process that uses oly edge removal. 3.2 Ajtai-Komlós-Szemerédi s r(3, k) < ck 2 / log k Erdős-Szekeres boud of r(s, t) ( ) ( s+t 2 s 2 implies that r(3, k) k+1 ) 2. Let s prove this boud i a alterative way, which will lead to the improved boud: if G has a vertex of degree k, the we are doe, otherwise G has average degree at most k (actually, maximum degree but we igore this), ad it is kow that a graph with average degree r 1 has a idepedet set of size at least /2t. Oe way to prove this fact, it to first observe that at least half the vertices have degree at most 2t ad the use a greedy algorithm that adds vertices to the idepedet set ad removes their eighbors. Note that whe we look for the idepedet set i the graph of average degree t (which i the proof for r(3, t) will be t = log ) we ever use the fact that G is triagle-free. It is easy to see that if we could show that a triagle-free graph with average degree t has a idepedet set of size 12t log t the we would get that ay triagle free graph has a idepedet set of size c log which would prove that r(3, k) ck 2 / log k. 3.3 AKS Ituitio 12t log t. If log t. So let s assume from We wat to prove that a triagle-free graph with average degree t satisfies α(g) log t 6, the we ca just use the basic result that α(g) 2t ow o that log t > 6. Suppose we fid a vertex of degree at least 2t. It makes sese to remove this vertex ad cotiue by iductio sice the average degree will sigificatly decrease. Ideed, the ew average degree would be at most t 4t 1 = t t 3t 1 12t = t 3t 1 t(1 3 ). While the umber of vertices wet dow by a factor of (1 1/), the average degree wet dow by a factor of (1 3/), so their ratio icreased by 1 + 2/. To get a feelig of the effect of oe such step, observe that after /2 such iteratios we will get a graph with /2 vertices ad average 9

10 degree less tha t/4. Sice (/2) 12(t/4) log(t/4) > 12t log t (here we use the assumptio that t is large) we see that we wi i this case, sice we ca apply iductio o the smaller subgraph. So suppose ow that all vertices are of degree at most 2t. Remember that the aive algorithm would pick a arbitrary vertex, add it to the idepedet set, remove its (at most) 2t eighbors, ad cotiue. This, of course, produces a idepedet set of size O(/t) as evideced by the uio of several K t+1. So the key idea is to show that if G is triagle free, the we ca pick a vertex v, so that after removig v it ad its d(v) eighbors 5, the average degree will go dow by 1 d(v)+1, that is, by the same factor the umber of vertices wet dow 6. As i the first case, to get a feelig of the effect of this step, suppose we applied it util /2 vertices are left. Sice we always had d 2t we icreased the idepedet set by (/2)/(2t) = /4t. Now, ad here comes the clicher, the graph spaed by the remaiig /2 vertices will have average degree t/2 (sice t ad always decreased by the same ratio). Hece, if we ca apply the same step till there are /4 vertices, we will add to the idepedet set at least (/4)/(2(t/2)) = /4t vertices. We see that i each iteratio we icrease the idepedet set by the same /4t. If we do this for log t iteratios (remember that we assume that log t > 6) we get the required idepedet set of size Ω((/t) log t). How do we fid a vertex v as above? We obviously wat the d vertices to be adjacet to as may edges as possible. The key observatio is that sice G is triagle-free, a edge caot be adjacet to two eighbors of v, hece the umber of edges icidet with the eighbors of v is exactly the sum of the degrees of v s eighbors. Sice the average degree is t it seems like the best we ca hope for, is for the sum of the degrees of v s eighbors to be td. Recappig, if we could fid such a vertex v, the removig it alog with its d eighbors would remove dt edges from G, thus givig us a ew graph o d 1 vertices with average degree ( t 2td t t(d + 1) t(d 1) = = t 1 d 1 ) d 1 d 1 d 1 ( < t 1 d 1 ) This is ot quite the t(1 d+1 ) that we wated, but this will tur out to suffice7. Hece we just eed to prove the followig: Lemma: Every graph with average degree t cotais a vertex v satisfyig x N(v) d(x) t d(v). 5 Of course, we add v to the idepedet set as before. 6 Note that this is ot doable i all graphs of average degree t, e.g. i the uio of K t+1 7 Sice 1 (d+1)/ 1 2/ we see that /t did t remai the same (as i the ituitive explaatio we had before) 1 (d 1)/ but istead wet dow by a factor 1 2/. However, assumig we always remove a vertex of degree t this meas that whe we are left with /2 vertices, the ratio /t will go dow by a factor (1 2/)(1 2/( t))(1 2/( 2t)) (1 2/(/2)) 1 2/t, so after the log t iteratios of the process we will lose oly a costat factor. Of course this is false if we remove vertices of degrees t, but the we ca claim that by the time we are left with /2 vertices we will have produced a idepedet set of size larger tha /2t.. 10

11 Proof: Set d 2 (v) = x N(v) d(x). The d 2(v) is the umber of 2-walks startig with v. Thus v d 2(v) gives the umber of ordered triples x, y, z so that (x, y), (y, z) are edges. This last quatity is just v d2 (v) t 2 = t( v d(v)). We thus get that v d 2(v) t( v d(v)) so there must be some v for which d 2 (v) t d(v). 3.4 AKS formal proof Let φ(, t) be the least iteger so that every triagle free graph of average degree t has a idepedet set of size φ(, t). We eed to prove that φ(, t) 12t log t. We apply iductio o, recallig that the result holds for log t 6. Also, if t /12 the 12t log t log t log, so we ca just use the simple boud r(3, k) k 2, which implies that every triagle-free graph has a idepedet set of size log (regardless of its average degree). We will thus also assume from ow o that t /12. (i) If we removed a vertex the we kow that the average degree became at most t(1 3/), so we eed to verify 8 1 that φ( 1, t(1 3/)) φ(, t), i.e. that 12t(1 3/) log(t(1 3/)) 12t log t. This is equivalet to showig that 1 1/ (log t + log(1 3/)) log t. 1 3/ Sice 1 1/ 1 3/ 1 + 2/, it is eough to show that 2 log t + (1 + 2/) log(1 3/) 0. Sice 6 we have log(1 3/) 6/ so (1 + 2/) log(1 3/) 10/, hece the iequality holds whe log t 6. (ii) Suppose we are i the secod case. We kow that if we removed a vertex of degree d, the d 2t ad the average degree wet dow by a factor (1 d 1 ). O the other had the umber of vertices wet dow by a factor of (1 d+1 ). We thus eed to verify that 1+φ((1 (d+1)/), t(1 (d 1)/)) φ(, t), that is, that (1 (d + 1)/) log(t(1 (d 1)/)) t(1 (d 1)/) 12t log t, wheever d 2t. As we assume that t /12 ad d 2t, we ifer that d /4, guarateeig that (1 d+1 )/(1 d 1 ) > 1 3. Thus, dividig both sides by /12t we eed to prove that wheever d 2t. The above follows from (1 3/) log(t(1 (d 1)/)) + 12t/ log t, 12t 3 log t log(1 (d 1)/). As oted above d /4, implyig that log(1 d 1 ) 2(d 1)/. Hece we eed to verify that 12t 3 log t + 2(d 1) which holds sice we assume that d 2t. 8 If the average degree is smaller tha t(1 3/) we of course oly gai more, so we ca assume that the average degree is exactly t(1 3/). This ca also be justified by the fact that log t t we cosider. 11 is decreasig for t t 0 which is the case

12 3.5 Improved boud for r(4, ) Erdős-Szekeres gives r(4, ) = O( 3 ) while probabilistic proofs give r(4, ) = Ω((/ log ) 2.5 ). We ca slightly improve the upper boud by usig the improved boud for r(3, ). We will eed the followig robust versio of the AKS boud. Lemma 1: If G has average degree t ad at most t 2 δ triagles, the α(g) > cδ(/t) log t. We ow use this lemma to prove that r(4, ) = O( 3 / log 2 ). Take a graph G o r = C 3 / log 2 vertices ad assume it is K 4 -free. If some vertex has degree at least r(3, ) the we are doe. I particular, this meas that G s average degree is at most t = r(3, ) 2 / log. For each edge (u, v) the vertices of N(u) N(v) must be idepedet, as a edge i this set will give a K 4 with u ad v. If for some u, v we have N(u) N(v) > the we are doe, so suppose this is ot the case. The each edges belogs to at most triagles, ad sice G has tr edges, G has at most tr = C 6 / log 3 triagles. This meas that G cotais at most rt 2 δ = (C 3 / log 2 )( 2 / log ) 2 δ triagles for some δ > 0 (ay δ < 1 2 will work) ad hece G has a idepedet set of size cδ r C t log t = cδ log log, so choosig C large eough gives a idepedet set of size. Proof: Pick every vertex with probability p. The whp (Cheroff) we will get at least p/2 vertices, (Markov) at most 2p 2 t edges (or less if the average degree is actually less tha t) ad (Markov) at most 3p 3 t 2 δ triagles. Let s pick p so that expected umber of triagles is about half the expected umber of vertices, i.e. p = 1/t 1 δ/2. The removig a vertex from each triagle will give us a triagle free graph G with (about) p vertices ad average degree (about) (2p 2 t)/(p/2) = 4pt. AKS theorem implies that G (ad thus also G) cotais a idepedet set of size c p 4pt log(4pt) = c t log(tδ/2 ) = cδ t log t. 4 Lecture Ajtai-Komlós-Szemerédi a-la Shearer We wat to prove that every -vertex triagle-free graph of average degree d has a idepedet set of size at least f(d). Let s try to do this by iductio o ad see what kid of f we ca pull. Let s guess that the f : [0, ) [0, 1] we will choose will be cotiuous, differetiable, satisfy f < 0 (sice icreasig average degree decreases α(g)) ad covex 9. We use d 2 (x) for sum of the degrees of x s eighbors, G x for the graph we get after removig x ad N(x), ad d x for the average degree of G x. Sice G is triagle free, we have e(g x ) = e(g) d(y) = 1 2 d d 2(x). (5) y N(x) 9 The trivial boud is f(d) = 1/d which is covex, so we expect the same here. 12

13 Sice we assume that f is covex, we have f(d x ) f(d) + (d x d)f (d). (6) For every vertex x, we have the trivial boud α(g) 1 + α(g x ). Averagig this iequality over all x, the usig iductio ad the (6) we deduce that α(g) 1+ 1 α(g x ) 1+ 1 f(d x )( d(x) 1) 1+ 1 (f(d)+(d x d)f (d))( d(x) 1) x which ca be rearraged (usig x d(x) = d) ito x 1 + f(d) f(d)(d + 1) + (d 2 + d d)f (d) + f (d) Sice d x ( d(x) 1) = 2e(G x ) we ca use (5) to simply the above ito 1 + f(d) f(d)(d + 1) + (d 2 + d d)f (d) + f (d) Cacellig the ±df (d) terms we get 1 + f(d) f(d)(d + 1) + (d 2 + d)f (d) + f (d) x d x ( d(x) 1). x (d 2d 2 (x)). x ( 2d 2 (x)). As i AKS, x d 2(x) is just the umber of 2-walks i G implyig that x d 2(x) d 2. Puttig this i the above expressio ad usig the fact that f (d) < 0 we fially get α(g) f(d) + ( 1 f(d)(d + 1) + f (d)(d d 2 ) ), so if we wat the iductio to work, we eed to fid a f so that the secod term above would vaish, ad that f will be cotiuous, differetiable, covex ad satisfy f (d) < 0. It ca be checked that f(d) = d l(d) d+1 with f(0) = 1, f(1) = 1/2 satisfies all these coditios. Furthermore, sice (d 1) 2 for every d 3 we have f(d) l(d)/d, we get the improved upper boud r(3, ) 2 / l. 4.2 Erdős-Hajal Cojecture/Theorem The lower boud for Ramsey umbers is obtaied by takig a radom graph. It is atural to ask if this is ecessary. Oe obvious property of radom graphs is that they are uiversal, that is, cotai a iduced copy of every fixed graph H. Let hom(g) deote the size of the largest homogeous subset of vertices i G. I other words, hom(g) = max{α(g), ω(g)}. The famous Erdős-Hajal cojecture is that if G is iduced H-free the hom(g) c for some c = c(h) > 0. While it is easy to see that this cojecture holds for may H (covice yourself it holds for H = P 2 ), it is ope already for H = C 5. The followig is still the best kow geeral boud. x 13

14 Theorem: If G is iduced H-free the hom(g) 2 c log for some c = c(h) > 0. We first prove the followig itermediate lemma. Lemma 1: Suppose G is iduced H free. The the followig holds for every δ > 0: there are two disjoit sets A, B satisfyig the followig: A δ ( h 2 ) ad B δ ( 2h ) Either every v A has has at most δ B eighbors i B or every v A has at least (1 δ) B eighbors i B. Proof: Partitio V (G) ito h sets V 1,..., V h of equal size /h ad let s try to embed H by placig i V (H) ito V i. We are boud to fail at some poit. Let us do the embeddig i the followig cotrolled way. As i previous embeddig lemmas, oce we pick vertex i, we are restricted to pick vertex j > i oly from a subset, deoted 10 by Aj i 1, of the previous vertices from which we could choose j (so iitially A 0 j = V j). So the pla is to pick vertex i so that for ay j > i the set of available vertices for vertex j, will ot shrik by more tha δ, i.e. we wat to make sure 11 that A i j δ Ai 1 j. Sice G is H free, we kow that at some poit the process will fail. Suppose we fail at step i for the first time. This meas that each vertex i A i 1 i has a problematic j > i, that is, for each vertex x A there is some j > i such that either (i, j) E(H) ad yet x has at most δ A i 1 eighbors i A i 1 j there must be A A i 1 i or (i, j) E(H) ad yet x has at least (1 δ) A i 1 j of size at least A i 1 i j eighbors i A i 1 j. Hece, /h δ h (/h 2 ) for which the same j is problematic. Settig B = A i 1 j we have thus foud the required sets A, B. Let us ow defie a family of graphs called Coraphs: K 1 is a cograph, ad the family is closed uder disjoit-uio ad complemetatio. Observe that we thus also get closure uder takig uio by addig a complete bipartite graph. Lemma 2: If G is a cograph 12 the α(g)ω(g). I particular, hom(g). Proof: The claim is true for K 1, so we use iductio o the legth of the sequece of operatios for buildig G. If last operatio was complemetatio the this is clear. So suppose last operatio was a disjoit uio of G 1 ad G 2. The iductio gives α(g 1 )ω(g 1 ) 1 ad α(g 2 )ω(g 2 ) 2. Sice ω(g) = max{ω(g 1 ), ω(g 2 )} ad α(g) = α(g 1 ) + α(g 2 ) we get as eeded. 10 A i 1 j ω(g)α(g) = ω(g)α(g 1 ) + ω(g)α(g 2 ) ω(g 1 )α(g 1 ) + α(g 2 )ω(g 2 ) =, meas the vertices available for j after i 1 iteratios 11 I particular we always maitai A i j 1, so that if the process ever stops we will obtai a copy of H. 12 It ca be show that G is a cograph iff G is iduced P 3-free. So this lemma proves the Erdős-Hajal Cojecture for P 3. 14

15 We are ow ready to prove Theorem 1. Proof: We wat to prove that every iduced H-free graph G o vertices has a homogeous set of size f() where f() = 2 c log with c = c(h) > 0. By Lemma 2 it is eough to fid a co-graph of size f(). The proof proceeds by iductio o. Assumig G is iduced H-free, let A, B be the sets from Lemma 1 where δ will be chose shortly. Apply iductio o A to fid a cograph o a set X of size at least f( A ) f(δ h (/h 2 )). If X f() the we are doe, so assume X < f(), ad remove from B all vertices adjacet to a vertex of X. We thus remove at most X δ B < f()δ B vertices from H. Thus, if we take δ = 1/2f() we are guarateed to be left with a set B of size at least B /2 δ h (/4h) δ h (/h 2 ), so that there are o edges betwee B ad X. We ow use iductio to fid a cograph of size at least f(δ h (/h 2 )) i B. Sice we made sure there are o edges betwee X ad B we have thus foud a cograph of size at least ( ) 2f(δ h (/h 2 )) = 2f 2 h (f() h )h 2, (7) where we used our choice of δ i the last equality. Hece we just eed to verify that there is a c = c(h) > 0 so that the fuctio f() = 2 c l satisfies ( ) 2f 2 h (f() h )h 2 f() (8) as (7) would the allow us to complete the proof by iductio o. If we plug f() = 2 c l ito (8) we see that it almost works, i the sese that o matter how large c is, we seem to lack a additive error i the expoet. Istead we prove that the fuctio 13 f() = e 1 h l M where M = satisfies (8), which is clearly eough for our eeds. Ideed ( ) 2f 2 h (f() h )h 2 = 2e 1 h ( l ) 2 h (f() h )h 2 M h l(2) + 2 l(h) h. = exp(l(2) + h 1 l() h l(2) h l(f()) 2 l(h) M) = exp(l(2) + 1 ( ) 1 l() h l(2) h l M 2 l(h) M) h h = exp(l(2) + 1 l() l() + Mh h l(2) 2 l(h) M) h = exp(l(2) + 1 l() l() M) h 13 Note that f(2) 2 so the claim holds for the base of the iductio. 15

16 so we just eed to verify that l(2) + 1 h l() l() h 1 l(), that is, that 1 h ( l() l() l()) l(2). But sice h 3 ad l(2) 1/3 it is eough to prove that x x x 1 which follows by x+ x x multiplyig both sides by x, that is, from the fact that ( x x x)( x + x x) x = 1 x + x x x. 5 Lecture Erdős-Szemerédi Theorem (Ramsey s Theorem for sparse graphs) Theorem: If G has ɛ 2 edges the hom(g) = Ω( log ɛ log(1/ɛ) ). Proof: We first recall that r(s, t) ( ) s+t 2 s 1. Suppose G has ɛ 2 edges. We ca clearly assume that G has maximum degree ɛ (check that you see why). We wish to fid a homogeous set of largest possible size s, where the exact value of s will be chose later. Let K be largest IS 14 i G. If k = K s, the we are doe, so suppose k < s. Sice G has maximum degree ɛ, there are at most ɛk edges coectig K to V \ K, hece at least ( k) /4 of the vertices of V \ K have at most 2ɛk eighbors i K. Call these vertices L. The choice of L implies that there are 2ɛk ( k ) ( i=i i < 2 k 2ɛk) ways to pick the eighborhood of a vertex from L i K. Thus, there must be a subset, call it L, of at least /8 ( ) ( k 2ɛk /8 s 2ɛs) vertices i L with the same eighborhood, call it K, i K. By the choice of L, we have K 2ɛk. Hece if L cotais a IS larger tha 2ɛs > 2ɛk, the addig this IS to K \ K would give a IS larger tha K, a cotradictio. By Erdős-Szekeres, if L also has o clique of size s the L r(s, 2ɛs) ( ) s+2ɛs 2ɛs. But we kow that L > /8 ( ) ( s 2ɛs, implyig that 8 s+2ɛs )( s 2ɛs 2ɛs) > /8. This iequality does ot hold whe s =, thus completig the proof. log 8ɛ log(1/ɛ) We ow observe that the above estimate is tight. G(, δ) oly gives hom(g) = O( log δ ). Actually, ω(g(, δ)) = O( log log log(1/δ) ) ad α(g(, δ)) = O( δ ). Takig δ = ɛ log(1/ɛ) we get a graph G satisfyig ω(g) = O( log log log(1/ɛ) ) ad α(g) = O( ɛ log(1/ɛ) ). Oly problem is that we have too may edges. So takig 1/ɛ disjoit copies of G we get a graph R of desity at most ɛ with α(r) = 1 log ɛ ω(g) = O( ɛ log(1/ɛ) ) ad ω(r) = α(g) = O( log ɛ log(1/ɛ) ). 14 It is (perhaps) temptig to thik that sice G is sparse we ca i fact always fid a IS of size Ω( easy to see that this is ot the case. 15 We assume that k = o() otherwise we have foud a IS of size c. log ɛ log(1/ɛ) ). It is 16

17 5.2 Beck s Theorem (Size-Ramsey umber of a path) Let us start by observig that s(k t ) = ( r(k t) ) 2. The part is trivial. Suppose G has the property that every 2-colorig has a moochromatic K t. We claim that χ(g) r(k t ). Ideed, if this is false the let R be a 2-colorig of K χ(g) without a moochromatic K t, ad color all the edges of G that coect color class i ad color class j usig the color R(i, j). It is easy to check that this colorig of E(G) does ot cotai a moochromatic K t. So we ifer that χ(g) r(k t ) which implies that G has at least ( r(k t) ) 2 edges. We kow that if G has bouded degree the r(g) = O() implyig that s(g) = O( 2 ). We will prove a theorem of Beck statig that s(p ) = O(). This ca be exteded to trees, but ot to graphs of bouded degree 3. It is true however that if G has bouded degree d the s(g) = O( 2 1/d ). We will eed a cosequece of Pósa s famous lemma, whose proof is differed to the ed. We use N(U) to deote the eighbors of U outside U, ad use P k to deote a path o k vertices. Lemma 1: If every U V (G), U u satisfies N(U) 2 U, the G cotais P 3u. We kow that every graph has a subgraph with mi-degree 1 2 d(g). Let e G(X) by umber of edges i G icidet with X. We will eed the followig geeralizatio. Lemma 2: Every G has a subgraph H i which e H (X) > 1 2d(G) X for every X V (H). Proof: If G itself does ot satisfy the coditio, the remove from G a set X satisfyig e G (X) 1 d(g) d(g) X 2d(G) X. The d(g \ X) X = d(g) implyig that cotiuig with this process we must evetually ed up with a o-empty subgraph satisfyig the coditio. Corollary 1: Suppose every X Y V (G) with Y < 3 X 3u satisfies 16 The every 2-colorig of G cotais P 3u. e(x, Y ) 1 d(g) X (9) 4 Proof: Let G be the graph spaed by the popular color. The d(g ) 1 2d(G). By Lemma 2, we ca fid a subgraph H i which every X V (H) satisfies e H (X) > 1 2 d(g ) X 1 d(g) X. (10) 4 We ow claim that H satisfies the coditio of Lemma 1. Take ay U V (H) of size at most u ad assume N(U) < 2 U. The (9) (with X = U ad Y = U N(U)) implies that e H (U) = e H (U, U N(U)) e G (U, U N(U)) 1 4 d(g) U which cotradicts (10). 16 For X Y, we use e(x, Y ) to deote e(x) + e(x, Y \ X) where e(x) is the umber of edges iside X. 17

18 Theorem: There is a -vertex sparse graph satisfyig (9) with u = α. Hece s(p ) = O(). Proof: Cosider G(, p) for some p = c/ where 4d < c < c = 5d will be chose later. Clearly, whp G has at most c edges. Also, whp d(g) 4d. Hece, if some X, Y violate (9) the they must satisfy e(x, Y ) d X. We ow prove that whp, G(, p) satisfies (9) with u = α. Fix some 1 s α, ad take X Y V (G) with 17 X = s, Y = 3s. The s=1 ( 5 ) P[e(X, Y ) d X ] p ds 2 s2 ds ( ) 5esc ds, 2d implyig that the probability of havig such a pair (of some size 1 s u = α) is at most α ( )( ) ( ) [ 5esc ds α (e ) ( e 2 2 ) ( ) ] 5esc d s [ α (e 3 3 ) ( ) ] 40s d s s 2s 2d s s 2 2d s 3, s=1 where we use the fact that c 5d. If d = 4 the we see that for small eough α > 0, the base is smaller tha 1/2 so the sum is o(1), implyig that whp G satisfies all the required properties. s=1 5.3 Pósa s rotatio-extesio lemma We ow prove Lemma 1. We start with Pósa s Lemma. Suppose P is a logest path i some graph G that starts from a give vertex x 0. Let A = A(x 0 ) be 18 the vertices of P which are edpoits of a (logest) path P which ca be obtaied from P by a sequece of rotatios 19. Let B = B(x 0 ) be the eighbors o P of the vertices i A, that is, the vertices that appear before/after some vertex of A o the path P. Clearly B 2 A. Actually, sice x k A we have B 2 A 1. Lemma 2 (Pósa s Lemma): I the otatio above, we have N(A) B. Proof: Sice P is a logest path, a vertex of a A caot have a eighbor ot i P, hece N(A) P, so we just eed to prove that if (a, b) is a edge the b A B. A importat observatio is that if P is a rotatio of P the all the vertices of P, save for x i, x i+1 ad x k, have the same pair of eighbors alog P ad P (but the order of these pairs might switch). Moreover, ote that x i+1, x k A ad x i B. Cosider ow the sequece of rotatios that eds up with a path P whose last vertex is a. If i oe of the rotatios alog the way, b was either x i, x i+1 or x k the b A B ad we are doe. If ot, the by the previous paragraph, the vertex appearig after b o P, call it b, was also a eighbour of b i P. It is ow easy to see that sice (a, b) is a edge, we ca perform aother rotatio that will place b at the ed of the path. But this meas that b A ad so b B. 17 Note that if a graph satisfies (9) wheever Y = 3 X the it also satisfies (9) wheever Y 3 X. 18 Strictly speakig we should have writte A = A(x 0, P ) sice A depeds o the specific path P that we chose to work with. 19 If P = x 0, x 1,..., x k the a rotatio of P is a path of the form x 0, x 1,..., x i, x k, x k 1,..., x i+1. 18

19 Proof (of Lemma 1): Let P be a logest path, ad suppose it starts at x 0. Settig A = A(x 0 ), B = B(x 0 ), we kow from Lemma 2 that N(A) B. But sice B 2 A 1 the assumptio o G imply that A > u (which already meas that P > u). Pick ay subset A A of size u. Assumptio o G, ad the fact that A N(A ) P, ow imply that P A N(A ) 3u. 5.4 Hamiltoicity of sparse radom graphs Lemma 1: If G G(, p) ad p A(x 0 ) > /4. 25 l() the with probability 1 o(1/) every vertex x 0 satisfies Proof: The probability that for some 1 k /4 a graph G G(, p) cotais a set of k vertices with o edge coectig it to aother set of size at least 3k is at most 20 /4 ( )( ) /4 /4 (1 p) k( 3k) 4k e pk/4 = [ 4 e p/4] k k 3k k=1 k=1 which is o(1/) whe p = 25 l()/. We claim that if the above holds, the G satisfies the assertio of the lemma. Ideed, suppose A(x 0 ) = k /4. By Pósa s Lemma we have N(A(x 0 )) B(x 0 ). But by the defiitio of B we have B(x 0 ) 2 A(x 0 ) = 2k, implyig that there is o edge betwee A(x 0 ) (whose size is k) ad V \ (A(x 0 ) B(x 0 )) (whose size is at least 3k). Lemma 2: If G G(, p) ad p 25 l() k=1 the whp, the followig holds for every vertex x 0 ; The logest path i G is loger tha the logest path i G \ x 0. I particular, G has a Hamilto path. Proof: Fix x 0. We first expose the edges i G \ x 0 which is just a istace of G( 1, p). By the previous lemmas with probability 1 o(1/), we have A(y 0 ) > ( 1)/4 for every vertex y 0 G\x 0. Let P be a logest path i G \ x 0 ad assume its first vertex is y 0. Suppose that A(y 0 ) > ( 1)/4. If we ow expose the edges betwee x 0 ad the rest of G we see that the probability of ot havig a edge from x 0 to A(y 0 ) is at most (1 25 l()/) ( 1)/4 = o(1/). Assumig oe such edge (x 0, v) exist, we ca obtai a path (i G) loger tha P by takig a path edig with v ad extedig it to x 0. Sice i both steps of exposig G we got the properties we wated with probability 1 o(1/) the claims follows. Sice the above holds with respect to x 0 with probability 1 o(1/) we get from a uio boud that it holds for all x 0 whp. Theorem: If G G(, 26 l()/) the whp G is Hamiltoia. Proof: Thik of G G(, 26 l()/) as the uio of idepedet G 1 G(, 25 l()/) ad G 2 G(, l /) (check that you see why we made do so). By Lemma 2, whp G 1 is hamiltoia, 20 It is clearly eough to cosider oly sets of size exactly 3k. 19

20 ad from Lemma 1, whp we have A(x 0 ) /4 for all x 0. Suppose both properties hold, let P be a Hamilto path ad suppose it starts with x 0. The the probability that G 2 cotais o edge betwee x 0 ad A(x 0 ) is (1 l /) /4 = o(1). Assumig such a edge (x 0, u) is preset we get a Hamilto cycle by takig a Hamilto path edig with u ad closig it with (x 0, u). Boud is tight i the sese that G(, 1 2 l()/) is ot eve coected whp. Actually, whp G(, 1 log +log log ±ω(1) 2 l()/) has isolated vertices. Exact threshold is aroud p =. Metio radom process ad mi-degree 2. 6 Lecture Depedet radom choice basic lemma Basic Lemma: Let a, d, m,, r be positive itegers. Let G = (V, E) be a graph with V = vertices ad average degree d = 2 E(G) /. If there is a positive iteger t such that d t ( ) (m ) t t 1 a, r the G cotais a subset U of at least a vertices such that every r vertices i U have at least m commo eighbors. Proof: Pick t vertices at radom (with repetitios) ad let A be the vertices adjacet to all the t chose vertices. The by liearity of expectatio ad covexity, we have E[ A ] = v (d(v)/) t = t v d(v) t 1 d t / t 1. O the other, if S is a set of r vertices that have less tha tha m commo eighbors, the the probability that S A is at most (m/) t. Hece, the expected umber of such sets i A, deoted by Y, is at most ( r) (m/) t. By liearity of expectatio ad the lemma s assumptio we have E[ A Y ] a, so if we remove a vertex from each set of r vertices with less tha m commo eighbors we get the required set. For what follows we will eed the followig embeddig lemma (give as home assigmet). Lemma 1: Suppose X, Y are vertex sets so that: every collectio of p vertices i X have at least b commo eighbors i Y. The G cotais a copy of every bipartite graph with vertex sets A, B satisfyig A X, B b ad where every vertex i B has at most p vertices i A. We ow derive the followig corollaries of the above two lemmas. Theorem: If H = (A, B, E) is a bipartite graph where every vertex i B has degree at most p the ex(, H) c 2 1/p. 20

21 Proof: By Lemma 1 we just eed to prove that for c large eough, there is a vertex set X of size at least h so that each subset of p vertices of its vertices have at least h commo eighbors. If G has c 2 1/p edges, the i the basic lemma we have d = 2c 1 1/p, r = p, m = h ad a = h, so it is eough 21 to fid a t so that (2c) t 1 t/p p t h t h. Cosiderig this differece, it is ow clear that we should take t = p, i which case the iequality becomes (2c) p h p h, so takig c = h makes sure the required iequality holds. This boud is believed to be tight. Note that there is a caveat i the above proof sice some vertices of the commo eighborhood of a p-tuple of vertices from X might be i X itself (resultig i choosig the same vertex twice). Thik how this ca be fixed (either by first pickig a bipartite subgraph or by replacig h with 2h). The above aalysis is tight i that if we have o( 2 1/p ) edges the the first term would be o(1) so we would have o chace of fidig a set of size h satisfyig the coditio. So i this case it was easy to guess the best boud that this proof ca give. We ow give a more ivolved example. Theorem: Let Q b deote the b-dimesioal Boolea cube. The r(q b ) 2 3b = Q b 3. Proof: As usual, we look at the popular color, that is, we ll prove that for large eough, every graph of average degree /2 cotais Q b. By Lemma 1, it is eough to fid a subset of size 2 b 1 so that every set of b vertices have at least 2 b 1 commo eighbors. Hece, i the basic lemma we have d = /2, r = b, m = 2 b 1 ad a = 2 b 1 so (1/2) t b (2 b 1 /) t 2 b 1. (11) Sice we kow that 2 b it makes sese to look for = 2 αb with the smallest α. We clearly eed to require the secod term be smaller tha the first, that is 2 αb t 2 αb2 αbt+bt i which case the boud will be basically the first term that is 2 αb t. So we require αb t b (so that the first term will be larger tha 2 b ) ad α(t + 1 b) b + t/b. This last requiremet implies that t should be of order b so we set t = βb. The we eed α 1 + β ad α 1+β/b β 1+1/b = 1+o(1) β 1+o(1). We look for the case whe both bouds meet that is whe 1 + β = 1 β 1 i.e. β = 2. For the sake of simplicity set β = 3 2 (so t = 3 2 b) ad α = 3 (so = 23 ). It is ow easy to check that this choice is valid i (11). We will soo improve this to r(q b ) b2 2b. It is cojectured that r(q b ) = O(2 b ). Theorem: If G has ɛ 2 edges the G has a 1-subdivisio of K b with b = ɛ 3/2. 21 We cosider here a expressio which is smaller tha the right had side expressio i the lemma. We will also do so i later proofs. 21

22 Proof: It is easy to see that if U = b ad every 2 vertices i U have at least ( b 2) commo eighbors, the we ca costruct a 1-subdivisio of K b. Takig d = (2ɛ), r = 2, a = b, m = b 2 i the basic lemma the situatio is ( ) ( ) b (2ɛ) t 2 t b. 2 If we wat the secod term to be smaller tha the first the b. It is clear that b should also deped o ɛ so let us parameterize b = (2ɛ) α. Agai, we clearly eed the secod term to be smaller tha the first so we require (2ɛ) t 2 (2ɛ) t2α, ad the goal is to fid the smallest α for which there is a t so the the above holds. I this case, we will get at least 1/2 of the term. We kow that (2ɛ) t 1/ (sice, agai, b ) implyig that α 3/2. It is ow easy to check that takig α = 3/2 (i.e. b = (2ɛ) 3/2 log ) ad choosig t so that (2ɛ) t = 1/ (i.e. t = log 2 log(1/ɛ) ) is a valid choice above. Above boud ca be improved to b = ɛ which is tight (cosider G(, ɛ)). 6.2 Variat of basic method We will ow revisit the topic of Ramsey umber of bouded degree graphs. For simplicity, we will oly cosider bipartite graphs. We wat to prove the followig improved boud for Ramsey umbers of bipartite graphs of bouded degree. Theorem 1: Suppose H is a -vertex bipartite graph of maximum degree b. If G has desity ɛ ad N 64b/ɛ b vertices the H G. We ote that settig ɛ = 1/2 we get the if H is a -vertex bipartite graph of maximum degree b the r(h) = O(b2 b ). It is kow 22 that there are -vertex bipartite graphs of maximum degree b satisfyig r(h) 2 cb so the above above upper boud is (almost) tight. We also get for the cube Q b that r(q b ) = O(b2 2b ) which is better tha the 2 3b boud we proved earlier. Suppose ɛ = 1/2 (we ca usually assume this sice oe of the colors has desity 1/2). If we wat to use the Basic Lemma, the we eed to fid a subset of size = Ω(N) where every b-tuple of vertices have = Ω(N) commo eighbours. The basic lemma caot give us this sice we get Θ() Θ( d )Θ(1). I fact, this stroger versio of the basic lemma is false! This calls for a relaxed versio, i which we allow some of the subsets to violate the coditio. Lemma 1: Suppose ɛ > 0 ad b. If N 16b/ɛ b ad G is a N N bipartite graph with ɛn 2 edges, the it cotais a subset U of size at least that cotais less tha ( U b ) /(2b) b b-sets with less tha commo eighbors. 22 The proof of this statemet is very similar to the oe we gave for ot ecessarily bipartite graphs. 22