The independent neighborhoods process

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1 The independent neighborhood proce Tom Bohman Dhruv Mubayi Michael Picollelli March 16, 2015 Abtract A triangle T (r) in an r-uniform hypergraph i a et of r + 1 edge uch that r of them hare a common (r 1)-et of vertice and the lat edge contain the remaining vertex from each of the firt r edge. Our main reult i that the random greedy triangle-free proce on n point terminate in an r-uniform hypergraph with independence number O((n log n) 1/r ). A a conequence, uing recent reult on independent et in hypergraph, the Ramey number r(t (r), K (r) ) ha order of magnitude r / log. Thi anwer quetion poed in [4, 10] and generalize the celebrated reult of Ajtai-Komló-Szemerédi [1] and Kim [9] to hypergraph. 1 Introduction An r-uniform hypergraph H (r-graph for hort) i a collection of r-element ubet of a vertex et V (H). Given r-graph G and H, the ramey number r(g, H) i the minimum n uch that every red/blue-edge coloring of the complete r-graph K n (r) := ( ) [n] r contain a red copy of G or a blue copy of H (often we will write K n for K n (r) ). Determining thee number for graph (r = 2) i known to be notoriouly difficult, indeed the order of magnitude (for fixed t) of r(k t, K ) i wide open when t 4. The cae t = 3 i one of the celebrated reult in graph Ramey theory: r(k 3, K ) = Θ( 2 / log ). (1) The upper bound wa proved by Ajtai-Komló-Szemerédi [1] a one of the firt application of the emi-random method in combinatoric (impler proof now exit due to Shearer [12, 13]). The lower bound, due to Kim [9], wa alo achieved by uing the emi-random or nibble method. More recently, the firt author [3] howed that a lower bound for r(k 3, K ) could alo be obtained by the triangle-free proce, which i a random greedy algorithm. Thi ettled a quetion of Spencer on the independence number of the triangle-free proce. Still more recently, Bohman-Keevah [6] and Fiz Pontivero-Griffith-Morri [8] have analyzed the triangle-free proce more carefully and improved the contant obtained o that the gap between the upper and lower bound for r(k 3, K ) i now aymptotically a multiplicative factor of 4. Department of Mathematical Science, Carnegie Mellon Univerity, Pittburgh, PA Reearch upported by NSF grant DMS Department of Mathematic, Statitic, and Computer Science, Univerity of Illinoi at Chicago, Chicago, IL Reearch upported by NSF grant DMS Department of Mathematic, California State Univerity San Marco, San Marco, CA

2 Given the difficulty of thee baic quetion in graph Ramey theory, one would expect that the correponding quetion for hypergraph are hopele. Thi i not alway the cae. Hypergraph behave quite differently for aymmetric Ramey problem, for example, there exit K (3) 4 -free 3- graph on n point with independence number of order log n, o r(k (3) 4, K(3) ) i exponential in unlike the graph cae. Conequently, to obtain r-graph reult parallel to (1), one mut conider problem r(g, K ) where G i much parer than a complete graph. A recent reult in thi vein due to Kotochka-Mubayi-Vertraëte [10] i that there are poitive contant c 1, c 2 with c 1 3/2 < r(c(3) (log ) 3/4 3, K(3) ) < c 2 3/2 where C (3) 3 i the looe triangle, compriing 3 edge that have pairwie interection of ize one and have no point in common. The author in [10] conjectured that r(c (3) 3, K(3) ) = o( 3/2 ) and the order of magnitude remain open. Another reult of thi type for hypergraph due to Phelp and Rödl [11] i that r(p (3) 2, K (3) ) = Θ( 2 / log ), where P (3) t i the tight path with t edge. Recently, the econd author and Cooper [7] prove that for fixed t 4, the behavior of thi Ramey number change and we have r(p (3) t, K (3) ) = Θ( 2 ); the growth rate for t = 3 remain open. Thee are the only nontrivial hypergraph reult of polynomial Ramey number, and in thi paper we add to thi lit with an extenion of (1). Definition 1. An r-uniform triangle T (r) i a et of r + 1 edge b 1,..., b r, a with b i b j = R for all i < j where R = r 1 and a = i (b i R). In other word, r of the edge hare a common (r 1)-et of vertice, and the lat edge contain the remaining point in all thee previou edge. When r = 2, then T (2) = K 3, o in thi ene T (r) i a generalization of a graph triangle. We may view a T (r) -free r-graph a one in which all neighborhood are independent et, where the neighborhood of an R ( ) V (H) r 1 i {x : R {x} H}. Frieze and the firt two author [4] proved that for fixed r 2, there are poitive contant c 1 and c 2 with c 1 r (log ) r/(r 1) < r(t (r), K (r) ) < c 2 r. They conjectured that the upper bound could be improved to o( r ) and believed that the log factor in the lower bound could alo be improved. Kotochka-Mubayi-Vertraëte [10] partially achieved thi by improving the upper bound to and believed that the log factor wa optimal. r(t (r), K (r) ) = O( r / log r) In thi paper we verify thi aertion by analyzing the T (r) -free (hyper)graph proce. Thi proce begin with an empty hypergraph G(0) on n vertice. Given G(i 1), the hypergraph G(i) i then formed by adding an edge e i elected uniformly at random from the r-et of vertice which neither form edge of G(i 1) nor create a copy of T (r) in the hypergraph G(i 1)+e i. The proce terminate with a maximal T (r) -free graph G(M) with a random number M of edge. Our main reult i the following: Theorem 1. For r 3 fixed the T (r) -free proce on n point produce an r-graph with independence number O ( (n log n) 1/r) with high probability. 2

3 Thi reult together with the aforementioned reult of Kotochka-Mubayi-Vertraëte give the following generalization of (1) to hypergraph. Corollary 2. For fixed r 3 there are poitive contant c 1 and c 2 with c 1 r log < r(t (r), K (r) r ) < c 2 log. Graph procee that iteratively add edge choen uniformly at random ubject to the condition that ome graph property i maintained have been ued to generate intereting combinatorial object in a number of context. In addition to the lower bound on the Ramey number r(k 3, K ) given by the triangle-free graph proce (dicued above), the H-free graph proce give the bet known lower bound on the Ramey number r(k t, K ) for t 4 fixed and the bet known lower bound on the Turán number for ome bipartite graph [5]. The proce that form a ubet of Z n by iteratively chooing element to be member of the et uniformly at random ubject to the condition that the et doe not contain a k-term arithmetic progreion produce a et that ha intereting propertie with repect to the Gower norm [2]. The T (r) -free (hyper)graph proce can be viewed a an intance of the random greedy hypergraph independent et proce. Let H be a hypergraph. An independent et in H i a et of vertice that contain no edge of H. The random greedy independent et proce form uch a et by tarting with an empty et of vertice and iteratively chooing vertice uniformly at random ubject to the condition that the et of choen vertice continue to be an independent et. We tudy the random greedy independent et proce for the hypergraph H T (r) which ha vertex et ( [n] ) r and edge et coniting of all copie of T (r) on vertex et [n]. Note that, ince an independent et in H T (r) give a T (r) -free r-graph on point et [n], the random greedy independent et proce on H T (r) i equivalent to the T (r) -free proce. Our analyi of the T (r) -free proce i baed on recent work on the random greedy hypergraph independent et proce due to Bennett and Bohman [2]. The remainder of the paper i organized a follow. In the following Section we etablih ome notation and recall the neceary fact from [2]. The proof of Theorem 1 i given in the Section that follow, modulo the proof of ome technical lemma. Thee lemma are proved in the final Section by application of the differential equation method for proving dynamic concentration. 2 Preliminarie Let H be a hypergraph on vertex et V = V (H). For each et of vertice A V, let N H (A) denote the neighborhood of A in H, the family of all et Y V \ A for which A Y H. We then define the degree of A in H to be d H (A) = N H (A). For a nonnegative integer a, we define a (H) to be the maximum of d H (A) over all A ( V a). Next, for a pair of (not necearily dijoint) et A, B V, we define the codegree of A and B to be the number of et X V \ (A B) for which A X, B X both lie in H. Recall that we define G(i) to be the r-graph produced through i tep of the T (r) -free proce. We let F i denote the natural filtration determined by the proce (ee [3], for example). We alo implify our notation omewhat and write N i (A) in place of N G(i) (A), d i (A) in place of d G(i) (A), etc., when appropriate. 3

4 The r-graph G(i) partition ( ) [n] r into three et E(i), O(i), C(i). The et E(i) i imply the et of i edge choen in the firt i tep of the proce. The et O(i) conit of the open r-et: all e ( ) n r \ E(i) for which G(i) + e i T (r) -free. The r-et in C(i) := ( ) [n] r \ (E(i) O(i)) are cloed. Finally, for each open r-et e O(i), we define the et C e (i) to conit of all open r-et f O(i) uch that the graph G(i) + e + f contain a copy of T (r) uing both e and f a edge. (That i, C e (i) conit of the open r-et whoe election a the next edge e i+1 would reult in e C(i + 1).) We now introduce ome notation in preparation for our application of the reult in [2]. Set ( ) ( ) n n r N := D := (r + 1) := N r r 1 D 1/r. Note that N i the ize of the vertex et of the hypergraph H T (r) and D i the vertex degree of H T (r) (in other word, every r-et in [n] i in D copie of T (r) ). The parameter i the caling for the length of the proce. Thi choice i motivated by the heuritic that E(i) hould be peudorandom; that i, E(i) hould reemble in ome way a collection of r-et choen uniformly at random (without any further condition). If thi i indeed the cae then the probability that a given r-et i open would be roughly ( ( ) i r ) D { ( ) i r 1 exp D} N N and a ubtantial number of r-et are cloed when roughly edge have been added. In order to dicu the evolution in more detail, we pa to a limit by introducing a continuou time variable t where t = t(i) = i/. The evolution of key parameter of the proce cloely follow trajectorie given by the function q(t) := exp { t r } and c(t) := q (t) = rt r 1 q(t). We introduce mall contant ζ, γ uch that ζ γ 1/r. (The notation α β here mean that α i choen to be ufficiently mall relative to β.) The point where we top tracking the proce i given by i max := ζ ND 1/r (log 1/r N) and t max := i max / = ζ log 1/r N. For i 0, let T i denote the event that the following etimate hold for all tep 0 i i : and for every open r-et e O(i) O(i) = ( q(t) ± N γ) N (2) C e (i) = ( c(t) ± N γ) D 1/r. (3) It follow from the reult of Bennett and Bohman that T imax hold with high probability. We now recall the reult of [2] in order to verify that thi i indeed the cae. Bennett and Bohman tudied the random greedy independent et proce applied to an ˆruniform, D-regular hypergraph H. A we dicu above, the T (r) -free proce i identical to the random greedy independent et proce on the hypergraph H T (r). Note that H T (r) i (r+1)-uniform, and o in our application of Bennett-Bohman we have ˆr = r + 1. Define the (ˆr 1)-codegree of a pair of ditinct vertice v, v in the hypergraph H to be the number of edge e, e H uch that v e \ e, v e \ e and e e = ˆr 1. We let Γ(H) be the maximum (ˆr 1)-codegree of H. 4

5 Theorem 3 (Theorem 1.1 of [2]). Let ˆr and ɛ > 0 be fixed. Let H be an ˆr-uniform, D-regular hypergraph on N vertice uch that D > N ɛ. If l (H) < D ˆr l ˆr 1 ɛ for l = 2,..., ˆr 1 (4) and Γ(H) < D 1 ɛ then the random greedy independent et algorithm produce an independent et I in H with ( ( ) 1 ) log N ˆr 1 I = Ω N (5) D with probability 1 exp { N Ω(1)}. Note that l (H T (r)) = Θ(n r l ) = Θ(nˆr 1 l ) and Γ(H T (r)) = 0, and therefore the work of Bennett- Bohman applie to the T (r) -free proce. We require more detailed information from [2]. Theorem 3 i proved by tracking key parameter of the proce, thee include the following, where we let I(i) be the independent et that ha been formed through i tep of the random greedy independent et algorithm: The ize of the et V (i) of vertice of H that remain available for incluion in the independent et after i vertice have been added to the independent et. Note that for the T (r) -free proce, the et of vertice that remain available for incluion in the independent et in H T (r) i preciely the collection of open edge O(i). For every vertex v available at tep i, the number d 2 (v, i) of available vertice u v with the property that there i ome edge e H uch that u, v e and e I(i) = ˆr 2. Note uch an available vertex v in the vertex et of H T (r) i an open edge in the T (r) -free proce, and the collection of vertice u that atify thee condition in H T (r) i the et of open edge in the T (r) -free proce in the et C e (i). Bound on V (i) and d 2 (v, i) are given in equation (8) and (9), repectively, of [2]. Thee bound immediately give the etimate (2) and (3) quoted above. Note that 2 = + 2 2, d 2 = d + 2 d 2, and the error function f v, f 2 can be bounded above by D to an arbitrarily mall contant, uniformly in t. Moreover, 2 (defined on page 11 of [2] and tranlated to our notation) i equal to rd 1/r t r 1 q which matche the main term in (3) a c(t) = rt r 1 q(t). We will alo make ue of the following fact regarding r-graph that appear a ubgraph of the T (r) -free proce. Lemma 4 (Lemma 5.1 of [2]). Fix a contant L and uppoe e 1,..., e L ( ) [n] r form a T (r) -free hypergraph. Then for all tep j i max, P [{e 1,..., e L } E(j)] = (j/n) L (1 + o(1)). Note that the fact that T imax hold with high probability doe not prove that the independence number of G(M) i O ( (n log n) 1/r) with high probability. Thi i the main reult in thi work; it i proved below. Before commencing with the detail of the proof, we briefly oberve that the deired bound on the independence number of G(M) can be viewed a a peudorandom property of the 5

6 r-graph G(i). Indeed, if G(i) reemble a collection of r-et choen uniformly at random then the expected number of independent et of ize k would be ( ) ( ( k i { )} n 1 ( k n = exp Θ (k log n) Θ (i r)) kr n r. If the proce lat through i = Θ(ND 1/r (log 1/r N)) = Θ(n r 1+1/r log 1/r n) tep then we would anticipate an independence number of O ( (n log n) 1/r). In the remainder of the paper we make thi heuritic calculation rigorou. 3 Independence number: Proof of Theorem 1 We expand the lit of contant given in the previou ection by introducing large contant κ and W, and mall contant ɛ uch that 1 κ ζ 1 W ε γ. (6) In the coure of the argument we introduce dynamic concentration phenomena that will tated in term of the error function f(t) := exp {W (t r + t)}. Define the contant λ := κ γ 2, and then let k := κ(n log n) 1/r and l := λ(n log n) 1/r, noting that a γ i mall, k 2l. Our aim i to how that the independence number of G(i max ) i at mot k with high probability. To do o, we will how that provided κ( i uitably large, w.h.p. for every tep 0 i i max, every k-element et of vertice ha at leat Ω q(t) ( k r) ) open r-et. A equation (2) etablihe (1 + o(1))q(t)n open r-et in total w.h.p., the probability that T imax hold and a given k-et remain independent over all i max tep i then at mot i max i=1 ( 1 Ω ( )) q(t)k r = q(t)n ( 1 Ω ( κ r log n n r 1 )) imax { } = exp ζκ r Ω(n 1/r log 1+1/r n), where our O( ), Ω( ), Θ( ) notation doe not uppre any contant that appear in (6). Since { } n k = exp κ O(n 1/r log 1+1/r n), thi uffice by the union bound, provided κ i uitably large with repect to r and ζ. There i a ignificant obtacle to proving that every et of k vertice contain the right number of open r-et. Since the forbidden r-graph T (r) conit of an (r 1)-et (R) along with an edge (a) contained in it neighborhood, it follow that all r-et within the neighborhood of an (r 1)-et in G(i) mut be cloed. (That i, if A ( ) ( [n] r 1 then Ni ) (A) r C(i)). So a et of k vertice that ha a large interection with the neighborhood of an (r 1)-et doe not have the right number of open r-et. To overcome thi obtacle, we extend the argument in [3] for bounding the independence number of the triangle-free proce. Our argument ha two tep: 6

7 1. We apply the differential equation method for etablihing dynamic concentration to how that unle a certain bad condition occur, a pair of dijoint l-et will have the right number of open r-et that are contained in the union of the pair of l-et and interect both l-et, that i about q(t) [ ( ) ( 2l r 2 l ) ( r ] open r-et. Note that 2l ) ( r 2 l ) r > 1 k 3( r), ay, a γ i mall. 2. We then argue that w.h.p., every k-et contain a (dijoint) pair of l-et which i good, i.e., for which the bad condition doe not occur. We formalize thi with the notion of r-et which are open with repect to a pair of dijoint l-et. Definition 2. Fix a dijoint pair A, B ( [n]) l. The topping time τa,b i the minimum of i max and the firt tep i for which there exit a (r 1)-et X uch that N i (X) A, N i (X) B, and N i (X) (A B) k/n 2ε. Definition 3. For each tep i 0, we ay that an r-et e A B i open with repect to the pair A, B in G(i) if e A, e B, and either e O(i) or e O(i 1) C(i) and i = τ A,B. Let Q A,B (i) count the number of r-et which are open with repect to the pair A, B in G(i). Lemma 5. With high probability, for every dijoint pair A, B ( [n]) l and all tep 0 i τa,b, ( )] l Q A,B (i) = ( q(t) ± f(t) ) n ε [( ) 2l 2 r r. (7) Lemma 6. With high probability, for every tep 0 i < i max and every et K ( [n]) k, there exit a pair of dijoint l-et A, B contained in K for which τ A,B > i. Lemma 5 and 6, repectively, complete tep 1 and 2 of the proof outlined above. The bad condition for a pair A, B of dijoint l-et i the event that we have reached the topping time τ A,B ; that i, the bad condition i that there i ome (r 1)-et whoe neighborhood interect both A and B and ha large interection with A B. Note that if i < τ A,B then Q A,B i equal to the number of open r-et that are contained in A B and interect both A and B. Thu, Lemma 5 ay that if we do not have the bad condition then we have the right number of uch et. Lemma 6 then ay that every k-et contain a pair dijoint pair A, B of l-et for which the bad condition doe not hold. Taken together, Lemma 5 and 6 yield that w.h.p., for every tep 0 i < i max, every k-et contain at leat q(t)(1 + o(1))[ ( ) ( 2l r 2 l ) r ] = Ω (q(t) ( k r) ) open r-et, a required. We now prove Lemma 6 modulo the proof of Lemma 7 which bound the maximum degree of an (r 1)-et. Lemma 5 and 7 are proved in the next Section. Proof of Lemma 6. We require a bound on the maximum degree of (r 1)-et of vertice. For each tep i 0 let D i denote the event that r 1 (G(i)) ε(n log n) 1/(r 1). Lemma 7. T imax D imax hold with high probability. 7

8 The proof of Lemma 7 i given in the next Section. Fix a tep 0 i < i max, and a et K ( [n]) k. Note that, by Lemma 7, we may aume that Di hold. We alo note that the maximum co-degree of a pair of et A, B ( [n] r 1) i at mot 5r with high probability. Thi follow from Lemma 4 and the union bound: ( ( ) ) ( )( ) ( ) [n] n n i 10r Pr A, B with co-degree 5r n 5r r 1 r 1 r 1 N = n 8 3r+o(1) = o(1). (8) Given thee two fact (i.e. thee degree and co-degree bound for (r 1)-et), the remainder of the proof i determinitic. To begin, define the et Claim 1. X < 2n 2ε. X := { ( ) } [n] X : N i (X) K k/n 2ε. r 1 Proof. Suppoe Y X with Y = 2n 2ε. Let N = Y Y (N i(y ) K). By incluionexcluion, k N Y (k/n 2ε ) Y 2 5r 2k 20rn 4ε, a contradiction a ε i mall and k = n 1/r+o(1). Next, we dicard from K the vertice which are common neighbor of (r 1)-et in X : let and K good := K \ K bad. Then ay, for large n. K bad := {v K : X, Y X with X Y and v N i (X) N i (Y )} K bad X 2 5r 20rn 4ε < γ 2 (n log n)1/r, We find dijoint l-ubet A, B of K good a follow, noting K good 2l + (γ/2)(n log n) 1/r. For each ubet Y X, let N(Y) = N i (Y ) K good. Y Y Now, chooe a maximal ubet X X ubject to N(X ) l. If X = X, then let A, B be l-et atifying N(X ) A K good and B K good \ A. Otherwie, pick any et X X \ X, o l < N(X {X }) < l + ε(n log n) 1/r ; let A N(X {X }) and B K good \ N(X {X }) be l-et. Oberve now that if X = X, then N i (X) B = for all X X. Otherwie, if X X {X }, N i (X) B =, but if X X \ (X {X }) then N i (X) A = a we are working within K good. In either cae, for every (r 1)-et X for which N i (X) (A B) k/n 2ε hold, either N i (X) A = or N i (X) B =, and τ A,B > i follow. 8

9 4 Dynamic Concentration In thi ection we prove Lemma 5 and 7. Both of thee tatement aert dynamic concentration of key parameter of the T (r) -free proce. We apply the differential equation method for proving dynamic concentration, which we now briefly ketch. Suppoe we have a combinatorial tochatic proce baed on a ground et of ize n that generate a natural filtration F 0, F 1,.... Suppoe further that we have a equence of random variable A 0, A 1,... and that we would like to prove a dynamic concentration tatement of the form A i T i + E i for all 0 i m(n) with high probability, (9) where T 0, T 1,... i the expected trajectory of the equence of random variable A i and E 0, E 1,... i a equence of error function. (One i often intereted in proving a lower bound on A i in conjunction with (9). The argument for proving thi i eentially the ame a the upper bound argument that we dicu here.) We often make thi tatement in the context of a limit that we define in term of a continuou time t given by t = i/ where i the time caling of the proce. The limit of the expected trajectory i determined by etting T i = f(t)s(n) where S = S(n) i the order caling of the random variable A i. Given thee aumption we hould have E [A i+1 A i F i ] = T i+1 T i = [f(t + 1/) f(t)] S f (t) S. Thu the trajectory i determined by the expected one-tep change in A i. We prove (9) by applying fact regarding the probability of large deviation in martingale with bounded difference. In particular, we conider the equence D i = A i T i E i. Note that if we et T 0 = A 0 (which i often the natural initial condition) then D 0 = E 0. If we can etablih that the equence D i i a upermartingale and E 0 i ufficiently large then it hould be unlikely that D i i ever poitive, and (9) follow. In order to complete uch a proof we how that the equence D i i a upermartingale, a fact that i ometime called the trend hypothei (ee Wormald [14]). The trend hypothei will often impoe a condition that the equence of error function E i i growing ufficiently quickly (i.e. the derivative of the limit of error function i ufficiently large). We then how that the one-tep change in D i are bounded in ome way (thi i ometime called the boundedne hypothei). Thi put u in the poition to apply a martingale inequality. In order to get good bound from the martingale inequality one generally need to make E 0 large. In thi ection we appeal to the following pair of martingale inequalitie (ee [3]). For poitive real b, B, the equence A 0, A 1,... i aid to be (b, B)-bounded if A i b A i+1 A i + B for all i 0. Lemma 8. Suppoe b B/10 and 0 < a < bm. If A 0, A 1,... i a (b, B)-bounded ubmartingale, then P [A m A 0 a] exp { a 2 /3bmB }. Lemma 9. Suppoe b B/10 and 0 < a < bm. If A 0, A 1,... i a (b, B)-bounded upermartingale, then P [A m A 0 + a] exp { a 2 /3bmB }. 9

10 Our application of thee Lemma make ue of topping time. Formally peaking, a topping time i imply a poitive integer-valued random variable τ for which {τ n} F n. In other word, τ i a topping time if the event τ n i determined by the firt n tep of the proce. We conider the topped proce (D i τ ), where x y := min{x, y}, in the place of the equence D 0, D 1,.... Our topping time τ i the firt tep in the proce when any condition on ome hort lit of condition fail to hold, where the condition D i 0 i one of the condition in the lit. Note that, ince the variable (D i τ ) doe not change once we reach the topping time τ, we can aume that all condition in the lit hold when we are proving the trend and boundedne hypothee. Alo note that if the topping time τ i imply the minimum of i max and the firt tep for which D i > 0 then {D imax τ > 0} contain the event { i i max : D i > 0}. 4.1 Proof of Lemma 7. For each et A ( [n] r 1) and tep i 0, let OA (i) := {e O(i) : A e}, and Q A (i) = O A (i). We define equence of random variable Y + A (i) := q(t) n Q A(i) + f(t) n 1 ε, Y A (i) := q(t) n Q A(i) f(t) n 1 ε, Z A (i) := d i (A) t D 1/r n f(t)q(t) 1 n 1/r ε, Finally, we define the topping time τ to be the minimum of ( n r), the firt tep i where Ti fail, or where any of Y + A (i) < 0, Y A (i) > 0, or Z A(i) > 0 hold for ome A ( [n] r 1). To prove Lemma 7, we how that for each A ( [n] r 1), P [ Y + A (i max τ) < 0 ] = o(n (r 1) ), (10) P [ Y A (i max τ) > 0 ] = o(n (r 1) ), and (11) P [Z A (i max τ) > 0] = o(n (r 1) ). (12) Conider the event τ i max. Thi event i the union of the event that T imax fail and the event that there exit A ( ) [n] r 1 uch that Y + A (i max τ) < 0 or Y A (i max τ) > 0 or Z A (i max τ) > 0. Since T imax hold with high probability, it follow from (10) (12) and the union bound that w.h.p. τ > i max. In particular, Z A (i) 0 for all (r 1)-et A and tep 0 i i max. It then follow ince ζ min{1/w, ε} implie that we may bound f(t max ) < n ε/2, ay that we have r 1 (G(i max )) t max D 1/r n + f(t max )n 1/r ε/2 = ζ O((n log n) 1/r ) ε(n log n) 1/r, for n ufficiently large. (We remark in paing that the bound on Y ± A (i) given when i < τ are neceary for our proof of the bound on Z A (i).) For the remainder of thi argument, fix a et A ( [n] r 1). We firt prove (10) and (11). Claim 2. For n ufficiently large, the variable Y + A (0),..., Y + A (i max τ) form an (O(n/), O(n 1 1 2r ))- bounded ubmartingale, and the variable Y A (0),..., Y A (i max τ) form an (O(n/), O(n 1 1 2r ))- bounded upermartingale. Proof. We begin by fixing a tep 0 i i max, and we aume that i < τ. Throughout we write t = t(i), and note t(i + 1) = t + 1 and that 1 = D 1/r /N = Θ(n 1 1/r r ). 10

11 To aid the calculation to follow, we begin by etimating the quantity Ξ := f(t + 1 ) f(t). Since f(t) = exp(w t r + W t), f (t) and f (t) are product of f(t) with polynomial in t. A ζ max{1/w, ε}, t max i polylogarithmic in n, and n i large, we have the crude bound f(t) n ɛ/2 and f (t) n o(1) f (t). Thu, by Taylor Theorem, Ξ f ( (t) = O maxt t max f (t ) ( ) f ) (t) 2 = o. (13) Oberve now that we may write Y ± A (i + 1) Y ± A (i) = (q(t + 1 ) q(t)) n (Q A (i + 1) Q A (i)) ± Ξ n 1 ε. (Note that thi tand for the pair of equation in which each ± i replaced with + or with, repectively.) We begin by etablihing the boundedne claim: it i routine to verify that c(t) and c (t) are bounded over the real, implying q(t + 1 ) q(t) c(t) 1 = O( 2 ), (14) and o 0 ( q(t + 1 ) q(t) ) ( n ) n O. A we have the bound f (t) = n ε/2+o(1) and (13), we have Ξ n 1 ε = o(n/), and the lower bound in the boundedne claim follow. To etablih the upper bound, it remain to bound Q A (i) Q A (i + 1). Conider the next edge e i+1 O(i) and oberve that Q A (i) Q A (i + 1) = ( {e i+1 } C ei+1 (i) ) O A (i). We bound C ei+1 (i) O A (i) by conidering five cae depending on e i+1 A : Cae 1: e i+1 A = 0. Let f O A (i) C ei+1 (i): then f = A {v} for ome vertex v, and ince G(i) + e i+1 + f contain a copy of T (r), v e i+1 mut hold. (Recall that every pair of edge in T (r) either hare exactly one or r 1 vertice.) In thi cae, C ei+1 (i) O A (i) e i+1 = r. Cae 2: e i+1 A = r 1. In thi cae, we may write e i+1 = A {u 1 }. Now, let f = A {v} O A (i) C ei+1 (i): ince f e i+1 = A and f C ei+1 (i), there mut exit vertice u 2,..., u r 1 N i (A) o that {u 1,..., u r 1, v} E(i). A then v N i ({u 1,..., u r 1 }), we may bound the number of uch choice of v (and hence of f) in thi cae above by r 1 (G(i)) r 1 ζ r 1 (n log n) (r 1)/r. (Note the bound on the maximum degree follow a Z A (i) 0 ince i < τ.) Cae 3: e i+1 A = 1. Write A = {x 1,..., x r 1 }, where we take e i+1 A = {x 1 }. Let f = A {v} C ei+1 (i) O A (i), and uppoe v / e i+1 (a there are at mot r 1 uch v), o f e i+1 = {x 1 }. Conider a copy of T (r) in G(i)+e i+1 +f uing both e i+1 and f a edge: without lo of generality, we may aume that one of e i+1, f map to the edge b 1 of T (r), the other to the edge a. 11

12 If e i+1 map to b 1, then the (r 1)-et e i+1 \ {x 1 } map to the common interection B of b 1,..., b r. Conequently v N i (e i+1 \ {x 1 }) mut hold, and o there are at mot r 1 (G(i)) uch r-et f C ei+1 (i) O A (i). Otherwie, if e i+1 map to the edge a and f map to b 1, then {x 2,..., x r 1, v} map to the common interection B. Thu, for each u e i+1 \{x 1 } we have {u, x 2,..., x r 1, v} E(i), implying v N i ({u, x 2,..., x r 1 }) and (a e i+1 i fixed), there are again at mot r 1 (G(i)) uch choice of f. Thu, in thi cae we have C ei+1 (i) O A (i) r 1 (G(i)) = n 1/r+o(1). Cae 4: 1 < e i+1 A = r 2. Let f = A {v} O A (i) C ei+1 (i). Since f e i+1 A e i+1 > 1, f e i+1 = r 1 mut hold, implying v e i+1 and o O A (i) C ei+1 (i) r a in Cae 1. Cae 5: 2 e i+1 A r 3. In thi cae, C ei+1 (i) O A (i) = 0, a every f O A (i) atifie 1 f e i+1 r 2. From the cae above it follow that Q A (i) Q A (i + 1) n (r 1)/r+o(1), and combining the above bound, it follow that the equence Y ± A (0),..., Y ± A (i max τ) are (O(n/), O(n 1 1 2r ))-bounded. We turn now to the ub- and upermartingale claim: all expectation calculation to follow are implicitly conditioned on the hitory of the proce up to tep i, and we recall that we aume i < τ. For each open r-et f O A (i), we have f / O A (i + 1) if and only if e i+1 C f (i) {f}. Thu, E [ Y ± A ((i + 1)) Y ± A (i)] = (q(t + 1 ) q(t)) n + f O A (i) C f (i) + 1 O(i) ± Ξ n 1 ε. To etablih the ubmartingale claim, conider the following chain of inequalitie: f O A (i) C f (i) + 1 O(i) (q(t) f(t)n ε ) n (c(t) N γ ) D 1/r (q(t) + N γ ) N = (1 N γ + f(t)n ε ) q(t) + N γ (c(t) N γ ) n. ( 1 2q(t) 1 f(t)n ε) (c(t) N γ ) n ( c(t) 2c(t)q(t) 1 f(t)n ε N γ) n ( c(t) (2c(t)q(t) 1 + 1) f(t)n ε) n. The firt inequality follow from the bound given by (2) and (3) on the event T i and a Y A (i) 0, ince i < τ. In the econd and fourth inequalitie we bounded N γ < f(t)n ε, valid a f(t) 1 and ε γ. Thu, applying thi bound and (14) give E [ Y + A (i + 1) Y + A (i)] Ξ n 1 ε (2c(t)q(t) 1 + 1)f(t) n1 ε Ξ n 1 ε (2c(t)q(t) 1 + 2)f(t) n1 ε 12 ( ) 1 O 2

13 = ( (1 + o(1))f (t) (2c(t)q(t) 1 + 2)f(t) ) n1 ε by (13). Since f (t) = (W rt r 1 + W )f(t) and 2c(t)q(t) 1 = 2rt r 1, thi final bound i nonnegative for large n a W i large, and o Y + A (0),..., Y + A (i max τ) form a ubmartingale. We imilarly bound E [Q A (i) Q A (i + 1)] above to etablih the upermartingale claim: a 1 < N γ D 1/r for large n, and a T i hold and Y + A (i) 0, f O A (i) C f (i) + 1 O(i) (q(t) + f(t)n ε ) n (c(t) + 2N γ ) D 1/r (q(t) N γ ) N = (1 + N γ + f(t)n ε ) q(t) N γ (c(t) + 2N γ ) n ( 1 + 4q(t) 1 f(t)n ε) (c(t) + 2N γ ) n ( c(t) + (4c(t)q(t) 1 + 4)f(t)n ε) n. In addition to the bound N γ f(t)n ε ued above, in the econd inequality, we bounded q(t) N γ q(t)/2, and in the final we bounded 2N γ (1 + 4q(t) 1 f(t)n ε ) 4f(t)n ε a q(t) 1 f(t)n ε 1 which hold a 2W ζ r < ɛ and n i large. Thu, E [ Y A (i + 1) Y A (i)] Ξ n 1 ε + (4c(t)q(t) 1 + 4)f(t) n1 ε Ξ n 1 ε + (4c(t)q(t) 1 + 5)f(t) n1 ε ( ) 1 + O 2 = ( (1 + o(1))f (t) + (4c(t)q(t) 1 + 5)f(t) ) n1 ε, and again, a W i large, thi i trictly negative for n ufficiently large. Y A (0),..., Y A (i max τ) form a upermartingale, completing the proof. Thu, the equence Since Q A (0) = n r + 1, Y + A (0) = r 1 + n1 ε and Y A (0) = r 1 n1 ε. Applying Lemma 8 and 9, repectively, we have { ( )} P [ Y + A (i max τ) < 0 ] exp Ω = exp { n 1 < exp { n 1 4r n 2 2ε n ζ log1/r N n 1 1 2r ) 2r 2ε+o(1)} (valid for large n a ε i mall), and an identical calculation yield We have etablihed (10) and (11). It remain to prove (12). } P [ Y A (i max τ) > 0 ] exp 13 { n 1 4r }.

14 Claim 3. The variable Z A (0),..., Z A (i max τ) form a (2n/N, 2)-bounded upermartingale. Proof. We begin by fixing a tep 0 i i max, and we aume that i < τ. Throughout we write t = t(i). Let f 1 (t) = f(t)q(t) 1 = exp((w + 1)t r + W t), and let Ξ 1 := f 1 (t + 1 ) f 1 (t). By the ame reaoning given in Claim 2, we may bound f 1 (t) < n ε/2, ay, for large n, and f 1 (t) no(1) f 1 (t), and o Ξ 1 f 1 (t) ( = O maxt <t max f 1 ) ( ) (t ) f 2 = o 1 (t). (15) Next, we oberve that Z A (i + 1) Z A (i) = d i+1 (A) d i (A) n N Ξ 1 n 1/r ε. The boundedne claim then follow for n ufficiently large a 0 d A (i + 1) d A (i) 1 and a a 1 = D 1/r /N = Θ(n 1 1/r /N). Ξ 1 n 1/r ε n ε/2+o(1) n 1/r ε 1 < n/n Turning to the upermartingale condition, oberve that d i+1 (A) = d i (A) + 1 if and only if e i+1 lie in the et of open r-et counted by Q A (i). Conditioned on the hitory of the proce up to tep i, it follow that E [Z A (i + 1) Z A (i)] = Q A(i) O(i) n N Ξ 1 n 1/r ε (q(t) + f(t)n ε ) n (q(t) N γ ) N = N γ + f(t)n ε (q(t) N γ ) n N Ξ 1 n 1/r ε n N Ξ 1 n 1/r ε (N γ + f(t)n ε ) 2q(t) 1 n N Ξ 1 n 1/r ε = (2q(t) 1 N γ + 2f 1 (t)n ε ) n N Ξ 1 n 1/r ε 4f 1 (t) n ε n N Ξ 1 n 1/r ε (16) Note that the firt inequality hold a T i and Y + A (i) 0 ince i < τ, the econd a q(t) N γ q(t)/2 ince ζ γ, and the final a N γ f(t) n ε, ince f(t) 1 and ε γ. Noting that for large n, D n r 1 /r r and o 1 n 1 1/r /(rn), by (15) we have Ξ 1 n 1/r ε = (1 + o(1)) f 1 (t) n 1/r ε (1 + o(1)) W f 1(t) n 1 1/r n 1/r ε rn > W 2r f 1(t) n ε n N. Thu, ince we aume W i large, the upermartingale condition follow now from (16). 14

15 Finally, to how (12), we apply Lemma 9 to yield { ( P [Z A (i max τ) > 0] exp Ω { = exp = exp n n 2/r 2ε N ζ log1/r N } n 2/r 2ε n 1 (r 1)/r+o(1) { n 1/r 2ε o(1)} which uffice a ε i mall. Thi complete the proof of Lemma 7. )} 4.2 Proof of Lemma 5 We begin by letting and we note that S = Θ(k r ). ( ) ( ) 2l l S = S(n) = 2, r r We fix a pair A, B of dijoint l-element ubet of [n], and define the following equence of random variable: for each tep i 0, let X + (i) = q(t) S Q A,B (i) + f(t) Sn ε, and X (i) = q(t) S Q A,B (i) f(t) Sn ε. We next define the topping time τ to be the minimum of τ A,B and the firt tep i for which X + (i) 0, X (i) 0, or the event T i fail to hold. Claim 4. The equence X + (0),..., X + (i max τ ) form a (O(k r /), O(k r 1 /n 4ε ))-bounded ubmartingale, and the equence X (0),..., X (i max τ ) form a (O(k r /), O(k r 1 /n 4ε ))-bounded upermartingale. Proof. We fix a tep 0 i i max, and we uppoe that i < τ. Throughout we write t = t(i), and note t(i + 1) = t + 1 and that 1 = D 1/r /N = Θ(n 1 1/r r ). To aid the calculation to follow, we begin by etimating the quantity Ξ := f(t + 1 ) f(t). Recall equation (13): Ξ f ( (t) = O maxt t max f (t ) ( ) f ) (t) 2 = o. Oberve that we may write X ± (i + 1) X ± (i) = (q(t + 1 ) q(t)) S (Q A,B (i + 1) Q A,B (i)) ± Ξ Sn ε. (A above, thi tand for the pair of equation in which each ± i replaced with + or with, repectively.) We begin by etablihing the boundedne claim: by (14) and a S = Θ(k r ), we have 0 ( q(t + 1 ) q(t) ) ( ) k r S O. 15

16 Next, bounding f (t) n ε/2+o(1), Ξ Sn ε n ε/2+o(1) kr In order to etablih the boundedne part of the claim, it remain to bound the quantity Q A,B (i + 1) Q A,B (i). Let O A,B (i) denote the et of r-et that are open with repect to the pair A, B in G(i), and let O τ denote the et of all open r-et whoe election a e i+1 would reult in τ A,B = i+1. Now, if e i+1 O τ, then Q A,B (i + 1) Q A,B (i) = 0 by definition, and, otherwie, we have Q A,B (i + 1) Q A,B (i) = O A,B (i) (C ei+1 (i) {e i+1 }). It uffice, then, to bound the quantity C e (i) O A,B (i) for all e O(i) \ O τ : fix uch an open r-et e. Now, for any f C e (i) O A,B (i), there i a copy T r,f of T (r) in the graph G(i) + e + f uing both e and f a edge. Up to iomorphim, there are only three poibilitie for the pair (e, f) in that copy: (e, f) map to (b 1, b 2 ), or to (b 1, a), or to (a, b 1 ). We treat thee three cae eparately. Cae 1: (e, f) map to (b 1, b 2 ). In thi cae, the r 1 vertice that map to the et R lie entirely in e, and f i the union of thoe r 1 vertice along with another vertex lying in A B. Thu, we may bound the total number of uch f above by rk. Cae 2: (e, f) map to (b 1, a). Let R = e f, the et of r 1 vertice hared by all edge b j in thi copy of T (r). Since f map to a, it follow that f N i (R ), and a f O A,B (i), we know f A B, f A, and f B. Conequently, N i (R ) interect both A and B: ince e / O τ, it follow that N i (R ) (A B) (k/n 2ε ) mut hold. Thu, by firt electing R e, which then identifie the ole vertex in e f, and then electing the r 1 vertice compriing f e from N i (R ) (A B), we can therefore bound the total number of uch open r-et f above by r(k/n 2ε ) r 1. Cae 3: (e, f) map to (a, b 1 ). There exit an (r 1)-et R A B and a vertex v e o that f = R {v} and o that e \ {v} N i (R ). To bound the number of uch f, it uffice to bound the number of (r 1)-et R A B for which N i (R ) contain (r 1) vertice from e. To that end, fix a vertex v e and let H v denote the (r 1)-uniform hypergraph on (A B) \ e whoe edge are the (r 1)-ubet X for which N i (X) e \ {v}. We claim that r 2 (H v ) < 4r. Suppoe to the contrary that thi doe not hold: then there exit an (r 2)-et Y (A B) \ e and vertice x 1, x 2,..., x 4r (A B) \ (Y e) o that for each for each vertex u e \ {v}, {u} Y {x j } E(i) for 1 j 4r. It follow from Lemma 4 that uch a configuration doe not appear in G(i). Indeed, a thi configuration pan 6r 3 vertice and ha 4r(r 1) edge, the probability that uch a configuration appear i at mot n 6r 3 ( i N ) 4r(r 1) = n 6r 3 4(r 1)2 +o(1) = o(1). 16

17 It follow that H v < 4r ( k r 2), and thu the total number of uch open r-et f a above i le than 4r 2 k r 2. A ε i mall and a k = n 1/r+o(1), it follow that for large n we have C e (i) O A,B (i) rk + r (k/n 2ε ) r 1 + 4r 2 k r 2 = O(k r 1 /n 2ε(r 1) ), and a r 3 we conclude that 0 Q A,B (i + 1) Q A,B (i) = O(k r 1 /n 4ε ). Thu, it follow that the equence X ± (0),..., X ± (i max τ ) are (O(k r /), O(k r 1 /n 4ε ))-bounded a claimed. We now turn to the ub- and upermartingale claim, and we remark that all expectation and probability calculation to follow are implicitly conditioned on the hitory of the proce up to tep i. We begin by bounding the expected value of Q A,B (i+1) Q A,B (i). Recall that we aume i < τ A,B and that O τ O(i) conit of the open r-et whoe election a e i+1 would yield τ A,B = i + 1. We claim that O τ 4n 2ε k (17) To ee thi, let R := { ( ) } [n] X : N i (X) (A B) k/(2n 2ε ). r 1 Then R < 4n 2ε, which can be argued a follow. Suppoe by way of contradiction that S R with S = 4n 2ε. Let N = Y S (N i(y ) (A B)). By incluion-excluion and the fact that Lemma 4 implie that the co-degree of any pair of (r 1)-et i at mot 5r (ee (8)), we have k 2l = A B N S k/(2n 2ε ) S 2 5r 2k 80rn 4ε, a contradiction a ε i mall and k = n 1/r+o(1). To deduce (17) it uffice to oberve that each open r-et e O τ can be written e = {v} X for ome vertex v A B and (r 1)-et X atifying N i (X) (A B) k/n 2ε 1 (and thu X R). Conditioning on the event e i+1 / O τ then yield E [Q A,B (i + 1) Q A,B (i)] = by linearity of expectation. Conequently, E [ X ± (i + 1) X ± (i) ] = (q(t + 1 ) q(t)) S + e O A,B (i) e O A,B (i) e O A,B (i) C e (i) \ O τ O(i) C e (i) \ O τ O(i) ± Ξ Sn ε. To etablih the ubmartingale claim, we note firt that a r 3 and ε γ 1/r, from (17) we have O τ = n 1/r+2ε+o(1) < N γ D 1/r. Now, a i < τ, T i and X (i) 0 hold, we have ( C e (i) \ O τ q(t) f(t) ) O(i) n ε S (c(t) 2N γ )D 1/r (q(t) + N γ )N 17

18 = (1 N γ + f(t)n ε ) q(t) + N γ (c(t) 2N γ ) S ( 1 2q(t) 1 f(t)n ε) (c(t) 2N γ ) S ( c(t) 2c(t)q(t) 1 f(t)n ε 2N γ) S ( c(t) (2c(t)q(t) 1 + 1)f(t)n ε) S. Note that thee bound follow for large n ince f(t) 1 and ε γ imply N γ f(t)n ε /2. Applying thi and (14) give E [ X + (i + 1) X + (i) ] ( ) Ξ Sn ε (2c(t)q(t) 1 + 1)f(t) Sn ε 1 O 2 Ξ Sn ε (2c(t)q(t) 1 + 2)f(t) Sn ε = ( (1 + o(1))f (t) (2c(t)q(t) 1 + 2)f(t) ) Sn ε by (13). Since f (t) = (W rt r 1 + W )f(t) and 2c(t)q(t) 1 = 2rt r 1, thi final bound i nonnegative for large n a W i large, and o X + (0),..., X + (i max τ) form a ubmartingale. Turning to the upermartingale claim, we take a imilar approach and begin by noting a T i hold and X + (i) 0, ( C e (i) \ O τ q(t) + f(t) ) O(i) n ε S (c(t) + N γ )D 1/r (q(t) N γ )N e O A,B (i) = (1 + N γ + f(t)n ε ) q(t) N γ (c(t) + N γ ) S ( 1 + 2q(t) 1 f(t)n ε) (c(t) + N γ ) S ( c(t) + (2c(t)q(t) 1 + 1)f(t)n ε) S. The upermartingale condition then follow in eentially the ame way a the ubmartingale condition above. Now, a X + (0) = Sn ε, X (0) = Sn ε, S = Θ(k r ) and i max = n o(1), it follow from Claim 4 and Lemma 8 and 9 that P [ X + (i max τ ) 0 ] { ( )} S 2 n 2ε { exp Ω = exp k n 2ε o(1)}. n o(1) Similarly, we have k r kr 1 n 4ε P [ X (i max τ ) 0 ] { exp k n 2ε o(1)}. Since there are fewer than n 2k = exp{2k log n} choice of the pair of et A and B, Lemma 5 follow from the union bound. 18

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