Multicolor Sunflowers

Transcription

1 Multicolor Sunflower Dhruv Mubayi Lujia Wang October 19, 2017 Abtract A unflower i a collection of ditinct et uch that the interection of any two of them i the ame a the common interection C of all of them, and C i maller than each of the et. A longtanding conjecture due to Erdő and Szemerédi olved recently in [9, 7], ee alo [23]) wa that the maximum ize of a family of ubet of [n] that contain no unflower of fixed ize k > 2 i exponentially maller than 2 n a n. We conider the problem of determining the maximum um and product of k familie of ubet of [n] that contain no unflower of ize k with one et from each family. For the um, we prove that the maximum i k 2 k 1)2 n for all n k 3, and for the k = 3 cae of the product, we prove that the maximum i ) o1) 2 3n. We conjecture that for all fixed k 3, the maximum product i 1/8 + o1))2 kn. =0 1 Introduction Throughout the paper, we write [n] = {1,..., n}, 2 [n] = {S : S [n]} and ) [n] = {S : S [n], S = }. A family A 2 [n] i -uniform if further A ) [n]. A unflower or trong -ytem) with k petal i a collection of k et S = {S 1,..., S k } uch that S i S j = C for all i j, and S i \ C for all i [k]. The common interection C i called the core of the unflower and the et S i \ C are called the petal. In 1960, Erdő and Rado [11] proved a fundamental reult regarding the exitence of unflower in a large family of et of uniform Department of Mathematic, Statitic, and Computer Science, Univerity of Illinoi, Chicago, IL 60607, mubayi@uic.edu. Reearch partially upported by NSF grant DMS and DMS Department of Mathematic, Statitic, and Computer Science, Univerity of Illinoi, Chicago, IL 60607, lwang203@uic.edu. 1

2 ize, which i now referred to a the unflower lemma. It tate that if A i a family of et of ize with A >!k 1), then A contain a unflower with k petal. Later in 1978, Erdő and Szemerédi [12] gave the following upper bound when the underlying et ha n element. Theorem 1 Erdő, Szemerédi [12]). There exit a contant c uch that if A 2 [n] with A > 2 n c n then A contain a unflower with 3 petal. In the ame paper, they conjectured that for n ufficiently large, the maximum number of et in a family A 2 [n] with no unflower with three petal i at mot 2 ɛ) n for ome abolute contant ɛ > 0. Thi conjecture, often referred to a the weak unflower lemma, i cloely related to the algorithmic problem of matrix multiplication [1] and remained open for nearly forty year. Recently, thi wa ettled via the polynomial method by Ellenberg Gijwijt [9] and Croot Lev Pach [7] ee alo Nalund Sawin [23]). A natural way to generalize problem in extremal et theory i to conider verion for multiple familie or o-called multicolor or cro-interecting problem. Beginning with the famou Erdő Ko Rado theorem [10], which tate that an interecting family of k-element ubet of [n] ha ize at mot n 1 k 1), provided n 2k, everal generalization were proved for multiple familie that are cro-interecting. In particular, Hilton [16] howed in 1977 that if t familie A 1,..., A t [n]) k are cro interecting meaning that Ai A j for all A i, A j ) A i A j ) and if n/k t, then t A i t n 1 k 1). On the other hand, reult of Pyber [24] in 1986, that were later lightly refined by Matumoto and Tokuhige [20] and Bey [2], howed that if two familie A [n]) k, B [n] ) l are cro-interecting and n max{2k, 2l}, then A B ) n 1 n 1 k 1 l 1). Thee are the firt reult about bound on um and product of the ize of cro-interecting familie. More general problem were conidered recently, for example for cro t-interecting familie i.e. pair of et from ditinct familie have interection of ize at leat t) and r-cro interecting familie any r et have a nonempty interection where each et i picked from a ditinct family) and labeled croing interecting familie, ee [4, 14, 15]. A more ytematic tudy of multicolored extremal problem with repect to the um of the ize of the familie) wa initiated by Keevah, Sak, Sudakov, and Vertraëte [17], and continued in [3, 18]. Cro-interecting verion of Erdő problem on weak -ytem for the product of the ize of two familie) were proved by Frankl and Rödl [13] and by the firt author and Rödl [22]. In thi note, we conider multicolor verion of unflower theorem. Quite urpriingly, thee baic quetion appear not to have been tudied in the literature. Definition 2. Let A i A i 2 [n] for i = 1,..., k. Then A i ) k i a unflower with k petal if there exit C [n] uch that A i A j = C for all i j, and A i \ C, for all i [k]. Say that A i ) k i unflower-free if it contain no unflower with k petal. For any k familie that are unflower-free, the problem of upper bounding the ize of any 2

3 ingle family i unintereting, ince there i no retriction on a particular family. So we are intereted in the um and product of the ize of thee familie. Given integer n and k, let Fn, k) = {A i ) k : A i 2 [n] for i [k] and A i ) k i unflower-free}, Sn, k) = max A i ) k Fn,k) A i and P n, k) = max A i ) k Fn,k) k A i. Our two main reult are harp or nearly harp etimate on Sn, k) and P n, 3). Theorem 1 or [7, 9, 23]) we obtain that By Sn, 3) 2 2 n + 2 n c n. Indeed, if A + B + C i larger than the RHS above then A B C > 2 n c n by the pigeonhole principle and we find a unflower in the interection which contain a unflower. Our firt reult remove the lat term to obtain an exact reult. Theorem 3. For n k 3 k 2 Sn, k) = k 1)2 n =0 The problem of determining P n, k) eem more difficult than that of determining Sn, k). Our bound for general k are quite far apart, but in the cae k = 3 we can refine our argument to obtain an aymptotically tight bound. Theorem 4. P n, 3) = ) o1) 2 3n. We conjecture that a imilar reult hold for all k 3. Conjecture 1. For each fixed k 3, P n, k) = ) o1) 2 kn. In the next two ection we give the proof of Theorem 3 and 4. 2 Sum In order to prove Theorem 3, we firt deal with -uniform familie and prove a tronger reult. Given a unflower S = A i ) k, define it core ize to be cs) = C, where C = A i A j, i j. 3

4 Lemma 5. Given integer 1 and c with 0 c 1, let n be an integer uch that n c + k c). For i = 1,..., k, let A i ) [n] uch that Ai ) k contain no unflower with k petal and core ize c. Then Furthermore, thi bound i tight. A i k 1). Proof. Randomly take an ordered partition of [n] into k + 2 part X 1, X 2,..., X k+2 uch that X 1 = n c + k c)), X 2 = c, and X i = c for i = 3,..., k + 2, with uniform probability for each partition. For each partition, contruct the bipartite graph G = {A i : i = 1,..., k} {X 2 X j : j [3, k + 2]}, E) where a pair {A i, X 2 X j } E if and only if X 2 X j A i. If there exit a perfect matching in G, then we will get a unflower with k petal and core ize c, ince X 2 will be the core. Thi how that G ha matching number at mot k 1. Then König theorem implie that the random variable EG) atifie EG) k 1)k. 1) Another way to count the edge of G i through the following expreion: EG) = k+2 χ {X2 X j A i }, j=3 where χ S i the characteritic function of the event S. Taking expectation and uing 1) we obtain k+2 E k 1)k. 2) j=3 χ {X2 X j A i } By linearity of expectation, k+2 E = j=3 χ {X2 X j A i } k+2 P X 2 X j A i ) = j=3 k+2 j=3 A A i P A = X 2 X j ). Since the partition of [n] i taken uniformly, for any j with 3 j k + 2, the et X 2 X j cover all poible -ubet of [n] with equal probability. Hence for any A A i, we have PA = X 2 X j ) = 1 n ). So we have k+2 E j=3 χ {X2 X j A i } = k+2 n = ) 1 j=3 A A i A i k n ). 4

5 Hence by 2), A i k 1). The bound hown above i tight, ince we can take A 1 = A 2 =... = A k 1 = ) [n], and A k =. Now we ue thi lemma to prove Theorem 3. Proof of Theorem 3. Recall that n k 3 and we are to how that n Sn, k) = k 1)2 n =n k+2 We firt how the lower bound by the following example: Let A i = 2 [n] for i = 1..., k 1 and A k = { } {S [n] : S n k + 2}. To ee that A i ) k i unflower-free, notice that any unflower ue a et from A k. The empty et doe not lie in any unflower. So if a et of ize at leat n k + 2 appear in a unflower S with k petal, it require at leat k 1 other point, but then the total number of point in S i at leat n + 1, a contradiction. To ee the upper bound, given familie A i ) k Fn, k), we define A i, = A i ) [n] for each i [k] and integer [0, n]. Thi give a partition of each family A i into n + 1 ubfamilie. Since A i ) k i unflower-free, o i A i,) k for all [0, n]. Now, for each = 1, 2,..., n k + 1, let c = 1. Then 0 c 1, and c + k c) = 1 + k n. Therefore, by Lemma 5, k A i, k 1) n ) for all [n k + 1]. For > n k + 1, the trivial bound for thi um i k n ). So we get, n n A i = A i, = A i, =0 Thi complete the proof. = =0 k + 0 n k+1 A i,0 + n k+1 =1 n n k 1) =0 =1 = k 1)2 n A i, + n k 1) ) k 2 = k 1)2 n n =n k+2 =0 n ) + n =n k+2 n =n k+2 n =n k+2 ). n A i, k )

6 3 Product From the bound on the um of the familie that do not contain a unflower, we deduce the following bound on the product by uing the AM-GM inequality. Corollary 6. Fix k 3. A n, ) o1) 2 kn P n, k) k ) 1 k + o1)) 2 kn. k Proof. The upper bound follow from Theorem 3 and the AM-GM inequality, k k A i A i k For the lower bound, we take ) k 1 + o1)) ) k k 1)2n = 1 + o1)) k A 1 = A 2 = {S [n] : 1 S} {[2, n]}, A 3 = {S [n] : 1 / S} {S [n] : S n 1}, ) k 1 k 2 kn. and A 4 = A 5 =... = A k = 2 [n]. A unflower with k petal mut ue three et from A 1, A 2, and A 3, call them A 1, A 2, A 3 repectively. Thee three et form a unflower with three petal. If any of thee et i of ize at leat n 1, then it will be impoible to form a 3-petal unflower with the other two et. So by their definition, we have 1 A 1 A 2, but 1 / A 3, which implie A 1 A 2 A 1 A 3, a contradiction. So A i ) k i unflower-free. The ize of thee familie are A 1 = A 2 = 2 n 1 + 1, A 3 = 2 n 1 + n and A i = 2 n for i 4. Thu, a required. k A i = ) o1) 2 kn For any poitive integer k we have k 1 k )k < 1/e, o Corollary 6 implie the upper bound 1/e + o1))2 kn for all k 3. For k = 3, we will improve the factor in the upper bound from 2/3) 3 = to our conjectured value of The main part of our proof i Lemma 7 below, which prove a much better bound than Sn, 3) = 2 + o1))2 n for the um of three unflower-free familie under the aumption that all of them contain a poitive proportion of et. Lemma 7. For all ɛ > 0 there exit n 0 = n 0 ɛ) > 0 uch that the following hold for n > n 0. Let A i 2 [n] with A i ɛ2 n for i [3], and uppoe that A i ) 3 i unflower-free. Then ) 3 A 1 + A 2 + A ɛ 2 n. 6 k

7 Lemma 7 immediately implie Theorem 4 by the AM-GM inequality a hown below. Proof of Theorem 4. Let ɛ 0, 1/8), n 0 be obtained from Lemma 7 and n > n 0. Suppoe there i an i, uch that A i < ɛ2 n. Then 3 A i < ɛ2 n 2 n 2 n < n. So we may aume that A i ɛ2 n for all i. Thu, by the AM-GM inequality and Lemma 7, 3 ) A1 + A 2 + A 3 3 A i 3 which i the bound ought. In the ret of thi ection we prove Lemma ɛ ) 3 ) 1 2 3n < ɛ 2 3n 3.1 Proof of Lemma 7 We begin with the following lemma, which ue idea imilar to thoe ued in the proof of Lemma 2.1 of [17]. Lemma 8. Let k 3, A 1,... A k be familie of ubet of [n] that are unflower-free. For any real number ɛ > 0, if A i ɛ2 n for all i, then there exit δ = δɛ) > 0 and k familie B 1,..., B k uch that the following hold. B i δ2 n for i = 1,..., k, k A i k B i + ɛ 2) 2 n, B i ) k i unflower-free, B 1,..., B k form a laminar ytem, that i, either B i B j =, B i B j, or B j B i for all i j. Proof. The familie A i, i = 1,..., k form a collection of ubet of 2 [n], hence they induce a partition of 2 [n] into at mot 2 k part. More preciely, the dijoint part ome may be empty) in thi partition are X I = i I A i i [k]\i A c i, where I [k]. Take δ = ɛ/k2 k ). For each I [k], if X I < δ2 n, update the A i by deleting X I from all A i that contain it, that i, all A i with i I. At the end of thi proce, let the reulting familie be A i, i = 1..., k. Now, all X I that are nonempty have ize at leat δ2 n. 7

8 For each original A i, the familie A i A j, j [k] \ {i} induce a partition on it into at mot 2 k 1 part. So, after the above deletion tep the remaining A i ha ize at leat A i ɛ2 n 2 k 1 δ2 n = ɛ2 n 2 k 1 ɛ k2 k 2n = 1 1 ) ɛ2 n. 2k If X I < δ2 n, it i deleted from all I of the A i that contain it. Hence, the total number of deleted part with repetition i at mot n ) k i = k2 k 1. i So, we have A i A i + k2 k 1 δ2 n = ɛ A i + 2 2) n. Two familie A and B are aid to be croing if all three of A B, A \ B and B \ A are nonempty. For each pair of croing familie A i and A j, replace A i and A j by A i A j and A i A j. Call the reulting familie B 1,..., B k. Notice firt that at the end of the proce which terminate after at mot k 2) tep, becaue it increae the number of incluion related pair at every tep), the familie B 1,..., B k contain no croing pair, hence form a laminar ytem. Secondly, the um of the ize of the familie remain the ame, ince X Y + X Y = X + Y for all et X, Y. Hence we get A i ɛ A i + 2 2) n = ɛ B i + 2 2) n. Next, notice that all part of the partition induced by A i, i = 1,..., k have ize at leat δ2 n. Moreover, the tep of replacing two croing familie by their interection and union only create new familie that conit of the union of nonempty part. Thi yield that B i δ2 n for all i [k]. Finally, we claim that B i ) k i unflower-free. The familie A i )k are certainly unflowerfree becaue A i A i for all i and A i ) k i unflower-free. So we are left to how that the tep of removing croing pair do not introduce unflower. Suppoe we have familie C i ) k, w.l.o.g, the croing pair C 1, C 2 are replaced by C 1 C 2 and C 1 C 2, and uppoe that C i, i = 1,..., k with C 1 C 1 C 2, C 2 C 1 C 2 and C i C i, i 3 i a unflower in the reulting familie. Then, w.l.o.g, C 2 i in C 2. Thu we find that C i, i = 1,..., k i alo a unflower in C i ) k. Thi complete the proof. We will ue the following lemma which follow from well-known propertie of binomial coefficient we omit the tandard proof). 8

9 Lemma 9. For each δ > 0, there exit a real number α = αδ) and integer n 0 uch that for n > n 0, every family A of ubet of [n] with ize A δ2 n contain a et S with S [n/2 α n, n/2 + α n]. Further, for each γ 0, δ), there exit a β = βγ), uch that all but at mot γ2 n element in A have ize in [n/2 β n, n/2 + β n]. Now we have all the neceary ingredient to prove Lemma7. Proof of Lemma 7. Let δ = ɛ/3 2 3 ) = ɛ/24) a in the proof of Lemma 8. By Theorem 1, we have A 1 A 2 A 3 2 n c n < δ2 n for large enough n. Apply Lemma 8 to obtain familie B i, i = 1, 2, 3 uch that B i δ2 n for i = 1, 2, 3, 3 A i 3 B i + ɛ 2) 2 n, B 1, B 2, B 3 ) i unflower-free, B 1, B 2, B 3 form a laminar ytem. Moreover, ince A 1 A 2 A 3 < δ2 n, the interection of all three familie i deleted from all three of them in the proce of forming B i which yield B 1 B 2 B 3 =. The ret of the proof i devoted to howing the claim below. Claim 10. B 1 + B 2 + B ɛ ) 2 n. 2 Proof. The laminar ytem formed by the three familie with an empty common interection fall into the following three type. Let {A, B, C} = {B 1, B 2, B 3 } and a := A, b := B, and c := C. Cae I. A, B, C are mutually dijoint. In thi cae, trivially we have a + b + c 2 n which i even better than what we need. Cae II. A B and A C =. Since C δ2 n, we may pick an S C with S [n/2 α n, n/2 + α n] by Lemma 9. Now for each ubet T S, conider the ubfamily of B defined by B T = {B B : B S = T }. Clearly, thee ubfamilie form a partition of B, i.e. B = T S B T. Now we define a new family derived from B T B T = {B \ T : B B T }. There i a naturally defined bijection between B T and B T, o B T = B T. Claim. b 1 + ɛ 2 )2n 1. 9

10 Proof. We firt how that B T \ { } i an interecting family if T S. Indeed, uppoe there are dijoint nonempty et B 1, B 2 B T, then we find a unflower coniting of B 1 T B A, B 2 T B and S C. So B T 2n S 1 + 1, which yield the following upper bound for B : b = T S B T = T S B T = T S B T + B S 2 S 1)2 n S 1 + 1) + 2 n S = 2 n S 2 n S n S 2 n n/2+α n 2 n n/2+α n) n n/2 α n) = 2 n n/2+α n 2 n/2 α n n/2+α n 1 + ɛ ) 2 n 1, 2 where the lat inequality hold for large enough n. Since A C =, the Claim implie that a + b + c 2 n ɛ ) 2 n 1 = ɛ ) 2 n. 2 Cae III. A B C) and B C =. We firt fix γ = min{δ, ɛ/12}, find β = βγ) a in Lemma 9. Then all but at mot γ2 n ɛ/12)2 n et in each family are of ize in [n/2 β n, n/2 + β n]. Hence we have a + b + c A β + B β + C β + ɛ 4 2n, where F β = {F F : n/2 β n F n/2 + β n}. It remain to how that 3 A β + B β + C β 2 + ɛ ) 2 n. 4 We may aume A = A β, B = B β and C = C β ; our tak i to prove a+b+c 3/2 + ɛ/4) 2 n. Conider a pair of et B, C) B C which atifie the following two condition: B C [n], B \ C and C \ B. Let A = B C = B C) B C. Then A / A, otherwie A, B, C together form a unflower. Hence the number of uch A i at mot 2 n a. We claim that for each uch A, there are at mot 1 + ɛ/4)2 n 1 pair B, C) B C with the two propertie above uch that A = B C. Indeed, for a given A, we firt partition it into two ordered part X 1, X 2 with X 2 here X 2 correpond to B C). There are 2 A 1 way to do o. Next we count the number of uch pair B, C) uch that 10

11 B C = X 1 and B C = X 2. Thi number at mot 1/2 of the number of ordered partition of [n] \ A into two nonempty part. The ratio 1/2 come from the fact that for each ordered bipartition [n] \ A = X 3 X 4, if X 3 X 1, X 4 X 1 ) B C), then we cannot alo have X 4 X 1, X 3 X 1 ) B C), becaue B and C are dijoint. So only half of the ordered bipartition could actually become deired pair. Conequently, the number of uch pair B, C) i 2 n A 2)/2 = 2 n A 1 1. The total number B, C) that give the ame A i therefore at mot 2 A 1)2 n A 1 1) = 2 n 1 2 A 2 n A n 1 2 n/2 β n 2 n n/2+β n) ɛ ) 2 n 1. 4 Here we ue the aumption that A = A β, which implie A [n/2 β n, n/2 + β n], and n i large enough. Thi yield bc 2 n a) 1 + ɛ ) 2 n n+1 4 where the error term 3 n+1 arie from the number of pair B, C) B C uch that either B C = [n], B C or C B. If 2 n a) ɛ/4) 2 n 1 < 3 n+1, then bc < 10/ɛ)3 n+1 and thi contradict b, c δ2 n. Therefore bc 2 n a) 1 + ɛ ) 2 n n+1 2 n a) 1 + ɛ ) 2 n Conequently, we have a 2 n bc 1 + ɛ/2)2 n 1. By the ame argument ued for the proof of the Claim in Cae II, we can how that b 1 + ɛ/2)2 n 1 and c 1 + ɛ/2)2 n 1. Now we obtain a + b + c 2 n bc 3 + b + c = fb, c) 1 + ɛ/2)2n ɛ ) 2 n 4 where the lat inequality follow by maximizing the function fb, c) ubject to the contraint b, c I = [δ2 n, 1 + ɛ/2)2 n 1 ]. Indeed, etting b f = c f = 0 we conclude that the extreme point occur at the boundary of I I. In fact, the maximum i achieved at b, c) = 1+ɛ/2)2 n, 1+ɛ/2)2 n ), and f1+ɛ/2)2 n, 1+ɛ/2)2 n ) = 3/2+ɛ/4)2 n a claimed above. 4 Concluding remark The definition of unflower can be generalized a follow: Let 0 t k and A i A i 2 [n] for i [k]. Then A i ) k i a t-unflower if 11

12 A i A j = C for all i j, and A i \ C hold for at leat t indice i [k]. Note that a t + 1)-unflower i a t-unflower but the convere need not hold. Let Fn, k, t) = {A i ) k : A i 2 [n] for i [k] and A i ) k i t-unflower-free}, Sn, k, t) = max A i ) k Fn,k,t) A i and P n, k, t) = max Uing the idea in thi paper, one can how that for each 0 t < k, t 2 Sn, k, t) = k 1)2 k +. The detail can be found in [21]. =0 A i ) k Fn,k,t) k A i. By the monotonicity of the function P n, k, t) in t, Theorem 4 implie for each fixed 0 t 3, ) 1 P n, 3, t) = 8 + o1) 2 3n. The cae t = 0 i particularly intereting. Let P n, k) = P n, k, 0), p n, k) = P n, k)/2 kn and pn, k) = P n, k)/2 kn. A pointed out by a referee, it i eay to how that p n, k) i monotone increaing a a function of n for each fixed k 3, while pn, k) i not. Indeed, given a collection of optimal familie A i ) k for P n, k), we can contruct k familie of ubet of [n+1] that are 0-unflower-free with the product of their ize at leat 2 k P n, k) a follow. We double each A i in the following way to get new familie: B i = A i {A {n + 1} : A A i }, i [k]. Clearly, k B i = k 2 A i = 2 k P n, k) and it i an eay exercie to how that B i ) k contain no 0-unflower. Since p n, k) 1, we conclude that p k) := lim n p n, k) exit. Clearly p 3) = 1/8, and in general 1/8 p k) 1 1/k) k < 1/e. Further, for a fixed k 4, if one can how that there exit a ingle value n 0 uch that p n 0, k) > 1/8, then by the monotonicity of p n, k) and P n, k) P n, k), Conjecture 1 would be diproved. Our approach for Sn, k) i imply to average over a uitable family of partition. It can be applied to a variety of other extremal problem, for example, it yield ome reult about cro interecting familie proved by Borg [5]. It alo applie to the ituation when the number of color i more than the ize of the forbidden configuration. In particular, the proof of Lemma 5 yield the following more general tatement. Lemma 11. Given integer 1, 1 t k and 0 c 1, let n be an integer uch that n c + t c). For i = 1,..., k, let A i ) [n] uch that {Ai } k contain no unflower with t petal and core ize c. Then, { t 1)k n ) A i m, if c + t c) n c + k c) t 1) n ), if n c + k c), 12

13 where m = n c)/ c). Note that both upper bound can be harp. For the firt bound, when c = 0, m = t < k and n = m, let each A i conit of all -et omitting the element 1. A unflower with t = m petal and core ize c = 0 i a perfect matching of [n]. Since every perfect matching ha a et containing 1, there i no unflower. Clearly i A i = k ) n 1 = t 1)k/m) n ). For the econd bound, we can jut take t 1 copie of ) [n] to achieve equality. Another general approach that applie to the um of the ize of familie wa initiated by Keevah Sak Sudakov Vertraëte [17]. We ued the idea behind thi approach in Lemma 7. Both method can be ued to olve certain problem. For example, a pointed out to u by Benny Sudakov, the approach in [17] can be ued to prove the k = 3 cae of Theorem 3 and perhap other cae too). Acknowledgment. We thank a referee for very helpful comment. Reference [1] N. Alon, A. Shpilka, C. Uman, On Sunflower and Matrix Multiplication, Computational Complexity, ), [2] C. Bey, On Cro-Interecting Familie of Set, Graph and Combinatoric, ), [3] B. Bollobá, P. Keevah, B. Sudakov, Multicolored extremal problem, Journal of Combinatorial Theory, Serie A ), [4] P. Borg, Interecting and Cro-Interecting Familie of Labeled Set, The Electronic Journal of Combinatoric, 15, #N9, [5] P. Borg, The maximum um and the maximum product of ize of cro-interecting familie, European Journal of Combinatoric ), [6] S. Boyd, L. Vandenberghe, Convex Optimization, Cambridge Univerity Pre 2004), 243. [7] E. Croot, V. Lev, and P. Pach. Progreion-free et in Z n ). URL: are exponentially mall [8] M. Deza, Une Propriete Extremale de Plan Projectif Fini Dan une Clae de Code Equidetant, Dicrete Mathematic, ), [9] J. Ellenberg, D. Gijwijt, Large ubet of F n q with no three-term arithmetic progreion, 2016). URL: [10] P. Erdő, C. Ko, R. Rado,Interection Theorem for Sytem of Finite Set, Quarterly Journal of Mathematic: Oxford Journal, 12 1) 1961),

14 [11] P. Erdő, R. Rado, Interection Theorem for Sytem of Finite Set, Journal of London Mathematical Society, Second Serie 35 1) 1960), [12] P. Erdő, E. Szemerédi, Combinatorial Propertie of Sytem of Set, Journal of Combinatorial Theory, Serie A 24 3) 1978), [13] P. Frankl, V. Rödl, Forbidden Interection, Tranaction of the American Mathematical Society, 300 1) 1987), [14] P. Frankl, N. Tokuhige, On r-cro Interecting Familie of Set, Combinatoric, Probability and Computing ), [15] P. Frankl, S.J. Lee, M. Sigger, N. Tokuhige, An Erdő-Ko-Rado Theorem for Cro t-interecting Familie, Journal of Combinatorial Theory, Serie A ), [16] A.J.W. Hilton, An Interection Theorem for a Collection of Familie of Subet of a Finite Set, Journal of London Mathematical Society, ), [17] P. Keevah, M. Sak, B. Sudakov, J. Vertraëte, Multicolor Turán problem, Advance in Applied Mathematic ), [18] P. Keevah, B. Sudakov, Set ytem with retricted cro-interection and the minimum rank of incluion matrice, Siam. J. Dicrete Math. Vol. 18, No ), [19] J.H. van Lint, A Theorem on Equiditant Code, Dicrete Mathematic, ), [20] M. Matumoto, N. Tokuhige, The Exact Bound in the Erdő-Rado-Ko Theorem for Cro-Interecting Familie, Journal of Combinatorial Theory, Serie A ), [21] D. Mubayi, L. Wang, Multicolor Sunflower, arxiv. [22] D. Mubayi, V. Rödl, Specified Interection, Tranaction of the American Mathematical Society ), No. 1, [23] E. Nalund, W.F. Sawin, Upper bound for unflower-free et, 2016). URL: [24] L. Pyber, A New Generalization of the Erdő-Ko-Rado Theorem, Journal of Combinatorial Theory, Serie A ),