Upper Bounds for Tura n Numbers. Alexander Sidorenko

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1 jounal of combinatoial theoy, Seies A 77, (1997) aticle no. TA Uppe Bounds fo Tua n Numbes Alexande Sidoenko Couant Institute of Mathematical Sciences, New Yok Univesity, 251 Mece Steet, New Yok, New Yok Communicated by the Managing Editos Received Octobe 12, 1994 A system of -element subsets (blocks) ofann-element set X n is called a Tua n (n, k, )-system if evey k-element subset of X n contains at least one of the blocks. The Tua n numbe T(n, k, ) is the minimum size of such a system. We pove uppe estimates: (1+o(1)) ln T(n, +1, ) 2 \ n as n, + T(n, k, ) (c #+o(1)) 2 \ k Academic Pess \ n + as n,, k=(#+o(1)), #>1. 1. INTRODUCTION The poblem of detemining T(n, k, ) was posed by Paul Tua n [15]. In genealization of the ealie esult of Mantel [8] fo =2, k=3, Tua n [14] completely solved the case =2 in 1941: T(n, :+1, 2)=mn& m(m+1) 2 Thee ae simple fomulae fo small n: T(n, :+1, )={ 3& Copyight 1997 by Academic Pess All ights of epoduction in any fom eseved. : fo m n : m+1. T(n, :+1, )=n&: if 1 n : &1, n &3: if #0 mod 2, &1 n : 3 3&4 ; 3n& \3&1 if #1 mod 2, 134 &1 n : 3+1 3&3.

2 UPPER BOUNDS FOR TURA N NUMBERS 135 The fist fomula is well-known; the second was obtained in [12] (its poof also can be found in [13]). Scho nheim [9] and, independently, Katona, Nemetz, and Simonovits [6] showed that T(n, k, ) n n& T(n&1, k, ). (1) Indeed, when we omit an element and all the blocks that contain this element, the Tua n (n, k, )-system educes to a Tua n (n&1, k, )-system. By omitting one element in n possible ways, we get n such subsystems; each of them has at least T(n&1, k, ) blocks. Evey block of the (n, k, )-system appeas in n& subsystems. Hence (n&) T(n, k, )nt(n&1, k, ). It follows fom inequality (1) that the atio T(n, k, )( n ) is nondeceasing. Thus thee exists the limit T(n, k, ) t(k, )= lim n \ n, + and the following inequality holds fo any n: T(n, k, ) + \n t(k, ). (2) The values t(k, ) wee found only fo =2 (except the tivial case k=). Edo s [3] offeed a ewad fo detemining t(k, ) just fo a single pai (k, ) with k>>2. De Caen [1] poved a geneal lowe bound which is cuently the best: T(n, k, ) n&k+1 n&+1 } k \ k + \n +. The pesent pape is devoted to the uppe bounds. Ou main esults (Theoems 1, 2, and 3) ae contained in Section RECURRENT INEQUALITIES Given a Tua n(n&1, k, )-system A and an (n&1, k&1, )-system B, one may constuct an (n, k, )-system in the following way (cf. [10]). Fist, the system B is tansfomed into a system B+v by adding a new element

3 136 ALEXANDER SIDORENKO v to evey block. Then the union of A and B+v is a Tua n (n, k, )- system. Thus we get T(n, k, )T(n&1, k, )+T(n&1, k&1, &1). (3) Inequalities (1) and (3) imply Theefoe, which esults in T(n, k, )=T(n, k, )&n& T(n, k, ) n n T(n, k, )&T(n&1, k, )T(n&1, k&1, ). T(n, k, ) \ n + T(n&1, k&1, &1) \ n&1, &1+ t(k, )t(k&1, &1). (4) Obviously, the union of disjoint Tua n (n$, k$, )- and (n", k", )-systems is a Tua n(n$+n", k$+k"&1, )-system. Moeove, the union of l disjoint Tua n systems with paametes (n i, : i +1, ) (whee, 2,..., l) is a Tua n system with paametes (n 1 +n 2 +}}}+n l,: 1 +: 2 +}}}+: l +1, ). This yields the inequality T \ : l l n i, : : i +1, + : l T(n i, : i +1, ). (5) Fo fixed integes : 1, : 2,..., : l and abitay n, one may select a patition n=n 1 (n)+n 2 (n)+ }}} n l (n) so that In this case, (5) implies n i (n) lim n n = t(: i+1, ) &1(&1) l t(: j=1 j+1, ). &1(&1) t \ : l : i +1, + \ : l &1(&1)+ &(&1) t(: i +1, ) (6) and, consequently, l &1 \ : : i+ t \ l : : i +1, max (: + i ) &1 t(: i +1, )., 2,..., l

4 UPPER BOUNDS FOR TURA N NUMBERS 137 This means that (k&1) &1 t(k, t) does not incease when k inceases and is fixed. Fo instance, let C(:, )=max[: &1 t(:+1, ), (:&1) &1 t(:, )]. Any intege geate than o equal to : 2 &3:+2 can be epesented as a sum of tems equal to : o :&1. Hence C(:, ) t(k, ) fo k: 2 &3:+3. (7) (k&1) &1 We will use inequalities (6) and (7) in the end of Section UPPER BOUNDS Accoding to (2), any uppe bound on t(k, ) yields an uppe bound on T(n, k, ). Thus we concentate on the bounds fo t(k, ). We fist conside the case k=+1 and then (k&) The Case k=+1 A lowe bound fo this case was poven in [11, 1]: T(n, +1, ) 1 } t(+1, ) 1. n& n&+1 \ n +, An uppe estimate t(+1, )C- was found in The estimate is a consequence of inequality (4) and the following inequality poven in [10]: t(2+1, 2) + \2 2&2. (8) In 1983, Kim and Roush [7] obtained a much bette bound: 1+2 ln t(+1, ). Thei esult was impoved by Fankl and Ro dl in 1985 [5]: ln +O(1) t(+1, ). (9)

5 138 ALEXANDER SIDORENKO We show below that the constuction descibed by Fankl and Ro dl yields the estimate t(+1, ) 1 l + \ 1&1 l+ (10) with an abitay intege l. The substitution l=(ln )(1+o(1)) in (10) poduces (9). Fo small, a stonge esult was obtained in [2]: t(2+1, 2) &2. (11) This gives the best known estimate fo t(5, 4) and t(7, 6). In the case =4, inequality (11) is still bette than (8) and (10) but aleady can be impoved. The main esult of this wok is a new uppe bound fo lage : Theoem 1. t(+1, ) \1 2 +o(1) + ln. (12) Befoe pesenting ou constuction, we need to descibe the constuctions fom woks [10, 5, 2] that yield inequalities (8), (9), and (11). Constuction 1 [10]. Most of the known constuctions of Tua n systems have a elatively small numbe of the classes of equivalent elements. In a typical constuction, the set of elements is patitioned into a fixed numbe of goups, and the fact whethe elements fom a block depends only on the goups they belong to. In contast, Constuction 1 is based on the odeing of the elements. We enumeate them as 1, 2,..., n. A subset of 2 elements, i 1 <i 2 <}}}<i 2, is a block of the system if the Boolean vecto ((i 1 +1) mod 2, (i 2 +2) mod 2,..., (i 2 +2) mod 2) has exactly zeos and ones. Using induction on, one may check that this is indeed a Tua n (n,2+1, 2)-system. Its size equals [( 2 )2&2 +o(1)]( n ) 2 which yields (8). Constuction 2 [2]. We associate n elements with the lines (ows and columns) of an (wn2x_wn2x)-matix M whose enties ae zeos and ones. We say that a submatix of M is even if the numbe of its ows, the numbe of its columns and the sum of its enties ae even numbes. A set of 2 lines of M is a block of the system if (i) all of these lines ae ows, o (ii) all ae columns, o (iii) the submatix induced by these lines is even. It is easy to see that the esulting system is a Tua n (n, 2+1, 2)-system. Its size

6 UPPER BOUNDS FOR TURA N NUMBERS 139 depends on the matix M. Let the enties of M be independent andom vaiables equal to zeo o one with pobability 1 2. Since any 2i ows and any 2(&i) columns (with 1i&1) induce an even submatix with pobability 1 2, the expected numbe of blocks is This yields (11). \ wn2x \ Wn2X = 1 2_\ wn2x 2 & : \ + + \ Wn2X 2 wn2x 2i +& +1 2 : = _ } 1 2 +O \ 1 n+&\ n 2+. +\ Wn2X 2(&i)+ i=0 \ wn2x 2i +\ 2(&i)+ Wn2X Constuction 3 [5]. We fix integes, l, and n so that n#0 mod l, and divide n elements into l equal goups A 0, A 1,..., A. Fo a subset BA 0 _ A 1 _ }}} _A, we denote by d(b) the numbe of indices i # [0, 1,..., ] that satisfy A i & B=<. We also set w(b)= : i=0 i A i &B. We denote by A k the family of all k-element subsets of A 0 _ A 1 _ }}} _A. Let B j be a subfamily which consists of blocks B # A that satisfy (w(b)+ j) mod l # [0, 1,..., d(b)]. (13) We notice that B j is a Tua n (n, +1, )-system. Indeed, fo any C # A +1, thee ae n&d(c) indices i such that A i & B{<, and thus at least one such index can be found among (w(c)+j) mod l, (w(c)+j&1) mod l,..., (w(c)+j&d(c)) mod l. The subset B=C"x, whee x # A i & B, satisfies (13) because w(b)# (w(c)&i) mod l and d(b)d(c). Let A i $=[B # A : B & A i =<]. We wish to estimate min[ B 0, B 1,..., B ]. Each B # A belongs to exactly d(b)+1 families among B 0, B 1,..., B. Thus : j=0 B j = : B#A (d(b)+1)= A + A$ 0 + A$ 1 + }}} + A$ = + \n +l } \ (()l)n + _ 1&l } \ 1+1 l+ &\ n (14) +

7 140 ALEXANDER SIDORENKO and min[ B 0, B 1,..., B ] 1 l This implies (10). : j=0 B j _1 l + \ 1&1 l+ &\ n +. In ode to obtain the estimate (12), we combine the main featues of Constuctions 13. Moe pecisely, we deive ou constuction fom Constuction 3 by emoving some supefluous blocks. This opeation is quite geneal and may be applied to any Tua n system povided that the numbe of equivalence classes of elements is small in compaison with the total numbe of elements. Removing supefluous blocks. Conside a system of -element blocks. Its automophism is a pemutation of the elements which peseves the set of blocks. The automophism goup geneates an equivalence elation on the elements as well as on the blocks. We denote the classes of equivalent elements by A 1, A 2,..., A l. Any equivalence class B of blocks coesponds to some intege patition =b 1 +b 2 +}}}+b l such that B=B(b 1, b 2,..., b l )=[B: B&A 1 =b 1, B&A 2 =b 2,..., B & A l =b l ]. Suppose, the consideed system is a Tua n (n, +1, )-system. We claim that one may omit at least half of the blocks fom evey existing class B(b 1, b 2,..., b l ) wheneve b 1 2, b 2 2,..., b l 2. As a esult, we will have a Tua n (n, +1, )-system of a smalle size. Indeed, suppose fist that =2l and b 1 =b 2 =}}}=b l =2. By analogy with Constuction 2, we intoduce an l-dimensional matix M whee the ith dimension coesponds to the equivalence class A i. Blocks of the class B(2, 2,..., 2) coespond to (2_2_ } } } _2)-submatices of M. We say that a block b # B(2, 2,..., 2) is even o odd if the sum of the enties of the coesponding submatix is such. Similaly to Constuction 2, we may omit all odd blocks of class B(2, 2,..., 2). Now let >2l and integes b 1, b 2,..., b l satisfy b 1 +b 2 +}}}+b l =, b i 2 fo evey i. By analogy with Constuction 1, we linealy ode the elements within each class A i. Fo evey block B # B(b 1, b 2,..., b l ), we define its 2-pojection as a (2l)-element subset which includes the two maximal elements fom each intesection B & A i. With espect to the l-dimensional matix M, this 2-pojection can be odd o even. We omit those blocks B # B(b 1, b 2,..., b l ) which have odd 2-pojections. If the enties of M ae independent andom vaiables equal to zeo o one with pobability 1 2, the expected numbe of omitted blocks is 1 2 B(b 1, b 2,..., b l ).

8 UPPER BOUNDS FOR TURA N NUMBERS 141 Constuction 4. Now we apply the descibed pocedue to Constuction 3. We select a function and denote f : A 0 _A 1 _}}}_A [0, 1] D=[D # A 2l : D&A 0 = D&A 1 = }}} = D&A =2]. Fo any D # D, we set q(d)= : x 0 # D&A 0, x 1 #D&A 1,..., x # D&A f(x 0, x 1,..., x ). (15) The sum in (15) consists of 2 l tems. We claim that fo any C # A 2l+1 that satisfies C & A i =3 and C & A k =2 with evey k # [0, 1,..., ]"[i], thee exists x # C & A i such that q(c"x) is even. Indeed, let C & A i = [x, y, z]. Among the 3 } 2 l tems of the sum q(c"x)+q(c"y)+q(c"z), evey tem appeas twice. Thus the sum is even, and one of the values q(c"x), q(c"y) oq(c"z) must be even. We linealy ode the elements within each goup A i and denote E=[B # A : B&A i 2 fo i=0, 1,..., ]. Fo any B # E, we fom a (2l)-element subset D(B) by taking the two maximal elements fom each B & A i with i=0, 1,...,. Now we omit a block B #(B j &E) fom B j if q(d(b)) is odd. We claim that the emaining system, B j $=[B # B j : B E o q(d(b)) is even], is a Tua n(n,+1, )-system. Indeed, conside any C # A +1. By the agument we used in Constuction 3, thee is an index i such that C & A i {< and (C"x)#B j fo any x # A i.if(c"x)#efo x # A i, then C & A i 3 and C & A k 2 fo evey k # [0, 1,..., ]"[i]. As we showed above, one may choose an element x among the thee maximal elements of C & A i such that q(d(c"x)) is even, so (C"x)#B j $. Now we know that B j $ is a Tua n (n, +1, )-system fo each j, and ou aim is to estimate min[ B$ 0, B$ 1,..., B$ ]. Obviously, : j=0 B j $ = : j=0 B j & [B # E : q(d(b)) is odd]. Let f in (15) be a andom function whose values fo diffeent sets of aguments ae independent; each value is 0 o 1 with pobability 1 2. With this function, the pobability that q(d(b)) is odd equals 1 2 fo any B # E.

9 142 ALEXANDER SIDORENKO Hence, the expectation of [B # E : q(d(b)) is odd] is 1 2 E. Thus thee exists a specific function f so that [B # E : q(d(b)) is odd] 1 E. 2 (Such a function f, if needed, could be constucted explicitly.) Theefoe, By applying (14), we get : j=<0 : j=0 B j $ : j=0 B j & 1 2 E. B$ j _1 2 +l} \ 1&1 l+ &\ n _\ n +& & E &. We set A i "=[B # A : B&A i 1]. Obviously, and Now we estimate \ n + & E : A i " i=0 A" = i \(()l)n + + \ (1l)n 1 + } \ (()l)n &1 + min[ B$ 0, B$ 1,..., B$ ] = &l(&1)n+\ (()l)n \1+ + &l(&1)n+\ \1+ 1&1 l+ \ n +. 1 : B j $ l _1 2l + \ 1&1 l+ &\ n : 2l j=0 i=0 _1 2l + \ 1&1 l+ &\ n \ 1+ = _1 2l +1 2\ 3+ &l(&1)n+\ 1&1 l+ &\ n +. A i " &l(&1)n+\ 1&1 l+ \ n +

10 UPPER BOUNDS FOR TURA N NUMBERS 143 So we get with abitay l. We choose t(+1, ) 1 2l +1 2\ 3+ +\ 1&1 l+ l= (1+*()) ln, whee the function * satisfies the conditions *() 0 and *()ln as. Since (1&1l) exp (&l), we finally get t(+1, ) 1 2l +1 2\ 3+ + exp \ & l+ = 1 2l +1 2\ exp[&*() ln] (1+o(1)) ln = 2 as. Remak. The coefficient 1 2 in the ight-hand side of (12) is inheited fom the genealized vaiant of the Tua n poblem and could be impoved. Given m, n,, we conside m disjoint goups of elements, X 1, X 2,..., X m, whee X 1 = X 2 = }}} = X m =n. Let T=T m (n, +1, ) be the minimum numbe of sets B 1, B 2,..., B T such that 1. B 1 = B 2 = }}} = B T =m. 2. B i & X j = fo evey, 2,..., T, j=1, 2,..., m. 3. Fo any set C of size m+1 that satisfies C & X j with j=1, 2,..., m, thee exists B i /C. In paticula, T 1 (n, +1, )=T(n, +1, ). One may show that thee exists the limit and t * ()= lim m, n m T m (n, +1, ) + <\n 1 =t 2 * (2)t (3) } }}. * 1 The coefficient 2 in the ight-hand side of (12) can be eplaced with lim t (). If lim * t ()=0, we would get t(+1, )=o((ln )). *

11 144 ALEXANDER SIDORENKO Unfotunately, we ae unable to impove the uppe bound, t * () 1 2, even assuming that is lage The Geneal Case In the case when k&1 is a multiple of &1, Tua n [16] showed that &1 t(k, ) k&1+ \&1. (16) Sidoenko [10] poved (16) fo abitay k and. Fankl and Ro dl [5] used pobabilistic aguments to obtain the estimate a(a+4+o(1)) ln t(+a, ) \ a+ as, (17) whee a is a constant. We descibe thei constuction and modify it to pove uppe bounds on t(k, ) with (k&). Constuction 5 [5]. We fix intege paametes, k, l, N. Fo evey -element subset of the set X=[1, 2,..., N], we assign at andom one of the colos 1, 2,..., l. Fo any k-element subset YX, the pobability that some specific colo is not used on the -element subsets of Y is (1&1l) ( k ) < exp [&(1l)( k )]. Thus the pobability p of the event E Y that not all l colos ae used on Y is less than l } exp [&(1l)( k )]. Let d denote the numbe of those k-element subsets Y$X fo which the events E Y and E Y$ ae dependent: We equie 4pd<1; that is, k&1 d= : i= \ 4l } exp _&1 l \ k k&1 +& } : k i+\ N&k k&i+. i= \ k i+\ N&k <1. (18) k&i+ This condition allows us to apply the Lova sz local lemma (see [4]) which states that thee exists a coloing whee none of the events E Y occus. With this coloing, the family A i of subsets of colo i is a Tua n (N, k, )-system. We choose one of the families A 1, A 2,..., A l which has the minimal size and denote it by A. Obviously, A (1l)( N ). Now we divide a lage set of n=mn elements into N equal goups Z 1, Z 2,..., Z N. Let B be the family of -element subsets BZ 1 _ Z 2 _ }}} _Z N such that

12 UPPER BOUNDS FOR TURA N NUMBERS 145 (i) B=[x 1, x 2,..., x ] whee x i # Z ji, j 1 < j 2 <}}}<j,[j 1, j 2,..., j ] # A; o (ii) thee is Z j fo which B & Z j 2. The size of this family is at most B m } A +N 2+\ mn&2 \m &2 + m l \ N + +(&1) 2N \ mn + \1 l +(&1) 2N +\ mn +. Obviously, B is a Tua n (mn, k, )-system. Thus (18) implies t(k, ) 1 l +(&1) 2N. (19) Fankl and Ro dl [5] used the paametization k=+a, l= w( )[a(a+4) ln ]x, a N=a+2 whee a is a constant. In this case, (18) holds and (19) yields (17). Constuction 6. In the case when k& inceases as, thee is a bette choice of paametes. In ou modification of Constuction 5, a=k& is not a constant anymoe. We equie k+3 and select paametes as follows: N= (&1) \ k + [(k&)! 2&k ] 1(k&), l=\ \ k + (k&+1) ln \ k +&. _(&1) Fist, we have to check the validity of (18): k&1 : i= \ 4l } exp _&1 l \ k +& } (N&1)k& (k&)! k i+\ N&k k&i+ < \ N k&1 k&+ : i= \ k i+ < \ k&+ N 2k < (N&1)k& 2 k, (k&)! 4 2 k (&1)(k&+1) ln _(&1) \ k +& <1.

13 146 ALEXANDER SIDORENKO Now we exploit (19): t(k, ) (k&+1) ln \ k +& _(&1) \ k + &(k&+1) ln _ (&1) \ k +& + (2 (k&)!) 1(k&) \ k +. (20) If k&log 2 then (2 (k&)!) 1(k&) =o[(k&)ln( k )]. Thus (20) yields the following esult. Theoem 2. If k+log 2 then (1+o(1))(k&+1) ln + t(k, ) \k \ k + as. (21) If k #>1 as, we estimate ln( k )=[# ln #&(#&1) ln(#&1)+o(1)]. In this case, Theoem 2 implies Theoem 3. If #>1 and k=(#+o(1)) then t(k, ) (c #+o(1)) 2 \ k + whee c # =(#&1)[# ln #&(#&1) ln(#&1)+o(1)]. as, (22) If k is much lage than, one may get futhe impovements in (21) by using (7) with :=( 2 ) and applying (22) to estimate t(:, ) and t(:+1, ). Tua n (fo instance, see wok [16]) conjectued that (16) tuns the equality wheneve k&1 is a multiple of &1. A counteexample with k=13, =4 was found in [10]. By combining inequality (6) with l=2s&1, : 1 =: 2 =}}}=: l =2s and inequality (8), we get the counteexample k=4s 2 &2s+1, =2s,(k&1)(&1)=2s fo any s2. Moeove, Theoem 3 shows that this conjectue fails fo any atio (k&1)(&1) when is sufficiently lage. REFERENCES 1. D. de Caen, Extension of a theoem of Moon and Mose on complete subgaphs, As Combin. 16 (1983), 510.

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