On an antiramsey type result


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1 On an antiramsey type result Noga Alon, Hanno Lefmann and Vojtĕch Rödl Abstract We consider antiramsey type results. For a given coloring of the kelement subsets of an nelement set X, where two kelement sets with nonempty intersection are colored differently, let inj (k, n) be the largest size of a subset Y X, such that the kelement subsets of Y are colored pairwise differently. Taking the minimum over all colorings, i.e. inj(k, n) = min {inj (k, n)}, it is shown that for every positive integer k there exist positive constants c k, c k > 0 such that for all integers n, n large, the following inequality holds c k (ln n) k 2k n 2k inj(k, n) c k (ln n) k 2k n 2k. Introduction In recent years some interest was drawn towards the study of antiramsey type results, cf. for example [ESS 75], [Ra 75], [Al 83], [SS 84], [GRS 89]. In particular, they include topics like Canonical Ramsey Theory and spectra of colorings. Some of these results show a close connection to the theory of Sidonsequences, cf. [BS 85], [So 90], and recently they turned out to be also fruitful in determining bounds for canonical Ramsey numbers, cf. [LR 90]. The general problem is: given a graph or hypergraph, where the edges are colored with certain restrictions on the coloring like, for example, edges of the same color cannot intersect nontrivially, one is interested in the largest size of some totally multicolored subgraph of a given type. In this paper we study this question for colorings of complete uniform hypergraphs. Let : [X] k ω, where ω = {0,,...}, be a coloring of the kelement subsets of X. A subset Y X is called totally multicolored if the restriction of to [Y ] k is a onetoone coloring. For coloring edges in complete graphs, where the coloring is such that edges of the same color are not incident, Babai gave bounds for the largest totally multicolored complete subgraph K k : Department of Mathematics, Raymond and Beverly Sackler Faculty of Exact Sciences, Tel Aviv University, Tel Aviv, Israel Fakultät für Mathematik, SFB 343, Universität Bielefeld, W Bielefeld, Germany Emory University, Department of Mathematics and Computer Science, Atlanta, Georgia 30322, USA
2 Theorem [Ba 85] Let n be a positive integer. Let X be an nelement set and let : E(K n ) ω, where ω = {0,, 2,...}, be a coloring of the edges of the complete graph K n on n vertices, such that incident edges get different colors. Then there exists a totally multicolored complete subgraph K k, which satisfies k (2 n) 3. () Moreover, for n large, there exists a coloring : E(K n ) ω, where incident edges are colored differently, such that the largest totally multicolored complete subgraph K k satisfies k 8 (n ln n) 3. (2) Here we improve the lower bound in Theorem () and show that the upper bound is tight, up to a constant factor. Moreover, we generalize Babai s result to colorings of kuniform complete hypergraphs. Specifically, we prove Theorem 2 Let k be a positive integer with k 2. Then there exist positive constants c k, c k > 0 such that for all positive integers n with n n 0 (k) the following holds. Let X be a set with X = n and let : [X] k ω be a coloring, where (S) (T ) for all sets S, T [X] k with S T and S T. Then there exists a totally multicolored set Y X of size Y c k (ln n) 2k n k 2k. (3) Moreover, there exists a coloring : [X] k ω, where (S) (T ) for all different S, T [X] k with S T, such that every totally multicolored subset Y X satisfies Y c k (ln n) 2k n k 2k. (4) The proof relies heavily on probabilistic arguments. The lower bound is proved is Section 2 and the upper bound is established in Section 3. 2 The proof of the lower bound Let F = (V, E) be a hypergraph with vertex set V and edge set E. For a vertex x V let deg F (x) denote the degree of x in F, i.e. the number of edges e E containing x, and let 2
3 (F) = max {deg F (x) x V}. Komlós, Pintz, Spencer and Szemerédi: Our lower bound is based on the following result of Ajtai, Theorem 3 [AKPSS 82] Let F be an (r + )uniform hypergraph on N vertices. Assume that (i) F is uncrowded ( i.e. contains no cycles of length 2, 3 or 4 ) and (ii) (F) t r, where t t 0 (r), and (iii) N N 0 (r, t), then the maximum independent set of F has size α(f) 0.98 e 0 5 r N t (ln t) r. (5) We want to apply this Theorem with the parameters r+ = 2 k and t = N δ, where δ r (2r ). The proof given in [AKPSS 82] suggests that N 0 (r, t) should be at least t 4r+O(). Indeed, these calculations can be avoided, as the statement of Theorem (3) is strong enough to imply the same statement, where condition (iii) is dropped. Theorem 3 Let G be an (r + )uniform hypergraph on n vertices. Assume that (i) G is uncrowded and (ii) (G) t r, where t t 0 (r), then α(g) 0.98 e 0 5 r n t (ln t) r. (6) To see that Theorem (3) implies (the seemingly stronger) Theorem (3 ) consider a hypergraph G on n vertices, which satisfies the assumptions of Theorem (3 ). Let m be a positive integer with m > N 0(r, t) n and consider the hypergraph F m consisting of m vertex disjoint copies of G. Then F m has N = m n vertices and satisfies the assumptions (i), (ii) and (iii) of Theorem (3). We infer that m α(g) = α(f m ) 0.98 e r N t (ln t) r
4 and thus α(g) 0.98 e 0 5 r n t (ln t) r. We can now prove the lower bound in Theorem (2). The basic idea is to define a hypergraph on the set of vertices X whose edges are all unions of pairs of members of [X] k which have the same color. Then we choose an appropriate random subset of the set of vertices and show that with positive probability the induced hypergraph on this set does not contain too many edges, and contains only a small number of cycles of length 2,3 or 4. We can then delete these cycles and apply Theorem (3 ) to obtain the desired result. The actual calculations are somewhat complicated and are described below. A similar application of Theorem (3 ) is given in [RS 9]. Proof: Let : [X] k ω be a coloring, which satisfies the assumptions in Theorem (2). We may assume that X = n is divisible by k. This will not change the calculations asymptotically. Put r + = 2 k and consider an (r + )uniform hypergraph H with vertex set X and with U E(H), where E [X] r+, being an edge of H if and only if there exist different kelement subsets S, T U with (S) = (T ). Indeed, by assumption on the coloring the sets S and T are disjoint. Note that if Z X is an independent set of H, then Z is totally multicolored. Our aim is to find a large independent set in H. As there are at most n k of H satisfies kelement subsets of the same color, we infer that the number of edges E(H) ( n k) n k ( n k 2 For i = 2, 3, 4 we will further bound the number µ i of icycles in H. ) 2 k 2 (k )! nk+. (7) For distinct vertices x, y of H let deg H (x, y) be the number of edges of H containing both vertices x and y. Let {x, y} U E(H) (8) for some set U. Let S, T [X] k be such that S T = U and (S) = (T ). As S and T are by assumption disjoint sets, we may assume without loss of generality that one of the following possibilities happens: (i) either x S and y T 4
5 (ii) or {x, y} S. ( n 2 k The number of edges U E(H) satisfying (8) and (i) ((8) and (ii)) is bounded from above by ) (n 2 ( ( n k ) k 2) respectively). Summing, we obtain that n 2 deg H (x, y) k Using (9) one easily sees that n + k µ 3 c 3 n 3k µ 4 c 4 n 4k. n 2 < 2 k n k. (9) k 2 k! We will now discuss the 2cycles: for j = 2, 3,..., 2k let ν j be the number of (unordered) pairs of edges of H intersecting in a jelement set. We clearly have µ 2 = 2k j=2 ν j. Set p = n 2 4k. Let Y be a random subset of X with vertices chosen independently, each with probability p. Then P rob ( Y p n) > 0.9. (0) For i = 3, 4 let µ i (Y ) be the random variable counting the number of icycles no two of whose edges form a two cycle of the subgraph of H induced on a set Y. Similarily, for j = 2, 3,..., 2k let ν j (Y ) be the random variable counting the number of (unordered) pairs of edges of H induced on Y, which intersect in j vertices. Let E(µ i (Y )) and E(ν j (Y )) be the corresponding expected values. It is easy to see that for i = 3, 4 (and k 2). E(µ i (Y )) p (2k ) i c i n ki = o(p n) () In order to give an upper bound on E(ν j (Y )) we will first estimate ν j. Fix an edge S E(H) [X] 2k and fix nonnegative integers j 0 and j with j 0, j k. We will count the number of unordered pairs {T 0, T }, which satisfy (T 0 ) = (T ) (2) 5
6 and j 0 = S T 0, j = S T. (3) Assume first that j 0, j. Fixing the set T 0, there are at most 2k j0 j sets T satisfying (2) and (3). If we apply the same argument also when T is fixed we infer that the number of pairs {T 0, T }, which satisfy (2) and (3), is bounded from above by { } 2k 2k j0 n 2k 2k j n min, j 0 j k j 0 j j 0 k j Assume now that j 0 = 0 and j. Having fixed the set T there are at most ( n k 2) sets T 0, which satisfy (2) and (3). Therefore, the number of such pairs {T 0, T } is bounded from above by most 2k n j k j Setting j = j 0 + j 2 and summing this means, that for every edge S E(H) there are at n k. c j,k n k j 2 + c j,k n k j+. c j,k n k j 2 edges T E(H) such that S T = j, and hence with (7) we have ν j c j,k n k j 2 E(H) c j,k n 2k+ j 2, where c j,k is a constant depending on j and k only. Thus, for j = 2, 3,..., 2k it follows that E(µ 2 (Y )) = 2k j=2 2k j=2 E(ν j (Y )) p 4k j c j,k n 2k+ j 2 = o(p n). (4) 6
7 Summarizing (0), () and (4) and the fact that E( E(H) [Y ] 2k ) = p 2k E(H), (5) we infer that there exists a subset Y 0 X with Y 0 p n, µ i (Y 0 ) = o(p n) and ν j (Y 0 ) = o(p n) for i = 3, 4 and j = 2, 3,..., 2k and with E(H) [Y 0 ] 2k 2 p 2k E(H). We delete from Y 0 all vertices, which are contained in icycles of length i = 2, 3, 4 to obtain a subset Y Y 0 with Y p n, such that the subgraph induced on Y is uncrowded and has at most 2 p 2k E(H) edges. Finally, delete all vertices Y with degree bigger than 8 k p 2k E(H) p n. We obtain a subset Z Y with at least p n 2 ( o()) vertices such that the subgraph G of H induced on the set Z satisfies the assumptions of Theorem (3 ) with (G) t 2k = 8 k p2k E(H) p n 4 k! p2k n k, hence, t ( 4 k!) 2k p n k 2k. We apply Theorem (3 ) to the hypergraph G and obtain α(h) α(g) 0.98 e 0 5 r Z t ( o()) 0.49 e (ln t) r ( k! 2k This completes the proof of the lower bound. (4k ) (4k 2) ) 2k n k 2k (ln n) 2k. 3 The proof of the upper bound Let k, n be positive integers, where n is divisible by k. Let X be a set of vertices with X = n. A perfect kmatching on X is a collection of n k pairwise disjoint kelement subsets of X. 7
8 The number of perfect kmatchings of an nelement set, n divisible by k, is given by = ( n k) ( n k k n! (k!) n k n k!. ) ( n 2k ) (... k k) k n k! In the following we prove the upper bound given in Theorem (2). The basic idea is similar to the one used by Babai in [Ba 85], but there are several complications. Proof: Let X be a set with X = n, and assume that n is divisible by k. As long as k is fixed, this will not change the calculations asymptotically. Let m = cn k, where c = (k )! 8. Let M, M 2,..., M m be random perfect kmatchings, chosen uniformly randomly and independently from the set of all perfect kmatchings. Put H j = i<j M i, hence H j is the set of all kelement subsets occurring in one of the first (j ) perfect kmatchings M i, i < j. Define a coloring : [X] k ω in the following way: for j =, 2,..., m color the sets in M j \ H j with color j, and, in order to complete the coloring, color those sets in [X] k \ H m+ with new colors in an arbitrary way, compatible with the assumptions. (E.g., using a different color for each edge). Now let Y X be a subset of X with Y = l, where l > n k k+ but l = o(n k ). Our objective is to estimate the probability that Y is totally multicolored. Very roughly, this is done as follows. We show that with a high probability Y does not contain too many edges of any single matching, and hence it does not contain too many edges of H j for every j. If this occurs, then for each j, with a reasonably high probability Y does contain two edges of M j that do not lie in H j (and thus have the same color). This implies that with extremely high probability such two edges exist for some j, and hence Y is not totally multicolored. The actual proof is rather complicated, and is described in the following sequence of lemmas. Lemma For all integers j, t, j m and t 0, the following inequality holds Proof: l k Prob ( M j [Y ] k t) ( k n k )t. (6) The kmatchings M j above have been chosen at random and the set Y was fixed. In order to give an upper bound for the probability Prob ( M j [Y ] k t), view it the other way around: Let M j be fixed and let Y be chosen at random. This does not change the corresponding probabilities. Now, Y can be chosen in ( n) l ways. From Mj we can choose t kelement sets in ( n ) k t 8
9 ways and the remaining elements of Y in at most ( n kt l kt ) ways. This implies Prob ( M j [Y ] k t) ( n ) ( k t n kt ) l kt ( n ) l = n t ( l kt k n) l k t k n k. Let E i denote the event H i [Y ] k 6c k lk = 3 4 (k!) lk. (7) Notice, that Prob (E i ) Prob (E i+ ) for every i, i m. The following Lemma gives a lower bound for the probability that E m+ occurs: Lemma 2 For n large, Prob (E m+ ) 2 6c k lk. Proof: For i =, 2,..., m define random variables x i by x i = M i [Y ] k. Thus E m+ [Y ] k mi= x i. As the kmatchings are chosen independently, the x i s are independent random variables too. Therefore, m Prob ( H m+ [Y ] k t) Prob ( x i t) i= (t i ) m i=,t i 0, t i =t m Prob (x i t i ). (8) i= The number of sequences (t i ) m i= with t i 0 and m i= t i = t is given by the binomial coefficient ). By Lemma we know ( t+m t l k Prob (x i t i ) ( k n k )t i, hence with (8) we have Prob ( H m+ [Y ] k t) t + m l k ( t k n k )t e(t + m) t t t k nk e(t + m)lk ( tkn k ) t (9) l k 9
10 For t = 6c k lk and l = o(n k k ) the quotient e(t+m) lk tkn k Hence Prob (E m+ ) 2 6c k lk. occurring in (9) is less than /2 for n large. Next, define another random variable y j by y j = [M j ] 2 [[Y ] k \ H j ] 2. (20) Clearly, y j counts the number of those pairs of disjoint kelement sets i [Y ] k \ H j which have both elements in the kmatching M j (and hence have the same color). Let E(y j M) denote the conditional expected value of y j given M. Lemma 3 For positive integers n, n large, E(y j E j, M, M 2,..., M j ) > 30 k 2 l 2k. (2) n2k 2 Proof: As E j holds, (7) implies for n large that [Y ] k \ H j l 6c k k lk c l k, (22) where, say, c = 8 (k!). For every set S [Y ] at most k( l k ) kelement subsets of Y, which are not disjoint from S, hence the the number of sets {S, T } [[Y ] k \ H j ] 2, where S and T are disjoint, is for n large at least l 2 c l k (c l k k k ) > 30 (k!) 2 l2k. (23) Now, for given disjoint kelement sets S, T, the probability that both are in M j is given by Prob (S, T M j ) = = ( n k ( n k ( ) n ( k) n k ) k ) ( n k ) ( n k ( n k ) 2 k (k )! n k ) 2. 0
11 Therefore, E(y j E j, M, M 2,..., M j ) 30 k 2 l 2k n 2k 2. Lemma 4 For every positive integer j, j m, and n large Prob (y j = E j, M, M 2,..., M j ) > 3 k 2 l 2k. (24) n2k 2 Proof: First we claim that Prob (y j t E j, M, M 2,..., M j ) ( k n l k 2t+ ) (25) k for every positive integer t. This follows from the fact that t pairwise different twoelement sets imply that the underlying set has cardinality at least 2t +. Hence, y j t implies M j [Y ] k 2t +. By Lemma this gives Prob (y j t E j, M, M 2,..., M j ) Prob ( M j [Y ] k 2t + ) ( k n l k 2t+ ), k proving (25). For every positive integer i put p i = Prob (y j = i E j, M, M 2,..., M j ). Clearly, we have E(y j E j, M, M 2,..., M j ) = i<ω i p i and by (25) this implies E(y j E j, M, M 2,..., M j ) p + i ( k n i 2 l k 2i+ ) k p + O(( lk n k )3 ) l 2k p + o. (26) n 2k 2 The last inequality (26) follows from the fact that l = o(n k k ). By Lemma 3 this implies that, say, p > 3 k 2 l 2k n 2k 2 for n large. Let F j denote the event (y j = 0 and E j+ ). Then F F 2... F j is the event that E j and y i = 0 for i =, 2,..., j.
12 Let M be the set of all mutually exclusive events (E j, M, M 2,..., M j ) for which y i = 0, i =, 2,..., j, holds. Then clearly and thus Hence, by Lemma 4 we infer that Now we are ready to prove F F 2... F j = M(E j, M, M 2,..., M j ) Prob (y j = F F 2... F j ) M = Prob (y j = (E j, M M 2... M j )) M Prob (E j, M, M 2,..., M j ) Prob (y j = (E j, M, M 2,..., M j )) min M Prob (E j, M, M 2,..., M j ) = min M Prob (y j = E j, M, M 2,..., M j ). Prob (y j = F F 2... F j ) > Lemma 5 For every positive integer n, n large, Proof: we infer that 3 k 2 l 2k. (27) n2k 2 Prob (F F 2... F m ) < exp( 048 k (k!) l 2k ). (28) nk As m Prob (F F 2... F m ) = Prob (F ) Prob (F j F F 2... F j ), (29) j=2 Prob (F j F F 2... F j )) Prob (y j = 0 F F 2... F j ) Prob (y j F F 2... F j ) With (29) it follows that Prob (F F 2... F m ) < ( 3 k 2 l 2k )m n2k 2 exp( 3 k 2 m l 2k n 2k 2 ) = Prob (y j = F F 2... F j ) < 3 k 2 l 2k n 2k 2 by (27). exp( 048 k (k!) l 2k n k ) as m = 8 ((k )!) nk. 2
13 We finish the proof of Theorem 2 by bounding the probability that there exists a totally multicolored subset Y X with Y = l. If a fixed subset Y X is totally multicolored, then either F F 2... F m is true or for some j, j m, the event E j+ does not occur and therefore by (7) also not the event E m+. With Lemma 2 and Lemma 5 we infer that Prob (Y is totally multicolored ) < 2 6c k lk + exp( 048 k (k!) l 2k n k ), and doing this for all ( n l ) lelement subsets Y X implies < Prob (there exists Y X with Y = l and Y is totally multicolored ) n (exp( l 048 k (k!) l 2k 6c ) + 2 k lk ). (30) nk For l (049 k (k!)) 2k n k 2k (ln n) 2k (where the constant can be easily improved) expression (30) goes to 0 with n going to infinity. Thus for X = n and n large there exists a coloring : [X] k ω such that every totally multicolored subset Y X has size Y (049 k (k!)) 2k n k 2k (ln n) 2k. References [AKPSS 82] [Al 83] M. Ajtai, J. Komlós, J. Pintz, J. Spencer and E. Szemerédi, Extremal uncrowded hypergraphs, Journal of Combinatorial Theory Ser. A 32, 982, N. Alon, On a conjecture of Erdös, Simonovits and Sós concerning AntiRamsey Theorems, J. Graph Theory 7, 983, [Ba 85] L. Babai, An antiramsey Theorem, Graphs and Combinatorics, 985, [BS 85] [ESS 75] L. Babai and V. T. Sós, Sidon sets in groups and induced subgraphs of Cayley graphs, Europ. J. Combinatorics 6, 985, 04. P. Erdös, M. Simonovits and V. T. Sós, AntiRamsey theorems, in: Infinite and Finite Sets, ed. by Hajnal, Rado, Sós, Coll. Math. Soc. Janos Bolyai, NorthHolland, Amsterdam, 975,
14 [GRS 89] [LR 90] [Ra 75] [RS 9] R. L. Graham, B. L. Rothschild and J. Spencer, Ramsey theory, 2nd edition, John Wiley, New York, 989. H. Lefmann and V. Rödl, On canonical Ramsey numbers for complete graphs versus paths, 990, to appear in J. Comb. Theory (Series B). R. Rado, AntiRamsey theorems, in: Infinite and Finite Sets, ed. by Hajnal, Rado, Sós, Coll. Math. Soc. Janos Bolyai, NorthHolland, Amsterdam, 975, V. Rödl and E. Si najová, Note on independent sets of Steiner systems, preprint, 99. [SS 84] M. Simonovits and V. T. Sós, On restricted colorings of K n, Combinatorica 4, 984, 00. [So 90] V. T. Sós, An additive problem in different structures, 990, preprint. 4
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