On the size-ramsey numbers for hypergraphs. A. Dudek, S. La Fleur and D. Mubayi

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1 On the size-ramsey numbers for hypergraphs A. Dudek, S. La Fleur and D. Mubayi REPORT No. 48, 013/014, spring ISSN X ISRN IML-R /14- -SE+spring

2 On the size-ramsey number of hypergraphs Andrzej Dudek Steven La Fleur Dhruv Mubayi Vojtech Rödl May 14, 015 Abstract The size-ramsey number of a graph G is the minimum number of edges in a graph H such that every -edge-coloring of H yields a monochromatic copy of G. Size-Ramsey numbers of graphs have been studied for almost 40 years with particular focus on the case of trees and bounded degree graphs. We initiate the study of size-ramsey numbers for k-uniform hypergraphs. Analogous to the graph case, we consider the size-ramsey number of cliques, paths, trees, and bounded degree hypergraphs. Our results suggest that size-ramsey numbers for hypergraphs are extremely difficult to determine, and many open problems remain. 1 Introduction Given graphs G and H, say that H G if every -edge-coloring of H results in a monochromatic copy of G in H. Using this notation, the Ramsey number RG of G is the minimum n such that K n G. Instead of minimizing the number of vertices, one can minimize the number of edges. Define the size-ramsey number ˆRG of G to be the minimum number of edges in a graph H such that H G. More formally, ˆRG = min{ EH : H G}. The study of size-ramsey numbers was proposed by Erdős, Faudree, Rousseau and Schelp [5] in By definition of RG, we have K RG G. Since the complete graph on RG vertices has RG edges, we obtain the trivial bound RG ˆRG. 1 Department of Mathematics, Western Michigan University, Kalamazoo, MI 49008, andrzej.dudek@wmich.edu. Supported in part by Simons Foundation Grant #4471. Work partially done during a visit to the Institut Mittag-Leffler Djursholm, Sweden. Department of Mathematics and Computer Science, Emory University, Atlanta, GA 303, slafleu@emory.edu. Department of Mathematics, Statistics, and Computer Science, University of Illinois at Chicago, Chicago, IL 60607, mubayi@math.uic.edu. Supported in part by NSF grant DMS Department of Mathematics and Computer Science, Emory University, Atlanta, GA 303, rodl@mathcs.emory.edu. Supported in part by NSF grants DMS and DMS

3 Chvátal see, e.g., [5] showed that equality holds in 1 for complete graphs. In other words, RKn ˆRK n =. One of the first problems in this area was to determine the size-ramsey number of the n vertex path P n. Answering a question of Erdős [4], Beck [1] showed that ˆRP n = On. 3 Since ˆRG EG for any graph, Beck s result is sharp in order of magnitude. The linearity of the size-ramsey number of paths was generalized to bounded degree trees by Friedman and Pippenger [11] and to cycles by Haxell, Kohayakawa and Luczak [1]. Beck [] asked whether ˆRG is always linear in the size of G for graphs G of bounded degree. This was settled in the negative by Rödl and Szemerédi [18], who proved that there are graphs of order n, maximum degree 3, and size-ramsey number Ωnlog n 1/60. They also conjectured that for a fixed integer there is an ε > 0 such that Ωn 1+ε = max G ˆRG = On ε, where the maximum is taken over all graphs G of order n with maximum degree at most. The upper bound was recently proved by Kohayakawa, Rödl, Schacht, and Szemerédi [15]. For further results about the size-ramsey number see, e.g, the survey paper of Faudree and Schelp [8]. Somewhat surprisingly the size-ramsey numbers have not been studied for hypergraphs, even though classical Ramsey numbers for hypergraphs have been studied extensively since the 1950 s see, e.g., [7, 6], and more recently [3]. In this paper we initiate this study for k-uniform hypergraphs. A k-uniform hypergraph G k-graph for short on a vertex set V G is a family of k-element subsets called edges of V G. We write EG for its edge set. Given k-graphs G and H, say that H G if every -edge-coloring of H results in a monochromatic copy of G in H. Define the size-ramsey number ˆRG of a k-graph G as ˆRG = min{ EH : H G}. Results and open problems Motivated by extending the basic theory from graphs to hypergraphs, we prove results for cliques, trees, paths, and bounded degree hypergraphs..1 Cliques For every k-graph G, we trivially have RG ˆRG, k where RG is the ordinary Ramsey number of G. Our first objective was to generalize to 3-graphs, which shows that equality holds for graphs. It is fairly easy to obtain a lower 3 bound for ˆRK n that is quadratic in RK n 3, but we were only able to do slightly better.

4 Theorem.1 ˆRK 3 n n 96 3 RK n The following basic questions remain open. Question. Is = RK n k? ˆRK k n k. Question.3 For k 3 let N = RK n k. Define K k N to be the hypergraph obtained from K k N by removing one edge. Is it true that Kk N K n k? Clearly, the affirmative answer to the latter gives a negative answer to Question... Trees Given integers 1 l < k and n, a k-graph T k of order n with edge set {e 1,..., e m } is an l-tree, if for each j m we have e j 1 i<j e i l and e j 1 i<j e i e i0 for some 1 i 0 < j. We are able to give the following general upper bound for trees. Theorem.4 Let 1 l < k be fixed integers. Then ˆRT k = Onl+1. One can easily show that this bound is tight in order of magnitude when l = 1 see Section 4 for details. The situation for l is much less clear. Question.5 Let l < k be fixed integers. Is it true that for every n there exists a k-uniform l-tree T of order at most n such that ˆRT = Ωn l+1. Here is another related question pointed out by Fox [9]. Let us weaken the restriction on the edge intersection in the definition of T k k. Let T be a k-graph of order n with edge set {e 1,..., e m } such that for each j m we have e j 1 i<j e i l. Question.6 Let l < k be fixed integers. Is k ˆR T polynomial in n?.3 Paths Given integers 1 l < k and n l mod k l, we define an l-path P k to be the k-uniform hypergraph with vertex set [n] and edge set {e 1,..., e m }, where e i = {i 1k l + 1, i 1k l +,..., i 1k l + k} and m = n l k l. In other words, the edges are intervals of length k in [n] and consecutive edges intersect in precisely l vertices. The two extreme cases of l = 1 and l = k 1 are referred to as, respectively, loose and tight paths. Clearly every l-path is also an l-tree. Thus, by Theorem.4 we obtain the following result. ˆRP k = Onl+1. 4 Our first result shows that determining the size-ramsey number of a path P k for l k can easily be reduced to the graph case. 3

5 Proposition.7 Let 1 l k. Then, ˆRP k ˆRP n = On. Clearly, this result is optimal. Determining the size-ramsey number of a path P k for l > k seems to be a much harder problem. Here we will only consider tight paths l = k 1. By 4 we get ˆRP k n,k 1 = Onk. 5 The most complicated result of this paper is the following improvement of 5. Theorem.8 Fix k 3 and let α = k / k Then ˆRP k n,k 1 = Onk 1 α log n 1+α. The gap in the exponent of n between the upper and lower bounds for this problem remains quite large between 1 and k 1 α. We believe that the lower bound is much closer to the truth. Indeed, the following question still remains open. Question.9 Is ˆRP k n,k 1 = On? k k If true, then since ˆRP ˆRP n,k 1, this would imply the linearity of the size-ramsey number of all l-paths..4 Bounded degree hypergraphs Our main result about bounded degree hypergraphs is that their size-ramsey numbers can be superlinear. This is proved by extending the methods of Rödl and Szémerédi [18] to the hypergraph case. Theorem.10 Let k 3 be an integer. Then there is a positive constant c = ck such that for every n there is a k-graph G of order at most n with maximum degree k + 1 such that ˆRG = Ωnlog n c. There are several other problems to consider such as finding the asymptotic of the size- Ramsey number of cycles and many other classes of hypergraphs. In general, they seem to be very difficult. Therefore, this paper is the first step towards a better understanding of this concept. In the next sections we prove these result for cliques Section 3, trees Section 4, paths Section 5, and hypergraphs with bounded degree Section 6. 3 Cliques Proof of Theorem.1. We show that if H is a 3-graph with EH < n 96 n 4, then H K n RK n for

6 Induction on N = V H. If N < RK n 3, then there is a -coloring of K 3 N with no monochromatic K n 3. Since H K 3, this coloring yields a -coloring of H with no monochromatic K 3 n. Suppose that N RK n 3. Since EH < n 96 degu, v = {e EH : {u, v} e} < n EH = 1 3 N {u,v} V H 3 RK n 3. Otherwise, degu, v 1 3 N, there are u and v in V H with n 3 > EH, a contradiction. Let u and v be such that degu, v < n 3. Define H u as follows: and V H u = V H \ {v} EH u = {e : v / e EH} {{u, x, y} : {v, x, y} EH and {u, x, y} / EH}. Clearly, V H u = N 1 and EH u EH < n 3 RK n 96. By the inductive hypothesis there is a -coloring χ u of the edges of H u with no monochromatic K n 3. Let T = T 1 = N H u, v = {w V H : {u, v, w} EH}. Thus, T 1 V H u and T 1 < n 3. If there exists S 1 T 1 such that S 1 n 4 and H u[s 1 {u}] is monochromatic, then set T = T 1 \S 1. If there exists S T such that S n 4 and H u[s {u}] is monochromatic, then set T 3 = T \ S. We continue this process obtaining T = S 1 S S m U, where H u [S i {u}] is monochromatic, S i n 4, and H u[u {u}] contains only monochromatic cliques of order at most n 4. Now we define a -coloring χ of H. i If v / e, then χe = χ u e. ii If v e = {v, x, y} and u / e, then χe = χ u {u, x, y}. iii If {u, v} e = {u, v, x} and x S i, then e takes the opposite color to the color of H u [S i {u}]. iv If {u, v} e = {u, v, x} and x U, then color e arbitrarily. Now suppose that there is a monochromatic clique K = K 3 n in H. Such a clique must contain v. Now there are two cases to consider. If u / V K, then the subgraph of H u induced by V K {u} \ {v} is also a monochromatic copy of K n 3, a contradiction. Otherwise, u V K. Thus, V K \ {u, v} T and V K \ {u, v} = n. Observe that V K S i and V K U < n 4. But this yields a contradiction n = V K \ {u, v} < m + n 4 < n 3 n 4 + n 4 = n n, for n 4. 5

7 4 Trees First for convenience we recall the definition of a hypertree. Given integers 1 l < k and n, recall that a k-graph T k of order n with edge set {e 1,..., e m } is an l-tree, if for each j m we have e j 1 i<j e i l and e j 1 i<j e i e i0 for some 1 i 0 < j. k Proof of Theorem.4. Fix 1 l k. We are to show that ˆRT = Onl+1. Recall that a partial Steiner system St, k, N is a k-graph of order N such that each t-tuple is contained in at most one edge. Due to a result of Rödl [17] it is known that there is a constant N 0 = N 0 t, k such that for every N N 0 there is an S = St, k, N with the number of edges satisfying N N 9 10 t t k ES k 6 t t see also [14, 19, 0, 1] for similar results. It is easy to observe that for 1 s t every s-tuple is contained in at most N s k s t s edges. Fix 1 l < k. Let N = cn + l, where the constant c is defined as { c = max N 0 l + 1, k, 0 } k l l + 1 t s Let H be a Sl + 1, k, N satisfying 6. Observe that if l + 1 = k, then H can be viewed as a complete k-graph of order N. Clearly, EH = On l+1. It remains to show that for any T = T k tree, H T. Define a degree of a set U V H 1 U < k by degu = {e EH : e U} and for EH a minimum non-zero l-degree by δ l H = min{degu : U = l and U e for some e EH}. First observe that for any -coloring of the edges of H, there is a monochromatic subhypergraph F with δ l F n. Indeed, suppose that H is colored with blue and red colors. Assume by symmetry that the red hypergraph R has at least 1 EH edges. Set R 0 = R. If there exists U 0 V R 0 with deg R0 U 0 < n, then let R 1 = R 0 U 0 we remove U 0 and all incident to U 0 edges. Now we repeat the process. If there exists U 1 V R 1 with deg R1 U 1 < n, then let R = R 1 U 1. Continue this way to obtain hypergraphs R = R 0 R 1 R R m, where either δ l R m n or R m is empty hypergraph. But the latter cannot happen, since the number of removed edges from R is less than N N l + 1 N l + 1 n = l l + 1 N l n 9 l + 1 c 0 6 N l+1 k l+1 < 1 EH.

8 Figure 1: A star of order n with n 1 4 arms each of length. Now we greedily embed T into F = R m. At every step we have a connected sub-tree T i T. Assume that we already embedded i edges of T obtaining T i. Let U l be such that U e for some e ET i. Observe that there is always an edge f EF \ ET i such that f V T i = U. Indeed, if U = l, then this is true since deg F U n and V T i < n and every l + 1-tuple of vertices of F is contained in at most one edge in F. Otherwise, if U < l, first we find a set W V F \ V T i such that W = l U and U W is contained in an edge of F, and next apply the previous argument to U W. Thus, we can extend T i to T i+1, as required. As mentioned in the introduction, it would be interesting to decide whether Theorem.4 is tight up to the hidden constant. This is definitely the case for l = 1. Indeed, let T be a k-uniform star-like tree of order n defined as follows. Assume that n 1 k divides n 1. T consists of k arms P i each with two edges: EP i = {{v, w1 i, wi,..., wi k 1 }, {wi k 1, wi k,..., wi k }}, where 1 i n 1 k and all wi j vertices are pairwise different see Figure 1. Assume that H T and color e H by red if degree in H of every vertex in e is less than n 1 k ; otherwise e is blue. Since H T e and there is no red copy of T, there must be a blue copy of T. Every edge in such a copy has at least one vertex of degree at least n 1 k in H. Since T has n 1 vertex disjoint edges and every edge in H can intersect at most k 3 of those disjoint edges, 5 Paths ˆRT 1 3 n 1 k n 1 k = Ωn. In this section we prove Proposition.7 and Theorem.8. Proof of Proposition.7. Let H be a graph satisfying H P n and EH = On cf. 3. We construct a k-graph H as follows. Replace every vertex v V H by an l-tuple {v 1, v,..., v l } different for every v and each e = {v, w} EH by {v 1,..., v l, w 1,..., w l, x 1,..., x k l }, 7

9 where x 1,..., x k l are different for every edge e, too. Thus, H is a k-graph with V H = l V H + k l EH and EH = EH. Now color EH. This coloring uniquely defines a coloring of EH. Since H contains a monochromatic copy of P n, H also contains a monochromatic copy of P k. Consequently, H Pk and the proof is complete. We now turn to the main result of this section which we restate for convenience. Theorem.8 Fix k 3 and let α = k / k Then ˆRP k n,k 1 = Onk 1 α log n 1+α. First we prove an auxiliary result. In order to do it we state some necessary notation. Set 1 β =. + 1 k 1 For a graph G = V, E let T l G be the set of all cliques of order l and let t l = T l G. Let A V and B T k 1 G be a family of pairwise vertex-disjoint cliques. Define x A,B as the number of k-cliques of G which k 1 vertices form a vertex set of some B B and the remaining vertex is from V \ A B B V B. Similarly, let y A,B be the number of k-cliques in G which k 1 vertices form a vertex set of some B B and the remaining vertex is from A B B V B. Finally, let z C for C V be the number of k-cliques containing at least one vertex from C. Proposition 5.1 Let k 3 be an integer and let c = 1 3 3k. Then there exists a graph G = V, E of order n for sufficiently large n satisfying the following: i For every A V, A cn, and every B T k 1 G, B = cn, vertex disjoint k 1- cliques such that A B B V B = we have y A,B 1 k + 1 x A,B. ii For every C V, C k 1cn, z C t k 4k. iii The total number of k-cliques satisfies where ν = 3/ k dk k 1k. t k νn k 1 α log n 1+α, 8

10 Proof. It suffices to show that the random graph G Gn, p with p = dlog n/n β and d = 3000 satisfies a.a.s. 1 i - iii. Below we will use the following bounds on the tails of the binomial distribution Bin n, p for details, see, e.g., [13]: PrBin n, p 1 γex exp γ EX, 7 PrBin n, p 1 + γex exp γ 3 EX. 8 First we show that G a.a.s. satisfies i. Fix an A V and B T k 1 with B = cn. Observe that without loss of generality we may assume that A = cn. Note that x A,B Bin cnn cn k 1cn, p k 1. Thus, Ex A,B = c1 kcn p k 1 = d k 1 c1 kcn k 1β log n k 1β and 7 applied with γ = 1/ implies Pr x A,B Ex A,B exp 1 8 Ex A,B = exp dk 1 8 c1 kcn k 1β log n k 1β. 9 Now we bound from above the number of all possible choices for A and B. Clearly we have at most n cn choices for A. Observe that the number of choices for B can be bounded from above by the number of ways of choosing an ordered subset of vertices of size k 1cn. { Indeed, suppose that v 1,..., v k 1cn is such a choice. Then B can be defined as {v1,..., v k 1 }, {v k,..., v k },..., {v k 1cn k+1,..., v k 1cn } }. Thus we conclude that there are at most n kcn ways to choose A and B. Hence, by 9 Pr { x A,B Ex } A,B n kcn Pr x A,B Ex A,B A,B exp kcn log n dk 1 8 c1 kcn k 1β log n k 1β Similarly, since y A,B Bin cn kcn, p k 1, and since c = 1 3 3k 1 k3k+4, = o1. 10 Ey A,B = kc n p k 1 = d k 1 kc n k 1β log n k 1β. Ex A,B c1 kc = k + 1 k + 1 dk 1 n k 1β log n k 1β 3 dk 1 kc n k 1β log n k 1β = 3 Ey A,B. 1 An event E n occurs asymptotically almost surely, or a.a.s. for brevity, if lim n Pr E n = 1. 9

11 Inequality 8 applied with γ = 1/ yields Pr y A,B Ex A,B Pr y A,B 3 k + 1 Ey A,B Therefore, we deduce that Pr A,B { y A,B Ex A,B k + 1 Consequently, by 10 and 11 we get that a.a.s. exp 1 1 Ey A,B. } n kcn exp 1 1 Ey A,B = o1. 11 y A,B Ex A,B k + 1 x A,B k + 1 for any choice of A and B. This finishes the proof of i. For each vertex v V, let deg k v denote the number of k-cliques of G which contain v. In order to show that a.a.s. G also satisfies ii, we will first estimate deg k v for each v V. The standard application of 8 applied with Bin n 1, p and γ = 1/ with the union bound imply that a.a.s. the degree of every vertex v V G satisfies degv 3 dn1 β log n β. The number of k-cliques which contain v is equal to the number of k 1-cliques in the neighborhood of v. Therefore, in order to show ii it suffices to bound the number of k 1-cliques in any set of size at most 3 dn1 β log n β. Let S V with s = S = 3 dn1 β log n β. First we will decompose all k 1-tuples of S into linear k 1-uniform hypergraphs S 1, S,..., S m with s k 1 s m = 1 + o1 / k 1 and S i = 1 + o1 s k 1 for every 1 i m. That means that each k 1-tuple of S belongs to exactly one S i and each pair of elements of S appears in at most one k 1-tuple in S i. The existence of such a decomposition follows from a more general result of Pippenger and Spencer [16] see also [10]. Let s i be the random variable that counts the number of k 1-tuples of S i which appear as k 1-cliques of G. Observe that s i Bin S i, p k 1. Therefore for each i, s Es i = 1 + o1 k 1 p k 1 s = 1 + o1 pk 1 k 1k 9 = 1 + o1 d+k 1 n 1 β log n 1+β 4k 1k 10

12 and by 8 with γ = 1/ Pr s i 3 Es i exp 1 1 Es i exp 3 d+k 1 16k n 1 β log n 1+β. Consequently, the union bound over all subsets S V of size s and over all i for each 1 i m implies Pr { i} s i 3 Es n m exp 3 d+k 1 s 16k n 1 β log n 1+β S, i n s s k 3 exp 3 d+k 1 16k n 1 β log n 1+β = s k 3 exp s log n 3 d+k 1 16k n 1 β log n 1+β 3 = s k 3 exp n 1 β log n 1+β d 3 d+k 1 16k = o1, since s k 3 grows like a polynomial in n. Therefore it follows that a.a.s. deg k v = m i=1 s i m 3 Es i s k 3 3 Es i = νn k 1 β log n 1+α, 1 where ν = In a similar way one can show that where 3 k d k k 1k. 13 deg k v λn k 1 β log n 1+α, λ = Note that equation 1 gives the bound 1 k 1 d k k 1k. 14 t k νn k 1 β+1 log n 1+α = νn k 1 α log n 1+α, which proves part iii. Now we finish the proof of ii. number of k-cliques is a.a.s. at least Since each k-clique is counted exactly k times, the t k n k λnk 1 β log n 1+α = λ k nk 1 α log n 1+α

13 It follows now from 1 and 15 that given a set C V, C k 1cn, the number of k-cliques of G which intersect C is a.a.s. at most z C k 1cn νn k 1 β log n 1+α = ck 1kν λ λ k nk 1 α log n 1+α Finally observe that 13, 14 together with the choice of c yield that ck 1kν t k. λ implying condition ii, as required. ck 1kν λ 1 4k Now we are ready to prove main result of this section. Proof of Theorem.8. We show that there exists a k-graph H with H = On k 1 α log n 1+α such that any two-coloring of the edges of H yields a monochromatic copy of P k n,k 1. Let G be a graph from Proposition 5.1. Set V H = V G and let EH be the set of k-cliques in G. We prove that such H is a Ramsey k-graph for P k m,k 1 with m = cn, where c = 1 3 3k. Take an arbitrary red-blue coloring of the edges of H 0 = H and assume that there is no monochromatic P k m,k 1. We will consider the following greedy procedure which at each step finds a blue tight path of length i labeled as v 1, v,..., v i. 1 Let B = be the trash set of k 1-tuples and U = V H be the set of unused vertices and set i := 0. At any point in the process, if B = m, then stop. In this step i = 0. If possible, then choose a blue edge from U and label its vertices by v 1,..., v k and then set i := k. Otherwise, if not possible, stop. 3 In this step i k. Let v i k+1,..., v i 1, v i be the labels of the last k 1 vertices of the constructed blue path. If possible, select a vertex u U for which v i k+1,..., v i 1, v i, u form a blue edge. Label u as v i+1, set U := U \ {u} and i := i + 1. Repeat this step until no such u can be found. 4 In this step also i k. Let v i k+1,..., v i 1, v i be the labels of the last k 1 vertices of the constructed blue path which cannot be extended in a sense described in step 3. Remove these k 1 vertices from the path and set B := B {{v i k+1,..., v i 1, v i }} and i := i k + 1. After this removal there are two possibilities: i if i < k, then put back v 1,..., v i to U i.e. U := U {v 1,..., v i }, set i := 0, and return to step ; ii otherwise, return to step 3. This procedure will terminate under two circumstances: either B = m or no blue edge can be found in step. 1

14 First let us consider the case when B = m, that means, there are m vertex disjoint k 1-tuples in B. Denote by A the vertex set of the blue path which was obtained when B = m. Clearly, A < m, otherwise there would be a blue P k m,k 1. We are going to apply Proposition 5.1 with sets A and B. Notice that every edge of H which contains a k 1-tuple from B and the remaining vertex from V H \ A B B B must be colored red. This is because for a k 1-tuple to end up in B, there must have been no vertex u in step 3 that could extend the blue path. It also follows from step 3 that each k 1- tuple in B is contained in at least one blue edge. Thus, Proposition 5.1 i implies that y A,B 1 k+1 x A,B. That means that the number of red edges which contain a k 1-tuple from B and the remaining vertex from U is at least k + 1 times the number of blue edges with a k 1-tuple from B. Now remove all the blue edges from H which contain a k 1-tuple from B and denote such k-graph by H 1. Perform the above procedure on H 1. This will generate a new trash set B 1. Observe that B 1 B =, since every edge of H 1 which contains a k 1-tuple from B must be red. Again, if B 1 = m, then we use the same argument as above to find that the number of red edges in H 1 which contain a k 1-tuple from B 1 and the remaining vertex from U is at least k + 1 times the number of blue edges in H 1 with a k 1-tuple from B 1. Indeed, we can again apply the inequality from Proposition i. This is because y A,B1 is smaller than the number of all blue edges in H containing a k 1-tuple from B 1, while since we do not remove red edges x A,B1 remains same in both H 1 and H. Now remove the blue edges from H 1 which contain a k 1-tuple from B 1 obtaining a k-graph H. Keep repeating the procedure until it is no longer possible. At some point, we will run out of blue edges in H j for some j 1, and the procedure will terminate prematurely in step. In this case B j < m, A = 0 and U has no blue edges. However, there still may be some blue edges which contain a vertex from B B j V B. Proposition 5.1 ii applied for C = B B j V B implies that the number of such edges is at most z C t k 4k. Let x i A,B and yi A,B be the numbers corresponding to x A,B and y A,B obtained at the end of the procedure applied to H i. Thus, y i A,B 1 k + 1 xi A,B for each 0 i j 1. Let t R and t B denote the number of red and blue edges in H. Observe that t B 0 i j 1 y i A,B + z C 1 k i j 1 x i A,B + t k 4k. 16 Furthermore, since all sets B i are mutually disjoint, each red edge in H containing a k 1- tuple from some B i can be only counted at most k times. Thus, x i A,B k t R i j 1 13

15 Consequently, by 16 and 17, we get and so t k = t R + t B t R + t R 4k 1 4k k k + 1 t R + t k 4k k + 1 k + 1 t k > 1 t k. The conclusion is that there are more red edges than there are blue edges in H. If we reverse the procedure and look for a red path instead of a blue one, we will conclude that there are more blue edges than red edges. Since these two statements contradict each other, the only way to avoid both statements is if a monochromatic path exists. 6 Hypergraphs with bounded degree In this section we prove Theorem.10, which states that hypergraphs with bounded degree can have nonlinear size-ramsey numbers. Proof of Theorem.10. We modify an idea from Rödl and Szemerédi [18]. For simplicity we only present a proof for k = 3, which can easily be generalized to k 3. The hypergraph G will be constructed as the vertex disjoint union of graphs G i each of which is a tree with a path added on its leaves. Next we will describe the details of such construction. Set c = 1 5. We make no effort to optimize c and always assume that n is sufficiently large. Let t = log n log. log log n Consider a binary 3-tree B = V, E on t vertices rooted at vertex z see Figure. Denote by LB the set of all its leafs. Call the edge containing z the root edge. Observe that V B = t = t+1 1 < log n 18 recall that n is large enough and LB = t. Let ϕ by an automorphism of B. Since the root edge e is the unique edge with exactly one vertex of degree 1, ϕz = z. The other two vertices of e are permuted by ϕ. Consequently, ϕ permutes two vertices of every other edge. Hence, it is easy to observe that the order of the automorphism group of B satisfies AutB = t 1 = t 1 < t. Now consider a tight path P of length LB placed on the leaves LB in an arbitrary order. Considering labeled vertices of LB there are clearly LB! such paths. Label them by P i for i = 1,,..., LT!. Let B i be vertex disjoint copies of B and G i = B i P i, where V P i = LB i. 14

16 Figure : Binary 3-tree B on vertices and rooted at vertex z. Let ϕ be an isomorphism between G i and G j. Since the only vertices of degree 4 are on paths P i and P j, ϕp i = P j. Thus, ϕeb i = ϕeg i \ EP i = EG j \ EP j = EB j and so B i and B j are isomorphic. Thus, the number of pairwise non-isomorphic G i s is at least Set LB! AutB t! t t t e t t = t 3t > n. t t n q = V B and let G = G 1 G q, where all G 1,..., G q are pairwise non-isomorphic. We show that G is a desired hypergraph. Clearly, V G n. Furthermore, by 18, we get n V G = q V B V B 1 V B > n log n. Moreover, H = 4 and the independence number of G satisfies αg 8 n Indeed, let I V = V G be an independent set of size α = αg. We estimate the number of edges ei, V \ I between sets I and V \ I. First observe that ei, V \ I G V \ I 4n α. Next, since each triple between I and V \ I intersects one of the partition classes on vertices and δg = 1, δg I ei, V \ I = α. This implies that α 4n α 15

17 and so 19. Now we are ready to finish the proof and show that for any 3-graph with we have H G. Set d = log n 1 5 and EH 1 30 nlog n 1 5 and define V high V H as V high = {v V H : degv d} V low = V H \ V high. Clearly, V high n 10 ; for otherwise, EH > n 10 d 1 3 EH, a contradiction. Recall that G consists of q pairwise non-isomorphic copies of G i. We estimate the number of copies of G i s contained in a sub-hypergraph induced by V low. First fix an edge e in V low [H] and count the number of copies of G i s for which e is a root edge. Since degv d for each v V low, we get that this number is at most 3 d +4+ +t 1 d t d t log n 1 5 log n log log n = n 4 5, where the factor 3 counts the number of choices for the root vertex, the next factors count the number of possible B i s with e as a root, and the last factor counts the number of paths on the set of leafs. Thus, there is an i 0 such that G i0 appears in V low [H] at most n 4 5 EH q < n 4 5 nlog n 1 5 n log n = n 4 5 log n 6 5 times. Denote by F the sub-hypergraph consisting of root edges from all copies of G i0 in V low [H]. Thus, V F 3n 4 5 log n 6 5. Color edges in F together with edges incident to V high blue; otherwise red. Clearly, there is no red copy of G, since there is no red copy of G i0. Moreover, there is no blue copy of G, since every blue sub-hypergraph of order V G has an independent set of size at least V G V high V F > n log n n 10 3n 4 5 log n 6 5 = 9 10 n log n 3n 4 5 log n 6 5, which is strictly bigger than αg cf. 19. References [1] J. Beck, On size Ramsey number of paths, trees, and circuits. I, Journal of Graph Theory , no. 1, [], On size Ramsey number of paths, trees and circuits. II, Mathematics of Ramsey theory, Algorithms Combin., vol. 5, Springer, Berlin, 1990, pp

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