MINIMUM VERTEX DEGREE THRESHOLD FOR LOOSE HAMILTON CYCLES IN 3-UNIFORM HYPERGRAPHS

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1 MINIMUM VERTEX DEGREE THRESHOLD FOR LOOSE HAMILTON CYCLES IN 3-UNIFORM HYPERGRAPHS JIE HAN AND YI ZHAO Abstract. We show that for sufficiently large n, every 3-unifor hypergraph on n vertices with iniu vertex degree at least ( n 1 ( 3 4 n + c, where c = if n 4N and c = 1 if n N \ 4N, contains a loose Hailton cycle. This degree condition is best possible and iproves on the work of Buß, Hàn and Schacht who proved the corresponding asyptotical result. 1. Introduction The study of Hailton cycles is an iportant topic in graph theory. In recent years, researchers have worked on extending the classical theore of Dirac [7] on Hailton cycles to hypergraphs see recent surveys of [3, 0]. Given k, a k-unifor hypergraph (in short, k-graph consists of a vertex set V and an edge set E ( V k, where every edge is a k-eleent subset of V. For 1 l < k, a k-graph is a called an l-cycle if its vertices can be ordered cyclically such that each of its edges consists of k consecutive vertices and every two consecutive edges (in the natural order of the edges share exactly l vertices. (If we allow l = 0, then a 0-cycle is erely a atching and perfect atchings have been intensively studied recently, e.g. [1, 5, 9, 16, 15, 1, 7, 30, 31] In k-graphs, a (k 1-cycle is often called a tight cycle while a 1-cycle is often called a loose cycle. We say that a k-graph contains a Hailton l-cycle if it contains an l-cycle as a spanning subhypergraph. Note that a Hailton l-cycle of a k-graph on n vertices contains exactly n/(k l edges, iplying that k l divides n. Given a k-graph H with a set S of d vertices (where 1 d k 1 we define deg H (S to be the nuber of edges containing S (the subscript H is oitted if it is clear fro the context. The iniu d-degree δ d (H of H is the iniu of deg H (S over all d-vertex sets S in H. We refer to δ 1 (H as the iniu vertex degree and δ k 1 (H the iniu codegree of H Hailton cycles in hypergraphs. Confiring a conjecture of Katona and Kierstead [13], Rödl, Ruciński and Szeerédi [5, 6] showed that for any fixed k, every k-graph H on n vertices with δ k 1 (H n/+o(n contains a tight Hailton cycle. This is best possible up to the o(n ter. With long and involved arguents, the sae authors [8] iproved this to an exact result for k = 3. Loose Hailton cycles were first studied by Kühn and Osthus [18], who proved that every 3-graph on n vertices with δ (H n/4+o(n contains a loose Hailton cycle. Czygrinow and Date: Deceber 14, Matheatics Subject Classification. Priary 05C65, 05C45. Key words and phrases. Hailton cycle, hypergraph, absorbing ethod, regularity lea. The second author is partially supported by NSA grant H and NSF grant DMS

2 JIE HAN AND YI ZHAO Molla [6] recently iproved this to an exact result. The result of Kühn and Osthus [18] was generalized for arbitrary k and arbitrary l < k/ by Hàn and Schacht [10], and independently by Keevash et al. [14] for l = 1 and arbitrary k. Later Kühn, Mycroft, and Osthus [17] obtained an asyptotically sharp bound on codegree for Hailton l-cycles for all l < k. Hence, the proble of finding Hailton l-cycles in k-graphs with large codegree is asyptotically solved. Much less is known under other degree conditions. Recently Rödl and Ruciński [4] gave a sufficient vertex degree condition that guarantees a tight Hailton cycle in 3-graphs. Glebov, Person and Weps [8] gave a nontrivial vertex degree condition for tight Hailton cycles in k-graphs for general k. Neither of these results is best possible see ore discussion in Section 4. Recently Buß, Hàn, and Schacht [] studied the iniu vertex degree that guarantees a loose Hailton cycle in 3-graphs and obtained the following result. Theore 1.1. [, Theore 3] For all γ > 0 there exists an integer n 0 such that the following holds. Suppose H is a 3-graph on n > n 0 with n N and ( ( 7 n δ 1 (H > 16 + γ. Then H contains a loose Hailton cycle. In this paper we iprove Theore 1.1 as follows. Theore 1. (Main Result. There exists an n 1. N such that the following holds. Suppose that H is a 3-graph on n > n 1. with n N and ( ( n 1 3 δ 1 (H 4 n + c, (1.1 where c = if n 4N and c = 1 otherwise. Then H contains a loose Hailton cycle. The following construction shows that Theore 1. is best possible. It is slightly stronger than [, Fact 4]. Proposition 1.3. For every n N there exists a 3-graph on n vertices with iniu vertex degree ( ( n n +c 1, where c is defined as in Theore 1., and which contains no loose Hailton cycle. Proof. Let H 1 = (V 1, E 1 be the 3-graph on n N \ 4N vertices such that V 1 = A B 1 with A = n 4 1 and = 3n 4 + 1, and E 1 consists of all triples intersecting A. Note that δ 1 (H 1 = ( ( n 1 3n 4. Suppose that H1 contains a loose Hailton cycle C. There are n/ edges in C and every vertex in A is contained in at ost two edges in C. Since A = n, there is at least one edge of C whose vertices are copletely fro B. This is a contradiction since B is independent. So H 1 contains no loose Hailton cycle. Let H = (V, E be a 3-graph on n 4N vertices such that V = A B with A = n 4 1 and = 3 4 n + 1, and E consists of all triples intersecting A and those containing both b 1 and b, where b 1, b are two fixed vertices in B. Then δ 1 (H = ( ( n n + 1. Suppose that H contains a loose Hailton cycle C. There are n/ edges in C and every vertex in A is contained in at ost two edges 1 Throughout the paper, we write A B for A B when sets A, B are disjoint.

3 VERTEX DEGREE THRESHOLD FOR LOOSE HAMILTON CYCLES IN 3-GRAPHS 3 in C. Thus, there are at least two edges of C whose vertices are copletely fro B. But due to the construction, every two edges in B share two vertices so they cannot both appear in one loose cycle. This contradiction shows that H contains no loose Hailton cycle. As a typical approach of obtaining exact results, we distinguish the extreal case fro the nonextreal case and solve the separately. Definition 1.4. Given β > 0, a 3-graph H on n vertices is called β-extreal if there is a set B V (H, such that = 3n/4 and e(b βn 3. Theore 1.5 (Extreal Case. There exist β > 0 and n 1.5 N such that the following holds. Let n > n 1.5 be an even integer. Suppose that H is a 3-graph on n vertices satisfying (1.1. If H is β-extreal, then H contains a loose Hailton cycle. Theore 1.6 (Nonextreal Case. For any β > 0, there exist γ > 0 and n 1.6 N such that the following holds. Let n > n 1.6 be an even integer. Suppose that H is a 3-graph on n vertices satisfying δ 1 (H ( 7 16 γ ( n. If H is not β-extreal, then H contains a loose Hailton cycle. Theore 1. follows fro Theores 1.5 and 1.6 iediately by choosing β fro Theore 1.5 and letting n 1. = ax{n 1.5, n 1.6 }. Let us discuss our proof ideas here. The proof of Theore 1.5 is soewhat standard (though non-trivial. The proof of Theoreo 1.6 follows the approach in the previous work [, 10, 17, 5, 6, 8]. Roughly speaking, we use the absorbing ethod initiated by Rödl, Ruciński and Szeerédi, which reduces the task of finding a loose Hailton cycle to finding constantly any vertex-disjoint loose paths that covers alost all the vertices of the 3-graph. More precisely, we first apply the Absorbing Lea (Lea.1 and obtain a (short absorbing path P 0 which can absorb any saller proportion of vertices. Second, apply the Reservoir Lea (Lea. and find a sall reservoir set R whose vertices ay be used to connect any constant nuber of loose paths to a loose cycle. Third, apply the Path-tiling Lea (Lea.3 in the reaining 3-graph and obtain constantly any vertexdisjoint loose paths covering alost all the vertices. Fourth, connect these paths (including P 0 together by the reservoir R and get a loose cycle C. Finally we absorb the vertices not in V (C to P 0 and obtain the desired loose Hailton cycle. The Absorbing Lea and the Reservoir Lea are not very difficult and already proven in []. Thus the ain step is to prove the Path-tiling Lea, under the assuption δ 1 (H ( 7 16 γ ( n and that H is not β-extreal (in contrast, δ 1 (H ( γ( n is assued in []. As shown in [, 10], after applying the (weak Regularity Lea, it suffices to prove that the cluster 3-graph K can be tiled alost perfectly by soe particular 3-graph. For exaple, the 3-graph M given in [] has the vertex set [8] = {1,,..., 8} and edges 13, 345, 456, 678. Since it is hard to find an M-tiling directly, the authors of [] found a fractional M-tiling instead and converted it to an (integer M-tiling by applying the Regularity Lea again. In this paper we consider a uch sipler 3-graph Y with vertex set [4] and edges 13, 34, and obtain an alost perfect Y-tiling in K directly. Interestingly, Y-tiling was studied (via the codegree condition in the very first paper on loose Hailton cycles [18]. Throughout the paper, we often represent a set {v1, v,..., v k } as v 1 v v k.

4 4 JIE HAN AND YI ZHAO Coparing with the first exact result on Hailton cycles in hypergraphs [8], our proof is uch shorter because the Absorbing and Reservoir Leas in [8] are uch harder to prove. The rest of the paper is organized as follows: we prove Theore 1.6 in Section and Theore 1.5 in Section 3, and give concluding rearks in Section Notations. Given a vertex v and disjoint vertex sets S, T in a 3-graph H, we denote by deg H (v, S the nuber of edges that contain v and two vertices fro S, and by deg H (v, ST the nuber of edges that contain v, one vertex fro S and one vertex fro T. Furtherore, let deg H (v, S = ( S degh (v, S and deg H (v, ST = S T deg H (v, ST. Given not necessarily disjoint sets X, Y, Z V (H, we define E H (XY Z = {xyz E(H : x X, y Y, z Z}, { ( V (H E H (XY Z = xyz 3 \ E(H : x X, y Y, z Z e H (XY Z = E H (XY Z, and e H (XY Z = E H (XY Z. The subscript H is often oitted when it is clear fro the context. A loose path P = v 1 v v k+1 is a 3-graph on {v 1, v,..., v k+1 } with edges v i 1 v i v i+1 for all i [k]. The vertices v 1 and v k+1 are called the ends of P.. Proof of Theore 1.6 In this section we prove Theore 1.6 by following the sae approach as in []..1. Auxiliary leas and Proof of Theore 1.6. For convenience, we rephrase the Absorbing Lea [, Lea 7] as follows. 3 Lea.1 (Absorbing Lea. For any 0 < γ there exists an integer n.1 such that the following holds. Let H be a 3-graph on n > n.1 vertices with δ 1 (H 3( 13 n. Then there is a loose path P with V (P γ1 n such that for every subset U V \V (P with U γ1n 3 and U N there exists a loose path Q with V (Q = V (P U such that P and Q have the sae ends. We also need the Reservoir Lea [, Lea 6]. Lea. (Reservoir Lea. For any 0 < γ < 1/4 there exists an integer n. such that for every 3-graph H on n > n. vertices satisfying ( n δ 1 (H (1/4 + γ, there is a set R of size at ost γ n with the following property: for every k γn/1 3 utually disjoint pairs {a i, b i } i [k] of vertices fro V (H there are 3k vertices u i, v i, w i, i [k] fro R such that a i u i v i, v i w i b i H for all i [k]. The ain step in our proof of Theore 1.6 is the following lea, which is stronger than[, Lea 10]. }, 3 Lea 7 in [] assues that δ1 (H ( γ( n and returns V (P γ 7 n with U ( γ n. We siply take their γ7 as our γ 1 and thus γ

5 VERTEX DEGREE THRESHOLD FOR LOOSE HAMILTON CYCLES IN 3-GRAPHS 5 Lea.3 (Path-tiling lea. For any 0 < γ 3, α < 1 there exist integers p and n.3 such that the following holds for n > n.3. Suppose H is a 3-graph on n vertices with iniu vertex degree ( ( 7 n δ 1 (H 16 γ 3, then there are at ost p vertex disjoint loose paths in H that together cover all but at ost αn vertices of H unless H is 050γ 3 -extreal. Proof of Theore 1.6. Given β > 0, let γ = in{ β 4101, }. We choose n 1.6 = ax{n.1, n., n.3, 19(p + 1/(γ/3 9 }, where p is the constant returned fro Lea.3 with γ 3 = γ and α = (γ/3 3 /. Let n > n 1.6 be an even integer. Suppose that H = (V, E is a 3-graph on n vertices with δ 1 (H ( 7 16 γ ( n. Since γ > 3, we can apply Lea.1 with γ 1 = γ/3 and obtain an absorbing path P 0 with ends a 0, b 0. We next apply Lea. with γ = (γ/3 3 / to H[(V \ V (P 0 {a 0, b 0 }] and obtain a reservoir R. Let V = V \(V (P 0 R and n = V. Note that n n γ 1 n + γ n < γn/. The induced subhypergraph H = H[V ] satisfies δ 1 (H ( ( 7 n 16 γ γ ( 7 n (n > 16 γ ( n Applying Lea.3 to H with γ 3 = γ and α = (γ/3 3 /, we obtain at ost p vertex disjoint loose paths that cover all but at ost αn vertices of H, unless H is 050γ 3 -extreal. In the latter case, there exists B V such that B = 3 4 n and e(b 4100γ(n 3. Then we add 3 4 n 3 4 n < γn/ arbitrary vertices fro V \ B to B to get a vertex set B such that = 3 4n and e(b 4100γ(n 3 + γn ( n 1 < 4101γn 3 βn 3, which eans that H is β-extreal, a contradiction. In the forer case, denote these loose paths by {P i } i [p ] for soe p p, and their ends by {a i, b i } i [p ]. The choice of n 1.6 guarantees that p + 1 p + 1 γn/4. 3 We can thus connect {a i, b i+1 } 0 i p 1 {a p, b 0 } by using vertices fro R obtaining a loose cycle C. Since V \ C R + αn γ n + γ n γ1n, 3 we can use P 0 to absorb all unused vertices in R and uncovered vertices in V. The rest of this section is devoted to the proof of Lea.3... Proof of Lea.3. Following the approach in [], we will use the weak regularity lea which is a straightforward extension of Szeerédi s regularity lea for graphs [9]. Below we only state this lea for 3-graphs. Let H = (V, E be a 3-graph and let A 1, A, A 3 be utually disjoint non-epty subsets of V. We define e(a 1, A, A 3 to be the nuber of edges with one vertex in each A i, i [3], and the density of H with respect to (A 1, A, A 3 as d(a 1, A, A 3 = e(a 1, A, A 3 A 1 A A 3. Given ɛ > 0, the triple (V 1, V, V 3 of utually disjoint subsets V 1, V, V 3 V is called ɛ-regular if d(a 1, A, A 3 d(v 1, V, V 3 ɛ.

6 6 JIE HAN AND YI ZHAO for all triple of subsets of A i V i, i [3], satisfying A i ɛ V i. We say (V 1, V, V 3 is (ɛ, d-regular if it is ɛ-regular and d(v 1, V, V 3 d for soe d 0. It is iediate fro the definition that in an (ɛ, d-regular triple (V 1, V, V 3, if V i V i has size V i c V i for soe c ɛ, then (V 1, V, V 3 is (ax{ɛ/c, ɛ}, d ɛ-regular. Theore.4. [, Theore 14] For any t 0 0 and ɛ > 0, there exist T 0 and n 0 so that for every 3-graph H = (V, E on n > n 0 vertices, there exists a partition V = V 0 V 1 V t such that (i t 0 t T 0, (ii V 1 = V = = V t and V 0 ɛn, (iii for all but at ost ɛ ( t 3 sets i1 i i 3 ( [t] 3, the triple (Vi1, V i, V i3 is ɛ-regular. A partition as given in Theore.4 is called an (ɛ, t-regular partition of H. For an (ɛ, t-regular partition of H and d 0 we refer to Q = (V i i [t] as the faily of clusters and define the cluster hypergraph K = K(ɛ, d, Q with vertex set [t] and i 1 i i 3 ( [t] 3 is an edge if and only if (Vi1, V i, V i3 is (ɛ, d-regular. The following corollary shows that the cluster hypergraph inherits the iniu degree of the original hypergraph. Its proof is the sae as that of [, Proposition 15] after we replace 7/16 + γ by c (we thus oit the proof. Corollary.5. For c > d > ɛ > 0 and t 0 0 there exist T 0 and n 0 such that the following holds. Suppose H is a 3-graph on n > n 0 vertices which has iniu vertex degree δ 1 (H c ( n. Then there exists an (ɛ, t-regular partition Q with t 0 < t < T 0 such that the cluster hypergraph K = K(ɛ, d, Q has iniu vertex degree δ 1 (K (c ɛ d ( t. In 3-graphs, a loose path is 3-partite with partition sizes about,, for soe integer. Proposition.6 below shows that every regular triple with partition sizes,, contains an alost spanning loose path as a subhypergraph. In contrast, [, Proposition 5] (ore generally [10, Lea 0] shows that every regular triple with partition sizes 3, 3, contains constant any vertex disjoint loose paths. The proof of Proposition.6 uses the standard approach of handling regularity. d Proposition.6. Fix any ɛ > 0, d > ɛ, and an integer ɛ(d ɛ. Suppose that V (H = V 1 V V 3 and (V 1, V, V 3 is (ɛ, d-regular with V i = for i = 1, 3 and V =. Then there is a loose path P oitting at ost 8ɛ/d + 3 vertices of H. Proof. We will greedily construct the loose path P = v 1 v v k+1 such that v i V, v 4i+1 V 1 and v 4i+3 V 3 until V i \V (P < ɛ d V i for soe i [3]. For j [3], let Uj 0 = V j and Uj i = V j \ {v 1,..., v i 1 } for i [k]. In addition, we require that for i = 0,..., k, deg(v i+1, UU i r i (d ɛ U U i r, i (.1 where r i 1 od 4. We proceed by induction on i. First we pick a vertex v 1 V 1 such that deg(v 1, V V 3 (d ɛ V V 3 (thus (.1 holds for i = 0. By regularity, all but at ost ɛ V 1 vertices can be chosen as v 1. Suppose that we have selected v 1,..., v i 1. Without loss of generality, assue that v i 1 V 1. Our goal is to choose v i U i, v i+1 U i 3 such that (i v i 1 v i v i+1 E(H, (ii deg(v i+1, U i 1U i (d ɛ U i 1 U i.

7 VERTEX DEGREE THRESHOLD FOR LOOSE HAMILTON CYCLES IN 3-GRAPHS 7 In fact, the induction hypothesis iplies that deg(v i 1, U i 1 U3 i 1 (d ɛ U i 1 U3 i 1. Since U i = U i 1 \ {v i } and U3 i = U3 i 1, we have deg(v i 1, U i U i 3 (d ɛ U i 1 U i 1 3 U i 1 3 = ((d ɛ U i 1 1 U i 1 3. By regularity, at ost ɛ V 3 vertices in V 3 does not satisfy (ii. So, at least deg(v i 1, U i U i 3 ɛ V 3 U i 1 ((d ɛ U i 1 1 U i 1 3 ɛ V 3 U i 1 (. pairs of vertices can be chosen as v i, v i+1. Since U3 i 1 ɛ d V 3 and U i 1 ɛ d V 4 d ɛ (using d ɛ(d ɛ, the right side of (. is at least ( ɛ ( ɛ (d ɛ U i 1 1 d V 3 ɛ V 3 U i 1 = (d ɛ U i 1 d V 3 > 0, thus the selection of v i, v i+1 satisfying (i and (ii is guaranteed. To calculate the nuber of the vertices oitted by P = v 1 v v k+1, note that V 1 V (P = k+1, V V (P = k, and V 3 V (P = k+1. Our greedy construction of P stops as soon as V i \ V (P < ɛ d V i for soe i [3]. As V 1 = V 3 = = V /, one of the following three inequalities holds: k + 1 < ɛ d, ɛ k + 1 k < d, < ɛ d. Thus we always have k+1 < ɛ k+ d, which iplies that > ( 1 ɛ d or k > ( 1 ɛ d. Consequently, ( ( V (H \ V (P = 4 (k + 1 < ɛ = 8ɛ d d + 3. Let Y be the 3-graph on the vertex set [4] with edges 13, 34 (the unique 3- graph with four vertices and two edges. The following lea is the ain step in our proof of Lea.3. In general, given two (hypergraphs F and G, an F-tiling is a sub(hypergraph of G that consists of vertex disjoint copies of F. The F-tiling is perfect if it is a spanning sub(hypergraph of G. Lea.7 (Y-tiling Lea. For any γ > 0, there exists an integer n.7 such that the following holds. Suppose H is a 3-graph on n > n.7 vertices with ( ( 7 n δ 1 (H 16 γ, then there is a Y-tiling covering all but at ost 19 /γ vertices of H unless H is 10 γ-extreal. Now we are ready to prove Lea.3 using the sae approach as in []. Proof of Lea.3. Given 0 < γ 3, α < 1, let n.3 = ax{n 0, 4T 0 /ɛ} and p = T 0 /, where T 0 and n 0 are the constants returned fro Corollary.5 with c = 7 16 γ 3, d = γ 3 /, ɛ = αd 8+α, and t 0 = ax{n.7, 0 γ }. 3α Suppose that H is a 3-graph on n > n.3 vertices with δ 1 (H ( 7 16 γ 3 ( n. By applying Corollary.5 with the constants chosen above, we obtain an (ɛ, t-regular partition Q. The cluster hypergraph K = K(ɛ, d, Q satisfies δ 1 (K ( 7 16 γ 3 ( t. Let be the size of each cluster except V 0, then (1 ɛ n t n t. By Lea.7, either K is 10 (γ 3 -extreal, or there is a Y-tiling Y of K that covers all but at ost 19 /(γ 3 vertices of K. In the first case, there exists a set B V (K such

8 8 JIE HAN AND YI ZHAO that = 3t 4 and e(b 11 γ 3 t 3. Let B V (H be the union of the clusters in B. By regularity, e(b e(b 3 + ( t d 3 + ɛ 3 ( t ( n, where the right-hand side bounds the nuber of edges fro regular triples with high density, edges fro regular triples with low density, edges fro irregular triples and edges that are fro at ost two clusters. Since n t, ɛ < d < γ 3, and t < t 0 < γ 3, we get ( e(b 11 γ 3 t 3 n 3 t (n 3 t (n + d( + ɛ( t 3 t 3 t Note that B = 3t 4 hand, B = ( 3 n/t + n < 049γ 3 n 3. 3t 4 n t = 3n 4 iplies that B 3n 4. On the other ( 3t 3t (1 ɛ n ( 3t n t 4 ɛt t = 3n 4 ɛn, by adding at ost ɛn vertices fro V \ B to B, we get a set B V (H of size exactly 3n/4, with e(b e(b + ɛn n < 050γ 3 n 3. Hence H is 050γ 3 - extreal. In the second case, the union of the clusters covered by Y contains all but at ost 19 γ 3 + V 0 αn/4 + ɛn < 3αn/8 vertices (here we use t 0 γ 3α. We will apply Proposition.6 to each eber Y Y. Suppose that Y has the vertex set [4] with edges 13, 34. For i [4], let V i denote the corresponding cluster in H. We split V i, i =, 3, into two disjoint sets Vi 1 and Vi of equal sizes. Then the triples (V 1, V 1, V3 1 and (V 4, V, V3 are (ɛ, d ɛ-regular and of sizes,,. Applying Proposition.6 to these two triples with =, we find a loose path in each triple covering all but at ost 8(ɛ d ɛ + 3 = α + 3 vertices (here we need ɛ = αd 8+α. Since Y t/4, we obtain a path tiling that consists of at ost t/4 T 0 / = p paths and covers all but at ost (α + 3 t 4 + 3α 8 n α n + 3t + 3α 8 n < αn vertices. This copletes the proof..3. Proof of Y-tiling Lea (Lea.7. Fact.8. Let H be a 3-graph on vertices which contains no copy of Y, then e(h 1 3(. Proof. Since there is no copy of Y, then given any u, v V (H, we have that deg(uv 1, which iplies e(h 1 3( 1 = 1 3(. Proof of Lea.7. Fix γ > 0 and let n N be sufficiently large. Let H be a 3-graph on n vertices that satisfies δ 1 (H ( 7 16 γ( n. Fix a largest Y-tiling Y = {Y 1,..., Y } and let V i = V (Y i for i []. Let V = i [] V i and U = V (H \ V. Assue that U > 19 /γ otherwise we are done. Our goal is to find a set C of vertices in V of size at ost n/4 that covers alost all the edges, which iplies that H is extreal.

9 VERTEX DEGREE THRESHOLD FOR LOOSE HAMILTON CYCLES IN 3-GRAPHS 9 Let A i be the ( set of all edges with exactly i vertices in V, for i = 0, 1,, 3. Note that A 0 1 U 3 by Fact.8. We ay assue that U < 3 4n and consequently > n 16. (.3 Indeed, if U 3 4 n, then taking U U of size 3 4 n, we get that e(u e(u ( 1 U n < γn 3. Thus H is γ-extreal and we are done. Clai.9. A 1 ( U + 1 U. Proof. Let D be the set of vertices v V such that deg(v, U 4 U. First observe that every Y i Y contains at ost one vertex in D. Suppose instead, two vertices x, y V i are both in D. Since deg(x, U 4 U > U /, the link graph 4 of x on U contains a path u 1 u u 3 of length two. The link graph of y on U \{u 1, u, u 3 } has size at least 4 U 3 U > U /, so it also contains a path of length two, with vertices denoted by u 4, u 5, u 6. Note that xu 1 u u 3 and yu 4 u 5 u 6 span two vertex disjoint copies of Y. Replacing Y i in Y with the creates a larger Y-tiling, contradicting the axiality of Y. So we conclude that D. Consequently, A 1 D ( U + V \ D 4 U ( U U ( U + 1 U. Fix u U, i j [], we denote by L i,j (u the link graph of u induced on (V i, V j, naely the bipartite link graph of u between V i and V j. Let T 6 be the set of all triples uij, u U, i, j [] such that e(l i,j (u 6. Let T7 1 be the set of all triples uij, u U, i, j [] such that 5 e(l i,j (u = 7 and L i,j (u has a vertex cover of two vertices with one fro V i and the other fro V j. Let T 7 be the set of all triples uij, u U, i, j [] such that e(l i,j (u 7 and L i,j (u has a vertex cover of two vertices both fro V i or V j. Let T 7 3 be the set of all triples uij, u U, i, j [] such that e(l i,j (u 7 and L i,j (u contains a atching of size three. By the König Egervary theore, a bipartite graph either contains a atching of size three or a vertex cover of size two. Thus T 6, T7 1, T 7, T 7 3 for a partition of U ( []. In order to bound the sizes of T 6, T7 1, T 7, T 7 3, we need the following fact. Fact.10. (i H does not contain i j [] and six vertices u 1,..., u 6 U such that u 1,..., u 6 have the sae (labeled link graph on (V i, V j and u 1 ij T 7 3. (ii H does not contain distinct i, j, k [] and eight vertices u 1,..., u 8 U such that the following holds. First, u 1,..., u 4 share the sae link graph on (V i, V j, and u 5,..., u 8 share the sae link graph on (V i, V k. Second, u 1 ij T 7 with the vertex cover in V j and u 5 ik T 7 with the vertex cover in V k. Proof. To see Part (i, since there is a atching of size three in the (sae link graph of u 1,..., u 6, say, a 1 b 1, a b, a 3 b 3, then u 1 u a 1 b 1, u 3 u 4 a b and u 5 u 6 a 3 b 3 4 Given a 3-graph H with a vertex v, the link graph of v has the vertex set V (H \ {v} and edge set {e \ {v} : e E(H}. 5 We could have e(li,j (u 7 here; but since L i,j (u contains a vertex cover with one vertex fro V i and one vertex fro V j, we ust have e(l i,j (u = 7.

10 10 JIE HAN AND YI ZHAO Figure 1. The bipartite graph L i,j (u when the triple is in T7 1 or, where the dotted line could be present or not. T 7 span three copies of Y. Replacing Y i, Y j by the gives a Y-tiling larger than Y, a contradiction. To see Part (ii, assue that V i = {a, b, c, d}. Suppose that the vertex cover of L i,j (u 1 is {x 1, y 1 } V j and the vertex cover of L i,k (u 5 is {x, y } V k. Since u 1 ij T 7, at ost one pair fro {x 1, y 1 } {a, b} is not in L i,j (u 1. Analogously at ost one pair fro {x, y } {c, d} is not in L i,k (u 5. Thus, without loss of generality, we ay assue that x 1 a, y 1 b L i,j (u 1 and x c, y d L i,k (u 5. Since u 1,..., u 4 share the sae link graph on (V i, V j, u 1 u x 1 a, u 3 u 4 y 1 b span two copies of Y. Siilarly, u 5 u 6 x c and u 7 u 8 y d span two copies of Y. Replacing Y i, Y j, Y k by these four copies of Y gives a Y-tiling larger than Y, a contradiction. We now show that ost triples uij, u U, i, j [] are in T 1 7. Clai.11. (i T 7 3 ( 16 5, (ii T 7 756( + U, (iii T 6 γ ( n ( U +, (iv T7 1 ( U γn U. Proof. To see Part (i, by Fact.10 (i, given i, j [] and a bipartite graph on (V i, V j containing a atching of size three, at ost five vertices in U can share this link graph on (V i, V j. Since there are 16 (labeled bipartite graphs on (V i, V j, we get that T 7 3 ( To see Part (ii, let D denote the digraph on [] such that (i, j E(D if and only if at least eight vertices u 1,..., u 8 U share the sae link graph on (V i, V j such that u 1 ij T 7, and the vertex cover is in V i. We clai that deg D (i 1 for every i [] and consequently e(d. Suppose instead, there are i, j, k [] such that (j, i, (k, i E(D, then eight vertices of U share the sae link graph on (V i, V j, and (not necessarily different eight vertices of U share the sae link graph on (V i, V k. Thus we can pick four distinct vertices for each of (V i, V j and (V i, V k and obtain a structure forbidden by Fact.10 (ii, a contradiction. Note that there are (4 ( = 108 (labeled bipartite graphs on (Vi, V j with at least seven edges and a vertex cover of two vertices both fro V i or V j. Furtherore, fixing one of these bipartite graphs, if (i, j / E(D and (j, i / E(D, then, by the definition of D, at ost seven vertices in U share this link graph. Hence ( ( T U = U.

11 VERTEX DEGREE THRESHOLD FOR LOOSE HAMILTON CYCLES IN 3-GRAPHS 11 To see Part (iii, recall that A i is the set of all edges of H with exactly i vertices in V. Then ( 4 A 6 T T T T U. Together with T 6 + T7 1 + T 7 + T 7 3 = ( U, we get, ( A 7 U T 6 + T T U ( ( 7 U T 6 + ( ( < 7 U T 6 + ( U by Parts (i, (ii + 7 U. (.4 We know that u U deg(u = 3 A ( 0 + A 1 + A. Thus, by A 0 1 U 3, Clai.9 and (.4, we have ( ( ( ( U U deg(u U + 7 U T U u U ( ( ( U = + U + 30 U + 7 U T 6 + < 7 ( U U U + 7 ( ( 4 U T 6 + as U > = 7 ( ( n U T 6 +, (.5 16 where the last inequality is due to ( ( U + 4 U + 4 ( = U +4 ( = n. On the other hand, δ 1 (H ( 7 16 γ( n iplies that u U deg(u ( 7 16 γ ( n U. Together with (.5, this gives the desired bound for T 6. To see Part (iv, note that (.3 iplies that U < 3 4 n < = 1. By Parts (i (iii, we have ( (( ( ( n T7 1 U ( U γ U ( ( ( n U γ U 3 as U < 1 ( ( ( n n U γ U 19 as < n 4 ( > U γn U as U > 19 /γ. For a triple uij T 1 7, we call v 1 V i and v V j centers for u if {v 1, v } is the vertex cover of L i,j (u. Define G as the graph on the vertex set V such that two vertices v 1, v V are adjacent if and only if there are at least 16 vertices u U such that v 1, v are centers for u. Fact.1. For every i [], at ost one vertex v V i satisfies deg G (v > 0.

12 1 JIE HAN AND YI ZHAO Proof. Suppose to the contrary, soe V i = {a, b, c, d} satisfies deg G (a, deg G (b > 0. Let a N G (a, b N G (b and assue that a V j, b V k (it is possible to have a = b. Pick x V j \{a, b } and y V k \{a, b, x}. By the definition of G, we can find u 1,..., u 4, u 1,..., u 4 U such that a, a are centers for u l and b, b are centers for u l for l = 1,..., 4. This gives three copies of Y on axu 1 u, a cu 3 u 4, byu 1u iediately. So if j = k, then replacing Y i, Y j by the in Y gives a larger Y- tiling, a contradiction. If j k, then we have a b, and we get one ore copy of Y on b du 3u 4. Replacing Y i, Y j, Y k by these four copies of Y in Y gives a larger Y-tiling, a contradiction. Let C be the set of vertices v V such that deg G (v 7 and deg G (v for soe v N G (v, where N G (v denotes the neighborhood of v in G. Clai.13. (1 11 γ C. Proof. The upper bound follows fro Fact.1 iediately. To see the lower bound, we first show that ( e(g (1 10 γ. (.6 To see this, let M be the set of pairs i, j ( [] such that there are at ost 40 vertices u U satisfying that uij T7 1. By Clai.11 (iv, the nuber of triples uij T7 1 (u U, i j [] is at ost γn U. Thus ( M γn U U 40 γn U 3 U = 3γn < 3γ(16 < 10 γ. where the second last inequality follows fro (.3. Fix a pair i, j ( [] \ M. There are at least 41 = vertices u U satisfying that uij T7 1. Since V i V j contains 16 pairs of vertices, by the pigeonhole principle, soe pair of vertices v 1 V i, v V j are centers for at least 16 vertices u U, naely, v 1 v G. Thus (.6 follows. By Fact.1, there are at ost vertices with positive degree in G. For convenience, define V V as an arbitrary set of vertices that contains all the vertices with positive degree in G. Furtherore, for any integer t <, let D t V denote the set of vertices v such that deg G (v t. Let F (V \ D 1 denote the set of vertices v such that N G (v D 1. We have e(g t D t + ( 1( D t = ( 1 ( t 1 D t. Together with (.6, it gives D t 10 γ ( 1 t 1. By definition, each vertex v F satisfies deg G (v, and its neighborhood is contained in D 1 (thus the vertices in F have disjoint neighborhoods. This iplies that F D 1 /. Recall that C = V \ (D 6 F. Since D 6 and D are not necessarily disjoint, C D 6 F 10 ( 1 γ 10 ( 1 γ 7 ( (1 11 γ. as claied. Let I C be the set of all i [] such that V i C. Fact.1 and Clai.13 together iply that I C = C (1 11 γ. Let A = ( i I C V i \ C U. Clai.14. H[A] contains no copy of Y, thus e(a 1 3( n.

13 VERTEX DEGREE THRESHOLD FOR LOOSE HAMILTON CYCLES IN 3-GRAPHS 13 Proof. The first half of the clai iplies the second half by Fact.8. Suppose instead, H[A] contains a copy of Y, denoted by Y 0, on V 0. Since H[U] contains no copy of Y, V 0 ust intersect soe V i with i I C. Without loss of generality, suppose that V 1,..., V j contain the vertices of V 0 \ U for soe 1 j 4 (recall that V i = V (Y i for i []. Here we separate two cases. Case 1. For any i [j], V i V 0. For i [j], let {c i } = V i C, and suppose that d i V i \ (V 0 {c i }. For each i [j], since deg G (c i 7, we can pick distinct v i N G (c i \ (V 1 V j. By Fact.1, v 1,..., v j are contained in different ebers of Y (also different fro Y 1,..., Y j. Let v 1,..., v j be arbitrary vertices in these ebers of Y, respectively, which are different fro v 1,..., v j. For every i [j], since c i, v i are centers for at least 16 vertices of U, we find a set of four vertices u 1 i,..., u4 i U \ V 0 disjoint fro the previous ones such that c i, v i are centers for the. This is possible because V 0 U 4 j and the nuber of available vertices in U is thus at least 16 (4 j = 1 + j 4j. Note that for i [j], c i v i u1 i u i, d iv i u 3 i u4 i span two copies of Y. Together with Y 0, this gives j+1 copies of Y while using vertices fro j ebers of Y, contradicting the axiality of Y. Case. There exists i 0 [j], such that V i0 V 0 = 3. Note that j = 1 or in this case. Without loss of generality, assue that V 1 V 0 = 3. First assue that j = 1 (then V 0 U = 1. Let {c 1 } = V 1 C. By the definition of C, there exists c N G (c 1 such that deg G (c. Let c 3 c 1 be a neighbor of c in G. Assue that Y i, Y i3 Y contains c, c 3, respectively. By the definition of G, we can find u 1,..., u 6 U \ V 0 such that c 1, c are centers for u 1, u, and c, c 3 are centers for u 3, u 4, u 5, u 6. Thus, c 1 w 1 u 1 u, c w 3 u 3 u 4, c 3 w u 5 u 6 span three copies of Y, where w 1, w are two vertices in V i \{c } and w 3 V i3 \{c 3 }. Together with Y 0, it gives four copies of Y while using vertices fro three ebers of Y, contradicting the axiality of Y. Now assue that j =, that is, V 0 V = 1. We pick c, c 3, u 1,..., u 6 in the sae way as in the j = 1 case. If c V, then this gives four copies of Y by using vertices fro three ebers of Y, a contradiction. Otherwise, let {c 4 } = V C and pick c 5 N G (c 4 \ {c 1, c, c 3 } (this is possible because deg G (c 4 7. Suppose that Y i5 contains c 5. We pick four new vertices u 7,..., u 10 U for who c 4, c 5 are centers. Thus, we can for two copies of Y by using vertices fro Y, Y i5 and u 7,..., u 10. Together with the four copies of Y given in the previous case, we obtain six copies of Y while using vertices fro five ebers of Y, a contradiction. Note that the edges not incident to C are either contained in A or incident to soe V i, i / I C. By Clai.14, C is incident to all but at ost ( n 1 e(a γ < 1 ( n + 10 γ(4n 3 ( 1 < 10 γn 10 γ + 4 < 10 γn 3, edges, where the last inequality holds because U > 1 10 γ. Since C n/4, we can pick a set B V \ C of order 3 4 n. Then e(b < 10 γn 3, which iplies that H is 10 γ-extreal.

14 14 JIE HAN AND YI ZHAO In Clai.14 we proved that H[A] contains no copy of Y, where, by Clai.13, A = n 3( C n n γ (1 11 γ 3 4 n. We suarize this in a lea and will use it in our forthcoing paper [1]. Lea.15. For any γ > 0, there exists an integer n 0 such that the following holds. Suppose H is a 3-graph on n > n 0 vertices with ( ( 7 n δ 1 (H 16 γ, then there is a Y-tiling covering all but at ost 19 /γ vertices of H unless H contains a set of order at least (1 11 γ 3 4n that contains no copy of Y. 3. The Extreal Theore In this section we prove Theore 1.5. Let n be sufficiently large and H be a 3-graph on n vertices satisfying (1.1. Assue that H is β-extreal, naely, there is a set B V (H, such that = 3 4 n and e(b βn3. For the convenience of later calculations, we let ɛ 0 = 18β and derive that e(b < ɛ 0 ( 3. (3.1 Let us outline our proof here. We define two disjoint sets A, B V (H such that A consists of the vertices with high degree in B and B consists of the vertices with low degree in B. We will show that A A and B B (Clai 3.. To illustrate our proof ideas, suppose that we are in an ideal case with n 4N, A = A, and B = B. In this case we arbitrarily partition B into three sets B 1, B and B 3 of equal size, and find a labelling B 1 = {b 1,..., b n/4 }, B = {b 1,..., b n/4 } and B 3 = {b 1,..., b n/4 } such that Γ has large iniu degree, where Γ denotes the bipartite graph on (A, [n/4] in which xi Γ for x A and i [n/4] if and only if b i b i, b i b i N H(x. It is easy to find a Hailton cycle in Γ, which gives rise to a loose Hailton cycle in H. In our actual proof, we first build a short loose path P that covers all the vertices of V 0 := V (H\(A B (and soe vertices fro A and B such that A \ V (P / B \ V (P is at least 1/3 (Clai 3.6. This is possible because of the iniu degree condition (1.1 and the fact that V 0 is sall. We next extend P to a loose path Q such that A \ V (Q / B \ V (Q is about 1/3 and finally find a loose Hailton path on V (H \ Q by following the approach for the ideal case (Lea Classifying vertices. Let ɛ 1 = 8 ɛ 0 and A = V (H \ B.. Assue that the partition A and B satisfies that = 3 4n and (3.1. In addition, assue that e(b is the sallest aong all the partitions satisfying these conditions. We now define { ( } A := v V deg(v, B (1 ɛ 1, { ( } B := v V deg(v, B ɛ 1, V 0 = V \ (A B. Clai 3.1. A B iplies that B B, and B A iplies that A A.

15 VERTEX DEGREE THRESHOLD FOR LOOSE HAMILTON CYCLES IN 3-GRAPHS 15 Proof. First, assue ( that A B. Then there is soe u A which satisfies that deg(u, B ɛ ( 1. If there exists soe v B \ B, naely, deg(v, B > ɛ 1, then we can switch u and v and for a new partition A B such that B = and e(b < e(b, which contradicts the iniality of e(b. Second, assue that B A. Then soe u B satisfies that deg(u, B (1 ɛ 1 (. Siilarly, by the iniality of e(b, we get that for any vertex v A, deg(v, B (1 ɛ 1 (, which iplies that A A. Clai 3.. { A \ A, B \ B, A \ A, B \ B } ɛ1 64 and V 0 ɛ1 3. Proof. First assue that B \ B > ɛ1 64. By the definition of B and the assuption ɛ 1 = 8 ɛ 0, we get that e(b > 1 3 ɛ 1 ( ɛ1 64 > ɛ 1 64 ( ( = ɛ 0, 3 3 which contradicts (3.1. Second, assue that A \ A > ɛ1 64. Then by the definition of A, for any ( vertex v / A, we have that deg(v, B > ɛ 1 e(abb > ɛ 1 64 ɛ 1. So we get ( ( = ɛ 0 ( > 3ɛ 0. 3 Together with (3.1, this iplies that ( ( ( deg(b 3e(B + e(abb > 3(1 ɛ 0 + 6ɛ 0 = 3(1 + ɛ b B By the pigeonhole principle, there exists b B, such that ( ( 3n deg(b > (1 + ɛ 0 = (1 + ɛ ( 3n > 4, where the last inequality follows fro the assuption that n is large enough. This contradicts (1.1. Consequently, A \ A = A B B \ B ɛ 1 64, B \ B = A B A \ A ɛ 1 64, V 0 = A \ A + B \ B ɛ ɛ 1 64 = ɛ 1 3. We next show that we can connect any two vertices of B with a loose path of length two without using any fixed n 8 vertices of V. Clai 3.3. For every pair of vertices u, v B and every vertex set S V with S n/8, there exist a A \ S and b 1, b B \ S such that ub 1 a, ab v E(H. Proof. For any x B, by (1.1, we have that deg(x ( 3 4 n ( =. So by the definition of B, ( ( ( deg(x, AB deg(x deg(x, B (1 ɛ 1 = ɛ 1.

16 16 JIE HAN AND YI ZHAO By Clai 3., we get that deg(x, A B deg(x, AB + A \ A B + B \ B A ( ɛ 1 + ɛ ( 1 64 n ɛ 1. (3. Consider a bipartite graph G on A \ S and B \ S with pairs ab E(G if and only if uab, vab E(H. Since S n A 8, we have A \ S 6 and B \ S > (, so A \ S B \ S > 1 ( 6 > 8ɛ1. Consequently, ( e(g A \ S B \ S 4ɛ 1 1 A \ S B \ S > A \ S. Hence there exists a vertex a A \ S such that deg G (a. By picking b 1, b N G (a we finish the proof. 3.. Building a short path. Clai 3.4. Suppose that A B = q > 0. Then there exists a faily P 1 of vertex disjoint loose paths in B, where P 1 consists of one edge if q = 1 and n / 4N two edges e 1, e with e 1 e 1 if q = 1 and n 4N q disjoint edges if q Proof. Let A B = q > 0. Since A B, by Clai 3.1, we get B B, which iplies B = 3 4n + q. By Clai 3., we get that q = A B A \ A ɛ1 64. Hence for any vertex b in B, deg (b, B deg (b, B + B \ B ( B 1 ( ( ɛ 1 + q( B 1 < ɛ 1. (3.3 Now we assue that q = 1, so B 1 = 3 4 n. By (1.1, for any b B, ( ( n 1 deg (b, B 3 4 n [( ( n 1 B ] 1 + c = c, where c = 1 if n / 4N and c = otherwise. The n / 4N case is trivial since B actually contains at least B /3 > 1 edges. If n 4N, then we have deg(b, B. Assue that B does not contain the desired structure. Then any two distinct edges of B share exactly two vertices. Fix an edge e 0 = v 1 v v 3 of B and two vertices u, u B \ e 0. Then every edge of B containing u ust have its two other vertices in e 0. Since deg(u, B, the link graph of u contains at least two pairs of vertices of e 0. So does the link graph of u. We thus find a loose path of length two fro u to u because two distinct pairs on e 0 share exactly one vertex.

17 VERTEX DEGREE THRESHOLD FOR LOOSE HAMILTON CYCLES IN 3-GRAPHS 17 Second, assue that q > 1. In this case we construct q disjoint edges greedily. By (1.1 and B = 3 4 n + q, for any b B, ( ( n 1 deg (b, B 3 4 n [( ( n 1 B ] 1 + c ( B ( 1 3 > 4 n 3 (q 1 4 n, which iplies that e(b > 1 3 B (q 1 3 4n. Suppose we have found i < q disjoint ( edges of B. By (3.3, there are at ost 3(q 1 ɛ 1 edges of B intersecting these i edges. Hence, there are at least ( e(b 3(q 1 ɛ 1 1 ( 3 3 B (q 1 4 n 6(q 1ɛ 1 ( ( (q 1 6(q 1ɛ 1 3 = ( 3 [(q 1 9(q 1ɛ 1] edges not intersecting the existing i edges. This quantity is positive provided that ɛ 1 < q 1 9(q 1. Thus, ɛ 1 < 1 7 suffices since the iniu of q 1 9(q 1, q > 1 is 1 7 attained by q =. Reark 3.5. Clai 3.4 is the only place where the constant c fro (1.1 is used. The goal of this subsection is to prove the following clai. Clai 3.6. There exists a loose path P in H with the following properties: V 0 V (P, V (P ɛ1 4, B \ V (P 3 A \ V (P 1, both ends of P are in B. Proof. We split into two cases here. Case 1. A B. By Clai 3.1, A B iplies that B B, which iplies that V 0 A. Let q = A B. We first apply Clai 3.4 and find a faily P 1 of vertex disjoint loose paths on at ost 6q vertices of B. Next we put each vertex of V 0 into a loose path of length two with four vertices fro B (so in B such that these paths are pairwise vertex disjoint and also vertex disjoint fro the paths in P 1. Let V 0 = {x 1,..., x V0 }. Suppose that we have found loose paths for x 1,..., x i with i < V 0. Since A \ A = V 0 (A B, by Clai 3., we have Thus, and consequently at ost 3ɛ1 3 q + V 0 = A \ A ɛ 1. ( i + 6q < 4 V 0 + 6q 6( V 0 + q 3ɛ 1 3 ( 3ɛ1 ( 1 = 16 pairs of B intersect the existing. Since every graph on n 4 paths. By the definition of V 0, deg(x i+1, B > ɛ 1 (

18 18 JIE HAN AND YI ZHAO vertices and n edges contains two vertex disjoint edges, we can find two vertex disjoint pairs in the link graph of x i+1 in B. Denote by P the faily of the loose paths that we obtained so far. Now we want to glue paths of P together to a single loose path. For this purpose, we apply Clai 3.3 repeatedly to connect the ends of two loose paths while avoiding previously used vertices. This is possible because V (P 5 V 0 + 6q and at ost 3( V 0 + q 1 vertices will be used to connect the paths in P. By (3.4, the resulting loose path P satisfies V (P 8 V 0 + 1q 3 < 1 ɛ1 64 < ɛ 1 4. We next show that B \ V (P 3 A \ V (P 1. To prove this, we split into three cases according to the structure of P 1. Note that B = 3n 4 + q and A = n 4 V 0 q. First, assue that q > 1. Our construction shows that P 1 consists of q disjoint edges in B. So V (P A = V 0 + q 1 and V (P B = 4 V q + ( V 0 + q 1 = 6 V q. Thus, 3n B \ V (P = + q (6 V q 4 ( n 3 V 0 3q = 3 A \ V (P 1. 4 Second, assue that q = 1 and n 4N. Then P 1 consists of a loose path of length two or two disjoint edges. For the first case, we have that V (P A = V 0 and V (P B = 4 V 0 + V = 6 V Thus, B \ V (P = 3n ( n (6 V = 3 4 V = 3 A \ V (P 1. In the second case, we have that V (P A = V and V (P B = 4 V 0 + ( V = 6 V Thus, B \ V (P = 3n ( n (6 V = 3 4 V 0 1 = 3 A \ V (P 1. Third, assue that q = 1 and n / 4N, so P 1 contains only one edge. We have V (P A = V 0 and V (P B = 4 V 0 + V = 6 V Let n = 4k + with soe k Z, so A = k + 1, = 3k + 1, B = 3k + and A = k V 0. Thus, B \ V (P = 3k + (6 V = 3(k V 0 1 = 3 A \ V (P 1. Case. A B =. Note that A B = eans that B B. The difference fro the first case is that we do not need to construct P 1. First we will put every vertex in V 0 into a loose path of length two together with four vertices fro B. By Clai 3., B \ B ɛ1 and thus for any vertex x V 0, deg(x, B deg(x, B B \ B ( 1 ɛ 1 ( 64 ɛ ( 1. (3.5 3 Siilar as in Case 1, let V 0 = {x 1,..., x V0 } and suppose that we have found loose paths for x 1,..., x i with i < V 0. By Clai 3., V 0 ɛ1 3. Thus, we have 4i < 4 V 0 ɛ1 8 and consequently at ost ( ɛ1 8 ( B 1 ɛ1 4 pairs of B

19 VERTEX DEGREE THRESHOLD FOR LOOSE HAMILTON CYCLES IN 3-GRAPHS 19 intersect the existing i loose paths. Then by (3.5, we ay find two vertex disjoint pairs in the link graph of x i+1 in B. As in Case 1, we connect the paths that we obtained to a single loose path by applying Clai 3.3 repeatedly. The resulting loose path P satisfies that V (P = 5 V 0 + 3( V 0 1 < 8 ɛ1 3 = ɛ 1 4. We next show that B \ V (P 3 A \ V (P 1. Note that V (P A = V 0 1 and V (P B = 4 V 0 + ( V 0 1 = 6 V 0. Since B B, we have A A A = n 4 V 0. Thus, n B \ V (P = B (6 V V ( A + V 0 V = 3( A V (P A 1 = 3 A \ V (P Copleting a Hailton cycle. Let P be the loose path given by Clai 3.6. Suppose that B \ V (P = 3 A \ V (P l for soe integer l 1. Since P is a loose path, V (P is odd. Since V = A B V 0 and V 0 V (P, we have V (P + B \ V (P + A \ V (P = n. (3.6 Since n is even, it follows that B \ V (P + A \ V (P is odd, which iplies that l = 3 A \ V (P B \ V (P is odd. If l > 1, then we extend P as follows. Starting fro an end u of P (note that u B, we add an edge by using one vertex fro A and one fro B. This is guaranteed by Clai 3.3, which actually provides a loose path starting fro u. We repeat this l 1 ties. The resulting loose path P satisfies B \ V (P = 3 A \ V (P 1. We clai that V (P 3ɛ1 4 (thus Clai 3.3 can be applied repeatedly. Indeed, by (3.6 and V (P ɛ1 4, l = 3 A \ V (P B \ V (P = 4 A \ V (P (n V (P 4 A n + ɛ 1 4. Since A A + B \ B n 4 + ɛ1 64 fro Clai 3., we have l ɛ1. Since V (P = V (P + l 1, we derive that V (P 3ɛ1 4. Finally, since both ends of P are vertices in B, we extend P by one ore ABB edge fro each end, respectively. Denote the ends of the resulting path Q be x 0, x 1 A. Let A 1 = (A \ V (Q {x 0, x 1 } and B 1 = B \ V (Q. Note that we have B 1 = 3( A 1 1. By Clai 3., we have B 1 \ B B \ B ɛ1 64. Furtherore, B 1 B 3ɛ 1 4 ɛ ɛ 1 4 > (1 ɛ 1. (3.7 For a vertex v A 1, since deg(v, B ɛ 1 (, we have deg (v, B 1 deg(v, B + B 1 \ B ( B 1 1 ( ɛ 1 + ɛ ( ɛ 1 64 ( ( B1 < ɛ 1 < 3ɛ 1,

20 0 JIE HAN AND YI ZHAO where the last inequality follows fro (3.7. In addition, (3. and (3.7 iply that for any vertex v B 1, ( deg(v, A 1 B 1 deg(v, A B ɛ 1 < ɛ 1 < 4ɛ 1 A 1 B 1. We finally coplete the proof of Theore 1.5 by applying the following lea with X = A 1, Z = B 1, and ρ = 4ɛ 1. Lea 3.7. Let ρ > 0 be sufficiently sall and n be sufficiently large. Suppose that H be a 3-graph on n vertices with V (H = X Z such that Z = 3( X 1. Further, assue that for every vertex v X, deg(v, Z ρ ( Z and for every vertex v Z, deg(v, XZ ρ X Z. Then given any two vertices x 0, x 1 X, there is a loose Hailton path fro x 0 to x 1. To prove Lea 3.7, we follow the approach 6 in the proof of [6, Lea 3.4] given by Czygrinow and Molla, who applied a result of Kühn and Osthus [19]. A bipartite graph G = (A, B, E with A = = n is called (d, ɛ-regular if for any two subsets A A, B B with A, B ɛn, (1 ɛd e(a, B A B (1 + ɛd, and G is called (d, ɛ-super-regular if in addition (1 ɛdn deg(v (1 + ɛdn for every v A B. Lea 3.8. [19, Theore 1.1] For all positive constants d, ν 0, η 1 there is a positive ɛ = ɛ(d, ν 0, η and an integer N 0 such that the following holds for all n N 0 and all ν ν 0. Let G = (A, B, E be a (d, ɛ-super-regular bipartite graph whose vertex classes both have size n and let F be a subgraph of G with F = ν E. Choose a perfect atching M uniforly at rando in G. Then with probability at least 1 e ɛn we have (1 ηνn M E(F (1 + ηνn. Proof of Lea 3.7. Let ɛ = ɛ(1, 7/8, 1/8 be given by Lea 3.8 and ρ = (ɛ/ 4. Suppose that n is sufficiently large and H is a 3-graph satisfying the assuption of the lea. Let G be the graph of all pairs uv in Z such that deg(uv, X (1 ρ X. We clai that for any vertex v Z, deg G (v ρ Z. (3.8 Otherwise, soe vertex v Z satisfies deg G (v > ρ Z. satisfies deg H (uv, X > ρ X, we have deg H (v, XZ > ρ Z ρ X = ρ Z X, As each u / N G (v contradicting our assuption. Let = X 1. Arbitrarily partition Z into three sets Z 1, Z, Z 3, each of order. By (3.8 and Z = 3, we have δ(g[z 1, Z ], δ(g[z, Z 3 ] (1 3 ρ. It is easy to see that both G[Z 1, Z ] and G[Z, Z 3 ] are (1, ɛ-super-regular as ɛ = 4 ρ. For any x X, let Fx 1 := {zz E(G[Z 1, Z ] : xzz E(H} and let Fx := {zz E(G[Z, Z 3 ] : xzz E(H}. Since deg(x, Z ρ ( Z 5ρ, 6 We proved this lea by the absorbing ethod in the previous version of this anuscript.

21 VERTEX DEGREE THRESHOLD FOR LOOSE HAMILTON CYCLES IN 3-GRAPHS 1 we have Fx 1, Fx (1 3 ρ 5ρ 7 8. Let M 1 and M be perfect atchings chosen uniforly at rando fro G[Z 1, Z ] and G[Z, Z 3 ], respectively. By applying Lea 3.8 with ν 0 = 7/8 and η = 1/8, for any x X, with probability at least 1 e ɛ, we have M 1 E(Fx 1, M E(Fx (1 ην ( Thus there exist a atching M 1 in G[Z 1, Z ] and a atching M in G[Z, Z 3 ] such that (3.9 holds for all x X. Label Z 1 = {a 1,..., a }, Z = {b 1,..., b } and Z 3 = {c 1,..., c } such that M 1 = {a 1 b 1,..., a b } and M = {b 1 c 1,..., b c }. Let Γ be a bipartite graph on (X, [] such that xi E(Γ if and only if xa i b i, xb i c i E(H for x X and i []. For every i [], since a i b i, b i c i E(G, we have deg Γ (i (1 ρ X by the definition of G. On the other hand, by (3.9, we have deg Γ (x (1 ( = 64 for any x X. By a result of Moon and Moser [], Γ contains a Hailton path x 1 j 1 x j... x j x 0, where [] = {j 1,..., j } and X = {x 0, x 1,..., x }. Since for each i [], x i j i, x i+1 j i E(Γ iplies that x i a ji b ji, b ji c ji x i+1 E(H (with x +1 = x 0, we get a loose Hailton path of H: x 1 a j1 b j1 c j1 x a j b j c j x a j b j c j x Concluding rearks Let h l d (k, n denote the iniu integer such that every k-unifor hypergraph H on n vertices with iniu d-degree δ d (H contains a Hailton l-cycle (provided that k l divides n. In this paper we deterine h 1 1(3, n for sufficiently large n. Can we apply the sae approach to find other values h l d (k, n? In the forthcoing paper [11], we deterine h l k 1 (k, n for all l < k/, iproving the asyptotic results in [18, 10, 14]. The authors of [] conjectured that h 1 1(k, n is asyptotically attained by a siilar construction as the one supported h 1 1(3, n. At present we cannot verify this conjecture because it sees that our success on h 1 1(3, n coes fro the relation d = k instead of the assuption d = 1. It was conjectured in [3] that h k 1 d (k, n approxiately equals to the iniu d-degree threshold for perfect atchings in k-graphs, in particular, h k 1 1 (k, n = (1 (1 1/k k 1 + o(1 ( n k 1. This conjecture sees very hard because we do not even know the iniu d-degree threshold for perfect atchings in general. The key lea in our proof, Lea.7, shows that every 3-graph H on n vertices with δ 1 (H (7/16 o(1 ( n either contains an alost perfect Y-tiling or is in the extreal case. Naturally this raises a question: what is the iniu vertex degree threshold for a perfect Y-tiling? The corresponding codegree threshold was deterined in [18] (asyptotically and [4] (exactly. We deterine this iniu vertex degree threshold exactly in [1] (Czygrinow [3] independently proved a siilar result. Acknowledgeent We thank two referees for their valuable coents that iproved the presentation of this paper.