Minimum vertex degree threshold for loose Hamilton cycles in 3uniform hypergraphs
Abstract.
We show that for sufficiently large , every 3uniform hypergraph on vertices with minimum vertex degree at least , where if and if , contains a loose Hamilton cycle. This degree condition is best possible and improves on the work of Buß, Hàn and Schacht who proved the corresponding asymptotical result.
Key words and phrases:
Hamilton cycle, hypergraph, absorbing method, regularity lemma1991 Mathematics Subject Classification:
Primary 05C65, 05C45Jie Han]jhan22@gsu.edu Yi Zhao]yzhao6@gsu.edu
1. Introduction
The study of Hamilton cycles is an important topic in graph theory. In recent years, researchers have worked on extending the classical theorem of Dirac [7] on Hamilton cycles to hypergraphs – see recent surveys of [23, 20].
Given , a uniform hypergraph (in short, graph) consists of a vertex set and an edge set , where every edge is a element subset of . For , a graph is called an cycle if its vertices can be ordered cyclically such that each of its edges consists of consecutive vertices and every two consecutive edges (in the natural order of the edges) share exactly vertices. (If we allow , then a cycle is merely a matching and perfect matchings have been intensively studied recently, e.g. [1, 5, 9, 16, 15, 21, 27, 30, 31]) In graphs, a cycle is often called a tight cycle while a cycle is often called a loose cycle. We say that a graph contains a Hamilton cycle if it contains an cycle as a spanning subhypergraph. Note that a Hamilton cycle of a graph on vertices contains exactly edges, implying that divides .
Given a graph with a set of vertices (where ) we define to be the number of edges containing (the subscript is omitted if it is clear from the context). The minimum degree of is the minimum of over all vertex sets in . We refer to as the minimum vertex degree and the minimum codegree of .
1.1. Hamilton cycles in hypergraphs
Confirming a conjecture of Katona and Kierstead [13], Rödl, Ruciński and Szemerédi [25, 26] showed that for any fixed , every graph on vertices with contains a tight Hamilton cycle. This is best possible up to the term. With long and involved arguments, the same authors [28] improved this to an exact result for . Loose Hamilton cycles were first studied by Kühn and Osthus [18], who proved that every 3graph on vertices with contains a loose Hamilton cycle. Czygrinow and Molla [6] recently improved this to an exact result. The result of Kühn and Osthus [18] was generalized for arbitrary and arbitrary by Hàn and Schacht [10], and independently by Keevash et al. [14] for and arbitrary . Later Kühn, Mycroft, and Osthus [17] obtained an asymptotically sharp bound on codegree for Hamilton cycles for all . Hence, the problem of finding Hamilton cycles in graphs with large codegree is asymptotically solved.
Much less is known under other degree conditions. Recently Rödl and Ruciński [24] gave a sufficient vertex degree condition that guarantees a tight Hamilton cycle in graphs. Glebov, Person and Weps [8] gave a nontrivial vertex degree condition for tight Hamilton cycles in graphs for general . Neither of these results is best possible – see more discussion in Section 4.
Recently Buß, Hàn, and Schacht [2] studied the minimum vertex degree that guarantees a loose Hamilton cycle in 3graphs and obtained the following result.
Theorem 1.1.
[2, Theorem 3] For all there exists an integer such that the following holds. Suppose is a 3graph on with and
Then contains a loose Hamilton cycle.
In this paper we improve Theorem 1.1 as follows.
Theorem 1.2 (Main Result).
There exists an such that the following holds. Suppose that is a 3graph on with and
(1.1) 
where if and otherwise. Then contains a loose Hamilton cycle.
The following construction shows that Theorem 1.2 is best possible. It is slightly stronger than [2, Fact 4].
Proposition 1.3.
For every there exists a 3graph on vertices with minimum vertex degree , where is defined as in Theorem 1.2, and which contains no loose Hamilton cycle.
Proof.
Let be the 3graph on vertices such that ^{1}^{1}1Throughout the paper, we write for when sets , are disjoint. with and , and consists of all triples intersecting . Note that . Suppose that contains a loose Hamilton cycle . There are edges in and every vertex in is contained in at most two edges in . Since , there is at least one edge of whose vertices are completely from . This is a contradiction since is independent. So contains no loose Hamilton cycle.
Let be a 3graph on vertices such that with and , and consists of all triples intersecting and those containing both and , where are two fixed vertices in . Then . Suppose that contains a loose Hamilton cycle . There are edges in and every vertex in is contained in at most two edges in . Thus, there are at least two edges of whose vertices are completely from . But due to the construction, every two edges in share two vertices so they cannot both appear in one loose cycle. This contradiction shows that contains no loose Hamilton cycle. ∎
As a typical approach of obtaining exact results, we distinguish the extremal case from the nonextremal case and solve them separately.
Definition 1.4.
Given , a 3graph on vertices is called extremal if there is a set , such that and .
Theorem 1.5 (Extremal Case).
There exist and such that the following holds. Let be an even integer. Suppose that is a 3graph on vertices satisfying (1.1). If is extremal, then contains a loose Hamilton cycle.
Theorem 1.6 (Nonextremal Case).
For any , there exist and such that the following holds. Let be an even integer. Suppose that is a 3graph on vertices satisfying . If is not extremal, then contains a loose Hamilton cycle.
Theorem 1.2 follows from Theorems 1.5 and 1.6 immediately by choosing from Theorem 1.5 and letting .
Let us discuss our proof ideas here. The proof of Theorem 1.5 is somewhat standard (though nontrivial). The proof of Theoreom 1.6 follows the approach in the previous work [2, 10, 17, 25, 26, 28]. Roughly speaking, we use the absorbing method initiated by Rödl, Ruciński and Szemerédi, which reduces the task of finding a loose Hamilton cycle to finding constantly many vertexdisjoint loose paths that covers almost all the vertices of the 3graph. More precisely, we first apply the Absorbing Lemma (Lemma 2.1) and obtain a (short) absorbing path which can absorb any smaller proportion of vertices. Second, apply the Reservoir Lemma (Lemma 2.2) and find a small reservoir set whose vertices may be used to connect any constant number of loose paths to a loose cycle. Third, apply the Pathtiling Lemma (Lemma 2.3) in the remaining 3graph and obtain constantly many vertexdisjoint loose paths covering almost all the vertices. Fourth, connect these paths (including ) together by the reservoir and get a loose cycle . Finally we absorb the vertices not in to and obtain the desired loose Hamilton cycle.
The Absorbing Lemma and the Reservoir Lemma are not very difficult and already proven in [2]. Thus the main step is to prove the Pathtiling Lemma, under the assumption and that is not extremal (in contrast, is assumed in [2]). As shown in [2, 10], after applying the (weak) Regularity Lemma, it suffices to prove that the cluster 3graph can be tiled almost perfectly by some particular 3graph. For example, the 3graph given in [2] has the vertex set and edges .^{2}^{2}2Throughout the paper, we often represent a set as . Since it is hard to find an tiling directly, the authors of [2] found a fractional tiling instead and converted it to an (integer) tiling by applying the Regularity Lemma again. In this paper we consider a much simpler 3graph with vertex set and edges , and obtain an almost perfect tiling in directly. Interestingly, tiling was studied (via the codegree condition) in the very first paper on loose Hamilton cycles [18].
1.2. Notations
Given a vertex and disjoint vertex sets in a 3graph , we denote by the number of edges that contain and two vertices from , and by the number of edges that contain , one vertex from and one vertex from . Furthermore, let and . Given not necessarily disjoint sets , we define
, and . The subscript is often omitted when it is clear from the context.
A loose path is a 3graph on with edges for all . The vertices and are called the ends of .
2. Proof of Theorem 1.6
2.1. Auxiliary lemmas and Proof of Theorem 1.6
For convenience, we rephrase the Absorbing Lemma [2, Lemma 7] as follows.^{3}^{3}3Lemma 7 in [2] assumes that and returns with . We simply take their as our and thus .
Lemma 2.1 (Absorbing Lemma).
For any there exists an integer such that the following holds. Let be a 3graph on vertices with Then there is a loose path with such that for every subset with and there exists a loose path with such that and have the same ends.
We also need the Reservoir Lemma [2, Lemma 6].
Lemma 2.2 (Reservoir Lemma).
For any there exists an integer such that for every 3graph on vertices satisfying
there is a set of size at most with the following property: for every mutually disjoint pairs of vertices from there are vertices , from such that for all .
The main step in our proof of Theorem 1.6 is the following lemma, which is stronger than[2, Lemma 10].
Lemma 2.3 (Pathtiling lemma).
For any there exist integers and such that the following holds for . Suppose is a 3graph on vertices with minimum vertex degree
then there are at most vertex disjoint loose paths in that together cover all but at most vertices of unless is extremal.
Proof of Theorem 1.6.
Given , let . We choose , where is the constant returned from Lemma 2.3 with and . Let be an even integer.
Suppose that is a 3graph on vertices with . Since , we can apply Lemma 2.1 with and obtain an absorbing path with ends . We next apply Lemma 2.2 with to and obtain a reservoir . Let and . Note that . The induced subhypergraph satisfies
Applying Lemma 2.3 to with and , we obtain at most vertex disjoint loose paths that cover all but at most vertices of , unless is extremal. In the latter case, there exists such that and . Then we add arbitrary vertices from to to get a vertex set such that and
which means that is extremal, a contradiction. In the former case, denote these loose paths by for some , and their ends by . The choice of guarantees that . We can thus connect by using vertices from obtaining a loose cycle . Since , we can use to absorb all unused vertices in and uncovered vertices in . ∎
The rest of this section is devoted to the proof of Lemma 2.3.
2.2. Proof of Lemma 2.3
Following the approach in [2], we will use the weak regularity lemma which is a straightforward extension of Szemerédi’s regularity lemma for graphs [29]. Below we only state this lemma for 3graphs.
Let be a 3graph and let be mutually disjoint nonempty subsets of . We define to be the number of edges with one vertex in each , , and the density of with respect to () as
Given , the triple of mutually disjoint subsets is called regular if
for all triple of subsets of , , satisfying . We say () is ()regular if it is regular and for some . It is immediate from the definition that in an regular triple (), if has size for some , then () is regular.
Theorem 2.4.
[2, Theorem 14] For any and , there exist and so that for every 3graph on vertices, there exists a partition such that

,

and ,

for all but at most sets , the triple is regular.
A partition as given in Theorem 2.4 is called an regular partition of . For an regular partition of and we refer to as the family of clusters and define the cluster hypergraph with vertex set and is an edge if and only if is regular.
The following corollary shows that the cluster hypergraph inherits the minimum degree of the original hypergraph. Its proof is the same as that of [2, Proposition 15] after we replace by (we thus omit the proof).
Corollary 2.5.
For and there exist and such that the following holds. Suppose is a 3graph on vertices which has minimum vertex degree . Then there exists an regular partition with such that the cluster hypergraph has minimum vertex degree .
In 3graphs, a loose path is partite with partition sizes about for some integer . Proposition 2.6 below shows that every regular triple with partition sizes contains an almost spanning loose path as a subhypergraph. In contrast, [2, Proposition 25] (more generally [10, Lemma 20]) shows that every regular triple with partition sizes contains constant many vertex disjoint loose paths. The proof of Proposition 2.6 uses the standard approach of handling regularity.
Proposition 2.6.
Fix any , , and an integer . Suppose that and is regular with for and . Then there is a loose path omitting at most vertices of .
Proof.
We will greedily construct the loose path such that , and until for some . For , let and for . In addition, we require that for ,
(2.1) 
where . We proceed by induction on . First we pick a vertex such that (thus (2.1) holds for ). By regularity, all but at most vertices can be chosen as . Suppose that we have selected . Without loss of generality, assume that . Our goal is to choose such that

,

.
In fact, the induction hypothesis implies that . Since and , we have
By regularity, at most vertices in does not satisfy (ii). So, at least
(2.2) 
pairs of vertices can be chosen as . Since and (using ), the right side of (2.2) is at least
thus the selection of satisfying (i) and (ii) is guaranteed.
To calculate the number of the vertices omitted by , note that , , and . Our greedy construction of stops as soon as for some . As , one of the following three inequalities holds:
Thus we always have , which implies that or . Consequently,
Let be the 3graph on the vertex set with edges (the unique 3graph with four vertices and two edges). The following lemma is the main step in our proof of Lemma 2.3. In general, given two (hyper)graphs and , an tiling is a sub(hyper)graph of that consists of vertex disjoint copies of . The tiling is perfect if it is a spanning sub(hyper)graph of .
Lemma 2.7 (tiling Lemma).
For any , there exists an integer such that the following holds. Suppose is a 3graph on vertices with
then there is a tiling covering all but at most vertices of unless is extremal.
Proof of Lemma 2.3.
Given , let and , where and are the constants returned from Corollary 2.5 with , , , and .
Suppose that is a 3graph on vertices with . By applying Corollary 2.5 with the constants chosen above, we obtain an regular partition . The cluster hypergraph satisfies . Let be the size of each cluster except , then . By Lemma 2.7, either is extremal, or there is a tiling of that covers all but at most vertices of . In the first case, there exists a set such that and . Let be the union of the clusters in . By regularity,
where the righthand side bounds the number of edges from regular triples with high density, edges from regular triples with low density, edges from irregular triples and edges that are from at most two clusters. Since , , and , we get
Note that implies that . On the other hand,
by adding at most vertices from to , we get a set of size exactly , with . Hence is extremal.
In the second case, the union of the clusters covered by contains all but at most vertices (here we use ). We will apply Proposition 2.6 to each member . Suppose that has the vertex set with edges . For , let denote the corresponding cluster in . We split , , into two disjoint sets and of equal sizes. Then the triples and are regular and of sizes . Applying Proposition 2.6 to these two triples with , we find a loose path in each triple covering all but at most vertices (here we need ).
Since , we obtain a path tiling that consists of at most paths and covers all but at most
vertices. This completes the proof. ∎
2.3. Proof of tiling Lemma (Lemma 2.7)
Fact 2.8.
Let be a 3graph on vertices which contains no copy of , then .
Proof.
Since there is no copy of , then given any , we have that , which implies . ∎
Proof of Lemma 2.7.
Fix and let be sufficiently large. Let be a 3graph on vertices that satisfies . Fix a largest tiling and let for . Let and . Assume that – otherwise we are done.
Our goal is to find a set of vertices in of size at most that covers almost all the edges, which implies that is extremal.
Let be the set of all edges with exactly vertices in , for . Note that by Fact 2.8. We may assume that and consequently
(2.3) 
Indeed, if , then taking of size , we get that . Thus is extremal and we are done.
Claim 2.9.
.
Proof.
Let be the set of vertices such that . First observe that every contains at most one vertex in . Suppose instead, two vertices are both in . Since , the link graph^{4}^{4}4Given a 3graph with a vertex , the link graph of has the vertex set and edge set . of on contains a path of length two. The link graph of on has size at least , so it also contains a path of length two, with vertices denoted by . Note that