Minimum vertex degree thresholds for tiling complete partite graphs
Abstract.
Given positive integers , let be the complete 3partite 3uniform hypergraph with three parts of sizes . Let be a 3uniform hypergraph on vertices where is divisible by . We asymptotically determine the minimum vertex degree of that guarantees a perfect tiling, that is, a spanning subgraph of consisting of vertexdisjoint copies of . This partially answers a question of Mycroft, who proved an analogous result with respect to codegree for uniform hypergraphs for all . Our proof uses a latticebased absorbing method, the concept of fractional tiling, and a recent result on shadows for 3graphs.
Key words and phrases:
graph packing, hypergraph, absorbing method, regularity lemma2010 Mathematics Subject Classification:
Primary 05C70, 05C65Jie Han]jhan@ime.usp.br Chuanyun Zang]czang1@student.gsu.edu Yi Zhao]yzhao6@gsu.edu
1. Introduction
Given , an uniform hypergraph (in short, graph) consists of a vertex set and an edge set , that is, every edge is an element subset of . Given an 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 . The minimum 1degree is also referred as the minimum vertex degree.
Given two graphs and , an tiling (also known as packing) of is a collection of vertexdisjoint copies of in . An tiling is called a perfect tiling (or an factor) of if it covers all the vertices of . An obvious necessary condition for to contain an factor is . Given an integer that is divisible by , we define the tiling threshold to be the smallest integer such that every graph of order with contains an factor.
As a natural extension of the matching problem, tiling has been intensively studied in the past two decades (see survey [21]). Much work has been done on graphs (), see e.g., [10, 2, 19, 22]. In particular, Kühn and Osthus [22] determined , for any graph , up to an additive constant. Tiling problems become much harder for hypergraphs (). For example, despite efforts from many researchers [1, 6, 13, 17, 18, 23, 29, 31, 32], we still do not know the vertex degree threshold for a perfect matching in graphs for arbitrary .
Other than the matching problem, only a few tiling thresholds are known (see survey [34]) Let denote the complete 3graph on four vertices, and let denote the (unique) 3graph on four vertices with three edges. Recently Lo and Markström [25] proved that , and Keevash and Mycroft [16] determined the exact value of for sufficiently large . In [24], Lo and Markström proved that . Let be the unique 3graph on four vertices with two edges (this 3graph was denoted by in [5], and by in [14]). Kühn and Osthus [20] showed that , and Czygrinow, DeBiasio and Nagle [5] determined exactly for large . Recently Mycroft [27] determined asymptotically for many partite graphs including all complete partite graphs and loose cycles.
There are fewer tiling results on vertex degree conditions. Lo and Markström [25] determined and asymptotically, where denotes the complete partite graph with vertices in each part. Recently Han and Zhao [15] and independently Czygrinow [4] determined exactly for sufficiently large . In this paper we extend these results by determining asymptotically for all complete partite graphs , and thus partially answer a question of Mycroft [26].
Given , let and define
(1.1) 
Given positive integers , let be the complete 3partite 3graph with three parts of size , and , respectively.
Theorem 1.1 (Main Result).
For integers ,
Let us compare Theorem 1.1 with the corresponding result in [27], which states that
where is the smallest prime factor of . Not only is Theorem 1.1 more complicated, but also it contains a case where the coefficient of the threshold is irrational. In fact, as far as we know, all the previously known tiling thresholds had rational coefficients.
The lower bound in Theorem 1.1 follows from six constructions given in Section 2. Three of them are known as divisibility barriers and two are known as space barriers. Roughly speaking, the divisibility barriers, known as latticebased constructions, only prevent the existence of a perfect tiling; in contrast, the space barriers are ‘robust’ because they prevent the existence of an almost perfect tiling. Our last construction is based on the fact that if a perfect tiling exists, then every vertex is covered by a copy of , so we call it a covering barrier. Such a barrier has never appeared before – see concluding remarks in Section 5.
Our proof of the upper bound of Theorem 1.1 consists of two parts: one is on finding an almost perfect tiling in , and the other is on ‘finishing up’ the perfect tiling. Our first lemma says that contains an almost perfect tiling if the minimum vertex degree of exceeds those of the space barriers.
Lemma 1.2 (Almost Tiling Lemma).
Fix integers . For any and , there exists an integer such that the following holds. Suppose is a graph of order with , then there exists a tiling covering all but at most vertices.
The absorbing method, initiated by Rödl, Ruciński and Szemerédi [28], has been shown to be effective in finding spanning (hyper)graphs. Our absorbing lemma says that contains a small tiling that can absorb any much smaller set of vertices of if the minimum vertex degree of exceeds those of the divisibility barriers and the covering barrier.
Lemma 1.3 (Absorbing Lemma).
Fix integers . For any , there exists such that the following holds for sufficiently large . Suppose is a graph on vertices such that
Then there exists a vertex set with such that for any vertex set with and , both and have factors.
Proof of Theorem 1.1 (upper bound).
Let be integers and . Let be the constant returned by Lemma 1.3 and let be sufficiently large. Suppose that is a 3graph on vertices with , where
We apply Lemma 1.3 to and get a vertex set with and the described absorbing property. In particular, . Let . Then
Next we apply Lemma 1.2 on with in place of and get a tiling covering all but a set of at most vertices of . Since , . By the absorbing property of , there exists a factor on . Thus we get a factor of . ∎
Although this proof is a straightforward application of the absorbing method, there are several new ideas in the proofs of Lemmas 1.2 and 1.3. First, in order to show that almost every set has many absorbing sets, we use latticebased absorbing arguments developed recently by Han [12]. Second, in order to prove Lemma 1.2, we use the concept of fractional homomorphic tiling given by Buß, Hàn and Schacht [3]. Third, we need a recent result of Füredi and Zhao [9] on the shadows of 3graphs, which can be viewed as a vertex degree version of the wellknown KruskalKatona Theorem for 3graphs.
The rest of the paper is organized as follows. We prove the lower bound in Theorem 1.1 by six constructions in Section 2. We prove Lemma 1.3 in Section 3 and Lemma 1.2 in Section 4, respectively. Finally, we give concluding remarks in Section 5.
Notations. Throughout this paper we let be three integers and . When it is clear from the context, we write as for short. By we mean that for any there exists such that for any the following statement holds. When and , we simply write (and this should not be confused with and ). We omit the floor and ceiling functions when they do not affect the proof.
2. Extremal examples
In this section, we prove the lower bound in Theorem 1.1 by six constructions. Following the definition in [27], we say a 3partite 3graph is of type 0 if or . We say is of type if and .
Construction 2.1 (Space Barrier I).
Let and be two disjoint sets of vertices such that and . Let be the graph on whose edge set consists of all triples such that . Then . Since , we have and .
We claim that has no perfect tiling. Indeed, consider a copy of in . We observe that at least one color class of is a subset of – otherwise contains at least one vertex from each color class; since is complete, there is an edge in , contradicting the definition of . Hence a tiling in covers at most vertices, so it cannot be perfect.
Construction 2.2 (Space Barrier II).
Let and be two disjoint sets of vertices such that and . Let be the graph on whose edge set consists of all triples such that . Then . Since , we have and .
We claim that has no perfect tiling. Similarly as in the previous case, for any copy of in , at least two color classes of are subsets of . Hence a tiling in covers at most vertices, so it cannot be perfect.
Construction 2.3 (Divisibility Barrier I).
Let and be two disjoint sets of vertices such that and . Let be the graph on such that and are two complete graphs. Then .
We claim that has no perfect tiling. Indeed, each copy of must be a subgraph of or . Since , due to the choice of and , we have mod and therefore cannot divide both and . Hence has no perfect tiling.
Construction 2.4 (Divisibility Barrier II).
Suppose that is of type for some even . Let and be two disjoint sets of vertices such that and is odd, and if . Note that we can pick satisfying these conditions because in the interval , there are at least two consecutive odd numbers, therefore at least one of them is not divisible by . Let be the graph on whose edge set consists of all triples such that is even (0 or 2). Then .
We claim that has no perfect tiling. Consider a copy of in . Since every edge intersects in an even number of vertices and is complete, no color class of intersects both and . Moreover, either 0 or 2 color classes of are subsets of . Thus . If , then is divisible by . Since , there is no perfect tiling. Otherwise, either or is even. In either case, all of and are even and thus is even. Since is odd, has no perfect tiling.
Construction 2.5 (Divisibility Barrier III).
Suppose that is of type for some odd , let and be two disjoint sets of vertices such that and and . Let be the graph on whose edge set consists of all triples such that . Then .
We claim that has no perfect tiling. Consider a copy of in . Similarly as in the previous case, exactly one color class of is a subset of , which implies . Since , we have mod and thus mod . If contains a perfect tiling , then mod , contradicting our assumption on . Hence has no perfect tiling.
Construction 2.6 (Covering Barrier).
Let and suppose that is partitioned into such that and . Define a graph on whose edge set consists of all triples with and all triples in such that or . Therefore, .
It is easy to see that is not contained in any copy of , and hence not contained in any copy of with . Therefore, has no perfect tiling with .
Proof of Theorem 1.1 lower bound.
Given positive integers and , where , let be the tiling threshold. By Constructions 2.1 and 2.2, we have and . Furthermore, assume has type . First, by definition, is even if and only if , or , or is even. By Construction 2.4, we have in this case. Second, assume that is odd, then by Construction 2.5, we have . Finally assume that . If , by Construction 2.3, we have . If , then by Construction 2.6, we have . ∎
3. Proof of the Absorbing Lemma
3.1. Preparation
We need a simple counting result, which, for example, follows from the result of Erdős [7] on supersaturation. Given , let denote the complete partite graph whose th part has exactly vertices for all .
Proposition 3.1.
Given , , there exists such that the following holds for sufficiently large . Let be an graph on vertices with a vertex partition . Suppose and contains at least edges such that , . Then contains at least copies of whose th part is contained in for all .
Given a graph , its shadow is the set of the pairs that are contained in at least one edge of . We need a recent result of Füredi and Zhao [9] on the shadows of 3graphs. The union of two (overlapping) complete 3graphs of order about shows that Lemma 3.2 is (asymptotically) best possible.
Lemma 3.2.
[9] Given , let be sufficiently large. If is a graph on vertices with , then .
The next lemma says that for any 3graph, after a removal of a small portion of edges, any two vertices with a positive codegree in the remaining 3graph has a linear codegree in .
Lemma 3.3.
Given and an vertex 3graph , there exists a vertex set and a subhypergraph of such that the following holds

,

for any ,

for any pair of vertices .
Proof.
If an edge contains a pair with , then it is called weak, otherwise called strong. Let be the subhypergraph of induced on strong edges. Then (iii) holds. Let
Then (ii) holds. Note that the number of weak edges in is at most . If , then there are more than weak edges in , a contradiction. Thus (i) holds. ∎
We use the notion of reachability introduced by Lo and Markström [24, 25]. Given an graph of order , , , two vertices in an graph on vertices are reachable (in ) if and only if there are at least sets such that both and contain factors. In this case, we call a reachable set for and . A vertex set is closed in if every two vertices in are reachable in . For , let be the set of vertices that are reachable to .
We use the following two results from [25].
Proposition 3.4.
[25, Proposition 2.1] Given and positive integers and , there exists such that the following holds for sufficiently large . Let be an graph on vertices. Given an vertex graph and a vertex with , then . In other words, if are reachable in and , then are reachable in .
The following lemma is essentially [25, Lemma 4.2]. In fact, [25, Lemma 4.2] shows that the density of ’s containing both and in the part of size is positive. By averaging, this implies that the density of ’s containing both and in the part of size is positive.
Lemma 3.5.
[25] Let be positive integers and . Given , there exists such that the following holds for sufficiently large . For any vertex graph , two vertices are reachable if the number of pairs with is at least .
3.2. Auxiliary Lemmas
Given positive integers , let , and . We call an set an absorbing set for a set if and both and contain factors. Denote by the set of all absorbing sets for .
Our proof of the Absorbing Lemma is based on the following lemma.
Lemma 3.6.
Given , , and , there exists such that the following holds for all sufficiently large integers . Suppose is an vertex graph with the following two properties

For any , there are at least copies of containing it.

There exists with such that is closed in .
Then there exists a vertex set with and such that for any vertex set with and , both and contain factors.
Proof.
Let
There are two steps in our proof. In the first step, we build an absorbing family that can absorb any small portion of vertices in . In the second step, we put the vertices in into a family of copies of . Then is the desired absorbing set.
Fix a set . Let . We claim that there are at least absorbing sets for , namely, . Indeed, we first find a set such that and spans a copy of . By , there are at least
choices for because there are at most sets containing and another fixed vertex, and . For each , since is closed, there are at least reachable sets for and . We greedily choose pairwise disjoint sets – when choosing , we need to avoid the vertices in so there are at least choices for . Let , then . We claim that both and contain factors. Indeed, by the definition of reachability, each spans copies of and thus contains a factor. Furthermore, since spans a copy of and each spans copies of , also contains a factor. Thus is an absorbing set for . In total, we get at least
such sets, thus .
Now we build the family by standard probabilistic arguments. Choose a family of sets in by selecting each of the possible sets independently with probability . Then by Chernoff’s bound, with probability as , the family satisfies the following properties:
(3.1) 
Furthermore, the expected number of pairs of sets in that are intersecting is at most
Thus, by using Markov’s inequality, we derive that with probability at least ,
(3.2) 
Hence, there exists a family with the properties in (3.1) and (3.2). By deleting one member of each intersecting pair and removing sets that are not absorbing sets for any set , we get a subfamily consisting of pairwise disjoint sets. Let and thus . Since every set in is an absorbing set for some set , has a factor. For any set , by (3.1) and (3.2) above we have
(3.3) 
For any set of size and , we arbitrarily partition it into at most sets. By the definition of , each such set has at least absorbing sets in so we can find a distinct absorbing set in for each of the sets. As a result, contains a factor.
In the second step, by (), we greedily build , a collection of copies of that cover the vertices in . Indeed, assume that we have built copies of . Together with the vertices in , at most vertices have already been covered by . So for any vertex not yet covered, we find the desired copy of containing by (), because .
Let , we get the desired absorbing set with . ∎
So it remains to show that and hold in the 3graph . We first study the property . Throughout this subsection, let . Note that because .
Lemma 3.7.
For any , there exists such that the following holds for sufficiently large . Let be an vertex graph with