Minimum vertex degree thresholds for tiling complete -partite -graphs
Given positive integers , let be the complete 3-partite 3-uniform hypergraph with three parts of sizes . Let be a 3-uniform 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 vertex-disjoint 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 lattice-based absorbing method, the concept of fractional tiling, and a recent result on shadows for 3-graphs.
Key words and phrases:graph packing, hypergraph, absorbing method, regularity lemma
2010 Mathematics Subject Classification:Primary 05C70, 05C65
Jie Han]email@example.com Chuanyun Zang]firstname.lastname@example.org Yi Zhao]email@example.com
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 1-degree is also referred as the minimum vertex degree.
Given two -graphs and , an -tiling (also known as -packing) of is a collection of vertex-disjoint 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 ). Much work has been done on graphs (), see e.g., [10, 2, 19, 22]. In particular, Kühn and Osthus  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 ) Let denote the complete 3-graph on four vertices, and let denote the (unique) 3-graph on four vertices with three edges. Recently Lo and Markström  proved that , and Keevash and Mycroft  determined the exact value of for sufficiently large . In , Lo and Markström proved that . Let be the unique 3-graph on four vertices with two edges (this 3-graph was denoted by in , and by in ). Kühn and Osthus  showed that , and Czygrinow, DeBiasio and Nagle  determined exactly for large . Recently Mycroft  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  determined and asymptotically, where denotes the complete -partite -graph with vertices in each part. Recently Han and Zhao  and independently Czygrinow  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 .
Given , let and define
Given positive integers , let be the complete 3-partite 3-graph with three parts of size , and , respectively.
Theorem 1.1 (Main Result).
For integers ,
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 lattice-based 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 , 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 3-graph 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 lattice-based absorbing arguments developed recently by Han . Second, in order to prove Lemma 1.2, we use the concept of fractional homomorphic tiling given by Buß, Hàn and Schacht . Third, we need a recent result of Füredi and Zhao  on the shadows of 3-graphs, which can be viewed as a vertex degree version of the well-known Kruskal-Katona Theorem for 3-graphs.
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
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
We need a simple counting result, which, for example, follows from the result of Erdős  on supersaturation. Given , let denote the complete -partite -graph whose th part has exactly vertices for all .
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  on the shadows of 3-graphs. The union of two (overlapping) complete 3-graphs of order about shows that Lemma 3.2 is (asymptotically) best possible.
 Given , let be sufficiently large. If is a -graph on vertices with , then .
The next lemma says that for any 3-graph, after a removal of a small portion of edges, any two vertices with a positive codegree in the remaining 3-graph has a linear codegree in .
Given and an -vertex 3-graph , there exists a vertex set and a subhypergraph of such that the following holds
for any ,
for any pair of vertices .
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 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.
 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.
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.
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:
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 ,
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
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 3-graph . We first study the property . Throughout this subsection, let . Note that because .
For any , there exists such that the following holds for sufficiently large . Let be an -vertex -graph with