A note on the Tur\'an number of a Berge odd cycle
D\'aniel Gerbner

TL;DR
This paper improves upper bounds on the number of hyperedges in 3-uniform hypergraphs avoiding Berge odd cycles, extending previous results and providing new insights into Berge cliques.
Contribution
It presents tighter bounds on hyperedge counts for Berge odd cycles and introduces a more general theorem that enhances prior work.
Findings
Improved upper bounds for Berge odd cycle-free hypergraphs
New results on Berge cliques
Extension of previous bounds by F"uredi and "Ozkahya
Abstract
In this note we obtain upper bounds on the number of hyperedges in 3-uniform hypergraphs not containing a Berge cycle of given odd length. We improve the bound given by F\"uredi and \"Ozkahya in 2017. The result follows from a more general theorem. We also obtain some new results for Berge cliques.
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Taxonomy
TopicsCoding theory and cryptography · Analytic Number Theory Research · Limits and Structures in Graph Theory
A note on the Turán number of a Berge odd cycle
Dániel Gerbner
Alfréd Rényi Institute of Mathematics, Hungarian Academy of Sciences
P.O.B. 127, Budapest H-1364, Hungary.
Abstract
In this note we obtain upper bounds on the number of hyperedges in 3-uniform hypergraphs not containing a Berge cycle of given odd length. We improve the bound given by Füredi and Özkahya in 2017. The result follows from a more general theorem. We also obtain some new results for Berge cliques.
Keywords: Berge, hypergraph, cycle, Turán number
1 Introduction
We say that a hypergraph is a Berge copy of a graph (in short: is a Berge-) if and there is a bijection such that for any we have . This definition was introduced by Gerbner and Palmer [11], extending the well-established notion of Berge cycles and paths. Note that there are several non-uniform Berge copies of , and a hypergraph is a Berge copy of several graphs. A particular copy of defining a Berge- is called its core. Note that there can be multiple cores in a Berge-.
We denote by the largest number of hyperedges in an -uniform Berge--free hypergraph on vertices. There are several papers dealing with (e.g. [8, 14, 15, 16]) or in general (e.g. [9, 10, 11, 12, 20]). For a short survey on this topic see Subsection 5.2.2 in [13].
In this note we consider . In the case , this was first studied by Bollobás and Győri [2]. They showed . This bound was improved to by Ergemlidze, Győri and Methuku [5]. For cycles of any length, Győri and Lemons [15, 16] proved . The constant factors were improved by Jiang and Ma [18], and in the case is even by Gerbner, Methuku and Vizer [10]. In the 3-uniform case, Füredi and Özkahya [8] obtained better constant factors (depending on ). In the case is even, further improvements were obtained by Gerbner, Methuku and Vizer [10] and by Gerbner, Methuku and Palmer [9].
A closely related area is counting triangles in -free graphs. More generally, let denote the maximum number of copies of in an -free graph on vertices. After some sporadic results, the systematic study of these problems (often called generalized Turán problems) was initiated by Alon and Shikhelman [1]. Their connection to Berge hypergraphs was established by Gerbner and Palmer [12], who proved
[TABLE]
for any , and .
Counting triangles in -free graphs and counting hyperedges in Berge--free 3-uniform hypergraphs was handled together already by Bollobás and Győri [2] for , and by Füredi and Özkahya [8], who proved and . Their upper bound for is still the best known bound, but their other upper bound was improved to by Gerbner, Methuku and Vizer [10] in the case and by Gerbner, Methuku and Palmer [9] in the case .
In the case of forbidden cycles of any odd length, the number of triangles was first studied by Győri and Li [17], who proved111We note that the bound is incorrectly stated in their paper [17]. . It was improved independently by Füredi and Özkahya [8] and by Alon and Shikhelman [1]. The latter had the stronger bound . In the case , the current best bound is due to Ergemlidze and Methuku [6].
Füredi and Özkahya [8] obtained the currently best upper bound on the Berge version by showing
[TABLE]
where denotes the largest number of hyperedges in an -uniform Berge--free linear hypergraph on vertices. Recall that a linear hypergraph is one in which any two hyperedges share at most one vertex.
In this note we improve the bound (1.1). Recall that we have , thus we cannot hope for a huge improvement, especially as might be the largest of the three terms. Indeed, the best upper bound currently known is for all the three terms, but the dependence of the known upper bound in is the largest for (we will state these bounds after Theorem 1.2).
Recall that in case of , the two upper bounds obtained by Füredi and Özkahya [8] were and , and the Berge bound was improved in [10, 9] to match the generalized Turán bound. Our goal would be to do the same here and get rid of the terms in (1.1). We cannot achieve that, but we decrease these additional terms. Recall that the currently best bound for the generalized Turán problem is by Alon and Shikhelman [1]. Our new upper bound on is larger than that bound by . We wonder if it is an example of a more general phenomenon and similar bounds could be obtained for other graphs.
The way we use the linearity involves subdividing an edge , i.e. deleting it and adding and for a new vertex . Our method uses only the following two properties of : it can be obtained from by subdividing an edge and deleting a vertex from we obtain a path. In the next theorem we state our result in the most general form.
Theorem 1.1**.**
Let be a connected graph obtained from by subdividing an edge and be obtained from by deleting a vertex. Let be such that for every . Then we have
(i) ,
(ii) .
In the case we have and , the path on vertices. A theorem of Luo [19] shows , but what we need for the 3-uniform case is the Erdős-Gallai theorem [4] showing . Using this, (ii) of Theorem 1.1 gives if . We can improve this a little bit.
Theorem 1.2**.**
If , then
.
The bound in Theorem 1.2 is currently stronger than the bound given by (i) of Theorem 1.1 for and . However, an improvement on would immediately improve the bound in (i). Any significant improvement would make (i) stronger than Theorem 1.2 for .
The second inequality in Theorem 1.2 follows from known results. Füredi and Özkahya [8] proved , and Bukh and Jiang [3] obtained the strongest bound on the Turán number of even cycles by showing . As we do not have good lower bounds on , we cannot be sure that the first term is actually the larger term. However, if is the larger term, then our improvement on the upper bound of is more significant, as we changed the constant factor of that term from 12 to 1. Obviously we have , hence further improvement is impossible here.
We prove Theorem 1.1 by combining the ideas of [8] and [1] with the methods developed in [9, 10]. In the next section we state some lemmas needed for the proof. We give a new proof of a lemma by Gerbner, Methuku and Palmer [9], and we strengthen the lemma a little bit. This strengthens results on for some values of , and . In Section 3 we prove Theorems 1.1 and 1.2.
2 Lemmas
We say that a graph is red-blue if each of its edges is colored with one of the colors red and blue. For a red-blue graph , we denote by the subgraph spanned by the red edges and the subgraph spanned by the blue edges. For two graphs and we denote by the number of subgraphs of that are isomorphic to . Let .
Lemma 2.1** (Gerbner, Methuku, Palmer [9]).**
For any graph and integers and , there is a red-blue -free graph on vertices, such that .
Note that an essentially equivalent version was obtained by Füredi, Kostochka and Luo [7]. The proof of Lemma 2.1 relies on a lemma about bipartite graphs (hidden in the proof of Lemma 2 in [9]). If is a matching and is an edge in , then with a slight abuse of notation we say and .
Lemma 2.2**.**
Let be a finite bipartite graph with parts and and let be a largest matching in . Let denote the set of vertices in that are incident to . Then we can partition into and and partition into and such that for we have , and every neighbor of the vertices of is in .
Here we present a proof that is built on the same principle, but is somewhat simpler than the proof found in [9]. Before that, let us recall the well-known notion of alternating paths. Given a bipartite graph and a matching in it, a path in is called alternating if its first edge is not in , and then it alternates between edges in and edges not in , finishing with an edge not in . It is well-known and easy to see that deleting the edges of from and replacing them with the edges of that were not in , we obtain another matching, that is larger than .
Proof.
First we build a set in the following way. Let be the set of vertices in that are not incident to any edges of . Then in the first step we add to the set of vertices in that are neighbors of a vertex in , to obtain . In the second step we add to the vertices in that are connected to a vertex in by an edge in , to obtain . Similarly, in the th step, if is odd we add to the set of vertices in that are neighbors of a vertex in , while if is even, we add to the vertices in that are connected to a vertex in by an edge in (i.e. for some ), to obtain . After finitely many steps, does not increase anymore, let be the resulting set of vertices.
We claim that no vertex from can be in . Indeed, such a vertex could be reached by an alternating path from a vertex in that is not incident to , thus is not a largest matching, a contradiction.
Then let , , and . A vertex in cannot be connected to a vertex not in , as could be added to then. Similarly, for a vertex , has to be in , otherwise is in and then can be added to .
∎
Let us briefly describe how we can apply this lemma to obtain Lemma 2.1. We take a Berge--free -uniform hypergraph on vertices. Let be the set of hyperedges in and be the set of sub-edges of these hyperedges (by edge and sub-edge we always mean an edge of size two, i.e. a pair of vertices). We connect to if . Let denote this auxiliary bipartite graph. Let be an arbitrary largest matching and be the vertices of incident to the edges in . It is easy to see that the elements of form an -free graph which we call . Indeed, otherwise defines the bijection between a copy of and hyperedges in to form a Berge-.
Now we apply Lemma 2.2 to and . We define a red-blue coloring of by taking the edges of in to be the red edges, and the edges of in to be the blue edges. We have . As hyperedges in have all their neighbors in , they each contain a blue , which is distinct from the other blue -cliques obtained this way, showing .
Let us remark here that Lemma 2.2 also gives some information on the structure of . If there is that has a neighbor , then we could obtain another matching by changing the neighbor of to , i.e. and if , then . Then is replaced by . In this case the same partition of into and , and the partition of into and satisfies Lemma 2.2. This means for that we can delete the (red) edge and replace it with the edge , to obtain another -free graph.
If on the other hand the vertices in have all their neighbors in , then we could recolor the red edges to blue. Therefore, in we can delete an edge and add another edge so that the resulting graph is still -free. Let be the largest value of , where is an -vertex -free blue-red graph. Assume that each -vertex -free blue-red graph with is not monoblue and we cannot delete an edge and add another edge to so that the resulting graph is still -free. Then by the above, cannot be one of these graphs, thus . This is usually a negligible improvement, as we often do not even know the order of magnitude.
However, if , Gerbner, Methuku and Palmer [9] proved that , where is the monoblue Turán graph and is the monored Turán graph . We mention without going into the details that their proof also shows that for any other graphs we have . As we cannot delete an edge from and add another edge to obtain a -free graphs, we do have an improvement. For example, if and , then the result in [9] determines for . For , has 36 copies of and 37 edges. Therefore, (as is a lower bound on ), we have . With our new observation, we know .
3 Proof of Theorems 1.1 and 1.2
Let be a Berge--free -graph on vertices. We say that an edge with is -heavy if are contained together in exactly hyperedges. First we will build a linear subhypergraph in a greedy way: if we can find a hyperedge that does not share an edge with any hyperedge in , we add to , and then repeat this procedure. By definition, is linear. Let consist of the remaining hyperedges. Note that , and the remainder of the proof is for proving the needed upper bound on .
We build an auxiliary bipartite graph in the usual way: let be the set of hyperedges in and be the set of sub-edges of these hyperedges. We connect to if . We will let be a largest matching in , however, we do not choose arbitrarily. Let be an arbitrary largest matching in . Let be the set of vertices in that are incident to some edge of and denote the set of vertices in that are incident to some edge of . Now a hyperedge contains a sub-edge , at least one sub-edge shared with a hyperedge in , maybe some sub-edges that are matched to some other , and maybe some other sub-edges . We have the option to replace in the edge between and with any of the edges of between and an unused sub-edge of , to obtain another largest matching. We will build a largest matching , that contains the same vertices () from as .
For , we pick to be one of the sub-edges of (potentially we let ) in the following way: should share exactly one vertex with (where is a sub-edge that is also a sub-edge of a hyperedge in ) if possible. We go through the hyperedges greedily; as long as there is a hyperedge such that can be changed in this way, we execute the change (it is possible that cannot be changed originally, but later a sub-edge of that is becomes free to use, when is chosen to be different from ). This process finishes after finitely many (at most ) steps, as we change to at most once for every . After this, we rename the unchanged to .
The resulting matching has the following property: for every , shares a sub-edge with a hyperedge in , such that that either shares exactly one vertex with , or all the sub-edges of sharing exactly one vertex with are for some .
Now we can apply Lemma 2.2 to and to obtain . Let us call the elements of red edges and the elements of blue edges. Let be the graph consisting of all the red and blue edges. Then is obviously -free.
Let us now take a random partition of into and . For every , we look at . If the two vertices of are in one part, and all the other vertices of are in the other part, we keep , otherwise we delete it. Let denote the set of elements in that are not deleted (note that elements in are never deleted, thus are in ). Let be the graph consisting of the elements of that are connected by an edge in to an element of . Then is obviously -free, as it is a subgraph of .
Claim 3.1**.**
* is -free, where is any graph for which F can be obtained from by subdividing an edge of .*
Proof.
Let us assume we are given a copy of in such that is the edge that needs to be subdivided to obtain . Observe that there is no edge between and in , thus is in one of them, say . Let be a vertex of with , then , thus is not in .
We say that a hyperedge in is good if contains and for some and is not for any edge of . If there is a good hyperedge, then we build a Berge- with the following core: we subdivide with . For each edge of this core we assign except for (where we assign ) and (where we assign ). This way we obtain a Berge-, a contradiction.
shares at least one sub-edge with a hyperedge . If the sub-edge shares exactly one vertex with , then is good and we are done. Thus every sub-edge of shared with a hyperedge in has to contain none or both of and . In both cases, when we tried to change when constructing , we failed, because all such edges are matched to some other hyperedges of . In particular, is for some and for some . Observe that is in , thus has vertices from both parts and , hence cannot be in by the definition of . This implies is good, finishing the proof.
∎
The above claim implies has at most edges. For an arbitrary , the probability that is in is at least . Let be any subset of , then we have that the expected value of the number of hyperedges in is at least , thus there is a partition with .
There are red edges in , and there is a random partition where at least elements of are undeleted, hence there are at least red edges in . This implies . Hence there are at most red edges altogether. For the total number of edges in we can use the same argument: there is a random partition where at least hyperedges in are undeleted, thus for the defined by that partition, we have .
Observe that we have , hence we are done with the proof of (i).
Note that is not necessarily -free, but it is -free. Let be the number of blue edges in , then has at most red edges. An argument of Gerbner, Methuku and Vizer [10] bounds the number of -cliques in -free graphs with the given number of vertices and edges. For sake of completeness, we include the argument here.
Let be the degree of in . Obviously the neighborhood of every vertex in is -free. An -free graph on vertices contains at most copies of . Thus is contained in at most copies of in . If we sum, for each vertex, the number of ’s containing a vertex, then each is counted times. On the other hand as , we have . This gives that the number of blue ’s is at most . Thus we have
[TABLE]
The above inequality, together with Lemma 2.1 implies that , finishing the proof of (ii).
Now we show how to obtain the small improvement needed to prove Theorem 1.2. It is based on the proof of the upper bound on in [1]. If is odd, replace it by . As the stated upper bound is the same in both cases, obvious mononicity conditions show we can do this. Thus we can assume is even. When we take the random partition into and , first we take a random partition into sets of size 2, and then randomly put one vertex into and the other into . The obtained graph will be -free, and it is divided into two components, hence it has at most edges. The way we chose ensures the above sum is . Then we can go through every step of the remaining part of the proof to obtain the result we need, if for an arbitrary , the probability that is in is still at least . We will separate into cases according to the intersection of with the parts . In case the three vertices of are in three different ’s, the probability is . In case contains for some , there are two cases. If , then the probability is [math], otherwise it is . As happens with probability (having the condition that contains ), for every we have that the probability of being in if contains is .
This gives the first inequality of Theorem 1.2. As we have mentioned after the statement, the second inequality follows from earlier results, stated there.
Acknowledgements: Research supported by the National Research, Development and Innovation Office - NKFIH under the grants SNN 129364, KH 130371 and K 116769 and by the János Bolyai Research Fellowship of the Hungarian Academy of Sciences.
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