Extremal results for Bergehypergraphs
Abstract
Let be a graph and be a hypergraph both on the same vertex set. We say that a hypergraph is a Berge if there is a bijection such that for we have . This generalizes the established definitions of “Berge path” and “Berge cycle” to general graphs. For a fixed graph we examine the maximum possible size (i.e. the sum of the cardinality of each edge) of a hypergraph with no Berge as a subhypergraph. In the present paper we prove general bounds for this maximum when is an arbitrary graph. We also consider the specific case when is a complete bipartite graph and prove an analogue of the KőváriSósTurán theorem.
1 Introduction
Let be a graph and be a hypergraph both on the same vertex set. We say that the hypergraph is a Berge if there is a bijection such that for we have . In other words, given a graph we can construct a Berge by replacing each edge of with a hyperedge that contains it. Alternatively, a hypergraph is a Berge if each hyperedge in can be mapped to two vertices contained in such that the resulting graph is (where we are allowed to ignore extra isolated vertices). Note that when is a cycle, this definition corresponds to the established definitions of “Berge path” and “Berge cycle.” Note that for a fixed graph , a Berge is a class of hypergraphs.
We say that a hypergraph contains the graph if has a Berge as a subhypergraph. If contains no , then we say that is free. We would like to examine the maximum number of edges possible in a hypergraph that is free and has vertices. Throughout most of this paper we allow our hypergraphs to include multiple copies of the same hyperedge (multihyperedges). A hypergrpah is simple if there are no duplicate hyperedges. For ease of notation we consider a hypergraph as a family of edges. Thus means is a hyperedge of and is the number of hyperedges in .
In [12], Győri, Katona, and Lemons proved the following analogue of the ErdősGallai theorem [4] for Berge paths.
Theorem 1 (Győri, Katona, Lemons [12]).
Let be a path on vertices. Suppose is a free uniform hypergraph on vertices and . If , then
If , then
Győri and Lemons [9] proved that if an uniform hypergraph on vertices is free and , then it contains at most edges, which matches the bound found in the graph case (see the even cycle theorem of Bondy and Simonovits [2]). They also showed the surprising fact that the maximum number of edges in a free uniform vertex hypergraph is also which is significantly different from the graph case. Moreover, they proved
Theorem 2 (Győri, Lemons [9]).
Suppose is either free or free hypergraph on vertices. If every hyperedge in has size at least , then
The case when all hyperedges are of size was addressed by Győri and Lemons [7] and recently improved by Füredi and Özkahya [6].
Throughout the paper when considering nonuniform hypergraphs we are interested in the value of instead of . Note that when a hypergraph is uniform than the two parameters can easily be computed from each other. We examine the size of hypergraphs that are free when is a general graph and when is a complete bipartite graph.
Theorem 3.
Let be a graph and let be an free hypergraph on vertices. If every hyperedge in has size at least , then
Furthermore, there exists a free hypergraph such that each edge has size and
The KőváriSósTurán theorem [10] gives an upperbound on the Turán number (extremal number) of the complete bipartite graph.
Theorem 4 (Kővári, Sós, Turán [10]).
If is a free graph on vertices and , then there is a constant depending on and such that
We prove the following analogue for free hypergraphs.
Theorem 5.
If is a free hypergraph, and every hyperedge in has size at least , then
Furthermore, there exists a free hypergraph such that each edge has size and
The case when in Theorem 5 is of particular interest as is also a . In [8] Győri and Lemons examined the maximum possible value of the sum for a free hypergraph. Observe that this sum is closely connected to the sum as . Furthermore, note that the term in the sum is necessary as a hypergraph may have arbitrary many copies of an edge of size and still avoid a . Our third main theorem involves this parameter,
Theorem 6.
If is a free hypergraph, then
Furthermore, there exists a free hypergraph such that
This improves the results of Győri and Lemons [8] who showed that the leading term of the sum above is between and .
If a hypergraph is free for , then we say that has girth (in the Berge sense). Note that the property of being free implies that any two hyperedges intersect in at most one vertex. Lazebnik and Verstraëte [11] examined hypergraphs of girth where all edges are of size . Of particular interest is their result when ,
Theorem 7 (Lazebnik, Verstraëte [11]).
If is a hypergraph on vertices with girth and every hyperedge in has size and the maximum number of edges, then
2 free hypergraphs
Lemma 8.
Let be a graph and suppose is an free hypergraph. If every edge of has size at least , then .
Proof.
Let us construct a graph on the ground set of by embedding a (unique) edge into each hyperedge of . By definition if is free, then must be free and thus .
We can construct greedily. Take the hyperedges of in arbitrary order and for each hyperedge embed an edge that has not already been used in . If at some step we cannot find such an edge, then contains a complete graph of size equal to the size of the hyperedge. Obviously, such a complete graph contains a copy of ; a contradiction. ∎
The following useful antiRamsey lemma appears in Babai [1] (a generalization appears in Erdős, Nešetřil, and Rödl [5]). We include the very short proof for the sake of completeness.
Lemma 9.
A properly edgecolored complete graph on at least vertices contains a rainbow .
Proof.
Suppose we have a proper edgecoloring of the complete graph on vertices and let be the largest rainbow subcomplete graph. Then every vertex not in is adjacent to some by an edge with the same color as one of the colors appearing in . Thus we have
Therefore
∎
Lemma 10.
Let be a graph on vertices. If is an free hypergraph on vertices such that each hyperedge is of size at least , then
Proof.
Similar to the proof of Lemma 8 we will embed a matching of size at least
into each hyperedge in such that the collection of matchings are pairwise disjoint (i.e. no edge is in more than one matching).
Take the hyperedges of in arbitrary order and suppose we have reached a hyperedge where we cannot find a large enough matching to embed, i.e., we can only embed a matching of size less than . Then there is a set of more than vertices such that every edge in has already been used by a matching from previous hyperedges. If we color each of these matchings by a unique color corresponding to their hyperedge, then we have a proper edgecoloring of the complete graph on . As has at least vertices, by Lemma 9 we have that contains a rainbow which obviously contains a rainbow . However, this rainbow would correspond to a Berge in ; a contradiction.
Therefore, we can embed a matching of size into each hyperedge of . The hypergraph is on vertices, so there are at most total edges embedded into the hyperedges of . Thus . Rearranging terms gives the lemma. ∎
Proof of Theorem 3.
First we split into two parts. Let be the collection of hyperedges each of size less than and let be the remaining edges.
All hyperedges in are of size at least , so by Lemma 8 we have and .
Combining the two estimates above gives,
To complete the proof of Theorem 3 we need to construct a free hypergraph with edges of size .
Suppose is an vertex free graph with edges (e.g. a Turán graph). Let be a bipartite subgraph of with at least half of the total edges and let and be the partition classes of . Let us replace each vertex with a set of copies of to get an uniform hypergraph . Put and . Observe that no hyperedge of contains two vertices from nor does it contain vertices from two distinct s. Therefore, the vertices of a in can only be those of some and a single vertex from . However, for each vertex in there is only one hyperedge of containing it and , so these vertices cannot be part of a Berge in .
Therefore is a free uniform hypergraph on at most vertices and edges, i.e.,
∎
Before proving Theorem 5 we need an antiRamsey lemma similar to Lemma 9 but for complete bipartite graphs. This lemma was also proved by Cho, Defant, Sonneborn [3].
Lemma 11.
A properly edgecolored contains a rainbow .
Proof.
Suppose is properly edgecolored and let be the class of size and be the other class. Consider the largest rainbow that uses all vertices of . That is, there are vertices in such that all edges between them and are of distinct colors. It is enough to show that .
Obviously as the edgecoloring is proper. For every vertex not in there is a neighbor of in such that the color of the edge between them is one of the colors used in (otherwise we can add to ). Each vertex is adjacent to vertices in , so there are colors that appear in not on edges incident to . So at most can force many vertices in to not be in . Therefore, in total, there are at most vertices in that are not in . So has at most vertices, hence . ∎
Proof of Theorem 5.
First we partition the hyperedges of into two parts. Let be the collection of hyperedges each of size at least and let be the remaining hyperedges.
Claim 12.
Proof.
We would like to construct a simple graph by embedding into each hyperedge of a matching of size
such that the collection of these matchings are pairwise disjoint (i.e. no edge is used more than once). As before, we take the edges of in arbitrary order. Suppose that we have reached a hyperedge to which we cannot embed a matching of the desired size. This implies that contains a complete graph on many vertices. If we color the edges of this complete graph according to their corresponding hyperedge, then we have a proper edgecoloring of . In particular, contains a properly edgecolored . Thus, by Lemma 11, contains a rainbow which corresponds to a Berge in ; a contradiction.
For the same reason, must be free, thus, applying the KőváriSósTurán theorem (Theorem 4) we have
Furthermore, as is free, Lemma 8 gives
By construction, the number of edges in is
Rearranging terms we get
∎
To complete the proof of Theorem 5 we need to construct a free hypergraph with edges of size of the appropriate size. Let . If , then is a star. In this case, is linear in and it is easy to construct uniform hypergraphs with no Berge that satisfy the theorem. Now we suppose and let be a free graph. Let be a bipartite subgraph of with at least half of the total edges and let and be the partition classes of . Let us replace each vertex with a set of copies of to get a uniform hypergraph . Put and . Now it remains to confirm that has no Berge.
Suppose contains a Berge, then we can map the hyperedges of the Berge into a copy of the graph such that each edge is contained in the hyperedge that was mapped to it. Let and be the classes of . If and (or vice versa), then each vertex of must be in a distinct as each and each vertex of are contained in exactly one edge. In this case, contains a ; a contradiction.
Now either or contains a vertex and a vertex . The vertices and are contained in a hyperedge and as has no hyperedges containing two vertices in we must have that contains and . No hyperedge of contains vertices from and for , so and must both be contained in some . Every vertex of and must be contained in a hyperedge with or , thus each vertex of and must be in or . As the size of is , there must be at least one vertex in . There is exactly one hyperedge of that contains and any other vertex of . However, the degree of in the Berge is at least ; a contradiction. ∎
3 free hypergraphs
In this section we prove Theorem 6. We begin with a short proof of a weaker upper bound that uses the ideas in Section 2.
Proposition 13.
If is a free hypergraph, then
Proof.
We will construct a graph on the vertex set of by replacing each hyperedge with independent edges on the vertices of . We proceed through the hyperedges in any order and assume that we have colored each matching with a unique color. Note that we may assume all edges of are of size at least .
Suppose we are replacing the hyperedge with a matching of color red. Consider an arbitrary subset of vertices of such that no vertex is incident to a red edge. Some of these vertices may be contained in other hyperedges and thus some of the six pairs may already be included in some other matching. However, it is easy to see that if all six pairs are used in macthings, then must contain a . Thus there is some unused pair that can be colored red. Repeat this step until every subset of vertices of is incident to one red edge. At this point we have a matching of red edges in .
Now suppose that contains a , then it is colored properly by the embedding above. Therefore, by Lemma 11, the must contain a rainbow . This corresponds to a Berge in which is a contradition. Therefore must be free. By the KőváriSósTurán theorem (Theorem 4), we have that has at most edges. Thus,
which gives the proposition. ∎
For the proof of Theorem 6 we will need some simple Ramseytype results. Recall that every edgecoloring of contains a monochromatic triangle. Let denote a clique with an edge removed.
Lemma 14.

A edgecoloring of contains either a red or a blue matching of size .

A edgecoloring of contains either a red or a blue triangle or a blue matching of size .

A edgecoloring of contains either a red or a blue triangle.
Proof.
1. Suppose there is no red , then among four vertices there are at least two blue edges. If they are independent, we are done. Therefore, suppose the vertex is incident to two blue edges. Then, among the other four vertices there are at least two blue edges. Among these four blue edges there must be two that are independent.
2. If there is no blue triangle on vertices, then there is a red triangle . For every vertex not in the triangle, there is at most one red edge going to the triangle as otherwise we have a red . First suppose that and are each connected by red edges to different vertices of . In this case it is easy to find a blue matching of size (between and ). Therefore, we may suppose (without loss of generality) that and are connected by red edges to the same vertex in , say . So, either or is connected to by only blue edges. Considering that there is no red on , there is at least one blue edge among . Among these four blue edges there must be a blue triangle.
3. If there is no blue triangle on vertices, then there is a red triangle . For every vertex not in the triangle, there is at most one red edge going to the triangle as otherwise we have a red . Furthermore, there must be a blue edge among , say . There is a vertex among that is not connected by a red edge to or therefore we have a blue triangle. ∎
We are now ready to give a proof of Theorem 6.
Proof of Theorem 6.
We begin with the upper bound. We will construct a graph on the vertex set of similar to the previous section. For each hyperedge we will embed edges on the vertices of such that the collection of edges in consists of pairwise disjoint s and (up to three) s. Let us proceed through the hyperedges in arbitrary order. At hyperedge we embed edges disjointly from the previously embedded edges. To show that such an embedding is possible we need a simple claim.
Claim 15.
Before embedding edges into , on any vertices in there are at most edges already embedded.
Proof.
Let be a set of vertices in with at least edges. For convenience we say that the edges already embedded into are each colored corresponding to their hyperedge. Note that if two incident edges are embedded into and they have the same color, then the third edge that forms a triangle must also be embedded into and have the same color. must contain a triangle and the edges and .
First suppose the triangle is monochromatic with color . Then the other two edges must have two new colors and . Then it is easy to see that there is a path on three edges with colors . Therefore, the hyperedges corresponding to colors and and form a Berge.
Now, if there is no monochromatic triangle in , then the edgecoloring is proper. Therefore, the triangle is rainbow. Without loss of generality suppose that is color , is and is . If there is a fourth color on either of the other two edges, then there is a path on three edges with three different colors. It is easy to see that the three hyperedges corresponding to colors and , together with form a Berge. If only colors and are used then, must be color and must be color . Therefore, the hyperedges corresponding to colors and contain all vertices of and the hyperedge corresponding to color contains and . These three hyperedges and form a Berge. ∎
Now we show that the desired embedding is indeed possible. Suppose we are trying to embed edges into the hyperedge . At this point some edges may already be embedded into . Recall that we would like to embed previously unused edges into such that these edges form triangles and (up to three) independent edges. By Claim 15 there is no among the already embedded edges. So if has size , there is an available edge to embed. If has size , then by Lemma 14 there must be an available matching of size . If has size , then by Lemma 14 there must be an available triangle or matching of size . If has size at least , then by Lemma 14 there must be an available triangle. So embed the triangle and on the remaining vertices we can repeat the above procedure.
In this way we have embedded at least unique edges into each hyperedge of . The graph spanned by these edges is . Thus
Furthermore, for each hyperedge, these unique edges form a set of independent triangles with at most three independent edges.
Claim 16.
contains no .
Proof.
Suppose contains a . We will show that there is a Berge in the original hypergraph . As before, let us color the edges of according to the hyperedge of into which they were embedded. Let be the class of size and be the class of size .
Case 1. For every vertex in the two incident edges in are different colors. Observe that there is a color, say , such that only one edge incident to is color . Call this edge . The edge is of a different color, say . Observe that there are at most other vertices in connected to or by edges of color or . Let be the remaining vertex in . Clearly, and must be two new colors. Thus the cycle is rainbow which corresponds to a Berge; a contradiction.
Case 2. There is a vertex such that and are both the same color, say . In this case, the edge is present in the graph and has color , therefore there is only one such . Thus and are colored and , respectively. So there are at most other vertices in connected by an edge of color to or and at most other vertex in connected by an edge of color to . Let be the remaining vertex and observe that must be of color . The edge cannot be color , , or . If is not color , then the cycle is rainbow which corresponds to a Berge; a contradiction. Therefore, is color . Thus and are both in the hyperedge corresponding to color . This hyperedge together with the three distinct hyperedges containing the edges , , and form a Berge. ∎
By Claim 16 and the KőváriSósTurán theorem (Theorem 4), we have
This completes the proof of the upper bound.
Now it remains to prove the lower bound. Lazebnik and Verstraëte [11] constructed an vertex hypergraph with all hyperedges of size and no Berge, Berge, or Berge (i.e. girth in the Berge sense) and
many hyperedges (see Theorem 7).
Starting with the above construction, we replace each vertex with three copies of it to get a uniform hypergraph . Now we show that has no Berge. If it contains a Berge then there are four vertices and four hyperedges such that are in distinct hyperedges. If are copies of four distinct vertices in , then contains a Berge; a contradiction. Thus, at least two of are copies of the same vertex in . Without loss of generality there are two cases: either or in . In the first case it is easy to see that there is either a Berge or Berge in ; a contradiction. Similarily, in the second case we will get a Berge; a contradiction. Therefore has no Berge.
If has vertices, then the number of edges in is
Therefore,
∎
4 Linear hypergraphs
In the previous sections we restricted our attention to free hypergraphs with edges of size at least . This condition allowed our theorems to hold even for hypergraphs that have multiple copies of the same hyperedge (multihyperedges). A hypergraph with multihyperedges can have arbitrarily many copies of a hyperedge of size less than and remain free. Therefore, we need to restrict our attention to simple hypergraphs, i.e., hypergraphs with no multihyperedges.
In the proof of Lemma 8 the assumption on the size of the hyperdges is crucial. Indeed there is no such general statement comparing the number of edges of an free hypergraph and in this case. For example the complete uniform partite hypergraph is clearly free but contains many hyperedges.
Unfortunately, in general for arbitrary free simple hypergraphs we know very little. One additional condition which can help in this case is if the hypergraph is free, i.e., two edges intersect in at most one point (free hypergraphs are typically called linear). Given a free hypergraph we can form a graph by replacing each hyperedge with an unique edge contained in . Clearly the resulting graph will not contain as a subgraph. This gives the following simple observation,
Observation 17.
Let be a graph and let be an free and free hypergraph. If every hyperedge in has size at least , then .
In order to improve the bound, we can attempt to replace each hyperedge of with several graph edges. However, this must be done in such a way that does not contain as a subgraph. If is uniform with girth at least , we can create a graph by replacing each hyperedge with the three graph edges in . The resulting graph does not contain (otherwise is not free). In this case, using the fact that we get the upper bound from Theorem 7. We prove the following slightly more general result,
Proposition 18.
If is a uniform hypergraph with girth , then
Proof.
Let us form a graph by replacing each hyperedge with the three graph edges contained in . It remains to show that does not contain a cycle of length between and . The assumption that has girth implies that has no cycle of length less than (not even a triangle) such that all of its edges come from different hyperedges of .
Therefore, if is a cycle of length between and , then contains two edges from the same hyperedge of . These two edges must be incident, so call them and . Now we can replace these edges with edge to get a shorter cycle. If we repeat this for any pair of edges of in the same hyperedge of we end up with a shorter nontriangle cycle which has each edge from a different hyperedge of ; a contradiction. ∎
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