1 Introduction
The KőváriSósTurán theorem is one of the most famous results in extremal combinatorics [16, 8, 10]. The theorem states that the maximum number of edges in a free graph of order is . There are multiple known proofs of this theorem, including a standard doublecounting proof that uses Jensen’s inequality, as well as a proof that uses dependent random choice and Jensen’s inequality [2].
Past proofs of the KőváriSósTurán theorem have relied on convexity and Jensen’s inequality, or the power mean inequality which is a corollary of Jensen’s inequality. For a student to fully understand the proof, they would need to understand both convexity and Jensen’s inequality, which would require calculus background, or they would need to know an alternative proof of the power mean inequality that avoids using Jensen’s inequality.
In this paper, we prove the KőváriSósTurán theorem without using calculus or Jensen’s inequality. Instead we use a method based on Nivasch’s bounds on DavenportSchinzel sequences [19] and Alon et al.’s bounds on interval chains [1]. This new proof gives a simple way to teach the proof of the KőváriSósTurán theorem to students with no calculus background, and the same method can be used to prove a generalization of the KőváriSósTurán theorem for uniform hypergraphs.
In [19], Nivasch found upper bounds on the maximum possible lengths of DavenportSchinzel sequences using two different methods. Both methods gave the same bounds, but the first method was more like the proofs in past papers on DavenportSchinzel sequences, and the second method was similar to proofs about interval chains in [1]. The second method in [19] was much simpler than the first for proving bounds on DavenportSchinzel sequences. We imitate the second method here for graph and hypergraph Turán problems.
Let denote the maximum number of edges in an free uniform hypergraph on vertices. Let be the uniform hypergraph obtained from by adding new vertices and replacing every edge in with in . For example, if is the uniform hypergraph of order with edges, then . Mubayi and Verstraëte [18] proved that when is a uniform hypergraph in which all edges are pairwise disjoint.
In Section 2, we provide an elementary proof that , giving an alternative proof of the KőváriSósTurán theorem when is the uniform hypergraph of order with edges. As a corollary, this implies that when is a uniform hypergraph in which all edges are pairwise disjoint, generalizing the upper bound of Mubayi and Verstraëte. In Section 3, we discuss analogous results about dimensional 01 matrices that can be proved with similar methods.
2 The letter method
An ordered uniform hypergraph is a uniform hypergraph with a linear order on the vertices. We define a lettered uniform hypergraph as the structure obtained from labeling each edge of an ordered uniform hypergraph with a letter such that two edges can be labeled with the same letter only if they have the same greatest vertex. Given a uniform hypergraph , we say that a lettered uniform hypergraph is free if its underlying uniform hypergraph is free.
For any uniform hypergraph , let denote the maximum possible number of distinct letters in an free lettered uniform hypergraph on vertices in which every letter occurs at least times.
Lemma 2.1.
For all positive integers and uniform hypergraphs , we have .
Proof.
Start with a uniform free hypergraph with edges. Order the vertices of arbitrarily. For each vertex in in order from greatest to least, label the unlabeled edges adjacent to in any order with letters , only using each letter exactly times and deleting up to remaining edges adjacent to if does not divide the total number of edges in which is the greatest vertex. Observe that the new lettered hypergraph has at most distinct letters with every letter occurring exactly times, and it is free. ∎
When combined with Lemma 2.1, the next lemma will complete our proof of the generalization of the KőváriSósTurán theorem. We use Stirling’s bound in the proof of the next lemma, but it is not actually necessary. We explain after the proof how the use of Stirling’s bound can be replaced with an elementary onesentence argument.
Lemma 2.2.
For and a uniform hypergraph with , we have .
Proof.
Suppose for contradiction that there exists a free lettered uniform hypergraph on vertices with distinct letters in which every letter occurs at least times. Suppose that is sufficiently large so that . Delete edges of until every letter occurs exactly times.
For each subset of , define to be the number of edges in that contain all of the vertices in and a greater vertex in the ordering. Let be the number of subsets of with . The number of tuples of edges in that have the same least vertices is equal to , which is at most , or else would contain a copy of . This follows by the pigeonhole principle, since every tuple of edges in that have the same least vertices must have different letters on each edge.
Then and , where the last inequality follows from Stirling’s bound. However , a contradiction. ∎
Theorem 2.3.
For fixed and uniform hypergraph , we have .
The use of Stirling’s bound in Lemma 2.2 may seem to make the proof nonelementary, but it was unnecessary. All we need is that there exists some constant such that for all , and then we can replace each in the last proof with . Proving this for : it is clearly true for , and if we assume that , then we also have , and . Thus the whole proof is elementary.
Corollary 2.4.
If is a uniform hypergraph in which all edges are pairwise disjoint, then .
Proof.
If is a uniform hypergraph in which all edges are pairwise disjoint, then , so this bound follows from Theorem 2.3. ∎
The last corollary yields the bound of Mubayi and Verstraëte from [18] when .
3 01 matrices
Using the same method, we can get similar bounds for Turántype problems on dimensional 01 matrices. In order to state these results, we introduce more terminology. We say that dimensional 01 matrix contains dimensional 01 matrix if some submatrix of can be turned into by changing some number of ones to zeroes. Otherwise avoids . For any dimensional 01 matrix , define to be the maximum number of ones in a dimensional 01 matrix of sidelength that avoids .
As with the case of uniform hypergraphs, most of the past research on the topic of dimensional 01 matrices has focused on when . We mention several results for that have been generalized to higher values of . For example, Klazar and Marcus [15] proved that for every dimensional permutation matrix , generalizing the result of Marcus and Tardos [17]. Geneson and Tian [14] sharpened this bound by proving that for dimensional permutation matrices of sidelength , generalizing a result of Fox [6]. Geneson and Tian also proved that for every dimensional double permutation matrix , generalizing the upper bound in [12].
In order to state the next result, we define to be the dimensional 01 matrix obtained from the dimensional 01 matrix by stacking copies of with the same orientation in the direction of the new dimension. For example if is the matrix of all ones, then is the matrix of all ones.
Theorem 3.1.

For fixed and dimensional 01 matrix , .

For any dimensional 01 matrix with , we have . In particular, for any dimensional 01 matrix with at least two ones such that . Moreover, for any dimensional 01 matrix with at least three ones differing in the first coordinate such that .
Proof.
4 Concluding remarks
The standard doublecounting method used to prove the KőváriSósTurán theorem can also be used to prove the bounds in Theorem 2.3 and 3.1. We did not include this method since it uses convexity, and it gives the same bounds up to a constant factor as the letter method. Dependent random choice can also be used to obtain the same bounds for uniform hypergraphs up to a constant factor when , and it can be applied to a larger family of hypergraphs that contains the family of . The next lemma is a generalization of the dependent random choice lemma from [2] and [7]. In the next lemma, we call a vertex and a subset of vertices of a uniform hypergraph neighbors if there is some edge of that contains and all of the vertices of . For each vertex and set of vertices , we define to be the set of subsets of vertices that are neighbors with , and we define to be the set of subsets of vertices that are neighbors with every vertex in .
Lemma 4.1.
Let be a uniform hypergraph with vertices and edges. If there is a positive integer such that , then contains a subset of at least vertices such that every vertices in have at least common neighbors among the subsets of .
Proof.
Pick a set of subsets of vertices of , choosing subsets uniformly at random with repetition. Let , and let be the cardinality of . Then , where the secondtolast inequality used Jensen’s inequality.
Let
be the random variable for the number of subsets
of size with fewer than common neighbors among the subsets of vertices of. The probability that an arbitrary
subset is a subset of is , so .Thus by linearity of expectation, . Thus there exists a choice of for which the corresponding set of cardinality satisfies , so we can remove vertices from to produce a new subset so that all subsets of have at least common neighbors among the subsets of . ∎
We can use Lemma 4.1 to get upper bounds for a more general family of uniform hypergraphs that contains the family of . The next theorem describes one such family.
Theorem 4.2.
For any uniform hypergraph , let be the uniform hypergraph obtained by starting with vertices , making disjoint copies of for each subset of vertices of , and replacing each edge in each with edges of the form for each . For any uniform hypergraph with and any integers and , we have .
Note that , and that the letter method also works to show that for any integers and . It would be interesting to see if the letter method is useful for other Turántype problems, and what else can be said about in general.
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