Forbidden Berge Hypergraphs

A \emph{simple} matrix is a (0,1)-matrix with no repeated columns. For a (0,1)-matrix $F$, we say that a (0,1)-matrix $A$ has $F$ as a \emph{Berge hypergraph} if there is a submatrix $B$ of $A$ and some row and column permutation of $F$, say $G$, with $G\le B$. Letting $||A||$ denote the number of columns in $A$, we define the extremal function $Bh(m,{ F})=\max\{||A||\,:\, A \hbox{ is }m\hbox{-rowed simple matrix with no Berge hypergraph }F\}$. We determine the asymptotics of $Bh(m,F)$ for all $3$- and $4$-rowed $F$ and most $5$-rowed $F$. For certain $F$, this becomes the problem of determining the maximum number of copies of $K_r$ in a $m$-vertex graph that has no $K_{s,t}$ subgraph, a problem studied by Alon and Shinkleman.


Introduction
This paper explores forbidden Berge hypergraphs and their relation to forbidden configurations. Define a matrix to be simple if it is a (0,1)-matrix with no repeated columns. Such a matrix can be viewed as an element-set incidence matrix. Given two (0,1)matrices F and A, we say A has F as a Berge hypergraph and write F Î A if there is a submatrix B of A and a row and column permutation of F , say G, with G ≤ B. The paper of Gerbner and Palmer [15] introduces this concept to generalize the notions of Berge cycles and Berge paths in hypergraphs. Let F be k × ℓ. A Berge hypergraph associated with the object F is a hypergraph whose restriction to a set of k elements yields a hypergraph that 'covers' F . Berge hypergraphs are related to the notion of a pattern P in a (0,1)-matrix A which has been extensively studied and is quite challenging [14]. We say A has pattern P if there is a submatrix B of A with P ≤ B. The award winning result of Marcus and Tardos [18] concerns avoiding a pattern corresponding to a permutation matrix. Row and column order matter to patterns.
We use heavily the concept of a configuration; see [7]. We say A has a configuration F if there is a submatrix B of A and a row and column permutation of F , say G, with B = G. Configurations care about the 0's as well as the 1's in F but do not care about row and column order. In set terminology the notation trace can be used.
For a subset of rows S, define A| S as the submatrix of A consisting of rows S of A. Define [n] = {1, 2, . . . , n}. If F has k rows and A has m rows and F Î A then there is a k-subset S ⊆ [m] such that F Î A| S . For two m-rowed matrices A, B, use [A | B] to denote the concatenation of A, B yielding a larger m-rowed matrix. Define t · A = [A A · · · A] as the matrix obtained from concatenating t copies of A. Let A c denote the (0,1)-complement of A. Let 1 a 0 b denote the (a + b) × 1 vector of a 1's on top of b 0's. We use 1 a instead of 1 a 0 0 . Let K ℓ k denote the k × k ℓ simple matrix of all columns of ℓ 1's on k rows and let K k = [K 0 We are mainly interested in F consisting of a single forbidden Berge hypergraph F . When |F | = 1 and F = {F }, we write BAvoid(m, F ) and Bh(m, F ).
The main goal of this paper is to explore the asymptotic growth rate of Bh(m, F ) for a given k×ℓ F . Theorem 3.1 handles k = 3, Theorem 4.4 handles k = 4 and Theorem 5.1 handles k = 5 (modulo Conjecture 7.1). The results apply some of the proof techniques (and results) for Forbidden configurations [7]. We have some interesting connections with ex(m, K s,t ) (the maximum number of edges in a graph on m vertices with no complete bipartite graph K s,t as a subgraph) and ex(m, K n , K s,t ) [6] (the maximum number of subgraphs K n in a graph on m vertices with no complete bipartite graph K s,t as a subgraph). Two such results are Theorem 6.1 and Theorem 6.3. We also obtain in Theorem 6.5, that if F is the vertex-edge incidence matrix of a tree T , then Bh(m, F ) is Θ(m) analogous to ex(m, T ). Note that K k has two meanings in this paper that are hopefully clear by context namely as the complete graph on k vertices or as the matrix [K 0 k K 1 k K 2 k · · · K k k ]. We first make some easy observations. Remark 1.1 Let F, F ′ be two k×ℓ (0,1)-matrices satisfying F Î F ′ . Then Bh(m, F ) ≤ Bh(m, F ′ ).
The related extremal problem for forbidden configurations is as follows: When |F | = 1 and F = {F }, we write Avoid(m, F ) and forb(m, F ). There are striking differences between Bh(m, F ) and forb(m, F ) such as T heorem 6.5 for Berge hypergraphs and T heorem 6.9 for Forbidden configurations. Note that the two notions of Berge hypergraphs and Configurations coincide when F has no 0's.
Note that any forbidden Berge hypergraph F can be given as a family B(F ) of forbidden configurations by replacing the 0's of F by 1's in all possible ways. Define Isomorphism can reduce the required set of matrices to consider, for example B(I 2 ) which has 4 matrices satisfies: BAvoid(m, B(I 2 ) = BAvoid(m, 1 0 0 1 , A product construction is helpful here. Let A, B be m 1 × n 1 and m 2 × n 2 matrices respectively. We define A × B as the (m 1 + m 2 ) × n 1 n 2 matrix whose columns are obtained by placing a column of A on top of a column of B in all n 1 n 2 possible ways. This extends readily to p-fold products. Let I t = K 1 t denote the t × t identity matrix. In what follows you may assume p divides m since we are only concerned with asymptotic growth with respect to m The p-fold product p I m/p × I m/p × · · · × I m/p , is an m × m p /p p simple matrix. This corresponds to the vertex-edge incidence matrix of the complete p-partite hypergraph with parts V 1 , V 2 , . . . , V p each of size m/p so that {v 1 , v 2 , . . . , v p } is an edge if and only if v i ∈ V i for i = 1, 2, . . . , p. These products sometimes yield the asymptotically best (in growth rate) constructions avoiding F as a Berge hypergraph. Remark 1.4 Let F be a given k × ℓ (0,1)-matrix so that F Î I m/p × I m/p × · · · × I m/p (a p-fold product). Then Bh(m, F ) is Ω(m p ).
Sometimes the product may contain the best construction using the following idea from [5], that when given two matrices F, P where P is m-rowed then Thus Theorem 4.3 yields Bh(m, I 2 × I 2 ) is Θ(f (I 2 × I 2 , I m/2 × I m/2 )). The result Lemma 6.2 indicates that things must be more complicated for general s, t.
A shifting argument works nicely here. We let T i (A) denote the matrix obtained from A by attempting to replace 1's in row i by 0's . We do not replace a 1 by a 0 in row i and column j if the resulting column is already present in A otherwise we do replace the 1 by a 0. We have that T i (A) = A and if A is simple then T i (A) is simple.
). Either T * (T * (A)) contains fewer 1's than T * (a) or we have T i (T * (A)) = T * (A) for i = 1, 2, . . . , m. In the former case replace A by T * (A) and repeat. In the latter case let T (A) = T * (A). Since the number of 1's in A is finite, then the algorithm will terminate with our desired matrix T (A).
Typically T (A) is referred to as a downset since when the columns of T (A) are interpreted as a set system T then if B ∈ T and C ⊂ B then C ∈ T . Note that if T (A) has a column of sum k with 1's on rows S, then K k Î T (A)| S and moreover the copy of K k on rows S can be chosen with 0's on all other rows. An easy consequence is that for A ∈ BAvoid(m, F ) where F is k-rowed and simple then we may assume A has no columns of sum k.

General results
This section provides a number of results about Berge hypergraphs that are used in the paper. The following results from forbidden configurations were useful.
Theorem 2.2 [6] Let k, t be given. Then forb(m, [1 k Theorem 2.3 [3] Let F be a k-rowed simple matrix. Assume there is some pair of rows i, j so than no column of F contains 0's on rows i, j, there is some pair of rows i, j so than no column of F contains 1's on rows i, j and there is some pair of rows i, j so than no column of F contains I 2 on rows i, j. Then forb(m, F ) is O(m k−2 ). Definition 2.4 Let F be a k-rowed (0,1)-matrix. Define G(F ) as the graph on k vertices such that we join vertices i and j by an edge if and only if there is a column in F with 1's in rows i and j. Let ω(G(F )) denote the size of the largest clique in G(F ) and χ(G(F )) is the chromatic number of G(F ). Let α(G(F )) denote the size of the largest independent set in G(F ).
Proof: F is not a Berge hypergraph of the t-fold product I m/t × I m/t × · · · × I m/t . Theorem 2.7 Let k be given and assume m ≥ k − 1. Then Bh(m, I k ) = 2 k−1 .
Proof: The construction consisting of K k−1 with m − k + 1 rows of 0's added yields Bh(m, I k ) ≥ 2 k−1 . The largest m-rowed matrix which avoids I 1 = [1] as a Berge hypergraph is [0 m ] since it avoids I 1 as Berge hypergraphs. This proves the base case k = 1 and the following is the inductive step.
Let A ∈ BAvoid(m, I k ). Let B be obtained from A by removing any rows of 0's so that B is simple and every row of B contains a 1. If B has k − 1 rows then A = B ≤ 2 k−1 which is our bound. Assume B has at least k rows. Either B ≤ 2 k−1 in which case we are done or B > 2 k−1 > 2 k−2 and so by induction, B must contain I k−1 as a Berge hypergraph. Permute B to form the block matrix Then G must be the matrix of 0's or else I k Î B. Thus D is simple. Since all rows of B contain a 1, then E must have a 1. If E contains a 1 then I k−1 Î D and so D ≤ 2 k−2 . This gives While Theorem 2.7 establishes a constant bound for the Berge hypergraph I k , we can see that this follows from a result of Balogh and Bollobás [8]. Let I c k = K k−1 k denote the k × k (0,1)-complement of I k and let T k denote the k × k upper triangular (0,1)-matrix with a 1 in row i and column j if and only if i ≤ j. Theorem 2.8 [8] Let k be given.
A corollary of Koch and the first author [4] gives one way to apply this result. .
We apply this result to a forbidden Berge hypergraph F using the family B(F ) from (1) which contains the k × ℓ matrix of 1's. Noting that I c k+ℓ+1 contains a k × ℓ block of 1's and T k+ℓ contains a k × ℓ block of 1's we obtain the following.
The following Lemma (from standard induction in [7]) was quite useful for Forbidden Configurations.
Proof: Let A ∈ BAvoid(m, F ). If we delete row 1 of A, then the resulting matrix may have columns that appear twice. We may permute the columns of A so that which yields the desired bound by induction on m.
Lemma 2.12 Let A be a k-rowed (0,1)-matrix, not necessarily simple, with all row sums at least kt. Then t · I k Î A.
Proof: We use induction on k where the case k = 1 and I 1 = [1] is easy. Choose t columns from A containing a 1 in row 1 and remove them and row 1 resulting in a matrix A ′ . The row sums of A ′ will be at least (k − 1)t and so we may apply induction. Thus (t − 1) · I k Î A ′ and so we obtain t · I k Î A.
An interesting corollary is that if we have an m-rowed matrix A with all rows sums at least kt then t · I k Î A| S for all S ∈ [m] k .

Lemma 2.13 Let A be a given m-rowed matrix and let S be a family of subsets of [m]
with the property that |S| ≤ k for all S ∈ S. Let c be given. Then by deleting at most c m k + m k−1 + · · · + m 1 columns from A we can obtain a matrix A ′ so that for each S ∈ S, A ′ | S either has more than c columns with 1's on all the rows of S or has no columns with 1's on all the rows of S.
Proof: For each subset of S ∈ S, if the number of columns of A| S with 1's on the rows of S is at most c, then delete all such columns. Repeat. The number of deleted columns Assume H, K are simple and have column sums at most k. Also assume for each column Moreover each column γ of G will appear at least c times in A ′ | S and so if α is a column of K and γ is a column of G with α ≤ γ, then This is reminiscent of Lemma 2.20 but the rules for eliminating columns of small column sum (at most k) are slightly more strict. The following are two important applications. We use the notation K p \1 p to denote the matrix obtained from K p by deleting the column of p 1's.
Then Bh(m, H(p, k, t)) is Θ(m p ). Moreover if we add to H(p, k, t) any column not already present t times in H(p, k, t) to obtain F ′ , then Bh(m, F ′ ) is Ω(m p+1 ).
To apply Reduction Lemma 2.14, with G to be the first k − p columns of F and with K to be the remaining 1 Applying Lemma 2.11 repeatedly (in essence deleting the first p rows of G) we obtain Bh(m, G) = O(m k Bh(m, I k−p )) and so with Lemma 2.7 this yields Bh(m, The remaining remarks concerning adding a column to H(p, k, t) are covered in Lemma 2.17.
Note that Bh(m, H(k − 1, k, t)) follows from Theorem 2.2. There is a more general form of H(p, k, t) as follows.
Definition 2.16 Let A be a given (0,1)-matrix. Let S(A) denote the matrix of all columns α so that there exists a column γ of A with α ≤ γ and α = γ. Let Then H(p, k, t) is H((a 1 , a 2 , . . . , a s ), t) where s = p + 1 and a 1 = a 2 = · · · = a p = 1 and a p+1 = k − p. The upper bounds of Theorem 2.15 do not generalize but the second part of the proof continues to hold. is Ω(m s ).
The rest follows from Lemma 2.17.
The following monotonicty result seems obvious but note that monotonicity is only conjectured to be true for forbidden configurations.
Proof: Let F ′ be the matrix obtained from F by deleting rows of 0's, if any. Then for m ≥ k, A ∈ BAvoid(m, F ) if and only if A ∈ BAvoid(m, F ′ ). Now assume A ∈ BAvoid(m, F ′ ) with m ≥ k. Then form A ′ from A by adding a single row or 0's. Then The following allows F to have rows of 0's which contrasts with Reduction Lemma 2.14. Proof: Let A ∈ BAvoid(m, [F | t · I k ]. For any row in A of row sum r we may remove that row and the r columns containing a 1 on that row and the remaining (m − 1)rowed matrix is simple. In this way remove all rows with row sum at most tk + l and call the remaining simple matrix B and assume it has m ′ rows. Then A ≤ B + (tk + ℓ)(m − m ′ ). Suppose B contains F on some k-rows S ⊆ [m ′ ] k . Remove the columns containing F from B to obtain B ′ and now the rows of B ′ have row sum ≥ tk. By Lemma 2.12, t · I k is contained in B ′ | S . Consequently [F | t · I k ] is contained in B. This is a contradiction so we conclude that B ∈ BAvoid(m ′ , F ). Hence B ≤ Bh(m ′ , F ) ≤ Bh(m, F ) (by Lemma 2.19). We also know that B ≥ A − (tk + ℓ)m and so A ≤ Bh(m, F ) + (tk + ℓ)m for all A. A more general result would be the following.
Assume F 1 appears in the first k rows so that If F 2 Î B then F Î A and so we may assume F 2 Î B. Now the multiplicity of any column of B is at most 2 k . Thus B ≤ 2 k Bh(m, F 2 ) and so A ≤ F 1 + 2 k Bh(m − k, F 2 ) ≤ F 1 + 2 k Bh(m, F 2 ) by Lemma 2.19. Interchanging F 1 , F 2 yields the result.
We conclude a Berge hypergraph result much in the spirit of Gerbner and Palmer [15]. They maximized a different extremal function: essentially the number of 1's in a matrix in BAvoid(m, C 4 ).

Theorem 4.3 Bh(m, C 4 ) is Θ(m 3/2 )
Proof: The lower bound follows from [17]. It is straightforward to see that C 4 Î T 2 ×T 2 and then we apply Theorem 4.1 for the upper bound.
We give an alternative argument in Section 6 that handles F = I 2 × I s for s ≥ 2. Other 4-rowed Berge hypergraph cases are more straightforward. Let To verify that all 4-rowed matrices are handled we first note that Bh(m, 2 · 1 4 ) is Θ(m 4 ). Consider matrices F with 2 · 1 4 Î F . Then The column minimal simple (0,1)-matrices F with ω(G(F )) = 4 and with column sums at least 2 are 1 4 , K 2 4 , H 5 , H 6 and H 7 . Since H 6 Î H 5 it suffices to drop H 5 from the list. Now assume ω(G(F )) ≤ 3 and so 1 4 , K 2 4 , H 6 or H 7 Î F . Also assume 2 · 1 3 Î F . Let 3, 4 be the rows so that no column has 1's in both rows 3, 4. Three columns of sum 3 in F either force ω(G(F )) = 4 or we have a column of sum 3 repeated. So F has at most 2 (different) columns of sum 3 and F Î H (2, 4, t). Now assume F Î H(2, 4, t) but 2·1 2 , 1 3 or G 2 Î F . Then G(F ) (from Definition 2.4) has no 3-cycle nor a repeated edge and so G(F ) is a subgraph of K 2,2 or K 1,3 . In the latter case, F Î H(1, 4, t). Then Bh(m, F ) is O(m). In the former case, F Î H ((2, 2), t) and so Theorem 4.3 applies.
For the following you may note that forb(m, F ) is m 2 + 2m − 1 [7]. Proof: Let A ∈ BAvoid(m, H 8 ). Assume that A is a downset by Lemma 1.5. Let A ′ = r(A). Since H 8 has column sums 2 then Bh(m, H 8 ) ≤ A ′ + m + 1. If A ′ has a column of column sum 4 (or more), then H 8 Î A ′ since H 8 has only 4 rows and is simple. If A ′ has a column of sum 3 say with 1's on rows 1,2,3, then we find [K 3 3 K 2 3 ] in those 3 rows. If A ′ has a column of column sum 3, say with 1's in rows 1,2,3 then we cannot have a column with a 1 in row 1 and a 1 in row 4 else H 8 Î A ′ using the fact that A is a downset (using the columns with 1's in rows 1,4 and the column with 1's in rows 2,3). If A ′ has only columns of sum 2 then we deduce that A ′ ≤ m − 1 and so Bh(m, H 8 ) ≤ 2m.
The construction to achieve the bound is to take the m − 1 columns of sum 2 that have a 1 in row 1 as well as all columns of sum 0 or 1. We conclude that Bh(m, H 8 ) = 2m. Proof: Proceed as above. Let A ∈ BAvoid(m, H 2 ). Assume that A is a downset by Lemma 1.5. Let A ′ =r(A) then Bh(m, H 2 ) ≤ A ′ + m + 1 since H 2 has column sum 2. If A ′ has a column of column sum 4 (or more), then H 2 Î A ′ since H 2 has only 4 rows and is simple. If A ′ has a column of sum 3 say with 1's on rows 1,2,3, then we find [K 3 3 K 2 3 ] in those 3 rows. If A ′ has such a column of column sum 3, then A ′ cannot have a column with a 1 in row 1 and a 1 in row 4 else F Î A using the fact that A is a downset (using the columns with 1's in rows 1,2 and the column with 1's in rows 1,3 and the column with 1's in rows 1,4). Thus the number of columns of sum 3 is at most ⌊m/3⌋.
Let t be the number of columns of sum 3. If m = 3t, then we can include all columns of sum 2 that are in the downset of the columns of sum 3. All other columns of sum 2 have their 1's in the m − 3t rows disjoint from those of the 1's in the columns of sum 3. The columns of sum 2, when interpreted as a graph, cannot have a vertex of degree 3 else H 2 Î A. So the number of columns of sum 2 is at most m − 3t for m − 3t ≥ 3 and 0 otherwise. This yields an upper bound.
A construction to achieve our bound is to simply take ⌊m/3⌋ columns of sum 3 each having their 1's on disjoint sets of rows and then, for each column of sum 3, add 3 columns of sum 2 whose 1's lie in the rows occupied by the 1's of the column of sum 3.

× ℓ Berge hypergraphs
First we give the 5-rowed classification which requires Conjecture 7.1 to be true. To verify that all 5-rowed matrices are handled we first note that Bh(m, 2 · 1 5 ) is Θ(m 5 ). Consider matrices F with 2 · 1 5 Î F . Then Now assume 2 · 1 2 Î F and χ(G(F )) ≤ 2 and so the columns of sum 2 of F form a bipartite graph G(F ) and there are no columns of larger sum. The graph G(F ) is either a tree in which case Bh(m, F ) is O(m) by Theorem 6.5 or if there is a cycle it must be C 4 and so Bh(m, F ) is Ω(m 3/2 ). But G(F ) is a subgraph of K 2,3 and so we may apply Theorem 6.1 (and Theorem 2.20) to obtain Bh(m, F ) is Ω(m 3/2 ). If 2 · 1 or 1 2 Î F then Bh(m, F ) is Ω(m). The only F with 2 · 1 1 Î F and 1 2 Î F satisfies F Î [I 5 | t · 0 5 ] for some t.
If we attempted the classification for 6-rowed F then we would need bounds such as Bh(m, I 1 × I 2 × I 3 ) and Bh(m, I 2 × I 2 × I 2 ).

Berge hypergraphs from graphs
Let G be a graph and let F be the vertex-edge incidence graph. This sections explores some connections for Berge hypergraphs F with extremal graph theory results.
The first results provides a strong connection with ex(m, K s,t ) and the related problem ex(m, T, H) (the maximum number of subgraphs T in an H-free graph on m vertices). Then we consider the case G is a tree (or forest). Finally we connect the largest clique size ω(G) with Bh(m, F ). Theorem 6.1 Let F = I 2 × I t be the vertex-edge incidence matrix of the complete bipartite graph K 2,t . Then Bh(m, F ) is Θ(ex(m, K 2,t )) which is Θ(m 3/2 ).
Proof: It is immediate that Bh(m, F ) is Ω(ex(m, K 2,t )) since the vertex-edge incidence matrix A of a graph on m vertices with no subgraph K 2,t has A ∈ BAvoid(m, F ). Now consider A ∈ BAvoid(m, F ). Applying Lemma 1.5, assume T i (A) = A for all i and so, when columns are viewed as sets, the columns form a downset. Thus for every column γ of A of column sum r, we have that there are all 2 r columns α in A with α ≤ γ. Assume for some column α of A of sum 2 that there are 2 t−1 columns γ of A with α ≤ γ. But the resulting set of columns have the Berge hypergraph 1 2 × I t by Theorem 2.7 and then, using the downset idea, will contain the Berge hypergraph F . Thus for a given column α of sum 2, there will be at most 2 t−1 − 1 columns γ of A with α < γ. Thus A ≤ (2 t−1 )p where p is the number of columns of sum 2 in A. We have p ≤ ex(m, K 2,t ) which proves the upper bound for Bh(m, F ).
Results of Alon and Shikhelman [1] are surprisingly helpful here. They prove very accurate bounds. For fixed graphs T and H, let ex(m, T, H) denote the maximum number of subgraphs T in an H-free graph on m vertices. Thus ex(m, K 2 , H) = ex(m, H). The following is their Lemma 4.4. The lower bound for s = 3 can actually be obtained from the construction of Brown [9]. The lower bounds for larger s have also been obtained by Kostochka, Mubayi and Verstraëtte [16]. We can use this directly in analogy to Theorem 6.1.
Proof: Let A ∈ BAvoid(m, I 3 × I t ). Applying Lemma 1.5, assume T i (A) = A for all i and so, when columns are viewed as sets, the columns form a downset. Thus for every column γ of A of column sum r, we have that there are all 2 r columns α in A with α ≤ γ. Let G be the graph associated with the columns of sum 2 and so a column of sum r corresponds to K r in G. In particular the number of columns of sum 3 is bounded by ex(m, K 3 , K 3,t ) since each column of sum 3 yields a triangle K 3 . Assume for some column α of A of sum 3 that there are 2 t−1 columns γ of A with α ≤ γ. But the resulting set of columns have the Berge hypergraph 1 3 × I t by Theorem 2.7 and then, using the downset idea, will contain the Berge hypergraph I 3 × I t . Thus for a given column α of sum 3, there will be at most 2 t−1 − 1 columns γ of A with α < γ. Thus A ≤ (2 t−1 )p + |E(G)| where p is the number of columns of sum 3 in A. We have p ≤ ex(m, K 3 , K 3,t ). This yields A ≤ 2 t−1 ex(m, K 3 , K 3,t ) + ex(m, K 3,t ). Now the standard inequalities yield ex(m, K 3,t ) is O(m 5/3 ) and combined with Lemma 6.2 we obtain the upper bound. The lower bound would follow from taking construction of Θ(m 3−(3/t) ) triples as columns of sum 3 from Lemma 6.2.
Thus for some choices r, s, t, ex(m, K r , K s,t ) grows something like Ω(m r−ǫ ) which shows we can take many columns of sum r and still avoid K s,t , i.e. Bh(m, K s,t ) grows very large.
Theorem 6.5 Let F be the vertex-edge incidence k × (k − 1) matrix of a tree (or forest) T on k vertices. Then Bh(m, F ) is Θ(m).
Proof: We generalize the result for trees/forests in graphs. It is known that if a graph G has all vertices of degree k − 1, then G contains any tree/forest on k vertices as a subgraph. We follow that argument but need to adapt the ideas to Berge hypergraphs. Let A ∈ BAvoid(m, F ) with A being a downset. We will show that A ≤ 2 k−1 m. If A has all rows sums at least 2 k−1 + 1 then we can establish the result as follows. If we consider the submatrix A r formed by those columns with a 1 in row r, then I k−1 is a Berge hypergraph contained in the rows [m]\r of A r (by Theorem 2.7). Thus the vertex corresponding to row r in G(A) has degree at least k − 1. Then G(A) has a copy of the tree/forest T and since A is a downset, F Î A, a contradiction. If A has some rows of sum at most 2 k−1 , then we use induction on m. Assume row r of A has row sum t ≤ 2 k−1 . Then we may delete that row and the t columns with 1's in row r and the resulting (m − 1)-rowed matrix A ′ is simple with A = A ′ + t. By induction A ′ ≤ 2 k−1 (m − 1) and this yields A ≤ 2 k−1 m.
This argument does not extend to other graphs since when we find our desired Berge hypergraph I ℓ under 1's on row r, we cannot control which rows of A contain a Berge hypergraph I ℓ . The following results shows the large gap between Berge hypergraph results and forbidden configurations results.
The following matrices will be used in our arguments. Case 3: H 9 ≺ F Because F has column sums 2, then H 9 ≺ F implies H 9 ×0 k−6 ≺ F and so forb(m, F ) is Ω(m k−1 ). Because F is simple then forb(m, F ) is O(m k−1 ) [7].
This concludes the case k ≥ 5.

Conjecture and Problems
We have used the following conjecture in Theorem 5.1.
What are the equivalent difficult cases for larger number of rows. The conjecture would yield Bh(m, 1 2 × C 4 ) is Θ(m 3 ) by Lemma 2.11 but we do not predict Bh(m, 1 1 × I 2 × I 3 ). For k = 6, we believe that F = I 2 × I 2 × I 2 will be quite challenging given an old result of Erdős [10].