$3$-uniform hypergraphs without a cycle of length five

In this paper we show that the maximum number of hyperedges in a $3$-uniform hypergraph on $n$ vertices without a (Berge) cycle of length five is less than $(0.254 + o(1))n^{3/2}$, improving an estimate of Bollob\'as and Gy\H{o}ri. We obtain this result by showing that not many $3$-paths can start from certain subgraphs of the shadow.


Introduction
A hypergraph H = (V, E) is a family E of distinct subsets of a finite set V . The members of E are called hyperedges and the elements of V are called vertices. A hypergraph is called 3-uniform if each member of E has size 3. A hypergraph H = (V, E) is called linear if every two hyperedges have at most one vertex in common.
A Berge cycle of length k 2, denoted Berge-C k , is an alternating sequence of distinct vertices and distinct edges of the form v 1 , h 1 , v 2 , h 2 , . . . , v k , h k where v i , v i+1 ∈ h i for each i ∈ {1, 2, . . . , k − 1} and v k , v 1 ∈ h k . (Note that if a hypergraph does not contain a Berge-C 2 , then it is linear.) This definition of a hypergraph cycle is the classical definition due to Berge. More generally, if F = (V (F ), E(F )) is a graph and Q = (V (Q), E(Q)) is a hypergraph, then we say Q is Berge-F if there is a bijection φ : E(F ) → E(Q) such that e ⊆ φ(e) for all e ∈ E(F ). In other words, given a graph F we can obtain a Berge-F by replacing each edge of F with a hyperedge that contains it.
Given a family of graphs F, we say that a hypergraph H is Berge-F-free if for every F ∈ F, the hypergraph H does not contain a Berge-F as a subhypergraph. The maximum possible number of hyperedges in a Berge-F-free 3-uniform hypergraph on n vertices is the Turán number of Berge-F, and is denoted by ex 3 (n, F). When F = {F } then we simply write ex 3 (n, F ) instead of ex 3 (n, {F }).
Determining ex 3 (n, {C 2 , C 3 }) is basically equivalent to the famous (6, 3)-problem. This was settled by Ruzsa and Szemerédi in their classical paper [23], showing that n 2− c √ log n < ex 3 (n, {C 2 , C 3 }) = o(n 2 ) for some constant c > 0. An important Turán-type extremal result for Berge cycles is due to Lazebnik and Verstraëte [21], who studied the maximum number of hyperedges in an r-uniform hypergraph containing no Berge cycle of length less than five (i.e., girth five). They showed the following.
The systematic study of the Turán number of Berge cycles started with the study of Berge triangles by Győri [15], and continued with the study of Berge five cycles by Bollobás and Győri [1] who showed the following.
The following example of Bollobás and Győri proves the lower bound in Theorem 2.
Bollobás-Győri Example. Take a C 4 -free bipartite graph G 0 with n/3 vertices in each part and (1 + o(1))(n/3) 3/2 edges. In one part, replace each vertex u of G 0 by a pair of two new vertices u 1 and u 2 , and add the triple u 1 u 2 v for each edge uv of G 0 . It is easy to check that the resulting hypergraph H does not contain a Berge cycle of length 5. Moreover, the number of hyperedges in H is the same as the number of edges in G 0 .
In this paper, we improve Theorem 2 as follows.
Roughly speaking, our main idea in proving the above theorem is to analyze the structure of a Berge-C 5 -free hypergraph, and use this structure to efficiently bound the number of paths of length 3 that start from certain dense subgraphs (e.g., triangle, K 4 ) of the 2-shadow. This bound is then combined with the lower bound on the number of paths of length 3 provided by the Blakley-Roy inequality [2]. We prove Theorem 3 in Section 2.
Ergemlidze, Győri and Methuku [3] considered the analogous question for linear hypergraphs and proved that ex 3 (n, {C 2 , C 5 }) = n 3/2 /3 √ 3 + o(n 3/2 ). Surprisingly, even though their lower bound is the same as the lower bound in Theorem 2, the linear hypergraph that they constructed in [3] is very different from the hypergraph used in the Bollobás-Győri example discussed above -the latter is far from being linear. In [3], the authors also strengthened Theorem 1 by showing that ex 3 (n, Recently, ex 3 (n, C 4 ) was studied in [5]. See [6] for results on the maximum number of hyperedges in an r-uniform hypergraph of girth six.
Győri and Lemons [16,17] generalized Theorem 2 to Berge cycles of any given length and proved bounds on ex r (n, C 2k+1 ) and ex r (n, C 2k ). These bounds were improved by Füredi andÖzkahya [9], Jiang and Ma [19], Gerbner, Methuku and Vizer [11]. Recently Füredi, Kostochka and Luo [7] started the study of the maximum size of an n-vertex r-uniform hypergraph without any Berge cycle of length at least k. This study has been continued in [8,18,20,4].

Notation
We introduce some important notations and definitions used throughout the paper.
• Length of a path is the number of edges in the path. We usually denote a path • For convenience, an edge {a, b} of a graph or a pair of vertices a, b is referred to as ab. A hyperedge {a, b, c} is written simply as abc.
• For a hypergraph H (or a graph G), for convenience, we sometimes use H (or G) to denote the edge set of the hypergraph H (or G respectively). Thus the number of edges in H is |H|.
• Given a graph G and a subset of its vertices S, let the subgraph of G induced by S be denoted by G[S].
• For a hypergraph H, the neighborhood of v in H is defined as • For a hypergraph H and a pair of vertices u, v ∈ V (H), let codeg(v, u) denote the number of hyperedges of H containing the pair {u, v}.

Proof of Theorem 3
Let H be a hypergraph on n vertices without a Berge 5-cycle and let G = ∂H be the 2-shadow of H. First we introduce some definitions.

Definition 4.
A pair xy ∈ ∂H is called thin if codeg(xy) = 1, otherwise it is called fat. We say a hyperedge abc ∈ H is thin if at least two of the pairs ab, bc, ac are thin.
Definition 5. We say a set of hyperedges (or a hypergraph) is tightly-connected if it can be obtained by starting with a hyperedge and adding hyperedges one by one, such that every added hyperedge intersects with one of the previous hyperedges in 2 vertices.
Definition 6. A block in H is a maximal set of tightly-connected hyperedges.
Definition 7. For a block B, a maximal subhypergraph of B without containing thin hyperedges is called the core of the block.
Let K 3 4 denote the complete 3-uniform hypergraph on 4 vertices. A crown of size k is a set of k 1 hyperedges of the form abc 1 , abc 2 , . . . , abc k . Below we define 2 specific hypergraphs: • Let F 1 be a hypergraph consisting of exactly 3 hyperedges on 4 vertices (i.e., K 3 4 minus an edge).
• For distinct vertices a, b, c, d and o, let F 2 be the hypergraph consisting of hyperedges oab, obc, ocd and oda.
Lemma 8. Let B be a block of H, and let B be a core of B. Then B is either ∅, K 3 4 , F 1 , F 2 or a crown of size k for some k 1.
Proof. If B = ∅, we are done, so let us assume B = ∅. Since B is tightly-connected and it can be obtained by adding thin hyperedges to B, it is easy to see that B is also tightly-connected. Thus if B has at most two hyperedges, then it is a crown of size 1 or 2 and we are done. Therefore, in the rest of the proof we will assume that B contains at least 3 hyperedges.
If B contains at most 4 vertices then it is easy to see that B is either K 3 4 or F 1 . So assume that B has at least 5 vertices (and at least 3 hyperedges). Since B is not a crown, there exists a tight path of length 3, say abc, bcd, cde. Since abc is in the core, one of the pairs ab or ac is fat, so there exists a hyperedge h = abc containing either ab or ac. Similarly there exists a hyperedge f = cde and f contains ed or ec. If h = f then B ⊇ F 2 . However, it is easy to see that F 2 cannot be extended to a larger tightly-connected set of hyperedges without creating a Berge 5-cycle, so in this case B = F 2 . If h = f then the hyperedges h, abc, bcd, cde, f create a Berge 5-cycle in H, a contradiction. This completes the proof of the lemma. Edge Decomposition of G = ∂H. We define a decomposition D of the edges of G into paths of length 2, triangles and K 4 's such as follows: Let B be a block of H and B be its core.
If B = ∅, then B is a crown-block {abc 1 , abc 2 , . . . , abc k } (for some k 1); we partition ∂B into the triangle abc 1 and paths ac i b where 2 i k. If B = ∅, then our plan is to first partition ∂B \ ∂B. If abc ∈ B \ B, then abc is a thin hyperedge, so it contains at least 2 thin pairs, say ab and bc. We claim that the pair ac is in ∂B. Indeed, ac has to be a fat pair, otherwise the block B consists of only one hyperedge abc, so B = ∅ contradicting the assumption. So by Observation 9, ac has to be a pair in ∂B. For every abc ∈ B \ B such that ab and bc are thin pairs, add the 2-path abc to the edge decomposition D. This partitions all the edges in ∂B \ ∂B into paths of length 2. So all we have left is to partition the edges of ∂B.
• If B is a crown {abc 1 , abc 2 , . . . , abc k } for some k 1, then we partition ∂B into the triangle abc 1 and paths ac i b where 2 i k.
Clearly, by Lemma 8 we have no other cases left. Thus all of the edges of the graph G are partitioned into paths of length 2, triangles and K 4 's.

Observation 10.
(a) If D is a triangle that belongs to D, then there is a hyperedge h ∈ H such that D = ∂h.
(b) If abc is a 2-path that belongs to D, then abc ∈ H. Moreover ac is a fat pair.
(c) If D is a K 4 that belongs to D, then there exists F = K 3 4 ⊆ H such that D = ∂F . Let α 1 |G| and α 2 |G| be the number of edges of G that are contained in triangles and 2-paths of the edge-decomposition D of G, respectively. So (1 − α 1 − α 2 ) |G| edges of G belong to the K 4 's in D.
Claim 11. We have, Proof. Let B be a block with the core B. Recall that for each hyperedge h ∈ B \ B, we have added exactly one 2-path or a triangle to D. Moreover, because of the way we partitioned ∂B, it is easy to check that in all of the cases except when B = K 3 4 , the number of hyperedges of B is the same as the number of elements of D that ∂B is partitioned into; these elements being 2-paths and triangles. On the other hand, if B = K 3 4 , then the number of hyperedges of B is 4 but we added only one element to D (namely K 4 ).
This shows that the number of hyperedges of H is equal to the number of elements of D that are 2-paths or triangles plus the number of hyperedges which are in copies of K 3 4 in H, i.e., 4 times the number of K 4 's in D. Since α 1 |G| edges of G are in 2-paths, the number of elements of D that are 2-paths is α 1 |G| /2. Similarly, the number of elements of D that are triangles is α 2 |G| /3, and the number of Combining this with the discussion above finishes the proof of the claim.
The link of a vertex v is the graph consisting of the edges {uw | uvw ∈ H} and is denoted by L v .
Proof. First let us notice that there is no path of length 5 in L v . Indeed, otherwise, there exist vertices v 0 , v 1 , . . . , v 5 such that vv i−1 v i ∈ H for each 1 i 5 which means there is a Berge 5-cycle in H formed by the hyperedges containing the pairs vv First we will prove that there is no path of length 12 in G v . Let us assume by we can conclude that there is a subsequence u 0 , u 1 , . . . , u 6 of v 0 , v 1 , . . . , v 12 and a sequence of distinct hyperedges In the first case the hyperedges f 1 , h 1 , h 2 , h 3 , f 2 , and in the second case the hyperedges f 2 , h 4 , h 5 , h 6 , f 3 form a Berge 5-cycle in H, a contradiction.
Therefore, there is no path of length 12 in G v , so by the Erdős-Gallai theorem, the number of edges in G v is at most 12−1 2 |N (v)| < 6 |N (v)|, as required.

Relating the hypergraph degree to the degree in the shadow
Let d and d G be the average degrees of H and G respectively. Suppose there is a vertex v of H, such that d(v) < d/3. Then we may delete v and all the edges incident to v from H to obtain a graph H whose average degree is more than 3(nd/3 − d/3)/(n − 1) = d. Then it is easy to see that if the theorem holds for H , then it holds for H as well. Repeating this procedure, we may assume that for every vertex v of H, d(v) d/3. Therefore, by (1), we may assume that the degree of every vertex of G is at least d/6. Proof. Let B P = {xy | x ∈ M, y ∈ M , xy ∈ G} be a bipartite graph, clearly |B P | = |P|.
Let d B (x) denote the degree of a vertex x in the graph B.
Proof. Let yx 1 , yx 2 , . . . , yx k ∈ B be the edges of B incident to y. For each 1 j k let f j ∈ H be a hyperedge such that vx j ⊂ f j . For each yx i ∈ B clearly there is a hyperedge yx i w i ∈ H \ E. We claim that for each 1 i k, w i ∈ M . It is easy to see that w i ∈ N (v) or w i ∈ M (because vx i w i is a 2-path in G). Assume for a contradiction that w i ∈ N (v), then since yx i w i / ∈ E we have, codeg(x i , w i ) 3. Let f ∈ H be a hyperedge such that vw i ⊂ f . Now take j = i such that x j = w i . If f j = f then since codeg(x i , w i ) 3 there exists a hyperedge h ⊃ x i w i such that h = f and h = x i w i y, then the hyperedges f, h, x i w i y, yx j w j , f j form a Berge 5-cycle. So f j = f , therefore f j = f i . Similarly in this case, there exists a hyperedge h ⊃ x i w i such that h = f i and h = x i w i y, therefore the hyperedges f i , h, x i w i y, yx j w j , f j form a Berge 5-cycle, a contradiction. So we proved that w i ∈ M for each 1 i k. Proof. If d B (w i ) = 1 for all 1 i k then we are done, so we may assume that there is Let f, h ∈ H be hyperedges with w i x ∈ h and xv ∈ f . If there are j, l ∈ {1, 2, . . . , k} \ {i} such that x, x j and x l are all different from each other, then clearly, either f = f j or f = f l , so without loss of generality we may assume f = f j . Then the hyperedges f, h, w i x i y, yw j x j , f j create a Berge cycle of length 5, a contradiction. So there are no j, l ∈ {1, 2, . . . , k} \ {i} such that x, x j and x l are all different from each other. Clearly this is only possible when k < 4 and there is a j ∈ {1, 2, 3} \ {i} such that x = x j . Let l ∈ {1, 2, 3} \ {i, j}. If f j = f l then the hyperedges f j , h, w i x i y, yw l x l , f l form a Berge 5-cycle. Therefore f j = f l . So we proved that d B (w i ) = 1 implies that k = 3 and for {j, l} = {1, 2, 3} \ {i}, we have f j = f l . So if d B (w i ) = 1 and d B (w j ) = 1 we have f j = f l and f i = f l , which is impossible. So d B (w j ) = 1. So we proved that if for any 1 i k, d B (w i ) = 1 then k = 3 and all but at most 2 of the vertices in {w 1 , w 2 , w 3 } have degree 1 in the graph B, as desired.
We claim that for any i = j where d B (w i ) = d B (w j ) = 1 we have w i = w j . Indeed, if there exists i = j such that w i = w j then w i x j and w i x i are both adjacent to w i in the graph B which contradicts to d B (w i ) = 1. So using the above claim, we conclude that the set {w 1 , w 2 , . . . , w k } contains at least k − 2 distinct elements with each having degree one in the graph B, so we can set S y to be the set of these k − 2 elements. (Then of course ∀w i ∈ S y we have d B (w i ) = 1.) Now we have to prove that for each z = y we have S y ∩S z = ∅. Assume by contradiction that w i ∈ S z ∩ S y for some z = y. That is, there is some hyperedge uw Clearly either f j = f i or f l = f i so without loss of generality we can assume f j = f i . Then it is easy to see that the hyperedges f j , x j w j y, yx i w i , w i zx i , f i are all different and they create a Berge 5-cycle (x j w j y = yx i w i because x j = w i ).
For each x ∈ M with d B (x) = k 3, let S x be defined as in Claim 16. Then the average of the degrees of the vertices in S x ∪{x} in B is (k+|S x |)/(k−1) = (2k−2)(k−1) = 2. Since the sets S 3) are disjoint, we can conclude that average degree of the set M is at most 2. Therefore 2 |M | |B|. So by (2) we have 2 |M | |B| > |P| − 48d G (V ), which completes the proof of the lemma.
Claim 18. We may assume that the maximum degree in the graph G is less than 160 √ n when n is large enough.
Proof. Let v be an arbitrary vertex with d G (v) = Cd for some constant C > 0. Let P be the set of the good 2-paths starting from the vertex v. Then applying Lemma 15 with M = N (v) and M = {y | vxy ∈ P}, we have |P| < 2 |M | + 48d G (v) < 2n + 48 · Cd. Since the minimum degree in G is at least d/6, the number of (ordered) 2-paths starting from v is at least d(v) · (d/6 − 1) = Cd · (d/6 − 1). Notice that the number of (ordered) bad 2-paths starting at v is the number of 2-paths vxy such that x, y ∈ N (v). So by Lemma 13, this is at most 2 · 8 |N (v)| = 16Cd, so the number of good 2-paths is at least Cd · (d/6 − 17). So |P| Cd · (d/6 − 17). Thus we have Cd · (d/6 − 17) |P| < 2n + 48Cd.
Theorem 2 implies that so d 3 √ 2 √ n + 13.5. So combining this with the fact that C < 36, we have d G (v) = Cd < 108 √ 2 √ n + 486 < 160 √ n for large enough n.
Combining Lemma 15 and Claim 18, we obtain the following. Claim 21. The number of (ordered) good 3-paths in G is at least nd 3 G − C 0 n 3/2 d G for some constant C 0 > 0 (for large enough n).
Proof. First we will prove that the number of (ordered) 3-walks that are not good 3-paths is at most 5440n 3/2 d G .
For any vertex x ∈ V (H) if a path yxz is a bad 2-path then zy is an edge of G, so the number of (ordered) bad 2-paths whose middle vertex is x, is at most 2 times the number of edges in G[N (x)], which is less than 2 · 8 |N (x)| = 16d G (x) by Lemma 13. The number of 2-walks which are not 2-paths and whose middle vertex is x is exactly d G (x). So the total number of (ordered) 2-walks that are not good 2-paths is at most Notice that, by definition, any (ordered) 3-walk that is not a good 3-path must contain a 2-walk that is not a good 2-path. Moreover, if xyz is a 2-walk that is not a good 2-path, then the number of 3-walks in G containing it is at most d G (x) + d G (z) < 320 √ n (for large enough n) by Claim 18. Therefore, the total number of (ordered) 3-walks that are not good 3-paths is at most 17nd G · 320 √ n = 5440n 3/2 d G . By the Blakley-Roy inequality, the total number of (ordered) 3-walks in G is at least nd 3 G . By the above discussion, all but at most 5440n 3/2 d G of them are good 3-paths, so letting C 0 = 5440 completes the proof of the claim.
Claim 22. Let {a, b, c} be the vertex set of a triangle that belongs to D. (By Observation 10 (a) abc ∈ H.) Then the number of good 3-paths whose first edge is ab, bc or ca is at most 8n + C 1 √ n for some constant C 1 and for large enough n.
For each x ∈ {a, b, c}, let P x be the set of good 2-paths xuv where u ∈ S x . Let S x = {v | xuv ∈ P x }. For each {x, y} ⊂ {a, b, c}, let P xy be the set of good 2-paths xuv and yuv where u ∈ S xy . Let S xy = {v | xuv ∈ P xy }.
Let {x, y} ⊂ {a, b, c} and z = {a, b, c} \ {x, y}. Notice that each 2-path yuv ∈ P xy (xuv ∈ P xy ), is contained in at most one good 3-path zyuv (respectively zxuv) whose first edge is in the triangle abc. Indeed, since u ∈ S xy , xyuv (respectively yxuv) is not a good 3-path. Therefore, the number of good 3-paths whose first edge is in the triangle abc, and whose third vertex is in S xy is at most |P xy |. The number of paths in P xy that start with the vertex x is less than 2 S xy + 7680 √ n, by Lemma 19. Similarly, the number of paths in P xy that start with the vertex y is less than 2 S xy + 7680 √ n. Since every path in P xy starts with either x or y, we have |P xy | < 4 S xy + 15360 √ n. Therefore, for any {x, y} ⊂ {a, b, c}, the number of good 3-paths whose first edge is in the triangle abc, and whose third vertex is in S xy is less than 4 S xy + 15360 √ n. In total, the number of good 3-paths whose first edge is in the triangle abc and whose third vertex is in S ab ∪ S bc ∪ S ac is at most Let x ∈ {a, b, c} and {y, z} = {a, b, c} \ {x}. For any 2-path xuv ∈ P x there are 2 good 3-paths with the first edge in the triangle abc, namely yxuv and zxuv. So the total number of 3-paths whose first edge is in the triangle abc and whose third vertex is in by Lemma 19. Now we will prove that every vertex is in at most 2 of the sets S a , S b , S c , S ab , S bc , S ac . Let us assume by contradiction that a vertex v ∈ V (G) \ {a, b, c} is in at least 3 of them. We claim that there do not exist 3 vertices u a ∈ N (a) \ {b, c}, u b ∈ N (b) \ {a, c} and u c ∈ N (c) \ {a, b} such that xu x v is a good 3-path for each x ∈ {a, b, c}. Indeed, otherwise, consider hyperedges h a , h a containing the pairs au a and u a v respectively (since au a v is a good 2-path, note that h a = h a ), and hyperedges h b , h b , h c , h c containing the pairs bu b , u b v, cu c , u c v respectively. Then either h a = h b or h a = h c , say h a = h b without loss of generality. Then the hyperedges h a , h a , h b , h b , abc create a Berge 5-cycle in H, a contradiction, proving that it is impossible to have 3 vertices u a ∈ N (a) \ {b, c}, u b ∈ N (b) \ {a, c} and u c ∈ N (c) \ {a, b} with the above mentioned property. Without loss of generality let us assume that there is no vertex u a ∈ N (a) \ {b, c} such that au a v is a good 2-path -in other words, v / ∈ S a ∪ S ab ∪ S ac . However, since we assumed that v is contained in at least 3 of the sets S a , S b , S c , S ab , S bc , S ac , we can conclude that v is contained in all 3 of the sets S b , S c , S bc , i.e., there are vertices u b ∈ S b , u c ∈ S c , u ∈ S bc such that vu b b, vu c c, vub, vuc are good 2-paths. Using a similar argument as before, if vu ∈ h, vu b ∈ h b and vu c ∈ h c , without loss of generality we can assume that h = h b , so the hyperedges abc,h,h b together with hyperedges containing uc and u b b form a Berge 5-cycle in H, a contradiction.
So we proved that This together with (4) and (5), we get that the number of good 3-paths whose first edge is in the triangle abc is at most for C 1 = 92160 and large enough n, finishing the proof of the claim.
Claim 23. Let P = abc be a 2-path and P ∈ D. (By Observation 10 (b) abc ∈ H.) Then the number of good 3-paths whose first edge is ab or bc is at most 4n + C 2 √ n for some constant C 2 > 0 and large enough n.
Proof. First we bound the number of 3-paths whose first edge is ab.
For each x ∈ {a, b}, let P x be the set of good 2-paths xuv where u ∈ S x , and let S x = {v | xuv ∈ P x }. The set of good 3-paths whose first edge is ab is P a ∪ P b , because the third vertex of a good 3-path starting with an edge ab can not belong to N (a) ∩ N (b) by the definition of a good 3-path.
We claim that |S a ∩ S b | 160 √ n. Let us assume by contradiction that and these hyperedges together with abc, would form a Berge 5-cycle in H, a contradiction. We claim that there are j, l ∈ {0, 1, . . . , k} such that a j = a l , otherwise there is a vertex x such that x = a i for each 0 i k. Then xv i ∈ G for each 0 i k, so we get that d G (x) > k > 160 √ n which contradicts Claim 18. So there are j, l ∈ {0, 1, . . . , k} such that a j = a l and a j v j b j , a l v l b l ∈ H. By observation 10 (b), there is a hyperedge h = abc such that ac ⊂ h. Clearly either a j / ∈ h or a l / ∈ h. Without loss of generality let a j / ∈ h, so there is a hyperedge h a with aa j ⊂ h a = h. Let h b ⊃ b j b, then the hyperedges abc, h, h a , a j v j b j , h b form a Berge 5-cycle, a contradiction, proving that |S a ∩ S b | 160 √ n. Notice that |S a | + |S b | = |S a ∪ S b | + |S a ∩ S b | n + 160 √ n. So by Lemma 19, we have |P a | + |P b | 2(|S a | + |S b |) + 2 · 7680 √ n 2(n + 160 √ n) + 2 · 7680 √ n = 2n + 15680 √ n for large enough n. So the number of good 3-paths whose first edge is ab is at most 2n + 15680 √ n. By the same argument, the number of good 3-paths whose first edge is bc is at most 2n + 15680 √ n. Their sum is at most 4n + C 2 √ n for C 2 = 31360 and large enough n, as desired.
Claim 24. Let {a, b, c, d} be the vertex set of a K 4 that belongs to D. Let F = K 3 4 be a hypergraph on the vertex set {a, b, c, d}. (By Observation 10 (c) F ⊆ H.) Then the number of good 3-paths whose first edge belongs to ∂F is at most 6n + C 3 √ n for some constant C 3 > 0 and large enough n.
Proof. First, let us observe that there is no Berge path of length 2, 3 or 4 between distinct vertices x, y ∈ {a, b, c, d} in the hypergraph H \ F , because otherwise this Berge path together with some edges of F will form a Berge 5-cycle in H. This implies, that there is no path of length 3 or 4 between x and y in G \ ∂F , because otherwise we would find a Berge path of length 2, 3 or 4 between x and y in H \ F . Let S = {u ∈ V (H) \ {a, b, c, d} | ∃{x, y} ⊂ {a, b, c, d}, u ∈ N (x) ∩ N (y)}. For each x ∈ {a, b, c, d}, let S x = N (x) \ (S ∪ {a, b, c, d}). Let P S be the set of good 2-paths xuv where x ∈ {a, b, c, d} and u ∈ S. Let S = {v | xuv ∈ P S }. For each x ∈ {a, b, c, d}, let P x be the set of good 2-paths xuv where u ∈ S x , and let S x = {v | xuv ∈ P x }.
Let v ∈ S . By definition, there exists a pair of vertices {x, y} ⊂ {a, b, c, d} and a vertex u, such that xuv and yuv are good 2-paths.
Suppose that zu v is a 2-path different from xuv and yuv where z ∈ {a, b, c, d}. If u = u then z / ∈ {x, y} so there is a Berge 2-path between x and y or between x and z in H \ F , which is impossible. So u = u . Either z = x or z = y, without loss of generality let us assume that z = x. Then zu vux is a path of length 4 in G \ ∂F , a contradiction. So for any v ∈ S there are only 2 paths of P a ∪ P b ∪ P c ∪ P d ∪ P S that contain v as an end vertex -both of which are in P S -which means that We claim that S a and S b are disjoint. Indeed, otherwise, if v ∈ S a ∩ S b there exists x ∈ S a and y ∈ S b such that vxa and vyb are paths in G, so there is a 4-path axvyb between vertices of F in G \ ∂F , a contradiction. Similarly we can prove that S a , S b , S c and S d are pairwise disjoint. This shows that the sets S , S a , S b , S c and S d are pairwise disjoint. So we have |S ∪ S a ∪ S b ∪ S c ∪ S d | = |S | + |S a | + |S b | + |S c | + |S d | .
By Lemma 19, we have |P a |+|P b |+|P c |+|P d | 2(|S a |+|S b |+|S c |+|S d |)+4·7680 √ n. Combining this inequality with (6), we get |P S | + |P a | + |P b | + |P c | + |P d | 2 |S | + 2(|S a | + |S b | + |S c | + |S d |) + 4 · 7680 √ n. (8) Combining (7) with (8) we get for large enough n. Each 2-path in P S ∪ P a ∪ P b ∪ P c ∪ P d can be extended to at most three good 3-paths whose first edge is in ∂F . (For example, auv ∈ P a can be extended to bauv, cauv and dauv.) On the other hand, every good 3-path whose first edge is in ∂F must contain a 2-path of P a ∪ P b ∪ P c ∪ P d ∪ P S as a subpath. So the number of good 3-paths whose first edge is in ∂F is at most 3 |P a ∪ P b ∪ P c ∪ P d ∪ P S | = 3(|P S | + |P a | + |P b | + |P c | + |P d |) which is at most 6n + C 3 √ n by (9), for C 3 = 92160 and large enough n, proving the desired claim.