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

TL;DR

This paper establishes an improved upper bound on the maximum number of hyperedges in a 3-uniform hypergraph on n vertices that avoids cycles of length five, advancing understanding of hypergraph cycle constraints.

Contribution

It provides a tighter upper bound on hyperedges in 3-uniform hypergraphs without 5-cycles, refining previous estimates by Bollobás and Győri.

Findings

Maximum hyperedges less than (0.254 + o(1))n^{3/2}

Fewer 3-paths originate from certain subgraphs of the shadow

Improved bounds on cycle-free hypergraph configurations

Abstract

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.

Figures

Equations

$${\mathrm{ex}}_3(n,\{C_2,C_3,C_4\})=\frac{1}{6}{n}^{3/2}+o(n^{3/2}).$$

$$(1+o(1))\frac{n^{3/2}}{3\sqrt{3}}\le e{x}_3(n,C_5)\le \sqrt{2}{n}^{3/2}+4.5n.$$

$${\mathrm{ex}}_3(n,C_5)<(1+o(1))0.254n^{3/2}.$$

$$N(v)=\{x\in V(H)\setminus \{v\}\mid v,x\in h\text{ for some }h\in E(H)\}.$$

$$\left|H\right|=\left(\frac{{\alpha }_1}{3}+\frac{{\alpha }_2}{2}+\frac{2(1-{\alpha }_1-{\alpha }_2)}{3}\right)\left|G\right|.$$

$$\frac{d_G(v)}{2}\le d(v)\le 2d_G(v).$$

$$\left|B\right|\ge \left|B_{\mathcal{P}}\right|-\left|\partial E\right|\ge \left|\mathcal{P}\right|-48\left|N(v)\right|=\left|\mathcal{P}\right|-48d_G(v).$$

$$C\overline{d}\cdot (\overline{d}/6-17)\le \left|\mathcal{P}\right|<2n+48C\overline{d}.$$

$$\left|H\right|=n\overline{d}/3\le \sqrt{2}{n}^{3/2}+4.5n,$$

$$4(\left|S_{ab}^{\prime }\right|+\left|S_{bc}^{\prime }\right|+\left|S_{ac}^{\prime }\right|)+46080\sqrt{n}.$$

$$4(\left|S_a^{\prime }\right|+\left|S_b^{\prime }\right|+\left|S_c^{\prime }\right|)+46080\sqrt{n},$$

$$2\left|S_a^{\prime }\cup S_b^{\prime }\cup S_c^{\prime }\cup S_{ab}^{\prime }\cup S_{bc}^{\prime }\cup S_{ac}^{\prime }\right|\ge \left|S_a^{\prime }\right|+\left|S_b^{\prime }\right|+\left|S_c^{\prime }\right|+\left|S_{ab}^{\prime }\right|+\left|S_{bc}^{\prime }\right|+\left|S_{ac}^{\prime }\right|$$

$$8\left|S_a^{\prime }\cup S_b^{\prime }\cup S_c^{\prime }\cup S_{ab}^{\prime }\cup S_{bc}^{\prime }\cup S_{ac}^{\prime }\right|+92160\sqrt{n}<8n+C_1\sqrt{n}$$

$$\left|{\mathcal{P}}_a\right|+\left|{\mathcal{P}}_b\right|\le 2(\left|S_a^{\prime }\right|+\left|S_b^{\prime }\right|)+2\cdot 7680\sqrt{n}\le 2(n+160\sqrt{n})+2\cdot 7680\sqrt{n}=2n+15680\sqrt{n}$$

$$\left|{\mathcal{P}}_S\right|\le 2\left|S^{\prime }\right|.$$

$$\left|S^{\prime }\cup S_a^{\prime }\cup S_b^{\prime }\cup S_c^{\prime }\cup S_d^{\prime }\right|=\left|S^{\prime }\right|+\left|S_a^{\prime }\right|+\left|S_b^{\prime }\right|+\left|S_c^{\prime }\right|+\left|S_d^{\prime }\right|.$$

$$\left|{\mathcal{P}}_S\right|+\left|{\mathcal{P}}_a\right|+\left|{\mathcal{P}}_b\right|+\left|{\mathcal{P}}_c\right|+\left|{\mathcal{P}}_d\right|\le 2\left|S^{\prime }\right|+2(\left|S_a^{\prime }\right|+\left|S_b^{\prime }\right|+\left|S_c^{\prime }\right|+\left|S_d^{\prime }\right|)+4\cdot 7680\sqrt{n}.$$

$$\left|{\mathcal{P}}_S\right|+\left|{\mathcal{P}}_a\right|+\left|{\mathcal{P}}_b\right|+\left|{\mathcal{P}}_c\right|+\left|{\mathcal{P}}_d\right|\le 2\left|S^{\prime }\cup S_a^{\prime }\cup S_b^{\prime }\cup S_c^{\prime }\cup S_d^{\prime }\right|+30720\sqrt{n}<2n+30720\sqrt{n},$$

$$\frac{{\alpha }_1}{3}\left|G\right|(8n+C_1\sqrt{n})+\frac{{\alpha }_2}{2}\left|G\right|(4n+C_2\sqrt{n})+\frac{(1-{\alpha }_1-{\alpha }_2)}{6}\left|G\right|(6n+C_3\sqrt{n})\le$$

$$\le \left|G\right|n\left(\frac{8{\alpha }_1}{3}+2{\alpha }_2+(1-{\alpha }_1-{\alpha }_2)\right)+(C_1+C_2+C_3)\sqrt{n}\left|G\right|=$$

$$=\frac{n^2{\overline{d}}_G}{2}\left(\frac{5{\alpha }_1+3{\alpha }_2+3}{3}\right)+(C_1+C_2+C_3)\frac{n^{3/2}{\overline{d}}_G}{2}.$$

$$n{\overline{d}}_G^3-C_0{n}^{3/2}{\overline{d}}_G\le \frac{n^2{\overline{d}}_G}{2}\left(\frac{5{\alpha }_1+3{\alpha }_2+3}{3}\right)+(C_1+C_2+C_3)\frac{n^{3/2}{\overline{d}}_G}{2}.$$

$${\overline{d}}_G^2\le \left(\frac{5{\alpha }_1+3{\alpha }_2+3}{6}\right)n+\frac{(C_1+C_2+C_3)}{2}\sqrt{n}+C_0\sqrt{n}.$$

$${\overline{d}}_G^2\le \left(\frac{5{\alpha }_1+3{\alpha }_2+3}{6}\right)n\left(1+\frac{(C_1+C_2+C_3)+2C_0}{\sqrt{n}}\right).$$

$${\overline{d}}_G\le \sqrt{1+\frac{C_4}{\sqrt{n}}}\sqrt{\frac{5{\alpha }_1+3{\alpha }_2+3}{6}}\sqrt{n}<\left(1+\frac{C_4}{2\sqrt{n}}\right)\sqrt{\frac{5{\alpha }_1+3{\alpha }_2+3}{6}}\sqrt{n},$$

$$\left|H\right|\le \frac{{\alpha }_1}{3}\left|G\right|+\frac{{\alpha }_2}{2}\left|G\right|+\frac{2(1-{\alpha }_1-{\alpha }_2)}{3}\left|G\right|=\frac{4-2{\alpha }_1-{\alpha }_2}{6}\frac{n{\overline{d}}_G}{2}.$$

$$\left|H\right|\le \left(1+\frac{C_4}{2\sqrt{n}}\right)\frac{(4-2{\alpha }_1-{\alpha }_2)}{12}\sqrt{\frac{5{\alpha }_1+3{\alpha }_2+3}{6}}{n}^{3/2},$$

$${\mathrm{ex}}_3(n,C_5)\le (1+o(1))\frac{(4-2{\alpha }_1-{\alpha }_2)}{12}\sqrt{\frac{5{\alpha }_1+3{\alpha }_2+3}{6}}{n}^{3/2}.$$

$${\mathrm{ex}}_3(n,C_5)\le (1+o(1))\frac{4-2/3}{12}\sqrt{\frac{5}{6}}{n}^{1.5}<(1+o(1))0.2536n^{3/2}.$$

Full text

We obtain this result by showing that not many $3$-paths can start from certain subgraphs of the shadow.

1 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 $r$-uniform is each member of $E$ has size $r$. A hypergraph $H=(V,E)$ is called linear if every two hyperedges have at most one vertex in common.

A Berge cycle of length $k\geq 2$, denoted Berge-$C_{k}$, is an alternating sequence of distinct vertices and distinct edges of the form $v_{1},h_{1},v_{2},h_{2},\ldots,v_{k},h_{k}$ where $v_{i},v_{i+1}\in h_{i}$ for each $i\in\{1,2,\ldots,k-1\}$ and $v_{k},v_{1}\in h_{k}$. (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 $\mathcal{Q}=(V(\mathcal{Q}),E(\mathcal{Q}))$ is a hypergraph, then we say $\mathcal{Q}$ is Berge-F if there is a bijection $\phi:E(F)\rightarrow E(\mathcal{Q})$ such that $e\subseteq\phi(e)$ for all $e\in E(F)$. 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 $\mathcal{F}$, we say that a hypergraph $\mathcal{H}$ is Berge-$\mathcal{F}$-free if for every $F\in\mathcal{F}$, the hypergraph $\mathcal{H}$ does not contain a Berge-$F$ as a subhypergraph. The maximum possible number of hyperedges in a Berge-$\mathcal{F}$-free hypergraph on $n$ vertices is the Turán number of Berge-$\mathcal{F}$, and is denoted by $\mathrm{ex}_{3}(n,\mathcal{F})$. When $\mathcal{F}=\{F\}$ then we simply write $\mathrm{ex}_{3}(n,F)$ instead of $\mathrm{ex}_{3}(n,\{F\})$.

Determining $\mathrm{ex}_{3}(n,\{C_{2},C_{3}\})$ is basically equivalent to the famous $(6,3)$-problem. This was settled by Ruzsa and Szemerédi in their classical paper [23], showing that $n^{2-\frac{c}{\sqrt{\log n}}}<\mathrm{ex}_{3}(n,\{C_{2},C_{3}\})=o(n^{2})$ 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.

Theorem 1** (Lazebnik, Verstraëte [21]).**

We have

[TABLE]

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.

Theorem 2** (Bollobás, Győri [1]).**

We have,

[TABLE]

The following construction 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.

Theorem 3**.**

We have,

[TABLE]

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 $ex_{3}(n,\{C_{2},C_{5}\})=n^{3/2}/{3\sqrt{3}}+o(n^{3/2})$. 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 $\mathrm{ex}_{3}(n,\{C_{2},C_{3},C_{4}\})\sim\mathrm{ex}_{3}(n,\{C_{2},C_{4}\}).$ Recently, $\mathrm{ex}_{3}(n,C_{4})$ 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 and Özkahya [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].

General results for Berge-$F$-free hypergraphs have been obtained in [12, 13, 10] and the Turán numbers of Berge-$K_{2,t}$ and Berge cliques, among others, were studied in [24, 22, 11, 14, 10].

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 $v_{0},v_{1},\ldots,v_{k}$, simply as $v_{0}v_{1}\ldots v_{k}$.

• •

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 $\left\lvert{H}\right\rvert$.

• •

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 $H$, let $\partial H=\{ab\mid ab\subset e\in E(H)\}$ denote its 2-shadow graph.

• •

For a hypergraph $H$, the neighborhood of $v$ in $H$ is defined as

[TABLE]

• •

For a hypergraph $H$ and a pair of vertices $u,v\in V(H)$, let $\mathrm{codeg}(v,u)$ denote the number of hyperedges of $H$ containing the pair $\{u,v\}$.

2 Proof of Theorem 3

Let $H$ be a hypergraph on $n$ vertices without a Berge $5$-cycle and let $G=\partial H$ be the $2$-shadow of $H$. First we introduce some definitions.

Definition 4**.**

A pair $xy\in\partial H$ is called thin if $\mathrm{codeg}(xy)=1$, otherwise it is called fat.

We say a hyperedge $abc\in H$ 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 $K_{4}^{3}$ denote the complete $3$-uniform hypergraph on $4$ vertices. A crown of size $k$ is a set of $k\geq 1$ hyperedges of the form $abc_{1},abc_{2},\ldots,abc_{k}$. Below we define $2$ specific hypergraphs:

• •

Let $F_{1}$ be a hypergraph consisting of exactly $3$ hyperedges on $4$ vertices (i.e., $K_{4}^{3}$ 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 $\mathcal{B}$ be a core of $B$. Then $\mathcal{B}$ is either $\emptyset,K_{4}^{3},F_{1},F_{2}$ or a crown of size $k$ for some $k\geq 1$.

Proof.

If $\mathcal{B}=\emptyset$, we are done, so let us assume $\mathcal{B}\not=\emptyset$. Since $B$ is tightly-connected and it can be obtained by adding thin hyperedges to $\mathcal{B}$, it is easy to see that $\mathcal{B}$ is also tightly-connected. Thus if $\mathcal{B}$ 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 $\mathcal{B}$ contains at least $3$ hyperedges.

If $\mathcal{B}$ contains at most $4$ vertices then it is easy to see that $\mathcal{B}$ is either $K_{4}^{3}$ or $F_{1}$. So assume that $\mathcal{B}$ has at least $5$ vertices (and at least $3$ hyperedges). Since $\mathcal{B}$ 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 $h\not=abc$ containing either $ab$ or $ac$. Similarly there exists a hyperedge $f\not=cde$ and $f$ contains $ed$ or $ec$. If $h=f$ then $\mathcal{B}\supseteq F_{2}$. 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 $\mathcal{B}=F_{2}$. If $h\not=f$ then the hyperedges $h,abc,bcd,cde,f$ create a Berge $5$-cycle in $H$, a contradiction. This completes the proof of the lemma. ∎

Observation 9**.**

Let $B$ be a block of $H$ and let $\mathcal{B}$ be the core of $B$. If $\mathcal{B}=\emptyset$ then the block $B$ is a crown, and if $\mathcal{B}\not=\emptyset$ then every fat pair of $B$ is contained in $\partial\mathcal{B}$.

Edge Decomposition of $G=\partial H$. We define a decomposition $\mathcal{D}$ 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 $\mathcal{B}$ be its core.

If $\mathcal{B}=\emptyset$, then $B$ is a crown-block $\{abc_{1},abc_{2},\ldots,abc_{k}\}$ (for some $k\geq 1$); we partition $\partial B$ into the triangle $abc_{1}$ and paths $ac_{i}b$ where $2\leq i\leq k$.

If $\mathcal{B}\not=\emptyset$, then our plan is to first partition $\partial B\setminus\partial\mathcal{B}$. If $abc\in B\setminus\mathcal{B}$, 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 $\partial{\mathcal{B}}$. Indeed, $ac$ has to be a fat pair, otherwise the block $B$ consists of only one hyperedge $abc$, so $\mathcal{B}=\emptyset$ contradicting the assumption. So by Observation 9, $ac$ has to be a pair in $\partial{\mathcal{B}}$. For every $abc\in B\setminus\mathcal{B}$ such that $ab$ and $bc$ are thin pairs, add the $2$-path $abc$ to the edge decomposition $\mathcal{D}$. This partitions all the edges in $\partial B\setminus\partial\mathcal{B}$ into paths of length $2$. So all we have left is to partition the edges of $\partial\mathcal{B}$.

• •

If $\mathcal{B}$ is a crown $\{abc_{1},abc_{2},\ldots,abc_{k}\}$ for some $k\geq 1$, then we partition $\partial B$ into the triangle $abc_{1}$ and paths $ac_{i}b$ where $2\leq i\leq k$.

• •

If $\mathcal{B}=F_{1}=\{abc,bcd,acd\}$ then we partition $\partial\mathcal{B}$ into $2$-paths $abc,bdc$ and $cad$.

• •

If $\mathcal{B}=F_{2}=\{oab,obc,ocd,oda\}$ then we partition $\partial\mathcal{B}$ into $2$-paths $abo,bco,cdo$ and $dao$.

• •

Finally, if $\mathcal{B}=K_{4}^{3}=\{abc,abd,acd,bcd\}$ then we partition $\partial\mathcal{B}$ as $K_{4}$, i.e., we add $\partial\mathcal{B}=K_{4}$ as an element of $\mathcal{D}$.

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 $\mathcal{D}$, then there is a hyperedge $h\in H$ such that $D=\partial h$.

• (b)

If $abc$ is a $2$-path that belongs to $\mathcal{D}$, then $abc\in H$. Moreover $ac$ is a fat pair.

• (c)

If $D$ is a $K_{4}$ that belongs to $\mathcal{D}$, then there exists $F=K_{4}^{3}\subseteq H$ such that $D=\partial F$.

Let $\alpha_{1}\left\lvert{G}\right\rvert$ and $\alpha_{2}\left\lvert{G}\right\rvert$ be the number of edges of $G$ that are contained in triangles and $2$-paths of the edge-decomposition $\mathcal{D}$ of $G$, respectively. So $(1-\alpha_{1}-\alpha_{2})\left\lvert{G}\right\rvert$ edges of $G$ belong to the $K_{4}$’s in $\mathcal{D}$.

Claim 11**.**

We have,

[TABLE]

Proof.

Let $B$ be a block with the core $\mathcal{B}$. Recall that for each hyperedge $h\in B\setminus\mathcal{B}$, we have added exactly one $2$-path or a triangle to $\mathcal{D}$.

Moreover, because of the way we partitioned $\partial\mathcal{B}$, it is easy to check that in all of the cases except when $\mathcal{B}=K_{4}^{3}$, the number of hyperedges of $\mathcal{B}$ is the same as the number of elements of $\mathcal{D}$ that $\partial\mathcal{B}$ is partitioned into; these elements being $2$-paths and triangles. On the other hand, if $\mathcal{B}=K_{4}^{3}$, then the number of hyperedges of $\mathcal{B}$ is $4$ but we added only one element to $\mathcal{D}$ (namely $K_{4}$).

This shows that the number of hyperedges of $H$ is equal to the number of elements of $\mathcal{D}$ that are $2$-paths or triangles plus the number of hyperedges which are in copies of $K_{4}^{3}$ in $H$, i.e., $4$ times the number of $K_{4}$’s in $\mathcal{D}$. Since $\alpha_{1}\left\lvert{G}\right\rvert$ edges of $G$ are in $2$-paths, the number of elements of $\mathcal{D}$ that are $2$-paths is $\alpha_{1}\left\lvert{G}\right\rvert/2$. Similarly, the number of elements of $\mathcal{D}$ that are triangles is $\alpha_{2}\left\lvert{G}\right\rvert/3$, and the number of $K_{4}$’s in $\mathcal{D}$ is $(1-\alpha_{1}-\alpha_{2})\left\lvert{G}\right\rvert/6$. 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 $\{uw\mid uvw\in H\}$ and is denoted by $L_{v}$.

Claim 12**.**

$\left\lvert{L_{v}}\right\rvert\leq 2\left\lvert{N(v)}\right\rvert$.

Proof.

First let us notice that there is no path of length $5$ in $L_{v}$. Indeed, otherwise, there exist vertices $v_{0},v_{1},\ldots,v_{5}$ such that $vv_{i-1}v_{i}\in H$ for each $1\leq i\leq 5$ which means there is a Berge $5$-cycle in $H$ formed by the hyperedges containing the pairs $vv_{1},v_{1}v_{2},v_{2}v_{3},v_{3}v_{4},v_{4}v$, a contradiction. So by the Erdős-Gallai theorem $\left\lvert{L_{v}}\right\rvert\leq\frac{5-1}{2}\left\lvert{N(v)}\right\rvert$, proving the claim. ∎

Lemma 13**.**

Let $v\in V(H)$ be an arbitrary vertex, then the number of edges in $G[N(v)]$ is less than $8\left\lvert{N(v)}\right\rvert$.

Proof.

Let $G_{v}$ be a subgraph of $G$ on a vertex set $N(v)$, such that $xy\in G_{v}$ if and only if there exists a vertex $z\not=v$ such that $xyz\in H$. Then each edge of $G[N(v)]$ belongs to either $L_{v}$ or $G_{v}$, so $\left\lvert{G[N(v)]}\right\rvert\leq\left\lvert{L_{v}}\right\rvert+\left\lvert{G_{v}}\right\rvert$. Combining this with Claim 12, we get $\left\lvert{G[N(v)]}\right\rvert\leq\left\lvert{G_{v}}\right\rvert+2\left\lvert{N(v)}\right\rvert$. So it suffices to prove that $\left\lvert{G_{v}}\right\rvert<6\left\lvert{N(v)}\right\rvert$.

First we will prove that there is no path of length $12$ in $G_{v}$. Let us assume by contradiction that $P=v_{0},v_{1},\ldots,v_{12}$ is a path in $G_{v}$. Since for each pair of vertices $v_{i},v_{i+1}$, there is a hyperedge $v_{i}v_{i+1}x$ in $H$ where $x\not=v$, we can conclude that there is a subsequence $u_{0},u_{1},\ldots,u_{6}$ of $v_{0},v_{1},\ldots,v_{12}$ and a sequence of distinct hyperedges $h_{1},h_{2},\ldots,h_{6}$, such that $u_{i-1}u_{i}\subset h_{i}$ and $v\notin h_{i}$ for each $1\leq i\leq 6$. Since $u_{0},u_{3},u_{6}\in N(v)$ there exist hyperedges $f_{1},f_{2},f_{3}\in H$ such that $vu_{0}\subset f_{1}$, $vu_{3}\subset f_{2}$ and $vu_{6}\subset f_{3}$. Clearly, either $f_{1}\not=f_{2}$ or $f_{2}\not=f_{3}$. 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 $\frac{12-1}{2}\left\lvert{N(v)}\right\rvert<6\left\lvert{N(v)}\right\rvert$, as required. ∎

2.1 Relating the hypergraph degree to the degree in the shadow

For a vertex $v\in V(H)=V(G)$, let $d(v)$ denote the degree of $v$ in $H$ and let $d_{G}(v)$ denote the degree of $v$ in $G$ (i.e., $d_{G}(v)$ is the degree in the shadow).

Clearly $d_{G}(v)\leq 2d(v)$. Moreover, $d(v)=\left\lvert{L_{v}}\right\rvert$ and $d_{G}(v)=\left\lvert{N(v)}\right\rvert$. So by Claim 12, we have

[TABLE]

Let $\overline{d}$ and $\overline{d}_{G}$ be the average degrees of $H$ and $G$ respectively.

Suppose there is a vertex $v$ of $H$, such that $d(v)<\overline{d}/3$. Then we may delete $v$ and all the edges incident to $v$ from $H$ to obtain a graph $H^{\prime}$ whose average degree is more than $3(n\overline{d}/3-\overline{d}/3)/(n-1)=\overline{d}$. Then it is easy to see that if the theorem holds for $H^{\prime}$, then it holds for $H$ as well. Repeating this procedure, we may assume that for every vertex $v$ of $H$, $d(v)\geq\overline{d}/3$. Therefore, by (1), we may assume that the degree of every vertex of $G$ is at least $\overline{d}/6$.

2.2 Counting paths of length $3$

Definition 14**.**

A $2$-path in $\partial H$ is called bad if both of its edges are contained in a triangle of $\partial H$, otherwise it is called good.

Lemma 15**.**

For any vertex $v\in V(G)$ and a set $M\subseteq N(v)$, let $\mathcal{P}$ be the set of the good $2$-paths $vxy$ such that $x\in M$. Let $M^{\prime}=\{y\mid vxy\in\mathcal{P}\}$ then $\left\lvert{\mathcal{P}}\right\rvert<2\left\lvert{M^{\prime}}\right\rvert+48d_{G}(v)$.

Proof.

Let $B_{\mathcal{P}}=\{xy\mid x\in M,y\in M^{\prime},xy\in G\}$ be a bipartite graph, clearly $\left\lvert{B_{\mathcal{P}}}\right\rvert=\left\lvert{\mathcal{P}}\right\rvert$. Let $E=\{xyz\in H\mid x,y\in N(v),\mathrm{codeg}(x,y)\leq 2\}$. By Lemma 13, $\left\lvert{E}\right\rvert\leq 2\cdot 8\left\lvert{N(v)}\right\rvert$ so the number of edges of $2$-shadow of $E$ is $\left\lvert{\partial E}\right\rvert\leq 48\left\lvert{N(v)}\right\rvert$. Let $B=\{xy\in B_{\mathcal{P}}\mid\exists z\in V(H),xyz\in H\setminus E\}$. Then clearly,

[TABLE]

Let $d_{B}(x)$ denote the degree of a vertex $x$ in the graph $B$.

Claim 16**.**

For every $y\in M^{\prime}$ such that $d_{B}(y)=k\geq 3$, there exists a set of $k-2$ vertices $S_{y}\subseteq M^{\prime}$ such that $\forall w\in S_{y}$ we have $d_{B}(w)=1$. Moreover, $S_{y}\cap S_{z}=\emptyset$ for any $y\not=z\in M^{\prime}$ (with $d_{B}(y),d_{B}(z)\geq 3$).

Proof.

Let $yx_{1},yx_{2},\ldots,yx_{k}\in B$ be the edges of $B$ incident to $y$. For each $1\leq j\leq k$ let $f_{j}\in H$ be a hyperedge such that $vx_{j}\subset f_{j}$. For each $yx_{i}\in B$ clearly there is a hyperedge $yx_{i}w_{i}\in H\setminus E$.

We claim that for each $1\leq i\leq k$, $w_{i}\in M^{\prime}$. It is easy to see that $w_{i}\in N(v)$ or $w_{i}\in M^{\prime}$ (because $vx_{i}w_{i}$ is a $2$-path in $G$). Assume for a contradiction that $w_{i}\in N(v)$, then since $yx_{i}w_{i}\notin E$ we have, $\mathrm{codeg}(x_{i},w_{i})\geq 3$. Let $f\in H$ be a hyperedge such that $vw_{i}\subset f$. Now take $j\not=i$ such that $x_{j}\not=w_{i}$. If $f_{j}\not=f$ then since $\mathrm{codeg}(x_{i},w_{i})\geq 3$ there exists a hyperedge $h\supset x_{i}w_{i}$ such that $h\not=f$ and $h\not=x_{i}w_{i}y$, 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 $f_{j}\not=f_{i}$. Similarly in this case, there exists a hyperedge $h\supset x_{i}w_{i}$ such that $h\not=f_{i}$ and $h\not=x_{i}w_{i}y$, 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 $w_{i}\in M^{\prime}$ for each $1\leq i\leq k$.

Claim**.**

For all but at most $2$ of the $w_{i}$’s (where $1\leq i\leq k$), we have $d_{B}(w_{i})=1$.

Proof.

If $d_{B}(w_{i})=1$ for all $1\leq i\leq k$ then we are done, so we may assume that there is $1\leq i\leq k$ such that $d_{B}(w_{i})\not=1$.

For each $1\leq i\leq k$, $w_{i}\in M^{\prime}$ and $x_{i}w_{i}\in\partial(H\setminus E)$ (because $x_{i}w_{i}y\in H\setminus E$), so it is clear that $d_{B}(w_{i})\geq 1$. So $d_{B}(w_{i})>1$. Then there is a vertex $x\in M\setminus\{x_{i}\}$ such that $w_{i}x\in B$. Let $f,h\in H$ be hyperedges with $w_{i}x\in h$ and $xv\in f$. If there are $j,l\in\{1,2,\ldots,k\}\setminus\{i\}$ such that $x,x_{j}$ and $x_{l}$ are all different from each other, then clearly, either $f\not=f_{j}$ or $f\not=f_{l}$, so without loss of generality we may assume $f\not=f_{j}$. 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 $j,l\in\{1,2,\ldots,k\}\setminus\{i\}$ 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 $j\in\{1,2,3\}\setminus\{i\}$ such that $x=x_{j}$. Let $l\in\{1,2,3\}\setminus\{i,j\}$. If $f_{j}\not=f_{l}$ 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 $d_{B}(w_{i})\not=1$ implies that $k=3$ and for $\{j,l\}=\{1,2,3\}\setminus\{i\}$, we have $f_{j}=f_{l}$. So if $d_{B}(w_{i})\not=1$ and $d_{B}(w_{j})\not=1$ 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 $1\leq i\leq k$, $d_{B}(w_{i})\not=1$ 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 $i\not=j$ where $d_{B}(w_{i})=d_{B}(w_{j})=1$ we have $w_{i}\not=w_{j}$. Indeed, if there exists $i\not=j$ 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 $\{w_{1},w_{2},\ldots,w_{k}\}$ 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 $\forall w_{i}\in S_{y}$ we have $d_{B}(w_{i})=1$.)

Now we have to prove that for each $z\not=y$ we have $S_{y}\cap S_{z}=\emptyset$. Assume by contradiction that $w_{i}\in S_{z}\cap S_{y}$ for some $z\not=y$. That is, there is some hyperedge $uw_{i}z\in H\setminus E$ where $u\in M$, moreover $u=x_{i}$ otherwise $d_{B}(w_{i})>1$. So we have a hyperedge $x_{i}w_{i}z\in H\setminus E$ for some $z\in M^{\prime}\setminus\{y\}$. Let $j,l\in\{1,2,\ldots,k\}\setminus\{i\}$ such that $j\not=l$. Recall that $x_{j}v\subset f_{j}$ and $x_{l}v\subset f_{l}$. Clearly either $f_{j}\not=f_{i}$ or $f_{l}\not=f_{i}$ so without loss of generality we can assume $f_{j}\not=f_{i}$. 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 ($x_{j}w_{j}y\not=yx_{i}w_{i}$ because $x_{j}\not=w_{i}$). ∎

For each $x\in M^{\prime}$ with $d_{B}(x)=k\geq 3$, let $S_{x}$ be defined as in Claim 16. Then the average of the degrees of the vertices in $S_{x}\cup\{x\}$ in $B$ is $(k+\left\lvert{S_{x}}\right\rvert)/(k-1)=(2k-2)(k-1)=2$. Since the sets $S_{x}\cup x$ (with $x\in M^{\prime}$, $d_{B}(x)\geq 3$) are disjoint, we can conclude that average degree of the set $M^{\prime}$ is at most $2$. Therefore $2\left\lvert{M^{\prime}}\right\rvert\geq\left\lvert{B}\right\rvert$. So by (2) we have $2\left\lvert{M^{\prime}}\right\rvert\geq\left\lvert{B}\right\rvert>\left\lvert{\mathcal{P}}\right\rvert-48d_{G}(V)$, which completes the proof of the lemma. ∎

Claim 17**.**

We may assume that the maximum degree in the graph $G$ is less than $160\sqrt{n}$ when $n$ is large enough.

Proof.

Let $v$ be an arbitrary vertex with $d_{G}(v)=C\overline{d}$ for some constant $C>0$. Let $\mathcal{P}$ be the set of the good $2$-paths starting from the vertex $v$. Then applying Lemma 15 with $M=N(v)$ and $M^{\prime}=\{y\mid vxy\in\mathcal{P}\}$, we have $\left\lvert{\mathcal{P}}\right\rvert<2\left\lvert{M^{\prime}}\right\rvert+48d_{G}(v)<2n+48\cdot C\overline{d}$. Since the minimum degree in $G$ is at least $\overline{d}/6$, the number of (ordered) $2$-paths starting from $v$ is at least $d(v)\cdot(\overline{d}/6-1)=C\overline{d}\cdot(\overline{d}/6-1)$. Notice that the number of (ordered) bad $2$-paths starting at $v$ is the number of $2$-paths $vxy$ such that $x,y\in N(v)$. So by Lemma 13, this is at most $2\cdot 8\left\lvert{N(v)}\right\rvert=16C\overline{d}$, so the number of good $2$-paths is at least $C\overline{d}\cdot(\overline{d}/6-17)$. So $\left\lvert{\mathcal{P}}\right\rvert\geq C\overline{d}\cdot(\overline{d}/6-17)$. Thus we have

[TABLE]

So $C\overline{d}(\overline{d}/6-65)<2n$. Therefore, $6C(\overline{d}/6-65)^{2}<2n$, i.e., $\overline{d}<6\sqrt{n/3C}+390$, so $\left\lvert{H}\right\rvert=n\overline{d}/3\leq 2n\sqrt{n/3C}+130n$. If $C\geq 36$ we get that $\left\lvert{H}\right\rvert\leq\frac{n^{3/2}}{3\sqrt{3}}+130n=\frac{n^{3/2}}{3\sqrt{3}}+O(n)$, proving Theorem 3. So we may assume $C<36$.

Theorem 2 implies that

[TABLE]

so $\overline{d}\leq 3\sqrt{2}\sqrt{n}+13.5$. So combining this with the fact that $C<36$, we have $d_{G}(v)=C\overline{d}<108\sqrt{2}\sqrt{n}+486<160\sqrt{n}$ for large enough $n$. ∎

Combining Lemma 15 and Claim 17, we obtain the following.

Lemma 18**.**

For any vertex $v\in V(G)$ and a set $M\subseteq N(v)$, let $\mathcal{P}$ be the set of good $2$-paths $vxy$ such that $x\in M$. Let $M^{\prime}=\{y\mid vxy\in\mathcal{P}\}$ then $\left\lvert{\mathcal{P}}\right\rvert<2\left\lvert{M^{\prime}}\right\rvert+7680\sqrt{n}$ when $n$ is large enough.

Definition 19**.**

A $3$-path $x_{0},x_{1},x_{2},x_{3}$ is called good if both $2$-paths $x_{0},x_{1},x_{2}$ and $x_{1},x_{2},x_{3}$ are good $2$-paths.

Claim 20**.**

The number of (ordered) good $3$-paths in $G$ is at least $n\overline{d}_{G}^{3}-C_{0}n^{3/2}\overline{d}_{G}$ 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 $5440n^{3/2}\overline{d}_{G}$.

For any vertex $x\in V(H)$ 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 $2\cdot 8\left\lvert{N(x)}\right\rvert=16d_{G}(x)$ 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 $\sum_{x\in V(H)}17d_{G}(x)=17n\overline{d}_{G}$.

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 $d_{G}(x)+d_{G}(z)<320\sqrt{n}$ (for large enough $n$) by Claim 17. Therefore, the total number of (ordered) $3$-walks that are not good $3$-paths is at most $17n\overline{d}_{G}\cdot 320\sqrt{n}=5440n^{3/2}\overline{d}_{G}$.

By the Blakley-Roy inequality, the total number of (ordered) 3-walks in $G$ is at least $n\overline{d}^{3}_{G}$. By the above discussion, all but at most $5440n^{3/2}\overline{d}_{G}$ of them are good $3$-paths, so letting $C_{0}=5440$ completes the proof of the claim. ∎

Claim 21**.**

Let $\{a,b,c\}$ be the vertex set of a triangle that belongs to $\mathcal{D}$. (By Observation 10 (a) $abc\in H$.) Then the number of good $3$-paths whose first edge is $ab,bc$ or $ca$ is at most $8n+C_{1}\sqrt{n}$ for some constant $C_{1}$ and for large enough $n$.

Proof.

Let $S_{abc}=N(a)\cap N(b)\cap N(c)$. For each $\{x,y\}\subset\{a,b,c\}$, let $S_{xy}=N(x)\cap N(y)\setminus\{a,b,c\}$. For each $x\in\{a,b,c\}$, let $S_{x}=N(x)\setminus(N(y)\cup N(z)\cup\{a,b,c\})$ where $\{y,z\}=\{a,b,c\}\setminus\{x\}$.

For each $x\in\{a,b,c\}$, let $\mathcal{P}_{x}$ be the set of good $2$-paths $xuv$ where $u\in S_{x}$. Let $S^{\prime}_{x}=\{v\mid xuv\in\mathcal{P}_{x}\}$. For each $\{x,y\}\subset\{a,b,c\}$, let $\mathcal{P}_{xy}$ be the set of good $2$-paths $xuv$ and $yuv$ where $u\in S_{xy}$. Let $S^{\prime}_{xy}=\{v\mid xuv\in\mathcal{P}_{xy}\}$.

Let $\{x,y\}\subset\{a,b,c\}$ and $z=\{a,b,c\}\setminus\{x,y\}$. Notice that each $2$-path $yuv\in\mathcal{P}_{xy}$ ($xuv\in\mathcal{P}_{xy}$), is contained in exactly one good $3$-path $zyuv$ (respectively $zxuv$) whose first edge is in the triangle $abc$. Indeed, since $u\in S_{xy}$, $xyuv$ (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 $\left\lvert{\mathcal{P}_{xy}}\right\rvert$. The number of paths in $\mathcal{P}_{xy}$ that start with the vertex $x$ is less than $2\left\lvert{S^{\prime}_{xy}}\right\rvert+7680\sqrt{n}$, by Lemma 18. Similarly, the number of paths in $\mathcal{P}_{xy}$ that start with the vertex $y$ is less than $2\left\lvert{S^{\prime}_{xy}}\right\rvert+7680\sqrt{n}$. Since every path in $\mathcal{P}_{xy}$ starts with either $x$ or $y$, we have $\left\lvert{\mathcal{P}_{xy}}\right\rvert<4\left\lvert{S^{\prime}_{xy}}\right\rvert+15360\sqrt{n}$. Therefore, for any $\{x,y\}\subset\{a,b,c\}$, 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 $4\left\lvert{S^{\prime}_{xy}}\right\rvert+15360\sqrt{n}$.

In total, the number of good $3$-paths whose first edge is in the triangle $abc$ and whose third vertex is in $S_{ab}\cup S_{bc}\cup S_{ac}$ is at most

[TABLE]

Let $x\in\{a,b,c\}$ and $\{y,z\}=\{a,b,c\}\setminus\{x\}$. For any $2$-path $xuv\in\mathcal{P}_{x}$ 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 $S_{a}\cup S_{b}\cup S_{c}$ is $2(\left\lvert{\mathcal{P}_{a}}\right\rvert+\left\lvert{\mathcal{P}_{b}}\right\rvert+\left\lvert{\mathcal{P}_{c}}\right\rvert)$, which is at most

[TABLE]

by Lemma 18.

Now we will prove that every vertex is in at most $2$ of the sets $S^{\prime}_{a},S^{\prime}_{b},S^{\prime}_{c},S^{\prime}_{ab},S^{\prime}_{bc},S^{\prime}_{ac}$. Let us assume by contradiction that a vertex $v\in V(G)\setminus\{a,b,c\}$ is in at least $3$ of them. We claim that there do not exist $3$ vertices $u_{a}\in N(a)\setminus\{b,c\}$, $u_{b}\in N(b)\setminus\{a,c\}$ and $u_{c}\in N(c)\setminus\{a,b\}$ such that $xu_{x}v$ is a good $3$-path for each $x\in\{a,b,c\}$. Indeed, otherwise, consider hyperedges $h_{a},h^{\prime}_{a}$ containing the pairs $au_{a}$ and $u_{a}v$ respectively (since $au_{a}v$ is a good $2$-path, note that $h_{a}\not=h_{a}^{\prime}$), and hyperedges $h_{b},h^{\prime}_{b},h_{c},h^{\prime}_{c}$ containing the pairs $bu_{b},u_{b}v,cu_{c},u_{c}v$ respectively. Then either $h_{a}^{\prime}\not=h_{b}^{\prime}$ or $h_{a}^{\prime}\not=h_{c}^{\prime}$, say $h_{a}^{\prime}\not=h_{b}^{\prime}$ without loss of generality. Then the hyperedges $h_{a},h_{a}^{\prime},h_{b}^{\prime},h_{b},abc$ create a Berge $5$-cycle in $H$, a contradiction, proving that it is impossible to have 3 vertices $u_{a}\in N(a)\setminus\{b,c\}$, $u_{b}\in N(b)\setminus\{a,c\}$ and $u_{c}\in N(c)\setminus\{a,b\}$ with the above mentioned property. Without loss of generality let us assume that there is no vertex $u_{a}\in N(a)\setminus\{b,c\}$ such that $au_{a}v$ is a good $2$-path – in other words, $v\notin S^{\prime}_{a}\cup S^{\prime}_{ab}\cup S^{\prime}_{ac}$. However, since we assumed that $v$ is contained in at least 3 of the sets $S^{\prime}_{a},S^{\prime}_{b},S^{\prime}_{c},S^{\prime}_{ab},S^{\prime}_{bc},S^{\prime}_{ac}$, we can conclude that $v$ is contained in all 3 of the sets $S^{\prime}_{b}$, $S^{\prime}_{c}$, $S^{\prime}_{bc}$, i.e., there are vertices $u_{b}\in S_{b},u_{c}\in S_{c},u\in S_{bc}$ such that $vu_{b}b,vu_{c}c,vub,vuc$ are good $2$-paths. Using a similar argument as before, if $vu\in h$, $vu_{b}\in h_{b}$ and $vu_{c}\in h_{c}$, without loss of generality we can assume that $h\not=h_{b}$, 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

[TABLE]

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

[TABLE]

for $C_{1}=92160$ and large enough $n$, finishing the proof of the claim. ∎

Claim 22**.**

Let $P=abc$ be a $2$-path and $P\in\mathcal{D}$. (By Observation 10 (b) $abc\in H$.) Then the number of good $3$-paths whose first edge is $ab$ or $bc$ is at most $4n+C_{2}\sqrt{n}$ for some constant $C_{2}>0$ and large enough $n$.

Proof.

First we bound the number of $3$-paths whose first edge is $ab$. Let $S_{ab}=N(a)\cap N(b)$. Let $S_{a}=N(a)\setminus(N(b)\cup\{b\})$ and $S_{b}=N(b)\setminus(N(a)\cup\{a\})$. For each $x\in\{a,b\}$, let $\mathcal{P}_{x}$ be the set of good $2$-paths $xuv$ where $u\in S_{x}$, and let $S_{x}^{\prime}=\{v\mid xuv\in\mathcal{P}_{x}\}$. The set of good $3$-paths whose first edge is $ab$ is $\mathcal{P}_{a}\cup\mathcal{P}_{b}$, because the third vertex of a good $3$-path starting with an edge $ab$ can not belong to $N(a)\cap N(b)$ by the definition of a good $3$-path.

We claim that $\left\lvert{S^{\prime}_{a}\cap S^{\prime}_{b}}\right\rvert\leq 160\sqrt{n}$. Let us assume by contradiction that $v_{0},v_{1},\ldots v_{k}\in S^{\prime}_{a}\cap S^{\prime}_{b}$ for $k>160\sqrt{n}$. For each vertex $v_{i}$ where $0\leq i\leq k$, there are vertices $a_{i}\in S_{a}$ and $b_{i}\in S_{b}$ such that $aa_{i}v_{i},bb_{i}v_{i}$ are good $2$-paths. For each $0\leq i\leq k$, the hyperedge $a_{i}v_{i}b_{i}$ is in $H$, otherwise we can find distinct hyperedges containing the pairs $aa_{i},a_{i}v_{i},v_{i}b_{i},b_{i}b$ and these hyperedges together with $abc$, would form a Berge $5$-cycle in $H$, a contradiction. We claim that there are $j,l\in\{0,1,\ldots,k\}$ such that $a_{j}\not=a_{l}$, otherwise there is a vertex $x$ such that $x=a_{i}$ for each $0\leq i\leq k$. Then $xv_{i}\in G$ for each $0\leq i\leq k$, so we get that $d_{G}(x)>k>160\sqrt{n}$ which contradicts Claim 17.

So there are $j,l\in\{0,1,\ldots,k\}$ such that $a_{j}\not=a_{l}$ and $a_{j}v_{j}b_{j},a_{l}v_{l}b_{l}\in H$. By observation 10 (b), there is a hyperedge $h\not=abc$ such that $ac\subset h$. Clearly either $a_{j}\notin h$ or $a_{l}\notin h$. Without loss of generality let $a_{j}\notin h$, so there is a hyperedge $h_{a}$ with $aa_{j}\subset h_{a}\not=h$. Let $h_{b}\supset b_{j}b$, then the hyperedges $abc,h,h_{a},a_{j}v_{j}b_{j},h_{b}$ form a Berge $5$-cycle, a contradiction, proving that $\left\lvert{S^{\prime}_{a}\cap S^{\prime}_{b}}\right\rvert\leq 160\sqrt{n}$.

Notice that $\left\lvert{S^{\prime}_{a}}\right\rvert+\left\lvert{S^{\prime}_{b}}\right\rvert=\left\lvert{S^{\prime}_{a}\cup S^{\prime}_{b}}\right\rvert+\left\lvert{S^{\prime}_{a}\cap S^{\prime}_{b}}\right\rvert\leq n+160\sqrt{n}$. So by Lemma 18, we have

[TABLE]

for large enough $n$. So the number of good $3$-paths whose first edge is $ab$ is at most $2n+15680\sqrt{n}$. By the same argument, the number of good $3$-paths whose first edge is $bc$ is at most $2n+15680\sqrt{n}$. Their sum is at most $4n+C_{2}\sqrt{n}$ for $C_{2}=31360$ and large enough $n$, as desired. ∎

Claim 23**.**

Let $\{a,b,c,d\}$ be the vertex set of a $K_{4}$ that belongs to $\mathcal{D}$. Let $F=K_{4}^{3}$ be a hypergraph on the vertex set $\{a,b,c,d\}$. (By Observation 10 (c) $F\subseteq H$.) Then the number of good $3$-paths whose first edge belongs to $\partial F$ is at most $6n+C_{3}\sqrt{n}$ 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 $x,y\in\{a,b,c,d\}$ in the hypergraph $H\setminus F$, 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 $G\setminus\partial F$, because otherwise we would find a Berge path of length $2,3$ or $4$ between $x$ and $y$ in $H\setminus F$.

Let $S=\{u\in V(H)\setminus\{a,b,c,d\}\mid\exists\{x,y\}\subset\{a,b,c,d\},u\in N(x)\cap N(y)\}$. For each $x\in\{a,b,c,d\}$, let $S_{x}=N(x)\setminus(S\cup\{a,b,c,d\})$. Let $\mathcal{P}_{S}$ be the set of good $2$-paths $xuv$ where $x\in\{a,b,c,d\}$ and $u\in S$. Let $S^{\prime}=\{v\mid xuv\in\mathcal{P}_{S}\}$. For each $x\in\{a,b,c,d\}$, let $\mathcal{P}_{x}$ be the set of good $2$-paths $xuv$ where $u\in S_{x}$, and let $S_{x}^{\prime}=\{v\mid xuv\in\mathcal{P}_{x}\}$.

Let $v\in S^{\prime}$. By definition, there exists a pair of vertices $\{x,y\}\subset\{a,b,c,d\}$ and a vertex $u$, such that $xuv$ and $yuv$ are good $2$-paths.

Suppose that $zu^{\prime}v$ is a $2$-path different from $xuv$ and $yuv$ where $z\in\{a,b,c,d\}$. If $u^{\prime}=u$ then $z\notin\{x,y\}$ so there is a Berge $2$-path between $x$ and $y$ or between $x$ and $z$ in $H\setminus F$, which is impossible. So $u\not=u^{\prime}$. Either $z\not=x$ or $z\not=y$, without loss of generality let us assume that $z\not=x$. Then $zu^{\prime}vux$ is a path of length $4$ in $G\setminus\partial F$, a contradiction. So for any $v\in S^{\prime}$ there are only $2$ paths of $\mathcal{P}_{a}\cup\mathcal{P}_{b}\cup\mathcal{P}_{c}\cup\mathcal{P}_{d}\cup\mathcal{P}_{S}$ that contain $v$ as an end vertex – both of which are in $\mathcal{P}_{S}$ – which means that $v\notin S_{a}^{\prime}\cup S_{b}^{\prime}\cup S_{c}^{\prime}\cup S_{d}^{\prime}$, so $S^{\prime}\cap(S_{a}^{\prime}\cup S_{b}^{\prime}\cup S_{c}^{\prime}\cup S_{d}^{\prime})=\emptyset$. Moreover,

[TABLE]

We claim that $S_{a}^{\prime}$ and $S_{b}^{\prime}$ are disjoint. Indeed, otherwise, if $v\in S_{a}^{\prime}\cap S_{b}^{\prime}$ there exists $x\in S_{a}$ and $y\in S_{b}$ such that $vxa$ and $vyb$ are paths in $G$, so there is a $4$-path $axvyb$ between vertices of $F$ in $G\setminus\partial F$, a contradiction. Similarly we can prove that $S_{a}^{\prime},S_{b}^{\prime},S_{c}^{\prime}$ and $S_{d}^{\prime}$ are pairwise disjoint. This shows that the sets $S^{\prime},S_{a}^{\prime},S_{b}^{\prime},S_{c}^{\prime}$ and $S_{d}^{\prime}$ are pairwise disjoint. So we have

[TABLE]

By Lemma 18, we have $\left\lvert{\mathcal{P}_{a}}\right\rvert+\left\lvert{\mathcal{P}_{b}}\right\rvert+\left\lvert{\mathcal{P}_{c}}\right\rvert+\left\lvert{\mathcal{P}_{d}}\right\rvert\leq 2(\left\lvert{S_{a}^{\prime}}\right\rvert+\left\lvert{S_{b}^{\prime}}\right\rvert+\left\lvert{S_{c}^{\prime}}\right\rvert+\left\lvert{S_{d}^{\prime}}\right\rvert)+4\cdot 7680\sqrt{n}.$ Combining this inequality with (6), we get

[TABLE]

Combining (7) with (8) we get

[TABLE]

for large enough $n$.

Each $2$-path in $\mathcal{P}_{S}\cup\mathcal{P}_{a}\cup\mathcal{P}_{b}\cup\mathcal{P}_{c}\cup\mathcal{P}_{d}$ can be extended to at most three good $3$-paths whose first edge is in $\partial F$. (For example, $auv\in\mathcal{P}_{a}$ can be extended to $bauv,cauv$ and $dauv$.) On the other hand, every good $3$-path whose first edge is in $\partial F$ must contain a $2$-path of $\mathcal{P}_{a}\cup\mathcal{P}_{b}\cup\mathcal{P}_{c}\cup\mathcal{P}_{d}\cup\mathcal{P}_{S}$ as a subpath. So the number of good $3$-paths whose first edge is in $\partial F$ is at most $3\left\lvert{\mathcal{P}_{a}\cup\mathcal{P}_{b}\cup\mathcal{P}_{c}\cup\mathcal{P}_{d}\cup\mathcal{P}_{S}}\right\rvert=3(\left\lvert{\mathcal{P}_{S}}\right\rvert+\left\lvert{\mathcal{P}_{a}}\right\rvert+\left\lvert{\mathcal{P}_{b}}\right\rvert+\left\lvert{\mathcal{P}_{c}}\right\rvert+\left\lvert{\mathcal{P}_{d}}\right\rvert)$ which is at most $6n+C_{3}\sqrt{n}$ by (9), for $C_{3}=92160$ and large enough $n$, proving the desired claim. ∎

2.3 Combining bounds on the number of $3$-paths

Recall that $\alpha_{1}\left\lvert{G}\right\rvert$, $\alpha_{2}\left\lvert{G}\right\rvert$, $(1-\alpha_{1}-\alpha_{2})\left\lvert{G}\right\rvert$ are the number of edges of $G$ that are contained in triangles, $2$-paths and $K_{4}$’s of the edge-decomposition $\mathcal{D}$ of $G$, respectively. Then the number of triangles, $2$-paths and $K_{4}$’s in $\mathcal{D}$ is $\alpha_{1}\left\lvert{G}\right\rvert/3$, $\alpha_{2}\left\lvert{G}\right\rvert/2$ and $(1-\alpha_{1}-\alpha_{2})\left\lvert{G}\right\rvert/6$ respectively. Therefore, using Claim 21, Claim 22 and Claim 23, the total number of (ordered) good $3$-paths in $G$ is at most

[TABLE]

[TABLE]

[TABLE]

Combining this with the fact that the number of good $3$-paths is at least $n\overline{d}_{G}^{3}-C_{0}n^{3/2}\overline{d}_{G}$ (see Claim 20), we get

[TABLE]

Rearranging and dividing by $n\overline{d}_{G}$ on both sides, we get

[TABLE]

Using the fact that $(5\alpha_{1}+3\alpha_{2}+3)/6\geq 1/2$, it follows that

[TABLE]

So letting $C_{4}=(C_{1}+C_{2}+C_{3})+2C_{0}$ we have,

[TABLE]

for large enough $n$. By Claim 11, we have

[TABLE]

Combining this with (10) we get

[TABLE]

for sufficiently large $n$. So we have

[TABLE]

The right hand side is maximized when $\alpha_{1}=0$ and $\alpha_{2}=2/3$, so we have

[TABLE]

This finishes the proof.

Acknowledgements

The research of the authors is partially supported by the National Research, Development and Innovation Office – NKFIH, grant K116769.