Super-pancyclic hypergraphs and bipartite graphs

TL;DR

This paper establishes Dirac-type conditions for hypergraphs to be hamiltonian and super-pancyclic, extending results to bipartite graphs and proving a longstanding conjecture on long cycles.

Contribution

It introduces new Dirac-type criteria for hypergraphs to be super-pancyclic and extends Jackson's results on bipartite graphs, including a proof of a 1981 conjecture.

Findings

Hypergraphs with few edges can be hamiltonian under certain conditions.

Conditions imply hypergraphs are super-pancyclic, containing Berge cycles for large vertex subsets.

Proved Jackson's conjecture on long cycles in 2-connected bipartite graphs.

Abstract

We find Dirac-type sufficient conditions for a hypergraph $\mathcal H$ with few edges to be hamiltonian. We also show that these conditions provide that $\mathcal H$ is {\em super-pancyclic}, i.e., for each $A \subseteq V(\mathcal H)$ with $|A| \geq 3$, $\mathcal H$ contains a Berge cycle with vertex set $A$. We mostly use the language of bipartite graphs, because every bipartite graph is the incidence graph of a multihypergraph. In particular, we extend some results of Jackson on the existence of long cycles in bipartite graphs where the vertices in one part have high minimum degree. Furthermore, we prove a conjecture of Jackson from 1981 on long cycles in 2-connected bipartite graphs.

Equations

$$\delta \ge n$$

$$A\subseteq X$$

$$y_1{x}_1\dots y_{i}x{y}^{\prime }{x}_i{y}_{i+1}{x}_{i+1}\dots y_{\ell }{x}_{\ell }{y}_1$$

$$y_1{x}_1\dots y_{i}x{y}_j{x}_{j-1}{y}_{j-1}\dots x_i{y}^{\prime }{x}_j{y}_{k+1}\dots y_{\ell }{x}_{\ell }{y}_1$$

$$|N(x_i)\cap V(C)|+|N(x_j)\cap V(C)|\le |V(C)\cap Y|+2.$$

$$I_1:=\{k\in \{i+1,\dots ,j-1\}:y_k\in N(x_j),y_{k+1}\in N(x_i)\}.$$

$$I_2:=\{k\in \{j+1,\dots ,\ell \}\cup \{1,\dots ,i-1\}:y_k\in N(x_i),y_{k+1}\in N(x_j)\}.$$

$$|N(x_i)\cap V(C)|+|N(x_j)\cap V(C)|$$

$$C^{\prime }:=y_{1}x{y}_j{x}_{j-1}{y}_{j-1}\dots y_2{x}_1{y}_{j+1}{x}_{j+2}\dots y_{\ell }{x}_{\ell }{y}_1.$$

$$\displaystyle|Y|\geq|N(x)\cup\Big{(}\bigcup_{y_{i}\in T}N(x_{i})\Big{)}|$$

$$=(t+1)\delta -\ell -(t-1)t-t+\ell$$

$$=3\delta +(t-2)\delta -t^2$$

$$\ge 3\delta +(t-2)(t+2)-t^2=3\delta -4.$$

$$|Y|\ge (t+1)\delta -\frac{t}{2}(\ell +2)-t+\ell .$$

$$|Y|\ge (t+1)\delta -\frac{t}{2}(\ell +2)-t+\ell$$

$$\ge (t+1)\delta -(\delta -1)(\frac{t}{2}-1)-2t$$

$$\ge (\frac{t}{2}+2)\delta -\frac{3}{2}t-1$$

$$=3\delta +(\frac{t}{2}-1)\delta -\frac{3}{2}t-1$$

$$\ge 3\delta +(\frac{t}{2}-1)4-\frac{3}{2}t-1$$

$$\ge 3\delta +\frac{1}{2}t-5$$

$$\ge 3\delta -4.$$

$$|Y|\ge (t+1)\delta -\frac{t-1}{2}(\ell +2)-\frac{\ell +2}{2}-t+\ell ,$$

$$y_{i+1}\dots x_k{y}^{\prime }x{y}_i{y}_{i-1}\dots y_{k+1}{x}_i{y}_{i+1}$$

$$y_{i+1}\dots x_k{y}^{\prime }{x}_q{y}_{q+1}\dots y_{i}x{y}_q{x}_{q-1}\dots y_{k+1}{x}_i{y}_{i+1}.$$

$$x_q{y}_{q+1}\dots y_k{x}_j{y}_{j+1}\dots y_{q}x{y}_j{y}_{j-1}\dots x_k{y}^{\prime }{x}_q.$$

$$y_{i+1}\dots x_k{y}^{\prime }{x}_q{y}_{q+1}\dots y_{i}x{y}_q{x}_{q-1}\dots y_{k+1}{x}_i{y}_{i+1}.$$

$$C^{\prime }:=y_{i+1}\dots y_k{x}_j{y}_{j+1}\dots y_{i}x{y}_j{x}_{j-1}\dots y_{k+1}{x}_i{y}_{i+1}.$$

$$|Y|\ge (t+1)\delta -(t-1)(\ell -s)-t-t+\ell .$$

$$|Y|\ge (t+1)\delta -(t-1)(\ell -s)-t-t+\ell$$

$$\ge (t+1)\delta -(t-2)(\delta -1)+s(t-1)-2t$$

$$=3\delta +(t-2)+s(t-1)-2t$$

$$\ge 3\delta +t-2+t-1-2t$$

$$=3\delta -3.$$

$$y_1{P}^{\prime }{y}^{\prime }{x}_1{y}_2\dots y_{\ell }{x}_{\ell }{y}_1$$

$$C^{\prime }:=y_1{P}_1\cup P_2{z}_2{y}_j{x}_{j-1}{y}_{j-1}\dots y_2{x}_1{y}^{\prime }{x}_{j+1}{y}_{j+2}\dots y_{\ell }{x}_{\ell }{y}_1$$

$$y_1{P}_1\cup P_2{z}_2{y}_j{x}_{j-1}{y}_{j-1}\dots y_2{x}_1{y}_k{x}_{k-1}\dots y_{j+2}{x}_{j+1}{y}_{k+1}{x}_{k+1}\dots y_{\ell }{x}_{\ell }{y}_1$$

$$y_1{P}_1\cup P_2{z}_2{y}_j{x}_{j-1}{y}_{j-1}\dots y_{k+1}{x}_1{y}_2\dots x_{k-1}{y}_k{x}_{j+1}{y}_{j+2}\dots y_{m}x{y}_1$$

Full text

We mostly use the language of bipartite graphs, because every bipartite graph is the incidence graph of a multihypergraph. In particular, we extend some results of Jackson on the existence of long cycles in bipartite graphs where the vertices in one part have high minimum degree. Furthermore, we prove a conjecture of Jackson from 1981 on long cycles in 2-connected bipartite graphs.

Mathematics Subject Classification: 05D05, 05C65, 05C38, 05C35.

Keywords: Berge cycles, extremal hypergraph theory, bipartite graphs.

1 Introduction

1.1 Cycles in bipartite graphs

For integers $n,m,$ and $\delta$ with $\delta\leq m$, we denote by ${\mathcal{G}}(n,m,\delta)$ the set of all bipartite graphs with partition $(X,Y)$ such that $|X|=n\geq 2,|Y|=m$ and for every $x\in X$, $d(x)\geq\delta$. Jackson [3, 4] proved the following theorem (among several other results) on the existence of long cycles in graphs ${\mathcal{G}}(n,m,\delta)$.

Theorem 1.1** (Jackson [3]).**

If a graph $G\in{\mathcal{G}}(n,m,\delta)$ satisfies $n\leq\delta$ and $m\leq 2\delta-2$, then it contains a cycle of length $2n$, i.e., a cycle that covers $X$.

The following two constructions with $m=2\delta-1$ show that Theorem 1.1 is best possible.

Example 1.2**.**

For $\delta=n$, let $G_{1}(n)\in{\mathcal{G}}(\delta,2\delta-1,\delta)$ be obtained from a copy of $K_{\delta,\delta-1}$ where every vertex in $X$ has an additional neighbor of degree 1. Then the longest cycle of $G_{1}(n)$ has length $2(n-1)$.

Example 1.3**.**

Fix positive integers $a\geq b$ such that $a+b=n$. Let $G_{2}(a,b)\in{\mathcal{G}}(n,2\delta-1,\delta)$ be the bipartite graph obtained from a copy $H_{1}$ of $K_{a,\delta}$ and a copy $H_{2}$ of $K_{b,\delta}$ by gluing together a vertex of $H_{1}$ in a part of size $\delta$ and a vertex of $H_{2}$ in a part of size $\delta$. Then the longest cycle of $G_{2}(a,b)$ has length $2a\leq 2(n-1)$.

Both examples above contain cut vertices. Jackson conjectured that if $G\in{\mathcal{G}}(n,m,\delta)$ is 2-connected, then we may relax the upper bound on $m$.

Conjecture 1.4** (Jackson [3]).**

Let $m,n,\delta$ be integers. If some graph $G\in{\mathcal{G}}(n,m,\delta)$ is 2-connected and satisfies

1. (i)

$m\leq 3\delta-5$* if $n\leq\delta$, or* 2. (ii)

$m\leq\lfloor\frac{2(n-\alpha)}{\delta-1-\alpha}\rfloor(\delta-2)+1$* if $n\geq\delta$,*

where $\alpha=1$ if $\delta$ is even and $\alpha=0$ if $\delta$ is odd, then $G$ contains a cycle of length $2\min(n,\delta)$.

If true, both (i) and (ii) would be best possible.

Example 1.5**.**

For (i), fix positive integers $n_{1}\geq n_{2}\geq n_{3}$ such that $n_{1}+n_{2}+n_{3}=n$. Let $G_{3}(n_{1},n_{2},n_{3})\in{\mathcal{G}}(n,3\delta-4,\delta)$ be the bipartite graph obtained from $K_{\delta-2,n_{1}}\cup K_{\delta-2,n_{2}}\cup K_{\delta-2,n_{3}}$ by adding two vertices $a$ and $b$ that are both adjacent to every vertex in the parts of size $n_{1},n_{2}$, and $n_{3}$. Then a longest cycle in $G_{3}(n_{1},n_{2},n_{3})$ has length $2(n_{1}+n_{2})\leq 2(n-1)$.

An extremal construction for (ii) is given in [3].

In this paper we prove part (i) of Conjecture 1.4.

Theorem 1.6**.**

Fix integers $n,m,\delta$ such that $n\leq\delta\leq m\leq 3\delta-5$. If $G\in{\mathcal{G}}(n,m,\delta)$ is 2-connected, then $G$ contains a cycle of length $2n$, i.e., a cycle that covers $X$.

We also prove the following extension of Theorem 1.1.

Theorem 1.7**.**

Let $\delta\geq n$ and $m\leq 2\delta-1$. If $G\in{\mathcal{G}}(n,m,\delta)$ does not contain a cycle of length $2n$, then either $G=G_{1}(n)$ in Example 1.2 or $G=G_{2}(a,b)$ for some $a$ and $b$ with $a+b=n$ in Example 1.3.

The setup of bipartite graphs with large degrees in one part is useful for finding long cycles in hypergraphs with high minimum degree.

1.2 Berge cycles in hypergraphs

A hypergraph ${\mathcal{H}}$ is a set of vertices $V({\mathcal{H}})$ and a set of edges $E({\mathcal{H}})$ such that each edge is a subset of $V({\mathcal{H}})$. Often we take $V({\mathcal{H}})=[n]$ where $[n]:=\{1,\ldots,n\}$ is the set of the first $n$ integers.

We consider hypergraphs without restrictions on edge cardinality (i.e., our hypergraphs may have edges of any size). The degree of a vertex $v$, denoted $d(v)$, is the number of edges that contain $v$. The minimum degree of a hypergraph ${\mathcal{H}}$ is denoted $\delta({\mathcal{H}}):=\min_{v\in V({\mathcal{H}})}d(v)$. The co-degree of a set of vertices $\{v_{1},\ldots,v_{\ell}\}$ is the number of edges that contain the set $\{v_{1},\ldots,v_{\ell}\}$.

Definition 1.8**.**

A Berge cycle of length $\ell$ in a hypergraph is a set of $\ell$ distinct vertices $\{v_{1},\ldots,v_{\ell}\}$ and $\ell$ distinct edges $\{e_{1},\ldots,e_{\ell}\}$ such that for every $i\in[\ell]$, $v_{i},v_{i+1}\in e_{i}$ with indices taken modulo $\ell$. The vertices $\{v_{1},\ldots,v_{\ell}\}$ are called the base vertices of the Berge cycle.

Furthermore, a Berge hamiltonian cycle in a hypergraph ${\mathcal{H}}$ is a Berge cycle whose set of base vertices is $V({\mathcal{H}})$.

Definition 1.9**.**

*Let ${\mathcal{H}}=(V({\mathcal{H}}),E({\mathcal{H}}))$ be a hypergraph. The **incidence graph *** of ${\mathcal{H}}$ is the bipartite graph $I({\mathcal{H}})$ with parts $(X,Y)$ where $X=V({\mathcal{H}})$, $Y=E({\mathcal{H}})$ such that for $e\in Y,v\in X,$ $ev\in E(I({\mathcal{H}}))$ if and only if the vertex $v$ is contained in the edge $e$ in ${\mathcal{H}}$.

If ${\mathcal{H}}$ has $n$ vertices, $m$ edges and minimum degree at least $\delta$, then $I({\mathcal{H}})\in{\mathcal{G}}(n,m,\delta)$. Furthermore, if $\{v_{1},\ldots,v_{\ell}\}$ and $\{e_{1},\ldots,e_{\ell}\}$ form a Berge cycle of length $\ell$ in ${\mathcal{H}}$, then $v_{1}e_{1}\ldots v_{\ell}e_{\ell}v_{1}$ is a cycle of length $2\ell$ in $I({\mathcal{H}})$, and vice versa.

Using incidence graphs, we also define 2-connectedness in hypergraphs.

Definition 1.10**.**

A hypergraph is 2-connected if its incidence graph is 2-connected.

Recall that Dirac’s Theorem states that every $n$-vertex graph with minimum degree at least $n/2$ is hamiltonian. This degree is much less than one needs to guarantee that an $n$-vertex hypergraph has a Berge hamiltonian cycle.

Example 1.11**.**

Let $V({\mathcal{H}})=V_{1}\cup V_{2}$ where $|V_{1}|=\lfloor(n+1)/2\rfloor$, $|V_{2}|=\lceil(n+1)/2\rceil$, $V_{1}\cap V_{2}=\{v\}$, and let $E({\mathcal{H}})$ consist of all sets of size at least 2 contained either in $V_{1}$ or in $V_{2}$. Then ${\mathcal{H}}$ has minimum degree $2^{\lfloor(n+1)/2\rfloor-1}-1$ but no Berge hamiltonian cycle, since the incidence graph has a cut vertex. If ${\mathcal{H}}^{\prime}$ is formed by the edges of ${\mathcal{H}}$ of size $n/4$, then this $n/4$-uniform hypergraph still has an exponential in $n$ minimum degree.

Example 1.12**.**

Let $V({\mathcal{H}})=V_{1}\cup V_{2}$ where $|V_{1}|=\lceil(n+2)/2\rceil$, $|V_{2}|=\lfloor(n-2)/2\rfloor$, $V_{1}\cap V_{2}=\emptyset$, and let $E({\mathcal{H}})=E_{1}\cup E_{2}$, where $E_{1}$ is the set of all subsets $A$ of $V({\mathcal{H}})$ of size $\lceil n/4\rceil$ such that $|V_{1}\cap A|=1$ (and $|V_{2}\cap A|=\lceil n/4\rceil-1$), and $E_{2}=\{V_{1}\}$. Then ${\mathcal{H}}$ has an exponential in $n$ minimum degree, high connectivity and positive codegree of each pair of the vertices. But again, ${\mathcal{H}}$ has no Berge hamiltonian cycle.

Thus without additional restrictions, the bounds on minimum degree, in terms of the number of vertices of a hypergraph, that guarantee a Berge hamiltonian cycle are far from linear. On the other hand, by translating Theorems 1.6 and 1.7 into the language of hypergraphs, we obtain the following lower bounds on the minimum degree in terms of the number of vertices and edges guaranteeing existence of hamiltonian cycles.

Theorem 1.13**.**

Fix integers $n,m,\delta$ such that

[TABLE]

If ${\mathcal{H}}$ is a 2-connected $n$-vertex hypergraph with $m$ edges and minimum degree at least $\delta$, then ${\mathcal{H}}$ has a hamiltonian Berge cycle.

Theorem 1.14**.**

Let $\delta\geq n$ and $\delta\geq(m+1)/2$. If an $n$-vertex hypergraph ${\mathcal{H}}$ with $m$ edges and minimum degree at least $\delta$ has no hamiltonian Berge cycle, then the incidence graph $I({\mathcal{H}})$ is either $G_{1}(n)$ in Example 1.2 or $G_{2}(a,b)$ for some $a$ and $b$ with $a+b=n$ in Example 1.3.

Examples 1.2 and 1.3 show the sharpness of Theorem 1.14, and Example 1.5 shows the sharpness of Theorem 1.13.

On the other hand, Condition (1) in a 2-connected $n$-vertex hypergraph ${\mathcal{H}}$ with $m$ edges and minimum degree at least $\delta$ implies more than simply a hamiltonian cycle in ${\mathcal{H}}$. We discuss this in the next section.

1.3 Pancyclic graphs and hypergraphs

An $n$-vertex graph is pancyclic if it contains a cycle of length $\ell$ for every $3\leq\ell\leq n$. There are many results on pancyclic graphs. For instance, Bondy [1] proved that every Hamiltonian graph with at least $n^{2}/4$ edges is either the balanced complete bipartite graph or is pancyclic. Later, Bondy [2] made the metaconjecture that every nontrivial condition that implies that a graph $G$ is hamiltonian also implies that $G$ is pancyclic. See [7] for more results on pancyclic graphs. In this paper, we consider similar notions for hypergraphs.

Definition 1.15**.**

A hypergraph ${\mathcal{H}}$ is pancyclic if it contains a Berge cycle of length $\ell$ for every $\ell\geq 3$. Furthermore, we say that ${\mathcal{H}}$ is super-pancyclic if for every $A\subseteq V({\mathcal{H}})$ with $|A|\geq 3$, ${\mathcal{H}}$ has a Berge cycle whose set of base vertices is $A$.

We can similarly define super-pancyclic bipartite graphs.

Definition 1.16**.**

A bipartite graph $G$ with partition $(X,Y)$ is $X$-super-pancyclic* if for every $X^{\prime}\subseteq X$ with $|X^{\prime}|\geq 3$, $G$ has a cycle $C$ with $V(C)\cap X=X^{\prime}$.*

An interesting result on pancyclic hypergraphs was proved by Lu and Wang [6]:

Theorem 1.17** (Lu, Wang [6]).**

Let $R$ be a finite set of integers. Then there exists some integer $n_{0}=n_{0}(R)$ such that for every $n\geq n_{0}$, each $n$-vertex hypergraph ${\mathcal{H}}$ with edge cardinalities in $R$ and minimum co-degree at least 1 is pancyclic. Furthermore, for each set of vertices $A$ with $|A|\geq n_{0}$, ${\mathcal{H}}$ contains a Berge cycle with the set of base vertices $A$.

Also, Jackson’s Theorem (Theorem 1.1) implies the following result on pancyclic hypergraphs.

Theorem 1.18** (Hypergraph version of Theorem 1.1).**

Suppose $\delta\geq n$ and $\delta\geq(m+2)/2$. Then every $n$-vertex hypergraph with $m$ edges and minimum degree at least $\delta$ is super-pancyclic.

One of the main goals of this paper is to extend this result to more general hypergraphs.

If $A$ is a subset of the part $X$ in a graph $G\in{\mathcal{G}}(n,m,\delta)$ and $C$ is a cycle in $G$ with $V(C)\cap X=A$ and $V(C)\cap Y=B$, then $G[A\cup B]$ is $2$-connected. Thus every $X$-super-pancyclic bipartite graph satisfies:

[TABLE]

We will show that this necessary condition for $G$ to be $X$-super-pancyclic is also sufficient when $m\leq 3\delta-5$.

Theorem 1.19**.**

Let $\delta\geq n$ and $m\leq 3\delta-5$. If $G\in{\mathcal{G}}(n,m,\delta)$ satisfies (2), then $G$ is $X$-super-pancyclic.

In terms of hypergraphs our result is as follows

Corollary 1.20** (Hypergraph version of Theorem 1.19).**

Let $\delta\geq n$ and $m\leq 3\delta-5$. If the incidence graph of an $n$-vertex hypergraph ${\mathcal{H}}$ with $m$ edges and minimum degree $\delta({\mathcal{H}})$ satisfies (2), then ${\mathcal{H}}$ is super-pancyclic.

It could be that the necessary condition (2) for ${\mathcal{H}}$ to be super-pancyclic is sufficient not only for $m\leq 3\delta-5$. It would be interesting to find the range of $m$ for which (2) yields the conclusion of Corollary 1.20.

We present the main proofs in the language of bipartite graphs. In Section 2, we introduce the notion of a tight pair and prove some properties of such pairs. Using these properties we prove Theorems 1.6 and 1.7 in Section 3. In Section 4, we prove Theorem 1.19 and the reduction to Corollary 1.20.

2 Tight pairs and their properties

Definition 2.1**.**

Let $G\in{\mathcal{G}}(n,m,\delta)$. A tight pair in $G$ is a pair $(C,x)$ where $C$ is a longest cycle in $G$ and $x\in X-C$ is such that $|N(x)\cap v(C)|$ is maximum over all pairs $(C^{\prime},x^{\prime})$ where $C^{\prime}$ is a longest cycle in $G$ and $x^{\prime}\in X-V(C^{\prime})$.

By definition, a graph $G\in{\mathcal{G}}(n,m,\delta)$ has a tight pair if and only if $G$ does not contain a cycle of length $2n$. In this section, $G$ is always a graph in ${\mathcal{G}}(n,m,\delta)$ and $(C,x)$ is a tight pair where $C=y_{1}x_{1}\ldots y_{\ell}x_{\ell}y_{1}$.

Claim 2.2**.**

If $y_{i}\in N_{C}(x)$, then $N(x_{i})-V(C)$ and $N(x)-V(C)$ are disjoint.

Proof.

Suppose $y^{\prime}\in(N(x_{i})\cap N(x))-V(C)$. Then the cycle

[TABLE]

contains $2(\ell+1)$ vertices, contradicting the choice of $C$.   $\Box$

Similarly, by reorienting the cycle, we also obtain $N(x_{i-1})-V(C)$ and $N(x)-V(C)$ are disjoint.

Claim 2.3**.**

If $y_{i},y_{j}\in N_{C}(x)$, then $N(x_{i})-V(C)$ and $N(x_{j})-V(C)$ are disjoint.

Proof.

Suppose $i<j$ and $y^{\prime}\in(N(x_{i})\cap N(x_{j}))-V(C)$. Then the cycle

[TABLE]

contains more vertices than $C$, a contradiction.   $\Box$

For $1\leq i,j\leq\ell$, let $C[i,j]$ denote the clockwise path segment of $C$ from $y_{i}$ to $y_{j}$ (where we take indices of the path modulo $\ell$ if $i>j$).

Definition 2.4**.**

For $1\leq i<j\leq\ell$ we say that vertices $x_{i}$ and $x_{j}$ are crossing in $C$ if there exists indices ${i^{\prime}},{j^{\prime}}\in[\ell]$ such that $y_{i^{\prime}}\in N(x_{i})$, $y_{j^{\prime}}\in N(x_{j}$), and either

• •

$i^{\prime}=j^{\prime}+1$* with $i+1\leq j^{\prime}\leq j-1$ (so $y_{i^{\prime}},y_{j^{\prime}}\in V(C[i+1,j])$, or*

• •

$j^{\prime}=i^{\prime}+1\mod\ell$* and $y_{i^{\prime}},y_{j^{\prime}}\in V(C[j+1,i])$.*

Note that we do not consider $\{i^{\prime},j^{\prime}\}=\{i,{i+1}\}$ or $\{i^{\prime},j^{\prime}\}=\{j,{j+1}\}$.

Lemma 2.5**.**

Let $1\leq i<j\leq\ell$. If $x_{i}$ and $x_{j}$ are not crossing, then

[TABLE]

Proof.

Let $P_{1}=y_{i+1}x_{i+1}\ldots y_{j}$. Define

[TABLE]

By pigeonhole principle, if $|N(x_{i})\cap V(P_{1})|+|N(x_{j})\cap V(P_{1})|\geq|V(P_{1})\cap Y|+2$ then $I_{1}$ is nonempty, which implies that $x_{i},x_{j}$ are crossing in $C$, a contradiction.

Similarly let $P_{2}=y_{j+1}x_{j+1}\ldots y_{\ell}x_{\ell}y_{1}\ldots x_{i-1}y_{i}$ and

[TABLE]

We obtain $|N(x_{i})\cap V(P_{2})|+|N(x_{j})\cap V(P_{2})|\leq|V(P_{2})\cap Y|+1$, otherwise $x_{i},x_{j}$ are crossing.

Hence

[TABLE]

$\Box$

Lemma 2.6**.**

If $2\leq|N(x)\cap V(C)|<|Y\cap V(C)|$ and $n=|X|\leq\delta$, then $m=|Y|\geq 3\delta-4$.

Proof.

Let $T$ be the set of vertices in $N(x)\cap V(C)$ and set $|T|=t$. Note all vertices of $T$ are in $Y$. By Claims 2.2 and 2.3, the neighborhoods all vertices in $\{x\}\cup\{x_{i}:y_{i}\in T\}$ are disjoint outside of $C$. Let $\ell-s$ be the maximum size of $|N(x_{i})\cap V(C)|$ for $x_{i}$ such that $y_{i}\in T$.

We will first show that $s>0$. If not, then without loss of generality, suppose $y_{1}\in T$ and ${V(C)\cap Y\subseteq N(x_{1})}$. For each $x_{j}$ such that $y_{j}\in T$ and $j>1$, we can form the cycle

[TABLE]

The cycle $C^{\prime}$ has the same length as $C$, and contains all of $C$ except $x_{j}$. By the choice of $(C,x)$, ${|N(x_{j})\cap V(C^{\prime})|\leq t}$. Thus, $|N(x_{j})\cap V(C)|\leq t$ for every $x_{j}$ such that $y_{j}\in T$ and $j>1$.

Since $t\leq|Y\cap V(C)|-1=\ell-1$ and $\ell\leq|X|-1\leq\delta-1$, we have $\delta\geq t+2$.

So

[TABLE]

Hence we may assume $s\geq 1$.

Case 1: There are no crossing vertices $x_{i},x_{j}$, with $y_{i},y_{j}\in T$. So by Lemma 2.5, for each $x_{i}$ and $x_{j}$, with $y_{i},y_{j}\in T$, $d_{C}(x_{i})+d_{C}(x_{j})\leq\ell+2$.

If $t$ is even, we arbitrarily pair up vertices in $T$ and apply Lemma 2.5 to obtain that the vertices in $T$ together have degree sum in $C$ at most $t(\ell+2)/2$. Therefore

[TABLE]

Recall $t\leq\ell-1$, $\ell\leq\delta-1$, and $\delta\geq t+2$ (implying $\delta\geq 4$). Therefore

[TABLE]

If $t$ is odd, fix $y_{i}\in T$ such that $d_{C}(x_{i})$ is minimum. By Lemma 2.5, $d_{C}(x_{i})\leq(\ell+2)/2$. The other $t-1$ vertices in $T$ can be arbitrarily paired up as in the previous case so that their total degree sum in $C$ is at most $(t-1)(\ell+2)/2$. Therefore

[TABLE]

as in the case of $t$ even. Thus $|Y|\geq 3\delta-4$.

Case 2: There are crossing vertices $x_{i}$ and $x_{j}$, such that $y_{i},y_{j}\in T$ for some $1\leq i<j\leq\ell$. Suppose that for some $i+1\leq k<j$, $y_{k+1}\in N(x_{i})$ and $y_{k}\in N(x_{j})$ (the other case is similar so we omit it).

We will show $N(x_{k})\cap N(x)\subset C$. Suppose not. Let $y^{\prime}\not\in C$ such that $y^{\prime}\in N(x_{k})\cap N(x)$. Then the cycle

[TABLE]

is longer than $C$, a contradiction.

Next, we will show $N(x_{k})\cap N(x_{q})\subset C$, for $x_{q}$ such that $y_{q}\in T$ and $q\neq k$. Suppose not. Then there is some vertex $y^{\prime}\not\in C$ such that $y^{\prime}\in N(x_{k})\cap N(x_{q})$. If $x_{q}\in C[j,i]$, then we have the cycle

[TABLE]

If $x_{q}\in C[i,k]$, then we have the longer cycle

[TABLE]

If $x_{q}\in C[k+1,j]$, then we have the longer cycle

[TABLE]

In any case we get a cycle with at least $2(\ell+1)$ vertices, a contradiction.

Next, we will show that $|N(x_{k})\cap V(C)|\leq t$. The cycle

[TABLE]

has $2\ell$ vertices, includes $x$, and includes all the vertices of $C$ except $x_{k}$. Since $(C,x)$ is a tight pair, $|N(x_{k})\cap V(C^{\prime})|\leq t$. So $|N(x_{k})\cap V(C)|\leq t$. Now, consider the vertices $x,x_{k}$ and all $x_{q}$ such that $y_{q}\in T.$

If $y_{k}\in T$, then we know $x_{k}$ has at most $t$ neighbors on $C$. Additionally, every other vertex $x_{q}$ such that $y_{q}\in N(x)$ has at most $\ell-s$ neighbors in $C$. So

[TABLE]

Since $s\geq 1$,

[TABLE]

If $y_{k}\not\in T$, then by the choice of $x$, $x_{k}$ has at most $t$ neighbors on $C$ and shares no neighbors outside $C$ with $x$ and $x_{q}$ such that $y_{q}\in T$. So $|Y|\geq(t+2)\delta-t(\ell-s)-t-t+\ell$. This is greater than $(t+1)\delta-(t-1)(\ell-s)-t-t+\ell$ which is at least $3\delta-3$.   $\Box$

Lemma 2.7**.**

If $n\leq\delta$ and $Y\cap V(C)\subseteq N(x)$, then for each $x_{i}\in X\cap V(C)$ and each $y\in N(w)-V(C)$, $w$ separates $y$ from $V(C)-w$.

Proof.

Without loss of generality, suppose $x_{1}\in C$ does not separate $y\in N(x_{1})$ from $V(C)-x_{1}$. Then $G-x_{1}$ contains a path $P$ from $y$ to $V(C)-x_{1}$. Let $z$ be the endpoint of $P$ in $V(C)$, say $z\in\{x_{i},y_{i}\}$ for some $1<i\leq\ell$. First suppose $x\notin V(P)$. If $z\neq y_{\ell}$, then either $y_{1}x_{1}yPzy_{i}x_{i-1}\ldots y_{2}xy_{i+1}x_{i+1}\ldots y_{\ell}x_{\ell}$ or $y_{1}x_{1}yPzx_{i-1}y_{i-1}\ldots y_{2}xy_{i+1}x_{i+1}\ldots y_{\ell}x_{\ell}$ is a longer cycle than $C$. If $z=y_{\ell}$, then $y_{2}x_{1}Py_{\ell}x_{\ell}y_{1}xy_{\ell-1}x_{\ell-2}\ldots y_{2}$ is a longer cycle. Otherwise, let $P^{\prime}$ be the segment of $P$ from $y$ to $x$. Then we instead take the cycle $y_{1}x_{1}yP^{\prime}xy_{2}x_{2}\ldots y_{\ell}x_{\ell}y_{1}$.   $\Box$

3 Proofs of Theorems 1.6 and 1.7

Proof of Theorem 1.6. Suppose a longest cycle of $G$ has length $2\ell$ where $\ell<n$ and fix a tight pair $(C,x)$. Say $C=y_{1}x_{1}\ldots y_{\ell}x_{\ell}y_{1}$, and set $t:=|N(x)\cap V(C)|$.

Case 1: $t\leq 1$. Because $G$ is 2-connected, there exist paths $P_{1}$ and $P_{2}$ from $x$ to $V(C)$ such that $P_{1}$ and $P_{2}$ are disjoint except at $x$, and both $P_{i}$ are internally disjoint from $V(C)$. Let $z_{1}$ and $z_{2}$ be the endpoints of $P_{1}$ and $P_{2}$ respectively that are not $x$. Among all such paths, choose $P_{1}$ and $P_{2}$ so that $|\{z_{1},z_{2}\}\cap Y|$ is maximum. Often we will use $P_{1}\cup P_{2}$ to refer to the combined path from $z_{1}$ to $z_{2}$.

Case 1.1: Either $z_{1}\in Y$ or $z_{2}\in Y$. Without loss of generality, let $z_{1}=y_{1}$. Note that this is exactly the case where $t=1$, because the second vertex in $P_{1}$ is a vertex of $X$ not in $V(C)$, so we may choose this vertex to be $x$ and $P_{1}$ to be the path $xy_{1}$. Since $t<2$, $|(P_{2}\cap Y)-V(C)|\geq 1$.

Say $z_{2}\in\{x_{j},y_{j}\}$ for some $1<j\leq\ell$. Also, $j\neq\ell$, otherwise we could replace the vertex $x_{\ell}$ in $C$ with the path $P_{1}\cup P_{2}$ to obtain a longer cycle.

By Claim 2.2, we have that $N(x_{1})$ and $N(x)$ do not intersect outside of $C$. Furthermore, we also claim that $N(x_{1})$ does not intersect $N(x_{j+1})$ outside of $C$. Suppose that ${y^{\prime}\in(N(x_{1})\cap N(x_{j+1}))-V(C)}$.

If $y^{\prime}\in V(P_{1}\cup P_{2})$, then let $P^{\prime}$ be the segment of $P_{1}\cup P_{2}$ from $y_{1}$ to $y^{\prime}$. Then

[TABLE]

is a cycle with at least $2(\ell+1)$ vertices, a contradiction.

So we may assume that $y^{\prime}\notin V(C)\cup V(P_{1}\cup P_{2})$. Then

[TABLE]

is a cycle which contains $(V(C)\cap Y)-y_{j+1}$ and $y^{\prime}\cup(V(P_{1}\cup P_{2})\cap Y)$. That is, $|V(C^{\prime})|\geq 2(\ell+1)$, a contradiction.

Next, we show that $x_{1}$ and $x_{j+1}$ are not crossing. Suppose first that for some $j+1<k\leq\ell$, $y_{k}\in N(x_{1})$ and $y_{k+1}\in N(x_{j+1})$. Then

[TABLE]

is at least as long as $X$, contains $y_{k}$ and $y_{k+1}$, and does not contain $x_{k}$. Similarly, if $x_{k+1}\in N(x_{1})$ and $x_{k}\in N(x_{j+1})$ for some $2\leq k\leq j$, then

[TABLE]

is a cycle with the same properties. Hence by the choice of $(C,x)$ we have that $N(x_{k})$ contains at most $t=1$ vertices in $C$. But $N(x_{k})\supseteq\{y_{k},y_{k+1}\}$, a contradiction.

Therefore by Lemma 2.5, $d_{C}(x_{1})+d_{C}(x_{j+1})\leq\ell+2$.

Note that

[TABLE]

To see this, suppose that $\{y^{\prime},y^{\prime\prime}\}\subseteq(N(x)\cap N(x_{j+1}))-V(C)$. If both $y^{\prime},y^{\prime\prime}\in V(P_{1}\cup P_{2})$, with say $y^{\prime}$ appearing before $y^{\prime\prime}$ in $P_{1}\cup P_{2}$, then let $P^{\prime}$ be the segment of $P_{1}\cup P_{2}$ from $y^{\prime}$ to $z_{2}$, oriented backwards (from $z_{2}$ to $y^{\prime}$). Then

[TABLE]

is a longer cycle than $C$. Otherwise, we may assume that $y^{\prime\prime}\notin V(P_{1}\cup P_{2})$. Then

[TABLE]

is a longer cycle.

Putting it all together, we get

[TABLE]

a contradiction.

Case 1.2: Both $z_{1}\in X$ and $z_{2}\in X$. Note that this implies $t=0$. For simplicity, let $z_{1}=x_{1}$ and $z_{2}=x_{j}$ for some $2\leq j\leq\ell$. We must have that $|(V(P_{1}\cup P_{2}))\cap Y|\geq 2$ since $G$ is bipartite.

We claim first that $N(x_{2})\cap N(x_{j+1})\subseteq V(C)$. Suppose that ${y^{\prime}\in(N(x_{2})\cap N(x_{j+1}))-C}$. If $y^{\prime}\in V(P_{1}\cup P_{2})$, then let $P_{1}^{\prime}$ be the segment of $P_{1}\cup P_{2}$ from $x_{1}$ to $y^{\prime}$, let $P_{2}^{\prime}$ be the segment of $P_{1}\cup P_{2}$ from $x_{j}$ to $y^{\prime}$, and let $P^{\prime}$ be the longer of the paths $P_{1}^{\prime}$ and $P_{2}^{\prime}$. Note that $P^{\prime}$ must contain $x$ and at least 2 vertices from $Y$. Then either

[TABLE]

is a cycle with at least $2(\ell-1+2)$ vertices, a contradiction.

Next, suppose $y^{\prime}\notin V(P_{1}\cup P_{2})$. Then the cycle

[TABLE]

contains at least $2(\ell-2+2+1)$ vertices. This proves that $N(x_{2})$ and $N(x_{j+1})$ are disjoint outside of $C$. The same proof as for (3) shows that also $N(x_{2})$ and $N(x)$ intersect at at most one vertex outside of $C$, and same for $N(x_{j+1})$ and $N(x)$.

Finally, we show that $x_{2}$ and $x_{j+1}$ are not crossing. Otherwise, if there exists some $j+1<k\leq\ell$, with $y_{k}\in N(x_{2})$ and $y_{k+1}\in N(x_{j+1})$. Then the cycle

[TABLE]

contains at least as many vertices as $C$. Furthermore $V(C)-\{y_{2},y_{j+1}\}\subset V(C^{\prime})$, and $x_{k}\notin V(C^{\prime})$. But $|N(x_{k})\cap C^{\prime}|\geq 2$, contradicting the choice of $(C,x)$. The case where $x_{k+1}\in N(x_{1})$ and $x_{k}\in N(x_{j+1})$ for some $2\leq k\leq j$ is similar so we omit the proof. By Lemma 2.5, this implies $d_{C}(x_{2})+d_{C}(x_{j+1})\leq\ell+2$.

Thus

[TABLE]

a contradiction.

Case 2: $2\leq t\leq\ell-1$. Apply Lemma 2.6 to obtain $n\geq 3\delta-4$, a contradiction.

Case 3: $t=\ell$. By Lemma 2.7, every vertex in $X\cap V(C)$ is a cut vertex of $G$, a contradiction.   $\Box$

Proof of Theorem 1.7. Suppose $G\in{\mathcal{G}}(n,2\delta-1,\delta)$ contains no cycle of length $2n$. If $n=2$ and $G$ has no cycles, then $G=G_{2}(1,1)$. If $n\geq 3$, since $m\leq 2\delta-1$, $G$ has a cycle. Let $(C,x)$ be a tight pair, and set $t:=|V(C)\cap N(x)|$.

Case 1: $t=0$. Since $|Y|=2\delta-1$, for every $x,x^{\prime}\in X$, $|N(x)\cap N(x^{\prime})|\geq 1$. In particular, for each $x_{i}\in V(C)\cap X$, $N(x_{i})\cap N(x)$ contains at least 1 vertex, say $y^{\prime}_{i}\in Y-V(C)$.

If for some $1\leq i\leq\ell$, $y^{\prime}_{i}\neq y^{\prime}_{i+1}$ (indices taken mod $\ell$), then the cycle $y_{1}x_{1}\ldots y_{i}x_{i}y^{\prime}_{i}xy^{\prime}_{i+1}x_{i+1}\ldots y_{\ell}x_{\ell}y_{1}$ contains more vertices than $C$. Otherwise, the cycle $C^{\prime}:=y_{1}x_{1}\ldots y_{i}x_{i}y^{\prime}_{i+1}x_{i+1}\ldots y_{\ell}x_{\ell}y_{1}$ is also a longest cycle of $G$, but $|V(C^{\prime})\cap N(x)|>0$, contradicting the choice of $(C,x)$ as a tight pair.

Case 2: $t=1$. Without loss of generality, let $N(x)=y_{1}$. We have that $N(x)-V(C)$ contains at least $\delta-1$ vertices. If $N(x)\cap N(x_{1})$ contains a vertex $y\in Y-V(C)$, then we get the longer cycle $y_{1}xyx_{1}y_{2}\ldots y_{\ell}x_{\ell}x_{1}$. So $N(x_{1})\cap N(x)=\{y_{1}\}$, and furthermore since ${|Y|=2\delta-1}$, ${Y=N(x_{1})\cup N(x)}$, and $Y\cap V(C)\subseteq N(x_{1})$. Note that by symmetry, $N(x_{\ell})=N(x_{1})$. Suppose that for some $1<i<\ell$, there exists $y\in(N(x)\cap N(x_{i}))-y_{1}$. Then we obtain the longer cycle $y_{1}x_{\ell}y_{\ell}\ldots x_{i+1}y_{i+1}x_{1}y_{2}\ldots y_{i}x_{i}yxy_{1}$. This shows that $N(x_{i})=N(x_{1})$ for all $1\leq i\leq\ell$.

By the choice of $(C,x)$ as a tight pair, $|N(x^{\prime})\cap V(C)|\leq 1$ for all $x^{\prime}\in X-V(C)$. Since ${Y=N(x_{1})\cup N(x)}$, ${N(x)-V(C)=N(x)-y_{1}}$. Hence the vertices of $V(C)$ induce a complete bipartite graph, as do the vertices of $V(G)-V(C)$. Furthermore, every $x\in X-V(C)$ contains exactly one neighbor in $V(C)$. It is easy to show that $G$ contains a cycle of length $2n$ unless for every $x\in X-V(C)$, $N(x)\cap V(C)=\{y_{1}\}$. Therefore $G$ is isomorphic to $G_{2}(n-\ell,\ell)$.

Case 3: $2\leq t\leq\ell-1$. If $\delta\geq 4$, then by Lemma 2.6 we get $|Y|\geq 3\delta-4$, a contradiction. Suppose $\delta=3$ and $G$ has no cycle of length $2n$. Since $G$ contains a cycle, $n\neq 2$ and so $n=3$. It is easy to check that $G=G_{1}(n)$ for the case $t\geq 2$.

Case 4: $t=\ell$. By Lemma 2.7, for any $x_{i}\in X\cap V(C)$ and each $y\in N(x_{i})-V(C)$, $x_{i}$ separates $y$ from $V(C)-x_{i}$. In particular, this implies that $y\notin N(x_{j})$ for any $j\neq i$. Since $|V(C)\cap Y|=\ell\leq n-1\leq\delta-1$, each $x_{i}$ has at least $\delta-\ell$ neighbors outside of $V(C)$ that are shared by no other $x_{j}$. Then $|Y|\geq\ell+(\ell+1)(\delta-\ell)=\ell\delta-\ell^{2}+\delta=\delta+\ell(\delta-\ell)\geq\delta+(\delta-1)(1)$, where equality holds only if $\ell=\delta-1$ (so $\delta=n$), and each vertex in $X$ has exactly one neighbor outside of $C$. That is, $G$ is isomorphic to $G_{1}(n)$.   $\Box$

4 Proofs of Theorem 1.19 and Corollary 1.20

Proof of Theorem 1.19. Fix $G\in{\mathcal{G}}(n,m,\delta)$ with partition $(X,Y)$. We will show that for every $X^{\prime}\subseteq X$ with $|X^{\prime}|\geq 3$, there exists a cycle $C$ in $G$ such that $V(C)\cap X=X^{\prime}$.

We proceed by induction on $|X^{\prime}|$. For the base case, suppose $|X^{\prime}|=3$. Let $Y^{\prime}\subseteq Y$ such that $G^{\prime}:=G[X^{\prime}\cup Y^{\prime}]$ is 2-connected. It is easy to check, via an ear-decomposition argument for instance, that a bipartite 2-connected graph has a cycle of length at least 6 unless one of its parts has size 2. But we have $|Y^{\prime}|\geq|X^{\prime}|\geq 3$. Hence for each $X^{\prime}\subset X$ with $|X^{\prime}|=3$, $G$ contains a cycle $C$ of length 6 with $V(C)\cap X=X^{\prime}$. This yields the following facts.

Claim 4.1**.**

(i) For each $x,x^{\prime}\in X$, $|N(x)\cap N(x^{\prime})|\geq 1$.

(ii) For any $x,x^{\prime},x^{\prime\prime}\in X$, ${|N(x)\cap(N(x^{\prime})\cup N(x^{\prime\prime}))|\geq 2}$.

Proof.

For (i), let $x^{\prime\prime}\in X-\{x,x^{\prime}\}$. Then $G$ contains a cycle, say $y_{1}xy_{2}x^{\prime}y_{3}x^{\prime\prime}y_{1}$, so ${y_{2}\in N(x)\cap N(x^{\prime})}$. For (ii), using the same cycle, we have $N(x)\cap(N(x^{\prime})\cup N(x^{\prime\prime}))\supseteq\{y_{1},y_{2}\}$.   $\Box$

For the induction step, fix $X^{\prime}\subseteq X$ with $k:=|X^{\prime}|\geq 4$ and consider the subgraph ${G_{X^{\prime}}=G[X^{\prime}\cup N(X^{\prime})]}$. We wish to show that $G_{X^{\prime}}$ contains a cycle using all of $X^{\prime}$. Suppose not. Then by induction hypothesis, for each $x\in X^{\prime}$, $G$ contains a cycle $C_{x}$ such that $V(C_{x})\cap X=X^{\prime}-x$. In particular, this cycle must be contained in $G_{X^{\prime}}$.

Among all $x\in X^{\prime}$, pick $x$ such that $|N(x)\cap V(C_{x})|$ is maximum. Let $C_{x}=y_{1}x_{1}\ldots y_{k-1}x_{k-1}y_{1}$ and $t:=|N(x)\cap V(C_{x})|$

Case 1: $t=0$. By Part (i) of Claim 4.1, $|N(x)\cap N(x_{i})|\geq 1$ for all $1\leq i\leq k-1$. If there exists some $y^{\prime}_{i}$, $y^{\prime}_{i+1}$ such that $y^{\prime}_{j}\in N(x)\cap N(x_{j})$ for $j\in\{i,i+1\}$ and $y^{\prime}_{i}\neq y^{\prime}_{i+1}$, then $G_{X^{\prime}}$ contains the cycle

[TABLE]

with $2(k-1-1+2)$ vertices. I.e., $V(C^{\prime})\cap X=X^{\prime}$.

Otherwise, suppose for every $i$, $N(x_{i})\cap N(x)=\{y^{\prime}\}$. Then $N(x)\cap(N(x_{1})\cup N(x_{2}))=\{y^{\prime}\}$, contradicting Part (ii) of Claim 4.1.

Case 2: $t=1$. Without loss of generality, say $y_{k-1}\in N(x)$. By Part (ii) of Claim 4.1, there exists some $y^{\prime}\in N(x)\cap(N(x_{k-2})\cup N(x_{k-1}))$ such that $y^{\prime}\neq y_{k-1}$. Since $t=1$, $y^{\prime}\notin V(C_{x})$. Then either

[TABLE]

is a cycle in $G_{X^{\prime}}$ which contains all of $X^{\prime}$.

Case 3: $2\leq t\leq k-2$. Apply Lemma 2.6 to $G^{\prime}_{X^{\prime}}$, $C_{x}$, and $x$ to obtain that $n\geq 3\delta-4$, a contradiction.

Case 4: $t=k-1$. We claim that for each $y\in V(G_{X^{\prime}})-V(C)$, $|N_{X^{\prime}}(y)|=1$. If not, then there exists some $y\notin V(C)$ and some $1\leq i<j\leq k$ such that $y\in N(x_{i})\cap N(x_{j})$. But this contradicts Theorem 2.7. Therefore every 2-connected subgraph of $G$ containing $X^{\prime}$ is a subgraph of $G[X^{\prime}\cup V(C)]$. But $|V(C)\cap Y|<|X^{\prime}|$, a contradiction.   $\Box$

Proof of Corollary 1.20. Fix an $n$-vertex hypergraph ${\mathcal{H}}$ with $\delta({\mathcal{H}})\geq n$ and at most $3n-5$ edges. Let $I({\mathcal{H}})$ be the incidence graph of ${\mathcal{H}}$.

Then $I({\mathcal{H}})\in{\mathcal{G}}(n,|E({\mathcal{H}})|,n)$ with $|E({\mathcal{H}})|\leq 3n-5$. Translating from the language of hypergraphs to bipartite graphs, for each $X^{\prime}\subseteq X$ with $|X^{\prime}|\geq 3$, there exists $Y^{\prime}\subseteq Y$ with $|Y^{\prime}|\geq|X^{\prime}|$ such that $G({\mathcal{H}})[X^{\prime}\cup Y^{\prime}]$ is 2-connected. By Theorem 1.19, $G({\mathcal{H}})$ is $X$-super-pancyclic; equivalently, ${\mathcal{H}}$ is super-pancyclic.   $\Box$

Concluding remarks.

1. In  [5], a significantly weaker version of Conjecture 1.4(ii) is proved.

2. Let the dual incidence graph of ${\mathcal{H}}$ be the bipartite graph $I({\mathcal{H}})$ with parts $(X,Y)$ where $X=E({\mathcal{H}})$, $Y=V({\mathcal{H}})$ such that for $e\in X,v\in Y,$ $ev\in E(I({\mathcal{H}}))$ if and only if the vertex $v$ is contained in the edge $e$ in ${\mathcal{H}}$.

By applying Theorem 1.19, we obtain the following.

Corollary 4.2**.**

Let ${\mathcal{H}}$ be a hypergraph with edge cardinalities at least $r$ such that $e({\mathcal{H}})\leq r$ and $|V({\mathcal{H}})|\leq 3r-5$. If the dual incidence graph of ${\mathcal{H}}$ satisfies (2), then for any set of edges $B$ with $|B|\geq 3$, ${\mathcal{H}}$ contains a Berge cycle using exactly the edges of $B$.

Acknowledgement. We thank Misha Lavrov for pointing our attention to Jackson’s Conjecture.