Some cyclic properties of $L_1$-graphs

TL;DR

This paper investigates the cycle properties of $L_1$-graphs, showing that non-Hamiltonian cycles can be extended to larger cycles, and explores related path extension properties.

Contribution

It extends previous work by demonstrating cycle extension properties in $L_1$-graphs, including non-Hamiltonian cycles and paths with specific endpoint conditions.

Findings

Non-Hamiltonian cycles can be extended to larger cycles containing all original vertices.

Not all $L_1$-graphs are pancyclic, contrary to some earlier assumptions.

Cycle extension results also apply to certain paths with non-adjacent endpoints.

Abstract

A graph $G$ is called an $L_1$-graph if $d(u)+d(v)\ge|N(u)\cup N(v)\cup N(w)|-1$ for every triple of vertices $u,v,w$ where $u$ and $v$ are at distance 2 and $w\in N(u)\cap N(v)$. Asratian et al. (1996) proved that all finite connected $L_1$-graphs on at least three vertices such that $|N(u)\cap N(v)|\ge2$ for each pair of vertices $u,v$ at distance 2 are Hamiltonian, except for a simple family $\mathcal{K}$ of exceptions. We show that not all such graphs are pancyclic, but that any non-Hamiltonian cycle in such a graph can be extended to a larger cycle containing all vertices of the original cycle and at most two other vertices. We also prove a similar result for paths whose endpoints do not have any common neighbors.

Equations

$$\mathcal{K}=\{G:K_{p,p+1}\subseteq G\subseteq K_p\vee \overline{K_{p+1}}\text{ for some }p\ge 2\},$$

$$d(u)+d(v)\ge |N(u)\cup N(v)\cup N(w)|.$$

$$G=K_1\vee \overline{K_2}\vee \underset{\text{at least two copies of }{K}_2}{\underbrace{K_2\vee \cdots \vee K_2}}\vee \overline{K_2}\vee K_1$$

$$W^{+}\cup \{v\}\text{ is an independent set.}$$

$$N(w_i^{+})\cap N(v)=N(w_i^{+})\cap W.$$

$$\lvert N(w_{i}^{+})\cap W\rvert=\lvert N(w_{i}^{+})\cap N(v)\rvert\geq\big{\lvert}N(w_{i})\setminus\bigl{(}N(w_{i}^{+})\cup N(v)\bigr{)}\big{\rvert}-1.$$

$$N(w_{i})\cap W^{+}\subseteq N(w_{i})\setminus\bigl{(}N(w_{i}^{+})\cup N(v)\cup\{v\}\bigr{)}.$$

$$|N(w_i)\cap W^{+}|\le |N(w_i^{+})\cap W|.$$

$$e(W^{+},W)=\sum _{i=1}^p|N(w_i)\cap W^{+}|\le \sum _{i=1}^p|N(w_i^{+})\cap W|=e(W^{+},W).$$

$$|N(w_i)\cap W^{+}|=|N(w_i^{+})\cap W|$$

$$N(w_{i})\setminus\bigl{(}N(w_{i}^{+})\cup N(v)\cup\{v\}\bigr{)}=N(w_{i})\cap W^{+}\subseteq W^{+}.\qed$$

$$w_i^{+}{w}_i^{-}\in E(G)\text{ for each }i=1,\dots ,p.$$

$$w_i^{+}{w}_i^{-}\in E(G)\text{ for each }i=k+1,\dots ,p.$$

$$|N(w_i)\cap W_1^{+}|\le |N(w_i)\cap W^{+}|=|N(w_i^{+})\cap W|=|N(w_i^{+})\cap W_1|$$

$$e(W_1^{+},W_1)=\sum _{i=1}^k|N(w_i)\cap W_1^{+}|\le \sum _{i=1}^k|N(w_i^{+})\cap W_1|=e(W_1^{+},W_1).$$

$$|N(w_i)\cap W_1^{+}|=|N(w_i)\cap W^{+}|,$$

$$|N(w_i^{+})\cap W|$$

$$\displaystyle\geq\big{\lvert}N(w_{i+1})\setminus\bigl{(}N(w_{i}^{+})\cup N(v)\bigr{)}\big{\rvert}-1$$

$$\ge |N(w_{i+1})\cap (W^{+}\cup \{v\})|-1$$

$$=|N(w_{i+1}^{+})\cap W|.$$

$$\displaystyle 1+e\bigl{(}\{w_{k+1}^{+},\dotsc,w_{p}^{+}\},\{w_{k+1},\dotsc,w_{p}\}\bigr{)}$$

$$=\sum _{i=k+1}^p|N(w_i)\cap W^{+}|$$

$$=\sum _{i=k+1}^p|N(w_i^{+})\cap W|$$

$$\displaystyle=e\bigl{(}\{w_{k+1}^{+},\dotsc,w_{p}^{+}\},W\bigr{)}.$$

$$|N(w_{k+1}^{+})\cap W|$$

$$=|N(w_j^{+})\cap N(v)|$$

$$\displaystyle\geq\big{\lvert}N(w_{i})\setminus\bigl{(}N(w_{j}^{+})\cup N(v)\bigr{)}\big{\rvert}-1$$

$$\ge |N(w_i)\cap W^{+}|$$

$$=|N(w_i^{+})\cap W|$$

$$\ge |N(w_k^{+})\cap W|.$$

$$|N(x)\cap N(w_p^{++})|\ge 2.$$

$$p=|N(w_1^{+})\cap N(w_2^{+})|$$

$$\ge |W^{+}\cup \{v,u\}|-1$$

$$=p+1,$$

$$p=|N(x^{+})\cap N(v)|$$

$$\ge |W^{+}\cup \{v,w_p^{++}\}|-1$$

$$=p+1,$$

Full text

We show that not all such graphs are pancyclic, but that any non-Hamiltonian cycle in such a graph can be extended to a larger cycle containing all vertices of the original cycle and at most two other vertices. We also prove a similar result for paths whose endpoints do not have any common neighbors.

1 Introduction

We use [9] for terminology and notation not defined here and consider simple graphs only. If $C$ is a cycle in a graph, then we use the notation $\vec{C}$ to denote the cycle with a given direction and

$C$

 

$\vec{\phantom{C}}$

for the reverse direction, and if $x$ is a vertex on the cycle then $x^{+}$ and $x^{-}$ denote the successor and predecessor of $x$, respectively, in the given direction. The same notation is used for paths. A cycle or a path in a finite graph $G$ is a Hamilton cycle or Hamilton path, respectively, if it contains all vertices of $G$, and a finite graph is Hamiltonian if it contains a Hamilton cycle. We also use the notation $e(X,Y)$, where $X$ and $Y$ are vertex sets, for the number of edges joining a vertex of $X$ with a vertex of $Y$.

A classic result on Hamiltonicity is the following by Dirac [13]: A finite graph $G$ with at least three vertices is Hamiltonian if $d(v)\geq\lvert V(G)\rvert/2$ for every vertex $v\in V(G)$. This was generalized by Ore [25] as follows: A finite graph $G$ with at least three vertices is Hamiltonian if $d(u)+d(v)\geq\lvert V(G)\rvert$ for every pair of non-adjacent vertices $u,v\in V(G)$. Graphs satisfying this condition are called Ore graphs, and there are many results on Hamiltonicity inspired by this theorem. Nara [23], among others, proved that the bound in Ore’s theorem can be improved under certain conditions:

Theorem A** (see e.g. Nara [23]).**

Let $G$ be a finite $2$-connected graph on at least three vertices such that $d(u)+d(v)\geq\lvert V(G)\rvert-1$ for every pair of non-adjacent vertices $u,v\in V(G)$. Then $G$ is Hamiltonian unless it belongs to the following set of exceptions:

[TABLE]

where $\vee$ denotes the join operation.

The above theorems only apply to graphs with large edge density ( $|E(G)|\geq\text{constant}\cdot|V(G)|^{2}$ ) and diameter 2. Asratian and Khachatryan pioneered a method to overcome this by using local structures of graphs. They generalized Ore’s theorem to cover sparse graphs with large diameter:

Theorem B** (Asratian–Khachatryan [4]).**

Let $G$ be a finite connected graph on at least three vertices such that for every triple $u,w,v$ with $d(u,v)=2$ and $w\in N(u)\cap N(v)$ the following property holds:

[TABLE]

Then $G$ is Hamiltonian.

A graph is called an $L_{i}$-graph if $d(u)+d(v)\geq\lvert N(u)\cup N(v)\cup N(w)\rvert-i$ for each triple of vertices $u,v,w$ with $d(u,v)=2$ and $w\in N(u)\cap N(v)$. Thus Theorem B can be reformulated as follows: all finite connected $L_{0}$-graphs on at least three vertices are Hamiltonian.

The class of $L_{1}$-graphs includes not only all $L_{0}$-graphs and thus all Ore graphs, but also all claw-free graphs – graphs that do not contain $K_{1,3}$ as an induced subgraph [3]. A related result on claw-free graphs is the following by Shi [27]: Any finite connected claw-free graph on at least three vertices such that $\lvert N(u)\cap N(v)\rvert\geq 2$ for each pair of vertices $u,v$ with $d(u,v)=2$ is Hamiltonian.

Every Hamiltonian graph $G$ is $1$-tough, that is, it contain no vertex set $S$ such that the subgraph $G-S$ contains more than $\lvert S\rvert$ components. All $L_{0}$-graphs and 2-connected claw-free graphs are 1-tough; for $L_{1}$-graphs we need a set of exceptions [3]: Any 2-connected $L_{1}$-graph is either 1-tough or lies in the set $\mathcal{K}$ defined above.

In [3], Asratian, Broersma, van den Heuvel, and Veldman proved the following local analogue of Theorem A, generalizing Theorem B (note that all $L_{0}$-graphs satisfy the $\lvert N(u)\cap N(v)\rvert\geq 2$ condition) and the result of Shi:

Theorem C** (Asratian et al. [3]).**

Let $G$ be a finite connected $L_{1}$-graph on at least three vertices such that $\lvert N(u)\cap N(v)\rvert\geq 2$ for each pair of vertices $u,v$ with $d(u,v)=2$. Then $G$ is Hamiltonian unless it belongs to the set $\mathcal{K}$.

Furthermore, it was proved in [3] that graphs satisfying these conditions have the property that every pair of vertices at distance at least three is connected by a Hamilton path.

Some other properties of $L_{1}$-graphs have been found. Saito [26] showed that all finite 2-connected $L_{1}$-graphs of diameter $2$ are Hamiltonian unless they belong to the set of exceptions $\mathcal{K}$, while Li and Schelp [22] showed that every finite 2-connected $L_{1}$-graph $G$ with minimum degree $\delta(G)\geq(\lvert V(G)\rvert-2)/3$ is Hamiltonian unless $G\in\mathcal{K}$. Furthermore, it was shown in [3] that every finite connected $L_{1}$-graph of even order has a perfect matching.

A finite graph $G$ is said to be pancyclic if it contains a cycle of each length from $3$ up to $\lvert V(G)\rvert$. Bondy [7] proved that all Ore graphs are pancyclic, except for complete bipartite graphs $K_{n,n}$, $n\geq 2$. He also made a metaconjecture that almost any nontrivial condition that implies Hamiltonicity also implies pancyclicity, though there may be a simple family of exceptional graphs. Aldred, Holton, and Min [1] proved that graphs satisfying the conditions of Theorem A are pancyclic, except for the graphs in the set $\mathcal{K}$, complete bipartite graphs $K_{n,n}$, and the cycle $C_{5}$.

An even stronger property is called cycle extendability, which means that any cycle that does not include all vertices of the graph can be extended to a new cycle containing a single new vertex in addition to all vertices of the original cycle. This notion was introduced by Hendry [18], who also proved that Ore graphs, with a relatively complicated set of exceptions, are cycle extendable. Without any exceptions, however, Bondy [8] had earlier proved that any cycle in an Ore graph that does not include all vertices can be extended to a larger cycle containing all vertices of the original cycle and at most two other vertices.

$L_{0}$-graphs (with the exception of the graphs $K_{n,n}$) have also been found to be pancyclic by Asratian and Sarkisian [5]. They further proved the following:

Theorem D** (Asratian–Sarkisian [6]).**

Let $G$ be a finite connected $L_{0}$-graph on at least three vertices. Then for each $\ell=4,\dotsc,\lvert V(G)\rvert$, unless $G=K_{n,n}$ for some $n\geq 2$, every vertex of $G$ lies on a cycle of length $\ell$, every edge of $G$ that does not lie on a triangle lies on a cycle of length $\ell$, and every pair of vertices at distance no less than three and at most $\ell$ is connected by a path with $\ell$ vertices.

In 2004, Diestel and Kühn [12] suggested a new concept for infinite locally finite graphs (infinite graphs with only finite vertex degrees), called Hamilton circles, which are analogues of Hamilton cycles in finite graphs. Let $G$ be an infinite locally finite graph. A ray in $G$ is a one-way infinite path. We define an equivalence relation on the set of rays in $G$ by saying that two rays are equivalent if they have a subray in the same component of $G-S$ for every finite vertex set $S$. The equivalence classes of this relation are called the ends of $G$, and can be seen as points at infinity. The Freudenthal compactification $\lvert G\rvert$ of $G$ is a topological space constructed by viewing $G$ as a 1-complex, and adding the ends of $G$ as additional points. Finally, a Hamilton circle in the Freudenthal compactification $\lvert G\rvert$ is a homeomorphic image of the unit circle that passes through every vertex and every end exactly once. For a more thorough exposition, see [10].

Diestel [11] launched the ambitious project of extending results on finite Hamilton cycles to Hamilton circles. Georgakopoulos [15] showed that if $G$ is the square of a 2-connected, infinite, locally finite graph, then $\lvert G\rvert$ has a Hamilton circle, extending Fleischner’s theorem [14] for finite graphs. Heuer [19] and Hamann et al. [17] showed that the Freudenthal compactification of every connected, locally connected, infinite, locally finite, claw-free graph has a Hamilton circle, extending Oberly–Sumner’s theorem [24].

Heuer [20] furthermore proved that the Freudenthal compactification of every claw-free, locally connected graph satisfying the conditions of Theorem B has a Hamilton circle. It is easy to see that for a triple $u,w,v$ with $d(u,v)=2$ and $w\in N(u)\cap N(v)$ in a claw-free graph, the inequality $d(u)+d(v)\geq\lvert N(u)\cup N(v)\cup N(w)\rvert$ is equivalent to the inequality $\lvert N(u)\cap N(v)\rvert\geq 2$. Thus the result of Heuer [20] can be reformulated as follows:

Theorem E**.**

Let $G$ be a locally finite, connected, claw-free graph on at least three vertices such that $\lvert N(u)\cap N(v)\rvert\geq 2$ for each pair of vertices $u,v$ with $d(u,v)=2$. Then $\lvert G\rvert$ has a Hamilton circle.

Kündgen, Li, and Thomassen [21] introduced another concept for infinite locally finite graphs: A closed curve in the Freudenthal compactification $\lvert G\rvert$ is called a Hamilton curve if it meets every vertex exactly once, but is allowed to meet the ends of $\lvert G\rvert$ multiple times. They showed that the condition of Theorem B implies the existence of a Hamilton curve.

In this article, which is partly based on the author’s master’s thesis [16], we investigate $L_{1}$-graphs in the same spirit as Theorem D, and show that they, unlike $L_{0}$-graphs, need not be pancyclic. However, we prove that if $G$ is a locally finite graph (not necessarily finite) satisfying the conditions of Theorem C, then

• •

any cycle $C$ in $G$ that does not contain all vertices of $G$ can be extended to a larger cycle containing all vertices of $C$ and at most two other vertices;

• •

for any pair of vertices $x,y$ with no common neighbors and any $x\mathord{-}y$-path $P$ in $G$ that does not include all vertices of $G$, there is a longer $x\mathord{-}y$-path containing all vertices of $P$ and at most two other vertices.

Furthermore we show that if $G$ is an infinite, locally finite graph satisfying the conditions of Theorem C, then $\lvert G\rvert$ has a Hamilton curve. Finally, we provide a characterization of all connected bipartite $L_{1}$-graphs.

2 Results

The main result of this paper is the following theorem:

Theorem 1**.**

Let $G$ be a connected, locally finite $L_{1}$-graph on at least three vertices such that $\lvert N(u)\cap N(v)\rvert\geq 2$ for each pair of vertices $u,v$ with $d(u,v)=2$. Then for every cycle $C_{n}$ of length $n$ in $G$ that does not contain all vertices of $G$, there is a cycle $C_{n+\ell}$ of length $n+\ell$, where $1\leq\ell\leq 2$, such that $V(C_{n})\subset V(C_{n+\ell})$, unless $n=\lvert V(G)\rvert-1$ and $G\in\mathcal{K}$.

Unlike for Theorem A, graphs satisfying the conditions of Theorem C need not be pancyclic, so Theorem 1 is best possible. The graph $K_{1}\vee\overline{K_{2}}\vee K_{2}\vee K_{2}\vee\overline{K_{2}}\vee K_{1}$ (see Fig. 1), for example, has 10 vertices and does not contain a 9-cycle. In general, the graph

[TABLE]

does not contain any cycle of length $\lvert V(G)\rvert-1$. Furthermore, the graph in Fig. 2 has 14 vertices and does not contain any cycle of length 11 or 13, and can be extended to an infinite family of graphs in the same way as above.

It is easy to see that every vertex in a graph satisfying the conditions of Theorem 1 lies on a cycle of length at most $4$. Thus we can draw the following conclusions:

Corollary 2**.**

Let $G$ be a finite connected $L_{1}$-graph on at least three vertices such that $\lvert N(u)\cap N(v)\rvert\geq 2$ for each pair of vertices $u,v$ with $d(u,v)=2$. Then for each vertex $x\in V(G)$ there is a number $r$ and a sequence of integers $n_{1},n_{2},\dotsc,n_{r}$, depending on $x$, such that $n_{1}\leq 4$, $n_{r}=\lvert V(G)\rvert$ (unless $G\in\mathcal{K}$, in which case $n_{r}=\lvert V(G)\rvert-1$), and $1\leq n_{i+1}-n_{i}\leq 2$ for each $i=1,\dotsc,r-1$, and a sequence of cycles $C_{n_{1}},C_{n_{2}},\dotsc,C_{n_{r}}$ of lengths $n_{1},n_{2},\dotsc,n_{r}$ respectively, such that $x\in V(C_{n_{1}})\subset V(C_{n_{2}})\subset\dotsb\subset V(C_{n_{r}})$.

Corollary 3**.**

Let $G$ be a finite connected $L_{1}$-graph on at least three vertices such that $\lvert N(u)\cap N(v)\rvert\geq 2$ for each pair of vertices $u,v$ with $d(u,v)=2$. Then for each vertex $x\in V(G)$ and each $m=4,\dotsc,\lvert V(G)\rvert$, the vertex $x$ lies on a cycle of length $m$ or $m-1$.

Using the same reasoning we also get the following:

Corollary 4**.**

Let $G$ be a connected, infinite, locally finite $L_{1}$-graph on at least three vertices such that $\lvert N(u)\cap N(v)\rvert\geq 2$ for each pair of vertices $u,v$ with $d(u,v)=2$. Then for each vertex $x\in V(G)$ and each $m\geq 4$, the vertex $x$ lies on a cycle of length $m$ or $m-1$.

We will also prove the following theorems:

Theorem 5**.**

Let $G$ be a connected, locally finite $L_{1}$-graph on at least three vertices such that $\lvert N(u)\cap N(v)\rvert\geq 2$ for each pair of vertices $u,v$ with $d(u,v)=2$, and let $x$ and $y$ be two adjacent vertices in $G$ with no neighbors in common. Then for every $x\mathord{-}y$-path $P_{n}$ with $n$ vertices in $G$ that does not contain all vertices of $G$, there is an $x\mathord{-}y$-path $P_{n+\ell}$ with $n+\ell$ vertices, $1\leq\ell\leq 2$, such that $V(P_{n})\subset V(P_{n+\ell})$, unless $n=\lvert V(G)\rvert-1$ and $G\in\mathcal{K}$.

Theorem 6**.**

Let $G$ be a connected, locally finite $L_{1}$-graph on at least three vertices such that $\lvert N(u)\cap N(v)\rvert\geq 2$ for each pair of vertices $u,v$ with $d(u,v)=2$, and let $x$ and $y$ be two vertices in $G$ with $d(x,y)\geq 3$. Then for every $x\mathord{-}y$-path $P_{n}$ with $n$ vertices in $G$ that does not contain all vertices of $G$, there is an $x\mathord{-}y$-path $P_{n+\ell}$ with $n+\ell$ vertices, $1\leq\ell\leq 2$, such that $V(P_{n})\subset V(P_{n+\ell})$.

Theorems 5 and 6 can be stated together as a single result by removing the requirement that $x$ and $y$ are adjacent from the formulation of Theorem 5, that is, $x$ and $y$ can be any pair of vertices without common neighbors.

The results in Theorems 5 and 6 are sharp; in the graph in Fig. 1 there are no $x\mathord{-}y$-paths with 9 vertices, and in the graph in Fig. 2 there are no $x\mathord{-}y$-paths with 11 or 13 vertices. Furthermore, the results cannot simply be extended to cover the case when $x$ and $y$ have neighbors in common; some counterexamples can be seen in Fig. 3.

Corollary 7**.**

Let $G$ be a finite connected $L_{1}$-graph on at least three vertices such that $\lvert N(u)\cap N(v)\rvert\geq 2$ for each pair of vertices $u,v$ with $d(u,v)=2$. Then for every pair of vertices $x,y\in V(G)$ with no neighbors in common, there is a number $r$ and a sequence of integers $n_{1},n_{2},\dotsc,n_{r}$, depending on $x$ and $y$, such that $n_{1}=d(x,y)+1$, $n_{r}=\lvert V(G)\rvert$ (unless $G\in\mathcal{K}$, in which case $n_{r}=\lvert V(G)\rvert-1$), and $1\leq n_{i+1}-n_{i}\leq 2$ for each $i=1,\dotsc,r-1$, and a sequence of $x\mathord{-}y$-paths $P_{n_{1}},P_{n_{2}},\dotsc,P_{n_{r}}$ with $n_{1},n_{2},\dotsc,n_{r}$ vertices, respectively, such that $V(P_{n_{1}})\subset V(P_{n_{2}})\subset\dotsb\subset V(P_{n_{r}})$.

Corollary 8**.**

Let $G$ be a finite connected $L_{1}$-graph on at least three vertices such that $\lvert N(u)\cap N(v)\rvert\geq 2$ for each pair of vertices $u,v$ with $d(u,v)=2$. Then for every pair of vertices $x,y\in V(G)$ with no neighbors in common and each $m=d(x,y)+1,\dotsc,\lvert V(G)\rvert$, there is an $x\mathord{-}y$-path with $m$ or $m-1$ vertices.

Corollary 9**.**

Let $G$ be a connected, infinite, locally finite $L_{1}$-graph on at least three vertices such that $\lvert N(u)\cap N(v)\rvert\geq 2$ for each pair of vertices $u,v$ with $d(u,v)=2$. Then for every pair of vertices $x,y\in V(G)$ with no neighbors in common and each $m\geq d(x,y)+1$, there is an $x\mathord{-}y$-path with $m$ or $m-1$ vertices.

The local nature of the $L_{1}$-condition allows us to easily extend Theorem C to Hamilton curves in infinite graphs.

Theorem 10**.**

Let $G$ be a connected, infinite, locally finite $L_{1}$-graph on at least three vertices such that $\lvert N(u)\cap N(v)\rvert\geq 2$ for each pair of vertices $u,v$ with $d(u,v)=2$. Then $\lvert G\rvert$ has a Hamilton curve.

We believe that Theorem 10 can be strengthened to the following, which would be a generalization of Theorem E:

Conjecture 11**.**

Let $G$ be a connected, infinite, locally finite $L_{1}$-graph on at least three vertices such that $\lvert N(u)\cap N(v)\rvert\geq 2$ for each pair of vertices $u,v$ with $d(u,v)=2$. Then $\lvert G\rvert$ has a Hamilton circle.

We end by characterizing all bipartite $L_{1}$-graphs.

Theorem 12**.**

Let $G$ be a connected, bipartite $L_{1}$-graph with maximum degree greater than $2$. Then either $G$ is a complete bipartite graph $K_{n,n}$, or $G$ is obtained from $K_{n,n}$ by removing a single vertex, edge, or perfect matching.

Note that a connected bipartite $L_{1}$-graph with maximum degree at most $2$ is either an even cycle or a finite or infinite path.

3 Proofs

In this section we prove our results.

Remark 1*.*

Let $uwv$ be a path in $G$ with $uv\notin E(G)$. Then the inequality $d(u)+d(v)\geq\lvert N(u)\cup N(v)\cup N(w)\rvert-1$ is equivalent to \lvert N(u)\cap N(v)\rvert\geq\big{\lvert}N(w)\setminus\bigl{(}N(u)\cup N(v)\bigr{)}\big{\rvert}-1.

Lemma 13**.**

If $G$ is a connected graph with at least three vertices such that $\lvert N(u)\cap N(v)\rvert\geq 2$ for each pair of vertices $u,v$ with $d(u,v)=2$, then $G$ is $2$-connected.

Lemma 14** **(see [3, thm. 5]111In [3], the result in Lemma 14

is only stated for finite graphs, but the same proof works for infinite, locally finite graphs as well.).

If $G$ is a $2$-connected $L_{1}$-graph, then either $G$ is $1$-tough or $G\in\mathcal{K}$.

3.1 Proof of Theorem 1

Assume that there is no cycle of length $n+1$ or $n+2$ containing the vertices of $C_{n}$. Specify a cyclic orientation $\vec{C}_{n}$ of $C_{n}$ and pick a vertex $v\in V(G)\setminus V(C_{n})$ such that $N(v)\cap V(C_{n})\neq\emptyset$. Set $W=N(v)\cap V(C_{n})$ and $p=\lvert W\rvert$. Let $w_{1},\dotsc,w_{p}$ be the vertices of $W,$ occurring on $\vec{C}_{n}$ in the order of their indices, and set $W^{+}=\{w_{1}^{+},\dotsc,w_{p}^{+}\}$. All indices are considered to be modulo $p$, so $w_{p+1}=w_{1}$.

Remark 2*.*

Note that any extension of $C_{n}$ that occurs in this proof contains either the vertex $v$ or a vertex of $M_{3}(v)$ (in Claims 1 and 2 it will always be the case that $v$ is included). This will be important in the proof of Theorem 10.

Claim 1**.**

The set $W^{+}\cup\{v\}$ is independent, $N(w_{i}^{+})\cap N(v)=N(w_{i}^{+})\cap W,$ $\lvert N(w_{i})\cap W^{+}\rvert=\lvert N(w_{i}^{+})\cap W\rvert$, and N(w_{i})\setminus\bigl{(}N(w_{i}^{+})\cup N(v)\cup\{v\}\bigr{)}\subseteq W^{+} for $i=1,\dotsc,p$.

Proof.

If there is an edge $vw_{i}^{+}\in E(G)$, then $G$ contains an $(n+1)$-cycle $w_{i}vw_{i}^{+}\vec{C}_{n}w_{i}$, and if there is an edge $w_{i}^{+}w_{j}^{+}\in E(G)$, then $G$ contains an $(n+1)$-cycle w_{i}vw_{j}\ooalign{C\cr{\raisebox{6.83331pt}{\hbox to0.0pt{\mkern 4.1mu\rotatebox[origin={c}]{180.0}{\vec{\phantom{C}}}\hss}}}}_{n}w_{i}^{+}w_{j}^{+}\vec{C}_{n}w_{i}. Thus

[TABLE]

Also, if \bigl{(}N(w_{i}^{+})\cap N(v)\bigr{)}\setminus V(C_{n})\neq\emptyset for some $1\leq i\leq p$, that is, if $w_{i}^{+}$ and $v$ have a common neighbor $u$ outside $C_{n}$, then $G$ contains an $(n+2)$-cycle $w_{i}vuw_{i}^{+}\vec{C}_{n}w_{i}$. Thus \bigl{(}N(w_{i}^{+})\cap N(v)\bigr{)}\setminus V(C_{n})=\emptyset, which means that

[TABLE]

Now for each $i=1,\dotsc,p$, we have $d(v,w_{i}^{+})=2$ and $w_{i}\in N(w_{i}^{+})\cap N(v)$, so by Remark 1,

[TABLE]

Obviously,

[TABLE]

Thus \lvert N(w_{i})\cap W^{+}\rvert\leq\big{\lvert}N(w_{i})\setminus\bigl{(}N(w_{i}^{+})\cup N(v)\bigr{)}\big{\rvert}-1. This and 3 together imply that

[TABLE]

We will now count the number of edges between $W^{+}$ and $W$ in two different ways:

[TABLE]

It follows for each $i=1,\dotsc,p$, that

[TABLE]

and that we have equality in 4, so

[TABLE]

Claim 2**.**

$w_{i}^{+}=w_{i+1}^{-}$* for $i=1,\dotsc,p$, that is, $n=2p$ and $v$ is adjacent to every second vertex of $C_{n}$.*

Proof.

Suppose that $v$ is not adjacent to every second vertex of the cycle $C_{n}$. Then $w_{i}^{+}\neq w_{i+1}^{-}$ for some $i$. Without loss of generality, assume that $w_{1}^{+}\neq w_{2}^{-}$, which means that $w_{2}^{-}\notin W^{+}$. This and 8 for $i=2$ imply that $w_{2}^{-}\in N(w_{2}^{+})$, because otherwise w_{2}^{-}\in N(w_{2})\setminus\bigl{(}N(w_{2}^{+})\cup N(v)\cup\{v\}\bigr{)}\subseteq W^{+}, a contradiction. Therefore $w_{2}^{-}w_{2}^{+}\in E(G)$. This in turn means that $w_{2}^{+}\neq w_{3}^{-}$, because otherwise there would be an $(n+1)$-cycle $w_{2}^{-}w_{2}^{+}w_{2}vw_{3}\vec{C}_{n}w_{2}^{-}$ (unless $p=1$, in which case recall that $w_{p+1}=w_{1}$ and skip this sentence). Repetition of this argument shows that $w_{i}^{+}\neq w_{i+1}^{-}$ for $i=1,\dotsc,p$, and that

[TABLE]

Now it is easy to see that $w_{1}^{+}w_{j}\notin E(G)$ for each $j\neq 1$, as otherwise there would be an $(n+1)$-cycle $w_{1}vw_{j}w_{1}^{+}\vec{C}_{n}w_{j}^{-}w_{j}^{+}\vec{C}_{n}w_{1}$ containing the vertices of $C_{n}$. This, together with 2, implies that $N(w_{1}^{+})\cap N(v)=\{w_{1}\}$. This contradicts the fact that $d(w_{1}^{+},v)=2$. Thus we can conclude that $w_{i}^{+}=w_{i+1}^{-}$ for each $i=1,\dotsc,p$, and that $n=2p$. ∎

Claim 3**.**

$n=\lvert V(G)\rvert-1$* and $G\in\mathcal{K}$.*

Proof.

We have concluded that $n=2p$ and that $N(v)$ contains every second vertex of $C_{n}$. Note that $p\geq 2$, as otherwise $N(w_{1}^{+})\cap N(v)=\{w_{1}\}$ by Claim 1, contradicting the conditions of the theorem. Suppose some vertex $w_{i}^{+}\in W^{+}$ has a neighbor $u$ outside $C_{n}$. Since $v$ was picked arbitrarily in the set $V(G)\setminus V(C_{n})$ such that $N(v)\cap V(C_{n})\neq\emptyset$, we can conclude that $u$ is adjacent to every second vertex of $C_{n}$ as well, that is, $N(u)\cap V(C_{n})=W^{+}$. But then there is an $(n+2)$-cycle $w_{1}vw_{2}w_{1}^{+}uw_{2}^{+}\vec{C}_{n}w_{1}$ containing the vertices of $C_{n}$, a contradiction, so no vertex outside $C_{n}$ is adjacent to any vertex in $W^{+}$. Thus $G$ is not 1-tough, so $G\in\mathcal{K}$ by Lemma 14. Also, since $G\in\mathcal{K}$ it follows that if $n<\lvert V(G)\rvert-1$ then there is a cycle of length $n+1$ or $n+2$ in $G$ containing the vertices of $C_{n}$. Thus $n=\lvert V(G)\rvert-1$. ∎

3.2 Proof of Theorem 5

Assume that there is no $x\mathord{-}y$-path with $n+1$ or $n+2$ vertices containing the vertices of $P_{n}$. Pick a vertex $v\in V(G)\setminus V(P_{n})$ such that $N(v)\cap V(P_{n})\neq\emptyset$. Since $x$ and $y$ have no neighbors in common, it follows that $\lvert N(v)\cap\{x,y\}\rvert\leq 1$. Without loss of generality we assume that $vy\notin E(G)$. Let $\vec{P}_{n}$ be $P_{n}$ directed from $x$ to $y$. Set $W=N(v)\cap V(P_{n})$ and $p=\lvert W\rvert$. Let $w_{1},\dotsc,w_{p}$ be the vertices of $W,$ occurring on $\vec{P}_{n}$ in the order of their indices, and set $W^{+}=\{w_{1}^{+},\dotsc,w_{p}^{+}\}$. The path $P_{n}$ together with the edge $xy$ of course forms a cycle, and for simplicity we define $z^{+}$ to be the successor of $z$ on this cycle, so $y^{+}=x$, etc. Also, all indices are considered to be modulo $p$, so $w_{p+1}=w_{1}$.

Claim 1**.**

The set $W^{+}\cup\{v\}$ is independent, $N(w_{i}^{+})\cap N(v)=N(w_{i}^{+})\cap W,$ $\lvert N(w_{i})\cap W^{+}\rvert=\lvert N(w_{i}^{+})\cap W\rvert$, and N(w_{i})\setminus\bigl{(}N(w_{i}^{+})\cup N(v)\cup\{v\}\bigr{)}\subseteq W^{+} for $i=1,\dotsc,p$.

Proof.

This follows using the same arguments as in the proof of Theorem 1. ∎

Claim 2**.**

$w_{1}=x$, $w_{p}=y^{-}$, and $w_{i}^{+}=w_{i+1}^{-}$ for $i=1,\dotsc,p-1$, that is, $n=2p$ and $v$ is adjacent to every second vertex of $P_{n}$.

Proof.

We will start by showing that $w_{i}^{+}=w_{i+1}^{-}$ for each $i=1,\dotsc,p-1$. Assume on the contrary that $w_{k}^{+}\neq w_{k+1}^{-}$ for some $k\leq p-1$, and furthermore assume that $k$ is the first such index, i.e., either $k=1$ or $k\geq 2$ and $w_{i}^{+}=w_{i+1}^{-}$ for every $i=1,\dotsc,k-1$. Then $w_{k+1}^{-}\notin W^{+}$. This and 8 for $i=k+1$ imply that $w_{k+1}^{-}\in N(w_{k+1}^{+})$, because otherwise w_{k+1}^{-}\in N(w_{k+1})\setminus\bigl{(}N(w_{k+1}^{+})\cup N(v)\cup\{v\}\bigr{)}\subseteq W^{+}, a contradiction. Therefore $w_{k+1}^{-}w_{k+1}^{+}\in E(G)$. This in turn means that $w_{k+1}^{+}\neq w_{k+2}^{-}$, because otherwise there would be an $x\mathord{-}y$-path $x\vec{P}_{n}w_{k+1}^{-}w_{k+1}^{+}w_{k+1}vw_{k+2}\vec{P}_{n}y$ with $n+1$ vertices (unless $k=p-1$, in which case skip this sentence). Repetition of this argument shows that $w_{i}^{+}\neq w_{i+1}^{-}$ for each $i=k,\dotsc,p-1$, and that

[TABLE]

Let $W_{1}=\{w_{1},\dotsc,w_{k}\}$ and $W_{1}^{+}=\{w_{1}^{+},\dotsc,w_{k}^{+}\}$. It is easy to see that $w_{i}^{+}w_{j}\notin E(G)$ for each $j>k$ and each $i\neq j$, as otherwise there would be an $x\mathord{-}y$-path $x\vec{P}_{n}w_{i}vw_{j}w_{i}^{+}\vec{P}_{n}w_{j}^{-}w_{j}^{+}\vec{P}_{n}y$ (if $i<j$) or $x\vec{P}_{n}w_{j}^{-}w_{j}^{+}\vec{P}_{n}w_{i}vw_{j}w_{i}^{+}\vec{P}_{n}y$ (if $i>j$) with $n+1$ vertices. This means that $N(w_{i}^{+})\cap W=N(w_{i}^{+})\cap W_{1}$ for each $i=1,\dotsc,k$. This, together with 7, means that

[TABLE]

for every $i=1,\dotsc,k$. We will now count the edges between $W_{1}^{+}$ and $W_{1}$ in two different ways:

[TABLE]

This means that we have equality in 11, so for every $i=1,\dotsc,k$

[TABLE]

which means that $w_{i}w_{j}^{+}\notin E(G)$ for all $i=1,\dotsc,k$ and $j=k+1,\dotsc,p$. But then 2 implies that $N(v)\cap N(w_{j}^{+})=\{w_{j}\}$ for every $j=k+1,\dotsc,p$. This contradicts the assumptions of the theorem, because the fact that $d(v,w_{j}^{+})=2$ implies that $\lvert N(v)\cap N(w_{j}^{+})\rvert\geq 2$. Thus we can conclude that $w_{i}^{+}=w_{i+1}^{-}$ for each $i=1,\dotsc,p-1$.

Now we can use an argument similar to the one in the beginning of this proof to show that $w_{1}=x$: If $w_{1}\neq x$ then $w_{1}^{-}\notin W^{+}$ (by assumption $vy\notin E(G)$, so no vertex on $x\vec{P}_{n}w_{1}$ is in $W^{+}$). This means that $w_{1}^{-}\in N(w_{1}^{+})$ by 8, so $w_{1}^{+}w_{1}^{-}\in E(G)$. Note also that $p\geq 2$, since otherwise $N(v)\cap N(w_{1}^{+})=\{w_{1}\}$, a contradiction as $d(v,w_{1}^{+})=2$. But now, since $w_{1}^{+}=w_{2}^{-}$, there is an $x\mathord{-}y$-path $x\vec{P}_{n}w_{1}^{-}w_{1}^{+}w_{1}vw_{2}\vec{P}_{n}y$ with $n+1$ vertices. This is a contradiction, so we can conclude that $w_{1}=x$. Also, since $y$ and $x=w_{1}$ are adjacent and have no neighbors in common and $y\neq v$, it follows that y\in N(w_{1})\setminus\bigl{(}N(w_{1}^{+})\cup N(v)\cup\{v\}\bigr{)}. Thus $y\in W^{+}$ by 8, so $w_{p}=y^{-}$, and $n=2p$. ∎

Claim 3**.**

$n=\lvert V(G)\rvert-1$* and $G\in\mathcal{K}$.*

Proof.

This follows using the same arguments as in the proof of Theorem 1. ∎

3.3 Proof of Theorem 6

Assume that there is no $x\mathord{-}y$-path with $n+1$ or $n+2$ vertices containing the vertices of $P_{n}$. Pick a vertex $v\in V(G)\setminus V(P_{n})$ such that $N(v)\cap V(P_{n})\neq\emptyset$. Since $d(x,y)\geq 3$, it follows that $\lvert N(v)\cap\{x,y\}\rvert\leq 1$. Without loss of generality we assume that $vy\notin E(G)$. Let $\vec{P}_{n}$ be $P_{n}$ directed from $x$ to $y$. Set $W=N(v)\cap V(P_{n})$ and $p=\lvert W\rvert$. Let $w_{1},\dotsc,w_{p}$ be the vertices of $W$, occurring on $\vec{P}_{n}$ in the order of their indices, and set $W^{+}=\{w_{1}^{+},\dotsc,w_{p}^{+}\}$.

Claim 1**.**

The set $W^{+}\cup\{v\}$ is independent, $N(w_{i}^{+})\cap N(v)=N(w_{i}^{+})\cap W,$ $\lvert N(w_{i})\cap W^{+}\rvert=\lvert N(w_{i}^{+})\cap W\rvert$, and N(w_{i})\setminus\bigl{(}N(w_{i}^{+})\cup N(v)\cup\{v\}\bigr{)}\subseteq W^{+} for $i=1,\dotsc,p$.

Proof.

This is proved exactly as Claim 1 in the proof of Theorem 5. ∎

Claim 2**.**

$w_{1}=x$* and $w_{i}^{+}=w_{i+1}^{-}$ for $i=1,\dotsc,p-1$, that is, $v$ is adjacent to every second vertex of $x\vec{P}_{n}w_{p}$.*

Proof.

This is proved exactly as Claim 2 in the proof of Theorem 5, without the last two sentences. ∎

Claim 3**.**

There exists a number $t$ such that $\lvert N(w_{i})\cap W^{+}\rvert=\lvert N(w_{i}^{+})\cap W\rvert=t$ for $i=1,\dotsc,p$.

Proof.

First, for $i=1,\dotsc,p-1$

[TABLE]

Now for any $k\in\{1,\dotsc,p-1\}$, $w_{k}^{+}w_{k+1}\in E(G)$, so

[TABLE]

Thus $G$ must contain some edge $w_{j}^{+}w_{i}$ with $i\leq k<j$. Now, by using 14 iteratively,

[TABLE]

We can thus conclude that $\lvert N(w_{k}^{+})\cap W\rvert=\lvert N(w_{k+1}^{+})\cap W\rvert$ for $k=1,\dotsc,p-1$. The rest of the claim now follows from Claim 1. ∎

Claim 4**.**

$N(w_{i}^{+})\subseteq V(P_{n})$* for $i=1,\dotsc,p-1$.*

Proof.

If $w_{i}^{+}u\in E(G)$ for some $i=1,\dotsc,p-1$ and some $u\notin V(P_{n})$, then repeating Claims 1 and 2 with $u$ instead of $v$, we get that $u$ is adjacent to every second vertex between $w_{i}^{+}$ and either $x$ or $y$. But it is then impossible that $ux\in E(G)$, since $x\vec{P}_{n}w_{i}$ has an odd number of vertices, which means that $u$ is adjacent to $y$ and, in particular, $w_{i+1}^{+}$. But then there is an $x\mathord{-}y$-path $x\vec{P}_{n}w_{i}vw_{i+1}w_{i}^{+}uw_{i+1}^{+}\vec{P}_{n}y$ with $n+2$ vertices, a contradiction. Thus $N(w_{i}^{+})\subseteq V(P_{n})$ for $i=1,\dotsc,p-1$. ∎

Claim 5**.**

$N(w_{i}^{+})=W$* for $i=1,\dotsc,p-1$ and $W\subseteq N(w_{p}^{+})$.*

Proof.

Note that when proving Claims 1, 2 and 3, every time we reached a contradiction by constructing a longer $x\mathord{-}y$-path, the new path contained the vertex $v$. Also, note that $p\geq 2$, as otherwise $N(w_{1}^{+})\cap N(v)=\{w_{1}\}$ by Claim 1, contradicting the conditions of the theorem. Now consider the path $P_{n}^{\prime}=xvw_{2}\vec{P}_{n}y$. Then Claims 1, 2 and 3 are valid for $P_{n}^{\prime}$ with $x^{+}$ instead of $v$ as the outside vertex, since otherwise we could construct an $x\mathord{-}y$-path containing all vertices of $V(P^{\prime}_{n})\cup\{x^{+}\}=V(P_{n})\cup\{v\}$. Note also that $t$ from Claim 3 has the property $t=\lvert N(v)\cap N(x^{+})\rvert$, so $t$ has the same value for $P_{n}$ and $v$ as for $P_{n}^{\prime}$ and $x^{+}$.

We shall now prove that $t=p$. Assume on the contrary that $t<p$ and let $W^{\prime}=N(x^{+})\cap V(P_{n}^{\prime})$. Since $x^{+}$ is adjacent to $t$ vertices in $W$, it is easy to see that Claim 2 for $P_{n}^{\prime}$ and $x^{+}$ implies that $W^{\prime}=\{w_{1},\dotsc,w_{t}\}$. It follows from Claim 3 for $P_{n}^{\prime}$ and $x^{+}$ that $w_{t}^{+}$ is adjacent to all vertices in $W^{\prime}$. But then $\{w_{1},\dotsc,w_{t+1}\}\subseteq N(w_{t}^{+})\cap W$, so $\lvert N(w_{t}^{+})\cap W\rvert\geq t+1$, contradicting Claim 3 for $P_{n}$ and $v$. We can conclude that $t=p$ and that $W\subseteq N(w_{i}^{+})$ for $i=1,\dotsc,p$.

Now assume that $N(w_{i}^{+})\neq W$ for some $i\in\{1,\dotsc,p-1\}$. Then Claim 4 implies that $w_{i}^{+}$ has a neighbor on $w_{p}^{++}P_{n}y$, since $W^{+}$ is independent. Now consider the path $P_{n}^{\prime\prime}=x\vec{P}_{n}w_{i}vw_{i+1}\vec{P}_{n}y$. As above, Claims 1, 2 and 3 are valid for $P_{n}^{\prime\prime}$ with $w_{i}^{+}$ instead of $v$ as the outside vertex and $t$ has the same value for $P_{n}$ and $v$ as for $P_{n}^{\prime\prime}$ and $w_{i}^{+}$. Thus $w_{p}^{++}\in W^{\prime\prime}=N(w_{i}^{+})\cap V(P_{n}^{\prime\prime})$ by Claim 2 for $P_{n}^{\prime\prime}$ and $w_{i}^{+}$. This means that $\lvert N(w_{p}^{+})\cap W^{\prime\prime}\rvert\geq\lvert W\cup\{w_{p}^{++}\}\rvert=t+1$, a contradiction. We can conclude that $N(w_{i}^{+})=W$ for $i=1,\dotsc,p-1$. ∎

We now know that $xw_{p}^{+}\in E(G)$, which means that $d(x,w_{p}^{++})\leq 2$. This will be used to get our final contradiction. If $xw_{p}^{++}\in E(G)$ then Claim 1 implies that $x^{+}w_{p}^{++}\in E(G)$, contradicting Claim 5. Thus $d(x,w_{p}^{++})=2$, which means that

[TABLE]

It follows from Claim 1 that $N(x)\subseteq W^{+}\cup N(x^{+})\cup N(v)\cup\{v\}$. Claim 5 shows that $N(x^{+})=W$, and we shall see that $N(v)\cap N(x)\subset W$ as well. Assume on the contrary that there is a vertex $u\in N(x)\cap N(v)\setminus V(P_{n})$. Then $u\notin N(w_{1}^{+})\cup N(w_{2}^{+})$ by Claim 1. Thus

[TABLE]

a contradiction. We can conclude that $N(x)\subseteq W\cup W^{+}\cup\{v\}$. Claim 5 implies that $N(w_{p}^{++})\cap W^{+}\subseteq\{w_{p}^{+}\}$, and together with Claim 1 it implies that $N(w_{p}^{++})\cap W\subseteq\{w_{p}\}$ since $w_{p}^{++}\notin W^{+}\cup N(v)\cup\{v\}$. Equation 17 now implies that $N(x)\cap N(w_{p}^{++})=\{w_{p},w_{p}^{+}\}$. But then

[TABLE]

our final contradiction. The Theorem follows.

3.4 Proof of Theorem 10

To prove that the conditions of Theorem 1 are sufficient to find a Hamilton curve, we will use the following theorem by Kündgen, Li, and Thomassen, along with an observation.

Theorem F** (Kündgen–Li–Thomassen [21]).**

The following are equivalent for any locally finite graph $G$.

For every finite vertex set $S$, $G$ has a cycle containing $S$. 2. 2.

$\lvert G\rvert$* has a Hamilton curve.*

Observation 15**.**

In the proof of Theorem 1, whenever we reach a contradiction by constructing a cycle $C_{n+\ell}$, the new cycle contains either the vertex $v$ or a vertex $u$ at distance at most $3$ from $v$ (see Remark 2). Thus, if $G$ satisfies the conditions of Theorem 1 and $v$ is a vertex adjacent to a cycle $C_{n}$ in $G$, then there is a cycle $C_{n+\ell}$ containing all vertices of $C_{n}$ and at least one additional vertex from the set $M_{3}(v)$, unless $G\in\mathcal{K}$.

Using Observations 15 and F it is straightforward to prove Theorem 10. First note that $G\notin\mathcal{K}$ since is infinite. Now for any finite vertex set $S$, pick a vertex $a$ and an integer $r$ such that $S\subseteq M_{r}(a)$, and let $C_{n}$ be a cycle through $a$ containing as many vertices as possible from the set $M_{r+3}(a)$. If $C_{n}$ does not contain all vertices of $S$, there is a vertex $v\in M_{r}(a)\setminus V(C_{n})$ with a neighbor on $C_{n}$, and by using Observation 15 we can find a cycle $C_{n+\ell}$ containing more vertices of $M_{r+3}(a)$, a contradiction. Thus $C_{n}$ contains all vertices of $S$. Now, using Theorem F we can conclude that $\lvert G\rvert$ has a Hamilton curve.

3.5 Proof of Theorem 12

For a non-regular graph $G$ with maximum degree at least three it is straightforward to use Observation 16 below to prove that $G$ is a subgraph of a complete bipartite graph $K_{n,n}$ with a single vertex or a single edge removed, by simply constructing the possible graphs vertex by vertex. Similarly one can prove, using Observation 17, that every regular, connected, bipartite $L_{1}$-graph with maximum degree at least three is either a complete bipartite graph $K_{n,n}$ or a subgraph of $K_{n,n}$ with a perfect matching removed. Theorem 12 follows. For details, see [16].

Observation 16** ([16]).**

Let $G$ be a bipartite $L_{1}$-graph and let $u$ and $v$ be two adjacent vertices in $G$. Then $\lvert d(u)-d(v)\rvert\leq 1$.

Observation 17** ([16]).**

Let $G$ be an $n$-regular bipartite $L_{1}$-graph and let $u$ and $v$ be two vertices at distance $2$ in $G$. Then $\lvert N(u)\cap N(v)\rvert\geq n-1$.

Acknowledgement

I would like to thank Armen Asratian and Carl Johan Casselgren for many helpful comments and fruitful discussions while preparing this work.