# L(2, 1)-Labeling Halin Graphs with Maximum Degree Eight

^{1}

^{2}

^{*}

## Abstract

**:**

## 1. Introduction

**Conjecture**

**1.**

- (1)
- If G is a Halin graph with $\Delta =8$, then $\lambda \left(G\right)\le 10$;
- (2)
- For every Halin graph G, it holds that $\lambda \left(G\right)\le \Delta +6$.

## 2. Structural Analysis

**Lemma**

**1.**

**(C1)**$k=4$, and there exists a vertex v adjacent to two 3-handles ${u}_{1}$ and ${u}_{2}$ such that $N\left({u}_{1}\right)=\{v,{x}_{1},{x}_{2}\}$ and $N\left({u}_{2}\right)=\{v,{x}_{3},{x}_{4}\}$.

**(C2)**$k=5$, and there exists a vertex v adjacent to a 3-handle ${u}_{1}$ and a 4-handle ${u}_{2}$ such that $N\left({u}_{1}\right)=\{v,{x}_{1},{x}_{2}\}$ and $N\left({u}_{2}\right)=\{v,{x}_{3},{x}_{4},{x}_{5}\}$.

**(C3)**$k=6$, and there exists a vertex v adjacent to two 4-handles ${u}_{1}$ and ${u}_{2}$ such that $N\left({u}_{1}\right)=\{v,{x}_{1},{x}_{2},{x}_{3}\}$ and $N\left({u}_{2}\right)=\{v,{x}_{4},{x}_{5},{x}_{6}\}$.

**(C4)**$k=4$, and there exists a vertex v adjacent to ${x}_{4}$ and a 4-handle u such that $N\left(u\right)=\{v,{x}_{1},{x}_{2},{x}_{3}\}$.

**(C5)**$k=3$, and there exists a ${5}^{-}$-vertex v adjacent to ${x}_{3}$ and a 3-handle u such that $N\left(u\right)=\{v,{x}_{1},{x}_{2}\}$.

**(C6)**$k=4$, and there exists a ${6}^{-}$-vertex v adjacent to ${x}_{1},{x}_{4}$ and a 3-handle u such that $N\left(u\right)=\{v,{x}_{2},{x}_{3}\}$.

**(C7)**$k=5$, and there exists a 7-vertex v adjacent to ${x}_{1},{x}_{4},{x}_{5}$ and a 3-handle u such that $N\left(u\right)=\{v,{x}_{2},{x}_{3}\}$.

**(C8)**$k=6$, and there exists a 8-vertex v adjacent to ${x}_{1},{x}_{4},{x}_{5},{x}_{6}$ and a 3-handle u such that $N\left(u\right)=\{v,{x}_{2},{x}_{3}\}$.

**(C9)**$k=7$, and there exists a vertex v adjacent to ${x}_{1},{x}_{4},{x}_{7}$ and two 3-handles ${u}_{1}$ and ${u}_{2}$ such that $N\left({u}_{1}\right)=\{v,{x}_{2},{x}_{3}\}$ and $N\left({u}_{2}\right)=\{v,{x}_{5},{x}_{6}\}$.

**(C10)**$k=8$, and there exists a vertex v adjacent to ${x}_{1},{x}_{4},{x}_{5},{x}_{8}$ and two 3-handles ${u}_{1}$ and ${u}_{2}$ such that $N\left({u}_{1}\right)=\{v,{x}_{2},{x}_{3}\}$ and ${N}_{G}\left({u}_{2}\right)=\{v,{x}_{6},{x}_{7}\}$.

**(C11)**$k\ge 4$, and there exists a $(k+1)$-handle u such that $N\left(u\right)=\{v,{x}_{1},{x}_{2},\dots ,{x}_{k}\}$.

**(C12)**$k\ge 7$, and there exists a vertex v adjacent to ${x}_{3},{x}_{4},\dots ,{x}_{k},w$ and a 3-handle u such that $N\left(u\right)=\{v,{x}_{1},{x}_{2}\}$ and $N\left(v\right)=\{u,{x}_{3},\dots ,{x}_{k},w\}$.

**(C13)**$k\ge 8$, and there exists a vertex v adjacent to ${x}_{3},{x}_{4},\dots ,{x}_{k-2},w$ and two 3-handles ${u}_{1}$ and ${u}_{2}$ such that $N\left({u}_{1}\right)=\{v,{x}_{1},{x}_{2}\}$, $N\left({u}_{2}\right)=\{v,{x}_{k-1},{x}_{k}\}$, and $N\left(v\right)=\{{u}_{1},{u}_{2},{x}_{3},\dots ,{x}_{k-2},w\}$.

**(C14)**$k=10$, and there exists a vertex v adjacent to ${x}_{3},{x}_{4},{x}_{7},{x}_{8},w$ and three 3-handles ${u}_{1},{u}_{2},{u}_{3}$ such that $N\left({u}_{1}\right)=\{v,{x}_{1},{x}_{2}\}$, $N\left({u}_{2}\right)=\{v,{x}_{5},{x}_{6}\}$, $N\left({u}_{3}\right)=\{v,{x}_{9},{x}_{10}\}$, and $N\left(v\right)=\{{u}_{1},{u}_{2},{u}_{3},{x}_{3},{x}_{4},{x}_{7},{x}_{8},w\}$.

**Proof.**

**Case 1.**$\beta =1$.

**Case 2.**$\beta =2$.

- $i=1$ and $j=m$. Then (C13) holds;
- $i=1$ and $j\le m-1$. Then ${z}_{j-1}$ and ${z}_{j+1}$ are leaves in T. If $d\left({y}_{2}\right)\le 6$, then (C6) holds. If $d\left({y}_{2}\right)=7$, then (C7) holds. If $d\left({y}_{2}\right)=8$, then $m=7$, we have $\left|\right\{{z}_{2},{z}_{3},\dots ,{z}_{s-1}\left\}\right|\ge 3$ or $\left|\right\{{z}_{s+1},{z}_{s+2},\dots ,{z}_{7}\left\}\right|\ge 3$. Thus, (C8) holds;
- $i\ge 2$ and $j\le m-1$. If $d\left({y}_{2}\right)\le 6$, then (C6) holds. If $d\left({y}_{2}\right)=7$, then (C7) holds. Otherwise, $d\left({y}_{2}\right)=8$. Note that $2\le j-i\le 4$. If $j-i=2$, then (C9) holds. If $j-i=3$, then (C10) holds. If $j-i=4$, that is $i=2$ and $j=6$, then (C8) holds.

**Case 3.**$\beta =3$.

**Case 4.**$\beta =4$.

## 3. Preliminary Results

**Lemma**

**2.**

**Lemma**

**3.**

**Proof.**

**Case 1.**$L\left({x}_{1}\right)=\{a,t\}$ where $t\in \{b,c,d\}$.

**Case 2.**$L\left({x}_{1}\right)=\{b,c\}$.

**Lemma**

**4.**

**Proof.**

**Case 1.**$q=p+1$, say $p=5$ and $q=6$.

**Case 2.**$q=p+2$, say $p=5$ and $q=7$.

**Lemma**

**5.**

**Proof.**

**Case 1.**$L\left({x}_{1}\right)=\{5,7\}$ and $L\left({x}_{2}\right)=\{4,5,6,7,8\}$.

**Case 2.**$L\left({x}_{1}\right)=\{5,6\}$, $\{4,5,6,7\}\subset L\left({x}_{2}\right)$, $L\left({x}_{4}\right)=\{b+1,b+2\}$, and $\{b,b+1,b+2,b+3\}\subset L\left({x}_{3}\right)$.

**Lemma**

**6.**

**Proof.**

**Lemma**

**7.**

- $\left|L\right({x}_{1}\left)\right|\ge 2$, $|L\left({x}_{2}\right)|,|L\left({x}_{3}\right)|,|L\left({x}_{4}\right)|\ge 5$, and $\left|L\right({x}_{5}\left)\right|\ge 3$.
- $L\left({x}_{2}\right)=L\left({x}_{3}\right)=L\left({x}_{4}\right)=S$ and $L\left({x}_{1}\right),L\left({x}_{5}\right)\subset S$.

**Proof.**

**Case 1.**$L\left({x}_{5}\right)$ contains ${a}_{1}$ or ${a}_{5}$, say ${a}_{1}\in L\left({x}_{5}\right)$.

- $L\left({x}_{1}\right)=\{{a}_{1},{a}_{2}\}$. If ${a}_{3}\in L\left({x}_{5}\right)$ or ${a}_{4}\in L\left({x}_{5}\right)$, we have a similar proof. Otherwise, $L\left({x}_{5}\right)=\{{a}_{1},{a}_{2},{a}_{5}\}$. Label ${x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5}$ with ${a}_{2},{a}_{4},{a}_{1},{a}_{3},{a}_{5}$, respectively;
- $L\left({x}_{1}\right)=\{{a}_{1},{a}_{5}\}$. With the similar reasoning, we have $L\left({x}_{5}\right)=\{{a}_{1},{a}_{2},{a}_{5}\}$, and ${x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5}$ can be labeled with ${a}_{5},{a}_{3},{a}_{1},{a}_{4},{a}_{2}$, respectively;
- $L\left({x}_{1}\right)=\{{a}_{2},{a}_{5}\}$. In view of the above discussion, we may assume that $L\left({x}_{5}\right)=\{{a}_{1},{a}_{3},{a}_{4}\}$. It suffices to label ${x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5}$ with ${a}_{2},{a}_{4},{a}_{1},{a}_{5},{a}_{3}$, respectively.

**Case 2.**$L\left({x}_{5}\right)=\{{a}_{2},{a}_{3},{a}_{4}\}$.

**Lemma**

**8.**

**Proof.**

**Case 1.**$L\left({x}_{1}\right)=\{{a}_{2},{a}_{5}\}$ and $L\left({x}_{2}\right)=S\backslash \left\{{a}_{7}\right\}$.

**Case 2.**$L\left({x}_{1}\right)=\{{a}_{2},{a}_{6}\}$ and $L\left({x}_{2}\right)=S\backslash \left\{{a}_{4}\right\}$.

**Case 3.**$L\left({x}_{1}\right)=L\left({x}_{7}\right)=\{{a}_{3},{a}_{6}\}$ and $L\left({x}_{2}\right)=L\left({x}_{6}\right)=S\backslash \left\{{a}_{1}\right\}$.

**Lemma**

**9.**

**Proof.**

- $a\in L\left(x\right)$. Labeling x with a, we define a list assignment ${L}^{\prime}$ for $y,z$ as follows: ${L}^{\prime}\left(y\right)=L\left(y\right)\backslash \{a,a+1\}$ and ${L}^{\prime}\left(z\right)=L\left(z\right)\backslash \{a,a+1\}$. Then $|{L}^{\prime}\left(y\right)|\ge 4-2=2$ and $|{L}^{\prime}\left(z\right)|\ge 5-2=3$. By Lemma 2, $y,z$ are ${L}^{\prime}$-$L(2,1)$-labelable;
- $a\in L\left(y\right)$ and $a\notin L\left(x\right)$. Labeling y with a, we define a list assignment ${L}^{\prime}$ for $x,z$ as follows: ${L}^{\prime}\left(x\right)=L\left(x\right)\backslash \{a+1\}$ and ${L}^{\prime}\left(z\right)=L\left(z\right)\backslash \{a,a+1\}$. Then $|{L}^{\prime}\left(x\right)|\ge 3-1=2$ and $|{L}^{\prime}\left(z\right)|\ge 5-2=3$. By Lemma 2, $x,z$ are ${L}^{\prime}$-$L(2,1)$-labelable;
- $a\in L\left(z\right)$ and $a\notin L\left(x\right)\cup L\left(y\right)$. Labeling z with a, we define a list assignment ${L}^{\prime}$ for $x,y$ as follows: ${L}^{\prime}\left(x\right)=L\left(x\right)\backslash \{a+1\}$ and ${L}^{\prime}\left(y\right)=L\left(y\right)\backslash \{a+1\}$. Then $|{L}^{\prime}\left(x\right)|\ge 3-1=2$ and $|{L}^{\prime}\left(y\right)|\ge 4-1=3$. By Lemma 2, $x,y$ are ${L}^{\prime}$-$L(2,1)$-labelable.

**Lemma**

**10.**

## 4. $L(2,1)$-Labeling

**Theorem**

**1.**

**Proof.**

**(C1)**Let $H=G-\{{x}_{1},{x}_{2},{x}_{3},{x}_{4}\}+\{y{u}_{1},{u}_{1}{u}_{2},{u}_{2}z\}$. Then H is a Halin graph with $\Delta \left(H\right)=8$ and $\left|H\right|<\left|G\right|$. By the induction hypothesis, H has an $L(2,1)$-labeling f with the label set B. Define a list assignment L for ${x}_{1},{x}_{2},{x}_{3},{x}_{4}$ as follows:

**(C2)**Let $H=G-{x}_{4}+{x}_{3}{x}_{5}$. Then H is a Halin graph with $\Delta \left(H\right)=8$ and $\left|H\right|<\left|G\right|$. By the induction hypothesis, H has an $L(2,1)$-labeling f with the label set B. Define a list assignment L for ${x}_{2},{x}_{3},{x}_{4}$ as follows:

**(C3)**Let $H=G-{x}_{2}+{x}_{1}{x}_{3}$. Then H is a Halin graph with $\Delta \left(H\right)=8$ and $\left|H\right|<\left|G\right|$. By the induction hypothesis, H has an $L(2,1)$-labeling f with the label set B. Define a list assignment L for ${x}_{2},{x}_{3},{x}_{4}$ as follows:

**(C4)**Let $H=G-{x}_{2}+{x}_{1}{x}_{3}$. Then H is a Halin graph with $\Delta \left(H\right)=8$ and $\left|H\right|<\left|G\right|$. By the induction hypothesis, H has an $L(2,1)$-labeling f with the label set B. Define a list assignment L for ${x}_{1},{x}_{2},{x}_{3}$ as follows:

**(C5)**Let $N\left(v\right)=\{u,{x}_{3},{t}_{1},\dots ,{t}_{l}\}$. Since $d\left(v\right)\le 5$, we see that $l\le 3$. Let $H=G-\{{x}_{1},{x}_{2}\}+\{{x}_{3}u,uy\}$. Then H is a Halin graph with $\Delta \left(H\right)=8$ and $\left|H\right|<\left|G\right|$. By the induction hypothesis, H has an $L(2,1)$-labeling f using B. Erasing the label of u, we define a list assignment L for ${x}_{1},{x}_{2},u$ as follows:

**(C6)**Let $N\left(v\right)=\{u,{x}_{1},{x}_{4},{t}_{1},\dots ,{t}_{m}\}$. Since $d\left(v\right)\le 6$, we see that $m\le 3$. Let $H=G-\{{x}_{2},{x}_{3}\}+\{{x}_{1}u,u{x}_{4}\}$. Then H is a Halin graph with $\Delta \left(H\right)=8$ and $\left|H\right|<\left|G\right|$. By the induction hypothesis, H admits an $L(2,1)$-labeling f using B. Erasing the label of u, we define a list assignment L for ${x}_{2},{x}_{3},u$ as follows:

**(C7)**Set $N\left(v\right)=\{u,{x}_{1},{x}_{4},{x}_{5},{t}_{1},{t}_{2},{t}_{3}\}$ because $d\left(v\right)=7$. Let $H=G-\{{x}_{2},{x}_{3}\}+\{{x}_{1}u,u{x}_{4}\}$. Then H is a Halin graph with $\Delta \left(H\right)=8$ and $\left|H\right|<\left|G\right|$. By the induction hypothesis, H admits an $L(2,1)$-labeling f using B. Let $f\left({x}_{1}\right)=a$, $f\left({x}_{4}\right)=b$, and $f\left({x}_{5}\right)=c$. Deleting the label of u, we define a list assignment L for ${x}_{2},{x}_{3},u$ as follows:

**(C8)**Set $N\left(v\right)=\{u,{x}_{1},{x}_{4},{x}_{5},{x}_{6},{t}_{1},{t}_{2},{t}_{3}\}$, and let $H=G-\{{x}_{2},{x}_{3}\}+\{{x}_{1}u,u{x}_{4}\}$. Then H is a Halin graph with $\Delta \left(H\right)=8$ and $\left|H\right|<\left|G\right|$. By the induction hypothesis, H admits an $L(2,1)$-labeling f using B such that ${x}_{1},{x}_{4},u,{x}_{5},$ ${x}_{6},y,v$ are labeled by $a,b,c,d,e,g,h$, respectively. Define a list assignment L for ${x}_{2}$ and ${x}_{3}$ as follows:

**(C9)**Let $H=G-\{{x}_{2},{x}_{3},{x}_{5},{x}_{6}\}+\{{x}_{1}{u}_{1},{u}_{1}{x}_{4},{x}_{4}{u}_{2},{u}_{2}{x}_{7}\}$. Then H is a Halin graph with $\Delta \left(H\right)=8$ and $\left|H\right|<\left|G\right|$. By the induction hypothesis, H admits an $L(2,1)$-labeling f using B such that ${u}_{1},{x}_{4},{u}_{2},{x}_{1},{x}_{7},v,y,z$ are labeled by $a,b,c,d,e,g,h,i$, respectively. Define a list assignment L for ${x}_{2},{x}_{3},{x}_{5},{x}_{6}$ as follows:

**(C10)**Without loss of generality, assume that $d\left(v\right)=8$ and $N\left(v\right)=\{{u}_{1},{u}_{2},{x}_{1},{x}_{4},$${x}_{5},{x}_{8},{t}_{1},{t}_{2}\}$. Let $H=G-\{{u}_{1},{u}_{2},{x}_{2},{x}_{3},\dots ,{x}_{7}\}+{x}_{1}{x}_{8}$. Then H is a Halin graph with $\Delta \left(H\right)\le 8$ and $\left|H\right|<\left|G\right|$. If $\Delta \left(H\right)\le 7$, then H is 10-$L(2,1)$-labelable by the result in [24]. If $\Delta \left(H\right)=8$, then H is also 10-$L(2,1)$-labelable by the induction hypothesis. Thus, H always admits an $L(2,1)$-labeling f using B such that $v,{x}_{1},{x}_{8},y,z,{t}_{1},{t}_{2}$ are labeled by ${i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5},{i}_{6},{i}_{7}$, respectively. Define a list assignment L as follows:

**(C10.1)**$a=0$. (If $d=10$, we have a similar argument.)

**(C10.2)**$b\le a+2$. (If $d\le c+2$, we have a similar discussion).

**(C10.3)**$b\ge a+3$ and $c\le b+2$.

**(C11)**Note that $4\le k\le 7$. Let $H=G-\{{x}_{1},{x}_{2},\dots ,{x}_{k}\}+\{yu,uz\}$. By the induction hypothesis or the result in [24], H has an $L(2,1)$-labeling f using B such that $u,y,z$ are labeled with $p,q,r$, respectively. Define a list assignment L for ${x}_{1},{x}_{2},\dots ,{x}_{k}$ as follows:

**(C11.1)**$k=4$.

**(C11.2)**$k=5$.

**(C11.3)**$k=6$.

- $L\left({x}_{1}\right)=\{{b}_{2},{b}_{5}\}$. Then $L\left({x}_{2}\right)=S\backslash \left\{{b}_{7}\right\}$. If $L\left({x}_{6}\right)=\{{b}_{2},{b}_{5}\}$, then $L\left({x}_{5}\right)=S\backslash \left\{{b}_{7}\right\}$, we label ${x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5},{x}_{6}$ with ${b}_{2},{b}_{6},{b}_{4},{b}_{1},{b}_{3},{b}_{5}$, respectively. If $L\left({x}_{6}\right)=\{{b}_{2},{b}_{6}\}$, or $L\left({x}_{6}\right)=\{{b}_{3},{b}_{6}\}$, then $L\left({x}_{5}\right)=S\backslash \left\{{b}_{4}\right\}$, or $L\left({x}_{5}\right)=S\backslash \left\{{b}_{1}\right\}$, we label ${x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5},{x}_{6}$ with ${b}_{5},{b}_{1},{b}_{4},{b}_{7},{b}_{3},{b}_{6}$, respectively;
- $L\left({x}_{1}\right)=\{{b}_{2},{b}_{6}\}$. Then $L\left({x}_{2}\right)=S\backslash \left\{{b}_{4}\right\}$. If $L\left({x}_{6}\right)=\{{b}_{2},{b}_{6}\}$, then $L\left({x}_{5}\right)=S\backslash \left\{{b}_{4}\right\}$, we label ${x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5},{x}_{6}$ with ${b}_{2},{b}_{7},{b}_{3},{b}_{5},{b}_{1},{b}_{6}$, respectively. If $L\left({x}_{6}\right)=\{{b}_{3},{b}_{6}\}$, then $L\left({x}_{5}\right)=S\backslash \left\{{b}_{1}\right\}$, we label ${x}_{1},{x}_{2},{x}_{3},{x}_{4},$${x}_{5},{x}_{6}$ with ${b}_{2},{b}_{7},{b}_{4},{b}_{1},{b}_{5},{b}_{3}$, respectively;
- $L\left({x}_{1}\right)=L\left({x}_{6}\right)=\{{b}_{3},{b}_{6}\}$. Then $L\left({x}_{2}\right)=L\left({x}_{5}\right)=S\backslash \left\{{b}_{1}\right\}$. We label ${x}_{1},{x}_{2},{x}_{3},{x}_{4},$${x}_{5},{x}_{6}$ with ${b}_{3},{b}_{7},{b}_{5},{b}_{1},{b}_{4},{b}_{6}$, respectively.

**(C11.4)**$k=7$.

**(C12)**Note that $6\le k\le 8$. Let $H=G-\{u,{x}_{1},{x}_{2},\dots ,{x}_{k}\}$$+\{yv,vz\}$. By the induction hypothesis or the result in [24], H has an $L(2,1)$-labeling f using B such that $v,w,y,z,{y}_{1},{y}_{2},$${z}_{1},{z}_{2}$ are labeled with $a,b,c,d,{c}_{1},{c}_{2},{d}_{1},{d}_{2}$, respectively. Label ${x}_{2}$ with some label in $\{a-1,a+1\}\backslash \left\{c\right\}$, say $a+1$. Then we label ${x}_{k}$ with $e\in B\backslash \{\underline{a},\underline{d},b,{d}_{1},{d}_{2}\}$, ${x}_{1}$ with $g\in B\backslash \{\underline{c},a,a+1,{c}_{1},{c}_{2}\}$, and u with $h\in B\backslash \{\underline{g},\underline{a},a+2,b,c,e\}$. Now we define a list assignment L for ${x}_{3},{x}_{4},\dots ,{x}_{k-1}$ as follows:

**(C13)**Note that $7\le k\le 9$. Let $H=G-\{{u}_{1},{u}_{2},{x}_{1},{x}_{2},\dots ,{x}_{k}\}$$+\{yv,vz\}$. By the induction hypothesis or the result in [24], H has an $L(2,1)$-labeling f using B such that $v,w,y,z,{y}_{1},{y}_{2},{z}_{1},{z}_{2}$ are labeled with $a,b,c,d,{c}_{1},{c}_{2},{d}_{1},{d}_{2}$, respectively. Label ${x}_{2}$ with some label in $\{a-1,a+1\}\backslash \left\{c\right\}$, say $a+1$, and ${x}_{k-1}$ with some label in $\{a-1,a+1\}\backslash \left\{c\right\}$, say $a-1$, Then we label ${x}_{1}$ with $g\in B\backslash \{\underline{c},\underline{a+1},{c}_{1},{c}_{2}\}$, ${x}_{k}$ with $h\in B\backslash \{\underline{d},\underline{a-1},{d}_{1},{d}_{2}\}$, ${u}_{1}$ with ${e}^{\prime}\in B\backslash \{\underline{e},\underline{a+1},a-1,b,c\}$, and ${u}_{2}$ with ${h}^{\prime}\in B\backslash \{\underline{h},\underline{a-1},a+1,b,d,{e}^{\prime}\}$.

**(C14)**Let $H=G-\{{u}_{1},{u}_{2},{u}_{3},{x}_{1},{x}_{2},\dots ,{x}_{k}\}$$+\{yv,vz\}$. By the induction hypothesis or the result in [24], H has an $L(2,1)$-labeling f using B such that $v,w,y,z,{y}_{1},{y}_{2},{z}_{1},{z}_{2}$ are labeled with $a,b,c,d,{c}_{1},{c}_{2},{d}_{1},{d}_{2}$, respectively. Similarly to the proof of the previous cases, we label, w.l.o.g., ${x}_{2},{x}_{5}$ with $a+1$, ${x}_{6},{x}_{9}$ with $a-1$, ${x}_{1}$ with g, ${x}_{10}$ with h, ${u}_{1}$ with ${g}^{\prime}$ and ${u}_{3}$ with ${h}^{\prime}$. Define a list assignment L for ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ as follows:

**Claim 1**${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ are L-${L}^{\ast}(2,1)$-labelable.

**Proof.**

**Case I.**$\beta <p$.

- $\alpha <\beta $. If $p\in L\left({x}_{4}\right)$, we label ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ with $\alpha ,p,p+1,\beta $, respectively. Otherwise, $p\notin L\left({x}_{4}\right)$. Then $p\notin L\left({x}_{3}\right)$ since $L\left({x}_{3}\right)\subset L\left({x}_{4}\right)$, and $p+1\in L\left({x}_{4}\right)$. This implies that at least one of $\beta $ and $p+1$ belongs to $L\left({x}_{3}\right)$ since $L\left({x}_{3}\right)\subset L\left({x}_{4}\right)$ and $|L\left({x}_{4}\right)\cap L\left({x}_{7}\right)|\ge 3$. Label ${x}_{7}$ with $\alpha $, ${x}_{8}$ with p, ${x}_{3}$ with some label $\gamma \in L\left({x}_{3}\right)\cap \{\beta ,p+1\}$, and ${x}_{4}$ with some label in $\{\beta ,p+1\}\backslash \left\{\gamma \right\}$;
- $\beta <\alpha <p$. If $p+1\in L\left({x}_{4}\right)$, we label ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ with $\alpha ,p+1,p,\beta $, respectively. Otherwise, $p+1\notin L\left({x}_{4}\right)$ and hence $p+1\notin L\left({x}_{3}\right)$, and $p\in L\left({x}_{4}\right)$. If $\beta \in L\left({x}_{3}\right)$, we label ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ with $\beta ,p,\alpha ,p+1$, respectively. Otherwise, $L\left({x}_{3}\right)=\{p,\alpha ,\gamma \}$ and $L\left({x}_{4}\right)=\{p,\alpha ,\beta ,\gamma \}$ for some $\gamma \in B$. If $\gamma <\alpha $, then we label ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ with $p,\gamma ,\beta ,p+1$, respectively. If $\gamma >p$, then we label ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ with $\alpha ,p,\beta ,p+1$, respectively;
- $\alpha >p+1$. If $p\in L\left({x}_{4}\right)$, we label ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ with $\alpha ,p,p+1,\beta $, respectively. Otherwise, $p\notin L\left({x}_{4}\right)$ and hence $p\notin L\left({x}_{3}\right)$, and $p+1\in L\left({x}_{4}\right)$. If $p+1\in L\left({x}_{3}\right)$, we label ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ with $p+1,\beta ,\alpha ,p$, respectively. Otherwise, $\beta \in L\left({x}_{3}\right)$, we label ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ with $\beta ,p+1,\alpha ,p$, respectively.

**Case II.**$\beta >p+1$.

- $\alpha <p$. If $p+1\in L\left({x}_{4}\right)$, then we label ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ with $\alpha ,p+1,\beta ,p$, respectively. Otherwise, $p+1\notin L\left({x}_{4}\right)$, and $p\in L\left({x}_{4}\right)$. If $\beta \in L\left({x}_{3}\right)$, then we label ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ with $\beta ,p,\alpha ,p+1$, respectively. Otherwise, $p\in L\left({x}_{3}\right)$, then we label ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ with $p,\beta ,\alpha ,p+1$, respectively;
- $p+1<\alpha <\beta $. If $p\in L\left({x}_{4}\right)$, then we label ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ with $\alpha ,p,\beta ,p+1$, respectively. Otherwise, $p\notin L\left({x}_{4}\right)$, and $p+1\in L\left({x}_{4}\right)$. If $\beta \in L\left({x}_{3}\right)$, then we label ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ with $\beta ,p+1,\alpha ,p$, respectively. Otherwise, $p+1\in L\left({x}_{3}\right)$, we label ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ with $p+1,\beta ,\alpha ,p$, respectively;
- $\alpha >\beta $. If $p+1\in L\left({x}_{4}\right)$, then we label ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ with $\alpha ,p+1,\beta ,p$, respectively. Otherwise, $p+1\notin L\left({x}_{4}\right)$, and $p\in L\left({x}_{4}\right)$. If $\beta \in L\left({x}_{3}\right)$, then we label ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ with $\beta ,p,\alpha ,p+1$, respectively. Otherwise, $p\in L\left({x}_{3}\right)$, we label ${x}_{3},{x}_{4},{x}_{7},{x}_{8}$ with $p,\beta ,\alpha ,p+1$, respectively. The completes the proof of Claim 1.

**Corollary**

**1.**

## 5. Concluding Remarks

**Problem**

**1.**

**Problem**

**2.**

## Author Contributions

## Funding

## Data Availability Statement

## Conflicts of Interest

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Qiu, H.; Che, Y.; Wang, Y.
*L*(2, 1)-Labeling Halin Graphs with Maximum Degree Eight. *Symmetry* **2023**, *15*, 50.
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**AMA Style**

Qiu H, Che Y, Wang Y.
*L*(2, 1)-Labeling Halin Graphs with Maximum Degree Eight. *Symmetry*. 2023; 15(1):50.
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**Chicago/Turabian Style**

Qiu, Haizhen, Yushi Che, and Yiqiao Wang.
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https://doi.org/10.3390/sym15010050