Trees whose even-degree vertices induce a path are antimagic

TL;DR

This paper proves that a specific class of trees, where even-degree vertices form a path, are antimagic, thus extending previous results and contributing to the understanding of the antimagic labeling conjecture.

Contribution

It establishes that trees with even-degree vertices inducing a path are antimagic, expanding the class of known antimagic trees.

Findings

Trees with even-degree vertices forming a path are antimagic.

Extends previous results by Liang, Wong, and Zhu.

Supports the broader conjecture that all trees are antimagic.

Abstract

An antimagic labeling a connected graph $G$ is a bijection from the set of edges $E(G)$ to $\{1,2,\dots,|E(G)|\}$ such that all vertex sums are pairwise distinct, where the vertex sum at vertex $v$ is the sum of the labels assigned to edges incident to $v$. A graph is called antimagic if it has an antimagic labeling. In 1990, Hartsfield and Ringel conjectured that every simple connected graph other than $K_2$ is antimagic; however, the conjecture remains open, even for trees. In this note we prove that trees whose vertices of even degree induce a path are antimagic, extending a result given by Liang, Wong, and Zhu [Discrete Math. 331 (2014) 9--14].

Figures

Equations

$$R_{m+2}(T^{\prime })=\{s_{T^{\prime }}(v)(\text{mod }m+2):v\in V(T^{\prime })\text{ and }{d}_{T^{\prime }}(v)\ge 1\}\subseteq \{0,1,\dots ,m+1\}.$$

$$L_{\varphi }(T_1)=[1,p+1]\cup [m-p+1,m].$$

$$R_{m+2}(T_1)=\{0,1,2,m+1\}.$$

$$R_{m+2}(T_{1})=\Big{(}[0,p+1]\setminus\{p\}\Big{)}\cup[m-p+2,m+1],$$

$$R_{m+2}(T_{1})=[0,p+1]\cup\Big{(}[m-p+1,m+1]\setminus\{m\}\Big{)}.$$

$$R_{m+2}(T_{1})=\Big{(}[0,p+1]\setminus\{p\}\Big{)}\cup\Big{(}[m-p+1,m+1]\setminus\{m-1\}\Big{)}.$$

$$T_1:=T[\{v_0,\dots ,v_{2p}\}\cup \{x_0,\dots ,x_{\ell }\}\cup \{y_0,\dots ,y_{\ell -1}\}].$$

$$L_{\varphi }(T_1)=[1,p+\ell +1]\cup [m-p-\ell +1,m],$$

$$s_{T_1}$$

$$R_{m+2}(T_{1})=[0,p+\ell]\cup\Big{(}[m-p-\ell+1,m+1]\setminus\{m-1\}\Big{)}$$

$$T_2:=T-E(T_1)=\bigcup _{v\in V(T_1)}{T}_2(v).$$

$$L_{\varphi }(T_2):=[p+\ell +2,m-p-\ell ].$$

Full text

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This project has received funding from the European Union’s Horizon 2020 research and innovation programme under the Marie Skłodowska-Curie grant agreement No 734922.

1 Introduction

All graphs considered in this work are finite, undirected and simple. Given a graph $G=(V(G),E(G))$ and a vertex $v\in V(G)$, we denote by $E_{G}(v)$ the set of edges incident to $v$ and by $d_{G}(v)=|E_{G}(v)|$, the degree of $v$ in $G$. We will just write $E(v)$ and $d(v)$ when $G$ is clear from context. A tree is a connected and acyclic graph, and a forest is a disjoint union of trees. Undefined terms in this work can be found in [2].

An (edge) labeling of a graph $G$ is a mapping from $E(G)$ to the set of nonnegative integers. A labeling $\phi$ of a connected graph $G$ is called antimagic if it is a bijection $\phi:E(G)\rightarrow\{1,2,\dots,|E(G)|\}$ such that all vertex sums are pairwise distinct, where the vertex sum $s(v)$ at vertex $v\in V(G)$ is defined as $s(v)=\sum_{e\in E(v)}\phi(e)$. A graph is called antimagic if it has an antimagic labeling.

In 1990, Hartsfield and Ringel [5] conjectured that every simple connected graph other than $K_{2}$ is antimagic. The conjecture has received much attention (see [4]); but it is widely open in general, even for trees. Nevertheless, several classes of trees are known to be antimagic (see [1, 3, 5, 6, 7, 8, 9, 10]).

Given a tree $T$, $V_{even}(T)$ (resp. $V_{odd}(T)$) denotes the set of even (resp. odd) degree vertices of $T$. Regarding trees such that $V_{even}$ induces a path, Liang, Wong, and Zhu [7] proved the following two theorems.

Theorem 1**.**

[7]* If $T$ is a tree such that $V_{even}(T)$ induces a path and $|V_{even}(T)|$ is odd, then $T$ is antimagic.*

Theorem 2**.**

[7]* Let $T$ be a tree such that $V_{even}(T)$ induces a path of length $2p$, $(v_{1},\dots,v_{2p})$. Let $v_{0}$ (resp. $v_{2p+1}$) be a neighbor of $v_{1}$ (resp. $v_{2p}$) different from $v_{2}$ (resp. $v_{2p-1}$). If $d(v_{p})\neq d(v_{2p+1})+1$ or $d(v_{p+1})\neq d(v_{0})+1$, then $T$ is antimagic.*

The aim of this note is to extend Theorem 2 to all cases, that is, to prove the antimagicness of trees such that $V_{even}(T)$ induces a path whenever $|V_{even}(T)|$ is even, obtaining as a consequence that trees whose even-degree vertices induce a path are antimagic.

2 Constructing an antimagic labeling

In the proof of the next theorem we follow and extend the main idea developed by Liang, Wong, and Zhu in [7]. We denote by $[a,b]$ the set of consecutive integers $\{a,a+1,\dots,b\}$, where $a\leq b$.

Theorem 3**.**

If $T$ is a tree such that $V_{even}(T)$ induces a path and $|V_{even}(T)|$ is even, then $T$ is antimagic.

Proof..

It is known that trees without vertices of degree $2$ are antimagic [6, 7], hence we may assume $|V_{even}(T)|=2p$, with $p\geq 1$. Let $P=(v_{0},v_{1},v_{2},\dots,v_{2p},v_{2p+1})$ be a path induced by $V_{even}(T)$ extended with endpoints in $V_{odd}(T)$, that is, $V_{even}(T)=\{v_{1},\dots,v_{2p}\}$ and $\{v_{0},v_{2p+1}\}\subseteq V_{odd}(T)$. For every $1\leq i\leq 2p+1$, we denote by $e_{i}$ the edge $v_{i-1}v_{i}$.

We prove the theorem by constructing an antimagic labeling $\phi$ of $T$ in two steps. The first step produces a labeling of a subtree of $T$ containing the path $P$ and satisfying a particular additional condition. This labeling will be extended to an antimagic labeling of $T$ at the second step.

Let $m=|E(T)|$. We will use the residues modulo $m+2$ to compare vertex sums: since vertex sums are distinct if they are distinct modulo $m+2$, it is enough to compare vertex sums whenever they are equal modulo $m+2$ in order to check that they all are pairwise distinct.

For each (not necessarily connected) subgraph $T^{\prime}$ of $T$, we set $L_{\phi}(T^{\prime})=\{\phi(e)\,:\,e\in E(T^{\prime})\}$ and $s_{T^{\prime}}(v)=\sum_{e\in E_{T^{\prime}}(v)}\phi(e)$ for every $v\in V(T^{\prime})$ such that $d_{T^{\prime}}(v)\geq 1$. Obviously, if $T^{\prime}=T$, then $s_{T^{\prime}}=s$. The set of all vertex sums modulo $m+2$ in $T^{\prime}$ will be denoted by $R_{m+2}(T^{\prime})$, that is,

[TABLE]

STEP I. The labeling of the tree $T_{1}$ constructed at this step will satisfy the following condition: all vertex sums in $T_{1}$ will be pairwise distinct modulo $m+2$ with at most one exception; moreover, if the vertex sums are equal modulo $m+2$ for a pair of vertices, then exactly one of them will be a leaf in $T$ and the vertex sums in $T_{1}$ for both vertices in the pair will be different.

As a starting point, let $T_{1}:=P$ and define

[math]

Hence,

[TABLE]

Moreover, $s_{T_{1}}(v_{0})=1$, $s_{T_{1}}(v_{2p+1})=p+1$, and for $1\leq i\leq 2p$, $s_{T_{1}}(v_{i})=m-p+i+1$. Next, we calculate the set $R_{m+2}(T_{1})$ according to the values of $p$ and to the degrees of $v_{0}$ and $v_{2p+1}$.

Case 1. $p=1$. In this case, $|V(T_{1})|=4$ and

[TABLE]

Hence, vertex sums at the vertices of $T_{1}$ are distinct.

Case 2. $p>1$. In such a case,

[TABLE]

and, only the residues of vertex sums at vertices $v_{0}$ and $v_{p+2}$ are equal. In fact, we have that $s_{T_{1}}(v_{0})\equiv 1\equiv s_{T_{1}}(v_{p+2})\pmod{m+2}$ (see an example in Figures 1(a) and 2(a)).

Now we distinguish two subcases.

1. 2.1.

At least one of the vertices $v_{0}$ or $v_{2p+1}$ is a leaf in $T$. Notice that by properly relabeling the vertices of $T_{1}$, we may assume $d_{T}(v_{0})=1$. Then, $s_{T_{1}}(v_{0})\equiv 1\equiv s_{T_{1}}(v_{p+2})\pmod{m+2}$, but $s_{T_{1}}(v_{0})=1<m+3=s_{T_{1}}(v_{p+2})$. 2. 2.2.

Neither $v_{0}$ nor $v_{2p+1}$ are leaves in $T$. In this case; for $1\leq i\leq\lfloor\small{\frac{p-1}{2}}\rfloor$, we swap the labels of the edges $e_{2i-1}$ and $e_{2i}$, that is,

[math]

Notice that the endpoints of the subpath of $T_{1}$ involved in the swaps are $v_{0}$ and $v_{k}$, where $k=p-2$, if $p$ is even; and $k=p-1$, if $p$ is odd. After the swaps, $s_{T_{1}}(v_{0})=m-p+1$; also it can be easily checked that $s_{T_{1}}(v_{k})=p-1$, if $p$ is even; and $s_{T_{1}}(v_{k})=p$, if $p$ is odd; and the vertex sum at any other vertex in $T_{1}$ remains unchanged. We distinguish cases depending on the parity of $p$.

1. (a)

$p$ odd. In this case we have that $s_{T_{1}}(v_{k})=p$, implying that the residues modulo $m+2$ at the vertices of $T_{1}$ are pairwise distinct. Concretely,

[TABLE] 2. (b)

$p$ even. Then, $s_{T_{1}}(v_{k})=p-1\equiv s_{T_{1}}(v_{2p})\pmod{m+2}$, and thus

[TABLE]

Notice that only $s_{T_{1}}(v_{k})$ and $s_{T_{1}}(v_{2p})$ have the same residue in $T_{1}$ (see an example in Figure 2(b)). Now, let $x_{0}=v_{2p+1}$ and let $P^{\prime}=(x_{0},x_{1},\dots,x_{\ell})$ be a maximal subpath of $T$ starting at $x_{0}$ and with vertices in $V_{odd}(T)$. Observe that, in such a case, $x_{\ell}$ is a leaf in $T$ and there exist vertices $y_{0},\dots,y_{\ell-1}\in V_{odd}(T)$ such that $x_{i}y_{i}\in E(T)$ (see an example in Figure 3). We update $T_{1}$ as the tree induced by the set of vertices of the paths $P$ and $P^{\prime}$, and $\{y_{0},\dots,y_{\ell-1}\}$:

[TABLE]

We define the labels of the new edges and update the label of the edge $e_{2p+1}$ as follows:

[math]

Thus, we have

[TABLE]

and

[TABLE]

Therefore, taking into account Equality (5),

[TABLE]

and only $s_{T_{1}}(x_{0})$ and $s_{T_{1}}(x_{\ell})$ have the same residue, concretely $p+\ell$. However, $s_{T_{1}}(x_{0})=m+p+\ell+2>p+\ell=s_{T_{1}}(x_{\ell})$, and hence all vertex sums in $T_{1}$ are different (see an example in Figure 3).

Notice that, in each of the above cases, $|L_{\phi}(T_{1})|=|E(T_{1})|$. Hence $\phi$, restricted to $E(T_{1})$, is a bijection from $E(T_{1})$ to $L_{\phi}(T_{1})$.

STEP II. Now, let $T_{2}$ be the forest obtained by removing all the edges of $T_{1}$. Each component of $T_{2}$ has exactly one vertex in $T_{1}$. Therefore, if $T_{2}(v)$ denotes the component of $T_{2}$ containing $v$,

[TABLE]

Clearly, $T_{2}(v)$ can be viewed as a directed rooted tree with root at $v$, where every edge is directed away from the root. Moreover, since every vertex of $T_{2}(v)$ different from $v$ has odd degree in $T$, each vertex in $T_{2}(v)$ has an even number of children in this rooted tree and, therefore, $|E(T_{2}(v))|$ is even for every $v\in V(T_{1})$. Hence, $|E(T_{2})|$ is even. If we set $\ell=0$ whenever $T_{1}=P$, then, by Equalities (1) and (6), the available labels for the edges of $T_{2}$ are

[TABLE]

As $L_{\phi}(T_{2})=[a,b]$, where $a+b=m+2$, and each $w\in V(T_{2})$ has an even number of children, we can label the edges of $T_{2}$ with integers in $L_{\phi}(T_{2})$ fulfilling this additional condition: if a vertex $w$ has an outgoing edge with label $t$ in the corresponding rooted tree of $T_{2}$, then $w$ has another outgoing edge with label $m+2-t$ (see an example in Figure 4). Concretely, if $|E(T_{2})|=2r$ for some integer $r$, then we set $E(T_{2})=E_{1}\cup\dots\cup E_{r}$, where $(E_{1},\dots,E_{r})$ is a partition of $E(T_{2})$ such that $E_{i}=\{f_{i}^{1},f_{i}^{2}\}$ contains exactly two outgoing edges from the same vertex in some rooted tree of the forest $T_{2}$, and we label the edges of $T_{2}$ in the following way:

[math]

Clearly, by the previous discussion, the labeling $\phi$ already constructed is a bijection from $E(T)$ to $[1,m]$. Finally, we just need to show that the vertex sums defined by $\phi$ in $T$ are pairwise distinct.

Observe that in $T_{2}$, the sum of the labels of the outgoing edges of $v$ is a multiple of $m+2$. Thus, the following two conditions hold.

For every $v\in V(T_{1})$, $s(v)\equiv s_{T_{1}}(v)\pmod{m+2}$. 2. 2.

For every $v\in V(T)\setminus V(T_{1})$, $s(v)\equiv\phi(f)\pmod{m+2}$, where $f$ is the incoming edge of $v$ in $T_{2}$.

Let $u,v\in V(T)$. We consider the following cases:

1. (a)

$u,v\in V(T)\setminus V(T_{1})$. Since $\phi:E(T)\rightarrow[1,m]$ is a bijection, Condition 2 implies that $s(u)\neq s(v)$. 2. (b)

$u,v\in V(T_{1})$. If $s_{T_{1}}(u)\not\equiv s_{T_{1}}(v)\pmod{m+2}$, then by Condition 1 we have that $s(u)\neq s(v)$. Otherwise, as we have seen in Cases 2.1 and 2.2(b) of Step I, one of these vertices, say $u$, is a leaf in $T$ and $s_{T_{1}}(u)<s_{T_{1}}(v)$. Therefore, we have that $s(u)=s_{T_{1}}(u)<s_{T_{1}}(v)\leq s(v)$, as we wanted to prove. 3. (c)

One of the vertices belongs to $V(T_{1})$ and the other to $V(T)\setminus V(T_{1})$. We can assume without loss of generality that $u\in V(T_{1})$ and $v\in V(T)\setminus V(T_{1})$. By Condition 1, $s(u)\pmod{m+2}\in R_{m+2}(T_{1})$. Moreover, by Condition  2, $s(v)\pmod{m+2}\in L_{\phi}(T_{2})$. By Equalities (2), (3), (4), (7), and (8), we have $R_{m+2}(T_{1})\cap L_{\phi}(T_{2})=\emptyset$. Hence, $s(u)\neq s(v)$.

Thus, the theorem holds. ∎

The next result follows from Theorems 1 and 3.

Corollary 1**.**

If $T$ is a tree such that $V_{even}(T)$ induces a path, then $T$ is antimagic.

Acknowledgments. A. Lozano is supported by the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation programme (grant agreement ERC-2014-CoG 648276 AUTAR); M. Mora is supported by projects Gen. Cat. DGR 2017SGR1336, MINECO MTM2015-63791-R, and H2020-MSCA-RISE project 734922-CONNECT; and C. Seara is supported by projects Gen. Cat. DGR 2017SGR1640, MINECO MTM2015-63791-R, and H2020-MSCA-RISE project 734922-CONNECT.