Edge k-Product Cordial Labeling of Trees

Abstract

The concepts of k-product cordial labeling and edge product cordial labeling were introduced in 2012 and further explored by various researchers. Building on these ideas, we define a new concept called ‘edge k-product cordial labeling’ as follows: For a graph $G=\left(V\right(G),E(G\left)\right)$, which does not have isolated vertices, an edge labeling $f:E\left(G\right)\to \left\{0,1,\dots ,k-1\right\}$, where $k\ge 2$ is an integer, is said to be an edge k-product cordial labeling of G if it induces a vertex labeling $f^{*}:V\left(G\right)\to \left\{0,1,\dots ,k-1\right\}$ defined by $f^{*}\left(v\right)={\prod }_{uv\in E\left(G\right)}f\left(uv\right)(modk)$, which satisfies $\left|e_f\left(i\right)-e_f\left(j\right)\right|\le 1$ and $\left|v_{f^{*}}\left(i\right)-v_{f^{*}}\left(j\right)\right|\le 1$ for $i,j\in \left\{0,1,\dots ,k-1\right\}$, where $e_f\left(i\right)$ and $v_{f^{*}}\left(i\right)$ denote the number of edges and vertices, respectively, having label i for $i=0,1,\dots ,k-1$. In this paper, we study the edge k-product cordial behavior of trees, a comet, and a double comet.

1. Preliminaries

The concept of graph labeling has experienced significant popularity over the past six decades, owing to its practical applications. A groundbreaking paper addressing graph labeling problems was published by Rosa [1]. Subsequently, numerous papers on various graph labeling methods have been published, and Gallian’s survey [2] elegantly categorizes and organizes these diverse labeling methods published by various mathematicians all over the world. ‘Cordial labeling’ is one of the popular labelings introduced by Cahit [3]. Inspired by this notion, ‘product cordial labeling’ was proposed in [4]. In 2012, this concept was extended further, and a new concept called ‘k-product cordial labeling’ [5] was introduced. In the same year, Vaidya et al. [6] introduced a variation of product cordial labeling called ‘edge product cordial labeling’. In this variant, the roles of vertices and edges in ‘product cordial labeling’ are interchanged. Since then, some more results on ‘edge product cordial labeling’ have been published by the same authors; see [6,7,8,9]. Building on this, Thamizharasi et al. [10] demonstrated the existence of ‘edge product cordial labeling’ of regular digraphs in 2015. In the subsequent years, Prajapati and Aboshady et al. [11,12,13,14] contributed additional results on ‘edge product cordial labeling’. Motivated by the concepts of ‘k-product cordial labeling’ and ‘edge product cordial labeling’ and the established results, we put forth a new labeling, namely ‘edge k-product cordial labeling’, which extends the concept of edge product cordial labeling by expanding the set of labels from $\{0,1\}$ to $\{0,1,2,ߪ,k-1\}$. Let $G=(V,E)$ be a graph without isolated vertices and $k\ge 2$. Let $f:E\to \{0,1,\dots ,k-1\}$ be an edge labeling. The induced vertex labeling $f^{*}:V\to \{0,1,\dots ,k-1\}$ is defined by

$$f^{*}\left(v\right)\equiv \prod _{uv\in E}f\left(uv\right)(modk).$$

$f^{*}$ is called the induced labeling of f. For convenience, let ${\mathbb{Z}}_k=\{0,1,\dots ,k-1\}$ be the complete residue system modulo k. Also, an edge is called i-edge if it is labeled by i; and a vertex is called j-vertex if its induced label is j. Let $e_f\left(i\right)$ and $v_{f^{*}}\left(i\right)$ denote the number of i-edges and i-vertices, respectively, for $i\in {\mathbb{Z}}_k$. f is said to be an edge k-product cordial labeling of G if $|e_f\left(i\right)-e_f\left(j\right)|\le 1$ and $|v_{f^{*}}\left(i\right)-v_{f^{*}}\left(j\right)|\le 1$ for $i,j\in {\mathbb{Z}}_k$. Also, G is called an edge k-product cordial graph.

Let L be a set of labels. For an edge labeling $f:E\to L$ of a graph $G=(V,E)$, if $|e_f\left(i\right)-e_f\left(j\right)|\le 1$ for all $i,j\in L$, then we say that the edges of the graph labeled by labels in Levenly (under f).

In this paper, we use this notation, these concepts, and the following definitions unless otherwise stated.

Definition 1.

For $n\ge 1$, let $P_n=u_1\dots u_n$ be the path of order n. For $m\ge 1$, let the vertex set and the edge set of the star graph $K_{1,m}$ be $\left\{c\right\}\cup \left\{v_i\right|1\le i\le m\}$ and $\left\{c{v}_i\right|1\le i\le m\}$, respectively. For $m,n\ge 1$, let the graph $C(n,m)=P_n\cup K_{1,m}$ with identifying c with $u_n$. Such a graph is called a comet [15].

Definition 2.

For $M\ge m\ge 1$, let the graph $DC(n,M,m)=P_n\cup K_{1,M}\cup K_{1,m}$ with the vertex set and the edge set $\left\{u_i\right|1\le i\le n\}\cup \left\{v_i\right|1\le i\le M\}\cup \left\{w_i\right|1\le i\le m\}$ and $\left\{u_i{u}_{i+1}\right|1\le i\le n-1\}\cup \left\{u_n{v}_i\right|1\le i\le M\}\cup \left\{u_1{w}_i\right|1\le i\le m\}$, respectively. Such a graph is called a double comet [16].

Notation and concepts, which are not defined in this paper, are referred to in [17]. All graphs considered here are simple and connected.

We use the next section to show the structure of the subtree induced by all edges labeled by 0 of an edge k-product cordial tree. In the consecutive sections, we investigate the edge k-product cordial behavior of comet and double comet graphs for $k=3,4$, and 5.

2. Properties of Edge $\mathit{k}$-Product Cordial Trees

Lemma 1.

Let $f:E\to {\mathbb{Z}}_k$ be an edge labeling of a tree $T=(V,E)$, where $k\ge 2$. Then $v_{f^{*}}\left(0\right)\ge e_f\left(0\right)+1$, where $f^{*}$ is the induced labeling of f.

Proof. 

Let $E_0=\{e\in E|f\left(e\right)=0\}$ and $T_0=T\left[E_0\right]$, the edge induced subgraph of T. Let $\omega$, p, and q be the number of components, the order and the size of $T_0$, respectively. Since $T_0$ is a forest, $p=q+\omega$. Since all its vertices are 0-vertices, $v_{f^{*}}\left(0\right)\ge p=q+\omega \ge e_f\left(0\right)+1$. □

Corollary 1.

Let f be an edge k-product cordial labeling of a tree T of order n, where $k\ge 2$. If $E_0$ is the set of 0 edges under f, then $T_0=T\left[E_0\right]$ is a tree and

$$v_{f^{*}}\left(0\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{k}}\hfill & if n\equiv 0(modk);\hfill \\ ⌊{\displaystyle \frac{n}{k}}⌋+1\hfill & if n\not\equiv 0(modk),\hfill \end{array}\right.e_f\left(0\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{k}}-1\hfill & if n\equiv 0(modk);\hfill \\ ⌊{\displaystyle \frac{n}{k}}⌋\hfill & if n\not\equiv 0(modk).\hfill \end{array}\right.$$

Proof. 

For each i$(0\le i\le k-1)$, we have

$$\begin{array}{cc}\hfill v_{f^{*}}\left(i\right)& =\left\{\begin{array}{cc}{\displaystyle \frac{n}{k}}\hfill & if n\equiv 0(modk);\hfill \\ ⌊{\displaystyle \frac{n}{k}}⌋ \mathrm{or} ⌊{\displaystyle \frac{n}{k}}⌋+1\hfill & if n\not\equiv 0(modk),\hfill \end{array}\right.\hfill \\ \hfill e_f\left(i\right)& =\left\{\begin{array}{cc}{\displaystyle \frac{n}{k}}-1 \mathrm{or} {\displaystyle \frac{n}{k}}\hfill & if n\equiv 0(modk);\hfill \\ {\displaystyle \frac{n-1}{k}}=⌊{\displaystyle \frac{n}{k}}⌋\hfill & if n\equiv 1(mod)k\hfill \\ ⌊{\displaystyle \frac{n}{k}}⌋\mathrm{or}⌊{\displaystyle \frac{n}{k}}⌋+1\hfill & if n\not\equiv 0,1(modk).\hfill \end{array}\right.\hfill \end{array}$$

By Lemma 1, we obtain this corollary. □

Example 1.

Suppose $n=7$ and $k=3$. Then T contains 7 vertices and 6 edges. We keep all notation defined in the proof of Lemma 1 and Corollary 2. For any edge 3-product cordial labeling f, $e_f\left(0\right)=e_f\left(1\right)=e_f\left(2\right)=2$ and $v_{f^{*}}\left(i\right)=2,3$ for $i=0,1,2$. From the proof of Lemma 1, $v_{f^{*}}\left(0\right)\ge e_f\left(0\right)+\omega$. Since $2\le v_{f^{*}}\left(0\right)\le 3$ and $e_f\left(0\right)+\omega =2+\omega \ge 3$, $\omega =1$ and hence $v_{f^{*}}\left(0\right)=3$.

Lemma 2.

Let G be a graph of order k. If k is a prime, then G is not an edge k-product cordial graph.

Proof. 

Let f be an edge k-product cordial labeling of G. Then $v_{f^{*}}\left(i\right)=1$ for $i\in {\mathbb{Z}}_k$. In particular, $v_{f^{*}}\left(0\right)=1$. Suppose there is a 0-edge, then it induces two 0-vertices, which is a contradiction. Suppose there are no 0-edges. Since k is a prime, there is no induced 0-vertex, which is also a contradiction. □

Corollary 2.

Let T be a tree of order k. If k is a prime or $k=4$, then T is not an edge k-product cordial.

Proof. 

When k is a prime, by Lemma 2, T is not edge k-product cordial. For $k=4$, let f be an edge 4-product cordial labeling of T. Then $e_f\left(i\right)=0,1$ and $v_{f^{*}}\left(i\right)=1$ for each i. Let u be the 0-vertex. By Lemma 2, there is no 0-edge in T. Thus, $e_f\left(i\right)=1$ for $1\le i\le 3$. This results in $v_{f^{*}}\left(2\right)=2$, which is a contradiction. □

3. Edge 3-Product Cordial Trees

In order to establish the edge 3-product cordial behavior of the comet, we prove the following two lemmas.

Lemma 3.

The path graph $P_n$ is an edge 3-product cordial for $n\ge 4$.

Proof. 

We define an edge labeling $f_n$ for $P_n$ recursively, and this labeling is represented by $(a_1,\dots ,a_{n-1})$ if $f_n\left(u_i{u}_{i+1}\right)=a_i$, $1\le i\le n-1$. We will use this representation for a labeling $f_n$ of $P_n$ in the whole paper.

The edge 3-product cordial labeling for $P_n$, $4\le n\le 9$, is shown in

Table 1. Edge 3-product cordial labeling for $P_n,4\le n\le 9$.

n	${\mathit{f}}_{\mathit{n}}$	${\mathit{e}}_{{\mathit{f}}_{\mathit{n}}}$(0)	${\mathit{e}}_{{\mathit{f}}_{\mathit{n}}}$(1)	${\mathit{e}}_{{\mathit{f}}_{\mathit{n}}}$(2)	${\mathit{v}}_{{\mathit{f}}_{\mathit{n}}^{*}}$(0)	${\mathit{v}}_{{\mathit{f}}_{\mathit{n}}^{*}}$(1)	${\mathit{v}}_{{\mathit{f}}_{\mathit{n}}^{*}}$(2)
4	(1,2,0)	1	1	1	2	1	1
5	(1,1,2,0)	1	2	1	2	2	1
6	(1,2,2,1,0)	1	2	2	2	2	2
7	(1,2,2,1,0,0)	2	2	2	3	2	2
8	(1,2,2,1,1,0,0)	2	3	2	3	3	2
9	(1,2,2,1,1,2,0,0)	2	3	3	3	3	3

.

For $n\ge 10$, define $f_n$ recurrently as $f_n=f_{m-i+6}=(1,2,2,1,a_1,\dots ,a_{m-i-1},0,0)$, where $0\le i\le 5$ and $m\ge 9$. Note that $a_1=1$. Clearly, $e_{f_{m-i+6}}\left(j\right)=e_{f_{m-i}}\left(j\right)+2$, $v_{f_{m-i+6}^{*}}\left(j\right)=v_{f_{m-i}^{*}}\left(j\right)+2$, $0\le j\le 2$, where $f_{m-i}=(a_1,\dots ,a_{m-i-1})$. Thus,

$$\begin{array}{cccccc}\hfill e_{f_n}\left(0\right)=e_{f_n}\left(1\right)=e_{f_n}\left(2\right)& =⌊{\displaystyle \frac{n}{3}}⌋,\hfill & \hfill v_{f_n^{*}}\left(0\right)-1=v_{f_n^{*}}\left(1\right)=v_{f_n^{*}}\left(2\right)& & \hfill =⌊{\displaystyle \frac{n}{3}}⌋& \mathrm{if} n\equiv 1(mod3);\hfill \\ \hfill e_{f_n}\left(0\right)=e_{f_n}\left(1\right)-1=e_{f_n}\left(2\right)& =⌊{\displaystyle \frac{n}{3}}⌋,\hfill & \hfill v_{f_n^{*}}\left(0\right)-1=v_{f_n^{*}}\left(1\right)-1=v_{f_n^{*}}\left(2\right)& & \hfill =⌊{\displaystyle \frac{n}{3}}⌋& \mathrm{if} n\equiv 2(mod3);\hfill \\ \hfill e_{f_n}\left(0\right)+1=e_{f_n}\left(1\right)=e_{f_n}\left(2\right)& ={\displaystyle \frac{n}{3}},\hfill & \hfill v_{f_n^{*}}\left(0\right)=v_{f_n^{*}}\left(1\right)=v_{f_n^{*}}\left(2\right)& & \hfill ={\displaystyle \frac{n}{3}}& \mathrm{if} n\equiv 0(mod3).\hfill \end{array}$$

Hence, $f_n$ is an edge 3-product cordial labeling of $P_n$. Note that $f_n^{*}\left(u_1\right)=1$ and $f^{*}\left(u_n\right)=0$. □

From [18], we have the following result.

Lemma 4.

The star graph $K_{1,m}$ is an edge 3-product cordial for $m\ge 3$.

Definition 3.

Suppose G and H are the two edge-disjoint graphs with edge labelings g and h, respectively. We say that ϕ is a combination of g and h (or combine g with h) if

$$\varphi \left(x\right)=\left\{\begin{array}{cc}g\left(x\right)\hfill & if x\in E\left(G\right),\hfill \\ h\left(x\right)\hfill & if x\in E\left(H\right).\hfill \end{array}\right.$$

Theorem 1.

The comet graph $C(n,m)$ is edge 3-product cordial for $n\ge 3$ and $m\ge 2$.

Proof. 

Note that $C(n,m)=P_n\cup K_{1,m}$. Let $c=u_n$. We label $P_n$ by $f_n$, which was defined in the proof of Lemma 3. Define labeling $g_m,m=2,3,4$, for $K_{1,m}$ as follows:

• Suppose $n\equiv 1(mod3)$.
  If $m=2$, then $g_2\left(c{v}_1\right)=1$, $g_2\left(c{v}_2\right)=2$.
  If $m=3$, then $g_3\left(c{v}_1\right)=1$, $g_3\left(c{v}_2\right)=2$, $g_3\left(c{v}_3\right)=0$.
  If $m=4$, then $g_4\left(c{v}_1\right)=1$, $g_4\left(c{v}_2\right)=2$, $g_4\left(c{v}_3\right)=0$, $g_4\left(c{v}_4\right)=1$.
• Suppose $n\equiv 2(mod3)$.
  If $m=2$, then $g_2\left(c{v}_1\right)=0$, $g_2\left(c{v}_2\right)=2$.
  If $m=3$, then $g_3\left(c{v}_1\right)=1$, $g_3\left(c{v}_2\right)=2$, $g_3\left(c{v}_3\right)=0$.
  If $m=4$, then $g_4\left(c{v}_1\right)=1$, $g_4\left(c{v}_2\right)=2$, $g_4\left(c{v}_3\right)=0$, $g_4\left(c{v}_4\right)=2$.
• Suppose $n\equiv 0(mod3)$.
  If $m=2$, then $g_2\left(c{v}_1\right)=1$, $g_2\left(c{v}_2\right)=0$.
  If $m=3$, then $g_3\left(c{v}_1\right)=1$, $g_3\left(c{v}_2\right)=2$, $g_3\left(c{v}_3\right)=0$.
  If $m=4$, then $g_4\left(c{v}_1\right)=1$, $g_4\left(c{v}_2\right)=2$, $g_4\left(c{v}_3\right)=0$, $g_4\left(c{v}_4\right)=0$.

Let $\varphi$ be the combination of $f_n$ and $g_m$. We can check that $e_{\varphi }\left(1\right)\ge e_{\varphi }\left(2\right)\ge e_{\varphi }\left(0\right)$ and $e_{\varphi }\left(1\right)-e_{\varphi }\left(0\right)\le 1$; $v_{{\varphi }^{*}}\left(0\right)\ge v_{{\varphi }^{*}}\left(1\right)\ge v_{{\varphi }^{*}}\left(2\right)$ and $v_{{\varphi }^{*}}\left(0\right)-v_{{\varphi }^{*}}\left(2\right)\le 1$.

Note that $C(n,m)=C(n,m-3)\cup K_{1,3}$ with the common vertex $u_n$. If $\varphi$ is an edge 3-product cordial labeling of $C(n,m-3)$, $m\ge 5$, then combine $\varphi$ for $C(n,m-3)$ and $g_3$ for $K_{1,3}$ to obtain an edge 3-product cordial labeling for $C(n,m)$.

This completes the proof. □

Example 2.

Here is an example to illustrate the proof of Theorem 1. Suppose $n=5$ and $m=7$. Firstly, we label $C(5,4)$. According to the labeling defined in the proof above, we have the following labeling (Figure 1):

Here we can see that $e_{\varphi }\left(0\right)=2$, $e_{\varphi }\left(1\right)=e_{\varphi }\left(2\right)=3$; $v_{{\varphi }^{*}}\left(0\right)=v_{{\varphi }^{*}}\left(1\right)=v_{{\varphi }^{*}}\left(2\right)=3$. Now, we combine the labeling $g_3$ of $K_{1,3}$ to the labeling ϕ of $C(5,4)$. We have (see Figure 2)

Here we can see that the number of 0-edges is 3, the number of i-edges is 4, $i=1,2$, and the number of j-vertices is 4, $0\le j\le 2$.

Remark 1.

By Lemma 3, $C(n,0)\cong P_n$ when $n\ge 4$ and $C(n,1)\cong P_{n+1}$ when $n\ge 3$ are edge 3-product cordial graphs. Again by Lemma 4, $C(1,m)\cong K_{1,m}$ for $m\ge 3$ and $C(2,m)\cong K_{1,m+1}$ for $m\ge 2$ are edge 3-product cordial graphs.

Also, note that under the labeling defined in Lemma 3, the vertex $u_1$ is always a 1-vertex and $u_n$ is always 0-vertex.

Consequently, if $C(n,m)$ is not isomorphic to $P_1$, $P_2$, or $P_3$, then $C(n,m)$ admits an edge 3-product cordial labeling ϕ such that $\varphi \left(u_1\right)=1$ and $\varphi \left(u_n\right)=0$.

Theorem 2.

The double comet graph $DC(n,M,m)$ is edge 3-product cordial for $n\ge 2$ and $M\ge m\ge 2$.

Proof. 

Let $S_1=K_{1,M}$ with the center $u_n$ and $S_2=K_{1,m}$ with the center $u_1$. Then $DC(n,M,m)=P_n\cup S_1\cup S_2$.

For $2\le q\le 5$, we label the edges of $S_2=K_{1,q}$ and the selected edges of $S_1$ by 0,1,2 evenly as shown below and denote this labeling by $\alpha$.

• When $q=2$, $\alpha \left(u_1{w}_1\right)=\alpha \left(u_1{w}_2\right)=1$, $\alpha \left(u_n{v}_M\right)=\alpha \left(u_n{v}_{M-1}\right)=0$ and $\alpha \left(u_n{v}_{M-2}\right)=\alpha \left(u_n{v}_{M-3}\right)=2$.
• When $q=3$, $\alpha \left(u_1{w}_1\right)=1$, $\alpha \left(u_1{w}_2\right)=\alpha \left(u_1{w}_3\right)=2$, $\alpha \left(u_n{v}_M\right)=\alpha \left(u_n{v}_{M-1}\right)=0$ and $\alpha \left(u_n{v}_{M-2}\right)=1$.
• When $q=4$, $\alpha \left(u_1{w}_1\right)=\alpha \left(u_1{w}_2\right)=1$, $\alpha \left(u_1{w}_3\right)=\alpha \left(u_1{w}_4\right)=2$, $\alpha \left(u_n{v}_M\right)=\alpha \left(u_n{v}_{M-1}\right)=0$.
• When $q=5$, $\alpha \left(u_1{w}_1\right)=\alpha \left(u_1{w}_2\right)=\alpha \left(u_1{w}_3\right)=1$, $\alpha \left(u_1{w}_4\right)=\alpha \left(u_1{w}_5\right)=2$, $\alpha \left(u_n{v}_M\right)=\alpha \left(u_n{v}_{M-1}\right)=\alpha \left(u_n{v}_{M-2}\right)=0$ and $\alpha \left(u_n{v}_{M-3}\right)=2$.

Note that ${\alpha }^{*}\left(u_1\right)=1$ and ${\alpha }^{*}\left(u_n\right)=0$. Also, $v_{{\alpha }^{*}}\left(0\right)=v_{{\alpha }^{*}}\left(1\right)=v_{{\alpha }^{*}}\left(2\right)+1$.

Now, consider $m=4p+q$, where $2\le q\le 5$ and $p\ge 0$. We split $S_2$ into $K_{1,q}$ and $K_{1,4p}$ with the common vertex $u_1$. We label $K_{1,q}$, and the selected edges of $S_1$ by $\alpha$ as defined above, and all the edges of $K_{1,4p}$ by 1 and 2 evenly. Again, we label the edges of an unlabeled subgraph of $S_1$, which is isomorphic to $K_{1,2p}$ by 0. We denote this labeling by $\alpha$. Then we have ${\alpha }^{*}\left(u_1\right)=1$ and ${\alpha }^{*}\left(u_n\right)=0$; $v_{{\alpha }^{*}}\left(0\right)=v_{{\alpha }^{*}}\left(1\right)=v_{{\alpha }^{*}}\left(2\right)+1$; and all $e_{\alpha }\left(i\right)$ are the same for $0\le i\le 2$.

Here, the unlabeled subgraph, say H, of $DC(n,M,m)$ is isomorphic to $C(n,M-2p-ϵ)$, where $ϵ=2,3,4$, which depends on q. That is,

• When $q=2$, $H\cong C(n,M-2p-4)$ if $M-2p-4\ge 0$.
• When $q=3$, $H\cong C(n,M-2p-3)$.
• When $q=4$, $H\cong C(n,M-2p-2)$.
• When $q=5$, $H\cong C(n,M-2p-4)$.

If $M-2p-ϵ\ge 2$, then by Theorem 1 there exists an edge 3-product cordial labeling $\varphi$ for H. Note that $v_{{\alpha }^{*}}\left(0\right)-1=v_{{\alpha }^{*}}\left(1\right)-1=v_{{\alpha }^{*}}\left(2\right)$, ${\alpha }^{*}\left(u_1\right)=1$, ${\alpha }^{*}\left(u_n\right)=0$ and ${\varphi }^{*}\left(u_1\right)=1$. When we combine $\alpha$ with $\varphi$, the number of j-vertices ($0\le j\le 2$) are $v_{{\varphi }^{*}}\left(0\right)+v_{{\alpha }^{*}}\left(0\right)-1$, $v_{{\varphi }^{*}}\left(1\right)+v_{{\alpha }^{*}}\left(1\right)-1$ and $v_{{\varphi }^{*}}\left(2\right)+v_{{\alpha }^{*}}\left(2\right)$. So the combination of $\alpha$ and $\varphi$ results in an edge 3-product cordial labeling for $DC(n,M,m)$.

The detailed labeling for receiving edge 3-product cordial of $DC(n,M-2p-ϵ)$ for $M-2p-ϵ\le 1$ are moved to Appendix A. This completes the proof. □

Example 3.

The following Figure 3 illustrates the proof of Theorem 2.

We combine α and ϕ to get an edge 3-product cordial labeling for $DC(7,8,6)$.

4. Edge 4-Product Cordial Trees

We begin this section with the necessary condition on the number of vertices and leaves for a tree to admit an edge 4-product cordial labeling.

Theorem 3.

Let T be a tree with n vertices, and S be the set of all the leaves of T. If T is an edge 4-product cordial graph, then

$$\left|S\right|\ge \left\{\begin{array}{cc}⌊{\displaystyle \frac{n}{4}}⌋\hfill & if n\equiv 0,1(mod4),\hfill \\ ⌊{\displaystyle \frac{n}{4}}⌋-1\hfill & if n\equiv 2,3(mod4).\hfill \end{array}\right.$$

Moreover, the bound is sharp.

Proof. 

Let f be an edge 4-product cordial labeling of T, and $E_i$ be the set of edges labeled by i, where $i=0,2$. Let $T_i=T\left[E_i\right]$. By Corollary 1, $T_0$ is a tree and $v_{f^{*}}\left(0\right)=e_f\left(0\right)+1=\left|V\left(T_0\right)\right|$. This implies that each vertex in $V\left(T\right)\setminus V\left(T_0\right)$ is not labeled by 0.

Now, consider forest $T_2$. If $v\in V\left(T_2\right)$, then $f^{*}\left(v\right)=0,2$; and if $v\notin V\left(T_2\right)$, then $f^{*}\left(v\right)\ne 2$. If $v\in V\left(T_2\right)$ and $f^{*}\left(v\right)=0$, then $v\in V\left(T_0\right)\cap V\left(T_2\right)=A_0$. If $f^{*}\left(v\right)=2$, then ${deg}_{T_2}\left(v\right)=1$.

Let $u_1{v}_1,\dots ,u_q{v}_q$ be the edges in $E_2$ such that $u_i\in V\left(T_0\right)$ and $v_i\in V\left(T_2\right)$, $1\le i\le q$. Note that $v_i$ are distinct, but $u_i$ may not be distinct. Then, there are $e_f\left(2\right)-q$ edges in $E_2$, and their end vertices are labeled by 2. Then $v_{f^{*}}\left(2\right)=q+2(e_f\left(2\right)-q)=2e_f\left(2\right)-q$, equivalently $q=2e_f\left(2\right)-v_{f^{*}}\left(2\right)$.

(1)

If $n\equiv 0(mod4)$, then $v_{f^{*}}\left(i\right)={\displaystyle \frac{n}{4}}$ for all i and $e_f\left(0\right)={\displaystyle \frac{n}{4}}-1$. Thus, $e_f\left(2\right)={\displaystyle \frac{n}{4}}=e_f\left(1\right)=e_f\left(3\right)$. Therefore, $q={\displaystyle \frac{n}{4}}$.

(2)

If $n\equiv 1(mod4)$, then $e_f\left(i\right)=\lfloor {\displaystyle \frac{n}{4}}\rfloor$ for all i, and $v_{f^{*}}\left(0\right)=\lfloor {\displaystyle \frac{n}{4}}\rfloor +1$. Therefore, $v_{f^{*}}\left(i\right)=\lfloor {\displaystyle \frac{n}{4}}\rfloor$ for $i=1,2,3$. Hence, $q=\lfloor {\displaystyle \frac{n}{4}}\rfloor$.

(3)

If $n\equiv 2,3(mod4)$, then $q=2e_f\left(2\right)-v_{f^{*}}\left(2\right)\ge 2⌊{\displaystyle \frac{n}{4}}⌋-(⌊{\displaystyle \frac{n}{4}}⌋+1)=⌊{\displaystyle \frac{n}{4}}⌋-1$.

We merge the tree $T_0$ into a vertex r to receive the resultant graph $T^{\prime }$. Then $T^{\prime }$ is a rooted tree with root r and ${deg}_{T^{\prime }}\left(r\right)=q$. So $T^{\prime }$ has at least q leaves, which are also the leaves of T. Since S is the set of all leaves of T and T contains at least q leaves, $\left|S\right|\ge q$. Hence,

$$\left|S\right|\ge \left\{\begin{array}{cc}⌊{\displaystyle \frac{n}{4}}⌋\hfill & \mathrm{if} n\equiv 0,1(mod4),\hfill \\ ⌊{\displaystyle \frac{n}{4}}⌋-1\hfill & \mathrm{if} n\equiv 2,3(mod4).\hfill \end{array}\right.$$

□

The following remark demonstrates that the bound in Theorem 3 is sharp.

Remark 2.

If $P_n$ is an edge 4-product cordial, then

$$n\le \left\{\begin{array}{cc}9\hfill & if n\equiv 0,1(mod4),\hfill \\ 15\hfill & if n\equiv 2,3(mod4).\hfill \end{array}\right.$$

Clearly, $P_4$, $P_3$, and $P_2$ are not edge 4-product cordial graphs. An edge 4-product cordial labeling for $P_n$, $5\le n\le 15$, is shown in

Table 2. Edge 4-product cordial labeling for $P_n,5\le n\le 15$.

n	Label the Edges in Order	v(0)	v(1)	v(2)	v(3)
5	2,0,3,1	2	1	1	1
6	2,0,2,3,1	2	1	2	1
7	2,0,2,1,3,3	2	1	2	2
8	2,0,2,1,3,3,1	2	2	2	2
9	2,0,0,2,1,3,3,1	3	2	2	2
10	2,0,0,2,1,3,3,1,3	3	2	2	3
11	2,0,0,2,3,3,1,1,3,1	3	3	2	3
14	2,0,0,0,2,1,3,3,1,1,3,3,2	4	3	4	3
15	2,0,0,0,2,1,3,3,1,1,3,3,1,2	4	3	4	4

.

Corollary 3.

$P_n$ is an edge 4-product cordial if and only if $n\in \{5,6,7,8,9,10,11,14,15\}$.

Note that in the following lemmas and corollary, all the induced vertex labelings work in the complete residues class modulo 4, ${\mathbb{Z}}_4$.

Lemma 5.

Suppose $S_1$ and $S_2$ are the trees such that $V\left(S_1\right)\cap V\left(S_2\right)=\left\{x\right\}$. Let the order of $S_1$ and $S_2$ be a and b, respectively, such that $2n\le a\le b\le 2n+1$ for some positive integer n. Let $f:E\left(S_1\right)\to \{0,2\}$ and $g:E\left(S_2\right)\to \{1,3\}$ be the edge labeling, which satisfy the following conditions:

(1) 

$0\le e_f\left(2\right)-e_f\left(0\right)\le 1$ and $0\le v_{f^{*}}\left(2\right)-v_{f^{*}}\left(0\right)\le 1$;

(2) 

$|e_g\left(1\right)-e_g\left(3\right)|\le 1$ and $|v_{g^{*}}\left(1\right)-v_{g^{*}}\left(3\right)|\le 1$;

(3) 

$g^{*}\left(x\right)=\ell$ only if $v_{g^{*}}\left({\ell }_1\right)\le v_{g^{*}}(\ell )$, where $\{\ell ,{\ell }_1\}=\{1,3\}$.

Let ϕ be the combination of f and g. Then ϕ is an edge 4-product cordial labeling of $S_1\cup S_2$.

Proof. 

Suppose the order of $S_1$ is $2n$. Then the order of $S_2$ is either $2n$ or $2n+1$. Thus, by conditions 1 and 2, we obtain $|e_{\varphi }\left(i\right)-e_{\varphi }\left(j\right)|\le 1$ for all $i\ne j$.

Since $f^{*}\left(x\right)=0$ or 2, we have ${\varphi }^{*}\left(x\right)=f^{*}\left(x\right)g^{*}\left(x\right)=f^{*}\left(x\right)$. Thus, the number of 0-vertices and 2-vertices does not change, and they are equal to n.

If the order of $S_2$ is $2n+1$ and $g^{*}\left(x\right)=1$, then $v_{g^{*}}\left(1\right)=n+1$ and $v_{g^{*}}\left(3\right)=n$. Hence, $v_{{\varphi }^{*}}\left(1\right)=n$ and $v_{{\varphi }^{*}}\left(3\right)=n$.

If the order of $S_2$ is $2n+1$ and $g^{*}\left(x\right)=3$, then $v_{g^{*}}\left(1\right)=n$ and $v_{g^{*}}\left(3\right)=n+1$. Hence, $v_{{\varphi }^{*}}\left(1\right)=n$ and $v_{{\varphi }^{*}}\left(3\right)=n$.

If the order of $S_2$ is $2n$, then $v_{g^{*}}\left(1\right)=v_{g^{*}}\left(3\right)=n$. Hence, $v_{{\varphi }^{*}}\left(1\right)=n-1$ and $v_{{\varphi }^{*}}\left(3\right)=n$ if $g^{*}\left(x\right)=1$; $v_{{\varphi }^{*}}\left(1\right)=n$ and $v_{{\varphi }^{*}}\left(3\right)=n-1$ if $g^{*}\left(x\right)=3$.

Suppose the order of $S_1$ is $2n+1$. Then the order of $S_2$ is $2n+1$. By conditions 1 and 2, the number of i-edges are n under $\varphi$ for $0\le i\le 3$. In this case $v_{g^{*}}(\ell )=n+1$ and $v_{g^{*}}\left({\ell }_1\right)=n$, where $g^{*}\left(x\right)=\ell$ and $\{\ell ,{\ell }_1\}=\{1,3\}$. Similarly, we have $v_{{\varphi }^{*}}\left(2\right)=v_{{\varphi }^{*}}\left(1\right)=v_{{\varphi }^{*}}\left(3\right)=n$ and $v_{{\varphi }^{*}}\left(0\right)=n+1$. □

Lemma 6.

Let $g:E\left(T\right)\to \{1,3\}$ be an edge labeling of tree T of order n, which satisfies the condition (2) of Lemma 5, then $n\not\equiv 2(mod4)$.

Proof. 

Let $E_3$ be the set of all 3-edges in T, and let $T_3=T\left[E_3\right]$. Since any 3-vertex must be incident to three edges, all 3-vertices are in $T_3$. Also, each 3-vertex is of odd degree in $T_3$. Thus, $v_{g^{*}}\left(3\right)$ is even. Hence, $n=v_{g^{*}}\left(3\right)+v_{g^{*}}\left(1\right)\equiv v_{g^{*}}\left(1\right)(mod2)$. If $v_{g^{*}}\left(1\right)$ is odd, then $n\not\equiv 2(mod4)$. If $v_{g^{*}}\left(1\right)$ is even, then $v_{g^{*}}\left(1\right)=v_{g^{*}}\left(3\right)$. Hence, $n\equiv 0(mod4)$. □

By Lemma 5, we have the following corollary.

Corollary 4.

Suppose $S_1$ and $S_2$ are the trees such that $V\left(S_1\right)\cap V\left(S_2\right)=\left\{x\right\}$. Let the order of $S_1$ be $4k+1$ or $4k+2$ and the order of $S_2$ be $4k+2$, where $k\ge 1$. Let $f:E\left(S_1\right)\to \{0,2\}$ and $g:E\left(S_2\right)\to \{1,3\}$ be the edge labeling, which satisfy the following conditions:

(a) 

$0\le e_f\left(2\right)-e_f\left(0\right)\le 1$ and $0\le v_{f^{*}}\left(2\right)-v_{f^{*}}\left(0\right)\le 1$;

(b) 

$g^{*}\left(x\right)=\ell$;

(c) 

$|e_g\left(1\right)-e_g\left(3\right)|\le 1$ and $v_{g^{*}}(\ell )-v_{g^{*}}\left({\ell }_1\right)=2$, where $\{\ell ,{\ell }_1\}=\{1,3\}$.

Let ϕ be the labeling of $S_1\cup S_2$ by combining f and g. Then ϕ is an edge 4-product cordial labeling of $S_1\cup S_2$.

A vertex that satisfies the condition (3) in Lemma 5 or the conditions (b) and (c) in Corollary 4 is called a major vertex under g.

Lemma 7.

If $n\not\equiv 2(mod4)$ and $n\ge 3$, then there exists a labeling $g:E\left(P_n\right)\to \{1,3\}$ that satisfies the condition (2) of Lemma 5. If $n\equiv 2(mod4)$ and $n\ge 2$, then there exists a labeling $g:E\left(P_n\right)\to \{1,3\}$ that satisfies the condition (c) of Corollary 4.

Proof. 

We label the edges of a path $P_n$ by 1,3,3,1 evenly and denote this required labeling by g. □

Lemma 8.

If $m+n\not\equiv 2(mod4)$ and $n\ge 3$, $m\ge 1$, then there exists a labeling $g:E\left(C\right(n,m\left)\right)\to \{1,3\}$ that satisfies the condition (2) of Lemma 5. If $m+n\equiv 2(mod4)$ and $n\ge 3$, $m\ge 2$, then there exists a labeling $g:E\left(C\right(n,m\left)\right)\to \{1,3\}$ that satisfies the condition (c) of Corollary 4. Moreover, $u_1$ is the major vertex under g.

Proof. 

We will define a labeling $g:E\left(C\right(,m\left)\right)\to \{1,3\}$ by the following approach. We first suitably label the edge $u_n{v}_i$ for $1\le i\le m$. And then label the edge of the path $u_n{u}_{n-1}\cdots u_1$. There are four cases. We put the details in Appendix B. Hence, we have the theorem. □

Now, we consider the comet graph $C(n,m)$, which has $m+1$ leaves. From Theorem 3, we have

$$n\le \left\{\begin{array}{cc}3m+4\hfill & \mathrm{if} m+n\equiv 0(mod4);\hfill \\ 3m+5\hfill & \mathrm{if} m+n\equiv 1(mod4);\hfill \\ 3m+10\hfill & \mathrm{if} m+n\equiv 2(mod4);\hfill \\ 3m+11\hfill & \mathrm{if} m+n\equiv 3(mod4).\hfill \end{array}\right.$$

When $n=1,2$, we have $C(n,m)\cong K_{1,n+m-1}$, which is a star. It is easy to check that $K_{1,n+m-1}$ is edge 4-product cordial when $n+m\ge 5$.

Theorem 4.

For $n\ge 3$, the comet graph $C(n,m)$ is edge 4-product cordial if and only if

$$n\le \left\{\begin{array}{cc}3m+4\hfill & if m+n\equiv 0(mod4);\hfill \\ 3m+5\hfill & if m+n\equiv 1(mod4);\hfill \\ 3m+10\hfill & if m+n\equiv 2(mod4);\hfill \\ 3m+11\hfill & if m+n\equiv 3(mod4).\hfill \end{array}\right.$$

Proof. 

The necessary part is shown in the discussion above. Now we have to show the sufficient part. Let $N=⌊{\displaystyle \frac{m+n}{4}}⌋$. We can check that $N\le m+1$ when $m+n\equiv 0,1(mod4)$; and $N\le m+2$ when $n+m\equiv 2,3(mod4)$.

We split the graph $C(n,m)$ into two subgraphs, $S_1$ and $S_2$, with a common vertex x. Then, we define the labelings f and g for $S_1$ and $S_2$, respectively, such that f and g satisfy all the conditions of Lemma 5 or Corollary 4. The details are referred to Appendix C.

Hence, we have the theorem. □

Example 4.

Edge 4-product labelings for $C(9,2)$, $C(8,3)$, and $C(5,6)$ are shown in Figure 4.

Example 5.

Edge 4-product labelings for $C(10,2)$, and $C(5,7)$ are provided in Figure 5.

Example 6.

Edge 4-product labelings for $C(16,2)$ and $C(17,2)$ are shown in Figure 6.

Remark 3.

Suppose T is a tree of order $4N+k$, which admits an edge 4-product cordial labeling f, where $0\le k\le 3$. Let $E_i$ be the set of edges labeled by i, where $i=0,2$ and $H=T[E_2\cup E_0]$. By Theorem 3 we obtain the following results.

(A) 

If $k=0,1$, then $q=N$. Thus, H is a tree that has at least N leaves.

(B) 

If $k=2,3$, then $q\ge N-1$. Clearly, $q\le N+1$. Also, we have $e_f\left(0\right)=N$ and $v_{f^{*}}\left(0\right)=N+1$.

B1. 

Suppose $q=N+1$. Clearly, $e_f\left(2\right)=N+1$ and $v_{f^{*}}\left(2\right)=N+1$. Then H is a tree of order $2N+2$ that has at least $N+1$ leaves.

B2. 

Suppose $q=N$. Recall that $q=2e_f\left(2\right)-v_{f^{*}}\left(2\right)$. Since $v_{f^{*}}\left(2\right)\le N+1$, we have $e_f\left(2\right)\le N+{\displaystyle \frac{1}{2}}$. Thus, $e_f\left(2\right)=N$ and $v_{f^{*}}\left(2\right)=N$. Then H is a tree of order $2N+1$ that has at least N leaves.

B3. 

Suppose $q=N-1$. Since $v_{f^{*}}\left(2\right)\le N+1$, we have $e_f\left(2\right)=N$ and $v_{f^{*}}\left(2\right)=N+1$. Then H is a disjoint union of a tree $T^{\prime }$ of order $2N$ with $P_2$. Moreover, $T^{\prime }$ has at least $N-1$ leaves.

Theorem 5.

Suppose $n\ge 2$, $M\ge m\ge 2$ and $M+m+n=4N+k$, where $0\le k\le 3$.

A. 

When $k=0,1$. The graph $DC(n,M,m)$ is an edge 4-product cordial if and only if $M\ge N-1$.

B. 

When $k=2,3$. The graph $DC(n,M,m)$ is an edge 4-product cordial if and only if $M\ge N-2$.

Proof. 

Suppose there is an edge 4-product cordial labeling f for $T=DC(n,M,m)$. Let H be the edge-induced subgraph defined in Remark 3.

(A)

Suppose $k=0,1$. By Remark 3, H is a tree of order $2N+k$ that has at least N leaves. Suppose $m\le M\le N-2$. Then $N={\displaystyle \frac{M+m+n-k}{4}}\le {\displaystyle \frac{2M+n-k}{4}}\le {\displaystyle \frac{2N-4+n-k}{4}}$. Thus, $n\ge 2N+k+4\ge 2N+4$. Since H has at least N leaves, we have $H\cong DC(n,M_1,m_1)$, where $M_1+m_1\ge N$. But the order of H is at least $3N+4$, which is a contradiction. Thus, if $M\le N-2$, then $DC(n,M,m)$ is not an edge 4-product cordial graph.

(B)

Suppose $k=2,3$. Then $e_f\left(0\right)=N$ and $v_{f^{*}}\left(0\right)=N+1$.

• Suppose $q=N+1$. Then H is a subtree of $DC(n,M,m)$ of order $2N+2$ that has $N+1$ leaves. Suppose $m\le M\le N-1$. Similarly to Case A, we obtain a contradiction. Thus, if $M\le N-1$, then $DC(n,M,m)$ is not an edge 4-product cordial graph.
• Suppose $q=N$. Then H is a subtree of $DC(n,M,m)$ of order $2N+1$ that has N leaves. Suppose $m\le M\le N-2$. Similarly to Case A, we obtain a contradiction. Thus, if $M\le N-2$, then $DC(n,M,m)$ is not an edge 4-product cordial graph.
• Suppose $q=N-1$. Then H is a disjoint union of a tree $T^{\prime }$ of order $2N$ with $P_2$. Moreover, $T^{\prime }$ has at least $N-1$ leaves. Thus, $T^{\prime }$ must be a comet $C(n_1,m_1)$ such that $n_1+m_1=2N$ and $m_1\ge N-2$. Thus, if $M\le N-3$, then $DC(n,M,m)$ is not an edge 4-product cordial.

Consequently, if $DC(n,M,m)$ is an edge 4-product cordial, then $M\ge N-1$ when $n+M+m\equiv 0,1(mod4)$; and $M\ge N-2$ when $n+M+m\equiv 2,3(mod4)$.

For the sufficient part, we split the graph $DC(n,M,m)$ into two subgraphs $S_1$ and $S_2$ with one or two common vertices. We label $S_1$ by 0 and 2, and $S_2$ by 1 and 3, respectively, such that these labelings induce an edge 4-product cordial labeling for $DC(n,M,m)$.

Since $M,n\ge 2$, we have $M+n\ge {\displaystyle \frac{M+m+n}{2}}+{\displaystyle \frac{n}{2}}\ge {\displaystyle \frac{4N+k}{2}}+1=2N+1+{\displaystyle \frac{k}{2}}$. This guarantees that the comets $S_1$ and $S_2$ defined below are well-defined. The details are referred to Appendix D.

This completes the proof. □

Example 7.

Edge 4-product labelings for $DC(14,3,2)$ and $DC(15,2,2)$ are shown in Figure 7.

5. Edge 5-Product Cordial Trees

In order to prove the main theorems, first we prove the following lemma. Note that, by Lemma 2, the path $P_5$ is not an edge 5-product cordial.

Lemma 9.

The path graph $P_n$ is an edge 5-product cordial for $n\ge 3$ and $n\ne 5$.

Proof. 

We define an edge labeling $f_n$ for $P_n$ recurrently, and this labeling is represented by $(a_1,\dots ,a_{n-1})$ if $f_n\left(u_i{u}_{i+1}\right)=a_i$, $1\le i\le n-1$.

We present the edge 5-product cordial labeling for $P_n$, $3\le n\le 12$, except $n=5$ in

Table 3. Edge 5-product cordial labeling for $P_n,3\le n\le 12$, except $n=5$.

n	${\mathit{f}}_{\mathit{n}}$	${\mathit{v}}_{{\mathit{f}}_{\mathit{n}}^{*}}$(0)	${\mathit{v}}_{{\mathit{f}}_{\mathit{n}}^{*}}$(1)	${\mathit{v}}_{{\mathit{f}}_{\mathit{n}}^{*}}$(2)	${\mathit{v}}_{{\mathit{f}}_{\mathit{n}}^{*}}$(3)	${\mathit{v}}_{{\mathit{f}}_{\mathit{n}}^{*}}$(4)
3	(3,2)	0	1	1	1	0
4	(1,2,4)	0	1	1	1	1
6	(3,2,1,4,0)	2	1	1	1	1
7	(1,3,2,1,4,0)	2	2	1	1	1
8	(1,1,3,3,4,2,0)	2	2	1	2	1
9	(1,2,4,1,4,3,2,0)	2	2	2	1	2
10	(1,3,3,4,4,2,2,1,0)	2	2	2	2	2
11	(1,3,3,4,4,2,2,1,0,0)	3	2	2	2	2
12	(1,3,3,4,4,2,2,1,1,0,0)	3	3	2	2	2

.

For $n\ge 13$, we define $f_n$ recurrently as $f_n=f_{m-i+10}=(1,3,3,4,4,2,2,1,a_1,\dots ,a_{m-i-1},0,0)$, where $0\le i\le 9$ and $m\ge 12$. When $m-i\ge 7$, we have $a_1=1$ and $a_{m-i-1}=0$. Thus, $e_{f_{m-i+10}}\left(j\right)=e_{f_{m-i}}\left(j\right)+2$, $v_{f_{m-i+10}^{*}}\left(j\right)=v_{f_{m-i}^{*}}\left(j\right)+2$, $0\le j\le 4$.

• For $m-i=3$, $v_{f_{13}^{*}}\left(0\right)=3$, $v_{f_{13}^{*}}\left(1\right)=3$, $v_{f_{13}^{*}}\left(2\right)=2$, $v_{f_{13}^{*}}\left(3\right)=3$, $v_{f_{13}^{*}}\left(4\right)=2$.
• For $m-i=4$, $v_{f_{14}^{*}}\left(0\right)=3$, $v_{f_{14}^{*}}\left(1\right)=3$, $v_{f_{14}^{*}}\left(2\right)=3$, $v_{f_{14}^{*}}\left(3\right)=3$, $v_{f_{14}^{*}}\left(4\right)=2$.
• For $m-i=5$, $v_{f_{15}^{*}}\left(j\right)=3$ for all $0\le j\le 4$.
• For $m-i=6$, $v_{f_{16}^{*}}\left(0\right)=4$ and $v_{f_{16}^{*}}\left(j\right)=3$ for all $1\le j\le 4$.

Thus, $f_n$ is an edge 5-product cordial labeling of $P_n$ for $n\ge 6$. Moreover, $f_3$ and $f_4$ are the required labelings for $P_3$ and $P_4$, respectively.

This completes the proof. □

Theorem 6.

The comet graph $C(n,m)$ is an edge 5-product cordial for $n\ge 3$ and $m\ge 2$, except $(n,m)=(3,2)$.

Proof. 

Let $m=5k+r$, where $0\le r\le 4$. In order to obtain an edge 5-product cordial labeling of $C(n,m)$ for $n\ge 3$ and $m\ge 2$, except for $(n,m)=(3,2)$, we split $C(n,m)$ into two subgraphs, $K_{1,5k}$ and $C(n,r)$, with a common vertex $u_n$. Note that when $k=0$, $K_{1,5k}$ does not appear; when $r=0$, $C(n,r)\cong P_n$. For the last case, it has been proved in Lemma 9.

First, we label $P_n$ by using $f_n$, which is defined below. For $1\le r\le 4$, we label $K_{1,r}$ to balance the number of i-edges and i-vertices, as shown in

Table 4. Edge labeling for $P_n,3\le n\le 12$ and $K_{1,r},1\le r\le 4$.

n	${\mathit{f}}_{\mathit{n}}$	Priority for Numbers Added to ${\mathit{K}}_{1,\mathit{r}}$
3	(1,3)	4,2,0,1
4	(3,2,1)	4,0,1,2
5	(3,2,1,4)	0,1,2,3
6	(3,2,1,4,0)	1,2,3,4
7	(1,3,2,1,4,0)	2,3,4,0
8	(1,1,3,3,4,2,0)	2,4,0,3
9	(1,2,4,1,4,3,2,0)	3,0,1,2
10	(1,3,3,4,4,2,2,1,0)	0,1,2,3
11	(1,3,3,4,4,2,2,1,0,0)	1,2,3,4
12	(1,3,3,4,4,2,2,1,1,0,0)	2,3,4,0

.

For $n\ge 13$, we label $f_n=f_{m+10}=(1,3,3,4,4,2,2,1,f_m,0,0)$, where $m\ge 3$. Then the difference between the number of i-edges and j-vertices does not change for all $0\le i,j\le 4$. Hence, according to the table above, we have an edge 5-product cordial labeling for $C(n,r)$, where $n\ge 3$ and $1\le r\le 4$.

For $k\ge 1$, we label the edges of $K_{1,5k}$ by 0, 1, 2, 3, and 4 evenly and denote this labeling by $\varphi$. Thus, $\varphi$ is an edge 5-product labeling for $C(n,5k+r)$ for $n\ge 3$ and $5k+r\ge 2$, except for $(n,m)=(3,2)$. By Corollary 2, $C(3,2)$ is not an edge 5-product cordial graph. This completes the proof. □

Example 8.

For the comet $C(10,13)$, we separate it into two edge-disjoint graphs, $C(10,3)$ and $K_{1,10}$. From the table above, we label $P_{10}$ as 1, 3, 3, 4, 4, 2, 2, 1, 0, and label the edges of $K_{1,3}$ by 0, 1, and 2. The resulting labeling is an edge labeling for $C(10,3)$. The numbers of 1-and 2-edges are 3, and the numbers of 0-, 3- and 4-edges are 2. The numbers of 0-, 1-, and 2- vertices are 3, and the numbers of 3- and 4- vertices are 2.

Finally, we label the edges of $K_{1,10}$ evenly by $0,1,2,3,4$. We obtain three 0-vertices and two i-vertices, where $1\le i\le 4$. The centers of $K_{1,10}$ and $K_{1,3}$ will be merged; thus, we obtain an edge 5-product cordial labeling ϕ for $C(10,13)$. We can check that $e_{\varphi }\left(0\right)=4$, $e_{\varphi }\left(1\right)=5$, $e_{\varphi }\left(2\right)=5$, $e_{\varphi }\left(3\right)=4$, $e_{\varphi }\left(4\right)=4$; $v_{{\varphi }^{*}}\left(0\right)=5$, $v_{{\varphi }^{*}}\left(1\right)=5$, $v_{{\varphi }^{*}}\left(2\right)=5$, $v_{{\varphi }^{*}}\left(3\right)=4$, $v_{{\varphi }^{*}}\left(4\right)=4$.

Theorem 7.

The double comet graph $DC(n,M,m)$ is an edge 5-product cordial for $n\ge 2$ and $M\ge m\ge 2$.

Proof. 

Let $S_1\cong K_{1,M}$ with center $u_n$ and $S_2\cong K_{1,m}$ with center $u_1$. Then $DC(n,M,m)=P_n\cup S_1\cup S_2$.

We define an edge labeling $\alpha$ for $S_2$ and the selected edges of $S_1$ by 0,1,2,3,4 evenly. Consider $m=3p+q$, where $0\le q\le 2$. First, we assume $p\ge 1$.

• For $q=0$, we label the edges of $S_2$ by 1, 2, 3 evenly and $p,p$ edges of $S_1$ by $0,4$, respectively.
• For $q=1$, we label $p-1,p,p,2$ edges of $S_2$ by $1,2,3,4$, respectively, and $p,p-2,1$ edges of $S_1$ by $0,4,1$, respectively.
• For $q=2$, we label $p,p,p,2$ edges of $S_2$ by $1,2,3,4$, respectively, and $p,p-2$ edges of $S_1$ by $0,4$, respectively.

Note that ${\alpha }^{*}\left(u_1\right)=1$. Also $e_{\alpha }\left(i\right)=p$ for all i and $v_{{\alpha }^{*}}\left(0\right)-1=v_{{\alpha }^{*}}\left(1\right)-1=v_{{\alpha }^{*}}\left(j\right)$ for $2\le j\le 4$.

Now the unlabeled edges form $C(n,M-2p+ϵ)$, where

$$ϵ=\left\{\begin{array}{cc}0\hfill & \mathrm{if} q=0,\hfill \\ 1\hfill & \mathrm{if} q=1,\hfill \\ 2\hfill & \mathrm{if} q=2.\hfill \end{array}\right.$$

Hence, $M-2p+ϵ\ge p+q+ϵ\ge 2$ except for $M=m=3$. By Theorem 6, there is an edge 5-product cordial labeling for $C(n,M-2p+ϵ)$ for $M-2p+ϵ\ge 2$.

For $M=m=3$, $C(n,1)\cong P_{n+1}=u_1\dots u_n{v}_3$. We label this $P_{n+1}$ by the labeling $f_{n+1}$ defined in Lemma 9. Now we check the number of i-vertices.

Before labeling $P_{n+1}$, we have ${\alpha }^{*}\left(v_1\right)=0={\alpha }^{*}\left(u_n\right)$, ${\alpha }^{*}\left(v_2\right)=4$, ${\alpha }^{*}\left(w_1\right)=1$, ${\alpha }^{*}\left(w_2\right)=2$, ${\alpha }^{*}\left(w_3\right)=3$ and ${\alpha }^{*}\left(u_1\right)=1$. Suppose $n+1\ge 7$. After labeling $P_{n+1}$, the vertex $u_n$ is still 0-vertex and the vertex $u_1$ changes from 1-vertex to $f_{n+1}^{*}\left(u_1\right)$. Thus ${\alpha }^{*}\left(u_n\right)=0$ and ${\alpha }^{*}\left(u_1\right)=1$ do not count towards the number of 0-vertices and 1-vertices. Thus, the combined labeling is an edge 5-product cordial labeling for $DC(n,3,3)$.

For $n+1=3,4,5$, the required labeling is shown in the following example.

Suppose $m=2$. If $M\ge 3$, then $\alpha \left(u_1{w}_1\right)=2$, $\alpha \left(u_1{w}_2\right)=3$, $\alpha \left(u_n{v}_1\right)=0$, $\alpha \left(u_n{v}_2\right)=1$ and $\alpha \left(u_1{v}_3\right)=4$. The unlabeled edges form $C(n,M-3)$. The argument is similar to the cases above. If $M=2$, let $\alpha \left(u_1{w}_1\right)=2$, $\alpha \left(u_1{w}_2\right)=3$, $\alpha \left(u_n{v}_1\right)=1$, $\alpha \left(u_n{v}_2\right)=4$ and $\alpha \left(u_1{u}_3\right)=0$. The unlabeled edges form $P_{n-1}$. If $n=2$, then we have an edge 5-product cordial labeling for $DC(2,2,2)$. If $n=3$, then label $u_1{u}_2$ by 1. We have an edge 5-product cordial labeling for $DC(3,2,2)$. If $n\ge 4$, then the labeling is the same as $M=m=3$. □

Example 9.

An edge 5-product cordial labelings for $DC(n,3,3)$, where $n=2,3,4,5$ are shown in Figure 8.

6. Conclusions

The notion of edge k-product cordial labeling was introduced only in the year 2025, and the authors showed that star, bistar, and path unions of star graphs admit edge k-product cordial labeling. They also investigated the edge k-product cordial behavior of the shadow and the splitting graph of a star. In this work, we further explore the relationship between the number of edges and vertices labeled with 0 in edge k-product cordial trees and investigate the edge k- product cordiality of trees of order k. Also, we establish the edge k-product cordial properties of comet and double comet trees for $k=3,4$, and 5. It is noted that edge k-product cordial labeling is a recent concept, and only a limited study has been carried out. Future researchers have ample scope to identify the families of graphs that admit or do not admit edge k-product cordial labeling, and also to investigate the edge k-product cordial behavior of a larger number of standard graphs.