On SD-Prime Labelings of Some One Point Unions of Gears

Abstract

Let $G=\left(V\right(G),E(G\left)\right)$ be a simple, finite and undirected graph of order n. Given a bijection $f:V\left(G\right)\to \{1,\dots ,n\}$, and every edge $uv$ in $E\left(G\right)$, let $S=f\left(u\right)+f\left(v\right)$ and $D=\left|f\right(u)-f(v\left)\right|$. The labeling f induces an edge labeling $f^{\prime }:E\left(G\right)\to \{0,1\}$ such that for an edge $uv$ in $E\left(G\right)$, $f^{\prime }\left(uv\right)=1$ if $gcd(S,D)=1$, and $f^{\prime }\left(uv\right)=0$ otherwise. Such a labeling is called an SD-prime labeling if $f^{\prime }\left(uv\right)=1$ for all $uv\in E\left(G\right)$. We provide SD-prime labelings for some one point unions of gear graphs.

1. Introduction

Let $G=\left(V\right(G),E(G\left)\right)$ (or simply $G=(V,E)$ for short, if there is no ambiguity) be a simple, finite and undirected graph of order $\left|V\right|=n$ and size $\left|E\right|=m$. For convenience, we shall use $[a,b]$ to denote the set of integers from a to b, where $a\le b$. All notation not defined in this paper can be found in [1].

Since the introduction of the concept of graph labeling by Rosa in 1967 [2], thousands of papers on various graph labeling problems have been published. It is interesting to note that the field of graph labelings is becoming more significant with more applications in other topics of graph theory being found. For example, the friendly index of a graph is related to eigenvalues, bipartition width, max-cut problem and isoperimetric problem of a graph [3].

One of the graph labeling problems that attracts great interest of researchers is the concept of prime graphs introduced in 1982 [4].

Definition 1.

A bijection $f:V\to [1,n]$ induces an edge labeling $f^{\prime }:E\to \{0,1\}$ such that for an edge $uv$ in G, $f^{\prime }\left(uv\right)=1$ if $gcd\left(f\right(u),f(v\left)\right)=1$, and $f^{\prime }\left(uv\right)=0$ otherwise. Such a labeling is called a prime labeling if $f^{\prime }\left(uv\right)=1$ for all $uv\in E$. We say G is a prime graph if it admits a prime labeling.

Since then, many papers that attempt to determine the primality of graphs are published (for example, [5,6,7,8,9,10,11]). Although a wheel graph of order $2n+1$ is prime if and only if n is even [12], it was proved that the one point union of cycles and that of wheels at the center vertex are all prime graphs [13]. However, we are unaware of any vertex labelings in which the induced edge labels are associated to the sum and difference of the incident end-vertex labels.

In view of this, Lau and Shiu [14] introduced a variant of prime graph labeling, known as SD-prime labeling, as defined below.

Given a bijection $f:V\to [1,n]$, and every edge $uv$ in E, one can associate two integers $S=f\left(u\right)+f\left(v\right)$ and $D=\left|f\right(u)-f(v\left)\right|$.

Definition 2.

A bijection $f:V\to [1,n]$ induces an edge labeling $f^{\prime }:E\to \{0,1\}$ such that for an edge $uv$ in G, $f^{\prime }\left(uv\right)=1$ if $gcd(S,D)=1$, and $f^{\prime }\left(uv\right)=0$ otherwise. Such a labeling is called an SD-prime labeling if $f^{\prime }\left(uv\right)=1$ for all $uv\in E$. We say G is SD-prime if it admits an SD-prime labeling.

In [15], Lau et al. provided more discussion on SD-prime labelings. This new concept leads to new research direction and possible future applications. In [16], one may find that the one point union of various families of standard graphs have been investigated under various labeling problems (see, for example, [17,18,19]). However, we are not aware of the study on one point union of gear graphs under any graph labeling problems. Thus, in this paper, we aim to determine the SD-primality of the various families of one point union of gear graphs. The following results follow directly from the definitions.

Corollary 1.

Every spanning subgraph of an SD-prime graph is also an SD-prime graph.

Corollary 2.

Suppose G admits an SD-prime labeling, then G also admits a prime labeling.

The following results are obtained in [14].

Theorem 1

([14], Theorem 1). Let f be an SD-prime labeling of G, then $f\left(u\right)$ and $f\left(v\right)$ have different parity for each edge $uv\in E$.

Theorem 2

([14], Theorem 3). Let G be an SD-prime graph of order greater than 1, then G is a spanning subgraph of either $K_{m,m}$ or $K_{m,m+1}$ for some $m\ge 1$.

Theorem 3

([14], Theorem 2.2). A graph G of order n is SD-prime if and only if G is bipartite and that there exists a bijection $f:V\to [1,n]$ such that for each edge $uv$ of G, $f\left(u\right)$ and $f\left(v\right)$ are of different parity and $gcd\left(f\right(u),f(v\left)\right)=1$.

2. Preliminary

For $n\ge 2$, let $G_n$ be the graph obtained from a wheel graph $W_{2n}$ of order $2n+1$ by deleting n spokes where no two of the spokes are consecutive. Such a graph is called a gear graph (or gear, for short). The core (center vertex) of the wheel is also called the core of the gear. From ([14], Theorem 2.13) we know that $G_n$ is SD-prime for all $n\ge 2$. In the paper, we want to study the SD-primality of some one point unions of gears.

Let $H_i$ be a graph and $v_i\in V\left(H_i\right)$ be fixed, $1\le i\le t$. A one point union of $H_i$, $1\le i\le t$, is a graph obtained from the disjoint union of $H_i$ by merging all $v_i$ into a single vertex which is called the merged vertex. So, there will be many different one point unions of $H_i$’s.

There are three types of vertices in a gear. We shall say that a noncore vertex of degree 2 is of Type 1; a noncore vertex of degree 3 is of Type 2; a core vertex is of Type 3.

Let $G_{n_i}$ be a gear, $1\le i\le t$ and $t\ge 2$. Suppose $0\le k\le l\le t$. Let $v_i$ be a Type 1 vertex in $G_{n_i}$ for $1\le i\le k$ (this case does not occur if $k=0$), a Type 2 vertex for $k+1\le i\le l$ (this case does not occur if $l=k$) and a Type 3 vertex for $l+1\le i\le t$ (this case does not occur if $l=t$). The graph $G(n_1,\dots ,n_k;n_{k+1},\dots ,n_l;n_{l+1},\dots ,n_t)$ is a one point union of the gear $G_{n_i}$’s by merging all $v_i$ into a vertex, say v. When one of the sequences $n_1,\dots ,n_k$; $n_{k+1},\dots ,n_l$ or $n_{l+1},\dots ,n_t$ is empty, we will denote it by ⌀. We shall also denote $G(n_1,\dots ,n_k;n_{k+1},\dots ,n_t;⌀)$ by $G(n_1,\dots ,n_k;n_{k+1},\dots ,n_t)$ and $G(n_1,\dots ,n_t;⌀;⌀)$ by $G(n_1,\dots ,n_t)$, for short.

Theorem 4.

If $G(n_1,\cdots ,n_k;n_{k+1},\cdots ,n_l;n_{l+1},\cdots ,n_t)$ is SD-prime, then $l=k+1$ or k.

Proof. 

Clearly, G is bipartite. We color each vertex of G by white or by black so that no two adjacent vertices receive the same color. Suppose the merged vertex is white.

Consider the subgraphs $G_{n_i}$ for $1\le i\le k$. All the Type 1 and Type 3 vertices are colored by white. So, there are $n_i+1$ white vertices and $n_i$ black vertices.

Consider the subgraphs $G_{n_i}$ for $k+1\le i\le l$. All the Type 1 and Type 3 vertices are colored by black. So, there are $n_i$ white vertices and $n_i+1$ black vertices.

Consider the subgraphs $G_{n_i}$ for $l+1\le i\le t$. All the Type 1 and Type 3 vertices are colored by white. So, there are $n_i+1$ white vertices and $n_i$ black vertices.

To obtain G, we identify t white vertices as the merged vertex. Thus, there are

$$1-t+\sum _{i=1}^k(n_i+1)+\sum _{i=k+1}^l{n}_i+\sum _{i=l+1}^t(n_i+1)=(k-l+1)+\sum _{i=1}^t{n}_i\mathrm{white}\mathrm{vertices}\mathrm{and}$$

$$\sum _{i=1}^k{n}_i+\sum _{i=k+1}^l(n_i+1)+\sum _{i=l+1}^t{n}_i=l-k+\sum _{i=1}^t{n}_i\mathrm{black}\mathrm{vertices}.$$

By Theorem 2 we have $\left|2\right(l-k)-1|\le 1$. It is equivalent to $0\le 2(l-k)\le 2$. Thus, we have Theorem 4. □

Corollary 3.

If $G(n_1,\cdots ,n_k;⌀;n_{k+1},\cdots ,n_t)$ is SD-prime, then all the Type 1 and Type 3 vertices are labeled by odd numbers.

Proof. 

From the proof of Theorem 4, we know that all the Type 1 and Type 3 vertices are colored white. Since the number of white vertices is one more than that of black vertices, white vertices are labeled by odd numbers. □

Corollary 4.

If $G(n_1,\cdots ,n_k;n_{k+1};n_{k+2},\cdots ,n_t)$ is SD-prime, then the merged vertex is labeled by even numbers.

Proof. 

For this case, from the proof of Theorem 4, the number of white vertices is one less than that of black vertices. Hence, the merged vertex is labeled by even number. □

The above theorem implies that at most one Type 2 vertex can be merged to obtain an SD-prime one point union of gears. In the following sections, we shall consider some simple cases. For the gear $G_{n_i}$, denote its vertex set by

$$\begin{array}{cc}\hfill V\left(G_{n_i}\right)& =\left\{c_i\right\}\cup \left\{v_{i,j}\right|1\le j\le 2n_i\}\mathrm{and}\hfill \\ \hfill E\left(G_{n_i}\right)& =\left\{c_i{v}_{i,2k}\right|1\le k\le n_i\}\cup \left\{v_{i,j}{v}_{i,j+1}\right|1\le j\le 2n_i\},\hfill \end{array}$$

where $v_{i,2n_i+1}=v_{i,1}$.

3. Merging of Type 1 Vertices

In this section, we only consider $G(n_1,\cdots ,n_t)$, when $t=2,3$. Before the discussion, we define two sequences first.

For $l\ge 2$, we define a sequence $s_l$ of 4 elements by

$$s_l=\left\{\begin{array}{cc}(6l+1,6l+2,6l+3,6l+4)\hfill & \mathrm{if}l\equiv 1,2,3(mod5);\hfill \\ (6l+1,6l+4,6l+3,6l+2)\hfill & \mathrm{if}l\equiv 4(mod5);\hfill \\ (6l+5,6l+4,6l+3,6l+2)\hfill & \mathrm{if}l\equiv 0(mod5).\hfill \end{array}\right.$$

We also define an order pair $t_l$ by

$$t_l=\left\{\begin{array}{cc}(6l+5,6l+6)\hfill & \mathrm{if}l\not\equiv 0(mod5);\hfill \\ (6l+1,6l+6)\hfill & \mathrm{if}l\equiv 0(mod5).\hfill \end{array}\right.$$

Let $S={\left(s_l\right)}_{l\ge 2}=(s_2,s_3,\cdots ,)$ and $T={\left(t_l\right)}_{l\ge 2}$.

Lemma 1.

Let sequences S and T be defined above. Then,

(1) 

each pair of consecutive terms in S is of different parity and coprime,

(2) 

each pair of consecutive terms in T is of different parity and coprime.

Proof. 

 

(1)

Clearly, (1) is true when the two numbers come from the same subsequence $s_l$ for some l.

Suppose $l\equiv 1(mod5)$. The last term of $s_{l-1}$ is $6(l-1)+2$ and the first term of $s_l$ is $6l+1$. Now, $(6l-4,6l+1)=(6l-4,5)=(l-4,5)=1$.

Suppose $l\equiv 2,3,4(mod5)$. The last term of $s_{l-1}$ is $6(l-1)+4$ and the first term of $s_l$ is $6l+1$. Clearly, $(6l-2,6l+1)=(6l-2,3)=(1,3)=1$.

Suppose $l\equiv 0(mod5)$. The last term of $s_{l-1}$ is $6(l-1)+2$ and the first term of $s_l$ is $6l+5$. Now, $(6l-4,6l+5)=(6l-4,9)$. Since $(6l-4,3)=1$, $(6l-4,9)=1$.

(2)

Clearly, (2) is true when the two numbers come from the same subsequence $t_l$ for some l.

Suppose $l\not\equiv 0(mod5)$. The last term of $t_{l-1}$ is $6(l-1)+6$ and first term of $t_l$ is $6l+5$. Clearly, $(6l,6l+5)=(6l,5)=(l,5)=1$.

Suppose $l\equiv 0(mod5)$. The last term of $t_{l-1}$ is $6(l-1)+6$ and first term of $t_l$ is $6l+1$. Clearly, $(6l,6l+1)=(6l,1)=1$.

□

Theorem 5.

$G(n_1,n_2)$ is SD-prime for any $n_2\ge n_1\ge 2$.

Proof. 

For simplicity, we replace $v_{1,x}$ by $u_x$, and $v_{2,y}$ by $v_y$, where $1\le x\le n_1$ and $1\le y\le n_2$. Let f be a labeling of G.

When $n_1=2$ and $n_2\ge 2$, we let $f\left(c_1\right)=3$, $f\left(c_2\right)=1$, $f\left(u_1\right)=7=f\left(v_1\right)$, $f\left(u_2\right)=4$, $f\left(u_3\right)=9$, $f\left(u_4\right)=8$, $f\left(v_{2n_2-2}\right)=2$, $f\left(v_{2n_2-1}\right)=5$ and $f\left(v_{2n_2}\right)=6$. When $n_2=2$, we have Figure 1a. When $n_2\ge 3$, labels in $[10,2n_2+5]$ need to be assigned. For this case, we simply label the vertices from $v_2$ to $v_{2n_2-3}$ by 10 to $2n_2+5$ consecutively. By Theorem 3, we see that f is an SD-prime labeling. Figure 1b shows the SD-prime labeling for $G(2,3)$ according to the construction above.

Now, we assume that $n_1\ge 3$.

Case 1:

Suppose $n_1=2k$ for some $k\ge 2$. We let $f\left(c_1\right)=3$, $f\left(c_2\right)=1$, $f\left(u_1\right)=7=f\left(v_1\right)$, $f\left(u_2\right)=4$, $f\left(u_{2n_1-1}\right)=9$, $f\left(u_{2n_1}\right)=8$, $f\left(v_2\right)=10$, $f\left(v_3\right)=11$, $f\left(v_4\right)=12$, $f\left(v_{2n_2-2}\right)=2$, $f\left(v_{2n_2-1}\right)=5$ and $f\left(v_{2n_2}\right)=6$ (see Figure 2a).

Now, there are $4k-4$ and $2n_2-7$ uncolored vertices in $G_{n_1}$ and $G_{n_2}$, respectively. We label consecutively the vertices from $u_3$ to $u_{2n_1-2}$ by the first $4k-4$ terms of S and from $v_5$ to $v_{n_1+2}$ by the first $2k-2$ terms of T, respectively, where S and T are the sequences in Lemma 1.

Note that $(f\left(u_{2n_1-2}\right),f\left(u_{2n_1-1}\right))=(6k+4,9)=1$ or $(f\left(u_{2n_1-2}\right),f\left(u_{2n_1-1}\right))=(6k+2,9)=1$, and the vertex labeled by $6l+3$ is not adjacent to the core $c_1$ for every l. So, f satisfies the requirement of SD-prime labeling for $G\left(n_1\right)$. Up to now, we have used labels from 1 to $6k+6=3n_1+6$. Finally, we label the unlabeled vertices from $v_{n_1+3}$ to $v_{2n_2-3}$ by $3n_1+7$ to $2n_1+2n_2+1$ consecutively, if $2n_2\ge n_1+6$. Note that, $(f\left(v_{2n_2-3}\right),f\left(v_{2n_2-2}\right))=(2n_1+2n_2+1,2)=1$. By Theorem 3, we see that f is an SD-prime labeling of G.

For the remaining case, i.e., $2n_2\le n_1+5$. Since $n_2\ge n_1$, $n_1\le 5$. Only $n_2=n_1=4$ is a case. For this case, we have the labeling:

Case 2:

Suppose $n_1=2k+1$ for some $k\ge 1$. We let $f\left(c_1\right)=3$, $f\left(c_2\right)=1$, $f\left(u_1\right)=7=f\left(v_1\right)$, $f\left(u_2\right)=8$, $f\left(u_3\right)=5$, $f\left(u_4\right)=4$, $f\left(u_{2n_1-1}\right)=9$, $f\left(u_{2n_1}\right)=10$, $f\left(v_2\right)=6$, $f\left(v_3\right)=11$, $f\left(v_4\right)=12$ and $f\left(v_{2n_2}\right)=2$ (see Figure 2b). When $k=1$, the graph $G\left(n_1\right)$ is labeled completely. For this case, we simply label the vertices from $v_5$ to $v_{2n_2-1}$ by 13 to $2n_2+7$ consecutively.

So, we assume that $k\ge 2$. Similar to Case 1, we may label the vertices from $u_5$ to $u_{2n_1-2}$ and from $v_5$ to $v_{2n_2-1}$ using the labels in $[13,2n_1+2n_2+1]$.

Following that, we consider $G=G(n_1,n_2,n_3)$, where $n_3\ge n_2\ge n_1\ge 2$. We always merge vertices $v_{1,1}$, $v_{2,1}$ and $v_{3,1}$ as v and let $f\left(c_1\right)=3$, $f\left(c_2\right)=9$, $f\left(c_3\right)=1$ and $f\left(v\right)=7$, where f is an expected SD-prime labeling of G.

Theorem 6.

$G(2,2,n_3)$ is SD-prime for any $n_3\ge 2$.

Proof. 

The following are SD-prime labelings for $G(2,2,2)$ and $G(2,2,3)$.

For $n_3\ge 4$, the labeling f for $G_{n_3}$ is listed in the following table.

The labeling f for the two $G_2$’s are same as the two $G_2$’s in $G(2,2,3)$ (see Figure 3b).

It is easy to check that f is an SD-prime labeling. □

Theorem 7.

$G(n_1,n_2,n_3)$ is SD-prime for any $n_3\ge n_2\ge 3$, $n_1\ge 2$ and $n_1+n_2+n_3\ge 41$.

Proof. 

Similar to the proof of Theorem 5, we let $S_1$ be the sequence obtained from S by removing subsequences $s_4$ and $s_{13}$, where S is defined in Lemma 1. Observe that the subsequences $(s_3,s_5)=(19,20,21,22,35,32,33,34)$ and $(s_{12},s_{14})=(73,74,75,76,85,88,87,86)$ have adjacent terms that are coprime. Hence, each pair of consecutive terms in $S_1$ is of different parity and coprime.

Case 1:

Suppose $n_1=2k_1$ and $n_2=2k_2$ for some $k_1\ge 1$ and $k_2\ge 2$. The initial labeling for some vertices are listed as follows:

When $k_1+k_2=3$, then $G_{n_1}$ and $G_{n_2}$ are already labeled. For this case, the set of unassigned labels is $[13,24]\cup [29,78]\cup [83,2n_3+13]$. Here, we may simply label the remaining vertices of $G_{n_3}$ consecutively.

Suppose $k_1+k_2\ge 4$. We label consecutively the vertices from $v_{1,3}$ to $v_{1,2n_1-2}$ and then $v_{2,7}$ to $v_{2,2n_2-2}$ by the first $4(k_1+k_2-3)$ terms of $S_1$ and from $v_{3,5}$ to $v_{3,n_1+n_2-4}$ by the first $2(k_1+k_2-4)$ terms of T, respectively, where T is the sequence defined in Lemma 1. Note that, when $k_1+k_2=4$, no vertices of $G_{n_3}$ are labeled at this step.

After this step, let R be the set of remaining labels arranged in natural order. Namely,

$$R=\left\{\begin{array}{cc}[17,2(n_1+n_2+n_3)+1]\backslash s_{13}\hfill & \mathrm{if}{k}_1+k_2=4;\hfill \\ t_{k_1+k_2-2}\cup t_{k_1+k_2-1}\cup \hfill & \\ \cup [6k_1+6k_2+1,2(n_1+n_2+n_3)+1]\backslash s_{13}\hfill & \mathrm{if}5\le k_1+k_2\le 12;\hfill \\ t_{k_1+k_2-2}\cup t_{k_1+k_2-1}\cup t_{k_1+k_2}\cup \hfill & \\ \cup [6k_1+6k_2+7,2(n_1+n_2+n_3)+1]\hfill & \mathrm{if}{k}_1+k_2\ge 13.\hfill \end{array}\right.$$

By part (2) of Lemma 1, we see that each pair of consecutive terms in R is of different parity and coprime.

So, we may label the vertices from $v_{3,n_1+n_2-5}$ to $v_{3,2n_3-3}$.

Hence, we obtain an SD-prime labeling for this case.

Case 2:

Suppose $n_1=2k_1$ and $n_2=2k_2+1$ for some $k_1\ge 1$ and $k_2\ge 1$.

If $k_2=1$, then $n_1=2$ and $n_2=3$. The labeling is defined as follows:

Suppose $k_2\ge 2$. The initial labeling for some vertices are listed as follows:

Similar to Case 1, we label consecutively the vertices from $v_{1,3}$ to $v_{1,2n_1-2}$ and then $v_{2,5}$ to $v_{2,2n_2-2}$ by the first $4(k_1+k_2-2)$ terms of $S_1$ and from $v_{3,7}$ to $v_{3,n_1+n_2-1}$ by the first $2(k_1+k_2-3)$ terms of T, respectively. The rest is similar to Case 1.

Case 3:

Suppose $n_1=2k_1+1$ and $n_2=2k_2$ for some $k_1\ge 1$ and $k_2\ge 2$. The initial labeling for some vertices are listed as follows:

This case is similar to Case 2.

Case 4:

Suppose $n_1=2k_1+1$ and $n_2=2k_2+1$ for some $k_1\ge 1$ and $k_2\ge 1$. The initial labeling for some vertices are listed as follows:

Similar to Case 1, we label consecutively the vertices from $v_{1,5}$ to $v_{1,2n_1-2}$ and then $v_{2,5}$ to $v_{2,2n_2-2}$ by the first $4(k_1+k_2-2)$ terms of $S_1$ and from $v_{3,5}$ to $v_{3,n_1+n_2-4}$ by the first $2(k_1+k_2-3)$ terms of T, respectively. The rest is similar to Case 1.

□

Example 1.

Below is the SD-prime labeling of $G(12,14,15)$ following the construction in Case 1 of the proof of Theorem 7.

Example 2.

Below is the SD-prime labeling of $G(11,15,15)$ following the construction in Case 4 of the proof of Theorem 7.

The labeling for $G_{n_3}=G_{15}$ is the same as in Example 1.

Corollary 5.

$G(n_1,n_2,n_3)$ is SD-prime for any $n_3\ge n_2\ge 3$, $n_1\ge 2$ and $n_1+n_2+n_3=40$.

Proof. 

Basically, the proof is the same as that of Theorem 7. The difference is that 82 is not an available label. So, we have to change the initial labeling of each case in the proof of Theorem 7.

Here, we only show the first case below. Other cases are similar.

Suppose $n_1=2k_1$ and $n_2=2k_2$ for some $k_1\ge 1$ and $k_2\ge 2$. The initial labeling for some vertices are listed as follows:

The rest assignment is same as that in the proof of Theorem 7.

Note that, in all cases, 79 is the largest unassigned label. Follow the labeling shown in the proof of Theorem 7, 79 is labeled at $v_{3,2n_3-1}$. □

From Case 2 of the proof of Theorem 7 we have

Corollary 6.

$G(2,3,n_3)$ is SD-prime for any $n_3\ge 3$.

For the remaining cases, we may assume $n_1+n_2\ge 6$. Hence, $9\le n_1+n_2+n_3\le 39$.

Theorem 8.

Suppose $6\le n_1+n_2\le 8$, then $G(n_1,n_2,n_3)$ is SD-prime for $n_3\ge n_2\ge n_1\ge 2$.

Proof. 

We label $G_{n_1}$ and $G_{n_2}$ as follows:

Case 1:

$n_1+n_2=6$. We have $3\le n_3\le 33$.

When $n_3=3$. This implies that $n_1=n_2=3$. For this case, following is an SD-prime labeling for $G(3,3,3)$:

Now, we assume that $n_3\ge 4$.

When $(n_1,n_2)=(3,3)$. We label $G_{n_1}$ and $G_{n_2}$ as follows:

When $(n_1,n_2)=(2,4)$. We label $G_{n_1}$ and $G_{n_2}$ as follows:

The initial labeling for some vertices of $G_{n_3}$ are as follows:

The unassigned labels in $[22,2n_3+13]$, if any, can be labeled from $v_{3,8}$ to $v_{3,2n_3-1}$ consecutively.

Case 2:

$n_1+n_2=7$.

When $(n_1,n_2)=(3,4)$.

The labeling for $G_{n_2}$ is as in Case 1.

When $(n_1,n_2)=(2,5)$. We label $G_{n_1}$ as in Case 1 and label $G_{n_2}$ as

The initial labeling for $G_{n_3}$ is as in Case 1. The unassigned labels in $[24,2n_3+15]$ can be labeled from $v_{3,8}$ to $v_{3,2n_3-1}$ consecutively.

Case 3:

$n_1+n_2=8$.

Suppose $(n_1,n_2)=(4,4)$. If $n_3=4$, we label $G_{n_2}$ as in Case 1.

The labeling for $G_{n_1}$ and $G_{n_3}$ are as follows:

If $n_3\ge 5$, the labeling for $G_{n_1}$ is as follows:

The labeling for $G_{n_2}$ is as in Case 1.

When $(n_1,n_2)=(3,5)$. We label $G_{n_1}$ as

The labeling for $G_{n_2}$ is as in Case 2.

When $(n_1,n_2)=(2,6)$. We label $G_{n_1}$ as in Case 1 and label $G_{n_2}$ as

The initial labeling for $G_{n_3}$ is as follows:

The unassigned labels in $[28,2n_3+17]$, if any, can be labeled from $v_{3,10}$ to $v_{3,2n_3-1}$ consecutively.

It is easy to see that all above labelings are SD-prime labelings. □

Now, the remaining case is $n_1+n_2\ge 9$ and $n_1+n_2+n_3\le 39$. This implies that $n_1+n_2\le 26$ and $n_1+n_2+n_3\ge 14$.

Theorem 9.

Suppose $9\le n_1+n_2\le 26$ and $14\le n_1+n_2+n_3\le 39$. $G(n_1,n_2,n_3)$ is SD-prime for $n_3\ge n_2\ge n_1\ge 2$.

Proof. 

Let $S_2$ be the subsequence of S by removing $s_2$ and $s_4$, and $T_2$ be the subsequence of T by removing $t_2$, where S and T are defined in Lemma 1. That is, $S_2=19,20,21,22;35,34,33,32;s_6,s_7,\cdots$ and $T_2=23,24;t_4,\cdots$. Clearly, each pair of consecutive terms in $S_2$ or $T_2$ is of different parity and coprime.

Note that $n_1\le 13$ and $n_2\ge 5$

Case 1:

Suppose $n_1=2k_1$ and $n_2=2k_2$ for some $k_1\ge 1$ and $k_2\ge 3$. In this case, $5\le k_1+k_2\le 13$. The initial labeling for some vertices in $G_{n_1}$ and $G_{n_2}$ are listed as follows:

We label consecutively the vertices from $v_{2,11}$ to $v_{2,2n_2-2}$ and then $v_{1,3}$ to $v_{2,2n_1-2}$, if any, by the first $4(k_1+k_2-4)$ terms of $S_2$. It is easy to see that the last term of $s_l$ is not a multiple of 17 for $3\le l\le 12$. The maximum used label is at most $6(k_1+k_2-1)+5\le 2(n_1+n_2+n_3)-1$. So, we may label $G_{n_3}$ initially as follows:

We label consecutively the vertices from $v_{3,5}$ to $v_{3,n_1+n_2-2}$ by the first $2(k_1+k_2-3)$ terms of $T_2$. The rest is similar to Case 1 of the proof of Theorem 7. Since $n_1+n_2-2\le 2n_3-2$, $f\left(v_{3,2n_3-2}\right)=2(n_1+n_2+n_3)$. The above labeling is SD-prime if $n_1+n_2+n_3\not\equiv 1(mod3)$.

If $n_1+n_2+n_3\equiv 1(mod3)$, then $n_1+n_2+n_3\in \{16,19,22,25,28,31,34,37\}$. We swap the labels of $v_{3,2n_3-1}$ and $v_{2,9}$. One may check that the resulting labeling is SD-prime.

Case 2:

Suppose $n_1=2k_1$ and $n_2=2k_2+1$ for some $k_1\ge 1$ and $k_2\ge 2$. In this case, $4\le k_1+k_2\le 12$. The initial labeling for some vertices in $G_{n_1}$ and $G_{n_2}$ are listed as follows:

We label consecutively the vertices from $v_{2,9}$ to $v_{2,2n_2-2}$ and then $v_{1,3}$ to $v_{1,2n_1-2}$, if any, by the first $4(k_1+k_2-3)$ terms of $S_2$. Note that $n_1+1\le n_2\le n_3$. The maximum used label is at most $6(k_1+k_2)+4=3(n_1+n_2)+1\le 2(n_1+n_2+n_3)$ when $k_1+k_2\not\equiv 0(mod5)$. So, we may label $G_{n_3}$ initially as follows:

We label consecutively the vertices from $v_{3,7}$ to $v_{3,n_1+n_2-1}$ by the first $2(k_1+k_2-3)$ terms of $T_2$. The rest is similar to Case 1. Since $n_1+n_2-1\le 2n_3-2$, $f\left(v_{3,2n_3-2}\right)=2(n_1+n_2+n_3)$, the above labeling is SD-prime if $n_1+n_2+n_3\not\equiv 1(mod3)$.

When $n_1+n_2+n_3\equiv 1(mod3)$. We swap the labels of $v_{3,2n_3-1}$ and $v_{2,7}$. Similar to Case 1, one may check that the resulting labeling is SD-prime.

When $k_1+k_2\equiv 0(mod5)$, there are only two cases: $k_1+k_2=5,10$.

Suppose $k_1+k_2=5$. After labeling $G_{n_1}$ and $G_{n_2}$, the maximum used labels is 35. Since $2k_2+1\ge 2k_1$, $k_2\ge 3$, we have $n_1+3\le n_2\le n_3$ so that $2(n_1+n_2+n_3)+1\ge 37$. Hence, the labeling method is as the case when $k_1+k_2\not\equiv 0(mod5)$.

Suppose $k_1+k_2=10$. After labeling $G_{n_1}$ and $G_{n_2}$, the maximum used labels is 65. In this case, we have $2(n_1+n_2+n_3)+1\ge 65$. When $2(n_1+n_2+n_3)+1>65$, the labeling method is as the case when $k_1+k_2\not\equiv 0(mod5)$. We now deal with $2(n_1+n_2+n_3)+1=65$. It implies that $(n_1,n_2,n_3)=(10,11,11)$. Actually, it is not a case according to the previous construction so we list a required labeling as follows:

Case 3:

Suppose $n_1=2k_1+1$ and $n_2=2k_2$ for some $k_1\ge 1$ and $k_2\ge 3$. In this case, $4\le k_1+k_2\le 12$. The initial labeling for some vertices in $G_{n_1}$ and $G_{n_2}$ are listed as follows:

We label consecutively the vertices from $v_{2,7}$ to $v_{2,2n_2-2}$ and then $v_{1,5}$ to $v_{1,2n_1-2}$, if any, by the first $4(k_1+k_2-3)$ terms of $S_2$. Note that $n_1+1\le n_2\le n_3$. The rest is as Case 2.

Case 4:

Suppose $n_1=2k_1+1$ and $n_2=2k_2+1$ for some $k_1\ge 1$ and $k_2\ge 2$. In this case, $4\le k_1+k_2\le 12$. The initial labeling for some vertices in $G_{n_1}$ and $G_{n_2}$ are listed as follows:

We label consecutively the vertices from $v_{2,9}$ to $v_{2,2n_2-2}$ and then $v_{1,5}$ to $v_{1,2n_1-2}$, if any, by the first $4(k_1+k_2-3)$ terms of $S_2$. The maximum used label is at most $6(k_1+k_2)+4=3(n_1+n_2)-2\le 2(n_1+n_2+n_3)-2$. So, we may label $G_{n_3}$ initially as follows:

The rest is similar to Case 1.

□

Example 3.

Following is an SD-prime labeling of $G(4,5,7)$ according to the construction in the proof of Theorem 9:

Conjecture 1.

$G(n_1,\cdots ,n_t)$ is SD-prime for $t\ge 4$.

4. Merging of Type 1 and Exactly One Type 2 Vertices

In this section, we assume that $k=t-1$. For convenience, we use $a^{\left[n\right]}$ to denote a sequence of length n whose terms are a.

Theorem 10.

For $t\ge 2$, $G=G(3^{[t-1]};3)$ is SD-prime.

Proof. 

Without loss of generality, we assume that $v_{t,2}$ is merged with $v_{i,1}$, $1\le i\le t-1$. Now, we define a labeling f (see Figure 4) for G by

$$\begin{array}{cc}\hfill f\left(c_t\right)& =7,f\left(v_{t,j}\right)=j\mathrm{for}1\le j\le 6;\mathrm{and}\hfill \\ \hfill f\left(c_i\right)& =6i+4,f\left(v_{i,1}\right)=2,f\left(v_{i,2}\right)=6i+3,f\left(v_{i,3}\right)=6i+2,\hfill \\ \hfill f\left(v_{i,j}\right)& =6i+j+1\mathrm{for}1\le i\le t-1,j=4,5,6.\hfill \end{array}$$

It is easy to check that f is an SD-prime labeling. □

Theorem 11.

For $a,b\ge 2$, $G=G(a;b)$ is SD-prime.

Proof. 

Same as the proof of Theorem 5, we use $u_x$ to replace $v_{1,x}$ and $v_y$ to replace $v_{2,y}$, where $1\le x\le a$ and $1\le y\le b$. Suppose $u_1$ is merged with $v_2$. Let f be a labeling of G. We always set $f\left(c_1\right)=2$, $f\left(c_2\right)=1$ and $f\left(u_1\right)=4=f\left(v_2\right)$. We assign the labels to other vertices as follows.

Case 1:

Suppose $b\not\equiv 2(mod3)$. Let

$$\begin{array}{cc}\hfill f\left(u_i\right)& =2b+i+1\mathrm{for}2\le i\le 2a,\hfill \\ \hfill f\left(v_1\right)& =3,f\left(v_j\right)=j+2\mathrm{for}3\le j\le 2b.\hfill \end{array}$$

Here, $(f\left(v_1\right),f\left(v_{2b}\right))=(3,2b+2)=1$ since $b\not\equiv 2(mod3)$.

Case 2:

Suppose $a=2=b$. Let

$$\begin{array}{cc}\hfill f\left(u_2\right)& =9,f\left(u_3\right)=8,f\left(u_4\right)=3,\hfill \\ \hfill f\left(v_1\right)& =5,f\left(v_3\right)=7,f\left(v_4\right)=6.\hfill \end{array}$$

Case 3:

Suppose $a=2$ and $b\equiv 2(mod3)$, $b\ge 5$. Let

$$\begin{array}{cc}\hfill f\left(u_2\right)& =3,f\left(u_3\right)=8,f\left(u_4\right)=9,\hfill \\ \hfill f\left(v_3\right)& =5,f\left(v_4\right)=6,f\left(v_5\right)=7,\hfill \\ \hfill f\left(v_1\right)& =2b+5,f\left(v_j\right)=j+4\mathrm{for}6\le j\le 2b.\hfill \end{array}$$

Case 4:

Suppose $a=3$ and $b\equiv 2(mod3)$. Let

$$\begin{array}{cc}\hfill f\left(u_2\right)& =3,f\left(u_3\right)=8,f\left(u_4\right)=5,f\left(u_5\right)=6,f\left(u_6\right)=7,\hfill \\ \hfill f\left(v_1\right)& =2b+7,f\left(v_j\right)=j+6\mathrm{for}3\le j\le 2b.\hfill \end{array}$$

Case 5:

Suppose $a\ge 4$ and $b\equiv 2(mod3)$. Let

$$\begin{array}{cc}\hfill f\left(u_2\right)& =3,f\left(u_{2a}\right)=9,f\left(u_{2a-1}\right)=10,f\left(u_{2a-2}\right)=7,\hfill \\ \hfill f\left(u_{2a-3}\right)& =6,f\left(u_{2a-4}\right)=5,f\left(u_{2a-5}\right)=8,\hfill \\ \hfill f\left(u_i\right)& =2b+i+7\mathrm{for}3\le i\le 2a-6(\mathrm{this}\mathrm{line}\mathrm{can}\mathrm{be}\mathrm{ignored}\mathrm{when}a=4),\hfill \\ \hfill f\left(v_1\right)& =9+2b,f\left(v_j\right)=j+8\mathrm{for}3\le j\le 2b.\hfill \end{array}$$

Now, $(f\left(u_2\right),f\left(u_3\right))=(3,2b+10)=(3,2)=1$.

It is easy to check the coprimality of other adjacent vertices. By Theorem 3, f is an SD-prime labeling of G. □

Theorem 12.

For $b\ge 2$, $G=G(2,2;b)$ is SD-prime.

Proof. 

When $b=2,3$, we have the following SD-prime labelings:

For $b\ge 4$, we merge vertices $v_{1,1}$, $v_{2,1}$ and $v_{3,2}$ and let $f\left(v_{1,1}\right)=f\left(v_{2,1}\right)=f\left(v_{3,2}\right)=8$, $f\left(c_1\right)=4$, $f\left(c_2\right)=2$ and $f\left(c_3\right)=1$.

We define the labeling f by the following table.

The labeling of the two $G_2$’s are same as shown in Figure 5b.

Clearly, f is an SD-prime labeling of $G(2,2;b)$. □

Theorem 13.

For $n_2\ge n_1$, $G=G(n_1,n_2;2)$ is SD-prime.

Proof. 

When $n_2=2$. We have proved in Theorem 12. So, we assume that $n_2\ge 3$. We merge vertices $v_{1,1}$, $v_{2,1}$ and $v_{3,2}$ and let $f\left(v_{1,1}\right)=f\left(v_{2,1}\right)=f\left(v_{3,2}\right)=8$, $f\left(c_1\right)=4$, $f\left(c_2\right)=2$ and $f\left(c_3\right)=1$.

Case 1:

Suppose $n_1\not\equiv 0(mod3)$. We define the labeling f by the following table.

Since $n_1\not\equiv 0(mod3)$, we have $2n_1+12\equiv 1,2(mod3)$. So, $(3,2n_1+12)=1$ and $(f\left(v_{2,6}\right),f\left(v_{2,7}\right))=1$ when $n_2\ge 4$.

Case 2:

Suppose $n_1\equiv 0(mod3)$. We list the labeling as in the following table.

Clearly, $(f\left(v_{2,4}\right),f\left(v_{2,5}\right))=(3,2n_1+10)=(3,10)=1$.

By Theorem 3, f is an SD-prime labeling of G for each case. □

Example 4.

By the construction from the proof of Case 1 above, we have an SD-prime labeling of $G(2,3;2)$ (Figure 6):

Theorem 14.

For $n_2\ge n_1$, $n_2\ge 3$ and $b\ge 3$, $G=G(n_1,n_2;b)$ is SD-prime.

Proof. 

We always merge vertices $v_{1,1}$, $v_{2,1}$ and $v_{3,2}$ and let $f\left(v_{1,1}\right)=f\left(v_{2,1}\right)=f\left(v_{3,2}\right)=8$, $f\left(c_1\right)=4$, $f\left(c_2\right)=2$ and $f\left(c_3\right)=1$.

Case 1: Suppose $b\not\equiv 0(mod3)$ and $b+n_1+1\not\equiv 0(mod3)$. This implies that $b\ge 4$. We define the labeling f by the following table.

Clearly, f satisfies the last two conditions of Theorem 3 for all vertices in $G_b$. Since $b\not\equiv 0(mod3)$, $(3,2b+12)=(3,2b)=1$. Thus, $(f\left(v_{1,4}\right),f\left(v_{1,5}\right))=(9,2b+12)=1$. Since $b+n_1+1\not\equiv 0(mod3)$, $(f\left(v_{2,6}\right),f\left(v_{2,7}\right))=(3,2b+2n_1+8)=(3,2(b+n_1+1))=1$.

Suppose $b\not\equiv 0(mod3)$ and $b+n_1+1\equiv 0(mod3)$. This implies that $b\ge 4$ and $n_1\ge 3$. We define the labeling f by the following table.

Same as the Case 1, f satisfies the last two conditions of Theorem 3 for all vertices in $G_b$ and $G_{n_1}$. Since $b+n_1\equiv 2(mod3)$, $(f\left(v_{2,4}\right),f\left(v_{2,5}\right))=(3,2b+2n_1+6)=1$.

Suppose $b\equiv 0(mod3)$ and $n_1\not\equiv 2(mod3)$. This implies that $n_1\ge 3$. We define the labeling f by the following table.

Since $b\equiv 0(mod3)$, $(3,2b+14)=(3,14)=1$. Thus, $(f\left(v_{1,6}\right),f\left(v_{1,7}\right))=(9,2b+14)=1$. Since $n_1\not\equiv 2(mod3)$, $(f\left(v_{2,6}\right),f\left(v_{2,7}\right))=(3,2b+2n_1+8)=(3,2n_1-1))=1$. Others are similar to the previous cases, we omit the arguments.

Case 2:

Suppose $b\equiv 0(mod3)$ and $n_1\equiv 2(mod3)$. If $n_2=3$, then $n_1=2$. We have the following labeling.

We define the labeling f by the following table for $n_2\ge 4$.

Since $n_1\equiv 2(mod3)$, $(f\left(v_{2,8}\right),f\left(v_{2,9}\right))=(3,2b+2n_1+10)=(3,5)=1$. If others are similar to the previous cases, we omit the arguments.

By Theorem 3, f is an SD-prime labeling of G for each case. □

Conjecture 2.

$G(n_1,\cdots ,n_k;b)$ is SD-prime, where $k\ge 3$ and $b\ge 2$.

5. Merging of Type 3 Vertices

We study the case $k=l=0$ first. We shall obtain some ad hoc results.

Theorem 15.

For $n\ge 2$, $G=G(⌀;⌀;3^{\left[n\right]})$ is SD-prime.

Proof. 

Let f be a labeling of G defined by $f\left(c_i\right)=1$ and

for $1\le i\le n$. Now, $(f\left(v_{i,1}\right),f\left(v_{i,6}\right))=(6i-4,6i-1)=(6i-4,3)=(-4,3)=1$ and $(f\left(v_{i,4}\right),f\left(v_{i,3}\right))=(6i+1,6i-2)=(3,6i-2)=(3,-2)=1$. Hence, by Theorem 3, f is an SD-prime labeling of G. □

Theorem 16.

For $2\le m\le 5$, $G=G(⌀;⌀;4^{\left[m\right]})$ is SD-prime.

Proof. 

Let f be a labeling of G defined by

$f\left(c_i\right)=1$ and $f\left(v_{i,j}\right)=1+j+8(i-1)$ for $1\le i\le m$, $1\le j\le 8$. In this case,

$$(f\left(v_{i,1}\right),f\left(v_{i,8}\right))=(8i-6,8i+1)=(8i-6,7)=(i-6,7)=1.$$

So, we have the theorem. □

Remark 1.

If we extend the labeling f in the proof of Theorem 16 to $m=6$, then $f\left(v_{6,8}\right)=49$. Here, $(f\left(v_{6,8}\right),f\left(v_{6,1}\right))=(49,42)=7$. For this case, we may swap the labels of $v_{6,8}$ and $v_{1,2}$. Then, f is an SD-prime labeling of $G(⌀;⌀;4^{\left[6\right]})$. Now, by the same way, we may extend f up to $m=12$. For the 13-th $G_4$, we may use the same f and swap the labels of $v_{13,8}$ and a suitable labeled vertex. So, that we believe that f can be extended to $G(⌀;⌀;4^{\left[13\right]})$ and then to $G(⌀;⌀;4^{\left[19\right]})$, and so on. So, we make the following conjecture.

Conjecture 3.

$G(⌀;⌀;4^{\left[m\right]})$ is SD-prime for $m\ge 13$.

Theorem 17.

Let v be the merged vertex of $H=G(n_1,\cdots ,n_k;n_{k+1})$ with the following conditions, where $k\ge 1$,

(1)

${\displaystyle \sum _{i=1}^{k+1}}{n}_i\equiv 0(mod3)$,

(2)

there is an SD-prime labeling h of H such that $h\left(v\right)=2^a$ for some $a\ge 1$.

Then, $G=G(n_1,\cdots ,n_k;n_{k+1};3^{\left[n\right]})$ is SD-prime, for $n\ge 1$.

Proof. 

Let $N=1+2{\displaystyle \sum _{i=1}^{k+1}}{n}_i$. By the assumption, all cores of $G_3$’s are merged to v. Now, we want to extend h to G by assigning labels in $[N+1,N+6n]$. Let the vertices of the i-th $G_3$ be $x_{i,j}$, $1\le i\le n$, $1\le j\le 6$. For $1\le i\le n$, let

Then, $(h\left(x_{i,5}\right),h\left(x_{i,6}\right))=(6i+N-2,6i+N-5)=(3,6i+N-5)=(3,2)=1$ and $(h\left(x_{i,3}\right),h\left(x_{i,2}\right))=(6i+N,6i+N-3)=(3,6i+N-3)=(3,1)=1$. Hence, h is extended to be an SD-prime labeling of $G(n_1,\cdots ,n_k;n_{k+1};3^{\left[n\right]})$ for $n\ge 1$. □

Theorem 18.

For $n\ge 1$ and $1\le k\le 4$, $G=G(⌀;⌀;3^{\left[n\right]},4^{\left[3k\right]})$ is SD-prime.

Proof. 

From Remark 1, we label $G(⌀;⌀;4^{\left[3k\right]})$ first, $1\le k\le 4$. Note that the merged vertex is labeled by 1. Then, shift the original labeling f defined in the proof of Theorem 15 by $24k$. By the same argument in the proof of Theorem 15, we have the theorem. □

To illustrate the proof of the above theorem, we provide the following example.

Example 5.

Consider $G=G(⌀;⌀;3^{\left[n\right]},4^{\left[3\right]})$, where $n\ge 1$. We use the labeling defined in the proof of Theorem 16 to label the three $G_4$’s. Here, labels in $[1,25]$ are occupied.

Let $v_{i,j}$ be vertex of the i-th $G_3$, $1\le i\le n$. We let

Now, $(f\left(v_{i,1}\right),f\left(v_{i,6}\right))=(6i+20,6i+23)=(6i+20,3)=(20,3)=1$ and $(f\left(v_{i,4}\right),f\left(v_{i,3}\right))=(6i+25,6i+22)=(3,6i+22)=(3,22)=1$. Hence, by Theorem 3, f is an SD-prime labeling of G.

6. Conclusions

In this paper, to consider SD-prime labelings for one point union of gear graphs is an initial work. We have found SD-prime labelings for a few one point union of gear graphs. Besides the two conjectures proposed in Section 3 and Section 4, there are many other cases that we can consider. For example, one may consider $G(n_1,\cdots ,n_k;⌀;n_{k+1})$ when $n_{k+1}\ge 5$; or $G(n_1,\cdots ,n_k;n_{k+1};n_{k+2})$, where $n_i\ge 2$.