# Graphs with Strong Proper Connection Numbers and Large Cliques

^{*}

## Abstract

**:**

## 1. Introduction

## 2. (Strong) Proper Connection and Clique Number

**Proposition 2**

**([12]).**

**Proposition 3**

**([18]).**

**Lemma 1.**

**Theorem 1.**

**Proof.**

**Theorem 2.**

- (i)
- $4\le n\le 5$, $G[V\left(G\right)\backslash V\left(Q\right)]\cong 2{K}_{1}$ and ${N}_{Q}\left({u}_{1}\right)={N}_{Q}\left({u}_{2}\right)=\left\{v\right\}$.
- (ii)
- $n\ge 6$, $G[V\left(G\right)\backslash V\left(Q\right)]\cong 2{K}_{1}$ and ${N}_{Q}\left({u}_{1}\right)={N}_{Q}\left({u}_{2}\right)=\left\{v\right\}$.

**Proof.**

**Case 1.**$F\cong {K}_{2}$. Since G is connected, it follows that $max\{{d}_{Q}\left({u}_{1}\right),{d}_{Q}\left({u}_{2}\right)\}\ge 1$. Note that G is traceable, and we have $pc\left(G\right)=2$ by Proposition 1. The following edge-coloring $\theta $ with two colors makes G strongly properly connected: color ${u}_{1}{u}_{2}$ and all edges of $E\left(Q\right)$ with 1, and color all edges of $E\left[V\right(F),V(Q\left)\right]$ with 2. Thus, $spc\left(G\right)=2$.

**Case 2.**$F\cong 2{K}_{1}$. Since G is connected, it follows that $min\{{d}_{Q}\left({u}_{1}\right),{d}_{Q}\left({u}_{2}\right)\}\ge 1$. Assume that ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)=\xd8$. Observe that G is traceable, and we have $pc\left(G\right)=2$ by Proposition 1. Assign an edge-coloring $\theta $ with two colors to G as follows: color all edges of $E\left(Q\right)$ with 1 and all edges of $E\left[V\right(F),V(Q\left)\right]$ with 2. It is clear that G is strongly properly connected with the above edge-coloring. Hence, $spc\left(G\right)=2$.

**Theorem 3.**

- (i)
- $G[V\left(G\right)\backslash V\left(Q\right)]\cong {P}_{3}$, where ${u}_{1}{u}_{2},{u}_{2}{u}_{3}\in E\left(G\right)$, ${d}_{Q}\left({u}_{2}\right)=0$, $min\{{d}_{Q}\left({u}_{1}\right),{d}_{Q}\left({u}_{3}\right)\}=1$, ${N}_{Q}\left({u}_{1}\right)\cup {N}_{Q}\left({u}_{3}\right)=V\left(Q\right)$ and ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{3}\right)=\xd8$.
- (ii)
- $n=6$, $G[V\left(G\right)\backslash V\left(Q\right)]\cong {K}_{2}+{K}_{1}$, where ${u}_{1}{u}_{2}\in E\left(G\right)$, and ${N}_{Q}\left({u}_{1}\right)={N}_{Q}\left({u}_{2}\right)={N}_{Q}\left({u}_{3}\right)=\left\{v\right\}$.
- (iii)
- $n\ge 7$, $G[V\left(G\right)\backslash V\left(Q\right)]\cong {K}_{2}+{K}_{1}$, where ${u}_{1}{u}_{2}\in E\left(G\right)$, and ${N}_{Q}\left({u}_{1}\right)={N}_{Q}\left({u}_{2}\right)={N}_{Q}\left({u}_{3}\right)=\left\{v\right\}$.
- (iv)
- $G[V\left(G\right)\backslash V\left(Q\right)]\cong {K}_{2}+{K}_{1}$, where ${u}_{1}{u}_{2}\in E\left(G\right)$, ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right)=\left\{v\right\}$, $min\{{d}_{Q}\left({u}_{1}\right),{d}_{Q}\left({u}_{2}\right)\}={d}_{Q}\left({u}_{3}\right)=1$ and ${d}_{Q}\left({u}_{1}\right)+{d}_{Q}\left({u}_{2}\right)\ge 3$.
- (v)
- $G[V\left(G\right)\backslash V\left(Q\right)]\cong {K}_{2}+{K}_{1}$, where ${u}_{1}{u}_{2}\in E\left(G\right)$, ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{3}\right)\ne \xd8$, ${N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right)\ne \xd8$, ${d}_{Q}\left({u}_{1}\right)={d}_{Q}\left({u}_{2}\right)=1$, ${d}_{Q}\left({u}_{3}\right)=2$ and ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)=\xd8$.
- (vi)
- $n=5$, $G[V\left(G\right)\backslash V\left(Q\right)]\cong 3{K}_{1}$ and ${N}_{Q}\left({u}_{1}\right)={N}_{Q}\left({u}_{2}\right)={N}_{Q}\left({u}_{3}\right)=\left\{v\right\}$.
- (vii)
- $n\ge 6$, $G[V\left(G\right)\backslash V\left(Q\right)]\cong 3{K}_{1}$ and ${N}_{Q}\left({u}_{1}\right)={N}_{Q}\left({u}_{2}\right)={N}_{Q}\left({u}_{3}\right)=\left\{v\right\}$.
- (viii)
- $G[V\left(G\right)\backslash V\left(Q\right)]\cong 3{K}_{1}$, $|({N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right))\cup ({N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{3}\right))\cup ({N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}$$\left({u}_{3}\right)\left)\right|=1$, ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right)\ne \xd8$ and ${d}_{Q}\left({u}_{1}\right)+{d}_{Q}\left({u}_{2}\right)+{d}_{Q}\left({u}_{3}\right)\ge 4$.
- (ix)
- $G[V\left(G\right)\backslash V\left(Q\right)]\cong 3{K}_{1}$, $|({N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right))\cup ({N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{3}\right))\cup ({N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}$$\left({u}_{3}\right)\left)\right|=2$, ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right)\ne \xd8$ and ${d}_{Q}\left({u}_{1}\right)+{d}_{Q}\left({u}_{2}\right)+{d}_{Q}\left({u}_{3}\right)=5$.

**Proof.**

**Case 1.**$F\cong {K}_{3}$. Observe that G is traceable, and so $pc\left(G\right)=2$ by Proposition 1. The following edge-coloring $\theta $ with two colors induces a strong proper coloring of G: color all edges of $E\left(F\right)$ and $E\left(Q\right)$ with 1, and color all edges of $E\left[V\right(F),V(Q\left)\right]$ with 2. Thus, $spc\left(G\right)=2$.

**Case 2.**$F\cong {P}_{3}$, where ${u}_{1}{u}_{2},{u}_{2}{u}_{3}\in E\left(G\right)$. Assume that $min\{{d}_{Q}\left({u}_{1}\right),{d}_{Q}\left({u}_{2}\right),{d}_{Q}\left({u}_{3}\right)\}\ge 1$. Note that G is traceable, and we have $pc\left(G\right)=2$ by Proposition 1. Assign a strong proper coloring $\theta $ with two colors to G as follows: $\theta \left({u}_{1}{u}_{2}\right)=1$; $\theta \left({u}_{2}{u}_{3}\right)=2$; and color all edges of $E\left(Q\right)$ with 1 and all edges of $E\left[V\right(F),V(Q\left)\right]$ with 2. Hence, $spc\left(G\right)=2$.

**Case 3.**$F\cong {K}_{2}+{K}_{1}$, where ${u}_{1}{u}_{2}\in E\left(G\right)$. Since G is connected, we obtain ${d}_{Q}\left({u}_{3}\right)\ge 1$. We distinguish the following three subcases.

**Subcase 3.1.**${d}_{Q}\left({u}_{3}\right)=1$. Let ${N}_{Q}\left({u}_{3}\right)=\left\{v\right\}$. Since $diam\left(G\right)=2$, we have ${d}_{Q}\left({u}_{1}\right)\ge 1$, ${d}_{Q}\left({u}_{2}\right)\ge 1$ and $v\in {N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)$. Assume that ${d}_{Q}\left({u}_{1}\right)={d}_{Q}\left({u}_{2}\right)=1$. This implies $n\ge 6$. If $n=6$, then $G\cong {G}_{2}$, where ${G}_{2}$ is displayed in Figure 1. Thus, $pc\left(G\right)=spc\left(G\right)=3$. Now we consider $n\ge 7$. Let $V\left(Q\right)=\{{w}_{1},{w}_{2},\dots ,{w}_{n-4},v\}$. Assign an edge-coloring $\theta $ with two colors to G as follows: $\theta \left({u}_{1}{u}_{2}\right)=\theta \left({u}_{1}v\right)=\theta \left({u}_{2}v\right)=\theta \left({w}_{i}v\right)=\theta \left({w}_{j}{w}_{n-4}\right)=1$ for $1\le i\le n-5$ and $2\le j\le n-5$, $\theta \left({u}_{3}v\right)=\theta \left({w}_{n-4}v\right)=\theta \left({w}_{1}{w}_{n-4}\right)=\theta \left({w}_{1}{w}_{2}\right)=2$, and color the remaining edges arbitrarily with 1 and 2. It is easy to verify that $\theta $ is a proper-path coloring of G. Thus, $pc\left(G\right)=2$. If G is strongly properly connected with an edge-coloring $\theta $, then $\theta \left({u}_{1}v\right)\ne \theta \left({u}_{3}v\right)\ne \theta \left({w}_{1}v\right)$, and so $spc\left(G\right)\ge 3$. Assign an edge-coloring ${\theta}^{\prime}$ with three colors to G as follows: ${\theta}^{\prime}\left({u}_{1}{u}_{2}\right)={\theta}^{\prime}\left({u}_{1}v\right)={\theta}^{\prime}\left({u}_{2}v\right)=1$, ${\theta}^{\prime}\left({u}_{3}v\right)=2$, and color all edges of $E\left(Q\right)$ with 3. We can check that G is strongly properly connected with the above edge-coloring. Hence, $spc\left(G\right)=3$.

**Subcase 3.2.**${d}_{Q}\left({u}_{3}\right)=2$. Let ${N}_{Q}\left({u}_{3}\right)=\{u,v\}$. Since $diam\left(G\right)=2$, we obtain ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{3}\right)\ne \xd8$ and ${N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right)\ne \xd8$. Observe that G is traceable, and we have $pc\left(G\right)=2$ by Proposition 1.

**Subcase 3.3.**${d}_{Q}\left({u}_{3}\right)\ge 3$. Note that G is traceable, and we obtain $pc\left(G\right)=2$ by Proposition 1. Assume that ${d}_{Q}\left({u}_{1}\right)={d}_{Q}\left({u}_{2}\right)=1$ and ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)=\xd8$. Let ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{3}\right)=\left\{u\right\}$ and ${N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right)=\left\{v\right\}$. Assign a strong proper coloring $\theta $ with two colors to G as follows: $\theta \left({u}_{1}{u}_{2}\right)=\theta \left({u}_{1}u\right)=\theta \left({u}_{2}v\right)=1$, $\theta \left({u}_{3}u\right)=\theta \left({u}_{3}v\right)=2$, and color all edges of $E\left(Q\right)$ with 2 and the remaining edges with 1. Thus, $spc\left(G\right)=2$.

**Case 4.**$F\cong 3{K}_{1}$. Since $diam\left(G\right)=2$, it follows that ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)\ne \xd8$, ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{3}\right)\ne \xd8$ and ${N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right)\ne \xd8$. This case is demonstrated by the following three subcases.

**Subcase 4.1.**$|({N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right))\cup ({N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{3}\right))\cup ({N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right))|=1$. This implies that $|{N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right)|=1$. Let ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right)=\left\{v\right\}$. Assume that ${d}_{Q}\left({u}_{1}\right)={d}_{Q}\left({u}_{2}\right)={d}_{Q}\left({u}_{3}\right)=1$. Then, $spc\left(G\right)\ge pc\left(G\right)\ge 3$ by Proposition 2. If $n=5$, then $G\cong {K}_{1,4}$. Hence, $pc\left(G\right)=spc\left(G\right)=4$. Now we consider $n\ge 6$. Let $V\left(Q\right)=\{v,{w}_{1},{w}_{2},\dots ,{w}_{n-4}\}$. Assign an edge-coloring $\theta $ with three colors to G as follows: $\theta \left({u}_{1}v\right)=1$; $\theta \left({u}_{2}v\right)=2$; $\theta \left({u}_{3}v\right)=3$; $\theta \left({w}_{1}{w}_{n-4}\right)=3$ if n is even, $\theta \left({w}_{1}{w}_{n-4}\right)=2$ if n is odd; color the sequence ${w}_{1}v{w}_{2}{w}_{3}\cdots {w}_{n-4}$ alternately with 1 and 2 starting with $\theta \left({w}_{1}v\right)=1$; and color the remaining edges arbitrarily with 1 and 2. It is not difficult to check that $\theta $ is a proper-path coloring of G. Thus, $pc\left(G\right)=3$. Suppose G has a strong proper coloring $\theta $, we have $\theta \left({u}_{1}v\right)\ne \theta \left({u}_{2}v\right)\ne \theta \left({u}_{3}v\right)\ne \theta \left({w}_{1}v\right)$, and so $spc\left(G\right)\ge 4$. On the other hand, there exists a strong proper coloring ${\theta}^{\prime}$ of G with four colors, as follows: ${\theta}^{\prime}\left({u}_{1}v\right)=1$, ${\theta}^{\prime}\left({u}_{2}v\right)=2$, ${\theta}^{\prime}\left({u}_{3}v\right)=3$, and color all edges of $E\left(Q\right)$ with 4. Therefore, we have $spc\left(G\right)=4$.

**Subcase 4.2.**$|({N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right))\cup ({N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{3}\right))\cup ({N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right))|=2$. Since $diam\left(G\right)=2$, we obtain ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right)\ne \xd8$, and say $v\in {N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right)$. Without a loss of generality, we consider $|{N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)|=2$, and say $u\in ({N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right))\backslash \left\{v\right\}$. Assign an analogous edge-coloring to that presented in Subcase 4.1 to G that satisfies ${d}_{Q}\left({u}_{1}\right)+{d}_{Q}\left({u}_{2}\right)+{d}_{Q}\left({u}_{3}\right)\ge 4$. Obviously, G is properly connected, and so $pc\left(G\right)=2$.

**Subcase 4.3.**$|({N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right))\cup ({N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{3}\right))\cup ({N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right))|\ge 3$, and let $\{{w}_{1},{w}_{2},{w}_{3}\}\subseteq ({N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right))\cup ({N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{3}\right))\cup ({N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right))$. Up to isomorphism, we only need to consider the following two cases.

**Theorem 4.**

- (i)
- $G[V\left(G\right)\backslash V\left(Q\right)]\cong {P}_{3}$, where ${u}_{1}{u}_{2},{u}_{2}{u}_{3}\in E\left(G\right)$, ${d}_{Q}\left({u}_{1}\right)={d}_{Q}\left({u}_{3}\right)=0$, and ${d}_{Q}\left({u}_{2}\right)=1$.
- (ii)
- $G[V\left(G\right)\backslash V\left(Q\right)]\cong {P}_{3}$, where ${u}_{1}{u}_{2},{u}_{2}{u}_{3}\in E\left(G\right)$, ${d}_{Q}\left({u}_{1}\right)={d}_{Q}\left({u}_{3}\right)=0$, and ${d}_{Q}\left({u}_{2}\right)\ge 2$.
- (iii)
- $G[V\left(G\right)\backslash V\left(Q\right)]\cong {P}_{3}$, where ${u}_{1}{u}_{2},{u}_{2}{u}_{3}\in E\left(G\right)$, $min\{{d}_{Q}\left({u}_{1}\right),{d}_{Q}\left({u}_{3}\right)\}=1$, ${d}_{Q}\left({u}_{2}\right)=0$, ${N}_{Q}\left({u}_{1}\right)\cup {N}_{Q}\left({u}_{3}\right)\ne V\left(Q\right)$, and ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{3}\right)=\xd8$.
- (iv)
- $5\le n\le 6$, $G[V\left(G\right)\backslash V\left(Q\right)]\cong {K}_{2}+{K}_{1}$, where ${u}_{1}{u}_{2}\in E\left(G\right)$, ${N}_{Q}\left({u}_{1}\right)={N}_{Q}\left({u}_{3}\right)=\left\{v\right\}$, and ${d}_{Q}\left({u}_{2}\right)=0$.
- (v)
- $n\ge 7$, $G[V\left(G\right)\backslash V\left(Q\right)]\cong {K}_{2}+{K}_{1}$, where ${u}_{1}{u}_{2}\in E\left(G\right)$, ${N}_{Q}\left({u}_{1}\right)={N}_{Q}\left({u}_{3}\right)=\left\{v\right\}$, and ${d}_{Q}\left({u}_{2}\right)=0$.
- (vi)
- $G[V\left(G\right)\backslash V\left(Q\right)]\cong {K}_{2}+{K}_{1}$, where ${u}_{1}{u}_{2}\in E\left(G\right)$, ${d}_{Q}\left({u}_{2}\right)\ge 1$, ${N}_{Q}\left({u}_{1}\right)={N}_{Q}\left({u}_{3}\right)=\left\{v\right\}$, ${N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right)=\xd8$, and ${N}_{Q}\left({u}_{1}\right)\cup {N}_{Q}\left({u}_{2}\right)\ne V\left(Q\right)$.
- (vii)
- $5\le n\le 6$, $G[V\left(G\right)\backslash V\left(Q\right)]\cong 3{K}_{1}$, ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right)=\xd8$, $|{N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)|=1$, and ${d}_{Q}\left({u}_{1}\right)={d}_{Q}\left({u}_{2}\right)={d}_{Q}\left({u}_{3}\right)=1$.
- (viii)
- $n=6$, $G[V\left(G\right)\backslash V\left(Q\right)]\cong 3{K}_{1}$, ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right)=\xd8$, $|{N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)|=1$, ${d}_{Q}\left({u}_{1}\right)={d}_{Q}\left({u}_{2}\right)=1$, and ${d}_{Q}\left({u}_{3}\right)=2$.
- (ix)
- $n\ge 7$, $G[V\left(G\right)\backslash V\left(Q\right)]\cong 3{K}_{1}$, ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right)=\xd8$, $|{N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)|=1$, ${d}_{Q}\left({u}_{1}\right)={d}_{Q}\left({u}_{2}\right)=1$, and ${d}_{Q}\left({u}_{3}\right)\ge 1$.
- (x)
- $G[V\left(G\right)\backslash V\left(Q\right)]\cong 3{K}_{1}$, ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right)=\xd8$, $|{N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)|=1$, ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{3}\right)\ne \xd8$, ${d}_{Q}\left({u}_{1}\right)=2$, and ${d}_{Q}\left({u}_{2}\right)={d}_{Q}\left({u}_{3}\right)=1$.

**Proof.**

**Case 1.**$diam\left(G\right)=3$. We distinguish the following four subcases by analyzing the structure of F.

**Subcase 1.1.**$F\cong {K}_{3}$. Note that G is traceable, and we have $pc\left(G\right)=2$ by Proposition 1. Assign an edge-coloring $\theta $ with two colors to G as follows: color all edges of $E\left(F\right)$ and $E\left(Q\right)$ with 1, and color all edges of $E\left[V\right(F),V(Q\left)\right]$ with 2. It is obvious that $\theta $ is a strong proper coloring of G, and so $spc\left(G\right)=2$.

**Subcase 1.2.**$F\cong {P}_{3}$, where ${u}_{1}{u}_{2},{u}_{2}{u}_{3}\in E\left(G\right)$. Assume that ${d}_{Q}\left({u}_{1}\right)={d}_{Q}\left({u}_{3}\right)=0$. Suppose ${d}_{Q}\left({u}_{2}\right)=1$, and let ${N}_{Q}\left({u}_{2}\right)=\left\{v\right\}$. Then, $spc\left(G\right)\ge pc\left(G\right)\ge 3$ by Proposition 2. Now we define a strong proper coloring $\theta $ of G with three colors as follows: $\theta \left({u}_{1}{u}_{2}\right)=1$, $\theta \left({u}_{2}{u}_{3}\right)=2$, $\theta \left({u}_{2}v\right)=3$, and color all edges of $E\left(Q\right)$ with 1. Thus, $pc\left(G\right)=spc\left(G\right)=3$. Suppose ${d}_{Q}\left({u}_{2}\right)\ge 2$, and let $u,v\in {N}_{Q}\left({u}_{2}\right)$. Assign an edge-coloring $\theta $ with two colors to G as follows: $\theta \left({u}_{1}{u}_{2}\right)=\theta \left({u}_{2}u\right)=\theta \left(vw\right)=1$ for any $w\in V\left(Q\right)\backslash \{u,v\}$, $\theta \left({u}_{2}{u}_{3}\right)=\theta \left({u}_{2}v\right)=\theta \left(uv\right)=\theta \left(uw\right)=2$ for any $w\in V\left(Q\right)\backslash \{u,v\}$, and color the remaining edges arbitrarily with 1 and 2. We can check that G is properly connected with the above edge-coloring, and so $pc\left(G\right)=2$. If G is strongly properly connected with an edge-coloring $\theta $, then $\theta \left({u}_{1}{u}_{2}\right)\ne \theta \left({u}_{2}{u}_{3}\right)\ne \theta \left({u}_{2}u\right)$. Thus, $spc\left(G\right)\ge 3$. Assign a strong proper coloring ${\theta}^{\prime}$ with three colors to G as follows: ${\theta}^{\prime}\left({u}_{1}{u}_{2}\right)=1$, ${\theta}^{\prime}\left({u}_{2}{u}_{3}\right)=2$, and color all edges of $E\left[V\right(F),V(Q\left)\right]$ with 3 and all edges of $E\left(Q\right)$ with 1. Thus, $spc\left(G\right)=3$.

**Subcase 1.3.**$F\cong {K}_{2}+{K}_{1}$, where ${u}_{1}{u}_{2}\in E\left(G\right)$. Since G is connected, we have ${d}_{Q}\left({u}_{3}\right)\ge 1$ and $max\{{d}_{Q}\left({u}_{1}\right),{d}_{Q}\left({u}_{2}\right)\}\ge 1$. Without a loss of generality, let ${d}_{Q}\left({u}_{1}\right)\ge 1$. Assume that ${d}_{Q}\left({u}_{2}\right)=0$. Since $diam\left(G\right)=3$, it follows that ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{3}\right)\ne \xd8$, and let $v\in {N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{3}\right)$.

**Subcase 1.4.**$F\cong 3{K}_{1}$. Since $diam\left(G\right)=3$, it follows that $min\left\{\right|{N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)|,|{N}_{Q}$

**Case 2.**$diam\left(G\right)\ge 4$. Since $diam\left(G\right)\ge 4$, it follows that $F\cong {P}_{3}$ or $F\cong {K}_{2}+{K}_{1}$. Assume that $F\cong {P}_{3}$, where ${u}_{1}{u}_{2},{u}_{2}{u}_{3}\in E\left(G\right)$. Since $diam\left(G\right)\ge 4$, we have $min\{{d}_{Q}\left({u}_{1}\right),{d}_{Q}\left({u}_{3}\right)\}=0$, $max\{{d}_{Q}\left({u}_{1}\right),{d}_{Q}\left({u}_{3}\right)\}\ge 1$, and ${d}_{Q}\left({u}_{2}\right)=0$. Without a loss of generality, let ${d}_{Q}\left({u}_{1}\right)\ge 1$ and ${d}_{Q}\left({u}_{2}\right)={d}_{Q}\left({u}_{3}\right)=0$. Note that G is traceable, and we have $pc\left(G\right)=2$ by Proposition 1. The following edge-coloring $\theta $ with two colors makes G strongly properly connected: color ${u}_{1}{u}_{2}$ and all edges of $E\left(Q\right)$ with 2, and color the remaining edges with 1. Thus, $spc\left(G\right)=2$.

## 3. Rainbow Connection and Clique Number

**Theorem 5.**

**Proof.**

**Theorem 6.**

- (i)
- $G[V\left(G\right)\backslash V\left(Q\right)]\cong {P}_{3}$, where ${u}_{1}{u}_{2},{u}_{2}{u}_{3}\in E\left(G\right)$, ${d}_{Q}\left({u}_{1}\right)={d}_{Q}\left({u}_{3}\right)=0$, and ${d}_{Q}\left({u}_{2}\right)=1$.
- (ii)
- $G[V\left(G\right)\backslash V\left(Q\right)]\cong {K}_{2}+{K}_{1}$, where ${u}_{1}{u}_{2}\in E\left(G\right)$, ${d}_{Q}\left({u}_{2}\right)=0$, ${d}_{Q}\left({u}_{1}\right)={d}_{Q}\left({u}_{3}\right)=1$, and ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{3}\right)\ne \xd8$.
- (iii)
- $G[V\left(G\right)\backslash V\left(Q\right)]\cong 3{K}_{1}$, ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)\cap {N}_{Q}\left({u}_{3}\right)=\xd8$, $|{N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)|=1$, and ${d}_{Q}\left({u}_{1}\right)={d}_{Q}\left({u}_{2}\right)={d}_{Q}\left({u}_{3}\right)=1$.
- (iv)
- $G[V\left(G\right)\backslash V\left(Q\right)]\cong {P}_{3}$, where ${u}_{1}{u}_{2},{u}_{2}{u}_{3}\in E\left(G\right)$, ${d}_{Q}\left({u}_{1}\right)\ge 1$, and ${d}_{Q}\left({u}_{2}\right)={d}_{Q}\left({u}_{3}\right)=0$.
- (v)
- $G[V\left(G\right)\backslash V\left(Q\right)]\cong {K}_{2}+{K}_{1}$, where ${u}_{1}{u}_{2}\in E\left(G\right)$, ${d}_{Q}\left({u}_{1}\right)\ge 1$, ${d}_{Q}\left({u}_{2}\right)=0$, ${d}_{Q}\left({u}_{3}\right)\ge 1$, and ${N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{3}\right)=\xd8$.

**Proof.**

**Case 1.**$diam\left(G\right)=3$. We have $rc\left(G\right)\ge diam\left(G\right)=3$. We distinguish the following four subcases by analyzing the structure of F.

**Subcase 1.1.**$F\cong {K}_{3}$. The following edge-coloring $\theta $ with three colors makes G rainbow-connected: $\theta \left({u}_{1}{u}_{2}\right)=\theta \left({u}_{2}{u}_{3}\right)=\theta \left({u}_{1}{u}_{3}\right)=1$, and color all edges of $E\left(Q\right)$ with 3 and all edges of $E\left[V\right(F),V(Q\left)\right]$ with 2. Thus, $rc\left(G\right)=3$.

**Subcase 1.2.**$F\cong {P}_{3}$, where ${u}_{1}{u}_{2},{u}_{2}{u}_{3}\in E\left(G\right)$. Assume that ${d}_{Q}\left({u}_{1}\right)={d}_{Q}\left({u}_{3}\right)=0$. Suppose ${d}_{Q}\left({u}_{2}\right)=1$, and say ${N}_{Q}\left({u}_{2}\right)=\left\{u\right\}$. If an edge-coloring $\theta $ is a rainbow coloring of G, then $\theta \left({u}_{1}{u}_{2}\right)\ne \theta \left({u}_{2}{u}_{3}\right)\ne \theta \left({u}_{2}u\right)\ne \theta \left(uv\right)$, where $v\in V\left(Q\right)\backslash \left\{u\right\}$. Hence, $rc\left(G\right)\ge 4$. Allocate a rainbow coloring ${\theta}^{\prime}$ with four colors to G as follows: ${\theta}^{\prime}\left({u}_{1}{u}_{2}\right)=1$, ${\theta}^{\prime}\left({u}_{2}{u}_{3}\right)=2$, ${\theta}^{\prime}\left({u}_{2}u\right)=3$, and color all edges of $E\left(Q\right)$ with 4. Thus, $rc\left(G\right)=4$. Suppose ${d}_{Q}\left({u}_{2}\right)\ge 2$, and say $u,v\in {N}_{Q}\left({u}_{2}\right)$. The following edge-coloring $\theta $ with three colors makes G rainbow-connected: $\theta \left({u}_{1}{u}_{2}\right)=\theta \left({u}_{2}u\right)=1$, $\theta \left({u}_{2}{u}_{3}\right)=\theta \left({u}_{2}v\right)=2$, and color the remaining edges with 3. Hence, $rc\left(G\right)=3$.

**Subcase 1.3.**$F\cong {K}_{2}+{K}_{1}$, where ${u}_{1}{u}_{2}\in E\left(G\right)$. Since G is connected, we obtain ${d}_{Q}\left({u}_{3}\right)\ge 1$ and $max\{{d}_{Q}\left({u}_{1}\right),{d}_{Q}\left({u}_{2}\right)\}\ge 1$. Without a loss of generality, let ${d}_{Q}\left({u}_{1}\right)\ge 1$.

**Subcase 1.4.**$F\cong 3{K}_{1}$. Since $diam\left(G\right)=3$, it follows that $min\left\{\right|{N}_{Q}\left({u}_{1}\right)\cap {N}_{Q}\left({u}_{2}\right)|,|{N}_{Q}$

**Case 2.**$diam\left(G\right)\ge 4$. We obtain $rc\left(G\right)\ge diam\left(G\right)\ge 4$. Since $diam\left(G\right)\ge 4$, it follows that $F\cong {P}_{3}$ or $F\cong {K}_{2}+{K}_{1}$. Assume that $F\cong {P}_{3}$, where ${u}_{1}{u}_{2},{u}_{2}{u}_{3}\in E\left(G\right)$. Since $diam\left(G\right)\ge 4$, we have $min\{{d}_{Q}\left({u}_{1}\right),{d}_{Q}\left({u}_{3}\right)\}=0$, $max\{{d}_{Q}\left({u}_{1}\right),{d}_{Q}\left({u}_{3}\right)\}\ge 1$, and ${d}_{Q}\left({u}_{2}\right)=0$. Without a loss of generality, let ${d}_{Q}\left({u}_{1}\right)\ge 1$ and ${d}_{Q}\left({u}_{2}\right)={d}_{Q}\left({u}_{3}\right)=0$. Allocate a rainbow coloring $\theta $ with four colors to G as follows: color ${u}_{1}{u}_{2}$ with 2 and ${u}_{2}{u}_{3}$ with 1, and color all edges of $E\left[V\right(F),V(Q\left)\right]$ with 3 and all edges of $E\left(Q\right)$ with 4. Therefore, $rc\left(G\right)=4$.

## Author Contributions

## Funding

## Institutional Review Board Statement

## Informed Consent Statement

## Data Availability Statement

## Conflicts of Interest

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**MDPI and ACS Style**

Ma, Y.; Zhang, X.; Xue, Y.
Graphs with Strong Proper Connection Numbers and Large Cliques. *Axioms* **2023**, *12*, 353.
https://doi.org/10.3390/axioms12040353

**AMA Style**

Ma Y, Zhang X, Xue Y.
Graphs with Strong Proper Connection Numbers and Large Cliques. *Axioms*. 2023; 12(4):353.
https://doi.org/10.3390/axioms12040353

**Chicago/Turabian Style**

Ma, Yingbin, Xiaoxue Zhang, and Yanfeng Xue.
2023. "Graphs with Strong Proper Connection Numbers and Large Cliques" *Axioms* 12, no. 4: 353.
https://doi.org/10.3390/axioms12040353