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Article

# Graphs with Strong Proper Connection Numbers and Large Cliques

College of Mathematics and Information Science, Henan Normal University, Xinxiang 453007, China
*
Author to whom correspondence should be addressed.
Axioms 2023, 12(4), 353; https://doi.org/10.3390/axioms12040353
Received: 22 February 2023 / Revised: 27 March 2023 / Accepted: 28 March 2023 / Published: 3 April 2023
(This article belongs to the Special Issue Graph Theory and Discrete Applied Mathematics)

## Abstract

:
In this paper, we mainly investigate graphs with a small (strong) proper connection number and a large clique number. First, we discuss the (strong) proper connection number of a graph G of order n and $ω ( G ) = n − i$ for $1 ⩽ i ⩽ 3$. Next, we investigate the rainbow connection number of a graph G of order n, $d i a m ( G ) ≥ 3$ and $ω ( G ) = n − i$ for $2 ⩽ i ⩽ 3$.
MSC:
05C15; 05C40

## 1. Introduction

We only consider graphs that are undirected, simple, finite, and connected in this paper. For terminology and notation that are not defined here, we refer to .
In 2008, Chartrand et al.  introduced the concept of rainbow connection. For an edge-colored graph G, if each pair of vertices is connected by a rainbow path, where its edges are assigned different colors, then G is said to be rainbow-connected. An edge-coloring that makes G rainbow-connected is said to be a rainbow coloring of G. The rainbow connection number of G, denoted by $r c ( G )$, is the smallest number of colors that are needed to make G rainbow-connected. Obviously, $r c ( G ) = 1$ if and only if G is complete, and $r c ( G ) ≥ d i a m ( G )$. As a natural generalization of the rainbow connection number, the concept of the vertex rainbow connection number was presented by Krivelevich et al. , and the concept of the total rainbow connection number was introduced by Liu et al. . There are abundant research results on this topic. In , Schiermeyer proved that a connected graph G with n vertices has $r c ( G ) < 4 n δ ( G ) + 1 + 4$. Huang et al.  provided upper bounds of the rainbow connection number of outerplanar graphs with small diameters. In , Li et al. studied the vertex rainbow connection numbers of some graph operations. Ma et al.  investigated the total rainbow connection numbers of some special graphs. The reader should also consult  for a survey and  for a monograph.
Inspired by the concept of rainbow connection, Borozan et al.  proposed the concept of proper connection, and Andrews et al.  presented the concept of strong proper connection. A path is called a proper path in an edge-colored graph if its adjacent edges are assigned distinct colors. An edge-colored graph G is said to be properly connected if any two vertices are connected by a proper path, and G is said to be strongly properly connected if every pair of vertices is connected by a proper geodesic. An edge-coloring $θ$ of graph G is called a proper-path coloring if it makes G properly connected, and $θ$ is called a strong proper coloring if it makes G strongly properly connected. The proper connection number of G, denoted by $p c ( G )$, is the smallest number of colors that are needed to make G properly connected. The strong proper connection number of G, denoted by $s p c ( G )$, is the smallest number of colors that are needed to make G strongly properly connected. From these definitions, it is easy to establish that $p c ( G ) = s p c ( G ) = 1$ if and only if G is complete. In [13,14], Huang et al. presented an upper bound for the proper connection number of a graph in terms of the bridge-block tree of the graph and investigated the proper connection number of the complement of a graph. Li et al.  used dominating sets to study the proper connection number of a graph. Ma and Zhang  characterized all connected graphs of size m with (strong) proper connection number $m − 4$. For more details, we refer the reader to a survey .
Some results regarding the (vertex) rainbow connection numbers of graphs with a large clique number are available; see [18,19]. These results motivated us to consider the (strong) proper connection numbers of graphs with a large clique number. In this paper, we mainly discuss the (strong) proper connection number of a graph G of order n and $ω ( G ) = n − i$ for $1 ⩽ i ⩽ 3$. Moreover, we also investigate the rainbow connection number of a graph G of order n, $d i a m ( G ) ≥ 3$ and $ω ( G ) = n − i$ for $2 ⩽ i ⩽ 3$.

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

In this section, we investigate graphs with a small (strong) proper connection number and a large clique number. We first introduce some definitions that will be used later.
A Hamiltonian path in a graph G is a path containing every vertex of G. A graph with a Hamiltonian path is called a traceable graph. Recall that a clique of a graph is a set of mutually adjacent vertices, and that the maximum size of a clique of graph G, i.e., the clique number of G, is denoted $ω ( G )$. For a connected graph G, we say Q is a subgraph of G which induces a maximum clique and $V ( F ) = V ( G ) \ V ( Q )$. We say $N Q ( u )$ is the set of neighbors of u in Q and $d Q ( u ) = | N Q ( u ) |$. Additionally, we say $E [ V ( F ) , V ( Q ) ]$ is the set of edges of G between vertices of $V ( F )$ and vertices of $V ( Q )$. Next, we present the following three useful propositions.
Proposition 1 ().
Let G be a non-complete graph. If G is traceable, then $p c ( G ) = 2$.
Proposition 2 ().
For a non-trivial connected graph G that contains a bridge, if b is the maximum number of bridges incident with a vertex in G, then $s p c ( G ) ≥ p c ( G ) ≥ b$.
Proposition 3 ().
Let G be a connected graph of order n and size m. If $n − 1 2 + 1 ≤ m ≤ n 2 − 1$, then $r c ( G ) = 2$.
As an immediate consequence of Proposition 3, we have the following Lemma.
Lemma 1.
Let G be a connected graph of order n and size m. If $n − 1 2 + 1 ≤ m ≤ n 2 − 1$, then $p c ( G ) = s p c ( G ) = 2$.
Theorem 1.
Let G be a connected graph of order n. If $ω ( G ) = n + 1 − i$ for $i ∈ { 1 , 2 }$, then $p c ( G ) = s p c ( G ) = i$.
Proof.
If $i = 1$, then $ω ( G ) = n$, which implies that G is a complete graph. Thus, $p c ( G ) = s p c ( G ) = 1$. If $i = 2$, then $ω ( G ) = n − 1$. Since G is connected, we obtain $| E ( G ) | ≥ n − 1 2 + 1$, and so $n − 1 2 + 1 ≤ | E ( G ) | ≤ n 2 − 1$. Hence, $p c ( G ) = s p c ( G ) = 2$ by Lemma 1. □
Theorem 2.
Let G be a connected graph of order $n ≥ 4$ and $ω ( G ) = n − 2$. Let Q be a maximum clique of G and $V ( G ) \ V ( Q ) = { u 1 , u 2 }$. Then, either $p c ( G ) = s p c ( G ) = 2$ or one of the following holds:
(i)
$4 ≤ n ≤ 5$, $G [ V ( G ) \ V ( Q ) ] ≅ 2 K 1$ and $N Q ( u 1 ) = N Q ( u 2 ) = { v }$.
(ii)
$n ≥ 6$, $G [ V ( G ) \ V ( Q ) ] ≅ 2 K 1$ and $N Q ( u 1 ) = N Q ( u 2 ) = { v }$.
Moreover, we have $p c ( G ) = s p c ( G ) = 3$ for (i), $p c ( G ) = 2$, and $s p c ( G ) = 3$ for (ii).
Proof.
Let $F = G [ V ( G ) \ V ( Q ) ]$ and let $θ$ be an edge-coloring of G. We prove this theorem by analyzing the structure of F.
Case 1. $F ≅ K 2$. Since G is connected, it follows that $m a x { d Q ( u 1 ) , d Q ( u 2 ) } ≥ 1$. Note that G is traceable, and we have $p c ( G ) = 2$ by Proposition 1. The following edge-coloring $θ$ with two colors makes G strongly properly connected: color $u 1 u 2$ and all edges of $E ( Q )$ with 1, and color all edges of $E [ V ( F ) , V ( Q ) ]$ with 2. Thus, $s p c ( G ) = 2$.
Case 2. $F ≅ 2 K 1$. Since G is connected, it follows that $m i n { d Q ( u 1 ) , d Q ( u 2 ) } ≥ 1$. Assume that $N Q ( u 1 ) ∩ N Q ( u 2 ) = Ø$. Observe that G is traceable, and we have $p c ( G ) = 2$ by Proposition 1. Assign an edge-coloring $θ$ with two colors to G as follows: color all edges of $E ( Q )$ with 1 and all edges of $E [ V ( F ) , V ( Q ) ]$ with 2. It is clear that G is strongly properly connected with the above edge-coloring. Hence, $s p c ( G ) = 2$.
Assume that $N Q ( u 1 ) ∩ N Q ( u 2 ) ≠ Ø$ and $d Q ( u 1 ) = d Q ( u 2 ) = 1$. Without a loss of generality, let $v ∈ N Q ( u 1 ) ∩ N Q ( u 2 )$. If $n = 4$, then $G ≅ K 1 , 3$. Hence, $p c ( G ) = s p c ( G ) = 3$. If $n = 5$, then $G ≅ G 1$, where $G 1$ is obtained by adding two pendant edges to a vertex of $K 3$. Thus, $p c ( G ) = s p c ( G ) = 3$. Now we consider $n ≥ 6$. Let $V ( Q ) = { v , w 1 , w 2 , … , w n − 3 }$. Define an edge-coloring $θ$ of G with two colors as follows: $θ ( u 1 v ) = θ ( w 1 w n − 3 ) = 1$; $θ ( u 2 v ) = θ ( v w n − 4 ) = 2$; color the sequence $v w 1 w 2 ⋯ w n − 3 v$ alternately with 1 and 2 starting with $θ ( v w 1 ) = 1$; 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 $p c ( G ) = 2$. If $θ$ is a strong proper coloring of G, then $θ ( u 1 v ) ≠ θ ( u 2 v ) ≠ θ ( v w 1 )$, and thus $s p c ( G ) ≥ 3$. On the other hand, we define a strong proper coloring $θ ′$ of G with three colors as follows: $θ ′ ( u 1 v ) = 1$, $θ ′ ( u 2 v ) = 2$, and color all edges of $E ( Q )$ with 3. Thus, $s p c ( G ) = 3$.
Assume that $N Q ( u 1 ) ∩ N Q ( u 2 ) ≠ Ø$ and $m a x { d Q ( u 1 ) , d Q ( u 2 ) } ≥ 2$. Without a loss of generality, let $v ∈ N Q ( u 1 ) ∩ N Q ( u 2 )$ and $d Q ( u 1 ) ≥ 2$. Observe that G is traceable, and we obtain $p c ( G ) = 2$ by Proposition 1. Assign an edge-coloring $θ$ with two colors to G as follows: $θ ( u 1 v ) = 1$; $θ ( u 2 v ) = θ ( u 1 w ) = 2$ for any $w ∈ N Q ( u 1 ) \ { v }$; and color the remaining edges with 1. It is clear that $θ$ is a strong proper coloring of G. Hence, $s p c ( G ) = 2$. □
Theorem 3.
Let G be a connected graph of order $n ≥ 5$, $d i a m ( G ) = 2$, and $ω ( G ) = n − 3$. Let Q be a maximum clique of G and $V ( G ) \ V ( Q ) = { u 1 , u 2 , u 3 }$. Then, either $p c ( G ) = s p c ( G ) = 2$ or one of the following holds:
(i)
$G [ V ( G ) \ V ( Q ) ] ≅ P 3$, where $u 1 u 2 , u 2 u 3 ∈ E ( G )$, $d Q ( u 2 ) = 0$, $m i n { d Q ( u 1 ) , d Q ( u 3 ) } = 1$, $N Q ( u 1 ) ∪ N Q ( u 3 ) = V ( Q )$ and $N Q ( u 1 ) ∩ N Q ( u 3 ) = Ø$.
(ii)
$n = 6$, $G [ V ( G ) \ V ( Q ) ] ≅ K 2 + K 1$, where $u 1 u 2 ∈ E ( G )$, and $N Q ( u 1 ) = N Q ( u 2 ) = N Q ( u 3 ) = { v }$.
(iii)
$n ≥ 7$, $G [ V ( G ) \ V ( Q ) ] ≅ K 2 + K 1$, where $u 1 u 2 ∈ E ( G )$, and $N Q ( u 1 ) = N Q ( u 2 ) = N Q ( u 3 ) = { v }$.
(iv)
$G [ V ( G ) \ V ( Q ) ] ≅ K 2 + K 1$, where $u 1 u 2 ∈ E ( G )$, $N Q ( u 1 ) ∩ N Q ( u 2 ) ∩ N Q ( u 3 ) = { v }$, $m i n { d Q ( u 1 ) , d Q ( u 2 ) } = d Q ( u 3 ) = 1$ and $d Q ( u 1 ) + d Q ( u 2 ) ≥ 3$.
(v)
$G [ V ( G ) \ V ( Q ) ] ≅ K 2 + K 1$, where $u 1 u 2 ∈ E ( G )$, $N Q ( u 1 ) ∩ N Q ( u 3 ) ≠ Ø$, $N Q ( u 2 ) ∩ N Q ( u 3 ) ≠ Ø$, $d Q ( u 1 ) = d Q ( u 2 ) = 1$, $d Q ( u 3 ) = 2$ and $N Q ( u 1 ) ∩ N Q ( u 2 ) = Ø$.
(vi)
$n = 5$, $G [ V ( G ) \ V ( Q ) ] ≅ 3 K 1$ and $N Q ( u 1 ) = N Q ( u 2 ) = N Q ( u 3 ) = { v }$.
(vii)
$n ≥ 6$, $G [ V ( G ) \ V ( Q ) ] ≅ 3 K 1$ and $N Q ( u 1 ) = N Q ( u 2 ) = N Q ( u 3 ) = { v }$.
(viii)
$G [ V ( G ) \ V ( Q ) ] ≅ 3 K 1$, $| ( N Q ( u 1 ) ∩ N Q ( u 2 ) ) ∪ ( N Q ( u 1 ) ∩ N Q ( u 3 ) ) ∪ ( N Q ( u 2 ) ∩ N Q$
, $N Q ( u 1 ) ∩ N Q ( u 2 ) ∩ N Q ( u 3 ) ≠ Ø$ and $d Q ( u 1 ) + d Q ( u 2 ) + d Q ( u 3 ) ≥ 4$.
(ix)
$G [ V ( G ) \ V ( Q ) ] ≅ 3 K 1$, $| ( N Q ( u 1 ) ∩ N Q ( u 2 ) ) ∪ ( N Q ( u 1 ) ∩ N Q ( u 3 ) ) ∪ ( N Q ( u 2 ) ∩ N Q$
, $N Q ( u 1 ) ∩ N Q ( u 2 ) ∩ N Q ( u 3 ) ≠ Ø$ and $d Q ( u 1 ) + d Q ( u 2 ) + d Q ( u 3 ) = 5$.
Moreover, we have $p c ( G ) = 2$ and $s p c ( G ) = 3$ for (i), (iii), (iv), (v), (viii), and (ix); $p c ( G ) = s p c ( G ) = 3$ for (ii); $p c ( G ) = s p c ( G ) = 4$ for (vi); and $p c ( G ) = 3$ and $s p c ( G ) = 4$ for (vii).
Proof.
Let $F = G [ V ( G ) \ V ( Q ) ]$ and let $θ$ be an edge-coloring of G. We prove this theorem by analyzing the structure of F.
Case 1. $F ≅ K 3$. Observe that G is traceable, and so $p c ( G ) = 2$ by Proposition 1. The following edge-coloring $θ$ with two colors induces a strong proper coloring of G: color all edges of $E ( F )$ and $E ( Q )$ with 1, and color all edges of $E [ V ( F ) , V ( Q ) ]$ with 2. Thus, $s p c ( G ) = 2$.
Case 2. $F ≅ P 3$, where $u 1 u 2 , u 2 u 3 ∈ E ( G )$. Assume that $m i n { d Q ( u 1 ) , d Q ( u 2 ) , d Q ( u 3 ) } ≥ 1$. Note that G is traceable, and we have $p c ( G ) = 2$ by Proposition 1. Assign a strong proper coloring $θ$ with two colors to G as follows: $θ ( u 1 u 2 ) = 1$; $θ ( u 2 u 3 ) = 2$; and color all edges of $E ( Q )$ with 1 and all edges of $E [ V ( F ) , V ( Q ) ]$ with 2. Hence, $s p c ( G ) = 2$.
Assume that $m i n { d Q ( u 1 ) , d Q ( u 2 ) , d Q ( u 3 ) } = 0$. Since $d i a m ( G ) = 2$, it follows that $d Q ( u 1 ) ≥ 1$, $d Q ( u 2 ) = 0$, $d Q ( u 3 ) ≥ 1$, and $N Q ( u 1 ) ∪ N Q ( u 3 ) = V ( Q )$. Observe that G is traceable, and we obtain $p c ( G ) = 2$ by Proposition 1. Next, we only consider the strong proper connection number of graph G under this assumption.
Suppose $N Q ( u 1 ) ∩ N Q ( u 3 ) = Ø$ and $m i n { d Q ( u 1 ) , d Q ( u 3 ) } = 1$. Without a loss of generality, let $d Q ( u 1 ) = 1$ and $N Q ( u 1 ) = { v }$. If there exists a strong proper coloring $θ$ of G with two colors, then $θ ( u 1 u 2 ) ≠ θ ( u 2 u 3 )$. Without a loss of generality, let $θ ( u 1 u 2 ) = 1$ and $θ ( u 2 u 3 ) = 2$. Since $u 2 u 1 v$ is the unique $u 2 − v$ geodesic and $u 2 u 3 w$ is the unique $u 2 − w$ geodesic for any $w ∈ N Q ( u 3 )$, it follows that $θ ( u 1 v ) = 2$ and $θ ( u 3 w ) = 1$. Note that $u 1 v w$ is the unique $u 1 − w$ geodesic for any $w ∈ N Q ( u 3 )$, and so $θ ( v w ) = 1$. There is no proper geodesic between $u 3$ and v, which is a contradiction. Thus, $s p c ( G ) ≥ 3$. Assign an edge-coloring $θ ′$ with three colors to G as follows: $θ ′ ( u 1 u 2 ) = θ ′ ( u 3 w ) = 1$ for any $w ∈ N Q ( u 3 )$, $θ ′ ( u 2 u 3 ) = θ ′ ( u 1 v ) = 2$, and color all edges of $E ( Q )$ with 3. Obviously, $θ ′$ is a strong proper coloring of G, and so $s p c ( G ) = 3$.
Suppose $N Q ( u 1 ) ∩ N Q ( u 3 ) = Ø$ and $m i n { d Q ( u 1 ) , d Q ( u 3 ) } ≥ 2$. Let $N Q ( u 1 ) = { v 1 , v 2 , … , v t }$ and $N Q ( u 3 ) = { w 1 , w 2 , … , w k }$, where $t + k = n − 3$. Assign an edge-coloring $θ$ with two colors to G such that G is strongly properly connected: $θ ( u 1 u 2 ) = θ ( v 1 w 1 ) = θ ( u 3 w 1 ) = θ ( u 3 w i ) = θ ( v 2 w i ) = 1$ for $2 ≤ i ≤ k$, $θ ( u 2 u 3 ) = θ ( v 1 w k ) = θ ( u 1 v 1 ) = θ ( u 1 v j ) = θ ( w 1 v j ) = 2$ for $2 ≤ j ≤ t$, and color the remaining edges arbitrarily with 1 and 2. Hence, $s p c ( G ) = 2$.
Suppose $N Q ( u 1 ) ∩ N Q ( u 3 ) ≠ Ø$, and say $v ∈ N Q ( u 1 ) ∩ N Q ( u 3 )$. This implies that $m i n { d Q ( u 1 ) , d Q ( u 3 ) } ≥ 2$. Color $u 1 u 2$, $u 2 u 3$, $u 1 v$ and all edges of $E ( Q )$ with 1, and color the remaining edges with 2. Clearly, G is strongly properly connected with the above edge-coloring, and so $s p c ( G ) = 2$.
Case 3. $F ≅ K 2 + K 1$, where $u 1 u 2 ∈ E ( G )$. Since G is connected, we obtain $d Q ( u 3 ) ≥ 1$. We distinguish the following three subcases.
Subcase 3.1. $d Q ( u 3 ) = 1$. Let $N Q ( u 3 ) = { v }$. Since $d i a m ( G ) = 2$, we have $d Q ( u 1 ) ≥ 1$, $d Q ( u 2 ) ≥ 1$ and $v ∈ N Q ( u 1 ) ∩ N Q ( u 2 )$. Assume that $d Q ( u 1 ) = d Q ( u 2 ) = 1$. This implies $n ≥ 6$. If $n = 6$, then $G ≅ G 2$, where $G 2$ is displayed in Figure 1. Thus, $p c ( G ) = s p c ( G ) = 3$. Now we consider $n ≥ 7$. Let $V ( Q ) = { w 1 , w 2 , … , w n − 4 , v }$. Assign an edge-coloring $θ$ with two colors to G as follows: $θ ( u 1 u 2 ) = θ ( u 1 v ) = θ ( u 2 v ) = θ ( w i v ) = θ ( w j w n − 4 ) = 1$ for $1 ≤ i ≤ n − 5$ and $2 ≤ j ≤ n − 5$, $θ ( u 3 v ) = θ ( w n − 4 v ) = θ ( w 1 w n − 4 ) = θ ( w 1 w 2 ) = 2$, and color the remaining edges arbitrarily with 1 and 2. It is easy to verify that $θ$ is a proper-path coloring of G. Thus, $p c ( G ) = 2$. If G is strongly properly connected with an edge-coloring $θ$, then $θ ( u 1 v ) ≠ θ ( u 3 v ) ≠ θ ( w 1 v )$, and so $s p c ( G ) ≥ 3$. Assign an edge-coloring $θ ′$ with three colors to G as follows: $θ ′ ( u 1 u 2 ) = θ ′ ( u 1 v ) = θ ′ ( u 2 v ) = 1$, $θ ′ ( u 3 v ) = 2$, and color all edges of $E ( Q )$ with 3. We can check that G is strongly properly connected with the above edge-coloring. Hence, $s p c ( G ) = 3$.
Assume that $m i n { d Q ( u 1 ) , d Q ( u 2 ) } = 1$ and $d Q ( u 1 ) + d Q ( u 2 ) ≥ 3$. Without a loss of generality, let $d Q ( u 1 ) ≥ 2$ and $d Q ( u 2 ) = 1$. Observe that G is traceable, and we have $p c ( G ) = 2$ by Proposition 1. If G is strongly properly connected with an edge-coloring $θ$, then $θ ( u 2 v ) ≠ θ ( u 3 v ) ≠ θ ( w v )$, where $w ∈ V ( Q ) \ N Q ( u 1 )$. Hence, $s p c ( G ) ≥ 3$. Define an edge-coloring $θ ′$ of G with three colors such that G is strongly properly connected: $θ ′ ( u 1 u 2 ) = θ ′ ( u 1 v ) = θ ′ ( u 2 v ) = 1$, and color all edges of $E ( Q )$ with 3 and the remaining edges with 2. Thus, $s p c ( G ) = 3$.
Assume that $m i n { d Q ( u 1 ) , d Q ( u 2 ) } ≥ 2$. Note that G is traceable, and so $p c ( G ) = 2$ by Proposition 1. Assign a strong proper coloring $θ$ with two colors to G as follows: $θ ( u 1 u 2 ) = θ ( u 1 v ) = θ ( u 2 v ) = 1$, and color all edges of $E ( Q )$ with 1 and the remaining edges with 2. Hence, $s p c ( G ) = 2$.
Subcase 3.2. $d Q ( u 3 ) = 2$. Let $N Q ( u 3 ) = { u , v }$. Since $d i a m ( G ) = 2$, we obtain $N Q ( u 1 ) ∩ N Q ( u 3 ) ≠ Ø$ and $N Q ( u 2 ) ∩ N Q ( u 3 ) ≠ Ø$. Observe that G is traceable, and we have $p c ( G ) = 2$ by Proposition 1.
Assume that $d Q ( u 1 ) = d Q ( u 2 ) = 1$ and $N Q ( u 1 ) ∩ N Q ( u 2 ) ≠ Ø$. Let $u ∈$$N Q ( u 1 ) ∩ N Q ( u 2 )$. There exists a strong proper coloring $θ$ of G with two colors as follows: $θ ( u 1 u 2 ) = θ ( u 1 u ) = θ ( u 2 u ) = θ ( u 3 v ) = 1$, $θ ( u 3 u ) = 2$, and color all edges of $E ( Q )$ with 2. Thus, $s p c ( G ) = 2$.
Assume that $d Q ( u 1 ) = d Q ( u 2 ) = 1$ and $N Q ( u 1 ) ∩ N Q ( u 2 ) = Ø$. Let $N Q ( u 1 ) = { u }$, $N Q ( u 2 ) = { v }$ and $V ( Q ) = { w 1 , w 2 , … , w n − 5 , u , v }$. If there exists a strong proper coloring $θ$ of G with two colors, then $θ ( u 1 u ) ≠ θ ( u u 3 )$. Without a loss of generality, let $θ ( u 1 u ) = 1$ and $θ ( u u 3 ) = 2$. Since $u 1 u w 1$ is the unique $u 1 − w 1$ geodesic, it follows that $θ ( u w 1 ) = 2$. Note that $u 2 v u 3$ is the unique $u 2 − u 3$ geodesic, and so $θ ( u 2 v ) ≠ θ ( v u 3 )$. We first consider $θ ( u 2 v ) = 1$ and $θ ( v u 3 ) = 2$. Since $u 2 v w 1$ is the unique $u 2 − w 1$ geodesic, we have $θ ( v w 1 ) = 2$. There is no proper geodesic between $u 3$ and $w 1$, which is a contradiction. Next, we consider $θ ( u 2 v ) = 2$ and $θ ( v u 3 ) = 1$. Note that $u 2 v w 1$ is the unique $u 2 − w 1$ geodesic, so we obtain $θ ( v w 1 ) = 1$. There is no proper geodesic between $u 3$ and $w 1$, which is a contradiction. Hence, $s p c ( G ) ≥ 3$. Allocate a strong proper coloring $θ ′$ with three colors to G as follows: $θ ′ ( u 1 u 2 ) = θ ′ ( u 1 u ) = θ ′ ( u 2 v ) = 1$, $θ ′ ( u u 3 ) = θ ′ ( v u 3 ) = 2$, and color all edges of $E ( Q )$ with 3. Thus, $s p c ( G ) = 3$.
Assume that $d Q ( u 1 ) + d Q ( u 2 ) ≥ 3$ and $N Q ( u 1 ) ∩ N Q ( u 2 ) ≠ Ø$. Without a loss of generality, let $d Q ( u 1 ) ≥ 2$ and $w ∈ N Q ( u 1 ) ∩$$N Q ( u 2 )$. Consider $u ∈ N Q ( u 1 ) ∩ N Q ( u 2 )$ or $v ∈ N Q ( u 1 ) ∩ N Q ( u 2 )$. Without a loss of generality, let $u ∈ N Q ( u 1 ) ∩ N Q ( u 2 )$. The following edge-coloring $θ$ with two colors makes G strongly properly connected: $θ ( u 1 u ) = θ ( u 2 u ) = θ ( u 1 u 2 ) = 1$, $θ ( u 3 u ) = 2$, and color all edges of $E ( Q )$ with 2 and the remaining edges with 1. Hence, $s p c ( G ) = 2$. Consider $u ∉ N Q ( u 1 ) ∩ N Q ( u 2 )$ and $v ∉ N Q ( u 1 ) ∩ N Q ( u 2 )$. Then, $m i n { d Q ( u 1 ) , d Q ( u 2 ) } ≥ 2$. Without a loss of generality, let $u ∈ N Q ( u 1 )$ and $v ∈ N Q ( u 2 )$. Assign an edge-coloring $θ$ with two colors to G as follows: $θ ( u 1 w ) = θ ( u 2 w ) = θ ( u 1 u 2 ) = θ ( u 1 u ) = θ ( u 3 v ) = 1$, $θ ( u 3 u ) = θ ( u 2 v ) = 2$, and color all edges of $E ( Q )$ with 2 and the remaining edges with 1. It is not difficult to verify that $θ$ is a strong proper coloring of G, and so $s p c ( G ) = 2$.
Assume that $d Q ( u 1 ) + d Q ( u 2 ) ≥ 3$ and $N Q ( u 1 ) ∩ N Q ( u 2 ) = Ø$. Without a loss of generality, let $d Q ( u 1 ) ≥ 2$, $u ∈ N Q ( u 1 )$, and $v ∈ N Q ( u 2 )$. There exists an edge-coloring $θ$ with two colors such that G is strongly properly connected, as follows: $θ ( u 1 u 2 ) = θ ( u 2 v ) = θ ( u 3 u ) = 1$, $θ ( u 1 u ) = θ ( u 3 v ) = 2$, and color all edges of $E ( Q )$ with 2 and the remaining edges with 1. Hence, $s p c ( G ) = 2$.
Subcase 3.3. $d Q ( u 3 ) ≥ 3$. Note that G is traceable, and we obtain $p c ( G ) = 2$ by Proposition 1. Assume that $d Q ( u 1 ) = d Q ( u 2 ) = 1$ and $N Q ( u 1 ) ∩ N Q ( u 2 ) = Ø$. Let $N Q ( u 1 ) ∩ N Q ( u 3 ) = { u }$ and $N Q ( u 2 ) ∩ N Q ( u 3 ) = { v }$. Assign a strong proper coloring $θ$ with two colors to G as follows: $θ ( u 1 u 2 ) = θ ( u 1 u ) = θ ( u 2 v ) = 1$, $θ ( u 3 u ) = θ ( u 3 v ) = 2$, and color all edges of $E ( Q )$ with 2 and the remaining edges with 1. Thus, $s p c ( G ) = 2$.
Assume that either $d Q ( u 1 ) = d Q ( u 2 ) = 1$ and $N Q ( u 1 ) ∩ N Q ( u 2 ) ≠ Ø$, or $d Q ( u 1 ) + d Q ( u 2 ) ≥ 3$. An analogous edge-coloring to that presented in Subcase 3.2 induces a strong proper coloring of G with $s p c ( G ) = 2$.
Case 4. $F ≅ 3 K 1$. Since $d i a m ( G ) = 2$, it follows that $N Q ( u 1 ) ∩ N Q ( u 2 ) ≠ Ø$, $N Q ( u 1 ) ∩ N Q ( u 3 ) ≠ Ø$ and $N Q ( u 2 ) ∩ N Q ( u 3 ) ≠ Ø$. This case is demonstrated by the following three subcases.
Subcase 4.1. . This implies that . Let $N Q ( u 1 ) ∩ N Q ( u 2 ) ∩ N Q ( u 3 ) = { v }$. Assume that $d Q ( u 1 ) = d Q ( u 2 ) = d Q ( u 3 ) = 1$. Then, $s p c ( G ) ≥ p c ( G ) ≥ 3$ by Proposition 2. If $n = 5$, then $G ≅ K 1 , 4$. Hence, $p c ( G ) = s p c ( G ) = 4$. Now we consider $n ≥ 6$. Let $V ( Q ) = { v , w 1 , w 2 , … , w n − 4 }$. Assign an edge-coloring $θ$ with three colors to G as follows: $θ ( u 1 v ) = 1$; $θ ( u 2 v ) = 2$; $θ ( u 3 v ) = 3$; $θ ( w 1 w n − 4 ) = 3$ if n is even, $θ ( w 1 w n − 4 ) = 2$ if n is odd; color the sequence $w 1 v w 2 w 3 ⋯ w n − 4$ alternately with 1 and 2 starting with $θ ( w 1 v ) = 1$; and color the remaining edges arbitrarily with 1 and 2. It is not difficult to check that $θ$ is a proper-path coloring of G. Thus, $p c ( G ) = 3$. Suppose G has a strong proper coloring $θ$, we have $θ ( u 1 v ) ≠ θ ( u 2 v ) ≠ θ ( u 3 v ) ≠ θ ( w 1 v )$, and so $s p c ( G ) ≥ 4$. On the other hand, there exists a strong proper coloring $θ ′$ of G with four colors, as follows: $θ ′ ( u 1 v ) = 1$, $θ ′ ( u 2 v ) = 2$, $θ ′ ( u 3 v ) = 3$, and color all edges of $E ( Q )$ with 4. Therefore, we have $s p c ( G ) = 4$.
Assume that $d Q ( u 1 ) + d Q ( u 2 ) + d Q ( u 3 ) ≥ 4$. Without a loss of generality, let $d Q ( u 1 ) ≥ 2$, and say $u ∈ N Q ( u 1 ) \ { v }$. Let $V ( Q ) = { u , v , w 1 , w 2 , … , w n − 5 }$ with $n ≥ 6$. The following edge-coloring $θ$ with two colors makes G properly connected: $θ ( u 1 v ) = θ ( u 2 v ) = θ ( u v ) = 1$, $θ ( u 3 v ) = 2$, color the sequence $v w 1 w 2 ⋯ w n − 5 u u 1$ alternately with 2 and 1 starting with $θ ( v w 1 ) = 2$, and color the remaining edges arbitrarily with 1 and 2. Thus, $p c ( G ) = 2$. Suppose G has a strong proper coloring $θ$, we have $θ ( u 1 v ) ≠ θ ( u 2 v ) ≠ θ ( u 3 v )$, and so $s p c ( G ) ≥ 3$. On the other hand, there exists a strong proper coloring $θ ′$ of G with three colors, as follows: $θ ′ ( u 1 u ) = θ ′ ( u 2 v ) = 1$, $θ ′ ( u 3 v ) = 2$, $θ ′ ( u 1 v ) = 3$, and color all edges of $E ( Q )$ with 3 and the remaining edges with 1. Hence, $s p c ( G ) = 3$.
Subcase 4.2. . Since $d i a m ( G ) = 2$, we obtain $N Q ( u 1 ) ∩ N Q ( u 2 ) ∩ N Q ( u 3 ) ≠ Ø$, and say $v ∈ N Q ( u 1 ) ∩ N Q ( u 2 ) ∩ N Q ( u 3 )$. Without a loss of generality, we consider , and say $u ∈ ( N Q ( u 1 ) ∩ N Q ( u 2 ) ) \ { v }$. Assign an analogous edge-coloring to that presented in Subcase 4.1 to G that satisfies $d Q ( u 1 ) + d Q ( u 2 ) + d Q ( u 3 ) ≥ 4$. Obviously, G is properly connected, and so $p c ( G ) = 2$.
Assume that $d Q ( u 1 ) + d Q ( u 2 ) + d Q ( u 3 ) = 5$. Suppose that there exists a strong proper coloring $θ$ of G with two colors. Note that $u 1 v u 3$ is the unique $u 1 − u 3$ geodesic, and $u 2 v u 3$ is the unique $u 2 − u 3$ geodesic. Without a loss of generality, let $θ ( u 1 v ) = θ ( u 2 v ) = 1$ and $θ ( u 3 v ) = 2$. Since $u 3 v w$ is the unique $u 3 − w$ geodesic, where $w ∈ V ( Q ) \ { u , v }$, it follows that $θ ( v w ) = 1$. In order to have a proper geodesic connecting $u 2$ and w, we have $θ ( u 2 u ) ≠ θ ( u w )$. Similarly, for the sake of having a proper geodesic between $u 1$ and $u 2$, we obtain $θ ( u 1 u ) ≠ θ ( u 2 u )$. Then, $θ ( u w ) = θ ( u 1 u )$, and so there is no proper geodesic connecting $u 1$ and w, which is a contradiction. Thus, $s p c ( G ) ≥ 3$. Now we assign a strong proper coloring $θ ′$ with three colors to G as follows: $θ ′ ( u 1 u ) = θ ′ ( u 1 v ) = θ ′ ( u 2 v ) = 1$, $θ ′ ( u 2 u ) = θ ′ ( u 3 v ) = 2$, and color all edges of $E ( Q )$ with 3. Hence, $s p c ( G ) = 3$.
Assume that $d Q ( u 1 ) + d Q ( u 2 ) + d Q ( u 3 ) ≥ 6$. Suppose $d Q ( u 3 ) = 1$. This implies that $m a x { d Q ( u 1 ) , d Q ( u 2 ) } ≥ 3$. Without a loss of generality, we consider $d Q ( u 1 ) ≥ 3$, and say $w ∈ N Q ( u 1 ) \ { u , v }$. The following edge-coloring $θ$ with two colors makes G strongly properly connected: $θ ( u 1 u ) = θ ( u 1 v ) = θ ( u 2 v ) = 1$, $θ ( u 2 u ) = θ ( u 3 v ) = θ ( u 1 w ) = 2$, and color all edges of $E ( Q )$ with 1 and the remaining edges with 2. Thus, $s p c ( G ) = 2$. Suppose $d Q ( u 3 ) ≥ 2$. Let $z ∈ N Q ( u 3 ) \ { v }$, where $u = z$ is possible. Assign an edge-coloring $θ$ with two colors to G as follows: $θ ( u 1 u ) = θ ( u 1 v ) = θ ( u 2 v ) = θ ( u 3 z ) = 1$, $θ ( u 2 u ) = θ ( u 3 v ) = 2$, and color all edges of $E ( Q )$ with 2 and the remaining edges with 1. Obviously, $θ$ is a strong proper coloring of G, and so $s p c ( G ) = 2$.
Subcase 4.3. , and let ${ w 1 , w 2 , w 3 } ⊆ ( N Q ( u 1 ) ∩ N Q ( u 2 ) ) ∪ ( N Q ( u 1 ) ∩ N Q ( u 3 ) ) ∪ ( N Q ( u 2 ) ∩ N Q ( u 3 ) )$. Up to isomorphism, we only need to consider the following two cases.
Let ${ u 1 w 1 , u 1 w 2 , u 2 w 1 , u 2 w 3 , u 3 w 2 , u 3 w 3 } ⊆ E [ V ( F ) , V ( Q ) ]$. Assign an edge-coloring $θ$ with two colors to G such that G is strongly properly connected: $θ ( u 1 w 1 ) = θ ( u 2 w 3 ) = θ ( u 3 w 2 ) = 1$, $θ ( u 1 w 2 ) = θ ( u 2 w 1 ) = θ ( u 3 w 3 ) = 2$, and color all edges of $E ( Q )$ with 2 and the remaining edges with 1. Hence, $p c ( G ) = s p c ( G ) = 2$.
Let ${ u 1 w 1 , u 1 w 2 , u 1 w 3 , u 2 w 1 , u 2 w 2 , u 2 w 3 , u 3 w 1 } ⊆ E [ V ( F ) , V ( Q ) ]$. The following edge-coloring $θ$ with two colors makes G strongly properly connected: $θ ( u 1 w 2 ) = θ ( u 2 w 3 ) = θ ( u 3 w 1 ) = 1$, $θ ( u 1 w 1 ) = θ ( u 2 w 1 ) = θ ( u 2 w 2 ) = 2$, and color all edges of $E ( Q )$ with 2 and the remaining edges with 1. Thus, $p c ( G ) = s p c ( G ) = 2$. □
Theorem 4.
Let G be a connected graph of order $n ≥ 5$, $d i a m ( G ) ≥ 3$, and $ω ( G ) = n − 3$. Let Q be a maximum clique of G and $V ( G ) \ V ( Q ) = { u 1 , u 2 , u 3 }$. Then, either $p c ( G ) = s p c ( G ) = 2$ or one of the following holds:
(i)
$G [ V ( G ) \ V ( Q ) ] ≅ P 3$, where $u 1 u 2 , u 2 u 3 ∈ E ( G )$, $d Q ( u 1 ) = d Q ( u 3 ) = 0$, and $d Q ( u 2 ) = 1$.
(ii)
$G [ V ( G ) \ V ( Q ) ] ≅ P 3$, where $u 1 u 2 , u 2 u 3 ∈ E ( G )$, $d Q ( u 1 ) = d Q ( u 3 ) = 0$, and $d Q ( u 2 ) ≥ 2$.
(iii)
$G [ V ( G ) \ V ( Q ) ] ≅ P 3$, where $u 1 u 2 , u 2 u 3 ∈ E ( G )$, $m i n { d Q ( u 1 ) , d Q ( u 3 ) } = 1$, $d Q ( u 2 ) = 0$, $N Q ( u 1 ) ∪ N Q ( u 3 ) ≠ V ( Q )$, and $N Q ( u 1 ) ∩ N Q ( u 3 ) = Ø$.
(iv)
$5 ≤ n ≤ 6$, $G [ V ( G ) \ V ( Q ) ] ≅ K 2 + K 1$, where $u 1 u 2 ∈ E ( G )$, $N Q ( u 1 ) = N Q ( u 3 ) = { v }$, and $d Q ( u 2 ) = 0$.
(v)
$n ≥ 7$, $G [ V ( G ) \ V ( Q ) ] ≅ K 2 + K 1$, where $u 1 u 2 ∈ E ( G )$, $N Q ( u 1 ) = N Q ( u 3 ) = { v }$, and $d Q ( u 2 ) = 0$.
(vi)
$G [ V ( G ) \ V ( Q ) ] ≅ K 2 + K 1$, where $u 1 u 2 ∈ E ( G )$, $d Q ( u 2 ) ≥ 1$, $N Q ( u 1 ) = N Q ( u 3 ) = { v }$, $N Q ( u 2 ) ∩ N Q ( u 3 ) = Ø$, and $N Q ( u 1 ) ∪ N Q ( u 2 ) ≠ V ( Q )$.
(vii)
$5 ≤ n ≤ 6$, $G [ V ( G ) \ V ( Q ) ] ≅ 3 K 1$, $N Q ( u 1 ) ∩ N Q ( u 2 ) ∩ N Q ( u 3 ) = Ø$, , and $d Q ( u 1 ) = d Q ( u 2 ) = d Q ( u 3 ) = 1$.
(viii)
$n = 6$, $G [ V ( G ) \ V ( Q ) ] ≅ 3 K 1$, $N Q ( u 1 ) ∩ N Q ( u 2 ) ∩ N Q ( u 3 ) = Ø$, , $d Q ( u 1 ) = d Q ( u 2 ) = 1$, and $d Q ( u 3 ) = 2$.
(ix)
$n ≥ 7$, $G [ V ( G ) \ V ( Q ) ] ≅ 3 K 1$, $N Q ( u 1 ) ∩ N Q ( u 2 ) ∩ N Q ( u 3 ) = Ø$, , $d Q ( u 1 ) = d Q ( u 2 ) = 1$, and $d Q ( u 3 ) ≥ 1$.
(x)
$G [ V ( G ) \ V ( Q ) ] ≅ 3 K 1$, $N Q ( u 1 ) ∩ N Q ( u 2 ) ∩ N Q ( u 3 ) = Ø$, , $N Q ( u 1 ) ∩ N Q ( u 3 ) ≠ Ø$, $d Q ( u 1 ) = 2$, and $d Q ( u 2 ) = d Q ( u 3 ) = 1$.
Moreover, we have $p c ( G ) = 2$ and $s p c ( G ) = 3$ for (ii), (iii), (v), (vi), (viii), (ix), and (x) and $p c ( G ) = s p c ( G ) = 3$ for (i), (iv), and (vii).
Proof.
Let $F = G [ V ( G ) \ V ( Q ) ]$, and let $θ$ be an edge-coloring of G. We prove this theorem by the following two cases.
Case 1. $d i a m ( G ) = 3$. We distinguish the following four subcases by analyzing the structure of F.
Subcase 1.1. $F ≅ K 3$. Note that G is traceable, and we have $p c ( G ) = 2$ by Proposition 1. Assign an edge-coloring $θ$ with two colors to G as follows: color all edges of $E ( F )$ and $E ( Q )$ with 1, and color all edges of $E [ V ( F ) , V ( Q ) ]$ with 2. It is obvious that $θ$ is a strong proper coloring of G, and so $s p c ( G ) = 2$.
Subcase 1.2. $F ≅ P 3$, where $u 1 u 2 , u 2 u 3 ∈ E ( G )$. Assume that $d Q ( u 1 ) = d Q ( u 3 ) = 0$. Suppose $d Q ( u 2 ) = 1$, and let $N Q ( u 2 ) = { v }$. Then, $s p c ( G ) ≥ p c ( G ) ≥ 3$ by Proposition 2. Now we define a strong proper coloring $θ$ of G with three colors as follows: $θ ( u 1 u 2 ) = 1$, $θ ( u 2 u 3 ) = 2$, $θ ( u 2 v ) = 3$, and color all edges of $E ( Q )$ with 1. Thus, $p c ( G ) = s p c ( G ) = 3$. Suppose $d Q ( u 2 ) ≥ 2$, and let $u , v ∈ N Q ( u 2 )$. Assign an edge-coloring $θ$ with two colors to G as follows: $θ ( u 1 u 2 ) = θ ( u 2 u ) = θ ( v w ) = 1$ for any $w ∈ V ( Q ) \ { u , v }$, $θ ( u 2 u 3 ) = θ ( u 2 v ) = θ ( u v ) = θ ( u w ) = 2$ for any $w ∈ V ( Q ) \ { u , v }$, 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 $p c ( G ) = 2$. If G is strongly properly connected with an edge-coloring $θ$, then $θ ( u 1 u 2 ) ≠ θ ( u 2 u 3 ) ≠ θ ( u 2 u )$. Thus, $s p c ( G ) ≥ 3$. Assign a strong proper coloring $θ ′$ with three colors to G as follows: $θ ′ ( u 1 u 2 ) = 1$, $θ ′ ( u 2 u 3 ) = 2$, and color all edges of $E [ V ( F ) , V ( Q ) ]$ with 3 and all edges of $E ( Q )$ with 1. Thus, $s p c ( G ) = 3$.
Assume that $m i n { d Q ( u 1 ) , d Q ( u 3 ) } = 0$ and $m a x { d Q ( u 1 ) , d Q ( u 3 ) } ≥ 1$. Without a loss of generality, let $d Q ( u 3 ) = 0$ and $d Q ( u 1 ) ≥ 1$. Since $d i a m ( G ) = 3$, it follows that $d Q ( u 2 ) ≥ 1$. Note that G is traceable, and we have $p c ( G ) = 2$ by Proposition 1. The following edge-coloring $θ$ with two colors makes G strongly properly connected: $θ ( u 1 u 2 ) = 1$, $θ ( u 2 u 3 ) = 2$, and color all edges of $E ( Q )$ with 2 and all edges of $E [ V ( F ) , V ( Q ) ]$ with 1. Hence, $s p c ( G ) = 2$.
Assume that $d Q ( u 1 ) ≥ 1$ and $d Q ( u 3 ) ≥ 1$. Since $d i a m ( G ) = 3$, it follows that $d Q ( u 2 ) = 0$ and $N Q ( u 1 ) ∪ N Q ( u 3 ) ≠ V ( Q )$. Observe that G is traceable, and we have $p c ( G ) = 2$ by Proposition 1. Now, we only consider the strong proper connection number of graph G under this assumption.
Suppose $N Q ( u 1 ) ∩ N Q ( u 3 ) = Ø$ and $m i n { d Q ( u 1 ) , d Q ( u 3 ) } = 1$. Without a loss of generality, we consider $d Q ( u 1 ) = 1$, and say $N Q ( u 1 ) = { u }$. If there exists a strong proper coloring $θ$ of G with two colors, then $θ ( u 1 u 2 ) ≠ θ ( u 2 u 3 )$. Without a loss of generality, let $θ ( u 1 u 2 ) = 1$ and $θ ( u 2 u 3 ) = 2$. Note that $u 2 u 1 u$ is the unique $u 2 − u$ geodesic, and $u 2 u 3 v$ is the unique $u 2 − v$ geodesic for any $v ∈ N Q ( u 3 )$; then, $θ ( u 1 u ) = 2$ and $θ ( u 3 v ) = 1$. Since $u 1 u v$ is the unique $u 1 − v$ geodesic for any $v ∈ N Q ( u 3 )$, we have $θ ( u v ) = 1$. There is no proper geodesic between $u 3$ and u, which is a contradiction. Thus, $s p c ( G ) ≥ 3$. On the other hand, we assign a strong proper coloring $θ ′$ with three colors to G as follows: $θ ′ ( u 1 u 2 ) = θ ′ ( u 3 v ) = 1$ for any $v ∈ N Q ( u 3 )$, $θ ′ ( u 2 u 3 ) = θ ′ ( u 1 u ) = 2$, and color all edges of $E ( Q )$ with 3. Hence, $s p c ( G ) = 3$.
Suppose $N Q ( u 1 ) ∩ N Q ( u 3 ) = Ø$ and $m i n { d Q ( u 1 ) , d Q ( u 3 ) } ≥ 2$. Let $N Q ( u 1 ) = { w 1 , w 2 , … , w t }$ and $N Q ( u 3 ) = { v 1 , v 2 , … , v k }$, where $t + k < n − 3$. Assign an edge-coloring $θ$ with two colors to G as follows: $θ ( u 1 u 2 ) = θ ( w 1 v 1 ) = θ ( u 3 v 1 ) = θ ( u 3 v i ) = θ ( w 2 v i ) = 1$ for $2 ≤ i ≤ k$, $θ ( u 2 u 3 ) = θ ( w 1 v k ) = θ ( u 1 w 1 ) = θ ( u 1 w j ) = θ ( v 1 w j ) = 2$ for $2 ≤ j ≤ t$, $θ ( v 1 w ) = 2$ and $θ ( w 1 w ) = 1$ for any $w ∈ V ( Q ) \ { N Q ( u 1 ) ∪ N Q ( u 3 ) }$, and color the remaining edges arbitrarily with 1 and 2. It is clear that $θ$ is a strong proper coloring of G, and so $s p c ( G ) = 2$.
Suppose $N Q ( u 1 ) ∩ N Q ( u 3 ) ≠ Ø$, and let $v ∈ N Q ( u 1 ) ∩ N Q ( u 3 )$. Consider $d Q ( u 1 ) = d Q ( u 3 ) = 1$. Color $u 1 u 2$ and all edges of $E ( Q )$ with 1, and color $u 2 u 3$, $u 1 v$ and $u 3 v$ with 2. Obviously, the above edge-coloring makes G strongly properly connected. Thus, $s p c ( G ) = 2$. Consider $m i n { d Q ( u 1 ) , d Q ( u 3 ) } = 1$ and $m a x { d Q ( u 1 ) , d Q ( u 3 ) } ≥ 2$. Without a loss of generality, let $d Q ( u 1 ) = 1$ and $d Q ( u 3 ) ≥ 2$. Assign a strong proper coloring $θ$ with two colors to G as follows: $θ ( u 1 u 2 ) = θ ( u 2 u 3 ) = θ ( u 3 v ) = 1$, $θ ( u 1 v ) = θ ( u 3 w ) = 2$ for any $w ∈ N Q ( u 3 ) \ { v }$, and color all edges of $E ( Q )$ with 1. Hence, $s p c ( G ) = 2$. Consider $m i n { d Q ( u 1 ) , d Q ( u 3 ) } ≥ 2$. Allocate a strong proper coloring $θ$ with two colors to G as follows: $θ ( u 1 u 2 ) = θ ( u 2 u 3 ) = θ ( u 3 v ) = 1$, and color all edges of $E ( Q )$ with 1 and the remaining edges with 2. Thus, $s p c ( G ) = 2$.
Subcase 1.3. $F ≅ K 2 + K 1$, where $u 1 u 2 ∈ E ( G )$. Since G is connected, we have $d Q ( u 3 ) ≥ 1$ and $m a x { d Q ( u 1 ) , d Q ( u 2 ) } ≥ 1$. Without a loss of generality, let $d Q ( u 1 ) ≥ 1$. Assume that $d Q ( u 2 ) = 0$. Since $d i a m ( G ) = 3$, it follows that $N Q ( u 1 ) ∩ N Q ( u 3 ) ≠ Ø$, and let $v ∈ N Q ( u 1 ) ∩ N Q ( u 3 )$.
Suppose $d Q ( u 1 ) = d Q ( u 3 ) = 1$. If $n = 5$, then $G ≅ G 3$, where $G 3$ is displayed in Figure 2. Hence, $p c ( G ) = s p c ( G ) = 3$. If $n = 6$, then $G ≅ G 4$, where $G 4$ is shown in Figure 2. Thus, $p c ( G ) = s p c ( G ) = 3$. Now, we consider $n ≥ 7$. Let $V ( Q ) = { w 1 , w 2 , … , w n − 4 , v }$. Assign an edge-coloring $θ$ with two colors to G as follows: $θ ( u 1 u 2 ) = θ ( u 3 v ) = θ ( w n − 4 w 1 ) = 1$, $θ ( u 1 v ) = θ ( w 2 v ) = 2$, color the sequence $v w 1 w 2 ⋯ w n − 4 v$ alternately with 1 and 2 starting with $θ ( v w 1 ) = 1$, and color the remaining edges arbitrarily with 1 and 2. We can verify that $θ$ is a proper-path coloring of G. Thus, $p c ( G ) = 2$. If G has a strong proper coloring $θ$, then $θ ( u 1 v ) ≠ θ ( u 3 v ) ≠ θ ( v w 1 )$, and so $s p c ( G ) ≥ 3$. On the other hand, there exists a strong proper coloring $θ ′$ of G with three colors: assign 1 to $u 1 u 2$ and $u 3 v$, assign 2 to $u 1 v$, and assign 3 to all edges of $E ( Q )$. Therefore, $s p c ( G ) = 3$.
Suppose $d Q ( u 1 ) = 1$ and $d Q ( u 3 ) ≥ 2$. Note that G is traceable, and we have $p c ( G ) = 2$ by Proposition 1. Allocate an edge-coloring $θ$ with two colors to G as follows: $θ ( u 1 v ) = θ ( u 3 w ) = 1$ for any $w ∈ N Q ( u 3 ) \ { v }$, $θ ( u 1 u 2 ) = θ ( u 3 v ) = 2$, and color all edges of $E ( Q )$ with 2. Obviously, $θ$ is a strong proper coloring of G, and so $s p c ( G ) = 2$.
Suppose $d Q ( u 1 ) ≥ 2$ and $d Q ( u 3 ) = 1$. Observe that G is traceable, and we obtain $p c ( G ) = 2$ by Proposition 1. The following edge-coloring $θ$ with two colors makes G strongly properly connected: $θ ( u 1 v ) = θ ( u 1 w 1 ) = 1$ for any $w 1 ∈ N Q ( u 1 ) \ { v }$, $θ ( u 1 u 2 ) = θ ( u 3 v ) = 2$, and color all edges incident with v in $E ( Q )$ with 1 and the remaining edges with 2. Hence, $s p c ( G ) = 2$.
Suppose $d Q ( u 1 ) ≥ 2$ and $d Q ( u 3 ) ≥ 2$. Note that G is traceable, and we have $p c ( G ) = 2$ by Proposition 1. Define a strong proper coloring $θ$ of G with two colors as follows: $θ ( u 1 v ) = 1$, $θ ( u 1 u 2 ) = θ ( u 3 v ) = 2$, and color all edges of $E ( Q )$ with 2 and the remaining edges with 1. Thus, $s p c ( G ) = 2$.
Assume that $d Q ( u 2 ) ≥ 1$. Since $d i a m ( G ) = 3$, it follows that $m i n { | N Q ( u 1 ) ∩ N Q ( u 3 ) |$,
$| N Q ( u 2 ) ∩ N Q ( u 3 ) | } = 0$. Suppose $m a x { | N Q ( u 1 ) ∩ N Q ( u 3 ) | , | N Q ( u 2 ) ∩ N Q ( u 3 ) | } ≥ 1$. Without a loss of generality, we consider $| N Q ( u 2 ) ∩ N Q ( u 3 ) | } = 0$ and , and say $v ∈ N Q ( u 1 ) ∩ N Q ( u 3 )$. Observe that G is traceable, and we have $p c ( G ) = 2$ by Proposition 1. Now, we only consider the strong proper connection number of graph G under this supposition.
We first consider . The following edge-coloring $θ$ with two colors makes G strongly properly connected: color $u 1 u 2$, $u 3 v$ and all edges of $E ( Q )$ with 2, and color the remaining edges with 1. Hence, $s p c ( G ) = 2$.
Next, we consider . Let $d Q ( u 1 ) ≥ 2$. Assign a strong proper coloring $θ$ with two colors to G: color $u 1 v$ and all edges of $E ( Q )$ with 1, and color the remaining edges with 2. Hence, $s p c ( G ) = 2$. Let $d Q ( u 3 ) ≥ 2$. Define a strong proper coloring $θ$ of G with two colors as follows: color $u 1 u 2$, $u 3 v$ and all edges of $E ( Q )$ with 2, and color the remaining edges with 1. Thus, $s p c ( G ) = 2$. Let $d Q ( u 1 ) = d Q ( u 3 ) = 1$ and $N Q ( u 1 ) ∪ N Q ( u 2 ) = V ( Q )$. Allocate an edge-coloring $θ$ with two colors to G: color $u 1 u 2$, $u 1 v$ and all edges of $E ( Q )$ with 1, and color the remaining edges with 2. We can check that G is strongly properly connected with the above edge-coloring, and so $s p c ( G ) = 2$. Let $d Q ( u 1 ) = d Q ( u 3 ) = 1$ and $N Q ( u 1 ) ∪ N Q ( u 2 ) ≠ V ( Q )$. If $θ$ is a strong proper coloring of G, then $θ ( u 1 v ) ≠ θ ( u 3 v ) ≠ θ ( v w )$, where $w ∈ V ( Q ) \ { N Q ( u 1 ) ∪ N Q ( u 2 ) }$. Thus, $s p c ( G ) ≥ 3$. On the other hand, there exists an edge-coloring $θ ′$ with three colors such that G is strongly properly connected: color $u 1 v$ with 1 and all edges of $E ( Q )$ with 3, and color the remaining edges with 2. Hence, $s p c ( G ) = 3$.
Suppose $m a x { | N Q ( u 1 ) ∩ N Q ( u 3 ) | , | N Q ( u 2 ) ∩ N Q ( u 3 ) | } = 0$. Observe that G is traceable, and we have $p c ( G ) = 2$ by Proposition 1. Assign an edge-coloring $θ$ with two colors to G as follows: color $u 1 u 2$ and all edges of $E ( Q )$ with 2, and color all edges of $E [ V ( F ) , V ( Q ) ]$ with 1. It is clear that $θ$ is a strong proper coloring of G, and so $s p c ( G ) = 2$.
Subcase 1.4. $F ≅ 3 K 1$. Since $d i a m ( G ) = 3$, it follows that $m i n { | N Q ( u 1 ) ∩ N Q ( u 2 ) | , | N Q$
$( u 1 ) ∩ N Q ( u 3 ) | , | N Q ( u 2 ) ∩ N Q ( u 3 ) | } = 0$. Assume that $m a x { | N Q ( u 1 ) ∩ N Q ( u 2 ) | , | N Q ( u 1 ) ∩ N Q ( u 3 ) | , | N Q ( u 2 ) ∩ N Q ( u 3 ) | } = 0$. The following edge-coloring $θ$ with two colors makes G strongly properly connected: color all edges of $E ( Q )$ with 2, and color all edges of $E [ V ( F ) , V ( Q ) ]$ with 1. Thus, $p c ( G ) = s p c ( G ) = 2$. Assume that $m a x { | N Q ( u 1 ) ∩ N Q ( u 2 ) | ,$
$| N Q ( u 1 ) ∩ N Q ( u 3 ) | , | N Q ( u 2 ) ∩ N Q ( u 3 ) | } ≥ 1$. Without a loss of generality, we consider , and say $u ∈ N Q ( u 1 ) ∩ N Q ( u 2 )$.
Suppose $d Q ( u 1 ) = d Q ( u 2 ) = 1$. If $θ$ is a strong proper coloring of G, then $θ ( u 1 u ) ≠ θ ( u 2 u ) ≠ θ ( u w )$, where $w ∈ V ( Q ) \ { u }$. Hence, $s p c ( G ) ≥ 3$. On the other hand, there exists a strong proper coloring $θ ′$ of G with three colors, as follows: $θ ′ ( u 1 u ) = 1$, $θ ′ ( u 2 u ) = 2$, and color all edges of $E ( Q )$ with 3 and the remaining edges with 1. Thus, $s p c ( G ) = 3$. Next, we discuss the proper connection number of G. If $n = 5$, then $G ≅ G 3$, where $G 3$ is displayed in Figure 2. Hence, $p c ( G ) = 3$. We consider $n = 6$. If $d Q ( u 3 ) = 1$, then $G ≅ G 5$. Thus, $p c ( G ) = 3$. If $d Q ( u 3 ) = 2$, then $G ≅ G 6$. Hence, $p c ( G ) = 2$. The graphs $G 5$ and $G 6$ are shown in Figure 3. Now, we consider $n ≥ 7$. Let $V ( Q ) = { u , v , w 1 , w 2 , … , w n − 5 }$ and $v ∈ N Q ( u 3 )$. Assign an edge-coloring $θ$ with two colors to G as follows: $θ ( u 1 u ) = θ ( u 3 v ) = θ ( u w n − 6 ) = 1$; $θ ( u 2 u ) = θ ( w n − 5 v ) = 2$; color $w n − 6 v$ with 1 for $n = 7$ and $w n − 6 v$ with 2 for $n ≥ 8$; color the sequence $u v w 1 w 2 ⋯ w n − 5 u$ alternately with 2 and 1 starting with $θ ( u v ) = 2$; 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 $p c ( G ) = 2$.
Suppose $m a x { d Q ( u 1 ) , d Q ( u 2 ) } ≥ 2$. Without a loss of generality, let $d Q ( u 1 ) ≥ 2$, and say $w ∈ N Q ( u 1 ) \ { u }$. We first consider $N Q ( u 1 ) ∩ N Q ( u 3 ) = Ø$. The following edge-coloring $θ$ with two colors makes G strongly properly connected: $θ ( u 1 u ) = 1$, $θ ( u 1 w ) = θ ( u 2 u ) = 2$, color all edges of $E ( Q )$ with 1 and all edges incident with $u 3$ in $E [ V ( F ) , V ( Q ) ]$ with 2, and color the remaining edges with 1. Thus, $p c ( G ) = s p c ( G ) = 2$.
Next, we consider $N Q ( u 1 ) ∩ N Q ( u 3 ) ≠ Ø$ and say $w 1 ∈ N Q ( u 1 ) ∩ N Q ( u 3 )$. Let $d Q ( u 1 ) = 2$ and $d Q ( u 2 ) = d Q ( u 3 ) = 1$. The following edge-coloring $θ$ with two colors makes G properly connected: color all edges of $E ( Q )$ with 1 and the remaining edges with 2. Hence, $p c ( G ) = 2$. If there exists a strong proper coloring $θ$ of G with two colors, then $θ ( u 1 u ) ≠ θ ( u 2 u )$. Without a loss of generality, let $θ ( u 1 u ) = 1$ and $θ ( u 2 u ) = 2$. Since $u 2 u w 1 u 3$ is the unique $u 2 − u 3$ geodesic, it follows that $θ ( u w 1 ) = 1$ and $θ ( u 3 w 1 ) = 2$. Note that $u 1 w 1 u 3$ is the unique $u 1 − u 3$ geodesic, and thus $θ ( u 1 w 1 ) = 1$. Since $u 2 u v$ is the unique $u 2 − v$ geodesic and $u 3 w 1 v$ is the unique $u 3 − v$ geodesic, we obtain $θ ( u v ) = θ ( w 1 v ) = 1$, where $v ∈ V ( Q ) \ { u , w 1 }$. There is no proper geodesic connecting $u 1$ and v, which is a contradiction. Hence, $s p c ( G ) ≥ 3$. On the other hand, we assign a strong proper coloring $θ ′$ with three colors to G as follows: $θ ′ ( u 1 u ) = θ ′ ( u 3 w 1 ) = 1$, $θ ′ ( u 2 u ) = θ ′ ( u 1 w 1 ) = 2$, and color all edges of $E ( Q )$ with 3. Therefore, $s p c ( G ) = 3$. Let $d Q ( u 1 ) ≥ 3$. The following edge-coloring $θ$ of G with two colors makes G strongly properly connected: $θ ( u 1 u ) = θ ( u 1 w 1 ) = 1$, $θ ( u 2 u ) = θ ( u 3 w 1 ) = θ ( u 1 w ) = 2$, where $w ∈ N Q ( u 1 ) \ { u , w 1 }$, and color the remaining edges with 1. Thus, $p c ( G ) = s p c ( G ) = 2$. Let $m a x { d Q ( u 2 ) , d Q ( u 3 ) } ≥ 2$. Without a loss of generality, we consider $d Q ( u 2 ) ≥ 2$. Define an edge-coloring $θ$ of G with two colors as follows: $θ ( u 1 u ) = θ ( u 3 w 1 ) = θ ( u 2 z ) = 1$, where $z ∈ N Q ( u 2 )$, $θ ( u 1 w 1 ) = θ ( u 2 u ) = 2$, and color all edges of $E ( Q )$ with 2 and the remaining edges with 1. Obviously, $θ$ is a strong proper coloring of G, and so $p c ( G ) = s p c ( G ) = 2$.
Case 2. $d i a m ( G ) ≥ 4$. Since $d i a m ( G ) ≥ 4$, it follows that $F ≅ P 3$ or $F ≅ K 2 + K 1$. Assume that $F ≅ P 3$, where $u 1 u 2 , u 2 u 3 ∈ E ( G )$. Since $d i a m ( G ) ≥ 4$, we have $m i n { d Q ( u 1 ) , d Q ( u 3 ) } = 0$, $m a x { d Q ( u 1 ) , d Q ( u 3 ) } ≥ 1$, and $d Q ( u 2 ) = 0$. Without a loss of generality, let $d Q ( u 1 ) ≥ 1$ and $d Q ( u 2 ) = d Q ( u 3 ) = 0$. Note that G is traceable, and we have $p c ( G ) = 2$ by Proposition 1. The following edge-coloring $θ$ with two colors makes G strongly properly connected: color $u 1 u 2$ and all edges of $E ( Q )$ with 2, and color the remaining edges with 1. Thus, $s p c ( G ) = 2$.
Assume that $F ≅ K 2 + K 1$, where $u 1 u 2 ∈ E ( G )$. Since $d i a m ( G ) ≥ 4$, we have $m i n { d Q ( u 1 ) , d Q ( u 2 ) } = 0$, $m a x { d Q ( u 1 ) , d Q ( u 2 ) } ≥ 1$, and $d Q ( u 3 ) ≥ 1$. Without a loss of generality, let $d Q ( u 1 ) ≥ 1$, $d Q ( u 2 ) = 0$ and $N Q ( u 1 ) ∩ N Q ( u 3 ) = Ø$. Observe that G is traceable, and we obtain $p c ( G ) = 2$ by Proposition 1. Assign a strong proper coloring $θ$ with two colors to G as follows: color $u 1 u 2$ and all edges of $E ( Q )$ with 2, and color the remaining edges with 1. Hence, $s p c ( G ) = 2$. □

## 3. Rainbow Connection and Clique Number

Kemnitz and Schiermeyer  considered the rainbow connection number of graph G of order n, $d i a m ( G ) = 2$, and $ω ( G ) = n − i$ for $2 ⩽ i ⩽ 3$. In this section, we investigate the rainbow connection number of graph G of order n, $d i a m ( G ) ≥ 3$, and $ω ( G ) = n − i$ for $2 ⩽ i ⩽ 3$.
Theorem 5.
Let G be a connected graph of order n, $d i a m ( G ) ≥ 3$, and $ω ( G ) = n − 2$. Let Q be a maximum clique of G and $V ( G ) \ V ( Q ) = { u 1 , u 2 }$. Then, $r c ( G ) = 3$.
Proof.
Let $F = G [ V ( G ) \ V ( Q ) ]$ and let $θ$ be an edge-coloring of G. Since $d i a m ( G ) ≥ 3$, we have $r c ( G ) ≥ d i a m ( G ) ≥ 3$. Assume that $F ≅ K 2$. Since $d i a m ( G ) ≥ 3$, we obtain $m a x { d Q ( u 1 ) , d Q ( u 2 ) } ≥ 1$ and $m i n { d Q ( u 1 ) , d Q ( u 2 ) } = 0$. The following edge-coloring $θ$ with three colors makes G rainbow-connected: color $u 1 u 2$ with 1 and all edges of $E [ V ( F ) , V ( Q ) ]$ with 2, and color all edges of $E ( Q )$ with 3. Thus, $r c ( G ) = 3$.
Assume that $F ≅ 2 K 1$. Since G is a connected graph with $d i a m ( G ) ≥ 3$, it follows that $d Q ( u 1 ) ≥ 1$, $d Q ( u 2 ) ≥ 1$ and $N Q ( u 1 ) ∩ N Q ( u 2 ) = Ø$. Assign an edge-coloring $θ$ with three colors to G as follows: assign 1 to all edges that are incident with $u 1$, assign 2 to all edges that are incident with $u 2$, and assign 3 to all edges of $E ( Q )$. It is not difficult to check that G is rainbow-connected with the above edge-coloring, and so $r c ( G ) = 3$. □
Theorem 6.
Let G be a connected graph of order n, $d i a m ( G ) ≥ 3$, and $ω ( G ) = n − 3$. Let Q be a maximum clique of G and $V ( G ) \ V ( Q ) = { u 1 , u 2 , u 3 }$. Then, either $r c ( G ) = 3$, or $r c ( G ) = 4$ if and only if one of the following holds.
(i)
$G [ V ( G ) \ V ( Q ) ] ≅ P 3$, where $u 1 u 2 , u 2 u 3 ∈ E ( G )$, $d Q ( u 1 ) = d Q ( u 3 ) = 0$, and $d Q ( u 2 ) = 1$.
(ii)
$G [ V ( G ) \ V ( Q ) ] ≅ K 2 + K 1$, where $u 1 u 2 ∈ E ( G )$, $d Q ( u 2 ) = 0$, $d Q ( u 1 ) = d Q ( u 3 ) = 1$, and $N Q ( u 1 ) ∩ N Q ( u 3 ) ≠ Ø$.
(iii)
$G [ V ( G ) \ V ( Q ) ] ≅ 3 K 1$, $N Q ( u 1 ) ∩ N Q ( u 2 ) ∩ N Q ( u 3 ) = Ø$, , and $d Q ( u 1 ) = d Q ( u 2 ) = d Q ( u 3 ) = 1$.
(iv)
$G [ V ( G ) \ V ( Q ) ] ≅ P 3$, where $u 1 u 2 , u 2 u 3 ∈ E ( G )$, $d Q ( u 1 ) ≥ 1$, and $d Q ( u 2 ) = d Q ( u 3 ) = 0$.
(v)
$G [ V ( G ) \ V ( Q ) ] ≅ K 2 + K 1$, where $u 1 u 2 ∈ E ( G )$, $d Q ( u 1 ) ≥ 1$, $d Q ( u 2 ) = 0$, $d Q ( u 3 ) ≥ 1$, and $N Q ( u 1 ) ∩ N Q ( u 3 ) = Ø$.
Proof.
Let $F = G [ V ( G ) \ V ( Q ) ]$, and let $θ$ be an edge-coloring of G. We prove this theorem by the following two cases.
Case 1. $d i a m ( G ) = 3$. We have $r c ( G ) ≥ d i a m ( G ) = 3$. We distinguish the following four subcases by analyzing the structure of F.
Subcase 1.1. $F ≅ K 3$. The following edge-coloring $θ$ with three colors makes G rainbow-connected: $θ ( u 1 u 2 ) = θ ( u 2 u 3 ) = θ ( u 1 u 3 ) = 1$, and color all edges of $E ( Q )$ with 3 and all edges of $E [ V ( F ) , V ( Q ) ]$ with 2. Thus, $r c ( G ) = 3$.
Subcase 1.2. $F ≅ P 3$, where $u 1 u 2 , u 2 u 3 ∈ E ( G )$. Assume that $d Q ( u 1 ) = d Q ( u 3 ) = 0$. Suppose $d Q ( u 2 ) = 1$, and say $N Q ( u 2 ) = { u }$. If an edge-coloring $θ$ is a rainbow coloring of G, then $θ ( u 1 u 2 ) ≠ θ ( u 2 u 3 ) ≠ θ ( u 2 u ) ≠ θ ( u v )$, where $v ∈ V ( Q ) \ { u }$. Hence, $r c ( G ) ≥ 4$. Allocate a rainbow coloring $θ ′$ with four colors to G as follows: $θ ′ ( u 1 u 2 ) = 1$, $θ ′ ( u 2 u 3 ) = 2$, $θ ′ ( u 2 u ) = 3$, and color all edges of $E ( Q )$ with 4. Thus, $r c ( G ) = 4$. Suppose $d Q ( u 2 ) ≥ 2$, and say $u , v ∈ N Q ( u 2 )$. The following edge-coloring $θ$ with three colors makes G rainbow-connected: $θ ( u 1 u 2 ) = θ ( u 2 u ) = 1$, $θ ( u 2 u 3 ) = θ ( u 2 v ) = 2$, and color the remaining edges with 3. Hence, $r c ( G ) = 3$.
Assume that $m i n { d Q ( u 1 ) , d Q ( u 3 ) } = 0$ and $m a x { d Q ( u 1 ) , d Q ( u 3 ) } ≥ 1$. Without a loss of generality, let $d Q ( u 3 ) = 0$ and $d Q ( u 1 ) ≥ 1$. Since $d i a m ( G ) = 3$, we have $d Q ( u 2 ) ≥ 1$. Define an edge-coloring $θ$ of G with three colors as follows: $θ ( u 1 u 2 ) = 1$, $θ ( u 2 u 3 ) = 2$, and color all edges of $E [ V ( F ) , V ( Q ) ]$ with 1 and all edges of $E ( Q )$ with 3. We can check that G is rainbow-connected with the above edge-coloring, and so $r c ( G ) = 3$.
Assume that $d Q ( u 1 ) ≥ 1$ and $d Q ( u 3 ) ≥ 1$. Since $d i a m ( G ) = 3$, it follows that $d Q ( u 2 ) = 0$ and $N Q ( u 1 ) ∪ N Q ( u 3 ) ≠ V ( Q )$. The following edge-coloring $θ$ with three colors makes G rainbow-connected: $θ ( u 1 u 2 ) = 1$, $θ ( u 2 u 3 ) = 2$, assign 3 to all edges of $E ( Q )$, assign 2 to the edges of $E [ V ( F ) , V ( Q ) ]$ which are incident with $u 1$, and assign 1 to the edges of $E [ V ( F ) , V ( Q ) ]$ which are incident with $u 3$. Thus, $r c ( G ) = 3$.
Subcase 1.3. $F ≅ K 2 + K 1$, where $u 1 u 2 ∈ E ( G )$. Since G is connected, we obtain $d Q ( u 3 ) ≥ 1$ and $m a x { d Q ( u 1 ) , d Q ( u 2 ) } ≥ 1$. Without a loss of generality, let $d Q ( u 1 ) ≥ 1$.
Assume that $d Q ( u 2 ) = 0$. Since $d i a m ( G ) = 3$, we have $N Q ( u 1 ) ∩ N Q ( u 3 ) ≠ Ø$, and say $u ∈ N Q ( u 1 ) ∩ N Q ( u 3 )$. Suppose $d Q ( u 1 ) = d Q ( u 3 ) = 1$. If there exists a rainbow coloring $θ$ of G with three colors, then $θ ( u 2 u 1 ) ≠ θ ( u 1 u ) ≠ θ ( u u 3 )$. Without a loss of generality, let $θ ( u 2 u 1 ) = 1$, $θ ( u 1 u ) = 2$ and $θ ( u u 3 ) = 3$. In order to have a rainbow path connecting $u 2$ and v for any $v ∈ V ( Q ) \ { u }$, let $θ ( u v ) = 3$. There is no rainbow path between $u 3$ and v, which is a contradiction. Thus, $r c ( G ) ≥ 4$. On the other hand, the following edge-coloring $θ ′$ with four colors makes G rainbow-connected: $θ ′ ( u 2 u 1 ) = 1$, $θ ′ ( u 1 u ) = 2$, $θ ′ ( u u 3 ) = 3$, and color all edges of $E ( Q )$ with 4. Hence, $r c ( G ) = 4$. Suppose $m a x { d Q ( u 1 ) , d Q ( u 3 ) } ≥ 2$. We first consider $d Q ( u 1 ) ≥ 2$, and say $v ∈ N Q ( u 1 ) \ { u }$. Assign an edge-coloring $θ$ with three colors to G as follows: $θ ( u 2 u 1 ) = 1$, $θ ( u 1 u ) = 2$, $θ ( u 3 u ) = θ ( u 1 v ) = 3$, and color the remaining edges with 2. It is obvious that G is rainbow-connected with the above edge-coloring, and so $r c ( G ) = 3$. Next, we consider $d Q ( u 3 ) ≥ 2$, and say $w ∈ N Q ( u 3 ) \ { u }$. Define a rainbow coloring $θ$ of G with three colors as follows: $θ ( u 2 u 1 ) = 1$, $θ ( u 1 u ) = θ ( u 3 w ) = 2$, $θ ( u 3 u ) = 3$, and color all edges of $E ( Q )$ with 3 and the remaining edges with 2. Thus, $r c ( G ) = 3$.
Assume that $d Q ( u 2 ) ≥ 1$. Since $d i a m ( G ) = 3$, we obtain $m i n { | N Q ( u 1 ) ∩ N Q ( u 3 ) | , | N Q$
$( u 2 ) ∩ N Q ( u 3 ) | } = 0$. Suppose $m a x { | N Q ( u 1 ) ∩ N Q ( u 3 ) | , | N Q ( u 2 ) ∩ N Q ( u 3 ) | } ≥ 1$. Without a loss of generality, let and . Let $u ∈ N Q ( u 1 ) ∩ N Q ( u 3 )$ and $v ∈ N Q ( u 2 )$. The following edge-coloring $θ$ with three colors makes G rainbow-connected: $θ ( u 1 u ) = θ ( u 2 v ) = 1$, $θ ( u 3 u ) = 2$, and color the remaining edges with 3. Hence, $r c ( G ) = 3$. Suppose $m a x { | N Q ( u 1 ) ∩ N Q ( u 3 ) | , | N Q ( u 2 ) ∩ N Q ( u 3 ) | } = 0$. Let $w ∈ N Q ( u 1 ) , v ∈ N Q ( u 2 )$ and $u ∈ N Q ( u 3 )$, where $w = v$ is possible. Allocate an edge-coloring $θ$ with three colors to G: $θ ( u 1 u 2 ) = θ ( u 3 u ) = 1$, $θ ( u 1 w ) = θ ( u 2 v ) = 2$, and color the remaining edges with 3. We can verify that G is rainbow-connected with the above edge-coloring, and so $r c ( G ) = 3$.
Subcase 1.4. $F ≅ 3 K 1$. Since $d i a m ( G ) = 3$, it follows that $m i n { | N Q ( u 1 ) ∩ N Q ( u 2 ) | , | N Q$
$( u 1 ) ∩ N Q ( u 3 ) | , | N Q ( u 2 ) ∩ N Q ( u 3 ) | } = 0$. Assume that $m a x { | N Q ( u 1 ) ∩ N Q ( u 2 ) | , | N Q ( u 1 ) ∩ N Q ( u 3 ) | , | N Q ( u 2 ) ∩ N Q ( u 3 ) | } = 0$. Let $u ∈ N Q ( u 1 ) , v ∈ N Q ( u 2 )$ and $w ∈ N Q ( u 3 )$. The following edge-coloring $θ$ with three colors makes G rainbow-connected: $θ ( u 1 u ) = θ ( v w ) = θ ( v z ) = 1$; $θ ( u v ) = θ ( u 3 w ) = θ ( u z ) = 2$; $θ ( u 2 v ) = θ ( u w ) = θ ( w z ) = 3$ for any $z ∈ V ( Q ) \ { u , v , w }$; and color the remaining edges with 1. Thus, $r c ( G ) = 3$.
Assume that $m a x { | N Q ( u 1 ) ∩ N Q ( u 2 ) | , | N Q ( u 1 ) ∩ N Q ( u 3 ) | , | N Q ( u 2 ) ∩ N Q ( u 3 ) | } ≥ 1$. Without a loss of generality, let , and say $u ∈ N Q ( u 1 ) ∩ N Q ( u 2 )$. Suppose $d Q ( u 1 ) = d Q ( u 2 ) = d Q ( u 3 ) = 1$. If an edge-coloring $θ$ is a rainbow coloring of G, then $θ ( u 1 u ) ≠ θ ( u 2 u ) ≠ θ ( u v ) ≠ θ ( u 3 v )$, where ${ v } = N Q ( u 3 )$. Thus, $r c ( G ) ≥ 4$. On the other hand, we define a rainbow coloring $θ ′$ of G with four colors as follows: $θ ′ ( u 1 u ) = 1 , θ ′ ( u 2 u ) = 2 , θ ′ ( u 3 v ) = 3$, and color all edges of $E ( Q )$ with 4. Hence, $r c ( G ) = 4$. Suppose $m a x { d Q ( u 1 ) , d Q ( u 2 ) } ≥ 2$. Without a loss of generality, let $d Q ( u 1 ) ≥ 2$, and say $w ∈ N Q ( u 1 ) \ { u }$. Assign an edge-coloring $θ$ with three colors to G: $θ ( u 1 u ) = θ ( u 2 u ) = 1$; $θ ( u 1 w ) = θ ( u 3 v ) = 2$, where $v ∈ N Q ( u 3 )$ and $v = w$ is possible; and color the remaining edges with 3. Obviously, the edge-coloring $θ$ is a rainbow coloring of G, and so $r c ( G ) = 3$. Suppose $d Q ( u 3 ) ≥ 2$, and say $v 1 , v 2 ∈ N Q ( u 3 )$. The following edge-coloring $θ$ with three colors makes G rainbow-connected: $θ ( u 1 u ) = θ ( u 3 v 1 ) = 1$, $θ ( u 2 u ) = θ ( u 3 v 2 ) = 2$, and color the remaining edges with 3. Thus, $r c ( G ) = 3$.
Case 2. $d i a m ( G ) ≥ 4$. We obtain $r c ( G ) ≥ d i a m ( G ) ≥ 4$. Since $d i a m ( G ) ≥ 4$, it follows that $F ≅ P 3$ or $F ≅ K 2 + K 1$. Assume that $F ≅ P 3$, where $u 1 u 2 , u 2 u 3 ∈ E ( G )$. Since $d i a m ( G ) ≥ 4$, we have $m i n { d Q ( u 1 ) , d Q ( u 3 ) } = 0$, $m a x { d Q ( u 1 ) , d Q ( u 3 ) } ≥ 1$, and $d Q ( u 2 ) = 0$. Without a loss of generality, let $d Q ( u 1 ) ≥ 1$ and $d Q ( u 2 ) = d Q ( u 3 ) = 0$. Allocate a rainbow coloring $θ$ 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 $E [ V ( F ) , V ( Q ) ]$ with 3 and all edges of $E ( Q )$ with 4. Therefore, $r c ( G ) = 4$.
Assume that $F ≅ K 2 + K 1$, where $u 1 u 2 ∈ E ( G )$. Since $d i a m ( G ) ≥ 4$, it follows that $m i n { d Q ( u 1 ) , d Q ( u 2 ) } = 0$, $m a x { d Q ( u 1 ) , d Q ( u 2 ) } ≥ 1$, and $d Q ( u 3 ) ≥ 1$. Without a loss of generality, let $d Q ( u 1 ) ≥ 1$, $d Q ( u 2 ) = 0$, and $N Q ( u 1 ) ∩ N Q ( u 3 ) = Ø$. The following edge-coloring $θ$ with four colors makes G rainbow-connected: $θ ( u 1 u 2 ) = 1$, $θ ( u 1 u ) = 2$, and $θ ( u 3 v ) = 3$, where $u ∈ N Q ( u 1 )$ and $v ∈ N Q ( u 3 )$, and color the remaining edges with 4. Hence, $r c ( G ) = 4$. □

## Author Contributions

Conceptualization, Y.X.; Methodology, Y.M.; Writing—original draft, X.Z.; Writing—review and editing, X.Z. and Y.X.; Supervision, Y.M.; Funding acquisition, Y.M. All authors have read and agreed to the published version of the manuscript.

## Funding

This work was supported by the Foundation of Henan Educational Committee (22A110003).

Not applicable.

Not applicable.

## Data Availability Statement

Data sharing not applicable. No new data were created or analyzed in this study. Data sharing is not applicable to this article.

## Conflicts of Interest

The authors declare no conflict of interest.

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Figure 1. The graph $G 2$ with a strong proper coloring.
Figure 1. The graph $G 2$ with a strong proper coloring.
Figure 2. The graphs $G 3$ and $G 4$ with a strong proper coloring.
Figure 2. The graphs $G 3$ and $G 4$ with a strong proper coloring.
Figure 3. The graphs $G 5$ and $G 6$ with a proper-path coloring.
Figure 3. The graphs $G 5$ and $G 6$ with a proper-path coloring.
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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

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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

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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

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