Strong conflict-free connection of graphs

TL;DR

This paper introduces the concept of strong conflict-free connection in graphs, providing bounds on the number of colors needed and characterizations for specific graph classes, including cubic graphs.

Contribution

It establishes bounds on the strong conflict-free connection number for graphs with triangles and characterizes graphs with specific strong conflict-free connection numbers.

Findings

For graphs with t triangles, scfc(G) ≤ m - 2t.

Characterization of graphs with scfc(G) in {1, m-3, m-2, m-1, m}.

Complete characterization of cubic graphs with scfc(G)=2.

Abstract

A path $P$ in an edge-colored graph is called \emph{a conflict-free path} if there exists a color used on only one of the edges of $P$. An edge-colored graph $G$ is called \emph{conflict-free connected} if for each pair of distinct vertices of $G$ there is a conflict-free path in $G$ connecting them. The graph $G$ is called \emph{strongly conflict-free connected }if for every pair of vertices $u$ and $v$ of $G$ there exists a conflict-free path of length $d_G(u,v)$ in $G$ connecting them. For a connected graph $G$, the \emph{strong conflict-free connection number} of $G$, denoted by $\mathit{scfc}(G)$, is defined as the smallest number of colors that are required in order to make $G$ strongly conflict-free connected. In this paper, we first show that if $G_t$ is a connected graph with $m$ $(m\geq 2)$ edges and $t$ edge-disjoint triangles, then $\mathit{scfc}(G_t)\leq m-2t$, and the equality holds if and only if $G_t\cong S_{m,t}$. Then we characterize the graphs $G$ with $scfc(G)=k$ for $k\in \{1,m-3,m-2,m-1,m\}$. In the end, we present a complete characterization for the cubic graphs $G$ with $scfc(G)=2$.

Full text

Keywords: strong conflict-free connection coloring (number); characterization; cubic graph

AMS subject classification 2010: 05C15, 05C40, 05C75.

1 Introduction

All graphs mentioned in this paper are simple, undirected and finite. We follow book [2] for undefined notation and terminology. Coloring problems are important topics in graph theory. In recent years, there have appeared a number of colorings raising great concern due to their wide applications in real world. We list a few well-known colorings here. The first of such would be the rainbow connection coloring, which is stated as follows. A path in an edge-colored graph is called a rainbow path if all the edges of the path have distinct colors. An edge-colored graph is called ($strongly$) rainbow connected if there is a ($shortest$ and) rainbow path between every pair of distinct vertices in the graph. For a connected graph $G$, the ($strong$) rainbow connection number of $G$ is defined as the smallest number of colors needed to make $G$ ($strongly$) rainbow connected, denoted by ($\mathit{src}(G)$) $\mathit{rc}(G)$. These concepts were first introduced by Chartrand et al. in [6].

Inspired by the rainbow connection coloring, the concept of proper connection coloring was independently posed by Andrews et al. in [1] and Borozan et al. in [3], the only difference from ($strong$) rainbow connection coloring is that distinct colors are only required for adjacent edges instead of all edges on the ($shortest$) path. For an edge-colored connected graph $G$, the smallest number of colors required to give $G$ a ($strong$) proper connection coloring is called the (strong) proper connection number of $G$, denoted by $(\mathit{spc}(G))$ $\mathit{pc}(G)$.

The hypergraph version of conflict-free coloring was first introduced by Even et al. in [10]. A hypergraph $H$ is a pair $H=(X,E)$ where $X$ is the set of vertices, and $E$ is the set of nonempty subsets of $X$, called hyperedges. The coloring was motivated to solve the problem of assigning frequencies to different base stations in cellular networks, which is defined as a vertex-coloring of $H$ such that every hyperedge contains a vertex with a unique color.

Later on, Czap et al. in [8] introduced the concept of conflict-free connection coloring of graphs, motivated by the earlier hypergraph version. A path in an edge-colored graph $G$ is called a conflict-free path if there is a color appearing only once on the path. The graph $G$ is called conflict-free connected if there is a conflict-free path between each pair of distinct vertices of $G$. For a connected graph $G$, the minimum number of colors required to make $G$ conflict-free connected is defined as the conflict-free connection number of $G$, denoted by $\mathit{cfc}(G)$. For more results, the reader can be referred to [4, 5, 6, 15].

In this paper, we focus on studying the strong conflict-free connection coloring which was introduced by Ji et al. in [13], where only computational complexity was studied. An edge-colored graph is called *strongly conflict-free connected *if there exists a conflict-free path of length $d_{G}(u,v)$ for every pair of vertices $u$ and $v$ of $G$. For a connected graph $G$, the strong conflict-free connection number of $G$, denoted $\mathit{scfc}(G)$, is the smallest number of colors that are required to make $G$ strongly conflict-free connected.

The paper is organized as follows. In Section 2, we give some preliminary results. In Section 3, we show that if $G_{t}$ is a connected graph with $m$ $(m\geq 2)$ edges and $t$ edge-disjoint triangles, then $\mathit{scfc}(G_{t})\leq m-2t$, and the equality holds if and only if $G_{t}\cong S_{m,t}$. In Section 4, we characterize the graphs $G$ with $scfc(G)=k$ for $k\in\{1,m-3,m-2,m-1,m\}$. In the last section, we completely characterize the cubic graphs $G$ with $scfc(G)=2$.

2 Preliminaries

In this section, we present some results which will be used in the sequel. In [13], the authors obtained the following computational complexity result.

Theorem 2.1

[13]* For a connected graph $G$ and integer $k\geq 2$, deciding whether $\mathit{scfc}(G)\leq k$ is NP-complete.*

They also showed the following result.

Theorem 2.2

[13] For a graph $G$, $\mathit{rc}(G)=2$ if and only if $diam(G)=2$ and $\mathit{scfc}(G)=2$.

Note that from [6], one has that $rc(G)=2$ if and only if $src(G)=2$. The following result is obvious.

Theorem 2.3

For a tree $T$, $\mathit{scfc}(T)=\mathit{cfc}(T)$. Therefore, for a path $P_{n}$ on $n$ vertices, $\mathit{scfc}(P_{n})=\lceil\log_{2}n\rceil$; for a star $S_{m}$ with $m$ edges, $\mathit{scfc}(S_{m})=m$.

The authors in [6] obtained the strong rainbow connection number for a wheel graph $W_{n}$, where $n$ is the degree of the central vertex.

Theorem 2.4

[6]* For $n\geq 3$, let $W_{n}$ be a wheel. Then $\mathit{src}(W_{n})=\lceil\frac{n}{3}\rceil$.*

For a complete bipartite graph $K_{s,t}$, they also got the following result.

Theorem 2.5

[6] For integers $s$ and $t$ with $1\leq s\leq t$, $\mathit{src}(K_{s,t})=\lceil\sqrt[s]{t}\rceil$.

From the above results, we get that

Theorem 2.6

$\mathit{scfc}(W_{n})=\lceil\frac{n}{3}\rceil$.

Proof. Note that for a graph $G$ with diameter 2, a strong rainbow path (of length 2) of $G$ is a strong conflict-free path of $G$, and vice versa. Since $diam(W_{n})=2$, then $\mathit{scfc}(W_{n})=\mathit{src}(W_{n})$. So, $\mathit{scfc}(W_{n})=\lceil\frac{n}{3}\rceil$ from Theorem 2.4. $\Box$

Theorem 2.7

For integers $s$ and $t$ with $1\leq s\leq t$, $\mathit{scfc}(K_{s,t})=\lceil\sqrt[s]{t}\rceil$.

Proof. Since $diam(K_{s,t})=2$, from Theorem 2.5 we have that $\mathit{scfc}(K_{s,t})=\lceil\sqrt[s]{t}\rceil$. $\Box$

Proposition 2.8

Let $C_{n}$ be a cycle of order $n$ and let $P_{n}$ be a spanning subgraph of $C_{n}$. Then $\mathit{scfc}(C_{n})\leq\mathit{scfc}(P_{n})$.

Proof. Let $P_{n}=e_{1}e_{2}\cdots e_{n-1}$ be a path with $n$ vertices and let $u$ and $v$ be the ends of $P_{n}$. We know that $\mathit{scfc}(P_{n})=\lceil\log_{2}n\rceil$ by Theorem 2.3. Now we first give a coloring for $P_{n}$: color the edge $e_{i}$ with color $x+1$, where $2^{x}$ is the largest power of 2 that divides $i$. One can see that $\lceil\log_{2}n\rceil$ is the largest number in the coloring by Theorem 2.3. Clearly, the color $\lceil\log_{2}n\rceil$ only occurs once. Thus, we color the edge $uv$ with $\lceil\log_{2}n\rceil$ in $C_{n}$ if there is only one color occurring once; otherwise, we color the edge $uv$ with $\lceil\log_{2}n\rceil-1$. Consequently, the coloring is a strong conflict-free connection coloring of $C_{n}$. $\Box$

Remark: The proposition does not hold for general graphs. Here is a counterexample. Let $G=C_{6}$ with the edge set $\{v_{1}v_{2},v_{2}v_{3},v_{3}v_{4},v_{4}v_{5},v_{5}v_{6},v_{6}v_{1}\}$. So $\mathit{scfc}(G)=2$. Let $G^{\prime}=C_{6}+v_{1}v_{3}$. Then $\mathit{scfc}(G^{\prime})=3$.

Theorem 2.9

If $C_{n}$ is a cycle with $n$ $(n\geq 3)$ vertices, then

$\mathit{scfc}(C_{n})=\lceil\log_{2}n\rceil-1$* or $\lceil\log_{2}n\rceil$.*

Proof. By Proposition 2.8 and Theorem 2.3, one can see that $\mathit{scfc}(C_{n})\leq\lceil\log_{2}n\rceil$. It remains to handle with the lower bound. We first consider the case that $diam(C_{n})=\frac{n}{2}$ for $n=2k$ $(k\in\mathbb{Z}^{+})$. Hence, $\mathit{scfc}(C_{n})\geq\lceil\log_{2}(\frac{n}{2}+1)\rceil=\lceil\log_{2}(n+2)\rceil-1\geq\lceil\log_{2}n\rceil-1$. We then consider the case that $diam(C_{n})=\frac{n-1}{2}$ for $n=2k+1$ $(k\in\mathbb{Z}^{+})$. Thus, $\mathit{scfc}(C_{n})\geq\lceil\log_{2}(\frac{n-1}{2}+1)\rceil=\lceil\log_{2}(n+1)\rceil-1\geq\lceil\log_{2}n\rceil-1$. Consequently, $\mathit{scfc}(C_{n})=\lceil\log_{2}n\rceil-1$ or $\lceil\log_{2}n\rceil$. $\Box$

Theorem 2.9 implies the following corollary.

Corollary 2.10

Let $G$ be a connected graph with $m$ edges and let $C$ be a cycle in $G$. Then $\mathit{scfc}(G)\leq m-|C|+\lceil\log_{2}|C|\rceil$.

Proof. By Theorem 2.9, $\mathit{scfc}(C)\leq\lceil\log_{2}|C|\rceil$. If we color the edges of $C$ with $\lceil\log_{2}|C|\rceil$ colors to make $C$ strongly conflict-free connected, and color each of the remaining $m-|C|$ edges with a fresh color, then we can verify that $G$ is strongly conflict-free connected. Consequently, $\mathit{scfc}(G)\leq m-|C|+\lceil\log_{2}|C|\rceil$. $\Box$

A graph $G$ is called $k$-$\mathit{cfc}$-$critical$ if $\mathit{cfc}(G)=k$ and for any proper subgraph $G^{\prime}$ of $G$, $cfc(G^{\prime})<k$.

Theorem 2.11

[13] Let $Q_{k}$ be the graph obtained from two copies of $K_{1,k-1}$ with $k\geq 2$ by identifying a leaf vertex in one copy with a leaf vertex in the other copy. Then $Q_{k}$ is $k$-$\mathit{cfc}$-critical.

3 Upper and lower bounds

At first, let us look at trees.

Theorem 3.1

Let $T$ be a tree of order $n$. Then

\max\{\lceil\log_{2}(diam(T)+1)\rceil,\Delta(T)\}$$\leq\mathit{scfc}(T)\leq n-1.

Proof. Clearly, it is a strong conflict-free coloring that colors the edges of $T$ with distinct colors, and so the upper bound holds. For the lower bound, let $P$ be a path of length $diam(T)$, which needs at least $\lceil\log_{2}(diam(T)+1)\rceil$ colors by Theorem 2.3. Meanwhile, since a strong conflict-free connection coloring of a tree must be a proper edge-coloring, it is obvious that $\mathit{scfc}(T)\geq\chi^{\prime}(T)=\Delta(T)$. Now we show that the upper bound is sharp. Let $H=(V(H),E(H))$ be a star. Then $\mathit{scfc}(H)=|E(H)|$ by Theorem 2.3. The lower bound is sharp by Theorem 2.3. $\Box$

Before we show the following theorem, we first define the notion of $t$-parallel paths. Let $G$ be a connected graph and let $u$, $v$ be two vertices of $G$. If there are $t$ paths between $u$ and $v$ in $G$, where the degree of internal vertices of the paths is 2, then we call the paths $t$-parallel paths.

Theorem 3.2

Let $G$ be a connected graph and let $v$, $u$ be two vertices of $G$ with $d(u,v)\geq 2$. If one of the following conditions holds, then $\mathit{scfc}(G)\geq 3$.

There exist a cut-vertex $w$ which splits $G$ into at least three components by deleting $w$. 2. 2.

There exists a path $P$ of length at least $4$ between $u$ and $v$, where the edges of the path are bridges. 3. 3.

There exist 2-parallel paths between $u$ and $v$, where the length of one path is $2$ and the length of the other one is $3$. 4. 4.

There exist 5-parallel paths between $u$ and $v$.

Proof.

1. Let $C_{1},C_{2},\cdots,C_{m}$ $(m\geq 3)$ be the components when deleting $v$ from $G$. We choose a vertex $u_{i}$ which is adjacent to $v$ in each component $C_{i}$. Clearly, each pair of $u_{i}$ and $u_{j}$ contains the only path, and it contains $v$. Consequently, the subgraph of $G$ induced by $\{u_{1},u_{2},\cdots,u_{m},v\}$ is a star $S_{m}$. By Theorem 2.3, we have $\mathit{scfc}(S_{m})=m$. Because every pair of $u_{i}$ and $u_{j}$ has the same path in $S_{m}$ and in $G$, we have that $\mathit{scfc}(G)\geq\mathit{scfc}(S_{m})=m\geq 3$.

2. Let $P=v_{1}v_{2}\cdots v_{t}$ $(t\geq 4)$. Since every edge of $P$ is a bridge, each pair of vertices $v_{i}$, $v_{j}$ contain the same path in $P$ and in $G$. Hence, we have $\mathit{scfc}(G)\geq\mathit{scfc}(P)\geq 3$.

3. Since the lengths of the two paths are 2 and 3, there is a 5-cycle in $G$. Clearly, $\mathit{scfc}(G)\geq 3$.

4. Since $d(u,v)\geq 2$, every path between $u$ and $v$ has a length at least 2. If we assign a coloring with 2 colors for the paths, then there always exist at least two internal vertices of the paths which do not contain a strong conflict-free path. Consequently, $\mathit{scfc}(G)\geq 3$. $\Box$

We now define a graph class. Let $S_{m}$ be a star with $m$ leaves $v_{1},v_{2},\cdots,v_{m}$. We denote by $S_{m,t}$ the graph $S_{m}+\{v_{i_{1}}v_{i_{2}},v_{i_{3}}v_{i_{4}},\cdots,v_{i_{t-1}}v_{i_{t}}\}$ $(v_{i_{j}}\neq v_{i_{k}}|j,k\in[t],i_{k},i_{j}\in[m])$.

Theorem 3.3

If $G_{t}$ is a connected graph with $m$ $(m\geq 2)$ edges and $t$ edge-disjoint triangles, then $\mathit{scfc}(G_{t})\leq m-2t$, and the equality holds if and only if $G_{t}\cong S_{m,t}$.

Proof. Clearly, $\mathit{scfc}(K_{3})=1$. Now we first give a coloring of $G_{t}$: color each triangle with a distinct color, that is, the three edges of each triangle receive a same color, and color each of the remaining $m-3t$ edges with a distinct color. Let $P$ be a strong conflict-free path for any pair of vertices $u$ and $v$ in $G$. Clearly, $P$ contains at most one edge from each triangle. Otherwise, it will produce a contradiction. Thus, $G_{t}$ is strongly conflict-free connected. So $\mathit{scfc}(G_{t})\leq m-2t$.

We now show that the equality holds if and only if $G_{t}\cong S_{m,t}$.

Claim 1. $\mathit{scfc}(S_{m,t})=m-2t$.

Proof of Claim 1. Clearly, $\mathit{scfc}(S_{m,t})\leq m-2t$. It remains to show the other round. Note that every pendant edge needs a distinct color and every triangle needs a fresh color. Assume that we color some triangle with one color used on some pendant edge. Then the shortest path is not a conflict-free path between the leaf incident with the pendant edge and one vertex of degree two. Also, if we provide the $t$ triangles with $t-1$ colors, there exist two triangle with the same color. There would also not exist a strong conflict-free path between the vertices of the two triangles. Consequently, $\mathit{scfc}(S_{m,t})\geq m-2t$.

Claim 2. Every edge is a cut-edge except the edges of triangles.

Proof of Claim 2. Assume that there is a cycle $C$ $(|C|\geq 3)$ except the $t$ triangles. By Theorem 2.9, we know that $\mathit{scfc}(C)\leq\lceil\log_{2}{|C|}\rceil$. Now we provide a coloring: color every triangle with a distinct color and color $C$ with $\lceil\log_{2}{|C|}\rceil$ fresh colors, and the remaining edges are colored by $m-|E(C)|-3t$ fresh colors. Clearly, $G_{t}$ is strongly conflict-free connected. So, $\mathit{scfc}(G_{t})\leq m-2t+\lceil\log_{2}{|C|}\rceil-|C|\leq m-2t-1$, a contradiction.

Claim 3. Each triangle contains at least two vertices of degree two in $G_{t}$.

Proof of Claim 3. Assume that there is only one vertex of degree two in a triangle $A$, say $A=v_{1}v_{2}v_{3}v_{1}$. Without loss of generality, let $u_{1}v_{1}$ and $u_{2}v_{2}$ be two edges. We will consider the following three cases:

Case 1. Both $u_{1}v_{1}$ and $u_{2}v_{2}$ are not contained in triangles. In order to find out a contradiction, we provide a coloring $c$: assign each triangle with a distinct color; assign both $u_{1}v_{1}$ and $u_{2}v_{2}$ with a fresh same color; the remaining $m-2-3t$ edges are colored by $m-2-3t$ fresh colors. We only need to check $u_{1}$-$u_{2}$ paths. By Claim 2, there is no other cycle except the $t$ triangles. So $u_{1}v_{1}v_{2}u_{2}$ is the unique path which is strongly conflict-free connected. Clearly, $G_{t}$ is strongly conflict-free connected. Hence, $\mathit{scfc}(G_{t})\leq(m-2-3t)+1+t=m-2t-1$, a contradiction.

Case 2. $u_{1}v_{1}$ and $u_{2}v_{2}$ are contained in different triangles. Let $X_{1}$ contain $u_{1}v_{1}$ and let $X_{2}$ contain $u_{2}v_{2}$. We now provide a coloring: assign $X_{1}$ and $X_{2}$ with the same color; assign the other triangles with $t-2$ fresh colors; each of the remaining edges is colored by a fresh color. Clearly, $G_{t}$ is strongly conflict-free connected, a contradiction.

Case 3. One of $u_{1}v_{1}$ and $u_{2}v_{2}$ is contained in a triangle. Without loss of generality, let $u_{1}v_{1}$ be contained in a triangle $X_{3}$. We color $X_{3}$ and $u_{2}v_{2}$ with the same color, the coloring of remaining edges is the same as Case 2. Also, this is a strong conflict-free connection coloring, a contradiction. Completing the proof of Claim 3.

Let $C(G_{t})$ be the graph induced by all the cut-edges of $G_{t}$.

Claim 4. $C(G_{t})$ is a tree.

Proof of Claim 4. Assume $C(G_{t})$ is not connected. Let $H_{1}$ and $H_{2}$ be two components with $C(G_{t})=H_{1}\cup H_{2}$. There exists one leaf $r_{1}$ in $H_{1}$ and one leaf $r_{2}$ in $H_{2}$ which are contained in the same triangle, say $r_{1}vr_{2}r_{1}$. Otherwise, $G_{t}$ is not connected. But both $d(r_{1})\geq 3$ and $d(r_{2})\geq 3$, which contradicts Claim 3.

Claim 5. $diam(C(G_{t}))\leq 2$.

Proof of Claim 5. Assume that $diam(C(G_{t}))=k\geq 3$. Let $P=v_{0}v_{1}\cdots v_{k}$ be a path of length $k$. Then we provide a coloring $c:$ $E(G)\mapsto[m-2t-k+\lceil\log_{2}(k+1)\rceil]$ of $G_{t}$: assign the edges of $P$ with $\lceil\log_{2}k\rceil$ colors to make $P$ strongly conflict-free connected by Theorem 2.3; assign each of the $t$ triangles with a fresh color; assign each of the remaining $m-3t-k$ edges with a fresh color. Clearly, $G_{t}$ is strongly conflict-free connected, a contradiction.

Clearly, from the above Claims we can deduce that $G_{t}\cong S_{m,t}$. $\Box$

4 Graphs with large or small $\mathit{scfc}$ numbers

In this section, we characterize the connected graphs $G$ of size $m$ with $\mathit{scfc}(G)=k$ for $k\in\{1,m-3,m-2,m-1,m\}$.

Theorem 4.1

For a nontrivial connected graph $G$, $\mathit{scfc}(G)=1$ if and only if $G$ is a complete graph.

Proof. Suppose that $G$ is a complete graph. Clearly, we have that $\mathit{scfc}(G)=1$. Conversely, suppose that $\mathit{scfc}(G)=1$. Assume that $G$ is not complete. Then there exists a pair of vertices $u,v$ with $d(u,v)\geq 2$. So, $\mathit{scfc}(G)\geq 2$, a contradiction. Thus, $G$ must be a complete graph. $\Box$

We now present an observation which will be used in the sequel.

Observation 4.2

Let $G$ be a connected graph with $\mathit{scfc}(G)=|E(G)|-k$ and let $H$ be a connected graph with $\mathit{scfc}(H)\leq|E(H)|-k-1$. Then $G$ does not contain a copy of $H$.

Proof. Assume, to the contrary, that $G$ contains a copy of $H$. Then, we give a coloring for $G$ as follows: assign the edges of $H$ with $|E(H)|-k-1$ colors to make $H$ strongly conflict-free connected, and then assign each of the remaining $m-|E(H)|$ edges of $G$ with a fresh color. Clearly, $G$ is strongly conflict-free connected. Consequently, $\mathit{scfc}(G)\leq E(G)-|E(H)|+|E(H)|-k-1\leq E(G)-k-1$, a contradiction. $\Box$

The following are two useful lemmas which will help to prove our latter theorems.

Lemma 4.3

Let $G$ be a connected graph with size $m$ and $\mathit{scfc}(G)=m-k$. Then

$diam(G)-\lceil\log_{2}(diam(G)+1)\rceil\leq k$.

Proof. Let $P$ be the path of length $diam(G)$. Now we provide a coloring with $m+\lceil\log_{2}{diam(G)+1}\rceil-diam(G)$ colors: assign the edges of $P$ with $\lceil\log_{2}{diam(G)+1}\rceil$ colors to make $P$ strongly conflict-free connected; assign each of the remaining $m-diam(G)$ edges a fresh color. Clearly, $G$ is strongly conflict-free connected. Since $\mathit{scfc}(G)=m-k$, then we have that $m-k\leq m+\lceil\log_{2}{(diam(G)+1)}-diam(G)\rceil$. Since $\lceil\log_{2}{(diam(G)+1)}-diam(G)\rceil$ is monotone decreasing, then $-diam(G)+\lceil\log_{2}{(diam(G_{t})+1)}\rceil\geq-k$. Consequently, $diam(G)-\lceil\log_{2}(diam(G)+1)\rceil\leq k$. $\Box$

Lemma 4.4

Let $G$ be a connected graph with size $m$ and $\mathit{scfc}(G)=m-k$, and let $C$ be a cycle of $G$. Then

$|C|-\lceil\log_{2}|C|\rceil\leq k$.

Proof. It is clear that $\mathit{scfc}(C)\leq\lceil\log_{2}|C|\rceil$ by Theorem 2.9. Then we give a coloring as follows: assign the edges of $C$ with $\lceil\log_{2}|C|\rceil$ colors to make $C$ strongly conflict-free connected and assign each of the remaining $m-|C|$ edges with a fresh color. We can easily verify that the coloring is a strong conflict-free coloring, a contradiction. Consequently, $|C|-\lceil\log_{2}|C|\rceil\leq k$. $\Box$

Theorem 4.5

Let $G$ be a nontrivial connected graph of size $m$. Then $\mathit{scfc}(G)=m$ if and only if $G\cong S_{m}$.

Proof. Suppose that $\mathit{scfc}(G)=m$. Assume that there is a cycle $C$ in $G$. Then $\mathit{scfc}(G)\leq m-|C|+\lceil\log_{2}|C|\rceil$ by Corollary 2.10, which is a contradiction. Hence, $G$ is a tree. Let $u$ and $v$ be two vertices with $d_{G}(u,v)\geq 3$ in $G$. Assume that $P$ is a path of length $d_{G}(u,v)$ between $u$ and $v$. Then we provide a coloring for $G$: assign the edges of $P$ with $\lceil\log_{2}(d_{G}(u,v)+1)\rceil$ colors to make $P$ strongly conflict-free connected; assign each of the remaining edges with a fresh color. Clearly, $G$ is strongly conflict-free connected by the edge-coloring with $m-d_{G}(u,v)+\lceil\log_{2}(d_{G}(u,v)+1)\rceil$ colors, a contradiction. Thus, $G\cong S_{m}$. $\Box$

Before proving the theorem below, we define some graph-classes. Let $S_{m}$ be a star with $m\ (\geq 2)$ edges and let $u$ be a leaf of $S_{m}$. We define a graph by $\Gamma_{m+1}=(V(S)\cup\{v\},E(S)\cup\{uv\})$ and we denote by $P_{n}$ a path of length $n$.

Lemma 4.6

If $G\in\{P_{3},P_{4},\Gamma_{m}\}$, then $\mathit{scfc}(G)=m-1$.

Proof. It is clear that $\mathit{scfc}(P_{3})=2$ and $\mathit{scfc}(P_{4})=3$ by Theorem 2.3, and $\mathit{scfc}(\Gamma_{m})\geq\Delta(\Gamma_{m})=m-1$ by Theorem 3.1. Then for the upper bound we give a coloring: assign each of the $m-1$ edges of $S_{m-1}$ with a fresh color and choose one color from the used colors except the color assigned to the edge incident with $u$. Clearly, $G$ is strongly conflict-free connected. Consequently, $\mathit{scfc}(\Gamma_{m})=m-1$. $\Box$

Theorem 4.7

Let $G$ be a connected graph of size $m$. Then $\mathit{scfc}(G)=m-1$ if and only if $G\in$ $\{P_{3}$, $P_{4}$, $\Gamma_{m}\}$.

Proof. The necessity holds by Lemma 4.6. On the contrary, suppose that $\mathit{scfc}(G)=m-1$. We first claim that $G$ is a tree. Assume that $G$ is not a tree. Let $C$ $(|C|\geq 3)$ be a cycle of $G$. We have that $\mathit{scfc}(C)\leq|C|-2$ by Corollary 2.10, and it is not true by Observation 4.2.

Suppose that $diam(G)\geq 5$ in $G$. Clearly, $diam(G)-\lceil\log_{2}(diam(G)+1)\rceil>1$ by Lemma 4.3, a contradiction. So $diam(G)\leq 4$. Suppose $diam(G)=4$. Let $P_{4}=v_{1}v_{2}v_{3}v_{4}v_{5}$ be a path with $\mathit{scfc}(P_{4})=3$. If $G=P_{4}$, then it is true. Assume that there is another vertex $w$ adjacent to $v_{i}$ of $P_{4}$, denote this structure by $R$. It is clear to see that $R$ can be colored by three colors to make it strongly conflict-free connected. Thus, $\mathit{scfc}(R)=|E(R)|-2$, and $R\nsubseteq G$ by Observation 4.2. Consequently, $G\cong P_{4}$.

Suppose that $diam(G)=3$. Let $P_{3}=v_{1}v_{2}v_{3}v_{4}$ be a path with $\mathit{scfc}(P_{3})=2$. If $G=P_{3}$, then it is true. Assume that there are two vertices $x,y$ adjacent to $v_{2}$, $v_{3}$ of $P_{3}$, respectively, denote by $L$ this structure. It is easy to check that $\mathit{scfc}(L)\leq 3$. So $L\nsubseteq G$ by Observation 4.2. Without loss of generality, let $d(v_{2})=t$ $(\geq 3)$ and $d(v_{3})=2$. Obversely, $\mathit{scfc}(G)\geq t$ by Theorem 3.1. Now we assign each of the edges incident with $v_{2}$ by a fresh color and assign the remaining edge $e$ by the color used on some edge not adjacent to $e$. Clearly, $G$ is strongly conflict-free connected. So, $G\in\{P_{3},\Gamma_{m}\}$. Suppose that $diam(G)=2$. Then $G\cong S_{n}$ with $\mathit{scfc}(G)=m$, a contradiction. Completing the proof. $\Box$

Theorem 4.8

$\mathit{scfc}(G)=m-2$* if and only if $G\in\{C_{3},C_{4},C_{5},P_{5},A_{1},A_{2},\cdots,A_{6}\}$.*

Proof. Suppose that $\mathit{scfc}(G)=m-2$. Then $diam(G)\leq 5$ by Lemma 4.3. Let $C$ be a cycle of $G$. Then $|C|\leq 5$ by Lemma 4.4. We now distinguish the following cases (The graphs are demonstrated in Figure 1).

Case 1: $G$ is a tree.

$(i)$ Suppose that $diam(G)=5$. Let $P_{5}=v_{1}v_{2}v_{3}v_{4}v_{5}v_{6}$ be a path. If $G=P_{5}$, then it is true by Theorem 2.3. If $G\neq P_{5}$, then we construct a graph $H$ by adding an edge $uv_{i}$ $(i\in\{2,3,4,5\})$ to $P_{5}$. It is easy to get $\mathit{scfc}(H)\leq 3$. Hence, $H\nsubseteq G$ by Observation 4.2. Consequently, $G\cong P_{5}$.

$(ii)$ Suppose that $diam(G)=4$. Let $P_{4}=v_{1}v_{2}v_{3}v_{4}v_{5}$ be a path for which $\mathit{scfc}(P_{4})=3$, and thus, $G\neq P_{4}$. We construct a graph $H_{1}$ by adding two vertices $u_{1}$, $u_{2}$ and connecting them to $v_{2}$ and $v_{3}$, respectively. Then $H_{1}\nsubseteq G$ by Observation 4.2 since $\mathit{scfc}(H_{1})=|E(H_{1})|-3$. We construct another graph $H_{2}$ by adding a vertex $u_{1}$ and connecting it to $v_{2}$ and adding another vertex $u_{2}$ and connecting it to $v_{4}$, which means that $\mathit{scfc}(H_{2})=|E(H_{2})|-2$. We construct $H_{2}^{\prime}$ by adding an edge $w_{1}v_{2}$ in $P_{4}$. We have that $\mathit{scfc}(H_{2}^{\prime})=|E(H_{2}^{\prime})|-2$ by easy calculation. Let $P=u_{1}u_{2}u_{3}$ be a path. We construct a graph $H_{3}$ by identifying $u_{1}$ with $v_{3}$ of $P_{4}$ (if $u_{1}$ is identified with other vertices of $P_{4}$, then it contradicts that $diam(G)=4$). Clearly, $\mathit{scfc}(H_{3})=|E(H_{3})|-3$. Thus, $H_{3}\nsubseteq G$ by Observation 4.2. Let $x_{1}x_{2}$ be an edge and we construct $H_{4}$ by identifying $x_{1}$ with $v_{3}$ of $H_{2}$. Clearly, $\mathit{scfc}(H_{4})=|E(H_{4})|-3$. Thus, $H_{4}\nsubseteq G$ by Observation 4.2. Consequently, $G$ can contain $H_{2}$ and $H_{2}^{\prime}$ but not $H_{3}$ and $H_{4}$.

Now we show that $G\in\{A_{2},A_{3},A_{4}\}$. Clearly, it is true for $G=A_{4}=H_{2}$. $A_{2}$ is constructed by identifying one end of each of $l$ new edges with $v_{2}$ in $P_{4}$. Clearly, $\mathit{scfc}(A_{2})\geq\Delta(A_{2})$ by Theorem 3.1. Then we give a coloring of $A_{2}$: first, assign each of the $\Delta(A_{2})$ edges incident with $v_{2}$ by a fresh color, and assign the remaining two edges with used colors except the color used on $v_{2}v_{3}$. Clearly, it is a strong conflict-free connection coloring. So $\mathit{scfc}(A_{2})=|E(A_{2})|-2$. Similarly, $\mathit{scfc}(A_{3})=|E(A_{3})|-2$. Consequently, $G\in\{A_{2},A_{3},A_{4}\}$.

$(iii)$ Suppose that $diam(G)=3$. By above similar manner, we have that $\mathit{scfc}(G)=m-2$ if and only if $G\in\{A_{5},A_{6}\}$. Suppose $diam(G)=2$. Then the only graph is a star $S_{m}$, which is a contradiction with $\mathit{scfc}(S_{m})=m$.

Case 2: There is at least one cycle $C$ in $G$.

$(i)$ Suppose that $C=v_{1}v_{2}v_{3}v_{4}v_{5}v_{1}$. If $G=C$, then it is true by $\mathit{scfc}(C)=3$. Assume that there is an edge $v_{i}v_{j}$ $(i,j\in[5])$ in $G$. Then there is a subgraph of $G$, say $C_{5}^{\prime}=C+v_{i}v_{j}$. Then $C_{5}^{\prime}\nsubseteq G$ by Observation 4.2 since $\mathit{scfc}(C_{5}^{\prime})=3$. We construct the graph $C^{+}_{5}$ by adding $u$ and connecting it to vertex $v_{i}$ $(i\in[5])$ of $C_{5}$. Obviously, $C^{+}_{5}\nsubseteq G$ by Observation 4.2 since $\mathit{scfc}(C^{+}_{5})=3$. So $G\cong C_{5}$.

$(ii)$ Suppose that $C=v_{1}v_{2}v_{3}v_{4}v_{1}$. We construct the graph $C_{4}^{\prime}$ by adding one edge $v_{2}v_{3}$ in $C_{4}$. Then $C_{4}^{\prime}\nsubseteq G$ by Observation 4.2 since $\mathit{scfc}(C_{4}^{\prime})=2$. We then construct another graph $C^{+}_{4}$ by adding one vertex $w$ and connecting it to $v_{i}$ $(i\in[4])$. Obviously, $C^{+}_{4}\nsubseteq G$. Consequently, $G\cong C_{4}$.

$(iii)$  Suppose that $C=v_{1}v_{2}v_{3}v_{1}$. Let $P=u_{1}u_{2}u_{3}$ be a path. We construct a graph by identifying $u_{1}$ with $v_{i}$ $(i\in[3])$, denote it by $H_{1}$. Clearly, $H_{1}\nsubseteq G$ by Observation 4.2 since $\mathit{scfc}(H_{1})\leq 2$. Let $P^{\prime}=u_{1}u_{2}$ and $P^{\prime\prime}=w_{1}w_{2}$ be two paths. We construct the graph $H_{2}$ by identifying $u_{1}$ with $v_{i}$ and identifying $w_{1}$ with $v_{j}$ $(i,j\in[3]$ and $i\neq j)$. Clearly, $\mathit{scfc}(H_{2})=2$. So $H_{1}\nsubseteq G$ by Observation 4.2. Finally we construct $H_{3}$ by identifying one vertex $v_{i}$ with $u_{1}$ of $P^{\prime}$. Clearly, $\mathit{scfc}(H_{3})=2$. Therefore, $G$ does not contain $H_{1}$ and $H_{2}$ but $G$ contain $H_{3}$. Clearly, the only graph class must be $A_{1}$ and $\mathit{scfc}(A_{1})=|E(A_{1})|-2$. Consequently, $G\in\{C_{3},A_{1}\}$. $\Box$

Lemma 4.9

Let $G$ be a connected graph of size $m$. If $G\in\{B_{1},B_{2},\cdots,B_{25}\}$, then $\mathit{scfc}(G)=m-3$.

Proof. The graphs are demonstrated in Figure 2. For $G\in\{B_{1},\cdots,B_{7},B_{9},B_{11},$ $\cdots,B_{14},B_{17},B_{18},B_{19},B_{21},B_{23},B_{24},B_{25}\}$, we can easily check that $\mathit{scfc}(G)=m-3$. For $G\in\{B_{8},B_{10}\}$, there is a triangle in $B_{8}$ and $B_{10}$, respectively. Then clearly $\mathit{scfc}(G)\geq\Delta(G)-1$. We give a coloring for $B_{8}$: color the triangle by 1 and color each of the edges incident with the vertices of the triangle by a fresh color, and color the remaining edge by a color used on the edges not adjacent to it. Clearly, $\mathit{scfc}(B_{8})=|E(B_{8})|-3$. Similarly, $\mathit{scfc}(B_{10})=|E(B_{10})|-3$. For $B_{15}$, it can be obtained by identifying one leaf of $P_{3}$ with one leaf of $S_{t}$ $(t\geq 3)$. Then we have that $\mathit{scfc}(S_{t})\geq t$. We give a coloring of $B_{15}$: color the edges of $S_{t}$ with $t$ colors and choose two colors used on two leaves of $S_{t}$ to color the remaining two edges. Clearly, it is a strong conflict-free connection coloring. Thus, $\mathit{scfc}(B_{15})=|E(B_{15})|-3$. Similarly, we can easily check that $\mathit{scfc}(G)=m-3$ for $G\in\{B_{20},B_{22},B_{23},B_{25}\}$. $\Box$

Theorem 4.10

Let $G$ be a connected graph with $m\ (m\geq 4)$ edges. Then $\mathit{scfc}(G)=m-3$ if and only if $G\in\{B_{1},B_{2},\cdots,B_{25}\}$.

Proof. The sufficiency holds by Lemma 4.9. Now we consider the necessity. We first have $diam(G)\leq 6$ by Lemma 4.3. Let $C$ be a cycle in $G$. Then $|C|\leq 6$ by Lemma 4.4. Then we consider the following two cases.

Case 1. Suppose that there is at least one cycle $C$ in $G$.

(i) Assume that $|C|=6$. Then $C\nsubseteq G$ by Observation 4.2 since $\mathit{scfc}(C)=2$. Thus, $|C|\leq 5$.

(ii) Assume that $|C|=5$. Let $C^{\prime}$ be a graph by adding a chord to $C$. We can easily check that $\mathit{scfc}(C^{\prime})=|E(C^{\prime})|-4$. So, $C^{\prime}\nsubseteq G$ by Observation 4.2. We construct $C^{\prime\prime}$ by adding a leaf vertex to $C$, for which $\mathit{scfc}(C^{\prime\prime})=|E(C^{\prime\prime})|-3$. Then we construct $C^{\prime\prime\prime}$ by adding two leave vertices to $C$. But, $\mathit{scfc}(C^{\prime\prime\prime})=|E(C^{\prime\prime\prime})|-4$. Let $P$ be a path of length 2. We construct $\bar{C}$ by identifying a vertex of $C$ with an end of $P$, for which $\mathit{scfc}(\bar{C})=|E(\bar{C})|-4$. Consequently, $G\cong C^{\prime\prime}=B_{1}$.

(iii)  Suppose that $C=v_{1}v_{2}v_{3}v_{4}v_{1}$. Let $P=u_{1}u_{2}u_{3}$ be a path. Then we construct $H_{1}$ by identifying $u_{1}$ in $P$ with $v_{4}$ in $C$. Clearly, $\mathit{scfc}(H_{1})\leq 2=|E(H_{1})|-4$. Hence, $H_{1}\nsubseteq G$ by Observation 4.2. Let $w_{1}w_{2}$ be an edge. We then construct $H_{2}$ by choosing arbitrarily $v_{i}$ $(i\in[4])$ and identifying $v_{i}$ with $w_{1}$. Then we get $\mathit{scfc}(H_{2})=|E(H_{2})|-3$. Let $s_{1}s_{2}$ be an edge. We construct $H_{3}$ by identifying $v_{1}$, $v_{2}$ of $C_{4}$ with $s_{1}$, $w_{1}$, respectively. Then clearly $\mathit{scfc}(H_{3})=|E(H_{3})|-4$=2, and thus $H_{3}\nsubseteq G$. We construct $H_{4}$ by identifying $s_{1}$ and $w_{1}$ with $v_{1}$ and $v_{3}$ of $C$, respectively. Clearly, $\mathit{scfc}(H_{4})=|E(H_{4})|-3$.

Hence, $G$ can contain the copies of $H_{2}$ and $H_{4}$ but not the copies of $H_{1}$ and $H_{3}$. Clearly, suppose that we construct $H_{5}$ by adding one pendant vertex to $v_{1}$ in $H_{4}$. Clearly, $\mathit{scfc}(H_{5})=|E(H_{5})|-4$. Then it does not hold for $H_{5}$. Obviously, $G\in\{B_{2},B_{3}\}$. We construct $H_{6}$ by adding a chord to $C_{4}$. Then we have that $\mathit{scfc}(H_{5})=|E(H_{5})|-3$. At last, we construct $H_{6}$ by adding a leaf vertex to connect it to a vertex of $H_{5}$. Clearly, $\mathit{scfc}(H_{6})=|E(H_{6})|-4$. Consequently, $G\in\{B_{2},B_{3},B_{4}\}$.

(iv) Suppose that $C=v_{1}v_{2}v_{3}v_{1}$. Let $P=u_{1}u_{2}u_{3}u_{4}u_{5}$. Clearly, $\mathit{scfc}(P)=3$ and $\mathit{scfc}(C)=1$. We construct a graph $H_{1}$ by identifying $u_{1}$ with $v_{1}$. Clearly, the coloring by assigning each edge $e$ $\in E(C)$ and $u_{3}u_{4}$ with color 2 and assigning $v_{1}u_{2}$ and $u_{4}u_{5}$ with color 1 and assigning $u_{2}u_{3}$ with color 3 is a strong conflict-free connection coloring. So $\mathit{scfc}(H_{1})\leq 3$. By Observation 4.2 $G$ does not contain any copy of $H_{1}$. Then we can use Observation 4.2 repeatedly, and eventually get that $G\in\{B_{5},B_{6},B_{7},B_{8},B_{9},B_{10}\}$.

Case 2. Suppose that $G$ is a tree. By the same arguments, we know that $G\in\{B_{11},B_{12},B_{13}\}$ if $diam(G)=6$; $G\in\{B_{14},\cdots,B_{18}\}$ if $diam(G)=5$; $G\in\{B_{19},\cdots,B_{24}\}$ if $diam(G)=4$; $G=B_{25}$ if $diam(G)=3$. But $\mathit{scfc}(G)\leq m-1$ when $diam(G)\leq 2$ by Theorems 4.7 and 4.5. $\Box$

5 Cubic graphs with $\mathit{scfc}$-number 2

In this section, we will characterize the cubic graphs $G$ with $\mathit{scfc}(G)=2$. We first discuss the relation between the strong conflict-free connection number and the strong proper connection number for cubic graphs.

We need the following definition.

Definition 5.1

A forced 2-path in a graph $G$ is a path $xyz$ such that $xz\notin E(G)$ and $xyz$ is the unique 2-path connecting $x$ and $z$. A $k$-path $P=u_{0}u_{1}\cdots u_{k}$ in a graph $G$ is called forced, if each 2-path $u_{i}u_{i+1}u_{i+2}$ is forced and $P$ is a path between $u_{0}$ and $u_{k}$, for $i=0,1,\cdots,k-2$. A cycle of a graph $G$ is called a forced cycle if any two successive edges of the cycle form a forced 2-path in $G$. An edge $e$ in a graph $G$ is called a forced edge if $e$ is not included in a cycle of length at most 4.

If $uv$ is a forced edge in $G$ and $vw$ is an edge adjacent to $uv$, then $uvw$ is a forced 2-path in $G$. The following two results follow directly from the definition.

Lemma 5.2

Let $P=u_{1}u_{2}\cdots u_{k}$ be a forced path in $G$ with $\mathit{scfc}(G)=2$. Then the adjacent edges of $P$ are colored by distinct colors for every strong conflict-free connection coloring with 2 colors.

Lemma 5.3

Let $C=u_{1}u_{2}\cdots u_{k}u_{1}$ be a forced cycle of length $k$ in $G$ with $\mathit{scfc}(G)=2$. Then the adjacent edges of $C$ are colored by distinct colors for every strong conflict-free connection coloring with 2 colors and $k$ is even.

Now we define some graph-classes. A $k$-ladder, denoted by $L_{k}$, is defined to be the product graph $P_{k}\Box K_{2}$, where $P_{k}$ is the path on $k$ vertices $(see\ Figure\ \ref{ladder})$. The Möbius ladder $M_{2k}$ is the graph obtained from $L_{k}$ by adding two new edges $s_{1}t_{k}$ and $t_{1}s_{k}$ $(see\ Figure\ \ref{mobius})$.

Lemma 5.4

$\mathit{scfc}(C_{k}\Box K_{2})=2$* if and only if $k$ equals $3$, $4$ or $6$.*

Proof. Let $k\geq 4$. Clearly, The graph has a forced cycle. Since $\mathit{scfc}(C_{k}\Box K_{2})=2$, we have that $k=4$ or $6$ by Lemma 5.3. When $k=3$, we define a $2$-edge-coloring $c$: for every edge $e$ in the triangles, $c(e)=1$; for the remaining edges $e$, $c(e)=2$. Clearly, the coloring is a strong conflict-free connection coloring for $C_{3}\Box K_{2}$. When $k=4,6$, we define a $2$-edge-coloring: assign alternate colors on the edges of $s_{1}s_{2}\cdots s_{k}s_{1}$ and $t_{1}t_{2}\cdots t_{k}t_{1}$ with colors 1 and 2 such that $c(s_{1}s_{2})\neq c(t_{1}t_{2})$, and all the remaining edges are colored by 1. One can easily check that this coloring is a strong conflict-free connection coloring. $\Box$

Lemma 5.5

$\mathit{scfc}(M_{2k})=2$* if and only if $3\leq k\leq 7$.*

Proof. It is clear to see that $\mathit{scfc}(M_{2k})\geq 2$ for every $k\geq 3$ since $M_{2k}$ is not a complete graph. First, when $k\geq 8$, clearly for the pair of vertices $s_{2}$ and $s_{6}$ there is only one shortest path connecting them, which is $P^{\prime}=s_{2}s_{3}s_{4}s_{5}s_{6}$. For every pair of vertices in $P$, there is only one shortest path in $M_{2k}$ connecting them. So we have that $\mathit{scfc}(M_{2k})\geq\mathit{scfc}(P^{\prime})=3$. For the graph $M_{2k}$ with $k\in\{4,6\}$, we define a $2$-edge-coloring $c$: for $i\in\{1,3,5\}$, $c(s_{i}s_{i+1})=c(t_{i}t_{i+1})=c(s_{i}t_{i})=1$; for the remaining edges $e$, $c(e)=2$. For the graph $M_{2k}$ with $k\in\{3,5,7\}$, we define a $2$-edge-coloring $c$: for $i\in\{1,3,5\}$, $c(s_{i}s_{i+1})=c(t_{i+1}t_{i+2})=1$; for $i\in\{1,2\cdots,k\}$, $c(s_{i}t_{i})=c(s_{k}t_{1})=1$; for the remaining edges $e$, $c(e)=2$. It is easy to check that every pair of vertices are connected by a strong conflict-free path under the above $2$-edge-colorings. $\Box$

In order to be more convenient to handle with the following theorem, let us start with some explanations. Let $G$ be a cubic graph, and let $c:$ $E(G)\mapsto\{1,2\}$ be a strong conflict-free connection coloring of $G$. Let $P=(u=)v_{1}v_{2}\cdots v_{t-1}v_{t}(=v)$ be a strong conflict-free path between $u$ and $v$. Suppose that there exists a 2-path $v_{i}v_{i+1}v_{i+2}$ in $P$ such that $c(v_{i}v_{i+1})=c(v_{i+1}v_{i+2})$. Then there must exist another 2-path $v_{i}v_{i+1}^{\prime}v_{i+2}$ with $c(v_{i}v_{i+1}^{\prime})\neq c(v_{i+1}^{\prime}v_{i+2})$ to replace $v_{i}v_{i+1}v_{i+2}$ since there exists a strong conflict-free path for the pair of $v_{i}$ and $v_{i+2}$. Then $v_{i}v_{i+1}^{\prime}v_{i+2}$ is called* a replacement*. Furthermore, suppose that $c(v_{i-1}v_{i})=c(v_{i}v_{i+1}^{\prime})$. Then there must also exist a replacement $v_{i-1}v_{i}^{\prime}v_{i+1}^{\prime}$ with $c(v_{i-1}v_{i}^{\prime})\neq c(v_{i}^{\prime}v_{i+1}^{\prime})$ for $v_{i-1}v_{i}v_{i+1}^{\prime}$. Continue the operation. If there does not exist a replacement sharing the same edges with $P$, then the sequence of replacements is called a finite replacement of $P$. Otherwise, the the sequence of replacements is called an infinite replacement of $P$.

Theorem 5.6

Let $G$ be a cubic graph with $G\ncong U$. If $\mathit{scfc}(G)=2$, then $\mathit{spc}(G)=2$.

Proof. Let $c:$ $E(G)\mapsto[2]$ be a strong conflict-free connection coloring of $G$. Let $P=(u=)v_{1}v_{2}\cdots v_{t-1}v_{t}(=v)$ be an arbitrary strong conflict-free path between $u$ and $v$. For every pair of $v_{i}$ and $v_{i+2}$ $(i\in[t])$, if $c(v_{i}v_{i+1})\neq c(v_{i+1}v_{i+2})$, then $P$ is a strong proper path. Suppose that there exists $v_{i}v_{i+1}v_{i+2}$ $(i\in[t-2])$ in $P$ such that $c(v_{i}v_{i+1})=c(v_{i+1}v_{i+2})$. If there exist a finite replacement for $P$, then there is a strong proper path for every pair of vertices in $G$. Suppose that the replacement is an infinite one for $P$ $(see\ Figure\ \ref{scfc=spc})$.

We denote $G[V^{\prime}]$ by $W$, where $V^{\prime}=\{v_{i-3},v_{i-2},v_{i-1},v_{i},v_{i+1},v_{i+2},v^{\prime}_{i},v^{\prime}_{i+1}\}$, and we say that $W$ is an attachment of path $P$. Then we first show the following claims.

Claim 1. For every strong conflict-free connection coloring $c$, $c(v_{i-3}v_{i-2})\neq c(v_{i-2}v_{i-1})=c(v_{i-2}v^{\prime}_{i-1})$ and $c(v_{i+1}v_{i+2})\neq c(v_{i}v_{i+1})=c(v^{\prime}_{i}v_{i+1})$.

Proof of Claim 1: Without loss of generality, suppose that $c(v_{i-3}v_{i-2})=c(v_{i-2}v_{i-1})$. Then $v_{i_{3}}v_{i-2}v^{\prime}_{i-1}$ is a unique shortest path between $v_{i-3}$ and $v^{\prime}_{i-1}$ since $G$ is a cubic graph with $d(v^{\prime}_{i-1})=3$. It contradicts that $c(v_{i-3}v_{i-2})=c(v_{i-2}v_{i-1})$ for the coloring $c$.

Claim 2. There is at most one attachment in $P$. Furthermore, let $C$ be a cycle. Then there are at most two attachments in $C$.

Proof of Claim 2: Assume that there are two attachments in $P$. Since $P$ is a shortest path, every subpath of $P$ is shortest. Hence, there is no strong conflict-free path between the attachments by Claim 1. Suppose that there are three attachments in $C$. Then $|C|\geq 12$, a contradiction by Claim 1. Completing the proof of Claim 2.

If the path $P$ with an attachment is not contained in a cycle, then there exist at least two cut-edges since $G$ is a cubic graph. Clearly, $\mathit{scfc}(G)\geq 3$ by Claim 1. If we identify $v_{i-3}$ with $v_{i+2}$, then $G=M_{6}$ with $\mathit{spc}(M_{6})=2$ by Lemma 5.7. Now we handle with the case that $P$ with an attachment is contained in a shortest cycle $C$. Clearly, $|C|\geq 6$, otherwise, $P$ does not contain an attachment. Suppose $|C|=6$. Then there are two vertices $u_{1},u_{2}$ except the vertices of the attachment in $C$. If $u_{1}$ and $u_{2}$ are not adjacent to the same neighbor, then every pair of edges incident with $u_{1}$ is a forced 2-path. Hence, there need at least three colors, a contradiction. Let $x$ be a common neighbor of $u_{1}$ and $u_{2}$, where $u_{2}$ is adjacent to $v_{i+1}$. Let $y$ be a neighbor of $x$, and $z$ be another neighbor of $y$ except $x$. Thus, $v_{i+1}u_{2}xyz$ is a unique forced path for the pair $v_{i+1},z$. Then it is not a strong conflict-free path by Lemma 5.2. Suppose $|C|=7$. Let $u_{1},u_{2},u_{3}$ be three vertices except the vertices of the attachment in $C$. If each of $u_{1},u_{2},u_{3}$ is in a triangle, then $G\cong U$ ($see\ Figure\ \ref{U}$). If one of $u_{1},u_{2},u_{3}$ is in a triangle, then there exists a unique forced 4-path for a pair of vertices in $C$, a contradiction. Suppose that $C=v_{1}v_{2}v_{3}v_{4}u_{1}u_{2}u_{3}u_{4}v_{1}$, and suppose further that there are two attachments in $C$. Then $G\cong L_{2}$ ($see\ Figure\ \ref{L}$) with an edge-coloring such that $\mathit{scfc}(G)=\mathit{spc}(G)=2$. Suppose that $u_{1},u_{2},u_{3},u_{4}$ are in triangles. Then $G\cong L_{3}$ ($see\ Figure\ \ref{L}$) such that $\mathit{scfc}(G)=\mathit{spc}(G)=2$. Otherwise, there will exist a unique forced 4-path for a pair of vertices in $C$, a contradiction. Suppose that at most one triangle contains two of the vertices $u_{1},u_{2},u_{3},u_{4}$, without loss of generality, say $u_{1},u_{2}$. Suppose further that $u_{3}u_{4}$ is a forced edge. Then $c(v_{1}u_{4})\neq c(u_{4}u_{3})\neq c(u_{4}x)$, where $x$ is a neighbor of $u_{4}$ except $v_{1},u_{3}$, a contradiction. Then suppose that $u_{3},u_{4}$ are contained a 4-cycle $C^{\prime}$. Clearly, there provides a unique forced 4-path for the pair of $v_{2}$ and one vertex in $C^{\prime}$ except $u_{3},u_{4}$, a contradiction. Suppose $9\leq|C|\leq 10$. Then there is a unique forced 4-path for some pair of vertices in $G$. Hence, $\mathit{scfc}(G)\geq 3$, a contradiction. Assume $|C|\geq 11$. Then there exists a unique shortest path of length 5 between $v_{i-3}$ and $v_{i+2}$, a contradiction by Claim 1. $\Box$

Now we only need to check whether 2 is the strong conflict-free connection number of $G$ with $\mathit{spc}(G)=2$ by Theorem 5.6.

Theorem 5.7

[12] Let $G$ be a cubic graph without forced edges. Suppose further that $G\neq K_{4}$. Then $\mathit{spc}(G)=2$ if and only if $G\in\{C_{3}\Box K_{2},C_{2k}\Box K_{2},M_{2k}\}$ for some $k\geq 2$.

Combining Theorem 5.6, Theorem 5.7, Lemma 5.4 and Lemma 5.5, we have the following result.

Lemma 5.8

Let $G$ be a cubic graph without forced edges. Then $\mathit{scfc}(G)=2$ if and only if $G\in\{C_{l}\Box K_{2},M_{2k}\}$ for $l\in\{3,4,6\}$ and for $k$ with $3\leq k\leq 7$.

Let $F_{0}(k)$ be the cubic graph which is obtained from $L_{k}$ by adding two new vertices $x$ and $y$ and adding five new edges $xy,xs_{1},xt_{1},ys_{k},yt_{k}$ ($see\ Figure\ \ref{F0(k)}$).

Lemma 5.9

$\mathit{scfc}(F_{0}(k))=2$* with $k\geq 2$ if and only if $k\in\{2,4\}$.*

Proof. When $k\geq 3$, the cycle $xs_{1}s_{2}\cdots s_{k}y$, say $C$, is a forced one in $F_{0}(k)$. Then we have that $k=4$ by Lemma 5.3. When $k=2$, we define an edge-coloring $c$ for $F_{0}(k)$: $c(xy)=2$; $c(xs_{1})=c(xt_{1})=c(ys_{k})=c(yt_{k})=1$; $c(s_{i}s_{i+1})=c(t_{i}t_{i+1})=c(s_{i}t_{i})=1$ for even $i\in[k]$; for all the remaining edges, $c(s_{i}s_{i+1})=c(t_{i}t_{i+1})=c(s_{i}t_{i})=2$ for odd $i\in[k]$. We can easily check that every pair of vertices have a strong conflict-free path connecting them. Since $F_{0}(k)>1$, we have that $\mathit{scfc}(F_{0}(k))=2$ for $k=2$ or $4$. $\Box$

We now introduce a family $\mathcal{H}$ of graphs which are demonstrated in Figure 8.

$\mathcal{H}=\{F_{0}^{*}(k),\hat{K_{4}},\hat{D_{3}},\tilde{K_{3,3}},\tilde{Q_{3}},F_{1}(k)\}$ $(k\in\mathbb{N})$

Theorem 5.10

[12] Let $G$ be a cubic graph with exactly one forced edge. Then $\mathit{spc}(G)=2$ if and only if $G=F_{0}(k)$ for some even $k\geq 4$, or $G$ is obtained from $H_{1}$ and $H_{2}$ by identifying the pendent edges to a single edge, where $H_{i}\in\{\hat{K_{4}},\hat{D_{3}}\}$ for $i=1,2$.

Lemma 5.11

Let $G$ be a cubic graph. If $G$ is obtained from $H_{1}$ and $H_{2}$ by identifying the pendent edges to a single edge, where $H_{i}\in\{\hat{K_{4}},\hat{D_{3}}\}$ for $i=1,2$, then $\mathit{scfc}(G)$=2 if and only if $H_{i}=\hat{K_{4}}$ for $i=1,2$.

Proof. Suppose $\mathit{scfc}(G)$=2. Let $H_{1}=\hat{D_{3}}$ ($see\ Figure\ \ref{H}$). If $G$ is constructed by identifying the pendent edge of $H_{1}$ with $H_{2}\in\{\hat{K_{4}},\hat{D_{3}}\}$, then there is a forced 4-path $t_{2}t_{1}ys_{1}s_{2}$, a contradiction by Lemma 5.2. Clearly, $\mathit{scfc}(G)$=2 when $H_{1}=H_{2}=\hat{K_{4}}$ under the edge-coloring in Figure 9.

$\Box$

Combining Lemma 5.9, Lemma 5.11, Theorem 5.6 and Theorem 5.10, we get the following result.

Lemma 5.12

Let $G$ be a cubic graph with exactly one forced edge. Then $\mathit{scfc}(G)=2$ if and only if $G\cong F_{0}(k)$ for $k\in\{2,4\}$ or $G\cong N$.

Before proceeding, we need one more definition.

Definition 5.13

Let $G$ be a connected graph. The forced graph of $G$ is obtained from $G$ by replacing each forced edge $uv$ (if any) by two pendant edges $uu^{\prime}$ and $vv^{\prime}$, where $u^{\prime}$ and $v^{\prime}$ are two new vertices with respect to the forced edge $uv$. Each component of the forced graph of $G$ is called a forced branch of $G$, and the new pendant edge $uu^{\prime}$ in the forced branch is called a forced link of $G$. For each forced edge $uv$ of $G$, we call $uu^{\prime}$ and $vv^{\prime}$ the twin links corresponding to the forced edge $uv$. In the case that a forced link $uu^{\prime}$ and its twin link $vv^{\prime}$ are contained in a common forced branch of $G$, we say that $uu^{\prime}$ is a selfish link.

Theorem 5.14

[12] Let $G$ be a cubic graph containing at least two forced edges, and let $H_{1},H_{2},\cdots,H_{r}$ be the forced branches of $G$. Then $\mathit{spc}(G)=2$ if and only if $H_{i}\in\mathcal{H}$ for $i=1,2,\cdots,r$, and there are $2$-$SPC$ (strong proper connection number being $2$) patterns $p_{1},p_{2},\cdots,p_{r}$ of $H_{1},H_{2},\cdots,H_{r}$, respectively, such that each pair of twin links receive the same color.

Lemma 5.15

Let $G$ be a cubic graph containing at least two forced edges, and let $H_{1},H_{2},\cdots,H_{r}\in\mathcal{H}$ be the forced branches of $G$. Then $\mathit{scfc}(G)$=2 if and only if $G\in\mathcal{L}$, demonstrated in Figure 10.

Proof. Let $G^{*}$ be the forced graph of $G$. According to the choices of $H$, we distinguish the following cases.

Suppose that $H_{i}\in\{\hat{K_{4}},\hat{D_{3}},\tilde{K_{3,3}},\tilde{Q_{3}}\}$. Clearly, there is no selfish link for $H_{i}$. Otherwise, there is no forced edge in $G$. Suppose that $G^{*}$ is constructed by $H_{1},H_{2}\in\{\hat{K_{4}},\hat{D_{3}}\}$. Then there is only one forced edge, a contradiction. Suppose that $G^{*}$ is constructed by $H_{1}\in\{\tilde{K_{3,3}},\tilde{Q_{3}}\}$ and $H_{2},H_{3}\in\{\hat{K_{4}},\hat{D_{3}}\}$. Then there is a forced 5-path between the two forced edges, a contradiction. Suppose that $G^{*}$ is constructed by $H_{1},H_{2}\cong\tilde{Q_{3}}$. Clearly, the two forced edges are contained in a forced cycle $C_{8}$ in $G$, which induces a forced 4-path, a contradiction. If $G^{*}$ is constructed by $H_{1},H_{2}\cong\tilde{K_{3,3}}$ or $H_{1}\cong\tilde{K_{3,3}},H_{2}\cong\tilde{Q_{3}}$, then $G\cong L_{2}$ or $L_{4}$ with $\mathit{scfc}(G)=2$ by the edge-coloring in Figure 10. Suppose that $H_{i}\in\{F_{0}^{*}(k),F_{1}(k)\}$. If $G^{*}\cong F_{0}^{*}(k)$, then there is at most one forced edge in $G$, a contradiction. If $G$ is constructed by identifying the pendent edges of $H_{i}\in\{F_{0}^{*}(k),F_{1}(k)\}$ to a single edge, then we can check that $G\cong L_{1}$. In the remaining case, $G\cong L_{3}$. $\Box$

Finally, Combining Theorem 5.6, Lemma 5.15 and Theorem 5.14, we have our main theorem of this section.

Theorem 5.16

Let $G$ be a cubic graph. Then $\mathit{scfc}(G)=2$ if and only if

$G\in\{L_{1},L_{2},L_{3},L_{4},N,C_{l}\Box K_{2},M_{2r},F_{0}(k)\}$,

where $l\in\{3,4,6\}$, $3\leq r\leq 7$ and $k\in\{2,4\}$.