Strong Edge Coloring of Generalized Petersen Graphs

Abstract

A strong edge coloring of a graph G is a proper edge coloring such that every color class is an induced matching. In 2018, Yang and Wu proposed a conjecture that every generalized Petersen graph $P(n,k)$ with $k\ge 4$ and $n>2k$ can be strong edge colored with (at most) seven colors. Although the generalized Petersen graph $P(n,k)$ is a kind of special graph, the strong chromatic index of $P(n,k)$ is still unknown. In this paper, we support the conjecture by showing that the strong chromatic index of every generalized Petersen graph $P(n,k)$ with $k\ge 4$ and $n>2k$ is at most 9.

1. Introduction

All graphs in this paper are finite and simple. We denote the minimum and maximum degree of vertices in G by $\delta \left(G\right)$ and $\Delta \left(G\right)$, respectively. $V\left(G\right)$ is the vertex set of graph G and $E\left(G\right)$ is the edge set of graph G. The distance between $v_1$ and $v_2$, denote as $dist(v_1,v_2)$, is the length of shortest path from $v_1$ to $v_2$. Let $dist(e_i,e_j)=min\left\{dist(u,v)\right|u\in e_i,v\in e_j\}$ denote the distance of two edges $e_i$ and $e_j$. We use $(b,k)$ to denote the greatest common division between two integers b and k.

A strong edge coloring of a graph G is a mapping $\varphi :E\left(G\right)⟶\{1,2,\cdots ,k\}$ such that any two edges $e_i$ and $e_j$ with distance at most one are colored by different colors, which is $\varphi \left(e_i\right)\ne \varphi \left(e_j\right)$. The minimum number of colors k required in a strong edge coloring of G is called the strong chromatic index denoted by ${\chi }_s^{\prime }\left(G\right)$, and the coloring is a strong k edge coloring. The concept of strong edge coloring was introduced by Fouquet and Jolivet [1]. A significant amount of interesting papers were devoted to strong edge coloring of graphs.

In 1985, Erdos and Nešetřil [2] conjectured that $\frac{5}{4}{\Delta }^2\left(G\right)$ when $\Delta \left(G\right)$ is even and $\frac{5}{4}{\Delta }^2\left(G\right)-\frac{1}{2}\Delta \left(G\right)+\frac{1}{4}$ when $\Delta \left(G\right)$ is odd.

Andersen [3] and Horák, Qing and Trotter [4] independently showed that ${\chi }_s^{\prime }\left(G\right)\le 10$ for any graph G with $\Delta \left(G\right)=3$. Cranston verified that ${\chi }_s^{\prime }\left(G\right)\le 22$ for graphs with $\Delta \left(G\right)=4$ in [5] and Huang, Santana and Yu [6] improved the upper bound to 21. Molloy and Reed [7] proved that ${\chi }_s^{\prime }\left(G\right)\le 1.998{\Delta }^2\left(G\right)$. Recently, the coefficient was improved to $1.835$ by Bonamy, Perrett and Postle [8]. Wang, Shiu, Wang and Chen [9] proved that every planar graph with a maximum degree of 4 can be strong edge colored with at most 19 colors.

In 1993, Brualdi and Massey [10] conjectured that every bipartite graph $G(A,B)$ can be strong edge colored with at most $\Delta \left(A\right)\Delta \left(B\right)$ colors.

Steger and Yu [11] confirmed that the chromatic index of any bipartite subcubic graph is at most 9. Nakprasit [12] confirmed the upper conjecture for $(2,\Delta )$-bipartite graphs. Then, Bensmail, Lagoutte and Valicov [13] proved that if G is a $(3,\Delta )$-bipartite graph, then ${\chi }_s^{\prime }\left(G\right)\le 4\Delta$. Recently, Huang, Yu and Zhou [14] confirmed the upper conjecture for $(3,\Delta )$-bipartite graphs.

Generalized Petersen graph$P(n,k)$ has $2n$ vertices where $v_1,v_2,\cdots v_n$ are denoted as outer vertices and $u_1,u_2,\cdots ,u_n$ are denoted as inner vertices. $E\left(P(n,k)\right)=E_1\bigcup E_2\bigcup E_3$ where $E_1$ is the set of edges $v_i{v}_{i+1}$ for $i\in [1,n]$, named as outer edges; $E_2$ is the set of edges $u_i{u}_{i+k}$ for $i\in [1,n]$, named as inner edegs; $E_3$ is the set of edges $v_i{u}_i$ for $i\in [1,n]$, named as called leg edges. Figure 1 is an illustration of generalized Petersen graph $P(n,k)$ with $n=9$ and $k=4$.

In 2018, Wu and Yang [15] conjectured that every generalized Petersen graph $P(n,k)$ with $n>2k$ and $k\ge 4$ can be strong edge colored with 7 colors. In the same paper, they showed some upper bound of strong chromatic index of $P(n,k)$ with $k\in \{1,2,3\}$.

In this paper, we prove following theorem:

Theorem 1.

Every generalized Petersen graph $P(n,k)$ with $n>2k$ and $k\ge 4$ can be strong edge colored with at most 9 colors.

The remainder of this paper is organized as follows. In Section 2, we give some lemmas. In Section 3, we show how to color generalized Petersen graphs $P(n,k)$ with $n>2k$ and $k\ge 4$ with 9 colors.

2. Lemmas

Lemma 1.

If $P(n,k)$ is a generalized Petersen graph with $n>2k$, $k\ge 4$ and $n=ak+b$, then all inner vertices $u_1,u_2,\cdots ,u_n$ are covered by either one cycle when $(b,k)=1$ or b cycles when $k=db$.

Proof. 

Since $P(n,k)$ is a generalized Petersen graph, all inner vertices are covered by some vertex disjoint cycles which are denoted as $C_1,C_2,\cdots ,C_k$. By symmetry, it is obvious that $|C_1|=|C_2|=\cdots =|C_k|$ and $k=\frac{xn}{k}$ where x is the minimum positive integer satisfying $x·b\equiv 0$ (mod k).

• If $(b,k)=1$, then $\left|C\right|=\frac{kn}{k}=n$. The cycle C cover all inner vertices;
• If $k\equiv 0$ (mod b), then $x=d$ and $|C_i|=\frac{dn}{k}=\frac{d(ak+b)}{k}=da+1$. The corresponding number of cycles which cover all inner vertices is $\frac{n}{|C_i|}=\frac{n}{da+1}=\frac{b(da+1)}{ad+1}=b$.

 ☐

Lemma 2.

If C is a cycle with $E\left(C\right)=\{e_1,e_2\cdots ,e_n\}$, $|L_{e_i}|\ge 3$ for $i\in [1,n]$ and $|L_{e_{n-1}}|=4$, $|L_{e_n}|=5$, then cycle C can be strong edge colored properly.

Proof. 

We greedily color edges from $e_1$ to $e_{n-2}$ by some color in their list. Since $|L_{e_i}|\ge 3$, it’s easy to know $|A_{e_i}|\ge |L_{e_i}|-2=1$ for $i\in [1,n-2]$. Hence, the former $n-2$ edges can be well strong edge colored. Next, we check and show how to color $e_{n-1}$ and $e_n$. For $e_{n-1}$, $A_{e_{n-1}}=L_{e_{n-1}}-\varphi \left(e_{n-2}\right)-\varphi \left(e_{n-3}\right)-\varphi \left(e_1\right)$. Since $|L_{e_{n-1}}|=4$, then $|A_{e_{n-1}}|\ge 1$, which means $e_{n-1}$ can be colored by some color in $A_{e_{n-1}}$. By the same argument, we can color $e_n$ by a color from $A_{e_n}$. ☐

Lemma 3.

If C is a cycle with $E\left(C\right)=\{e_1,e_2\cdots ,e_n\}$, $|L_{e_i}|\ge 3$ for $i\in [1,k-2]\bigcup [k+1,n-1]$, $|L_{e_{n-1}}|=4$, $|L_{e_n}|=5$, and $|L_{e_{k-1}}|=4$, $|L_{e_k}|=2$, then cycle C can be strong edge colored properly.

Proof. 

By the argument of Lemma 2, we only need to show how to color $e_k$, and we have following two cases.

• If $A_{e_k}=L_{e_k}-\varphi \left(e_{k-1}\right)-\varphi \left(e_{k-2}\right)\ne \varnothing$, then $e_k$ can be colored by some color in $A_{e_k}$;
• If $A_{e_k}=L_{e_k}-\varphi \left(e_{k-1}\right)-\varphi \left(e_{k-2}\right)=\varnothing$, then we know $L_{e_k}=\varphi \left(e_{k-1}\right)+\varphi \left(e_{k-2}\right)$. Next, we show how to color $e_k$ by recoloring $e_{k-1}$. Since, $|L_{e_{k-1}}|=4$, then $|A_{e_{k-1}}|=|L_{e_{k-1}}-\varphi \left(e_{k-2}\right)-\varphi \left(e_{k-3}\right)|=2$. W.l.o.g, $A_{e_{k-1}}=\{c,d\}$ and $\varphi \left(e_{k-2}\right)=a$, $\varphi \left(e_{k-3}\right)=b$, $\varphi \left(e_{k-1}\right)=c$. By assumption, we can recolor $e_{k-1}$ by d, and color $e_k$ by c.

 ☐

Lemma 4.

If C is a cycle with $E\left(C\right)=\{e_1,e_2\cdots ,e_n\}$, $|L_{e_i}|\ge 3$ for $i\in [1,n-2]$ and $|L_{e_{n-1}}|=5$, $|L_{e_n}|=4$, then C can be strong edge colored properly.

Proof. 

By Lemma 2, we only need to show how to color $e_n$ and have following two cases.

• If $A_{e_n}=L_{e_n}-\varphi \left(e_{n-1}\right)-\varphi \left(e_{n-2}\right)-\varphi \left(e_1\right)-\varphi \left(e_1\right)-\varphi \left(e_2\right)\ne \varnothing$, then we can color $e_n$ by some color in $A_{e_n}$;
• If $A_{e_n}=L_{e_n}-\varphi \left(e_{n-1}\right)-\varphi \left(e_{n-2}\right)-\varphi \left(e_1\right)-\varphi \left(e_1\right)-\varphi \left(e_2\right)=\varnothing$, then we know $L_{e_n}=\varphi \left(e_{n-1}\right)+\varphi \left(e_{n-2}\right)+\varphi \left(e_1\right)+\varphi \left(e_1\right)+\varphi \left(e_2\right)$. Since $|L_{e_{n-1}}|=5$, and $|A_{e_{n-1}}|=|L_{e_{n-1}}-\varphi \left(e_{n-2}\right)-\varphi \left(e_{n-3}\right)-\varphi \left(e_1\right)|\ge 2$, then we can recolor edge $e_{n-1}$ by another color of $A_{e_{n-1}}$. W.l.o.g, $A_{e_{n-1}}=\{c,d\}$, $L_{e_n}=\{c,a,e,f\}$, $\varphi \left(e_{n-2}\right)=a$, $\varphi \left(e_{n-3}\right)=b$, $\varphi \left(e_1\right)=e$ and $\varphi \left(e_{n-1}\right)=c$. We can recolor $e_{n-1}$ by d, and color $e_n$ by c. All edges can be strong edge colored properly.

 ☐

Lemma 5.

If C is a cycle with $E\left(C\right)=\{e_1,e_2\cdots ,e_n\}$, $|L_{e_i}|\ge 3$ for $i\in [5,n-2]$, $|L_{e_1}|=4$, $|L_{e_2}|=2$ and $|L_{e_3}|=3$, $|L_{e_4}|=2$, $|L_{e_{n-1}}|=4$, $|L_{e_n}|=5$, then cycle C can be strong edge colored properly.

Proof. 

By Lemma 2, we only need to show how to color $e_4$ properly and have following two cases.

• If $L_{e_4}-\varphi \left(e_2\right)-\varphi \left(e_3\right)\ne \varnothing$, then we can color $e_4$ by some color of $A_{e_4}$;
• If $L_{e_4}-\varphi \left(e_2\right)-\varphi \left(e_3\right)=\varnothing$, then we consider the available colors of $e_3$.

  -

  If $L_{e_3}-\varphi \left(e_2\right)-\varphi \left(e_1\right)-\varphi \left(e_3\right)\ne \varnothing$, then we can recolor $e_3$ by some color of $A_{e_3}$. In this case, we have $L_{e_4}-\varphi \left(e_2\right)-\varphi \left(e_3\right)\ne \varnothing$, which means we can color $e_4$ by another color differ from $\varphi \left(e_3\right)$;

  -

  If $L_{e_3}-\varphi \left(e_2\right)-\varphi \left(e_1\right)-\varphi \left(e_3\right)=\varnothing$, then we can recolor $e_1$ by a color differ from $\varphi \left(e_2\right)$, recolor $e_3$ by the former color of $e_1$ and color edge $e_4$ by the former color of $e_3$.

 ☐

Lemma 6.

If C is a cycle with $E\left(G\right)=\{e_1,e_2\cdots ,e_n\}$, $|L_{e_i}|\ge 3$ for $i\in [1,j-3]\bigcup [j+1,n-1]$, $|L_{e_{j-2}}|=5$, $|L_{e_{j-1}}|=3$, $|L_{e_j}|=2$, and $|L_{e_{n-1}}|=4$, $|L_{e_n}|=5$, then cycle C can be strong edge colored properly.

Proof. 

By Lemma 2, we only need to show how to color $e_j$, and we have following two cases.

• If $L_{e_j}-\varphi \left(e_{j-1}\right)-\varphi \left(e_{j-2}\right)\ne \varnothing$, then we can color $e_j$ by some color of $A_{e_j}$;
• If $L_{e_j}-\varphi \left(e_{j-1}\right)-\varphi \left(e_{j-2}\right)=\varnothing$. Since $|L_{e_{j-2}}|=5$, then we have $|L_{e_{j-2}}-\varphi \left(e_{j-1}\right)-\varphi \left(e_{j-3}\right)-\varphi \left(e_{j-4}\right)|\ge 2$ hold. Hence, we can recolor $e_{j-2}$ by another color which differ from $\varphi \left(e_{j-2}\right)$. After recounting $A_{e_j}$, we have $L_{e_j}-\varphi \left(e_{j-1}\right)-\varphi \left(e_{j-2}\right)\ne \varnothing$ holds. Next, we can color $e_j$ by the former color of $e_{j-2}$ and all edges can be strong edge colored properly.

 ☐

3. Coloring

Theorem 2.

If G is a generalized Petersen graph with $n>2k$, $k\ge 4$ and $n\equiv 0$ (mod k), then ${\chi }_s^{\prime }\left(G\right)\le 9$.

Proof. 

We prove this theorem by following progresses.

First, we pre-color all inner edges by following progresses.

• If $a\equiv 0$ (mod 3), then we color edges $u_i{u}_{i+k}$, $u_{i+k}{u}_{i+2k}$, …, $u_{i+(a-2)k}{u}_{i+(a-1)k}$ with “1, 2, 3” in order for $i\in \{1,k\}$;
• If $a\equiv 1$ (mod 3), then we color edges $u_i{u}_{i+k}$, $u_{i+k}{u}_{i+2k}$, …, $u_{i+(a-2)k}{u}_{i+(a-1)k}$ for $i\in \{1,k\}$ with “1, 2, 3, 4, 1, 2, 3, …, 1, 2, 3” in order;
• If $a\equiv 2$ (mod 3), then we color edges $u_i{u}_{i+k}$, $u_{i+k}{u}_{i+2k}$, …, $u_{i+(a-2)k}{u}_{i+(a-1)k}$ for $i\in \{1,k\}$ with “1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, …, 1, 2, 3” in order.

Second, we pre-color all leg-edges by following four cases, and show that every generalized Petersen graph $P(n,k)$ can be strong edge colored with 9 colors in each cases.

Case 1. If k is even and a is even, then we pre-color all leg edges by following four steps.

• Step 1: Color $e_{L_{(i-1)k+j}}$ with color “6” for {i|$i\in [1,a]$ and i is odd} and {j|$j\in [1,k]$ and j is odd};
• Step 2: Color $e_{L_{(i-1)k+j}}$ with color “7” for {i|$i\in [1,a]$ and i is odd} and {j|$j\in [1,k]$ and j is even};
• Step 3: Color $e_{L_{(i-1)k+j}}$ with color “8” for {i|$i\in [1,a]$ and i is even} and {j|$j\in [1,k]$ and j is odd};
• Step 4: Color $e_{L_{(i-1)k+j}}$ with color “9” for {i|$i\in [1,a]$ and i is even} and {j|$j\in [1,k]$ and j is even}.

In Case 1, we have three subcases separately as $a\equiv 0$ (mod 3), $a\equiv 2$ (mod 3) and $a\equiv 2$ (mod 3). For the sake of description, we only consider subcase $a\equiv 0$ (mod 3), and the other two subcases can be done by same arguments.

By upper steps, we have that all leg edges are colored by $\{6,7,8,9\}$ and inner edges are colored by $\{1,2,3\}$. Hence, all leg edges can be distinguished from all inner edges. Next, we only need to show that all leg edges and inner edges can be well strong edge colored separately.

For two leg edges, we have following two cases.

• If two leg edges are $e_{L_{ik+j}}$ and $e_{L_{ik+j+1}}$, then they can be distinguished by two different colors of {6, 7};
• If two leg edges are $e_{L_{ik+j}}$ and $e_{L_{(i+1)k+j}}$, then they can be distinguished by two different color sets {6, 7} and {8, 9}.

For any two inner edges with distance at most 1, since they were colored in some order, it is obvious that all inner edges can be strong edge colored properly.

By upper arguments, all leg edges and inner edges have been pre-strong edge colored properly. It’s easy to count and get $|A_{e_{ik+1}}|=4$, $|A_{e_{ik+2}}|=5$, …, $|A_{e_{ik+k-2}}|=5$, $|A_{e_{ik+k-1}}|=4$ and $|A_{e_{ik+k}}|=2$ for $i\in \{0,a-1\}$. By Lemma 2 and Lemma 3, all outer edges can be strong edge colored properly.

Hence, every generalized Petersen graph $P(n,k)$ with $n>2k$, $k\ge 4$, $n=ak$, k is even and a is even can be strong edge colored properly with at most 9 colors.

Case 2. k is even and a is odd. First, we pre-color all leg edges by following six steps.

• Step 1: Color $e_{L_{(i-1)k+j}}$ with color “6”, for {i|$i\in [1,a-1]$ and i is odd}, and {j|$j\in [1,k]$ and j is odd};
• Step 2: Color $e_{L_{(i-1)k+j}}$ with color “7”, for {i|$i\in [1,a-1]$ and i is odd}, and {j|$j\in [1,k]$ and j is even};
• Step 3: Color $e_{L_{(i-1)k+j}}$ with color “8”, for {i|$i\in [1,a-1]$ and i is even}, and {j|$j\in [1,k]$ and j is odd};
• Step 4: Color $e_{L_{(i-1)k+j}}$ with color “9”, for {i|$i\in [1,a-1]$ and i is even}, and {j|$j\in [1,k]$ and j is even};
• Step 5: Color $e_{L_{(a-1)k+j}}$ with color “7” for {j|$j\in [1,k]$ and j is odd};
• Step 6: Color $e_{L_{(a-1)k+j}}$ with color “8”, for {j|$j\in [1,k]$ and j is even}.

By similar argument of Case 1, we only consider the subcase $a\equiv 1$ (mod 3). By counting $A_{e_i}$ for $i\in [0,ak]$, we have

• $|A_{e_{ik+1}}|=4$, $|A_{e_{ik+2}}|=5$, …, $|A_{e_{ik+k-2}}|=5$, $|A_{e_{ik+k-1}}|=4$ and $|A_{e_{ik+k}}|=2$ for $i\in [0,a-3]\bigcup \{a-1\}$;
• $|A_{e_{(a-2)k+1}}|=4$, $|A_{e_{(a-2)k+2}}|=\cdots |A_{e_{(a-1)k-2}}|=5$, $|A_{e_{(a-1)k-1}}|=4$ and $|A_{e_{(a-1)k}}|=3$.

By Lemmas 2 and 3, all outer edges can be strong edge colored properly. Hence, every generalized Petersen graph $P(n,k)$ with $n>2k$, $k\ge 4$, $n=ak$, k is even and a is odd can be strong edge colored with 9 colors.

Case 3. k is odd and a is even. First, we pre-color all leg edges by following four steps.

• Step 1: Color $e_{L_{ik+j}}$ with color “6”, for {i|$i\in [1,a-1]$ and i is odd}, and {j|$j\in [1,a]$ and j is odd};
• Step 2: Color $e_{L_{ik+j}}$ with color “7”, for {i|$i\in [1,a-1]$ and i is odd}, and {j|$j\in [1,a]$ and j is even};
• Step 3: Color $e_{L_{ik+j}}$ with color “7”, for {i|$i\in [1,a-1]$ and i is even}, and {j|$j\in [1,k]$ and j is odd};
• Step 4: Color $e_{L_{ik+j}}$ with color “6”, for {i|$i\in [1,a-1]$ and i is even}, and {j|$j\in [1,k]$ and j is even}.

By similar argument of Case 1, we only consider the subcase $a\equiv 1$ (mod 3). By counting $A_{e_i}$ for $i\in [0,a-1]$, we have $|A_{e_{ik+1}}|=|A_{e_{ik+2}}|=\cdots =|A_{e_{ik+k-1}}|=5$, $|A_{e_{ik+k}}|=4$ hold.

By Lemma 2, all outer edges can be strong edge colored properly. Hence, generalized Petersen graph $P(n,k)$ with $n>2k$, $k\ge 4$, $n=ak$, k is odd and a is even can be strong edge colored with 9 colors.

Case 4. k is odd and a is odd. First, we pre-color all leg edges by following three steps.

• Step 1: Color $e_{L_{ik+j}}$ with color “6”, for {i|$i\in [1,a-2]$ and i is odd}, and {j|$j\in [1,a]$ and j is odd};
• Step 2: Color $e_{L_{ik+j}}$ with color “7”, for {i|$i\in [1,a-2]$ and i is odd}, and {j|$j\in [1,a]$ and j is even};
• Step 4: Color $e_{L_{ik+j}}$ with color “7”, for {i|$i\in [1,a-2]$ and i is even}, and {j|$j\in [1,k]$ and j is odd};
• Step 5: Color $e_{L_{ik+j}}$ with color “6”, for {i|$i\in [1,a-2]$ and i is even}, and {j|$j\in [1,k]$ and j is even};
• Step 6: Color $e_{L_{(a-1)k+j}}$ with color “8”, for {j|$j\in [1,k]$ and j is odd};
• Step 7: Color $e_{L_{(a-1)k+j}}$ with color “9”, for {j|$j\in [1,k]$ and j is even}.

By similar argument of Case 1, we only consider the subcase $a\equiv 2$ (mod 3). By counting $A_{e_i}$ for $i\in [0,ak]$, we have

• $|A_{e_1}|=4$, $|A_{e_2}|=\cdots =|A_{e_{k-1}}|=5$, $|A_{e_k}|=4$;
• $|A_{e_{ik+1}}|=\cdots =|A_{e_{ik+k-1}}|=5$, $|A_{e_{ik+k}}|=4$ for $i\in [1,a-3]$;
• $|A_{e_{(a-2)k+1}}|=\cdots =|A_{e_{(a-1)k-2}}|=5$, $|A_{e_{(a-1)k-1}}|=4$ and $|A_{e_{(a-1)k}}|=2$;
• $|A_{e_{(a-1)k+1}}|=4$, $|A_{e_{(a-1)k+2}}|=\cdots =|A_{e_{(a-1)k+k-2}}|=5$, $|A_{e_{(a-1)k+k-1}}|=4$ and $|A_{e_{ak}}|=2$.

By Lemma 2 and Lemma 3, all outer edges can be strong edge colored properly. Hence, every generalized Petersen graph $P(n,k)$ with $n>2k$, $k\ge 4$, $n=ak$, k is odd and a is odd can be strong edge colored with 9 colors.

By upper four cases, we have Theorem 2 hold. ☐

Theorem 3.

If G is a generalized Petersen graph with $n>2k$, $k\ge 4$, $a\ge 2$ and $n=ak+b$, then ${\chi }_{sl}^{\prime }\left(G\right)\le 9$.

Proof. 

To prove the theorem, we divide the proof into following eight cases. Case 1: k is even, a is even and b is even; Case 2: k is even, a is even and b is odd; Case 3: k is even, a is odd and b is even; Case 4: k is even, a is odd and b is odd; Case 5: k is odd, a is even and b is even; Case 6: k is odd, a is even and b is odd; Case 7: k is odd, a is odd and b is odd; Case 8: k is odd, a is odd and b is even. First, we pre-color all inner edges by following progresses.

• If $a-1\equiv 0$ (mod 3), then we color edges by following four steps.

  -

  Color edges $u_i{u}_{i+k}$, $u_{i+k}{u}_{i+2k}$, …, $u_{i+(a-2)k}{u}_{i+(a-1)k}$ for $i\in \{1,k\}$ with “1, 2, 3” in order;

  -

  Color edges $u_{i+(a-1)k}{u}_{i+ak}$ for $i\in \{1,b\}$ with “4”;

  -

  Color edges $u_i{u}_{(a-1)k+b+i}$ for $i\in \{1,b\}$ with “5”;

  -

  Color edges $u_{ak+i}{u}_{k-b+i}$ for $i\in \{1,b\}$ with “5”.

• If $a-1\equiv 1$ (mod 3), then we color edges by following four steps.

  -

  Color edges $u_i{u}_{i+k}$, $u_{i+k}{u}_{i+2k}$, …, $u_{i+(a-2)k}{u}_{i+(a-1)k}$ for $i\in \{1,k\}$ with “1, 4, 2, 3, 1, 2, 3, …, 1, 2, 3” in order;

  -

  Color edges $u_{i+(a-1)k}{u}_{i+ak}$ for $i\in \{1,b\}$ with “4”;

  -

  Color edges $u_i{u}_{(a-1)k+b+i}$ for $i\in \{1,b\}$ with “5”;

  -

  Color edges $u_{ak+i}{u}_{k-b+i}$ for $i\in \{1,b\}$ with “5”.

• If $a-1\equiv 2$ (mod 3), then we color edges by following four steps.

  -

  Color edges $u_i{u}_{i+k}$, $u_{i+k}{u}_{i+2k}$, …, $u_{i+(a-2)k}{u}_{i+(a-1)k}$ for $i\in \{1,k\}$ with “1, 4, 2, 1, 3, 2, 1, …, 2, 1, 3” in order;

  -

  Color edges $u_{i+(a-1)k}{u}_{i+ak}$ for $i\in \{1,b\}$ with “4”;

  -

  Color edges $u_i{u}_{(a-1)k+b+i}$ for $i\in \{1,b\}$ with “5”;

  -

  Color edges $u_{ak+i}{u}_{k-b+i}$ for $i\in \{1,b\}$ with “5”.

Next, we color all leg edges by following eight cases, and show every generalized Petersen graph $P(n,k)$ can be strong edge colored with 9 colors in each case.

Case 1. k is even, a is even and b is even. First, we pre-color all leg edges $e_{L_i}$ for $i\in \{1,n\}$ by following six steps.

• Step 1: Color $e_{L_{(i-1)k+j}}$ with color “6”, for {i|$i\in [1,a]$ and i is odd}, and {j|$j\in [1,a]$ and j is odd};
• Step 2: Color $e_{L_{(i-1)k+j}}$ with color “7”, for {i|$i\in [1,a]$ and i is odd}, and {j|$j\in [1,a]$ and j is even};
• Step 3: Color $e_{L_{(i-1)k+j}}$ with color “8”, for {i|$i\in [1,a]$ and i is even}, and {j|$j\in [1,k]$ and j is odd};
• Step 4: Color $e_{L_{(i-1)k+j}}$ with color “9”, for {i|$i\in [1,a]$ and i is even}, and {j|$j\in [1,k]$ and j is even};
• Step 5: Color $e_{L_{ak+j}}$ with color “7”, for {j|$j\in [1,k]$ and j is odd};
• Step 6: Color $e_{L_{ak+j}}$ with color “8”, for {j|$j\in [1,k]$ and j is even}.

It is obvious that all inner edges and all leg edges can be distinguished separately, and all inner edges can be distinguished from all leg edges. Hence, we only need to color all outer edges properly. Similarly, we only show the case $a-1\equiv 1$ (mod 3), and the other two cases $a-1\equiv 0$ (mod 3) and $a-1\equiv 2$ (mod 3) can be done by same arguments.

If $a-1\equiv 0$ (mod 3), then we know that

• $|A_{e_1}|=4$, $|A_{e_2}|=|A_{e_3}|=\cdots =|A_{e_{k-2}}|=5$, $|A_{e_{k-1}}|=4$ and $|A_{e_k}|=2$;
• $|A_{e_{ik+1}}|=4$, $|A_{e_{ik+2}}|=|A_{e_{ik+3}}|=\cdots =|A_{e_{(i+1)k-2}}|=5$, $|A_{e_{(i+1)k-1}}|=4$ and $|A_{e_{(i+1)k}}|=2$ for $i\in [0,a-2]$;
• $|A_{e_{(a-1)k+1}}|=4$, $|A_{e_{(a-1)k+2}}|=|A_{e_{(a-1)k+3}}|=\cdots =|A_{e_{(a-1)k+b-2}}|=5$, $|A_{e_{(a-1)k+b-1}}|=4$, $|A_{e_{(a-1)k+b}}|=|A_{e_{(a-1)k+b+1}}|=\cdots =|A_{e_{ak-2}}|=5$, $|A_{e_{ak-1}}|=4$ and $|A_{e_{ak}}|=3$;
• $|A_{e_{ak+1}}|=4$, $|A_{e_{ak+2}}|=|A_{e_{ak+3}}|=\cdots =|A_{e_{ak+b-2}}|=5$, $|A_{e_{ak+b-1}}|=4$ and $|A_{e_{ak+b}}|=3$.

By Lemma 2 and Lemma 3, all outer edges can be strong edge colored properly. The other two subcases $a-1\equiv 1$ (mod 3) and $a-1\equiv 2$ (mod 3) can be done by similar counting progresses and arguments. For sake of describe, we omit the proof of the left two subcases.

Case 2. k is even, a is even and b is odd. First, we pre-color all leg edges by following six steps.

• Step 1: Color $e_{L_{(i-1)k+j}}$ with color “6”, for {i|$i\in [1,a]$ and i is odd}, and {j|$j\in [1,k]$ and j is odd};
• Step 2: Color $e_{L_{(i-1)k+j}}$ with color “7”, for {i|$i\in [1,a]$ and i is odd}, and {j|$j\in [1,k]$ and j is even};
• Step 3: Color $e_{L_{(i-1)k+j}}$ with color “8”, for {i|$i\in [1,a]$ and i is even}, and {j|$j\in [1,k]$ and j is odd};
• Step 4: Color $e_{L_{(i-1)k+j}}$ with color “9”, for {i|$i\in [1,a]$ and i is even}, and {j|$j\in [1,k]$ and j is even};
• Step 5: Color $e_{L_{ak+j}}$ with color “2”, for {j|$j\in [1,b]$ and j is odd};
• Step 6: Color $e_{L_{ak+j}}$ with color “8”, for {j|$j\in [1,b]$ and j is even}.

For sake of describe, we only show the subcase $a-1\equiv 1$ (mod 3), and the other two subcases can be done by similar arguments. If $a-1\equiv 1$(mod 3), then by upper coloring progress, we know that

• $|A_{e_{ik+1}}|=4$, $|A_{e_{ik+2}}|=|A_{e_{ik+3}}|=\cdots =|A_{e_{(i+1)k-2}}|=5$, $|A_{e_{(i+1)k-1}}|=4$ and $|A_{e_{(i+1)k}}|=2$ for $i\in [0,a-2]$;
• $|A_{e_{(a-1)k+1}}|=4$, $|A_{e_{(a-1)k+2}}|=\cdots =|A_{e_{(a-1)k+b-1}}|=5$, $|A_{e_{(a-1)k+b}}|\ge 3$, $|A_{e_{(a-1)k+b+1}}|=\cdots |A_{e_{ak-2}}|=$ 5, $|A_{e_{ak-1}}|\ge 3$ and $|A_{e_{ak}}|=3$;
• $|A_{e_{ak+1}}|=4$, $|A_{e_{ak+2}}|=\cdots =|A_{e_{ak+b-2}}|=5$, $|A_{e_{ak+b-1}}|=4$ and $|A_{e_{ak+b}}|=2$.

By Lemma 2 and Lemma 3, all outer edges can be strong edge colored properly. We are done.

Case 3. k is even, a is odd and b is even. First, we pre-color all leg edges by following six steps.

• Step 1: Color $e_{L_{(i-1)k+j}}$ with color “6”, for {i|$i\in [1,a-1]$ and i is odd}, and {j|$j\in [1,a]$ and j is odd};
• Step 2: Color $e_{L_{(i-1)k+j}}$ with color “7”, for {i|$i\in [1,a-1]$ and i is odd}, and {j|$j\in [1,a]$ and j is even};
• Step 3: Color $e_{L_{(i-1)k+j}}$ with color “8”, for {i|$i\in [1,a-1]$ and i is even}, and {j|$j\in [1,k]$ and j is odd};
• Step 4: Color $e_{L_{(i-1)k+j}}$ with color “9”, for {i|$i\in [1,a-1]$ and i is even}, and {j|$j\in [1,k]$ and j is even};
• Step 5: Color $e_{L_{(a-1)k+j}}$ with color “7”, for {j|$j\in [1,k]$ and j is odd};
• Step 6: Color $e_{L_{(a-1)k+j}}$ with color “6”, for {j|$j\in [1,k]$ and j is even};
• Step 7: Color $e_{L_{ak+j}}$ with color “8”, for{j|$j\in [1,b]$ and j is odd};
• Step 8: Color $e_{L_{ak+j}}$ with color “9”, for{j|$j\in [1,b]$ and j is even}.

It is easy to see that the following progress, counting available colors for each outer edge, of three subcases $a-1\equiv 0$ (mod 3), $a-1\equiv 1$ (mod 3), $a-1\equiv 2$ (mod 3) are similar. Hence, we only show the counting progress of subcase $a-1\equiv 2$ (mod 3), and the other two subcases can be done by a same argument.

If $a-1\equiv 2$ (mod 3), then by upper coloring progress, we know that

• $|A_{e_{ik+1}}|=4$, $|A_{e_{ik+2}}|=|A_{e_{ik+3}}|=\cdots =|A_{e_{(i+1)k-2}}|=5$, $|A_{e_{(i+1)k-1}}|=4$ and $|A_{e_{(i+1)k}}|=2$ for $i\in [0,a-2]$;
• $|A_{e_{(a-1)k+1}}|=3$, $|A_{e_{(a-1)k+2}}|=|A_{e_{(a-1)k+3}}|=\cdots =|A_{e_{(a-1)k+b-1}}|=5$, $|A_{e_{(a-1)k+b}}|=4$, $|A_{e_{(a-1)k+b+1}}|=\cdots =|A_{e_{ak-2}}|=5$, $|A_{e_{ak-1}}|=4$, $|A_{e_{ak}}|=2$;
• $|A_{e_{ak+1}}|=4$, $|A_{e_{ak+2}}|=\cdots =|A_{e_{ak+b-2}}|=5$, $|A_{e_{ak+b-1}}|=4$, $|A_{e_{ak+b}}|=2$.

By Lemma 2 and Lemma 3, all outer edges can be strong edge colored properly. We are done.

Case 4. k is even, a is odd and b is odd. First, we pre-color all leg edges by following six steps.

• Step 1: Color $e_{L_{(i-1)k+j}}$ with color “6”, for {i|$i\in [1,a]$ and i is odd}, and {j|$j\in [1,k]$ and j is odd};
• Step 2: Color $e_{L_{(i-1)k+j}}$ with color “7”, for {i|$i\in [1,a]$ and i is odd}, and {j|$j\in [1,k]$ and j is even};
• Step 3: Color $e_{L_{(i-1)k+j}}$ with color “8”, for {i|$i\in [1,a-1]$ and i is even}, and {j|$j\in [1,k]$ and j is odd};
• Step 4: Color $e_{L_{(i-1)k+j}}$ with color “9”, for {i|$i\in [1,a-1]$ and i is even}, and {j|$j\in [1,k]$ and j is even};
• Step 5: Color $e_{L_{ak+j}}$ with color “8”, for{j|$j\in [1,b]$ and j is odd};
• Step 6: Color $e_{L_{ak+j}}$ with color “9”, for{j|$j\in [1,b]$ and j is even}.

It is easy to see that the following progress, counting available colors for each outer edge, of three subcases $a-1\equiv 0$ (mod 3), $a-1\equiv 1$ (mod 3), $a-1\equiv 2$ (mod 3) are similar. Hence, we only show the counting progress of subcase $a-2\equiv 2$ (mod 3), and the other two subcases can be done by a same argument.

If $a-1\equiv 2$ (mod 3), then by upper coloring progress, we know that

• $|A_{e_{ik+1}}|=4$, $|A_{e_{ik+2}}|=|A_{e_{ik+3}}|=\cdots =|A_{e_{(i+1)k-2}}|=5$, $|A_{e_{(i+1)k-1}}|=4$ and $|A_{e_{(i+1)k}}|=2$ for $i\in [0,a-2]$;
• $|A_{e_{(a-1)k+1}}|=4$, $|A_{e_{(a-1)k+2}}|=\cdots =|A_{e_{(a-1)k+b-1}}|=5$, $|A_{e_{(a-1)k+b}}|\ge 3$, $|A_{e_{(a-1)k+b+1}}|=\cdots =|A_{e_{ak-2}}|=$ 5, $|A_{e_{ak-1}}|=4$, $|A_{e_{ak}}|\ge 2$;
• $|A_{e_{ak+1}}|=4$, $|A_{e_{ak+2}}|=\cdots =|A_{e_{ak+b-2}}|=5$, $|A_{e_{ak+b-1}}|=4$, $|A_{e_{ak+b}}|\ge 2$.

By Lemma 2 and Lemma 3, all outer edges can be strong edge colored properly. We are done.

Case 5. k is odd, a is even and b is even. First, we pre-color all leg edges by following six steps.

• Step 1: Color $e_{L_{(i-1)k+j}}$ with color “6”, for {i|$i\in [1,a]$ and i is odd}, and {j|$j\in [1,k]$ and j is odd};
• Step 2: Color $e_{L_{(i-1)k+j}}$ with color “7”, for {i|$i\in [1,a]$ and i is odd}, and {j|$j\in [1,k]$ and j is even};
• Step 3: Color $e_{L_{(i-1)k+j}}$ with color “7”, for {i|$i\in [1,a]$ and i is even}, and {j|$j\in [1,k]$ and j is odd};
• Step 4: Color $e_{L_{(i-1)k+j}}$ with color “6”, for {i|$i\in [1,a]$ and i is even}, and {j|$j\in [1,k]$ and j is even};
• Step 5: Color $e_{L_{ak+j}}$ with color “6”, for{j|$j\in [1,b]$ and j is odd};
• Step 6: Color $e_{L_{ak+j}}$ with color “7”, for{j|$j\in [1,b]$ and j is even}.

It is easy to see that the following progress, counting available colors for each outer edge, of three subcases $a-1\equiv 0$ (mod 3), $a-1\equiv 1$ (mod 3), $a-1\equiv 2$ (mod 3) are similar. Hence, we only show the counting progress of subcase $a-1\equiv 2$ (mod 3), and the other two subcases can be done by a same argument.

If $a-1\equiv 2$ (mod 3), then by upper coloring progress, we know that

• $|A_{e_1}|=5$, $|A_{e_2}|=\cdots =|A_{e_{k-1}}|=5$, $|A_{e_k}|=4$;
• $|A_{e_{ik+1}}|=4$, $|A_{e_{ik+1}}|=\cdots =|A_{e_{(i+1)k-1}}|=5$, $|A_{e_{(i+1)k}}|=4$, for $i\in \{1,a-2\}$;
• $|A_{e_{(a-1)k+1}}|=\cdots =|A_{e_{(a-1)k+b-1}}|=5$, $|A_{e_{(a-1)k+b}}|=4$, $|A_{e_{(a-1)k+b+1}}|=\cdots =|A_{e_{ak-1}}|=5$, $|A_{e_{ak}}|=4$;
• $|A_{e_{ak+1}}|=\cdots =|A_{e_{ak+b-1}}|=5$, $|A_{e_{ak+b}}|=4$.

By Lemma 3 and Lemma 4, all outer edges can be strong edge colored properly. We are done.

Case 6. k is odd, a is even and b is odd. First, we pre-color all leg edges by following six steps.

• Step 1: Color $e_{L_{(i-1)k+j}}$ with color “6”, for {i|$i\in [1,a-1]$ and i is odd}, and {j|$j\in [1,k]$ and j is odd};
• Step 2: Color $e_{L_{(i-1)k+j}}$ with color “7”, for {i|$i\in [1,a-1]$ and i is odd}, and {j|$j\in [1,k]$ and j is even};
• Step 3: Color $e_{L_{(i-1)k+j}}$ with color “7’, for {i|$i\in [1,a-1]$ and i is even}, and {j|$j\in [1,k]$ and j is odd};
• Step 4: Color $e_{L_{(i-1)k+j}}$ with color “6”, for {i|$i\in [1,a-1]$ and i is even}, and {j|$j\in [1,k]$ and j is even};
• Step 5: Color $e_{L_{(a-1)k+j}}$ with color “8”, for {j|$j\in [1,k]$ and j is odd};
• Step 6: Color $e_{L_{(a-1)k+j}}$ with color “9”, for {j|$j\in [1,k]$ and j is even};
• Step 7: Color $e_{L_{ak+j}}$ with color “7”, for{j|$j\in [1,b]$ and j is odd};
• Step 8: Color $e_{L_{ak+j}}$ with color “8”, for{j|$j\in [1,b]$ and j is even}.

It is easy to see that the following progress, counting available colors for each outer edge, of three subcases $a-1\equiv 0$ (mod 3), $a-1\equiv 1$ (mod 3), $a-1\equiv 2$ (mod 3) are similar. Hence, we only show the counting progress of subcase $a-1\equiv 2$ (mod 3), and the other two subcases can be done by a same argument.

If $a-1\equiv 1$ (mod 3), then by upper coloring progress, we know that

• $|A_{e_{ik+1}}|=\cdots =|A_{e_{(i+1)k-1}}|=5$, $|A_{e_{(i+1)k}}|=4$, for $i\in [0,a-3]$;
• $|A_{e_{(a-2)k+1}}|=\cdots =|A_{e_{(a-1)k-2}}|=5$, $|A_{e_{(a-1)k-1}}|=4$, $|A_{e_{(a-1)k}}|=2$;
• $|A_{e_{(a-1)k+1}}|=4$, $|A_{e_{(a-1)k+2}}|=\cdots =|A_{e_{(a-1)k+b-1}}|=5$, $|A_{e_{(a-1)k+b}}|=4$, $|A_{e_{(a-1)k+b+1}}|=\cdots =|A_{e_{ak-2}}|=5$, $|A_{e_{ak-1}}|=4$, $|A_{e_{ak}}|=3$;
• $|A_{e_{ak+1}}|=\cdots |A_{e_{ak+b-2}}|=5$, $|A_{e_{ak+b-1}}|=4$, $|A_{e_{ak+b}}|=3$.

By Lemma 2 and Lemma 3, all outer edges can be strong edge colored properly. We are done.

Case 7. k is odd, a is odd and b is odd. First, we pre-color all leg edges by following six steps.

• Step 1: Color $e_{L_{(i-1)k+j}}$ with color “6”, for {i|$i\in [1,a]$ and i is odd}, and {j|$j\in [1,k]$ and j is odd};
• Step 2: Color $e_{L_{(i-1)k+j}}$ with color “7”, for {i|$i\in [1,a]$ and i is odd}, and {j|$j\in [1,k]$ and j is even};
• Step 3: Color $e_{L_{(i-1)k+j}}$ with color “7”, for {i|$i\in [1,a]$ and i is even}, and {j|$j\in [1,k]$ and j is odd};
• Step 4: Color $e_{L_{(i-1)k+j}}$ with color “6”, for {i|$i\in [1,a]$ and i is even}, and {j|$j\in [1,k]$ and j is even};
• Step 5: Color $e_{L_{ak+j}}$ with color “7”, for {j|$j\in [1,k]$ and j is odd};
• Step 6: Color $e_{L_{ak+j}}$ with color “8”, for {j|$j\in [1,k]$ and j is even}.

It is easy to see that the following progress, counting available colors for each outer edge, of three subcases $a-1\equiv 0$ (mod 3), $a-1\equiv 1$ (mod 3), $a-1\equiv 2$ (mod 3) are similar. Hence, we only show the counting progress of subcase $a-1\equiv 2$ (mod 3), and the other two subcases can be done by a same argument.

If $a-1\equiv 1$ (mod 3), then by upper coloring progress, we know that

• $|A_{e_{ik+1}}|=\cdots =|A_{e_{(i+1)k-1}}|=5$, $|A_{e_{(i+1)k}}|=4$, for $i\in [0,a-2]$;
• $|A_{e_{(a-1)k+1}}|=\cdots =|A_{e_{(a-1)k+b-1}}|=5$, $|A_{e_{(a-1)k+b}}|=4$, $|A_{e_{(a-1)k+b+1}}|=\cdots =|A_{e_{ak-1}}|=5$, $|A_{e_{ak}}|=3$;
• $|A_{e_{ak+1}}|=4$, $|A_{e_{ak+2}}|=\cdots =|A_{e_{ak+b-2}}|=5$, $|A_{e_{ak+b-1}}|=4$, $|A_{e_{ak+b}}|=3$;

By Lemma 2, all outer edges can be strong edge colored properly. We are done.

Case 8. k is odd, a is odd and b is even. First, we pre-color all leg edges by following six steps.

• Step 1: Color $e_{L_{(i-1)k+j}}$ with color “6”, for {i|$i\in [1,a-1]$ and i is odd}, and {j|$j\in [1,a]$ and j is odd};
• Step 2: Color $e_{L_{(i-1)k+j}}$ with color “7”, for {i|$i\in [1,a-1]$ and i is odd}, and {j|$j\in [1,a]$ and j is even};
• Step 3: Color $e_{L_{(i-1)k+j}}$ with color “7”, for {i|$i\in [1,a-1]$ and i is even}, and {j|$j\in [1,k]$ and j is odd};
• Step 4: Color $e_{L_{(i-1)k+j}}$ with color “6”, for {i|$i\in [1,a-1]$ and i is even}, and {j|$j\in [1,k]$ and j is even};
• Step 5: Color $e_{L_{(a-1)k+j}}$ with color “9”, for {j|$j\in [1,k]$ and j is odd};
• Step 6: Color $e_{L_{(a-1)k+j}}$ with color “8”, for {j|$j\in [1,k]$ and j is even};
• Step 7: Color $e_{L_{ak+j}}$ with color “6”, for{j|$j\in [1,b]$ and j is odd};
• Step 8: Color $e_{L_{ak+j}}$ with color “7”, for{j|$j\in [1,b]$ and j is even}.

It is easy to see that the following progress, counting available colors for each outer edge, of three subcases $a-1\equiv 0$ (mod 3), $a-1\equiv 1$ (mod 3), $a-1\equiv 2$ (mod 3) are similar. Hence, we only show the counting progress of subcase $a-1\equiv 2$ (mod 3), and the other two subcases can be done by a same argument.

If $a-1\equiv 0$ (mod 3), then by upper coloring progress, we know that

• $|A_{e_{ik+1}}|=\cdots =|A_{e_{(i+1)k-1}}|=5$, $|A_{e_{(i+1)k}}|=4$, for $i\in [0,a-3]$;
• $|A_{e_{(a-2)k+1}}|=\cdots =|A_{e_{(a-1)k-2}}|=5$, $|A_{e_{(a-1)k-1}}|=4$, $|A_{e_{(a-1)k}}|=2$;
• $|A_{e_{(a-1)k+1}}|=4$, $|A_{e_{(a-1)k+2}}|=\cdots =|A_{e_{(a-1)k+b-1}}|=5$, $|A_{e_{(a-1)k+b}}|=4$,$|A_{e_{(a-1)k+b+1}}|=\cdots =|A_{e_{ak-2}}|=$ 5, $|A_{e_{ak-1}}|=4$, $|A_{e_{ak}}|=2$;
• $|A_{e_{ak+1}}|=4$, $|A_{e_{ak+2}}|=\cdots =|A_{e_{ak+b-1}}|=5$, $|A_{e_{ak+b}}|=4$.

By Lemma 2, all outer edges can be strong edge colored properly. We are done.

After proving Theorem 2 and Theorem 3, it is easy to know that Theorem 1 is correct. ☐

4. Conclusions

By showing how to use 9 colors to color generalized Petersen graphs $P(n,k)$ with $n>2k$ and $k\ge 4$, we got our main result.