^{1}

^{*}

^{1}

^{1}

A coloring
of
*G* is
*d-distance* if any two vertices at distance at most d from each other get different colors. The minimum number of colors in
*d-distance* colorings of
*G* is its
*d-distance*
*chromatic number*, denoted by
* χ*
_{d}(
*G*). In this paper, we give the exact value of
*χ*
_{d}
(
*G*
) (d = 1, 2), for some types of generalized Petersen graphs
*P*(
*n, k*) where k = 1, 2, 3 and arbitrary
*n*.

Let G = (V, E) be simple graph. A vertex k-coloring of G is a mapping from V(G) to the set { 1 , 2 , ⋯ , k } such that any two adjacent vertices are mapped to different integers. The smallest integer k for which a k-coloring exists is called the chromatic number of G, denoted by c(G). The d-distance between two distinct vertices u and v, d(u, v) is the number of edges of the shortest path joining them. The d-distance k-coloring, also called distance (d, k)-coloring, is a k-coloring of the graph G, that is, any two vertices within distance d in G receive different colors. The d-distance chromatic number of G is exactly the chromatic number of G under the d-distance condition, denoted by c_{d}(G). For a simple graph G, the dth power of G, (G^{d} of G) is defined such that V ( G d ) = V ( G ) and two vertices u and v are adjacent in G^{d} if and only if the distance between u and v in G is at most d. Clearly, the following inequality is holds:

χ ( G ) = χ 1 ( G ) ≤ χ ( G 2 ) = χ 2 ( G ) ≤ χ ( G d ) = χ d ( G ) for d ≥ 2.

The theory of plane graph coloring has a long history, extending back to the middle of the 19^{th}century. In 1969, Florica Kramer and Horst Kramer [^{2}, as G ranges over all graphs with maximum degree d and girth g. Bonamy et al. [_{d}(G) of hexagonal lattice graph when d is odd and some value when d is even. Borodin and Ivanova [_{d}(G). Many papers have been devoted to it during the last decade, see for example [

In this paper, all graphs are finite, simple and undirected. For a graph G, we denote by V(G), E(G), d(u,v), ∆(G), diam(G), G^{d} and χ d ( G ) its vertex set, edge set, the distance between u and v which is the length of shortest path connecting them, the maximum vertex degree, the diameter of G, the power of G and d-distance coloring of G.

Theorem 1.1. [

1) | V | = d + 1 .

2) G is a path of length greater than d.

3) G is a cycle of length a multiple of (d + 1). □

Theorem 1.2. [

The path P n and the cycle C n :

1). χ d ( P n ) = min { n , d + 1 } .

2). χ d ( C n ) = { d + 1 : n ≡ 0 ( mod ( d + 1 ) ) , min { i + 1 ≥ d + 2 : n mod i ≤ n i } .

Theorem 1.3. [

χ d ( G ) ≤ Δ ( Δ − 1 ) d − 1 Δ − 2 + 1.

□

Lemma 1.1. [

1) If H a subgraph of G, then χ d ( H ) ≤ χ d ( G ) .

2) χ d ( G ) equals the order of G if and only if G is connected and diam ( G ) ≤ d . □

Definition 1.1. [

V ( P ( n , k ) ) = { a i , b i : 1 ≤ i ≤ n , 1 ≤ k ≤ [ n − 1 2 ] } ,

E ( P ( n , k ) ) = { a i a i + 1 , a i b i , b i b i + k : 1 ≤ i ≤ n }

□

We will call A ( n , k ) (respectively B ( n , k ) ) the outer (respectively inner) subgraph of P ( n , k ) . Note that we take the skip k ≤ ⌊ n − 1 2 ⌋ , because of the obvious isomorphism P ( n , k ) ≅ P ( n , n − k ) .

Our main results here are to establish the exact chromatic number χ d ( P ( n , k ) ) (d = 1, 2) for k = 1, 2, 3 and arbitrary n.

Theorem 2.1. χ 1 ( P ( n , 1 ) ) = { 2 : n even , 3 : n odd .

Proof. Let G = P ( n , 1 ) , observe from Definition 1.1, that Generalized Petersen Graphs composed of one outer cycle and several inner cycles dependent on k. So, when k = 1 there is one inner cycle, then G composed of two cycles of size n. There are two cases:

Case 1: n is even, immediately from Theorem 1.1 we have χ 1 ( C n ) = 2 (because d = 1) then

χ 1 ( G ) ≥ 2 (1)

We define a function f with colors in the set {1, 2} for a_{i} and b_{i} as follows:

f ( a i ) = { 1 : i odd , 2 : i even . , f ( b i ) = { 2 : i odd , 1 : i even .

Then

χ 1 ( G ) ≤ 2 (2)

By (1) and (2) we get χ 1 ( G ) = 2 .

Case 2: n is odd, from Theorem 1.2, we have χ 1 ( C n ) = 3 . Then

χ 1 ( G ) ≥ 3 (3)

We define a function f with colors in the set {1, 2, 3} for a_{i} and b_{i} as follows:

f ( a i ) = { 1 : i odd and i < n , 2 : i even , 3 : i = n . , f ( b i ) = { 2 : i odd and i < n , 1 : i even and i = n such that i ≠ n − 1 , 3 : i = n − 1.

So,

χ 1 ( G ) ≤ 3 . (4)

From (3) and (4), gets χ 1 ( G ) = 3 . As example see

Theorem 2.2: χ 2 ( P ( n , 1 ) ) = { 4 : n ≡ 0 ( mod 4 ) , 6 : n = 3 , 6 , 5 : otherwise .

Proof. Let G = P ( n , 1 ) . We have C 4 is an induced subgraph of G and from Theorem 1.2, gets χ 2 ( C 4 ) = 4 . So,

χ 2 ( G ) ≥ 4 (5)

We define a function f with colors in the set { 1 , 2 , 3 , 4 } for a_{i} and b_{i} as follows:

f ( a 1 ) = 1 , f ( b 1 ) = 3 , f ( a 2 ) = 2 , f ( b 2 ) = 4 .

By follow-up the coloring to the right, for a 3 there is only a single color as f ( a 3 ) = 3 .

So, for each vertex there is only a single color:

f ( b 3 ) = 1 , f ( a 4 ) = 4 , f ( b 4 ) = 2 , f ( a 5 ) = 1 ,

f ( b 5 ) = 3 , f ( a 6 ) = 2 , f ( b 6 ) = 4 .

Observe that we have a repeat of the same order of the colors for each 4-inner (4-outer) vertices. Consider G with n ≥ 4 . Assume that n = 4 q + r : 0 ≤ r < 4 for each j ∈ { 0 , 4 , ⋯ , 4 ( q − 1 ) } we define a subset S j of V(G) by S j = { a i , a i + 1 , a i + 2 , a i + 3 , b i , b i + 1 , b i + 2 , b i + 3 } then there is a function f with colors in the set { 1 , 2 , 3 , 4 } define as follows:

f ( a i ) = { 1 : i ≡ 1 ( mod 4 ) , 2 : i ≡ 2 ( mod 4 ) , 3 : i ≡ 3 ( mod 4 ) , 4 : i ≡ 0 ( mod 4 ) . , f ( b i ) = { 3 : i ≡ 1 ( mod 4 ) , 4 : i ≡ 2 ( mod 4 ) , 1 : i ≡ 3 ( mod 4 ) , 2 : i ≡ 0 ( mod 4 ) .

We have four cases according to the value of n modulo 4:

Case 1: r = 0. Then V ( G ) = ∪ j = 0 4 ( q − 1 ) S j . By function f is

χ 2 ( G ) ≤ 4 (6)

From (5) and (6), we get χ 2 ( G ) = 4 : n ≡ 0 ( mod 4 ) .

Case 2: r = 1. There are two leftover vertices in V ( G ) − ∪ S j = { a n , b n } . By function f we have f ( a n ) = 1 , f ( b n ) = 3 which is a contradiction with a 1 and b 1 . Moreover d ( a n , b n ) = 1 . So, each of a n and b n needs deferent color then χ 2 ( G ) > 4 . We define f 1 = f \ { a n , a n − 1 , a n − 2 , b n } ∪ f 2 , where f 2 is a function with colors in the set {3, 4, 5} define as follows:

f 2 ( v ) = { 4 : v = a n , 3 : v = a n − 1 , 5 : v = a n − 2 , b n .

Then we get χ 2 ( G ) = 5 = when n ≡ 1 ( mod 4 ) .

Case 3: For r = 2, we have two subcases:

Case 3.1: r = 2 and n > 6, a similar argument, there is a contradiction for a n , b n , a n − 1 , b n − 1 . Then, χ 2 ( G ) > 4 . We define

f 1 = f \ { a n , b n , a n − 1 , b n − 1 , a n − 2 , b n − 2 , a n − 3 , a n − 4 , b 1 , b 2 } ∪ f 2 .

f 2 is a function with colors in the set { 2 , 3 , 4 , 5 } define as follows:

f 2 ( v ) = { 2 : v = a n − 3 , b n − 1 , 3 : v = a n − 2 , b n , 4 : v = a n − 1 , b 1 , 5 : v = a n , a n − 4 , b n − 2 , b 2 .

Then we get χ 2 ( G ) = 5 when n ≡ 2 ( mod 4 ) , see

Case 3.2: r = 2 and n = 6. There are two cycles of order 6 and know that χ 2 ( C 6 ) = 3 . Without loss of generality, assuming that f ( a 1 ) = 1 . Then the vertices a_{2}, a_{3}, a_{5}, a_{6}, b_{1}, b_{2}, b_{6}, can’t take the color 1. Moreover, at most one of a_{4}, b_{3}, b_{4}, b_{5} can be coloring by 1. This implies that each color has only two vertices from P ( 6 , 1 ) . So, needs 6 colors for 12 vertices. Furthermore, χ 2 ( P ( 6 , 1 ) ) = 6 .

Case 4: For r = 3, there are two subcases:

Case 4.1: n ≥ 7. For a function coloring f there is a contradiction in a n and b n . Then χ 2 ( G ) > 4 . We define

f 1 = f \ { a n , b n , a n − 1 , b n − 2 } ∪ f 2

where f 2 is a function with colors in the set { 2 , 3 , 5 } define as follows:

f 2 ( v ) = { 2 : v = b n , 3 : v = a n − 1 , 5 : v = a n , b n − 2 .

Then, χ 2 ( G ) = 5 when n ≡ 3 ( mod 4 ) .

Case 4.2: n = 3. Then, diam ( P ( 6 , 1 ) ) = 2 . By Lemma 1.1, we get χ 2 ( G ) = 6 . □

Theorem 2.3: χ 1 ( P ( n , 2 ) ) = 3 .

Proof: Let G = P ( n , 2 ) . We have C 5 is an induced subgraph of G. Then by Theorem 1.2, is χ 1 ( C 5 ) = 3 . So,

χ 1 ( G ) ≥ 3 (7)

We define a function f as follows:

f ( a i ) = { 1 : i ≡ 1 ( mod 3 ) , 2 : i ≡ 2 ( mod 3 ) , 3 : i ≡ 0 ( mod 3 ) . , f ( b i ) = { 2 : i ≡ 1 ( mod 3 ) , 3 : i ≡ 2 ( mod 3 ) , 1 : i ≡ 0 ( mod 3 ) .

We have three cases according to the value of n modulo 3:

Case 1: r = 0. By definition f we have

χ 1 ( G ) ≤ 3 (8)

By (7) together with (8), gets χ 1 ( G ) = 3 when n ≡ 0 ( mod 3 ) .

Case 2: r = 1. Then there is a contradiction for a n . We define

f 1 = f \ { a 1 } ∪ f 2

where f 2 ( a 1 ) = 3 . This implies that χ 1 ( G ) = 3 when n ≡ 1 ( mod 3 ) .

Case 3: r = 2. There is a problem with colors the vertices { b n , b n − 1 } .

We define f 1 = f \ { a n , b n , a n − 1 , b n − 1 } ∪ f 2 , where f 2 is a function with colors in the set { 1 , 2 , 3 } define as follows:

f 2 ( v ) = { 1 : v = b n − 1 , 3 : v = a n , 2 : v = a n − 1 , b n .

By the last result together with (7), we get χ 2 ( G ) = 3 when n ≡ 2 ( mod 3 ) . □

Theorem 2.4: χ 2 ( P ( n , 2 ) ) = { 5 : n ≡ 0 ( mod 10 ) , 10 : n = 5 , 6 : otherwise .

Proof: Let G = P ( n , 2 ) . G including C 5 as an induced subgraph. We have diam ( C 5 ) = 2 . Then by Lemma 1.1, we get χ 2 ( C 5 ) = 5 . Furthermore

χ 2 ( G ) ≥ 5. (9)

We define a function f with colors in the set { 1 , 2 , 3 , 4 , 5 } for a i and b i as follows:

f ( a 1 ) = 2 , f ( a 2 ) = 2 , f ( a 3 ) = 3 , f ( b 1 ) = 4 , f ( b 3 ) = 5

Then for b 2 there are two cases:

Case a: f ( b 2 ) = 4 . (By coloring to the right). So, a 4 has only a single color as f ( a 4 ) = 1 and f ( b 4 ) = 5 . Follow-up coloring inner (outer) vertices each vertex will have only a single color as follows:

f ( b 5 ) = 2 , f ( a 5 ) = 4 , f ( b 6 ) = 2 , f ( a 6 ) = 3 , f ( b 7 ) = 1 , f ( a 7 ) = 5 ,

f ( b 8 ) = 1 , f ( a 8 ) = 4 , f ( b 9 ) = 3 , f ( a 9 ) = 2 , f ( b 10 ) = 3 , f ( a 10 ) = 5

By continue we will have a repeat of the same order of the colors for each 10-inner (10-outer) vertices.

Case b: f ( b 2 ) = 5 . By coloring to the left. So, we back to consider the Case 1.

We will consider G with n ≥ 10 . Assume that n = 10 q + r : 0 ≤ r < 10 . Now, for each j ∈ { 0 , 10 , ⋯ , 10 ( q − 1 ) } we define a subset S j of V(G) by

S j = { a j , a j + 1 , ⋯ , a j + 9 , b j , b j + 1 , ⋯ , b j + 9 }

Then there is a function f define as follows:

f ( a i ) = { 1 : i ≡ 1 , 4 ( mod 10 ) , 2 : i ≡ 2 , 9 ( mod 10 ) , 3 : i ≡ 3 , 6 ( mod 10 ) , 4 : i ≡ 5 , 8 ( mod 10 ) , 5 : i ≡ 7 , 0 ( mod 10 ) . , f ( b i ) = { 4 : i ≡ 12 ( mod 10 ) , 5 : i ≡ 3 , 4 ( mod 10 ) , 2 : i ≡ 5 , 6 ( mod 10 ) , 1 : i ≡ 7 , 8 ( mod 10 ) , 3 : i ≡ 0 , 9 ( mod 10 ) .

We have ten cases according to the value of n modulo 10:

Case 1: r = 0. Then V ( G ) = ∪ j = 0 10 ( q − 1 ) S j . Moreover, by define f we have

χ 2 ( G ) ≤ 5 (10)

From (9) and (10) we get χ 2 ( G ) = 5 when n ≡ 0 ( mod 10 ) .

Case 2: r = 1. There are two leftover vertices in V ( G ) − ∪ S j = { a n , b n } . By function f, f ( a n ) = 1 , f ( b n ) = 4 which is a contradiction with a 1 and b 1 , and d ( a n , b n ) = 1 . So each of a n and b n needs deferent color then χ 2 ( G ) > 5 . We define

f 1 = f \ { a n , b n , a n − 1 , a n − 2 , a n − 3 } ∪ f 2

where f 2 is a function with colors in the set { 2 , 4 , 5 , 6 } define as follows:

f 2 ( v ) = { 2 : v = a n − 1 , 4 : v = a n − 2 , 5 : v = a n , 6 : v = b n , a n − 3 .

Then we get χ 2 ( G ) = 6 when n ≡ 1 ( mod 10 ) .

Case 3: r = 2. There are four leftover vertices in V ( G ) − ∪ S j = { a n , b n , a n − 1 , b n − 1 } , which are a contradiction with { a 1 , a 2 , b 1 , b 2 } . This implies that χ 2 ( G ) > 5 .

Let

f 1 = f \ { b n , b n − 1 , a 1 , a 2 , a 3 , a 4 , a 6 } ∪ f 2

where f 2 is a function with colors in the set { 1 , 3 , 6 } define as follows:

f 2 ( v ) = { 1 : v = a 2 , 3 : v = a 1 , a 4 , 6 : v = a 3 , a 6 , b n , b n − 1 .

Then χ 2 ( G ) = 6 when n ≡ 2 ( mod 10 ) . See

Case 4: r = 3. By same argument there is a contradiction for b n , b n − 1 , b n − 2 . Which implies that χ 2 ( G ) > 5 . So, we define

f 1 = f \ { b n , b n − 1 , b n − 2 , a n − 3 , a n − 5 } ∪ f 2

where that f 2 is a function with colors in the set { 4 , 5 , 6 } define as follows:

f 2 ( v ) = { 4 : v = a n − 3 , 5 : v = b n − 2 , 6 : v = b n , b n − 1 , a n − 5 .

Then we get χ 2 ( G ) = 6 when n ≡ 3 ( mod 10 ) .

Case 5: r = 4. There is a contradiction for b n , b n − 1 , b n − 2 , b n − 3 , a n . We define

f 1 = f \ { a 1 , a n , a n − 3 , b n , b n − 1 , b n − 2 , b n − 3 } ∪ f 2

where f 2 is a function with colors in the set { 1 , 4 , 5 , 6 } define as follows:

f 2 ( v ) = { 1 : v = b n , b n − 1 , 4 : v = a n − 3 , 5 : v = a n , 6 : v = a 1 , b n − 2 , b n − 3 .

We get χ 2 ( G ) = 6 when n ≡ 4 ( mod 10 ) .

Case 6: When r = 5 we have two subcases:

Case 6.1: r = 5 and n > 5 . The contradiction is for b n , b n − 1 , b n − 3 , a n − 1 , a n . We will need at least three new deferent colors for them, then χ 2 ( G ) > 5 . We define

f 1 = f \ { a n , b n , a n − 1 , b n − 1 , a n − 2 , b n − 3 , a n − 4 , b n − 4 , a n − 5 , b n − 6 , b n − 7 , a 2 , a 3 } ∪ f 2

where f 2 is a function with colors in the set { 1 , 2 , 3 , 4 , 5 , 6 } define as follows:

f 2 ( v ) = { 1 : v = a n − 2 , a n − 5 , 2 : v = a n , 3 : v = a n − 1 , a 2 , b n − 4 , 4 : v = a n − 4 , 5 : v = b n − 3 , 6 : v = a 3 , b n , b n − 1 , b n − 6 , b n − 7 .

Then χ 2 ( G ) = 6 when n ≡ 5 ( mod 10 ) . See

Case 6. 2: r = 5 and n = 5. We have diam(G) = 2. So, Lemma 1.1, gets χ 2 ( G ) = 10 .

Case 7: r = 6. A contradiction for b n , a n − 1 . We define f 1 = f \ { b 1 , b n } ∪ f 2 and f 2 is a function with color 6. So, f 2 ( b 1 ) = f 2 ( b n ) = 6 , gets χ 2 ( G ) = 6 when n ≡ 6 ( mod 10 ) .

Notice when n = 6 we have the same argument but q = 0 , so the vertices will take the sequence of colors for (outer, inner)vertices as follows (1, 2, 3, 1, 2, 3, 4, 4, 5, 5, 6, 6).

Case 8: r = 7. A contradiction is only for b n . Let f 1 = f \ { b n } ∪ f 2 . Where f 2 ( b n ) = 6 . Moreover, χ 2 ( G ) = 6 when n ≡ 7 ( mod 10 ) . Also when n = 7 we get χ 2 ( G ) = 6 by the same condition with sequence of colors (outer, inner) vertices as following (1, 2, 3, 1, 4, 3, 5, 6, 4, 5, 5, 2, 2, 6).

Case 9: r = 8. A contradiction is for b n − 1 , b n , a n , then χ 2 ( G ) > 5 . We define f 1 = f \ { a n , b n , b n − 1 , a n − 2 , a n − 3 } ∪ f 2 with f 2 is a function with colors in the set { 3 , 4 , 6 } define as follows:

f 2 ( v ) = { 3 : v = b n , b n − 1 , 4 : v = a n − 2 , 6 : v = a n , a n − 3 .

And so, gets χ 2 ( G ) = 6 when n ≡ 8 ( mod 10 ) .

Also notice when n = 8 we have the same argument but q = 0 , so the vertices will take the sequence of colors for (outer, inner) vertices as follows (1, 2, 3, 1, 6, 4, 5, 6, 4, 4, 5, 5, 2, 2, 3, 3).

Case 10: r = 9. There is a contradiction for b n − 1 , a n − 1 , a n then χ 2 ( G ) > 5 . Let us define f 1 = f \ { a n , b n , a n − 1 , b n − 1 , a n − 2 , a n − 4 } ∪ f 2 , with f 2 is a function with colors in the set { 3 , 4 , 5 , 6 } define as follows:

f 2 ( v ) = { 3 : v = a n , 4 : v = a n − 2 , 5 : v = a n − 1 , 6 : v = b n , b n − 1 , a n − 4 .

Furthermore, χ 2 ( G ) = 6 when n ≡ 9 ( mod 10 ) .

As before when n = 9 we get χ 2 ( G ) = 6 , by the same condition with sequence of colors (outer, inner) vertices as follows (1, 2, 3, 1, 6, 3, 4, 5, 3, 4, 4, 5, 5, 2, 2, 1, 6, 6).

Finally, we conclude that:

χ 2 ( P ( n , 2 ) ) = { 5 : n ≡ 0 ( mod 10 ) , 10 : n = 5 , 6 : otherwise .

Theorem 2.5: χ 1 ( P ( n , 3 ) ) = { 2 : n ≡ 0 ( mod 2 ) , 3 : n ≡ 1 ( mod 2 ) .

Proof. Let G = P ( n , 3 ) . There are two cases:

Case 1: n ≡ 0 ( mod 2 ) . From Theorem 1.2, we have χ 1 ( C n ) = 2 . Then

χ 1 ( G ) ≥ 2 (11)

We define a function f with colors in the set {1, 2} for a i and b i ( 1 ≤ i ≤ n ),

f ( a i ) = { 1 : i odd , 2 : i even . , f ( b i ) = { 2 : i odd , 1 : i even .

Then

χ 1 ( G ) ≤ 2 (12)

From (11) and (12), gets χ 1 ( G ) = 2 .

Case 2: n ≡ 1 ( mod 2 ) . From Theorem 1.2, we have χ 1 ( C n ) = 3 . Moreover,

χ 1 ( G ) ≥ 3 (13)

Let f : V ( G ) → { 1 , 2 , 3 } for a i and b i , where

f ( a i ) = { 1 : i odd , i = n − 1 such that i ≠ n , n − 2 , 2 : i even , i = n , n − 2 such that i ≠ n − 1 , n − 3 , 3 : i = n − 3.

f ( b i ) = { 1 : i even , i < n − 2 , 2 : i odd , i < n − 2 , 3 : i = n , n − 1 , n − 2.

Then

χ 1 ( G ) ≤ 3 . (14)

From (13) and (14) is χ 1 ( G ) = 3 . □

Theorem 2.6: χ 2 ( P ( n , 3 ) ) = { 4 : n ≡ 0 ( mod 4 ) , 6 : n = 7 or 2 ( mod 4 ) , 5 : otherwise .

Proof. Let G = P ( n , 3 ) . K ( 1 , 3 ) is an induced subgraph of G and χ 2 ( K ( 1 , 3 ) ) = 4 . This implies that

χ 2 ( G ) ≥ 4 . (15)

Without loss of generality, we define a function f as follows: f ( a 1 ) = 1 , f ( a 2 ) = 2 , f ( a 3 ) = 3 , f ( b 2 ) = 4 . By follow-up the coloring to the right, for b 1 there are two cases f ( b 1 ) = 4 or f ( b 1 ) = 3 .

Case 1: f ( b 1 ) = 4 . Then, there are two cases for b 3 , f ( b 3 ) = 4 or f ( b 3 ) = 1 . So, if f ( b 3 ) = 4 then absolutely f ( a 4 ) = 1 and f ( b 1 ) = 3 . For coloring a 5 we need another color because it has the four colors as neighbors. If f ( b 3 ) = 1 , then f ( a 4 ) = 1 and f ( b 4 ) = 2 . Furthermore, for coloring a 5 we need another color. So to avoiding the fifth color we have to take the second case.

Case 2: f ( b 1 ) = 3 . There are two cases for b 3 , f ( b 3 ) = 4 or f ( b 3 ) = 1 , we have two subcases:

Case 2.1: f ( b 3 ) = 4 . Absolutely f ( a 4 ) = 1 and f ( b 4 ) = 4 or f ( b 4 ) = 2 . If we take f ( b 4 ) = 4 then f ( a 5 ) = 2 , f ( b 5 ) = 3 , but we need another color for a 6 . Also if we take f ( b 4 ) = 2 then we need new color for a 5 .

Case 2.2: f ( b 3 ) = 1 . For each vertex there is only a single color: f ( a 4 ) = 4 , f ( b 4 ) = 2 , f ( a 5 ) = 1 , f ( b 5 ) = 3 , f ( a 6 ) = 2 , f ( b 6 ) = 4 . Observe that, we have a repeat of the same order of the colors for each (4-outer) and (4-inner) vertices as respectively for colors { 1 , 2 , 3 , 4 } and { 3 , 4 , 1 , 2 } . Consider G with n ≥ 4 . Assume that n = 4 q + r : 0 ≤ r < 4 for each j ∈ { 0 , 4 , ⋯ , 4 ( q − 1 ) } , we define a subset S j of V(G) by S j = { a i , a i + 1 , a i + 2 , a i + 3 , b i , b i + 1 , b i + 2 , b i + 3 } then there is a function f define as follows:

f ( a i ) = { 1 : i ≡ 1 ( mod 4 ) , 2 : i ≡ 2 ( mod 4 ) , 3 : i ≡ 3 ( mod 4 ) , 4 : i ≡ 0 ( mod 4 ) . , f ( b i ) = { 3 : i ≡ 1 ( mod 4 ) , 4 : i ≡ 2 ( mod 4 ) , 1 : i ≡ 3 ( mod 4 ) , 2 : i ≡ 0 ( mod 4 ) .

We have four cases according to the value of n modulo 4:

Case 2.2.1: r = 0. Then V ( G ) = ∪ j = 0 4 ( q − 1 ) S j . By function f we have

χ 2 ( G ) ≤ 4 (16)

From (15) and (16) we get χ 2 ( G ) = 4 : n ≡ 0 ( mod 4 ) .

Case 2.2.2: r = 1. Then there are two leftover vertices in V ( G ) = ∪ j = 0 4 ( q − 1 ) S j = { a n , b n } , by function f we get f ( a n ) = 1 , f ( b n ) = 3 which is a contradiction with a 1 and a 3 . So each of a n and b n needs deferent color then χ 2 ( G ) > 4 . We define

f 1 = f \ { a n , a n − 1 , a n − 2 , a n − 3 , b 1 , b n , b n − 1 , b n − 2 , b n − 3 } ∪ f 2

where f 2 is a function with colors in the set {2, 3, 4, 5} define as follows:

f 2 ( v ) = { 2 : v = a n − 1 , b n − 3 , 3 : v = a n , b n − 2 , 4 : v = a n − 2 , b n , 5 : v = a n − 3 , b n − 1 , b 1 .

Then gets χ 2 ( G ) = 5 when n ≡ 1 ( mod 4 ) .

Case 2.2.3: r = 2. Here, we will consider χ 2 ( P ( 10 , 3 ) ) , {we delete the details of the general case because they are too long}.

We have χ 2 ( P ( 10 , 3 ) ) ≥ 5 . Suppose χ 2 ( P ( 10 , 3 ) ) = 5 . It is easy to prove that each color can be given at most to four vertices. This implies that each color has exactly four vertices. {If drawing P 1 ( 10 , 3 ) as following form: (outer cycle, inner cycle) respectively, b 1 b 4 b 7 b 10 b 3 b 6 b 9 b 2 b 5 b 8 , a 1 a 4 a 7 a 10 a 3 a 6 a 9 a 2 a 5 a 8 such that b i a i ∈ E ( P 1 ( 10 , 3 ) ) we gets the same graph ( P ( 10 , 3 ) , i.e., P 1 ( 10 , 3 ) ≅ P ( 10 , 3 ) }. Furthermore, no more three vertices from (outer cycle, inner cycle) respectively, can be take the same color.

Assume that there are five sets of colors, D_{1}, D_{2}, D_{3}, D_{4}, D_{5}, i.e., f(v) = i if and only v ∈ D i ( 1 ≤ i ≤ 5 ). We will study the cases for one of D_{i}. If D_{i} contain r vertices of outer cycle and q vertices of inner cycle, then we called D_{i} is (r-outer, q-inner). Without loss of generality, we consider D_{1}. Thus, we distinguish two cases:

Case a: D_{1} is (3-outer, 1-inner).

(This Case is similar by symmetry to D_{1} is (1-outer, 3-inner).

Let’s start with a 1 then we have (up to isomorphism) D 1 = { a 1 , a 4 , a 7 , b 9 } , a 2 ∈ D 2 , and a 3 ∈ D 3 . Thus, b 2 ∈ D 4 or b 2 ∈ D 5 and b 3 ∈ D 4 or b 3 ∈ D 5 . We have two cases:

Case a.1: b 2 ∈ D 4 and b 3 ∈ D 5 or b 2 ∈ D 5 and b 3 ∈ D 4 . Two cases are similar by symmetry. Let b 2 ∈ D 4 and b 3 ∈ D 5 . Then b 6 ∈ D 2 , a 5 ∈ D 5 , b 5 ∈ D 3 , b 8 ∈ D 2 , b 4 ∈ D 4 . Then b 10 ∉ D i , ( 1 ≤ i ≤ 5 ) , a contradiction with our hypothesis, χ 2 ( P ( 10 , 3 ) ) = 5 .

Case a.2: b 2 and b 3 are belonging to the same set, let b 2 , b 3 ∈ D 4 . There are two subcases a 5 ∈ D 2 or a 5 ∈ D 5 :

Case a.2.1: a 5 ∈ D 2 . Then b 6 ∈ D 5 , b 10 ∈ D 2 , a 6 ∈ D 3 , b 5 ∈ D 5 , b 7 ∈ D 5 , b 4 ∈ D 4 , b 1 ∈ D 3 . So, b 8 ∉ D i , ( 1 ≤ i ≤ 5 ) , a contradiction with χ 2 ( P ( 10 , 3 ) ) = 5 .

Case a.2.2: a 5 ∈ D 5 . Then b 5 ∈ D 3 , b 6 ∈ D 2 . This implies that a 6 ∉ D i , ( 1 ≤ i ≤ 5 ) , again gets a contradiction with χ 2 ( P ( 10 , 3 ) ) = 5 .

Case b: D_{1} is (2-outer, 2-inner).

Assume that a 1 ∈ D 1 . We have three cases to choose the second vertex from outer cycle.

Case b.1: a 4 ∈ D 1 . (We have the same result if we take a 8 ∈ D 1 ). Just one of b 6 , b 9 can belongs to D 1 . So, D 1 has three vertices and that means a contradiction with our proof that each set is from size 4.

Case b.2: a 5 ∈ D 1 . (We have the same result if we take a 7 ∈ D 1 ). Then D 1 = { a 1 , a 5 , b 7 , b 9 } , a 2 ∈ D 2 , a 3 ∈ D 3 , a 4 ∈ D 4 , b 3 ∈ D 5 . Also, b 4 ∈ D 2 or b 4 ∈ D 5 .

Case b.2.1: b 4 ∈ D 2 . Then b 10 ∈ D 4 , b 6 ∈ D 2 , a 6 ∈ D 3 , b 5 ∈ D 5 , a 7 ∈ D 5 , b 1 ∈ D 3 , b 8 ∈ D 4 . Thus, b 2 ∉ D i , ( 1 ≤ i ≤ 5 ) , a contradiction with χ 2 ( P ( 10 , 3 ) ) = 5 .

Case b.2.2: b 4 ∈ D 5 . Then b 1 ∈ D 3 , a 10 ∈ D 4 , b 10 ∈ D 2 , b 5 ∈ D 5 , b 2 ∈ D 4 . Thus, we get b 6 ∉ D i , ( 1 ≤ i ≤ 5 ) , a contradiction with χ 2 ( P ( 10 , 3 ) ) = 5 .

Case b.3: a 6 ∈ D 1 . Then no vertex in inner cycle can take the color 1. We get a contradiction with our proof that each set is from size 4.

Finally, we conclude that χ 2 ( P ( 10 , 3 ) ) > 5 . To prove that χ 2 ( P ( 10 , 3 ) ) ≤ 6 , we take a function f : V ( G ) → { 1 , 2 , 3 , 4 , 5 , 6 } as follows:

f ( v ) = { 1 : v = a 1 , a 5 , a 8 , b 3 , 2 : v = a 2 , a 6 , a 9 , b 4 , 3 : v = a 3 , a 7 , b 1 , 4 : v = a 4 , a 10 , b 2 , 5 : v = b 5 , b 6 , b 7 , 6 : v = b 8 , b 9 , b 10 .

Then we get χ 2 ( G ) = 6 when n ≡ 2 ( mod 4 ) . See

Case 2.2.4: r = 3. we have two subcases:

Case 2.2.4.1: r = 3 and n > 7. The contradiction in a n , b n − 1 , b n − 2 and b n . We define

f 1 = f \ { a n , b n , b n − 1 , b n − 2 } ∪ f 2

where f 2 is a function with colors in the set {4, 5}, define as follows:

f 2 ( v ) = { 4 : v = a n , 5 : v = b n , b n − 1 , b n − 2 .

Then we get χ 2 ( G ) = 5 when n ≡ 3 ( mod 4 ) for n > 7.

Case 2.2.4.2: r = 3 and n = 7. In this case we have C 5 induced subgraph from P ( 7 , 3 ) . Furthermore, χ 2 ( P ( 7 , 3 ) ) ≥ 5 . Let take the cycle a 1 a 2 b 2 b 5 b 1 and give it the fife color as follows: f ( a 1 ) = 1 f ( a 2 ) = 2 , f ( b 2 ) = 3 , f ( b 5 ) = 4 , f ( b 1 ) = 5 , so for a 3 there are two cases f ( a 3 ) = 4 or 5.

Case 2.2.4.2.a: f ( a 3 ) = 4 . Then for b 3 we have two choices 1 or 5. For the first choice f ( b 3 ) = 1 we get f ( a 4 ) = 3 , f ( b 4 ) = 2 , f ( a 5 ) = 1 . But for a 6 there are two colors 2 or 5. If f ( a 6 ) = 5 , then we will need a new color for b 6 . Also, if f ( a 6 ) = 2 then f ( b 6 ) = 5 . Obviously, we need a new color for b 7 . For second choice f ( b 3 ) = 5 then f ( a 4 ) = 1 or f ( a 4 ) = 3 . If f ( a 4 ) = 1 we have for b 4 two colors 2 or 3 if we take the color 2 then needs a new color for the vertices a 5 . Also, if we take the color 3 we will need a new color for a 6 because a 5 can only take the color 2. If f ( a 4 ) = 3 then f ( b 4 ) = 2 , f ( a 5 ) = 1 , f ( a 6 ) = 2 . Moreover, we will need a new color for b 6 .

Case 2.2.4.2.b: f ( a 3 ) = 5 then for b 3 we have two choices 1 or 4. For f ( b 3 ) = 1 we get f ( a 4 ) = 3 , f ( b 4 ) = 2 , f ( a 5 ) = 1 , f ( a 6 ) = 5 Then we need a new color for b 6 . For second choice f ( b 3 ) = 4 then f ( a 4 ) = 1 or f ( a 4 ) = 3 . If f ( a 4 ) = 1 , then we have for b 4 two colors 2 or 3. If we take the color 2 we will need a new color for the vertices a 5 . Also, if we take the color 3 we will need a new color for a 7 . If f ( a 4 ) = 3 , then f ( b 4 ) = 2 ,so we will need a new color for b 7 . We conclude that for all the cases, needs six colors. Furthermore, χ 2 ( P ( 7 , 3 ) ) > 5 . To prove that χ 2 ( P ( 7 , 3 ) ) ≤ 6 , we take a function f : V ( G ) → { 1 , 2 , 3 , 4 , 5 , 6 } as follows:

f ( a 1 ) = f ( a 5 ) = f ( b 3 ) = 1 , f ( a 2 ) = f ( a 6 ) = f ( b 4 ) = 2 , f ( b 1 ) = f ( a 3 ) = 3 ,

f ( b 2 ) = f ( a 4 ) = f ( a 7 ) = 4 , f ( b 5 ) = f ( b 7 ) = 5 , f ( b 6 ) = 6 .

Finally, we get χ 2 ( P ( 7 , 3 ) ) = 6 .

The authors would like to thank the referees for their careful reading of the paper and their helpful comments.

Shaheen, R., Kanaya, Z. and Jakhlab, S. (2017) d-Distance Coloring of Generalized Petersen Graphs P(n, k). Open Journal of Discrete Mathematics, 7, 185-199. https://doi.org/10.4236/ojdm.2017.74017