The square chromatic number of the torus

Discrete Mathematics 339 (2016) 447–456

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The square chromatic number of the torus A.G. Chegini a , Morteza Hasanvand a , E.S. Mahmoodian a,∗ , Farokhlagha Moazami b a

Department of Mathematical Sciences, Sharif University of Technology, P.O. Box 11155–9415, Tehran, Iran

b

Cyberspace Research Institute, Shahid Beheshti University, G.C., P.O. Box 1983963113, Tehran, Iran

article

info

Article history: Received 28 January 2015 Received in revised form 1 September 2015 Accepted 2 September 2015

abstract The square of a graph G denoted by G2 , is the graph with the same vertex set as G and edges linking pairs of vertices at distance at most 2 in G. The chromatic number of the square of the Cartesian product of two cycles was previously determined for some cases. In this paper, we determine the precise value of χ ((Cm Cn )2 ) for all the remaining cases. We show |V ((C C )2 )|

Keywords: 2-distance colouring Cartesian product Torus

that for all ordered pairs (m, n) except for (7, 11) we have χ ((Cm Cn )2 ) = ⌈ α((Cm Cn )2 ) ⌉, m n where α(G) denotes the independent number of G. This settles a conjecture of Sopena and Wu (2010). We also show that the smallest integer k such that χ ((Cm Cn )2 ) ≤ 6 for every m, n ≥ k is 10. This answers a question of Shao and Vesel (2013). © 2015 Published by Elsevier B.V.

1. Introduction A k-distance colouring of a graph is a colouring of its vertices such that any two vertices at distance at most k receive distinct colours. By the kth power of a graph G denoted by Gk , we mean the graph with the same set of vertices in which two vertices are adjacent when their distance in G is at most k. For a given graph G, it is of interest to find χk (G), the minimum number of colours necessary to have a k-distance colouring of G. Note that χk (G) = χ (Gk ), where χ stands for the ordinary chromatic number. The k-distance colouring of a graph was defined by Florica Kramer and Horst Kramer in 1969 [5,6] and has been studied extensively throughout the literature, for a survey see [7]. The Cartesian product of two graphs G and H, denoted by GH, is a graph with vertex set V (G) × V (H ) where two vertices (u, u′ ) and (v, v ′ ) are adjacent when either u = v and u′ is adjacent with v ′ in H, or u′ = v ′ and u is adjacent with v in G. For u ∈ V (H ) let Gu = G{u} and for v ∈ V (G) let Hv = {v}H. We call Gu the uth column and Hv the v th row of GH. For convenience let Tm,n stand for Cm Cn , and let α(G) stand for the size of an independent set in G with maximum cardinality. It is clear that for every graph G

χ (G) ≥

|V (G)| · α(G)

(1)

Sopena and Wu [10] proposed the following conjectures. Conjecture 1 ([10]). If m, n ≥ 3, then χ (Tm2 ,n ) = ⌈

2 )| |V (Tm ,n 2 ) α(Tm ,n

⌉.

Conjecture 2 ([10]). There exists some constant c such that if m, n ≥ c, then χ (Tm2 ,n ) ≤ 6. B.M. Kim, et al. in [4] have worked on the conjecture above, while Shao and Vesel [9] discovered a colouring showing that Conjecture 2 holds for c = 40. More formally, they proved the following theorem.

∗

Corresponding author. E-mail addresses: [email protected] (A.G. Chegini), [email protected] (M. Hasanvand), [email protected] (E.S. Mahmoodian), [email protected] (F. Moazami). http://dx.doi.org/10.1016/j.disc.2015.09.003 0012-365X/© 2015 Published by Elsevier B.V.

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A.G. Chegini et al. / Discrete Mathematics 339 (2016) 447–456

Theorem A ([9]). If m, n ≥ 40, then χ (Tm2 ,n ) ≤ 6. Also, Shao and Vesel [9] proposed the following question. Question 1 ([9]). What is the smallest c such that if m, n ≥ c, then χ (Tm2 ,n ) ≤ 6? Jamison and Matthews [2], Mahmoodian and Mousavi [8], also Sopena and Wu [10] independently proved the following theorem. Theorem B. For all m and n, χ (Tm2 ,n ) = 5 if and only if both m and n are multiple of 5. The exact values of χ (Tm2 ,n ) are determined in some papers for infinitely many other cases of (m, n), see [3,8–10]. Here we

find χ (Tm2 ,n ) for all remaining cases which were not known. Also we show that Conjecture 1 holds for all m and n except when (m, n) = (7, 11) and also we show that the optimal value of c in Question 1 is 10. The number c in Question 1 is equal to 10. 2. General results We state a theorem from [3] which is a generalization of Theorem B. They state this theorem with different mathematical language and notation. An important corollary of this theorem will be used to prove Conjecture 1. We give a proof for one side of this theorem which can be instrumental in the proof of its corollary. |V (G2 )|

Theorem C ([3]). Let G = Cn1 Cn2  · · · Cnk . Then χ (G2 ) = 2k + 1 if and only if α(G2 ) = 2k + 1. Proof. (H⇒) Let V (G2 ) = {(x1 , x2 , . . . , xk ) | 0 ≤ xi ≤ ni − 1}. Assume that A0 is an arbitrary independent set of G2 . Let e+i = (0, . . . , 0, 1, 0, . . . , 0) and e−i = (0, . . . , 0, −1, 0, . . . , 0), where for each i, 1 ≤ i ≤ k, the ith coordinate is equal to +1 or −1. For each x ∈ A0 let Ax = {x} ∪ {x + e±i | 1 ≤ i ≤ k}, where addition is taken modulo ni . Now let A = ∪x∈A0 Ax . A is a collection of pairwise disjoint sets each of size 2k + 1. Therefore, (2k + 1)α(G2 ) ≤ |V (G2 )|. So

χ (G2 ) ≥

|V (G2 )| ≥ 2k + 1. α(G2 )

(2) |V (G2 )|

By the hypothesis χ (G2 ) = 2k + 1, hence α(G2 ) = 2k + 1. For (⇐H) see [3].  |V (G2 )|

Corollary 1. Let G = Cn1 Cn2  · · · Cnk , if χ (G2 ) ≤ 2k + 2 then ⌈ α(G2 ) ⌉ = χ (G2 ). |V (G2 )|

Proof. If χ (G2 ) = 2k + 1 then the statement follows from Theorem C. If χ (G2 ) = 2k + 2 then by (1) 2k + 2 = χ (G2 ) ≥ α(G2 ) |V (G2 )|

and by (2) and Theorem C, we have α(G2 ) > 2k + 1.



Let x and y be two integers, and S (x, y) = {α x + β y | α, β are nonnegative integers}. Sylvester has shown the following lemma. Lemma A ([1,11]). Let x and y be relatively prime integers greater than 1. Then n ∈ S (x, y) for all n ≥ (x − 1)(y − 1). By applying Sylvester’s Lemma, one can observe that S (5, 6) = N \ {1, 2, 3, 4, 7, 8, 9, 13, 14, 19}. Theorem 1. If m, n ∈ S (5, 6), then χ (Tm2 ,n ) ≤ 6. Proof. The following patterns are proper 6-colourings of T52,5 , T52,6 , T62,5 , and T62,6 , respectively. 1 2 A= 3 4 5

3 4 5 2 6

5 1 6 3 4 T52,5

2 3 4 5 1

4 5 1 6 3

1 2 B= 3 4 5

3 4 5 2 6

5 6 1 3 4 T52,6

2 3 4 6 1

6 1 2 5 3

4 5 6 1 2

1 2 3 C = 4 5 6

3 4 5 6 1 2 T62,5

5 6 1 2 3 4

2 3 4 5 6 1

4 5 6 1 2 3

1 2 3 D= 4 5 6

3 4 5 6 1 2

5 6 1 2 3 4 T62,6

2 3 4 5 6 1

6 1 2 3 4 5

4 5 6 . 1 2 3



A.G. Chegini et al. / Discrete Mathematics 339 (2016) 447–456

A C

Also, by considering

B D

449

2 we obtain a proper colouring for the graph T11 ,11 with 6 colours. By an appropriate

combination of these patterns we can colour the vertices of Tm2 ,n where m = 5α + 6β , n = 5α ′ + 6β ′ . So, for every m, n ∈ S (5, 6), we have a 6-colouring of the graph Tm2 ,n .  Remark 1. It should also be mentioned that Shao and Vesel in [9], by a proof similar to the proof of Theorem 1 have shown this result for all m, n ≥ 40. By Theorem B, Corollary 1, and Theorem 1 we have:

• χ(Tm2 ,n ) = 5, if m and n both are multiple of 5. • χ(Tm2 ,n ) = 6, when m, n ∈ S (5, 6), and at least one of them is not a multiple of 5. Sopena and Wu [10] obtained the exact value of χ (Tm2 ,n ) for all n ≥ 3, where m = 3, 4, 5, 6. In the next section we state results for the exact values of χ (Tm2 ,n ), for all the remaining cases. This will settle Conjecture 1 completely. 3. Filling the gaps By Theorem B, if at least one of m or n is not a multiple of 5, then χ (Tm2 ,n ) ≥ 6. So in these cases, if we want to show

that χ(Tm2 ,n ) = 6, all we need is to present a 6-colouring. By Theorem 1 we only need to find the value of χ (Tm2 ,n ) for m ∈ {7, 8, 9, 13, 14, 19}. 3.1. T72,n Theorem 2. If n ̸∈ {7, 8, 9, 11, 13, 14, 19} and n ≥ 6 then χ (T72,n ) = 6. Proof. Consider the following 6-colouring patterns for T72,n , where n = 6, 10, 15, and 17, respectively. 1 3 5 1 3 6 4

2 4 6 2 4 1 5

3 5 1 3 5 2 6

4 6 2 4 6 3 1

5 1 3 5 1 4 2

6 2 4 6 2 5 3

1 3 5 1 3 6 4

2 4 6 2 4 1 5

3 5 1 3 5 2 6

1 2 4 6 1 3 4

T72,6 1 3 5 1 3 6 4

2 4 6 2 4 1 5

3 5 1 3 5 2 6

1 2 4 6 1 3 4

6 3 5 2 4 5 2

5 4 1 3 6 1 3

1 2 6 4 5 2 4

6 3 5 2 1 3 5

2 4 1 3 6 4 1

5 6 2 4 5 2 3

4 1 3 2 4 1 5

6 2 4 6 5 2 3

T72,10 5 6 3 2 4 6 2

3 4 5 1 3 5 1

6 1 2 4 6 2 4

2 5 6 3 5 1 3

1 3 4 1 2 4 5

4 6 2 5 6 3 2

3 5 1 3 4 1 6

1 2 4 6 5 2 4

6 3 5 2 1 3 5

2 4 1 3 6 4 1

5 6 2 4 5 2 3

3 2 1 5 6 2 1

5 4 6 2 3 4 6

2 1 5 4 1 5 3

4 6 2 3 6 2 1

5 3 1 5 4 3 6

1 2 4 6 2 5 4

3 6 5 1 3 6 2

T72,15 1 3 5 1 3 6 4

2 4 6 2 4 1 5

3 5 1 3 5 2 6

1 2 4 6 1 3 4

5 6 3 5 4 6 2

3 4 2 1 3 5 1

2 1 5 6 2 4 6

4 6 3 4 1 3 5

T72,17 For any integer n ≥ 6 where n ̸∈ {7, 8, 9, 11, 13, 14, 19} there exist positive integers r1 , r2 , r3 , r4 such that n = 6r1 + 10r2 + 15r3 + 17r4 . To see this, for n ≥ 80 apply Sylvester’s Lemma A and for n < 80 it can be easily checked. First two columns in all of the above four patterns are identical. So, by taking ri copies of each and aligning them (and horizontally joining one after the other) we can construct a 6-colouring for T72,n , where n = 6r1 + 10r2 + 15r3 + 17r4 . 

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A.G. Chegini et al. / Discrete Mathematics 339 (2016) 447–456 Table 1 Independent sets for T72,n of size 7, 8, 9, 10 and 12, respectively.

• ◦ ◦ • ◦ ◦ ◦

◦ ◦ ◦ ◦ ◦ ◦ ◦

◦ • ◦ ◦ • ◦ ◦

◦ ◦ ◦ ◦ ◦ ◦ •

◦ ◦ • ◦ ◦ ◦ ◦

◦ ◦ ◦ ◦ ◦ • ◦

• ◦ ◦ • ◦ ◦ ◦

◦ ◦ ◦ ◦ ◦ ◦ ◦

◦ • ◦ ◦ • ◦ ◦

◦ ◦ ◦ ◦ ◦ ◦ •

◦ ◦ • ◦ ◦ ◦ ◦

◦ ◦ ◦ ◦ • ◦ ◦

• ◦ ◦ • ◦ ◦ ◦

• ◦ ◦ ◦ ◦ ◦ ◦

◦ ◦ ◦ ◦ ◦ ◦ ◦

◦ ◦ • ◦ ◦ • ◦

◦ • ◦ ◦ ◦ • ◦

◦ ◦ ◦ • ◦ ◦ ◦

◦ ◦ ◦ ◦ ◦ ◦ •

• ◦ ◦ • ◦ ◦ ◦

◦ ◦ ◦ ◦ ◦ ◦ ◦

◦ • ◦ ◦ • ◦ ◦

◦ ◦ ◦ ◦ ◦ ◦ ◦

• ◦ ◦ • ◦ ◦ ◦

◦ ◦ ◦ ◦ ◦ ◦ ◦

◦ • ◦ ◦ ◦ • ◦

◦ ◦ ◦ • ◦ ◦ ◦

• ◦ ◦ ◦ ◦ ◦ ◦

◦ ◦ ◦ ◦ ◦ ◦ ◦

◦ ◦ • ◦ ◦ • ◦

◦ • ◦ ◦ ◦ • ◦

◦ ◦ ◦ ◦ ◦ ◦ ◦

◦ ◦ ◦ • ◦ ◦ ◦

• ◦ ◦ ◦ • ◦ ◦

◦ ◦ ◦ ◦ ◦ ◦ •

◦ ◦ • ◦ ◦ ◦ ◦

◦ • ◦ ◦ • ◦ ◦

◦ ◦ ◦ ◦ ◦ ◦ ◦

◦ ◦ ◦ ◦ ◦ • ◦

◦ ◦ ◦ ◦ ◦ • ◦

By computing α(T72,n ), we will show that χ (T72,n ) = 7, for n ∈ {7, 8, 9, 13, 14, 19} (see Table 1). Lemma 1. If n = 5k + r and n ≥ 6 then α(T72,n ) ≥ 6k + r. Proof. The following patterns illustrate some of those desired sets. Now by induction assume that we have an independent set S for the graph T72,n of size 6(k − 1) + r, where n = 5(k − 1) + r. We note that in the above construction we have only two patterns for the first three columns. Our objective is to add 5 more columns containing 6 independent vertices, between the second and the third columns, such that the first three columns keep their earlier patterns. This can be done as follows

• ◦ ◦ • ◦ ◦ ◦

◦ ◦ ◦ ◦ ◦ ◦ ◦

◦ • ◦ ◦ ◦ • ◦

◦ ◦ ◦ • ◦ ◦ ◦

• ◦ ◦ ◦ ◦ ◦ ◦

◦ ◦ • ◦ ◦ • ◦

◦ ◦ ◦ ◦ ◦ ◦ ◦

◦ • ◦ ◦ • ◦ ◦

And by induction the statement follows.

• ◦ ◦ • ◦ ◦ ◦

◦ • ◦ ◦ • ◦ ◦

◦ ◦ ◦ ◦ ◦ ◦ ◦

◦ ◦ ◦ ◦ ◦ ◦ •

◦ ◦ • ◦ ◦ ◦ ◦

• ◦ ◦ ◦ • ◦ ◦

◦ ◦ ◦ ◦ ◦ ◦ ◦

◦ • ◦ ◦ . ◦ • ◦



Let I and ai be the number of vertices in the ith column. We note that the sequence (a1 , . . . , an ) satisfies the following definition. be an independent set in T72,n

Definition 1. A sequence s = (a1 , . . . , an ), where 0 ≤ ai ≤ 2, and i ∈ Zn , is called a toric sequence if 1. (ai , ai+1 ) ̸= (2, 2) and 2. (ai−1 , ai , ai+1 ) ̸= (2, 1, 2) or (1, 2, 1). For every toric sequence s = (a1 , . . . , an ) we define w(s) :=

n

i=1

ai as the weight of the sequence s. And we denote

Mn = max{w(s)|s is a toric sequence of length n}. Our objective is to compute the exact value of Mn , then by using this we can compute α(T72,n ). Lemma 2. For every integer n = 5k + r, we have Mn ≤ 6k + r. Proof. Assume that (a1 , a2 , . . . , an ) is a toric sequence with the maximum weight. If it contains a subsequence (ai−2 , ai−1 , ai , ai+1 , ai+2 ) = (2, 0, 2, 0, 2) for some i ∈ Zn , then one can change it into another toric sequence with (ai−2 , ai−1 , ai , ai+1 , ai+2 ) = (1, 1, 2, 0, 2) without having reduced its weight. Hence, for all i ∈ Zn , by a simple observation one can assume that among (ai−2 , ai−1 , ai , ai+1 , ai+2 ) we have twice the value 2. Assume that xj is the number of the value j in the sequence (a1 , a2 , . . . , an ), for j = 0, 1, 2. By a double-counting on the   . Since the sequence (a1 , a2 , . . . , an ) number of the value 2 in every (ai−2 , ai−1 , ai , ai+1 , ai+2 ) one can deduce that x2 ≤ 2n 5 is toric, if ai = 2, then 0 ∈ {ai−1 , ai+1 }; thus, x0 ≥

 x2  2

. Hence:

x  2 w((a1 , a2 , . . . , an )) = 2x2 + x1 = n + x2 − x0 ≤ n + x2 − 2 x  n 2 = n+ ≤n+ = 6k + r .  2

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451

Theorem 3. Let n = 5k + r, if n ≥ 3 and n ̸= 5 then α(T72,n ) = 6k + r. Proof. Assume that I is a maximum independent set of T72,n . Let bi denote the number of vertices of the ith column in the

set I. (b1 , . . . , bn ) is a toric sequence and α(T72,n ) = w((b1 , . . . , bn )) ≤ Mn . So, for every n ≥ 6, α(T72,n ) = 6k + r. Also, it is

an obvious and known fact that α(T72,3 ) = 3 and α(T42,7 ) = 4. In [10] it is shown that α( for χ(

T32,3

) = 9 and χ (



) = 5. Also in [10], Sopena and Wu have shown that if 3 ≤ m ≤ n then χ (Tm2 ,n ) ≤ 7, except ) = χ (T42,4 ) = 8. By using this fact and Theorem 3 the following corollary is proved.

T72,5

T32,5

Corollary 2. For every n ∈ {7, 8, 9, 13, 14, 19} we have

 χ(

T72,n

)=

|V (T72,n )|



α(T72,n )

= 7.

Theorem 3 shows that α(T72,11 ) = 13 so ⌈ is not true at least in one case.

|V (T72,11 )| α(T72,11 )

⌉ = 6. But we have the following remark which shows that Conjecture 1

Remark 2. We have χ (T72,11 ) = 7. Proof. This conclusion can be obtained by a Maple program using the following commands with(GraphTheory): with(SpecialGraphs): with(LinearAlgebra): A := AdjacencyMatrix(CartesianProduct(CycleGraph(7), CycleGraph(11))): G := UnderlyingGraph(Graph([‘$ ‘(1 .. 77)], A2 + A − 4∗IdentityMatrix(77))); kernelopts(printbytes = false): ChromaticNumber(G); Note that if A is the adjacency matrix of T7,11 then A2 + A − 4I is the adjacency matrix of T72,11 .



3.2. T82,n Theorem 4. If n ≥ 9 then χ (T82,n ) = 6. Proof. Consider the following pattern 1 2 3 1 2 4 3 5

3 6 5 4 6 5 1 2

5 2 4 3 2 1 3 5 1 4 2 6 4 5 6 1 8×5

6 5 4 6 3 1 2 4

1 2 3 1 2 4 6 3

4 6 5 4 6 3 1 5

3 5 1 4 2 6 3 5 1 4 5 2 4 3 2 6 8×6

2 3 1 2 3 6 5 1

6 5 4 6 . 5 1 2 4

The first five columns of it give a 6-colouring for T82,5 and the last six columns a 6-colouring for T82,6 . By combining the above

subpatterns, we find a proper colouring for the graph T82,5α+6β with 6 colours. The remaining cases are n = 9, 13, 14, 19. For these cases consider the following patterns 1 2 3 1 2 3 4 5

3 4 5 6 4 1 2 6

5 1 2 3 5 6 3 4

6 3 4 1 2 4 1 2

1 2 5 6 3 5 6 3 T82,9

4 6 1 2 4 1 2 5

2 3 4 5 6 3 4 1

6 1 2 3 1 2 5 3

4 5 6 4 , 5 6 1 2

1 2 3 1 2 3 4 5

3 4 5 6 4 1 2 6

2 6 1 2 3 6 5 1

5 3 4 5 1 2 3 4

6 1 2 6 4 5 1 2

3 4 5 1 3 6 4 5

1 2 3 4 5 2 3 6 T82,13

4 5 6 2 1 4 5 2

6 1 4 5 3 6 1 3

5 2 3 1 4 5 2 4

3 4 5 6 2 3 6 1

6 1 2 3 1 4 5 2

4 5 6 4 , 5 6 1 3

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A.G. Chegini et al. / Discrete Mathematics 339 (2016) 447–456

1 2 3 1 2 3 4 5

3 4 5 6 4 1 2 6

2 6 1 2 3 6 5 1

5 3 4 5 1 2 3 4

6 1 2 3 4 5 1 2

3 4 6 1 2 3 4 5

1 2 3 5 6 1 2 6

4 5 1 2 3 4 5 3

2 3 6 4 5 2 6 1

6 4 5 1 6 3 4 5

3 1 2 3 4 5 1 2

5 6 4 5 1 2 3 4

2 3 1 2 3 4 5 6

4 5 6 4 , 5 6 1 3

2 3 6 5 1 2 6 1

5 1 4 3 6 5 3 4

3 2 5 1 2 4 1 6

1 4 3 6 5 3 2 5

6 5 2 4 1 6 4 3

4 3 6 5 3 2 5 1

T82,14 1 2 3 1 2 3 4 5

3 4 5 6 4 1 2 6

2 6 1 2 3 6 5 1

5 3 4 5 1 2 3 4

6 1 2 3 4 5 1 2

3 4 5 1 2 3 4 5

1 2 3 4 5 1 2 6

4 5 1 2 3 4 5 3

2 1 4 2 6 1 3 6

5 6 3 1 4 5 2 4

3 4 5 6 2 3 6 1

6 1 2 3 1 4 5 2

4 5 6 4 . 5 6 1 3



T82,19

In [8] it is shown that χ (T82,8 ) = ⌈

|V (T82,8 )| α(T82,8 )

⌉ = 7.

3.3. T92,n

Theorem 5. If n ≥ 10 then χ (T92,n ) = 6. Proof. Consider the following pattern 1 2 3 1 4 2 3 5 4

5 2 4 4 3 6 6 1 2 2 4 5 5 6 1 1 3 4 4 5 2 2 1 3 3 6 5 9×5

3 5 4 6 3 5 1 6 2

6 1 3 2 4 6 3 5 4

5 2 6 1 5 2 4 1 3

1 2 4 4 3 6 5 1 2 3 4 5 6 2 1 1 3 4 5 6 2 2 4 3 6 5 1 9×6

3 5 4 6 3 5 1 6 2

.

For every integer n ≥ 10 where n ̸= 13, 14, 19 then n ∈ S (5, 6). So, by combining the aforementioned subpatterns, we can obtain a proper colouring for T92,5α+6β with 6 colours. For the cases n = 13, 14, 19 consider the following patterns 1 2 3 1 2 3 4 5 6

3 5 4 6 5 1 6 2 4

2 6 1 2 3 4 5 3 1

4 3 5 4 6 2 1 6 5

6 1 2 3 1 5 3 4 2

3 4 6 5 2 4 6 5 1

5 2 1 4 6 3 1 2 6 T92,13

1 6 5 3 1 2 5 4 3

4 3 2 6 5 4 6 1 5

2 5 1 4 3 1 2 3 6

3 4 6 5 2 6 5 4 1

5 1 2 3 1 4 3 6 2

4 6 5 4 6 , 5 2 1 3

A.G. Chegini et al. / Discrete Mathematics 339 (2016) 447–456

1 2 3 1 2 3 1 2 3

4 5 6 4 5 6 4 5 6

2 3 1 2 3 1 2 3 1

5 6 4 5 6 4 5 6 4

1 2 3 1 2 3 1 2 3

4 5 6 4 5 6 4 5 6

2 3 1 2 3 1 2 3 1

5 6 4 5 6 4 5 6 4

1 2 3 1 2 3 1 2 3

4 5 6 4 5 6 4 5 6

2 3 1 2 3 1 2 3 1

5 6 4 5 6 4 5 6 4

3 1 2 3 1 2 3 1 2

6 4 5 6 4 , 5 6 4 5

4 5 3 4 6 3 1 2 6

1 2 6 5 1 2 4 3 5

6 3 4 2 3 5 6 1 4

2 5 1 6 4 1 2 5 3

1 4 2 3 5 6 3 4 6

3 6 5 4 1 2 5 1 2

453

T92,14 1 2 3 1 2 3 4 5 6

3 5 4 6 5 1 6 2 4

2 6 1 2 3 4 5 3 1

4 3 5 4 6 2 1 6 5

1 2 6 1 5 3 4 2 3

5 4 3 2 4 6 5 1 6

2 1 5 6 3 1 2 3 4

3 6 2 1 5 4 6 5 1

4 1 3 2 6 3 4 6 5

2 5 6 1 4 5 2 3 1

6 3 4 5 2 6 1 5 4

5 1 2 3 1 4 3 6 2

4 6 5 4 6 . 5 2 1 3



T92,19 In [8] it is shown that χ (T92,9 ) = ⌈

|V (T92,9 )| α(T92,9 )

⌉ = 7.

2 3.4. T13 ,n 2 Theorem 6. If n ≥ 13 then χ (T13 ,n ) = 6.

Proof. Consider the following pattern 1 2 3 1 2 3 4 5 1 2 3 4 5

4 5 6 4 5 6 2 3 4 5 6 2 3

2 5 3 3 6 1 1 4 2 2 5 3 3 6 1 1 4 2 5 3 6 6 1 4 2 5 3 3 6 1 1 4 2 5 3 6 6 1 4 13 × 6

6 4 5 6 4 5 1 2 6 4 5 1 2

1 2 3 1 2 6 3 5 1 2 3 4 5

4 2 5 5 3 6 6 1 2 4 5 3 3 6 1 1 4 2 2 5 6 4 1 3 3 2 5 5 6 1 1 4 2 2 5 3 3 1 4 13 × 5

3 4 5 6 4 5 1 . 2 6 4 5 1 2

2 For n ∈ S (5, 6), combination of the above subpatterns gives a proper colouring for T13 ,5α+6β with 6 colours. For the case n = 13, 14, 19 consider the following patterns

6 1 5 3 4 6 2 5 3 4 5 1 2

5 2 4 6 2 1 3 4 6 1 2 4 3

4 6 3 1 5 4 6 1 2 5 3 6 1

3 1 5 2 6 3 2 5 4 6 1 2 5

6 2 4 3 1 5 4 3 1 2 5 3 4

1 3 6 5 4 2 1 6 5 3 4 6 2

4 5 2 1 6 3 5 2 4 6 1 5 3 2 T13 ,13

2 6 4 3 5 1 4 3 1 2 3 4 1

3 1 5 2 4 6 2 5 6 4 5 2 6

5 2 3 6 1 5 3 4 2 1 6 3 4

6 4 1 5 3 4 6 1 5 3 2 5 1

2 5 6 4 2 1 5 2 4 6 1 4 3

4 3 2 1 5 3 4 , 6 1 2 3 6 5

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1 2 3 1 2 3 1 2 3 1 2 3 4

3 5 4 6 5 4 6 5 4 6 5 1 6

2 6 1 2 3 1 2 3 1 2 3 4 5

1 4 3 5 4 6 5 4 6 5 1 6 3

6 5 2 6 1 2 3 1 2 3 4 5 2

4 3 1 4 3 5 4 6 5 1 6 3 1

5 2 6 5 2 6 1 2 3 4 5 2 6

1 4 3 1 4 3 5 4 6 2 1 4 3

2 6 5 2 6 1 2 3 1 5 3 6 5

3 1 4 3 5 4 6 5 4 6 2 1 4

5 2 6 1 2 3 1 2 3 1 5 3 6

4 3 5 4 6 5 4 6 5 4 6 2 1

6 1 2 3 1 2 3 1 2 3 1 5 3

5 4 6 5 4 6 5 , 4 6 5 4 6 2

5 6 3 1 4 5 6 1 5 4 3 1 2

3 1 2 5 6 3 2 4 3 1 2 5 6

4 5 6 3 2 4 5 6 2 5 6 3 1

6 2 4 1 5 6 3 1 4 3 1 4 5

1 3 5 6 3 1 2 5 6 2 5 6 2

5 6 1 2 4 5 6 3 1 4 3 1 4

2 T13 ,14

1 2 4 5 6 1 3 4 5 6 2 4 3

4 5 1 2 3 4 5 6 2 4 1 5 6

2 3 6 4 5 6 2 3 1 5 6 3 1

6 4 5 1 2 3 1 5 4 3 2 4 5

3 1 2 3 6 5 4 6 2 1 5 6 2

4 5 6 4 1 2 3 1 5 6 4 3 1

6 3 1 5 3 6 5 2 4 3 1 2 5

1 2 4 6 2 1 4 3 6 2 5 6 3

2 4 3 5 6 3 1 4 5 6 2 5 6

1 5 6 1 2 4 5 6 2 3 1 4 3

6 2 4 3 5 6 2 3 1 5 6 2 5

3 1 5 6 4 3 1 5 6 2 3 1 4

5 6 3 1 2 5 6 . 2 3 1 5 6 2



2 T13 ,19

2 3.5. T14 ,n 2 Theorem 7. If n ≥ 14 then χ (T14 ,n ) = 6.

Proof. Consider the following pattern 1 4 3 6 2 5 3 4 1 5 2 6 3 5

2 5 1 4 3 6 1 5 2 6 3 4 1 6

4 6 2 5 1 4 2 6 3 4 1 2 5 3

5 6 1 3 2 4 4 5 6 6 1 3 2 4 5 3 6 2 5 1 4 4 2 3 1 6 5 2 3 1 5 4 2 3 1 6 6 2 3 1 4 5 14 × 8

2 5 1 4 6 3 5 1 2 6 3 4 1 6

3 6 2 5 1 4 2 6 3 4 1 5 2 4

1 2 3 4 5 6 3 1 2 6 4 5 2 3 1 5 6 4 3 1 2 . 4 5 6 1 2 3 5 6 4 2 3 1 6 4 5 3 1 2 5 6 4 14 × 3

By Sylvester’s Theorem, for every integer n ≥ 14, n ∈ S (3, 8). So, by combining the aforementioned subpatterns, we can 2 obtain a proper colouring for the graph T14  ,3r1 +8r2 with 6 colours. 2 3.6. T19 ,n 2 Theorem 8. If n ≥ 19 then χ (T19 ,n ) = 6.

Proof. Consider the left pattern that is shown below. Using an appropriate combination of its subpatterns, we can obtain a 2 proper colouring for the graph T19 ,5α+6β with 6 colours. For every integer n ≥ 20 we have n ∈ S (5, 6). For the case n = 19

A.G. Chegini et al. / Discrete Mathematics 339 (2016) 447–456

455

consider the right pattern below. 1 3 5 2 4 1 3 5 2 4 1 3 5 2 6 1 3 2 4 1 2 4 1 3 5 4 2 3 1 5 4 2 3 1 5 4 6 3

2 3 4 4 1 6 6 2 3 3 4 5 5 6 1 2 3 4 4 1 6 6 2 3 3 4 5 5 6 1 2 3 4 4 1 6 6 2 3 3 4 5 5 1 2 4 3 6 6 2 1 1 4 5 5 6 2 19 × 5 6 3 5 6 2 1 6 5 4 6 2 1 6 4 2 6 1 2 4

5 2 1 6 3 5 2 1 6 3 5 2 4 1 3 5 4 6 3

2 4 1 3 5 4 2 3 1 5 4 3 5 1 3 4 5 3 1

1 3 5 2 4 1 3 5 2 4 1 3 5 2 6 1 3 2 4 3 6 2 4 6 3 1 4 6 2 1 6 4 2 6 1 2 6 5

2 4 6 3 5 2 4 6 3 5 2 4 6 3 5 4 6 1 5

3 4 5 5 6 1 1 2 3 4 5 6 6 1 2 3 4 5 5 6 1 1 2 3 4 5 6 6 1 2 3 4 5 5 6 1 1 2 3 4 5 6 2 1 4 3 6 5 5 2 1 4 3 6 6 1 2 19 × 6 1 5 3 1 2 5 6 2 3 4 5 2 1 3 4 5 3 4 2

6 2 4 6 3 1 4 5 1 6 3 4 6 5 1 2 6 1 3

6 2 4 1 3 6 2 4 1 3 , 6 2 4 1 3 2 4 5 3 4 1 5 2 4 6 2 3 4 2 1 5 2 4 6 3 4 2 5

2 6 3 1 5 3 1 6 5 3 4 6 1 3 2 1 5 3 1

3 5 2 4 6 2 5 4 1 6 2 3 4 6 5 4 2 6 4

6 4 1 5 3 1 6 2 3 4 1 5 2 1 3 6 1 5 2

5 2 3 6 2 4 3 1 5 2 3 4 6 5 4 2 3 4 1

3 1 5 4 1 5 2 4 6 1 5 2 1 3 6 1 5 2 6

2 4 6 2 3 6 1 3 2 4 6 3 5 4 2 3 6 1 5

6 5 3 1 5 2 4 6 1 5 2 4 6 1 5 4 2 3 4

3 1 2 4 6 3 5 2 3 6 1 5 2 3 6 1 5 6 2

4 6 5 3 2 4 6 1 4 2 3 6 1 5 2 3 4 1 5

1 2 4 6 5 1 2 3 6 5 4 2 3 6 4 5 2 3 6

3 5 1 2 4 3 5 4 2 3 1 5 4 2 3 1 6 4 2

4 6 3 5 6 2 1 6 5 4 . 6 3 1 5 6 2 3 1 5



2 T19 ,19

4. Conclusion We summarize all the results in a table in which the exact value of χ (Tm2 ,n ), for each m and n, is given. (m, n)

χ (Tm2 ,n )

Reference

(3, 3) (3, 5) (3, 2k + 1), k ≥ 3 (3, 2k)

9 8 7 6

[8,10] [8,10] [10] [10]

(4, 4) (4, 3k ± 1), k ≥ 2 (4, 3k)

8 7 6

[8,10] [10] [10]

(5, 7) (5k, n), 5 - n, n ̸= 7 (5k, 5k′ )

7 6 5

[10] [10] [8] (continued on next page)

456

A.G. Chegini et al. / Discrete Mathematics 339 (2016) 447–456

(m, n)

χ (Tm2 ,n )

Reference

(6, n)

6

[8,10]

(7, n), n ∈ {7, 8, 9, 11, 13, 14, 19} (7, n), n ̸∈ {7, 8, 9, 11, 13, 14, 19}, n ≥ 6

7 6

Corollary 2 and Remark 2 Theorem 2

(8, 8) (8, n), n ≥ 9

7 6

[8] Theorem 4

(9, 9) (9, n), n ≥ 10

7 6

[8] Theorem 5

(m, n), m, n ≥ 10, 5 - m, 5 - n

6

Sections 2 and 3

Also we have answered Conjecture 1 as follows. Theorem 9. Conjecture 1 is true for all m, n ≥ 3, except for (m, n) = (7, 11).



Acknowledgements 2 2 The 6-colouring of T13 ,13 and T19,19 was found by using the supercomputer of the school of Mathematics of Institute for Research in Fundamental Sciences (IPM). This work was done while Farokhlagha Moazami had a Postdoctoral Fellowship at Sharif University of Technology, supported by a grant from the National Elites Foundation. She is grateful to professor E.S. Mahmoodian for his support. This research was partially supported by a grant from the INSF.

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