Strong Edge-Coloring for Cubic Halin Graphs

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國立臺灣大學 數學系 預印本 Department of Mathematics, National Taiwan University

www.math.ntu.edu.tw/ ~ mathlib/preprint/2011- 10.pdf

Strong Edge-Coloring for Cubic Halin Graphs

Gerard Jennhwa Chang and Daphne Der-Fen Liu

November 4, 2011

Strong Edge-Coloring for Cubic Halin Graphs

Gerard Jennhwa Chang

and Daphne Der-Fen Liu

1Department of Mathematics, National Taiwan University, Taipei 10617, Taiwan

2Taida Institute for Mathematical Sciences, National Taiwan University, Taipei 10617, Taiwan

3National Center for Theoretical Sciences, Taipei Office, Taipei, Taiwan

4Department of Mathematics, California State University, Los Angeles, USA

November 4, 2011

Abstract

A strong edge-coloring of a graph G is a function that assigns to each edge a color such that two edges within distance two apart must receive different colors. The minimum number of colors used in a strong edge-coloring is the strong chromatic indexof G. Lih and Liu [14] proved that the strong chromatic index of a cubic Halin graph, other than two special graphs, is 6 or 7. It remains an open problem to determine which of such graphs have strong chromatic index 6. Our article is devoted to this open problem. In particular, we disprove a conjecture of Shiu, Lam and Tam [18] that the strong chromatic index of a cubic Halin graph with characteristic tree a caterpillar of odd leaves is 6.

1 Introduction

The coloring problem considered in this article has restrictions on edges within dis- tance two apart. The distance between two edges e and e⁰ in a graph is the minimum

∗E-mail: gjchang@math.ntu.edu.tw. Supported in part by the National Science Council under grant NSC98-2115-M-002-013-MY3.

†Corresponding author. Email: dliu@calstatela.edu.

k for which there is a sequence e0, e₁, . . . , e_k of distinct edges such that e = e0, e⁰ = ek, and ei−1 shares an end vertex with ei for 1 ≤ i ≤ k. A strong edge-coloring of a graph is a function that assigns to each edge a color such that any two edges within distance two apart must receive different colors. A strong k-edge-coloring is a strong edge-coloring using at most k colors. The strong chromatic index of a graph G, denoted by χ⁰_s(G), is the minimum k such that G admits a strong k-edge-coloring.

Strong edge-coloring was first studied by Fouquet and Jolivet [8, 9] for cubic planar graphs. A trivial upper bound is that χ⁰_s(G) ≤ 2∆² − 2∆ + 1 for any graph G of maximum degree ∆. Fouquet and Jolivet [8] established a Brooks type upper bound χ⁰_s(G) ≤ 2∆²− 2∆, which is not true only for G = C5 as pointed out by Shiu and Tam [19]. The following conjecture was posed by Erd˝os and Neˇsetˇril [5, 6] and revised by Faudree, Schelp, Gy´arf´as and Tuza [7]:

Conjecture 1. For any graph G of maximum degree ∆,

χ⁰_s(G) ≤ ( 5

4∆², if ∆ is even;

5

4∆²−¹₂∆ + ¹₄, if ∆ is odd.

Faudree, Schelp, Gy´arf´as and Tuza [7] also asked whether χ⁰_s(G) ≤ 9 if G is cubic planar. If this upper bound is proved to be true, it would be the best possible. For graph with maximum degree ∆ = 3, Conjecture 1 was verified by Andersen [1] and by Hor´ak, Qing and Trotter [12] independently. For ∆ = 4, while Conjecture 1 says that χ⁰_s(G) ≤ 20, Hor´ak [11] obtained χ⁰_s(G) ≤ 23 and Cranston [4] proved χ⁰_s(G) ≤ 22.

The main theme of this paper is to study strong edge-coloring for the following planar graphs. A Halin graph G = T ∪ C is a plane graph consisting of a plane embedding of a tree T each of whose interior vertex has degree at least 3, and a cycle C connecting the leaves (vertices of degree 1) of T such that C is the boundary of the exterior face. The tree T and the cycle C are called the characteristic tree and the adjoint cycle of G, respectively. Strong chromatic index for Halin graphs was first considered by Shiu, Lam and Tam [18] and then studied in [19, 13, 14].

A caterpillar is a tree whose removal of leaves results in a path called the spine of the caterpillar. For k ≥ 1, let Gk be the set of all cubic Halin graphs whose characteristic trees are caterpillars with k + 2 leaves. For a graph G = T ∪ C in Gk, let P : v1, v₂, . . . , v_k be the spine of T and each vi is adjacent to a leaf ui for 1 ≤ i ≤ k with v1 (resp. vk) adjacent to one more leaf u0 = v0 (resp. uk+1 = vk+1). We draw G on the plane by putting the path v0P v_k+1 horizontally in the middle, and the pending

edges (leaf edges) viu_i, 1 ≤ i ≤ k. by either up or down edges vertically to P . See Figure 1 for an example of G8.

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v v v v v v v v v v

v v

v v v

v v v

u₀ = v0 v₁ v₂ v₃ v₄ v₅ v₆ v₇ v₈ v₉ = u9

u₁ u₂ u₆ u₇ u₈

u₃ u₄ u₅

Figure 1: The graph G2,3,3 in G8.

From this drawing, we associate to G with a list of positive integers (n1, n₂, . . . , nr), where ni is the number of maximum consecutive up or down edges, starting from the leftmost to the rightmost on P . We use Gn1,n2,...,nr to denote this graph. For instance the graph in Figure 1 is G2,3,3. Notice that n1+ n2+ . . . + nr = k. For a special case when these pending edges are all in the same direction (up or down), the graph Gk

is called the necklace and denoted by Nek in [18]. Notice that Gk is the only graph in Gk for k ≤ 3.

Observation 1. G_n1,n2,...,n^r ∼= Gn^r,...,n2,n1. Observation 2. G_n1,n2,...,nr,1 ∼= Gn1,n2,...,nr+1.

It is easy to see that χ⁰_s(G) ≥ 6 for any G ∈ Gk, k ≥ 1. Shiu, Lam and Tam [18]

obtained the following results:

•

χ⁰_s(Gk) =











9, m = 2;

8, m = 4;

7, m is even and m ≥ 6;

6, m is odd.

• If G ∈ Gk with k ≥ 4, then 6 ≤ χ⁰_s(G) ≤ 8.

• If G is a cubic Halin graph, then 6 ≤ χ⁰_s(G) ≤ 9.

Moreover, the authors [18] raised the following conjectures:

Conjecture 2. If G∈ Gk with k≥ 5, then χ⁰_s(G) ≤ 7.

Conjecture 3. If G∈ Gk with odd k ≥ 5, then χ⁰_s(G) = 6.

Conjecture 4. If G= T ∪ C is a Halin graph, then χ⁰_s(G) ≤ χ⁰_s(T ) + 4.

Faudree, Schelp, Gy´arf´as and Tuza [7] proved, for any tree T , it holds that χ⁰_s(T ) = maxuv∈E(T )(deg(u)+deg(v)−1). Conjecture 4 was confirmed by Lai, Lih and Tsai [13], who proved a stronger result that χ⁰_s(G) ≤ χ⁰_s(T ) + 3 for any Halin graph G = T ∪ C other than G2 and wheels Wn with n 6≡ 0 (mod 3), where Wn = K1,n∪ Cn. Note that χ⁰_s(W5) = χ⁰_s(K1,5) + 5; and χ⁰_s(G) = χ⁰_s(T ) + 4 for G = G2 or G = Wn with n 6≡ 0 (mod 3) and n 6= 5.

Conjecture 2 was confirmed by Lih and Liu [14], who proved a more general result that χ⁰_s(G) ≤ 7 is true for any cubic Halin graph other than G2 and G4. Hence, the strong chromatic index for any cubic Halin graph G 6= G2, G₄ is either 6 or 7.

It remains open to determine the cubic Halin graphs G with χ⁰_s(G) = 6 (or the ones with χ⁰_s(G) = 7). Our aim is to investigate this problem. In particular, we establish methods that can be used to study the graphs Gk. As a result, we discover counterexamples to Conjecture 3. We prove that for any k ≥ 7, there exists graph G ∈ Gk with χ⁰_s(G) = 7; and for any k 6= 2, 4, there exists G ∈ Gk (other than necklaces) with χ⁰_s(G) = 6.

2 Cubic Halin graphs G with χ

(G) = 6

This section gives some cubic Halin graphs with strong chromatic index 6. We begin with the development of several general transformation theorems for Halin graphs.

For a positive integer r, an r-tail of a tree T is a path P : v1, v₂, . . . , vr, v_r+1 in which v1 is not a leaf but all vertices in Li = {u 6∈ P : uvi ∈ E(T )} are leaves for 1 ≤ i ≤ r. For integer s < r, cutting Ps from T means deleting the vertices {v1, v₂, . . . , v_s−1} ∪1≤i≤sL_i from T , which results in a tree denoted by T Ps. Notice that vs becomes a leaf adjacent to vs+1 in T Ps.

Suppose P : v1, v₂, . . . , v_r, v_r+1 is an r-tail of the characteristic tree T of a Halin graph G = T ∪ C. For any j with 1 ≤ j ≤ r, the vertices in ∪1≤i≤jL_i form a consecutive portion on the adjoint cycle C. See Figure 2 for an example of a 4-tail.

For any two vertices in ∪2≤i≤sL_i, we may regard that they are on the same or different sides of L1. For instance, in Figure 2, u¹₃ and u²₃ are on the same side of L1, while u¹₂ and u²₂ are on different sides of L1. For s < r, the tree T Ps is the characteristic

tree of a new Halin graph, denoted by G Ps, whose adjoint cycle is obtained from C by replacing the segment {x} ∪1≤i≤sL_i∪ {y} by the path xvsy originally not in G, where x (respectively, y) is the vertex in C right before (respectively, after) ∪1≤i≤sL_i. See the dashed path for xv3y in Figure 2.

C ^{v v}

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v₄

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L₁ = {u¹₁, u²₁, u³₁} L₂ = {u¹₂, u²₂} L₃ = {u¹₃, u²₃} L₄ = {u¹₄}

Figure 2: A cutting 4-tail from T , resulting in G P3 with two new edges, v3x and v₃y, while vertices in {v1, v₂} ∪ L1∪ L2∪ L3 are all gone.

We denote a 4-cycle by (v1, v₂, v₃, v₄), which consists of the edges v4v₁, and viv_i+1 for i = 1, 2, 3.

Lemma 3. Suppose (x1, x₂, x₃, x₄) is a 4-cycle in a graph G in which each xi is adjacent to a vertex y_i not in the 4-cycle for 1 ≤ i ≤ 4. If χ⁰_s(G) = 6, then for every strong 6-edge-coloring f of G we have

(i) f(x1y₁) = f(x3y₃) and f(x2y₂) = f(x4y₄), and (ii) f(y3y₄) = f(x1x₂) whenever y3 is adjacent to y₄.

Proof. Part (i) follows from that for each i the edges on the 4-cycle (x1, x₂, x₃, x₄) together with the edges xiy_i and xi+1y_i+1 use all the 6 colors, where x5y₅ = x1y₁.

Part (ii) follows from that the edges on the 4-cycle (x3, y₃, y₄, x₄) together with the two edges x1x₄, x2x₃ use all the 6 colors. See Figure 3 for an illustration.

We now consider the cutting tail operation for the characteristic tree of a cubic Halin graph G = T ∪ C. We shall study the conditions for which such an operation preserves the fact that χ⁰_s(G) = 6.

t t

t t

t t

t t

y₂ y₁

x₂ x₁

x₃ x₄

y₃ y₄

2 1 3

4 5 6

b a

c

Figure 3: a, b, c are forced to be 2, 1, 3, respectively.

Theorem 4. Suppose P: v1, v₂, v₃, v₄ is a4-tail of the characteristic tree T of a cubic Halin graph G = T ∪ C, where L1 = {u0, u₁} and Li = {ui} for i ≥ 2. If u2 and u₃ are on the same side of L₁, then χ⁰_s(G) = 6 if and only if χ⁰_s(G P2) = 6.

Proof. (⇒) Suppose χ⁰_s(G) = 6. Let f be a strong 6-edge-coloring of G. Without loss of generality, we may assume that f(xu0) = 1, f(u0u₁) = 2, f(u1u₂) = 3, f(v1u₀) = 4, f(v1u₁) = 5, and f(v1v₂) = 6 as the bold faced numbers in Figure 4. It is then the case that f(v2u₂) = 1. Repeatedly applying Lemma 3, we have f(u2u₃) = 4, f(v2v₃) = 2, f(v3u₃) = 5, f(u3z) = 6 and f(v3v₄) = 3 (see Figure 4). In G P2, we use the old color for edges in G, and color the new edges xv2 and v2y by 1 and 4, respectively. It is easy to check that the new coloring is a strong 6-edge-coloring for G P2. Hence, χ⁰_s(G P2) = 6.

(⇐) Suppose χ⁰_s(G P2) = 6. Let f⁰ be a strong 6-edge-coloring of G P2. Without loss of generality, assume that the colors are as in Figure 4. We may delete the edges xv2 and v2y, and extend the coloring to G using the colors as in Figure 4.

This gives a strong 6-edge-coloring of G, so χ⁰_s(G) = 6.

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1 (1)

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Figure 4: A cutting G P2.

Corollary 5. χ⁰_s(Gn1,n2,...,nr) = 6 if and only if χ⁰_s(Gn1,n2,...,nr+2) = 6.

Theorem 6. Suppose P: v1, v₂, v₃, v₄, v₅ is a 5-tail of the characteristic tree T of a cubic Halin graph G= T ∪ C, where L1 = {u0, u₁} and Li = {ui} for i ≥ 2. Assume u₂ and u₃ are on different sides of L₁, while u₂ and u₄ are on the same side of L₁. If χ⁰_s(G P2) = 6, then χ⁰_s(G) = 6.

Proof. Let f⁰ be a strong 6-edge-coloring of G P2. By Lemma 3, without loss of generality, we may assume that f⁰(v3v₄) = 1, f⁰(v2v₃) = 2, f⁰(v2y) = 3, f⁰(v4u₄) = 4, f⁰(v4v₅) = f⁰(xv2) = 5 and f⁰(v3u₃) = f⁰(u4z) = 6, as the bold faced numbers shown in Figure 5. We delete the edges xv2 and v2y from G P2, and extend the coloring to G using the colors shown in Figure 5. This gives a strong 6-edge-coloring of G, so χ⁰_s(G) = 6.

C ^{v v}

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QQQ QQQ

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v₄ v SS

SS SS

S SS

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v₅ v

1 2

3 4 5

6 5

5 2 4 6

3 1 6

5

3

Figure 5: A cutting G P2.

Corollary 7. If χ⁰_s(Gn1,n2,...,nr,1) = 6, then χ⁰_s(Gn1,n2,...,nr,1,2) = 6.

Corollary 8. Assume χ⁰_s(Gn1,n2,...,nr

−1,1) = 6. Then χ⁰_s(Gn1,n2,...,nr

−1,F1,F2,...,F^t,1) = 6, where each F_i is either a single positive even integer, or a list of two integers, (1, t), for some odd integer t. In particular, χ⁰_s(Gn1,n2,...,nr−1,1∗m) = 6 for odd m, where 1 ∗ m stands for a sequence of m 1’s.

Proof. Assume χ⁰_s(Gn1,n2,...,n^r₋1,1) = 6. It suffices to prove the result for the case t = 1.

Assume F1 is a single even integer, F1 = nr. By Corollary 5 and Observation 2, 6 = χ⁰_s(Gn1,n2,...,nr

−1,1) = χ⁰_s(Gn1,n2,...,nr

−1,1+n^r) = χ⁰_s(Gn1,n2,...,nr

−1,nr,1).

Next, assume F1 is a list of two integers (1, t) for some odd t = 2s + 1. By Corollaries 7 and 5, and Observation 2,

6 = χ⁰_s(Gn1,n2,...,nr

−1,1,2) = χ⁰_s(Gn1,n2,...,n^r₋1,1,2(s+1)) = χ⁰_s(Gn1,n2,...,nr

−1,1,t,1).

Let k be an even integer. Although it is known [18] that χ⁰_s(Gk) > 6, there exist graphs G ∈ Gk with χ⁰_s(G) = 6. Figures 6 and 7 show two examples.

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v v v v v v v v

v v

v v

v v

3 1

5 1

2 6 1 2 3 5 2

4 2 6

3

5 3 1 4

4 6

Figure 6: A strong 6-edge-coloring for G2,2,2= G2,2,1,1.

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v

v

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v

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3 4 6 3 5 6

4 6 5

2

5 1 4

1 5 2 1 4

2 6 1 3 4 2 6 1 3

Figure 7: A strong 6-edge-coloring for G1?8.

By the results we have shown, one can verify that for every positive integer k 6= 2, 4, there exists G ∈ Gk (other than necklaces) with χ⁰_s(G) = 6. This is be- cause χ⁰_s(G1) = χ⁰_s(G1,1,1) = χ⁰_s(G2,2,1,1) = χ⁰_s(G1∗8) = 6, by Corollary 8, one gets χ⁰_s(G2,2,1∗m) = 6 for m even m ≥ 4, and χ⁰_s(G1∗n) = 6 for n 6= 2, 4, 6.

We end this section with another family of cubic Halin graphs with strong chro- matic index 6.

Theorem 9. χ⁰_s(G3?m) = 6 for any m 6= 2, 4, 5.

Proof. Since χ⁰_s(G3) = χ⁰_s(G5) = 6, so χ⁰_s(G1,3,1) = 6. By Corollary 5, we get χ⁰_s(G3,3,3) = 6. Hence, the result holds for m = 1, 3.

Assume m ≥ 6. If χ⁰_s(G2,3?(m−4),2) = 6, then by Corollary 5, χ⁰_s(G4,3?(m−4),4) = χ⁰_s(G1,3?(m−2),1) = χ⁰_s(G3?m) = 6. Hence, it is enough to find a strong 6-edge-coloring f for G2,3?(m−4),2.

In the following we let f(v0u₁) = 1, f(v0v₁) = 2, f(v0u₃) = 3, f(u1u₂) = 4, f(v1u₁) = 5, and f(v1v₂) = 6. Consequently, by Lemma 3, f(v2v₃) = f(u4u₅) = 1, f(u2u₆) = f(v7v₈) = 2, and f(u2v₂) = f(v4v₅) = 3. Since the color 3 has to be used in the 4-cycle (u6u₇v₇v₆), it must be the case that f(v7u₇) = 3.

Assume m is even. Let m − 4 = 2k. Define f(v4u₄) = 4, f(u3u₄) = 6, and the remaining by the following recursive process for 1 ≤ t ≤ 2k:

f(v3tv_3t+1) =

( f(v3t−2u_3t−2) if t is even;

f(v3t−3v_3t−2) if t is odd.

f(v3tu_3t) =

( f(u3t−2u_3t−1) if t is even;

f(v3t−2u_3t−2) if t is odd.

f(u3tu_3t+1) =

( f(v3t−1u_3t−1) if t is even;

f(u3t−2u_3t−1) if t is odd, t ≥ 3.

f(v3t+1u_3t+1) =

( f(v3t−2v_3t−1) if t is even;

f(v3t−1u_3t−1) if t is odd, t ≥ 3.

By Lemma 3, the colors for the remaining edges are fixed. It is not hard to see that f is a strong 6-edge-coloring for G2,3?(2k),2. See Figure 8 for an example.

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v v

v v v

v v v

v v

3 1

2

4 2 5 6 4

6 1 2 3 6 4 2 5 6 2

5 3 1 3 1

5 4 5 4 1

6 1 2 3 5

Figure 8: A strong 6-edge-coloring for G2,3,3,2.

Assume m is odd. Let m − 4 = 2k + 1. Let f(v3v₄) = 4, f(v3u₃) = 5, f(u3u₄) = 6, and f(u4v₄) = 2. For 2 ≤ t ≤ 2k, define f by the following recursive process:

f(v3tv_3t+1) =

( f(v3t−3v_3t−2) if t is even;

f(u3t−2u_3t−1) if t is odd.

f(v3tu_3t) =

( f(u3t−2u_3t−1) if t is even;

f(v3t−2u_3t−2) if t is odd.

f(u3tu_3t+1) =

( f(v3t−1u_3t−1) if t is even;

f(v3t−2v_3t−1) if t is odd.

f(v3t+1u_3t+1) =

( f(u3t−2v_3t−2) if t is even, t 6= 2;

f(u3t−1v_3t−1) if t is odd.

Note, for t = 2 in the last case above, f(v7u₇) = 3 is fixed as discussed at the beginning of the proof.

For t = 2k + 1, let f(v6k+3v_6k+4) = f(u6k+1v_6k+1), f(v6k+3u_6k+3) = f(u6k+1u_6k+2), f(u6k+3u_6k+4) = f(v6k+1v_6k+2), and f(v6k+4u_6k+4) = f(u6k+2v_6k+2).

Again, by Lemma 3, the colors for the remaining edges are fixed. It is not hard to see that f is a strong 6-edge-coloring for G2,3?(2k+1),2. See Figure 9 for an example.

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u u

u u u

u u u

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u u

3 1 2

4 2 5 6 4 5

2

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

5 3 1 3 1 3 6

5 2 5 6 1 6

6 1 4 2 5 3

Figure 9: A strong 6-edge-coloring for G2,3,3,3,2. In the next section, we will show that χ⁰_s(G3?m) = 7 for m = 2, 4, 5.

3 Cubic Halin graphs G with χ

(G) = 7

We present some cubic Halin graphs with strong chromatic index 7. In particular, we prove that for any k ≥ 7, there exists G ∈ Gk with χ⁰_s(G) = 7.

Let us start with an example, χ⁰_s(G2,2) = 7. Suppose to the contrary that χ⁰_s(G2,2) = 6. Choose a strong 6-edge-coloring f for G2,2. By Lemma 3 (i), f(v0u₃) = f(v4v₅). Since the color f(v0u₃) has to be used by the 4-cycle (u1, u₂, v₂, v₁), it is the case that f(v0u₃) = f(u2v₂), and so f(v4v₅) = f(u2v₂), a contradiction.

Lemma 10. Let G = G2,3,1,n⁴,n5...,nr with r ≥ 4. If f is a strong 6-edge-coloring for G, then f(u1v₁) = f(u6u_7+n4) = f(v7v₈).

Proof. Without loss of generality, assume that f(v0u₁) = 1, f(v0v₁) = 2, f(v0u₃) = 3, f(v1u₁) = 4, f(u1u₂) = 5, and f(v1v₂) = 6. See the bold faced numbers in Figure 10. Then f(v2u₂) = 3. By Lemma 3 (i), f(u2u₆) = 2, f(v2v₃) = f(u4u₅) = 1, and f(v4v₅) = 3. See the italic numbers in Figure 10.

Suppose f(v3u₃) = x and f(u3u₄) = y. Then x ∈ {4, 5} and y ∈ {4, 5, 6}. By Lemma 3, f(v5u₅) = x and f(v5v₆) = y. Let z be the only label in {4, 5, 6} − {x, y}.

Then {f(v6v₇), f(v6u₆)} = {1, z}, since v6v₇ and v6u₆ can not be labeled by 2, 3, x, y.

Let f(u6u_7+n4) = a. Then a 6∈ {1, 2, 3, 5, y, z}. Hence, it must be the case that a= x 6= 5, implying a = x = 4.

By Lemma 3 (i), f(u5u₇) = f(v3v₄) = c ∈ {5, 2}. If c = 5, then f(v6v₇) = 1 and f(v6u₆) = z = c = 5, which is impossible as f(u1u₂) = 5. Hence, c = 2. Then f(v7u₇) 6∈ {1, 2, x, y, z}, so f(v7u₇) = 3. Consequently, f(v7v₈) = b 6∈ {1, 2, 3, y, z}.

Therefore, b = x = 4. This completes the proof.

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v

v

v₀ v₁ v₂ v₃ v₄ v₅ v₆ v₇

u₁ u₂ u₆

u₇ u₃ u₄ u₅

1 2 3

4 5

6 3

2

1 3

1

a

b x

y

x y

c

Figure 10: In G2,3,1,n⁴,n5,...,nr, labels a and b are forced to be 4.

Theorem 11. The following graphs have strong chromatic index 7:

(a) G2,3,1,n⁴.

(b) G2,3,1,1,n5,n6,...,n^r with r≥ 5.

(c) G2,3,1,3,n⁵.

(d) G2,3,1,3,2,n6,n7,...,n^r with r≥ 6.

(e) G2,3,1,3,4,n6.

(f) G2,3,1,3,4,2,n⁷,n8,...,nr with r ≥ 6.

Proof. For each case in the following, we suppose to the contrary that the given graph has strong chromatic index 6. Let f be a strong 6-edge-coloring for G. We shall derive a contradiction for each case.

(a) By Corollary 5 we may assume that n4 ≤ 2. By Lemma 10, f(u6u_7+n4) = f(v7v₈), which contradicts the fact that the edges u6u_7+n4 and v7v₈are within distance two apart.

(b) Since n4 = 1, by Lemma 10, f(u6u₈) = f(v7v₈), which contradicts the fact that the edges u6u₈ and v7v₈ are distance two apart.

(c) By Corollary 5 we may assume that n5 ≤ 2. By Lemma 10 and Corollary 5, f(u6u₁₀) = f(v7v₈) = f(u9u_10+n5). For the case n5 = 1, this is a contradiction as u₆u₁₀ and u9u_10+n5 are of distance two apart. For the case n5 = 2, by Corollary 5, f(u6u₁₀) = f(v11v₁₂), and so f(v11v₁₂) = f(u9u₁₂), a contradiction.

The proofs for (d), (e), and (f) are similar. We leave the details to the reader.

An immediate consequence of Theorem 11 is that for every integer k ≥ 7, there exists G ∈ Gk with χ⁰_s(G) = 7. This gives counter examples to Conjecture 3.

We now give complete solutions for the value of χ⁰_s(G3?m). By Theorem 9 it is enough to verify that χ⁰_s(G3?m) = 7 for m = 2, 4, 5. Because χ⁰_s(G3,1) = χ⁰_s(G4) > 6, Corollary 5 implies that χ⁰_s(G3,3) > 6, so χ⁰_s(G3,3) = 7.

For m = 4, since χ⁰_s(G2,2) > 6, by Corollary 5, we get χ⁰_s(G4,4) = χ⁰_s(G1,3,3,1) > 6.

Use Corollary 5 twice again, we obtain χ⁰_s(G3,3,3,3) > 6, so χ⁰_s(G3,3,3,3) = 7.

For m = 5, by Theorem 11 (a), χ⁰_s(G2,3,1,1) = χ⁰_s(G2,3,2) = 7. This implies, by Corollary 5, χ⁰_s(G4,3,4) = χ⁰_s(G1,3,3,3,1) = χ⁰_s(G3?5) = 7.

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