Received: 15 January 2017 Revised: 27 October 2017 Accepted: 20 November 2017
DOI: 10.1002/jgt.22230
ARTICLE
Star chromatic index of subcubic multigraphs
Hui Lei1 Yongtang Shi1 Zi-Xia Song2
1 Center for Combinatorics and LPMC, Nankai Abstract University, Tianjin 300071, China The star chromatic index of a multigraph πΊ, denoted πβ²(πΊ), 2Department of Mathematics, University of π Central Florida, Orlando, FL 32816, USA is the minimum number of colors needed to properly color Correspondence the edges of πΊ such that no path or cycle of length four Zi-Xia Song, Department of Math- is bicolored. A multigraph πΊ is star π-edge-colorable if ematics, University of Central πβ² πΊ β€ π Florida, Orlando, FL 32816, USA. π ( ) . DvoΕΓ‘k, Mohar, and Ε Γ‘mal [Star chromatic Email: [email protected] index, J. Graph Theory 72 (2013), 313β326] proved that Contract grant sponsor: National Natural Sci- every subcubic multigraph is star 7-edge-colorable. They ence Foundation of China and Natural Science conjectured in the same article that every subcubic multi- Foundation of Tianjin; contract grant number: 17JCQNJC00300. graph should be star 6-edge-colorable. In this article, we ο¬rst prove that it is NP-complete to determine whether β² ππ (πΊ) β€ 3 for an arbitrary graph πΊ. This answers a ques- tion of Mohar. We then establish some structure results on β² subcubic multigraphs πΊ with πΏ(πΊ) β€ 2 such that ππ (πΊ) >π β² but ππ (πΊ β π£) β€ π for any π£ β π (πΊ), where π β{5, 6}. We ο¬nally apply the structure results, along with a sim- ple discharging method, to prove that every subcubic multi- graph πΊ is star 6-edge-colorable if πππ(πΊ) < 5β2, and star 5-edge-colorable if πππ(πΊ) < 24β11, respectively, where πππ(πΊ) is the maximum average degree of a multigraph πΊ. This partially conο¬rms the conjecture of DvoΕΓ‘k, Mohar, and Ε Γ‘mal.
KEYWORDS maximum average degree, star edge-coloring, subcubic multigraphs
1 INTRODUCTION
All multigraphs in this article are ο¬nite and loopless; and all graphs are ο¬nite and without loops or multiple edges. Given a multigraph πΊ,letπ βΆ πΈ(πΊ) β [π] be a proper edge-coloring of πΊ, where π β₯ 1 is an integer and [π]βΆ={1, 2, β¦ ,π}. We say that π is a star π-edge-coloring of πΊ if no path or
J Graph Theory. 2017;1β11. wileyonlinelibrary.com/journal/jgt Β© 2017 Wiley Periodicals, Inc. 1 2 LEI ET AL. cycle of length four in πΊ is bicolored under the coloring π; and πΊ is star π-edge-colorable if πΊ admits β² astarπ-edge-coloring. The star chromatic index of πΊ, denoted by ππ (πΊ), is the smallest integer π such that πΊ is star π-edge-colorable. The chromatic index of πΊ is denoted by πβ²(πΊ). As pointed out in [7], the deο¬nition of star edge-coloring of a graph πΊ is equivalent to the star vertex-coloring of its line graph πΏ(πΊ). Star edge-coloring of a graph was initiated by Liu and Deng [10], motivated by the vertex version (see [1,5,6,9,11]). Given a multigraph πΊ,weuse|πΊ| to denote the number of vertices, π(πΊ) the number of edges, πΏ(πΊ) the minimum degree, and Ξ(πΊ) the maximum degree of πΊ, respectively. For π£ π πΊ π π£ π π£ π£ πΊ any β ( ),let πΊ( ) and πΊ( ) denoteβ the degree and neighborhood of in , respectively. For π΄, π΅ βπ πΊ π π΄ π π π΄ π΅ π΄ π΅ π΅ π any subsets ( ),let πΊ( )βΆ= πβπ΄ πΊ( ), and let β βΆ= β .If ={ },wesimply write π΄βπ instead of π΄βπ΅.WeuseπΎπ and ππ to denote the complete graph and the path on π vertices, respectively. It is well-known [13] that the chromatic index of a graph with maximum degree Ξ is either Ξ or Ξ+1. However, it is NP-complete [8] to determine whether the chromatic index of an arbitrary graph with maximum degree Ξ is Ξ or Ξ+1. The problem remains NP-complete even for cubic graphs. A multigraph πΊ is subcubic if the maximum degree of πΊ is at most three. Mohar (private communica- β² tion with the second author) proposed that it is NP-complete to determine whether ππ (πΊ) β€ 3 for an arbitrary graph πΊ. We ο¬rst answer this question in the positive.
β² Theorem 1.1. It is NP-complete to determine whether ππ (πΊ) β€ 3 for an arbitrary graph πΊ. We prove Theorem 1.1 in Section 2. Theorem 1.2 below is a result of DvoΕΓ‘k et al. [7], which gives an upper bound and a lower bound for complete graphs.
Theorem 1.2 ( [7]). The star chromatic index of the complete graph πΎπ satisο¬es β β 2 2(1+π(1)) log π β² 2 2π(1 + π(1)) β€ ππ (πΎπ) β€ π . (log π)1β4
β² 1+π In particular, for every π>0, there exists a constant π such that ππ (πΎπ) β€ ππ for every integer π β₯ 1. πβ² πΎ The true order of magnitude of π ( π) is still unknown. From Theorem 1.2, an upper boundβ in β² π(1) log Ξ terms of the maximum degree for general graphs is also derived in [7], i.e. ππ (πΊ) β€ Ξ β 2 for any graph πΊ with maximum degree Ξ. In the same article, DvoΕΓ‘k et al. [7] also considered the star chromatic index of subcubic multigraphs. To state their result, we need to introduce one notation. A graph πΊ covers a graph π» if there is a mapping π βΆ π (πΊ) β π (π») such that for any π’π£ β πΈ(πΊ), π(π’)π(π£)βπΈ(π»), and for any π’ β π (πΊ), π is a bijection between ππΊ(π’) and ππ» (π(π’)).Theyproved the following. Theorem 1.3 ( [7]). Let πΊ be a multigraph.
β² (a) If πΊ is subcubic, then ππ (πΊ) β€ 7. β² (b) If πΊ is cubic and has no multiple edges, then ππ (πΊ) β₯ 4 and the equality holds if and only if πΊ covers the graph of 3-cube.
πβ² πΎ As observed in [7], π ( 3,3)=6and the Heawood graph is star 6-edge-colorable. No subcubic multigraphs with star chromatic index seven are known. DvoΕΓ‘k et al. [7] proposed the following con- jecture.
β² Conjecture 1.4. Let πΊ be a subcubic multigraph. Then ππ (πΊ) β€ 6. LEI ET AL. 3
As far as we know, not much progress has been made yet toward Conjecture 1.4. It was recently shown in [2] that every subcubic outerplanar graph is star 5-edge-colorable. A tight upper bound for trees was also obtained in [2]. We summarize the main results in [2] as follows. Theorem 1.5 ( [2]). Let πΊ be an outerplanar graph. Then β β 3Ξ(πΊ) (a) πβ²(πΊ) β€ if πΊ is a tree. Moreover, the bound is tight. π 2 (b) πβ²(πΊ) β€ 5 if Ξ(πΊ) β€ 3. π β β 3Ξ(πΊ) (c) πβ²(πΊ) β€ +12if Ξ(πΊ) β₯ 4. π 2
The maximum average degree of a multigraph πΊ, denoted πππ(πΊ), is deο¬ned as the maximum of 2π(π»)β|π»| taken over all the subgraphs π» of πΊ. We want to point out here that there is an error in the proof of Theorem 2.3 in a recent published article by Pradeep and Vijayalakshmi [Star chromatic index of subcubic graphs, Electronic Notes in Discrete Mathematics 53 (2016), 155β164]. Theorem β² 2.3 in [12] claims that if πΊ is a subcubic graph with πππ(πΊ) < 11β5, then ππ (πΊ) β€ 5. The error in the proof of Theorem 2.3 arises from ambiguity in the statement of Claim 3 in their article. From its proof given in [12] (on page 158), Claim 3 should be stated as βπ» does not contain a path π’π£π€, where either all of π’, π£, π€ are 2-vertices or all of π’, π£, π€ are light 3-vertices.β This new statement of Claim 3 does not imply that βa 2-vertex must be adjacent to a heavy 3-vertexβ in Case 2 of the proof of Theorem 2.3 (on page 162). It seems nontrivial to ο¬x this error in their proof. If Claim 3 in their article is true, using the technique we developed in the proof of Theorem 1.6(b), one can obtain a stronger result that every subcubic multigraph with πππ(πΊ) < 7β3 is star 5-edge-colorable. In this article, we prove two main results, namely Theorem 1.1 mentioned above and Theorem 1.6 below. Theorem 1.6. Let πΊ be a subcubic multigraph.
β² (a) If πππ(πΊ) < 2, then ππ (πΊ) β€ 4 and the bound is tight. β² (b) If πππ(πΊ) < 24β11, then ππ (πΊ) β€ 5. β² (c) If πππ(πΊ) < 5β2, then ππ (πΊ) β€ 6.
The rest of this article is organized as follows. We prove Theorem 1.1 in Section 2. Before we prove Theorem 1.6 in Section 4, we establish in Section 3 some structure results on subcubic multigraphs πΊ β² β² with πΏ(πΊ) β€ 2 such that ππ (πΊ) >πand ππ (πΊ β π£) β€ π for any π£ β π (πΊ), where π β{5, 6}. We believe that our structure results can be used to solve Conjecture 1.4.
2 PROOF OF THEOREM 1.1
First let us denote by SEC the problem stated in Theorem 1.1, and we denote by 3EC the following well-known NP-complete problem of Holyer [8]: Given a cubic graph πΊ,isπΊ 3-edge-colorable? Proof of Theorem 1.1. Clearly, SEC is in the class NP. We shall reduce 3EC to SEC. Let π» be an instance of 3EC. We construct a graph πΊ from π» by replacing each edge π = π’π€ β πΈ(π») with a copy of graph π»ππ, identifying π’ with π and π€ with π, where π»ππ is depicted in Figure 1. The size of πΊ is clearly polynomial in the size of π», and Ξ(πΊ)=3. 4 LEI ET AL.
FIGURE 1 Graph π»ππ
β² β² β² It suο¬ces to show that π (π») β€ 3 if and only if ππ (πΊ) β€ 3. Assume that π (π») β€ 3.Letπ βΆ πΈ(π») β {1, 2, 3} be a proper 3-edge-coloring of π».Letπβ be an edge coloring of πΊ obtained from π π π’π€ πΈ π» πβ ππ£π πβ π£ππ£π πβ π£ππ π π’π€ πβ π£ππ£π as follows: for each edge = β ( ),let ( 1)= ( 3 4)= ( 6 )= ( ), ( 1 2)= πβ π£ππ£π πβ π£ππ£π πβ π£ππ£π π π’π€ πβ π£ππ£π πβ π£ππ£π π π’π€ ( 3 7)= ( 4 8)= ( 5 6)= ( )+1, and ( 2 3)= ( 4 5)= ( )+2, where all colors here and henceforth are done modulo 3. Notice that πβ is a proper 3-edge-coloring of πΊ. Furthermore, it can be easily checked that πΊ has no bicolored path or cycle of length four under the coloring πβ. Thus β β² π is a star 3-edge-coloring of πΊ and so ππ (πΊ) β€ 3. β² β Conversely, assume that ππ (πΊ) β€ 3.Letπ βΆ πΈ(πΊ) β {1, 2, 3} be a star 3-edge-coloring of πΊ.Letπ π» πβ π π πβ ππ£π π π’π€ πΈ π» be an edge-coloring of obtained from by letting ( )= ( 1) for any = β ( ). Clearly, π π» π π’π€ πΊ πβ ππ£π πβ π£ππ is a proper 3-edge-coloring of if for any edge = in , ( 1)= ( 6 ). We prove this next. Let π = π’π€ be an edge of π». We consider the following two cases.
Case 1: πβ π£ππ£π πβ π£ππ£π ( 3 7)= ( 4 8). πβ π£ππ£π πβ π£ππ£π πΌ πΌ , , In this case, let ( 3 7)= ( 4 8)= , where β{1 2 3}. We may further assume πβ π£ππ£π π½ πβ π£ππ£π πβ π£ππ£π πΎ π½,πΎ , , πΌ that ( 3 4)= and ( 2 3)= ( 4 5)= , where { }={1 2 3}β . This is possible π π£π π π£π πβ πΊβ πβ because πΊβ ( 3)= πΊβ ( 4)=3and is a proper 3-edge-coloring of . Since is a star πΊβ πβ π£ππ£π πβ π£ππ£π πΌ πβ ππ£π πβ π£ππ π½ edge-coloring of , we see that ( 1 2)= ( 5 6)= and so ( 1)= ( 6 )= . Case 2: πβ π£ππ£π β πβ π£ππ£π ( 3 7) ( 4 8). πβ π£ππ£π πΌ πβ π£ππ£π π½ πβ π£ππ£π πΎ πΌ,π½,πΎ , , In this case, let ( 3 7)= , ( 4 8)= , ( 3 4)= , where { }={1 2 3}.Thisis possible because πΌ β π½ by assumption. Since πβ is a proper edge-coloring of πΊβ, we see that πβ π£ππ£π π½ πβ π£ππ£π πΌ πβ π£ππ£π πΌ πβ π£ππ£π ( 2 3)= and ( 4 5)= . One can easily check now that ( 1 2)= and ( 5 6)= π½ πβ ππ£π πβ π£ππ πΎ πβ πΊβ , and so ( 1)= ( 6 )= , because is a star edge-coloring of . πβ ππ£π πβ π£ππ π π» In both cases we see that ( 1)= ( 6 ). Therefore is a proper 3-edge-coloring of and so πβ²(π») β€ 3. This completes the proof of Theorem 1.1. β
3 PROPERTIES OF STAR π-CRITICAL SUBCUBIC MULTIGRAPHS
In this section, we establish some structure results on subcubic multigraphs πΊ with πΏ(πΊ) β€ 2 such β² β² that ππ (πΊ) >πand ππ (πΊ β π£) β€ π for any π£ β π (πΊ), where π β{5, 6}. For simplicity, we say that a β² β² multigraph πΊ is star π-critical if ππ (πΊ) >πand ππ (πΊ β π£) β€ π for any π£ β π (πΊ), where π β{5, 6}. Clearly, every star π-critical graph must be connected. Throughout the remainder of this section, let πΊ beastarπ-critical subcubic multigraph with πΏ(πΊ) β€ 2, and let π(π£) and π(π£) denote the neighborhood and degree of a vertex π£ in πΊ, respectively. Since every multigraph with maximum degree two is star 4-edge-colorable, we see that Ξ(πΊ)=3and LEI ET AL. 5
|πΊ| β₯ 3.Letπ₯ β π (πΊ) with π(π₯) β€ 2.Letπ» = πΊ β π₯ and let π βΆ πΈ(π») β [π] be a star π-edge- π» π , π’ π π» π π’ coloring of , where β{5 6}. For any β ( ),let ( ) denote the set of all colors suchβ that each is π’ π π΄βπ π» π π΄ π π used to color an edge incident with under the coloring . For any ( ),let ( )βΆ= πβπ΄ ( ). By abusing the notation we use π(π’π£) to denote the set of all colors on the edges between π’ and π£ under the coloring π if π’π£ β πΈ(π») is a multiple edge. Observation. If π(π₯)=2, then |π(π₯)| =2. Proof. Suppose that |π(π₯)| =1.Letπ(π₯)={π§}. Since πΊ is connected, we see that π(π§)=3.Let π(π§)={π₯,β π§ }. We obtain a star π-edge-coloring of πΊ by coloring the two edges between π₯ and π§ by two distinct colors in [π]βπ(π§β), a contradiction. β Lemma 3.1. Assume that π(π₯)=1.Letπ(π₯)={π¦}. The following are true.
(a) π(ππ» (π¦)) = [π] and |π(π¦)| =3. π π¦ πΊ π π¦ π π¦ β₯ π π π¦ π₯, π¦ ,π¦ (b) ( ) is an independent set in , ( 1)=3and ( 2) β3, where ( )={ 1 2} with π π¦ β₯ π π¦ ( 1) ( 2). π π¦ π π , π£ π π¦ π¦ π π¦π¦ π π£ |π π£ | β₯ (c) If ( 2)= β3, then for any β{1 2} and any β π» ( π)β , ( π)β ( ), ( ) 2, |π π¦ | |π π¦ | π π π¦ π π¦ π¦ ( 1) =3, ( 2) = β3,and [ 1]β© [ 2]={ }. π π¦ π π π€ π€ π¦ πΊ (d) If ( 2)=2, then =5and ( 1)=3, where 1 is the other neighbor of 2 in . β² β² (e) If π =6and for some π β{1, 2}, π(π¦π)βπ¦ has a vertex π£ with π(π£)=2, then π£π£ β πΈ(πΊ), π(π£ ) β² β² is an independent set in πΊ,andπ(π’)=3for any π’ β π(π£)βͺπ[π£ ], where π(π¦π)={π¦, π£, π£ }.
(f) If π =6and for some π β{1, 2},eachvertexofπ(π¦π)βπ¦ has degree three in πΊ, then either π(π£) β₯ 2 1 2 1 2 for any π£ β π(π¦π ) or π(π£) β₯ 2 for any π£ β π(π¦π ), where π(π¦π)={π¦,π π¦ ,π¦π }. π π π¦ π π£ β₯ π£ π π¦ π π£ β₯ π£ π π¦ (g) If =5and ( 2)=3, then either ( ) 2 for any β ( 1) or ( ) 2 for any β ( 2).
Proof. To prove Lemma 3.1(a), suppose that π(ππ» (π¦)) β [π]. Then coloring the edge π₯π¦ by a color in [π]βπ(ππ» (π¦)), we obtain a star π-edge-coloring of πΊ, a contradiction. Thus π(ππ» (π¦)) = [π] and so |π(π¦)| =3. π π¦ π₯, π¦ ,π¦ π π¦ β₯ π π¦ π¦ π¦ πΈ πΊ Next let ( )={ 1 2} with ( 1) ( 2) by Lemma 3.1(a). Suppose that 1 2 β ( ). Then |π π¦ | |π π¦ | π¦ π¦ ( 1) = ( 2) =3and all the edges incident to 1 or 2 are colored with distinct colors because π π π¦ π π₯π¦ π π¦ π¦ π πΊ ( π» ( )) = [ ]. Now coloring the edge by color ( 1 2) yields a star -edge-coloring of ,a π¦ π¦ πΈ πΊ |π π π¦ | π β₯ π π¦ π π¦ β₯ π contradiction. Thus 1 2 β ( ). Since ( π» ( )) = 5, we see that ( 1)=3and ( 2) β 3 β₯ 2. This proves Lemma 3.1(b). π π¦ π π¦ π¦ To prove Lemma 3.1(c), since ( 2)= β3, we see that all the edges incident to 1 or 2 are colored with distinct colors because π(ππ» (π¦)) = [π]. Suppose that for some π β{1, 2}, there exists a vertex π£ β ππ» (π¦π)βπ¦ such that π(π¦π¦π)βπ(π£). Then we obtain a star π-edge-coloring of πΊ by coloring the edge π₯π¦ by a color in π(π¦ππ£), a contradiction. Thus for any π β{1, 2} and any π£ β ππ» (π¦π)βπ¦, π(π¦π¦π)βπ(π£). |π π£ | β₯ πΊ π π¦ π π¦ π¦ |π π¦ | Hence ( ) 2. Since Ξ( )=3, we see that [ 1]β© [ 2]={ }. We next show that ( 1) =3 |π π¦ | π and ( 2) = β3. |π π¦ | < |π π¦ | π π¦ π π¦ π¦,β π¦ π π¦β Suppose that ( 1) 3. Then ( 1) =2because ( 1)=3.Let ( 1)={ 1}. Then ( 1)= π’ π¦β πΊ π π¦βπ’ π π¦π¦ πΌ,π½ 3 by Observation. Let be the other neighbor of 1 in . Then ( 1 )= ( 1).Let be two distinct π¦ π¦β πΌ,π½ π π’ π π π» πβ colors on the parallel edges 1 1. Then β ( ) because is a star -edge-coloring of .Let be π¦ π¦β πΌ π π¦ π π’ β π the edge between 1 and 1 with color .If ( 2)β ( ) β , then we obtain a star -edge-coloring of πΊ πβ π π¦ π π’ π¦π¦ πΌ π₯π¦ by recoloring the edge by a color in ( 2)β ( ), 1 by color , and coloring the edge by color π π¦π¦ π π¦ βππ’ π |π π¦ | π π¦β ( 1), a contradiction. Thus ( 2) ( ) and so =5. Clearly, ( 2) = β3=2.Let 2 be the π¦ πΊ π π¦π¦ π π¦β πΊ other neighbor of 2 in . Then ( 2)β ( 2). We obtain a star 5-edge-coloring of by recoloring π¦π¦ πΌ,π½ π π¦β π¦π¦ π π¦π¦ π₯π¦ π π¦π¦ the edge 2 by a color in { }β ( 2), 1 by color ( 2), and coloring the edge by color ( 1), 6 LEI ET AL.
|π π¦ | |π π¦ | π |π π¦ | π a contradiction. Thus ( 1) =3. By symmetry, ( 2) =3if =6. Clearly, ( 2) = β3=2 if π =5. This proves Lemma 3.1(c). π π¦ It remains to prove Lemma 3.1(d), (e), (f), and (g). Notice that if ( 2)=2, then by Lemma 3.1(b), π π¦ β₯ π π π π¦ π ( 2)=2 β3. Thus =5and so ( 2)= β3. For each proof of Lemma 3.1(d), (e), and (f), π π¦ π π π¦ π¦, π§ ,π§ π π¦ π¦, π€ let ( 2)= β3. By Lemma 3.1(c), we may assume that ( 1)={ 1 2} and ( 2)={ 1} if π π π¦ π¦, π€ ,π€ π π π¦ π π¦ π¦ =5(and ( 2)={ 1 2} if =6), where [ 1]β© [ 2]={ }. By Lemma 3.1(a), we may π π¦π¦ π π¦π¦ π π¦ π§ π π¦ π§ π π¦ π€ π π¦ π€ further assume that ( 1)=1, ( 2)=2, ( 1 1)=3, ( 1 2)=4, ( 2 1)=5(and ( 2 2)=6 π π π§ π π§ π π€ π π π€ π π€ when =6). By Lemma 3.1(c), 1β ( 1)β© ( 2) and 2β ( 1) if =5(and 2β ( 1)β© ( 2) if π =6). π π π€ β€ π π€ We next prove Lemma 3.1(d). Clearly, =5. Suppose that ( 1) 2. By Lemma 3.1(c), ( 1)=2 π π€ , π€β π π€ π π€ π€β π€β and ( 1)={2 5}.Let β ( 1) with ( 1 )=2. Notice that is not necessarily diο¬erent π§ π§ π π» π π€β π π€β from 1 or 2. Since is a star edge-coloring of , 5β ( ).If3β ( ), then we obtain a star πΊ π₯π¦ π¦ π€ 5-edge-coloring of by coloring the edge by color 5 and recoloring the edge 2 1 by color 3, a π π€β π π€β π€β π§ ,π§ πΊ contradiction. Thus 3β ( ), and similarly, 4β ( ). Hence β{ 1 2} because Ξ( )=3.We πΊ π₯π¦ π¦π¦ obtain a star 5-edge-coloring of by coloring the edge by color 2, recoloring the edge 2 by color π¦ π€ 5, and 2 1 by color 1, a contradiction. π π π¦ , , π π¦ , , To prove Lemma 3.1(e), since =6, we see that ( 1)={1 3 4} and ( 2)={2 5 6}. By symme- π π£ π§ π£β² π§ π π§ π π§ π§ π§ πΈ πΊ try, we may assume that =1, = 1, and = 2. Then 1β ( 1)β© ( 2). Suppose that 1 2 β ( ). π π§ π§ π¦ π§ , π π§ π₯π¦ Then ( 1 2)=1. Now recoloring the edge 1 1 by a color in {5 6}β ( 2) and coloring the edge πΊ π§ π§ πΈ πΊ π π§ by color 3, we obtain a star 6-edge-coloring of , a contradiction. Thus 1 2 β ( ).Let ( 1)= π¦ ,π§β π π§ π§β π§ π§ πΈ πΊ π§β β π§ , , π π§β π π§ β { 1 1}. Then ( 1 1)=1. Since 1 2 β ( ), we see that 1 2.If{2 5 6}β( ( 1)βͺ ( 2)) β , π¦ π§ , , π π§β π π§ π¦ π¦ then recoloring the edge 1 1 by a color in {2 5 6}β( ( 1)βͺ ( 2)), 1 by color 3, and color- π₯π¦ πΊ , , βππ§β π π§ ing by color 1 yields a star 6-edge-coloring of , a contradiction. Thus {2 5 6} ( 1)βͺ ( 2) π π§β π π§ π§βπ§ πΈ πΊ πΊ π π§ , ,πΌ πΌ and so ( 1)= ( 2)=3. Clearly, 1 2 β ( ) because Ξ( )=3.Let ( 2)={1 4 }, where β , , π π π§ π¦ β πΊ {2 5 6}. Suppose that ( ( 2)β 1) [6]. Then we obtain a star 6-edge coloring of by recoloring π¦ π§ π½ π π π§ π¦ π¦ π§ πΌ π¦π¦ π½ β the edge 1 2 by a color, say ,in[6]β ( ( 2)β 1), 1 1 by color , 1 by color 3 if 3 or color 4 if π½ π₯π¦ π π π§ π¦ |π π§ | =3, and ο¬nally coloring by color 1, a contradiction. Thus ( ( 2)β 1)=[6]and so ( 2) =3. π π§ π¦ ,π§1,π§2 π§1π§2 πΈ πΊ π π§1 π π§2 π π§1 π π§2 Let ( 2)={ 1 2 2}. Clearly, 2 2 β ( ) because ( 2)βͺ ( 2)=[6], and ( 2)= ( 2)=3, as desired. This proves Lemma 3.1(e). π π§ π¦1 π§ π¦2 To prove Lemma 3.1(f), we may assume that =1, 1 = 1, and 2 = 1. Suppose that there π§1 π π§ π§1 π π§ π π§1 π π§1 |π π§ | |π π§ | exist vertices 1 β ( 1) and 2 β ( 2) such that ( 1)= ( 2)=1. Then ( 1) = ( 2) =3 π π§ π¦ ,π§1,π§2 π π§ π¦ ,π§1,π§2 π π§2 π π§2 by Lemma 3.1(a). Let ( 1)={ 1 1 1} and ( 2)={ 1 2 2}. Then ( 1)= ( 2)=3 by π π§ π§1 , , π π§2 β π§ π§1 Lemma 3.1(a). Assume ο¬rst that ( 1 1)=1.If{2 5 6}β ( 1) β , then recoloring the edge 1 1 , , π π§2 π» π π§ by a color in {2 5 6}β ( 1), we obtain a star 6-edge-coloring of with 1β ( 1), contrary to π π§2 , , π π§ π§ π§1 Lemma 3.1(c). Thus ( 1)={2 5 6}. Clearly, 3β ( 2), otherwise recoloring the edge 1 1 by color π» π π§ π π§ , , 4 yields a star 6-edge coloring of with 1β ( 1), contrary to Lemma 3.1(c). Thus ( 2)={1 3 4}. , , π π§2 β π§ π§1 , , π π§2 Then {2 5 6}β ( 2) β . Now recoloring the edge 2 2 by a color in {2 5 6}β ( 2), we obtain a π» π π§ β , , π π§ π§1 β star 6-edge-coloring of with ( 2) {1 3 4}, a contradiction. Thus ( 1 1) 1. By symmetry, π π§ π§1 β π π§ π§2 π π§ π§2 π π§2 π» ( 2 2) 1. By Lemma 3.1(c), ( 1 1)= ( 2 2)=1. Clearly, 3β ( 1) because has no bicolored , π π§2 β πΊ path of length four. If {5 6}β ( 1) β , then we obtain a star 6-edge-coloring of by recoloring the π§ π§1 π§ π¦ , π π§2 π₯π¦ edge 1 1 by color 3, 1 1 by a color in {5 6}β ( 1), and coloring by color 3, a contradiction. Thus π π§2 , , π π§2 , , π§ π§1 π§ π¦ ( 1)={1 5 6}. Similarly, ( 2)={1 5 6}. Now recoloring the edges 1 1 by color 3, 1 1 by color π§1π§ π¦ π§ π¦π¦ π₯π¦ 4, 2 2 by color 4, 1 2 by color 2, 1 by color 3, and ο¬nally coloring the edge by color 1, we obtain a star 6-edge-coloring of πΊ, a contradiction. This proves Lemma 3.1(f). π¦1 π π¦ π¦1 π π¦ It remains to prove Lemma 3.1(g). Suppose that there exist vertices 1 β ( 1) and 2 β ( 2) π π¦1 π π¦1 |π π¦ | |π π¦ | π π¦ π¦,1 π¦ ,π¦2 such that ( 1)= ( 2)=1. Then ( 1) = ( 2) =3by Lemma 3.1(a). Let ( 1)={ 1 1} LEI ET AL. 7
π π¦ π¦,1 π¦ ,π¦2 π π¦ π π¦ π π¦ π¦1 π π¦ π¦1 and ( 2)={ 2 2}. By Lemma 3.1 (a), ( 1)βͺ ( 2)=[5].If ( 1 1)= ( 2 2), then we obtain πΊ π₯π¦ π π¦ π¦1 π π¦ π π¦ a star 5-edge-coloring of by coloring the edge by color ( 1 1) because ( 1)βͺ ( 2)=[5],a π π¦ π¦1 β π π¦ π¦1 π π¦ π π¦ π π¦ π¦1 π π¦ contradiction. Thus ( 1 1) ( 2 2). Since ( 1)βͺ ( 2)=[5], we see that either ( 1 1)β ( 2) or π π¦ π¦1 π π¦ π π¦ π¦1 π π¦ π₯π¦ π π¦ π¦1 ( 2 2)β ( 1). We may assume that ( 1 1)β ( 2). But then coloring the edge by color ( 1 1) yields a star 5-edge-coloring of πΊ, a contradiction. This completes the proof of Lemma 3.1. β
Lemma 3.2. Assume that π(π₯)=2.Letπ(π₯)={π§, π€} with |π(π§)| β€ |π(π€)|.
(a) If π§π€ β πΈ(πΊ), then π =5, |π(π§)| = |π(π€)| =3and π(π£) β₯ 2 for any π£ β π(π§)βͺπ(π€). (b) If π§π€ β πΈ(πΊ), then |π(π€)| =3or π =5, |π(π€)| = |π(π§)| =2,andπ(π€)=π(π§)=3. (c) If π(π§)=2 and π§βπ€ β πΈ(πΊ), then π =5, |π(π§β)| = |π(π€)| =3,andπ(π’)=3 for any π’ β (π[π€]βͺπ[π§β])β{π₯,} π§ , where π§β is the other neighbor of π§ in πΊ. (d) If π =6 and π(π§)=2, then π§βπ€ β πΈ(πΊ), |π(π§β)| = |π(π€)| =3,andforanyπ£ β(π(π€)βͺ π(π§β))β{π₯,} π§ , π(π£)=3and π(π’) β₯ 2 for any π’ β π(π£), where π(π§)={π₯,β π§ }. (e) If π =5and π(π§)=2, then |π(π§β)| = |π(π€)| =3,and|π(π£)| β₯ 2 for any π£ β π(π€)βͺπ(π§β), where π(π§)={π₯,β π§ }.
Proof. Assume that π§π€ β πΈ(πΊ). Since πΊ is connected, we see that |π(π€)| =3.Letπ(π€)= {π₯, π§, π€β}. We ο¬rst show that |π(π§)| =3 and π(π§)β©π(π€)={π₯}. Suppose that |π(π§)| =2 or |π(π§)| =3and π§π€β β πΈ(πΊ). Then |π(π€)βͺπ(π€β)| β€ 4 when π(π§π€)βπ(π€β) and |π(π€)βͺπ(π€β)| β€ 3 when π(π§π€)βπ(π€β). We obtain a star π-edge-coloring of πΊ by coloring the edge π₯π€ by a color, say πΌ,in[π]β(π(π€)βͺπ(π€β)) and then coloring π₯π§ by color π(π€π€β) if π(π§π€)βπ(π€β) or a color in [π]β(π(π€)βͺπ(π€β)βͺ{πΌ}) if π(π§π€)βπ(π€β), a contradiction. Thus |π(π§)| =3and π§β β π€β, where π(π§)={π₯, π€,β π§ }. We next show that π =5. Suppose that π =6. Then π(π§π€)βπ(π§β), otherwise we obtain a star 6-edge-coloring of πΊ by coloring the edge π₯π§ by a color, say πΌ,in[6]β(π(π§β)βͺπ(π€)) and π₯π€ by a color in [6]β(π(π€)βͺπ(π€β)βͺ{πΌ}). We then obtain a star 6-edge-coloring of πΊ by coloring the edge π₯π€ by a color, say π½,in[6]β(π(π€β)βͺπ(π§)) and π₯π§ by a color in [6]β(π(π§β)βͺ{π(π€π€β),π½}),a contradiction. Hence π =5. By Lemma 3.1(b), we see that π(π£) β₯ 2 for any π£ β π(π§)βͺπ(π€).This proves Lemma 3.2(a). To prove Lemma 3.2(b), by Lemma 3.1(a), |π(π€)| β₯ |π(π§)| β₯ 2. We are done if |π(π€)| =3.Sowe may assume that |π(π€)| =2. Then |π(π§)| =2because |π(π§)| β€ |π(π€)|.Letπ§β and π€β be the other neighbor of π§ and π€, respectively. If π€π€β or π§π§β is not a multiple edge (say the former) or π =6, then we obtain a star π-edge-coloring of πΊ by coloring the edge π₯π§ by a color, say πΌ,in[π]β(π(π§β)βͺ{π(π€π€β)}) and π₯π€ by a color in [π]β(π(π€β)βͺ{πΌ}), a contradiction. Thus |π(π€π€β)| = |π(π§π§β)| =2and π =5.We see that π(π€)=π(π§)=3. We next prove Lemma 3.2(c). Since π(π§)=2, we see that π§π€ β πΈ(πΊ) by Lemma 3.2(a). Then |π(π€)| =3by Lemma 3.2(b). Since π₯π§β β πΈ(πΊ), by Lemma 3.2(b) again, |π(π§β)| =3.Letπ(π€)= π₯,β π§ ,π€β π π§β π§,β π§ ,π€ π π π { } and ( )={ 1 }. We ο¬rst show that =5. Suppose that =6. Then can be extended to be a star 6-edge-coloring of πΊ by coloring the edge π₯π€ by color π(π§π§β) if π(π§π§β)βπ(π€β) or a color πΌ in [6]β(π(π§β)βͺπ(π€β)) if π(π§π§β)βπ(π€β), and then coloring the edge π₯π§ by a color in [6]β(π(π§β)βͺ{πΌ,π(π€π€β)}), a contradiction. Thus π =5. We next show that π(π€β)=3. Suppose that π(π€β) β€ 2.Ifπ(π§π§β)βπ(π€β), then we obtain a star 5-edge-coloring of πΊ by coloring the edge π₯π€ by color π(π§π§β) and π₯π§ by a color in [5]β(π(π§β)βͺπ(π€)), a contradiction. Thus π(π§π§β)βπ(π€β) and so |π(π€β)βͺπ(π§β)| β€ 4. We obtain a star 5-edge coloring of πΊ by coloring the edge π₯π€ by a color, say πΌ,in 8 LEI ET AL.
π π§β π π€β π₯π§ π π§β πΌ π π§β [5]β( ( )βͺ ( )) and by a color in [5]β( ( )βͺ{ }), a contradiction. By symmetry, ( 1)=3. This proves Lemma 3.2(c). To prove Lemma 3.2(d), since π =6, by Lemma 3.2(a,c), we see that π€π§,β π€π§ β πΈ(πΊ).By |π π§β | |π π€ | π π€ π₯, π€ ,π€ π π§β π§,β π§ ,π§β Lemma 3.2(b), ( ) = ( ) =3.Let ( )={ 1 2} and ( )={ 1 2}. Then π π§π§β π π€ π π€ πΊ ( )β ( 1)βͺ ( 2), otherwise we obtain a star 6-edge-coloring of by coloring the edge π₯π€ by color π(π§π§β) and π₯π§ by a color in [6]β(π(π§β)βͺπ(π€)), a contradiction. We next show that π π€ π π€ π π€ π π€ β π₯π€ ( 1)βͺ ( 2)=[6]. Suppose that ( 1)βͺ ( 2) [6]. Now coloring the edge by a color, say πΌ π π€ π π€ π₯π§ π π€π€ π π€π€ π π§β ,in[6]β( ( 1)βͺ ( 2)), and then coloring by color ( 1) if ( 1)β ( ) or a color in π π§β π π€ πΌ π π€π€ π π§β πΊ [6]β( ( )βͺ ( )βͺ{ }) if ( 1)β ( ), we obtain a star 6-edge-coloring of , a contradiction. π π€ π π€ π π€ π π€ π π§β π π§β Thus ( 1)βͺ ( 2)=[6]and so ( 1)= ( 2)=3. By symmetry, ( 1)= ( 2)=3. Finally, for π’ π π§β π π§β π π€ π π€ π π’ β₯ any β ( 1)βͺ ( 2)βͺ ( 1)βͺ ( 2), by Lemma 3.1(e), ( ) 2. This proves Lemma 3.2(d). It remains to prove Lemma 3.2(e). By Lemma 3.2(a), π§π€ β πΈ(πΊ). We may assume that π€π§β β πΈ(πΊ) by Lemma 3.2(c). By Lemma 3.2(b), |π(π§β)| = |π(π€)| =3. Suppose that there exists a vertex π£ β π(π€)βͺπ(π§β) with π(π£)=1. We may assume that π£ β π(π€).Letπ(π€)={π₯, π£,β π€ }. Then π(π§π§β)β π(π£)βͺπ(π€β), otherwise we recolor the edge π€π£ by color π(π§π§β). Now we obtain a star 5-edge-coloring of πΊ by coloring the edge π₯π€ by a color, say πΌ,in[5]β(π(π£)βͺπ(π€β)) and π₯π§ by a color in [5]β(π(π§β)βͺ {πΌ}), a contradiction. Thus |π(π£)|, |π(π€β)| β₯ 2. By symmetry, |π(π’)| β₯ 2 for any π’ β π(π§β). This completes the proof of Lemma 3.2. β
Corollary 3.3. Let πΊ be a subcubic multigraph that is star 6-critical. Let π£ β π (πΊ) be a vertex with π π£ π£ ,π£ ,π£ π π£ β₯ π π£ β₯ π π£ ( )={ 1 2 3} and ( 1) ( 2) ( 3)=2. The following are true.
(a) π£ ,π£ ,π£ are pairwise distinct, π π£ and π£β π£ ,π£ , where π π£ π£,β π£ . 1 2 3 ( 1)=3 3 β{ 1 2} ( 3)={ 3} (b) If π π£β , then π π£ ,andπ π’ β₯ for each π’ π π£ π π£ , ( 3)=2 ( 2)=3 ( ) 2 β ( 1)βͺ ( 2) π π£ π π£ π π£ πΊ π’ π π£ π£ (c) If ( 2)=2, then every vertex of ( 2)βͺ ( 3) has degree three in ,andforany β ( 1)β , π(π’) β₯ 2. π π£ π£β π π£ π π£β π , (d) If ( 2)=3and there exists a vertex π β ( π) with ( π )=1for some β{1 2}, then for any π’ π π£ π π’ β ( 3βπ), ( )=3.
π£ ,π£ ,π£ Proof. To prove Corollary 3.3(a), we ο¬rst show that 1 2 3 are pairwise distinct. By Observation, π£ π£ ,π£ π£ π£ π£β π£ π£β,π£β 3 is distinct from 1 2. Suppose that 1 = 2.Let 1 be the other neighbor of 1, where 1 3 are not necessarily distinct. Let π βΆ πΈ(πΊβπ£) β [6] be a star edge-coloring of πΊβπ£. We obtain a star 6-edge- πΊ π£π£ πΌ,π½ π π£β π π£ π£β coloring of by coloring the edges 1 by two distinct colors, say ,in[6]β( ( 1)βͺ{ ( 3 3)}) π£π£ π π£β πΌ,π½ π£ ,π£ ,π£ and 3 by a color in [6]β( ( 3)βͺ{ }), a contradiction. This proves that 1 2 3 are pairwise π£β π£ ,π£ π π£ π π£ β distinct. By Lemma 3.2(a), 3 β{ 1 2}. We next show that ( 1)=3. Suppose that ( 1) 3. Then π π£ π π£ π£ πΈ πΊ π£β ( π)=2for all β[3]. By Lemma 3.2(a), we see that 1 2 β ( ).Let π be the other neighbor π£ π , π£β,π£β,π£β π πΈ πΊ π£ β of π for all β{1 2}, where 1 2 3 are not necessarily distinct. Let βΆ ( β ) [6] be a star πΊ π£ πΊ π£π£ edge-coloring of β . We obtain a star 6-edge-coloring of by coloring the edge 1 by a color, say πΌ π π£β π π£ π£β ,π π£ π£β π£π£ π½ π π£β πΌ,π π£ π£β π£π£ ,in[6]β( ( 1)βͺ{ ( 2 2) ( 3 3)}), 2 by a color, say ,in[6]β( ( 2)βͺ{ ( 3 3)}), and 3 by π π£β πΌ,π½ π π£ a color in [6]β( ( 3)βͺ{ }), a contradiction. Thus ( 1)=3. This proves Corollary 3.3(a). By Lemma 3.2(d), Corollary 3.3(b) is true. By Lemma 3.2(d) and Lemma 3.1(e), Corollary 3.3(c) β is true. Finally, by Lemma 3.1(b,e) applied to π£π , Corollary 3.3(d) is true. β
4 PROOF OF THEOREM 1.6
We are now ready to prove Theorem 1.6. LEI ET AL. 9
β² FIGURE 2 AgraphπΊ with πππ(πΊ)=2and ππ (πΊ)=5
To prove Theorem 1.6(a), let πΊ be a subcubic multigraph with πππ(πΊ) < 2. Then πΊ must be a simple graph. Notice that a simple graph πΊ has πππ(πΊ) < 2 if and only if πΊ is a forest. Now applying β² Theorem 1.5(a) to every component of πΊ, we see that ππ (πΊ) β€ 4. This bound is sharp in the sense β² that there exist graphs πΊ with πππ (πΊ)=2and ππ (πΊ) > 4. One such example from [2] is depicted in Figure 2. We next proceed the proof of Theorem 1.6(c) by contradiction. Suppose the assertion is false. Let πΊ β² be a subcubic multigraph with πππ(πΊ) < 5β2 and ππ (πΊ) > 6. Among all counterexamples we choose πΊ so that |πΊ| is minimum. By the choice of πΊ, πΊ is connected, star 6-critical, and πππ(πΊ) < 5β2. π π΄ π£ π πΊ π π£ π π |π΄ | π π΄ For all β[3],let π ={ β ( )βΆ πΊ( )= } and π = π for all β[3]. By Lemma 3.1(a), 1 πΊ π π΄ βπ΄ πΊβ πΊ π΄ πππ πΊβ < is an independent set in and πΊ( 1) 3.Let = β 1. Then ( ) 5β2. We see that π πΊβ π πΊ π π π π < π π π <π π π΄ π΄ β 2 ( )=2 ( )β2 1 =2 2 +3 3 β 1 5( 2 + 3)β2 and so 3 2 +2 1. Thus 1 βͺ 2 β .By β β Lemma 3.1(b), πΏ(πΊ ) β₯ 2. We say that a vertex π£ β π (πΊ ) with ππΊβ (π£)=2is good if ππΊ(π£)=3; bad if ππΊ(π£)=2and π£ is adjacent to another vertex of degree two in πΊ; and fair if ππΊ(π£)=2and π£ is not bad. We shall apply the discharging method below to obtain a contradiction. β β 5 For each vertex π£ β π (πΊ ),letπ(π£)βΆ=ππΊβ (π£)β be the initial charge of π£. Then β π(π£)= 2 π£βπ (πΊ ) 2π(πΊβ)β 5 |πΊβ| =(2π +3π β π )β 5 (π + π )=(π β π β2π )β2 < 0, because π <π +2π . 2 2 3 1 2 2 3 3 2 1 3 2 1 β Notice that for each π£ β π (πΊ ), π(π£)=2β5β2=β1β2if ππΊβ (π£)=2, and π(π£) = 3 β 5β2 = 1β2 if β ππΊβ (π£)=3.Letπ₯ β π (πΊ ) be a vertex with ππΊβ (π₯)=3such that π₯ is adjacent to exactly π‘ β₯ 1 vertices β of degree two in πΊ . We claim that π‘ β€ 2 and π‘ =1when ππΊβ (π₯) has a bad vertex. Clearly, ππΊβ (π₯) has at most two good vertices by Lemma 3.1(f), and at most two fair vertices by Corollary 3.3(a). By Corollary 3.3(b), ππΊβ (π₯) has at most one bad vertex, and if such a bad vertex exists, then ππΊβ (π₯) has no good or fair vertex. Finally, ππΊβ (π₯) has at most one fair vertex and one good vertex simultane- ously by Corollary 3.3(c,d). Thus π‘ β€ 2 and π‘ =1when ππΊβ (π₯) has a bad vertex, as claimed. We will redistribute the charges of π₯ according to the following discharging rule: β β (R): For each π₯ β π (πΊ ) with ππΊβ (π₯)=3and exactly π‘ β₯ 1 neighbors of degree two in πΊ , π₯ sends 1 β₯ 1 charges to each of its neighbors of degree two in πΊβ. 2π‘ 4 Let πβ be the new charge of πΊβ after applying the above discharging rule. We see that for any β β β π£ β π (πΊ ) with ππΊβ (π£)=3, π (π£) β₯ 0. We next show that for any π£ β π (πΊ ) with ππΊβ (π£)=2, β β π (π£) β₯ 0.Letπ£ β π (πΊ ) be a vertex with ππΊβ (π£)=2. By Observation and Lemma 3.1(a), |ππΊβ (π£)| = β 2.LetππΊβ (π£)={π’,} π€ .Ifπ£ is good, then ππΊβ (π’)=ππΊβ (π€)=3by Lemma 3.1(c). Thus π (π£) β₯ π(π£) + 1β4 + 1β4 = 0. Next, suppose that π£ is bad. We may assume that π’ is bad. By Lemma 3.2(d), ππΊβ (π€)=3. By the above claim, π£ is the only (bad) vertex of degree two of ππΊβ (π€). By the discharging β rule (R), π (π£) β₯ π(π£)+1β2=0. Finally, suppose that π£ is a fair vertex. Then ππΊ(π’)=ππΊ(π€)=3. By Lemma 3.1(b), neither π’ nor π€ is good. Thus ππΊβ (π’)=ππΊβ (π€)=3. By the discharging rule (R), 10 LEI ET AL.
πβ π£ β₯ π π£ πβ π£ β₯ π£ π πΊβ π π£ β ( ) ( ) + 1β4 + 1β4 = 0. This proves thatβ ( ) 0 for anyβ β ( ) with πΊβ ( )=2. Thus πβ π£ β₯ πβ π£ π π£ < π£βπ (πΊβ) ( ) 0, contrary to the fact that π£βπ (πΊβ) ( )= π£βπ (πΊβ) ( ) 0. This completes the proof of Theorem 1.6(c). The proof of Theorem 1.6(b) is similar to the proof Theorem 1.6(c). For its completeness, we include its proof here because the discharging part is diο¬erent and more involved. Suppose the assertion is false. Let πΊ be a subcubic multigraph with πππ(πΊ) < 24β11 and πΊ is not star 5-edge-colorable. Among all counterexamples we choose πΊ so that |πΊ| is minimum. By the choice of πΊ, πΊ is connected and star 5- critical. Clearly, πππ(πΊ) < 24β11. For all π β[3],letπ΄π ={π£ β π (πΊ)βΆ ππΊ(π£)=π} and let ππ = |π΄π| π π΄ πΊ π π΄ βπ΄ πΊβ πΊ π΄ for all β[3]. By Lemma 3.1(a), 1 is an independent set in and πΊ( 1) 3.Let = β 1. πππ πΊβ < π πΊβ π πΊ π π π π < π π Then ( ) 24β11. We see that 2 ( )=2 ( )β2 1 =2 2 +3 3 β 1 24( 2 + 3)β11 and π < π π π΄ π΄ β πΏ πΊβ β₯ π£ so 9 3 2 2 +11 1. Thus 1 βͺ 2 β . By Lemma 3.1(b), ( ) 2. We say that a vertex β β π (πΊ ) with ππΊβ (π£)=2is good if ππΊ(π£)=3; bad if ππΊ(π£)=2and π£ is adjacent to another vertex β of degree two in πΊ; and fair if ππΊ(π£)=2and π£ is not bad. Let π΅ βΆ= {π£ β π (πΊ )βΆ ππΊβ (π£)=2}.We πΊβ π΅ πΎ πΎ π next claim that every component of [ ] is isomorphic to 1, 2,or 3. π πΊβ π΅ π₯ ,π₯ , ,π₯ π β₯ Suppose not. Let be a longest path in [ ] with vertices 1 2 β¦ π in order, where 4 π π₯ π₯ πΈ πΊ π π₯ π₯ π β or =3and 1 3 β ( ) or =2and 1 2 is a multiple edge. Clearly, 2 by Observation and π π₯ ,π₯ ,π₯ Lemma 3.1 (a). Suppose that =3. By Lemma 3.2(a), none of 1 2 3 is a fair or bad vertex. Thus π₯ ,π₯ ,π₯ π β₯ all of 1 2 3 must be good, contrary to Lemma 3.1(b). Hence 4. By Lemma 3.2(e), none of the π π₯ π₯ π₯ π₯ π₯ vertices of are bad. If 2 is fair, then both 1 and 3 must be good because neither 1 nor 3 are bad. π₯ π π₯ π₯ π₯ By Lemma 3.1(c) applied to 2, πΊβ ( 4)=3, a contradiction. Thus 2 is good. Similarly, 3 is good. π₯ π₯ By Lemma 3.1(g, c), 1 is neither good nor fair, and so 1 must be bad, a contradiction. Thus every πΊβ π΅ πΎ πΎ π component of [ ] is isomorphic to 1, 2 or 3. π£ π πΊβ We shall apply the discharging method to obtain aβ contradiction. For each vertex β ( ), 24 β 24 β let π(π£)βΆ=ππΊβ (π£)β be the initial charge of π£. Then β π(π£)=2π(πΊ )β |πΊ | =(2π + 11 π£βπ (πΊ ) 11 2 π π π 3π β π )β 24 (π + π )= 9 3β2 2β11 1 < 0, because 9π < 2π +11π . Notice that for each π£ β 3 1 11 2 3 11 3 2 1 β π (πΊ ), π(π£) = 2 β 24β11 = β2β11 if ππΊβ (π£)=2and π(π£) = 3 β 24β11 = 9β11 if ππΊβ (π£)=3.We will redistribute the charges of vertices in πΊβ according to the following discharging rule: β β (R): For each π₯ β π (πΊ ) with ππΊβ (π₯)=3and exactly π‘ β₯ 1 neighbors of degree two in πΊ , π₯ sends 9 β₯ 3 charges to each of its neighbors of degree two in πΊβ. 11π‘ 11 πβ πΊβ Let be the new charge of after applying the above dischargingβ rule. We see that for π£ π πΊβ π π£ πβ π£ β₯ πβ π΅ πβ π£ β₯ any β ( ) with πΊβ ( )=3, ( ) 0. We next show that ( )βΆ= π£βπ΅ ( ) 0.Bythe π πΊβ π΅ πΎ πΎ π above claim, each component of [ ] is isomorphic to 1, 2,or 3. Thus each endpoint π π of (with the endpoint of 1 βcounted twice) receivesβ at least 3β11 charge from its neighbor π πΊβ π΅ πβ π πβ π£ β₯ π π£ β₯ in ( )β and so ( β)βΆ= π£βπ (π ) ( ) π£βπ (π ) ( ) + 3β11 + 3β11 β6β11 + 3β11 + πβ π΅ πβ π β₯ π πΊβ π΅ π 3β11 = 0. Hence ( )=β π βπΊβ[π΅] ( β) 0, where β [ ] denotes that is a component πΊβ π΅ πβ π£ πβ π£ πβ π΅ β₯ ofβ [ ]. We seeβ that π£βπ (πΊβ) ( )= π£βπ (πΊβ)βπ΅ ( )+ ( ) 0, contrary to the fact that πβ π£ π π£ < π£βπ (πΊβ) ( )= π£βπ (πΊβ) ( ) 0. Remark. Kerdjoudj, Kostochka, and Raspaud [3] considered the list version of star edge-colorings of simple graphs. They proved that every subcubic graph is star list-8-edge-colorable, and further proved the following. Theorem 4.1 ( [3]). Let πΊ be a subcubic simple graph.
(a) If πππ(πΊ) < 7β3, then πΊ is star list-5-edge-colorable. (b) If πππ(πΊ) < 5β2, then πΊ is star list-6-edge-colorable. LEI ET AL. 11
ACKNOWLEDGMENTS The authors would like to thank one anonymous referee for many helpful comments, in particular, πΎ for pointing out that the Heawood graph is star 5-edge-colorable and 4 with one subdivided edge has star chromatic index 6, and bringing Reference [3] to our attention. Yongtang Shi would like to thank Bojan Mohar for his helpful discussion and for mentioning the complexity problem on star edge- coloring during his visit to Simon Fraser University. The authors would like to thank Rong Luo for his helpful comments.
ORCID Zi-Xia Song http://orcid.org/0000-0001-6183-0110
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How to cite this article: Lei H, Shi Y, Song Z-X. Star chromatic index of subcubic multigraphs. J Graph Theory. 2017;00:1β11. https://doi.org/10.1002/jgt.22230