Information Processing Letters 112 (2012) 767–771

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Information Processing Letters

www.elsevier.com/locate/ipl

The linear arboricity of planar graphs with maximum degree at least 5 ✩

Hong-Yu Chen a, Jian-Ming Qi b,∗a School of Sciences, Shanghai Institute of Technology, Shanghai, 201418, Chinab Department of Mathematics and Physics, Shanghai Dianji University, Shanghai, 201306, China

a r t i c l e i n f o a b s t r a c t

Article history:Received 12 March 2012Accepted 14 June 2012Available online 26 June 2012Communicated by Jinhui Xu

Keywords:Combinatorial problemsPlanar graphLinear arboricityCycle

Let G be a planar graph with maximum degree �(G) � 5. It is proved that la(G) = � �(G)2 �

if G has no intersecting 4-cycles and intersecting 5-cycles.© 2012 Elsevier B.V. All rights reserved.

1. Introduction

In this paper, all graphs are finite, simple and undi-rected, and we follow [3] for the terminologies and nota-tions not defined here. For a real number x, �x� is the leastinteger not less than x and �x� is the largest integer notlarger than x. Let G be a graph. We use V (G), E(G), �(G)

and δ(G) (or simply V , E , � and δ) to denote the vertexset, the edge set, the maximum degree and the minimumdegree of G , respectively. A k-, k+- or k−-vertex is a vertexof degree k, at least k, or at most k, respectively. A k-cycleis a cycle of length k. Two cycles are said to be intersectingif they are incident with a common vertex.

A linear forest is a graph in which each component is apath. A map ϕ from E(G) to {1,2, . . . , t} is called a t-linearcoloring if the induced subgraph of edges having the samecolor α is a linear forest for 1 � α � t . The linear arboricityla(G) of a graph G defined by Harary [9] is the minimumnumber t for which G has a t-linear coloring.

✩ This work was partially supported by National Natural Science Foun-dation of China (10971121) and the Key Development Project 12C104,12C401, 10XKJ01 from Shanghai Dianji University.

* Corresponding author.E-mail address: qijianming1981@gmail.com (J.-M. Qi).

0020-0190/$ – see front matter © 2012 Elsevier B.V. All rights reserved.http://dx.doi.org/10.1016/j.ipl.2012.06.007

Akiyama, Exoo, and Harary [1] conjectured that la(G) =��(G)+1

2 � for any regular graph G . The conjecture is equiv-alent to the following conjecture.

Conjecture A. For any graph G, ��(G)2 �� la(G) � ��(G)+1

2 �.

This conjecture was confirmed for planar graphs, see[12] and [16], and the linear arboricity has been deter-mined for some graphs, see [1,2,7,8,11,13,15]. Wu [12]proved that for a planar graph G with maximum degree�(G) � 13, la(G) = ��(G)

2 �. Recently, Cygan et al. [6] im-proved the result to �(G) � 9, and they posed the follow-ing conjecture.

Conjecture B. If G is a planar graph with �(G) � 5, thenla(G) = ��(G)

2 �.

Wu et al. [14] proved that if G is a planar graph with�(G) � 7 and without i-cycles for some i ∈ {4,5}, thenla(G) = ��(G)

2 �. In [17], it is proved that if G is a planargraph with �(G) � 5 and without 4-cycles, then la(G) =��(G)

2 �. Tan et al. [10] showed that if G is a planar graphwith �(G) � 5 and without intersecting 4-cycles and in-tersecting i-cycles for some i ∈ {3,6}. In this paper, wegeneralize these results and prove that if G is a planar

768 H.-Y. Chen, J.-M. Qi / Information Processing Letters 112 (2012) 767–771

graph with �(G) � 5 and without intersecting 4-cycles andintersecting 5-cycles, then la(G) = ��(G)

2 �.In the paper, we always assume that a planar graph G

has been embedded in the plane. Let G be a planar graphand F (G) be the face set of G . For f ∈ F (G), the degreeof f , denoted by d( f ), is the number of edges incidentwith it, where each cut-edge is counted twice. A k-, k+-or k−-face is a face of degree k, at least k, or at most k,respectively.

Given a t-linear coloring ϕ and a vertex v of G , wedenote by C i

ϕ(v) the set of colors appears i times at v ,

where i = 0,1,2. Then |C0ϕ(v)| + |C1

ϕ(v)| + |C2ϕ(v)| = t and

|C1ϕ(v)| + 2|C2

ϕ(v)| = d(v). Let x be a vertex of G , denoteϕ(x) = (ϕ(xy1), . . . ,ϕ(xyk)), where vertices y1, . . . , yk aredistinct neighbors of x. For any two vertices u and v ,let Cϕ(u, v) = C2

ϕ(u) ∪ C2ϕ(v) ∪ (C1

ϕ(u) ∩ C1ϕ(v)), that is,

Cϕ(u, v) is the set of colors that appear at least two timesat u and v . A monochromatic path is a path of whoseedges receive the same color. For two different edges e1and e2 of G , they are said to be in the same color compo-nent, denoted by e1 ↔ e2 if there is a monochromatic pathof G connecting them. Furthermore, if two ends of ei areknown, that is, ei = xi yi (i = 1,2), then x1 y1 ↔ x2 y2 de-notes more accurately that there is a monochromatic pathfrom x1 to y2 passing through the edges x1 y1 and x2 y2in G (that is, y1 and x2 are internal vertices in the path).Otherwise, we use x1 y1 � x2 y2 (or e1 � e2) to denote thatsuch monochromatic path passing through them does notexist. Note that x1 y1 ↔ x2 y2 and x1 y1 ↔ y2x2 are dif-ferent. Let xy be an edge of G , xy ↔ (v, i) denote thatx and v have a monochromatic path of color i betweenthem through y. (u, i) ↔ (v, i) denote that u and v have amonochromatic path of color i between them.

Let v be a vertex with d(v) = d, denote by f1, f2, . . . , fdthe faces incident with v in a clockwise order, and letv1, v2, . . . , vd be neighbors of v , where vi is incident withf i, f i+1, i = 1,2, . . . ,d − 1, and vd is incident with fdand f1.

2. Main result and its proof

Theorem 1. If G is a planar graph with �(G) � 5 and with-out intersecting 4-cycles and intersecting 5-cycles, then la(G) =��(G)

2 �.

Proof. According to [4], if G is a planar graph with�(G) � 7 and without adjacent 4-cycles, then la(G) =��(G)

2 �. According to [12] and [16], Conjecture A is truefor all planar graphs. Henceforth, to prove Theorem 1, weonly need to prove that a planar graph with �(G) = 6 andwithout intersecting 4-cycles and intersecting 5-cycles hasa 3-linear coloring. Let G = (V , E) be a minimal counterex-ample to the theorem in terms of the number of verticesand edges, and uv be an edge of G . Then G − uv has a3-linear coloring ϕ . We first show some known propertiesof G (see [5]).

(a) |Cϕ(u, v)| = 3.(b) If there is a color i such that i ∈ C1

ϕ(u) ∩ C1ϕ(v), then

(u, i) ↔ (v, i).

(c) For any uv ∈ E(G), dG(u) + dG(v) � 8.(d) δ(G) � 2.(e) Any two 3−-vertices are not adjacent.(f) Any 3-face is incident with three 4+-vertices, or at

least two 5+-vertices.(g) Any 5-vertex has no neighbors of degree 2.

Claim 1. (See [5].) Every vertex is adjacent to at most two2-vertices. Moreover, suppose that a vertex v is adjacent to two2-vertices x, y, let x′ and y′ be the other neighbors of x, y, re-spectively. Then x′v, y′v /∈ E(G).

Claim 2. (See [5].) If a vertex u is adjacent to two 2-verticesv, w and incident with a 3-face uxyu, then min{d(x),d(y)} � 4.

Claim 3. (See [5].) If a 5-vertex u is incident with a 3-face uvxusuch that d(v) = 3, then all neighbors of u except v are 4+-vertices.

Claim 4. A 4+-vertex v is incident with at most � d(v)2 � 3-faces.

By the choice of G , Claim 4 is obvious.

Claim 5. (See [10].) G has no (4,4,5−)-face.

Claim 6. G has no configurations depicted in Fig. 1(a)–(e) wherethe vertices marked by • have no other neighbors in G and in(a)–(c), d(v) = 6.

Proof of Claim 6. G has no configurations depicted inFig. 1(a)–(c) (see [5]).

Now, suppose that G has a configuration as depictedin Fig. 1(d). By the minimality of G , G ′ = G − u1 v has a3-linear coloring ϕ .

Suppose C2ϕ(v) �= ∅. Without loss of generality, assume

ϕ(v) = {1,1}. Then ϕ(u1) = {2,2,3,3} by (a) and (b). Thenwe can recolor u1u2 with 1, u2 v with ϕ(u1u2) and coloru1 v with 1. So ϕ is extendable to a 3-linear coloring of G ,a contradiction.

Suppose C2ϕ(v) = ∅. Without loss of generality, as-

sume ϕ(v) = {1,2} and ϕ(u3 v) = 1, ϕ(u2 v) = 2. Thenϕ(u1) = {1,2,3,3} and (v, i) ↔ (u1, i) for i = 1,2 by(a) and (b). If ϕ(u1u2) = 1, ϕ(u1u3) = 2, then {1,2} ∈C2

ϕ(u2), {1,2} ∈ C2ϕ(u3) since (v,1) ↔ (u1,1) and (v,2) ↔

(u1,2). Thus ϕ(u2) = ϕ(u3) = {1,1,2,2,3}. We can re-color u2 v with 3 and color u1 v with ϕ(u2 v). Hence, weobtain a 3-linear coloring of G , a contradiction. Supposeϕ(u1u2) = 2, ϕ(u1u3) = 1. There is one color i appearsone time at u2 since d(u2) = 5. Obviously, i �= 2. Thenrecolor u2 v with i and color u1 v with ϕ(u2 v). So, weobtain a 3-linear coloring of G , a contradiction. Supposeϕ(u1u2) = ϕ(u1u3) = 3. Assume i ∈ C1

ϕ(u2). Then i �= 2since (v,2) ↔ (u1,2). We can recolor u2 v with i and coloru1 v with ϕ(u2 v). Thus we can obtain a 3-linear coloringof G , a contradiction shows that G has no configuration asdepicted in Fig. 1(d).

H.-Y. Chen, J.-M. Qi / Information Processing Letters 112 (2012) 767–771 769

Fig. 1. The forbidden configurations in G .

Suppose that G has a configuration as depicted inFig. 1(e). By the minimality of G , G ′ = G −uv has a 3-linearcoloring ϕ . Without loss of generality, assume ϕ(vt) = 1.By (a) and (b), C0

ϕ(u) = ∅, C1ϕ(u) = {1} and vt ↔ (u,1).

If ϕ(uw) �= 1 and ϕ(uz) �= 1, then ϕ(w w ′) = ϕ(xw) = 1,otherwise, we can recolor uw with 1, and color uv withϕ(uw). Similarly, ϕ(zz′) = ϕ(zy) = 1. Since d(x) = 5, thereis one color i appears one time at x. If i �= 1, then we canrecolor xw with i, uw with 1 and color uv with ϕ(uw).So i = 1. Then ϕ(xu) �= 1. We can recolor xu with 1 andcolor uv with ϕ(xu). If one of ϕ(uw) and ϕ(uz) is 1, with-out loss of generality, assume ϕ(uw) = 1, then 1 ∈ C2

ϕ(w)

since vt ↔ (u,1), and 1 ∈ C2ϕ(z), 1 ∈ C2

ϕ(y), 1 ∈ C2ϕ(x), oth-

erwise, we can recolor uz or uy or ux with 1, and color uvwith ϕ(uz) or ϕ(uy) or ϕ(ux). So ϕ(yz) = 1 and there isone color i (i �= 1) appears one time at y since d(y) = 5.Then we can recolor yz with i, uz with 1, and color uvwith ϕ(uz). Thus ϕ is extendable to a 3-linear coloringof G , a contradiction. Hence G has no configuration as de-picted in Fig. 1(e). �

By Euler’s formula |V | − |E| + |F | = 2, we have∑v∈V

(2d(v) − 6

) +∑f ∈F

(d( f ) − 6

)

= −6(|V | − |E| + |F |) = −12 < 0.

We define the initial charge function ch(x) of x ∈ V ∪ Fto be ch(v) = 2d(v) − 6 if v ∈ V and ch( f ) = d( f ) − 6 if

f ∈ F . It follows that∑

x∈V ∪F ch(x) < 0. Now, we designappropriate discharging rules and redistribute weights ac-cordingly. Note that any discharging procedure preservesthe total charge of G . If we can define suitable dischargingrules to charge the initial charge function ch to the finalcharge function ch′ on V ∪ F , such that ch′(x) � 0 for allx ∈ V ∪ F , then we get an obvious contradiction.

For convenience, A k-face with consecutive verticesv1, v2, . . . , vk along its boundary in some direction is of-ten said to be (d(v1),d(v2), . . . ,d(vk))-face, and we use(d(v1),d(v2), . . . ,d(vk)) → (c1, c2, . . . , ck) to denote thatthe vertex vi sends the face the amount of charge ci fori = 1,2, . . . ,k. Our discharging rules are defined as follows.

R1. Every 2-vertex receives 1 from each of its incident 6+-vertices.

R2. Suppose that f = v1 v2 v3 is a 3-face.

(3−,5,5

) →(

0,3

2,

3

2

),

(3−,5,6+) →

(0,

6

5,

9

5

),

(3−,6+,6+) →

(0,

3

2,

3

2

),

(4,4,6+) →

(7,

7,

5)

,

8 8 4

770 H.-Y. Chen, J.-M. Qi / Information Processing Letters 112 (2012) 767–771

Fig. 2. The illustration of Case 1 and Case 2.

(4,5+,5+) →

(9

10,

21

20,

21

20

),

(5+,5+,5+) → (1,1,1).

R3. Suppose that f = v1 v2 v3 v4 is a 4-face.(3−,5+,3−,5+) → (0,1,0,1),(3−,5+,4+,5+) → (0,1,0,1),

(4+,4+,4+,4+) →

(1

2,

1

2,

1

2,

1

2

).

R4. Suppose that f = v1 v2 v3 v4 v5 is a 5-face.

(3−,5+,3−,5+,5+) →

(0,

1

3,0,

1

3,

1

3

),

(3−,5+,4+,4+,5+) →

(0,

1

4,

1

4,

1

4,

1

4

),

(4+,4+,4+,4+,4+) →

(1

5,

1

5,

1

5,

1

5,

1

5

).

R5. For a 6+-vertex v , v receives d( f )−6n from each of

its incident 7+-face f (n denotes the number of 6+-vertices incident with f ).

In the following, we will check ch′(x) � 0 for all x ∈V ∪ F . First, note that our discharging rules are just de-signed such that ch′( f ) � 0 for all f ∈ F and ch′(v) � 0for all 2-vertices v ∈ V . So we only need to check thatch′(v) � 0 for all 3+-vertices in G .

Let v be a 3+-vertex of G . If d(v) = 3, according to ourdischarging rules, no charge is discharged to or from v ,that is, ch′(v) = ch(v) = 0. If d(v) = 4, then v is inci-dent with at most two 3-faces by Claim 4. If v is inci-dent with two 3-faces, then it is not incident with 4-facesand at most one of the other two faces is a 5-face. Soch′(v) � ch(v) − 7

8 × 2 − 14 = 0 by R2 and R4. If v is inci-

dent with one 3-face, then at least one 6+-face is incidentwith v . So ch′(v) � ch(v)− 7

8 − 12 − 1

4 > 0 by R2, R3 and R4.If v is not incident with 3-faces, then v is incident with atleast two 6+-faces. So ch′(v) � ch(v) − 1

2 − 14 > 0 by R3

and R4.If d(v) = 5, then v is incident with at most three

3-faces by Claim 4. If v is incident with three 3-faces, thenit is not incident with 4-faces and at most one of the othertwo faces is a 5-face. Obviously, two of the three 3-facesare adjacent. Assume the two adjacent 3-faces are f1

and f2. If f1 and f2 are incident with a common 3-vertex,then by Claim 6(d), at most one of f1 and f2 is a (3,5,5)-face, and by Claim 3, the third 3-face is not incident witha 3-vertex. So ch′(v) � ch(v) − 3

2 − 2120 − 6

5 − 14 = 0. Other-

wise, at most one of the three 3-faces is incident with a3-vertex by Claim 3. So ch′(v) � ch(v)− 21

20 × 2 − 32 − 1

3 > 0by R2 and R4. If v is incident with two 3-faces, then vis incident with at least two 6+-faces. So ch′(v) � ch(v) −32 × 2 − 1 = 0 by R2, R3 and R4. If v is incident with one3-face, then v is incident with at least two 6+-faces. Soch′(v) � ch(v)− 3

2 − 1 − 13 > 0 by R2, R3 and R4. If v is not

incident with 3-faces, then v is incident with at most one4-face and at most one 5-face. So ch′(v) � ch(v)−1− 1

3 > 0by R3 and R4.

Suppose d(v) = 6. By Claim 1, v is adjacent to at mosttwo 2-vertices.

Suppose v is adjacent to two 2-vertices. Then v is inci-dent with at most two 3-faces by Claim 1 and the 3-facesare not incident with 3−-vertices by Claim 2. If v is in-cident with two 3-faces, then v is incident with at leasttwo 6+-faces. So ch′(v) � ch(v) − 2 − 5

4 × 2 − 1 − 13 > 0

by R1, R2, R3 and R4. If v is incident with one 3-face,then v is incident with at least three 6+-faces. So ch′(v) �ch(v) − 2 − 5

4 − 1 − 13 > 0 by R1, R2, R3 and R4. If v is not

incident with 3-faces, then ch′(v) � ch(v) − 2 − 1 − 13 > 0

by R1, R2, R3 and R4.Suppose v is adjacent to one 2-vertex. Then v is inci-

dent with at most three 3-faces by Claim 4. If v is incidentwith three 3-faces, then v is incident with at least two6+-faces and incident with no 4-faces, and it must be thefollowing cases (see Fig. 2).

Case 1. In this case, at most one of f3 and f5 is a (3,5,6)-face by Claim 6(e) and f6 is not a 5−-face, otherwise, thereare intersecting 4-cycles or intersecting 5-cycles. If f6 is a6-face, then f2 and f4 are 6+-faces. So ch′(v) � ch(v) −1 − 3

2 − max{ 95 + 3

2 , 32 × 2} > 0 by R1, R2 and R4. If f6 is

a 7+-face, then neither f2 nor f4 is a 4-face and at mostone of them is a 5-face. So ch′(v) � ch(v) + d( f6)−6

d( f6)−1 − 1 −32 − max{ 9

5 + 32 + 1

3 , 32 × 2 + 1

3 } > 0 by R1, R2, R4 and R5.

Case 2. In this case, at most two 3-faces of f1, f2, f4 areincident with 3−-vertices by Claim 6(a)(b)(c). So ch′(v) �ch(v) − 1 − max{ 9

5 × 2 + 2120 + 1

3 , 54 × 2 + 9

5 + 13 } > 0 by R1,

R2 and R4.

H.-Y. Chen, J.-M. Qi / Information Processing Letters 112 (2012) 767–771 771

If v is incident with at most two 3-faces, then v is in-cident with at least two 6+-faces. So ch′(v) � ch(v) − 1 −95 × 2 − 1 − 1

3 > 0.Suppose v is not adjacent to 2-vertices. Then v is in-

cident with at most three 3-faces by Claim 4. So ch′(v) �ch(v) − max{ 9

5 × 3 + 13 , 9

5 × 2 + 1 + 13 , 9

5 × 1 + 1 + 13 } > 0.

Hence we complete the proof of the theorem. �According to Theorem 1, we can draw a conclusion

which generalizes related known results.

Corollary. If G is a planar graph with �(G) � 5 and with-out intersecting 4-cycles and intersecting i-cycles for some i ∈{3,5,6}, then la(G) = ��(G)

2 �.

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