edge choosability of planar graphs without short cycles

Science in China Ser. A Mathematics 2005 Vol. 48 No. 11 1531—1544 1531

Edge choosability of planar graphs withoutshort cycles

WANG Weifan

School of Mathematics and Physics, Zhejiang Normal University, Jinhua 321004, China (email: [email protected])

Received September 28, 2004; revised June 10, 2005

Abstract In this paper we prove that if G is a planar graph with Δ = 5 and without 4-cycles

or 6-cycles, then G is edge-6-choosable. This consequence together with known results show that,

for each fixed k ∈ {3, 4, 5, 6}, a k-cycle-free planar graph G is edge-(Δ + 1)-choosable, where Δ

denotes the maximum degree of G.

Keywords: planar graph, coloring, choosability, cycle.

DOI: 10.1360/022004-049

1 Introduction

We only consider simple graphs in this paper unless otherwise stated. A planegraph is a particular drawing of a planar graph in the Euclidean plane. For aplane graph G, we denote its vertex set, edge set, face set, minimum degree,and maximum degree by V (G), E(G), F (G), δ(G), and Δ(G), respectively. Wesay that a graph G is k-cycle-free if G contains no cycles of a length k.

An edge-k-coloring of a graph G is a mapping φ from E(G) to the set of colors{1, 2, · · · , k} such that φ(x) �= φ(y) for any adjacent edges x and y of G. Thegraph G is edge-k-colorable if it has an edge-k-coloring. The chromatic indexχ′(G) is the smaller integer k such that G is edge-k-colorable. The mappingL is said to be an edge assignment for the graph G if it assigns a list L(e) ofpossible colors to each edge e of G. If G has some proper edge coloring φ suchthat φ(e) ∈ L(e) for all edges e, then we say that G is edge-L-colorable or φ isan edge-L-coloring of G. We call G edge-k-choosable if it is edge-L-colorablefor every edge assignment L satisfying |L(e)| = k for all edges e. The edgechoice number or list chromatic index χ′

l(G) of G is the smallest k such that Gis edge-k-choosable.

The well-known List Coloring Conjecture is stated as follows:

Conjecture 1. If G is a multigraph, then χ′l(G) = χ′(G).

The conjecture has been proved for a few special cases, such as bipartitemultigraphs[1], complete graphs of odd order[2], multicircuits[3], outerplanargraphs[4], and graphs with Δ(G) � 12 which can be embedded in a surfaceof nonnegative characteristic[5].

Copyright by Science in China Press 2005

1532 Science in China Ser. A Mathematics 2005 Vol. 48 No. 11 1531—1544

The following weaker conjecture on an edge list coloring was proposed byVizing (see ref. [6]).

Conjecture 2. Every graph G is edge-(Δ(G) + 1)-choosable.

An earlier result of Harris[7] shows that χ′l(G) � 2Δ(G)−2 if G is a graph with

Δ(G) � 3. This implies Conjecture 2 for the case Δ(G) = 3. In 1999, Juvan,Mohar, and Skrekovski[8] settled the case for Δ(G) = 4. Conjecture 2 has alsobeen confirmed for other special cases such as complete graphs[2], graphs withgirth at least 8Δ(G)(ln Δ(G) + 1.1)[6], and planar graphs with Δ(G) � 9[9].Wang and Lih[10] proved that a planar graph G with Δ(G) �= 5 and withouttwo 3-cycles sharing a common vertex is edge-(Δ(G) + 1)-choosable. Supposethat G is a planar graph without k-cycles for some fixed integer 3 � k � 6.Then it was shown that Conjecture 2 holds if G satisfies one of the followingconditions: (i) either k = 3, or k = 4 and Δ(G) �= 5[11]; (ii) k = 5[12]; (iii) k = 6and Δ(G) �= 5[13] .

In this paper, we will prove that a planar graph G with Δ(G) = 5 andwithout 4-cycles or 6-cycles is edge-6-choosable, which settles two unsolvedcases in refs. [11] and [13]. Hence the following theorem follows immediatelyfrom our conclusion and the known results.

Theorem 1. If G is a k-cycle-free planar graph for some fixed k ∈ {3, 4, 5, 6},then G is edge-(Δ(G) + 1)-choosable.

Let us introduce some notations and definitions. Let G be a plane graph.We use dG(x) (for short, d(x)) to denote the degree of a vertex or a face xof G. A vertex (or face) of degree k is called a k-vertex (or k-face). For avertex x ∈ V (G) (or a face x ∈ F (G)), let mk(x) denote the number of k-faces incident (or adjacent) to x for k � 3. Let n3(v) denote the number of3-vertices adjacent to v. A 3-vertex v is called bad if it and its three neighborsinduce a subgraph K4. For f ∈ F (G), we use b(f) to denote the boundary off and write f = [u1u2 · · ·un] if u1, u2, . . . , un are the boundary vertices of f ina clockwise order. A face f of degree n is called simple if b(f) forms a cycle oflength n. Obviously, every face of degree at most 5 is simple. We sometimesuse (d1, d2, · · · , dn) to represent a cycle (or a face) whose (boundary ) verticesare of degree sequence d1, d2, · · · , dn in a clockwise order in the graph G.

A subgraph C of G is called a cluster if C consists of a non-empty minimalset of 3-faces in G such that no other 3-face is adjacent to a member of thesets. Let t(C) denote the number of 3-faces in C. If t(C) = k, then C is calleda k-cluster. A face f and a cluster C are said to be adjacent to each other iff /∈ C and f is adjacent to a member of C. By definition, every 3-face belongsto exactly one of clusters of G.

Let H1 denote the graph consisting of a 4-cycle x1x2x3x4x1 with a chordx2x4. Let H2 denote the graph obtained from H1 by adding a pendant vertexyi at the vertex xi for each i = 1, 2, 3. Let H3 denote the graph by adding theedge x1x3 to H1, namely H3 is the complete graph K4. Let H4 denote a graph


Edge choosability of planar graphs without short cycles 1533

consisting of two 3-cycles xyzx and xvux with a unique common vertex x. LetH5 denote a graph obtained by adding a chord u2u6 to a 6-cycle u1u2 · · ·u6u1.The graphs H2, H4, and H5 are depicted in Fig. 1.

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Fig. 1

2 6-Cycle-free planar graphs

Lemma 2. Let G be a plane graph with Δ(G) = 5 and without 6-cycles.Then G contains one of the following configurations:

(A1) An edge xy with d(x) + d(y) � 7;

(A2) a (3, 5, 3, 5)-cycle or a (4, 4, 4, 4)-cycle;

(A3) a (4, 4, 4, 5)-cycle with a chord;

(A4) a subgraph H2 with d(y1) = d(y3) = 3 and d(x2) = d(x4) = 4;

(A5) a bad 3-vertex.

Proof. Suppose that the lemma is false. Let G be a counterexample withthe fewest vertices. Thus G is a connected plane graph with Δ(G) = 5, without6-cycles, and without (A1)-(A5). In particular, it follows from the lack of (A1)that δ(G) � 3. Moreover, the following configurations are excluded from G:

(C1) A simple 6-face;

(C2) a 5-face adjacent to a 3-face;

(C3) two adjacent 4-faces sharing a single edge;

(C4) a 4-face adjacent to two non-adjacent 3-faces;

(C5) a 4-face having only one common edge with two adjacent 3-faces;

(C6) a 3-face adjacent to three mutually non-adjacent 3-faces;

(C7) a 3-vertex incident to three 3-faces;

(C8) a 5-vertex incident to at least four 3-faces.

Euler’s formula |V (G)|− |E(G)|+ |F (G)| = 2 can be rewritten as (6|E(G)|−10|V (G)|)+(4|E(G)|−10|F (G)|) = −20.It follows from

∑

v∈V (G)

d(v) =∑

f∈F (G)

d(f) =

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2|E(G)| that∑

v∈V (G)

(3d(v) − 10) +∑

f∈F (G)

(2d(f) − 10) = −20. (1)

Let w denote the charge function defined on V (G)∪F (G) by w(v) = 3d(v)−10if v ∈ V (G) and w(f) = 2d(f) − 10 if f ∈ F (G). So the sum of the charges forall vertices and faces is −20. We are going to redistribute these charges, notchanging their sum, so that the new charge w′(x) becomes non-negative for allx ∈ V (G) ∪ F (G). Thus a contradiction is produced below and henceforth theproof is complete.

0 �∑

x∈V (G)∪F (G)

w′(x) =∑

x∈V (G)∪F (G)

w(x) = −20.

Our discharging rules are defined as follows:

(R1) Every face f of degree at least 7 sends w(f)/m3(f) to each adjacent3-face f ′ through each edge in b(f) ∩ b(f ′).

(R2) Each 4-vertex v sends 13

to each incident 4-face, and (2− 13m4(v))/m3(v)

to each incident 3-face.

(R3) Each 5-vertex v sends 13

to each adjacent 3-vertex, 1 to each incident4-face, and (2 − 1

3n3(v) − m4(v))/m3(v) to each incident 3-face.

Let τ(x → y) denote the amount transferred out of an element x into anotherelement y according to the above rules. We have some properties below.

(P1) If f is a face of degree at least 7 and f ′ is a 3-face adjacent to f , thenτ(f → f ′) � (2d(f) − 10)/d(f) = 2 − (10/d(f)) � 4

7by (R1). Moreover, if

m3(f) � d(f) − 1, then τ(f → f ′) � (2d(f) − 10)/(d(f) − 1) � 23.

(P2) Suppose that v is a 4-vertex and f is a 3-face incident to v. Thenw(v) = 2 and m3(v) � 4. If m3(v) = 4, then τ(v → f) � 1

2by (R2). If

m3(v) = 3, then m4(v) = 0 by (C4) and hence τ(v → f) � 23. If m3(v) = 2,

then m4(v) = 0 by (C4) and (C5), thus τ(v → f) � 1. If m3(v) = 1, thenm4(v) � 2 by (C3) and τ(v → f) � 2 − 2 · 1

3= 4

3by (R2).

(P3) Suppose that v is a 5-vertex and f is a 3-face incident to v. Thenw(v) = 5 and m3(v) � 3 by (C8). First assume that m3(v) = 3. Both (C4) and(C5) imply that m4(v) = 0. Since G contains no (A1) and (A2), v is adjacentto at most two 3-vertices. If n3(v) = 1, then τ(v → f) � (5 − 1

3)/3 = 14

9by

(R3). If n3(v) = 2, then τ(v → f) � (5 − 2 · 13)/3 = 13

9by (R3). Next assume

that m3(v) = 2. We see that m4(v) � 1 by (C3)–(C5) and n3(v) � 3. Thenτ(v → f) � (5−1−3 · 1

3)/2 = 3

2by (R3). Finally assume that m3(v) = 1. Then

m4(v) � 2 and n3(v) � 4 by (C3) so that τ(v → f) � 5− 2− 4 · 13

= 53

by (R3).

In order to check that the new charge function w′ is nonnegative on V (G) ∪F (G), we start with an arbitrary vertex v ∈ V (G). If d(v) = 3, then w(v) = −1.Since G contains no (A1), v is adjacent to three 5-vertices, each of which givesv 1

3by (R3). Thus w′(v) = −1 + 3 · 1

3= 0. If d(v) = 4, then it follows from



(C3) that m4(v) � 2 and thus v sends at most 23

to its incident 4-faces. Notingw(v) = 2, we have w′(v) � 0 by (R2). If d(v) = 5, then n3(v) � 5, andm4(v) � 2 by (C3). Therefore, (R3) and the fact that w(v)−m4(v)− 1

3n3(v) �

5 − 2 − 53

> 0 imply that w′(v) � 0.

Now let f ∈ F (G). If d(f) = 5, then w′(f) = w(f) = 0. If d(f) � 6, thenw(f) = 2d(f) − 10 � 2 and hence w′(f) � 0 by (R1). Suppose that d(f) = 4,then w(f) = −2. If f is incident to a 3-vertex u, then two neighbors of u inb(f) are of degree 5 by (A1), each of which sends 1 to f by (R3). Thereforew′(f) � −2 + 1 + 1 = 0. So suppose that every vertex in b(f) is of degree atleast 4. Since G contains no (A2), f is incident to at least one 5-vertex. By(R2) and (R3), w′(f) � −2 + 1 + 3 · 1

3= 0.

Suppose that d(f) = 3. Then w(f) = −4, and f is contained in a uniquecluster C. Instead of proving w′(f) � 0, we show

w′(C) =∑

f∈Cw′(f) = β(C) +

∑

f∈Cw(f) = β(C) − 4t(C) � 0,

equivalently β(C) � 4t(C), where β(C) represents the sum of charges transferredinto C from all its adjacent faces and incident vertices by (R1)–(R3). As a directconsequence of definition, every adjacent face of C is of degree at least 4. Thisfact will be frequently used in the following proof. Since G contains neitherbad 3-vertices nor 6-cycles, it follows that t(C) � 4. In the sequel, we use fe todenote the adjacent face of C sharing the common edge e with some face of C.

Assume that t(C) = 1, that is C consists of a 3-face f . Suppose that f =[x1x2x3] with d(x1) � d(x2) � d(x3). Then every adjacent face of f is ofdegree 4 or at least 7 by (C1) and (C2). First assume that d(x1) = 3. Thusd(x2) = d(x3) = 5 by (A1). If f is adjacent to at least two faces of degree atleast 7, then β(C) � 2 · 13

9+ 2 · 4

7= 4 2

63by (P1) and (P3). So assume that

f is adjacent to at least two 4-faces. Since x1 is a 3-vertex and G has no twoadjacent 4-faces, we may suppose that d(fx2x3) = d(fx3x1) = 4 and d(fx1x2) � 7.Since G contains no (C4), x3 is incident to the unique 3-face, i.e. f . As fx3x1

is incident to the unique 3-vertex x1 by (A2), x3 is adjacent to at most three3-vertices. We deduce that τ(x3 → f) � 2 and thus β(C) � 2 + 13

9+ 4

7= 4 1

63.

Next assume that d(x1) � 4. If f is adjacent to at least two faces of degree atleast 7, then β(C) � 3 · 1 + 2 · 4

7= 4 1

7. Otherwise, each of x1, x2, x3 gives f at

least 43

and hence β(C) � 3 · 43

= 4.

Assume that t(C) = 2. Let C = {[ux1x2], [ux2x3]} such that d(u) � d(x2). Wesee d(u) � 4 by (A1). Since G contains no 6-cycles, every face adjacent to C is ofdegree at least 7. First assume that d(u) = 5. If d(x2) = 3, then both x1 and x3

are 5-vertices by (A1) to make β(C) � 2 · 139

+ 139

+ 139

+4 · 47

= 8 463

. If d(x2) = 4,then d(x1), d(x3) � 4 and hence β(C) � 2 · 1 + 1 + 1 + 2 · 13

9+ 4 · 4

7= 9 11

63.

If d(x2) = 5, then at least one of x1, x3 is of degree at least 4 by (A2), andthus β(C) � 2 · 13

9+ 2 · 13

9+ 1 + 4 · 4

7= 9 4

63. Now suppose that d(u) = 4.

By (A1) and the assumption that d(u) � d(x2), we have d(x2) = 4. Again,since G has no (A1)–(A3), both x1 and x3 are 5-vertices. Therefore β(C) �

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2 · 1 + 2 · 1 + 139

+ 139

+ 4 · 47

= 9 1163

.

Assume that t(C) = 3. Let C = {[ux1x2], [ux2x3], [ux3x4]} such that d(x2) �d(x3). Since G contains no (A5), d(u) � 4 and x1 �= x4. Each face adjacent toC is of degree at least 7. We consider two cases, depending on the size of d(u).First assume that d(u) = 5. If d(x2) = 3, then d(x1) = d(x3) = 5 and d(x4) � 4by (A1) and (A2). Thus β(C) � 1 + 6 · 13

9+ 5 · 4

7= 12 11

21. If d(x2) = 4, then

d(x1), d(x3) � 4. When d(x3) = 4, we derive d(x1) = d(x4) = 5 by (A3) so thatβ(C) � 2+2+5· 13

9+5· 4

7= 14 5

63. When d(x3) = 5, β(C) � 2+1+5· 13

9+5· 4

7= 13 5

63.

If d(x2) = 5, then d(x3) = 5 and β(C) � 7 · 139

+ 5 · 47

= 12 6163

. Now assume thatd(u) = 4. It follows that d(xi) � 4 for all i = 1, 2, 3, 4, and at least one of d(x2)and d(x3) is equal to 5. Without loss of generality, we suppose d(x2) = 5. Ifd(x3) = 5, then β(C) � 4 · 13

9+ 1 + 1 + 2 + 5 · 4

7= 12 40

63. If d(x3) = 4, then

d(x1) = d(x4) = 5 by (A3), and β(C) � 2 + 2 + 4 · 139

+ 5 · 47

= 12 4063

.

Assume that t(C) = 4. In view of (C7) and (C8), we may suppose thatC = {[ux1x2], [ux2x3], [ux3x4], [ux4x1]} such that d(x1) = min

1�i�4{d(xi)}, as

shown in Fig. 2. It is easy to derive that d(x1) � 4 by (A1). If some xi isof degree 5, then it is obvious that n3(xi) � 2. When n3(xi) � 1, τ(xi →C) � 2 · (5 − 1

3) · 1

3= 28

9. When n3(xi) = 2, we derive m3(xi) = 2 and thus

τ(xi → C) � 5 − 2 · 13

= 133. Thus we always have τ(xi → C) � 28

9. So, if

d(x1) = 5, then β(C) � 2 + 4 · 289

+ 4 · 47

= 16 4663

. Assume that d(x1) = 4. Thend(xi) = 5 for i = 2, 3, 4 by (A3). Since G contains no (A4), at least one of x2, x4

is not adjacent to any 3-vertex, say x2. It follows that τ(x2 → C) � 2 · 53

= 103.

Moreover, since fx1x2 is adjacent to fx4x1 , we have m3(fx1x2) � d(fx1x2)− 1 andm3(fx4x1) � d(fx4x1) − 1. Thus τ(fx1x2 → C) � 2

3and τ(fx4x1 → C) � 2

3by

(P1), and therefore β(C) � 2 + 2 + 2 · 289

+ 103

+ 2 · 47

+ 2 · 46

= 16 263

.

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Lemma 3. Let L be an assignment of colors to the edges of H1 thatsatisfies |L(x2x4)| = 4, |L(x1x4)| = 3, and |L(x1x2)| = |L(x2x3)| = |L(x3x4)| =2. Then H1 is edge-L-colorable.

Proof. If L(x1x2)∩L(x3x4) �= ∅, we first color x1x2 and x3x4 with the samecolor from L(x1x2)∩L(x3x4). Then we color x2x3, x4x1, and x2x4 successively.If L(x2x3)∩L(x4x1) �= ∅, we give a similar coloring. If there exists c ∈ L(x3x4)\L(x2x4), we color x3x4 with c, then color x2x3, x1x2, x4x1, and x2x4 successively.



If L(x2x3)\L(x2x4) �= ∅, or L(x1x2)\L(x2x4) �= ∅, a coloring can be constructedsimilarly. So suppose that L(x1x2) ∪ L(x2x3) ∪ L(x3x4) ⊆ L(x2x4). SinceL(x4x1) ∩ L(x2x3) = ∅ and |L(x2x3)| + |L(x4x1)| = 5 > 4 = |L(x2x4)|, thereexists b ∈ L(x4x1) \L(x2x4), implying b /∈ L(x1x2). We color x4x1 with b, thencolor x3x4, x2x3, x1x2, and x2x4 successively.

Lemma 4. Let L be an assignment of colors to the edges of H1 that satisfies|L(x1x2)| = |L(x2x3)| = |L(x2x4)| = 3 and |L(x3x4)| = |L(x4x1)| = 2. Then H1

is edge-L-colorable.

Proof. If L(x3x4) = L(x4x1), we first color x2x4 with some color fromL(x2x4) \L(x3x4). Afterwards, every edge in the 4-cycle x1x2x3x4x1 admits atleast two colors to choose from and it can be colored properly. If L(x3x4) ∩L(x4x1) = ∅, we color L(x2x4) with a color from L(x2x4) \L(x3x4), then colorx4x1, x1x2, x2x3, and x3x4, successively. So we may suppose that L(x3x4) ={1, 2}, L(x4x1) = {1, 3}, and L(x2x4) = {1, 2, 3}. If there exists c ∈ L(x2x4) \L(x2x3), we color x2x4 with c, then color x3x4, x4x1, x1x2, and x2x3 successively.Thus assume that L(x2x3) = {1, 2, 3} and similarly L(x1x2) = {1, 2, 3}. Wecolor x2x3 and x4x1 with 1, x3x4 and x1x2 with 2, and x2x4 with 3.

Lemma 5. Let L be an assignment of colors to the edges of H2 that satisfies|L(x2x4)| = 5, |L(x1x2)| = |L(x2x3)| = 4, |L(x1x4)| = |L(x3x4)| = 3, and|L(x1y1)| = |L(x2y2)| = |L(x3y3)| = 2. Then H2 is edge-L-colorable.

Proof. If there exists a color a ∈ L(x1y1) ∩ L(x2y2), we color x1y1 andx2y2 with a, and x3y3 with b ∈ L(x3y3). Then the subgraph H2 − {y1, y2, y3} isisomorphic to H1. We define a reduced list L′ by L′(e) = L(e) \ {a} for eachedge e ∈ {x1x2, x1x4, x2x4}, L′(x3x4) = L(x3x4)\{b}, and L′(x2x3) = L(x2x3)\{a, b}. It is easy to check that |L′(x2x4)| � 4, |L′(x1x2)| � 3, and |L′(e)| � 2 fore ∈ {x1x4, x2x3, x3x4}. By Lemma 3, H2 − {y1, y2, y3} is edge-L′-colorable andhence H2 is edge-L-colorable. So we may suppose that L(x1y1) ∩ L(x2y2) = ∅,and similarly L(x3y3) ∩ L(x2y2) = ∅.

If there exists a ∈ L(x1y1) \ (L(x1x2) ∪ L(x1x4)), we first color x1y1 with a,then give a proper coloring for x2y2 and x3y3. Now the remaining subgraph isisomorphic to H1 with a reduced list L′ satisfying Lemma 3. So we supposethat L(x1y1) ⊆ L(x1x2)∩L(x1x4), and similarly, L(x3y3) ⊆ L(x2x3)∩L(x3x4),and L(x2y2) ⊆ L(x1x2) ∩ L(x2x3).

If L(x2y2) �⊆ L(x2x4), we color x2y2 with a color from L(x2y2)\L(x2x4), thencolor properly x1y1 and x3y3. Now each of the edges on the 4-cycle x1x2x3x4x1

has at least two colors to choose from and hence it can be colored properly.Finally we color x2x4 to establish an edge-L-coloring of H2. Henceforth wefurthermore suppose L(x2y2) ⊆ L(x2x4).

In view of the previous discussion, we assume that L(x1y1) = {1, 2}, L(x2y2) ={3, 4}, L(x1x2) = {1, 2, 3, 4}, 1, 2 ∈ L(x1x4), and 3, 4 ∈ L(x2x4). Obviously, weonly need to consider the following cases.

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Assume that L(x3y3) = {1, 2}. So we further put L(x2x3) = {1, 2, 3, 4}. Ifthere exists a ∈ L(x2x4) ∩ {1, 2}, we color x1y1, x3y3, x2x4 with a and x2y2

with 3. Since each of the edges on the 4-cycle x1x2x3x4x1 has at least twocolors to choose from and hence it can be colored properly. So suppose that1, 2 /∈ L(x2x4). Since 1, 2 ∈ L(x1x4) ∩ L(x3x4), we color x1x4 and x2x3 with1, x1y1 and x3y3 with 2, x2y2 with 3, then color properly x3x4, x1x2 and x2x4,successively.

Assume that L(x3y3) = {5, 6}. We derive that L(x2x3) = {3, 4, 5, 6} in thiscase. Since |L(x2x4)| = 5 and 3, 4 ∈ L(x2x4), at least one of 1,2,5,6 does notbelong to L(x2x4), say 1 /∈ L(x2x4). If 1 ∈ L(x3x4), we color x1x2 and x3x4

with 1, x1y1 with 2, x2y2 with 3, x2x3 with 4, x3y3 with 5, then color x1x4

and x2x4 successively. If 1 /∈ L(x3x4), then we color x1x4 with 1, x1y1 with 2,x2y2 with 3, x1x2 with 4, x2x3 with 5, x3y3 with 6, then color x3x4 and x2x4

successively.

Assume that L(x3y3) = {1, 5}. We see that L(x2x3) = {1, 3, 4, 5}, 1, 2 ∈L(x1x4), and 1, 5 ∈ L(x3x4). If 1 ∈ L(x2x4), we first color x1y1, x3y3, x2x4 with1, and x2y2 with 3. Afterwards the 4-cycle x1x2x3x4x1 can be colored properly.So assume that 1 /∈ L(x2x4). We color x1x4 and x2x3 with 1, x1y1 with 2, x2y2

with 3, x1x2 with 4, x3y3 with 5, then color x3x4 and x2x4 successively.

Lemma 6. Let L be an assignment of colors to the edges of H3 that satisfies|L(x1x4)| = |L(x2x4)| = |L(x3x4)| = 4, |L(x1x2)| = 3, and |L(x2x3)| =|L(x1x3)|= 2. Then H3 is edge-L-colorable.

Proof. If L(x1x3) �= L(x2x3), we color x1x3 with a color in L(x1x3) \L(x2x3). Then H3 − x1x3 is isomorphic to H1 with a reduced list L′ suchthat |L′(e)| � 3 for each e ∈ {x1x4, x2x4, x3x4} and |L′(e)| � 2 for each e ∈{x1x2, x2x3}. By Lemma 4, H3 − x1x3 is edge-L′-colorable and hence H3 isedge-L-colorable. So suppose L(x1x3) = L(x2x3). Since |L(x1x2)| = 3, we maycolor x1x2 with a color from L(x1x2) \ L(x1x3). Again, note that H3 − x1x2 isisomorphic to H1 with a reduced list L′ such that |L′(x3x4)| = 4, |L′(e)| � 3for e ∈ {x1x4, x2x4}, and |L(e)| = 2 for e ∈ {x1x3, x2x3}. Thus H3 − x1x2 isedge-L′-colorable by Lemma 3, and hence H3 is edge-L-colorable.

Lemma 7[8]. Every graph G with Δ(G) � 4 is edge-5-choosable.

To show that a planar graph G with Δ(G) = 5 and without 6-cycles is edge-6-choosable, we prove a stronger result as follows.

Theorem 8. If G is a planar graph with Δ(G) � 5 and without 6-cycles,then G is edge-6-choosable.

Proof. The proof is carried out by induction on |V (G)| + |E(G)|. Thetheorem holds trivially for |V (G)|+ |E(G)| � 5. Let G be a planar graph withΔ(G) � 5 and without 6-cycles such that |V (G)| + |E(G)| � 6. By Lemma 7,we may suppose that Δ(G) = 5. Let L be an edge assignment of G such that



|L(e)| = 6 for each e ∈ E(G). By Lemma 2, we consider five cases as follows.

Case 1. G contains an edge xy with d(x) + d(y) � 7. The inductionhypothesis implies that G − xy has an edge-L-coloring. Since there exist atmost five edges adjacent to xy and |L(xy)| = 6, xy can be colored properly.

Case 2. There is a 4-cycle C = x1x2x3x4x1 such that either d(xi) = 4for all i = 1, 2, 3, 4, or d(x1) = d(x3) = 3 and d(x2) = d(x4) = 5. Let Hbe the subgraph of G obtained by deleting the edges on C. By the inducionhypothesis, H has edge-L-coloring φ. Define an edge assignment L′ of C suchthat L′(e) = L(e) \ {φ(e′)|e′ ∈ E(H) is adjacent to e in G} for each edgee ∈ E(C). It is easy to inspect that |L′(e)| � 2. Thus C is edge-L′-colorableand hence G is edge-L-colorable.

Case 3. G contains a 4-cycle C = x1x2x3x4x1 with a chord e∗ such thatd(xi) = 4 for i = 1, 2, 3 and d(x4) = 5. Let H = G−(E(C)∪{e∗}) and note thatE(C) ∪ {e∗} induces a subgraph H1. By the induction hypothesis, H admitsan edge-L-coloring φ. For each edge e ∈ E(C) ∪ {e∗}, we define a reducedlist L′(e), similar to Case 2. If e∗ = x1x3, we derive that |L′(x1x3)| � 4,|L′(e)| � 3 for e ∈ {x1x2, x2x3}, and |L′(e)| � 2 for e ∈ {x1x4, x3x4}. ByLemma 3, E(C) ∪ {e∗} is edge-L′-colorable. If e∗ = x2x4, we have |L′(e)| � 3for e ∈ {x1x2, x2x3, x2x4}, and |L′(e)| � 2 for e ∈ {x1x4, x3x4}. By Lemma 4,E(C) ∪ {e∗} is edge-L′-colorable.

Case 4. G contains a subgraph H2 with d(y1) = d(y3) = 3 and d(x2) =d(x4) = 4. Let H = G − E(H2) and suppose that φ is an edge-L-coloringof H by the induction hypothesis. Again, we define a reduced list L′(e) foreach edge e ∈ E(H2). It is easy to check that |L′(x2x4)| � 5, |L′(e)| � 4for e ∈ {x1x2, x2x3}, |L′(e)| � 3 for e ∈ {x1x4, x3x4}, and |L′(e)| � 2 fore ∈ {x1y1, x2y2, x3y3}. By Lemma 5, E(H2) is edge-L′-colorable, thus G isedge-L-colorable.

Case 5. G contains a bad vertex v with three neighbors x, y, z such that{v, x, y, z} induces the subgraph K4, as shown in Fig. 3. We suppose d(x) =d(y) = d(z) = 5 since, otherwise, the argument can be reduced to Case 1.Let H = G − v and let φ denote an edge-L-coloring of H by the induc-tion hypothesis. For each t ∈ {x, y, z}, we define L′(vt) = L(vt) \ {φ(e)|e ∈E(H) is adjacent to vt in G}. So |L′(vt)| � 6 − 4 = 2. If either one of|L′(vx)|, |L′(vy)|, |L′(vz)| is greater or equal to 3, or L′(vx), L′(vy), L′(vz) arenot all identical, we can give a proper coloring for vx, vy, vz. Thus assume thatL′(vx) = L′(vy) = L′(vz) = {p, q}. We write C(x) = {φ(xx′)|x′ ∈ NG(x) \{v, y, z}}, C(y) = {φ(yy′)|y′ ∈ NG(y) \ {v, x, z}}, and C(z) = {φ(zz′)|z′ ∈NG(z) \ {v, x, y}}. Here NG(u) denotes the set of neighbors of a vertex u in G.It is evident that |C(x)| = |C(y)| = C(z) = 2.

If either C(x) ∩ C(y) �= ∅ or (C(x) ∪ C(y)) \ L(xy) �= ∅, we delete thecolors of edges xy, yz and zx and then define an edge assignment L′ as follows:L′(vt) = L(vt) \ C(t) for each t ∈ {x, y, z}, L′(st) = L(st) \ (C(s) ∪ C(t))

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for each edge st ∈ {xy, yz, zx}. It is easy to inspect that |L′(vt)| � 4 for allt ∈ {x, y, z}, |L′(xy)| � 6 − 2 − 2 + 1 = 3, |L′(yz)| � 2, and |L′(zx)| � 2. ByLemma 6, these uncolored edges can be colored properly. For other situations,we have a similar proof. Thus, without loss of generality, we may assume thatC(x) = {1, 2}, C(y) = {3, 4}, C(z) = {5, 6}, φ(xy) = a, φ(yz) = b, andφ(zx) = c. It follows that L(vx) = {1, 2, a, c, p, q}, L(vy) = {3, 4, a, b, p, q},L(vz) = {5, 6, b, c, p, q}. Furthermore we assume that L(xy) = {1, 2, 3, 4, a, a′},L(yz) = {3, 4, 5, 6, b, b′}, and L(zx) = {1, 2, 5, 6, c, c′}.

If a′ /∈ {b, c}, we recolor xy with a′, and color vx with a, vy with p, and vzwith q. Thus we suppose that a′ ∈ {b, c}, similarly, b′ ∈ {a, c}, and c′ ∈ {a, b}.We further suppose that a′ = b. (If a′ = c, we can give a similar proof.) Ifb′ = a, we color xy with b, yz and vx with a, vy with p, and vz with q. Soassume that b′ = c. If c′ = b, we color xz and vy with b, yz with c, vx with p,and vz with q. If c′ = a, we color xz and vy with a, xy and vz with b, and yzand vx with c. The proof of the theorem is complete.

3 4-Cycle-free planar graphs

Lemma 9. Let G be a plane graph with Δ(G) = 5 and without 4-cycles.Then G contains one of the following configurations:

(B1) An edge xy with d(x) + d(y) � 7;

(B2) a subgraph H4 with all vertices of degree 4 in G;

(B3) a subgraph H5 with either d(ui) = 4 for all i = 1, 2, · · · , 5, or d(u4) = 3and d(ui) = 4 for i = 1, 2, 6.

Proof. Suppose that the lemma is false. Let G be a counterexample withthe fewest vertices. Thus G is a connected plane graph with Δ(G) = 5, without4-cycles, and without (B1)–(B3). It is easy to derive that δ(G) � 3 by (B1).

Euler’s formula |V (G)|−|E(G)|+ |F (G)| = 2 can be rewritten as a new form:∑

v∈V (G)

(2d(v) − 6) +∑

f∈F (G)

(d(f) − 6) = −12. (2)

Define the charge function w by w(v) = 2d(v) − 6 if v ∈ V (G) and w(f) =d(f) − 6 if f ∈ F (G). Thus

∑{w(x)| x ∈ V (G) ∪ F (G)} = −12.

For a vertex v ∈ V (G) and for i = 4, 5, we use mi3(v) to denote the number

of (4, 4, i)-faces incident to v. We carry out the following discharging rules.

(r1) Every 4-vertex v gives 1 to each incident (4, 4, 4)-face, 34

to each incident(4, 4, 5)-face, (2 − m4

3(v) − 34m5

3(v))/m5(v) to each incident 5-face.

(r2) Every 5-vertex v gives 32

to each incident 3-face, and (4− 32m3(v))/m5(v)

to each incident 5-face.

Let w′ denote the resultant charge function. If we can show that w′(x) � 0for all x ∈ V (G) ∪ F (G), then a contradiction will be produced and the proof



is complete.

Let v ∈ V (G). Since G contains no 4-cycles, G contains no 4-faces and twoadjacent 3-faces. Thus m3(v) � �d(v)

2. If d(v) = 3, then w′(v) = w(v) = 0. If

d(v) = 4, then w(v) = 2 and m3(v) � 2. By (r1), we transfer, in total, at most2 from v to its incident 3-faces, thus w′(v) � 0. If d(v) = 5, then w(v) = 4 andm3(v) � 2. (r2) implies that v transfers at most 3 to its incident 3-faces, andhence w′(v) � 0.

Let f ∈ F (G). Then d(f) �= 4. Assume that d(f) = 3. Then w(f) = −3. Iff is incident to a 3-vertex, then two other boundary vertices of f are of degree5 by (B1). Thus w′(f) = −3 + 3

2+ 3

2= 0. Suppose that f is not incident

to any 3-vertex. If f is a (4, 4, 4)-face, then w′(f) = −3 + 3 · 1 = 0 by (r1).If f is a (4, 4, 5)-face, then w′(f) = −3 + 3

2+ 2 · 3

4= 0 by (r1) and (r2). If

f is a (4, 5, 5)-face, then w′(f) = −3 + 2 · 32

= 0. If f is a (5, 5, 5)-face, thenw′(f) = −3 + 3 · 3

2= 3

2.

Assume that d(f) = 5. Then w(f) = −1. Let v be a boundary vertex of f . Ifd(v) = 5 and m3(v) � 1, then τ(v → f) � 5

8by (r2). If d(v) = 5 and m3(v) = 2,

then τ(v → f) � 13

by (r2). If d(v) = 4 and m3(v) = 0, then τ(v → f) � 12

by (r1). If d(v) = 4 and m3(v) = 1, then τ(v → f) � 13. If d(v) = 4 and v is

incident to a (4, 4, 4)-face and a (4, 4, 5)-face, then τ(v → f) � 18. If d(v) = 4

and v is incident to at least one (4, 5, 5)-face, then τ(v → f) � 12. If d(v) = 4

and v is incident to two (4, 4, 5)-faces, then τ(v → f) � 14.

If f is incident to at least three 5-vertices, then w′(f) � −1 + 3 · 13

= 0 bythe above analysis. From now on, suppose that f is incident to at most two5-vertices. This implies that f is incident to at most one 3-vertex by (B1). Wewrite f = [y1y2y3y4y5].

If f is incident to a 3-vertex, say y1, then d(y2) = d(y5) = 5 and d(y3) =d(y4) = 4. If m3(y3) � 1 or m3(y4) � 1, then w′(f) � −1+3 · 1

3= 0. Otherwise,

it follows that f is adjacent to a 3-face [y3y4z]. Since G contains no 4-cycles,z /∈ b(f). Since G contains no (B3), it follows that d(z) = 5 and τ(yi → f) � 1

4

for i = 3, 4. Thus w′(f) � −1 + 2 · 13

+ 2 · 14

= 16.

Suppose that d(yi) = 4 for all i = 1, 2, . . . , 5. If f is adjacent to at most three3-faces, then f is incident to at least three vertices each of which is incident toat most one 3-face. Hence w′(f) � −1 + 3 · 1

3= 0. If f is adjacent to exactly

four 3-faces, e.g., [y1y2z1], [y2y3z2], [y3y4z3], and [y4y5z4], then each of y1 andy5 is incident to one 3-face. We assert that d(zi) = 5 for all i = 1, 2, 3, 4 since,otherwise, G would contain a subgraph satisfying (B3). Thus τ(yi → f) � 1

3

for i = 1, 5, τ(yj → f) � 14

for j = 2, 3, 4, and w′(f) � −1 + 2 · 13

+ 3 · 14

= 512

.If f is adjacent to five 3-faces, then each yi is incident to two (4, 4, 5)-faces andthus w′(f) � −1 + 5 · 1

4= 1

4.

Suppose that d(y1) = 5 and d(yi) = 4 for i = 2, 3, 4, 5. First assume thatm3(y1) � 1, which implies that m3(y2) � 1 or m3(y5) � 1. It is easy to seethat τ(y1 → f) � 5

8, hence w′(f) � −1 + 5

8+ 1

3+ 1

8= 1

12. Next assume that

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m3(y1) = 2. It follows that at least one of the edges y1y2 or y5y1 lies on theboundary of some 3-face. Without loss of generality, suppose that f is adjacentto a 3-face [y1y2z1]. Since G contains no (B3), d(z1) = 5, which implies thatτ(y2 → f) � 1

2. When m3(y5) � 1, we have w′(f) � −1 + 1

2+ 2 · 1

3= 1

6.

When m3(y5) = 2, f is adjacent to a 3-face [y5y1z2]. Then d(z2) = 5 andτ(y5 → f) � 1

2, so w′(f) � −1 + 1

3+ 2 · 1

2= 1

3.

Suppose that d(y1) = d(y2) = 5 and d(yi) = 4 for i = 3, 4, 5. If m3(yi) � 1for some i ∈ {3, 4, 5}, then w′(f) � −1+3 · 1

3= 0. Suppose that m3(yi) = 2 for

all i = 3, 4, 5. Thus f is adjacent to four 3-faces [y2y3z1], [y3y4z2], [y4y5z3], and[y5y1z4]. Since G contains no (B2), one of z2 and z3 is of degree 5. Suppopsethat z2 is such a vertex. Thus τ(y3 → f) � 1

4, τ(y4 → f) � 1

8, and w′(f) �

−1 + 14

+ 18

+ 2 · 13

= 124

.

Suppose that d(y1) = d(y3) = 5 and d(yi) = 4 for i = 2, 4, 5. It is easy to seethat τ(yi → f) � 1

8for all i = 2, 4, 5, and thus w′(f) � −1 + 2 · 1

3+ 3 · 1

8= 1

24.

Lemma 10. Let L be an assignment of colors to the edges of H5 such that|L(e)| = 3 for e ∈ {u1u2, u2u3, u2u6} and |L(e)| = 2 for all other edges e. ThenH5 is edge-L-colorable.

Proof. If there is some color a ∈ L(u2u6) \ (L(u5u6) ∪ L(u6u1)), we as-sign a to u2u6. Note that H5 − u2u6 forms a 6-cycle u1u2 · · ·u6u1 such thateach edge has a reduced list of colors of length at least 2. Thus this cycle canbe edge-colored properly. Suppose that L(u2u6) ⊆ L(u5u6) ∪ L(u6u1). Since|L(u2u6)| = 3 and |L(u5u6)| = |L(u6u1)| = 2, we deduce L(u5u6) �= L(u6u1). IfL(u2u6) �= L(u1u2), we color u2u6 with a color a ∈ L(u2u6)\L(u1u2), then coloru6u1, u5u6, u4u5, u3u4, u2u3, and u1u2, successively. If L(u2u6) �= L(u2u3), wecolor u2u6 with a color a ∈ L(u2u6)\L(u2u3), then color u6u1, u1u2, u5u6, u4u5, u3u4

and u2u3, successively. The coloring is available since L(u6u1)\{a} �= L(u5u6)\{a}.

Now we may assume that L(u2u6) = L(u1u2) = L(u2u3) = {1, 2, 3}. Weconsider two subcases as follows.

Assume that L(u6u1)∪L(u5u6) = {1, 2, 3}, say L(u5u6) = {1, 2} and L(u6u1) ={2, 3}. If 1 ∈ L(u3u4), we color u1u2, u3u4 and u5u6 with 1, u6u1 and u2u3 with2, u2u6 with 3, and u4u5 with a color a ∈ L(u4u5) \ {1}. If 2 ∈ L(u3u4),we have a similar coloring. So suppose 1, 2 /∈ L(u3u4). This implies thatthere is a color a ∈ L(u3u4) \ {1, 2, 3}. We color u3u4 with a, then coloru4u5, u5u6, u6u1, u2u6, u1u2 and u2u3 successively.

Assume that L(u6u1)∪L(u5u6) �= {1, 2, 3}. If L(u5u6) = {1, 2} and L(u6u1) ={3, 4}, we color u6u1 with 4, then color u5u6, u4u5, u3u4, u2u3, u2u6 and u1u2,successively. If L(u6u1) = {1, 2} and L(u5u6) = {3, 4}, we color u6u1 and u2u3



with 1, then color u3u4, u4u5, u5u6, u2u6 and u1u2, successively.

Lemma 11. Let L be an assignment of colors to the edges of H5 such that|L(u2u6)|=4, |L(u1u2)|= |L(u6u1)|= 3, and |L(e)| = 2 for all other edges e. Then H5 isedge-L-colorable.

Proof. If there exists a color a ∈ L(u2u6) \ (L(u2u3) ∪ L(u5u6)), we coloru2u6 with a. Note that every edge e in the 6-cycle C = u1u2 · · ·u6u1 ad-mits at least two colors to choose from. Thus E(C) can be colored prop-erly. If there exists b ∈ L(u2u3) \ L(u1u2), we color u2u3 with b, then coloru3u4, u4u5, u5u6, u2u6, u6u1 and u1u2, successively. If L(u5u6)\L(u6u1) �= ∅, wehave a similar proof. Thus we may suppose that L(u2u6) = L(u2u3)∪L(u5u6),L(u2u3) ∩ L(u5u6) = ∅, L(u2u3) ⊆ L(u1u2), and L(u5u6) ⊆ L(u6u1). Thesefacts imply that L(u1u2) �= L(u6u1). Let us color u1u2 with a color fromL(u1u2)\L(u6u1), then color u2u3, u3u4, u4u5, u5u6, u2u6 and u6u1, successively.

Theorem 12. If G is a planar graph with Δ(G) � 5 and without 4-cycles,then G is edge-6-choosable.

Proof. The proof is carried out by induction on |V (G)| + |E(G)|. Thetheorem holds trivially for |V (G)|+ |E(G)| � 5. Let G be a planar graph withΔ(G) � 5 and without 4-cycles such that |V (G)|+|E(G)| � 6. We may supposeΔ(G) = 5 by Lemma 7. Let L be an edge assignment of G such that |L(e)| = 6for each e ∈ E(G). According to Lemma 9, we need to handle three cases.

If G contains (B1), the proof is similar to Case 1 in Theorem 8.

If G contains (B2), that is a subgraph H4 such that d(x) = d(y) = d(z) =d(u) = d(v) = 4, then H = G−E(H4) has an edge-L-coloring φ by the inductionhypothesis. For each edge e ∈ E(H4), we define a list L′(e) = L(e) \ {φ(e′)|e′ ∈E(H) is adjacent to e in G}. It is easy to check that |L′(e)| � 4 for e ∈{xu, xv, xy,

xz} and |L′(e)| � 2 for e ∈ {uv, yz}. If |L′(yz)| � 3, we can color uv, xu, xv, xy, xz

and yz successively. If |L′(yz)| = 2, we can color xy with a color from L′(xy) \L′(yz), then color xu, uv, xv, xz, and yz successively.

Suppose now that G contains (B3), that is a subgraph H5 such that eitherd(ui) = 4 for all i = 1, 2, · · · , 5, or d(u4) = 3 and d(ui) = 4 for i = 1, 2, 6.Let H = G − E(H5) and let φ be an edge-L-coloring of H by the inductionassumption. For each edge e ∈ E(H5), we define similarly a reduced list L′(e).If d(ui) = 4 for all i = 1, 2, · · · , 5, then |L′(e)| � 3 for e ∈ {u1u2, u2u3, u2u6}

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and |L′(e)| � 2 for e ∈ {u3u4, u4u5, u5u6, u6u1}. By Lemma 10, E(H5) is edge-L′-colorable and hence G is edge-L-colorable. If d(u4) = 3 and d(ui) = 4 fori = 1, 2, 6, then |L′(e)| = 2 for all e ∈ {u2u3, u3u4, u4u5, u5u6}, |L(u1u2)| =|L(u6u1)| = 3, and |L(u2u6)| = 4. Lemma 11 asserts that E(H5) is edge-L′-colorable. Thus G is edge-L-colorable.

Acknowledgements This work was supported partially by the National Natural Science Founda-

tion of China (Grant No.10471131) and the Natural Science Foundation of Zhejiang Province (Grant

No. M103094)

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