planar graphs without 4-cycles are acyclically 6-choosable

Planar Graphs Without4-Cycles Are Acyclically6-Choosable

Weifan Wang and Min Chen

DEPARTMENT OF MATHEMATICS, ZHEJIANGNORMAL UNIVERSITY, ZHEJIANG

JINHUA 321004, CHINAE-mail: [email protected]; [email protected]

Received June 5, 2007; Revised October 23, 2008

Published online 6 April 2009 in Wiley InterScience (www.interscience.wiley.com).DOI 10.1002/jgt.20381

Abstract: A proper vertex coloring of a graph G= (V,E) is acyclic if Gcontains no bicolored cycle. A graph G is acyclically L-list colorable if for agiven list assignment L={L(v)|v∈V}, there exists a proper acyclic coloring� of G such that �(v)∈L(v) for all v∈V. If G is acyclically L-list colorablefor any list assignment with |L(v)|≥k for all v∈V, then G is acyclicallyk-choosable. In this paper we prove that every planar graph G without4-cycles is acyclically 6-choosable. � 2009 Wiley Periodicals, Inc. J Graph Theory 61:

307–323, 2009

Keywords: acyclic coloring; choosability; planar graph; cycle

1. INTRODUCTION

Let G be a graph with the vertex set V (G) and the edge set E(G). A proper k-coloringof a graph G is a mapping � from V (G) to the set of colors {1,2, . . . ,k} such

Contract grant sponsor: NSFC; Contract grant number: 10771197.

Journal of Graph Theory� 2009 Wiley Periodicals, Inc.

307

308 JOURNAL OF GRAPH THEORY

that �(x) �=�(y) for every edge xy of G. A proper vertex coloring of a graph isacyclic if there is no bicolored cycle in G. The acyclic chromatic number, denotedby �a(G), of a graph G is the smallest integer k such that G has an acyclick-coloring.

The acyclic colorings of graphs were introduced by Grünbaum in [6] and studied byMitchem [9], Albertson and Berman [1], and Kostochka [7]. In 1979, Borodin [2] provedGrünbaum’s conjecture that every planar graph is acyclically 5-colorable. This result isbest possible in the sense that there exist infinite planar graphs G such that �a(G)=5.Such an example of 4-regular planar graph was first obtained by Grünbaum [6]. In 1976,Kostochka and Mel’nikov [8] further constructed examples of bipartite 2-degenerateplanar graphs of acyclic chromatic number 5. Borodin et al. [5] improved this boundfor planar graphs with a given girth by showing that every planar graph of girth atleast 7 is acyclically 3-colorable and every planar graph of girth at least 5 is acyclically4-colorable. A graph is 1-planar if it can be drawn on the plane in such a way thatevery edge crosses at most one other edge. It was shown in [4] that every 1-planargraph is acyclically 20-colorable.

We say that L is an assignment for the graph G if it assigns a list L(v) of possiblecolors to each vertex v of G. If G has a proper coloring � such that �(v)∈L(v) for allvertices v, then we say that G is L-colorable or � is an L-coloring of G. The graph Gis k-choosable if it is L-colorable for every assignment L satisfying |L(v)|≥k for allvertices v. A graph G is acyclically L-list colorable if for a given list assignment Lthere is an acyclic coloring � of the vertices such that �(v)∈L(v). If G is acyclicallyL-list colorable for any list assignment with |L(v)|≥k for all v∈V (G), then G isacyclically k-choosable. The acyclic list chromatic number of G, �la(G), is the smallestinteger k such that G is acyclically k-choosable.

Thomassen [13] proved that every planar graph is 5-choosable, whereas Voigt [14]presented an example of a planar graph which is not 4-choosable. Borodin et al. [3]first investigated the acyclical list coloring of planar graphs to prove that every planargraph is acyclically 7-choosable. They also put forward the following challengingconjecture:

Conjecture 1 (Borodin et al. [3]). Every planar graph is acyclically 5-choosable.

By studying the average degree of graphs, Montassier et al. [10] confirmedConjecture 1 for planar graphs of girth at least 5. Further, it was proved in [12] thatevery planar graph without 4- and 5-cycles or without 4- and 6-cycles is acyclically5-choosable. Other sufficient conditions for a planar graph to be acyclically 4-choosablewere established in [11].

The following weaker conjecture appeared in [12]:

Conjecture 2 (“Domaine de la Solitude 2000” Conjecture). Every planar graphwithout 4-cycles is acyclically 4-choosable.

In this paper, we prove the following result:

Theorem 1. Every planar graph without 4-cycles is acyclically 6-choosable.

Journal of Graph Theory DOI 10.1002/jgt

PLANAR GRAPHS WITHOUT 4-CYCLES 309

2. NOTATION

Only simple graphs are considered in this paper. A plane graph is a particular drawingof a planar graph in the Euclidean plane. For a plane graph G, we denote its face setby F(G). For x ∈V (G)∪F(G), let dG(x), or simply d(x), denote the degree of x inG. A k-vertex, k−-vertex, or k+-vertex is a vertex of degree k, at most k, or at leastk. Similarly, we can define k-face, k−-face, k+-face, in a similar way. We say thattwo cycles or faces are adjacent if they share at least one common edge, respectively.We use [u1u2 · · ·un] to denote a face f if u1,u2, . . . ,un are the boundary vertices off in clockwise order. Repeated occurrences of a vertex are allowed. The degree ofa face is the number of edge-steps in its boundary walk. Note that each cut-edge iscounted twice. Let NG(v) (or N (v)) denote the set of neighbors of a vertex v in G. Forx ∈V (G)∪F(G) and i≥2, let ni (x) denote the number of adjacent/incident i-verticesto x . For j ≥3, let m j (v) denote the number of j-faces incident to a vertex v.

We say that a 3-face f = [v1v2v3] is an (a1,a2,a3)-face if the degree of the vertexvi is ai for i=1,2,3. A 3-vertex v is light if it is incident to a 3-face. A 3-face is badif it is a (3,3,7+)-face. Let b(v) denote the number of bad 3-faces incident to a vertexv. If a vertex v is adjacent to a 3-vertex u such that the edge uv is not incident toany 3-face, then we call u a pendent 3-vertex of v. A pendent light 3-vertex is a lightand pendent 3-vertex. A bad pendent light 3-vertex is a pendent light 3-vertex whichis incident to a bad 3-face. Let p3(v) and p∗

3(v) denote the number of pendent light3-vertices and the number of bad pendent light 3-vertices of a vertex v, respectively.Thus, p∗

3(v)≤ p3(v).Suppose that f = [xyz] is a (4,4,6+)-face, i.e., d(x)=d(y)=4 and d(z)≥6. Let

f ′ = [xzuvw] be the adjacent face of f sharing a common edge xz with f . If d(u)=d(v)=3, we say that f ′ is a heavy 5-face.

3. STRUCTURAL PROPERTIES

The proof of Theorem 1 is proceeded by contradiction. We suppose that G is aminimal counterexample (i.e., with the smallest vertex number) to the theorem whichis embedded in the plane. Thus, G is connected. We first investigate the structuralproperties of G, then use Euler’s formula and the discharging technique to derive acontradiction.

Lemma 2. The minimal counterexample G to Theorem 1 satisfies the following:

(B1) G contains no 1-vertices.(B2) No 2-vertex is adjacent to a 5−-vertex.(B3) If a 3-vertex v is adjacent to a 3-vertex u, then v is not adjacent to a 5−-vertex

except u.(B4) Let v be a 4−-vertex. Then p3(v)=0.(B5) Let v be a 5−-vertex. Then p3(v)≤1.(B6) Let v be a 6-vertex. Then:

(B6.1) n2(v)≤1; moreover, if n2(v)=1, then p3(v)=0;(B6.2) p3(v)≤4; moreover, if p3(v)=4, then m3(v)=0.



The proof of Lemma 2 is similar to that of Lemma 1 in [12].

Lemma 3. Suppose that v∈V (G) is a 7-vertex. Then the following hold:

(A1) If 3≤n2(v)≤5, then n2(v)+ p3(v)≤6.(A2) If m3(v)=1 and n2(v)≥2, then n2(v)+ p3(v)≤4.(A3) If at least one of the following conditions holds, then b(v)=0:

(A3.1) n2(v)=4;(A3.2) n2(v)=3 and m3(v)=2;(A3.3) n2(v)=m3(v)=2 and p3(v)=1.

(A4) n2(v)≤5; moreover, if n2(v)=5, then p3(v)=0.(A5) If m3(v)=0 and p3(v)=2, then n2(v) �=4.

Proof. Let L be a list assignment such that |L(x)|=6 for all x ∈V (G). Given a7-vertex v, let v1,v2, . . . ,v7 denote the neighbors of v. If vi is a 2-vertex, we use uito denote the neighbor of vi different from v. If vi is a pendent light 3-vertex, we usexi and yi to denote the neighbors of vi different from v such that [vi xi yi ] is a 3-face.For simplicity, we write that n2=n2(v), p3= p3(v), m3=m3(v), etc. We make use ofcontradiction to show (A1)–(A4).

(A1) Suppose to the contrary that n2+ p3>6. Since d(v)=7, it follows easily thatn2+ p3=7. Without loss of generality, we assume that v1,v2, . . . ,vn2 are 2-verticesand vn2+1,vn2+2, . . . ,v7 are pendent light 3-vertices. By the minimality of G, G−{v,v1,v2, . . . ,v7} admits an acyclic L-coloring �. It is obvious that �(xi ) �=�(yi ) for alli≥n2+1. Let S={u1,u2, . . . ,un2 , xn2+1, yn2+1, . . . , x7, y7}. Note that |L(v)|=6 and|S|=n2+2×(7−n2)=14−n2≤11. Thus, there exists a color c∈L(v) which appearsat most once in the set S. We may color v with c, then color vi with a color differentfrom c,�(ui ) for i=1,2, . . . ,n2, and vi with a color different from c,�(xi ),�(yi ) fori=n2+1, . . . ,7.

(A2) Suppose that [vv6v7] is a 3-face. Assume to the contrary that n2+ p3≥5.It is easy to show that v6 and v7 cannot be 2-vertices or pendent light 3-verticesof v. So n2+ p3=5. Suppose that v1, . . . ,vn2 are 2-vertices and vn2+1, . . . ,v5 arependent light 3-vertices. By the minimality of G, G−{v,v1,v2, . . . ,v5} admitsan acyclic L-coloring �. Clearly, �(v6) �=�(v7). Let S1={u1,u2, . . . ,un2} andS2={xn2+1, yn2+1, . . . , x5, y5}. Since |S1∪S2|=|S1|+|S2|=n2+2×(5−n2)=10−n2≤8, there exists a color c∈L(v)\{�(v6),�(v7)} which is assigned to at most twoelements s, t ∈ S1∪S2. Without loss of generality, suppose �(s)=�(t)=c. We assignc to v.

If s, t ∈ S1, say s=ui and t=u j , we color vi with a∈L(vi )\{�(v6),�(v7),c} and v jwith a color in L(v j )\{�(v6),�(v7),c,a}.

If s∈ S1 and t ∈ S2, say s=ui and t= x j , we color vi with a∈L(vi )\{�(v6),�(v7),c}and v j with a color in L(v j )\{�(v6),�(v7),�(y j ),c,a}.

If s, t ∈ S2, say s= xi and t= x j , we color vi with a∈L(vi )\{�(v6),�(v7),�(yi ),c}and v j with a color in L(v j )\{�(v6),�(v7),�(y j ),c,a}.

Each of other uncolored vertices vi ’s is assigned a color different from that of itsneighbors.



(A3) Suppose to the contrary that b(v)≥1. Let [vv6v7] be a bad 3-face with d(v6)=d(v7)=3. Let z6∈N (v6)\{v,v7} and z7∈N (v7)\{v,v6}. We consider the followingthree cases:

(3.1) n2=4. Let d(vi )=2 for i=1,2,3,4. By the minimality of G, G−{v,v1,v2,v3,v4} has an acyclic L-coloring �. Thus, �(v6) �=�(v7). If �(v5) /∈{�(v6),�(v7)}, then there exists a color c∈L(v)\{�(v5),�(v6),�(v7)} appearing atmost once in the set {u1,u2,u3,u4}, say �(u1)=c. We color v with c, v1 with a colorin L(v1)\{c,�(v5),�(v6),�(v7)}, and vi , for i=2,3,4, with a color different from thatof its neighbors. If �(v5)∈{�(v6),�(v7)}, say �(v5)=�(v6), we recolor v6 with a colordifferent from that of v5,v7, z6, z7 and then reduce to the previous case.

(3.2) n2=3 and m3=2. Assume that [vv4v5] is another 3-face incident to v except[vv6v7], and d(v1)=d(v2)=d(v3)=2. We note that the two 3-faces incident to v

cannot share a common edge since G contains no 4-cycles. Let � be an acyclicL-coloring of G−{v,v1,v2,v3,v6,v7}. Then, �(v4) �=�(v5). There exists a color c∈L(v)\{�(v4),�(v5)} appearing at most once in the set {u1,u2,u3, z6, z7}. We color v

with c.If �(ui )=c for some 1≤ i≤3, we color vi with a color different from that of

v,v4,v5,v6 with a color different from that of v,v4,v5, z6, z7, and v7 with a colordifferent from that of v,v4,v5,v6, z7.

If �(z j )=c for j =6 or 7, we color v j with a color different from that ofv,v4,v5, z6, z7.

In every possible case, other uncolored vertex vi is assigned a color different fromthat of its neighbors.

(3.3) n2=m3=2 and p3=1. Assume that d(v1)=d(v2)=2, v3 is a pendent light3-vertex, and [vv4v5] is another 3-face incident to v except [vv6v7]. Let � be an acyclicL-coloring of G−{v,v1,v2,v3,v6,v7}. Then, �(v4) �=�(v5) and �(x3) �=�(y3). Thereexists c∈L(v)\{�(v4),�(v5)} which appears at most once in {u1,u2, x3, y3, z6, z7}. Wecolor v with c. If �(x3)=c, we color v3 with a color different from that of v,v4,v5, y3.Other discussion is similar to Case (3.2).

(A4) If n2≥6, the proof is similar to that of (C7) in [12]. Suppose that n2=5and p3≥1. Let d(vi )=2 for i=1,2, . . . ,5, and assume that v6 is a pendent light3-vertex. Let � be an acyclic L-coloring of G−{v,v1,v2, . . . ,v6}. We color v with acolor c∈L(v)\{�(v7)} which appears at most once in {u1,u2, . . . ,u5, x6, y6}. With thesimilar argument, we can give an acyclic coloring for all vertices v1,v2, . . . ,v6.

(A5) The proof is analogous to that of (A2). �

Lemma 4. Suppose that v∈V (G) is an 8-vertex. Then:

(Q1) n2(v)≤6.(Q2) If m3(v)=1, then n2(v)≤5.(Q3) If n2(v)+ p3(v)=8, then p3(v)≥4.

Proof. The proof is similar to that of (A1) or (A4) in Lemma 3. �

Lemma 5. Suppose that f = [v1v2v3] is a 3-face with d(v1)≤d(v2)≤d(v3). Then:

(F1) d(v1)≥3.(F2) If d(v1)=d(v2)=3, then d(v3)≥7.



(F3) If d(v1)=3, then v1 is adjacent to at least two 6+-vertices.(F4) If d(v1)=d(v2)=d(v3)=4, then m3(vi )=1 for each i=1,2,3.(F5) If d(v1)=d(v2)=4, d(v3)=5, and v3 is incident to another 3-face, then

p3(v3)=0.

Proof. Suppose that N (v1)\{v2,v3}={x1, . . . , xl}, N (v2)\{v1,v3}={y1, . . . , ys}, andN (v3)\{v1,v2}={z1, . . . , zk}. So, k≥s≥ l≥1.(F1) Obvious.(F2) The proof is similar to (C8.2) in Lemma 1 in [12].(F3) In this case, l=1. Let N (x1)\{v1}={u1,u2, . . . ,uq}. Suppose to the contrary

that at least two of v2,v3, x1 are of degree at most 5, which implies that d(v2)≤5.If d(v3)≤5, then s,k≤3 and G−v1 has an acyclic L-coloring �. Thus, �(v2) �=�(v3).

If �(x1) /∈{�(v2),�(v3)}, we properly color v1. If �(x1)=�(v2), we color v1 with a colordifferent from that of x1,v3, y1, . . . , ys . If �(x1)=�(v3), we have a similar argument.

If d(x1)≤5, i.e., s≤3 and q≤4, then G−v1 has an acyclic L-coloring �. If �(x1) �=�(v3), we have a similar discussion. Suppose that �(x1)=�(v3). Let L(v1)={1,2, . . . ,6}.If there exists a color c∈L(v1)\{�(v2),�(v3),�(u1),�(u2), . . . ,�(uq )}, then we colorv1 with c. Otherwise, we may suppose q=4 and �(ui )= i for i=1,2,3,4, �(v2)=5,and �(v3)=�(x1)=6. If L(x1) �= L(v1), then we recolor x1 with c∈L(x1)\L(v1), andcolor v1 with a color different from that of x1,v2,v3. If L(x1)= L(v1), then we recolorx1 with 5 and then reduce to the former case.

(F4) In this case, l=s=k=2. To derive a contradiction, we suppose, without lossof generality, that m3(v1)=2, that is, v1 is incident to another 3-face [v1x1x2]. LetL(v1)={1,2, . . . ,6}. By the minimality of G, G−v1 has an acyclic L-coloring �. Then�(v2) �=�(v3) and �(x1) �=�(x2). If the colors of v2,v3, x1, x2 are mutually distinct, weproperly color v1. Otherwise, we consider the following cases by symmetry:

Assume that �(x1)=�(v3) and �(x2) �=�(v2). We color v1 with a color inL(v1)\{�(x1),�(x2),�(v2),�(z1),�(z2)}.

Assume that �(x1)=�(v3)=1 and �(x2)=�(v2)=2. If there exists c∈L(v1)\{1,2,�(y1),�(y2),�(z1),�(z2)}, then we color v1 with c. Otherwise, we may suppose that�(y1)=3, �(y2)=4, �(z1)=5, and �(z2)=6. Then we recolor v2 with a color differentfrom 1,2,3,4 and then reduce to the previous case.

(F5) Assume to the contrary that v3 is incident to another 3-face [v3z1z2] and isadjacent to a pendent light 3-vertex z3 such that [z3u1u2] is a 3-face. Thus, l=s=2and k=3. Let L(v3)={1,2, . . . ,6} and � be an acyclic L-coloring to G−{v3, z3}. Wewrite �=|{�(v1),�(v2),�(z1),�(z2)}|, which is the number of distinct colors used onv1,v2, z1, z2. Obviously, 2≤�≤4, since �(v1) �=�(v2) and �(z1) �=�(z2).

If �=3, say �(v1)=�(z1)=1, �(v2)=2, and �(z2)=3, then there exists a colorc∈L(v3)\{�(v1),�(v2),�(z2),�(x1),�(x2)} which appears at most once on {u1,u2},e.g., �(u1)=c. We color v3 with c, and then color z3 with a color different fromc,�(u2),1,2,3.

Suppose that �=2, that is {�(v1),�(v2)}={�(z1),�(z2)}. If there exists a color c∈L(v3)\{�(v1),�(v2),�(x1),�(x2),�(y1),�(y2)}, then we color v3 with c, and then colorz3 with a color different from that of v3,u1,u2, z1, z2. Otherwise, we assume that�(v1)=�(z1)=1, �(v2)=�(z2)=2, {�(x1),�(x2)}={3,4}, and {�(y1),�(y2)}={5,6}.We can recolor v1 with 5 and go back to the former case.



Finally, we suppose that �=4, say �(v1)=1, �(v2)=2, �(z1)=3, �(z2)=4. Ifthere exists c∈{5,6}\{�(x1),�(x2)}, then we color v3 with c. Then we color z3 inthis way: when c /∈{�(u1),�(u2)}, we color properly z3. When c∈{�(u1),�(u2)}, say�(u1)=c, we color z3 with a color different from c,�(u2),2,3,4. If there exists c∈{5,6}\{�(y1),�(y2)}, we have a similar proof. Now we assume that {�(x1),�(x2)}={�(y1),�(y2)}={5,6}. We recolor v1 with a color different from 1,2,5,6 and reduceto the previous case. �

4. PROOF OF THEOREM 1

Let G be a counterexample to Theorem 1 with the least vertex number. Let L be a listassignment with |L(v)|=6 for all v∈V (G). Thus, G satisfies Lemmas 2–5.

Using Euler’s formula |V (G)|−|E(G)|+|F(G)|=2 and the relation∑

v∈V (G) d(v)=∑f ∈F(G) d( f )=2|E(G)|, we can derive the following identity:

∑v∈V (G)

(2d(v)−6)+ ∑f ∈F(G)

(d( f )−6)=−12. (1)

We define a weight function w by w(v)=2d(v)−6 for each vertex v∈V (G), andw( f )=d( f )−6 for each face f ∈F(G). It follows from identity (1) that the totalsum of weights is equal to −12. We design appropriate discharging rules and redis-tribute weights accordingly. Once the discharging is finished, a new weight func-tion w′ is produced. However, the total sum of weights is kept fixed when thedischarging is in process. Nevertheless, after the discharging is complete, the newweight function w′(x)≥0 for all x ∈V (G)∪F(G). This leads to the following obviouscontradiction:

0≤ ∑x∈V (G)∪F(G)

w′(x)= ∑x∈V (G)∪F(G)

w(x)=−12. (2)

Our discharging rules are as follows:(R1) Suppose that f = [v1v2v3] is a 3-face with d(v1)≤d(v2)≤d(v3). We use

(d(v1),d(v2),d(v3))→ (c1,c2,c3) to denote that the vertex vi gives to f the amount ofweight ci for i=1,2,3. For a boundary edge e of f , let fe denote the adjacent face off with e as a common edge. We set

(3,3,7+)→ ( 1730 ,1730 ,

2815 ),

(3,4,6+)→ ( 1115 ,23 ,

85 ),

(3,5,6+)→ ( 1115 ,5348 ,

9380 ),

(3,6+,6+)→ ( 1115 ,1715 ,

1715 ),

(4,4,4) or (5+,5+,5+)→ (1,1,1),



(4,4,5)→ ( 34 ,34 ,

32 ),

(4,5,5)→ ( 1724 ,5548 ,

5548 ),

(4,5,6+)→ ( 23 ,98 ,

2924 ),

(4,6+,6+)→ ( 23 ,76 ,

76 ),

(4,4,6+)→

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

( 23 ,23 ,

53 ) if fv1v3 and fv2v3 are heavy 5-faces,

( 23 ,1724 ,

138 ) if fv1v3 is a heavy 5-face but fv2v3 is not,

( 1724 ,23 ,

138 ) if fv2v3 is a heavy 5-face but fv1v3 is not,

( 1724 ,1724 ,

1912 ) if neither fv2v3 nor fv1v3 is a heavy 5-face.

(R2) Every 5-vertex v gives 34 to each adjacent pendent light 3-vertex.

(R3) Every 6+-vertex v gives 1 to each adjacent 2-vertex, 1730 to each bad pendent

light 3-vertex, and 1115 to each other pendent light 3-vertex.

(R4) Every 4+-vertex v gives 1/(5−n2( f )−n3( f )) to each incident 5-face f .For x, y∈V (G)∪F(G), let �(x→ y) denote the amount of weights transferred from

x to y according to the above rules. Let �in(x) (respectively, �out (x)) denote the totalsum of weights transferred into (respectively, out of) x .

Claim 1. Suppose that f = [v1v2 · · ·v5] is a 5-face. Let i ∈{1,2, . . . ,5}.(1) If 4≤d(vi )≤5, then �(vi → f )≤ 1

3 .(2) If d(vi )≥6, then �(vi → f )≤ 1

2 .(3) If d(vk)≥4 for k=1,2,3 and min{d(v1),d(v3)}≤5, then every 4+-vertex in the

boundary of f gives to f at most 14 .

Proof. (1) It suffices to show that n2( f )+n3( f )≤2 by (R4). Assume, w.l.o.g., thati=1. First, it follows from (B2) that neither v2 nor v5 is of degree 2. If both v2 andv5 are 4+-vertices, we are done. If d(v2)=d(v5)=3, then v3 and v4 are 4+-vertices by(B3). So suppose, w.l.o.g., that d(v2)=3 and d(v5)≥4. Again, it follows from (B2) or(B3) that v3 cannot be a 2-vertex or a 3-vertex. Thus, n2( f )+n3( f )≤2.

(2) (B2) and (B3) imply that f is incident to at most three 3−-vertices. Thus (2)holds by (R4).

(3) It suffices to prove that n2( f )+n3( f )≤1. We assume, w.l.o.g., that d(v3)≤5.By (B2), d(v4)≥3. If d(v4)≥4, we are done. If d(v4)=3, then d(v5)≥4 by (B3). Theproof of the claim is complete. �

Remark 1. If the boundary of a 5-face f contains two consecutive 4+-vertices, thenn2( f )+n3( f )≤2 by (B2) and (B3). Thus, every boundary 4+-vertex gives to f atmost 1

3 by (R4).



Since G contains no 4-cycles, G has no 4-faces and adjacent 3-faces. Thus:

Claim 2. For a vertex v∈V (G), m3(v)≤d(v)/2�.Let f ∈F(G). Then d( f ) �=4. The proof is divided into the following cases.

Case a. If d( f )≥6, then it is trivial that w′( f )=w( f )=d( f )−6≥0.

Case b. Let d( f )=3 and so w( f )=−3. Let f = [xyz] such that d(x)≤d(y)≤d(z).By (F1), d(x)≥3. If we can show that �in( f )≥3, then it is evident that w′( f )=w( f )+�in( f )≥−3+3=0. We have three subcases as follows:

(b1) Assume that d(x)=3. If d(y)=3, then (F2) asserts that d(z)≥7, i.e., f is a(3,3,7+)-face. Thus, by (R1), �in( f )= 17

30 ×2+ 2815 =3. If d(y)=4, then d(z)≥6 by

(F3), i.e., f is a (3,4,6+)-face. Thus, �in( f )= 1115 + 2

3 + 85 =3 by (R1). If d(y)=5, then

d(z)≥6 by (F3) and hence f is a (3,5,6+)-face. Consequently, �in( f )= 1115 + 53

48+ 9380=3

by (R1). If d(y)≥6, then f is a (3,6+,6+)-face and therefore �in( f )= 1115 + 17

15 ×2=3by (R1).

(b2) Assume that d(x)=4. If f is a (4,4,4)-face, then �in( f )=1×3=3 by (R1). If fis a (4,4,5)-face, then �in( f )= 3

4 ×2+ 32 =3. If f is a (4,5,5)-face, then �in( f )= 17

24 +5548 ×2=3. If f is a (4,5,6+)-face, then �in( f )= 2

3 + 98 + 29

24 =3. If f is a (4,6+,6+)-face, then �in( f )= 2

3 +2× 76 =3.

Now suppose that f is a (4,4,6+)-face, i.e., d(x)=d(y)=4 and d(z)≥6. If fxz andfyz are heavy 5-faces, then �in( f )= 5

3 +2× 23 =3 by (R1). If exactly one of fxz and

fyz is a heavy 5-face, then �in( f )= 23 + 17

24 + 138 =3. If neither fxz nor fyz is a heavy

5-face, then �in( f )= 1724 ×2+ 19

12 =3.(b3) Assume that d(x)≥5. As f is a (5+,5+,5+)-face, �in( f )=1×3=3 by (R1).

Case c. Let d( f )=5 and so w( f )=−1. It is easy to see by (B2) and (B3) that5−n2( f )−n3( f )≥2, thus w′( f )≥−1+1/(5−n2( f )−n3( f ))×(5−n2( f )−n3( f ))=0by (R4).

Let v∈V (G). Let v1,v2, . . . ,vd(v) denote the neighbors of v drawn in the plane inclockwise order. For 1≤ i≤d(v)−1, let fi denote the incident face of v with vvi andvvi+1 as two boundary edges, and fn the incident face of v with vvd(v) and vv1 as twoboundary edges. We see that d(v)≥2 by (B1).

Case 1. d(v)=2.

Then w(v)=−2. Note that v is adjacent to two 6+-vertices by (B2). Therefore,w′(v)≥−2+1×2=0 by (R3).

Case 2. d(v)=3.

We have that w(v)=0, and m3(v)≤1 by Claim 2. If m3(v)=0, then w′(v)=w(v)=0.If m3(v)=1, let f1= [v1v2v] be a 3-face. So v is a pendent light 3-vertex of v3. By(B4), d(v3)≥5. If [vv1v2] is bad, i.e., a (3,3,7+)-face, then d(v3)≥6 by (B3) andhence w′(v)≥− 17

30 + 1730 =0 by (R1) and (R3). Thus, assume that f1 is not bad. When



d(v3)≥6, w′(v)≥− 1115 + 11

15 =0 by (R1) and (R3). When d(v3)=5, w′(v)≥− 1115 + 3

4 =160 by (R1) and (R2).In what follows, we suppose that d(v)≥4. In order to show that w′(v)=w(v)−

�out (v)≥0, it suffices to prove that �out (v)≤w(v)=2d(v)−6.

Case 3. d(v)=4.

Thenw(v)=2, andm3(v)≤2 by Claim 2. We see that n2(v)=0 by (B2) and p3(v)=0by (B4). If m3(v)≤1, then �out (v)≤1+3× 1

3 =2 by (R1) and Claim 1. So we supposethat m3(v)=2, and let f1= [vv1v2] and f3= [vv3v4] be two 3-faces. By (F4), at mostone of v1,v2 is of degree 4, and at most one of v3,v4 is of degree 4. Thus, fi (i=1,3)is not a (4,4,4)-face and therefore gets from v at most 3

4 by (R1). Clearly, n3(v)≤2by (F2). We have some subcases below:

(3.1) Assume that n3(v)=0. Set �1=min{d(v1),d(v4)} and �2=min{d(v2),d(v3)}.If �1≤5 and �2≤5, then v gives at most 1

4 to each of f2, f4 by Claim 1. Thus,�out (v)≤2× 3

4 +2× 14 =2.

If �1≥6 and �2≥6, then both f1 and f3 are (4,6+,6+)-faces, thus �out (v)≤2× 23 +

2× 13 =2 by (R1) and Claim 1.

Now suppose, w.l.o.g., that �1≤5 and �2≥6. It follows that d(v2)≥6 and d(v3)≥6.Noting that �(v→ fi )≤ 17

24 for i=1,3 and �(v→ f4)≤ 14 , we have �out (v)≤2× 17

24 +13 + 1

4 =2.(3.2) Assume that n3(v)=1 and let d(v1)=3. We see that d(v2)≥6 by (F3) and

n4(v)≤1 by (F4). This implies that f1 is a (3,4,6+)-face, thus v gives 23 to f1.

If d(v3)≤5, then �(v→ f2)≤ 14 by Claim 1, thus �out (v)≤ 2

3 + 34 + 1

3 + 14 =2.

Suppose that d(v3)≥6. If d(v4)≥5, then �(v→ f3)≤ 23 , thus �out (v)≤2× 2

3 +2× 13 =

2. Assume that d(v4)=4, i.e., f3 is a (4,4,6+)-face. If d( f2) �=5, it is easy to see that�out (v)≤ 2

3 + 1724 + 1

3 = 4124 . So suppose that f2= [v3vv2xy] is a 5-face. If at least one of

x and y is a 4+-vertex, then �(v→ f2)≤ 14 by (R4), and hence �out (v)≤ 2

3 + 34 + 1

3 + 14 =

2. Thus, suppose that d(x)≤3 and d(y)≤3. By (B2), we further derive that d(x)=d(y)=3, i.e., f2 is a heavy 5-face adjacent to f3. Thus, v gives 2

3 to f3 by (R1) and�out (v)≤2× 2

3 +2× 13 =2.

(3.3) Assume that n3(v)=2. Then one of v1,v2 and one of v3,v4 are 3-verticesby (B3). Furthermore, (F3) implies that n4(v)=n5(v)=0, namely both f1 and f3 are(3,4,6+)-faces, thus each of them receives from v at most 2

3 by (R1). Consequently,�out (v)≤2× 2

3 +2× 13 =2.

Case 4. d(v)=5.

We see thatw(v)=4, n2(v)=0 by (B2), p3(v)≤1 by (B5), andm3(v)≤2 by Claim 2.If m3(v)=0, then �out (v)≤5× 1

3 + 34 =2 5

12 by (R2) and Claim 1.If m3(v)=1, then �out (v)≤4× 1

3 + 32 + 3

4 =3 712 by (R1), (R2), and Claim 1.

Now suppose that m3(v)=2, and f1= [vv1v2] and f3= [vv3v4] are two 3-faces. Ifv5 is not a pendent light 3-vertex of v, then �out (v)≤2× 3

2 +3× 13 =4 by (R1) and

Claim 1. Assume that v5 is a pendent light 3-vertex of v. Then v gives 34 to v5 by (R2).

Since p3(v)=1, f1 and f3 are not (4,4,5)-faces by (F5). Thus, v gives at most 5548 to

each of f1 and f3 by (R1).



If neither f1 nor f3 is a (4,5,5)-face, then �out (v)≤2× 98 +3× 1

3 + 34 =4 by (R1),

(R2), and Claim 1. Otherwise, we may assume that f1 is a (4,5,5)-face by symmetry.If d(v3)≥4, then f2 gets from v at most 1

4 by Claim 1. Therefore, �out (v)≤2× 5548 +

2× 13 + 1

4 + 34 =323

24 . If d(v3)=3, then f3 must be a (3,5,6+)-face, which yields that�out (v)≤ 55

48 + 5348 +3× 1

3 + 34 =4.

Case 5. d(v)=6.

We see that w(v)=6, m3(v)≤3 by Claim 2, n2(v)≤1 and if n2(v)=1 then p3(v)=0by (B6.1), p3(v)≤4 and if p3(v)=4 then m3(v)=0 by (B6.2). In view of the size ofm3(v), we divide the proof into four cases below:

(5.1) Suppose that m3(v)=0. If n2(v)=0, then �out (v)≤6× 12 +4× 11

15 =51415 by (R3)

and Claim 1. If n2(v)=1, then p3(v)=0 and thus �out (v)≤6× 12 +1=4.

(5.2) Suppose that m3(v)=1. Let f1= [vv1v2] be a 3-face. By (B6.2), p3(v)≤3.If n2(v)=1, then p3(v)=0 by (B6.1), which implies that �out (v)≤ 5

3 +5× 12 +1=51

6by (R1), (R3), and Claim 1. So suppose that n2(v)=0. We have the following twopossibilities:

First assume that f1 is a (4,4,6)-face. Then �(v→ fi )≤ 13 for i=2,6 by Remark 1.

If p3(v)≤2, then �out (v)≤ 53 +2× 1

3 +3× 12 +2× 11

15 =5 310 . Now suppose that p3(v)=3.

Then at least one of v3 and v6 is a pendent light 3-vertex of v, assuming that v3 issuch a vertex. If f2 is not a heavy 5-face, then �(v→ f1)≤ 13

8 by (R1), so �out (v)≤138 +2× 1

3 +3× 12 +3× 11

15 =5119120 . If f2 is a heavy 5-face, then v3 is incident to a

(3,3,7+)-face, i.e., v3 is a bad pendent light 3-vertex of v, which gets 1730 from v by

(R3). Consequently, �out (v)≤ 53 +2× 1

3 +3× 12 + 17

30 +2× 1115 =513

15 .Next assume that f1 is not a (4,4,6)-face. Then v gives to f1 at most 8

5 . If p3(v)≤2,then �out (v)≤ 8

5 +5× 12 +2× 11

15 =51730 . Suppose that p3(v)=3. If f1 is not a (3,4,6)-

face, then v gives to f1 at most 2924 and hence �out (v)≤ 29

24 +5× 12 +3× 11

15 =5109120 .

Suppose that f1 is a (3,4,6)-face. Note that at least one of f2 and f6 gets from v atmost 1

3 by Remark 1. If p∗3(v)≥2, then �out (v)≤ 8

5 + 13 +4× 1

2 +2× 1730 + 11

15 =5.8. Ifp∗3(v)≤1, then it is easy to show that v sends 1

2 to each of at most three incident faces,thus �out (v)≤ 8

5 +2× 13 +3× 1

2 +3× 1115 =529

30 .(5.3) Suppose that m3(v)=2. Let f1= [vv1v2] and f ′ be 3-faces, where f ′ ∈{ f3, f4}.

It follows that p3(v)≤2 and v gives at most 13 to at least two incident 5+-faces by

(B3), (B4), (F1), and Remark 1. If n2(v)=1, then p3(v)=0, and thus �out (v)≤2×53 +2× 1

3 +2× 12 +1=6. If p3(v)≤1, then �out (v)≤2× 5

3 +2× 12 +2× 1

3 + 1115 =511

15 . Sosuppose that n2(v)=0 and p3(v)=2. There are three possibilities as follows (up tosymmetry):

(5.3.1) Assume that f1 and f ′ are both (4,4,6)-faces.(5.3.1a) Let f ′ = f3= [vv3v4]. Both v5 and v6 are pendent light 3-vertices. By

Claim 1 and Remark 1, �(v→ f2)≤ 14 , �(v→ f4)≤ 1

3 , �(v→ f6)≤ 13 . Obviously, f2 is

not a heavy 5-face by (B3), which means that neither f1 nor f3 gets 53 from v by

(R1). If �(v→ f5)≤ 13 , then �out (v)≤2× 13

8 + 14 +3× 1

3 +2× 1115 =529

30 . If �(v→ f5)= 12 ,

then at least one of v5 and v6 is a bad pendent light 3-vertex of v. Thus, �out (v)≤2× 13

8 + 14 +2× 1

3 + 12 + 11

15 + 1730 =529

30 .



(5.3.1b) Let f ′ = f4= [vv4v5]. By Remark 1, �(v→ fi )≤ 13 for i=2,3,5,6. Both

v3 and v6 are pendent light 3-vertices of v. If at least one of v3 and v6 is bad, then�out (v)≤2× 5

3 +4× 13 + 11

15 + 1730 =529

30 . If neither v3 nor v6 is bad, then it is easy toderive that none of f2, f3, f5, f6 is a heavy 5-face. Thus, �(v→ fi )≤ 19

12 for i=1,4 by(R1) and �out (v)≤2× 19

12 +4× 13 +2× 11

15 =52930 .

(5.3.2) Assume that f1 is a (4,4,6)-face, but f ′ is not.(5.3.2a) Let f ′ = f3= [vv3v4]. Then �(v→ fi )≤ 1

3 by Remark 1 for i=2,6. Firstsuppose that f2 is a heavy 5-face, say f2= [v2vv3xy] such that d(v3)=d(x)=3. By (B3),we see that d(v4)≥6. Thus f3 is a (3,6,6+)-face and �(v→ f4)≤ 1

3 by Remark 1. Hence-forth, �out (v)≤ 5

3+ 1715+3× 1

3+ 12+2× 11

15=52330 .Next suppose that f2 is not a heavy 5-face.

(a1) Suppose that f6 is a heavy 5-face. This implies that v6 is bad. It follows that�(v→ f1)= 13

8 and �(v→v6)= 1730 by (R1) and (R3).

(a1.1) v5 is bad. If d(v4)≥4, then �(v→ f4)≤ 13 , so that �out (v)≤ 13

8 + 85 +3× 1

3 +12 +2× 17

30 =5103120 . If d(v4)=3, then d(v3)≥4 by (F2) and �(v→ f2)≤ 1

4 by Claim 1.Thus, �out (v)≤ 13

8 + 85 +2× 1

2 + 13 + 1

4 +2× 1730 =5113

120 .(a1.2) v5 is not bad. We see that �(v→ f5)≤ 1

3 since f5 is incident to exactly three4+-vertices. If �(v→ f4)≤ 1

3 , then �out (v)≤ 138 + 8

5 +4× 13 + 11

15 + 1730 =5103

120 . If �(v→f4)= 1

2 , it is easy to deduce that f3 is a (3,6,6+)-face and �(v→ f3)= 1715 . Thus,

�out (v)≤ 138 + 17

15 +3× 13 + 1

2 + 1115 + 17

30 =5 67120 .

(a2) Suppose that f6 is not a heavy 5-face. Then, �(v→ f1)= 1912 .

If �(v→vi )= 1730 for i=5,6, then �out (v)≤ 19

12 + 85 +2× 1

2 +2× 13 +2× 17

30 =55960 .

Assume that �(v→v5)= 1730 and �(v→v6)= 11

15 . If �(v→ f4)≤ 13 or �(v→ f5)≤ 1

3 , then�out (v)≤ 19

12+ 85+3× 1

3+ 12+ 17

30+ 1115=559

60 . So suppose that �(v→ f4)=�(v→ f5)= 12 .

It follows that d(v4)=3 and f3 is a (3,6,6+)-face. Thus, �out (v)≤ 1912 + 17

15 +2× 1

3+2× 12+ 17

30+ 1115=541

60 .If �(v→v5)= 11

15 and �(v→v6)= 1730 , the proof is similar to the previous case.

Assume that �(v→v5)=�(v→v6)= 1115 . It is easy to see that �(v→ f5)≤ 1

3 . If�(v→ f4)≤ 1

3 , then �out (v)≤ 1912 + 8

5 +4× 13 +2× 11

15 =55960 . If �(v→ f4)= 1

2 , then f3 isa (3,6,6+)-face, thus �out (v)≤ 19

12 + 1715 +3× 1

3 + 12 +2× 11

15 =54160 .

(5.3.2b) Let f ′ = f4= [vv4v5]. Note that �(v→ fi )≤ 13 for i=2,6 and for at least

one i ∈{3,5}.(b1) Assume that �(v→ f4)= 8

5 . Then f4 is a (3,4,6)-face.If �(v→ f1)= 5

3 , then both v3 and v6 are bad, therefore �out (v)≤ 53 + 8

5 +3× 13 + 1

2 +2× 17

30 =5.9.Suppose that �(v→ f1)= 13

8 . Without loss of generality, we assume that f2 is aheavy 5-face. This implies that v3 is bad. If v6 is bad, we have �out (v)≤ 13

8 + 85 +3×

13 + 1

2 +2× 1730 =5103

120 . If v6 is not bad, then it is easy to show that v gives at most 13 to

each of f3 and f5. Thus, �out (v)≤ 138 + 8

5 +4× 13 + 11

15 + 1730 =57

8 .Suppose that �(v→ f1)= 19

12 . If �(v→ fi )≤ 13 for i=3,5, then �out (v)≤ 19

12 + 85 +4×

13 +2× 11

15 =55960 . If exactly one of f3 and f5 gets 1

2 from v, then v3 or v6 is bad. Thus,�out (v)≤ 19

12 + 85 +3× 1

3 + 12 + 11

15 + 1730 =559

60 .



(b2) Assume that �(v→ f4)≤ 2924 . If �(v→ f1)= 5

3 , then both v3 and v6 are bad. Thus,�out (v)≤ 5

3 + 2924 +3× 1

3 + 12 +2× 17

30 =5 61120 . Otherwise, �out (v)≤ 13

8 + 2924 +3× 1

3 + 12 +

2× 1115 =5.8.

(5.3.3) Assume that neither f1 nor f ′ is a (4,4,6)-face.We note that v gives at most 8

5 to each incident 3-face, and at most 13 to at least

one incident face of degree greater than 3 by (F2) and Remark 1. If �(v→ f1)≤ 2924

and �(v→ f ′)≤ 2924 , then �out (v)≤2× 29

24 +3× 12 + 1

3 +2× 1115 =553

60 . Otherwise, we onlyneed to handle the following two possibilities:

(5.3.3.1) Suppose that �(v→ f1)= 85 and �(v→ f ′)≤ 29

24 . This means that f1 is a(3,4,6)-face. If at least two incident faces obtain from v at most 1

3 , then �out (v)≤85 + 29

24 +2× 13 +2× 1

2 +2× 1115 =5113

120 . Assume that v gives 12 to exactly three incident

5-faces. It implies that f ′ = f3= [vv3v4], d(v1)=d(v4)=3, and d(v2)=4. Since f6 isincident to three 3−-vertices, v6 must be bad. Thus, �out (v)≤ 8

5 + 2924 + 1

3 +3× 12 + 11

15 +1730 =5113

120 .(5.3.3.2) Suppose that �(v→ f1)=�(v→ f ′)= 8

5 . So both f1 and f ′ are (3,4,6)-faces. We consider two cases as follows:

(5.3.3.2a) Assume that f ′ = f3= [vv3v4]. By symmetry, we consider the followingsubcases:

(a1) Assume that d(v2)=d(v3)=3 and d(v1)=d(v4)=4. By Remark 1, �(v→ fi )≤ 13

for i=4,6 and �(v→ f2)≤ 13 by (B3) and (B4). If �(v→ f5)≤ 1

3 , we have �out (v)≤2×85 +4× 1

3 +2× 1115 =6. If �(v→ f5)= 1

2 , then at least one of v5 and v6 is bad, therefore�out (v)≤2× 8

5 +3× 13 + 1

2 + 1730 + 11

15 =6.(a2) Assume that d(v2)=d(v4)=3 and d(v1)=d(v3)=4. By Remark 1, �(v→ fi )≤

13 for i=2,6. If v gives at most 1

3 to each of f4 and f5, then �out (v)≤2× 85 +4× 1

3 +2× 11

15 =6. If �(v→ f4)= 12 and �(v→ f5)≤ 1

3 , then it is easy to derive that v5 is bad.Similarly, if �(v→ f5)= 1

2 and �(v→ f4)≤ 13 , then at least one of v5 and v6 is bad. In

these two cases, we have �out (v)≤2× 85 +3× 1

3 + 12 + 17

30 + 1115 =6. If �(v→ f4)=�(v→

f5)= 12 , then both v5 and v6 must be bad. Therefore, �out (v)≤2× 8

5 +2× 13 +2× 1

2 +2× 17

30 =6.

(a3) Assume that d(v1)=d(v4)=3 and d(v2)=d(v3)=4. (B3) asserts that �(v→f2)≤ 1

4 , and v1,v4 cannot be adjacent to any 3-vertex. This implies that at least oneof f4, f5, f6 gets from v at most 1

3 . If v5 and v6 are both bad, then �out (v)≤2×85 +2× 1

2 + 13 + 1

4 +2× 1730 =511

12 . If neither v5 nor v6 is bad, then �(v→ fi )≤ 13 for all

i=4,5,6, and �out (v)≤2× 85 +3× 1

3 + 14 +2× 11

15 =51112 . If exactly one of v5,v6 is bad,

then two of f4, f5, f6 get from v at most 13 each, thus �out (v)≤2× 8

5 + 12 +2× 1

3 + 14 +

1115 + 17

30 =51112 .

(5.3.3.2b) Assume that f ′ = f4= [vv4v5]. We assume that d(v1)=3 and d(v2)=4.By symmetry, it suffices to consider the following two subcases:

(b1) d(v5)=3 and d(v4)=4. Then �(v→ fi )≤ 13 for i=2,3. We again note that both

v1 and v5 cannot be adjacent to any 3-vertex. This implies that either f5 or f6 receivesat most 1

3 from v. If v6 is bad, then �out (v)≤2× 85 +3× 1

3 + 12 + 17

30 + 1115 =6. Otherwise,

�(v→ fi )≤ 13 for i=5,6, thus �out (v)≤2× 8

5 +4× 13 +2× 11

15 =6.



(b2) d(v4)=3 and d(v5)=4. We see that �(v→ fi )≤ 13 for i=2,5 by Remark 1.

If �(v→ f3)≤ 13 , then �(v→ f3)+�(v→v3)≤ 1

3 + 1115 = 16

15 . If �(v→ f3)= 12 , then v3

must be bad, thus �(v→ f3)+�(v→v3)= 12 + 17

30 = 1615 . Similarly, we always have �(v→

f6)+�(v→v6)≤ 1615 . Consequently, �out (v)≤2× 8

5 +2× 13 +2× 16

15 =6.(5.4) Suppose that m3(v)=3. Let f1= [vv1v2], f3= [vv3v4], and f5= [vv5v6] be

3-faces. Clearly, p3(v)=0 and n2(v)=0. Moreover, (F2) and Remark 1 imply that atleast two of f2, f4, f6 get from v at most 1

3 each. If �(v→ fi )≤ 13 for =2,4,6, then

�out (v)≤3× 53 +3× 1

3 =6. Now, w.l.o.g., assume that �(v→ f2)= 12 and �(v→ fi )≤ 1

3for i=4,6. Suppose that f2= [vv2xyv3]. Then v2,v3, and exactly one of x, y are3-vertices, e.g., d(x)=3. By (B3), d(v1)≥6 and hence f1 is a (3,6,6+)-face. Therefore,�out (v)≤2× 5

3 + 1715 +2× 1

3 + 12 =519

30 .

Case 6. d(v)≥7.

For simplicity, we again write that mi =mi (v), p3= p3(v), n2=n2(v), etc. Thefollowing useful fact follows directly from definitions:

Claim 3. n2+ p3+2m3≤d(v).

A 5-face f incident to v is called weak if v gives to f exactly 12 . Let m

′5 denote the

number of weak 5-faces incident to v.

Claim 4. If n2+ p3+2m3=d(v)−k, where k∈{0,1}, then m′5≤ p∗

3 +2k.

Proof. We first suppose that k=0, that is, n2+ p3+2m3=d(v). Let fi = [vvi xyvi+1]be a weak 5-face incident to v. Then fi is incident to two 3-vertices and one 3−-vertex. We assume, w.l.o.g., that d(y)=d(vi+1)=3 and 2≤d(vi )≤3. By (B4), fi+1=[vi+1vvi+2 . . .] cannot be a 3-face, i.e., vi+1 does not lie in the boundary of any 3-faceincident to v. Since n2+ p3+2m3=d(v), vi+1 is a bad pendent light 3-vertex. Thisshows that every weak 5-face incident to v must be incident to a bad pendent light3-vertex of v. Conversely, it is easy to see by Remark 1 that every bad pendent light3-vertex lies in the boundary of at most one weak 5-face. Thus, we conclude thatm′

5≤ p∗3 .

Next, suppose that k=1, i.e., n2+ p3+2m3=d(v)−1. There is a vertex vi that isneither a 2-vertex nor a pendent light 3-vertex and that does not lie in the boundaryof 3-faces incident to v. According to the previous proof, the total number of weak5-faces incident to v, different from fi−1 and fi , is at most p∗

3 , thus the numberof all weak 5-faces incident to v is at most p∗

3 +2. This completes the proof of theclaim. �

In view of Claim 4, the argument is divided into two cases.(6.1) n2+ p3+2m3=d(v)−k for k=0,1.Using (R1), (R3), and Claim 4, we have

�out (v)≤ 2815m3+n2+ 17

30 p∗3 + 11

15 (p3− p∗3)+ 1

2m′5+ 1

3 (m5−m′5)

= 2815m3+(d(v)−k− p3−2m3)+ 11

15 p3− 16 p

∗3 + 1

3m5+ 16m

′5



≤ d(v)−k− 215m3− 4

15 p3− 16 p

∗3 + 1

3 (d(v)−m3)+ 16 (p

∗3 +2k)

= 43d(v)− 2

3k− 715m3− 4

15 p3.

(6.1.1) k=0. Then n2+ p3+2m3=d(v) and

�out (v)≤ 43d(v)− 7

15m3− 415 p3. (∗)

(6.1.1.1) If d(v)≥9, then �out (v)≤ 43d(v)≤2d(v)−6=w(v) by (∗).

(6.1.1.2) Assume that d(v)=8. If m3≥2, then �out (v)≤ 43 ·8− 7

15 ·2=91115<

10=w(v).Suppose thatm3=1. If p3≥1, then �out (v)≤ 4

3 ·8− 715 − 4

15 =91415 . If p3=0, it follows

that n2=d(v)− p3−2m3=8−2=6, which contradicts (Q2).Suppose that m3=0. Then n2+ p3=8. By (Q3), p3≥4. Thus, �out (v)≤ 4

3 ×8− 415 ×

4=9.6.(6.1.1.3) Assume that d(v)=7. Then w(v)=8, and m3≤3 by Claim 2. If m3=3,

then �out (v)≤ 43 ×7− 7

15 ×3=71415 by (∗). Thus, assume that m3≤2.

(6.1.1.3a) Suppose that m3=0. Then n2+ p3=7. If p3≥5, then �out (v)≤ 43 ×7−

415 ×5=8 by (∗). If p3≤1, then n2≥6, contradicting (A4). If 2≤ p3≤4, then 3≤n2≤5,contradicting (A1).

(6.1.1.3b) Suppose that m3=1. Then n2+ p3=5. If p3≥4, then �out (v)≤ 43 ×7−

715 − 4

15 ×4=7.8 by (∗). If p3≤3, then n2≥2, contradicting (A2).(6.1.1.3c) Suppose that m3=2. Then n2+ p3=3.If p3≥2, then �out (v)≤ 4

3 ×7− 715 ×2− 4

15 ×2=71315 by (∗).

If p3=0, then n2=3. By (A3.2), v is not incident to any bad 3-face, i.e., b(v)=0.Note that no heavy 5-face is incident to v in this case, which implies that v sends atmost 8

5 to every incident 3-face by (R1). Furthermore, it is easy to observe that m′5=0

by Remark 1, thus �out (v)≤3×1+2× 85 +5× 1

3 =71315 .

If p3=1, then n2=2, and b(v)=0 by (A3.3). There are at least three incident5+-faces receiving at most 1

3 from v by Remark 1. Thus, �out (v)≤2+2× 85 +3× 1

3 +2× 1

2 + 1115 =714

15 .(6.1.2) k=1. Then n2+ p3+2m3=d(v)−1 and

�out (v)≤ 43d(v)− 2

3 − 715m3− 4

15 p3. (∗∗)

(6.1.2.1) If d(v)≥8, then �out (v)≤ 43d(v)− 2

3 ≤2d(v)−6=w(v) by (∗∗).(6.1.2.2) Suppose that d(v)=7. Then w(v)=8 and m3≤3. If m3=3, then �out (v)≤

43 ×7− 2

3 − 715 ×3=7 4

15 by (∗∗). Thus, assume that m3≤2.(6.1.2.2a) Suppose that m3=2. Then n2+ p3=7−1−2m2=2.If p3≥1, then �out (v)≤ 4

3 ×7− 23 − 7

15 ×2− 415 ×1=7 7

15 . If p3=0, then n2=2 andm′

5≤2 by Claim 4, thus �out (v)≤2×1+2× 2815 +2× 1

2 +3× 13 =711

15 .(6.1.2.2b) Suppose that m3=1. Then n2+ p3=7−1−2m2=4. If p3≥1, then

�out (v)≤ 43 ×7− 2

3 − 715 ×1− 4

15 ×1=71415 . If p3=0, then n2=4 and m′

5≤2 by Claim 4.By (A3.1), b(v)=0. Thus, �out (v)≤4×1+ 8

5 +2× 12 +4× 1

3 =71415 .



(6.1.2.2c) Suppose that m3=0. Then n2+ p3=7−1=6. If p3≥3, then �out (v)≤43 ×7− 2

3 − 415 ×3=713

15 by (∗∗). If p3≤1, then n2=5 or n2=6, contradicting (A4). Ifp3=2, then n2=4, contradicting (A5).(6.2) n2+ p3+2m3≤d(v)−2.Using (R1) and (R3), we obtain

�out (v)≤ 2815m3+n2+ 11

15 p3+ 12m5

≤ 2815m3+(d(v)−2− p3−2m3)+ 11

15 p3+ 12 (d(v)−m3)

= 32d(v)−2− 19

30m3− 415 p3 (∗∗∗)

(6.2.1) If d≥8, then �out (v)≤ 32d(v)−2≤2d(v)−6=w(v) by (∗∗∗).

(6.2.2) Suppose that d(v)=7. Then w(v)=8. If m3≥1, then �out (v)≤ 32 ×7−2−

1930 ×1=713

15 by (∗∗∗). So suppose that m3=0.If p3≥2, then �out (v)≤ 3

2 ×7−2− 415 ×2=729

30 by (∗∗∗).Assume that p3=1. Then n2≤7−2−1=4. If n2≤3, then �out (v)≤3+ 11

15 +7×12 =7 7

30 . If n2=4, then it is easy to show that m′5≤2, thus �out (v)≤4+ 11

15 +5× 12 +

2× 13 =7.9.

Assume that p3=0. Then n2≤7−2=5. If n2≤4, then �out (v)≤4+7× 12 =7.5.

Assume that n2=5. It is easy to see that m′5≤4 and therefore �out (v)≤5+4× 1

2 +3×13 =8.

ACKNOWLEDGMENTS

The authors would like to thank the referee for his/her valuable suggestions to improvethis work. This research was supported partially by NSFC (No. 10771197).

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planar graphs without 4-cycles are acyclically 6-choosable

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