Steinberg ’s Conjecture and near-colorings

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

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Steinberg ’s Conjecture and near-colorings

Gerard J. Chang, Fr´ ed´ eric Havet, Mickael Montassier, Andr´ e Raspaud

June 20, 2011

Steinberg’s Conjecture and near-colorings ^∗

Gerard J. Chang

Frédéric Havet

Mickael Montassier

André Raspaud

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

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

3National Center for Theoretical Sciences, Taipei, Taiwan

4Projet MASCOTTE – INRIA Sophia-Antipolis, 2004 route des lucioles BP93, 06902 Sophia-Antipolis Cedex, France

5LaBRI – Univerity of Bordeaux, 351 cours de la libéation, 33405 Talence Cedex, France

June 20, 2011

Abstract

LetF be the family of planar graphs without cycles of length 4 and 5. Steinberg’s Conjecture (1976) that says every graph ofF is 3-colorable remains widely open. Focusing on a relaxation proposed by Erd˝os (1991), many studies proved the conjecture for some subfamilies ofF. For example, Borodin et al. proved that every planar graph without cycles of length 4 to 7 is 3- colorable. In this note we propose to relax the problem not on the family of graphs but on the coloring by considering near-colorings. A graph G = (V, E) is said to be (i, j, k)-colorable if its vertex set can be partitioned into three sets V1, V2, V3 such that the graphs G[V1], G[V2], G[V3] induced by the sets V1, V2, V3 have maximum degree at most i, j, k respectively. Under this terminology, Steinberg’s Conjecture says that every graph ofF is (0, 0, 0)-colorable. A result of Xu (2008) implies that every graph ofF is (1, 1, 1)-colorable. Here we prove that every graph of F is (2, 1, 0)-colorable and (4, 0, 0)-colorable.

1 Introduction

In 1976, Appel and Haken proved that every planar graph is 4-colorable [2, 3], and as early as 1959, Grötzsch [20] showed that every planar graph without 3-cycles is 3-colorable. As proved by Garey, Johnson and Stockmeyer [19], the problem of deciding whether a planar graph is 3-colorable is NP- complete. Therefore, some sufficient conditions for planar graphs to be 3-colorable were stated. In 1976, Steinberg [24] raised the following:

Steinberg’s Conjecture ’76 Every planar graph without 4- and 5-cycles is 3-colorable.

There were then no progress in this direction until Erd˝os (1991) proposed the following relax- ation of Steinberg’s Conjecture:

Erd˝os’ relaxation ’91 Determine the smallest value of k, if it exists, such that every planar graph without cycles of length from 4 to k is 3-colorable.

Abbott and Zhou [1] proved that such a k does exist, with k ≤ 11. This result was later on improved to k ≤ 10 by Borodin [4], to k ≤ 9 by Borodin [5] and Sanders and Zhao [22], to k ≤ 8

∗Supported by ANR/NSC projects ANR-09-blan-0373-01 and NSC99-2923-M-002-007-MY3.

†E-mail: gjchang@math.ntu.edu.tw.

‡E-mail: Frederic.Havet@sophia.inria.fr.

§E-mail: mickael.montassier@labri.fr.

¶E-mail: andre.raspaud@labri.fr.

by Salavatipour [21]. The best known bound for such a k is 7 which was proved by Borodin, Glebov, Raspaud and Salavatipour [10].

This approach was at the origin of sufficient conditions of 3-colorability of subfamilies of planar graphs where some families of cycles are forbidden. See for examples [8, 9, 12, 13, 14, 15, 16, 17, 25].

A graph G is called improperly (d₁, d₂, . . . , d_k)-colorable, or simply (d1, d₂, . . . , d_k)-colorable, if the vertex set of G can be partitioned into subsets V₁, V₂, . . . , V_ksuch that the graph G[V_i] induced by Vi has maximum degree at most di for 1≤ i ≤ k. This notion generalizes those of proper k- coloring (when d1= d2= . . . = dk = 0) and d-improper k-coloring (when d1= d2= . . . = dk= d≥ 0). Under this terminology, the Four Color Theorem says that every planar graph is (0, 0, 0, 0)- colorable. Eaton and Hull [18] and independently Škrekovski [23] proved that every planar graph is 2-improperly 3-colorable (in fact, 2-improperly 3-choosable), i.e. (2, 2, 2)-colorable.

In this note we focus on near-colorings and Steinberg’s Conjecture. LetF be the family of planar graphs without cycles of length 4 and 5. We prove:

Theorem 1 Every graph ofF is (2, 1, 0)-colorable and (4, 0, 0)-colorable.

The remaining of the paper is dedicated to the proof of this theorem.

2 General setting for (s

, s

, s

)-colorability of F

The proof of the main theorem is done by reducible configurations and discharging procedure. Sup- pose the theorem is not true. Let G = (V, E, F ) be a counterexample with the minimum order embedded in the plane. We apply a discharging procedure to reach to a contradiction.

We first assign to each vertex v and face f of G a charge ω such that ω(v) = 2d(v)− 6 and ω(f ) = r(f )− 6, where d(v) and r(f) denote the degree of the vertex v and the length of the face f respectively. By Euler’s Formula|V | − |E| + |F | = 2 and formula∑

v∈V d(v) = 2|E| =

∑

f∈Fr(f ), we have: ∑

v∈V

ω(v) +∑

f∈F

ω(f ) =−12 < 0.

We then redistribute the charges according to some discharging rules. During the process, no charges are created or disappear. Let ω^∗be the new charge on each vertex and face after the proce- dure. It follows that:

∑

v∈V

ω(v) +∑

f∈F

ω(f ) =∑

v∈V

ω^∗(v) +∑

f∈F

ω^∗(f ).

However, we will show that under some structural properties of G the new charge on each vertex and face is non-negative. This leads to the following obvious contradiction

−12 = ∑

v∈V

ω(v) +∑

f∈F

ω(f ) = ∑

v∈V

ω^∗(v) +∑

f∈F

ω^∗(f ) > 0

implying that no counterexample can exist.

Establishing structural properties is essential in the proof of the theorem. Although the properties for (2, 1, 0)-coloring and for (4, 0, 0)-coloring are not the same, they share some common part. In this section, we derive lemmas for a general setting. Suppose s1≥ s2≥ s3≥ 0 and s = s1+s2+s3. In this section we assume that G is a minimum counterexample inF that is not (s1, s2, s3)-colorable.

A vertex of degree k (resp. at least k, at most k) will be called k-vertex (resp. k⁺-vertex, k⁻- vertex). A similar notation will be used for cycles and faces. A k-neighbor (resp. k⁺-neighbor, k⁻-neighbor) of some vertex u is a neighbor of u which is a k-vertex. An (a, b, c)-face is a 3-face uvw such that d(u) = a, d(v) = b and d(w) = c. In addition, a⁻(resp. a⁺) will mean d(u)≤ a

(resp. d(u)≥ a) and ∗ will mean any degree. For example, a (3, 4⁻,∗)-face is a 3-face uvw such that d(u) = 3, d(v)≤ 4 and w has no restriction on its degree. A pendent 3-face of a vertex v is a 3-face not containing v but is incident to a 3-vertex adjacent to v. In the following we will color the vertices of the graphs by partitioning the vertex set into V1, V₂, V₃ such that each Viinduces a subgraph of maximum degree at most si. Coloring a vertex with color i means adding the vertex into V_i. We will say that we nicely color a vertex if we color it by i and at most max{0, si− 1} of its neighbors are colored by i. We say that we properly color a vertex if we color it by a color not used by its neighbors. Properly colored vertices are nicely colored. When the colored neighbors of an uncolored vertex v use at most two colors, in particular when v has at most two colored neighbors, we can always color v properly by using the third color not used by its neighbors. We will use this frequently. As an easy consequence, every vertex of G has degree at least 3.

First, since G has no 4-cycles, we have the following:

Observation 2 Two 3-faces may not share an edge. If a k-vertex v is incident to α 3-faces and has β pendent 3-faces, then 2α + β≤ k.

Next, three useful lemmas.

Lemma 3 Let v be an (s + 2)⁻-vertex of G. If G− v has an (s1, s2, s3)-coloring such that all neighbors of v are nicely colored, then G is (s1, s2, s3)-colorable.

PROOF. For 1≤ i ≤ 3, if we cannot assign color i to v, then v has at least si+ 1 neighbors colored by i. It follows that v has degree at least∑3

i=1(si+ 1) = s + 3, a contradiction. 2 Lemma 4 Graph G contains no (s + 2)⁻-vertex v adjacent only to 4⁻-vertices, each 4-neighbor of which is adjacent some 3-neighbor of v.

PROOF. Suppose to the contrary that G contains such a (s + 2)⁻-vertex v. By the minimality of G, the graph G^′obtained from G by deleting v and all of its neighbors admits an (s1, s2, s3)-coloring.

We first color all 4-neighbors of v properly, and then color all 3-neighbors of v properly. Then all neighbors of v are nicely colored. Thus, by Lemma 3, G is (s1, s2, s3)-colorable, a contradiction.2 Lemma 5 The three neighbors x1, x2, x3of a 3-vertex v of G use different colors in an (s1, s2, s3)- coloring of G− v. Moreover, assume xiis colored by i, we have d(xi) ≥ si+ 3 for 1 ≤ i ≤ 3.

Furthermore, if si> 0 and xiis adjacent to xj, then either d(xi) > si+ 3 or d(xj) > sj+ 3.

PROOF. If x1, x2, x3do not use three distinct colors, then we can properly color v, a contradiction.

Hence w.l.o.g. we can assume that xiis colored by i for 1≤ i ≤ 3.

Suppose for a contradiction that some d(xi)≤ si+ 2 for some i. Then si ≥ 1 as d(xi) ≥ 3.

If xiis nicely colored by i, then we color v by i and this extends the coloring to G, a contradiction.

Hence, xihas at least sineighbors colored by i. Since xi has an uncolored neighbor v, there is at least one color different from i not used by its neighbors. We then color v by i and recolor xiby the unused color. This extends the coloring to G, a contradiction.

Suppose for a contradiction that xi is adjacent to xj, but d(xi) = si+ 3 and d(xj) = sj+ 3.

Let k be the color distinct from i and j. Since G has no 4-cycle, xkis not adjacent to xiand xj. As above, xi(resp. xj) has si(resp. sj) neighbors colored by i (resp. j) and another colored neighbor x^′_i (resp. x^′_j) other than xj (resp. xi). If x^′_iis colored by j, then we may color v by i and recolor x_i by k to get an (s₁, s₂, s₃)-coloring of G, a contradiction. Hence, x^′_iis colored by k. Similarly, x^′_j is also colored by k. Then we may color v by i, recolor x_i by j and recolor x_j by i to get an (s₁, s₂, s₃)-coloring of G (notice that si > 0), again a contradiction. Hence, d(xi) > s_i+ 3 or

d(xj) > sj+ 3. 2

3 (2, 1, 0)-colorability of F

In this section we prove that every graph inF is (2, 1, 0)-colorable, namely we consider the case (s₁, s₂, s₃) = (2, 1, 0) for which s = s1+ s₂+ s₃= 3.

3.1 Reducible configurations for (2, 1, 0)-coloring

We first establish structural properties of G. More precisely, we prove that some ‘configurations’, i.e. subgraphs, are ‘reducible’, i.e cannot appear in G because it is a minimum counterexample. Lots of this configuartions are depicted in Figure 1.

A light 5-vertex is a 5-vertex incident to a (3, 5, 5)-face f and adjacent to three 3-vertices not in f . A poor (3, 5, 5)-face is a (3, 5, 5)-face incident to a light 5-vertex. If a 3-vertex is incident to a 3-face, then its neighbor not incident to this 3-face is said to be its outer neighbor.

As already mentioned we have the following.

(C1) G contains no 2⁻-vertices.

The two following claims come from Lemma 4 with s = 3.

(C2) G contains no 5-vertex adjacent to five 3-vertices.

(C3) G does not contain 5-vertices v incident to a (3, 4, 5)-face f and adjacent to three 3-vertices not in f .

(C4) G contains no non-light 5-vertex incident to a poor (3, 5, 5)-face and a (3, 5⁻, 5)-face, and adjacent to a 3-vertex not in these faces.

Proof. Suppose to the contrary that G contains such a 5-vertex v. Let uvw be the poor (3, 5, 5)- face, rvs be the (3, 5⁻, 5)-face with d(u) = d(r) = 3, and x be the neighbor of v not in these faces. Vertex w is light and thus is adjacent to three 3-vertices distinct from u, say w1, w2, w3. By the minimality of G, the graph G− {u, v, w, w1, w2, w3, r, x} admits a (2, 1, 0)-coloring. Now we extend this coloring as follows. We may assume that, if s is colored by 1, then it has at most one neighbor colored by 1, otherwise we can properly recolor it. Then we color r and x properly. If s, r, x use different colors, then we color v with 1; otherwise we color v properly. We then color u, w1, w2, w3 properly. It follows that all neighbors of w are nicely colored. By Lemma 3, G is (2, 1, 0)-colorable, a contradiction. 2

(C5) G does not contain a poor (3, 5, 5)-face incident to two light 5-vertices.

Proof. Suppose to the contrary that G contains a poor (3, 5, 5)-face uvw with light vertices v and w. For x∈ {v, w}, let x1, x2, x3be the three neighbors of x not in{u, v, w}. By the minimality of G, the graph G− {u, v, w, w1, w2, w3, v1, v2, v3} admits a (2, 1, 0)-coloring. We extend the coloring to{v1, v₂, v₃} by coloring each of them properly. If v1, v₂, v₃use three distinct colors, then we color v with 1, and properly otherwise. After this, we color u, w1, w₂, w₃properly. It follows that all neighbors of w are nicely colored. By Lemma 3, G is (2, 1, 0)-colorable, a contradiction. 2 Let v be a 3-vertex adjacent to three vertices y1, y₂, y₃. Consider G− v. By Lemma 5, the colors 1, 2, and 3 appear on the neighbors of v. Moreover the vertex colored with 1 (resp. 2, 3) has degree at least 5 (resp. 4, 3). Thus (C6) and (C7) follow.

(C6) G does not contain 3-vertices adjacent to two 3-vertices.

(C7) If uvw is a (3, 4, 4)-face with d(u) = 3, then the outer neighbor of u has degree at least 5.

Now, if the three vertices y1, y₂, y₃satisfy d(y1) = 3, d(y₂) ≤ 4 and d(y2) ≤ d(y3), then y1

(resp. y₂, y₃) is colored with 3 (resp. 2, 1) and has degree 3 (resp. 4, at least 5). By the last sentence of Lemma 5, the vertices y₁, y₂are non-adjacent; moreover if d(y₃) = 5, then y3is not adjacent to y1or y2. Thus (C8), (C9), and (C10) follow.

(C8) G does not contain (3, 3, 4⁻)-faces.

(C9) If uvw is a (3, 3, 5)-face with d(u) = 3, then the outer neighbor of u has degree at least 5.

(C10) If uvw is a (3, 4, 5)-face with d(u) = 3, d(v) = 4 and d(w) = 5, then the outer neighbor of u has degree at least 4.

(C10)

(C2) (C3)

(C5) (C6)

(C8) (C9)

(C4)

(C7)

Figure 1: Reducible configurations (C2)-(C10). Black dots represent vertices all neighbours of which are drawn in the figure; the white dots represent vertices that can have nondepicted neighbours.

Dashed lines represent edges that may possibly not exist.

3.2 Discharging procedure for (2, 1, 0)-coloring

We now apply a discharging procedure to reach to a contradiction. The discharging rules are as follows:

R1. Every 4-vertex gives ¹₂to each pendent 3-face.

R2. Every 5⁺-vertex gives 1 to each pendent 3-face.

R3. Every 4-vertex gives 1 to each incident 3-face.

R4. Every non-light 5-vertex gives 2 to each incident poor (3, 5, 5)-face.

R5. Every 5-vertex gives ³₂to each incident non-poor (3, 5, 5)-face or (3, 4, 5)-face.

R6. Every 5-vertex gives 1 to each other incident 3-face.

R7. Every 6⁺-vertex gives 2 to each incident 3-face.

Let v be a k-vertex with k≥ 3 by (C1).

Case k = 3.The discharging procedure does not involves 3-vertices. Hence ω^∗(v) = ω(v) = 0.

Case k = 4. Initially ω(v) = 2. Vertex v gives 1 to each of the α incident 3-faces by R3 and¹₂ to each of the β pendent 3-faces by R1. By Observation 2, ω^∗(v)≥ 2 − (α +¹₂β)≥ 2 −¹₂· 4 = 0.

Case k = 5. Initially ω(v) = 4. Assume v is not incident to any 3-face. By (C2), v is adjacent to at most four 3-vertices and so has at most four pendent 3-faces. By R2, ω^∗(v)≥ 4 − 4 · 1 = 0.

Assume v is incident to exactly one 3-face f . If v is a non-light 5-vertex and f is a poor (3, 5, 5)- face, then v has at most two pendent 3-faces by definition. By R4 and R2, ω^∗(v)≥ 4 − 2 − 2 · 1 = 0.

If f is a non-poor (3, 5, 5)-face, then v has at most two pendent 3-faces by definition. By R5 and R2, ω^∗(v) ≥ 4 − ³₂− 2 · 1 > 0. If f is a (3, 4, 5)-face, then v has at most two pendent 3-faces by (C3). By R5 and R2, ω^∗(v)≥ 4 −³₂− 2 · 1 > 0. If f is a 3-face of other type, then by R6 and R2 ω^∗(v)≥ 4 − 1 − 3 · 1 = 0.

Assume v is incident to exactly two 3-faces f1and f2. If v gives twice at most ³₂ to the 3-faces, then ω^∗(v)≥ 4 − 2 ·³₂− 1 = 0. So we may assume that f1or f2, say f1, is a poor (3, 5, 5)-face. If f2is a (3, 5⁻, 5)-face, then v has no pendent 3-faces by (C4) and ω^∗(v)≥ 4 − 2 − 2 = 0. If f2is a 3-face of other type, then v may have a pendent 3-face and ω^∗(v)≥ 4 − 2 − 1 − 1 = 0 by R6.

Case k≥ 6. Initially ω(v) = 2k − 6. Vertex v gives 2 to each of the α incident 3-faces by R7 and 1 to each of the β pendent 3-faces by R2. By Observation 2, ω^∗(v) ≥ 2k − 6 − 2α − β ≥ 2k− 6 − k = k − 6 ≥ 0.

Let f be a k-face.

Case k = 3.Initially ω(f ) =−3. By (C8), f is not a (3, 3, 4⁻)-face.

Let f = uvw be a (3, 3, 5)-face so that d(u) = d(v) = 3 and d(w) = 5. By (C9) the outer neighbor of u (resp. v) has degree at least 5 and so gives at least 1 to f by R2. By R6, w gives 1 to f . It follows that ω^∗(f ) =−3 + 2 · 1 + 1 = 0.

Let f = uvw be a (3, 3, 6⁺)-face so that d(u) = d(v) = 3 and d(w) ≥ 6. By (C6), the outer neighbor of u (resp. v) has degree at least 4 and so gives at least ¹₂ to f by R1. By R7, w gives 2 to f . It follows that ω^∗(f ) =−3 + 2 ·¹₂+ 2 = 0.

Let f = uvw be a (3, 4, 4)-face so that d(u) = 3 and d(v) = d(w) = 4. By (C7) the outer neighbor of u has degree at least 5 and so gives 1 to f by R2. Vertices v (resp. w) give 1 to f by R3.

Hence ω^∗(f ) =−3 + 1 + 2 · 1 = 0.

Let f = uvw be a (3, 4, 5)-face so that d(u) = 3, d(v) = 4 and d(w) = 5. By (C10), the outer neighbor of u has degree at least 4 and so gives at least¹₂ to f by R1. Vertices v and w give each 1 and³₂to f respectively by R3 and R5. Hence ω^∗(f ) =−3 +¹₂+ 1 + ³₂= 0.

Let f = uvw be a (3, 4, 6⁺)-face so that d(u) = 3, d(v) = 4 and d(w)≥ 6. By R3 and R7, vertices v and w give each 1 and 2 to f respectively. Hence ω^∗(f ) =−3 + 1 + 2 = 0.

Let f = uvw be a (3, 5, 5)-face so that d(u) = 3, d(v) = d(w) = 5. Assume f is poor and v is light. By (C5) w cannot be light. Hence ω^∗(f ) =−3 + 1 + 2 = 0 by R4 and R6. Assume f is not poor. Then ω^∗(f ) =−3 + 2 · ³₂ = 0 by R5.

Let f = uvw be a (3, 5⁺, 6⁺)-face so that d(u) = 3, d(v)≥ 5, d(w) ≥ 6. Vertices v and w give each at least 1 and 2 respectively by R6-7. Hence ω^∗(f )≥ −3 + 1 + 2 = 0.

Let f = uvw be a (4⁺, 4⁺, 4⁺)-face. Each incident vertex gives at least 1 to f by R3-7. Hence ω^∗(f )≥ −3 + 3 · 1 = 0.

Case k≥ 4. Faces of length 4 and 5 do not exist by hypothesis. Faces of length at least 6 are not involved in the discharging procedure. Hence ω^∗(f ) = ω(f ) = r(f )− 6 ≥ 0.

It follows that every vertex and face has a non-negative charge as required. This completes the proof.

4 (4, 0, 0)-colorability of F

In this section we prove that every graph ofF is (4, 0, 0)-colorable, namely we consider the case of (s₁, s₂, s₃) = (4, 0, 0) for which s = s1+ s₂+ s₃= 4.

4.1 Reducible configurations for (4, 0, 0)-coloring

In this section we study structural properties of G and establish a number of reducible configuarions.

See Figure 3.

A bad 8-vertex is a 8-vertex v incident to three (3, 3, 8)-faces and to a (3, 8,∗)-face f = uvw with d(u) = 3, d(v) = 8, where the vertex w is called the sponsor of f and f is a bad face of v. See Figure 2.

sponsor v

u w

f

Figure 2: A bad 8-vertex v whose bad face is uvw with sponsor w. (Drawing conventions are the same as in Figure 1.)

(C1’) G contains no 2⁻-vertices.

(C2’) For 8≤ k ≤ 10, a k-vertex cannot be incident to exactly k − 5 (3, 3, k)-faces and adjacent to k 3-vertices.

Proof. Suppose v is a k-vertex incident to exactly k− 5 (3, 3, k)-faces and adjacent to 10 − k other 3-vertices not in these (3, 3, k)-faces. By the minimality of G, the graph G^′obtained from G by deleting v and all its neighbors admits a (4, 0, 0)-coloring. We color properly and sequentially all neighbors of v. Since each (3, 3, k)-face contains at most one vertex colored by 1, color 1 appears at most k− 5 + 10 − k = 5 times on the neighbors of v. If it appears less than 5 times, we can

color v with 1, a contradiction. Hence color 1 appears exactly 5 times, once in each (3, 3, k)-face and on all the 10− k other 3-vertices. For each (3, 3, k)-face vxy with d(x) = d(y) = 3, where x is colored by 1, y is colored by 2 or 3. In the case of y is colored by 3, if the outer neighbor of y is colored by 1 (resp. 2), then we can recolor y by 2 (resp. 1). Then we can color v with 3 to obtain a (4, 0, 0)-coloring of G, a contradiction. 2

(C3’) Every 3-vertex of G is adjacent to at least one 7⁺-vertex.

Proof. This follows from the fact that the degree sequence for the three neighbors of a 3-vertex is lexicographically at least (7, 3, 3) by Lemma 5. 2

(C4’) If uvw is a (3, 3, 7)-face with d(u) = 3, then the outer neighbor of u has degree at least 4.

Proof. Suppose to the contrary that G has a (3, 3, 7)-face uvw with d(u) = d(v) = 3 and d(w) = 7, but the outer vertex x of u has d(x) = 3. By Lemma 5, the degree sequence for the three neighbors of u is lex-graphically at least (7, 3, 3). Hence w is colored by 1 and v is colored by 2 or 3. This contradicts the last sentence of Lemma 5 as w is adjacent to v. 2

(C5’) The sponsor w of a bad 8-vertex v has degree at least 8 and is not a bad 8-vertex.

Proof. Suppose to the contrary that the bad 8-vertex v is incident to three (3, 3, 8)-faces x1x₂v, y₁y₂v and z1z₂v and to a (3, 8,∗)-face uvw with d(u) = 3 and 3 ≤ d(w) ≤ 7 or w a bad 8-vertex.

By the minimality of G, the graph G^′ = G− {v, x1, x₂, y₁, y₂, z₁, z₂, u} admits a (4, 0, 0)-coloring.

We can assume that w is nicely colored; otherwise, if d(w)≤ 7, then we can recolor it properly, and if w is a bad 8-vertex, then we can recolor properly all its colored neighborhood and then color w nicely. Now we color properly and sequentially x1, x2, y1, y2, z1, z2, u, and we assign color 1 to v (color 1 appears at most 4 times on the neighbors of v). This extends the (4, 0, 0)-coloring to G, a contradiction. 2

4.2 Discharging procedure for (4, 0, 0)-coloring

We now apply a discharging procedure to reach a contradiction. The discharging rules are as follows:

R1’. For 4≤ k ≤ 6, every k-vertex gives ¹₂to each pendent 3-face.

R2’. Every 7⁺-vertex gives 1 to each pendent 3-face.

R3’. For 4≤ k ≤ 6, every k-vertex gives 1 to each incident 3-face.

R4’. Every 7⁺-vertex gives 1 to each incident (4⁺, 4⁺, 4⁺)-face.

R5’. Every non-bad7⁺-vertex gives 2 to each incident (3, 4⁺, 4⁺)-face; every bad 8-vertex gives 1 to its bad 3-face.

R6’. Every 7-vertex gives 2 to each incident (3, 3, 7)-face.

R7’. For k≥ 8, every k-vertex gives 3 to each incident (3, 3, k)-face.

Let v be a k-vertex with k≥ 3 by (C1’). Initially ω(v) = 2k − 6.

Case k = 3. The discharging procedure does not involves 3-vertices. Hence ω^∗(v) = ω(v) = 0.

Case 4≤ k ≤ 6. Vertex v gives 1 to each of the α incident 3-faces by R3’ and¹₂to each of the β pendent 3-faces by R1’. By Observation 2, ω^∗(v)≥ 2k−6−(α+¹₂β)≥ 2k−6−¹₂k = ³₂k−6 ≥ 0.

Case k = 7. Vertex v gives 2 to each of the α^′ incident (3, 3⁺, 4⁺)-faces by R5’-6’, 1 to each of the α^′′incident (4⁺, 4⁺, 4⁺)-faces by R4’, and 1 to each of the β pendent 3-faces by R2’. By Observation 2, ω^∗(v)≥ 2k − 6 − (2α^′+ α^′′+ β)≥ 2k − 6 − k = k − 6 > 0.

Case k ≥ 8. For the case when v is a bad 8-vertex, v gives 3 to each incident (3, 3, 8)-face by R7’ and 1 to the bad 3-face by R5’. Hence ω^∗(v) = 2· 8 − 6 − 3 · 3 − 1 = 0.

(C5’)

(C2’) (C2’)

(C2’)

(C3’) (C4’)

(C5’)

Figure 3: The reducible configurations (C2’)-(C5’). (Drawing conventions are the same as in Fig- ure 1.)

Now assume that v is not a bad 8-vertex. By R7’, R5’, R4’ and R2’, v gives 3 to each of the α^′ incident (3, 3, k)-faces, 2 to each of the α^′′ incident (3, 4⁺, 4⁺)-faces, 1 to each of the α^′′′

incident (4⁺, 4⁺, 4⁺)-faces, and 1 to each of the β pendent 3-faces. By Observation 2, ω^∗(v) = 2k− 6 − (3α^′+ 2α^′′+ α^′′′+ β)≥ 2k − 6 − ⌊^3k₂⌋ = ⌈^k₂⌉ − 6 ≥ 0 except for the cases (1) k = 10 with α^′ = 5, (2) k = 9 with α^′ = 4 and β = 1, (3) k = 8 with α^′ = 3 and β = 2 (note that the bad 8-vertex case, i.e. α^′ = 4 or α^′ = 3 with α^′′ = 1, is excluded). The exceptional cases give a k-vertex, 8 ≤ k ≤ 10, with exactly k − 5 (3, 3, k)-faces and adjacent only to 3-vertices, a contradiction to (C2’).

Let f be a k-face.

Case k = 3.Initially ω(f ) =−3.

Let f = uvw be a (a1, a2, a3)-face with 3 ≤ a1 ≤ 6, 3 ≤ a2 ≤ 6 and 3 ≤ a3 ≤ 6. By (C3’), the outer neighbor of each 3-vertex incident to f has degree at least 7 and gives each at least 1 to f by R2’. By R3’, each d-vertex with 4 ≤ d ≤ 6 incident to f gives 1 to f. It follows that ω^∗(f ) =−3 + 3 = 0.

Let f = uvw be a (3, 3, 7)-face so that d(u) = d(v) = 3 and d(w) = 7. By (C4’) the outer neighbor of u (resp. v) has degree at least 4 and so gives at least¹₂to f by R1’. By R6’, w gives 2 to f . It follows that ω^∗(f ) =−3 + 2 ·¹₂+ 2 = 0.

Let f = uvw be a (3, 3, 8⁺)-face so that d(u) = d(v) = 3 and d(w)≥ 8. By R7’, w gives 3 to f . It follows that ω^∗(f ) =−3 + 3 = 0.

Let f = uvw be a (3, 4⁺, 7⁺)-face so that d(u) ≥ 3, d(v) ≥ 4 and d(w) ≥ 7. By R3’-5’, vertices v and w gives at least 3 to f and so ω^∗(f ) =−3 + 3 = 0, except for the case when f is a bad 3-face with the pair v, w being either two bad 8-vertices or a bad 8-vertex and a 6⁻-vertex. But these two exceptional cases are impossible by (C5’).

Finally, let f = uvw be a (4⁺, 4⁺, 4⁺)-face. Every incident vertex gives at at least 1 to f by R3’-4’. Hence ω^∗(f )≥ 0.

It follows that every vertex and face has a non-negative charge as required. This completes the proof.

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