On a conjecture by Plummer and Toft

On a Conjecture byPlummer and Toft

Mirko Hornak and Stanislav Jendrol’DEPARTMENT OF GEOMETRY AND ALGEBRA

P. J. SAFARIK UNIVERSITYJESENNA 5, 041 54 KOSICE, SLOVAKIA

E-mail: [email protected], [email protected]

Received August 27, 1996; accepted August 29, 1998

Abstract: The cyclic chromatic number χc(G) of a 2-connected plane graph Gis the minimum number of colors in an assigment of colors to the vertices of Gsuch that, for every face-bounding cycle f of G, the vertices of f have differentcolors. Plummer and Toft proved that, for a 3-connected plane graph G, under theassumption ∆∗(G) ≥ 42, where ∆∗(G) is the size of a largest face of G, it holdsthatχc(G) ≤ ∆∗(G)+4. They conjectured that, ifG is a 3-connected plane graph,then χc(G) ≤ ∆∗(G)+2. In the article the conjecture is proved for ∆∗(G) ≥ 24.c© 1999 John Wiley & Sons, Inc. J Graph Theory 30: 177–189, 1999

Keywords: plane graph, cyclic coloring, cyclic chromatic number

1. INTRODUCTION

In this article we consider 2-connected plane graphs without loops and multipleedges. Let G be such a graph and let x be a vertex or a face of G. The degree of x,in symbol deg(x), is the number of edges incident with x. By ∆∗(G) we denotethe maximum face degree of G.

A cyclic coloring of a plane graph G is such a coloring of the vertices of G that,if two vertices are incident with a common face, they receive different colors. Leta k-coloring of a graph G be a coloring of G using k colors. The cyclic chromaticnumber (introduced by Ore and Plummer [15]) of a graphG, denoted χc(G), is theminimum k such that G has a cyclic k-coloring. Obviously, ∆∗(G) ≤ χc(G). For

Correspondence to: S. Jendrol’Contract grant sponsor: Slovak VEGAContract grant number: 1/4377/97

c© 1999 John Wiley & Sons, Inc. CCC 0364-9024/99/030177-13

178 JOURNAL OF GRAPH THEORY

the graph P of a 3-sided prism it holds χc(P ) = 6 = ∆∗(P ) + 2. In a recent bookJensen and Toft [14] have posed the following two questions (see Problem 2.5):

(1) Is χc(G) ≤ 32∆∗(G)?

(2) If G is 3-connected, is χc(G) ≤ ∆∗(G) + 2?

Question 1 is still open. For a recent progress in solving it, consult the articleBorodin et al. [6], where it is proved that χc(G) ≤ 9

5∆∗(G).The present article is devoted to Question 2. Plummer and Toft [16] proved

that χc(G) ≤ ∆∗(G) + 9 in general and, among other similar results, χc(G) ≤∆∗(G)+4 if ∆∗(G) ≥ 42. In the same article (see also [14]) they have conjectured(PTC) thatχc(G) ≤ ∆∗(G)+2. The validity of PTC forGwith ∆∗(G) = 3 followsfrom the Five Color Theorem by Heawood [11]; moreover, the Four Color Theoremby Appel and Haken [2] yields a stronger statement, namelyχc(G) ≤ ∆∗(G)+1 =4. For ∆∗(G) = 4, PTC has been proved by Borodin [4]. For ∆∗(G) ≥ 24 it isknown that χc(G) ≤ ∆∗(G) + 3—the result has been obtained independently byBorodin [5] and Hornak and Jendrol’ [12]. Recently, rapid progress in the areacan be seen: Borodin and Woodall [7] proved PTC for ∆∗(G) ≥ 61. Further, theyproved that χc(G) ≤ ∆∗(G) + 1 if ∆∗(G) ≥ 122; Enomoto et al. [9] improvedthis to ∆∗(G) ≥ 60. Graphs of pyramids (i.e., plane embeddings of wheels) showthat the upper bound ∆∗(G)+1 is best possible. We found a Lebesgue type result,which enabled us to show in [13] that PTC is true provided that ∆∗(G) ≥ 40.Borodin and Woodall [8] proved that χc(G) ≤ ∆∗(G) + 3 if ∆∗(G) ≥ 21.

The aim of this article is to make a further step in tackling PTC—to prove thefollowing.

Theorem. For any 3-connected plane graphG,χc(G) ≤ max{∆∗(G)+2, 26}.

PTC is open for graphs G with 5 ≤ ∆∗(G) ≤ 23. The best known estimates forχc(G) in these cases are:

∆∗(G) χc(G) Reference5 8 Borodin et al.[6]6 107 12 Borodin [3]8 139 15

10 1711 1912 20 Hornak and Jendrol’ [12]

13, 14, 15, 16 21 [12], Borodin [5]17, 18 ∆∗(G) + 519, 20 ∆∗(G) + 4

21, 22, 23 ∆∗(G) + 3 Borodin and Woodall [8]

ON A CONJECTURE BY PLUMMER AND TOFT 179

2. PRELIMINARIES

Let G be a 3-connected graph with at least five vertices and let xy be an edge ofG. By G(x → y) we denote the graph formed by adding to G− x all the edges yzsuch that the vertex z /= y is adjacent (inG) to x, but not to y. Evidently, the graphsG(x → y) andG(y → x) are isomorphic, and the edgexy is said to be contractible,if the graphG(x → y) is 3-connected. ByG\xy we denote the graph formed fromG − xy by smoothing all (possibly arisen) 2-vertices. If G is a plane graph, thenboth G(x → y) and G\xy are also plane graphs, and ∆∗(G(x → y)) ≤ ∆∗(G).

Two vertices x, y of G are defined to be cyclically adjacent if they are incidentwith a common face of G. Let the cyclic degree of a vertex x, in symbol cd(x), bethe number of vertices that are cyclically adjacent to x. The cyclic neighborhoodof a vertex x is the set of all vertices that are cyclically adjacent to x.

If a vertex or a face of G is of degree n, it is said to be an n-vertex or ann-face, respectively. Let x be an n-vertex incident with faces f1, . . . , fn in anatural order (when rotating around x) and let deg(fi) = di, i = 1, . . . , n; we willrefer to x as to a (d1, . . . , dn)-vertex. For the cyclic degree of x we clearly havecd(x) =

∑ni=1(di − 2).

For integers p, q, we denote by [p, q] the interval of all integers between p and q(inclusively).

Proposition 1. Any partial cyclic k-coloring of a graph G in which the set ofuncolored vertices consists only of either

(i) a vertex x with cd(x) ≤ k − 1 or(ii) cyclically adjacent vertices x, y with cd(x) ≤ k − 1 and cd(y) ≤ k

may be extended to a cyclic k-coloring of G.

Proof. The cyclic neighborhood of x has cardinality cd(x); this provides anextension in the case (i). In the case (ii), we color first y and then x.

We need a result of Halin [10].

Lemma 1. Let G be a 3-connected graph with at least five vertices. Then any3-vertex of G is incident with a contractible edge.

From a result of Ando et al. [1], Lemma 2 follows.

Lemma 2. Let G be a 3-connected graph with at least five vertices. If a vertexx of G with deg(x) ≥ 4 is incident with no contractible edge, then it is adjacent tothree 3-vertices.

First we will show that a counterexample to our theorem, which is minimalwith respect to the number of edges, cannot contain certain configurations. Someof them are in Figs. 1(a)–(f) and 2(a), where encircled numbers are degrees ofcorresponding vertices and 4+ means that a corresponding vertex is of degree ≥ 4.(An empty circle corresponds to a vertex of an arbitrary degree.)


FIGURE 1.

Lemma 3. LetG be a 3-connected plane graph with a smallest number of edgessuch that χc(G) > max{∆∗(G) + 2, 26}. Then G contains none of the followingconfigurations:1. a configuration from Figs. 1(a)–(f), where, in Fig. 1(e), deg(f1) ≤ ∆∗(G) − 1,and, in Fig. 1(f), deg(f1) + deg(f2) ≤ 2∆∗(G) − 1;2. a configuration from Fig. 2(a), where cd(x) ≤ ∆∗(G)+1 and deg(fi) ≥ 5, i =1, 3;3. a k-vertex x, k ≥ 4, with cd(x) ≤ ∆∗(G) + 1, which is either


FIGURE 2.

(i) incident with a contractible edge xy or(ii) adjacent to a 3-vertex y with cd(y) ≤ ∆∗(G) + 2;

4. a 3-vertex x with cd(x) ≤ ∆∗(G) + 1.

Proof. (For simplicity we will write ∆∗ instead of ∆∗(G).) Suppose that Gpossesses some of the configurations above. By modifying G, we will define a3-connected plane graph G′ with |E(G′)| < |E(G)| and ∆∗(G′) ≤ ∆∗. Thenχc(G′) ≤ m′ := max{∆∗(G′) + 2, 26} ≤ m := max{∆∗ + 2, 26} and thereexists a cyclic m′-coloring ϕ of G′. This coloring will serve as a basis for findinga cyclic m-coloring ψ of G, which is in contradiction with χc(G) > m. All valuesof ψ not defined explicitly are ‘‘inherited’’ from ϕ, i.e., ψ(v) := ϕ(v) for (some)vertices v ∈ V (G) ∩ V (G′).1. If G has a configuration of Fig. 1 we define G′ := G\x1x2 for figures (a)–(e), and G′ := G(x1 → x0) for figure (f), where the face f1 (incident with thecontracted edge x0x1) has been chosen in such a way that deg(f1) ≤ ∆∗ − 1. LetC(fk) be the set of colors used in the coloring ϕ for the vertices of the face fk.

(a)–(b): If ϕ(y) ∈ C(fk) for some k ∈ {1, 2}, then we can extend ϕ to ψ bycoloring firstx3−k and nextxk, because in such a case at most ∆∗+1 colors are usedin the cyclic neighborhood of each of x1, x2 at the time when it is colored. (Notethatϕ(y) is used twice in the cyclic neighborhood ofxk because of 3-connectednessof G.) On the other hand, provided that ϕ(y) /∈ C(f1) ∪ C(f2), we may recolor zwith ϕ(y) to create the situation above.

(c): If ϕ(z1) /∈ C(f2), putting ψ(x2) := ϕ(z1), we have at least one color forx1. If ϕ(z1) ∈ C(f2), we remove the color from y2, define ψ(x2) := ϕ(y2), andthen color first x1 and next y2. (We have f1 /= f2, since G is 3-connected.)

(d): Since ϕ(w1) /= ϕ(w2), we may assume that ϕ(w1) /= ϕ(z). Also, we mayassume that ϕ(w1) ∈ C(f2), because otherwise we may recolor y2 with ϕ(w1).Now, color x1, then color x2.

(e): Take simply ψ := ϕ.(f): We have deg(f1) ≤ ∆∗ − 1. Thus, if {ϕ(y1), ϕ(y2)} ∩ C(f1) /= ∅, at

least one color is available for the vertex x1. If ϕ(yk) /∈ C(f1) ∪ C(f2) forsome k ∈ {1, 2}, we may recolor the vertex z with ϕ(yk) to obtain the preceding


situation. Finally, if {ϕ(y1), ϕ(y2)} ∩ C(f1) = ∅ and {ϕ(y1), ϕ(y2)} ⊆ C(f2),we may remove the color from x2 and then color first x1 and next x2.2. In this case, we form G′ by replacing the configuration from Fig. 2(a) by thatfrom Fig. 2(b), where the edge e = y1y2 is added if and only if deg(f2) ≥ 4, andmake use of Proposition 1.(i).3.(i) After setting G′ := G(y → x) we employ Proposition 1.(i).3.(ii) By Lemma 1, the vertex y is incident with a contractible edge yz. By 3.(i),we may assume that z /= x. We define G′ := G(y → z), remove the color fromthe vertex x, and use Proposition 1.(ii).4. Lemma 1 yields a contractible edge xy—this allows us to use Proposition 1.(i).

3. PROOF OF THEOREM

The proof is by induction on the number of edges of G. Evidently, there is nothingto prove if |E(G)| ≤ 40, because in such a case |V (G)| ≤ 26—the assumption|V (G)| ≥ 27 would yield |E(G)| ≥ d3·27

2 e = 41. Suppose that our Theorem isnot true and G = (V,E, F ) is a counterexample to it with a minimum number ofedges. With respect to the results mentioned in the introduction, we know that then∆∗ = ∆∗(G) ≥ 24. Euler’s formula for G may be rewritten as

∑v∈V

2(3 − deg(v)) +∑f∈F

(6 − deg(f)) = 12,

−12 =∑v∈V

(2 deg(v) − 6) +∑f∈F

∑v∈f

deg(f) − 6deg(f)

=∑v∈V

(2 deg(v) − 6) +∑v∈V

∑f3v

deg(f) − 6deg(f)

=∑v∈V

2 deg(v) − 6 +

∑f3v

deg(f) − 6deg(f)

.

Thus, if we define the charge of a vertex v ∈ V by

c(v) := 2 deg(v) − 6 +∑f3v

deg(f) − 6deg(f)

,

then the sum of all the charges of vertices in G is negative; in this way we obtaina charge mapping c ∈ QV . We will create a new charge mapping c′ ∈ QV as aredistribution of c, i.e.,

∑v∈V

c′(v) =∑v∈V

c(v) = −12.


This new charge mapping leads to a contradiction, since it is possible to find adecomposition D of V such that

∑v∈D

c′(v) ≥ 0 for any D ∈ D. (∗)

We will denote by D(v) the set of D containing v ∈ V . In fact, nontrivial (con-taining more than one element) sets in D consist of 3-vertices only. To obtain thenew charge mapping c′, the charges of c are redistributed by the following twoRules, where, in Rule 1, V (f) is the set of vertices incident with a face f andV3(f) := {v ∈ V (f) : deg(v) = 3}:

Rule 1. Let f be a 3-face such that V3(f) /= ∅ /= V (f) − V3(f). Then anyvertex y ∈ V (f) − V3(f) sends to any vertex z ∈ V3(f) the amount t(y, z) =−c(z)/|V (f) − V3(f)|.

Rule 2. If z is a (4, 4,∆∗)-vertex, any of its neighbors y of degree ≥ 4 sends to zthe amount t(y, z) = −c(z) = 6

∆∗ .Observe that the notation t(y, z) used in Rules is correct—if vertices y, z are

adjacent and deg(y) ≥ 4,deg(z) = 3, then y can send an amount to z by at mostone of Rules 1, 2, and, from Lemma 3.4, there is at most one 3-face incident withyz for which Rule 1 can be applied.

If v ∈ V is a (d1, . . . , dn)-vertex adjacent to vertices v1, . . . , vn in such a waythat the edge vvi is incident with faces of degrees di and di+1 (indices are takenmodulo n), then

c(v) = 2n−6+n∑

i=1

di − 6di

=n∑

i=1

(3 − 6

di− 6n

)=

n∑i=1

(3 − 3

di− 3di+1

− 6n

).

Setting

γ(d1, . . . , dn) := 2n− 6 +n∑

i=1

di − 6di

,

σ(k, l,m) := 3 − 3k

− 3l

− 6m

p(v, vi) := σ(di, di+1, n)

yields

c(v) = γ(d1, . . . , dn) =n∑

i=1

σ(di, di+1, n) =n∑

i=1

p(v, vi).

Thus, the charge of the vertex v may be counted in a natural way as the sumof partial charges p(v, vi) corresponding to the oriented edges

−→vvi leaving v, i =

1, . . . , n. If we define the new partial charge of−→vvi by p′(v, vi) := p(v, vi) −

t(v, vi), then the new charge of the vertex v is c′(v) =∑n

i=1 p′(v, vi).


In our analysis, two properties of the function γ(d1, . . . , dn) are needed. Clearly,

γ(d1, . . . , dn) = γ(dπ(1), . . . , dπ(n)) for any permutation π of [1, n].

The next Lemma provides a lower bound for the charge of a vertex v incidentwith faces of degrees d1, . . . , dq, d

′q+1, . . . , d

′n, where d′

i ≥ di and d′n ≥ di for

i ∈ [q + 1, n− 1] and the weight of v,∑q

i=1 di +∑n

i=q+1 d′i, is at least w.

Lemma 4. For a nonnegative integer q and for positive integers n, d1, . . . , dn−1,w with n ≥ q + 2 and w ≥ ∑n−1

i=1 di + max{di : i ∈ [q + 1, n − 1]} letSq(d1, . . . , dn−1;w) denote the set of all integer sequences (d1, . . . , dq, d′

q+1, . . . ,

d′n) in which d′

i ≥ di and d′n ≥ di for all i ∈ [q + 1, n − 1] and

∑qi=1 di +∑n

i=q+1 d′i ≥ w. Then the minimum of γ(d1, . . . , dq, d

′q+1, . . . , d

′n) overSq(d1, . . . ,

dn−1;w) is equal to γ(d1, . . . , dn−1, w − ∑n−1i=1 di).

Proof. We have γ(d1, . . . , dq, d′q+1, . . . , d

′n) = 3n − 6 − 6

∑qi=1

1di

− 6∑n

i=q+11d′

i

. Let us decrease d′i to di while increasing d′

n by d′i − di for any

i ∈ [q + 1, n − 1] with d′i > di. Since for positive integers a1, a2, a3, a4 with

a1 + a2 = a3 + a4 and a1 < min{a3, a4} it holds −( 1a1

+ 1a2

) < −( 1a3

+1a4

), putting d′′n := d′

n +∑n−1

i=q+1(d′i − di) we obtain γ(d1, . . . , dn−1, d

′′n) ≤

γ(d1, . . . , dq, d′q+1, . . . , d

′n). Here the inequality turns into equality if and only

if d′i = di for all i ∈ [q + 1, n − 1], or, equivalently, d

′′n = d′

n. In such a case,d′

n ≥ w − ∑n−1i=1 di ≥ max{di : i ∈ [q + 1, n − 1]} ≥ 1 and, as the sequence

{− 1m}∞

m=1 is increasing, the proof follows.It is easy to find an upper bound for the amount t(y, z) sent from y to z. Namely,

if y sends something to z by Rule 1, then z is a (3, k, l)-vertex. By Lemma 3.4,cd(z) = k + l − 3 ≥ ∆∗ + 2, so that min{k, l} ≥ 5. Hence, from Lemma 4 withq = 1, d1 = 3, d2 = 5, andw = ∆∗+8, we obtainγ(3, k, l) ≥ γ(3, 5,∆∗) = −1

5−6

∆∗ ≥ −15 − 6

24 = − 920 . Thus, we have either t(y, z) = −c(z) = −γ(3, k, l) ≤ 9

20or t(y, z) = −1

2c(z) ≤ 940 . If y makes a transfer to z with respect to Rule 2, then

t(y, z) = 6∆∗ ≤ 1

4 .Now we are going to define the sets D(v), v ∈ V , in such a way that (*) is

fulfilled. Suppose that v is a (d1, . . . , dn)-vertex adjacent to vertices v1, . . . , vn

and that, for any i ∈ [1, n], the edge vvi is incident with di-face fi and di+1-facefi+1.

(1). If n = 3, then, without loss of generality, d1 ≤ d2 ≤ d3. (We may choose asf1 any face incident with v and we may choose also one of two possible orientationsto rotate around v.)

(11). If d1 = 3, then, by Lemma 3.4, cd(v) = d2 + d3 − 3 ≥ ∆∗ + 2, henced2 ≥ 5.

(111). If max{deg(v1),deg(v3)} ≥ 4, then, because of Rule 1, c′(v) = 0 and wemay set D(v) := {v}.


(112). Now suppose that deg(v1) = deg(v3) = 3; then c′(v) = c(v) and c′(vi) =c(vi), i = 1, 3. If d′

1 is the degree of the face adjacent to f1 along the edgev1v3, then, by Lemma 3.1. (a), min{d′

1, d2, d3} ≥ 6. We are going to show thatc′(v) + c′(v1) + c′(v3) ≥ 0, which will allow us to define D(v) := {v, v1, v3}.

(1121). The assumption d2 = 6 forces (see Lemma 3.1. (f)) the vertex v3 to bea (3,∆∗,∆∗)-vertex. In such a case c′(v) + c′(v1) + c′(v3) = −3 + 4 · ∆∗−6

∆∗ =1 − 24

∆∗ ≥ 0.

(1122). d2 ≥ 7.

(11221). If d′1 = 6, then, by Lemma 3.1.(f), d2 = d3 = ∆∗ and c′(v) + c′(v1) +

c′(v3) = 1 − 24∆∗ ≥ 0.

(11222). For d′1 ≥ 7, we have d := min{d′

1, d2} ≥ 7.

(112221). If d ∈ [7, 11], the remaining (besides d) terms of the triple (d′1, d2, d3)

are at least ∆∗ + 5 − d (by Lemma 3.4), so that c′(v) + c′(v1) + c′(v3) = 3 −12d′1

− 12d2

− 12d3

≥ 3 − 12d − 24

∆∗+5−d . Using ∆∗ ≥ 24, it is easy to see that the last

expression is lower bounded by 3 − 127 − 24

∆∗−2 ≥ 97 − 24

22 > 0.

(112222). d ∈ [12,∆∗] implies c′(v) + c′(v1) + c′(v3) ≥ 3 − 36d ≥ 0.

(12). d1 = 4.

(121). If d2 = 4, then, by Lemma 3.4, d3 = ∆∗. Letm be the number of neighborsof v of degree ≥ 4.

(1211). If m = 0, then c′(v) = c(v) = − 6∆∗ . Again by Lemma 3.4, v1 is a

(4, 4,∆∗)-vertex and fi = vv1v′4−iv4−i, i = 1, 2.

(12111). If there is i ∈ {1, 2} with deg(v′4−i) = 3, then, by Lemmas 3.1. (c) and

3.1. (d), the face adjacent to fi along the edge v′4−iv4−i is of degree k ≥ 6 and

c′(v′4−i) = c(v′

4−i) = c′(v4−i) = c(v4−i) = 32 − 6

k − 6∆∗ . Since c′(v1) ≥ c(v1) =

− 6∆∗ , the estimate

∑x∈V (fi) c

′(x) ≥ 2 · (− 6∆∗ ) + 2 · (3

2 − 6k − 6

∆∗ ) ≥ 3 − 126 −

2424 = 0 enables us to set D(v) := V (fi).

(12112). If min{deg(v′2),deg(v′

3)} ≥ 4, then, by Rule 2, c′(v1) = c(v1) −2c(v1) = 6

∆∗ and, with respect to c′(v) + c′(v1) = 0, we define D(v) := {v, v1}.Note that Lemma 3.1. (c) guarantees that for any two (4, 4,∆∗)-verticesu1, u2 withno neighbor of degree ≥ 4 we have eitherD(u1) = D(u2) orD(u1)∩D(u2) = ∅.

(1212). If m ≥ 1, we may suppose that D(u) is already determined for all(4, 4,∆∗)-vertices u having no neighbor of degree ≥ 4. If v is in no such D(u),we may put D(v) := {v}, because of c′(v) = c(v) −mc(v) = 6(m−1)

∆∗ ≥ 0.

(122). For d2 ≥ 5, we assume that D(u) is already defined for all (4, 4,∆∗)-vertices. If v does not belong to any such D(u), we set D(v) := {v}—we have,by Lemma 3.4, d2 + d3 ≥ ∆∗ + 4 and, by Lemma 4, c′(v) = c(v) ≥ γ(4, 5,∆∗ − 1) = 3

10 − 6∆∗−1 ≥ 3

10 − 623 > 0.


(13). For d1 ≥ 5, analogously as above, c′(v) = c(v) ≥ γ(5, 5,∆∗ − 2) ≥35 − 6

22 > 0 and D(v) := {v}.

(2). For n ≥ 4, showing that c′(v) ≥ 0 will allow us to set D(v) := {v}.

(21). n = 4.

(211). cd(v) ≤ ∆∗ + 1: By Lemma 3.3.(i) there is no contractible edge incidentwith v, hence, by Lemma 2, v is adjacent to at least three 3-vertices. By Lemmas 3.1.(e) and 3.4, there is no i ∈ [1, 4] such that di = di+1 = 3. Because cd(v) ≤ ∆∗+1,no face incident with v is of degree ∆∗ Now, by Lemmas 3.1. (e), 3.2, and 3.4,all faces incident with v are of degree ≥ 4; then c(v) = 6 − 6

∑4i=1

1di

≥ 6 −6

∑4i=1

14 = 0. On the other hand, by Lemma 3.3.(ii), t(v, vi) = 0, i = 1, 2, 3, 4,

and c′(v) = c(v) ≥ 0.

(212). If cd(v) ≥ ∆∗ + 2, then m := |{i ∈ [1, 4] : deg(fi) = 3}| ≤ 2.

(2121). m = 2.

(21211). Suppose that v is incident with two adjacent 3-faces, say, f2 and f3.Then, by Lemmas 3.1. (e) and 3.4, deg(vi) ≥ 4, i = 1, 2, 3. From Lemma 4 weknow that c(v) ≥ γ(3, 3, 4,∆∗) = 1

2 − 6∆∗ ≥ 1

4 and, as t(v, vi) = 0, i = 1, 2, 3,and t(v, v4) ≤ 1

4 (v can send something to v4 only by Rule 2), c′(v) ≥ 0.

(21212). Now let v be incident with two nonadjacent 3-faces, say, f2 and f4. Wewill show that p′(v, v1) + p′(v, v2) ≥ 0; the inequality p′(v, v3) + p′(v, v4) ≥ 0can be proved similarly. We have p(v, v1) + p(v, v2) =

∑2i=1 σ(di, di+1, 4) =

1 − 3d1

− 3d3

, which is, because of d1 + d3 ≥ ∆∗ + 4, lower bounded by 1 − 34 −

3∆∗ ≥ 1

8 . Thus, it is sufficient to analyze the case t(v, v1)+ t(v, v2) > 0, for whichmin{deg(v1),deg(v2)} = 3.

(212121). If exactly one of v1, v2 is of degree 3, say, deg(v1) = 3 < deg(v2),then, by Lemmas 3.1. (e) and 3.4, d1 ≥ 5 and d3 = ∆∗; in this case t(v, v1) =−1

2c(v1) ≤ 940 , t(v, v2) = 0, and p′(v, v1) + p′(v, v2) = 1 − 3

d1− 3

∆∗ + 12c(v1) ≥

1 − 35 − 3

24 − 940 = 1

20 .

(212122). Now let deg(v1) = deg(v2) = 3. Let d′2 be the degree of the face

adjacent to f1 along the edge v1v2. Then, by Lemma 3.4, d1 + d′2 ≥ ∆∗ + 5.

We may assume that d1 ≤ d3. Since t(v, v1) + t(v, v2) = −c(v1) − c(v2) =−2 + 6

d1+ 12

d′2

+ 6d3

≤ −2 + 12d1

+ 12d′2, we may suppose that min{d1, d

′2} ≤ 11. We

have to estimate p′(v, v1) + p′(v, v2) = 3 − 9d1

− 12d′2

− 9d3

.

(2121221). If d1 ≤ 11, then, with respect to min{d′2, d3} ≥ ∆∗ + 4 − d1, p′(v,

v1) + p′(v, v2) ≥ 3 − 9d1

− 21∆∗+4−d1

≥ 3 − 9d1

− 2128−d1

≥ 3 − 95 − 21

23 > 0.

(2121222). If d′2 ≤ 11, then d3 ≥ d1 ≥ ∆∗+5−d′

2. Since d′2 ≥ 6 by Lemma 3.1.

(b), p′(v, v1)+p′(v, v2) ≥ 3− 12d′2− 18

∆∗+5−d′2

≥ 3− 12d′2− 18

29−d′2

≥ 3− 126 − 18

23 > 0.


(2122). m = 1: Without loss of generality, assume that d2 = 3. Set r := |{i ∈{1, 2} : deg(vi) = 3}|.(21221). If r = 0, then t(v, vi) = 0, i = 1, 2, and t(v, vi) ≤ 1

4 , i = 3, 4 (here onlyRule 2 can be applied). From Lemma 4, it follows c(v) ≥ γ(3, 4, 4,∆∗ − 1) =1 − 6

∆∗−1 ≥ 1723 , so that c′(v) ≥ 17

23 − 2 · 14 > 0.

(21222). For r = 1, we may assume that deg(v1) = 3. Then t(v, v2) = 0 and, byLemma 3.4, d1 ≥ 5. With respect to d4 ≥ 4 then t(v, v4) = 0 and

∑4i=1 t(v, vi) ≤

940 + 1

4 = 1940 . On the other hand, by Lemma 4, c(v) ≥ γ(3, 4, 5,∆∗ − 2) =

1310 − 6

∆∗−2 ≥ 1310 − 6

22 >1940 .

(21223). If r = 2, then, by Lemma 3.4, d1, d3 ≥ 5. As in (21212), we havep′(v, v1) + p′(v, v2) ≥ 0. Thus, it will be sufficient to prove that p′(v, v3) +p′(v, v4) ≥ 0. It holds p(v, v3) + p(v, v4) = 3 − 3

d1− 3

d3− 6

d4≥ 3 − 2 · 3

5 − 64 =

310 , hence, we are done if

∑4i=3 t(v, vi) ≤ 1

4 < 310 . However, the mentioned

sum can be greater than 14 only if d1 = d3 = ∆∗ and d4 = 4; in such a case

p′(v, v3) + p′(v, v4) = 3 − 2 · 3∆∗ − 6

4 − 2 · 6∆∗ > 0.

(2123). Ifm = 0, then, by Lemma 4, c(v) ≥ γ(4, 4, 4,∆∗ −2) = 32 − 6

∆∗−2 ≥ 2722

and c′(v) ≥ 2722 − 4 · 1

4 > 0.

(22). n = 5.

(221). For cd(v) ≤ ∆∗ + 1, we know from Lemma 3.3.(i) that there is no con-tractible edge incident with v, hence, by Lemma 2, at least three from amongneighbors of v are of degree 3. Therefore, by Lemmas 3.1. (e) and 3.4, there is noi ∈ [1, 5] with di = di+1 = 3. We are going to show that p′(v, vi) > 0 for anyi ∈ [1, 5]. If deg(vi) ≥ 4, then p′(v, vi) = p(v, vi) ≥ 3 − 3

3 − 34 − 6

5 > 0. Fordeg(vi) = 3 and min{di, di+1} ≥ 4, we have p′(v, vi) ≥ (3 − 2 · 3

4 − 65) − 1

4 > 0.Finally, suppose that deg(vi) = 3 and, without loss of generality, di = 3. ByLemma 3.3.(ii), cd(vi) ≥ ∆∗ + 3. Therefore, if fi shares the edge vi−1vi with ad′

i-face, then d′i + di+1 ≥ ∆∗ + 6 and p′(v, vi) = σ(3, di+1, 5) + γ(3, d′

i, di+1) =95 − 6( 1

d′i

+ 1di+1

) − 3di+1

≥ 95 − 6(1

6 + 1∆∗ ) − 3

6 ≥ 310 − 6

24 > 0. Thus, c′(v) is a

sum of positive summands.

(222). cd(v) ≥ ∆∗ + 2.

(2221). Suppose first there is a 3-face incident with v, say, f2. Without loss ofgenerality, assume that it is incident with the maximum number, say q, of 3-vertices,from among all 3-faces incident with v.

(22211). If q = 2, then, by Lemma 3.4, min{d1, d3} ≥ 5, and, by Lemma 4,c(v) ≥ γ(3, 3, 5, 5,∆∗ − 4) = 13

5 − 6∆∗−4 ≥ 23

10 , so that c′(v) ≥ 2310 − 5 · 9

20 > 0.

(22212). For q = 1 assume, without loss of generality, that deg(v1) ≥ 4. Then,by Lemmas 3.1. (e) and 3.4, d1 = ∆∗ and d3 ≥ 5. Also, we have t(v, v1) = 0 andt(v, v2) ≤ 9

40 .


(222121). For d4 = d5 = 3, we obtain deg(v4) ≥ 4 and t(v, v4) = 0, so thatc′(v) ≥ (3 − 6

∆∗ − 6d3

) − ( 940 + 2 · 9

20) ≥ 3 − 624 − 6

5 − 98 > 0.

(222122). If max{d4, d5} ≥ 4, then c′(v) ≥ (7− 6∆∗ −∑5

i=36di

)−( 940 +3 · 9

20) ≥7 − 6

24 − 65 − 6

3 − 64 − 63

40 > 0.

(22213). If q = 0, then t(v, vi) = 0, i = 1, 2.

(222131). If v is incident with at most three 3-faces, then, by Lemma 4, c(v) ≥γ(3, 3, 3, 4,∆∗ − 1) = 3

2 − 6∆∗−1 ≥ 57

46 and, since v can send out something onlyby Rule 2, c′(v) ≥ 57

46 − 3 · 14 > 0.

(222132). Ifv is incident with four 3-faces, then c′(v) = c(v) = γ(3, 3, 3, 3,∆∗) =1 − 6

∆∗ ≥ 34 .

(2222). If min{d1, . . . , d5} ≥ 4, then p(v, vi) = 95 − 3

di− 3

di+1≥ 9

5 − 2 · 34 = 3

10

and p′(v, vi) ≥ 310 − 1

4 = 120 so that c′(v) ≥ 1

4 .

(23). n = 6.

(231). Suppose first cd(v) ≤ ∆∗ + 1. If deg(vi) ≥ 4, then p′(v, vi) = p(v, vi) ≥3 − 2 · 3

3 − 66 = 0. For deg(vi) = 3, we may proceed as in (221) to see that

p′(v, vi) > 0. We conclude that c′(v) is a sum of nonnegative summands.

(232). If cd(v) ≥ ∆∗ + 2, Lemma 4 yields c(v) ≥ γ(3, 3, 3, 5, 5,∆∗ − 5) =185 − 6

∆∗−5 ≥ 31295 and c′(v) ≥ 312

95 − 6 · 920 > 0.

(24). The assumption n ≥ 7 leads to p(v, vi) ≥ 157 − 3

di− 3

di+1> 0 for all i ∈

[1, n]. If t(v, vi) > 0 and min{di, di+1} = 3, then max{di, di+1} ≥ 5 (Lemma3.4) and p(v, vi) = σ(di, di+1, n) ≥ 8

7 − 35 = 19

35 , so that p′(v, vi) ≥ 1935 − 9

20 > 0.If t(v, vi) > 0 and min{di, di+1} = 4, then p(v, vi) ≥ 15

7 − 2 · 34 = 9

14 andp′(v, vi) ≥ 9

14 − 14 > 0. Thus, c′(v) is a sum of positive summands.

4. CONCLUDING REMARK

Note that besides our Theorem, it is possible to prove that χc(G) is upper boundedby max{∆∗(G) + 3, 23} and by max{∆∗(G) + 4, 21}. For that, an analogue ofLemma 3 and the Discharging Method with Rule 1 can be used.

ACKNOWLDEGEMENTS

The authors wish to thank to anonymous referees for their helpful comments, whichaided to improve the presentation of the article.


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On a conjecture by Plummer and Toft

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