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Article

Sufficient Conditions of 6-Cycles Make Planar Graphs DP-4-Colorable

by
Kittikorn Nakprasit
1,
Watcharintorn Ruksasakchai
2,* and
Pongpat Sittitrai
1
1
Department of Mathematics, Faculty of Science, Khon Kaen University, Khon Kaen 40002, Thailand
2
Department of Mathematics, Statistics and Computer Science, Faculty of Liberal Arts and Science, Kasetsart University, Kamphaeng Saen Campus, Nakhon Pathom 73140, Thailand
*
Author to whom correspondence should be addressed.
Mathematics 2022, 10(15), 2762; https://doi.org/10.3390/math10152762
Submission received: 28 June 2022 / Revised: 23 July 2022 / Accepted: 29 July 2022 / Published: 4 August 2022
(This article belongs to the Special Issue Domination, Independence and Distances in Graphs)

Abstract

:
In simple graphs, DP-coloring is a generalization of list coloring and thus many results of DP-coloring generalize those of list coloring. Xu and Wu proved that every planar graph without 5-cycles adjacent simultaneously to 3-cycles and 4-cycles is 4-choosable. Later, Sittitrai and Nakprasit showed that if a planar graph has no pairwise adjacent 3-, 4-, and 5-cycles, then it is DP-4-colorable, which is a generalization of the result of Xu and Wu. In this paper, we extend the results on 3-, 4-, 5-, and 6-cycles by showing that every planar graph without 6-cycles simultaneously adjacent to 3-cycles, 4-cycles, and 5-cycles is DP-4-colorable, which is also a generalization of previous studies as follows: every planar graph G is DP-4-colorable if G has no 6-cycles adjacent to i-cycles where i { 3 , 4 , 5 } .
MSC:
05C10; 05C15

1. Introduction

The concept of list coloring was introduced independently by Vizing [1] and Erdős et al. [2]. A k-list assignment L of a graph G assigns for each vertex v in G a list L ( v ) of k colors. An L-coloring is a proper coloring c such that c ( v ) L ( v ) for each v in V ( G ) . A graph G is L-colorable if G has an L-coloring. If G is L-colorable for any k-list assignment L, then G is said to be k-choosable.
DP-coloring is a generalization of list coloring. Dvořák and Postle [3] introduced the concept of DP-coloring and they called it correspondence coloring. Later on, it is called DP-coloring by Bernshteyn et al. [4].
Assume L is an assignment of a graph G. H is a cover of G if it admits all the following properties:
(i)
Its vertex set V ( H ) is v V ( G ) ( { v } × L ( v ) ) = { ( v , c ) : v V ( G ) , c L ( v ) } ;
(ii)
H [ { v } × L ( v ) ] is a complete graph for every v V ( G ) ;
(iii)
The set E H ( { u } × L ( u ) , { v } × L ( v ) ) is a matching (empty matching is allowable) for each u v E ( G ) .
(iv)
If u v E ( G ) , then there are no edges of H connect { u } × L ( u ) and { v } × L ( v ) .
An independent set in a cover H of a graph G with size | V ( G ) | is called an ( H , L ) -coloring of G. If every cover H with any k-assignment L of a graph G admits an ( H , L ) -coloring for G , then we say that G is DP-k-colorable. The minimum k in which a graph G is DP-k-colorable is called the DP-chromatic number of G and denoted by χ D P ( G ) .
If edges on H are defined to match exactly identical colors between L ( u ) and L ( v ) for each u v E ( G ) , then G admits an ( H , L ) -coloring is equivalent to G is L-colorable. Consequently, DP-coloring is a generalization of list coloring. Furthermore, this implies that χ D P ( G ) χ l ( G ) .
Dvořák and Postle [3] proved that for every planar graph G, χ D P ( G ) 5 , which extends a seminal result by Thomassen [5] on list coloring. Meanwhile, Voigt [6] constructed an example of a non-4-choosable planar graph (and thus, not DP-4-colorable). It motivates the investigation to obtain sufficient conditions for being DP-4-colorable of planar graphs. Kim and Ozeki [7] proved that every planar graph is DP-4-colorable if it does not contain k-cycles for each k = 3 , 4 , 5 , 6 . Kim et al. [8] proved that every planar graph is DP-4-colorable if it contains neither 7-cycles nor butterflies. In [9], Kim and Yu proved that every planar graph is DP-4-colorable if it does not contain triangles adjacent to 4-cycles, which extends the result on 3- and 4-cycles. In 2019, Liu and Li [10] improved the previous result of Kim and Yu [9] by relaxing the condition of one triangle into two triangles. Chen et al. [11] showed that every planar graph that contains no 4-cycles adjacent to k-cycles where k = 5 , 6 is DP-4-colorable. Liu et al. [12] extended the result of Kim and Ozeki [7] on 3-, 5-, and 6-cycles by proving that every planar graph contains no k-cycles adjacent to triangles is DP-4-colorable. Xu and Wu [13] proved that every planar graph, which contains no 5-cycles adjacent simultaneously to 3-cycles and 4-cycles is 4-choosable. Recently, Sittitrai and Nakprasit [14] showed that every planar graph that contains no pairwise adjacent 3-, 4-, and 5-cycle is DP-4-colorable which generalizes the result of Xu and Wu [13].
In this work, the results on 3-, 4-, 5-, and 6-cycles are extended by the result on Theorem 1, which generalizes the aforementioned results by Chen et al. [11] and Liu et al. [12].
Theorem 1.
Every planar graph without 6-cycles simultaneously adjacent to 3-cycles, 4-cycles, and 5-cycles is DP-4-colorable.
Then we have the following two Corollaries. Moreover, some results on [11,12] are some part of Corollary 1 for i = 4 and i = 3 , respectively.
Corollary 1.
Every planar graph without 6-cycles adjacent to i-cycles is DP-4-colorable for each i { 3 , 4 , 5 } .
Corollary 2.
Every planar graph without 6-cycles simultaneously adjacent to i-cycles and j-cycles is DP-4-colorable for each i , j { 3 , 4 , 5 } and i j .

2. Preliminaries

First, some notations and definitions are introduced in this section. Let G be a plane graph. The vertex set, edge set, and face set of the graph G are denoted, respectively, by V ( G ) , E ( G ) , and F ( G ) . We use B ( f ) to denote the boundary of a face f . Two faces f and g are adjacent if B ( f ) and B ( g ) are adjacent. A wheel  W n is a graph of n vertices formed by connecting all vertices of an ( n 1 ) -cycle (these vertices are called external vertices) to a single vertex (hub). A k-vertex, k + -vertex, and k -vertex is a vertex of degree k, at least k, and at most k , respectively. Similar notation is applied to cycles and faces.
Note that some faces may appear several times in the order. If a face is incident to at least two 5 + -vertices (respectively, exactly one 5 + -vertex, no 5 + -vertices), it is called rich (semi-rich, poor, respectively).
A semi-rich 5-face is a proper semi-rich 5-face if each incident edge with two endpoints of degree 4 is on the boundary of a 3-face, otherwise it is called an improper semi-rich 5-face.
A bounded face is an extreme face if it has a vertex incident to the unbounded face. An inner face is a bounded face but is not an extreme face.
An edge u v is a chord in an embedding cycle C if u , v V ( C ) but u v is not in E ( C ) . If a chord is inside C , then it is called an internal chord, otherwise it is called an external chord. A graph C ( m , n ) is obtained from a cycle x 1 x 2 x m + n 2 with an internal chord x 1 x m . For example, cycles u v w and v w x y z form C ( 3 , 5 ) . A graph C ( l , m , n ) is obtained from a cycle x 1 x 2 x l + m + n 4 with internal chords x 1 x l and x 1 x l + m 2 . The previous definition can be extended similarly to a graph C ( m , n , p , q ) . The graphs i n t ( C ) and e x t ( C ) are induced by vertices inside and outside a cycle C , respectively. A separating cycle C is a cycle with non-empty i n t ( C ) and e x t ( C ) .
Let A denote the family of planar graphs without 6 cycle simultaneously adjacent 3-, 4-, and 5-cycle.
To prove that every planar graph without 6-cycles simultaneously adjacent to 3-cycles, 4-cycles, and 5-cycles is DP-4-colorable, we prove a stronger result as follows.
Theorem 2.
If G A with a precolored 3-cycle, then the precoloring can be extended to be a DP-4-coloring of G.

3. Structures

Let G be a minimal counterexample to Theorem 2 with respect to the order | V ( G ) | . Then, (i) G A and (ii) G is a minimal graph with a precoloring of a 3-cycle that cannot be extended to be a DP-4-coloring in G . Some tools in [14] are used to deal with graphs satisfying (ii). We assume that G contains a 3-cycle since every planar graph without 3-cycles is DP-4-colorable [9].
Thus we let C 0 be a 3-cycle in G that is precolored.
Lemma 1
(Lemma 3.1 in [14]). G has no separating 3-cycles (See the proof in Lemma A1).
It follows from Lemma 1 that we may assume C 0 to be the boundary of the unbounded face of G .
Lemma 2
(Lemma 3.3 in [14]). Each vertex in i n t ( C 0 ) has degree at least four (See the proof in Lemma A3).
Lemma 3.
The following statements hold.
(i)
A bounded 6 -face has its boundary as a cycle.
(ii)
If a bounded k 1 -face f and a bounded k 2 -face g with k 1 + k 2 8 are adjacent, then B ( f ) B ( g ) = C ( k 1 , k 2 ) .
(iii)
Let a bounded 3-face f and a bounded 4-face g be adjacent. If f or g is adjacent to a bounded 3-face h, then B ( f ) B ( g ) B ( h ) is a 6-cycle with two internal chords.
Proof. 
(i)
Clearly, a boundary of a 5 -face is a cycle. Consider a bounded 6-face f . A boundary closed walk is in a form of u v w x y w u if B ( f ) is not a cycle. By Lemma 2, u or x has degree at least 4 . It follows that u v w or x y w is a separating 3-cycle, contrary to Lemma 1.
(ii)
It suffices to show that B ( f ) and B ( g ) share exactly two vertices.
If B ( f ) = u v w , B ( g ) = v w x and u = x , then f or g is the unbounded face, a contradiction.
If B ( f ) = u v w , B ( g ) = v w x y and u = x or y , then d ( w ) = 2 or d ( v ) = 2 , which contradicts Lemma 2.
If B ( f ) = u v w , B ( g ) = v w x y z and u = x or z , then d ( w ) = 2 or d ( v ) = 2 , which contradicts Lemma 2. If B ( f ) = u v w , B ( g ) = v w x y z and u = y , then v y z or w x y is a separating 3-cycle, which contradicts Lemma 1.
If B ( f ) = s t u v , B ( g ) = u v w x and s = w , then d ( v ) = 2 , which contradicts Lemma 2. If B ( f ) = s t u v , B ( g ) = u v w x and s = x , then u t x or v w x is a separating 3-cycle, which contradicts Lemma 1. The remaining cases are similar.
(iii)
Lemma 3 (ii) yields that B ( f ) B ( g ) is a 5-cycle with one chord. Similar to the proof of Lemma 3 (ii), one can show that B ( h ) and B ( f ) B ( g ) share exactly two vertices. This yields a desired result.
Lemma 4.
If C is a 6-cycle and has a triangular chord, then C has only one chord. Moreover, every 6-cycle has at most one internal chord.
Proof. 
Let C be a 6-cycle t u v x y z and let t v be its triangular chord. Suppose to the contrary that C has at least two chords. Since C is adjacent to a 3-cycle t u v and a 5-cycle u v x y z , it suffices to show that C is adjacent to a 4-cycle. By symmetry, we assume another chord e of C is u x , u y , t x , t y , or x z .
If e = u x , then C is adjacent to a 4-cycle t u x v .
If e = u y , then C is adjacent to a 4-cycle u v x y .
If e = t x , then C is adjacent to a 4-cycle t u v x .
If e = t y , then C is adjacent to a 4-cycle v x y t .
If e = x z , then C is adjacent to a 4-cycle t v x z .
Thus, C has exactly one chord. Note that C has a triangular chord if C has at least two internal chords. It follows that every 6-cycle has at most one internal chord. □
A cluster in a plane graph G is a subgraph of G consisting of 3-cycles from a minimal set of bounded 3-faces such that they are not adjacent to other bounded 3-faces outside the set. A k-cluster is formed by k bounded 3-faces. An adjacent face of an i-cluster H i is a face that is adjacent to some bounded 3-face in H i . Since G A , one can observe that every cluster in G is a 4 -cluster where a 4-cluster is isomorphic to W 5 .
Lemma 5.
The following statements hold.
(i)
If a 4-face f is adjacent to an inner 3-face g, then f is not adjacent to other inner 3-faces and f is not adjacent to any 4-faces.
(ii)
If an inner 3-face f is adjacent to a 5-face g , then f and g are not adjacent to any 4-faces.
(iii)
Every adjacent face of a 2-cluster is a 6 + -face or the unbounded 3-face D.
(iv)
Every adjacent face of a 3 + -cluster is a 7 + -face or the unbounded 3-face D.
Proof. 
(i)
Let f be a 4-face adjacent to an inner 3-face g and another face h.
Suppose to the contrary that h is an inner 3-face or a 4-face.
If h is an inner 3-face, then B ( f ) B ( g ) B ( h ) is a 6-cycle with two internal chords by Lemma 3 (iii), contrary to Lemma 4.
If h is a 4-face, then Lemma 3 (ii) yields a 6-cycle from B ( f ) B ( h ) , which is adjacent to a 5-cycle from B ( f ) B ( g ) , a 4-cycle from B ( f ) , and a 3-cycle from B ( g ) , contrary to G A .
(ii)
Let an inner 3-face f and a 5-face g be adjacent. Lemma 3 (ii) yields that B ( f ) B ( g ) contains a 6-cycle. Thus, f or g is not adjacent to any 4-faces since G A .
(iii)
Let f and g be bounded 3-faces in a 2-cluster H 2 and let h be a bounded face adjacent to f. By the definition, h is not a bounded 3-face.
If h is a 4-face, then Lemma 3 (iii) yields that B ( f ) B ( h ) B ( g ) contains a 6-cycle with two internal chords, contrary to Lemma 4.
If h is a 5-face, then it follows from Lemmas 3 (i) and (ii) that a 6-cycle from B ( f ) B ( h ) is adjacent to a 5-cycle from B ( h ) , a 4-cycle from B ( f ) B ( g ) , and 3-cycle from B ( f ) , contrary to G A .
Thus, h is a 6 + -face or the unbounded face.
(iv)
Let f 1 , f 2 , and f 3 be the bounded 3-faces of 3 + -cluster H 3 in a consecutive order.
By similar arguments as in the proof of (iii), it follows that H 3 cannot be adjacent to a bounded 5 -face.
Let H 3 be adjacent to a 6-face f 4 . By Lemma 3 (ii) and an argument similar to its proof, one can show that H 3 is a 5-cycle with two chords. Since B ( f 4 ) is a 6-cycle by Lemma 3 (i), we have a 6-cycle adjacent to a 3-, a 4-, and a 5-cycle in H 3 , contrary to G A .
If H 3 is adjacent to a 6-face f 4 , then by Lemma 3 (ii), a 6-cycle B ( f 4 ) is adjacent to a 3-, a 4-, and a 5-cycle, which are in H 3 , contrary to G A .
For Corollary 3 (i), it is proved by the fact that every 5 + -vertex is not adjacent to four consecutive bounded 3-faces. Thus, each 5 + -vertex has at least two 4 + -faces. For Corollary 3 (ii), it is proved by Lemmas 5 (iii) and (iv) that each 3-face in H 2 + is not adjacent to a 5-face. Thus, each 5 + -vertex has at least three 4 + -faces.
Corollary 3.
Let v be a k-vertex in G where v V ( C 0 ) and k 5 . It follows that:
(i)
v is incident to at most k 2 bounded 3-faces;
(ii)
v is incident to at most k 3 bounded 3-faces, if v has an incident 5-face.
Proof. 
If v is incident to k 1 bounded 3-faces, then there are four consecutive bounded faces forming a 4-cluster that is not a wheel, contrary to G A . This proves (i). It follows from Lemmas 5 (iii) and (iv) that each 3-face in a 2 + -cluster is not adjacent to a 5-face. Thus, each 5 + -vertex incident to a 5-face must be incident to at least three 4 + -faces. This proves (ii). □
Lemma 6
(Lemma 3.6 in [14]). C ( l 1 , , l k ) is defined to be a cycle C = x 1 x m with k internal chords such that x 1 is their common endpoint and V ( C ) V ( C 0 ) = . Suppose x 2 or x m is not the endpoint of any chords in C . If d ( x 1 ) k + 3 , then some i { 2 , 3 , , m } satisfies d ( x i ) 5 (See the proof in Lemma A4).
Lemma 7.
Let a 4-vertex v be incident to bounded faces f 1 , , f 4 in cyclic order and let F = B ( f 1 ) B ( f 2 ) , where V ( F ) V ( C 0 ) = . If ( d ( f 1 ) , d ( f 2 ) ) = ( 3 , 3 ) or ( 3 , 5 ) , then there is a vertex w V ( F ) { v } such that d ( w ) 5 .
Proof. 
If ( d ( f 1 ) , d ( f 2 ) ) = ( 3 , 3 ) , it follows from Lemma 3 (ii) that F = C ( 3 , 3 ) . Moreover, F has exactly one chord, otherwise there is a separating 3-cycle, which contradicts Lemma 1.
If ( d ( f 1 ) , d ( f 2 ) ) = ( 3 , 5 ) , it follows from Lemma 3 (ii) that F = C ( 3 , 5 ) . Moreover, F has exactly one chord by Lemma 4.
The proof is complete by Lemma 6. □
Lemma 8.
Let v be a 5-vertex with incident bounded faces f 1 , , f 5 in a cyclic order. Let F = B 1 B 2 B 3 where B i denote B ( f i ) and V ( F ) V ( C 0 ) = . If ( d ( f 1 ) , d ( f 2 ) , d ( f 3 ) ) = ( 5 , 3 , 5 ) , then there exists w V ( F ) { v } such that d ( w ) 5 .
Proof. 
Let B 1 = x 1 x 2 x 3 x 4 x 5 , B 2 = x 1 x 5 x 6 , and B 3 = x 1 x 6 x 7 x 8 x 9 , where x 1 = v . It follows from Lemma 3 (ii) that B 1 B 2 is a C ( 3 , 5 ) and B 2 B 3 is a C ( 3 , 5 ) . Suppose to the contrary that F is not a C ( 5 , 3 , 5 ) . Then, there is i { 2 , 3 , 4 } and j { 7 , 8 , 9 } such that x i = x j . If i = 2 , then a 6-cylcle x 1 x 5 x 6 x 7 x 8 x 9 has a triangular chord x 1 x 6 and a chord x 1 x j , contrary to Lemma 4. If i = 2 , then a 6-cylcle x 1 x 5 x 6 x 7 x 8 x 9 , has a triangular chord x 1 x 6 and a chord x 5 x j , contrary to Lemma 4.
Suppose that i = 3 . Note that a 6-cycle C = x 1 x 5 x 6 x 7 x 8 x 9 is adjacent to a 3-cycle x 1 x 5 x 6 and a 5-cycle x 1 x 6 x 7 x 8 x 9 . It suffices to show that C is adjacent to a 4-cycle to get a contradiction. If x 3 = x 7 , then C is adjacent to a 4-cycle x 1 x 2 x 7 x 6 . If x 3 = x 8 , then C is adjacent to a 4-cycle x 1 x 2 x 8 x 9 . If x 3 = x 9 , then C is adjacent to a 4-cycle x 1 x 5 x 4 x 9 .
Thus, F = C ( 5 , 3 , 5 ) . By Lemma 6, it remains to show that x 2 or x m is not an endpoint to a chord in C , say x 1 x 2 x 9 . Suppose C has a chord e = x 2 x i , otherwise the desired condition is obtained. If x 2 x 9 E ( G ) , then we have separating 3-cycle x 1 x 2 x 9 , contrary to Lemma 1. By Lemma 4, we have i { 4 , 5 , 6 } . Then, x i = x 7 or x 8 . By Lemma 4, x 9 is not adjacent to x 6 or x 7 . Thus, a chord of C cannot have x 9 as its endpoint. □
Corollary 4.
Let v be a 4-vertex incident to bounded faces f 1 , , f 4 in cyclic order, where f 1 is an inner 5-face, f 2 is an inner 3-face, f 3 is an inner 5-face, and f 4 is an arbitrary face. If f 3 is a poor 5-face, then f 1 is a rich 5-face or an improper semi-rich 5-face.
Proof. 
Let B 1 = x 1 x 2 x 3 x 4 x 5 , B 2 = x 1 x 5 x 6 , and B 3 = x 1 x 6 x 7 x 8 x 9 , where x 1 = v . Let f 3 be a poor 5-face. Then, x 1 , x 6 , x 7 , x 8 , and x 9 are 4-vertices. By Lemma 7, x 5 is a 5 + -vertex. If x 2 , x 3 , or x 4 is a 5 + -vertex, then f 1 is a rich 5-face. Now suppose that x 2 , x 3 , and x 4 are 4-vertices. If f 4 is a not a 3-face, then f 1 is an improper semi-rich 5-face. If f 4 is a 3-face, then x 2 is a 5 + -vertex by considering f 1 and f 4 into Lemma 7, a contradiction. □

4. Discharging Process

In this section, we use the discharging procedure to get a contradiction and complete the proof of Theorem 2.
For each vertex and bounded face x V ( G ) F ( G ) , let an initial charge of x be μ ( x ) = d ( x ) 4 and let μ ( D ) = d ( D ) + 4 = 7 where D is the unbounded face. By Euler’s Formula, x V F μ ( x ) = 0 . Let μ * ( x ) be the charge after the discharge procedure of x V F . To get a contradiction, we prove that μ * ( x ) 0 for each x V ( G ) F ( G ) and μ * ( D ) > 0 .
Let w ( x f ) be the transferred charge from x to a face f where x is a vertex or a face.
The discharging rules:
(R1) Let v be a 5-vertex where v V ( C 0 ) and f be an incident 3-face of v.
w ( v f ) = 1 2 , if v is incident to some 5 - faces , 1 7 , if v is not incident to any 5 - faces and f is not adjacent to any incident 3 - faces of v , 3 7 , if v is not incident to any 5 - faces and f is adjacent to exactly one incident 3 - face of v .
(R2) Let v be a 6 + -vertex where v V ( C 0 ) and f be an incident 3-face of v.
w ( v f ) = 2 3 , if v is incident to some 5 - faces , 1 2 , if v is not incident to any 5 - faces .
Let g be a k-face with k incident vertices, say v 1 , v 2 , , v k in cyclic order, and with k adjacent faces, say f 1 , f 2 , , f k in cyclic order. Let f i be incident to v i and v i + 1 (i is taken modulo k).
(R3) Let g be a 4-face.
w ( g f i ) = 1 3 if f i is an inner 3-face.
(R4) Let g be a 5-face.
w ( g f i ) = 1 5 if f i is a 4-face.
  • Let g be an inner poor 5-face.
    w ( g f i ) = 1 5 if f i is an inner 3-face.
  • Let g be an inner proper semi-rich 5-face.
    w ( g f i ) = 1 3 if f i is an inner 3-face where both v i and v i + 1 are 4-vertices.
  • Let g be an inner rich 5-face or an inner improper semi-rich 5-face.
    w ( g f i ) = 1 6 , if f i is an inner 3 - face where exactly one of v i and v i + 1 is a 4 - vertex . 1 3 , if f i is an inner 3 - face where both v i and v i + 1 are 4 - vertices .
  • Let g be an extreme 5-face.
    w ( g f i ) = 2 3 if f i is an inner 3-face.
(R5) Let g be a k-face where k 6 .
w ( g f i ) = θ ( f i ) + χ ( f i + 1 ) θ ( f i + 1 ) + χ ( f i 1 ) θ ( f i 1 ) , if f i is a 4 - face , 0 , otherwise . where χ ( f i ) = 1 2 , if f i is not a 4 - face , 0 , otherwise . and θ ( f i ) = d ( g ) 4 d ( g ) for each i { 1 , 2 , , n } .
(R6) The unbounded face D incident to a vertex v receives charge μ ( v ) from v but gives 1 to each of its intersecting 3-faces and 5-faces.
It follows from (R6) that μ * ( v ) = 0 for every v V ( C 0 ) . By this, we consider only a vertex v such that v V ( C 0 ) .
CASE 1:v is a 5-vertex.
  • v is incident to some 5-faces.
    Then, v has at most two incident 3-faces by Corollary 3. Thus, μ * ( v ) μ ( v ) 2 × 1 2 = 0 by (R1).
  • v is not incident to any 5-faces.
    It follows from Corollary 3 that v is incident to at most three 3-faces. Then, v has at most two incident 3-faces, which are adjacent to exactly one incident 3-face of v. Thus, μ * ( v ) μ ( v ) 2 × 3 7 1 7 = 0 by (R1).
CASE 2:v is a 6 + -vertex.
  • v is incident to some 5-faces.
    It follows from Corollary 3 that v is incident to not more than d ( v ) 3 of 3-faces. Thus, μ * ( v ) μ ( v ) ( d ( v ) 3 ) × 2 3 = ( d ( v ) 4 ) ( 2 d ( v ) 3 2 ) = d ( v ) 3 2 0 by (R2) and d ( v ) 6 .
  • v is not incident to any 5-faces.
    It follows from Corollary 3 that v is incident to at most d ( v ) 2 3-faces. Thus, μ * ( v ) μ ( v ) ( d ( v ) 2 ) × 1 2 = ( d ( v ) 4 ) ( d ( v ) 2 1 ) = d ( v ) 2 3 0 by (R2) and d ( v ) 6 .
For a 3-face in an i-cluster H i , we consider the total of charges in the same cluster. That is μ ( H i ) = i and we show that μ * ( H i ) 0 instead.
CASE 3:f is a 3-face in an i-cluster, say H i where | V ( H i ) V ( C 0 ) | 1 .
  • If | V ( H 1 ) V ( C 0 ) | 1 , then μ * ( H 1 ) μ ( H 1 ) + 1 = 0 by (R6).
  • If | V ( H 2 ) V ( C 0 ) | = 1 , then each adjacent face of H 2 is a 6 + -face by Lemma 5 (iii). Thus, μ * ( H 2 ) μ ( H 2 ) + 1 + 4 × 1 3 > 0 by (R5) and (R6).
  • If | V ( H 2 ) V ( C 0 ) | 2 , then each 3-face in H 2 is an extreme 3-face. Thus, μ * ( H 2 ) μ ( H 2 ) + 2 × 1 = 0 by (R6).
  • If | V ( H 3 ) V ( C 0 ) | = 1 , then each adjacent face of H 3 is a 7 + -face by Lemma 5 (iv). Thus, μ * ( H 3 ) μ ( H 3 ) + 1 + 5 × 3 7 > 0 by (R5) and (R6).
  • If | V ( H 3 ) V ( C 0 ) | = 2 , then H 3 is adjacent to at least four 7 + -faces by Lemma 5 (iv). Moreover, there are at least two extreme 3-faces in H 3 . Thus, μ * ( H 3 ) μ ( H 3 ) + 2 + 4 × 3 7 > 0 by (R5) and (R6).
  • If | V ( H 3 ) V ( C 0 ) | = 3 , then each 3-face in H 3 is an extreme 3-face. Thus, μ * ( H 3 ) μ ( H 3 ) + 3 × 1 = 0 by (R6).
  • If | V ( H 4 ) V ( C 0 ) | = 1 , then there are two extreme 3-faces in H 4 and each adjacent face of H 4 is a 7 + -face by Lemma 5 (iv). If each vertex in V ( H 4 ) V ( C 0 ) is a 4-vertex, we have μ * ( H 4 ) μ ( H 4 ) + 2 + 2 × 9 14 + 2 × 6 7 > 0 by (R5) and (R6). Otherwise, there is a vertex in V ( H 4 ) V ( C 0 ) , which is not a 4-vertex, then we have μ * ( H 4 ) μ ( H 4 ) + 2 + 6 × 3 7 > 0 by (R1), (R2), (R5), and (R6).
  • If | V ( H 4 ) V ( C 0 ) | = 2 , then there are at least three extreme 3-faces in H 4 . Moreover, H 3 is adjacent to at least three 7 + -faces by Lemma 5 (iv). Thus, μ * ( H 4 ) μ ( H 4 ) + 3 × 1 + 3 × 3 7 > 0 by (R5) and (R6).
  • If | V ( H 4 ) V ( C 0 ) | = 3 , then each 3-face in H 4 is an extreme 3-face. Thus, μ * ( H 4 ) μ ( H 4 ) + 4 × 1 = 0 by (R6).
CASE 4:f is an inner 3-face in a 1-cluster.
Let v 1 , v 2 , v 3 be three incident vertices in cyclic order and f 1 , f 2 , f 3 be three adjacent faces in cyclic order. Moreover, let f i be incident to v i and v i + 1 (i is taken modulo 3) (See Figure 1).
Subcase 4.1:f is not adjacent to any 5-faces.
Thus, μ * ( f ) μ ( f ) + 3 × 1 3 = 0 by (R3) and (R5).
Next, we consider that f is adjacent to some 5-faces in Subcases 4.2 to 4.5. It follows from Lemma 5 (ii) and the assumption of Case 4 that f is not adjacent to a 4 -faces.
Subcase 4.2: An inner 3-face f is adjacent to some extreme 5-faces.
WLOG, let f 1 be an extreme 5-face. Then, w ( f 1 f ) = 2 3 by (R4).
  • f i is not an inner 5-face where i = 2 or 3.
    Then, f i is an extreme 5-face or a 6 + -face. Thus, w ( f i f ) 1 3 by (R4) and (R5). Therefore, μ * ( f ) μ ( f ) + 2 3 + 1 3 = 0 .
  • f 2 and f 3 are inner 5-faces.
    - If v i is a 5 + -vertex for some i { 1 , 2 , 3 } , then w ( v i f ) = 1 2 by (R1) and (R2). Thus, μ * ( f ) μ ( f ) + 2 3 + 1 2 > 0 .
    - If v i is a 4-vertex for each i { 1 , 2 , 3 } , then w ( f 2 f ) 1 5 and w ( f 3 f ) 1 5 by (R4). Thus, μ * ( f ) μ ( f ) + 2 3 + 2 × 1 5 > 0 .
We now consider the cases that each adjacent 5-face of f is not an extreme 5-face.
Subcase 4.3:f is a poor 3-face.
It follows from Lemma 7 that f i is not a poor 5-face for each i { 1 , 2 , 3 } . Thus, μ * ( f ) μ ( f ) + 3 × 1 3 = 0 by (R4) and (R5).
Subcase 4.4:f is a semi-rich 3-face.
Let v 1 be a 5 + -vertex. By symmetry, we only consider two following cases.
  • f 2 is a poor 5-face.
    Then w ( f 2 f ) 1 5 by (R4). Note that if f i is an improper semi-rich 5-face, a rich 5-face, or a 6 + -face where i { 1 , 3 } , then w ( f i f ) 1 6 by (R4) and (R5).
    - If f i is a 5-face for i = 1 or 3, then f i is an improper semi-rich 5-face or a rich 5-face by Corollary 4. It follows that w ( v 1 f ) 1 2 by (R1) and (R2). Thus, μ * ( f ) μ ( f ) + 2 × 1 6 + 1 5 + 1 2 > 0 .
    - If f 1 and f 3 are 6 + -faces, then w ( v 1 f ) 1 7 by (R1) and (R2). Thus, μ * ( f ) μ ( f ) + 2 × 1 3 + 1 5 + 1 7 > 0 .
  • f 2 is a 5 + -face but not a poor 5-face.
    Then w ( f 2 f ) 1 3 by (R4) and (R5). If f 1 and f 3 are 6 + -faces, then μ * ( f ) μ ( f ) + 3 × 1 3 = 0 by (R5). If v 1 is a 6 + -vertex and f 1 or f 3 is a 5-face, then μ * ( f ) μ ( f ) + 1 3 + 2 3 = 0 by (R2). Thus, it remains to check the case that f 1 or f 3 is a 5-face and v 1 is a 5-vertex. Note that w ( v 1 f ) 1 2 by (R1).
    - If f 1 and f 3 are 5-faces, then f i is a rich 5-face for i = 1 or 3 by Lemma 8. It follows that w ( f i f ) = 1 6 by (R4). Thus, μ * ( f ) μ ( f ) + 1 3 + 1 2 + 1 6 = 0 .
    - If f 1 is a 5-face and f 3 is a 6 + -face, then w ( f 3 f ) 1 3 by (R5). Thus, μ * ( f ) μ ( f ) + 2 × 1 3 + 1 2 = 0 .
Subcase 4.5:f is an inner rich 3-face.
Let v 1 and v 2 be 5 + -vertices. Recall that f 1 , f 2 , and f 3 are inner 5 + -faces and at least one of them is a 5-face. By symmetry, we only consider two following cases.
  • f 1 is a 5-face or f 2 and f 3 are 5-faces.
    That makes v 1 and v 2 incident to some 5-faces. Then, w ( v 1 f ) 1 2 and w ( v 2 f ) 1 2 by (R1) and (R2). Thus, μ * ( f ) μ ( f ) + 2 × 1 2 = 0 .
  • f 1 is a 6 + -face and either f 2 or f 3 is a 6 + -face.
    WLOG, let f 2 be a 5-face. That makes v 2 incident to some 5-faces. Then w ( f 1 f ) 1 3 and w ( f 3 f ) 1 3 by (R5) and w ( v 2 f ) 1 2 by (R1) and (R2). Thus, μ * ( f ) μ ( f ) + 2 × 1 3 + 1 2 > 0 .
CASE 5:f is a 3-face in a 2-cluster H 2 where | V ( H 2 ) V ( D ) | = 0 .
Let H 2 be a 4-cycle v 1 v 2 v 3 v 4 with a chord v 1 v 3 . Let f 1 , f 2 , f 3 , f 4 be four adjacent faces of H 2 in cyclic order. Moreover, let f i be incident to v i and v i + 1 (i is taken modulo 4) (See Figure 2). It follows from Lemma 5 (iii) that f 1 , f 2 , f 3 , and f 4 are 6 + -faces. By symmetry, we only consider two following cases.
  • v 1 and v 3 are 4-vertices.
    Then w ( f i H 2 ) 1 2 for i { 1 , 2 , 3 , 4 } by (R5). Thus, μ * ( H 2 ) μ ( H 2 ) + 4 × 1 2 = 0 .
  • v 1 is a 5 + -vertex and v 3 is a 4 + -vertex.
    Then w ( v 1 H 2 ) 2 × 3 7 by (R1) and (R2), and w ( f i H 2 ) 1 3 for i { 1 , 2 , 3 , 4 } by (R5). Thus, μ * ( H 2 ) μ ( H 2 ) + 4 × 1 3 + 2 × 3 7 > 0 .
CASE 6:f is a 3-face in a 3-cluster H 3 where | V ( H 3 ) V ( D ) | = 0 .
Let H 3 be a 5-cycle v 1 v 2 v 3 v 4 v 5 with two chords v 1 v 3 and v 1 v 4 . Let f 1 , f 2 , f 3 , f 4 , f 5 be five adjacent faces of H 3 in cyclic order. Moreover, let f i be incident to v i and v i + 1 (i is taken modulo 5). Note that f 1 and f 5 may be the same face (See Figure 3). It follows from Lemma 5 (iv) that f 1 , f 2 , f 3 , f 4 , and f 5 are 7 + -faces. By symmetry, we only consider the two following cases.
  • v 3 and v 4 are 4-vertices.
    Then w ( f 1 H 3 ) 3 7 and w ( f 5 H 3 ) 3 7 by (R5), w ( f 2 H 3 ) 9 14 and w ( f 4 H 3 ) 9 14 by (R5), and w ( f 3 H 3 ) = 6 7 by (R5). Thus, μ * ( H 3 ) μ ( H 3 ) + 2 × 3 7 + 2 × 9 14 + 6 7 = 0 .
  • v 3 is a 5 + -vertex and v 4 is a 4 + -vertex.
    Then w ( v 3 H 3 ) 2 × 3 7 by (R1) and (R2), and w ( f i H 3 ) 3 7 for i { 1 , 2 , 3 , 4 , 5 } by (R5). Thus, μ * ( H 3 ) μ ( H 3 ) + 7 × 3 7 = 0 .
CASE 7:f is a 3-face in a 4-cluster H 4 where | V ( H 4 ) V ( D ) | = 0 .
Let H 4 be the wheel W 5 where v 5 is a hub and v 1 , v 2 , v 3 , and v 4 are external vertices in cyclic order. Let f 1 , f 2 , f 3 , f 4 be four adjacent faces of H 4 in cyclic order. Moreover, let f i be incident to v i and v i + 1 (i is taken modulo 4) (See Figure 4). By Lemma 5 (iv), f 1 , f 2 , f 3 , and f 4 are 7 + -faces. Moreover, at least two vertices in { v 1 , v 2 , v 3 , v 4 } are 5 + -vertices by Lemma 7. By symmetry, we only consider the three following cases.
  • v 1 and v 2 are 5 + -vertices and v 3 and v 4 are 4-vertices.
    Then w ( f 1 H 4 ) 3 7 , w ( f 2 H 4 ) 9 14 , w ( f 4 H 4 ) 9 14 , w ( f 3 H 4 ) = 6 7 by (R5), w ( v 1 H 4 ) 2 × 3 7 and w ( v 2 H 4 ) 2 × 3 7 by (R1) and (R2). Thus, μ * ( H 4 ) μ ( H 4 ) + 2 × 9 14 + 6 7 + 5 × 3 7 > 0 .
  • v 1 and v 3 are 5 + -vertices and v 2 and v 4 are 4-vertices.
    Then w ( f i H 4 ) 9 14 for i { 1 , 2 , 3 , 4 } by (R5), and w ( v 1 H 4 ) 2 × 3 7 and w ( v 3 H 4 ) 2 × 3 7 by (R1) and (R2). Thus, μ * ( H 4 ) μ ( H 4 ) + 4 × 9 14 + 4 × 3 7 > 0 .
  • v 1 , v 2 , and v 3 are 5 + -vertices and v 4 is a 4 + -vertex.
    Then w ( f i H 4 ) 3 7 for i { 1 , 2 , 3 , 4 } by (R5) and w ( v i H 4 ) 2 × 3 7 for i { 1 , 2 , 3 } by (R1) and (R2). Thus, μ * ( H 4 ) μ ( H 4 ) + 10 × 3 7 > 0 .
CASE 8:f is a 4-face adjacent to an inner 3-face, say h.
Since h is an inner 3-face, we have | B ( f ) B ( D ) | 2 where D is the unbounded 3-face. Consequently, there are at least two adjacent faces of f, which are not h and D. Moreover, they are 5 + -faces by Lemma 5 (i). Thus μ * ( f ) μ ( f ) 1 3 + 2 × 1 5 > 0 by (R3), (R4), and (R5).
CASE 9:f is a 5-face.
  • Let f be adjacent to some 4-faces.
    Then f is not adjacent to any 3-faces by Lemma 5 (ii). Thus, μ * ( f ) μ ( f ) 5 × 1 5 = 0 by (R4).
  • Let f be an inner poor 5-face.
    Then μ * ( f ) μ ( f ) 5 × 1 5 = 0 by (R4).
  • Let f be an inner semi-rich 5-face.
    - If f is a proper semi-rich 5-face, then B ( f ) has three edges with two 4-endpoints. Thus, μ * ( f ) μ ( f ) 3 × 1 3 = 0 by (R4).
    - If f an improper semi-rich 5-face, then B ( f ) has at most two edges with two 4-endpoints and at most two edges with exactly one 5 + -endpoint. Thus, μ * ( f ) μ ( f ) 2 × 1 3 2 × 1 6 = 0 by (R4).
  • Let f be an inner rich 5-face.
    Then f has at least two incident 5 + -vertices.
    If two incident 5 + -vertices are not adjacent in B ( f ) , then B ( f ) has at most one edge with two 4-endpoints. Thus, μ * ( f ) μ ( f ) 1 3 4 × 1 6 = 0 by (R4). It remains to consider the case that f has exactly two incident 5 + -vertices and they are adjacent in B ( f ) . Then B ( f ) has two edges with two 4-endpoints and two edges with exactly one 5 + -endpoint. Thus, μ * ( f ) μ ( f ) 2 × 1 3 2 × 1 6 = 0 by (R4).
  • Let f be an extreme 5-face.
    Then f has at most an adjacent inner 3-face. Thus, μ * ( f ) μ ( f ) + 1 3 × 2 3 = 0 by (R4) and (R6).
CASE 10:f is an m-face where m 6 .
Then, by (R5) we have w ( f f i ) ( 1 2 χ ( f i ) ) θ ( f i ) + χ ( f i + 1 ) θ ( f i + 1 ) + χ ( f i 1 ) θ ( f i 1 ) .
μ * ( f ) = μ ( f ) i = 1 m w ( f f i ) μ ( f ) i = 1 m ( ( 1 2 χ ( f i ) ) θ ( f i ) + χ ( f i + 1 ) θ ( f i + 1 ) + χ ( f i 1 ) θ ( f i 1 ) ) = μ ( f ) i = 1 m ( θ ( f i ) 2 χ ( f i ) θ ( f i ) + 2 χ ( f i ) θ ( f i ) ) = m 4 m ( m 4 m ) = 0 .
CASE 11: The unbounded face D.
Let the number of intersecting 3-faces and 5-faces of D be denoted by f . Let E ( C 0 , V ( G ) C 0 ) denote the set of edges between V ( G ) C 0 and C 0 where this set has size e ( C 0 , V ( G ) C 0 ) . Then by (R6),
μ * ( D ) = 3 + 4 + v C 0 ( d ( v ) 4 ) f = 1 + v C 0 ( d ( v ) 2 ) f = 1 + e ( C 0 , V ( G ) C 0 ) f .
So we may consider that D sends charge 1 to each edge e E ( C 0 , V ( G ) C 0 ) . So each intersecting 3-face and 5-face contains at least two edges in E ( C 0 , V ( G ) C 0 ) . It follows that e ( C 0 , V ( G ) C 0 ) f 0 . Thus, μ * ( D ) > 0 .
This completes the proof.

5. Conclusions

We prove that every planar graph without 6-cycles simultaneously adjacent to 3-cycles, 4-cycles, and 5-cycles is DP-4-colorable. This result is a special case of two following open problems.
1. Every planar graph without i-cycles simultaneously adjacent to j-cycles, k-cycles, and l-cycles is DP-4-colorable for { i , j , k , l } = { 3 , 4 , 5 , 6 } .
2. Every planar graph without 3-, 4-, 5-, and 6-cycles that are pairwise adjacent is DP-4-colorable.

Author Contributions

Conceptualization, K.N. and P.S.; investigation, W.R.; methodology, P.S. and K.N.; validation, K.N. and W.R.; writing—original draft preparation, W.R.; writing—review and editing, K.N. and W.R.; supervision, K.N.; funding acquisition, P.S., K.N. and W.R. All authors have read and agreed to the published version of the manuscript.

Funding

NSRF via the Program Management Unit for Human Resources and Institutional Development, Research and Innovation [Grant number B05F640106], and National Research Council of Thailand (NRCT) [Grant number N41A640141].

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

This research has received funding support from NSRF via the Program Management Unit for Human Resources and Institutional Development, Research, and Innovation [Grant number B05F640106]. The first author was supported by National Research Council of Thailand (NRCT) [Grant number N41A640141]. The third author has received scholarship under the Post-Doctoral Training Program from Khon Kaen University, Thailand. The authors take this opportunity to thank the funders for the support.

Conflicts of Interest

The authors declare no conflict of interest.

Appendix A

Lemma A1
([14]). G has no separating 3-cycles.
Proof. 
Suppose to the contrary that G contains C 0 , which is a separating 3-cycle. Consider a 3-cycle C, which is precolored. Note that C and C 0 may be different. By symmetry, one may assume V ( C ) V ( C 0 ) i n t ( C 0 ) . By minimality, a precoloring can be extended from C to V ( C 0 ) i n t ( C 0 ) . After C 0 is colored, one can extend the coloring of C 0 to e x t ( C 0 ) . In this way, we obtain a DP-4-coloring of G, a contradiction. □
To prove Lemmas A3 and A4, Ref. [14] gave the definition of residual list assignment and Lemma A2 as follows.
Let G be a graph with a list assignment L and let H be its cover. Let F be an induced subgraph of G and G = G F . A restriction of L on G is a list assignment, say L such that L ( u ) = L ( u ) for every vertex u in G .
If a graph H = H [ { { v } × L ( v ) : v V ( G ) } ] , then we say H is a restriction of H on G . Assume G has an ( H , L ) -coloring such that I is an independent set in H with | I | = | V ( G ) | | V ( F ) | .
Define a residual list assignment  L * of F to be
L * ( x ) = L ( x ) u x E ( G ) { c L ( x ) : ( u , c ) ( x , c ) E ( H ) and ( u , c ) I }
for every x V ( F ) .
Define residual cover H * to be H [ { { x } × L * ( x ) : x V ( F ) } ] .
Lemma A2.
Let I be an ( H , L ) -coloring of G . It follows that a residual cover H * becomes a cover of F with a list assignment L * . Additionallay, F is ( H * , L * ) -colorable implies G is ( H , L ) -colorable.
Proof. 
The first part follows immediately from the definitions of a cover and a residual cover.
Suppose that F is ( H * , L * ) -colorable. Consequently, H * has an independent set I * with the size | I * | = | F | . The definition of residual cover implies that no edges connect between H * and I . Furthermore, I and I * are disjoint. Put them together, we have I = I I * is an independent set of H such that | I | = ( | V ( G ) | | V ( F ) | ) + | V ( F ) | = | V ( G ) | . So we can conclude that G is ( H , L ) -colorable as desired. □
Lemma A3
([14]). Each vertex in i n t ( C 0 ) has degree of at least four.
Proof. 
Suppose otherwise that G has a vertex v of degree less than 4 . Let L be a 4-assignment in G and H be a cover of G in which G has no ( H , L ) -coloring. By minimality, we have G = G x with an ( H , L ) -coloring where L (respectively, H ) is a restriction of L (respectively, H) on G . Thus, there is an independent set I with | I | = | G | in H . Let L * be a residual list assignment. Since d ( x ) 3 and | L ( v ) | = 4 , it follows that | L * ( v ) | 1 . It is obvious that { ( v , c ) } with c L * ( v ) is an independent set in G [ { v } ] . It follows that G [ { v } ] is ( H * , L * ) -colorable. Lemma A2 yields that G is ( H , L ) -colorable. This contradiction completes the proof. □
Lemma A4.
Assume C ( l 1 , , l k ) is a cycle C = v 1 v m with k internal chords that share an endpoint v 1 with V ( C ) V ( C 0 ) = . Suppose v m is not an endpoint of a chords in C . If d ( v 1 ) k + 3 , then there exists v i V ( C ) { v 1 } such that d ( v i ) 5 .
Proof. 
Let v m be not an endpoint of a chord in C . Suppose otherwise that d ( v i ) 4 for each v i V ( C ) { v 1 } . Assume G has a 4-assignment L with a cover H in which G has no ( H , L ) -coloring. By minimality, G = G { v 1 , , v m } has an ( H , L ) -coloring where L (respectively, H ) is a restriction of L (respectively, H) in G . Thus an independent set I in H with | I | = | G | exists.
Let L * be a residual list assignment on F . From | L ( v ) | = 4 for every v V ( G ) , it follows that | L * ( v 1 ) | 3 and | L * ( v ) | 3 for each vertex v V ( C ) such that v 1 v is an edge whereas | L * ( v i ) | 2 for each remaining vertex v i in V ( C ) . Assume H * is a residual cover of F . Recall that v m is not an endpoint of a chord in C . It follows that there exists a color c in L * ( v 1 ) with | L * ( v m ) { c : ( v 1 , c ) ( v m , c ) E ( H * ) } | 2 . Greedily coloring v 2 , v 3 , , v m sequently, we have an independent set I * where its size | I * | = m = | F | . It follows that F is ( H * , L * ) -colorable. By Lemma A2, we have G is ( H , L ) -colorable, which is a contradiction. □

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Figure 1. The configuration in CASE 4.
Figure 1. The configuration in CASE 4.
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Figure 2. The configuration in CASE 5.
Figure 2. The configuration in CASE 5.
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Figure 3. The configuration in CASE 6.
Figure 3. The configuration in CASE 6.
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Figure 4. The configuration in CASE 7.
Figure 4. The configuration in CASE 7.
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Nakprasit, K.; Ruksasakchai, W.; Sittitrai, P. Sufficient Conditions of 6-Cycles Make Planar Graphs DP-4-Colorable. Mathematics 2022, 10, 2762. https://doi.org/10.3390/math10152762

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Nakprasit K, Ruksasakchai W, Sittitrai P. Sufficient Conditions of 6-Cycles Make Planar Graphs DP-4-Colorable. Mathematics. 2022; 10(15):2762. https://doi.org/10.3390/math10152762

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Nakprasit, Kittikorn, Watcharintorn Ruksasakchai, and Pongpat Sittitrai. 2022. "Sufficient Conditions of 6-Cycles Make Planar Graphs DP-4-Colorable" Mathematics 10, no. 15: 2762. https://doi.org/10.3390/math10152762

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