1. Introduction
The graphs discussed are simple, planar graphs. Vertex coloring stands out as a significant area of study within graph theory. A graph
G with
list assignment allocates a color set
to each
. Considering
to be a proper coloring of
G, which ensures that
differs from
across every edge
in
G. If
in respect to every vertex
v, then
is defined as a
list coloring of
G, which proposed by was Vizing et al. [
1,
2] independently. Considering each list assignment with a size more than
k, if
G has a list coloring, then we could define
G as
k-choosable. The least amount
k is the
choosable number, represented by
. Thomassen [
3] demonstrated the 5-choosability of planar graph, while not all planar graphs are 4-choosable, as proven by Voigt [
4]. Thus, many experts focus on identifying sufficient conditions under which planar graphs have the property of 4-choosability.
In list coloring, vertex identification cannot be used due to the different lists on vertices. To address this limitation, Dvořák et al. [
5] proposed DP-coloring, which enabled them to resolve a long-standing conjecture by Borodin [
6], namely that every planar graph
G without cycles of lengths 4 to 8 is 3-choosable. Let
be two disjoint vertex sets of
G, and
denote the collection of edges with one end-vertex in
X and the other end-vertex in
Y. The following definition of DP-coloring was subsequently provided by Bernshteyn et al. [
7].
Definition 1. Consider that a graph denoted as G along with its respective list assignment L. qualifies as a cover for when it meets these criteria:
, where and ;
for each vertex w in G, the collection makes up a clique;
for every edge , is a matching set (which may be empty), denoted by ;
when , is an empty set.
For simplicity, we denote
.
Figure 1b,c are two distinct covers of
with
for each vertex of
.
Definition 2. Considering denoted as a cover of . Then, the following two conditions are equivalent:
- (i)
includes an independent subset I with size ;
- (ii)
G has a coloring ψ on satisfying across each vertex v. This coloring is said to -coloring.
A graph is called DP-k-colorable when it has an -coloring for every L with and every cover . The minimum such k is the DP-chromatic number, represented as .
For every
, distinct matchings between
and
correspond to different covers of
G. If the matching between
and
is
, an
-coloring of
G becomes equivalent to a list coloring of
G. It indicates that DP-coloring serves as a generalization of list coloring, leading to the conclusion that
. From
Figure 1,
has an independent set
; hence,
has an
-coloring with vertices
colored 1 and vertices
colored 2, and this corresponding coloring is also a 2-list coloring of
. However,
does not have an
-coloring for
and has no 4-independent set. So,
. The fact that
and
marks an important difference between DP-coloring and list coloring.
Recently, DP-coloring has gained significant attention. Dvořák et al. [
5] showed that
does not exceed 5, and it can be reduced to 3 for planar graphs with a girth of at least 5. Li et al. [
8] proved that any planar graphs without intersecting 5-cycles is DP-4-colorable. Li et al. [
9] discussed the DP-4-colorability of planar graphs with some restrictions on cycles. For more results about DP-colorability, the readers can see in [
10,
11,
12,
13,
14,
15,
16,
17,
18]. Kim and Ozeki [
12] improved results about 4-choosability, any planar graph that avoids a
k-cycle (where
k is in the set
) is DP-4-colorable. Moreover, Farzad [
19] confirmed the 4-choosability for every planar graph lacking 7-cycles. One naturally asks whether every planar graph lacking 7-cycles has the property of DP-4-colorability. For this purpose, Kim et al. [
20] demonstrated the subsequent finding.
Theorem 1. Any planar graph that lacks both butterflies (see Figure 2b) and 7-cycles is DP-4-colorable. On the other hand, there is another way to study the sufficient conditions for DP-4-colorability, that is, forbidden adjacent cycles. For more details, the reader can refer to [
11,
12,
13,
16,
17,
18,
21].
Theorem 2. Any planar graph that does not have -cycles adjacent to -cycles is DP-4-colorable, where .
In a planar graph
G, a
cluster refers to a subgraph formed by a maximal collection of 3-faces, ensuring that no other 3-face shares an edge with any face within this collection. When such a cluster contains
k 3-faces, it is designated as
k-cluster (see
Figure 3).
Adjacent cycles (faces) means that they share no less than one edge. A face
f is said to be
adjacent to one cluster
H if there exists a face within
H that shares adjacency with
f. A
Near-bow-tie is a graph isomorphic to the configuration depicted in
Figure 2a where
v is a 7-vertex. Inspired by Theorems 1 and 2, the subsequent result is formulated.
Theorem 3. Consider a planar graph G without a Near-bow-tie as an induced subgraph. If G lacks 7-cycles that are adjacent to 4- or 5-cycles, then it is DP-4-colorable.
If a graph G is free of 7-cycles that are adjacent to either 4-cycles or 5-cycles, then G may contain 7-cycles and a butterfly. Additionally, if the graph G lacks the Near-bow-tie as an induced subgraph, then it may still contain a butterfly. Therefore, Theorem 3 generalizes Theorem 1.
A 3-cycle
C is considered
bad provided that it is isomorphic to the bordering of the outer face of
in
Figure 3; all other 3-cycles are
good. Let
denote a portion of graph
G. Therefore,
G is DP-
k-colorable, provided that every DP-
k-coloring restricted on
K could be expanded to cover the whole
G. To demonstrate Theorem 3, a stronger result is established as outlined below.
Theorem 4. Consider a planar graph G without a Near-bow-tie as an induced subgraph. If G lacks 7-cycles that are adjacent to 4- or 5-cycles, then each DP-4-coloring restricted on one good 3-cycle within G can be extended to the entire graph.
At the conclusion of this section, we will present some definitions and notations. The notation is defined as , where m is a positive integer. A k-vertex represents a node whose degree is k, while a (or )-vertex represents a node whose degree is no less (or no more) than k. A face labeled as -face corresponds to one k-face where is -vertex for . If , then u is incident with f. The notation denotes the number of edges associated with cycle (or face) C of G.
2. Preliminaries
This section focuses on the corresponding coloring on some small subgraph, which is crucial to the proof of Theorem 4.
Lemma 1. Consider L as a list assignment on . If for each , the size of the list is equal to 2, then has an -coloring except when the subgraph of induced by is a union of two 3-cycles.
Moreover, if for each and for some , then has an -coloring for any cover of .
Proof. Assume
for every index
. If the subgraph of
generated by
is a union of two 3-cycles, see
Figure 4a, say
for each
, then we cannot find a 3-independent set in this cover; hence,
does not have an
-coloring for this cover. Thus, we assume that the subgraph of
induced by
is not a union of two 3-cycles. Without loss of generality, suppose
forms a 2-path in this cover, but
. Then,
is a 3-independent set of
; see
Figure 4b. Hence, we obtain an
-coloring of
for this cover.
If there is one , for some , then we can identify a 3-independent set of for any cover of . Consequently, an -coloring of can be achieved for any cover of . □
Lemma 2. Consider L as a list assignment on , where comprises two 3-faces for . If for each and , then we can assign colors to each using at most one color from , so that has an -coloring for any given cover of .
Proof. Assuming and is valid for our analysis. We assume that there is a color, say , which cannot be chosen. This means if we choose in , then by Lemma 1, has two available colors and the corresponding vertices of in induce two triangles, say for every . In this case, is a 3-cycle of . Then, for each and for certain values i and j where , the set is an independent set in . Thus, each color in other than 1 can be chosen so that has an -coloring. For vertex , we can discuss a similar method. □
Lemma 3. Consider be a 2-path with a list assignment L on P, such that for every . If , then we can assign colors to , such that has at least available colors.
Proof. If there is a vertex in such that is not adjacent to any vertices in , then we have completed this section. Thus, assume each vertex in is adjacent to a vertex of , and the same holds for . Given that , it follows that there exists a vertex in that is adjacent to at least one vertex from both and . Without loss of generality, assume , where , . Consequently, can be colored with color a and with c, so that has at least available colors. □
Lemma 4. Let be the graph depicted in Figure 3 with a list assignment L on . If for every 3-vertex of , for some and for other three vertices in , then we can color u with all colors but at most two of , ensuring that has an -coloring for any given cover of . Proof. Due to symmetry, we can assume that . Thus, for each and for each .
For 2-path , if , then by Lemma 3, we can color such that has available colors, denoted by . After such coloring, has at least two available colors and has at least three available colors for each . For , by Lemma 2, we can color with at least colors such that can be colored. This means we can color with all colors in , except for at most two, ensuring that has an -coloring for any given cover of , which completes this section.
Thus, . Without loss of generality, we assume , . From the above discussion, we only need to prove the case that after coloring and , has exactly two available colors. Without loss of generality, we assume for each . For , by Lemma 2, we can color with 2 or 4 such that can be colored. For the same reason, for , by Lemma 2, we can color with 1 or 3 such that can be colored. This means we can color with all colors in , except for at most two, ensuring that has an -coloring for any given cover of , which completes this section. □
Lemma 5. Consider as the 7-cluster shown in Figure 3, along with a list assignment L. Let . If there is at most one v in S for which , while for remaining vertices in S, and for every vertex v in except S, then admits an -coloring for any given cover of . Proof. By symmetry, for each , assume that and . Furthermore, for each . Consequently, set , , . Suppose otherwise that does not have an -coloring for some cover of .
Suppose the subgraph induced by on vertices and contains two triangles in since ; otherwise, by Lemma 1, we can color and such that has two available colors. Then, by Lemma 2, we can color . Therefore, this results in an -coloring of , leading to a contradiction. Let be the two triangles for and .
Since does not have an -coloring, we have the following proposition, Proposition 1, constructed easily using Lemma 1. □
Proposition 1. If are colored, then the subgraph of induced by edge between is two triangles, where denotes the available colors in after coloring .
By Proposition 1 and Lemma 1, there are four triangles between () in . Let be the four triangles in for each . We assume that . Assume for some . By symmetry, we only need to discuss two cases: or .
If , then by Proposition 1, for 3-independent set and for 3-independent set ; contrary to this, is a matching set.
If , then by Proposition 1, for some , say . In this case, for 3-independent set , , for some . The independent set , , contrary to , is a matching set.
Lemma 6. Consider as the 7-cluster shown in Figure 3, along with a list assignment L. If for and for the other vertices of , then we can color with all colors but at most one in , ensuring that maintains an -coloring for any given cover of . Proof. Without loss of generality, we assume that for each , and for each . For any given cover of , by contradiction, we assume that there are two colors that cannot be chosen in , say . Since , by Lemma 5, there is an -coloring for any given cover of , a contradiction. This means we can color with all colors, but only use at most one in , ensuring that maintains an -coloring for any cover of . □
The following definition is due to Sittitrai and Nakprasit [
22].
Definition 3. Let represent a cover of , with as an induced subgraph within G. The following definitions apply:
- (i)
A list assignment is considered a restriction of L on if for every vertex of .
- (ii)
A graph is termed a restriction of on if . According to the cover definition, .
Assuming that has an -coloring with an independent set , the residual list assignment for is defined as follows:
and
for every . The residual cover is established by .
Lemma 7. ([
22])
. Let represent a cover of with as an induced subgraph of G. If maintains an - coloring, then the residual cover serves as a cover for , where represents the residual list assignment. Additionally, if is colorable with respect to , then G is also -colorable. For the remainder of this paper, is regarded as an induced subgraph of , where for every vertex v of G. By Lemma 7, to establish that G is -colorable, it suffices to demonstrate that maintains an -coloring and maintains an -coloring.
3. Reducible Configurations
Consider
as a counterexample to Theorem 4, where the number of vertices is minimized, and within that constraint, the number of edges is also minimized, with
representing a good 3-cycle. The proof presented in [
12] establishes that any planar graph devoid of 3-cycles is DP-4-colorable. Therefore, it is reasonable to assume that
G includes a 3-cycle. Indeed, we can always identify a good 3-cycle within
G by Lemma 5. Let
L represent a 4-list assignment on the vertex set
, and
as the outer cycle of
G. Let
be any cover of
and
the
-coloring restricted on
. The following lemma holds true when
G lacks any 7-cycle that is adjacent to 4- or 5-cycles.
Lemma 8. G contains no k-cluster for ; moreover, the graphs of Figure 3 are all k-clusters of G, where . Consider C representing a cycle within G. The notation (or ) is employed to indicate the collections of vertices lying inside (or outside) the cycle C. A cycle C is referred to as separating if it divides the vertex set such that both and include vertices. A vertex u in G is classified as internal if . Additionally, a face f is classified as internal if all vertex of f is internal.
Lemma 9. The statements below are all valid:
- (i)
;
- (ii)
G does not have any good 3-cycle that is separating;
- (iii)
the degree of every internal vertex in G is at least 4.
Proof. (i) Assume, for the purpose of contradiction, that . Since is a good 3-cycle, the -coloring restricted on would also serve as an -coloring for G, which contradicts the initial assumption.
(ii) Suppose C is a separating good 3-cycle. Given the property of minimality in G, we can extend to ext(C), say . This extension can then be restricted to C and further extended to create an -coloring for . Consequently, by merging and , we obtain an -coloring for G, contrary to the assumption.
(iii) Let v represent an internal -vertex. Given that G is minimal, the -coloring restricted on can be applied to . Since , v has at least one available color for any cover of . This means that G has an -coloring, a contradiction. □
Lemma 10. No two internal -faces in G can share a common edge.
Proof. Assume, for the sake of contradiction, that () are two internal (4, 4, 4)-faces. Denote . By the minimality of G, the graph maintains an -coloring extended by on , and , are the residual list assignment and the residual cover of . Since each vertex of is a 4-vertex, for every and . By Lemmas 2 and 7, we can obtain an -coloring for G, which contradicts the original assumption. □
A cluster is
special if it is one of
in
Figure 3 with an internal (4, 4, 4)-face
(or
by symmetry if the cluster is
); otherwise non-special. An internal
-vertex
v located inside a cluster
H is classified as
i-type with respect to
H if it is connected to precisely
i edges of
H. Moreover, if
H is a special cluster, then this
-vertex
v is referred to as
i-type good; otherwise, it is considered
i-type bad. The following result is obvious from the fact that
G contains no 7-cycle adjacent to a 4- or 5-cycle.
Lemma 11. - (i)
A 1-cluster is adjacent to no more than a 4-face.
- (ii)
The face of G adjacent to i-cluster should be -face for each .
- (iii)
The face of G adjacent to 2- or 3-cluster should be 4-face or -face; moreover, s 2-cluster is adjacent to no more than two non-adjacent 4-faces, where these two 4-faces are adjacent to different 3-face of this 2-cluster. Similarly, a 3-cluster is adjacent to no more than one 4-face, which is incident to one 4-type 4-vertex to this 3-cluster.
Lemma 12. No internal 5-vertex of G can belong to two clusters which are special.
Proof. Let v represent an internal 5-vertex, and let its neighborhood be . Assume, for the purpose of contradiction, that v is a vertex of two special clusters . Without loss of generality, assume the internal (4, 4, 4)-faces of and are and , respectively. Define and as the graphs induced by vertex set and , respectively. Clearly, is isomorphic to . Let and , denote the residual list assignment and the residual cover of . Due to the minimality property of , maintains an -coloring extended from . Notably, and are both at least 3 for each , while . Applying Lemma 2 to , we can color v with at least two colors in , ensuring that maintains an -coloring for every cover of . Applying Lemma 2 to subsequently yields an -coloring for . By Lemma 7, this implies that G maintains an -coloring, contrary to the assumption. □
Internal 6-vertex
v is classified as
special if it qualifies as a 4-type good vertex to an internal
-cluster
and as a 2-type good vertex in association with a
-cluster
, where
(see
Figure 5a). Internal 6-vertex
v is
poor if it is a 3-type good vertex to an internal 6-cluster
and a 3-type good vertex to a
-cluster
, where
(see
Figure 5b).
Lemma 13. Consider v as either a special or poor 6-vertex contained in and (see Figure 5a,b). Let , with , representing the residual list assignment and residual cover of . Then, if v is a special 6-vertex, and if v is a poor 6-vertex. Proof. Since v is either a special or poor 6-vertex, is a special cluster. Denote to represent the (4, 4, 4)-face of . Denote , with , as the residual list assignment and residual cover for . This implies that and for each .
If v is a special 6-vertex, then . If v is a poor 6-vertex, then . In either case, by Lemma 2, we can precolor v using all colors in except for at most one, allowing the vertices to be colored as well. This means when the vertex v is classified as a special 6-vertex and when v is a poor 6-vertex. □
We use a recursive method to define the poor 7-vertex, the special 7-vertex, and the special 8-vertex.
Definition 4. Consider v as a 7-vertex of G, which is 4-type good to one internal 6-cluster and 3-type good to one internal 6-cluster .
- (i)
v is a
poor -vertex
if contains two 3-type vertices, each of which is in {3-type good 5-vertex, 3-type poor 6-vertex};
- (ii)
v is a
poor -vertex
if contains one 3-type poor -vertex and one 3-type vertices which is in {3-type good 5-vertex, 3-type poor 6-vertex, 3-type poor -vertex }.
Then, v is called a
3-type poor 7-vertex
if v is a 3-type poor -vertex, where .
Let 3-type good 5-vertex, 3-type poor 6-vertex, 3-type poor 7-vertex}.
Definition 5. Consider v as a 7-vertex of G which is 4-type good to one internal -cluster and 3-type good to one internal 6-cluster .
- (i)
v is a
special -vertex
if contains one vertex of and one vertex of {4-type good 5-vertex, 4-type special 6-vertex};
- (ii)
v is a
special -vertex
if contains one vertex of and one special -vertex.
Then, v is called a
4-type special 7-vertex
if v is a 4-type special -vertex, where .
Let 4-type good 5-vertex, 4-type special 6-vertex, 4-type special 7-vertex}.
Definition 6. Consider v as an 8-vertex of G which is 4-type good to one internal 7-cluster and 4-type good to one internal -cluster .
- (i)
v is a
special -vertex
if contains two vertices which are in ;
- (ii)
v is a
special -vertex
if contains one special -vertex and one vertex of 4-type special -vertex .
Then, v is called a
special 8-vertex
if v is a special -vertex, where .
By Definition 6, in order to distinguish the above clusters and , is called type I and is called type II.
Lemma 14. Consider v as a 3-type vertex to an internal cluster K. Let , with and as the residual list assignment and residual cover on , respectively. If , then .
Proof. If
v is a 5-vertex, then the cardinality of
is at least 2. If
v is a 3-type poor 6-vertex, then
by Lemma 13. Then,
v is a poor 7-vertex. By Definition 4, assuming
v is a poor
-vertex,
. We use the symbol shown in
Figure 5c, where
and
are 6-clusters. Let
and
be (4, 4, 4)-faces in
and
, respectively. Since
v is poor,
is also internal, and
. Let
, and let
and
denote the residual list assignment and residual cover of
. We use the induction hypothesis on
m. If
, then
is a 3-type 5- or poor 6-vertex, by the above discussion,
,
. Clearly,
for each
. Lemma 4 provides a minimum of two colors available in
to color
. This means
. By the induction hypothesis, it is assumed that the lemma is valid for
.
Next, we prove that the lemma is valid if .
By Definition 4, each of is a 5-, poor 6-, or -vertex , using Lemma 13 and the induction hypothesis, and . Then, using Lemma 4, we have completed this section. □
Lemma 15. Consider v as a 4-type vertex to an internal cluster K. Let with and representing the residual list assignment and residual cover of .
- (i)
If , then the cardinality of is at least 3;
- (ii)
If v is a special 8-vertex and K is type II, then .
Proof. (i) Clearly,
if
v is a 5-vertex. If
v is a 4-type special 6-vertex, then
by Lemma 13. Then,
v is a special 7-vertex. By Definition 5, assuming
v is a special
-vertex,
. We use the labeling shown in
Figure 5d, where
and
is an internal 6-cluster. Then, the (4, 4, 4)-face corresponding to the cluster
is denoted by
and
. Let
, with
and
representing the residual list assignment and residual cover of
. Then, for each vertex
,
. Since
, by Lemma 14,
.
Next, we use the induction hypothesis on m.
If , then is a good 5- or special 6-vertex. According to Lemma 13, we have . Without loss of generality, we can take for each . By Lemma 4, for , there is an -coloring, say , with . Similarly, for , there is an -coloring, say , with ; for , there is an -coloring, say , with . Hence, .
By the induction hypothesis, it is assumed that the lemma is valid for .
Next we prove that the lemma is valid if . By Definition 5, is a special -vertex. By the induction hypothesis, . By Lemma 4 and the similar reasoning for the case , we have .
(ii) Since v is a special 8-vertex, v is a special -vertex, . We use the induction hypothesis on m. If , then v is a special -vertex, K is an internal -cluster and v is also a vertex of internal 7-cluster , and contains two vertices in , say . Let , with and representing the residual list assignment and residual cover of . Let be the (4, 4, 4)-face in . Then, for each vertex , . Since , by Lemma 15(i), , . By Lemma 6, we can assign colors to v using all but at most one color, ensuring that maintains an -coloring for any given cover of . This means .
By the induction hypothesis, it is assumed that the lemma is valid for .
Next, we prove that the lemma is valid if . By Definition 6, assuming that u is a special -vertex and w is a vertex from the 4-type special -vertex . By the induction hypothesis, the cardinality of both and is at least 3. By Lemma 6, we have . □
Lemma 16. If v is a poor 7-vertex in G, then it cannot be a special 7-vertex of G.
Proof. Since v is a poor 7-vertex, we will consider v as a good 4-type vertex in relation to the internal 6-cluster and as a good 3-type vertex to internal 6-cluster . Clearly, . Suppose, on the contrary, that v is also a special 7-vertex. This means . Let , with and representing the residual list assignment and residual cover for . Given the minimal property of G, has an -coloring that extends . Since , there are three available colors in to choose, so that can be colored by Lemma 15. Since , there are two available colors in to choose, so that can be colored by Lemma 14. Since , we can precolor v so that can be colored. By Lemma 7, G is -coloring for any given cover of , leading to a contradiction. □
Lemma 17. Consider v as a 7-vertex of G, where it is 4-type to a 7-cluster and 3-type to a -cluster . Denote represent a (4, 4, 4)-face within and . If the special 8-vertex and is type II}, then cannot be special.
Proof. Assume, to the contrary, that is special. In this case, let represent the internal (4, 4, 4)-face within with . Since the special 8-vertex and is type II}, is internal. Define , and let and be the residual list assignment and the residual cover of . In this setup, and . By Lemma 15, . Clearly, and . According to Lemma 2, v has a minimum of two colors that can be used from , allowing for the coloring of the set . Then, by Lemma 5, has an -coloring for any cover of . Therefore, maintains an -coloring. By the minimality of G, also maintains an -coloring. By Lemma 7, G maintains an -coloring, a contradiction. □
Lemma 18. Consider as an internal special 6-cluster with a (4, 4, 4)-face , shown in Figure 3. If for each , then and the special 8-vertex, and is type II}.
Proof. By Lemma 10, . Denote with and as the residual list assignment and residual cover of . By the minimal property of G, maintains an -coloring that extends . Suppose otherwise that or is a special 8-vertex with type II. Then, by Lemma 15, . Since for each , by Lemma 14, . Note that for each . By Lemma 4, has an -coloring. According to Lemma 7, G admits an -coloring, which is a contradiction. □
Lemma 19. Consider as an internal 7-cluster (see Figure 3) and . - (i)
If X contains two vertices in the special 8-vertex, and is type II}, then the third vertex of X is a -vertex;
- (ii)
X does not contain three vertices in the special 8-vertex, and is type II}.
Proof. (i) By symmetry, assume the special 8-vertex, and is type II}. Suppose, to the contrary, that .
Define with and as the residual list assignment and the residual cover of . By the minimal property of G, maintains an -coloring that extends , with . Note that for each . Since the special 8-vertex, and is type I}, by Lemma 15, , for each . By Lemma 5, an -coloring of can be achieved. Then, by Lemma 7, G admits an -coloring, a contradiction.
(ii) follows easily by Lemmas 5 and 15. □
4. Proof of Theorem 4
Consider V as the family of vertices of G and F as the family of faces. An initial charge is defined as for every , while , where is designated as the outer face of G. According to Euler’s Formula, the sum . Let represent the charge of after the discharging procedure. To reach a contradiction, the purpose of this is to demonstrate that for all and that .
The discharging rules are as follows:
- (R1)
Every 5- or -face distributes to every neighboring 3-face, while a k-face () distributes to every neighboring 3-face;
Consider v as a vertex of k-face f and let t represent the number of edges of f that are incident to v, such that these edges are not in any 3-faces. In cases where , f distributes to v; f distributes to v when ;
- (R2)
Every 2-cluster obtains from its 3-type -vertex and from its 2-type -vertex, which is on a 4-face adjacent to this cluster;
- (R3)
Every 3-cluster which is adjacent to 4-face obtains from its 3-type -vertex and from its 2-type -vertex; otherwise, 3-cluster obtains from its 3-type -vertex;
- (R4)
Every 4-cluster obtains the charge from its incident vertex
v, defined by
- (R5)
Every 5-cluster obtains the charge from its incident vertex
v defined by
- (R6)
Every 6-cluster
K obtains the charge from its incident vertex
v defined by
- (R7)
Every 7-cluster
K obtains the charge from its incident internal vertex
v defined by
- (R8)
receives from every incident vertex, and distributes 1 to every non-internal 3-face;
- (R9)
At last, each face excepting transfers its present charge to .
The following lemma can be easily obtained by discharging rules.
Lemma 20. Assuming u is a k-type vertex within a cluster, if , then u gives at most We begin by examining the final charge assigned to each vertex of G. Consider v as a vertex of G. If , then according to rule (R8), we have . Now, consider the situation in which v is not in . By applying Lemma 9 (iii), the minimal degree of G is 4.
. Suppose v is a k-type vertex of cluster K for some . If , then v does not participate in the discharge process, hence . If , then K is a -cluster and v must be contained in at least one -face. By (R2)–(R6), v gives or to K. By (R1), v obtains at least either if v is incident with a 4-face or otherwise. Therefore, if v is not incident with 4-face, then is not negative. Assuming v is incident to a single 4-face. It can be inferred from Lemma 11 that K is a 2-cluster. By (R2) and (R1), . If , then K is a -cluster. By (R2)–(R5), v gives a charge to K when K is a 2- or 3-cluster. If K is a 2- or 3-cluster, then v gives to K when v is incident with a 4-face adjacent to K according to (R2)–(R3). Given that G lacks 7-cycles that are adjacent to 4- or 5-cycles, and according to Lemma 11, the three other faces incident with v in clockwise direction are 4-, -, and -faces, respectively. By applying (R1), we have .
. Assume that v represents a k-type vertex of cluster K for some . By applying Lemma 20, it can be inferred that is at least , which yields when . If , then K is a -cluster and v is contained in two -faces. By (R3)–(R7), v gives at most to K. By (R1), v receives from each incident -face, yielding a total of . Therefore, we find that . If , then K is a -cluster. By rules (R2)–(R6), v contributes no more than 1 to K for . If v is part of only a single cluster, then it follows that . Thus, let us consider the scenario where v is a 2-type vertex of other cluster, referred to as L. By Lemma 11, v is contained in two -faces. According to rules (R2)–(R5), v contributes to L when L is a special 4-, 5-cluster. By Lemma 12, K is not a special cluster. According to rules (R2)–(R6), v contributes no more than to K. Then, . The case of can be proved similarly as .
. Assume v is contained in no more than three clusters. By (R2)–(R7), v contributes no more than 2 to its incident cluster. Then, when v is contained in only one cluster. If v is contained in three clusters, then it is a 2-type vertex for each cluster. According to Lemma 20, . Therefore, we consider the case where v belongs to exactly two clusters, say . Suppose firstly that v is a 2-type vertex for . According to Lemma 20, assume v is a 4-type vertex to and v gives more than to . By (R6)–(R7), v gives 2 to , where is an internal 6- or 7-cluster. By (R6) and (R7), either v is a good but not special 6-vertex, or is a special 6-cluster containing two vertices in . In the latter situation, by Lemma 18, v is also good but not a special 6-vertex. In either case, v is a non-special 6-vertex, while is internal and special. By the definition of the special 6-vertex, is not a special 4-, 5-cluster. By (R2)–(R5), v gives no charge to . In this case, . Now, let us consider the scenario where v acts as a 3-type vertex for . Due to the symmetric properties, v is necessarily classified as a 3-type vertex in relation to . According to rules (R2)–(R6), we can assume that is -cluster for each . By (R5)–(R6), assume v gives to . Then, is an internal 6-cluster and v is a 3-type good 6-vertex which is not poor. By the definition of a poor 6-vertex, is not a special 5- or 6-cluster. Hence, v is a 3-type bad 6-vertex to . By (R5)–(R6), v contributes at most to . Thus, we have .
. Then, v is contained in no more than three clusters and . By Lemma 20, we only need to discuss the case that v is contained in two clusters, say and , and v is classified as a 4-type vertex to and a 3-type vertex to . Since v gives at most to , we need only show the case that v gives at least 2 to cannot occurs by (R2)–(R7). Further, by rules (R2)–(R6), v contributes at most to . If v contributes to , then is an internal 7-cluster and contains two vertices in the special 8-vertex and is type II} by (R7). By Lemma 17, is not special. According to (R2)–(R6), v contributes at most to . Hence, we find that , and we complete this section. If v gives 2 to , then one of the following applies: (i) v is a non-special 7-vertex, while is classified as an internal 7-cluster, lacking two vertices of the special 8-vertex and is type II} by (R7); (ii) is an internal special 6-cluster that includes two vertices from by (R6). When (i) occurs, is not a 6-cluster for G and contains no Near-bow-tie flaw. Following rules (R2)–(R5), v contributes no more than 1 to . Therefore, we have , concluding our argument. When (ii) occurs, v is poor if is an internal and special 6-cluster. By the same rules (R2)–(R6), v contributes no more than 1 to , again leading to , which completes the proof.
. Hence, . By applying Lemma 20, suppose v belongs to at least two clusters. If v is present in four clusters, then according to Lemma 20. For the case where v belongs to three clusters, we have or by Lemma 20. Thus, suppose v is contained in just two clusters, say and . By Lemma 20, suppose v gives to and v is a 4-type vertex of . According to (R3)–(R7), is an internal 7-cluster and contains two vertices in the special 8-vertex and is type II}. We only need to prove cases that v gives 2 or to cannot occur. Suppose otherwise, by (R2)–(R7), one of the following applies: (i) is an internal and special 6-cluster and contains two vertices in ; (ii) is an internal 7-cluster and v is a non-special 8-vertex. By Definition 6, v would indeed be a special 8-vertex if is an internal 7-cluster, contradicting the assumption that v is a non-special 8-vertex. This makes the second case (ii) impossible. Therefore, is an internal special 6-cluster containing two vertices from . Let , where be the residual list assignment and the residual cover of . By the minimal property of G, admits -coloring extending with . By Lemmas 4 and 14, v has at least two available colors to color . Thus, by Lemmas 5 and 15, has an -coloring. By applying Lemma 7, an -coloring of G can be achieved, leading to a contradiction.
. Then we have . According to Lemma 20, , or .
Consider a face
in
G. By (R9),
. Now, we examine
in
G. Consider that
represents the number of 3-faces that have vertices in common with
and let
b denote the charge transferred to
from other faces by (R9). Define
s as the size of
. By Lemma 9 (i),
. If
, then
. If
, then
, and so we also find
. Using rules (R8) and (R9), we can express
as follows:
Since , for . If , then , which completes this section.
Next, we prove that . Define as the charge assigned to face f following the application of rules (R1)–(R8). Clearly, . To begin with, let us examine the scenario where . In this instance, f is not impacted by the discharging procedure. Thus, . If , then f is adjacent to no more than a 3-face for G contains no 7-cycle adjacent to 4- or 5-cycles. By (R1), . Now, assume f is a -face. Denote f to be adjacent to m 3-faces, meaning f has m edges shared with 3-faces. If f is a 6- or 7-face, then by (R1), . In the case of , we find by (R1). Finally, . If f is a 3-face, then according to Lemma 8, f could be a subgraph of an i-cluster where .
Let K be an l-cluster. Note that is not a face of K. This means . We define the initial and the middle charges of K after applying (R1)–(R8), denoted by and , as the sum of the initial and middle charges of l 3-faces in it. Note that the final charge of every 3-face is 0 by (R9). We check in the following.
. If K is not internal, then by (R8). Thus, assume that K is internal. Assume that is the face adjacent to K for with . If , then and for each . By (R1), each () gives at least to K and . Thus, . By (R1), .
. Assume that K is not internal. If , then by rule (R8), . So, suppose . According to Lemma 11(iii), K is adjacent to at least two -faces. Utilizing (R1) and (R8), .
Thus, K is internal. By Lemma 11 (iii), the face adjacent to K is a 4-face or an -face and there are no more than two 4-faces. If all faces adjacent to K are -faces, then by (R1), . Thus suppose one adjacent face is a 4-face. Utilizing (R2) and Lemma 9 (iii), each 3-type vertex of K gives to K. Then, according to (R1) and (R2), we find or .
. If , then K must be adjacent to at least two -faces. Utilizing (R1) and (R8), . If , then K is adjacent to a minimum of three -faces. By applying (R1), (R3) and (R8), it can be concluded that or . Thus . If the faces adjacent to K are all -faces, then by (R1) and (R3). Thus, according to Lemma 11(iii), there exists one adjacent 4-face which contains 4-type vertex of K. By applying (R1) and (R3), .
. By Lemma 8,
K is a graph of
shown in
Figure 3. By Lemma 11 (ii), faces adjacent to
K are
-faces. If
, where
, then
gives at least
t to
K when
,
to
K when
by (R8). If
, then there are at least
edges of
K adjacent to
-faces. By (R1) and (R8),
. If
,
by (R1) and (R8).
Now, suppose K is internal. Since has four 3-type -vertices and each of has two 3-type -vertices, or by (R1) and (R4). Suppose . If K is special, then by (R1), (R4) and Lemma 10, . Otherwise, .
. By Lemma 8,
K is a graph of
shown in
Figure 3. By Lemma 11(ii), faces adjacent to
K are
-faces. If
, then
gives 3 to
K by (R8). In this case, there are four edges of
K adjacent to
-faces. By (R1) and (R8),
. If
, then there are five edges of
K adjacent to
-faces. If
K is
, then
K has a 5-type 5-vertex and
. Assume
K is
. If
, then
when
is special; otherwise,
. If
, then
by (R1), (R5) and (R8).
Thus, assume K is internal. If , then K contains five 3-type bad -vertices and one 5-type 5-vertex. By (R1) and (R5), . Assume K is . If K is special, then K contains two 3-type good -vertices and one 2-type good -vertex. By (R1) and (R5), . Otherwise, .
. By Lemma 8,
K is the graph
shown in
Figure 3. By Lemma 11(ii), the faces adjacent to
K are
-faces. If
, then
gives 4 to
K by (R8). In this case, there are three edges of
K adjacent to
-faces. Due to symmetry, suppose that
. Then
is a 3-type
-vertex. By (R6),
gives at least
to
K. By (R1) and (R8),
. If
, then
contributes at least 2 to
K by (R8) and there are four edges of
K adjacent to
-faces. Suppose firstly,
K is special. Then,
K has at least two internal
-vertices. Applying (R1), (R6), and (R8), we find that
. Suppose secondly, that
K is not special. Then,
K contains one 4-type
-vertex and one 3-type bad
-vertex. Utilizing (R1), (R6), and (R8),
.
Thus, assume K is internal. If K is special, then K contains two 3-type good -vertices and one 4-type good -vertex by Lemma 10. If , then by Lemma 18. This means is a 4-type good -vertex and K contains two vertices of . In this case, by (R1) and (R6). Then, assume . By (R6), gives at least to K. By (R1) and (R6), . Suppose K is not special. In this situation, is classified as a 4-type -vertex for each . By (R1) and (R6), .
. By Lemma 8,
K is the graph
shown in
Figure 3. By Lemma 11(ii), faces adjacent to
K are
-faces. If
, then
gives 5 to
K by (R8) and according to Lemma 10,
K contains one 4-type
-vertex. If
, then
contributes 3 to
K and
K contains two 4-type
-vertices similarly. By (R1), (R7) and (R8),
or
. Thus assume
K is internal. By Lemma 10,
for each
. By Lemma 19(ii), there are at most two vertices of
in
the special 8-vertex and
K is type II}. If any vertex of
is not in
the special 8-vertex and
K is type II}, then
. If there is only one vertex in
the special 8-vertex and
K is type II}, then
. Thus assume there are two vertices in
the special 8-vertex and
K is type II}, say
. By Lemma 19(i),
. By (R1) and (R7),
.