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
of
k colors. An
L-coloring is a proper coloring
c such that
for each
v in
. 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 is
- (ii)
is a complete graph for every
- (iii)
The set is a matching (empty matching is allowable) for each .
- (iv)
If then there are no edges of H connect and .
An independent set in a cover H of a graph G with size is called an -coloring of G. If every cover H with any k-assignment L of a graph G admits an -coloring for 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 .
If edges on H are defined to match exactly identical colors between and for each then G admits an -coloring is equivalent to G is L-colorable. Consequently, DP-coloring is a generalization of list coloring. Furthermore, this implies that
Dvořák and Postle [
3] proved that for every planar graph
G,
, 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
. 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
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
and
, respectively.
Corollary 1. Every planar graph without 6-cycles adjacent to i-cycles is DP-4-colorable for each .
Corollary 2. Every planar graph without 6-cycles simultaneously adjacent to i-cycles and j-cycles is DP-4-colorable for each and .
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 and We use to denote the boundary of a face Two faces f and g are adjacent if and are adjacent. A wheel is a graph of n vertices formed by connecting all vertices of an -cycle (these vertices are called external vertices) to a single vertex (hub). A k-vertex, -vertex, and -vertex is a vertex of degree k, at least k, and at most 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 -vertices (respectively, exactly one -vertex, no -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 is a chord in an embedding cycle C if but is not in If a chord is inside then it is called an internal chord, otherwise it is called an external chord. A graph is obtained from a cycle with an internal chord For example, cycles and form A graph is obtained from a cycle with internal chords and The previous definition can be extended similarly to a graph The graphs and are induced by vertices inside and outside a cycle respectively. A separating cycle C is a cycle with non-empty and
Let denote the family of planar graphs without 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 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
Then, (i)
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
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 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 to be the boundary of the unbounded face of
Lemma 2 (Lemma 3.3 in [
14]).
Each vertex in has degree at least four (See the proof in Lemma A3). Lemma 3. The following statements hold.
- (i)
A bounded -face has its boundary as a cycle.
- (ii)
If a bounded -face f and a bounded -face g with are adjacent, then
- (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 is a 6-cycle with two internal chords.
Proof. - (i)
Clearly, a boundary of a -face is a cycle. Consider a bounded 6-face A boundary closed walk is in a form of if is not a cycle. By Lemma 2, u or x has degree at least It follows that or is a separating 3-cycle, contrary to Lemma 1.
- (ii)
It suffices to show that and share exactly two vertices.
If and then f or g is the unbounded face, a contradiction.
If and or then or which contradicts Lemma 2.
If and or then or which contradicts Lemma 2. If and then or is a separating 3-cycle, which contradicts Lemma 1.
If and then which contradicts Lemma 2. If and then or is a separating 3-cycle, which contradicts Lemma 1. The remaining cases are similar.
- (iii)
Lemma 3 (ii) yields that is a 5-cycle with one chord. Similar to the proof of Lemma 3 (ii), one can show that and 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 and let be its triangular chord. Suppose to the contrary that C has at least two chords. Since C is adjacent to a 3-cycle and a 5-cycle , it suffices to show that C is adjacent to a 4-cycle. By symmetry, we assume another chord e of C is or
If then C is adjacent to a 4-cycle
If then C is adjacent to a 4-cycle
If then C is adjacent to a 4-cycle
If then C is adjacent to a 4-cycle
If then C is adjacent to a 4-cycle
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 is a face that is adjacent to some bounded 3-face in . Since , one can observe that every cluster in G is a -cluster where a 4-cluster is isomorphic to .
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 then f and g are not adjacent to any 4-faces.
- (iii)
Every adjacent face of a 2-cluster is a -face or the unbounded 3-face D.
- (iv)
Every adjacent face of a -cluster is a -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 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 , which is adjacent to a 5-cycle from , a 4-cycle from , and a 3-cycle from , contrary to
- (ii)
Let an inner 3-face f and a 5-face g be adjacent. Lemma 3 (ii) yields that contains a 6-cycle. Thus, f or g is not adjacent to any 4-faces since .
- (iii)
Let f and g be bounded 3-faces in a 2-cluster 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 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 is adjacent to a 5-cycle from , a 4-cycle from , and 3-cycle from , contrary to
Thus, h is a -face or the unbounded face.
- (iv)
Let and be the bounded 3-faces of -cluster in a consecutive order.
By similar arguments as in the proof of (iii), it follows that cannot be adjacent to a bounded -face.
Let be adjacent to a 6-face . By Lemma 3 (ii) and an argument similar to its proof, one can show that is a 5-cycle with two chords. Since is a 6-cycle by Lemma 3 (i), we have a 6-cycle adjacent to a 3-, a 4-, and a 5-cycle in , contrary to
If is adjacent to a 6-face , then by Lemma 3 (ii), a 6-cycle is adjacent to a 3-, a 4-, and a 5-cycle, which are in , contrary to .
□
For Corollary 3 (i), it is proved by the fact that every -vertex is not adjacent to four consecutive bounded 3-faces. Thus, each -vertex has at least two -faces. For Corollary 3 (ii), it is proved by Lemmas 5 (iii) and (iv) that each 3-face in is not adjacent to a 5-face. Thus, each -vertex has at least three -faces.
Corollary 3. Let v be a k-vertex in G where and It follows that:
- (i)
v is incident to at most bounded 3-faces;
- (ii)
v is incident to at most bounded 3-faces, if v has an incident 5-face.
Proof. If v is incident to bounded 3-faces, then there are four consecutive bounded faces forming a 4-cluster that is not a wheel, contrary to . This proves (i). It follows from Lemmas 5 (iii) and (iv) that each 3-face in a -cluster is not adjacent to a 5-face. Thus, each -vertex incident to a 5-face must be incident to at least three -faces. This proves (ii). □
Lemma 6 (Lemma 3.6 in [
14]).
is defined to be a cycle with k internal chords such that is their common endpoint and . Suppose or is not the endpoint of any chords in If then some satisfies (See the proof in Lemma A4). Lemma 7. Let a 4-vertex v be incident to bounded faces in cyclic order and let , where If or , then there is a vertex such that
Proof. If , it follows from Lemma 3 (ii) that . Moreover, F has exactly one chord, otherwise there is a separating 3-cycle, which contradicts Lemma 1.
If , it follows from Lemma 3 (ii) that . 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 in a cyclic order. Let where denote and If then there exists such that
Proof. Let and , where It follows from Lemma 3 (ii) that is a and is a . Suppose to the contrary that F is not a . Then, there is and such that . If then a 6-cylcle has a triangular chord and a chord contrary to Lemma 4. If then a 6-cylcle , has a triangular chord and a chord contrary to Lemma 4.
Suppose that Note that a 6-cycle is adjacent to a 3-cycle and a 5-cycle . It suffices to show that C is adjacent to a 4-cycle to get a contradiction. If , then C is adjacent to a 4-cycle . If , then C is adjacent to a 4-cycle If , then C is adjacent to a 4-cycle
Thus, . By Lemma 6, it remains to show that or is not an endpoint to a chord in say Suppose C has a chord otherwise the desired condition is obtained. If , then we have separating 3-cycle contrary to Lemma 1. By Lemma 4, we have Then, or By Lemma 4, is not adjacent to or Thus, a chord of cannot have as its endpoint. □
Corollary 4. Let v be a 4-vertex incident to bounded faces in cyclic order, where is an inner 5-face, is an inner 3-face, is an inner 5-face, and is an arbitrary face. If is a poor 5-face, then is a rich 5-face or an improper semi-rich 5-face.
Proof. Let and , where Let be a poor 5-face. Then, and are 4-vertices. By Lemma 7, is a -vertex. If or is a -vertex, then is a rich 5-face. Now suppose that and are 4-vertices. If is a not a 3-face, then is an improper semi-rich 5-face. If is a 3-face, then is a -vertex by considering and 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 , let an initial charge of x be and let where D is the unbounded face. By Euler’s Formula, . Let be the charge after the discharge procedure of To get a contradiction, we prove that for each and
Let 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 and f be an incident 3-face of v.
(R2) Let v be a -vertex where and f be an incident 3-face of v.
Let g be a k-face with k incident vertices, say , in cyclic order, and with k adjacent faces, say , in cyclic order. Let be incident to and (i is taken modulo k).
(R3) Let g be a 4-face.
if is an inner 3-face.
(R4) Let g be a 5-face.
if is a 4-face.
Let g be an inner poor 5-face.
if is an inner 3-face.
Let g be an inner proper semi-rich 5-face.
if is an inner 3-face where both and are 4-vertices.
Let g be an inner rich 5-face or an inner improper semi-rich 5-face.
Let g be an extreme 5-face.
if is an inner 3-face.
(R5) Let g be a k-face where .
where and for each .
(R6) The unbounded face D incident to a vertex v receives charge from v but gives 1 to each of its intersecting 3-faces and 5-faces.
It follows from (R6) that for every By this, we consider only a vertex v such that
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, 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, by (R1).
CASE 2:v is a -vertex.
v is incident to some 5-faces.
It follows from Corollary 3 that v is incident to not more than of 3-faces. Thus, by (R2) and .
v is not incident to any 5-faces.
It follows from Corollary 3 that v is incident to at most 3-faces. Thus, by (R2) and .
For a 3-face in an i-cluster , we consider the total of charges in the same cluster. That is and we show that instead.
CASE 3:f is a 3-face in an i-cluster, say where .
If , then by (R6).
If , then each adjacent face of is a -face by Lemma 5 (iii). Thus, by (R5) and (R6).
If , then each 3-face in is an extreme 3-face. Thus, by (R6).
If , then each adjacent face of is a -face by Lemma 5 (iv). Thus, by (R5) and (R6).
If , then is adjacent to at least four -faces by Lemma 5 (iv). Moreover, there are at least two extreme 3-faces in . Thus, by (R5) and (R6).
If , then each 3-face in is an extreme 3-face. Thus, by (R6).
If , then there are two extreme 3-faces in and each adjacent face of is a -face by Lemma 5 (iv). If each vertex in is a 4-vertex, we have by (R5) and (R6). Otherwise, there is a vertex in , which is not a 4-vertex, then we have by (R1), (R2), (R5), and (R6).
If , then there are at least three extreme 3-faces in . Moreover, is adjacent to at least three -faces by Lemma 5 (iv). Thus, by (R5) and (R6).
If , then each 3-face in is an extreme 3-face. Thus, by (R6).
CASE 4:f is an inner 3-face in a 1-cluster.
Let
,
be three incident vertices in cyclic order and
,
be three adjacent faces in cyclic order. Moreover, let
be incident to
and
(
i is taken modulo 3) (See
Figure 1).
Subcase 4.1:f is not adjacent to any 5-faces.
Thus, 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 -faces.
Subcase 4.2: An inner 3-face f is adjacent to some extreme 5-faces.
WLOG, let be an extreme 5-face. Then, by (R4).
is not an inner 5-face where or 3.
Then, is an extreme 5-face or a -face. Thus, by (R4) and (R5). Therefore,
and are inner 5-faces.
- If is a -vertex for some , then by (R1) and (R2). Thus, .
- If is a 4-vertex for each , then and by (R4). Thus,
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 is not a poor 5-face for each . Thus, by (R4) and (R5).
Subcase 4.4:f is a semi-rich 3-face.
Let be a -vertex. By symmetry, we only consider two following cases.
is a poor 5-face.
Then by (R4). Note that if is an improper semi-rich 5-face, a rich 5-face, or a -face where , then by (R4) and (R5).
- If is a 5-face for or 3, then is an improper semi-rich 5-face or a rich 5-face by Corollary 4. It follows that by (R1) and (R2). Thus, .
- If and are -faces, then by (R1) and (R2). Thus, .
is a -face but not a poor 5-face.
Then by (R4) and (R5). If and are -faces, then by (R5). If is a -vertex and or is a 5-face, then by (R2). Thus, it remains to check the case that or is a 5-face and is a 5-vertex. Note that by (R1).
- If and are 5-faces, then is a rich 5-face for or 3 by Lemma 8. It follows that by (R4). Thus, .
- If is a 5-face and is a -face, then by (R5). Thus, .
Subcase 4.5:f is an inner rich 3-face.
Let and be -vertices. Recall that , and are inner -faces and at least one of them is a 5-face. By symmetry, we only consider two following cases.
is a 5-face or and are 5-faces.
That makes and incident to some 5-faces. Then, and by (R1) and (R2). Thus, .
is a -face and either or is a -face.
WLOG, let be a 5-face. That makes incident to some 5-faces. Then and by (R5) and by (R1) and (R2). Thus, .
CASE 5:f is a 3-face in a 2-cluster where .
Let
be a 4-cycle
with a chord
. Let
,
be four adjacent faces of
in cyclic order. Moreover, let
be incident to
and
(
i is taken modulo 4) (See
Figure 2). It follows from Lemma 5 (iii) that
,
, and
are
-faces. By symmetry, we only consider two following cases.
and are 4-vertices.
Then for by (R5). Thus, .
is a -vertex and is a -vertex.
Then by (R1) and (R2), and for by (R5). Thus, .
CASE 6:f is a 3-face in a 3-cluster where .
Let
be a 5-cycle
with two chords
and
. Let
,
be five adjacent faces of
in cyclic order. Moreover, let
be incident to
and
(
i is taken modulo 5). Note that
and
may be the same face (See
Figure 3). It follows from Lemma 5 (iv) that
,
, and
are
-faces. By symmetry, we only consider the two following cases.
and are 4-vertices.
Then and by (R5), and by (R5), and by (R5). Thus, .
is a -vertex and is a -vertex.
Then by (R1) and (R2), and for by (R5). Thus,
CASE 7:f is a 3-face in a 4-cluster where .
Let
be the wheel
where
is a hub and
,
and
are external vertices in cyclic order. Let
,
be four adjacent faces of
in cyclic order. Moreover, let
be incident to
and
(
i is taken modulo 4) (See
Figure 4). By Lemma 5 (iv),
,
, and
are
-faces. Moreover, at least two vertices in
are
-vertices by Lemma 7. By symmetry, we only consider the three following cases.
and are -vertices and and are 4-vertices.
Then , , , by (R5), and by (R1) and (R2). Thus, .
and are -vertices and and are 4-vertices.
Then for by (R5), and and by (R1) and (R2). Thus, .
, and are -vertices and is a -vertex.
Then for by (R5) and for by (R1) and (R2). Thus, .
CASE 8:f is a 4-face adjacent to an inner 3-face, say h.
Since h is an inner 3-face, we have 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 -faces by Lemma 5 (i). Thus 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, by (R4).
Let f be an inner poor 5-face.
Then by (R4).
Let f be an inner semi-rich 5-face.
- If f is a proper semi-rich 5-face, then has three edges with two 4-endpoints. Thus, by (R4).
- If f an improper semi-rich 5-face, then has at most two edges with two 4-endpoints and at most two edges with exactly one -endpoint. Thus, by (R4).
Let f be an inner rich 5-face.
Then f has at least two incident -vertices.
If two incident -vertices are not adjacent in , then has at most one edge with two 4-endpoints. Thus, by (R4). It remains to consider the case that f has exactly two incident -vertices and they are adjacent in Then has two edges with two 4-endpoints and two edges with exactly one -endpoint. Thus, by (R4).
Let f be an extreme 5-face.
Then f has at most an adjacent inner 3-face. Thus, by (R4) and (R6).
CASE 10:f is an m-face where .
Then, by (R5) we have
.
CASE 11: The unbounded face D.
Let the number of intersecting 3-faces and 5-faces of
D be denoted by
Let
denote the set of edges between
and
where this set has size
Then by (R6),
So we may consider that D sends charge 1 to each edge . So each intersecting 3-face and 5-face contains at least two edges in . It follows that . Thus,
This completes the proof.