choosability of planar graphs without adjacent short cycles
Abstract
A list assignment of a graph is a function that assigns a list of colors to each vertex . An coloring is a mapping that assigns a color to each vertex so that at most neighbors of receive color . A graph is said to be choosable if it admits an coloring for every list assignment with for all . In 2001, Lih et al. [6] proved that planar graphs without  and cycles are choosable, where . Later, Dong and Xu [3] proved that planar graphs without  and cycles are choosable, where .
There exist planar graphs containing cycles that are not choosable (Crown, Crown and Woodall, 1986 [1]). This partly explains the fact that in all above known sufficient conditions for the choosability of planar graphs the cycles are completely forbidden. In this paper we allow cycles nonadjacent to relatively short cycles. More precisely, we prove that every planar graph without cycles adjacent to  and cycles is choosable. This is a common strengthening of all above mentioned results. Moreover as a consequence we give a partial answer to a question of Xu and Zhang [11] and show that every planar graph without cycles is choosable.
Keyword: Planar graphs; Improper choosability; Cycle.
1 Introduction
All graphs considered in this paper are finite, loopless, and without multiple edges. A plane graph is a particular drawing of a planar graph in the Euclidean plane. For a graph , we use , , , and to denote its vertex set, edge set, order, size and minimum degree, respectively. For , denotes the set of neighbors of in . If there is no confusion about the context, we write for .
A coloring of is a mapping from to a color set such that for any adjacent vertices and . A graph is colorable if it has a coloring. Cowen, Cowen, and Woodall [1] considered defective colorings of graphs. A graph is said to be improper colorable, or simply, colorable, if the vertices of can be colored with colors in such a way that each vertex has at most neighbors receiving the same color as itself. Obviously, a coloring is an ordinary proper coloring.
A list assignment of is a function that assigns a list of colors to each vertex . An coloring with impropriety of integer , or simply an coloring, of is a mapping that assigns a color to each vertex so that at most neighbors of receive color . A graph is choosable with impropriety of integer , or simply choosable, if there exists an coloring for every list assignment with for all . Clearly, a choosable is the ordinary choosability introduced by Erdős, Rubin and Taylor [5] and independently by Vizing [10].
The concept of list improper coloring was independently introduced by Škrekovski [7] and Eaton and Hull [4]. They proved that every planar graph is choosable and every outerplanar graph is choosable. These are both improvement of the results showed in [1] which say that every planar graph is colorable and every outerplanar graph is colorable. Let denote the girth of a graph , i.e., the length of a shortest cycle in . The choosability of planar graph with given has been studied by Škrekovski in [9]. He proved that every planar graph is choosable if , choosable if , choosable if , and choosable if and . Recently, Cushing and Kierstead [2] proved that every planar graph is choosable. So it would be interesting to investigate the sufficient conditions of choosability of subfamilies of planar graphs where some families of cycles are forbidden. Škrekovski proved in [8] that every planar graph without cycles is choosable. Lih et al. [6] proved that planar graphs without  and cycles are choosable, where . Later, Dong and Xu [3] proved that planar graphs without  and cycles are choosable, where . Moreover, Xu and Zhang [11] asked the following question:
Question 1
Is it true that every planar graph without adjacent triangles is choosable?
Recall that there is a planar graph containing cycles that is not colorable [1]. Therefore, while describing choosability planar graphs, one must impose these or those restrictions on cycles. Note that in all previously known sufficient conditions for the choosability of planar graphs, the cycles are completely forbidden. In this paper we allow cycles, but disallow them to have a common edge with relatively short cycles.
The purpose of this paper is to prove the following
Theorem 1
Every planar graph without cycles adjacent to  and cycles is choosable.
Corollary 1
Every planar graph without cycles is choosable.
2 Notation
A vertex of degree (resp. at least , at most ) will be called a vertex (resp. vertex, vertex). A similar notation will be used for cycles and faces. A triangle is synonymous with a 3cycle. For , we use to denote the boundary walk of and write if are the boundary vertices of in cyclic order. For any , we let denote the neighbors of in a cyclic order. Let be the face with and as two boundary edges for , where indices are taken modulo . Moreover, we let denote the number of faces incident to and let denote the number of vertices adjacent to .
An face is called an face if the degree of the vertex is for . Suppose is a vertex incident to a face and adjacent to two vertices not on . If , then we call a light vertex. Otherwise, we call a soft vertex if . A vertex is called an vertex if it is either a vertex or a light vertex. Moreover, we say a face is an face if for each and is an vertex. Suppose is a vertex incident to two 3faces and . Let be the neighbour of not belonging to the faces. If and is a face, then we call a bad vertex.
For all figures in the following section, a vertex is represented by a solid circle when all of its incident edges are drawn; otherwise it is represented by a hollow circle. Moreover, we use a hollow square to denote an vertex.
3 Proof of Theorem 1
The proof of Theorem 1 is done by reducible configurations and discharging procedure. Suppose the theorem is not true. Let be a counterexample with the least number of vertices and edges embedded in the plane. Thus, is connected. We will apply a discharging procedure to reach a contradiction.
We first define a weight function on the vertices and faces of by letting if and if . It follows from Euler’s formula and the relation that the total sum of weights of the vertices and faces is equal to
We then design appropriate discharging rules and redistribute weights accordingly. Once the discharging is finished, a new weight function is produced. The total sum of weights is kept fixed when the discharging is in process. Nevertheless, after the discharging is complete, the new weight function satisfies for all . This leads to the following obvious contradiction,
and hence demonstrates that no such counterexample can exist.
3.1 Reducible configurations of
In this section, we will establish structural properties of . More precisely, we prove that some configurations are reducible. Namely, they cannot appear in because of the minimality of . Since does not contain a cycle adjacent to an cycle, where , by hypothesis, the following fact is easy to observe and will be frequently used throughout this paper without further notice.
Observation 1
does not contain the following structures:
(a) adjacent cycles;
(b) a cycle adjacent to a cycle;
(c) a cycle adjacent to a cycle.
Before showing Lemmas 27, we need to introduce some useful concepts, which were firstly defined by Zhang in [12].
Definition 1
For , let denote the subgraph of induced by . We simply write . Let be an arbitrary list assignment of , and be an coloring of . For each , let , and we call an induced assignment of from . We also say that can be extended to if admits an coloring.
Lemma 2
Suppose that contains the configuration , depicted in Figure 2. Let be an coloring of , where . Denote by an induced list assignment of . If for each , then can be extended to the whole graph .
Proof. Since for each , we can color each with a color properly. Note that . If there exists a color in which appears at most once on the set , then we assign such a color to . It is easy to check that the resulting coloring is an coloring and thus we are done. Otherwise, w.l.o.g., suppose , , and each color in appears exactly twice on the set . W.l.o.g., suppose .
By definition, we see that is either a vertex or a light vertex. We label two steps in the proof for future reference.
(i) If , then . We may assign color to and then recolor with a color in .
(ii) If is a light vertex, denote by the other two neighbors which are different from and . Erase the color of , color with , and recolor and with a color different from its neighbors. We can do this since by definition. Next, we will show how to extend the resulting coloring, denoted by , to . If , then color with a color in . Otherwise, we color with a color in . In each case, one can easily check that the obtained coloring of is an coloring.
Therefore, we complete the proof of Lemma 2.
Lemma 3
satisfies the following.
(B1) A vertex is adjacent to at most two vertices.
(B2) There is no face.
(B3) There is no face which is incident to two light vertices.
(B4) There is no vertex incident to a face and adjacent to two vertices not on .
(B5) There is no vertex incident to two faces and one face.
Proof. Let be a list assignment such that for all . We make use of contradiction to show (B1)(B5).

Suppose that is adjacent to three vertices and . Denote . By the minimality of , admits an coloring . Let be an induced list assignment of . It is easy to deduce that and for each . So for each , we assign the color to it. Now we observe that there exists a color in appearing at most once on the set . We color with such a color. The obtained coloring is an coloring of . This contradicts the choice of .

It suffices to prove that does not contain a face by (A3). Suppose is a face with . For each , let denote the other two neighbors of not on . Denote by the graph obtained from by deleting edge . By the minimality of , has an coloring . If , then itself is colorable and thus we are done. Otherwise, suppose . If is not an coloring of the whole graph , then without loss of generality, assume that and . Moreover, none of ’s neighbors except is colored with 1. First, we recolor each with a color in , where . We should point out that may be the same as , but it does not matter. Note that if at most two of are equal then the resulting coloring is an coloring and thus we are done. Otherwise, suppose that . Since and , we may further reassign color 1 to to obtain an coloring of . This contradicts the choice of .

Suppose is a face incident to two light vertices and . By definition, we see that each () is incident to two other vertices, denoted by and , which are not on . Let denote the graph obtained from by deleting edge . Obviously, has an coloring by the minimality of . Similarly, if , then itself is colorable and thus we are done. Otherwise, suppose . If is not an coloring of , then w.l.o.g., assume that and . Erase the color of and recolor with a color different from its neighbors. If , then color with a color in . Otherwise, color with . It is easy to verify that the resulting coloring is an coloring of , which is a contradiction.

Suppose that a vertex is incident to a face and adjacent to two vertices and . Let . By the minimality of , has an coloring . Let be an induced list assignment of . Obviously, for each and . By Lemma 2, can be extended to , which is a contradiction.

Suppose that a vertex is incident to two faces and one face such that for each , and is an vertex. Namely, is either a vertex or a light vertex. Let . By minimality, admits an coloring . Denote by an induced list assignment of . It is easy to verify that for each and . So we can color with for each . If there exists a color appearing at most once on the set , then we further assign color to and thus obtain an coloring of . Otherwise, each color in appears exactly twice on the set . Since is an vertex, we can apply versions of arguments (i) and (ii) in the proof of Lemma 2 to obtain an coloring of .
Lemma 4
Suppose that is a face. Then
(F1) .
(F2) cannot be a soft vertex.
Proof. (F1) Suppose to the contrary that . Let . By the minimality of , admits an coloring . Let be an induced list assignment of . Notice that , , and . First, we color with and color with . Then color with and with . One can easily check that the resulting coloring of is an coloring. This contradicts the assumption of .
(F2) Suppose to the contrary that is a soft vertex. By definition, has other two neighbors whose degree are both 3, say and . Observe that neither nor is on . Let . Obviously, admits an coloring . Let be an induced list assignment of . For each , we deduce that . Moreover, . We first color with and color with a color in . If at least one of and has the same color as , we can color with a color different from that of and . Otherwise, we can color with a color different from and . Therefore, we achieve an coloring of , which is a contradiction.
Lemma 5
There is no adjacent soft vertices.
Proof. Suppose to the contrary that and are adjacent soft vertices such that is a face and are vertices, which is depicted in Figure 3. By Observation 1(b), cannot be coincided with , where . Let . For each , we color and with a color in and , respectively. If , then color with . It is easy to see that there exists at least one color in which appears at most once on the set . So we may assign such a color to . Now suppose that . By symmetry, we may suppose that . This implies that . Thus, we can first color with and then assign a color in to .
Lemma 6
Suppose is a vertex incident to two faces and . Let be the neighbour of not belonging to and . Then the following holds.
(C1) If and are both faces, then .
(C2) If is a face and is a face, then .
(C3) and cannot be both faces.
Proof. In each of following cases, we will show that an coloring of can be extended to , which is a contradiction.

We only need to show that since by (A1). Suppose that is a vertex. Let . By the minimality of , has an coloring . Let be an induced list assignment of . It is easy to deduce that for each and . So we first color each with . Observe that there exists a color that appears at most once on the set . Therefore, we can color with to obtain an coloring of .

Suppose that , and and are both vertices. By definition, we see that is either a vertex or a light vertex, where . Let . By the minimality of , has an coloring . Let be an induced list assignment of . The proof is split into two cases in light of the conditions of .

Assume is a vertex. It is easy to calculate that for each and . By Lemma 2, can be extended to .

Assume is a light vertex. By definition, let denote the other two neighbors of not on . Recolor and with a color different from its neighbors. Next, we will show how to extend the resulting coloring to . Denote be the induced assignment of . Notice that for each . If , then by Lemma 2, can be extended to . Otherwise, we derive that . First we assign a color in to each , where . It is easy to see that there is at least one color, say , belonging to that appears at most once on the set . We assign such a color to . Then color with a color in but different from .


Suppose that and are both faces such that and and are vertices. Let . Obviously, has an coloring by the minimality of . Let be an induced list assignment of . We assert that satisfies that for each and . By Lemma 2, we can extend to the whole graph successfully.
Lemma 7
There is no face incident to two bad vertices.
Proof. Suppose to the contrary that there is a face incident to two bad vertices and , depicted in Figure 4. Let . By the minimality of , has an coloring . Let be an induced list assignment of . Since each has at most two neighbors in , we deduce that for each . So we first color each with a color . If , namely , then by Lemma 2 we may easy extend to , since for each . Otherwise, we deduce that there exists a color in that is the same as for some fixed . Color with and with a color firstly, where . For our simplicity, denote .
First, suppose that there is a color, say , appearing at most once on the set . We assign such a color to . If , the obtained coloring is obvious an coloring. Otherwise, assume that . Now we erase the color from . One may check that the resulting coloring, say , satisfies that each of has at least one possible color in . In other words, for each . Hence, by Lemma 2, we can easily extend to .
Now, w.l.o.g., suppose that , , and each color in appears exactly twice on the set . It implies that . We apply versions of discussion (i) and (ii) in the proof of Lemma 2. After doing that, one may check that now is colored with and is recolored with a new color, say . There are two cases left to discuss: if , namely the new color of is , then the obtained coloring is an coloring and thus we are done; otherwise, we uncolor . Again, it is easy to see that the resulting coloring, say , satisfies that for each . Therefore, we can easily extend to successfully by Lemma 2.
3.2 Discharging progress
We now apply a discharging procedure to reach a contradiction. Suppose that is adjacent to a vertex such that is not incident to any faces. We call a free vertex if and a pendant vertex if . For simplicity, we use to denote the number of free vertices adjacent to and to denote the number of pendant vertices of . Suppose that is a soft vertex such that is a face and . If the opposite face to via , i.e., , is of degree at least , then we call a weak vertex. We notice that every weak vertex is soft but not vice versa.
For and , let denote the amount of weights transferred from to . Suppose that is a 3face. We use to denote for . Our discharging rules are defined as follows:
(R1) Let be a face. We set
(R1.1)
(R1.2)
(R1.3)
(R1.4)
(R1.5)
(R2) Suppose that is a vertex incident to a face . Then
(R2.1) if and ;
(R2.2) otherwise.
(R3) Suppose that is a nonweak vertex incident to a face .
(R3.1) Assume . Then
(R3.1.1) if the opposite face to via is of degree ;
(R3.1.2) otherwise.
(R3.2) Assume and . Then
(R3.2.1) if at least one of and is a soft vertex;
(R3.2.2) otherwise.
(R3.3) Assume and . Then .
(R4) Every vertex sends to each pendant vertex and to each free vertex.
According to (R3), we notice that a weak vertex does not send any charge.
We first consider the faces. Let be a face.
Case . Initially . Let with . By (A1), . If , then by (A2). Together with (B2), we deduce that is either a face, a face, a face, a face or a face. It follows from (B3) and Lemma 7 that every possibility is indeed covered by rule (R1). Obviously, takes charge 4 in total from its incident vertices. Therefore, .
Case . Clearly, . Assume that is a face. By (A2), there are no adjacent vertices in . It follows that is incident to at most two vertices. By symmetry, we have to discuss three cases depending on the conditions of these vertices.

. By (F1), we deduce that at least one of and is of degree at least . Moreover, if one of and is a vertex, say , we claim that cannot be weak by definition and (B1). Hence, by (R2) and (R3).

and . Note that and are both vertices. Similarly, neither nor can be a weak vertex. It follows from (R3.3) and (R2) that each of and sends charge at least to . So if one of them is a vertex, say , then by (R2) we have that and thus gets in total from incident vertices of . Otherwise, suppose