1 Introduction
Given a simple graph with vertex set and edge set , we use to denote the set of neighbors of in and say that is the degree of in . A planar graph is a graph admitting a drawing in the plane with no crossing and typically we say such a drawing a plane graph. By we denote the face set of a plane graph , and for any face , we use to denote the degree of in , which is the number of edges that are incident with in (cutedges are counted twice). By , we denote the set of vertices incident with a face in a plane graph . A , , or vertex (resp. face) is a vertex (resp. face) of degree , at least , or at most , respectively.
A proper coloring of a graph is a coloring on using colors so that adjacent vertices receive distinct colors. If every vertex of degree at least two is incident with at least two colors, then we call this proper coloring a dynamic coloring. The minimum integer such that has a proper (resp. dynamic) coloring is the chromatic number (resp. dynamic chromatic number) of , denoted by (resp. ).
The wellknown four color theorem states that for every planar graph . In 2013, Kim, Lee, and Park [14] proved that for every planar graph and the equality holds if and only if , answering a conjecture of Chen et al. [10]. Furthermore, the same conclusion holds even for minorfree graphs, which was proved by Kim, Lee and Oum [16] in 2016. For other results on the dynamic coloring of graphs, we refer the reads to [1, 4, 8, 9, 10, 13, 18, 19, 20, 21, 24, 26].
Imaging that each vertex is assigned a list of distinct candidate colors, our goal is to color the vertices of so that every vertex receives color from its list assignment and the resulting coloring of is a proper (resp. dynamic) coloring. If we win for a given list assignment to , then is colorable (resp. dynamically colorable). Furthermore, if we win for every given list assignment to with for each , then is choosable (resp. dynamically choosable). The minimum integer so that is choosable (resp. dynamically choosable) is the list chromatic number (resp. dynamic list chromatic number) of , denoted by (resp. ).
Thomassen’s theorem [25] states that for every planar graph , and the sharpness of this upper bound was confirmed by Voigt [27], who constructed a planar graph with and . This reminds us that and are not always the same, even for planar graphs. Similarly, Esperet [12] showed that there is a planar bipartite graph with and , and moreover, there exists for every a bipartite graph with and . Hence the gap between (or ) and can be any large. For further interesting readings on the dynamic list coloring of graphs, we refer the readers to [4, 14, 15].
A subdivision of a graph is the graph derived from by inserting on each edge a new vertex of degree two, denoted by . One can see Figure 1 for an example of , which is 1planar.
Fact 1.
For any graph , and .
Proof.
Let be a mapping that maps a vertex of to the vertex of corresponding to it, and let be the set of new added 2vertices to while doing the 2subdivision. Let be an arbitrary list assigment on , where . We extend to an list on , i.e, for any . Since the two neighbors of a 2vertex of shall be colored with distinct colors in any dynamic coloring of , there is a dynamic coloring of so that , , and for any . Therefore, we construct an coloring of by letting for any . This implies that . The proof for is similar (we just proceed by fixing every list used in the privious proof to be ). ∎
Note that the equality in Fact 1 does not always hold. One easy example is the cycle on vertices. Since and it is known [2, 18, 21] that
we have
A graph is planar if it can be drawn in the plane so that each edge is crossed at most times. Specially, the 1planarity was initially introduced by Ringel [23] in 1965, who proved that for every 1planar graph and conjectured that every 1planar graph is 6colorable. This conjecture was solved by Borodin [7] in 1984, who also gave a new proof [6] in 1995. Due to the 1planar graph , the upper bound 6 for the chromatic number of the class of 1planar graphs is sharp. Since 2006, the list coloring of 1planar graphs was also investigated by many researchers including Albertson and Mohar [3], Wang and Lih [28]. In particular, the second group [28] proved that for every 1planar graph . Actually, the class of 1planar graphs is among the most investigated graph families within the socalled “beyond planar graphs”, see [11]. For those who want to know more about 1planar graphs, we refer them to a recent survey due to Kobourov, Liotta and Montecchiani [17].
Let be the class of graphs that are planar and nonplanar. By we denote the minimum integer so that for each . Similarly, we can define , , and . If with , then it is easy to see that the 2subdivision of is planar.
Pach and Tóth [22] showed that for each 2planar graph . This implies that each 2planar graph has a vertex of degree at most and thus . Since is a non1planar 2planar graph, and .
We now look back at Fact 1. If there is a 2planar graph with (resp. ), then is a 1planar graph with (resp. ). This implies
Fact 2.
and .
The aim of this paper is to give a reasonable upper bound, say 11, for (note that ). In other words, we prove the following
Theorem 1.
If is a 1planar graph, then .
2 Dynamically Minimal Graphs
A graph class is hereditary if is closed by taking subgraphs. A graph is dynamically minimal in a hereditary class if is not dynamically choosable and any graph with is dynamically choosable.
In this section, we use to stand the class of 1planar graphs, i.e., . Suppose that is a dynamically minimal graph in , It follows that is a 1planar graph with the smallest value of such that there is an list assignment to the vertices of such that is not dynamically colorable. Moreover, we assume that is a plane graph (i.e, a drawing of in the plane so that its 1planarity is satisfied) that has the minimum number of crossings.
The associated plane graph of a 1plane is the plane graph derived from by turning all crossings into new vertices of degree 4, and those 4vertices in are called false vertices. If a vertex of is not false, then it is a true vertex. A face of the plane graph is a false face if it is incident with at least one false vertex, and is a true face otherwise. Clearly, no two false vertices are adjacent in by the definition of the 1planarity and each face of is incident with at most false vertices.
In the following statements or the proofs of the propositions, stands for an arbitrary given hereditary graph class, and is the list assignment mentioned above.
Proposition 1.
If is a dynamically minimal graph in with , then .
Proof.
Suppose, to the contrary, that has an edge with . By the minimality of , is dynamically colorable. Let be a dynamic coloring of . Coloring from with a color different from the colors on and a neighbor of besides , we obtain a dynamic coloring of , a contradiction. ∎
Proposition 2.
If is a dynamically minimal graph in with , then no two vertices are adjacent in .
Proof.
Suppose, to the contrary, that has an edge with . Let , , , and . By the minimality of , has a dynamic coloring . Coloring with and with , we get a dynamic coloring of , a contradiction. ∎
Actually, Proposition 2 can be generalized to the following
Proposition 3.
If is a dynamically minimal graph in with , then each edge of is incident with at least one vertex.
Proof.
We first claim that if and , then . Suppose, to the contrary, that . Let and , where for each and for each . Let for each . Note that or may be 0, in which case or , respectively. By Proposition 2, , , and for each . By the minimality of , has a dynamic coloring . Color in this order with colors such that , where , for each , and if , or if . Note that . It is easy to see that this results in a dynamic coloring of , a contradiction.
We come back to the proof of Proposition 3. Suppose, to the contrary, that has an edge with . Let and . By the arguments in the first paragraph, and for each and . By the minimality of , has a dynamic coloring . Without loss of generality, assume that .
If , then we construct a dynamic coloring of by coloring and in order with and such that and , where and .
If , then we construct a dynamic coloring of by coloring and in order with and such that and , where and .
In each of the above two cases we win since and . So we have contradictions. ∎
Proposition 4.
If is a dynamically minimal graph in with and is a vertex incident with a triangle, then .
Proof.
Proposition 5.
If is a dynamically minimal graph in with and is a vertex incident with a false face of , then either is false or .
Proof.
Suppose, to the contrary, that is true and . Let be the false 3face that is incident with , where is a false vertex. Basically we assume crosses in at a point . Let . Since , by Proposition 3, and for each . If , then let . If , then let . In any case, we can see that is still 1planar, i.e, . Let be another neighbor of in that is not or or . By the minimality of , has a dynamic coloring . Extending to a dynamic coloring of by coloring with a color , where , we get a contradiction. Note that . ∎
Proposition 6.
If is a dynamically minimal graph in with and is a face of with , where , then both and are false.
Proof.
Suppose, to the contrary, that at least one of and is true. We divide the proof into two major cases.
First of all, we assume that both and are true. If , then let . If , then let . In any case, it is easy to see that is still 1planar, i.e, . By the minimality of , has a dynamic coloring . Let . By Proposition 3, any neighbor of in has degree at least since . Let be another neighbor of in that is not among , and let be another neighbor of in that is not among (such vertices exist since the number of the excluded vertices are at most ). Color with a color , where . Since , we obtain a dynamic coloring of , a contradiction.
On the other hand, we assume, by symmetry, that is true and is false. Basically we assume that crosses in at the point . If , then let , and otherwise let . The 1planarity of is easy to be confirmed (note that the crossing point in is removed by the deletion of , and if we have to add the edge , it can be drawn so that it is only crossed by in ). By the minimality of , has a dynamic coloring . Let . By Proposition 3, any neighbor of in has degree at least since . Let or be another neighbor of or in that is not among or , respectively. Color with a color , where . Since , we get a dynamic coloring of , a contradiction. ∎
3 Discharging: the Proof of Theorem 1
If Theorem 1 is false, then there is a dynamically minimal 1planar graph . For every element , we assign an initial charge . By the wellknown Euler formulae on the plane graph , we have
If there is a 4face in such that , and are false vertices, then we call a special face.
Initially, we define the following discharging rules (also see Figure 2) so that the charges are transferred among the elements in .

Every true 3face in receives from each of its incident vertices;

Every false 3face in receives from each of its incident vertices;

Every vertex incident with a special face sends to , from which the special vertex on receives ;

Every face in sends 1 to each of its incident special 2vertices if there are some ones;

After applying R1–R4, every face in redistributes its charge equitably to each of its incident nonspecial 2vertices or (special or nonspecial) vertices if there are some ones.
Let be the final charge of the element after discharging. Clearly
In the following, we show that for every by Claims 1, 2, 4, 5, and 8. This contradiction completes the proof of Theorem 1.
Claim 1.
Every face in has a nonnegative final charge.
Proof.
If is a true 3face (i.e., a triangle in ), then every vertex incident with is a vertex by Proposition 4, and thus by R1. If is a false 3face, then is incident with two vertices by Proposition 5, which implies by R2.
If is a nonspecial face, then no rule is valid for and thus . If is a special face, then by R3.
If is a 5face, then is incident with at most one 2vertex by Proposition 6. Therefore, the remaining charge of after R1–R4 are applied to it is at least , and thus has a nonnegative final charge by R5. ∎
Claim 2.
Every face is incident with at most two special vertices in . Therefore, every face in has a nonnegative final charge.
Proof.
Suppose, to the contrary, that is a 6face such that are special 2vertices and are false vertices. According to the definition of the special 2vertices, there are three vertices and such that (resp. and ) crosses (resp. and ) in at the crossing (resp. and ), see Figure 3(a). Pulling the vertex (resp. ) into the face of that is incident with the path (resp. ), we get another one 1planar drawing of with three less crossings, see Figure 3(b). This contradicts the initial assumption that the drawing of has the minimum number of crossings.
Therefore, every face in has charge at least after R1–R4 are applied to it, and thus has nonnegative final charge by R5.
Claim 3.
Let be a face in with and being five consecutive vertices on the boundary of such that is a vertex, is a nonspecial vertex or a (special or nonspecial) vertex, and are false vertices.
If is a vertex, then sends at least to ;
If is a vertex, then sends at least to .
Proof.
Let (resp. ) be the number of special vertices (resp. nonspecial vertices and special or nonspecial vertices) that are incident with .
(1) Suppose that is a vertex. By Proposition 6, there are at least false vertices in . This implies that
Therefore, sends to at least
by R4 and R5.
(2) Suppose that is a vertex. By Proposition 6, there are at least false vertices in . This implies that
Therefore, by R4 and R5, sends to at least
if .
On the other hand, if , then and , since . Hence would send at least to by R5. ∎
Claim 4.
Every vertex in has a nonnegative final charge.
Proof.
By Propositions 4 (applying it by choosing as ) and 5, every vertex in is not incident with a 3face in . By Proposition 6, the neighbors of in , say and , are both false vertices.
If is a special 2vertex, then receives 1 from its incident special 4face, say , by R3. If the other face incident with in is still a 4face, say , then there would be two edges in connecting to , one passing through the crossing and the other passing through the crossing . This contradicts the fact that is a simple graph. Therefore, is incident with a face, from which receives another 1 by R4. Hence .
So in the following, we assume that is a nonspecial 2vertex.
If is incident with a 4face, say , then since is nonspecial. Let (resp. ) be the vertices in such that (resp. ) passes through the crossing (resp. ). Since is a simple graph, , and moreover, and are vertices by Proposition 3. Therefore, is incident with a face that satisfies the condition of Claim 3(1). Since such a face would send at least 2 to by Claim 3(1),
If is incident with two faces and , then let and be edges of that pass through the crossings and , respectively, such that and are vertices on , where . By Proposition 3, there are at least two vertices among and . Therefore, either or satisfies the condition of Claim 3(1), or both and satisfy the condition of Claim 3(2). In each case receives at least 2 from and , and thus . ∎
Claim 5.
Every vertex in has a nonnegative final charge.
Proof.
By Propositions 4 and 5, every vertex in is not incident with a 3face in . By Proposition 6, the three neighbors of in , say and , are false vertices. Let and be the face that is incident with the path , and in .
Let be the edge of that passes through the crossings , where and . By Proposition 3, either or , say , is a vertex. If is a face, then it satisfies the condition of Claim 3(1) or Claim 3(2). This implies that receives at least 1 from , and thus . Hence we assume that is a 4face. Actually, is a special face now, from which receives by R3. This implies that . ∎
Claim 6.
No two special faces sharing a common vertex are adjacent in .
Proof.
Suppose, to the contrary, and are two adjacent special 4faces in so that . By the definition of the special face, and are vertices and is a false vertex. This implies that , contradicting Proposition 3. ∎
Claim 7.
If is a vertex and and are three consecutive faces that are incident with in , then totally sends to and at most .
Proof.
If there is only one special 4face among and , then by R1–R3, totally sends to and at most . If there are at least two special 4faces among and , then by Claim 6, they are and , and is a nonspecial face. In this case totally sends to and by R3. ∎
Claim 8.
Every vertex in has a nonnegative final charge.
Proof.
Since vertices of degree between 4 and 8 are not involved in the discharging rules, their final charges are the same with their initial charges, which are nonnegative. Suppose that is a vertex of degree .
If , then by R2.
If , then let be the faces in this order around . Let with be the charge that sends to and let , where the subscripts are taken modular . One can see that
where the second inequality holds by Claim 7. Hence
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