1 Introduction
A graph is colorable if its vertex set can be partitioned into a graph with maximum degree at most and and a graph with maximum degree at most . Choi, Liu, and Oum CLO have established that there exists exactly two minimal sets of forbidden cycle length such that every planar graph is colorable for some absolute constant .

planar graphs without odd cycles are bipartite, that is,
colorable. 
planar graphs without cycles of length , , and are colorable.
The aim of this paper is to improve this last result. Notice that forbidding cycles of length , , and as subgraphs or as induced subgraphs result in the same graph class. For every , we denote by the cycle on vertices. So we are interested in the class of free planar graph.
We will prove the following two theorems in the next two sections.
Theorem 1.
Every graph in is colorable.
Theorem 2.
For every , either every graph in is colorable, or deciding whether a graph in is colorable is NPcomplete.
In addition, we construct a graph in that is not colorable in Section 4. This graph and Theorem 2 imply the following.
Corollary 3.
Deciding whether a graph in is colorable is NPcomplete.
Since we deal with colorings for some , we denote by the letter the color of the vertices that induce the independent set and we denote by the letter the color of the vertices that induce the graph with maximum degree .
2 Proof of Theorem 1
The proof will be using the discharging method. For every plane graph , we denote by the set of vertices of , by the set of edges of , and by the set of faces of .
Let us define the partial order . Let be the number of vertices in . For any two graphs and , we have if and only if at least one of the following conditions holds:

and .

.
Note that the partial order is welldefined and is a partial linear extension of the subgraph poset.
We suppose for contradiction that is a graph in that is not colorable and is minimal according to . Let denote the number of vertices, the number of edges and the number of faces of . For every vertex , the degree of in is denoted by . For every face , the degree of , denoted , is the number of edges that are shared between this face and another face, plus twice the number of edges that are entirely in . More generally, when counting the number of edges of a certain type in a face, we will always count twice the edges that are only in this face. For all , let us call a vertex of of degree , at most , and at least a vertex, a vertex, and a vertex respectively. For all vertex , a neighbor, a neighbor, and a neighbor of is a neighbor of that is a vertex, a vertex, and a vertex respectively. For all , let us call a face of of degree , at most , and at least a face, a face, and a face respectively. For all set of vertices, an vertex is a vertex that belongs to , and an neighbor of a vertex is a neighbor of that belongs to . For all set of vertices, let denote the set of vertices induced by , and . For convenience, we will note for .
Let us first prove some results on the structure of , and then we will prove that cannot exist, thus proving the theorem.
Lemma 4.
is connected.
Proof.
If is not connected, then every connected component of is smaller than and thus admits a coloring. The union of these colorings gives a coloring of , a contradiction. ∎
Lemma 5.
has no vertex.
Proof.
Let be a vertex and be the neighbor of . The graph admits a coloring since . We get a coloring by assigning to the color distinct from the color of , a contradiction. ∎
Lemma 6.
Every vertex of has a neighbor.
Proof.
Let be a vertex with no neighbors. The graph admits a coloring since . If there is a neighbor of with no neighbor colored , then we color with . Thus, we can assume that every neighbor of that is colored has a neighbor colored in , and thus at most neighbors colored in . Also, we can assume that has at least one neighbor colored , since otherwise can be colored . Thus, has at most neighbors colored and can be colored , a contradiction. ∎
Lemma 7.
Every vertex with degree at least and at most has two neighbors.
Proof.
Suppose for contradiction that contains a vertex such that and such that has at most one neighbor. By Lemma 6, has exactly one neighbor . Let be the other neighbors of . Let be the graph obtained from by adding vertices , such that for every , is adjacent to and .
Notice that since . Moreover, every cycle of length in is associated a cycle of length or in . Therefore , so has a coloring.
If is colored , then every is colored , coloring with leads to a coloring of , a contradiction. Therefore is colored .
While at least one of the ’s has no neighbor colored in , we color it , and color the corresponding with if it was colored . By doing this, we keep a coloring of . We can thus assume that in , every that is colored has a neighbor colored and thus at most five neighbors colored . If at least one of the ’s is colored , then has at most five neighbors colored in , and assigning to gives a coloring of . Otherwise, every is colored , every is colored , and is colored . Thus we assign to to obtain a coloring of , a contradiction. ∎
Lemma 8.
No vertex is adjacent to a vertex.
Proof.
Let be a vertex adjacent to a vertex , let and be the other two neighbors of , and let be the other neighbor of . Let be the graph obtained from by adding five vertices , , , , and which form the cycle . It is easy to check that is in . By Lemmas 6 and 7, , , and are vertices in and thus are vertices in . Since is in but not in , , so . Therefore has a coloring.
Suppose that and are both colored . Then , , and are colored . We color with and with . The number of neighbors of (resp. ) colored in is at most the number of neighbors of (resp. ) colored in . Thus we have a coloring of , a contradiction. Now we assume without loss of generality that is colored . We color with the color of and we color with . The number of neighbors of (resp. , ) colored in is at most the number of neighbors of (resp. , ) colored in . Thus we have a coloring of , a contradiction. ∎
A special face is a face with three vertices and two nonadjacent vertices. See figure 1, left. A special configuration is three faces sharing a common vertex adjacent to three vertices, such that all the other vertices of these faces are vertices. See figure 1, right. We say special structure to speak indifferently about a special face or a special configuration.
Let us define a hypergraph whose vertices are the vertices of and the hyperedges correspond to the sets of vertices contained in the same special structure. For every vertex of , let denote the degree of in , that is the number of hyperedges containing .
Lemma 9.
Let be a special stucture, with the notations of Figure 1. Consider a coloring of .
We can change the color of the ’s, ’s and such that the ’s have no more neighbors colored than before, and for all , if is colored , then has a neighbor colored .
Proof.
If all of the ’s are colored , then there is noting to do. If they are all colored , then we assign to . If one of the ’s, say , is colored and another one, say , is colored , then and are colored and we assign to . Moreover, if is a special configuration and is colored , then is colored and we assign to . ∎
Lemma 10.
For every vertex in , .
Proof.
Let be a vertex that does not verify the lemma, i.e. such that . As is an vertex, . Let be a special structure incident to in . We use the notations of Figure 1, with say . The graph is smaller than , thus it admits a coloring. Since does not admit a coloring, is colored and is colored . By Lemma 9, we can assume that has a neighbor colored in each of its special structures distinct from . If is colored , then is colored , a contradiction. Thus is colored . If is a special face, or if is colored , then we assign to . If is a special configuration and is colored , then is colored and we assign to . In both cases, has at least neighbors colored . Thus has at most neighbors colored and we can assign to , a contradiction. ∎
Lemma 11.
Every component of has at least one vertex such that .
Proof.
Suppose the lemma is false, and let be a component of that does not verify the lemma. If has only one vertex, then this vertex is an vertex, which verifies . Therefore has at least one hyperedge, which corresponds to a special structure of . By Lemma 10, every vertex of verifies . We use the notations of Figure 1. The graph is smaller than , thus it admits a coloring. Since admits no coloring, and are colored . If is a special configuration and is colored , then and are colored and we can color and with . Otherwise, we can color with . Note that and , as well as if it exists and is colored , all have six neighbors colored , and by Lemma 9, we can assume that they all have at least one neighbor colored in each of their special structures besides .
If one of the ’s, say , has an additional neighbor colored , it verifies , a contradiction. Thus, for every , either is colored or has no neighbor colored outside of its special structures and at most one neighbor colored in each special structure besides .
We uncolor and all the ’s and ’s, and let equal to where , the ’s and the ’s are removed. By symmetry, we only consider the vertex . The following procedure either assigns to or ensures that has two neighbors colored in one of its special structures:

For each special structure containing and completely contained in , we use the notations of Figure 1, keeping the same vertex for , but changing the other ones for the vertices in , and do the following:

By Lemma 9, we can assume that every colored has a neighbor colored in each of its special structures that are completely contained in .

Suppose that one of the vertices of distinct from , say , has two neighbors colored in a special structure distinct from or a neighbor colored outside of its special structures. Since , has at most five neighbors colored outside of if is a special face, and at most four neighbors colored outside of if is a special configuration. We assign to and to . If exists and is colored , then we assign to , and otherwise we assign to . We end the procedure.

We uncolor the vertices of and remove them from .

For every vertex in colored , we apply the procedure with instead of . Now is colored or has two neighbors colored in the same special structure.

We add back to the vertices of . If is colored , then we give them color if they are adjacent to a vertex colored and we assign them otherwise, and we end the procedure. If is a special face and is colored , or if is a special configuration and and are colored , then we color and with , we color the other vertices with , and we end the procedure. Suppose is a special configuration, either or , say , is colored , and the other one is colored . We assign to , , and , and to , , , and , and we end the procedure. Now all of the ’s distinct from are colored . We color and (if it exists) with and we color the other vertices in with color .


Now in each special structure containing and completely contained in , all of the vertices distinct from is colored . We assign to and to all of the neighbors of .
Let us prove that the previous procedure terminates. It always calls itself iteratively on a graph with fewer vertices, thus the number of nested iterations is bounded by the order of the initial graph. Furthermore, each iteration of the procedure only does a bounded number of calls to the procedure (at most two). That proves that the procedure terminates.
In the end, if one of the ’s is colored , then it has at most five neighbors colored outside of if is a special face, and at most four neighbors colored outside of if is a special structure. If every is colored , then color with color and the other vertex of with color . Otherwise, assign to , and do the following:

If every is colored , then assign to the ’s and the ’s.

If is a special face and one of the ’s, say , is colored while the other one is colored , then assign to and to .

If is a special structure, then assign to the ’s, and for all , if is colored , then assign to , and if is colored then assign to .
In all cases, we get a coloring of , a contradiction. ∎
For each component of , we choose a vertex in such that as the root of . We then choose an orientation of the edges of such that the only vertices with no incoming edges are the roots (for example do a breadth first search from the root of each component). For each vertex , is said to sponsor all of the special faces that correspond to its outgoing edges in .
We are now going to give some weight on the vertices and faces of the graph. Initially, for all , every vertex has weight , and every face has weight . Thus every face and every vertex has nonnegative initial weight.
We apply the following discharging procedure.

Every vertex gives weight to each of its neighbors, to each special face it sponsors, and to the vertex of each special configuration it sponsors. Additionally, for every edge where and are vertices, and each give to each of the faces containing the edge , and more to the face containing if there is only one face containing .

For each vertex with degree at most in , gives to each of its neighbors. Moreover, it gives to each of its faces where it is adjacent to two vertices and where there are two vertices.

Each face gives to its vertices with degree at most that are consecutive to an vertex, for each time they appear consecutively to an vertex in the boundary of .

Each face gives to each of its vertices with no neighbor and to its vertices with a neighbor.

Each face gives to each of its vertices that belong to a face and have no neighbors, to each of its vertices that belong to a face and have a neighbor, to each of its vertices that do not belong to a face and have no neighbors, and to each of its vertices that do not belong to a face and have a neighbor.
Let be the initial weight distribution, and let be the final weight distribution, after the discharging procedure.
Lemma 12.
Every vertex verifies .
Proof.
Let be a vertex of degree . We have .

Suppose first that . The vertex gives to each of its neighbors and two times for each of its neighbors in Step 1, for a total of . As , we have , therefore if sponsors no special structure, then .
Suppose sponsors a special structure. If sponsors all of its special structures, then is the root of its component in , thus , and thus . If does not sponsor all of its special structures, then , and .

Suppose now that . By Lemma 7, has at least two neighbors. The vertex only gives weight in Step 2. Moreover, it gives at most to each of its neighbors plus for each pair of consecutive vertices in Step 2. If has only neighbors, then it receives in Step 1, and gives at most in Step 2, so . Suppose has at least one neighbor. Let be the number of neighbors of . The vertex receives in Step 1. It gives at most to the vertices and at most to the faces for a total of at most in step 2. It receives at least in Step 3. We have , since and .

Suppose that . By Lemma 7, has at least two neighbors, and by Lemma 8, has no neighbors. If has exactly two neighbors, then it receives in Step 1, gives in Step 2, and receives in Step 3, therefore . If has three neighbors, then receives in Step 1 and an additional in Step 3, and it gives at most in Step 2 unless it is in a special configuration, in which case it gives at most in Step 2 and receives in Step 1. Therefore if has three neighbors, then .

Suppose that . Note that cannot be in two faces since .
In all cases, receives over the procedure, and thus .
∎
Lemma 13.
Every face satisties .
Proof.
Let be a vertex of degree . We have .

If has no two consecutive vertices, then it gives at most to its small vertices over Steps 3 and 4, and does not actually give anything unless one of its vertices is an vertex, and thus gives at most overall.
If has two consecutive vertices and its three other vertices are vertices, then it receives in Step 1 and gives at most overall.
The only remaining case is when has, in this consecutive order, two vertices, an vertex, a vertex with degree at most , and another vertex. In this case, receives in Step 2, and gives over Steps 3 and 4.
In all cases, .

Suppose . Note that if there are two adjacent vertices in , then these two vertices are not in a face, otherwise there would be a cycle of length in . The face has an initial charge of , gives at most to its vertices that are adjacent to an vertex in and nothing to its other vertices. There can be at most four of these vertices. Therefore .

Suppose . Note that at most one pair of adjacent vertices is in a face, otherwise there would be a cycle of length in . The face has an initial charge of , gives at most to its vertices that are adjacent to an vertex in , and nothing to its other vertices. There can be at most five of these vertices, and at most two are given , the other being given at most . Therefore .

Suppose . Note that at most two pairs of adjacent vertices are in a face, otherwise there would be a cycle of length in . The face has an initial charge of , gives at most to its vertices that are adjacent to an vertex in , and nothing to its other vertices. There can be at most six of these vertices, at most four are given , and the others are given at most . Therefore .

Suppose . The face has an initial charge of , gives at most to its vertices that are adjacent to an vertex in , and nothing to its other vertices. There can be at most of these vertices, therefore .
∎
By Euler’s formula, since is connected by Lemma 4 and has at least one vertex, . The initial weight of the graph is . Therefore the initial weight of the graph is negative, thus the final weight of the graph is negative. Since by Lemmas 12 and 13, the final weight of every face and every vertex is nonnegative, we get a contradiction. This completes the proof of Theorem 1.
3 Proof of Theorem 2
Let be a fixed integer. Suppose that there exists a graph in that is not colorable. We consider such a graph that is minimal according to . By adapting the proofs of Lemmas 4, 5, and 6, we obtain that the minimum degree of is at least two and every vertex in is adjacent to a vertex. Suppose for contradiction that contains no vertex. We consider the discharging procedure such that the initial chage of every vertex is equal to its degree and every vertex gives to every adjacent vertex. Then the final charge of a vertex is at least , the final charge of a vertex with is at least , and the final charge of every remaining vertex is at least . This implies that the maximum average degree of is at least , which is a contradiction since is a planar graph with girth at least . Thus, contains a vertex adjacent to the vertices and .
By minimality of , is colorable, every coloring of is such that and get distinct colors, and the vertex in that is colored has exactly neighbors that are colored .
Consider the graph obtained from by adding three vertices , , and which form a path . Notice that is colorable and that in every coloring of is such that is colored and is adjacent to exactly one vertex colored . It is easy to see that is in .
We are ready to prove that deciding whether a graph in is colorable is NPcomplete. The reduction is from the NPcomplete problem of deciding whether a planar graph with girth at least is colorable EMOP:2013 . Given in instance of this problem, we construct a graph , as follows For every vertex in , we add copies of and we add an edge between and the vertex of each these copies. Notice that is in since is planar and every cyle of length at most is contained in a copy of which is in . Notice that a coloring of can be extended to a coloring of . Conversely, a coloring of induces a coloring of . So is colorable if and only if is colorable.
4 A graph in that is not colorable
Consider the graph depicted in Figure 3. Suppose for contradiction that admits a coloring such that all the neighbors of and are colored (the white vertices in the picture). Then the neighbors of those white vertices are colored . We consider the big vertices. Each of them is colored and is ajacent to two vertices colored . For every pair of adjacent red vertices, at least one of them is colored . Notice that every red vertex is adjacent to a big vertex. Since there are pairs of adjacent red vertices, there exists a big vertex that is adjacent to at least two red vertices colored . This big vertex is thus adjacent to four vertices colored , which is a contradiction.
In the graph depicted in Figure 3, every dashed line represent a copy of such that the extremities are and . Suppose for contradiction that this free planar graph admits a coloring. Each of the two drawn edges has at least one extremity colored . Thus, there exist two vertices and colored that are linked by copies of . Since at most neighbors of and at most neighbors of can be colored , one these copies of is such that all the neighbors of and are colored . This is contradiction proves Theorem 2.
Following the proof above, we see that if we remove the green parts in Figures 3 and 3, we obtain a planar graph with girth that is not colorable. A graph with such properties is already known MO14 , but this new graph is smaller ( vertices instead of ) and the proof of noncolorability is simpler.
References
 (1) I. Choi, CH. Liu S. Oum. Characterization of cycle obstruction sets for improper coloring planar graphs. http://mathsci.kaist.ac.kr/~sangil/pdf/2016balanced.pdf
 (2) L. Esperet, M. Montassier, P. Ochem, and A. Pinlou. A complexity dichotomy for the coloring of sparse graphs. J. Graph Theory 73(1) (2013), 85–102.
 (3) M. Montassier and P. Ochem. Nearcolorings: noncolorable graphs and NPcompleteness. Electron. J. Comb. 22(1) (2015), #P1.57.
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