Planar graphs without cycles and close triangles are colorable
Abstract.
For a set of nonnegative integers , a coloring of a graph is a partition of into such that for every , has maximum degree at most . We prove that all planar graphs without 4cycles and no less than two edges between triangles are colorable.
1. Introduction
The coloring of planar graphs has a long history. The wellknown Four Color Theorem, proved by Appel and Haken (see [1][2]) in the 1970s, states that all planar graphs are 4colorable. Determining whether an arbitrary planar graph is colorable is NPcomplete; much attention has been given to proving sufficient conditions under which planar graphs are 3colorable. The classic example is the theorem by Grötzch [9] showing that planar graphs without 3cycles are 3colorable.
Recently, the study of the coloring of planar graphs with 3 colors has taken a very interesting turn. Steinberg [17] in 1976 famously conjectured that planar graphs without 4cycles and 5cycles are 3colorable. Erdős asked for the constant such that planar graphs excluding cycles of lengths from to are colorable. Borodin, Glebov, Raspaud, and Salavatipour [4] showed that . After being open for almost 40 years, in a very recent paper [6], the Steinberg Conjecture was disproved by a counterexample. This surprising result suggests that the property of planar graphs being 3colorable may be more rare than was previously thought, and spurs the search for more classes of planar graphs that are 3colorable.
One interesting restriction that gives rise to classes of 3colorable planar graphs involves forbidding triangles that are close together. This idea is illustrated in the famous conjecture by Havel.
Conjecture 1.1 (Havel, 1969).
There is a constant (perhaps as small as ) such that any planar graph whose triangles are at distance at least from each other is colorable.
This conjecture was resolved by Dvořák, Král’ and Thomas [8] by showing the truth for any planar graph with , where is the length of the shortest path between the vertices of any two 3cycles. Clearly more work is needed to understand the constant , but in the meantime, there have been advances that combine the hypotheses of the Steinberg and the Havel conjectures. For example, Borodin and Glebov [3] showed that any planar graph without cycles and satisfying is colorable.
With the recent counterexample to Steinberg’s conjecture showing that it may be more difficult to find 3colorable planar graphs than originally thought, it becomes more interesting to investigate “nearly” 3colorable planar graphs. A graph is colorable if can be partitioned into nonempty subsets such that the maximum degree of is at most . In other words, there exists a coloring such that for each color , each vertex colored with has at most neighbors of the same color. Clearly, a graph is properly colorable if and only if it is colorable. In [7], it is shown that every planar graph is colorable.
There are many results in this area; we refer interested readers to [16]. As an illustration, the following is a list of results known for 5cyclefree planar graphs.
Theorem 1.2.
In [18], Wang and Xu proved that planar graphs without cycles are colorable (in fact, choosable), and constructed a noncolorable planar graph that has no cycles (and ). (See Figure 1.) Furthermore, although the result by Borodin and Glebov [3] (and other likewise results) forbids 5cycles and not 4cycles, their proof involves showing that there are no internal cycles in a minimal counterexample.
This motivates the study of the colorability of planar graphs without cycles (but perhaps with cycles) and satisfying . We conjecture that the following is true.
Conjecture 1.3.
If is a planar graph without 4cycles such that , then is 3colorable.
In this paper, we prove a relaxation of Conjecture 1.3. Let be the set of planar graphs with and no 4cycles.
Theorem 1.4.
If , then is colorable.

A coloring of a subgraph of missingsuperextends to if there exists a coloring of that extends with the property that whenever and , where is the set of neighbors of . We say that a subgraph is superextendable to if every coloring of superextends to . When we wish to specify , we will say is superextendable.
We need the following definition.

A 6cycle is missingbad if alternating vertices along the 6cycle are matched to the vertices of a triangle. The triangle is called an interior triangle of a bad cycle. (See Figure 2.) Otherwise, a 6cycle is good.
Our approach is to prove the following stengthening of Theorem 1.4.
Theorem 1.5.
For each , every triangle, 5cycle and good 6cycle in is superextendable.
Observe that the restriction to good 6cycles is necessary. Otherwise, the graph in Figure 2 is a counterexample: precolor the vertices of degree 3 on the 6cycle with color 1.
2. Preliminaries and Definitions
The advantage of proving the stronger theorem involving superextendable colorings was noted by Xu in [20]. Let a cycle in a plane graph be a separating cycle if the deletion of results in a disconnected graph. Let denote the interior of , and similarly the exterior, when the vertices of are deleted. If a proper coloring of a separating cycle can be extended to and individually, then the union of the two colorings is a proper coloring. However, this property would not hold for colorings of the two subgraphs; a vertex of precolored with color 1 may have two neighbors of color 1 in both and , so the union of the two colorings would contain a vertex of color 1 with four neighbors of color 1. The superextendable property allows us to combine colorings of and into a coloring of the entire graph.
In order to illustrate this more clearly, we must introduce some notation that will be used for the remainder of the paper. Our proof of Theorem 1.5 is by contradiction, so we will let for be a counterexample to the theorem of minimum order. That is, some fixed precoloring of a cycle of length 3 or 5 or a good cycle of length 6 in cannot be superextended to , and is the smallest graph with this property. Let denote the vertices of the cycle, and let .
We first observe that cannot be a separating cycle, analogous to Lemma 1 in [20]. Otherwise, can be superextended individually to and by the minimality of , and then the union of these two colorings would be a superextension of to , a contradiction. Hence we may assume that is drawn with as the exterior face.
Lemma 2.1.
does not contain separating triangles, 5cycles or good 6cycles.
Proof.
Suppose otherwise that contains a separating cycle , where the length of is 3 or 5, or is a good 6cycle. Let be the subgraph of induced by together with , and the subgraph of induced by together with . Note that is contained in . By the minimality of , superextends to a coloring of . Now restricted to is a precoloring of , and again applying the minimality of , it superextends to a coloring of . The union of these two colorings is a superextension of to , a contradiction. ∎
The lack of separating short cycles provides additional information about the structure of . The next lemma follows [20].
Lemma 2.2.
The cycle is chordless, and for nonadjacent , .
Proof.
The conclusion is trivial if ; suppose that or .
If has a chord and , then the chord separates into a 3cycle and a 4cycle, contradicting . If , then the chord would create either a 4cycle and a 5cycle, again contradicting , or a 3cycle and a 5cycle. In the second case, since , one of these cycles would have to be a separating cycle, contradicting Lemma 2.1.
Now consider nonadjacent in . Suppose that there exists where . If , then together with forms a 4cycle and a 5cycle, which is impossible. If , then together with forms either a 6cycle and a 4cycle, which is impossible, or two 5cycles. Neither of the 5cycles can be separating, but then , and it is easy to verify that such a graph is not a counterexample. Therefore and can have no such neighbor. ∎
Now we introduce some definitions we use in the rest of the paper. In a coloring of , a vertex is saturated if it is colored with and has two neighbors of color ; otherwise, it is called nicely colored (i.e., it is colored with or , or it is colored with but not saturated). If is nicely colored, then a neighbor of can be (re)colored with 1. If has at most three colored neighbors, then nicely recoloring means is either recolored with color 2 or 3 (if one of those colors is available), or has at most one neighbor with color 1, and remains color 1.
Lemma 2.3.
Suppose is a superextension of to , and . If is nicely recolored, then the extension remains a superextension.
Proof.
Since the color of is not changed to 1, can be color 1 only if , in which case must not have a neighbor of color 1 on . ∎
Suppose that is a vertex of a face . A neighbor of is an outer neighbor (with respect to ) if is not on . A vertex (or vertex) in is a vertex of degree at least (or at most) . A cycle is a cycle whose vertices’ degrees are , respectively. A vertex is triangular if it lies on a 3face, otherwise it is called nontriangular.
Let be a face in . If is adjacent to a vertex on , then is called a pendant face to , and and are pendant neighbors to each other.
A vertex on a face in is called special if its two neighbors on the face are vertices. If is a vertex with a special neighbor , we call the pendant 5face containing a pendant special 5face to . (See Figure 3.) A vertex in is potentially special if two of its neighbors are vertices in . Note that a special vertex is also potentially special.
3. Reducible Configurations
Lemma 3.1.
There are no vertices in .
Proof.
Let be a vertex in . Then superextends to by the minimality of . Now can be properly colored, a contradiction. ∎
The following lemma is a foundational one for our paper, and similar lemmas appear in other related results (see for example [14]). We include the proof for completeness.
Lemma 3.2.
Every 3vertex in is adjacent to at least one vertex or a vertex on .
Proof.
Let be a vertex not adjacent to a vertex of . By the minimality of , superextends to . Since this coloring cannot extend to , the colors must appear in , and the vertex colored with is neither nicely colored, nor can it be recolored with or . Hence is adjacent to vertices colored with and and to two vertices of color , thus . ∎
Lemma 3.3.
Suppose has been superextended to some subgraph of , and let be a vertex that has exactly two colored neighbors, and , both in . Then can be recolored with color , unless one of the following holds:

and are adjacent and , or

and are nonadjacent and one is a vertex.
Proof.
Suppose that cannot be recolored with . Then if we recolor with , some neighbor of (say that is colored with will have three neighbors of color . Note that color and must appear in , for otherwise we recolor with the absent color, so the degree of is at least . Assume further that . If , then and . As there are at most two different colors in , we can properly recolor (if its color is ) or remove the color of and then properly recolor and in order(if its color is or ). Now in both cases, we can recolor with . ∎
Observation 3.4.
Lemma 3.5.
Let be a vertex in . If is adjacent to special vertices and pendant  or faces, then .
Proof.
Suppose to the contrary, that there exists some vertex with . Consider . We know that , so superextends to a coloring of . By Lemma 3.3, the pendant or special vertices in can be recolored with color 1, leaving at most one vertex in with a different color. Now either color 2 or 3 is available to properly color , superextending to , a contradiction. ∎
Lemma 3.6.
Let be an internal cycle with chord . Let , and be adjacent to internal pendant , faces and special vertices. If and is not the interior triangle of any bad cycle, then , and .
Proof.
Suppose the statement is not true. Consider the graph formed by deleting and and identifying and into a single vertex .
We claim that . First of all, we do not create chords in of since both and are in . The only path of length 3 from to in goes through and , else there would be a separating 5cycle in or a triangle at vertex or . Hence no new triangles are created by identifying and . Further, since and are triangular in , neither nor can be triangular. Thus . It remains to show that has no 4cycles; such a 4cycle could be created by a path of length 4 in from to . For such a path to exist, there must be a 4cycle containing , which does not exist, or a separating 6cycle in . Since does not contain good separating 6cycles and is not the interior triangle of any bad cycle, must contain a 6cycle containing with a triangle inside, pendant to . Let be the pendant neighbor to . But since there are no separating triangles or 5cycles in , and the two triangular neighbors of are all degree 3, contradicting Lemma 3.2. Hence there is no such bad 6cycle, and .
By the minimality of , we know that there exists a superextension of to . We claim we can extend to a coloring that superextends to , a contradiction. Let for and . Recolor so that it is nicely colored. It remains to color and , and to verify that is adjacent to at most two neighbors of color when it is colored with .
If (or symmetrically 3), then we can properly color . Since has three nicely colored neighbors, it can be colored, completing the superextension to . Hence we may assume .
Suppose that either , or and and are the only neighbors of that are colored with 1. At least one of is nicely colored with 1; assume by symmetry that is nicely colored. Properly color , and now there is a color available for , again a contradiction. If , then can be recolored in this way, hence .
Now suppose that and has a third neighbor colored with 1. (Since was adjacent to in , cannot have more than three neighbors with color 1.) Observe that is a pendant neighbor of . Remove the color of , and by Lemma 3.3, recolor with color 1 the other neighbors of that are special vertices or pendant neighbors on  or faces. Since , has at most one neighbor colored from . Hence can be properly colored. If is not colored with , then we can color with and properly color . If is colored with , then and can be consecutively colored properly. Therefore . ∎
Lemma 3.7.
Let be a vertex with . Let be the face containing for . If and are both vertices that are on internal  or faces and both and are faces in , then .
Proof.
Suppose otherwise, that . Then we discuss the following two cases.
Case 1: One of and (say , by symmetry) is not on the interior triangle of a bad cycle. Let be the graph formed by identifying and in into vertex . First of all, we do not create chords of , for otherwise, the chord must be incident with , conradicting and in . Note that no new triangles can be created, else there would be a separating cycle in , contradicting Lemma 2.1. Since , and its neighbors are all nontriangular, hence . Also, contains no cyces, else there would be a path between and which implies a separating good cycle in , contradicting Lemma 2.1. Therefore, . By the minimality of , we know that there exists a superextension of to . We show that can be extended to a coloring of . Let , and let for . It remains to color to arrive at a contradiction. Recolor and with 1 by Lemma 3.3. If , then can be properly colored. If or (say 2), then must be colored , else we can color properly. In this case, we can properly recolor and , and then color with .
Case 2: Both and are on the interior faces of bad cycles. Let for such that and are on and and are on . Let . By the minimality of , there exists a superextension of to . We first claim that is saturated. For otherwise, color with , and then either and we color with and then properly, or and we color properly in order, a contradiction. Similarly, is also saturated. Furthermore, neither nor can be recolored, so their neighborhoods must have color set . Further, must be colored with , or we could recolor it by Lemma 3.3. We claim that must also be colored with . For otherwise, we can recolor with 1 and then recolor properly, a contradiction. But since all neighbors of are colored with , can be recolored with a different color so that can be nicely recolored, a contradiction again. ∎
Lemma 3.8.
Given a face in , the pendant neighbors of the 3vertices on either are in or have degree at least .
Proof.
Consider a 3face in , where and (see Figure 4). Assume to the contrary that the outer neighbor of has degree at most , but . Consider . Because , we know that there exists a superextension of to . Since and have degree at most 3 in , they can be nicely recolored. But now we extend to by properly coloring and then coloring , a contradiction. ∎
Lemma 3.9.
Let be a vertex with in the clockwise order. If is an internal face, then neither nor can be a vertex in .
Proof.
Without loss of generality, let be a vertex in . Let be the graph formed by identifying and into in . First of all, we do not create chords in of , for otherwise, the chord must be at , thus there is a path connecting two vertices on , which will create a separating or good cycle, a contradiction to Lemma 2.1. Note that no new triangles can be created, else there would be a separating cycle in , contradicting Lemma 2.1. Since , both and are nontriangular in , hence . None of is on a bad cycle, thus contains no cycles, else there would be a separating good cycle in , contradicting Lemma 2.1. Therefore, . By the minimality of , we know that there exists a superextension of to . We show that can be extended to a coloring of . Let , and let for . We claim this coloring extends to , a contradiction.
If (or ), then we properly recolor and and color with 1. If , then has color set , else we can color with the missing color. By symmetry, let be colored with and be colored with . We can recolor with either if the outer neighbor of is colored with , or otherwise. In either case, can be colored with 2, a contradiction. ∎
Lemma 3.10.
Suppose that is a face in , with , , and . Let the outer neighbors of be in clockwise order so that and are on the same face. Let be the outer neighbor of (See Figure 4).

At most one of (and symmetrically ) is potentially special (and hence at most one is special).

If and are potentially special, then either , or .
Proof.
Consider the graph formed by identifying vertices and into vertex , and deleting the vertex . Note all 3cycles in were 3cycles in , else there would be a separating 5cycle in , contradicting Lemma 2.1. Also, since was incident to a 3face, cannot be triangular, and hence is maintained in . We also claim that does not contain any 4cycles. Any such 4cycle would correspond to a path of length 4 in between and , and such a path would imply a separating 6cycle in ; such a 6cycle must be bad. But since is triangular, the 6cycle could not have another interior triangle, a contradiction. Hence , and by the minimality of , superextends to a coloring of . We show that can be extended to a coloring of when the hypotheses fail. Let , and let for all other . It remains to color to arrive at a contradiction.
(1) Assume first that and are both potentially special. Properly recolor , and properly recolor and . If (or symmetrically 3), then can be colored with 1, unless , , and are all colored with 1, in which case can be colored with ; even if , this would be a superextension of to , a contradiction. If , then and can be recolored with 1 by Lemma 3.3, and can be properly colored. Hence at most one of is potentially special. Repeating the proof with in place of shows that at most one of is potentially special.
Now suppose one of is potentially special (assume by symmetry), and is, as well. Properly recolor and . If and is colored with or , then can be colored with 1. Otherwise, and can be recolored with 1, and either 2 or 3 is available for .
(2) Now assume that and are potentially special, , and .
If , then recolor with 1. Now can be colored with or unless and neither is color 1. If , , or are not nicely colored with 1, then properly recolor them, and color 1 is now available for . If they are all nicely colored with 1, then can be recolored 1, unless , in which case can be recolored with . In either case, becomes available for , a contradiction.
If (or symmetrically 3), consider the color on . If , then properly recolor and , and now color 1 is available for . If , then recolor with 1 by Lemma 3.3. So can be colored with 3, unless is given color 3. In this case, consider . Since , the vertex can be nicely recolored, and we can color with 1, a contradiction. ∎
Lemma 3.11.
Let be a face in with vertices such that . Then either a neighbor of is in , or has at most two potentially special neighbors.
Proof.
Suppose that no neighbors of are in . Let , and be the outer neighbors of , labeled as in Figure 4. Let be the graph formed by identifying and in into a single vertex , and be graph formed by identifying and in into a single vertex . Let and let . Assume by symmetry that the number of potentially special vertices in is at most the number of potentially special vertices in . This implies that we will consider for this proof, but a similar argument would hold for if had more potentially special vertices.
Note that all 3cycles in were 3cycles in , else there would be a separating 5cycle in , contradicting Lemma 2.1. Also, since was incident to a 3face, is maintained in . We also claim that does not contain any 4cycles. Any 4cycle in would correspond to the contraction of the edges between two vertices in , and that would imply a separating 6cycle in ; such a 6cycle must be good, since the outer neighbors of cannot be triangular, but no such separating cycle exists. Thus , and by the minimality of , we know that superextends to a coloring of .
We claim that extends to a coloring of that superextends , a contradiction. Let for , and . It remains to assign colors to and . Let be the outer neighbor of . If contains at most one potentially special vertex, then by the minimality of , the result holds. Hence we may assume contains at least two potentially special vertices, and by symmetry, we may assume is potentially special.
Suppose first that . Recolor with 1 by Lemma 3.3. If , then can be colored with , leaving a color available for . If , then can be colored with 1 and a color is left for , unless and is improperly colored (and cannot be nicely recolored). This implies that is not potentially special, and hence is potentially special. Thus can be recolored with 1, and and can be colored with 2 and 3.
Otherwise, by symmetry we may assume that . Properly color , and then color 1 is available for unless either all of receives color 1, or some vertex in is not nicely colored with 1 and cannot be nicely recolored. In the former case, color 3 is available for . In the latter case, some neighbor of in is not potentially special. But then the other two vertices in must be potentially special, and they can be recolored with 1. This leaves color 3 available for . ∎
4. Discharging Procedure
We are now ready to present a discharging procedure that will complete the proof of the theorem. Let each vertex have an initial charge of , and each face in our fixed plane drawing of have an initial charge of . Recall that the length of is ; let . By Euler’s Formula, .
Let be the charge of after the discharge procedure. To lead to a contradiction, we shall prove that for all and .
Let a face with exactly one vertex in be an face, and a face with two or more vertices in be an face for . Note that by Lemma 2.2, no 3face contains three vertices of and no 5face contains four consecutive vertices of . Observe also that since , a vertex can be incident to at most one 3face.
We call a vertex good if it contains three consecutive neighbors that are neither special vertices on faces nor on internal pendant  or faces of , furthermore, they are the nontriangular neighbors when is on a face. Otherwise, it is bad. Extending this, a vertex in is good if it is a nontriangular vertex, a good vertex, or a vertex. We call a face in rich if it has one good vertex and two or more other vertices.
Below are the discharging rules:

If is a 4vertex and is an incident face in , then :

gives to when is a face, and to when is any other triangular face.

gives to when is a 5face and is nontriangular, and gives to when is a 5face and is a triangular vertex with no incident face.


If is a vertex with , then :

gives to each incident face in with exactly two vertices that are consecutive, gives to each incident face in that is not rich and has at least three vertices, and gives to each other incident face, unless is a bad vertex and the face is rich, in which case gives . In addition, gives to each of its pendant special 5faces in .

gives to pendant , faces, and faces in , respectively.

gives to incident ,  and other incident faces in , respectively (when ).

gives to incident , , and faces in , respectively (when ).


The initial charge of on is distributed as follows:

gets from each vertex , from each face.

gives to each face in , to each face in , to pendant  and faces in , and to its pendant special 5faces.

Lemma 4.1.
The face has a positive final charge.
Proof.
Let be the number of pendant faces and pendant special faces at , respectively. Assume that gets from faces. Let be the set of edges between and and let be its size. Then by (R3),
(1) 
We aim to balance the charge of 2 on each with the charge distributed to the incident and pendant faces; we can view this as sharing a charge of 2 for each with the faces.

If is on a face , then can give to , to a potential pendant face, and to a potential incident face.

If is adjacent to a pendant face, then it can give to the face and to each potential incident face.

If is neither on a face nor adjacent to any face, then it can give to each potential incident face and to a potential pendant face. In this case, would have a surplus of at least .
Observe first that pendant faces are collectively allocated from . Since each face in contains two edges in , it is allocated a charge of . Each face in contains two edges in , and it is allocated at least , unless it shares an edge with a face in . In that case, it is not adjacent to pendant face, so it gains . This implies that
Hence from (1), if is a  or cycle. When , , with equality only if and
This implies that each edge must be as in (a) or (b), and it is either the common edge of two faces and adjacent to a pendant face or the common edge of a face and a face and adjacent to a pendant face. Note that edges on faces in cannot be adjacent to pendant faces, so .
Let . Suppose that is on a face in . Then must be a vertex for ; otherwise and must be on the same face and thus implies cannot be on a 3face or adjacent to a pendant face in . But in this case there is a face, contradicting . Therefore .
Now if are vertices and in the same face