Threecoloring trianglefree graphs on surfaces V. Coloring planar graphs with distant anomalies^{†}^{†}thanks: 18 January 2015. A preprint of an earlier version of this paper appeared (under different title) as [11].
Abstract
We settle a problem of Havel by showing that there exists an absolute constant such that if is a planar graph in which every two distinct triangles are at distance at least , then is colorable. In fact, we prove a more general theorem. Let be a planar graph, and let be a set of connected subgraphs of , each of bounded size, such that every two distinct members of are at least a specified distance apart and all triangles of are contained in . We give a sufficient condition for the existence of a coloring of such that for every the restriction of to is constrained in a specified way.
1 Introduction
This paper is a part of a series aimed at studying the colorability of graphs on a fixed surface that are either trianglefree, or have their triangles restricted in some way. Here, we are concerned with coloring planar graphs. All graphs in this paper are finite and simple; that is, have no loops or multiple edges. All colorings that we consider are proper, assigning different colors to adjacent vertices. The following is a classical theorem of Grötzsch [17].
Theorem 1.1.
Every trianglefree planar graph is colorable.
There is a long history of generalizations that extend the theorem to classes of graphs that include triangles. An easy modification of Grötzsch’ proof shows that every planar graph with at most one triangle is colorable. Even more is true—every planar graph with at most three triangles is colorable. This was first claimed by Grünbaum [18], however his proof contains an error. This error was fixed by Aksionov [1] and later Borodin [5] gave another proof. There are infinitely many critical planar graphs with four triangles, but they were recently completely characterized by Borodin et al. [6].
As another direction of research, Grünbaum [18] conjectured that every planar graph with no intersecting triangles is colorable. This was disproved by Havel [19], who formulated a more cautious question whether there exists a constant such that every planar graph such that the distance between every two triangles is at least is colorable. In [20], Havel shows that if such a constant exists, then , and Aksionov and Mel’nikov [2] improved this bound to . Borodin [4] constructed a family of graphs that suggests that it may not be possible to obtain a positive answer to Havel’s question using local reductions only.
The answer to Havel’s question is known to be positive under various additional conditions (e.g., no cycles [8], no cycles adjacent to triangles [7], a distance constraint on cycles [9]), see the online survey of Montassier [21] for a more complete list. The purpose of this paper is to describe a solution to Havel’s problem.
Theorem 1.2.
There exists an absolute constant such that if is a planar graph and every two distinct triangles in are at distance at least , then is colorable.
Let us remark that our proof gives an explicit upper bound on the constant of Theorem 1.2, which however is not very good (, say) compared with the known lower bounds.
A natural extension of Havel’s question is whether instead of triangles, we could allow other kinds of distant anomalies, such as colorable subgraphs containing several triangles (the simplest one being a diamond, that is, without an edge) or even more strongly, prescribing specific colorings of some distant subgraphs. Similar questions have been studied for other graph classes. For example, Albertson [3] proved that if is a set of vertices in a planar graph that are precolored with colors and are at distance at least from each other, then the precoloring of can be extended to a 5coloring of . Furthermore, using the results of the third paper of this series [12], it is easy to see that any precoloring of sufficiently distant vertices of a planar graph of girth at least can be extended to a coloring of . We can even precolor larger connected subgraphs, as long as these precolorings can be extended locally to the vertices of at some bounded distance from the precolored subgraphs. Both for coloring planar graphs and coloring planar graphs of girth at least five this follows from the fact that the corresponding critical graphs satisfy a certain isoperimetric inequality [22].
The situation is somewhat more complicated for graphs of girth four. Firstly, as we will discuss in Section 4, there is a global constraint on colorings of plane graphs based on winding number, which implies that in graphs with almost all faces of length four, precoloring a subgraph may give restrictions on possible colorings of distant parts of the graph. For example, if we prescribed specific colorings of the triangles in Theorem 1.2, the resulting claim would be false, even though such precolorings extend locally. Secondly, nonfacial (separating) cycles are problematic as well and they need to be treated with care in many of the results of this series, see e.g. Theorem 2.2 below. Specifically, we cannot replace triangles in Theorem 1.2 by diamonds, even though this seems viable when considering only the winding number argument, as shown by the class of graphs (with many separating cycles) constructed by Thomas and Walls [23].
Thus, in our second result, we only deal with graphs without separating cycles, and we need to allow certain flexibility in the prescribed colorings of distant subgraphs. The exact formulation of the result (Theorem 5.1) is somewhat technical, and we postpone it till Section 5. Here, let us give just a special case covering several interesting kinds of anomalies. The pattern of a coloring is the set . That is, two colorings have the same pattern if they only differ by a permutation of colors.
Theorem 1.3.
There exists an absolute constant with the following property. Let be a plane graph without separating cycles. Let be a set of vertices of . Let be a set of cycles of . Let be a set of vertices of of degree at most . For each , let be a color. For each , let be a coloring of . Suppose that the distance between any two vertices or subgraphs belonging to is at least . If all triangles in belong to , then has a coloring such that

for every ,

has the same pattern on as for every , and

for every neighbor of a vertex .
Let us remark that forbidding separating cycles is necessary when the anomalies (except for triangles) and are considered, as shown by simple variations of the construction of Thomas and Walls [23]. On the other hand, there does not appear to be any principal reason to exclude cycles when only precolored single vertices are allowed.
Conjecture 1.4.
There exists an absolute constant with the following property. Let be a plane trianglefree graph, let be a set of vertices of and let be an arbitrary function. If the distance between every two vertices of is at least , then extends to a coloring of .
In Theorem 5.1, we show that Conjecture 1.4 is implied by the following seemingly simpler statement.
Conjecture 1.5.
There exists an absolute constant with the following property. Let be a plane trianglefree graph, let be a cycle bounding a face of and let be a vertex of . Let be a coloring of . If the distance between and is at least , then extends to a coloring of .
If an vertex planar trianglefree graph has bounded maximum degree, then we can select a subset of its vertices of size such that the distance between any two of vertices of is at least . If does not contain separating cycles, then by Theorem 1.3, we can color so that all vertices of have prescribed colors. By choosing the colors of vertices in , we obtain exponentially many 3colorings of . This solves a special case of a conjecture of Thomassen [24] that all trianglefree planar graphs have exponentially many 3colorings.
Corollary 1.6.
For every , there exists such that every planar trianglefree graph of maximum degree at most and without separating cycles has at least 3colorings.
The rest of the paper is structured as follows. In the next section, we state several previous results which we need in the proofs. In Section 3, we study the structure of graphs where no faces can be collapsed without decreasing distances between anomalies, showing that they contain long cylindrical quadrangulated subgraphs. In Section 4, we study the colorings of such cylindrical subgraphs. Finally, in Section 5, we prove a statement generalizing Theorems 1.2 and 1.3.
2 Previous results
We use the following lemma of Aksionov [1].
Lemma 2.1.
Let be a plane graph with at most one triangle, and let be either the null graph or a facial cycle of of length at most five. Assume that if has length five and has a triangle , then and are edgedisjoint. Then every coloring of extends to a coloring of .
We also need several results from previous papers of this series. Let be a graph and its subgraph. We say that is critical if and for every proper subgraph of that includes , there exists a coloring of that extends to a coloring of , but does not extend to a coloring of . The following claim is a special case of the general form of the main result of [13] (Theorem 4.1).
Theorem 2.2.
There exists an absolute constant with the following property. Let be a plane graph and a (not necessarily connected) subgraph of such that all triangles and all separating cycles in are contained in . If is critical, then , where the summation is over all faces of of length at least five.
The following is a simple corollary of Lemma 5.3 of [13].
Lemma 2.3.
Let be a trianglefree plane graph with outer face bounded by a cycle and with another face bounded by a cycle of length at least . If every cycle separating from in has length at least , then every coloring of the cycle bounding extends to a coloring of .
Finally, let us state a basic property of critical graphs.
Proposition 2.4.
Let be a graph and its subgraph such that is critical. If , and , then is critical.
3 Structure of graphs without collapsible faces
Essentially all papers dealing with colorability of trianglefree planar graphs first eliminate faces by identifying their opposite vertices, thus reducing the problem to graphs of girth . However, this reduction might decrease distances in the resulting graph, which constrains its applicability for the problems we consider. In this section, we give a structural result on graphs in that no face can be reduced.
Let be a cycle in a graph , and let . We say that the cycle is tight if has length four and the vertices of can be numbered in order such that for some integer the vertices are at distance exactly from , and the vertices are at distance exactly from . We say that a face is tight if it is bounded by an tight cycle.
Lemma 3.1.
Let be an integer, let be a graph, and let be a family of subsets of such that the distance between every two distinct sets of is at least . Let be a cycle in of length four and assume that for each pair of diagonally opposite vertices of , two distinct sets in are at distance at most in the graph obtained from by identifying and . Then there exists a unique set at distance at most from . Furthermore, is tight.
Proof.
Let the vertices of be in order. By hypothesis there exist sets , where is at distance from , such that , , , and . From the symmetry we may assume that and . The distance between and is at most , and thus . Let us set . If any is at distance at most from , then the distance between and is at most , and thus . It follows that is the unique element of at distance at most from .
Note that , and hence , because and are at distance at least . This and the inequality imply that . But there is a symmetry between and , and hence an analogous argument shows that . Thus for the vertices are both at distance from . If was at distance or less from , then and would be at distance at most , a contradiction. The same holds for by symmetry, and hence and are at distance from , as desired. ∎
Let be a graph, let and let be a cycle in . We say that is equidistant from if for some integer every vertex of is at distance exactly from . We will also say that is equidistant from at distance .
We say that a plane graph is a cylindrical quadrangulation with boundary faces and if the distinct faces and of are bounded by cycles and all other faces of have length four. The union of the cycles bounding and is called the boundary of . The cylindrical quadrangulation is a joint if , every cycle of separating from has length at least and the distance between and in is at least . If appears as a subgraph of another plane graph , we say that the appearance is clean if every face of except for and is also a face of . An cylindrical grid is the Cartesian product of a path with vertices and a cycle of length .
Lemma 3.2.
Let be a plane graph and let induce a connected subgraph of . Let be an equidistant cycle at some distance from . Let and . Let denote the set of vertices of at distance at least and at most from that are separated from by , including . Let denote the set of faces of which are separated from by and are incident with at least one vertex of . Assume that every face in is tight and every vertex in has degree at least three. Then contains a clean joint such that .
Proof.
For an integer , let . Note that for every , and . Choose the smallest integer for that there exists an equidistant cycle at distance from such that and . Such an integer exists, since satisfies the requirements for . Let be the maximum integer such that contains a clean cylindrical grid with boundary faces and as a subgraph such that , is bounded by and is bounded by an equidistant cycle at distance from . Such an integer exists, since (treated as a cylindrical grid) satisfies the requirements for .
We claim that , and thus satisfies the conclusion of the theorem. Suppose that . Removing splits the plane to two open sets, let be the one that does not contain . Observe that has no chord contained in , as otherwise there exists an equidistant cycle of length less than at distance from , contrary to the minimality of . Consider an edge and let be the face of incident with contained in . Then is at distance from , and thus is tight, i.e., the vertices on the boundary on may be denoted by in order and and are at distance from . Now let be the other neighbor of on , and let us repeat the same argument to the edge , obtaining a face boundary . If , then there exists a face incident with and contained in that is incident neither with nor with . However, since has no chord contained in , both neighbors of in are at distance at least from , contrary to the assumption that is tight. This proves that .
Hence, every has a unique neighbor contained in . Let be the subgraph of induced by . If contained a vertex whose degree in is one, then would have degree two in , contrary to the assumptions of this lemma. Hence, the minimum degree of is at least two, and thus contains a cycle . Note that is equidistant at distance from , and by the minimality of , it follows that , and thus . Therefore, for distinct vertices . We conclude that we can extend to a clean cylindrical grid by adding the vertices and edges of the faces of incident with and contained in , contrary to the maximality of . This finishes the proof. ∎
Next, we consider the case that some of the faces of the set are not tight, but instead are contained in a short separating cycle. A face is attached to a cycle if the boundary cycle of and intersect in a path of length two.
Lemma 3.3.
Let and be integers, and let . Let be a connected plane graph and let induce connected subgraphs of at distance at least from each other. Suppose that every equidistant cycle in at distance at least and at most from has length at most and that all vertices of at distance at least and at most from have degree at least three. Assume furthermore that every face of at distance at least and at most from has length four, and it is either tight or attached to a cycle of length at most that separates from . Then contains a clean joint such that every vertex of is at distance at least and at most from .
Proof.
Let be the set of all cycles in that separate from and contain a vertex at distance at least and at most from . For an integer such that , let denote the subgraph of induced by vertices at distance exactly from .
If and is a connected component of whose distance from every element of in is at least , then has minimum degree at least two.  (1) 
Proof.
Consider a vertex and any face incident with . The distance between and is at least and at most , and thus is a face. Since the distance between and every element of is at least two, is not attached to a cycle separating from ; hence, is tight. Therefore, the boundary of contains an edge of incident with . Since the same claim holds for each face incident with and since has degree at least three in , we conclude that has degree at least two in . ∎
Consider a cycle . Removing splits the plane to two open sets, denote by the one that does not contain . For , we write if is contained in the closure of , and we write for the subgraph of drawn in the closure of .
Consider cycles of the same length such that and no with satisfies . For , let denote the distance between and . If and , then contains a clean joint such that every vertex of is at distance at least and at most from .  (2) 
Proof.
Note that by the assumptions of the claim, no cycle in that separates from has length less than and the distance between and is at least . If each vertex of is at distance at most from , then by the assumptions of this lemma, every face of has length four, and thus we can set .
Therefore, assume that contains a vertex at distance more than from . Let and let be a connected component of contained in . Observe that every element of which intersects is at distance at most from , and thus its distance from is at least two. By (1), we conclude that contains a cycle . Note that is equidistant from at distance , and by the assumptions of this lemma, has length at most .
Consider any face at distance at most from that is separated from by . Note that is a face and its distance from every element of is at least two, and thus it is not attached to any cycle of . We conclude that each such face is tight. By Lemma 3.2, contains a clean joint as required. ∎
Let and . For , let and be chosen so that , every cycle in of length is at distance either at most or at least from , and subject to these conditions, is as large as possible.
Consider a fixed . If no cycle in has length and is at distance more than and less than from , then we have and . Otherwise, let be a cycle of length whose distance from satisfies and subject to that, is as small as possible; and, let be a cycle of length whose distance from satisfies and subject to that, is as large as possible. If , then (2) implies that the conclusion of this lemma holds, and thus we can assume that . Note that the distance of every cycle in of length from is at most , or between and (inclusive), or at least . Furthermore, , and thus .
It follows that . Let and let be a connected component of . Note that the distance between and every element of is at least two, and thus by (1), contains a cycle . Since is equidistant at distance from , the assumptions of this lemma imply that has length at most . Consider a face of at distance at least and at most from . Note that is a face and observe that the distance between and any element of is at least one. Therefore, is tight by the assumptions of this lemma. By Lemma 3.2, contains a clean joint such that every vertex of is at distance at least and at most from , as required. ∎
Let be a plane graph, let be an odd cycle in , let be one of the two connected open subsets of the plane bounded by , let be an edge of , let be the vertex of that is farthest (as measured in ) from and let be a vertex of such that either , or does not belong to the closure of . Let and be the paths in joining and , respectively, with . We say that is a petal with top if there exists a path in between and such that and are shortest paths from to and in , respectively.
Let be a connected subgraph of and consider a cycle which is equidistant at some distance from . The removal of splits the plane into two open sets, let be the one containing . For each , choose a path of length joining to . We can choose the paths so that for every , the paths and are either disjoint or intersect in a path ending in . Removing and the paths for splits to several parts; for each , let be the one whose boundary contains . Clearly, and are disjoint for distinct . Note that if is an edge of , the path intersects and is the common endpoint of and in , then is a petal with top . We call the collection a flower of with respect to .
Lemma 3.4.
Let and be integers, let be a connected plane graph and let be a subset of inducing a connected subgraph. Let be either the null graph, or a cycle in bounding a face of length at most . Suppose that every face of not bounded by at distance at least and at most from is tight, and all vertices belonging to at distance exactly from have degree at least three. Let be a petal with top for some and some edge of , such that is disjoint from the closure of and the distance between and is at most . Then, there exists a face of at distance at most from which is contained in , such that either has length other than or it is bounded by .
Proof.
We can assume that is minimal, i.e., there is no such that is a petal satisfying the assumptions of the lemma. Since is bounded by an odd cycle, there exists an odd face contained in . It suffices to consider the case that the distance between and is at least . Let be the subgraph of induced by vertices at distance exactly from that are contained in the closure of . Note that may intersect the boundary of only in the edge .
If , then forms a cut in that separates the rest of the boundary of from the vertices incident with . Observe that this implies that there exists a face contained in and incident with both and which either does not have length or is bounded by (let us remark that cannot have length unless it is bounded by , since it would not be tight). Hence, the conclusion of this lemma is satisfied.
Therefore, we can assume that . Consider a vertex . If any face incident with has length other than or is bounded by , then the conclusion of this lemma is satisfied. Hence, assume that this is not the case, and thus all such faces are tight. Furthermore, has degree at least three in . As in (1) in the proof of Lemma 3.3, we conclude that has degree at least two in . Hence, contains a cycle , which is equidistant at distance from . Let be a flower of with respect to . We can choose so that is a petal for every . There exists exactly one element such that is contained in the closure of . Every distinct from is a subset of . Since , it follows that each such petal is a proper subset of . This contradicts the minimality of . ∎
Consider a face in a plane graph , bounded by a closed walk going clockwise around . A pair for (where and ) is called an angle in , and is its tip.
Lemma 3.5.
For all integers , there exists an integer with the following property. Let be a connected plane graph. Let be a set of subsets of such that each induces a connected subgraph of with at most vertices and the distance between every two elements of is at least . Let be either the null graph or a cycle of length at most bounding a face of . Suppose that every face of not bounded by at distance at least and at most from any element is tight. Let , and assume that every triangle of is contained in . Furthermore, assume that for every separating cycle of , if one of the open regions of the plane bounded by is disjoint from and contains a vertex of exactly one , then the distance between and is less than . Let . If is critical and , then contains a clean joint whose vertices are at distance at least and at most from some element of . Furthermore, is vertexdisjoint from .
Proof.
Let , where is the constant from Theorem 2.2, and let . Let .
Without loss of generality, we can assume that if is nonnull, then it bounds the outer face of . If there exists such that either

is a separating cycle such that for at least two distinct , the open disk bounded by contains at least one vertex of both and , or

for some distinct , forms the boundary of a face of which is not outer and which contains ,
then choose so that is inclusionwiseminimal and let be the subgraph of drawn in the closure of . Otherwise, let and . Let consist of the sets such that . If there exists (necessarily unique by the assumptions of this lemma) such that is not disjoint from , then let , otherwise let .
For each , let consist of and of all separating cycles such that the open disk bounded by contains at least one vertex of .
For each , the boundary of the outer face of has only one component and . The distance in between and every vertex of is less than . Furthermore, for any distinct from , the subgraph is drawn in the outer face of .  (3) 
Proof.
Let be all separating cycles in such that for , there exists a vertex contained in the open disk bounded by , no other separating cycle of bounds an open disk containing , and . Consider distinct indices and such that . By the choice of , no set in distinct from has vertices in the disk bounded by , and by the assumptions of this lemma, the distance between and is less than . It follows that the distance between and each vertex of is less than . If say had a chord, then it would be contained in a union of two triangles and we would have by the assumptions of this lemma. Thus, neither nor has a chord, and we conclude that if the open disks bounded by and intersected, then their union would contain a disk bounded by a cycle contradicting the maximality of or . Therefore, the open disks bounded by and are disjoint. If a vertex is contained in the open disk bounded by , then , and thus each vertex of contributes at most vertices to .
Since is connected, either and , or each of the cycles , …, contains a vertex of . We conclude that is connected. Finally, consider a set distinct from . Since the distance between each vertex of and (or and ) is less than and the distance between and is at least , it follows that and are vertexdisjoint. If were not drawn in the outer face of , then would be contained either inside a face of or inside one of the cycles , …, , which contradicts the choice of . ∎
Let be the intersection of the outer faces of the graphs for , and let be the subgraph of contained in the closure of . Observe that is a subgraph of . Let and note that all triangles and separating cycles in are contained in . By Proposition 2.4, is critical.
A face of is poisonous if either it has length at least or its boundary is contained in . Since for each , and , the choice of and Theorem 2.2 imply that
, where the summation is over all poisonous faces of .  (4) 
Consider . We say that an angle in is contaminated if is poisonous and the distance between and in is at most . Since every contaminated angle contributes at least one toward the sum in (4), we deduce that there exists such that there are at most angles that are contaminated. Let us fix such .
We say that an integer such that is contaminated if there exists an contaminated angle in whose tip is at distance exactly from in . By the choice of and , we conclude that there exist integers and such that
and and no integer such that is contaminated.  (5) 
Our next claim bounds the length of equidistant cycles. Note that contains some vertices whose distance from in is at least . By the choice of and the assumptions that , the distance between elements of is at least and is connected.
If is a cycle in that is equidistant at distance from in , where , then .  (6) 
Proof.
Let be a flower of with respect to . Since , it follows that the outer face of has length at most , and thus at most elements of the flower contain an edge of in their closure.
Let us recall that at most angles are contaminated. If , then there exists such that contains no contaminated angle and the closure of does not contain any edge of . Observe that is a petal for some . By Lemma 3.4, there exists a face of contained in which is odd or bounded by , and the distance of from is at most . Since does not contain a contaminated angle, it follows that is not a face of .
Let be the face of that contains . Note that the distance between and is at most , and thus the boundary of is not contained in for any distinct from . Furthermore, by the choice of , the boundary of intersects in at most one vertex. Since no angle in is contaminated, we conclude that is not a face of and it is a face whose boundary forms a separating cycle in . By the construction of , there exists such that . By the choice of , we have . However, the distance between and in is less than by (3), and thus the distance between and in is less than . This is a contradiction. ∎
Consider a vertex whose distance from in is , where . Since is not contaminated by (5), all faces incident with are tight. Furthermore, , and since is critical, it follows that has degree at least three in . Let be the set of vertices of whose distance from is at most and that are not drawn in the outer face of . Let . Note that a vertex is at distance from in , where , if and only if and the distance between and in is . The conclusion of this lemma then follows by Lemma 3.3 applied to with and equal to an arbitrary vertex of at distance at least from . ∎
We also need a variant of Lemma 3.5 dealing with the case that .
Lemma 3.6.
For all integers and , there exists an integer with the following property. Let be a connected plane graph, a subset of at most vertices of inducing a connected subgraph and a cycle of length at most bounding a face of , such that the distance between and is at least . Suppose that every face of at distance at least and at most from is either tight or attached to a cycle separating from . Let and assume that every triangle in is contained in and that the distance of every separating cycle of from is at most . If is critical, then contains a clean joint vertexdisjoint from .
Proof.
Let , where is the constant from Theorem 2.2, and let . Let .
Without loss of generality, assume that bounds the outer face of . By Lemma 2.1, each separating cycle in bounds an open disk containing at least one vertex of , since is critical. Let be the boundary of the outer face of the subgraph of consisting of and all separating cycles. As in (3), is connected and . Let be the subgraph of drawn in the closure of the outer face of , let and note that is critical (by Proposition 2.4) and contains no separating cycles. Let us define a poisonous face and contamination as in the proof of Lemma 3.5.
Since and , the choice of and Theorem 2.2 implies that , where the summation is over all poisonous faces of . Consequently, there exist integers and such that