Painting and Correspondence Coloring of
Squares of Planar Graphs with
no 4cycles
Abstract
We study listcoloring of squares of planar graphs with no 4cycles. We show that if is such a graph, then . When is sufficiently large, we strengthen this bound to . Our bounds also hold for Alon–Tarsi number, paint number, and correspondence chromatic number. To complement these results, we show that 4cycles are unique in having this property. Specifically, let be a finite list of positive integers, with . For each constant , we construct a planar graph with no cycle with length in , but for which .
1 Introduction
The square\marginnotesquare[0cm], , of a graph is formed from by adding an edge for each pair of vertices, and , at distance two in . It is easy to check that , and this bound can be tight, as when is the 5cycle or the Petersen graph (here and denote the chromatic number and maximum degree). Even when is sufficiently large, there exist constructions showing that this upper bound on cannot be improved much. For example, when is the incidence graph of a projective plane, . However, for planar graphs, we have much better bounds on .
Recall that Euler’s formula implies that every planar graph is 5degenerate. Coloring vertices greedily in the reverse of this degeneracy order [9],[5, Theorem 4.9] shows that . Refinements of this approach have led to successive improvements of this upper bound, culminating with the result of Molloy and Salavatipour [10] that .
Every graph satisfies , and for planar graphs we might naively hope to prove a matching upper bound, or at least a bound of the form , for some constant . However, for each , Wegner constructed a planar graph with and ; see Figure 1 for his construction. So to prove a bound of the form , we must restrict to some proper subset of planar graphs.
Wang and Lih [12] conjectured that, for each , there exists such that if is a planar graph with girth at least and , then . This is true for , but false for ; see [3]. For each , there exists a planar graph with and with girth 6 such that . However, Dvořák et al. [7] proved a surprising complementary result: , whenever is a planar graph with girth 6 and sufficiently large. This work inspired analogous results for planar graphs with (i) girth 5 [2] and (ii) no 4cycles or 5cycles (though 3cycles are allowed) [6]^{1}^{1}1Here we only hit the highlights. For a more detailed history of this problem, we recommend the introduction of [8] and [5, Conjecture 4.7 ff.].. In each case the bound still holds (though the required lower bound on is larger).
The work above naturally leads to the following question. Exactly which cycle lengths can be forbidden from planar graphs to get a bound of the form ? For a set of positive integers, let denote the family of planar graphs having no cycles with length in .
Main Theorem.
For a finite set there exists a constant such that for all if and only if .
We prove the Main Theorem in two parts. Immediately below we give a construction that proves the “only if” part. In Section 2 we handle the “if” part, the case when . In fact, we prove the stronger statement that the vertices of every graph can be ordered so that each vertex is preceded in the order by at most of its neighbors in . Now the coloring result follows by coloring greedily. In Section 3, when is sufficiently large we strengthen our bound to , which is sharp. This bound also holds for paint number, Alon–Tarsi number, and correspondence chromatic number (all defined at the end of Section 2).
Lemma 1.1.
If , for some odd integer , then there does not exist a constant such that for every .
Proof.
Begin with a cycle and replace each edge with a copy of , so that the two vertices of degree replace and . The resulting graph, has maximum degree and has cycles only of lengths and . In every proper coloring of , each color class contains at most vertices of degree 2 in (by the Pigeonhole Principle). Since has 2vertices, . Given any constant , we can choose sufficiently large so that . ∎
2 Graphs with no 4cycles
Our goal in this section is to prove Theorem 2.1, below. First we need a few definitions. A vertex\marginnotevertex[0cm] (resp. vertex, vertex) is a vertex of degree equal to (resp. at least, at most) ; a neighbor\marginnoteneighbor[0cm], of a vertex , is an adjacent vertex. Analogously, we define face\marginnoteface[.3cm], face, and face. We write for the degree of a vertex and for the length of a face . We write \marginnote[0cm] to denote and for . We write for the set of neighbors of in . When the context could be unclear, we specify our meaning by using , , and . An order, , of is good for \marginnotegood for [0cm] if each vertex, , of is preceded in by at most vertices in . Following the approach of [4], we prove the degeneracy result below, which immediately implies the desired coloring bounds, by coloring greedily.
Theorem 2.1.
For every planar graph with no 4cycles, there exists a vertex order such that each vertex is preceded in by at most of its neighbors in .
Our proof of Theorem 2.1 is by discharging, with initial charge for each vertex and for each face . In the next section we discuss the discharging rules, but for now it is enough to note that we only need to give extra charge to 2vertices, 3vertices, and 3faces. Here we prove that certain configurations are reducible; that is, they cannot appear in a minimal counterexample. In each case we assume that our minimal counterexample contains such a configuration. We modify to get a smaller graph (that is also planar and without 4cycles), and which therefore has the desired vertex order, . Finally, we modify to get , a good vertex order for of . Each reducible configuration formalizes the intuition that every 2vertex, 3vertex, and 3face of must be near a vertex of high degree. This is useful, since has extra charge to share with nearby vertices and faces that need it.
Proof of Theorem 2.1. Suppose the theorem is false, and let be a counterexample that minimizes the number of vertices and, subject to that, the number of edges. A vertex is big if , and is small\marginnotebig, small[0cm] if . Note that , since otherwise each vertex has at most neighbors in , so every vertex order shows that is not a counterexample.
2.1 Reducible Configurations
Key Lemma.
For an edge in , if both and are not big, then at least one of and has at least two big neighbors.
Proof.
Suppose to the contrary that both and are not big, and that each has at most one big neighbor. By minimality, has a good order, . By deleting and from , we get a good order (for ) of . Since is not big and has at most one big neighbor, . By symmetry, . Thus, by appending and to the order, we get a good order for , which is a contradiction. ∎
Lemma 2.2.
If a face is incident with a vertex, then the other two vertices on must be big vertices.
Proof.
Let be a face that is incident with a vertex . Suppose to the contrary that is not big. By minimality, has a good order, which is a good order for of . Since is not big, . Thus, we can append to obtain a good order of , which is a contradiction. ∎
Lemma 2.3.
Every face that is incident with two vertices is also incident with a big vertex.
Proof.
Suppose that a face is incident with two vertices and a vertex . Applying the Key Lemma to shows that must be big. ∎
Lemma 2.4.
Every vertex has a big neighbor.
Proof.
Let be a vertex with neighbors . Suppose to the contrary that every is not big. Applying the Key Lemma to each edge shows that each must be a vertex. Consider the graph formed from by adding a path of length two between each pair of neighbors of . (Since each is not big, we have .) Since has fewer vertices, by minimality has a good order , and also is a good order for of . Since each neighbor of is small, . So appending to gives a good order for , which is a contradiction. ∎
Lemma 2.5.
If a face is incident with a vertex and at most one big vertex, then the neighbor of that is not on must be a big vertex.
Proof.
Let be a vertex on a face and let be the neighbor of that is not on . Suppose to the contrary that both and are not big. Applying the Key Lemma to edge shows that is a vertex. Consider the graph formed from by adding paths of length two between and and also between and . So, has fewer vertices than . By minimality, has a good order, , which also is a good order for of . Since has at most one big neighbor, . So appending to gives a good order for , which is a contradiction. ∎
2.2 Discharging
We use the initial charges for each vertex and for each face . Note that, by Euler’s formula, the sum of these initial charges is . Using the structural lemmas in Section 2.1, we redistribute this charge so that each vertex and face ends with nonnegative charge. However, this gives a contradiction, since a sum of nonnegatives is equal to . To redistribute charge, we use the following six discharging rules, applied in succession. (See Figure 2 for an illustration of the discharging rules.)

Each edge takes from each incident face and from each incident big vertex^{2}^{2}2A cutedge takes from its incident face..

If edge is incident to a 3face , then gives all its charge (received by (R1)) to . Otherwise, distributes its charge equally among incident vertices where .

Each big vertex gives to each neighbor.

Each vertex, vertex, and small vertex gives to each 2neighbor. If either is a vertex with at least two big neighbors or is a small vertex, then gives to each incident 3face that is incident with a vertex other than that is not big.

Assume vertices and are big and the edge lies on a 3face . If is a vertex, then all charge given from to (and vice versa) by (R3) continues on to . If is a vertex, then all charge given from to (and vice versa) by (R3) continues on to face .

If a vertex has an incident 3face with negative charge, then gives its excess charge to .
Now we show that each vertex and face ends with nonnegative charge, which yields the desired contradiction.
Each face ends with charge . Each edge receives charge by (R1) and gives it all away by (R2), so ends with 0. Consider a big vertex . For each of its neighbors , the charge that gives to by (R1) is and to by (R3) is , for a total of . So ends with ; this is nonnegative, since .
Consider a small vertex . Let \marginnote, [0cm] denote the number of 2neighbors of and the number of 3faces incident with that are not incident with two big neighbors of (that is, 3faces that get from ). By (R4), gives away . Suppose that has a 2neighbor. By the Key Lemma, has at least two big neighbors, so . Furthermore, Lemma 2.2 implies that . To see this, note that if gives charge to 3face , then gives no charge to , , and the other face incident to each of and . So, ends with at least ; this is positive, since . Now instead assume that has no 2neighbors. Since has no 4cycles, . Thus, ends with at least ; this is nonnegative since .
So, to complete the proof we only need to consider faces, vertices, 3vertices, and 4vertices.
Claim 2.6.
Every vertex that is on a face ends with nonnegative charge.
Proof.
By Lemma 2.2, both and must be big. By (R3), gets from each of and . And by (R5), gets another . So ends with . ∎
Claim 2.7.
Every vertex that is not on a face ends with nonnegative charge.
Proof.
Let and be the neighbors of a vertex . It suffices to show that gets total charge at least 1 from and , since by symmetry it also gets at least 1 from and , so ends with at least . Applying the Key Lemma to shows that either is big or is a vertex with two big neighbors. By (R1), gets from incident faces and by (R2) gives all this charge to . So we only need to show that gets at least from . If is a vertex that is not big, then gives to by (R4). If is big, then it gives charge by (R3), and gives edge an extra by (R1), and all this charge goes to by (R2). Thus, gets , as desired. ∎
Claim 2.8.
Every vertex ends with nonnegative charge.
Proof.
By Lemma 2.4, has a big neighbor .
First suppose that does not have a 2neighbor. If is not on a 3face, then by (R3) gives charge , and by (R2) edge gives charge . So ends with at least . So assume is on a 3face and is on a 3face for every big neighbor of . By Lemma 2.5, vertex has at least two big neighbors, say and . Since each of and must be on a 3face, and has only a single incident 3face, it must be . Now, by (R3) and (R5), gets at least . So ends (R5) with at least .
Now assume that has a 2neighbor , which gets from by (R4). Applying the Key Lemma to shows that has two big neighbors, and . If is a 3face, then each of and gives to , by (R3) and (R5). So ends with at least . If is not a 3face, then each of and gives to , by (R1) and (R2). So ends with at least . ∎
Claim 2.9.
Every vertex ends with nonnegative charge.
Proof.
Let and denote the numbers of 2neighbors and incident 3faces that get charge from by (R4).
Suppose has no neighbor. If gives no charge to incident faces by (R4), then gives no charge at all, so ends with at least . If does give charge to an incident face by (R4), then (R4) implies that has two big neighbors; by (R3), each big neighbor gives charge . Since has no 4cycles, gives charge to at most two faces. So ends with at least .
So assume has a neighbor, . Applying the Key Lemma to shows that has two big neighbors, and ; by (R3) each gives charge . If is not on a face, for some , then by (R2) gives charge . Thus, ends with at least . So we assume that each is on a 3face. Since has a 2neighbor (which is not on a 3face with , by Lemma 2.2), and has no 4cycles, has at most one incident 3face. Since and are both on 3faces, the 3face must be . Because and are both big, gives no charge to . So ends with at least . ∎
Claim 2.10.
Every face ends with nonnegative charge.
Proof.
Let be a face, where . By (R1) each of gets from its incident face, and by (R2) all of this charge goes to . If has two incident big vertices, then by (R1) edges get in total an additional . So ends with at least . If is a vertex, then and are both big, by Lemma 2.2, and we are done, as above. So assume that is a vertex, and is not big. If some is a small vertex or a 4vertex with two big neighbors (which, by assumption, are not both incident to ), then gives to by (R4), so ends with at least . So we assume that has at most one incident big vertex, and has no incident small vertex, and no incident 4vertex with two big neighbors. Applying the Key Lemma to shows that must have an incident big vertex. Otherwise and are each vertices with at most one big neighbor, a contradiction. Thus, we can assume that has exactly one incident big vertex, and has no incident 2vertex, small vertex, or 4vertex with two big neighbors.
So assume that is big and that and are each either a 3vertex or else a 4vertex with no big neighbor other than . Applying the Key Lemma to shows that must be a 3vertex. Furthermore, at least one of and is a 3vertex with a big neighbor not on ; by symmetry, assume this is . By (R3), and each give charge . Since edge is not on a face, by (R2) it gives charge . So finishes (R5) with at least ; by (R6) all of this charge continues on to . So ends with at least . ∎
This completes the proof of Theorem 2.1.
For completeness, we conclude this section with the definitions of Alon–Tarsi number, paint number and correspondence chromatic number, and the corollary that bounds these parameters for planar graphs with no 4cycles.
An eulerian digraph\marginnoteeulerian digraph[0cm] is one in which each vertex has indegree equal to outdegree. For a digraph , let and denote the numbers of eulerian subgraphs of in which the number of edges is even and odd, respectively. A digraph is Alon–Tarsi\marginnoteAlon–Tarsi[.5cm] if , and it is Alon–Tarsi\marginnoteAlon–Tarsi[0cm] if also each vertex has outdegree less than . An orientation\marginnoteorientation[0cm] of a graph is formed from by directing each edge toward one of its endpoints. The Alon–Tarsi number\marginnoteAlon–Tarsi number[0cm] of , denoted , is the smallest such that some orientation of is Alon–Tarsi. Note that every acyclic orientation is Alon–Tarsi, since ; the only eulerian subgraph of is the spanning edgeless graph. Suppose that has degeneracy , and is a vertex ordering witnessing this. By orienting each edge toward its endpoint that appears earlier in , we conclude that .
The paint number is defined using a twoplayer game. At round , one player (Lister) chooses a set of vertices and the other one (Painter) answers by coloring an independent subset of with color . The winning conditions depend on a fixed integer : Lister wins if he presents a vertex on rounds but Painter never colors it. Otherwise, Painter wins. The paint number\marginnotepaint number[0cm] is the smallest integer such that Painter has a winning strategy with parameter . This problem can be seen as a generalization of list coloring, where the lists are not all known at the beginning of the coloring process (take as the set of vertices whose lists contain color ). As shown by Schauz [11], each Alon–Tarsi graph is paintable. Thus, every degenerate graph satisfies .
Given a graph and a function , an correspondence assignment is given by a matching , for each , between and . We say that each vertex has available colors. A correspondence assignment is an correspondence assignment where for all . Given an correspondence assignment , a coloring\marginnotecoloring[0cm] is a function such that for each , and, for each edge , the pairs and are nonadjacent in . The correspondence chromatic number\marginnotecorrespondence chromatic number[0cm] of is the least integer such that, for every correspondence assignment of , graph admits a coloring. It is denoted by . Note that if is degenerate, then coloring greedily in some order witnessing this shows that . Thus, we have the following corollary of Theorem 1.
Corollary 2.11.
If is a planar graph with no 4cycles, then , , and .
3 Graphs with no 4cycles and large
Theorem 3.1.
There exists such that if is a plane graph with no 4cycles and with , then is choosable. In fact, .
Let and fix . We prove by contradiction that if is a plane graph with no 4cycles and with , then is choosable. (By plane graph\marginnoteplane graph[0cm], we mean a planar graph with a fixed embedding in the plane.) For ease of exposition, we present the proof only for choosability, although it also works for paintability and Alon–Tarsi orientations. Most of the reducible configurations rely only on degeneracy, though at one point we use the kernel lemma.
Assume the theorem is false and let be a counterexample that minimizes . Let be an assignment of lists of size to the vertices of such that has no coloring. Throughout Section 3 we prove several structural lemmas, which ultimately lead to a contradiction. We follow the same general approach as in [2], which considered planar graphs with girth at least 5; however, we need new ideas to handle the presence of triangles.
3.1 First Reducible Configurations
Lemma 3.2.
Graph is connected and has minimum degree at least .
Proof.
Note that is connected, since otherwise one of its components is a smaller counterexample. Now assume there exists a 1vertex . By the minimality of , we can color . Since , and has at most neighbors in , we can color with a color not used on its neighbors in , which is a contradiction. ∎
The next two lemmas essentially show that every vertex of must be near a vertex of high degree. To formalize this, we use the following terminology: a vertex is big\marginnotebig, small[0cm] if and small otherwise. Denote by and the sets of big and small vertices. To refine the set , we write for the set of small vertices with exactly big neighbors.
Remark 3.3.
In our figures in the rest of the paper, we draw small vertices as circles, and big vertices as squares. Further, we use black circles for vertices with all neighbors shown. So a white vertex could have more neighbors than those shown; in fact, it could also have edges (that are not drawn) to other vertices that are shown. For example, Figure 3 shows the configurations forbidden by Lemma 3.4.
Lemma 3.4.
For each edge , either or . Further, if , then .
Proof.
Assume to the contrary that some edge has . By minimality, we can color . We uncolor and . Since , both and have at most colored neighbors in . Since , we can find distinct available colors for and .
Suppose instead that and and . Again, by minimality we color , then uncolor and . Now has at most colored neighbors in , so has an available color. As before, we can color . This gives an coloring for , a contradiction. ∎
Lemma 3.5.
If is an edge with , then and have no common neighbor.
Proof.
Assume there exists a triangle with . By minimality, we can color . Both and have colored neighbors in . So and each have at least available colors, and thus we can color them both. ∎
Lemma 3.6.
Let be a triangle of such that some vertex has a common 2neighbor with . If either (a) or (b) and and have a common 2neighbor, then .
Proof.
Note that it suffices to handle case (b), since (a) is a subgraph of (b). Let and denote the 2neighbors of common with and . Assume that . Note that none of , , and are neighbors of , since has no 4cycle. By minimality, we color . For each , the number of colored neighbors in of is
Thus, and both have 2 available colors, so we can color them. Further, for each , the number of colored neighbors of is
Therefore, and both have available colors. So we can color them to get an coloring for , a contradiction. ∎
We combine Lemmas 3.4 and 3.6 to prove the reducibility of the bigger configuration shown in Figure 5.
Lemma 3.7.
Fix such that . Now cannot contain vertices that are consecutive neighbors of and that satisfy both conditions below; see Figure 5.

Each has degree two and has a common neighbor with .

For each , each vertex inside cycle is adjacent to .
Proof.
We assume that contains such a configuration and reach a contradiction, by showing that contains a configuration forbidden by Lemma 3.6. Note that all ’s are distinct, since contains no 4cycle.
Below when we write a statement about , we mean that it is true for each . Since , Lemma 3.4 implies that . Because are consecutive neighbors of , vertex is not adjacent to . Since has no 4cycle, has at most one common neighbor with . Thus . Define so that . If , then contains the second configuration in Lemma 3.6, a contradiction. If has a neighbor other than and , then call it ; now is adjacent to (by hypothesis 2), so is a 4cycle, a contradiction. Thus, is a 2vertex with . Now contains the first configuration in Lemma 3.6, again a contradiction. ∎
3.2 Outline of the proof
Recall that \marginnote, [0cm] is the set of small vertices, and is the set of small vertices with exactly big neighbors. Let \marginnote[0cm] denote the multigraph formed from by suppressing every vertex of degree in , and then contracting every edge between and . (Suppressing a vertex means deleting and adding an edge between its two neighbors.) Let \marginnote[0cm] denote the multigraph formed from by removing every loop, and let \marginnote[0cm] denote the underlying multigraph of , i.e., the multigraph formed from by deleting the minimal number of edges to remove all faces of length . Note that can have parallel edges. For example, suppose and have parallel edges, say and , in . If some vertices are embedded inside the cycle , then in vertices and still have parallel edges, with those same vertices embedded inside the cycle they bound. However, cannot have faces of length 2.
An region of \marginnoteregion of [0cm] is a set of consecutive faces of length sharing the same boundary , where and are distinct vertices of . Note that each of the faces in an region is constructed from some cycle of when we apply the construction rules above. By extension, an region of \marginnoteregion of [0cm] is the subgraph of induced by the vertices of these cycles, together with those lying on the inside of those cycles. (We often simply write region\marginnoteregion[0cm], when the specific value of is less important.) When is an region of , we denote by \marginnote[0cm] the set of vertices appearing on all faces of , excluding and .
To reach a contradiction, we prove the following two propositions.
Proposition 3.8.
contains an region of size at least .
Proposition 3.9.
does not contain any region for .
Our contradiction now comes quickly. These propositions give that . This inequality implies , contradicting the hypothesis .
3.3 Structure of Regions
We now classify each edge of based on its corresponding path in . An edge in corresponds to a path in \marginnotecorresponds to a path in [.4cm] if and for each , vertex disappeared when we constructed . (As we will see, always .)
Lemma 3.10.
Each edge of corresponds to a path in for which exactly one of the following six conditions holds (see Figure 6). If satisfies condition below (for some ), then we say that has type \marginnotetype [0cm]. If , then has one of types 1–4. If is a loop of , then has type . Finally if , then has type 1, 5, or 6.

.

and corresponds to a path in with .

and corresponds to a path in with and .

and corresponds to a path in with .

corresponds to a path in with .

corresponds to a path in with and .
Proof.
Due to the construction of , each edge in between and comes from a path in between and , and each internal vertex of is small. Further, every internal vertex has degree , except possibly the neighbors of whichever of and are big.
If and , then Lemma 3.4 shows that contains at most edges. So has type 1, 2, or 3. If , then has at most edges, so has either type or type . If , then has at most edges. If has only two edges, then, since is simple, , so in the construction of , we do not delete the middle vertex. Thus, has 1, 3, or 4 edges; so has type 1, 5, or 6. If is a loop, then and are either both big or both small, so has neither type nor . Since is loopless and simple, does not have type or . Since has no 4cycle, also does not have type . Therefore, every loop has type . ∎
In what follows, when referring to an edge with type , we use , , and as defined in the corresponding part of Lemma 3.10. This lemma implies the following facts about the structure of regions in .
Corollary 3.11.
Let be a region of . Now is the disjoint union of three sets such that for some , and is an independent set of vertices, each with a neighbor in each of and .
Proof.
Let be a region of . By definition, there exists such that every face of in has boundary . Therefore, in , the edges appearing in are either loops over or edges .
Note that is the set of all vertices of that disappear when we construct the edges of in . For each , define as the set of vertices of such that is contracted when constructing an edge of in . We also define as the set of vertices in that are suppressed when constructing an edge of in . By definition, we have
By Lemma 3.10, since , each edge between and in has type 1, 5, or 6, and each loop around has type . This ensures that and that contains only vertices of degree in . By Lemma 3.4, this implies that is an independent set.
It remains to show that these sets are pairwise disjoint. Assume that there is . Now and are both contracted when constructing . This requires that