1 Introduction
Colorings of graphs form one of the most fascinating and wellstudied topics in graph theory. Early results date back to 1852 [25] and since then several variants have been proposed and studied over the years (e.g., vertex colorings [2, 5, 11, 17, 29], edge colorings [6, 7, 27], total colorings [22, 28, 30]); for a survey refer to [23]. In the classical vertex coloring problem, the vertices of a given graph must be colored with a minimum number of colors, so that adjacent vertices have different colors. In other words, the monochromatic connected components (or monochromatic components, for short) must form an independent set, i.e., they must be single vertices. A notable result on this topic is the socalled color theorem [14], which states that four colors suffice to color the vertices of any planar graph such that adjacent vertices are of different colors; deciding whether a planar graph can be colored with three colors such that adjacent vertices are of different colors is a wellknown NPcomplete problem [13].
In this paper, we continue the study of a variant of the vertex coloring problem, where each monochromatic component may be formed by more than one vertex but we require its cardinality to be “small” [16]. Formally, for a graph and an integer we denote by the smallest integer such that there exists a coloring of the vertices of with colors so that any monochromatic component of has at most vertices. The classical vertex coloring is a restriction of this problem, in which one seeks for the smallest so that .
This problem is closely related to the wellknown HEX lemma [12], which in its simplest form states that in any coloring of a triangulated grid there exists either a monochromatic path from the left to the right side of the grid in one of the colors or a monochromatic path from the top side to the bottom side in the other color; see also [21]. As a consequence, not all (planar) graphs of bounded degree admit colorings in which all monochromatic components are of bounded size. Lovász [20] proved that any (not necessarily planar) cubic graph admits a coloring in which all monochromatic components are of size at most two, i.e., . Also, such graphs admit colorings in which one colorclass is an independent set, while the other one induces monochromatic components of size at most 750 [3].
Alon et al. [1] generalized these results to graphs with maximum degree 4 by proving that such graphs admit colorings in which all monochromatic components are of size at most . Haxell et al. [15] improved this bound to and also proved that every graph with maximum degree 5 can be colored so that all monochromatic components have size at most 20000, a bound which was later reduced to 1908 by Berke in his Ph.D. thesis [4]. Linial et al. [18] showed that there exist regular vertex graphs whose monochromatic components are of size in any coloring. Note that for regular graphs the corresponding order of magnitude ranges between and [1, 12]. Also, Linial et al. [18] proved that , where is planar or belongs to a minorclosed class.
As for three or more colors, Kleinberg et al. [16] and Alon at al. [1] constructed planar graphs that do not admit colorings so that each monochromatic component has bounded size. Esperet et al. [10] proved that there is a function so that every planar graph with maximum degree admits a coloring in which each monochromatic component has size at most , where . When a fixed number of colors is available and is of bounded treewidth, Linial et al. [18] proved that .
Note that there is a fairly rich literature on colorings of graphs where each monochromatic component is acyclic (see, e.g., [9]) or has a small diameter (see, e.g., [19]), or has small degree (see, e.g., [8]).
Our contribution. In this paper, we study  and colorings for two important subclasses of planar graphs: (i) Maximal outerplanar graphs, and (ii) Complete planar 3trees. Note that both these classes are meaningful subclasses of planar graphs which are of interest in computational complexity theory, since many NPcomplete graph problems are solvable in linear time on these graphs (see, e.g., [26]). We also note that both of these classes of graphs are of bounded treewidth, which implies that any graph belonging to them has and [18]. We present improvements upon these bounds:

For an outerplanar graph with maximum degree , Berke has shown that [4]. Observe that this upper bound drastically improves the general one by Esperet et al. [10] for the class of outerplanar graphs. We present an efficient dynamicprogramming based algorithm which for a given outerplanar graph determines the exact value of . We further show improved upper bounds for special subclasses of outerplanar graphs. We note that outerplanar graphs admit proper colorings; see, e.g., [24].

For a complete planar 3tree , we prove that and . For two colors we also give a corresponding lower bound of . However, our bounds do not transfer to general planar trees; Linial [18] presented infinitely many planar graphs, that are subgraphs of planar trees and meet the aforementioned upper bounds of and for  and colorings, respectively.
2 Preliminaries
Let be a graph. is connected if there is a path between any two vertices of . is connected, for , if the removal of vertices leaves connected. A connected graph is also called biconnected.
A planar drawing of is a drawing in the plane such that each vertex is drawn as a distinct point ; each edge is drawn as a simple curve connecting and and no two edges intersect in except at their common endpoints. divides the plane into topologically connected regions called faces; the unbounded face is called the outer face, all other faces are called internal faces. Each face is described by the circular sequence of vertices encountered while moving clockwise along its boundary. The description of the set of faces determined by a planar drawing of is called a planar embedding of . Graph together with a given planar embedding is an embedded planar graph.
An outerplanar graph is a planar graph that admits a planar embedding such that all vertices are on the outer face. A maximal outerplanar graph is an outerplanar graph to which no edge can be added while preserving outerplanarity (see Figure 0(a)).
The dual of an embedded planar graph is a graph that has a vertex for each face of and an edge between two facevertices if and only if the corresponding faces share an edge in the embedding of . The weak dual of is the subgraph of the dual of obtained by omitting the facevertex of the outer face of (see Figure 0(a)). The weak dual of an outerplanar graph is a forest in general. For a biconnected outerplanar graph the weak dual is a tree, and vice versa.
A complete planar tree is a graph formally defined as follows. A cycle is a planar tree with levels. A complete planar tree with levels is obtained from one with levels by inserting a vertex in every internal face and connecting it to all of its vertices (see Figure 0(b)).
3 Outerplanar graphs
In this section, we focus on outerplanar graphs. We start with a dynamic programming based algorithm which for a given maximal outerplanar graph determines the exact value of .
Theorem 1.
There exists a dynamic programming based algorithm which in polynomial time computes a coloring of a maximal outerplanar graph, such that the size of the largest monochromatic component is minimized.
Proof.
Let be a maximal outerplanar graph on vertices and assume without loss of generality that is embedded according to its outerplanar embedding. Since is maximal, is also biconnected, and it follows that the weak dual of is a tree. We further assume that is rooted at a leaf of it; note that has at least two leaves, since . Our algorithm is mainly based on a definition of equivalent colorings, which we use in a bottomup traversal of to maintain representative colorings from each class of equivalent colorings.
For a node of , we denote by the subgraph of whose weak dual is the subtree of rooted at . Consider a node of that is not the root of , that is, . Let be the parent of in . Let also and be the faces of corresponding to nodes and of . Finally, let be the edge of shared by and . We say that is the attachment edge of to , while its endpoints and are the poles of and . The attachment edge of root is any edge of face that is incident to the outer face (since is biconnected and is a leaf, this edge always exists).
Consider a coloring of with two colors, black and white, and let and be the sizes of the largest monochromatic components in black and white, respectively, that contain neither nor . Let and be the colors of and in , respectively. Let also and be the size of the monochromatic components containing and in , respectively. Since and are adjacent, it follows that if holds, then also holds. We say that two colorings and of are equivalent with respect to the attachment edge if and only if (i) has the same color in and , that is, , (ii) has the same color in and , that is, , (iii) the size of the monochromatic component containing is the same in and , that is, , (iv) the size of the monochromatic component containing is the same in and , that is, , (v) the size of the largest black monochromatic component that contains neither nor is the same in and , that is, , and (vi) the size of the largest white monochromatic component that contains neither nor is the same in and , that is, . This definition of equivalence determines a partition of the colorings of into a set of equivalence classes.
To store a representative from each class of equivalent colorings of we employ a table , in which entry
is true if and only if there exists a coloring of in which (i) the color of is , (ii) the color of is , (iii) the size of the monochromatic component containing is , (iv) the size of the monochromatic component containing is , (v) the size of the largest black monochromatic component that contains neither nor is , and (vi) the size of the largest white monochromatic component that contains neither nor is . Table is of size , since and due to Berke [4] holds.
When visiting a node of in the traversal of , we assume that for each of the two children and of in , we have computed a representative coloring from each class of equivalent colorings and that we have stored this information in and . To compute , we consider each pair of possible equivalence classes of colorings of and and check whether their combination yields a representative in some equivalence class of colorings of ; initially, all entries of are false. For each of and , consider the following entries:
If , then we can ignore this pair and proceed to a next one, as the vertex shared by and has different colors in the colorings of and , and hence cannot yield a valid coloring for . So, we can assume without loss of generality that . To simplify the presentation, we assume that the color of the shared vertex of and is black. The case where this vertex is colored white is symmetric.
We proceed by considering two cases: (i) , and (ii) . In Case (i), we first assume that and correspond to white and black colors, respectively. In this case, it holds that . Since we have assumed that the color of the shared vertex of and is black, it follows that is true. We cope with the case where and correspond to black and white colors, respectively, symmetrically. This completes our analysis for Case (i).
In Case (ii), we again consider two subcases, namely, either both and correspond to black or both to white. Note that these cases are not symmetric, as we have assumed that the color of the shared vertex of and is black. In the former case, it holds that and . Since we have assumed that the color of the shared vertex of and is black, it follows that is true. To complete the description of our case analysis, it remains to consider the more involved case where both and correspond to white. Since the color of the shared vertex of and is black, it follows that it may form a new monochromatic component (in black) that is not incident to the poles of . Hence, we set to true. This completes our analysis for Case (ii).
At the end of the bottomup traversal of , we search in for the entry which minimizes the maximum of , , and . The corresponding coloring of can be easily constructed by traversing topdown, and by following the choices performed during the bottomup visit. Note that for a node of the computation of table can be done in , since each of the tables and of the children and of are of size . Hence, the timecomplexity of our algorithm is , which is linear when has bounded degree. ∎
In the rest of this section, we present two sideresults that form improvements on the general upper bound of by Berke [4] for special subclasses of outerplanar graphs with maximum degree .
The first subclass of outerplanar graphs that we consider is the class of snowflakes. Intuitively, a snowflake is a biconnected maximal outerplanar graph whose weak dual is a complete binary tree. Formally, a snowflake of height is defined recursively as follows (see Figure 2 for an example). For , is a cycle, i.e., 3 vertices and 3 edges between them. For with , we take and extend it by attaching on each edge of the outer face a path of length 2. The new vertices are at height and the length of the outer face is doubled. When we add the set of vertices at height , for they are adjacent to two vertices that we call ancestors: One of them is at height , and is denoted by , while the other has height . The height of the snowflake is related to the maximum degree of any vertex, namely, where is any vertex at height .
Theorem 2.
For a snowflake of maximum degree , holds.
Proof.
Assume that has height . We color its vertices as follows: Initially, the vertices of height are colored with the same color. For a vertex of height , we consider the colors of its two ancestors: If they both have the same color, we give to vertex the opposite color, otherwise we assign to the color of vertex .
We claim that each monochromatic component is a path. By construction, the edges between monochromatic vertices are of the form , which we consider to be directed towards . Hence, they form monochromatic trees with direction in decreasing height. Note that if a vertex has two incoming edges, then its two ancestors both have different color than . So, we can conclude that the monochromatic components consist of at most two heightdecreasing paths that end in a common vertex. Since the paths have length at most , the component has size , which is equal to . That concludes the proof. ∎
The second subclass of outerplanar graphs that we consider is the class of outerpaths. Formally, an outerpath is a biconnected maximal outerplanar graph whose weak dual is a path. Before we proceed with the description of our coloring scheme, we introduce some necessary definitions and notation. Let be an outerpath (see Figure 3 for an example). We call spine vertices of the vertices that have degree at least four in and, for the sake of simplicity, we denote by and the two (unique) vertices of degree two of . Consider a spine vertex of () and denote by and the edges and of , respectively. We call fan the subgraph of induced by , , and by all the vertices such that edge is between and in counterclockwise (clockwise, resp.) order around if
is odd (even, resp.). Denote by
the number of vertices of . Observe that for each , for , and that and share the two spine vertices and for each . We say that a vertex is isolated if all its adjacent vertices have different color.The last tool that we need for our analysis is the following lemma, proven in several papers; see, e.g., [4]. Recall that a wheel on vertices is a graph formed by vertices connected in a cycle, called the rim of , and a socalled central vertex that has an edge, called spoke, towards each vertex of the rim.
Lemma 3 (Berke [4]).
For a planar wheel W on vertices, holds.
We are now ready to prove the last result of this section.
Theorem 4.
For an outerpath of maximum degree , holds.
Proof.
We color with a technique similar to the one used in Lemma 3 for a wheel. Note that the coloring of a wheel with vertices is obtained by coloring its center and of its neighbors black so that the remaining vertices, which are colored white, are partitioned into blocks of size or . For the case of an outerpath, we first color the spine vertices and then we use the idea of the wheel to color its fan vertices. More precisely, we do the following (see Figure 3). Without loss of generality, set the color of black.
Spine vertices: Starting from , consider pairs of consecutive spine vertices; color the vertices of each pair with the same color and alternate the color of two consecutive pairs. So, and are colored black, and are colored white, and are colored black, and so on. For the two vertices and of degree two do the following: Color white; if and have the same color, then color with the other color, otherwise color with the same color as .
Fan vertices: If there is no vertex to color. So, assume and assume without loss of generality that is colored black. Then, either or is white. Suppose that is white. Color black the unique vertex adjacent to both and . The remaining vertices of each fan are colored using the technique of Lemma 3.
We now prove that the subgraph induced by the fans , , , has with the following additional property: If is odd, then has vertex isolated, while if is even, has both vertices and isolated. The proof is by induction on the number of spine vertices of .
If , is composed of fan and . By construction, is black, is white, the other vertex adjacent to (call it ) is black, and is black (see Figure 3(a)). The other vertices of are colored according to the technique of Lemma 3. Observe that is isolated. Consider the subgraph . Since is isolated, . Since has vertices, has vertices. Now consider the graph obtained from by making vertices and coincident. is a wheel with vertices (see Figure 3(b)). Then . Since (in vertices and are both black and distinct), it follows that .
If , is composed of fans and (see Figure 3(c)). By construction, and are black, while and are white. Observe that , while . If has size three, is composed of fan plus vertex that is isolated. Then by the same argument as above. Otherwise, denote by and the two vertices adjacent to both and and to both and , respectively. Vertices and are colored black. Observe that and are isolated and that and are shared by and . Denote by () the subgraph of induced by (). We have that . Using the same argument as for the case for both and and considering the (worst) case where both and have degree , it follows that which is , and then .
For the inductive hypothesis, assume that the statement holds for , with . We prove that it also holds for . If is even, decompose in the two subgraphs and . The induction hypothesis holds for , then has both and isolated. Suppose that is white. Since is isolated, and have different color. By construction the (unique) vertex of fan adjacent to both and has the same color of . Observe that the vertices of adjacent to have color different from , and the vertices of adjacent to have color different from (recall that they are black). Then which is . If is odd, decompose in the subgraph and in the subgraph induced by fans and . Since is composed of an even number of fans, by induction it has both and isolated. Then, as before, . To complete the proof, we note by Lemma 3 follows, since each fan of can be regarded as a wheel. ∎
4 Complete Planar 3trees
In this section, we present  and colorings for complete planar trees. However, before we proceed with the description of our approach, we will first introduce some important properties of planar trees, which we use in Section 4.1.
Let be a complete planar tree with levels. By definition, is fully triangulated. Let , and be the vertices of on its outer face, called outer vertices of . With a slight abuse of notation, we write . As mentioned above, , with , is constructed from as follows: In every bounded face of we add a vertex in its interior and connect it to the three vertices of , splitting into three “smaller” triangular faces. We say that is at level and is the central vertex of . A vertex of maximum level is called a leafvertex and it is adjacent to three emptytriangles, that is, faces of . By definition each triangle of contains in its interior another complete planar tree with levels; if we refer to such a triangle as nonempty. Note that emptytriangles induce degenerate complete planar trees of levels.
Outer vertices and (with ) have common neighbors that form a socalled central path , where is the central vertex of , namely, is the path from the central vertex to the unique leafvertex that is a common neighbor of and . For , vertices , and of form a triangle, called sidetriangle. Note that sidetriangles contain in their interior complete planar trees with levels; sidetriangles are in principle nonempty, except for the case where . Vertices , and form an emptytriangle, called leaftriangle.
Property 5.
has vertices and triangular faces.
Property 6.
The number of neighbors for outer vertex is equal to , and the number of internal neighbors of all three outer vertices equals .
We are now ready to present the main results of this section. As a last tool, we further need the following result, which is a direct consequence of Lemma 3.
Corollary 7.
A (planar) graph that has as a subgraph a wheel on vertices has .
Note that Lemma 3 holds only for two colors. For three colors, it is not difficult to see that or . One can argue similarly for a double wheel. A double wheel on vertices consists of a wheel on vertices, and a second central vertex that is also adjacent to every vertex of the rim. It is not hard to see that or holds for a double wheel as well. Note that in such a coloring the two central vertices always have the same color. However, for the case where the two centers must have different colors, one can see, following an approach similar to the one of the proof of Lemma 3, that there exists a monochromatic component of size .
4.1 Colorings of Complete Planar 3trees
We are now ready to present our main results for complete planar trees. In Theorem 8 we focus on colorings, while Theorem 10 regards colorings.
Theorem 8.
For a complete planar tree on vertices, holds.
Proof.
To prove the upper bound, assume that has levels. We describe a recursive coloring algorithm (see Figure 5 for an example), which maintains the following simple invariant: The outer vertices of the graph to be colored are bicolored. To this end, we initially color vertices and with the color , and vertex with the color . Clearly, this conforms with our invariant. Then we color the central vertex and the three central paths for , with as their common endvertex, with . The induced subgraph with vertices , , and the vertices of the three central paths contains a monochromatic component of size , and splits into inner triangles. Two of the three leaftriangles have all three colors on their vertices but they contain no other vertex in their interior. Sidetriangles have their vertices colored with two colors (since two vertices belong to a central path and are colored with ). The algorithm recursively computes the colors of the vertices in the interior of the nonempty sidetriangles. Note that every sidetriangle is a complete planar tree with levels and its outer vertices have two colors, which conforms with our invariant. Hence, at each recursive step of our algorithm, a monochromatic component of size is created. By construction, each monochromatic component inside the sidetriangle is either not adjacent to the three outer vertices, or it has different color. This property ensures that no two monochromatic components of the same color, created at two different steps, are adjacent. Hence, the size of the largest monochromatic component is . In order to prove the lower bound, we need the following claim.
Claim 9.
Let be a complete colored planar tree with levels, where , and have all different colors. Then, there exists a monochromatic path of length at least , with one endpoint on the outer face of .
Proof.
For the claim trivially holds, so we assume that . Denote by a degenerate path consisting only of vertex , for . We shall prove that at least one of these three paths can be extended to a monochromatic path of length at least . Let be the central vertex of . Since , and have different colors, must have the same color with exactly one of the three outer vertices. Assume without loss of generality that has the same color as . This implies that we can extend monochromatic path by adding vertex , that is . Also, vertex forms a colored triangle together with outer vertices and of . Now is a complete planar tree with levels and vertices of its outer face have different colors. Furthermore, paths , and contain four vertices in total. Following a similar approach, from the central vertex of , one can have a complete planar tree with levels, whose outer vertices have different colors. Furthermore, the central vertex of extends one of the three monochromatic paths by one vertex. Repeating the above procedure times, we can create a chain of nested complete planar trees , , …, , where their outer vertices have different colors. At each step, the central vertex is added to one of the three monochromatic paths. Paths , and have a total of vertices. Hence, at least one of them has length at least , and the claim holds. ∎
From the above claim, in order to prove the lower bound, it suffices to find a subgraph of that is a complete planar tree with levels, whose outer vertices have all three colors. Consider outer vertex of and let be its neighboring vertices at level at most , where . Then the vertices , , and form a wheel with the center vertex . By Property 6, , and therefore wheel has vertices. Let be the color of vertex . We consider two cases depending on whether there exist two consecutive vertices along the rim of with colors and . Suppose first that this is not the case, and vertices of with color are not adjacent to vertices with color . This implies that if one identifies colors and to one color, the size of all monochromatic components in the derived coloring of remains the same. By Lemma 3, we have that contains a monochromatic component of size . For , it follows that , and therefore , contains a monochromatic component of size . In the second case, there exist two consecutive vertices on the cycle of with colors and . These two vertices, together with , form a nonempty triangle that is a complete planar tree with at least levels. Since the three outer vertices of are of different colors, the previous claim assures that there exists a monochromatic path of length at least . For , it follows that there exists a monochromatic path of length . This completes our proof. ∎
We now adjust the technique used in the proof of Theorem 8 to obtain the following result.
Theorem 10.
For a complete planar tree on vertices, the following holds:
Proof.
The general idea is to color the first levels with color , their neighbors with color and to use recursion. Assume that has levels. We color the three outer vertices plus the vertices of the first levels with color . This yields a monochromatic component of size (by Property 5). This coloring implies a “subdivision” of the remaining vertices into components, that are planar trees with levels. In each of the smaller trees, say , we use color for the neighbors of the outer vertices (outer vertices have already color ). By Property 6 we create components of size .
After this step, uncolored vertices form smaller components in the interior of complete planar trees that have fewer than levels and their outer vertices have color . Note that there exist nonmonochromatic triangles, however we claim that they are empty. Assume that there is an uncolored vertex inside a nonmonochromatic triangle . At least one of the vertices of has color and it is therefore an outer vertex of a tree . Hence is adjacent to this outer vertex, and therefore has already color , a contradiction.
The remaining components are complete planar trees with levels each. The outer vertices of have color and we proceed by coloring the neighbors of its outer vertices with color . Thus we create one monochromatic component with fewer than vertices. As before, uncolored vertices form smaller complete planar trees with even fewer levels than . Furthermore the property of empty nonmonochromatic triangles still holds, and we can recursively color the remaining uncolored components, by alternatively using colors and . At every recursive step, we create monochromatic components with fewer than vertices. Hence, , which is minimized for , that is, . So, we have:
where . Then:
By Property 5, it follows that and thus . ∎
Theorem 11.
For a complete planar tree on vertices, the following holds:
Proof.
The neighbors of each of the outer vertices form a wheel, each of which has size . Then, the lemma follows from Corollary 7. ∎
5 Conclusions and Open Problems
In this paper, we studied  and colorings and presented improved bounds on the size of the monochromatic components for internally triangulated outerplanar graphs and complete planar 3trees. Several questions remain open:

The study of other classes of graphs with bounded treewidth, which would give rise to improved bounds w.r.t. the ones of Linial et al. [18], is of interest.

Other classes of graphs that allow for an efficient computation of the largest monochromatic component are of importance.

We believe that the class of planar graphs would be interesting in this context, as it is neither minor closed nor of bounded treewidth. No results are known for the size of the largest monochromatic component (for  and colorings) of these graphs.
Acknowledgments
We thank Tamara Mchedlidze for useful discussions.
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