1 Introduction
Ramsey’s theorem [15] claims that any graph is Ramsey in the class of all complete graphs, i.e., for any graph and any number of colors there is a sufficiently large complete graph such that in any coloring of its edges in colors there is a monochromatic copy of . In general for graphs and , we write and say that arrows if any coloring of the edges of in colors contains a monochromatic copy of . We write and say that arrows if . There are classes of graphs that are Ramsey in their own class, meaning that for any graph in a class there is a graph such that
. Examples of such classes include bipartite graphs, graphs with a given clique number, and graphs of a given odd girth, see
[12, 13]. Here, we are concerned with Ramsey properties of the class of all planar graphs. We say that a planar graph is planar unavoidable if there is a planar graph such that , otherwise we call planar avoidable. Similarly, we define outerplanar unavoidable and outerplanar avoidable graphs. When , we write planar unavoidable instead of planar unavoidable, or, if clear from context, simply unavoidable. The complexity of the problem to edgecolor planar graphs with a given number of colors so that there is no monochromatic copy of a given graph was addressed by Broersma et al. [3]. A related problem of bounding local density of Ramsey graphs has been addressed for example in [16] and [10].A result of Gonçalves [8] states that any planar graph can be edgecolored in two colors so that each color class is an outerplanar graph.
Thus any planar unavoidable graph is outerplanar.
The Four Color Theorem [2] implies that any planar graph is a union of two bipartite graphs.
In general any graph that is colorable for is a union of at most bipartite graphs.
This shows that any planar unavoidable graph is bipartite and outerplanar and thus gives necessary conditions for planar unavoidability.
Next we give several sufficient conditions. Here, a generalized broom is a union of a path and a star such that they share only the center of the star.
Theorem 1.
If be a path, a cycle on vertices, a tree of radius at most , or a generalized broom, then is planar unavoidable. Moreover, if is a path, then it is outerplanar unavoidable.
The next result shows that not only odd cycles and nonouterplanar graphs are planar avoidable, but also some trees.
Theorem 2.
There is a planar avoidable tree of radius and an outerplanar avoidable tree of radius .
Moreover, any planar avoidable tree has at least vertices and there is a planar avoidable tree on vertices.
A result of Hakimi et al. [6], see also [1], states that any planar graph can be edgedecomposed into at most five star forests. Thus the planar unavoidable graphs for are precisely the star forests. Next we summarise our results for planar unavoidable graphs, for and .
Theorem 3.
If is planar unavoidable for , then is a forest. If is planar unavoidable, then is a caterpillar forest. There are  and planar avoidable trees of radius .
Moreover, there are  and planar avoidable trees on and vertices, respectively.
2 Definitions
We denote a complete graph, a path, and a cycle on vertices by and , respectively.
A complete bipartite graph with parts of sizes and is denoted by .
For an integer , , a ary tree is a rooted tree in which each vertex has at most children.
A perfect ary tree is a ary tree in which every nonleaf vertex has children and all leaf vertices have the same distance from the root.
For all other standard graph theoretic definitions, we refer the reader to the book of West [17].
Iterated Triangulation :
An iterated triangulation is a plane graph defined as follows: is a triangle, , is obtained from by inserting a vertex in each of the inner faces of and connecting this vertex with edges to all the vertices on the boundary of the respective face, see Figure 1.
We see that is a triangulation and each triangle of bounds a face of for some .
Universal outerplanar graph :
A universal outerplanar graph is defined as follows: is a triangle. An edge on the outer face is called an outer edge.
For , is an outerplanar graph that is a supergraph of obtained by introducing, for each outer edge , a new vertex and new edges: and . Then the set of outeredges of is
Triangulated Grid :
Let a triangulated grid be a graph with and if and only if either ( and ) or ( and ) or ( and ) or ( and ).
We define left, right, top, and bottom sides of the grid as subsets of vertices
, , , and respectively.
Fish:
A graph is called a fish and denoted if , where , and are each adjacent to each vertex in , induces a path in , and is an edge.
We call the set of spine vertices, is called the spine, are called ribs, , and the paths of length are called double ribs.
Sometimes we say that a fish hangs on an edge .
In an edgecolored fish, a double rib is called bicolored if there are different colors used on two edges of this double rib.
We will call two double ribs and , with , , identically bicolored, if the same color is used on both of the edges and , and a different color is used on both of the edges and .
Note that for any positive integers and and for any edge , there is a fish on in with spine vertices.
Indeed, consider an inner face in .
We can pick spine vertices such that , and such that is inserted in the face of , , .
3 Proof of Theorem 1
Theorem 1 follows immediately from the following lemmas.
The following proof closely resembles the Hexlemma [7].
Lemma 4.
Let be a neartriangulation with outer cycle , that is, is a planar graph with outer face boundary and each other face is bounded by a triangle. Let be vertices on in clockwise order dividing the edges of in four paths , respectively. If the edges of are colored red and blue, then either there is a blue path from to or a red path from to (or both).
Proof.
Suppose there is no blue path from to . Then the red graph contains a minimal edgecut separating and . A minimal edgecut in is a cycle in the dual graph . This cycle must contain the vertex corresponding to the outer face of . Since is a neartriangulation, it follows that any two consecutive edges of (except the two edges incident with ) correspond to two edges in that are incident with the same vertex. Thus the edges in corresponding to the edges in contain a red path path from to . ∎
Corollary 5.
Any path is planar unavoidable, even in a class of planar graphs of bounded degrees (in fact of maximum degree at most ).
Proof.
If the edges of the triangulated grid are colored red or blue, then there is a monochromatic by Lemma 4, where the paths correspond to the top, right, bottom, and the left sides of the grid. ∎
The above gives planar graphs of bounded maximum degree that arrow arbitrarily long paths, which however have large treewidth. Complementary, we can find planar graphs of treewidth that also arrow arbitrarily long paths, where however the maximum degree is large.
Lemma 6.
Any path is outerplanar unavoidable. In particular, for any positive integer , .
Proof.
We shall show that . Let and let it be edgecolored red and blue. We see that each edge of is on the outer face of for some , where as in the definition of the universal outer planar graph. Consider the unique outerplanar embedding of and for each edge , consider such that is on the outer face of . For a vertex let its rank be the least for which it is in the vertex set of . For each edge , we define graphs and such that , where and share only the edge and no vertices except for the endvertices of . We require in addition that contains as a subgraph, see Figure 3. Observe that among vertices of rank in there are two at distance in , .
For an edge in with endvertex of rank , we define the following.
Let be a longest red path in with last edge and last vertex .
Let be a longest blue path in with last vertex .
We shall write for two edges of if , or and .
Consider the edges in .
Assume that the endvertices of each such edge belong to the same blue component.
Then for any two vertices of rank , there is a blue path joining them.
Since there are two such vertices at distance at least in , we see that there is a blue path on edges, and we are done.
So, assume that is an outer edge of such that its endvertices belong to different blue components of .
Assume we constructed a sequence of edges of outer edges in respectively such that the endvertices of each of these edges belong to different blue components of .
Consider , we shall construct .
Let with of rank and let be two adjacent outer edges of that are incident to and , respectively.
Let and where has rank .
Note that either ( and ) or ( and ) are in different blue components in , otherwise and would have been in the same blue component.
See Figure 4 for illustrations.
Case 1. and are in different blue components.
Then is red and the path is a red path in of length ending in at vertex .
Then let .
We see that .
Case 2. and are in the same blue component and and are in different blue components.
Then is red and the path is a red path of length in ending with at vertex .
Since and are in the same blue component, there is a blue path of length , , between them in .
The union of and the blue path ending at in forms a blue path ending at in .
Let .
We see that .
We can continue in this manner until rank , i.e., we create a desired sequence . Note that and for , and whenever . As there are exactly edges in this sequence, there exists some with or , proving that there is a red or a blue path of length at least . ∎
Lemma 7.
Any generalized broom is planar unavoidable.
Proof.
Let be a union of and that share only their center vertices. Note that any generalized broom on at most vertices is a subgraph of . Let be sufficiently large, say . Consider colored red and blue. Since , we see that there is monochromatic path on edges in order in a twoedge colored , say is red. Consider a set of fishes hanging on respectively such that the spines of fishes from are pairwise disjoint and each fish has at least spine vertices. If at least one of these fishes contains a red star of size centered at a vertex of , we have a red . Otherwise, each fish in contains at least blue double ribs. The union of blue subgraphs of fishes from clearly contains a blue copy of . ∎
Lemma 8.
Any tree of radius is planar unavoidable.
Proof.
Let be a perfect ary tree of radius . Consider together with a fixed edge coloring in red and blue.
Claim.
If there is a red star and a blue star on edges in , , such that the stars have the same leafset , then there is a monochromatic copy of in .
Let denote the centers of and , respectively. Each vertex has at least neighbors in that are not neighbors of any vertex in . Hence, by pigeonhole principle is the center of a monochromatic star on edges, whose leaves have distance at least two to , see Figure 5. At least of these monochromatic stars are of the same color that together with either or form a monochromatic copy of . This proves the Claim.
Now consider any two adjacent vertices in , , and the set of their at least common neighbors in . By the Claim, we may assume that fewer than vertices of have a red edge to and a blue edge to , and fewer than vertices of have a blue edge to and a red edge to . Each of the remaining at least vertices in has its edges to and in the same color, and by pigeonhole principle we may assume that for at least of these vertices this the same color. We let denote this monochromatic copy of in .
Finally, consider two adjacent vertices in . Say that is blue. If for every vertex in we find a monochromatic in blue for some , then there is a blue copy of , as desired. So assume that for at least one vertex in all monochromatic for some are red; see Figure 6. Then in particular is red with vertices and . Moreover, for each the monochromatic is red. However, this gives a red copy of rooted at ; see Figure 6. ∎
Lemma 9.
A cycle is planar unavoidable. For , .
Proof.
Consider the graph consisting of a fish hanging on edge with spine vertices , and a vertex of degree three in each face of bounded by two spine vertices; see the left part of Figure 7. Note that . Consider any fixed edgecoloring of in red and blue.
First we claim that contains a monochromatic or a monochromatic inner face such that any two vertices of have a common neighbor in , not in , such that edges and have the same color. To this end, consider the spine vertices and the corresponding double ribs , . If contains no monochromatic , at most two double ribs are monochromatic – one red and one blue. Hence there are five consecutive spine vertices whose double ribs are bicolored. Assume, without loss of generality, that . Further assume that the edges and are red, so the edge is blue.
 Case 1: is red.

Then is blue. If the spine edge is blue, then bound an inner face with the desired properties ensured by the vertex that sends red edges to . So we may assume that is red. For the same reason, if is also red, then also is red, giving a red with vertices . So we may assume that is blue and hence is red. Now if is red, there is a red with vertices . So we may assume that is blue.
Symmetrically, we may assume that is blue, hence is red, and is blue. But now bound an inner face with the desired properties; see the right part of Figure 7.
 Case 2: is blue.

Then is red. Now if is red, we have a red with vertices . So we may assume that is blue. By symmetry we may also assume that is blue, hence is red, and is blue. But then we have a blue with vertices ; see the right part of Figure 7.
This proves the claim that contains a monochromatic or a monochromatic inner face , say in red, such that any two vertices of are joined by a blue in . In the former case we are done. In the latter case note that as is all red, any two vertices of are also joined by a red in . Now consider the vertex in whose three neighbors are the vertices of . As two of the three edges incident to have the same color, there are two vertices in that are joined by two distinct but identically colored ’s in . That is, there is a monochromatic copy of in . ∎
4 Proof of Theorem 2
Let be a tree of radius with root and all vertices of distance to having degree .
See Figure 8.
Let be a planar graph.
Let be a partition of such that each induces a linear forest in , , such a partition exists by a result of Poh [14].
Further, consider an orientation of with outdegree at most at each vertex, see [5].
(This orientation result also follows from [11].)
For color the edges in alternately red and blue along the paths in .
For each remaining directed edge of we have and for some .
Color red if and blue if .
See Figure 8.
Assume that there is a monochromatic copy of , say red.
Since the outdegree of each vertex in is at most , we see that each nonleaf vertex of has at least two incoming edges.
Due to the color alternation in each , at least one of the two incoming edges has its two endvertices in distinct parts.
In particular, the root is in or in .
Then at least one vertex at distance or from is in .
However, the vertices of have indegree at most in the red graph, a contradiction.
Let be a tree of radius with root and all vertices of distance to having degree .
See Figure 8.
Similarly, let be an outerplanar graph.
Let be a partition of such that each induces a linear forest in , , such a partition exists by a result of Cowen et al. [4].
Further, consider an orientation of with outdegree at most at each vertex, see [5, 11].
For color the edges in alternately red and blue along the paths in .
For each remaining directed edge of we have and for some .
Color red if and blue if .
See Figure 8.
Assume that there is a red copy of .
Since the outdegree of each vertex in is at most , each nonleaf vertex of has two incoming edges.
Thus the root is in and at least one of its neighbors is in , a contradiction.
The trees and have and vertices, respectively, and are illustrated in Figure 8. We know that every planar avoidable tree has at least vertices since it has radius at least three and it is not a generalized broom.
5 Proof of Theorem 3
A result of NashWilliams [11] implies that any planar graph can be edgedecomposed into at most three forests.
Thus any graph that is not a forest is planar avoidable.
Another result of Gonçalves [9] states that any planar graph can be edgecolored in four colors so that each color class is a forest of caterpillars.
Thus any graph that is not a caterpillar forest is planar avoidable.
For the remainder of the proof let be any planar graph. Let be a partition of the vertex set so that is a linear forest [14]. We shall define two colorings and of the edges of with three and four colors, respectively. To this end, consider the bipartite subgraphs of with partitions , , , and containing all edges of between respective parts. For each orient the edges of so that the outdegree at each vertex is no more than . (Such an orientation exists by [5, 11] as bipartite vertex planar graphs have no more than edges, by Euler’s formula.)
 Coloring :

For , color all edges in and all edges of that are oriented incoming at a vertex of in color .
 Coloring :

For , color all edges of that are oriented incoming at a vertex of in color . Further, color all edges in , , in color .
Next we show that a tree of radius with root and all vertices of distance to of degree at least (see Figure 8) is planar avoidable.
We claim that does not contain a monochromatic copy of .
In fact, if is any vertex with at least three incident edges of the same color , then must be a vertex in .
However, has maximum degree at most , while the vertices of degree at least in induce a subgraph of maximum degree at least .
Hence there is no monochromatic copy of in under coloring .
Finally, we show that a symmetric double star on vertices, i.e, a tree with two adjacent vertices of degree and four leaves (see Figure 8) is planar avoidable.
We claim that does not contain a monochromatic copy of .
First, color is a disjoint union of paths, and thus there is no copy of in color .
For color we see that, as before, only vertices in may have three incident edges of color .
However, as is an independent set in the subgraph of color , there is no copy of in that subgraph.
Hence there is no monochromatic copy of in under coloring , as desired. ∎
Let us remark that coloring shows that every graph in which the vertices of degree at least induce a subgraph of maximum degree at least is planar avoidable. Similarly, coloring shows that every graph with an oddlength path whose two endvertices have degree at least three each, is planar avoidable.
6 Conclusions
In this paper we initiated the study of Ramsey properties of planar graphs.
When two colors are considered, only some outerplanar bipartite graphs are unavoidable and even some trees are avoidable.
We showed that is unavoidable. The following questions remain open:
1. Are other even cycles unavoidable?
2. What is the smallest number of vertices in an avoidable tree?
All of our positive results, showing that some graphs are unavoidable, use the fact that the iterated triangulation arrows these graphs.
3. Is is true that for each planar unavoidable graph there is such that ?
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