Minimal Ordered Ramsey Graphs

12/25/2017 ∙ by Jonathan Rollin, et al. ∙ 0

An ordered graph is a graph equipped with a linear ordering of its vertex set. A pair of ordered graphs is Ramsey finite if it has only finitely many minimal ordered Ramsey graphs and Ramsey infinite otherwise. Here an ordered graph F is an ordered Ramsey graph of a pair (H,H') of ordered graphs if for any coloring of the edges of F in colors red and blue there is either a copy of H with all edges colored red or a copy of H' with all edges colored blue. Such an ordered Ramsey graph is minimal if neither of its proper subgraphs is an ordered Ramsey graph of (H,H'). If H=H' then H itself is called Ramsey finite. We show that a connected ordered graph is Ramsey finite if and only if it is a star with center being the first or the last vertex in the linear order. In general we prove that each Ramsey finite (not necessarily connected) ordered graph H has a pseudoforest as a Ramsey graph and therefore is a star forest with strong restrictions on the positions of the centers of the stars. In the asymmetric case we show that (H,H') is Ramsey finite whenever H is a so-called monotone matching. Among several further results we show that there are Ramsey finite pairs of ordered stars and ordered caterpillars of arbitrary size and diameter. This is in contrast to the unordered setting where for any Ramsey finite pair (H,H') of forests either one of H or H' is a matching or both are star forests (with additional constraints). Several of our results give a relation between Ramsey finiteness and the existence of sparse ordered Ramsey graphs. Motivated by these relations we characterize all pairs of ordered graphs that have a forest as an ordered Ramsey graph and all pairs of connected ordered graphs that have a pseudoforest as a Ramsey graph.

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1 Introduction

Graph Ramsey theory is concerned with the phenomenon that for any given graph there are graphs , called the Ramsey graphs of , such that for any -coloring of the edges of there is a copy of in with all its edges of the same color. For most graphs it is a challenging problem to determine all its Ramsey graphs exactly which is solved only for few classes of graphs like small matchings or stars [8, 11]. Therefore, particular properties of the set of all Ramsey graphs of and its members are studied. This line of research was initiated by fundamental work of Nešetřil and Rödl [32] and Burr, Erdős, and Lovász [11]. One of the most famous questions asks for the smallest number of vertices of graphs in called the Ramsey number and denoted . Determining the Ramsey number of complete graphs is a challenging problem on its own and no exact formula is known yet.

In this paper we study some structural questions from the Ramsey theory for ordered graphs. Here an ordered graph is a graph equipped with a linear ordering of its vertex set. An (ordered) subgraph of an ordered graph is a subgraph of the underlying graph of that inherits the ordering of vertices from . Analogously to the unordered setting we have the following definitions. An ordered graph is an ordered Ramsey graph of some pair of ordered graphs if for any coloring of the edges of in colors red and blue there is either a copy of with all its edges colored red or a copy of with all its edges colored blue. In this case we write to indicate this fact and let . If , then we write and . A fundamental relation between ordered and unordered Ramsey graphs is given in the following observation.

Observation 1.

Let and be an ordered graphs with underlying (unordered) graphs and . If is an ordered Ramsey graph of , then is a Ramsey graph of .

If is a complete graph, then also the reverse statement holds, otherwise it may fail. Figure 1 shows an example of an ordered graph and a Ramsey graph of which does not form an ordered Ramsey graph of in any ordering.

Figure 1: An ordered matching and an ordered Ramsey graph of (left). An (unordered) -cycle is a Ramsey graph of the underlying graph of , but no ordering of the vertices of yields an ordered Ramsey graph of (right). Here we show just a few possible orderings of . In [36] we show that any ordered Ramsey graph of contains a copy of (which is not possible for orderings of ).

We study the fundamental question whether the set has a finite number of minimal elements, where is minimal if for each proper subgraph of . In this case we call the pair Ramsey finite. The corresponding question in the unordered setting is studied intensively (see Theorems 6, 7, 8 below) but a full answer is not known. Our results indicate that any Ramsey finite pair of ordered graphs necessarily has sparse ordered Ramsey graphs. Motivated by this relations we first give several results on sparse ordered Ramsey graphs, which might be of independent interest.

Outline.

Next we present some basic definitions which are used throughout the paper. In particular some frequently used notions for ordered graphs are introduced. Then we present our results on sparse ordered Ramsey graphs, followed by our results on minimal ordered Ramsey graphs. A brief summary of previous work on Ramsey theory for ordered graphs is given at the end of Section 1. In Section 2 we prove our results and Section 3 contains concluding remarks and open questions.

Preliminary Remarks and Definitions.

Before stating our results we need to introduce some notions. For a positive integer we shall write . For a given (ordered) graph we refer to its vertex set by and to its edge set by . We consider the vertices of an ordered graph laid out along a horizontal line from left to right such that a vertex is to the left of a vertex if , and to the right if . For two sets , we write () if () for all and . For two subgraphs of , of we write () if (). An interval of an ordered graph is a set of consecutive vertices of , i.e., for any , , , with , we have .

A complete graph on vertices is denoted . A path on vertices is denoted and an ordered path is a monotone path if . A partial matching is a graph without any copy of . A right star is an ordered star with all its leaves to the right of its center and a left star is defined accordingly. A right star with leaves is denoted . Right and left stars with exactly two edges are also called bend. An ordered graph is a right caterpillar if it is connected and for some there are vertices in such that each edge in is of the form with and . For the segment of a right caterpillar is the subgraph of induced by . The defining sequence of is . Left caterpillars are defined accordingly. See Figure 2 for examples.

Figure 2: A monotone path (left), a right star (middle), and a right caterpillar with four segments (right).

Sparse Ordered Ramsey Graphs.

First, we state some known results on densities of (unordered) Ramsey graphs. Given a graph let denote its density and let denote its -density, provided that has at least two edges. For a pair of graphs let denote its Ramsey density. The Ramsey density is studied by Rödl and Ruciński [34], Kurek and Ruciński [25], and Mütze and Peter [29]. The exact value is known only for few classes of graphs, including stars and complete graphs. The following result is due to Rödl and Ruciński [34] (see also Nenadov and Steger [31]).

Theorem 1 ([34]).

Let be a graph with . Then for each .

This leaves to consider graphs with . Observe that if and only if is a forest. Recall that a partial matching is a graph without any copy of . More precisely we have if and only if is a partial matching and if and only if is a forest which is not a partial matching. Further observe that if and only if is a forest and if and only if each component of contains at most one cycle and is not a forest. Graphs of density at most are called pseudoforests. A proper pseudoforest is a graph of density exactly , that is, a pseudoforest that contains at least one cycle. It seems well-known which pairs of graphs have Ramsey density at most . We give a proof of the following result in Section 2.1 for completeness.

Lemma 1.1.

For each pair of graphs with at least one edge each we have

  1. if and only if is a pair of a forest and a star forest,

  2. and there is a pseudoforest in if and only if is a pair of a partial matching and a proper pseudoforest or both and are forests of stars and copies of , both with at least one copy of .

With Theorem 1 we obtain the following corollary, implicitly in [21, 34]. Note that for any matching there is a matching , that is, .

Corollary 2 ([21, 34]).

Let be a graph. Then there is a graph with if and only if each component of is either a star or a copy of .

Here we are interested in the corresponding parameter for a pair of ordered graphs. We completely determine the pairs of ordered graphs that have forests as ordered Ramsey graphs in the following theorem.

Theorem 3.

Let and be ordered graphs with at least one edge each. Then if and only if and are forests and one of the following statements holds.

  1. or is a partial matching.

  2. For one of or each component is a right star and for the other each vertex has at most one neighbor to the left.

  3. For one of or each component is a left star and for the other each vertex has at most one neighbor to the right.

  4. For one of or each component is a left or a right star and for the other each component is a monotone path.

Moreover contains a partial matching if and only if both and are partial matchings.

Further we characterize all pairs of connected ordered graphs that have pseudoforests as Ramsey graphs.

Theorem 4.

Let and be connected ordered graphs with at least one edge each. Then and there is an ordered pseudoforest in if and only if is a pair of and a connected ordered proper pseudoforest or both and form a monotone .

Corollary 5.

Let be a connected ordered graph. Then there is an ordered graph with if and only if is a left star, a right star, or a monotone .

Ramsey (In)Finiteness.

Recall that a graph is minimal if for each proper subgraph of . A pair of graphs is Ramsey finite if there are only finitely many minimal graphs in , and Ramsey infinite otherwise. The following theorems summarize some of the known main results in the unordered setting.

Theorem 6 ([34, 35], see [26]).

If a graph contains a cycle, then is Ramsey infinite.

Theorem 7 ([10],[26]).

Let be a forest without isolated vertices and let be graph with at least one cycle.

  1. If is a matching, then is Ramsey finite.

  2. If is not a matching, then is Ramsey infinite.

Theorem 8 ([19]).

Let and be forests without isolated vertices. Then there is a constant such that is Ramsey finite if and only if one of the following statements holds.

  1. At least one of or is a matching.

  2. Both and

    are vertex disjoint unions of a star with an odd number of edges and a matching.

  3. One of or is a vertex disjoint union of a matching and at least two stars with respectively edges, while the other is a vertex disjoint union of a matching on edges and a star with edges. Moreover , are odd, , and .

The asymmetric case for a pair of graphs containing a cycle is not completely resolved. Nešetřil and Rödl [33] prove that is Ramsey infinite if both and are -connected or both are of chromatic number at least while results in [12] show that it is sufficient to consider pairs of -connected graphs. Bollobás et al. [7] prove that is Ramsey infinite for each cycle if is -connected and contains no induced cycles of length at least , provided . We think that all Ramsey finite pairs of graphs are characterized in Theorem 8, see the discussion of the asymmetric case in Section 3.

As for unordered graphs, we call a pair of ordered graphs Ramsey finite if there are only finitely many minimal graphs in , and Ramsey infinite otherwise. Here an ordered graph is minimal if for each proper ordered subgraph of . In case we call itself Ramsey finite or infinite, respectively. We shall establish results similar to Theorems 6 and 7(a) while our results for ordered forests show that the ordering plays a significant role.

For wide families of ordered graphs we shall show that for a carefully chosen probability large random ordered graphs contain large minimal ordered Ramsey graphs. Note that random graphs

are usually defined with vertex set and can be considered as ordered graphs without modifications. Rödl and Ruciński [35] (see also [23]) determine for a given (unordered) graph the threshold probability that is a Ramsey graph of , provided that is not a forest whose components are stars or copies of only. They prove that for such a graph there are constants and such that if then the probability that is a Ramsey graph of tends to as (the -statement), while if then the probability that is a Ramsey graph of tends to as (the -statement). Note that the -statement immediately carries over to ordered graphs by Observation 1. We prove a corresponding -statement for ordered graphs. Our proof closely follows a recent new proof of the -statement for (unordered) graphs due to Nenadov and Steger [31] using the hypergraph container method. It would be interesting whether one can deduce the -statement for ordered graphs from the unordered -statement directly, without reformulating the proof.

Theorem 9.

Let be an ordered graph that is not a partial matching. There is a constant such that if then the probability that is an ordered Ramsey graph of tends to as .

Based on this theorem we use random graphs to prove the following result.

Theorem 10.

Let be an ordered graph. If for each ordered graph , then is Ramsey infinite.

Essentially this theorem yields that every Ramsey finite ordered graph has a pseudoforest as a Ramsey graph. Indeed by Theorems 1 and 10 we have that is a forest and there is some with , that is, is a pseudoforest. Using our results on sparse ordered Ramsey graphs from the first part of this article we obtain the following.

Theorem 11.

If is a Ramsey finite ordered graph, then each component of which is not a monotone is a right star or each such component is a left star.

A full characterization of Ramsey finite ordered graphs which are connected is given in Corollary 17 below. As in the unordered setting, we believe that Theorems 9 and 10 generalize to the asymmetric case, see the discussion in Section 3. Now we turn to pairs of ordered graphs where one is a forest and the other contains a cycle. In the unordered setting such a pair is Ramsey finite if and only if the forest is a matching [10, 26] (see Theorem 7 above). The following theorem gives a partial result for ordered graphs, similar to Theorem 7(a). Recall that is an interval of an ordered graph if for any , and with we have . An ordered graph with at least two vertices is loosely connected if for any any partition of the vertices of into two disjoint intervals and there is an edge with one endpoint in and the other endpoint in . See Figure 3.

Figure 3: Each ordered graph on the left side is loosely connected while each ordered graph on the right is not loosely connected. We see that a disconnected ordered graph might be loosely connected, and each ordered graph which is not loosely connected contains two vertices next two each other which are not “spanned” by any edge (the dark gray parts).

Further denotes the intervally disjoint union of ordered graphs and , that is, a vertex disjoint union of and where all vertices of are to the left of all vertices of . Note that each ordered graph without isolated vertices has a unique representation where is a loosely connected ordered graph, .

Theorem 12.

Let and be positive integers and let , be loosely connected ordered graphs. If is Ramsey finite for all , , then is Ramsey finite.

A monotone matching is an ordered matching of the form . Clearly is Ramsey finite for any ordered graph .

Corollary 13.

If is a monotone matching, then is Ramsey finite for each ordered graph  without isolated vertices.

Finally we consider pairs of ordered forests. A large part of the full characterization in the unordered setting (see Theorem 8) is due to Nešetřil and Rödl [33] who prove that each pair of (unordered) forests which are not star forests is Ramsey infinite. Their proof is based on the fact that each pair of (unordered) forests has Ramsey graphs of arbitrarily large girth. This in turn relies on the fact that for each (unordered) forest there is an integer such that each graph of chromatic number at least contains a copy of . This second fact is not true for ordered forests [1]. We think though that the first fact holds for ordered graphs as well.

Conjecture 1.

For each integer and any pair of ordered forests there is with .

If Conjecture 1 is true, then each pair of ordered forests where does not contain a forest has minimal Ramsey graphs of arbitrarily large (but finite) girth, and hence of arbitrarily large order.

Observation 2.

Let be a pair of ordered forests such that and for each integer there is with . Then is Ramsey infinite.

Here we focus on pairs of ordered forests which satisfy the second fact mentioned above, that is, there is an integer such that each graph of chromatic number at least contains a copy of . We call such an ordered forest -unavoidable. Ordered forest which are not -unavoidable are discovered in [1] and for some small such forests we show that they are Ramsey infinite in [36]. The proof from [33] can be easily adopted for -unavoidable ordered forests. So Conjecture 1 holds for -unavoidable ordered forests and we have the following theorem.

Theorem 14.

If and are -unavoidable ordered graphs and , then is Ramsey infinite.

This theorem leaves to consider pairs of -unavoidable ordered graphs with . We address such pairs of connected ordered graphs next and defer the study of such disconnected forests to future work. Recall that a right caterpillar is an ordered tree with segments being right stars with at least one edge each. Further if are the segments of a right caterpillar , then the defining sequence of is . A left or right caterpillar with defining sequence is called almost increasing if or (, , and ). See Figure 4.

Figure 4: Two almost increasing right caterpillars (left, middle) and two not almost increasing right caterpillars (right).
Theorem 15.

Let be a Ramsey finite pair of -unavoidable connected ordered graphs with at least two edges. Then is a pair of a right star and an almost increasing right caterpillar or a pair of a left star and an almost increasing left caterpillar.

Theorem 16.

Let be a pair of a right star and a right caterpillar or a pair of a left star and a left caterpillar, and let be the defining sequence of the caterpillar. If either or , then is Ramsey finite.

Unfortunately we do not resolve this case completely, see Conjecture 5 and the preceding discussion in Section 3. Nevertheless Theorems 3, 11, 14 and 16 yield the following result.

Corollary 17.

A connected ordered graph is Ramsey finite if and only if it is a left or a right star.

A summary of our results is given in Table 1.

Table 1: Summary of results on Ramsey finiteness of for ordered graphs and .

Ordered Ramsey Numbers.

Recently, Ramsey numbers were studied for ordered (hyper)graphs. The ordered Ramsey number of some ordered -uniform hypergraph is the smallest integer such that for any -coloring of the (hyper)edges of an ordered complete -uniform hypergraph on vertices there is a copy of with all edges of the same color. Due to several applications, mostly geometric Erdős-Szekeres type results, Ramsey numbers of so-called monotone (hyper)paths received particular attention [28]. For given positive integers and a monotone -uniform -hyperpath is an ordered -uniform hypergraph with edges , where each edge forms an interval in the vertex ordering and consists of the rightmost vertices in and the leftmost vertices in , . Building on previous results of Moshkovitz and Shapira [27], Cox and Stolee [17] prove that the ordered Ramsey number of such paths grows, as a function in the number of edges of , like a tower of height proportional to the maximum degree of (note that the maximum degrees of all sufficiently large monotone -uniform -hyperpaths coincide for fixed and ). In contrast to this, the Ramsey numbers of unordered hyperpaths, and more general of any hypergraph of bounded maximum degree, are linear in the size. Indeed for any uniformity and any positive integer there is a constant such that for each (unordered) -uniform hypergraph on vertices and of maximum degree at most its Ramsey number is at most  [14, 16]. In a similarly striking contrast to this result, Conlon et al. [15] and independently Balko et al. [3] prove the existence of ordered matchings with superpolynomial Ramsey numbers. On the other hand Conlon et al. [15] present results showing that for dense graphs the ordered Ramsey numbers behave similar to the unordered Ramsey numbers.

2 Proofs

2.1 Proof of Theorems 3 and 4

First we introduce several types of edge-colorings which we shall use to proof that some ordered forest or pseudoforest is not a Ramsey graph for certain pairs of ordered graphs. The distance of an edge and a vertex in some graph is the smallest number of edges in a path that contains and . Three vertices of an ordered graph form a bend if either is adjacent to and or is adjacent to and . An edge-coloring of an ordered graph is a

  • [wide]

  • with respect to if an edge is colored red if its distance to is odd and blue otherwise,

  • with respect to a partition if an edge is colored red if its left endpoint is in and blue otherwise,

  • with respect to if is a tree and an edge is colored red if its right endpoint is or the edge next to on the (unique) path to exists and forms a bend with , and is colored blue otherwise. See Figure 5 (left) for examples of such a coloring.

Figure 5: A bend-coloring of the edges of an ordered tree with respect to a vertex (left) in colors red (solid) and blue (dashed). There is no red monotone path on at least two edges (right, top) and in the blue component containing each vertex has at most one neighbor to the left (right, bottom).
Lemma 2.1.

Let be an ordered graph and let be an edge-coloring of .

  1. If is a star-coloring with respect to , then each monochromatic component is a star and all edges incident to are red.

  2. If is a bipartite-coloring, then there is no monochromatic copy of a monotone .

  3. If is a bend coloring, then there is no red copy of a monotone and for each blue component either each vertex in has at most one neighbor to the left in or each vertex in has at most one neighbor to the right in .

Proof.

The first two statements follow immediately from the definitions, so we only prove (c). Suppose that is a bend-coloring of some ordered tree with respect to . First consider a copy of a monotone in that contains some red edge , . Then either or forms a bend with the edge next to on the path to in . In both cases the other edge in neither has as its right endpoint nor forms a bend with the edge next to it on the path to in . See Figure 5 (right, top). Hence there is no red monotone path on two edges.

Next consider a blue component and the vertex in that has shortest distance to in (it might happen that ). Suppose that has some neighbor in with . If , then each edge in with is red (and is not in ). If and is the edge next to on the path to in , then (as is blue). Hence does not have any neighbor to the left in , since any such edge is colored red as and form a bend. Moreover each other vertex in has exactly one neighbor to the left in , since otherwise there is a path from (and hence from ) to some vertex in that contains a bend and hence a red edge. Hence each vertex in has at most one neighbor to the left in . See Figure 5 (right, bottom).

Similar arguments show that if has some neighbor to the left in , then each vertex in has at most one neighbor to the right in . ∎

First we prove Lemma 1.1 on unordered graphs with Ramsey density at most . In that proof we shall freely use the star-coloring adopted for unordered graphs. Clearly an analogous statement to Lemma 2.1(a) holds in this case.

Proof of Lemma 1.1.

Let be a forest of maximum degree and let be a star forest of maximum degree . One can see that for any -coloring of the edges of a sufficiently large -ary tree without blue copies of there is a copy of in some component of the red subgraph by a greedy embedding. Therefore and .

On the other hand consider a forest and a pair of graphs . If is not a forest, then since coloring all its edges red yields neither red copies of nor blue copies of (since contains at least one edge). If neither nor is a star forest then choose a root in each component of and color the edges of the components according to a star coloring with respect to their respective roots. Then there are neither red copies of nor blue copies of by Lemma 2.1(a) and . So in both cases there is no forest in and we have that .

Consider a pair of graphs. If one of or is not a pseudoforest, or one of or contains a cycle while the other is not a partial matching (that is, contains a copy of ), then clearly there is no pseudoforest in .

If is a partial matching and is a proper pseudoforest, then let be a vertex disjoint union of many copies of . For any coloring of the edges of either all edges in one of the copies of are blue or there is red copy of . Hence is a Ramsey graph of and . Moreover by part (a) and thus .

This leaves to consider pairs of ordered forests. If one of or is a star forest, then by part (a). So suppose that both and contain a copy of .

First assume that has a component which is not a star and not a . Then contains either a copy of or a copy of a graph obtained from by adding a pendant edge, that is, by adding a new vertex and an edge connecting to one the vertices of degree in . Consider some proper pseudoforest . We shall prove that .

If contains a copy of , then let denote the forest obtained from by removing a smallest set of edges which contains one edge from each cycle in . Color each component of according to some star coloring and color all edges in blue. Then by Lemma 2.1(a) the red edges in form a star forest and there is no blue copy of in . Hence there is no red copy of and no blue copy of and is not a Ramsey graph of . So there is no pseudoforest in in this case, as was arbitrary.

If contains a copy of with a pendant edge, then let denote the subgraph of formed by all the cycles in and let denote the subgraph of formed by all edges not in cycles. Each component of contains at most one vertex from . Color all edges in blue and color each component of according to a star coloring with respect to the unique vertex shared with , if it exists, and with respect to an arbitrary vertex otherwise. Then by Lemma 2.1(a) the red edges form a star forest while the blue edges form a vertex disjoint union of a star forest and cycles. Hence there is no red copy of and no blue copy of and . Again there is no pseudoforest in in this case, as was arbitrary.

It remains to consider pairs of forests of stars and copies of , both with at least one copy of . Let be a graph obtained from a -cycle with vertices by adding vertices and an edge for each , . Moreover let denote the largest degree among all vertices in and , let be a forest that is a Ramsey graph for a pair of stars on edges, and let be a forest that is a Ramsey graph for a pair of a star on edges and . Such forests exist by part (a). One can see that  [29] and that a suitable vertex-disjoint union of several copies of the graphs , , and forms a Ramsey graph of . Since such a union is a pseudoforest we have . Moreover by part (a) and thus .∎

Lemma 2.2.

Let and be ordered graphs. Then does not contain a pseudoforest in each of the following cases.

  1. One of and contains a cycle and the other is not a partial matching.

  2. One of or contains a vertex with two neighbors to the right and the other contains a vertex with two neighbors to the left.

  3. One of and contains a copy of a monotone and the other contains a copy of an ordered .

  4. One of and contains a copy of a monotone and the other contains a copy of a star on three edges that is neither a right star nor a left star .

Proof.

Let be a pseudoforest. For each of the cases we shall give a coloring of the edges of without red copies of or blue copies of .

Without loss of generality assume that contains a cycle and is not a partial matching. Choose a smallest set of edges which contains one edge from each of the cycles of . Then color all edges in blue and all the other edges of red. Since the edges in form a forest there is no red copy of and since the edges in form a matching there is no blue copy of . Hence and contains no pseudoforest.

Without loss of generality assume that contains a vertex with two neighbors to the right and contains a vertex with two neighbors to the left. We give a -coloring of the edges of without red copies of or blue copies of by induction on the number of edges of . Indeed such a coloring clearly exists if . If we distinguish two cases. First suppose that has some vertex of degree . Remove from and color the resulting pseudoforest inductively. If is to the left of its neighbor in then color red, otherwise color it blue. This coloring of contains neither red copies of nor blue copies of . Next assume that contains no vertex of degree , that is, is a vertex disjoint union of cycles. For each even cycle in color its edges alternatingly red and blue and for each odd cycle color both edges incident to its leftmost vertex blue and the remaining edges alternatingly red and blue. This coloring contains neither red copies of nor blue copies of . In both cases and hence contains no pseudoforest.

Without loss of generality assume that contains a copy of a monotone and contains a copy of an ordered . We shall give a coloring of with no red copies of and no blue copies of . Since and are connected we assume without loss of generality that is connected. We distinguish cases based on the ordering of .

First assume that forms a monotone . If is bipartite, then color its edges using a bipartite-coloring with respect to an arbitrary bipartition of . Then there is no red copy of and no blue copy of by Lemma 2.1(b). If is not bipartite, then we obtain a bipartite graph from by removing some edge from the (unique) odd cycle in . Note that for any bipartition of the endpoints of belong to the same part. Choose such a partition such that the endpoints of belong to . Color the edges in using the bipartite-coloring with respect to the partition formed by and . Further color blue. By Lemma 2.1(b) there is no red copy of and each blue copy of a monotone contains . Since the left endpoint of is in we have that all edges incident to this vertex to the left are colored red. Therefore there is no blue copy of a monotone and thus no blue copy of . In both cases and hence contains no pseudoforest.

Next assume that contain a copy of a monotone whose rightmost vertex has two neighbors to the left in . If is bipartite, then color its edges using a bipartite-coloring with respect to an arbitrary bipartition of . Then there is no red copy of and no blue copy of by Lemma 2.1(b). Otherwise consider the odd cycle in and edges , in such that is the rightmost vertex of and . Let denote the subgraph of formed by the union of all monotone paths in whose leftmost vertex is . Then is a tree, since it does not contain (and is the only cycle in ). Color all edges of blue. Then there is no blue copy of in . The remaining edges form a forest since does not contain . Consider a bipartition where each vertex shared with is in . Such a partition exists since either and are connected by a path in with an odd number of vertices or are in distinct components. Color the edges in using a bipartite-coloring with respect to the partition formed by and , see Figure 6 (left). Then in there are no monochromatic copies of a monotone by Lemma 2.1(b). In particular there is no red copy of in , since all edges in are blue. Moreover each blue edge in does not contain any vertex of since edges in may share only their right endpoint with . Hence each blue component of is either in or in and hence there is no blue copy of in . Altogether and hence contains no pseudoforest.

Figure 6: Colorings of proper pseudoforests. In the left all edges of a certain tree are colored blue and the remaining edges are coloring using a bipartite coloring. In the middle an edge is removed from a cycle and the remaining forest is colored with a bend coloring with respect to . In the right all edges of a cycle are colored blue and the remaining edges are colored using a bipartite coloring.

Next assume that contain a copy of a monotone whose leftmost vertex has two neighbors to the right. In this case contains no pseudoforest with arguments similar to the previous case.

Finally assume that does not contain any copy of a monotone . Then has some vertex with two neighbors to the left and another vertex with two neighbors to the right. If is a tree, then color its edges according to some bend-coloring with respect to an arbitrary vertex. Then there is no red copy of a monotone and no blue copy of by Lemma 2.1(c). Otherwise let denote the leftmost vertex of the cycle in and let and be its neighbors in that cycle. Let be a tree obtained y removing the edge from . Color with a bend-coloring with respect to , see Figure 6 (middle). Then in there is no red copy of a monotone and no blue copy of by Lemma 2.1(c). Note that any edge in with is colored blue, since the next edge on the path to is which does not form a bend with . Moreover any edge in with is colored blue. The coloring of gives a coloring of by coloring red. Then there is no blue copy of since there is no such copy in and no red copy of a monotone since edges , , and edges , , are blue. Altogether and hence contains no pseudoforest.

Without loss of generality assume that contains a copy of a monotone and contains a star on three edges whose center has one neighbor to the left and two neighbors to the right. Note that contains a copy of a monotone . If is bipartite, then color its edges using a bipartite-coloring with respect to an arbitrary bipartition of . Then there is no red copy of and no blue copy of by Lemma 2.1(b). Otherwise color all edges of that are contained in a cycle blue. The remaining edges form a forest . Choose a bipartition such that all vertices in the cycle in that have two neighbors to the right in that cycle are in and the other vertices of the cycle are in . Such a partition exists since the vertices of the cycle in are in different components of . Color the edges in using a bipartite-coloring with respect to the partition formed by and , see Figure 6 (right). Then in there are no monochromatic copies of a monotone by Lemma 2.1(b). In particular there is no red copy of in , since all edges not in are blue. Moreover each vertex which is left endpoint of at least two blue edges in is in . Such a vertex is not a right endpoint of any blue edge. This shows that there is no blue copy of in . Altogether and hence contains no pseudoforest.∎

Lemma 2.3.

Let and be ordered forests. Then does not contain a forest in each of the following cases.

  1. Both and contain a component that is not a star.

  2. One of or contains a vertex with two neighbors to the right and the other contains a vertex with two neighbors to the left.

  3. Both and contain a monotone path on two edges.

  4. One of and contains a copy of a monotone and the other contains a copy of an ordered .

Proof.

Let be a forest. For each of the cases we shall give a coloring of the edges of without red copies of or blue copies of .

For each component of color its edges according to a star-coloring with respect to some arbitrary vertex in that component. Then each color class forms a star forest by Lemma 2.1(a), that is, there is neither a monochromatic copy of nor of . Hence contains no forest.

This follows immediately from Lemma 2.2(b).

Color the edges of according to a bipartite-coloring with respect to an arbitrary bipartition. This coloring contains neither red copies of nor blue copies of by Lemma 2.1(b). Hence contains no forest.

This follows immediately from Lemma 2.2(c).∎

Proof of Theorem 3.

We shall prove that all pairs of ordered forests that do not have a forest as an ordered Ramsey graph are covered by Lemma 2.3. To this end we provide explicit constructions of ordered forests that are ordered Ramsey graphs for the remaining pairs.

First we shall show that each pair where contains a forest satisfies at least one of the cases of this theorem. Let be such a pair. Clearly and are forests, since any monochromatic subgraph of an edge-colored forest is a forest itself. If either or is a partial matching then Case (a) of this theorem holds. So assume that neither nor is a partial matching. Due to Lemma 2.3 (a) one of or is a star forest. Without loss of generality assume that is a star forest. Due to Lemma 2.3 (c) one of or does not contain a monotone .

First suppose that does not contain a monotone . Then each component of is a left or a right star. Due to Lemma 2.3 (b) the following holds. If each component of is a right star, then each vertex of has at most one neighbor to the left (as is not a partial matching). Thus and satisfy Case (b) of this theorem. Similarly, if each component of is a left star, then each vertex of has at most one neighbor to the right. Thus and satisfy Case (c) of this theorem. If contains a right star on two edges as well as a left star on two edges, then each component of is a monotone path. Thus and satisfy Case (d) of this theorem.

Now suppose that contains a monotone . Then neither contains a monotone nor any ordered due to Lemma 2.3 (c) and (d). Therefore each component of is a left or a right star. The same arguments as above show that and satisfy one of the Cases (b), (c), or (d) of this theorem.

Next consider two ordered forests and that satisfy Case (a), (b), (c), or (d) of this theorem. We shall show that there is a forest in and distinguish which case of this theorem holds. First of all suppose that and together contain some isolated vertices. Let and be obtained from respectively by removing these isolated vertices. Then there is a forest in if and only if there is a forest in . Indeed, if is an ordered forest in , then . Suppose that is an ordered forest in . Then we obtain an ordered forest in by adding isolated vertices to the left of all vertices in , to the right of all vertices in , as well as between any pair of consecutive vertices of . Similarly contains a partial matching if and only if contains a partial matching. For the remaining proof we assume that neither nor contains isolated vertices.

Without loss of generality assume that is a matching. Consider a complete ordered graph of order with vertices . Let , , and . Note that contains exactly copies of and each vertex of is contained in exactly copies of in . We shall construct an ordered graph that is a vertex disjoint union of copies of . For each let denote the copies of in containing . Choose disjoint ordered vertex sets of size each, . Let denote the ordered graph with vertex set , , where is an edge in if and only if and the edge is in , , . See Figure 7.

Figure 7: Disjoint copies of forming an ordered graph in for some matching .

Observe that is a vertex disjoint union of copies of and hence a forest. Moreover, if is a matching, then is a matching. We claim that . Consider a -coloring of the edges of . We shall show that there is either a red copy of or a blue copy of . To this end consider the edge-coloring of where an edge , , is colored red if there is at least one red edge between and in and blue otherwise. Due to the choice of there is either a red copy of or a blue copy of under . In either case there is a red copy of respectively a blue copy of