## 1 Introduction

Ramsey theory is devoted to the study of the minimum size of a system that guarantees the existence of a highly organized subsystem.
Given graphs and , their *Ramsey number* is the smallest such that any two-coloring of the edges of contains either as a red subgraph or as a blue subgraph of .
The case is called the *diagonal* case and in this case we use the abbreviation .

The growth rate of Ramsey numbers has been of interest to many researchers. In general, it is notoriously difficult to find tight estimates on Ramsey numbers. For example, despite many efforts, the best known bounds on

are essentially(1) |

obtained by Erdős and Szekeres [10], although some smaller term improvements are known. For a more comprehensive survey we can refer the reader to [8].

In this paper, we study ordered graphs.
An *ordered graph* on vertices is a graph whose vertex set is and it is ordered by the standard ordering of integers.
For an ordered graph , we use to denote its unordered counterpart.
An ordered graph on is an *ordered subgraph* of another ordered graph on if there exists a mapping such that for and also is an edge of whenever is an edge of .
Definitions that are often stated for unordered graphs, such as vertex degrees, degeneracy, colorings, and so on, have their natural analogues for ordered graphs. Note that, for every , there is a unique complete ordered graph .

Motivated by connections to classical results such as the Erdős–Szekeres theorem on monotone subsequences [10],
various researchers [2, 7] recently initiated the study of Ramsey numbers of ordered graphs.
Given two ordered graphs and , the *ordered Ramsey number* is defined as the smallest such that any two-coloring of the edges of contains either as a red ordered subgraph or as a blue ordered subgraph.

Observe that for any two ordered graphs and on and vertices, respectively, we have . Thus, by (1), the number is finite and, in particular, is at most exponential in the number of vertices for every ordered graph .

It is known that for dense graphs, there is not a huge difference in the growth rate of their ordered and unordered Ramsey numbers [2, 7]).
On the other hand, ordered Ramsey numbers of sparse ordered graphs behave very differently from their unordered counterparts.
For example, Ramsey numbers of *matchings* (that is, graphs with maximum degree ) are clearly linear in the number of their vertices. However, it was proved independently in [2, 7] that there exist ordered matchings vertices such that their diagonal ordered Ramsey numbers grow superpolynomially.

The bound from Theorem 1 is quite close to the truth, as Conlon, Fox, Lee and Sudakov [7] proved that for every ordered graph on vertices with degeneracy . In particular, we have if has its maximum degree bounded by a constant.

There has also been a keen interest in studying the off-diagonal ordered Ramsey numbers. Conlon, Fox, Lee and Sudakov [7] proved that there exist ordered matchings such that . On the other hand, the best known upper bound on is

which follows from the well-known result [1], which is tight [12]. Note that the first inequality only uses the fact that is an ordered subgraph of and does not utilize any special properties of ordered matchings such as its sparseness. Conlon, Fox, Lee and Sudakov [7] expect that the upper bound is far from optimal and posed the following problem.

###### Problem 2 ([7]).

Does there exist an such that any ordered matching on vertices satisfies ?

Problem 2 remains open, but there was some progress obtained by Rohatgi [15], who resolved some special cases of this problem.
In particular, he proved that if the edges of an ordered matching do not cross, then the ordered Ramsey number is almost linear.
The basic building block of the proof of this result is formed by so-called nested matchings.
For , a *nested matching* (or a *rainbow*) is the ordered matching on vertices with edges for every .

Rohatgi [15] determined the off-diagonal ordered Ramsey numbers of nested matchings up to a constant factor.

###### Proposition 3 ([15]).

For every , we have

He believed that the upper bound is far from optimal and posed the following conjecture, which he verified for .

###### Conjecture 4 ([15]).

For every , we have

The ordered graphs that do not contain as an ordered subgraph for some are known to be equivalent to so-called *-queue graphs* [11] and, in particular, 1-queue graphs correspond to *arched-leveled-planar graphs* [11].
As we will see, estimating the ordered Ramsey numbers is connected to extremal questions about -queue graphs.
In particular, there is a close connection to the problem of Dujmovic̀ and Wood [9] about determining the chromatic number of such graphs.

###### Problem 5 ([9]).

What is the maximum chromatic number of a -queue graph?

Dujmovic̀ and Wood [9] note that and they prove that the lower bound is attainable for .

## 2 Our results

In this paper, we also focus on off-diagonal ordered Ramsey numbers. In particular, we improve and generalize the bounds on and we disprove Conjecture 4. We also consider ordered Ramsey numbers for general connected ordered graphs and we introduce the concept of Ramsey goodness of ordered graphs. Finally, we pose several new open problems.

### 2.1 Nested matchings versus complete graphs

First, we improve the leading constant in the upper bound from Proposition 3 and thus show that the bound by Rohatgi is indeed not tight. However, we believe that our estimate can be improved as well.

###### Theorem 6.

For every , we have

Next, we disprove Conjecture 4 by showing for every . For , we determine exactly.

###### Theorem 7.

For every , we have

Moreover, and .

We prove the lower bound by constructing a specific red-blue coloring of the edges of that avoids a red copy of and a blue copy of . To determine and exactly, we use a computer-assisted proof based on SAT solvers. For more details about the use of SAT solvers for finding avoiding colorings computationally, we refer the reader to the bachelor’s thesis of the second author [14]. The utility we developed for computing ordered Ramsey numbers for small ordered graphs and is publicly available [13].

By performing the exhaustive computer search, we know that there are only 326 red-blue colorings of the edges of without a red copy of and a blue copy of . They all share the same structure except for 6 red edges that can be switched to blue while not introducing a blue triangle. Using the same computer search, we were able to find many red-blue colorings of the edges of without a red copy of and a blue copy of , some of which even had certain symmetry properties. There were no such symmetric colorings on 15 vertices, which suggests that the lower bound on might be further improved for larger values of .

Using the lower bounds from Theorem 7, we can address Problem 5 about the maximum chromatic number of -queue graphs. In particular, we can improve the lower bound by for any ; see Subsection 3.3 for a proof.

###### Corollary 8.

For every , the maximum chromatic number of -queue graphs is at least .

We recall that the maximum chromatic number of -queue graphs is [9]. We use this result to prove the exact formula for the off-diagonal ordered Ramsey numbers of nested matchings with two edges.

###### Theorem 9.

For every , we have .

For general nested matchings versus arbitrarily large complete graphs, we can determine the asymptotic growth rate of their ordered Ramsey numbers, generalizing the linear bounds from Proposition 3 and Theorem 6.

###### Theorem 10.

For every , we have

### 2.2 Ramsey goodness for ordered graphs

To obtain the lower bound for a connected graph on vertices, one might consider a simple construction that is usually attributed to Chvátal and Harary [5].
Take red cliques, each with vertices, and connect vertices in different red cliques by blue edges.
For some graphs , this lower bound is the best possible and such graphs are called *(Ramsey) -good*.
That is, a connected graph on vertices is -good if . We call a graph *good* if it is -good for all .
A famous result by Chvátal [6] states that all trees are good.

Studying -good graphs is a well-established area in extremal combinatorics.
Despite this, to the best of our knowledge, Ramsey goodness has not been considered for ordered graphs.
Motivated by our results from Subsection 2.1, we thus extend the definition of good graphs to ordered graphs and we attempt to characterize all good connected ordered graphs.
A connected ordered graph on vertices is *-good* if . A connected ordered graph is *good* if it is -good for all .

A generalization of the well-known Erdős–Szekeres theorem on monotone subsequences states that for every and every *monotone path * [4], which is an ordered path on vertices where edges connect consecutive vertices in .
In other words, any monotone path is good, which gives a first example of good ordered graphs.
Note, however, that not all ordered paths are good, which follows immediately from Theorem 1.

First, we prove some basic properties of good ordered graphs, some of them resembling their unordered counterparts. It can be shown similarly as in the unordered case that if a connected ordered graph is -good, then it is -good.

Let be an ordered graph containing an ordered cycle as an ordered subgraph. It is known that, for every cycle on vertices and for going to infinity, the Ramsey number grows superlinearly with [3, 12]. Since , the number is also superlinear in and thus the ordered graph cannot be good. We thus obtain the following result that limits good ordered graphs to ordered trees.

###### Proposition 11.

Every good ordered graph is an ordered tree.

In our attempt to determine which ordered trees are good, we discovered a class of good ordered trees, which significantly extends the example with monotone paths. In order to describe this new class, we need to introduce some notation.

An *ordered star graph* is an ordered graph on vertices such that the th vertex in the vertex ordering is adjacent to all other vertices and there are no other edges; see part (a) of Figure 1.
We call an ordered star *one-sided*, if or .

For any two ordered graphs and on and vertices, respectively, the *join* is an ordered graph on vertices constructed by identifying the leftmost vertex of with the rightmost vertex of ; see part (b) of Figure 1.
The join operation is associative and if and are both connected, then is connected as well.

The following result gives a construction of good ordered graphs based on the join operation.

###### Theorem 12.

For all , if a connected ordered graph is -good, then the ordered graphs , , , and are also -good.

Theorem 12 immediately implies that every ordered star graph is good.
More generally, it follows that all ordered trees from the following class are good.
An ordered graph is a *monotone caterpillar graph* if there exist positive integers such that or for each and
.
In other words, if can be obtained by performing joins on one-sided ordered star graphs; see part (b) of Figure 1.
Note that monotone paths and ordered stars are all monotone caterpillar graphs.

###### Corollary 13.

All monotone caterpillar graphs are good.

Computer experiments based on our SAT solver based utility [13] proved that all good ordered graphs up to 6 vertices are monotone caterpillar graphs. We believe that there are no other good ordered graphs; see Conjecture 17.

To get a better understanding of good ordered graphs, we prove an alternative characterization of monotone caterpillar graphs stated in terms of forbidden ordered subgraphs.

###### Proposition 14.

A connected ordered graph is a monotone caterpillar graph if and only if does not contain any of the four ordered graphs from Figure 2 as an ordered subgraph.

We note that if we assume that is an ordered tree, then we can leave out in Figure 2 and the characterization still holds.
It follows from Proposition 14 that monotone caterpillar graphs can be also characterized using nine forbidden *connected* ordered subgraphs.
It suffices to extend the two disconnected ordered graphs and from Figure 2 into connected ordered graphs using a simple case analysis.

###### Corollary 15.

An ordered tree is a monotone caterpillar graph if and only if it does not contain any ordered graph from Figure 3 as an ordered subgraph.

It follows from our computer-assisted proofs that none of the nine ordered graphs from Figure 3 is good.

### 2.3 Open problems

By Theorem 7, the value of does not match the value of the ordered Ramsey number of an -good ordered graph on vertices if already in the case . However, Theorem 9 implies that the two values are the same for and every . In other words, can be considered good, but the ordered graphs with cannot (although neither of these ordered graphs is connected and the definition of good ordered graphs is stated only for connected ordered graphs). We do not know where the truth lies for , although we have computationally verified that for . These results are closely connected to proper colorings of ordered graphs that do not contain nested matchings as ordered subgraphs; see Subsection 3.3, for an example. In particular, a related problem is to decide whether the maximum chromatic number of -queue graphs is at most .

###### Problem 16.

Is it true that for every positive integer ?

Despite our efforts, we still lack complete characterization of all ordered Ramsey good graphs. We exhaustively searched for good graphs with small number of vertices and we verified that there are no -good graphs with at most 6 vertices other than monotone caterpillar graphs. Based on our experimental results, we believe that all good ordered graphs are monotone caterpillar graphs and we pose the following conjecture.

###### Conjecture 17.

A connected ordered graph is good if and only if it is a monotone caterpillar graph.

## 3 Nested matchings versus complete graphs

This section contains proofs of all statements from Subsection 2.1, that is, all results about off-diagonal ordered Ramsey numbers of nested matchings versus complete graphs. In particular, we prove Theorems 6, 7, 10, and 9.

### 3.1 Proof of Theorem 6

To prove Theorem 6, that is, to show that for every , we first state a lemma about the number of edges in an ordered graph that does not contain as an ordered subgraph.

We will assume that an ordered graph on vertices is represented by a matrix.
The *matrix representation* of is an -matrix such that the entry of on position is 1 if and only if and is an edge of .
Sometimes we will not distinguish between and and we identify the edges of with positions of 1-entries in .
Also, given a red-blue coloring of the edges of , we define the *matrix representation of * to be the matrix representation of the ordered graph formed by red edges in .
In particular, red edges correspond to 1-entries and blue edges correspond to 0-entries in such matrix.

We state the following lemma about the maximum number of edges in an ordered graph without a copy of . This result was proved by Dujmovic̀ and Wood [9] in their study of queue graphs. Nevertheless, we include its proof for completeness and present it using the matrix representations, which are more suitable for representing the red-blue colorings of complete ordered graphs.

###### Lemma 18 ([9]).

For every , if an ordered graph on vertices with does not contain as an ordered subgraph, then the number of edges in is at most . Moreover, this upper bound is tight.

###### Proof.

Let be an ordered graph with the vertex set such that does not contain as an ordered subgraph.
Let be a matrix representation of .
For , we define the th *anti-diagonal* of to be the set of positions of such that , and ; see part (a) of Figure 4.
Note that there are exactly anti-diagonals and each one of them contains at most entries, as 1-entries in an anti-diagonal give a copy of the nested matching in .

It follows that there can be at most edges in . To obtain a stronger estimate, we take into account the fact that, for each , the th anti-diagonal contains only at most 1-entries. A similar estimate holds for the number of -entries of a th anti-diagonal with . Altogether, by summing the estimates for all anti-diagonals, we see that the number of edges of is at most

where we used the assumption .

This upper bound is tight, which can be seen by considering the ordered graph on vertices with all edges of length at most .
Here, the *length* of an edge is defined as .
Summing over the lengths of the edges of , we see that has exactly

edges. The ordered graph then does not contain as an ordered subgraph, as the longest edge in each copy of in has to have length at least . ∎

A more general construction to achieve this maximum number of edges in an ordered graph without a copy of is to lead pairwise disjoint “routes” in the matrix and set all entries of with positions in these routes to and all other entries of to ; see part (b) of Figure 4.
For , the th *route* in the matrix is the set of positions such that we have , , and or for every .
The third condition says that a route “goes” only to the right or down in , while the first two conditions only specify the start and the end of each route.
Note that a route can contain entries of that lie on or below the main diagonal of .
We say that an ordered graph is *covered* by a set of routes if every position in the matrix representation of that corresponds to an edge of is contained in some route from .

We note that a route in a matrix corresponds to a *queue* in a *linear queue layout* that is represented by , as defined in [9].
Here, we stick to routes in matrix representations, as we find it more convenient to visualize red-blue coloring of complete ordered graphs with matrices.

We prove that there is no copy of in the ordered graph that is covered by pairwise disjoint routes. This result will be used several times later and it can be shown that if all such routes lie above the main diagonal of , then we have an ordered graph without containing the maximum number of edges.

###### Lemma 19.

For every , every ordered graph that is covered by pairwise disjoint routes does not contain as an ordered subgraph.

###### Proof.

Suppose for contradiction that contains a copy of . Every edge of this copy then belongs to one of the pairwise disjoint routes. No two distinct edges of a nested matching can belong to the same route, since a copy of corresponds to two entries and of the matrix such that , whereas any two entries and of a route satisfy and . However, by the pigeonhole principle, there is at least one such route with at least two edges of , a contradiction. ∎

By Lemma 19, any ordered graph on vertices whose edges can be partitioned into pairwise disjoint routes does not contain . Moreover, if the routes lie above the main diagonal of its matrix representation, then has edges, which is tight by Lemma 18.

We are now ready to improve the upper bound from Proposition 3. Let us assume we have a red-blue coloring of the edges of with no red copy of and no blue copy of . Since there is no blue copy of , there cannot be any vertex of contained in at least blue edges, as otherwise there are only red edges between any two such neighbors of , which gives a red copy of in and thus also a red copy of in .

If every vertex of is contained in less than blue edges, then there is less than blue edges in . Therefore there are more than red edges. However, by Lemma 18, there can be at most red edges in , as otherwise we have a red copy of in . Therefore we obtain

which can be rewritten as

By solving this quadratic inequality for we arrive at the bound

For , the right side of the above expression is at most , which concludes the proof of Theorem 6.

### 3.2 Proof of Theorem 7

We show that, for every , we have . Moreover, we prove and . First, we prove the lower bound by showing that there exists a red-blue coloring of the edges of the complete ordered graph on vertices without a blue triangle and a red copy of .

Let be an integer.
The matrix representation of the coloring is illustrated in Figures 5 and 6.
We now describe the construction of formally by listing all its blue edges.
We note that this coloring is *symmetric*, that is, for all with , the edges and have the same color.

The blue edges in are decomposed into the following sets: the set , which forms a square in , the set , which corresponds to the L-shaped upper right corner of of width and height , and two sets and , which form two rectangles in . Finally, there are two single blue edges and . All the remaining edges of are red in .

We show that the red edges of can be covered by pairwise disjoint routes. Then it will follow from Lemma 19 that there is no red copy of the nested matching in . The set of routes covering the ordered graph formed by red edges in is constructed inductively with respect to . As the basis of the induction, we use the set of routes for that is illustrated in Figure 5. For , we use essentially the same routes we had for , we only elongate them. However, we additionally have to cover two new diagonals formed by entries on positions with and ; see Figure 6. Covering these two new diagonals by an st route is clearly possible and thus we can cover the whole ordered graph by pairwise disjoint routes. Note that some entries of the two new diagonals might be covered by the first routes, but this makes covering their entries by the st route only simpler.

To prove that does not contain a blue triangle for any , we consider the ordered graph formed by edges that are blue in . First, there is no blue triangle containing the edge , as in any such blue triangle there is another blue edge incident to vertex 3. However, all other edges containing vertex 3 are of the form for and there is no blue edge of the form . By symmetry, there is no blue triangle containing the edge

The edges from form a bipartite graph and thus there is no blue triangle with vertices in and any blue triangle in has to have an edge in . Since both sets and induce a bipartite graph, any blue triangle in contains at most one edge in and at most one edge in .

Consider a blue triangle with an edge from . By the definition of , this edge contains a vertex . Since there is at most one edge of in , there is an edge of that is not contained in . The vertex satisfies , as all blue edges with lie in . However, the only blue edge of this form is for and , which gives the edge and we already know that is not contained in a blue triangle. Thus there is no blue triangle with an edge in . By symmetry, there is also no blue triangle with an edge from and, altogether, contains no blue triangle.

It is likely that our construction can be modified to obtain stronger lower bounds on . However, the coloring is easy to describe for any and one can show that it does not contain the forbidden monochromatic ordered subgraphs without employing too complicated case analysis. We also note that some of the blue edges might be colored red without introducing a red copy of in the resulting coloring.

Now, we prove the rest of the statement of Theorem 7, that is, we show and . The lower bounds and are obtained using red-blue colorings and whose matrix representations can be seen in parts (a) and (b) of Figure 7, respectively. We prove that the colorings and do not contain a red copy of and , respectively, and that there is also no blue triangle.

The red edges of and can be partitioned into three and four, respectively, disjoint routes in the matrix representation (see Subsection 3.1 for the definition). It follows from Lemma 19 that and do not contain a red copy and , respectively, as an ordered subgraph. Note that all the routes lie above the main diagonal and thus both examples contain the maximum possible number of red edges by Lemma 18.

Proving that does not contain a blue triangle is a bit tedious and we omit it here, as we verified it using a computer. We can, however, use the fact that is symmetric to give a short explicit proof of the fact that contains no blue copy of .

Let and suppose for contradiction there exists a blue copy of in . It follows from the symmetry of that this copy is formed by vertices such that . Vertices do not have any blue neighbors in . The only blue neighbors of vertex 3 in are and the only blue neighbors of vertex 3 outside of are . However, there is no blue edge between vertices from , hence vertex 3 also cannot be a part of any blue triangle. Thus .

Blue neighbors of vertex 8 are and these vertices form a red clique. Similarly, blue neighbors of vertex 9 are and also form a red clique. Blue neighbors of 10 are two symmetric sets of vertices and . These six vertices also form a red clique. The only option left is , but there are no blue edges between these vertices, a contradiction. The coloring thus contains no blue triangle.

The proof of the upper bounds and is computer assisted. We performed an exhaustive computer search using our SAT-solver-based program [13] and verified that every red-blue coloring of the edges of the ordered complete graph on vertices contains either a red copy of or a blue triangle. Similarly, every red-blue coloring of the edges of contains either a red copy of or a blue triangle.

### 3.3 Proofs of Theorem 9 and Corollary 8

First, we prove Theorem 9 by deriving the exact formula for the ordered Ramsey number of the nested matching versus . The lower bound follows from a simple red-blue coloring of the edges of , where we partition the vertex set into consecutive cliques, each of size 3, and we color all edges between vertices from the same clique red and all other edges blue. Then it is easy to see that there is no red copy of and no blue copy of .

To show the upper bound , we first state an auxiliary result about ordered graphs that do not contain as an ordered subgraph for some . This result was proved by Dujmovic̀ and Wood [9].

###### Lemma 20 ([9]).

Every 1-queue graph is 3-colorable.

Now, let be a red-blue coloring of the edges of . Assume that does not contain a red copy of . The ordered graph formed by edges that are red in is a 1-queue graph, as does not contain a copy of . By Lemma 20, the ordered graph is 3-colorable and therefore we can partition its vertex set into 3 disjoint sets such that no two vertices from the same set are connected by an edge in .

By the pigeonhole principle, there is a set that contains at least vertices. Since there is no edge of between any two vertices from , we see that induces a blue copy of in , which gives the desired upper bound and finishes the proof of Theorem 9.

In the rest of the subsection, we prove Corollary 8 by showing that the maximum chromatic number of -queue graphs is at least for every .

###### Proof of Corollary 8.

For , let be a positive integer such that . We will show that . Since , there is a red-blue coloring of the edges of without a red copy of and a blue copy of . Let be the ordered subgraph of formed by red edges. Suppose for contradiction that the chromatic number of is less than . Then there is an independent set in of size . However, since there is no blue copy of , we have , a contradiction.

By Theorem 7, we have for every . Applying this estimate to the previous observation, we obtain . ∎

### 3.4 Proof of Theorem 10

Here, we prove Theorem 10 by showing for every .
We recall the famous *Turán’s theorem* [16], which states that, for any , every graph on vertices that does not contain as a subgraph has at most
edges.

Let be a red-blue coloring of the edges of , which does not contain a red copy of nor a blue copy of . We proceed along the lines of our proof for Proposition 6. By Lemma 18, there can be at most red edges in . Since the subgraph formed by blue edges does not contain a copy of , Turán’s theorem implies that there can be at most blue edges in . Thus the following inequality

is satisfied and it can be rewritten as

By solving this quadratic inequality, we get

and thus .

The lower bound can be obtained from a coloring on vertices formed by red cliques, each of size , such that any two vertices from different cliques form a blue edge. This coloring clearly contains no red copy of and no blue copy of as ordered subgraphs.

## 4 Ramsey goodness for ordered graphs

This section is devoted to proofs of results of about Ramsey goodness of ordered graphs that are stated in Subsection 2.2. That is, we prove Theorem 12 and Proposition 14.

### 4.1 Proof of Theorem 12

Here, we present the proof of Theorem 12 by showing that a join of a good ordered graph with a one-sided star is a good graph. Formally, we prove that, for all , if a connected ordered graph is -good, then the ordered graphs , , , and are also -good. We start with the following auxiliary result.

###### Lemma 21.

Let be a connected ordered graph such and set . Then every red-blue coloring of the edges of , where , contains either a blue copy of or there are vertices of such that each one of them is the rightmost vertex of some red copy of .

###### Proof.

Let be a red-blue coloring of the edges of , where . By the definition of and since , the coloring contains either a red copy of or a blue copy of as an ordered subgraph. In the latter case we are finished. In the former case, we can find a red copy of in and delete rightmost vertex from , obtaining a red-blue coloring of the edges of vertices. We can iterate this approach as long as the number of vertices is at least . For every , we either find a blue copy of or a vertex that is the rightmost vertex of some red copy of in coloring that is obtained by removing . At the end, the vertices satisfy the statement of the lemma. ∎

Assume that we have a connected -good ordered graph with vertices. It is sufficient to prove that the two ordered graphs , , where is an arbitrary integer larger than , are -good, as the remaining two cases from the statement of Theorem 12 follow by symmetry. Let us denote and and note that .

Let be a red-blue coloring of the edges of for . We will prove that contains either red copies of both and or a blue copy of as ordered subgraphs. Since is -good, we have

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