Off-diagonal ordered Ramsey numbers of matchings

For ordered graphs $G$ and $H$, the ordered Ramsey number $r_<(G,H)$ is the smallest $n$ such that every red/blue edge coloring of the complete graph on vertices $\{1,\dots,n\}$ contains either a blue copy of $G$ or a red copy of $H$, where the embedding must preserve the relative order of vertices. One number of interest, first studied by Conlon, Fox, Lee, and Sudakov, is the"off-diagonal"ordered Ramsey number $r_<(M, K_3)$, where $M$ is an ordered matching on $n$ vertices. In particular, Conlon et al. asked what asymptotic bounds (in $n$) can be obtained for $\max r_<(M, K_3)$, where the maximum is over all ordered matchings $M$ on $n$ vertices. The best-known upper bound is $O(n^2/\log n)$, whereas the best-known lower bound is $\Omega((n/\log n)^{4/3})$, and Conlon et al. hypothesize that $r_<(M, K_3) = O(n^{2-\epsilon})$ for every ordered matching $M$. We resolve two special cases of this conjecture. We show that the off-diagonal ordered Ramsey numbers for matchings in which edges do not cross are nearly linear. We also prove a truly sub-quadratic upper bound for random matchings with interval chromatic number $2$.


Introduction
A classical area of extremal combinatorics is Ramsey theory. Introduced by Ramsey [6] and popularized by Erdős and Szekeres [4], the Ramsey number of a graph G, commonly denoted by r(G), is the smallest n so that every edge bicoloring of the complete graph K n contains a monochromatic copy of G. Shrinking the sizable gap between the asymptotic upper/lower bounds on r(K n ) has been a major open problem for decades, spurring extensive work on a plethora of related questions in Ramsey theory.
One variant of Ramsey numbers which has recently received attention is the analogue for ordered graphs. An ordered graph on [n] is a graph on n vertices which are given distinct labels in {1, . . . , n}. Given an ordered graph G, the ordered Ramsey number of G, denoted by r < (G), is the smallest n so that every edge bicoloring of the ordered complete graph on n vertices contains a monochromatic copy of G which preserves the relative vertex ordering of G. As with the unordered case, one can define the off-diagonal ordered Ramsey number of two graphs G and H, denoted by r < (G, H), as the smallest n so that every edge bicoloring of the ordered complete graph on n vertices contains either an order preserving red copy of G or an order preserving blue copy of H.
The first systematic studies of ordered Ramsey numbers were conducted by Conlon, Fox, Lee, and Sudakov [3] and by Balko, Cibulka, Král, and Kynčl [2]. However, as pointed out by the authors of [3], a number of classic results in extremal combinatorics can be reinterpreted as statements about ordered Ramsey numbers. For instance, Erdős and Szekeres proved [4] that every sequence of at least (n − 1) 2 + 1 distinct numbers contains either an increasing subsequence of length n or a decreasing subsequence of length n. This result is implied by the bound r < (P n , K n ) ≤ (n − 1) 2 + 1, where P n is the n-vertex path imbued with the natural monotonic ordering: for any sequence of n distinct numbers x 1 , . . . , x n , color (i, j) red if x i < x j and blue otherwise.
Perhaps the simplest nontrivial family of ordered graphs from the perspective of ordered Ramsey theory is matchings, in which every vertex has degree 1. Conlon, Fox, Lee, and Sudakov provide a number of bounds for general matchings, for matchings satisfying certain properties, and for offdiagonal ordered Ramsey numbers involving matchings. Relevant to this paper is their work on bounding the largest possible value of r < (M, K 3 ), where M is a matching. They have the following result: Theorem 1.1 (Conlon, Fox, Lee, and Sudakov [3]). There are positive constants c 1 and c 2 such that for all even positive integers n, where the maximum is taken over all ordered matchings M on n vertices.
The upper bound in this theorem is in some sense trivial. Since every graph on n vertices embeds in the complete graph K n , and the ordered Ramsey number r < (K n , K 3 ) is equal to the Ramsey number r(n, 3), which has been asymptotically determined [1,5] to be Θ(n 2 / log n), it follows (as pointed out in [3]) that r < (M, K 3 ) = O(n 2 / log n) for a matching M on n vertices. However, this bound does not make use of any properties of matching graphs, only making use of the fact that every graph on n vertices can be embedded in K n . For this reason and perhaps other reasons, Conlon, Fox, Lee, and Sudakov hypothesize [3] that the upper bound can be improved to We contribute two results in the direction of this conjecture. We first look at the special case of ordered matchings where the edges do not cross. That is, for any two edges (i, j) and (k, l) with i < j and k < l, the intervals [i, j] and [k, l] are either disjoint or nested one inside the other. We call the matchings which satisfy this condition "parenthesis matchings", after the useful fact that these matchings correspond with balanced parenthesis sequences. Indeed, it is this correspondence which partially motivates our proof of the following theorem.
Theorem 2.8. For any > 0 there is a constant c such that every parenthesis matching M on n vertices has To state our second result, we must define the interval chromatic number of an ordered graph. Analogous to the chromatic number of an unordered graph, the interval chromatic number χ < (G) of a graph G is the minimum number of contiguous intervals into which the vertex set must be split so that each interval is an independent set in G.
Conlon, Fox, Lee, and Sudakov present a number of general results accompanied by much stronger specific results for matchings with small interval chromatic number [3]. In a similar spirit, we prove a sub-quadratic bound on r < (M, K 3 ) for random matchings with interval chromatic number 2.
Theorem 3.7. There is a constant c such that for every even n, if an ordered matching M on n vertices with interval chromatic number 2 is picked uniformly at random, then with high probability.
Observe that the statement is not probabilistic over bicolorings; rather, it is a true Ramsey-type result which applies to almost all matchings.

Roadmap
We outline the remainder of this paper. In Section 2, we achieve a nearly linear bound for matchings whose edges do not cross. In Section 3, we obtain a slightly sub-quadratic bound for random matchings with interval chromatic number 2. Finally, in Section 4 we outline possible directions for future research, describing a few of the many interesting questions about ordered Ramsey numbers which remain open.
Throughout the paper, we make no serious attempts to optimize constants.

Parenthesis Matchings
Earlier we defined "parenthesis matchings" as matchings for which the edges do not cross. We claim without proof that every parenthesis matching corresponds uniquely with a balanced parenthesis sequence-that is, a sequence of correctly matched open and close parentheses. The bijection is straightforward; each matched pair of parentheses corresponds with an edge in the matching. See Figure 1 for an example.
We start with perhaps the simplest nontrivial parenthesis matching, and work our way up to general parenthesis matchings. Define the nested matching graph N M k of size k to be the graph on Proof. The lower bound follows from a simple construction: color the ordered complete graph K 4k−2 such that {1, . . . , 2k − 1} and {2k, . . . , 4k − 2} form two red cliques, and all remaining edges are blue. Then there are no blue triangles, and no red edge (i, j) has |i − j| > 2k − 2, so there cannot be a red matching on 2k vertices.
For the upper bound, pick an arbitrary bicoloring of K 6k . Suppose the graph contains no blue copies of K 3 . If any vertex has blue degree at least 2k, then there is a red clique of size 2k, which must contain M . Otherwise, the number of blue edges is at most 6k 2 . Hence, the number of red edges is at least 12k 2 − 3k. Let E R be the set of red edges, and define a strict partial order on E R by (i, j) < (l, m) if l < i < j < m. We wish to show that there is a "chain" of edges e 1 , . . . , e k with e 1 < · · · < e k .
For the sake of contradiction, suppose the contrary, so every chain has length at most k − 1. Define a function L : E R → {1, . . . , k − 1} where L(e) is the longest chain ending at e. Observe that L −1 (n) is an "anti-chain" for each n ∈ [k − 1]. That is, for any e 1 , e 2 ∈ L −1 (n), we cannot have e 1 < e 2 nor e 2 < e 1 .
Applying the pigeonhole principle, fix some n such that |L −1 (n)| ≥ 12k. For 1 ≤ i ≤ 6k let a i be the minimum index j such that (i, j) ∈ |L −1 (n)|, and let b i be the maximum such index. Then It follows that there exist indices i < j with b i > a j . But then i < j < a j < b j , so edges (i, a i ) and (j, b j ) are comparable. This contradicts our claim that L −1 (n) is an anti-chain, so there must be a chain of length at least k. The edges in the chain comprise the red embedding of N M k into the graph.
We believe that the upper bound is far from optimal. In particular, we make the following conjecture.
Conjecture 2.2. For any positive integer k, The nested matching can be used to bound the corresponding ordered Ramsey numbers for a more general class of matchings. As we will build up more complex parenthesis matchings from simpler ones, we need a way to keep track of the growth of r < (M, K 3 ). One approach is the following lemma: is a balanced parenthesis sequence which induces some matching M , with Proof. Pick an arbitrary bicoloring of the complete graph on r < (N M k+t , K 3 ) vertices. Assume that there is no blue copy of K 3 . Then there is a red copy of N M k+t . Starting with the innermost edge of the matching and working outwards, delete as many matched pairs as necessary until there is space for a red copy of M k . Every deletion increases the number of inner vertices by at least one, so there will be space after at most r < (M k , K 3 ) steps. Save the current innermost matched pair (which will correspond to the parentheses around A k ), and continue deleting subsequent matches until there is space for a red copy of M k−1 (to the left of the saved match) and a red copy of M k+1 (to the right of the saved match). The number of deletions is at most max(r < (M k−1 , K 3 ), r < (M k+1 , K 3 )); save the new innermost match.
Repeating the above process k − 2 more times yields a complete red copy of M . Note that the process does not run out of matches, since only k matches are saved, and at most t matches are deleted.
In the above lemma, the Ramsey number of each matching M i is multiplied by a constant factor arising from the Ramsey number of a nested matching N M n . It is possible to decrease the dependence on the central matching M k , in exchange for larger constants on the remaining matchings and on the length of the matching.
Proof. Pick an arbitrary bicoloring of the ordered complete graph on t + 20(k + l + |M k |) vertices. Assume that there is no blue copy of K 3 . Let X denote the first 10(k + l + |M k |) vertices; let Y denote the next t vertices; and let Z denote the remaining 10(k + l + |M k |) vertices. Observe that Y contains a red copy of M k .
Suppose that there is a red copy of N M k+l in X ∪ Z, where the first k + l vertices are in X and the remaining k + l vertices are in Z. Then, just as in Lemma 2.3, we can start with the innermost matching and work outwards, deleting matchings to make space for red copies of M 1 , . . . , M k−1 and M k+1 , . . . , M 2k−1 . Only l matchings need be deleted, and by the end, the graph X ∪ Y ∪ Z contains a red copy of M . Now suppose the converse, so the maximum number of nested matchings from X to Z is less than k + l. As in Proposition 2.1, define the natural strict partial order on the red edges between X and Z. A set of nested edges forms a "chain", and the largest anti-chain contains no more than |X| + |Z| = 20(k + l + |M k |) red edges. We know that the red edges can be partitioned into less than k + l anti-chains, so the number of red edges between X and Z is at most 20(k + l + |M k |)(k + l), which we upper bound by 20(k + l + |M k |) 2 .
Thus, the number of blue edges between X and Z is at least 80(k + l + |M k |) 2 . Hence there must be a vertex v ∈ X with at least 8(k + l + |M k |) blue edges into Z. Since the graph was assumed to be blue K 3 -free, it follows that the set of blue neighbors of v forms a red clique of size 8(k + l + |M k |).
As |M | ≤ 8(k + l + |M k |), we conclude that the bicoloring contains a red copy of M .
Every parenthesis matching is in a bijection with an ordered, rooted tree. The above lemma allows us to bound the off-diagonal Ramsey number of the tree by the Ramsey numbers of all the branches off any path. Intuitively (and we will formalize the intuition later), this bound is strong on unbalanced trees and weak on well-balanced trees. For the latter case, we have the following simple lemma. While it is a special case of the above lemma aside from unimportant constant factors, we will use it for a different purpose (namely, well-balanced trees), so we state it separately for clarity.
Proof. Let t = r < (M, K 3 ) and let n be the number of vertices in matching M . Pick an arbitrary bicoloring of the ordered complete graph on [t + n + 1]. Suppose there are no blue triangles. Then there is a red copy of M in {2, . . . , t+1}. So if there is a red edge from 1 to any of {t+2, . . . , t+n+1}, we have found a red copy of M . Otherwise, every edge from 1 to {t + 2, . . . , t + n + 1} is blue, so {t + 2, . . . , t + n + 1} form a red clique of size n, which must contain the matching M .
With the above lemmas, we can prove a subquadratic bound on the Ramsey numbers of all balanced parenthesis matchings. Two convexity results are needed; we postpone their proofs to Appendix A.
In the following proof we'll use the bijection between parenthesis matchings on n vertices and ordered rooted trees of size s = n/2 + 1. The basic idea is to induct on tree size and decompose the tree into smaller trees by one of two methods, depending on the relative weights of the root's child subtrees.
Call an edge r-heavy if s child ≥ r · s parent , where s child is the size of the child subtree and s parent is the size of the parent subtree. If the inequality does not hold, call the edge r-light. Similarly call a vertex r-heavy or r-light if its parent edge is r-heavy or r-light, respectively.
If all children of the root are r-light for an appropriate choice of r (slightly less than 1), we apply the inductive hypothesis to each child separately, and use Lemma 2.5 to obtain a bound for the entire tree. Since every child subtree is a constant factor smaller than the entire tree, the lemma intuitively yields a sufficiently good recurrence.
If however the root has an r-heavy child, Lemma 2.5 does not suffice. Instead we trace a path of heavy edges from the root down, decomposing the tree into a number of branches, as well as possibly some subtrees at the tail end of the path. Here we use Lemma 2.4. We know that every branch is (1 − r)-light, so can afford to multiply the sum of Ramsey numbers of the branches by 20 in the lemma. We only know the tail subtrees to be r-light, which is why they are treated differently in the lemma.
Formalizing the above proof sketch requires some manipulation of inequalities and applications of Lemma 2.6 and Lemma 2.7. We work through these below. Proof. Let > 0. Set r = 1 − 23 −2/ , and set c = 23/(1 − r ). A parenthesis matching on n vertices uniquely corresponds with an ordered rooted tree of size s = n/2 + 1. We induct on the tree size s. If s = 1, the corresponding matching is the empty matching on 0 vertices, for which the claim is trivially true. Fix an ordered rooted tree of size s > 1, corresponding to a matching M . There are two cases which we will treat separately; either the tree root has an r-heavy child, or not.
Suppose that the tree root does not have an r-heavy child. Let s 1 , . . . , s k be the sizes of the child subtrees of the root. Let M 1 , . . . , M k be the matchings corresponding to the respective subtrees, and let t i = r < (M i , K 3 ) for each i ∈ [k]. With a slight abuse of notation, identifying the matchings with their parenthesis sequences, we have Lemma 2.5 provides the bound r < ((M i ), K 3 ) ≤ t i + 2s i + 1. Since the Ramsey number of a union of ordered graphs on disjoint intervals of vertices is subadditive, it follows that By the inductive hypothesis and Lemma 2.7 (using the assumption that every subtree is r-light), we have The last step follows since c was chosen to be sufficiently large.
The remaining case to consider is if the tree root has a heavy child. Then there is some path which starts at the root and consists entirely of heavy edges (possibly only one edge, or possibly more). Let ). By Lemma 2.5, the following bound holds for every matching M b i : By the inductive hypothesis, it follows that Reordering terms and absorbing the term 3 k i=k +1 s b i into the outer constant factor through the bound c ≥ 20, we get (1) To bound the first two terms of Equation 1, we observe that for each i ≤ k , subtree i is r-light, and therefore s b i ≤ rs. An application of Lemma 2.7, along with the bound cr + 23 ≤ c, gives For the remaining terms of Equation 1, observe that for any i > k , subtree i has an r-heavy sibling, so s b i is at most 1 − r times the parent's subtree size, and therefore at most (1 − r)s. We will use one of two approaches (below, A and B) depending on the cumulative weight of these subtrees. (3) Summing together the bounds from Equation 2 and Equation 3 and applying the most basic convexity bound, we get the desired bound B. If s h + k j=k +1 s b j < 23 −1/ s, then we are unable to bound s b i against s h + k j=k +1 s b j , but we know that the latter quantity is much smaller than s. So we instead use the weak bound (4) Now we combine Equation 2 with Equation 4, using the simple inequality (1 − x) 1+ + 23x 1+ ≤ 1 for x ∈ (0, 23 −1/ ), and obtain This completes the induction.

Random Matchings with χ < (M ) = 2
Recall that the interval chromatic number χ < (G) of an ordered graph G is the minimum number of contiguous intervals into which the vertex set must be split so that each interval is an independent set in G.
In this section, we show that for almost every matching M with interval chromatic number 2, the bound of O(n 2 ) on r < (M, K 3 ) can be beaten. More specifically, we exhibit a condition on M which is sufficient to guarantee an improved bound on r < (M, K 3 ), and then prove that a random matching with interval chromatic number 2 satisfies this condition with high probability.
The set of matchings on 2n vertices with interval chromatic number 2 is in bijection with the permutation group S n , and it is often notationally convenient to examine the permutation corresponding to a given matching. See Figure 2 for an example.  3.2. We say that a given permutation π ∈ S n contains an "exact pattern" ρ if ρ is an ordered subset of [n] and there are indices 1 ≤ i 1 < · · · < i k ≤ n, where k = |ρ|, such that π(i j ) = ρ(j) for all j.
We are interested in using exact patterns as a metric for the "intersection" of two permutations. Specifically, we make the following definition. Definition 3.3. Let π, σ ∈ S n be permutations. Define the "ordered intersection" of π and σ, denoted Int(π, σ), to be the largest k such that both π and σ share an exact pattern of length k.
In the theorem below, we do something slightly stronger than bounding the Ramsey number r < (M, K 3 ) for certain matchings M . Rather, we show that in a blue K 3 -free graph on 2n vertices, there is a tradeoff between finding a red copy of the matching M in the bipartite subgraph [1, n] ∪ [n + 1, 2n] and finding a large red clique (which of course contains every matching of that size) in [1, n] or symmetrically in [n + 1, 2n].
Proof. Fix a bicoloring C of the ordered complete graph on [2n], and suppose that it contains none of the hypothesized blue or red structures. Then in particular, for 1 ≤ i ≤ n + 1 − n 1/2+α , we know that there are no red copies of M between [1, n] and [n + i, n + i + n 1/2+α ).
The vertices v 1 (i), . . . , v f (i)−1 (i) demarcate f (i) blue segments in the adjacency matrix of [1, n] ∪ [n+1, n+2n 1/2+α ]. That is, for 1 ≤ j ≤ f (i) we have C(k, i+π(j)) is blue for all v j−1 (i) < k < v j (i) (where for convenience we set v 0 (i) = 0 and v f (i) (i) = n + 1). Treating C as an n × 2n 1/2+α matrix, each segment is in a distinct row, and the segments occupy distinct intervals of columns, covering a total of at least n − n 1/2+α columns. If any segment had length at least n 1/2+β , then some vertex would have n 1/2+β blue edges, so the coloring would contain either a blue triangle or a red K n 1/2+β . So henceforth we assume that every segment has length at most n 1/2+β . For each i ≤ n /4 let F (i) be the set of f (i) blue segments as defined above (see Figure 3 for an example). We seek to lower bound the number of blue edges in F (i) which are not contained in any F (i ) for i < i. So fix i < i ≤ n /4 . Suppose that there are k segments in F (i) which intersect with segments in F (i ).
Since each segment in F (i) is in a different row, as is each segment of F (i ), each of the k intersecting segments in F (i) intersects with a unique segment in F (i ). Suppose that s 1 , s 2 ∈ F (i) and t 1 , t 2 ∈ F (i ) where s 1 intersects t 1 and s 2 intersects t 2 . Then row(s 1 ) = row(t 1 ), and row(s 2 ) = row(t 2 ). And since the segments F (i) hit disjoint intervals of columns, as do the segments F (i ), we have columns(s 1 ) is "left" of columns(s 2 ) in the adjacency matrix if and only if columns(t 1 ) is "left" of columns(t 2 ). So the k intersecting segments define an exact pattern in both π +i , which describes the row indices of the segments F (i ), and π + i, which describes the row indices of the segments F (i). It follows that k is at most Int(π, π + i − i ), which is by assumption at most n (1− )(1/2+α) . Summing over all i < i, at most in (1− )(1/2+α) segments in F (i) intersect with previous segments.
We seek to show that for random permutations π and for any integer h, the intersection of π with the shifted permutation π + h is sublinear in the length of π with high probability. The general outline of the proof is as follows. We bound the expected number of long exact patterns contained in both π and π + h. To do so, we of course sum over all long exact patterns, splitting into two cases. If the exact pattern ρ has small intersection with ρ + h, we can straightforwardly obtain a good bound on the probability that ρ embeds into both permutations. However, if ρ has large intersection with ρ + h, we cannot do so. Instead we show that the number of such exact patterns is extremely small.
The following lemma formalizes the last step of the above outline.
Lemma 3.5. Fix positive integers n, k ≤ n, and h. Pick an exact pattern ρ of length k from [n] uniformly at random. Then the probability that the set intersection ρ ∩ (ρ + h) has size at least t, and there exists some permutation π ∈ S n such that ρ and ρ + h are both exact patterns in π, does not exceed 2 2k−t k k−t k! .
Proof. Observe that it is possible to pick an exact pattern uniformly at random by two independent choices: first, pick an unordered subset of [n] with size k. Second, pick some ordering for the subset. We will show that for any unordered subset U ⊆ [n] with size k such that |U ∩ (U + h)| ≥ t, if we pick an ordering on U uniformly at random and thereby induce an exact pattern ρ, then the probability that there exists a permutation π in which ρ and ρ + h are both exact patterns does not exceed 2 2k−t k k−t /k!. This will prove the lemma.
Fix any U ⊆ [n] with |U | = k and |U ∩ (U + h)| ≥ t. The number of elements a ∈ U such that a − h ∈ U does not exceed k − t, so U can be partitioned into k − t arithmetic progressions, each with common difference h.
Pick some permutation σ ∈ S k . This yields an ordering of U , in which the smallest element of U is placed in position σ(1), and so forth. Hence, an exact pattern ρ is induced. Suppose that the ordering is "compatible": that is, ρ and ρ + h are both exact patterns in some permutation π. Since ρ and ρ + h fix the order in π of the sets of elements U and U + h respectively, it must hold that U ∩ (U + h) has the same order in ρ and ρ + h. Pick any arithmetic progression {a + ih} m i=0 ⊆ U . We have that a + ih precedes a + (i + 1)h in ρ if and only if a + ih precedes a + (i + 1)h in ρ + h, or equivalently a + (i − 1)h precedes a + ih in ρ. So the arithmetic progression must either have a monotone increasing order or a monotone decreasing order in ρ.
The key observation was that for any a, b ∈ U where neither a nor b is the first term in its arithmetic progression, a precedes b in ρ if and only if a − h precedes b − h. We use this observation to bound the total number of compatible orderings. There are 2 k−t ways to assign a direction to each progression, either monotone increasing or monotone decreasing. Fix one such assignment, and suppose that m inc progressions are monotone increasing. There are at most 2 k ways to pick the subset of locations L inc ⊆ [k] to which the increasing-ordered progressions are assigned. It remains to pick an embedding of the increasing-ordered progressions in L inc , and an embedding of the decreasing-ordered progressions in [k] \ L inc . The two cases are symmetric, so we consider the increasing-ordered progressions.
For notational convenience, arbitrarily index the increasing-ordered progressions A 1 , . . . , A minc . Now define a map Φ : S |Linc| → L minc inc from embeddings of the increasing-ordered progressions into L inc (which are in bijection with the permutations S |Linc| ) to tuples (v 1 , . . . , v minc ), where v i is the index assigned to the first element of progression A i .
We claim that the restriction of Φ to compatible embeddings is injective. Pick two different compatible orderings of U , inducing exact patterns ρ 1 and ρ 2 , and assume for the sake of contradiction that Φ(ρ 1 ) = Φ(ρ 2 ). Suppose that j is the first index at which ρ 1 and ρ 2 differ. By assumption, the first term of each arithmetic progression has the same index in ρ 1 and ρ 2 . Therefore neither ρ 1 (j) nor ρ 2 (j) is a first term in its progression. Now observe that ρ 1 (j) precedes ρ 2 (j) in ρ 1 , but ρ 2 (j) precedes ρ 1 (j) in ρ 2 . Hence, ρ 1 (j) − h precedes ρ 2 (j) − h in ρ 1 , and in ρ 2 the opposite holds. However, ρ 1 (j) − h and ρ 2 (j) − h are both in the first j − 1 terms of ρ 1 , which are equal to the first j − 1 terms of ρ 2 . So one of the relative orderings is impossible! Contradiction, so the restriction of Φ is injective.
Thus, there are at most |L inc | minc ways to compatibly embed the increasing-ordered progressions into L inc , and similarly there are at most (k − |L inc |) k−t−minc ways to embed the decreasing-ordered progressions into [k] \ L inc . So the total number of compatible orderings is at most 2 k−t 2 k k k−t . Since the total number of orderings is k!, the result follows. Now we can prove our desired result on random permutations. Lemma 3.6. Fix some α > 0 and some positive integers n and h. If π ∈ S n is a permutation chosen uniformly at random, then The above lemma and Theorem 3.4 imply the main result of this section-a subquadratic bound on r < (M, K 3 ) for random matchings with interval chromatic number 2-as a corollary. So with high probability, every bicoloring of [2n] contains either a blue triangle or a red copy of M or a red clique of size at least 2m.

Future Work
Many open questions about the ordered Ramsey numbers of matchings remain. Most significant, perhaps, is the original question posed by Conlon, Fox, Lee, and Sudakov: does there exist some > 0 such that r < (M, K 3 ) ≤ n 2− for every ordered matching M on n vertices? Based on our Theorem 3.7, a number of natural intermediate questions arise. In particular, a reasonably modest step beyond random matchings with χ < (M ) = 2 would be the following: Conversely, we are curious how far from the truth the exponent 24 13 in our Theorem 3.7 is. It seems plausible that our argument can be optimized to produce a significantly better bound, and we do not know of any lower bounds for this class of matchings that come anywhere near this bound.
Regarding parenthesis matchings, we were unable to find a family for which r < (M, K 3 ) is superlinear, leaving a slight gap beneath our upper bound. Such a construction would be quite interesting to us.