Partitioning permutations into monotone subsequences

A permutation is $k$-coverable if it can be partitioned into $k$ monotone subsequences. Barber conjectured that, for any given permutation, if every subsequence of length $k+2 \choose 2$ is $k$-coverable then the permutation itself is $k$-coverable. This conjecture, if true, would be best possible. Our aim in this paper is to disprove this conjecture for all $k \ge 3$. In fact, we show that for any $k$ there are permutations such that every subsequence of length at most $(k/6)^{2.46}$ is $k$-coverable while the permutation itself is not.

A (possibly empty) subsequence of a permutation is monotone if it is either increasing or decreasing. A permutation is k-coverable if it can be partitioned into k monotone subsequences. If a permutation π is k-coverable, then so is every pattern of π. Barber [2] conjectured that, conversely, if every pattern of π of length at most k+2 2 is k-coverable, then so is π itself. This has been verified for k ≤ 2.
We say π is k-critical if every proper pattern of π is k-coverable while π itself is not. Barber's conjecture is equivalent to the assertion that critical permutations have length at most k+2 2 . A simple inductive argument shows that any k-critical permutation has length at least k+2 2 . So Barber's conjecture, if true, would be the best possible. In this paper, we show that the conjecture is false for every k ≥ 3. In fact the length of a k-critical permutation need not even be O(k 2 ) -we show that it can be as big as (k/6) 2.46 . Kézdy et al [6] showed that k-critical permutations cannot be arbitrarily long. In fact they showed that the length of a k-critical permutation is at most k O(k 6 ) . Feder and Hell [3] improved this to show that the length is at most k O(k 2 ) . We present a version of their argument. This exponential upper bound is currently the best known.
Wagner [8] showed that the problem of recognising k-coverable permutations is NPcomplete. Thus, unless NP = coNP, there is generally no simple reason why a permutation is not k-coverable.
Although this is a paper about permutations, the results all have graph-theoretic analogues. Given a permutation π, we can construct a graph -called a permutation graph -with vertex set [n] where i < j are adjacent if π(i) < π(j). Thus the problem of partitioning a permutation into increasing and decreasing subsequences is a special case of the problem of partitioning the vertex set of a graph into cliques and independent sets. This is called graph cocolouring [7]. One special property of permutation graphs is that they are perfect. As we will mention in the text, some of our results generalise to perfect graphs, and some to general graphs.
The plan of the paper is as follows. In Section 2 we make some simple remarks, and then in Section 3 we give our first examples that disprove Barber's conjecture. In Section 4, which is really the heart of the paper, we show how to construct long k-critical permutations for general k. In Section 5, we show that Barber's conjecture is true in the special case of separable permutations. In Section 6, we discuss upper bounds. And finally in Section 7 we mention some open problems.

Initial remarks
For completeness, we start with the following easy fact. It is equivalent to the assertion that permutation graphs are perfect, and is a special case of Dilworth's theorem. Lemma 1. Let π be a permutation, and suppose the length of the longest increasing subsequence of π is s. Then π can be partitioned into s decreasing subsequences.
Similarly, if the length of the longest decreasing subsequence of π is r, then π can be partitioned into r increasing subsequences.
Proof. By symmetry, it suffices to prove the first part.
For 1 ≤ t ≤ s, let D t be the set of i ∈ [n] such that the longest increasing subsequence of π ending at π(i) has length t. The D t certainly partition [n]. Moreover, if i < j and π(i) < π(j), then we can get an increasing subsequence ending at π(j) by appending π(j) to an increasing subsequence ending at π(i). So each D t must be decreasing, as needed.
Note that this result is sharp: we cannot hope to partition π into less than s decreasing subsequences.
We will now use the previous lemma to say something about partitions into monotone subsequences. For a permutation π and nonnegative integers r, s ∈ N, we say π is (r, s)coverable if π can be partitioned into r increasing sequences and s decreasing sequences. Thus π is k-coverable iff π is (r, s)-coverable for some r + s = k. We say π is (r, s)-critical if every proper pattern is (r, s)-coverable while π itself is not.
We write D(π) ⊆ N 2 for the set of pairs (r, s) such that π is not (r, s)-coverable.
We remark that the graph-theoretic analogue of Lemma 2 is true for perfect graphs by the same proof, but false for general graphs.
Let us write C(k) for the length of the longest k-critical permutation, and C(r, s) for the length of the longest (r, s)-critical permutation. We will later see that C(k) and C(r, s) are always finite.
We certainly have C(r, s) ≥ (r + 1)(s + 1), since any (r, s)-critical permutation is at least this long. Lemma 1 shows that we get equality C(r, s) = (r + 1)(s + 1) when r = 0 or s = 0. One can show that C(1, 1) = 4 = (1 + 1)(1 + 1) as follows. A permutation is (1, 1)-coverable iff its graph is a split graph -that is, a graph whose vertex set can be partitioned into a clique and an independent set. A graph is split iff it does not contain a 5-cycle, a 4-cycle, or the complement of a 4-cycle as an induced subgraph [4]. Finally, no permutation graph contains an induced 5-cycle, since it is not perfect, so C(1, 1) = 4.
Similarly, we have C(k) ≥ k+2 2 since any k-critical permutation is at least this long. It is easy to see that C(0) = 1 = 0+2 2 and C(1) = 3 = 1+2 2 . Jørgensen [5] characterised the graph analogues of 2-critical permutations, and from this characterisation we get C(2) = 6 = 2+2 2 . (This again uses the fact that permutation graphs do not contain induced long odd cycles.) As we will see, we never get equality in C(k) ≥ k+2 2 or C(r, s) ≥ (r + 1)(s + 1) except for the cases listed above. We now present some specific examples of critical permutations that are a bit longer than expected. We will later use these to construct infinite families of critical permutations, but the exact structure of these starting examples will be irrelevant. These examples and many more were found by computer search.
This permutation has a clear asymmetry to it: the longest decreasing subsequence has length 6, but the longest increasing subsequence has length 3. The latter property will be useful later.

Combining critical permutations
We will now explain the main idea of this paper: combining critical permutations to get new critical permutations. The next lemma is the simplest example of this.
Note that we would get equality above if we replaced C(r, s) with (r + 1)(s + 1) throughout.
Proof. By symmetry, it suffices to prove the first part.
Now consider a proper pattern of π ⊖ σ. Without loss of generality it is of the form τ ⊖ σ where τ is a proper pattern of π. By assumption, τ is (r 1 , s)-coverable. It remains to check that σ is (r 2 + 1, s)-coverable. Pick some term in σ. By assumption, σ minus this term is (r 2 , s)-coverable, so by putting in the term as an increasing subsequence of length 1, σ is (r 2 + 1, s)-coverable, as needed.
We now look more closely at direct sums, and in particular at D(π ⊕ σ). First note two properties of D(π) ⊆ N 2 : it is finite, and downward closed, in the sense that if r ′ ≤ r, s ′ ≤ s, and (r, s) ∈ D(π), then (r ′ , s ′ ) ∈ D(π). We will call a subset of N 2 with these two properties a downset. Our prototypical downset is the 'triangle' Given downsets A, B, we let A ⊕ B be the downset consisting of (r, s) such that whenever s = s 1 + s 2 , either (r, s 1 ) ∈ A or (r, s 2 ) ∈ B. We can think of this as merging A and B column by column. We have |A ⊕ B| = |A| + |B|. Similarly we let A ⊖ B consist of (r, s) such that whenever r = r 1 + r 2 , either (r 1 , s) ∈ A or (r 2 , s) ∈ B. Here A and B are merged row by row, and again |A ⊖ B| = |A| + |B|. The following lemma motivates these definitions.
Given a permutation π and a downset A, we say π is A-coverable if π is (r, s)-coverable for some (r, s) ∈ A. As before, we say π is A-critical if every proper pattern of π is Acoverable while π itself is not. Thus π is k-coverable iff π is T (k)-coverable, and k-critical iff π is T (k)-critical.
If π is A-critical and moreover A = D(π), then we say π is A-minimal. For example, π is T (k)-minimal iff π is k-critical and also (r, s)-coverable whenever r + s = k + 1. By inspection, π 12 is T (3)-minimal.
Let M(k) be the length of the longest T (k)-minimal permutation, so that M(k) ≤ C(k) and M(3) ≥ 12. We introduced the notion of A-minimal permutations in order for the following lemma to work.
Repeated application of Lemma 6 starting from M(3) ≥ 12 gives the following.

Lemma 8. For any positive integers a, b, c, d, we have
Note that, as before, the inequality becomes an equality if we replace S(r, s) with (r + 1)(s + 1).
There are three things to check. In each case, we use the fact that an increasing subsequence of π ⊗ τ is obtained by picking an increasing subsequence I of π and then an increasing subsequence of τ for every copy of τ corresponding to a point in I.
Firstly, we claim that π ⊗τ is (0, bd)-coverable. Simply pick b decreasing subsequences to cover π and use each one d times to cover the corresponding copies of τ .
Secondly, we claim that π ⊗ τ is not (ac − 1, bd − 1)-coverable. Suppose it were. Each copy of τ must either have at least c increasing subsequences or d decreasing subsequences passing through it. Thus we get two sets C and D which cover all of π. Because π is not (a − 1, b − 1)-coverable, by Lemma 1 either C has a decreasing subsequence of length a or D has an increasing subsequence of length b. In the first case, we get a contradiction as we used less than ac increasing subsequences to cover each point of C c times. The second case is analogous.
Finally, we claim that any proper subsequence of π ⊗ τ is (ac − 1, bd − 1)-coverable. Suppose we remove a term from the ith copy of τ . First pick a (0, b)-covering of π, and make d − 1 copies of each of these b subsequences. Then pick an (a − 1, b − 1)-covering of π minus the ith term, making 1 copy of each decreasing sequences and c copies of each increasing sequence. Now each copy of τ except the ith one has either d − 1 + 1 = d decreasing sequences through it, or c increasing seqeunces and d−1 decreasing sequences; either way it is fully covered. So far we have used (a − 1)c = ac − c increasing sequences and b(d − 1) + b − 1 = bd − 1 decreasing sequences, so we have c − 1 increasing sequences left. This is precisely enough to cover the the ith copy since it already has d−1 decreasing sequences through it and it is missing one term.
Proof. We first show that if τ is (r, s)-critical, then there exists an r + s-critical permutation containing τ as a pattern. It follows that C(r + s) ≥ C(r, s).
Let L be a skew sum of r copies of 1 . . . N, and let R be a direct sum of s copies of N . . . 1, where N > r + s. For any permutation σ, we claim that (L ⊕ σ) ⊖ R is r + s-coverable iff σ is (r, s)-coverable.
If σ is (r, s)-coverable, then we extend the r increasing subsequences to cover L and the s decreasing subsequences to cover R. Conversely, suppose (L ⊕ σ) ⊖ R is r + scoverable. Each of copy of 1 . . . N in L must have at least one increasing subsequences through it, since even r + s decreasing subsequences would not be enough. Moreover, the r copies cannot share any increasing subsequences, by construction. So we need at least r increasing subsequences, and similarly at least s decreasing subsequences. So (L ⊕ σ) ⊖ R must in fact be (r, s)-coverable, and hence σ is (r, s)-coverable.
For the second part, we show that if τ is k-critical, then there exists a k + 1-critical permutation containing τ as a proper pattern. The key is to observe that for any permutation σ, the permutation (1 . . . k + 2) ⊖ σ is k + 1-coverable iff σ is k-coverable. Indeed a k + 1-covering of (1 . . . k + 2) ⊖ σ must use one increasing sequence to cover 1 . . . k + 2, and then we are left to k-cover σ.
Thus as before we get a k + 1-critical permutation π containing τ as a pattern. We cannot have π = τ since τ itself is k + 1-coverable.
We remark that ⊕, ⊖, and ⊗ can be defined on arbitrary graphs. Thus all results in this section remain true in the more general setting of graph cocolouring, with the same proofs, except for Lemma 8 which only remains true for perfect graphs. If we instead consider the problem of partitioning the vertex set of a graph into two sets A and B such that A has no independent set of size r + 1 and B has no clique of size s + 1, then all the results in this section remain true, including Lemma 8, with similar proofs.

Separable permutations
We were able to construct long critical permutations using skew sums, direct sums, and tensor products, but only given some base cases. We now explain this phenomenon.
A permutation is said to be separable if it can be obtained from the length-one permutation (1) by repeatedly taking skew sums and direct sums. For example, 2 1 4 3 is separable but 3 1 4 2 is not. It can be shown that π ⊗ σ is separable if π and σ are, by inducting on π and using the right-distributivity (π ⊕ τ ) ⊗ σ = (π ⊗ σ) ⊕ (τ ⊗ σ), We will classify k-critical and (r, s)-critical separable permutations, showing that they all have lengths k+2 2 and (r + 1)(s + 1), respectively. This explains why our lemmas about ⊕, ⊖, and ⊗ alone are not enough to get non-trivial lower bounds on C(k) and C(r, s). The proof crucially relies on our more general notion of being critical with regard to a downset.
Lemma 11. Given a downset A, if a direct sum π ⊕ σ of permutations is A-critical, then A = B ⊕ C for some B, C such that π is B-critical and σ is C-critical.
Similarly, if a skew sum π ⊖ σ is A-critical, then A = B ⊖ C for some B, C such that π is B-critical and σ is C-critical.
Proof. By symmetry, it suffices to prove the first part.
Since π ⊕σ is not A-coverable, we have A ⊆ D(π ⊕σ) = D(π)⊕D(σ), using Lemma 4. It follows that there exist downsets B, C such that B ⊆ D(π), C ⊆ D(σ), and A = B ⊕C -the idea is to start from B 0 := D(π), C 0 := D(σ), and successively remove points (r, s) from them, starting from larger r. Now we claim that π is B-critical. We know that π is not B-coverable. Moreover, if B ⊆ D(τ ) for some proper pattern τ of π, then A = B ⊕ C ⊆ D(τ ⊕ σ) contradicting the assumption that every proper pattern of π ⊕ σ if A-coverable. So π is B-critical.
Similarly σ is C-critical, as needed.
Proof. The first claim follows from Lemma 11 and induction on π. The other claims are special cases of the first.
We can say a bit more here. Namely, if π is a separable permutation, then |D(π)| is the length of π, and π is A-critical iff A = D(π). This gives a procedure to generate all separable A-critical permutations: consider all ways to write A as B ⊕ C or B ⊖ C, and for each one recursively find the corresponding separable critical permutations.
We remark that if A ⊕ B is of the form {0 . . . r} × {0 . . . s} for some r, s, then so are A and B. Thus if we were only interested in (r, s)-critical separable permutations, then we would not need to consider general downsets. However, in order to understand k-critical separable permutations, we must consider general downsets.
A graph built from the single-vertex graph using the graph analogues of ⊕ and ⊖ is called a cograph. It is immediate that a graph is a cograph iff it corresponds to a separable permutation. Thus Corollary 12 characterises cocolourable cographs in terms of forbidden subgraphs.

Upper bounds
We now prove that C(r, s) and C(k) are always finite. The following combinatorial lemma will be key. It can be deduced from the sunflower lemma [1], but we will get a slightly better bound using an ad-hoc argument. The argument we present is essentially the same as the argument used to prove Theorem 3.1 in [3].
Lemma 13. Fix r, d ∈ N with r ≥ 2 and let U be any nonempty set. Suppose for every i ∈ U, we have a partial Boolean function P i : U \ {i} → [2]. Suppose the P i are all close in the sense that for any i, j ∈ U, there are at most d points k ∈ U \ {i, j} where P i (k) = P j (k).
Then there exists a least integer N = N(r, d) such that if |U| > N, then we can find a total Boolean function P : U → [2] such that for any S ⊆ U of size at most r + 1, there exists i ∈ U \ S such that P (k) = P i (k) for all k ∈ S.
Proof. We first explain why N(r, d) ≤ 4r d+1 . Fix an arbitrary i ∈ U. The idea is to extend P i to all of U and then try change it a bit to get the desired P . We show that this works when U is sufficiently large. Let p ∈ [2] be arbitrary. For any T ⊆ U \ {i} we get a Boolean function P T : U → [2] by taking P T (i) = p, and P T (j) = P (j) iff j ∈ T for j ∈ U \ {i}. We may assume that this P T does not satisfy the desired properties, i.e. that there exists S T ⊆ U of size at most r + 1 such that no P k agrees with P T on all of S T . We must have i ∈ S T or S T ∩ T = ∅ since P i agrees with itself. For j ∈ S T \ (T ∪ {i}), say T ∪ {j} is a child of T .
Let F p be the minimal collection of subsets of U \ {i} such that ∅ ∈ F p , and if T ∈ F p has size at most d, then all its children are also in F p . Let T p = T ∈Fp T , and let We claim that |T | ≤ 4r d+1 . It suffices to prove that |T p | < 2r d+1 . We have |T p | ≤ |F p |, since each new set in F p introduces at most one new element. By construction, F p has 1 element of size 0, at most r elements of size 1, etc, and at most r d+1 elements of size d + 1. So |F p | ≤ 1 + r + . . . r d+1 < 2r d+1 , as desired.
Finally, suppose T is not all of U. Pick j ∈ U \ T . Consider P j . Let p = P j (i). Let T ∈ F p be maximal such that P j (k) = P i (k) for all k ∈ T . Since P i and P j are close, T has size at most d. By definition of S T , we know that P j (k) = P T (k) for some k ∈ S T . Now this contradicts maximality of T , since the child T ∪ {k} of T is also in F p . This finishes the proof that N(r, h) ≤ 4r d+1 .
Let us now prove that N(r, h) ≤ (4r) d/2+1 . We may assume that, for some i, j, the functions P i and P j differ in exactly d points, since otherwise we could take d smaller. Fix such i and j and write D for the set where they differ. Given Q : D → [2], we have that Q differs with P i or P j in at least d/2 points. Say Q differs with P i in at least d/2 points. As before, for p ∈ [2] and T ⊆ U \ (D ∪ {i}), we define P T : U → [2] by P T (k) = Q(k) for k ∈ D, P T (i) = p, and P T (k) = P i (k) iff k ∈ T for k ∈ U \ (D ∪ {i}). The rest of the proof proceeds as before: for every Q and p we construct a tree of sets T where this time the maximum size allowed is d/2 + 1 instead of d + 1. The set T ⊆ U we end up with has size at most d + 2 + 2 d+1 (r + r 2 + . . . + r ⌊d/2⌋+1 ) ≤ 2 d+1 · 2r d/2+1 = (4r) d/2+1 , as needed.
In the other direction, we have N(r, d) ≥ r d/2 . To see this, let U be a rooted tree with d/2+1 layers where every non-leaf node has r children. For i ∈ U, define P i : U \{i} → [2] by taking P i (j) = 2 iff j lies on the path from the root to i. The P i are close since each one takes the value 2 at most d/2 times. Suppose there were a P : U → [2] with the desired properties. Then P must take the value 2 at the root, since all P i 's do, and 1 at all leaves. We get a contradiction by looking at the deepst vertex where P takes the value 2.
We now explain how Lemma 13 relates to permutations.
Proof. Let π be an (r, s)-critical permutation of length n. We wish to show that n ≤ N(r, rs). For any i ∈ [n], π minus its ith term is (r, s)-coverable, so pick an (r, s)-covering. Define P i : [n] \ {i} → [2] by P i (j) = 1 if j is in an increasing subsequences and 2 if it is in a decreasing subsequence. Given i, j ∈ [n], suppose there are 2rs + 1 points k ∈ [n] \ {i, j} where P i (k) = P j (k). Then we can find a set S ⊆ [n] \ {i, j} of size rs + 1 such that, without loss of generality, P i (k) = 1 and P j (k) = 2 for all k ∈ S. Now S is coverable by r increasing sequences, since P i comes from an (r, s)-covering, and also by s decreasing sequences, since P j comes from an (r, s)-covering. This is a contradiction. So the P i are all close.