Color avoidance for monotone paths
- Department of Mathematics, Emory University
- ORCID iD: 0009-0003-7545-6504
- Department of Mathematics, Emory University
- ORCID iD: 0009-0004-4842-2939
- More about Cosmin Pohoata
- Department of Mathematics, Massachusetts Institute of Technology
- More about Dmitrii Zakharov
Editorial introduction
Color avoidance for monotone paths, Discrete Analysis 2025:23, 14 pp.
The famous Erdős–Szekeres theorem states that every sequence a1,…,ars+1 of rs+1 real (and without loss of generality distinct) numbers has an increasing subsequence of length r+1 or a decreasing subsequence of length s+1. One way to prove it is to colour each pair of integers mn with m<n red if am<an and blue if am>an. This gives us a red/blue colouring of the complete graph Krs+1. Defining a path m1,…,mk to be monotone if m1<⋯<mk, we see that it is sufficient to find a red monotone path of length r+1 or a blue monotone path of length s+1. To see that such a path must exist, we assign to each integer m a pair of integers (p(m),q(m)), where p is the length of the longest red monotone path that finishes at m and q is the length of the longest blue monotone path that finishes at m. One readily sees that if m<m′, then p(m)<p(m′) or q(m)<q(m′) (since the edge from m to m′ can be appended to one or other of the longest monochromatic paths to m). It follows that all the pairs (p(m),q(m)) are distinct, and since there are rs+1 of them, it must be possible for p(m) to take the value r+1 or for q(m) to take the value s+1.
In 2015 Po-Shen Loh formulated a beautiful conjecture that is still open. Suppose that we 3-colour the edges of Kn. If we look for a monochromatic monotone path, then the argument given above generalizes straightforwardly, but what if instead we try to find a path that uses at most two colours, or equivalently that avoids one of the colours? With a bit of thought, one finds examples where the longest color-avoiding monotone path has length n2/3, and the open question is whether this bound is best possible. If one imitates the proof technique described above, then one arrives at the following equivalent question. Define a sequence ((ri,si,ti))ni=1 of triples of positive integers to be 2-increasing if whenever i<j, at least two of the inequalities ri<rj,si<sj and ti<tj hold. Is it true that n must be at most m3/2 for every 2-increasing sequence of length n with every ri,si and ti belonging to {1,2,…,m}? There is a trivial upper bound of m2 (the pairs (ri,si) must be distinct). Loh observed that the triangle-removal lemma can be used to yield an upper bound of o(m2), and Gowers and Long improved that to m2−c for a very small absolute constant c>0, but the problem is still wide open.
This paper concerns generalizations of some of these ideas to hypergraphs. For this we first need to say what we mean by a “path” in a k-uniform hypergraph. The two commonest definitions are of loose paths, where consecutive edges intersect in exactly one vertex, and tight paths, where consecutive edges intersect in k−1 vertices. When k=2, these definitions coincide, but for k>2 they are very different. In this paper tight paths are considered. More precisely, the authors look at tight monotone paths, which are paths where one has a monotone sequence v1<v2<⋯<vr+k−1 of vertices and the path consists of all edges of the form {vi,vi+1,…,vi+k−1}.
With that definition in place, one can ask questions analogous to the Erdős–Szekeres theorem and to the conjecture of Loh. To express these questions in a unified way, the authors define Ak(n;r,s) to be the number N of vertices you need if for every r-colouring of the edges of the complete k-uniform hypergraph with vertex set {1,2,…,N} there is a tight monotone path that uses at most s of the colours. For example, the Erdős–Szekeres theorem implies that A2(n;2,1)=(n−1)2+1, and Loh’s question turns out to be whether A2(n;3,2) is always at most (n−1)3/2+1.
The question of generalizing the Erdős–Szekeres theorem was almost completely solved by Guy Moshkovitz and Asaf Shapira, who gave upper and lower bounds for Ak(n;r,1), for every k≥3 and r≥2, that were both of the form a (k−2)-fold iterated exponential applied to a number of order of magnitude nr−1. That is, they more or less determined how large N needs to be if for every r-colouring of the complete k-uniform hypergraph with vertex set {1,2,…,N} there is a monochromatic tight monotone path of length n.
This paper considers the colour-avoiding version of the problem. The main result is rather surprising: the authors show that the bound one obtains for the colour-avoiding version is not just better but much better, in that it beats the monochromatic version by an exponential. More precisely, the authors provide an upper bound for Ak(n;r,s), showing that if n≥3 and s>r/2, then one needs not a (k−2)-fold iterated exponential but a (k−3)-fold iterated exponential, applied to a suitable power of n that depends on r and s only. Note that when k=3 this implies that we do not need any exponentials, and indeed the first thing the authors do is prove that A3(n;3,2) is at most n9.
The paper concludes with some appealing open problems. The most obvious is whether the upper bound provided in this paper can be matched by a lower bound, at least for the number of exponentials needed.