Order-isomorphic twins in permutations

Abstract

Let a 1 , . . . , a n be a permutation of [n]. Two disjoint order-isomorphic subsequences are called twins. We show that every permutation of [n] contains twins of length Ω(n 3/5 ) improving the trivial bound of Ω(n 1/2 ). We also show that a random permutation contains twins of length Ω(n 2/3 ), which is sharp.In this paper we regard permutations as sequences of symbols, devoid of any group-theoretic meaning. So, for us a permutation on a finite set Σ is a sequence of elements of Σ in which each element of Σ appears exactly once. We call a subsequence of a permutation subpermutation. For instance, 135642 is a permutation of [6], and 1562 is a subpermutation inside, which itself is a permutation of {1, 2, 5, 6}. We denote permutations by bold letters.Throughout the paper we consider only the permutations of finite sets of natural numbers. We say that permutations a = (a 1 , . . . , a L ) and b = (b 1 , . . . , b L ) are order-isomorphic if (a i < a j ) ⇐⇒ (b i < b j ). For example, 1562 is order-isomorphic to 1342.We call a pair of subpermutation a, b of c twins if a and b are order-isomorphic and disjoint (do not contain the same symbol). For example, 152 and 364 are twins in 135642 of length 3. We denote by t(n) the largest integer such that every permutation of [n] contains a pair of twins of length t(n).The problem of estimating t(n) was raised by Gawron [5], who observed that t(n) ≥ (n 1/2 − 1)/2 follows from the Erdős-Szekeres theorem, and that t(n) = O(n 2/3 ) follows from the first moment method. He further conjectured that t(n) = Ω(n 2/3 ). This is not known even for random permutations: the best result is due to Dudek, Grytczuk, Ruciński [4] who showed that a random permutation almost surely contains twins of length Ω(n 2/3 / log 1/3 n).In this short note, we give a first non-trivial lower bound on t(n), and remove the logarithmic factor from the Dudek-Grytczuk-Ruciński result.Theorem 1. For n ≥ 2, every permutation of [n] contains twins of length at least 1 8 n 3/5 . Theorem 2. A random permutation of [n] almost surely contains twins of length at least 1 80 n 2/3 , as n → ∞.In view of Gawron's result, Theorem 2 is sharp up to the constant factor.

On weak twins and up-and-down sub-permutations

“…We confirmed this conjecture in [7] (up to a logarithmic factor) for a random permutation. A refinement of our result (getting rid of the logarithmic factor) was then obtained by Bukh and Rudenko [4]. In the deterministic case, the tpnq ě Ωp ?…”

“…nq follows immediately from the famous result of Erdős and Szekeres [9] on monotone subsequences in permutations. Currently, the best lower bound tpnq " Ωpn 3{5 q is due to Bukh and Rudenko [4].…”

Variations on twins in permutations

“…Next we present the proof of an optimal bound that was also recently obtained by Bukh and Rudenko [6] and is based on some additional ideas provided by those authors.…”

“…The new idea, delivered by them was to replace estimation of the maximum degree of the auxiliary bipartite graph B by the estimation of the average degree (see the proofs below for details). Bukh and Rudenko wrote then a short note [6] which contains the improvement. In their version they switched to a less standard way of generating a random permutation by a Poisson point process on the unit square.…”