A new lower bound for deterministic pop-stack-sorting

Pop-stack-sorting was introduced by Avis and Newborn in [4]. It is a variation of the stack-sorting process first presented by Knuth in The Art of Computer Programming [14]. Pop-stack-sorting has been the subject of much research in recent years (see [2, 3, 5, 6, 7, 8, 10, 11, 12, 13]). For our purposes, a permutation is an ordered list of the elements of $\{1,2,...,n\}$, where each element appears once. We write permutations in the one-line notation, that is $12...n$. We consider the deterministic algorithm to which Defant refers as the function $\mathsf{Pop}$ [9]. The function $\mathsf{Pop}$ acts as follows: given a permutation $\sigma$, read the permutation from left to right and if there are any consecutive elements which are in decreasing order, reverse the order of those elements. For example, $\mathsf{Pop}(4\hskip 1.4457pt\underline{71}\hskip 1.4457pt\underline{83}\hskip 1.4457pt6\hskip 1.4457pt\underline{952})=4\hskip 1.4457pt\underline{17}\hskip 1.4457pt\underline{38}\hskip 1.4457pt6\hskip 1.4457pt\underline{259}$. Let $\mathcal{D}_{n}^{\mathsf{Pop}}$ be the average number of iterations of $\mathsf{Pop}$ that it takes to transform a uniformly randomly chosen permutation of length $n$ to the identity permutation $12...n$. In [9], Defant conjectured that

and asked, in that paper and at the Banff workshop on Analytic and Probabilistic Combinatorics in November 2022, if there exists $c\in\left(\frac{1}{2},1\right]$ such that

It was first proven in [16] that $c\leq 1$, and reproven in both [3, 1] using more direct arguments. The lower bound of $c=\frac{1}{2}$ was not proven by Defant in that paper, but we believe that his conjecture was based on an argument similar to a combination of Proposition 2 and Lemma 7.

In [15], Lichev gave a lower bound of $c=0.503$. We improve this bound by proving the following theorem.

For brevity, we adopt the notation $\sigma_{t}:=\mathsf{Pop}^{t}(\sigma)$ throughout the paper. Following the standard notation, the position of the element $k$ in the permutation $\sigma_{t}$ is denoted $\sigma_{t}^{-1}(k)$. An increasing run in $\sigma_{t}$ is a maximal set of elements $\{p_{1},p_{2},...,p_{k}\}$, $k\geq 2$, such that $p_{1}<p_{2}<\dots<p_{k}$ and $\sigma_{t}^{-1}(p_{1})=\sigma_{t}^{-1}(p_{2})-1=...=\sigma_{t}^{-1}(p_{k})-k+1.$ A decreasing run is defined similarly. Note that, by this definition, we do not refer to single elements as increasing (or decreasing) runs of length one. An element is in the interior of an increasing (or decreasing) run if it is an element of an increasing (or decreasing) run, but neither the first nor the last element of that run. We also refer to iterations of $\mathsf{Pop}$ simply as iterations, and the $t$th iteration as iteration $t$.

Defant noted without proof in [9] that in the image of a permutation under $\mathsf{Pop}$, there are no decreasing runs of length more than three.

Since all decreasing runs in a pop-stack-sorted permutation are of length two or three, and we are intensely concerned with the manner in which individual elements move, we name each possible way of moving or holding position for an element during an iteration after the first.

We say that $\mathsf{Pop}$ causes two elements to switch if they are in reverse order but not part of a decreasing run of length three. We say that $\mathsf{Pop}$ causes three elements to pivot if they are in a decreasing run of length three, where the outermost elements both move and the middle of the three elements holds its position. We call the middle element of a pivot the center of the pivot. If an element is in the interior of an increasing run, or at the start of an increasing run in position $1$, or the end of an increasing run in position $n$, it has the same position after $\mathsf{Pop}$ is applied, and we say that the element stops.

In any given iteration after the first, there are four ways to move: switch left, switch right, pivot left, pivot right, and two ways to hold position: be the center of a pivot, or stop.

Consider the example $\sigma=471836952$ from above. Then we have $\sigma_{1}=\mathsf{Pop}(4\hskip 1.4457pt\underline{71}\hskip 1.4457pt\underline{83}\hskip 1.4457pt6\hskip 1.4457pt\underline{952})=4\hskip 1.4457pt\underline{17}\hskip 1.4457pt\underline{38}\hskip 1.4457pt6\hskip 1.4457pt\underline{259}.$ Applying $\mathsf{Pop}$ again, we have $\sigma_{2}=\mathsf{Pop}(\underline{41}\hskip 1.4457pt\underline{73}\hskip 1.4457pt\underline{862}\hskip 1.4457pt5\hskip 1.4457pt9)=\underline{14}\hskip 1.4457pt\underline{37}\hskip 1.4457pt\underline{268}\hskip 1.4457pt5\hskip 1.4457pt9.$ During iteration 2, the pair 1 and 4 and the pair 3 and 7 switch and the elements 2, 6, and 8 pivot, with 6 as the center of the pivot. The only elements which hold their position during iteration 2 are 6, 5, and 9; 6 is the center of a pivot and both 5 and 9 stop. Note also that the element 6, which is the center of a pivot in iteration 2, stops during iteration 1.

We begin with two observations regarding elements which stop. Let $\sigma$ be a permutation of length $n\geq 3$.

The following lemma, which we refer to as the pop-stop lemma, gives some insight into the circumstances under which an element stops during iteration $t$. Such elements are important because pivots (after the first iteration) are always preceded by a stop. Fix some $1\leq k\leq n$ in a permutation $\sigma$ of length $n$, where $\sigma^{-1}(k)>k$ so that the sorting process eventually causes $k$ to move to the left. Intuitively, $k$ almost always moves to the left until it becomes a left-to-right maximum unless it encounters an element smaller than $k$ to its left and stops or it is the center of a pivot. However, both cases necessitate stops and therefore there must be smaller elements to the left of $k$, and such elements never move to the right of $k$. If $k$ stops during iteration $t$ where $t$ is large, there must be many elements which are smaller than $k$ to its left, or it must already be a left-to-right maximum. Thus, there exists some $t$ which depends on $k$ but not on $\sigma$ such that $k$ cannot stop during iteration $t$ unless it is a left-to-right maximum.

In order to be in the center of a pivot (after the initial iteration), an element must first stop. If an element is in position $i$ in $\sigma_{2i-4}$ and it is in the interior of an increasing run, the element has at least $\left\lceil\frac{2i-4+1}{2}\right\rceil=i-1$ smaller elements to its left, so the pop-stop lemma guarantees that if it stops, then it is a left-to-right maximum. Since the center of a pivot has a larger element to its left, $\sigma_{2i-4}(i)$ is not the center of a pivot, and neither are any other elements which are within $i-1$ positions from either end of $\sigma_{2i-4}$.

The most an element can move in any iteration after the first is two positions by pivoting. Thus, for every two iterations, an element can move four positions, however, after those two iterations one more position at the end of the permutation cannot be the center of a pivot. Thus, we split the permutation into fifths; during roughly the first $\frac{2}{5}n$ iterations, an element can move about $\frac{4}{5}n$ positions, but in the following iterations $\frac{1}{5}n$ positions can not be centers of pivots. This can be leveraged to yield the following theorem.

We note that this bound could be improved by a single iteration for some values of $n,k,$ and $i$ if one were to consider the possibilities of the distance to travel after $2i-4$ iterations mod 5, but such considerations give at most a change of one in the bound at the cost of a more complicated statement.

Lichev proved the following lemma ([15], Observation 1 and Observation 2).

This lemma combined with Theorem 6 yields Theorem 1 almost immediately.

We note that the bound of $\frac{2}{5}n$ is asymptotically the correct upper bound for the possible number of pivots to the right which any element can make. We present a family of permutations in which the element $n$ pivots $\frac{2}{5}n-1$ times and reaches the final position after $\frac{3}{5}n$ iterations.

We do not believe this bound is attainable for both the elements $1$ and $n$ in the same permutation. The following construction gives a permutation where each of $1$ and $n$ pivot roughly $\frac{3}{8}n$ times.

We believe that the pop-stop lemma can be used in future arguments to improve this lower bound in two ways.

First, for every two pivots in an “early” iteration which can be shown to be unlikely, the pop-stop lemma guarantees an “extra” pivot which is prevented later. For example, if it were possible to show that the average number of pivots in the first $n/5$ iterations for the element which must travel the furthest was $n/10$, the pop-stop lemma gives a corresponding $n/20$ guaranteed switches at the end.

Second, our arguments in the proof of Theorem 6 use the pop-stop lemma to argue that there are no pivots near the left or right of $\sigma_{t}$. However, The lemma also implies that elements near the top or the bottom of $\sigma_{t}$ cannot be the center of pivots. For example, in $\sigma_{2i-4}$, if the element $i$ is in the interior of an increasing run, it has at least $\left\lceil\frac{2i-3}{2}\right\rceil=i-1$ smaller elements to its left, so it cannot be the center of a pivot in $\sigma_{2i-3}$ and beyond, as a pivot requires that the element $i$ must stop in iteration $t$ and have an element smaller immediately to its right in $\sigma_{t}$. One can picture that the pop-stop lemma guarantees a square where no centers of pivots can occur that starts one element from the edge of the permutation (for the elements in the first and last positions and the elements $1$ and $n$ can never be the center of a pivot) and closes in one element in each direction every two iterations (see Figure 3). As Example 8 shows, it is not impossible for elements to pivot every time in the first $\frac{2}{5}n-1$ iterations, but using these closing windows at the top and bottom might be one approach to show that it is rare in a uniformly randomly chosen permutation.

The authors are grateful to Colin Defant for introducing us to the problem and for some helpful suggestions and corrections. We are also grateful to Kelly Debure for lending us some time.