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Can it be done in linear or at least subquadratic time?

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    $\begingroup$ Can you format your problem and give more details/context? Also, any bounds on $x_i$? Are they non-negative? Is weakly polynomial (i.e sth like $O(n \log M)$ where M is the maximum $x_i$ useful? $\endgroup$ Commented Aug 11, 2023 at 7:40
  • $\begingroup$ Can anyone find a sub-quadratic-time exact algorithm for the following (perhaps easier) decision problem? Given a sequence $x_1, x_2, \ldots, x_n$ of rationals in increasing order, do there exist $i, j$ with $i < j$ such that $x_j - x_i < \sqrt{j-i}$? $\endgroup$
    – Neal Young
    Commented Dec 13, 2023 at 16:04

2 Answers 2

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Getting an algorithm which runs in time $O(n \log n \log(1/\epsilon))$ for a $\epsilon$-approximation is pretty standard. We will require the following data structure, which I'll call a ``line container''.

Line container: There is a data structure that maintains a set $\mathcal{L}$ of lines $L: \mathbb{R} \to \mathbb{R}$, and supports the following operations, in amortized time $O(\log n)$.

  • Insert($L$): Insert a line $L$ into $\mathcal{L}$.
  • Query($x \in \mathbb{R}$): Return $\max_{L \in \mathcal{L}} L(x)$.

By binary search, it suffices to check for each real number $k > 0$, whether there is a pair $i < j$ such that $(x_j-x_i)^2/(j-i) > k$. This rearranges to $-2x_i \cdot x_j + x_i^2 + ki > kj - x_j^2$. This suggests the following algorithm.

  1. For $i = 1, 2, \dots, n-1$.
  2. Insert the line $L_i(x) = -2x_i \cdot x + x_i^2 + ki$ into $\mathcal{L}$.
  3. Check if Query($x_{i+1}$) $> k(i+1) - x_{i+1}^2$.

Sketch for construction of line container: For any set $\mathcal{L}$, the function $f(x) = \max_{L \in \mathcal{L}} L(x)$ is some upper convex hull-like object. When a line is inserted, binary search to find where it intersects the convex hull, and update the convex hull. This is efficient because a line intersects this convex hull in at most two points.

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  • $\begingroup$ 3 should be check if $\text{Query}(x_{i+1}) > k(i+1) - x^2_{i+1}$ ? $\endgroup$ Commented Aug 11, 2023 at 17:55
  • $\begingroup$ Yes, that's correct. I fixed the typo. $\endgroup$
    – yangpliu
    Commented Aug 11, 2023 at 20:32
  • $\begingroup$ I wonder if one can adopt parametric search to avoid that $\epsilon$. $\endgroup$
    – Chao Xu
    Commented Dec 12, 2023 at 16:36
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Here is a different algorithm that outputs a $1-\epsilon$ multiplicative approximation in time $O(n \log^2(n)/\epsilon)$, but works for more general settings, (if we want to find the maximum of $\frac{f(|x_j - x_i|)}{j-i}$ for any increasing function $f(t)$ such as $f(t) = t$). The computational model is that two real numbers can be compared in $O(1)$ time.

Subroutine: Create a data structure $D$ which takes in a contiguous interval $I = [a,b] = \{a, a+1, \ldots, b\}$ and outputs $i,j = \text{argmax}_{i,j \in I} |x_j - x_i|$.

Assume the subroutine exists for now. The idea is now to try all possible values of $j-i$ rounded to the nearest power of $(1+\epsilon)$. For every fixed power $(1+\epsilon)^k$, we can invoke the subroutine $O(n)$ times to find the largest value of $(x_j - x_i)^2$, conditioned on $j-i \approx (1+\epsilon)^k$. This gives the desired approx. guarantees.

In more detail, we iterate over $ t = 1, \lfloor (1+\epsilon) \rfloor, \lfloor(1+\epsilon)^2 \rfloor, \lfloor(1+\epsilon)^3\rfloor, \ldots, $. For a fixed power, we invoke the subroutine at most $n$ times on the intervals $[1, t+1], [2, t+2], \ldots, $. Now let $i^* < j^*$ denote the OPT pair and suppose $(1+\epsilon)^{t^*}$ is the nearest power larger than $j^*- i^*$. We return some pair $j' > i'$ when we query $[i^*, i^* + (1+\epsilon)^{t^*}]$ and $(x_j' - x_i')^2 \ge (x_j^* - x_i^*)^2$, $(j^* - i^*) \ge (1-\epsilon)(j' - i')$. In other words, we find a value that is at least $(1-\epsilon)$OPT.

Thus it remains to analyze the runtime of the subroutine. We show that we can construcut such a datstructure in $O(n \log n)$ time where each query times $O(\log n)$ time.

$D$ is simply a balanced binary tree over the interval ${1, \ldots, n}$ where each node stores the min and max $x_i$ over all $i$ that are `covered' by the node. To answer any given query interval $I$, we can break it down into $O(\log n)$ contiguous intervals $I = I_1 \cup I_2 \ldots$ (this is a standard procedure for ex. interval trees), query eeach of them, and just return the max - min over all $x$'s returned.

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