An Improved FPTAS for 0-1 Knapsack

TL;DR

This paper presents an improved fully polynomial-time approximation scheme (FPTAS) for the 0-1 knapsack problem, reducing the runtime closer to the theoretical lower bound through advanced algorithmic techniques.

Contribution

The paper introduces a faster FPTAS for 0-1 knapsack using a multi-level number-theoretic approach and a greedy lemma to minimize unnecessary computations.

Findings

Runtime improved to  O(n+(1/)^{9/4})

Approaches the quadratic lower bound more closely

Enhances efficiency over previous FPTAS by Chan (2018)

Abstract

The 0-1 knapsack problem is an important NP-hard problem that admits fully polynomial-time approximation schemes (FPTASs). Previously the fastest FPTAS by Chan (2018) with approximation factor $1+\varepsilon$ runs in $\tilde O(n + (1/\varepsilon)^{12/5})$ time, where $\tilde O$ hides polylogarithmic factors. In this paper we present an improved algorithm in $\tilde O(n+(1/\varepsilon)^{9/4})$ time, with only a $(1/\varepsilon)^{1/4}$ gap from the quadratic conditional lower bound based on $(\min,+)$-convolution. Our improvement comes from a multi-level extension of Chan's number-theoretic construction, and a greedy lemma that reduces unnecessary computation spent on cheap items.

Tables

Table 1. Table 1 : FPTASs for 0-1 knapsack

$O\left(n\log n+{\left(\frac{1}{\epsilon }\right)}^4\log \frac{1}{\epsilon }\right)$	Ibarra and Kim [6]	1975
$O\left(n\log \frac{1}{\epsilon }+{\left(\frac{1}{\epsilon }\right)}^4\right)$	Lawler [13]	1979
$O\left(n\log \frac{1}{\epsilon }+{\left(\frac{1}{\epsilon }\right)}^3{\log }^2\frac{1}{\epsilon }\right)$	Kellerer and Pferschy [11]	2004
$O\left(n\log \frac{1}{\epsilon }+{\left(\frac{1}{\epsilon }\right)}^{5/2}{\log }^3\frac{1}{\epsilon }\right)$ (randomized)	Rhee [15]	2015
$O\left(n\log \frac{1}{\epsilon }+{\left(\frac{1}{\epsilon }\right)}^{12/5}/2^{\Omega \left(\sqrt{\log \left(1/\epsilon \right)}\right)}\right)$	Chan [3]	2018
$O\left(n\log \frac{1}{\epsilon }+{\left(\frac{1}{\epsilon }\right)}^{9/4}/2^{\Omega \left(\sqrt{\log \left(1/\epsilon \right)}\right)}\right)$	This work
$O\left(\frac{1}{\epsilon }{n}^3\right)$	Textbook algorithm
$O\left(\frac{1}{\epsilon }{n}^2\right)$	Lawler [13]	1979
$O\left({\left(\frac{1}{\epsilon }\right)}^{2}n\log \frac{1}{\epsilon }\right)$	Kellerer and Pferschy [10]	1999
$\tilde{O}\left(\frac{1}{\epsilon }{n}^{3/2}\right)$ (randomized, Las Vegas)	Chan [3]	2018
$O\left(\left({\left(\frac{1}{\epsilon }\right)}^{4/3}n+{\left(\frac{1}{\epsilon }\right)}^2\right)/2^{\Omega \left(\sqrt{\log \left(1/\epsilon \right)}\right)}\right)$	Chan [3]	2018
$O\left(\left({\left(\frac{1}{\epsilon }\right)}^{3/2}{n}^{3/4}+{\left(\frac{1}{\epsilon }\right)}^2\right)/2^{\Omega \left(\sqrt{\log \left(1/\epsilon \right)}\right)}+n\log \frac{1}{\epsilon }\right)$	This work

Equations

$$f_{I}(x)=\max\Bigg{\{}\sum_{i\in J}p_{i}:\sum_{i\in J}w_{i}\leq x,\;\;J\subseteq I\Bigg{\}}$$

$$\epsilon \le {\delta }_1,{\delta }_i\le {\delta }_{i+1}/2,{\delta }_d\le 1/8$$

$${\Delta }_{i+1}:=[{\delta }_{i+1},8{\delta }_{i+1}]\cap \bigcup _{k\in \mathbb{Z}}k{\Delta }_i,i=1,2,\dots ,d-1.$$

$$\displaystyle|\Delta_{i}|=\Big{\lvert}[\delta_{i},8\delta_{i}]\cap\bigcup_{k\in\mathbb{Z}}k\Delta_{i-1}\Big{\rvert}\leq\sum_{x\in\Delta_{i-1}}8\delta_{i}/x\leq|\Delta_{i-1}|8\delta_{i}/\delta_{i-1}.$$

$$k_1{k}_2\cdots k_{i-1}\in [{\delta }_i/{\delta }_1,2{\delta }_i/{\delta }_1],\text{ for every 2≤i≤d}.$$

$$k_1{k}_2\cdots k_{d-1}j\in [p/(4{\delta }_1),p/(2{\delta }_1)],$$

$$\frac{p/(4\delta_{1})}{\big{(}\log(p/(4\delta_{1}))\big{)}^{O(d)}}=\Omega\Big{(}\frac{\delta_{1}^{-1}}{(\log\varepsilon^{-1})^{O(d)}}\Big{)},$$

$$I_{p}\,\,:=\bigcup_{\text{$K$ good for $p$}}\left[\frac{p-\varepsilon}{K},\frac{p}{K}\right]\,\,\,\,\subset\,\big{[}\delta_{1},4\delta_{1}\big{]}.$$

$$\lambda (I_p)\ge \frac{{\delta }_1^{-1}}{(\log {\epsilon }^{-1}{)}^{O(d)}}\cdot \frac{p-(p-\epsilon )}{\max K}\ge \frac{\epsilon }{(\log {\epsilon }^{-1}{)}^{O(d)}}.$$

$$s:=\frac{{\min }_p\lambda (I_p)}{4{\delta }_1-{\delta }_1}\ge \frac{\epsilon /{\delta }_1}{(\log {\epsilon }^{-1}{)}^{O(d)}},$$

$$1+{\log }_{1/(1-s)}|P|=O\left(\frac{\log |P|}{s}\right)=\frac{{\delta }_1}{\epsilon }(\log {\epsilon }^{-1}{)}^{O(d)}.$$

$$\sum _{x\in {\Delta }_1}\lceil |Q_x|/s\rceil \le |{\Delta }_1|+\sum _{x\in {\Delta }_1}|Q_x|/s=|{\Delta }_1|+m/s\le 2D.$$

$$2^{2^{d-1}}\le \frac{\sqrt{B}}{\sqrt{r}}<2^{2^d}=4^{2^{d-1}}.$$

$${\alpha }^{2^{d-1}}=\frac{\sqrt{B}}{\sqrt{r}}.$$

$$\delta_{i}:=\varepsilon\sqrt{Br}\Big{/}{\alpha^{2^{d-i}}},\,\,0\leq i\leq d.$$

$${\delta }_d=\frac{\epsilon \sqrt{Br}}{\alpha },{\delta }_1=\epsilon r$$

$$t:=\max\Big{\{}\alpha,\,\,\max_{j}|\Delta_{i}^{(j)}|\Big{/}\frac{\delta_{i}}{\varepsilon r}\Big{\}}=(\log\varepsilon^{-1})^{O(d)}$$

$$B_{i}:=Bt\Big{/}\alpha^{2^{d-i}},\,\,0\leq i\leq d.$$

$$|{\Delta }_i^{(j)}|\cdot {\delta }_i\le B_{i-1}\epsilon ,(1\le i\le d).$$

$$\tilde{O}({\epsilon }^{-1}(|P^{(j)}|B_0+{\epsilon }^{-1})B_0^{-0.01})+\sum _{1\le j\le d}\tilde{O}(|P^{(j)}|B_j/{\delta }_j)$$

$$\displaystyle=\$$

$$O({\epsilon }^{-2}{B}^{1/3}/2^{\Omega (\sqrt{\log (1/\epsilon )})}).$$

$$O({\epsilon }^{-3/2}{m}^{3/4}/2^{\Omega (\sqrt{\log (1/\epsilon )})}).$$

$$\pi (x)-\pi (x/2)\ge x/(C\log x),\text{for all x≥2,}$$

$$\pi(T_{k}/P)-\pi\big{(}T_{k}/(2P)\big{)}\geq\frac{T_{k}/P}{C\log(T_{k}/P)}\geq\frac{T_{k}/T_{k-1}}{C\log T_{k}}.$$

$$N_k\ge N_{k-1}\cdot \frac{T_k/T_{k-1}}{C\log T_k}\ge \frac{T_{k-1}}{(C\log T_{k-1}{)}^{k-1}}\cdot \frac{T_k/T_{k-1}}{C\log T_k}\ge \frac{T_k}{(C\log T_k{)}^k}.$$

$$\frac{N_d}{d!}\ge \frac{T_d}{(Cd\log T_d{)}^d}\ge \frac{T_d}{(C{\log }^2{T}_d{)}^d}\ge \frac{T_d}{(\log T_d{)}^{O(d)}},$$

Full text

An Improved FPTAS for 0-1 Knapsack

Ce Jin

1 Introduction

1.1 Background

In the 0-1 knapsack problem, we are given a set $I$ of $n$ items where each item $i\in I$ has weight $w_{i}$ and profit $p_{i}$, and we want to select a subset $J\subseteq I$ such that $\sum_{j\in J}w_{j}\leq W$ and $\sum_{j\in J}p_{j}$ is maximized.

The 0-1 knapsack problem is a fundamental optimization problem in computer science and is one of Karp’s 21 NP-complete problems [8]. An important field of study on NP-hard problems is to find efficient approximation algorithms. A $(1+\varepsilon)$-approximation algorithm (for a maximization problem) outputs a value $\mathrm{SOL}$ such that $\mathrm{SOL}\leq\mathrm{OPT}\leq(1+\varepsilon)\cdot\mathrm{SOL}$, where $\mathrm{OPT}$ denotes the optimal answer. The 0-1 knapsack problem is one of the first problems that were shown to have fully polynomial-time approximation schemes (FPTASs), i.e., algorithms with approximation factor $1+\varepsilon$ for any given $0<\varepsilon<1$ and running time polynomial in both $n$ and $1/\varepsilon$.

There has been a long line of research on finding faster FPTASs for the 0-1 knapsack problem, as summarized in Table 1. The first algorithm with only subcubic dependence on $1/\varepsilon$ was due to Rhee [15]. Very recently, Chan [3] gave an elegant algorithm for the 0-1 knapsack problem in deterministic $O(n\log\frac{1}{\varepsilon}+(\frac{1}{\varepsilon})^{5/2}/2^{\Omega(\sqrt{\log(1/\varepsilon)})})$ via simple combination of the SMAWK algorithm [1] and a standard divide-and-conquer technique. The speedup of superpolylogarithmic factor $2^{\Omega(\sqrt{\log(1/\varepsilon)})}$ is due to recent progress on $(\min,+)$-convolution [2, 16, 4]. Using an elementary number-theoretic lemma, Chan further improved the algorithm to $O(n\log\frac{1}{\varepsilon}+(\frac{1}{\varepsilon})^{12/5}/2^{\Omega(\sqrt{\log(1/\varepsilon)})})$ time, and obtained two new algorithms running in $\tilde{O}(\frac{1}{\varepsilon}n^{3/2})$ and $O((\frac{1}{\varepsilon})^{4/3}n+(\frac{1}{\varepsilon})^{2})/2^{\Omega(\sqrt{\log(1/\varepsilon)})})$ time respectively, which are faster for small $n$.

FPTASs on several special cases of 0-1 knapsack are also of interest. For the unbounded knapsack problem, where every item has infinitely many copies, Jansen and Kraft [7] obtained an $O(n+(\frac{1}{\varepsilon})^{2}\log^{3}\frac{1}{\varepsilon})$-time algorithm; the unbounded version can be reduced to 0-1 knapsack with only a logarithmic blowup in the problem size [5]. For the subset sum problem, where every item has $p_{i}=w_{i}$, Kellerer et al. [9] obtained an algorithm with $O(\min\{n/\varepsilon,n+(\frac{1}{\varepsilon})^{2}\log\frac{1}{\varepsilon}\})$ running time, which will be used in our algorithm as a subroutine. For the partition problem, which is a special case of the subset sum problem where $W=\frac{1}{2}\sum_{i\in I}w_{i}$, Mucha et al. [14] obtained an algorithm with a subquadratic $\tilde{O}(n+(\frac{1}{\varepsilon})^{5/3})$ running time.

On the lower bound side, recent reductions showed by Cygan et al. [5] and Künnemann et al. [12] imply that 0-1 knapsack and unbounded knapsack have no FPTAS in $O((n+\frac{1}{\varepsilon})^{2-\delta})$ time, unless $(\min,+)$-convolution has truly subquadratic algorithm [14]. It remains open whether 0-1 knapsack has a matching upper bound.

1.2 Our results

In this paper we present improved FPTASs for the 0-1 knapsack problem. Our results are summarized in the following two theorems.

Theorem 1.1**.**

There is a deterministic $(1+\varepsilon)$-approximation algorithm for 0-1 knapsack with running time $O(n\log\frac{1}{\varepsilon}+(\frac{1}{\varepsilon})^{9/4}/2^{\Omega(\sqrt{\log(1/\varepsilon)})})$.

Theorem 1.2**.**

For $n=O(\frac{1}{\varepsilon})$, there is a deterministic $(1+\varepsilon)$-approximation algorithm for 0-1 knapsack with running time O\Big{(}\big{(}n^{3/4}(\frac{1}{\varepsilon})^{3/2}+(\frac{1}{\varepsilon})^{2}\big{)}/2^{\Omega(\sqrt{\log(1/\varepsilon)})}\Big{)}.

Theorem 1.2 gives the current best time bound for $(\frac{1}{\varepsilon})^{2/3}\ll n\ll\frac{1}{\varepsilon}$, improving upon the previous $O((\frac{1}{\varepsilon})^{4/3}n+(\frac{1}{\varepsilon})^{2})/2^{\Omega(\sqrt{\log(1/\varepsilon)})})$ algorithm by Chan [3]. For $n\ll(\frac{1}{\varepsilon})^{2/3}$, Chan’s $\tilde{O}(\frac{1}{\varepsilon}n^{3/2})$ time randomized algorithm [3] remains the fastest.

For $n\gg\frac{1}{\varepsilon}$, Theorem 1.1 gives a better time bound, improving upon the previous $O(n\log\frac{1}{\varepsilon}+(\frac{1}{\varepsilon})^{12/5}/2^{\Omega(\sqrt{\log(1/\varepsilon)})})$ algorithm by Chan [3].

1.3 Outline of our algorithm

We give an informal overview of our improved algorithm for 0-1 knapsack.

Using a known reduction [3], it suffices to solve an easier instance of 0-1 knapsack where profits of all items satisfy $p_{i}\in[1,2]$. Here “solving an instance” means approximating the function f(x):=[\text{maximum total profit of items with at most x total weight}] for all $x\geq 0$, rather than for just a single point $x=W$. In this restricted case, simple greedy (sorting according to unit profits $p_{i}/w_{i}$) gives an additive error of at most $\max_{j}p_{j}=O(1)$, so it suffices to approximate the capped function $\min\{\varepsilon^{-1},f(x)\}$ with approximation factor $1+O(\varepsilon)$. Chan gave an algorithm that $(1+\varepsilon)$-approximates $\min\{B,f(x)\}$ in $\tilde{O}(n+\varepsilon^{-2}B^{1/2})$ time (implied by [3, Lemma 7]), which immediately implies an $\tilde{O}(n+\varepsilon^{-5/2})$ time FPTAS by setting $B=\varepsilon^{-1}$.

Greedy.

Now we explain how to use a greedy argument (described in detail in Section 5) to improve this algorithm to $\tilde{O}(n+\varepsilon^{-7/3})$ time. We sort all items (with $p_{i}\in[1,2]$) in non-increasing order of unit profits $p_{i}/w_{i}$, and divide them into three subsets $H,M,L$ (items with high, medium, low unit profits), where $H$ contains the top $\Theta(\varepsilon^{-1})$ items, and $L$ contains all items $i$ for which $p_{i}/w_{i}\leq(1-\varepsilon^{2/3})\cdot\min_{h\in H}\{p_{h}/w_{h}\}$, so there is a gap between the unit profits of $H$-items and $L$-items. Intuitively, there are sufficiently many $H$-items available, so it’s not optimal to include too many cheap $L$-items when the knapsack capacity is not very big. To be more precise, we prove that in any optimal solution we care about (i.e., having optimal total profit smaller than $\varepsilon^{-1}$), the total profit contributed by $L$-items cannot exceed $O(\varepsilon^{-2/3})$. Hence, for subset $L$ we only need to approximate up to $B=\Theta(\varepsilon^{-2/3})$ in $\tilde{O}(n+\varepsilon^{-2}B^{1/2})=\tilde{O}(n+\varepsilon^{-7/3})$ time. Subset $H$ has only $O(\varepsilon^{-1})$ items and can be solved using Chan’s $\tilde{O}(\varepsilon^{-4/3}n+\varepsilon^{-2})$ algorithm in $\tilde{O}(\varepsilon^{-7/3})$ time. To solve subset $M$, we round down the profit value $p_{i}$ for every item $i\in M$, so that the unit profit $p_{i}/w_{i}$ becomes a power of $(1+\varepsilon)$. Then there are $O(\varepsilon^{-1/3})$ distinct unit profit values in $M$. Items with the same unit profit can be solved together using the efficient FPTAS for subset sum by Kellerer et al. [9] in $\tilde{O}(n+\varepsilon^{-2})$ time. Finally we merge the results for $H,M,L$. The total time complexity is $\tilde{O}(n+\varepsilon^{-7/3})$.

Multi-level number-theoretic construction.

The above approach invokes two of Chan’s algorithms: an $\tilde{O}(n+\varepsilon^{-2}B^{1/2})$ algorithm (useful for small $B$) and an $\tilde{O}(\varepsilon^{-4/3}n+\varepsilon^{-2})$ algorithm (useful for small $n$). The key ingredient in these algorithms is a number-theoretic lemma: we can $(1+\varepsilon)$-approximate all profit values $p_{i}\in[1,2]$ by multiples of elements from a small set $\Delta\subset[\delta,2\delta]$ of size $|\Delta|=\tilde{O}(\frac{\delta}{\varepsilon})$ (small $|\Delta|$ can reduce the additive error incurred from rounding).

Chan obtained an $\tilde{O}(n+\varepsilon^{-2}B^{2/5})$ time algorithm using some additional tricks. First, evenly partition $\Delta$ into $r$ subsets $\Delta^{(1)},\dots,\Delta^{(r)}$, and divide the items into $P=P^{(1)}\cup\dots\cup P^{(r)}$ accordingly, so that profit values from $P^{(j)}$ are approximated by $\Delta^{(j)}$-multiples. To $(1+\varepsilon)$-approximate the profit function $f_{j}$ for each $P^{(j)}$, pick a threshold $B_{0}\ll B$, and return the combination of a $(1+\varepsilon)$-approximation of $\min\{f_{j},B_{0}\}$ and an $\varepsilon B_{0}$-additive-approximation of $\min\{f_{j},B\}$. Since the size of $\Delta^{(j)}$ is only $|\Delta|/r$, the latter function can be approximated faster when $r\gg 1$. Finally, merge $f_{j}$ over all $1\leq j\leq r$. By fine-tuning the parameters $r,\delta,B_{1}$, the time complexity is improved to $\tilde{O}(n+\varepsilon^{-2}B^{2/5})$.

Our new algorithm extends this technique to multiple levels. To $(1+\varepsilon)$-approximate the profit function $f_{j}$ for each $P^{(j)}$, we will pick $B_{0}\ll B_{1}\ll\dots\ll B_{d-1}\ll B_{d}\approx B$, and compute the $\varepsilon B_{i-1}$-additive-approximation of $\min\{f_{j},B_{i}\}$, for all $i\in[d]$. An issue of this multi-level approach is that, different levels have different optimal parameters $\delta_{i}$ and different $\Delta_{i}^{(1)},\dots,\Delta_{i}^{(r)}$, but we have to stick to the same partition of items $P=P^{(1)}\cup\dots\cup P^{(r)}$ over all levels. We overcome this issue by enforcing that $\Delta_{i}^{(j)}$ at level $i$ must be generated by multiples of elements from $\Delta_{i-1}^{(j)}$ at level $i-1$, so that $P^{(j)}$ can be approximated by $\Delta_{i}^{(j)}$-multiples for all levels. To achieve this, we need a multi-level version of the number-theoretic lemma. We will discuss this part in detail in Section 4.

Using this multi-level construction, we obtain algorithms in $\tilde{O}(n+\varepsilon^{-2}B^{1/3})$ time and $\tilde{O}(\varepsilon^{-3/2}n^{3/4}+\varepsilon^{-2})$ time. Combining these improved algorithms with the greedy argument previously described (the threshold which splits $M$ and $L$ needs to be adjusted accordingly), we obtain an algorithm in $\tilde{O}(n+\varepsilon^{-9/4})$ time as claimed in Theorem 1.1.

2 Preliminaries

Throughout this paper, $\log x$ stands for $\log_{2}x$, and $\tilde{O}(f)$ stands for $O(f\cdot\text{poly}\log(f))$.

We will describe our algorithm with approximation factor $1+O(\varepsilon)$, which can be lowered to $1+\varepsilon$ if we scale down $\varepsilon$ by a constant factor at the beginning.

We are only interested in the case where $n=O(\varepsilon^{-4})$. For greater $n$, Lawler’s $O(n\log\frac{1}{\varepsilon}+(\frac{1}{\varepsilon})^{4})$ algorithm [13] is already near-optimal. Hence we assume $\log n=O(\log\varepsilon^{-1})$.

Assume $0<w_{i}\leq W$ and $p_{i}>0$ for every item $i$. Then a trivial lower bound of the maximum total profit is $\max_{j}p_{j}$. At the beginning, we discard all items $i$ with $p_{i}\leq\frac{\varepsilon}{n}\max_{j}p_{j}$. Since the total profit of discarded items is at most $\varepsilon\max_{j}p_{j}$, the optimal total profit is only reduced by a factor of $1+O(\varepsilon)$. So we can assume that $\frac{\max_{j}p_{j}}{\min_{j}p_{j}}\leq\frac{n}{\varepsilon}$.

We adopt Chan’s terminology in presenting our algorithm. For a set $I$ of items, define the profit function

[TABLE]

over non-negative real numbers $x\geq 0$. Note that $f_{I}$ is a monotone (nondecreasing) step function. The complexity of a monotone step function refers to the number of its steps.

We say that a function $\tilde{f}$ approximates a function $f$ with factor $1+\varepsilon$ if $\tilde{f}(x)\leq f(x)\leq(1+\varepsilon)\tilde{f}(x)$ for all $x\geq 0$. We say that $\tilde{f}$ approximates $f$ with additive error $\delta$ if $\tilde{f}(x)\leq f(x)\leq\tilde{f}(x)+\delta$ for all $x\geq 0$. Our goal is to approximate $f_{I}$ with factor $1+O(\varepsilon)$ on the input item set $I$.

Let $I_{1},I_{2}$ be two disjoint subsets of items, and $I=I_{1}\cup I_{2}$. We have $f_{I}=f_{I_{1}}\oplus f_{I_{2}}$, where $\oplus$ denotes the $(\max,+)$-convolution, defined by $(f\oplus g)(x)=\max_{0\leq x^{\prime}\leq x}(f(x^{\prime})+g(x-x^{\prime}))$. If two non-negative monotone step functions $f,g$ are approximated with factor $1+\varepsilon$ by functions $\tilde{f},\tilde{g}$ respectively, then $f\oplus g$ is also approximated by $\tilde{f}\oplus\tilde{g}$ with factor $1+\varepsilon$.

For a monotone step function $f$ with range111Here range refers to the set of possible output values of the function. contained in $\{0\}\cup[A,B]$, we can obtain a function $\tilde{f}$ with complexity only $O(\varepsilon^{-1}\log(B/A))$ which approximates $f$ with factor $1+\varepsilon$, by simply rounding $f$ down to powers of $(1+\varepsilon)$. For our purposes, $B/A$ will be bounded by polynomial of $n$ and $1/\varepsilon$, hence we may always assume that the approximation results are monotone step functions with complexity $\tilde{O}(\varepsilon^{-1})$.

For an item set $I$ with the same profit $p_{i}=p$ for every item $i\in I$, the step function $f_{I}$ can be exactly computed in $O(n\log n)$ time by simple greedy: the function values are $0,p,2p,\dots,np$ and the $x$-breakpoints are $w_{1},w_{1}+w_{2},\,\dots,\,w_{1}+\dots+w_{n}$, after sorting all $w_{i}$’s in nondecreasing order. We say that a monotone step function is $p$-uniform if its function values are of the form $0,p,2p,\dots,lp$ for some $l$. We say that a $p$-uniform function is pseudo-concave if the differences of consecutive $x$-breakpoints are nondecreasing from left to right. In the previous case where all $p_{i}$’s are equal to $p$, $f_{I}$ is indeed $p$-uniform and pseudo-concave.

3 Chan’s techniques

In this section we review several useful lemmas by Chan [3].

3.1 Merging profit functions

Lemma 3.1** ([3, Lemma 2(i)]).**

Let $f_{1},\dots,f_{m}$ be monotone step functions with total complexity $O(n)$ and ranges contained in $\{0\}\cup[A,B]$. Then we can compute a monotone step function that approximates $f_{1}\oplus\dots\oplus f_{m}$ with factor $1+O(\varepsilon)$ and complexity $\tilde{O}(\frac{1}{\varepsilon}\log B/A)$ in $O(n)+\tilde{O}((\frac{1}{\varepsilon})^{2}m/2^{\Omega(\sqrt{\log(1/\varepsilon)})}\log B/A)$ time.

Remark 3.2**.**

Lemma 3.1 is proved using a divide-and-conquer method, which was also used previously in [10]. The speedup of superpolylogarithmic factor $2^{\Omega(\sqrt{\log(1/\varepsilon)})}$ is due to recent progress on $(\min,+)$-convolution [2, 16, 4].

Lemma 3.1 enables us to focus on a simpler case where all $p_{i}\in[1,2]$. For the general case, we divide the items into $O(\log\frac{\max_{j}p_{j}}{\min_{j}p_{j}})=O(\log\varepsilon^{-1})$ groups, each containing items with $p_{i}\in[2^{j},2^{j+1}]$ for some $j$ (which can be rescaled to $[1,2]$), and finally merge the profit functions of all groups by using Lemma 3.1 in $\tilde{O}(n+\varepsilon^{-2})$ time.

Assuming $\varepsilon^{-1}$ is an integer and every $p_{i}\in[1,2]$, we can round every $p_{i}$ down to a multiple of $\varepsilon$, introducing only a $1+\varepsilon$ error factor. Then there are only $m=O(\varepsilon^{-1})$ distinct values of $p_{i}$. For every value of $p_{i}$, the corresponding profit function $f_{i}$ is $p_{i}$-uniform and pseudo-concave, and can be obtained by simple greedy (as discussed in Section 2).

3.2 Approximating big profit values using greedy

When all $p_{i}$’s are small, simple greedy gives good approximation guarantee when the answer is big enough.

Lemma 3.3**.**

Suppose $p_{i}\in[1,2]$ for all $i\in I$. For $B=\Omega(\varepsilon^{-1})$, the function $f_{I}$ can be approximated with additive error $O(\varepsilon B)$ in $O(n\log n)$ time.

Proof 3.4**.**

Sort the items in nonincreasing order of unit profit $p_{i}/w_{i}$. Let $\tilde{f}$ be the monotone step function resulting from greedy, with function values $0,p_{1},p_{1}+p_{2},\dots,p_{1}+\dots+p_{n}$ and $x$-breakpoints $0,w_{1},w_{1}+w_{2},\dots,w_{1}+\dots+w_{n}$. It approximates $f_{I}$ with an additive error of $\max_{i}p_{i}\leq 2\leq O(\varepsilon B)$ for $B=\Omega(\varepsilon^{-1})$.

When every $p_{i}\in[1,2]$, we set $B=\varepsilon^{-1}$ and let $f_{H}$ denote the result from greedy (Lemma 3.3). Then we only need to obtain a function $f_{L}$ which $1+O(\varepsilon)$ approximates $\min\{f_{I},B\}$, and finally return $\max\{f_{L},f_{H}\}$ as a $1+O(\varepsilon)$ approximation of $f_{I}$ (because when $f_{I}(x)$ exceeds $B$, an additive error $O(\varepsilon B)$ implies $1+O(\varepsilon)$ approximation factor).

3.3 Approximation using $\Delta$-multiples of small set $\Delta$

For a set $\Delta$ of numbers, we say that $p$ is a $\Delta$-multiple if it is a multiple of $\delta$ for some $\delta\in\Delta$.

Lemma 3.5** ([3, Lemma 5]).**

Let $f_{1},\dots,f_{m}$ be monotone step functions with ranges contained in $[0,B]$. Let $\Delta\subset[\delta,8\delta]$. If every $f_{i}$ is $p_{i}$-uniform and pseudo-concave for some $p_{i}\in[1,2]$ which is a $\Delta$-multiple, then we can compute a monotone step function that approximates $\min\{f_{1}\oplus\dots\oplus f_{m},B\}$ with additive error $O(|\Delta|\delta)$ in $\tilde{O}(Bm/\delta)$ time.

Remark 3.6**.**

An intuition of Lemma 3.5 is as follows. When $p_{i}$’s are exact multiples of $\delta$, standard dynamic programming algorithm maintains a result array of length $B/\delta$, and adding a new $f_{i}$ to the result can be done in linear time (by exploiting the pseudo-concavity of $f_{i}$ using the SMAWK algorithm222The SMAWK algorithm [1] finds all row-minima in an $n\times n$ matrix $A$ satisfying the Monge property $A[i,j]+A[i+1,j+1]\leq A[i,j+1]+A[i+1,j]$ using only $O(n)$ queries. ). Now if the next $p_{i}$ to be considered is a multiple of $\delta^{\prime}\neq\delta$, we first have to round down the current results to multiples of $\delta^{\prime}$, introducing an additive error of $\delta^{\prime}$. We round our results for $|\Delta|-1$ times, so smaller $|\Delta|$ implies smaller overall additive error.

Corollary 3.7**.**

Let $f_{1},\dots,f_{m}$ be monotone step functions with ranges contained in $[0,B]$. If every $f_{i}$ is $p_{i}$-uniform and pseudo-concave for some $p_{i}\in[1,2]$, then we can compute a monotone step function that approximates $\min\{f_{1}\oplus\dots\oplus f_{m},B\}$ with factor $1+O(\varepsilon)$ in $\tilde{O}(\varepsilon^{-1}Bm)$ time.

Proof 3.8**.**

Assuming $\varepsilon^{-1}$ is an integer, adjust every $p_{i}$ down to the nearest multiple of $\varepsilon$, and adjust $f_{i}$ accordingly. This introduces a $1+\varepsilon$ overall error factor. Then use Lemma 3.5 with $\delta=\varepsilon,\Delta=\{\varepsilon\}$ to compute the desired function in $\tilde{O}(Bm\varepsilon^{-1})$ time.

4 Extending Chan’s number-theoretic construction

As mentioned in Section 1.3, the main results of this section are two approximation algorithms in $\tilde{O}(n+\varepsilon^{-2}B^{1/3})$ and $\tilde{O}(\varepsilon^{-3/2}n^{3/4}+\varepsilon^{-2})$ time respectively (the latter time bound assumes $n=O(1/\varepsilon)$). These results rely on Lemma 3.5.

4.1 Number-theoretic construction

To avoid checking boundary conditions, from now on we assume $\varepsilon>0$ is sufficiently small.

Our algorithm extends Chan’s technique by using a multi-level structure defined as follows.

Definition 4.1**.**

For fixed parameters $\delta_{1},\delta_{2},\dots,\delta_{d}$ satisfying condition

[TABLE]

and a finite real number set $\Delta_{1}\subset[\delta_{1},8\delta_{1}]$, a set tower $(\Delta_{1},\Delta_{2},\dots,\Delta_{d})$ generated by $\Delta_{1}$ is defined by recurrence333For a number $k$ and a set $A$ of numbers, $kA:=\{ka:a\in A\}$.

[TABLE]

We refer to $\Delta_{1}$ as the base set and $\Delta_{d}$ as the top set of this set tower. We also say that the base set $\Delta_{1}$ generates the top set $\Delta_{d}$.

If $\Delta_{d}^{\ast}$ is the top set generated by a singleton base set $\Delta_{1}^{\ast}=\{x\}$, then for every $y\in\Delta_{d}^{\ast}$ we say $x$ generates $y$.

We have the following simple facts about set towers.

Proposition 1**.**

If $x$ generates $y$ then $x\in\Delta_{1}$ implies $y\in\Delta_{d}$. Conversely, for every $y\in\Delta_{d}$, there exists $x\in\Delta_{1}$ which generates $y$, and for every $1\leq i\leq d$ there exists $z\in\Delta_{i}$ such that both $y/z$ and $z/x$ are integers.

Proposition 2**.**

For any $1\leq i\leq d$, $|\Delta_{i}|\leq 8^{i-1}(\delta_{i}/\delta_{1})|\Delta_{1}|$, and we can compute $\Delta_{i}$ in $\tilde{O}(8^{i-1}(\delta_{i}/\delta_{1})|\Delta_{1}|)$ time given $\Delta_{1}$ as input.

Proof 4.2**.**

For $2\leq i\leq d$, we have

[TABLE]

The proof of size upper bounds follows by induction. Elements of $\Delta_{i}$ can be generated straightforwardly within the time bound.

Lemma 4.3**.**

Let $T_{1},T_{2},\dots,T_{d}$ be positive real numbers satisfying $T_{1}\geq 2$ and $T_{i+1}\geq 2T_{i}$. There exist at least T_{d}\big{/}(\log T_{d})^{O(d)} integers $t$ satisfying the following condition: $t$ can be written as a product of integers $t=n_{1}n_{2}\cdots n_{d}$, such that $n_{1}n_{2}\cdots n_{i}\in(T_{i}/2,T_{i}]$ for every $1\leq i\leq d$.

The proof of Lemma 4.3 is deferred to Appendix A. Lemma 4.3 helps us prove the following fact, which is a multi-level extension of [3, Lemma 6].

Lemma 4.4**.**

For any parameters $\delta_{1},\dots,\delta_{d}$ satisfying condition (1), there exists a base set $\Delta_{1}$ of size $\frac{\delta_{1}}{\varepsilon}\cdot(\log\varepsilon^{-1})^{O(d)}$, such that every $p\in[1,2]$ can be approximated by a $\Delta_{d}$-multiple with additive error $O(\varepsilon)$, where $\Delta_{d}$ is the top set generated by $\Delta_{1}$.

This base set $\Delta_{1}$ can be constructed in $\tilde{O}(\varepsilon^{-1}\delta_{1}^{-1})$ time deterministically.

Proof 4.5**.**

Let $P=\{1,1+\varepsilon,1+2\varepsilon,\dots,1+\lfloor\frac{1}{\varepsilon}\rfloor\varepsilon\}$. It suffices to approximate every value $p\in P$ with additive error $\varepsilon$ using $\Delta_{d}$-multiples. For any $p\in P$ and $y\in\Delta_{d}\subset[\delta_{d},8\delta_{d}]$, $p$ is approximated with additive error $\varepsilon$ by a multiple of $y$ if and only if $y\in\bigcup_{j\in\mathbb{Z}}\left[\frac{p-\varepsilon}{j},\frac{p}{j}\right]$.

Our constructed base set $\Delta_{1}$ will satisfy $\Delta_{1}\subset[\delta_{1},4\delta_{1}]$. Suppose integers $k_{1},k_{2},\dots,k_{d-1}$ satisfy

[TABLE]

Then by Definition 4.1, for any $x\in\Delta_{1}\subset[\delta_{1},4\delta_{1}]$, we have $xk_{1}k_{2}\cdots k_{i-1}\in\Delta_{i}$ for every $2\leq i\leq d$.

For any integer $j$ satisfying

[TABLE]

the interval $[\frac{p-\varepsilon}{k_{1}k_{2}\cdots k_{d-1}j},\frac{p}{k_{1}k_{2}\cdots k_{d-1}j}]$ is contained in $[\delta_{1},4\delta_{1}]$.

We say an integer $K$ is good for $p$, if $K$ can be expressed as a product of integers $k_{1}k_{2}\cdots k_{d-1}j$ satisfying conditions (3) and (4). For such $K$, any $x\in[\frac{p-\varepsilon}{K},\frac{p}{K}]\cap\Delta_{1}$ generates an element $y=xk_{1}k_{2}\cdots k_{d-1}\in\Delta_{d}\cap[\frac{p-\varepsilon}{j},\frac{p}{j}]$ such that $p$ can be approximated by a multiple of $y$ with additive error $\varepsilon$.

By Lemma 4.3, the number of good integers $K$ for $p$ is at least

[TABLE]

and at most $p/(2\delta_{1})=O(\delta_{1}^{-1})$, by (4). Using conditions (3) and (4) we can compute all these $K$’s by simple dynamic programming. We denote the union of their associated intervals by

[TABLE]

Note that these intervals are disjoint since $p/(K+1)\leq(p-\varepsilon)/K$, so the total length of $I_{p}$ can be lower-bounded as

[TABLE]

We have seen that $p$ is approximated by a $\Delta_{d}$-multiple with additive error $\varepsilon$ as long as $\Delta_{1}\cap I_{p}\neq\emptyset$. We compute $I_{p}$ for every $p\in P$, and use the standard greedy algorithm (for Hitting Set problem) to construct a base set $\Delta_{1}\subset[\delta_{1},4\delta_{1}]$ which intersects with every $I_{p}$: in each round we find a point $x\in[\delta_{1},4\delta_{1}]$ that hits the most $I_{p}$’s, include $x$ into $\Delta_{1}$, and remove the $I_{p}$’s that are hit by $x$. In each round the number of remaining $I_{p}$’s decreases by

[TABLE]

so the solution size $|\Delta_{1}|$ is upper-bounded by

[TABLE]

To implement this greedy algorithm, we use standard data structures (for example, segment trees) that support finding $x$ which hits the most intervals, reporting an interval hit by $x$, removing an interval, all in logarithmic time per operation. The number of operations is bounded by the total number of small intervals, so the running time is at most $\tilde{O}(|P|\cdot\frac{p}{2\delta_{1}})=\tilde{O}(\delta_{1}^{-1}\varepsilon^{-1})$.

The following lemma evenly partitions the base set $\Delta_{1}$ into $r$ subsets $\Delta_{1}^{(1)},\dots,\Delta_{1}^{(r)}$, and partitions the profit values $P=\{p_{1},\dots,p_{m}\}$ into $P^{(1)}\cup\dots\cup P^{(r)}$, so that $P^{(j)}$ can be approximated by $\Delta_{d}^{(j)}$-multiples. An additional requirement is that $P^{(1)},\dots,P^{(r)}$ should have size $O(|P|/r)$ each.

Lemma 4.6**.**

Let $\delta_{1},\dots,\delta_{d}$ be parameters satisfying condition (1). Let $P=\{p_{1},\dots,p_{m}\}\subset[1,2]$ with $m=O(\varepsilon^{-1})$. Given a positive integer parameter $r=O(\min\{\frac{\delta_{1}}{\varepsilon},m\})$, there exist $r$ base sets $\Delta_{1}^{(1)},\Delta_{1}^{(2)},\dots,\Delta_{1}^{(r)}$ each of size $\frac{\delta_{1}}{\varepsilon r}\cdot(\log\varepsilon^{-1})^{O(d)}$, and a partition of $P=P^{(1)}\cup P^{(2)}\cup\cdots\cup P^{(r)}$ each of size $O(m/r)$, such that for every $1\leq j\leq r$, every $p\in P^{(j)}$ can be approximated by a $\Delta_{d}^{(j)}$-multiple with additive error $O(\varepsilon)$, where $\Delta_{d}^{(j)}$ is the top set generated by $\Delta_{1}^{(j)}$.

These $r$ base sets and the partition of $P$ can be computed by a deterministic algorithm in $\tilde{O}(\varepsilon^{-2}/r)$ time .

Proof 4.7**.**

First construct the base set $\Delta_{1}$ of size $\frac{\delta_{1}}{\varepsilon}(\log\varepsilon^{-1})^{O(d)}$ from Lemma 4.4 in $\tilde{O}(\delta_{1}^{-1}\varepsilon^{-1})=\tilde{O}(\varepsilon^{-2}/r)$ time, and compute the top set $\Delta_{d}$ that it generates. By Proposition 2, $|\Delta_{d}|\leq 8^{d-1}\frac{\delta_{d}}{\delta_{1}}|\Delta_{1}|\leq\frac{\delta_{d}}{\varepsilon}(\log\varepsilon^{-1})^{O(d)}$. Generate and sort all $\Delta_{d}$-multiples in interval $[1,2]$. For every $p\in P$, use binary search to find the $\Delta_{d}$-multiple $ky\leq p\,(y\in\Delta_{d})$ closest to $p$, and then add $p$ to the set $Q_{x}$, where $x\in\Delta_{1}$ is an element that generates $y$. ($Q_{x}$ is initialized as empty for every $x\in\Delta_{1}$.) Then remove every $x$ with $Q_{x}=\emptyset$ from $\Delta_{1}$, so that $|\Delta_{1}|\leq m$, while every $p\in P$ can still be approximated with $O(\varepsilon)$ additive error by a $\Delta_{d}$-multiple.

Let $D:=\max\{r,|\Delta_{1}|\}$ and let $s:=\lceil m/D\rceil$. For every $x\in\Delta_{1}$ we divide $Q_{x}$ evenly into small subsets each having size at most $s$. The total number of these small subsets is

[TABLE]

We merge these small subsets into $r$ groups, each having at most $\lceil 2D/r\rceil$ small subsets. Then, define set $P^{(j)}$ as the union of small subsets from the $j$-th group, and let base set $\Delta_{1}^{(j)}$ contain $x\in\Delta_{1}$ if any of these small subsets comes from $Q_{x}$. So $|\Delta_{1}^{(j)}|\leq\lceil 2D/r\rceil=\frac{\delta_{1}}{\varepsilon r}(\log\varepsilon^{-1})^{O(d)}$, and $|P^{(j)}|\leq s\cdot\lceil 2D/r\rceil=O(m/D)\cdot O(D/r)=O(m/r)$.

4.2 Approximation using set towers

We first prove a slightly improved version of Corollary 3.7. The only purpose of this lemma is to get rid of the $(\log\varepsilon^{-1})^{O(\log\log\varepsilon^{-1})}$ factor in the final running time.

Lemma 4.8**.**

Let $f_{1},\dots,f_{m}$ be monotone step functions with ranges contained in $[0,B]$ for some $1\leq B\leq O(\varepsilon^{-1})$. If every $f_{i}$ is $p_{i}$-uniform and pseudo-concave for some $p_{i}\in[1,2]$, then we can compute a monotone step function that approximates $\min\{f_{1}\oplus\dots\oplus f_{m},B\}$ with factor $1+O(\varepsilon)$ in $\tilde{O}(\varepsilon^{-1}(Bm+\varepsilon^{-1})/B^{0.01})$ time.

Proof 4.9**.**

Using Lemma 4.4 with parameters $d=1,\delta_{1}=\varepsilon B^{0.01}$, we get $\Delta\subset[\delta_{1},8\delta_{1}]$ with size $|\Delta|\leq\tilde{O}(\delta_{1}/\varepsilon)=\tilde{O}(B^{0.01})$, in $\tilde{O}(\varepsilon^{-2}/B^{0.01})$ time. Adjust every $p_{i}$ down to the nearest $\Delta$-multiple, and adjust $f_{i}$ accordingly. This introduces at most $1+O(\varepsilon)$ error factor. Then use Lemma 3.5 to compute a monotone step function $f_{H}$ that approximates $\min\{f_{1}\oplus\cdots\oplus f_{m},B\}$ with additive error $e=O(|\Delta|\delta_{1})=\tilde{O}(\varepsilon B^{0.02})$, in $\tilde{O}(B^{0.99}m\varepsilon^{-1})$ time.

Let $B_{L}:=e/\varepsilon$, and use Corollary 3.7 to compute a monotone step function $f_{L}$ that approximates $\min\{f_{1}\oplus\cdots\oplus f_{m},B_{L}\}$ with factor $1+O(\varepsilon)$ in only $\tilde{O}(B_{L}m\varepsilon^{-1})=\tilde{O}(B^{0.02}m\varepsilon^{-1})$ time.

Since $f_{H}$ approximates $\min\{f_{1}\oplus\cdots\oplus f_{m},B\}$ with additive error $\varepsilon B_{L}$, $\max\{f_{L},f_{H}\}$ is a $1+O(\varepsilon)$ approximation of $\min\{f_{1}\oplus\cdots\oplus f_{m},B\}$.

Now we can approximate the profit function $\min\{B,\bigoplus_{p_{k}\in P^{(j)}}f_{k}\}$ for each group $P^{(j)}$, using the multi-level approach described in Section 1.3.

Lemma 4.10**.**

Let $f_{1},\dots,f_{m}$ be given monotone step functions with ranges contained in $[0,B]$, and every $f_{k}$ is $p_{k}$-uniform and pseudo-concave for some $p_{k}\in[1,2]$. Assume $m=O(\varepsilon^{-1})$, $\Omega(\varepsilon^{-0.01})\leq B\leq O(\varepsilon^{-1})$. Let $r$ be a given positive integer parameter with $r=O(m),r=o(B)$.

We can set $d=O(\log\log\varepsilon^{-1})$ and choose $d$ parameters $\delta_{1},\dots,\delta_{d}$ satisfying condition (1), such that the following statement holds:

Let $P^{(1)}\cup\dots\cup P^{(r)}$ be the partition of set $P=\{p_{1},\dots,p_{m}\}$ returned by the algorithm in Lemma 4.6 with the above parameters. Then for any $1\leq j\leq r$, using the base set $\Delta_{1}^{(j)}$ from Lemma 4.6, we can compute a monotone step function that approximates $\min\{B,\bigoplus_{p_{k}\in P^{(j)}}f_{k}\}$ with factor $1+O(\varepsilon)$, in $(\varepsilon^{-2}/r^{0.01}+m\varepsilon^{-1}B^{1/2}/r^{3/2})(\log\varepsilon^{-1})^{O(d)}$ time.

Proof 4.11**.**

We can assume $B\geq 4r$, and define $d$ to be the unique positive integer such that

[TABLE]

Then $d=O(\log\log\frac{\sqrt{B}}{\sqrt{r}})=O(\log\log\varepsilon^{-1})$. Pick $\alpha\in[2,4)$ such that

[TABLE]

Define

[TABLE]

Then

[TABLE]

Note that $\delta_{d}=\varepsilon\sqrt{B}\cdot O(\sqrt{r})=\varepsilon\sqrt{B}\cdot o(\sqrt{B})=\varepsilon\cdot o(B)=o(1)$. Hence the parameters $\delta_{1},\dots,\delta_{d}$ satisfy condition (1) for sufficiently small $\varepsilon$.

The base set $\Delta_{1}^{(j)}$ from Lemma 4.6 has size $\frac{\delta_{1}}{\varepsilon r}(\log\varepsilon^{-1})^{O(d)}$. We compute the generated set tower $\Delta_{1}^{(j)},\Delta_{2}^{(j)},\dots,\Delta_{d}^{(j)}$. By Proposition 2, $|\Delta_{i}^{(j)}|\leq\frac{\delta_{i}}{\varepsilon r}(\log\varepsilon^{-1})^{O(d)}$. Let

[TABLE]

and define

[TABLE]

Then $B\leq B_{d}\leq B\cdot(\log\varepsilon^{-1})^{O(d)}$, and it’s easy to verify that

[TABLE]

For every $1\leq i\leq d$, adjust every $p_{k}\in P^{(j)}$ down to the nearest $\Delta_{i}^{(j)}$-multiple and adjust $f_{k}$ accordingly, which introduces a $1+O(\varepsilon)$ error factor. Then use Lemma 3.5 to obtain a monotone step function $g_{i}$ which approximates $\min\{\bigoplus_{p_{k}\in P^{(j)}}f_{k},B_{i}\}$ with additive error $O(|\Delta_{i}^{(j)}|\delta_{i})=O(\varepsilon B_{i-1})$ in $\tilde{O}(|P^{(j)}|B_{i}/\delta_{i})$ time.

Then we use Lemma 4.8 to obtain a monotone step function $g_{0}$ which approximates $\min\{\bigoplus_{p_{k}\in P^{(j)}}f_{k},B_{0}\}$ with $1+O(\varepsilon)$ factor, in $\tilde{O}(\varepsilon^{-1}(|P^{(j)}|B_{0}+\varepsilon^{-1})B_{0}^{-0.01})$ time. Notice that $B_{0}=rt$.

Finally, $\max\{g_{0},g_{1},g_{2},\dots,g_{d}\}$ is a $1+O(\varepsilon)$ approximation of $\min\{\bigoplus_{p_{k}\in P^{(j)}}f_{k},B_{d}\}$, where $B_{d}\geq B$. Overall running time is

[TABLE]

Now we merge the results from all $r$ groups, and obtain an approximation of the final result $\min\{f_{1}\oplus\dots\oplus f_{m},B\}$.

Lemma 4.12**.**

Let $f_{1},\dots,f_{m}$ be given monotone step functions with ranges contained in $[0,B]$, and every $f_{k}$ is $p_{k}$-uniform and pseudo-concave for some $p_{k}\in[1,2]$. Assume $m=O(1/\varepsilon),\Omega(\varepsilon^{-0.01})\leq B\leq O(\varepsilon^{-1})$. We can approximate $\min\{f_{1}\oplus\dots\oplus f_{m},B\}$ with factor $1+O(\varepsilon)$ in $O(\varepsilon^{-2}B^{1/3}/2^{\Omega(\sqrt{\log(1/\varepsilon)})})$ time.

Proof 4.13**.**

Assume $m\geq\varepsilon^{-1}$, by adding zero functions which do not change the answer.

Let $r=o(B)$ be a positive integer parameter to be determined later.

Using Lemma 4.6 and Lemma 4.10, we can get a partition of $\{p_{1},\dots,p_{m}\}=P^{(1)}\cup\dots\cup P^{(r)}$ and then get an $1+O(\varepsilon)$ approximation of $\min\{\bigoplus_{p_{k}\in P^{(j)}}f_{k},B\}$ for every $1\leq j\leq r$, in $r\cdot(\varepsilon^{-2}/r^{0.01}+m\varepsilon^{-1}B^{1/2}/r^{3/2})(\log\varepsilon^{-1})^{O(d)}=(r^{0.99}+\sqrt{B/r})\varepsilon^{-2}(\log\varepsilon^{-1})^{O(\log\log\varepsilon^{-1})}$ overall time.

Then we use Lemma 3.1 to merge all these $r$ functions in $\tilde{O}((\frac{1}{\varepsilon})^{2}r/2^{\Omega(\sqrt{\log(1/\varepsilon)})})$ time.

Setting $r=B^{1/3}2^{c\sqrt{\log(1/\varepsilon)}}$, where $c>0$ is some small enough constant, the total complexity is

[TABLE]

Lemma 4.14**.**

Let $I$ be a set of $m$ items with $p_{i}\in[1,2]$ for every $i\in I$, where $\Omega(\varepsilon^{-2/3})\leq m\leq O(\varepsilon^{-1})$. One can approximate $f_{I}$ with factor $1+O(\varepsilon)$ in $O(\varepsilon^{-{3/2}}m^{3/4}/2^{\Omega(\sqrt{\log(1/\varepsilon)})})$ time.

Proof 4.15**.**

Let $f_{1},\dots,f_{m}$ denote the profit functions of the $m$ items.

Let $r=o(m^{1/2})$ be a positive integer parameter to be determined later. Obtain a partition of $\{p_{1},\dots,p_{m}\}=P^{(1)}\cup\dots\cup P^{(r)}$ using Lemma 4.6. Let $B:=\max_{i}\sum_{p\in P^{(i)}}p\leq 2\max_{i}|P^{(i)}|=\Theta(m/r)$. Then $r=o(B)$. Use Lemma 4.10 to get an $1+O(\varepsilon)$ approximation of $\bigoplus_{p_{k}\in P^{(j)}}f_{k}=\min\{\bigoplus_{p_{k}\in P^{(j)}}f_{k},B\}$ for every $1\leq j\leq r$, in $r\cdot(\varepsilon^{-2}/r^{0.01}+m\varepsilon^{-1}B^{1/2}/r^{3/2})(\log\varepsilon^{-1})^{O(d)}=(\varepsilon^{-2}r^{0.99}+m^{3/2}\varepsilon^{-1}/r)(\log\varepsilon^{-1})^{O(\log\log\varepsilon^{-1})}$ overall time.

Then we use Lemma 3.1 to merge all these $r$ functions in $\tilde{O}((\frac{1}{\varepsilon})^{2}r/2^{\Omega(\sqrt{\log(1/\varepsilon)})})$ time.

Setting $r=m^{3/4}\varepsilon^{1/2}2^{c\sqrt{\log(1/\varepsilon)}}$, where $c>0$ is some small enough constant, the total complexity is

[TABLE]

Corollary 4.16** (restated Theorem 1.2).**

For $n=O(\frac{1}{\varepsilon})$, there is a deterministic $(1+\varepsilon)$-approximation algorithm for 0-1 knapsack in O\Big{(}\big{(}n^{3/4}(\frac{1}{\varepsilon})^{3/2}+(\frac{1}{\varepsilon})^{2}\big{)}/2^{\Omega(\sqrt{\log(1/\varepsilon)})}\Big{)} time.

Proof 4.17**.**

Divide the items into $O(\log\frac{n}{\varepsilon})$ groups, each containing items with $p_{i}\in[2^{j},2^{j+1}]$ for some $j$. Use Lemma 4.14 to solve each group, and merge them using Lemma 3.1.

5 Main algorithm

5.1 A greedy lemma

Our improved algorithm uses the following lemma, which gives an upper bound on the total profit of cheap items (with low $p_{i}/w_{i}$) in an optimal knapsack solution.

Lemma 5.1**.**

Let $H,L$ be two subsets of items with $p_{i}\in[1,2]$. Let $W=\sum_{h\in H}w_{h}$ and $q=\min_{h\in H}\frac{p_{h}}{w_{h}}$. Suppose $\max_{l\in L}\frac{p_{l}}{w_{l}}\leq q(1-\alpha)$ for some $0<\alpha<1$. Let $f=f_{H}\oplus f_{L},\tilde{f}=f_{H}\oplus\min\{\frac{2}{\alpha},f_{L}\}$. Then for every $x\leq W$, $f(x)=\tilde{f}(x)$.

Proof 5.2**.**

By greedy, $f(W)=\sum_{h\in H}p_{h}=\tilde{f}(W)$ clearly holds. Now consider $0\leq x<W$. Suppose $f_{L}(x^{\prime})+f_{H}(x-x^{\prime})$ achieves its maximum value at $x^{\prime}=w_{L}$, i.e., $f(x)=f_{L}(w_{L})+f_{H}(x-w_{L})$. It suffices to prove $f_{L}(w_{L})\leq\frac{2}{\alpha}$.

Let $J\subseteq H$ be a subset of items with total weight $w_{J}\leq x-w_{L}$ and total profit achieving optimal value $f_{H}(x-w_{L})$. Let $K\subseteq H\backslash J$ be a subset of items with total weight $w_{K}$, such that $w_{K}\leq w_{L}$, and $w_{K}+w_{i}>w_{L}$ for every remaining item $i\in H\backslash(J\cup K)$. Such $K$ can be constructed by a simple greedy algorithm.

Since $w_{J}+w_{K}\leq(x-w_{L})+w_{L}<W=\sum_{h\in H}w_{h}$, the remaining set $H\backslash(J\cup K)$ contains at least one item $h_{0}$. Hence, $w_{L}-w_{K}<w_{h_{0}}=p_{h_{0}}/\frac{p_{h_{0}}}{w_{h_{0}}}\leq 2/q$, and equivalently $qw_{K}>qw_{L}-2$.

Since $J\cup K$ is a subset of $H$ with total weight bounded by $x$, we have $f_{H}(x)\geq\sum_{k\in K}p_{k}+\sum_{j\in J}p_{j}$, and thus $f_{H}(x)-f_{H}(x-w_{L})=f_{H}(x)-\sum_{j\in J}p_{j}\geq\sum_{k\in K}p_{k}\geq qw_{K}>qw_{L}-2$.

Hence $qw_{L}-2<f_{H}(x)-f_{H}(x-w_{L})\leq f(x)-f_{H}(x-w_{L})=f_{L}(w_{L})\leq q(1-\alpha)w_{L}$, which shows that $q\alpha w_{L}\leq 2$. So $f_{L}(w_{L})\leq q(1-\alpha)w_{L}\leq qw_{L}\leq 2/\alpha$, which concludes the proof.

5.2 FPTAS for Subset Sum

We will use the efficient FPTAS for the subset sum problem by Kellerer et al. [9] as a subroutine in our algorithm.

Lemma 5.3** ([9], implicit).**

Let $I$ be a set of $n$ items and $W$ be a number. We can obtain a list $S$ of $O(\frac{1}{\varepsilon})$ numbers in $O(n+(\frac{1}{\varepsilon})^{2}\log\frac{1}{\varepsilon})$ time, such that for every $s\leq W$ that is the subset sum $s=\sum_{j\in J}w_{j}$ of some subset $J\subseteq I$, there exists $s^{\prime}\in S$ with $s-\varepsilon W\leq s^{\prime}\leq s$.

Remark 5.4**.**

This result wasn’t explicitly stated in [9], but can be easily seen from their analysis of the correctness of the FPTAS.

Corollary 5.5**.**

Let $I$ be a set of $n$ items with $p_{i}\in[1,2]$ and $p_{i}=w_{i}$ for every item $i\in I$. We can approximate $f_{I}$ with factor $1+O(\varepsilon)$ in $O(n\log n+\varepsilon^{-2}\log\frac{1}{\varepsilon}\log n)$ time.

Proof 5.6**.**

Notice that approximating $s$ with additive error $\varepsilon W$ implies approximation factor $1+O(\varepsilon)$ for $W/2\leq s\leq W$. So we simply apply Lemma 5.3 with $W=2^{j}$ for $0\leq j\leq 1+\log n$, and merge all obtained lists.

5.3 Improved algorithm

Lemma 5.7**.**

Let $I$ be a set of $n$ items with $p_{i}\in[1,2]$ for every $i\in I$. We can approximate $f_{I}$ with factor $1+O(\varepsilon)$ in $O(n\log\frac{1}{\varepsilon}+(\frac{1}{\varepsilon})^{9/4}/2^{\Omega(\sqrt{\log(1/\varepsilon)})})$ time.

Proof 5.8**.**

Let $B=\lceil\varepsilon^{-1}\rceil$ and assume $n\geq B$ (if $n<B$, we can simply apply Lemma 4.14). By Lemma 3.3, we can approximate $f_{I}$ with additive error $O(\varepsilon B)$ in $O(n\log\frac{1}{\varepsilon})$ time, so we only need to approximate $\min\{f_{I},B\}$ with factor $1+O(\varepsilon)$.

We sort the items by their unit profits $p_{i}/w_{i}$. Let set $H$ contain the top $B$ items with the highest unit profits. Define $q=\min_{h\in H}\frac{p_{h}}{w_{h}}$, and let $M$ be the set of remaining items $i$ with $q(1-\alpha)\leq\frac{p_{i}}{w_{i}}\leq q$, where parameter $0<\alpha<1$ is to be determined later. Let set $L$ contain the remaining items not included in $H$ or $M$.

Using Lemma 4.14, we can compute $\tilde{f}_{H}$ which approximates $f_{H}$ with factor $1+O(\varepsilon)$ in time $O(B^{3/4}\varepsilon^{-3/2}/2^{\Omega(\sqrt{\log(1/\varepsilon)})})=O(\varepsilon^{-9/4}/2^{\Omega(\sqrt{\log(1/\varepsilon)})})$.

Since $\max_{l\in L}\frac{p_{l}}{w_{l}}<q(1-\alpha)$, Lemma 5.1 states that $f_{H}\oplus f_{L}$ and $f_{H}\oplus\min\{2/\alpha,f_{L}\}$ agree when $x\leq W_{H}=\sum_{h\in H}w_{h}$. Since $(f_{H}\oplus f_{L})(W_{H})=\sum_{h\in H}p_{h}\geq B$, this implies $\min\{B,f_{H}\oplus f_{L}\}=\min\{B,f_{H}\oplus\min\{2/\alpha,f_{L}\}\}$. For every item $l\in L$, we round down $p_{l}$ to a power of $1+\varepsilon$, so that there are only $\log_{1+\varepsilon}2=O(\varepsilon^{-1})$ distinct values. This only multiplies the approximation factor by $1+\varepsilon$. Then we use Lemma 4.12 to compute an approximation of $\min\{2/\alpha,f_{L}\}$ with factor $1+O(\varepsilon)$ in $\tilde{O}(\varepsilon^{-2}(2/\alpha)^{1/3}/2^{\Omega(\sqrt{\log(1/\varepsilon)})})$ time. We merge it with $\tilde{f}_{H}$ and obtain an approximation of $\min\{f_{H}\oplus f_{L},B\}$ with factor $1+O(\varepsilon)$.

For every $m\in M$, we round down $p_{m}$ so that the unit profit $p_{m}/w_{m}$ becomes a power of $1+\varepsilon$. After rounding, the approximation factor is only multiplied by $1+\varepsilon$, and there are at most $\log_{1+\varepsilon}\frac{q}{q(1-\alpha)}=O(\alpha/\varepsilon)$ distinct unit profits in $M$. Let $M_{q}$ denote the set of items in $M$ with unit profit $q$. For each $q$, we use Lemma 5.5 to obtain a $1+\varepsilon$ approximation of the function $f_{M_{q}}$ in $O(|M_{q}|+\varepsilon^{-2})$ time. Then we use Lemma 3.1 to merge these functions and obtain a $1+\varepsilon$ approximation of $f_{M}$. The total time is $O(|M|\log n)+\tilde{O}(\alpha\varepsilon^{-3})$.

Finally we merge the functions and get an approximation of $\min\{B,f_{L}\oplus f_{H}\oplus f_{M}\}$ with factor $1+O(\varepsilon)$. The total time is $O(n\log\frac{1}{\varepsilon})+\tilde{O}(\alpha\varepsilon^{-3}+\varepsilon^{-2}(2/\alpha)^{1/3}/2^{\Omega(\sqrt{\log(1/\varepsilon)})})$, which is $O(n\log\frac{1}{\varepsilon}+\varepsilon^{-9/4}/2^{\Omega(\sqrt{\log(1/\varepsilon)})})$ if we choose $\alpha=\varepsilon^{3/4}/2^{c\sqrt{\log(1/\varepsilon)}}$ for a sufficiently small constant $c$.

Corollary 5.9** (restated Theorem 1.1).**

There is a deterministic $(1+\varepsilon)$-approximation algorithm for 0-1 knapsack with running time $O(n\log\frac{1}{\varepsilon}+(\frac{1}{\varepsilon})^{9/4}/2^{\Omega(\sqrt{\log(1/\varepsilon)})})$.

Proof 5.10**.**

Divide the items into $O(\log\frac{n}{\varepsilon})$ groups, each containing items with $p_{i}\in[2^{j},2^{j+1}]$ for some $j$. Use Lemma 5.7 to solve each group, and merge them using Lemma 3.1.

Appendix A Proof of Lemma 4.3

Theorem A.1** (Reminder of Lemma 4.3).**

Let $T_{1},T_{2},\dots,T_{d}$ be positive real numbers satisfying $T_{1}\geq 2$ and $T_{i+1}\geq 2T_{i}$. There exist at least T_{d}\big{/}(\log T_{d})^{O(d)} integers $t$ satisfying the following condition: $t$ can be written as a product of integers $t=n_{1}n_{2}\cdots n_{d}$, such that $n_{1}n_{2}\cdots n_{i}\in(T_{i}/2,T_{i}]$ for every $1\leq i\leq d$.

Proof A.2**.**

For every $1\leq k\leq d$, we say an ordered $k$-tuple $(p_{1},p_{2},\dots,p_{k})$ is valid if every $p_{i}$ is prime, and $p_{1}p_{2}\cdots p_{i}\in(T_{i}/2,T_{i}]$ for every $1\leq i\leq k$. Then the product $t=p_{1}p_{2}\cdots p_{d}$ of any valid $d$-tuple $(p_{1},\dots,p_{d})$ satisfies our condition. For any integer $t$, there are at most $d!$ different valid $d$-tuples with product $t$ (which could be obtained by permuting $t$’s prime factors). Let $N_{k}$ denote the number of valid $k$-tuples. Then it suffices to show $N_{d}/(d!)\geq T_{d}/(\log T_{d})^{O(d)}$.

By the prime number theorem and Bertrand-Chebyshev theorem, there exists a positive constant $C$ such that

[TABLE]

where $\pi(x)$ denotes the number of primes less than or equal to $x$. We will prove $N_{k}\geq T_{k}/(C\log T_{k})^{k}$ for all $1\leq k\leq d$ by induction.

First note that this statement is trivial for $k=1$. For $k\geq 2$, a valid $k$-tuple $(p_{1},\dots,p_{k})$ can be obtained by appending any prime p_{k}\in\big{(}T_{k}/(2P),T_{k}/P\big{]} to any valid $(k-1)$-tuple $(p_{1},\dots,p_{k-1})$ with product $P=p_{1}\cdots p_{k-1}\leq T_{k-1}$. The number of such primes $p_{k}$ is

[TABLE]

Summing over all valid $(k-1)$-tuples, we have

[TABLE]

Hence, $N_{d}\geq T_{d}/(C\log T_{d})^{d}$ by induction. Observe that $T_{d}\geq 2^{d}$ and we have

[TABLE]

which finishes the proof.