Chao Xu
Professor at UESTC.
Almost tight \(\ell\)-covering of \(\mathbb{Z}_n\)
Ke Shi
Chao Xu
Flash Memory Storage
state \(\in \{0,\ldots,\ell\}\)
cell stores a state
block \(\sim 10^6\) cells
Operations
increase a state
= single cell update
set a state to 0
= clear entire block
Abstract memory cell (AMC): stores a value in \(\N\)
- Read: read the number in the cell
- Update: increase the cell by a number in \(\{0,\ldots,\ell\}\)
\(\Z_n=\{0,1,2,\ldots,n-1\}\)
Store and rewrite a number in \(\Z_n\)
1 rewrite = 1 AMC update
Store a number in \(\Z_n\) use AMCs
Store a number in \(\Z_n\) use AMCs
store \(y\in \Z_n\) by storing \(x\in \N^t\)
\(s\in \Z_n^t\) a fixed vector
D(x) = s\cdot x \pmod n = y
\(2\)
\(4\)
\(3\)
\(1\)
\(D(x) = (1,2,3,4) ·(1,7,2,9)\pmod{12} =9\)
Example:
AMC storage
t=4
n=12
y=9
s=(1,7,2,9)
x
1 rewrite = 1 AMC update
\(D(x) = (1,2,3,4) ·(1,7,2,9)\pmod{12} =9\)
Example:
AMC storage
t=4
n=12
y=9
s=(1,7,2,9)
x
\(D(x) = (1,2,{\color{red}5},4) ·(1,7,2,9)\pmod{12} ={\color{red}1}\)
\(2\)
\(4\)
\(3\)
\(1\)
\(5\)
1 AMC update per rewrite is possible if and only if
\displaystyle \Z_n = \bigcup_{i\in s}\{ei\pmod{n} | e\leq \ell\}
x_j\gets x_j + e \Leftrightarrow y \gets y+e s_j\pmod{n}
y={\color{red}1}
1 rewrite = 1 AMC update
1 AMC update per rewrite is possible if and only if
\displaystyle \Z_n = \bigcup_{i\in s}\{ei\pmod{n} | e\leq \ell\}
\(t\) should be as small as possible to save space
For a fixed \(\ell\) and \(n\), how large must \(t\) be?
\(\ell\)-covering
S_i^{\ell} = \{0i,1i,2i,\ldots, \ell i\}
\(A\subseteq \mathbb{Z}_n\) is a \(\ell\)-covering of \(U\),
If \(U\subseteq \bigcup_{i\in A} S_i^{\ell}\).
\(U=\Z_n\) most of the time.
We omit \(\ell\) when it is clear
A segment of length \(\ell\) with slope \(i\).
Find a small \(\ell\)-covering \(A\subseteq \mathbb{Z}_n\)
\(n=7, \ell=3\)
\(S_1\)
\(S_3\)
\(S_4\)
\(\mathbb{Z}_n\)
\(S_2\)
\(S_5\)
\(S_6\)
\(5\)
\(4\)
\(6\)
\(0\)
\(1\)
\(3\)
\(5\)
\(0\)
\(1\)
\(4\)
\(5\)
\(0\)
\(3\)
\(2\)
\(6\)
\(0\)
\(2\)
\(4\)
\(6\)
\(0\)
\(1\)
\(3\)
\(2\)
\(0\)
\(1\)
\(3\)
\(2\)
\(4\)
\(5\)
\(6\)
\(0\)
\(S_0\)
\(0\)
\(n=7, \ell=3\)
\(\mathbb{Z}_n\)
\(1\)
\(4\)
\(5\)
\(0\)
\(3\)
\(2\)
\(6\)
\(0\)
\(1\)
\(3\)
\(2\)
\(4\)
\(5\)
\(6\)
\(0\)
\(A=\{3,4\}\)
\(A\)
\(S_3\)
\(S_4\)
Find a small \(\ell\)-covering \(A\subseteq \mathbb{Z}_n\)
\(n=7, \ell=3\)
\(\mathbb{Z}_n\)
\(1\)
\(4\)
\(5\)
\(0\)
\(3\)
\(2\)
\(6\)
\(0\)
\(1\)
\(3\)
\(2\)
\(4\)
\(5\)
\(6\)
\(0\)
\(f(n,\ell)\) is the size of the smallest \(\ell\)-covering of \(\mathbb{Z}_n\).
\(f(7,3)\leq |A| = 2\)
\left\lceil \frac{n-1}{\ell} \right\rceil \leq f(n,\ell) \leq n-1
\( 2 = \left\lceil \frac{7-1}{3} \right\rceil \leq \)
\(A=\{3,4\}\)
Find a small \(\ell\)-covering \(A\subseteq \mathbb{Z}_n\)
Main Questions
Combinatorial: What is the correct bound for \(f(n,\ell)\)?
Algorithmic: How to find a small \(\ell\)-covering?
\(f(n,\ell)=\left\lceil \frac{n-1}{\ell} \right\rceil\)?
No.
\(f(16,5)>3=\lceil \frac{16-1}{5}\rceil\)
Klove et. al. multiple papers in the 2010s: \(f(n,\ell)\) for \(\ell=2,3,4\)
Chen et. al. 2013: \(f(n,\ell) \leq \frac{n^{1+o(1)}}{\sqrt{\ell}}\)
Previous Work
\(f(n,\ell) = O(\frac{n}{\ell}\log n\log \log n)\)
if \(\ell\) is neither too small nor too large
n^{\Omega(\frac{1}{\log \log n})} \leq \ell \leq n^{1-\Omega(\frac{1}{\log \log n})}
Koiliaris & Xu 2019
Our results
\(f(n,\ell) = O(\frac{n}{\ell}\log n\log \log n)\)
A \(O(f(n,\ell)\log n)\) size \(\ell\)-covering can be found in time
\(\tilde{O}(\frac{n}{\ell}) + n^{o(1)}\)
There exist examples where
\(f(n,\ell)=\Omega(\frac{n}{\ell}\frac{\log n}{\log \log n})\)
\(f(n,\ell) = O(\frac{n}{\ell}\log n\log \log n)\)
* \(\ell\) is not too small and not too large.
*
A more refined analysis of \(\varphi(n,\ell)\).
A more refined analysis of covering.
\displaystyle\varphi(n,\ell)=\sum_{\substack{
x\leq\ell \\
\gcd(x,n)=1}}1
\varphi(n)=\varphi(n,n)
Set cover
Given \(\mathcal{S} \subseteq 2^U\).
A collection of sets in \(\mathcal{S}\) covers \(U\) is a set cover.
Formally, if \(\mathcal{S}'\subseteq \mathcal{S}\) such that \(U= \bigcup_{X\in \mathcal{S}'} X\), then \(\mathcal{S}'\) is a set cover of \(U\).
\(\ell\)-covering as set cover
\(U=\Z_n\)
\(S_i=\{ij\pmod{n}|j\leq\ell\}\)
\(\mathcal{S}=\{S_i|i\in\Z_n\}\)
An algebraic special case
Each elements covered by \(b\) sets
\(O\left(\frac{|\mathcal{S}|}{b}\log \ell\right)\) size set cover
(naively) apply to \(\ell\)-covering?
\begin{aligned}
&\Z_{n,d}&&=&&\{x|\gcd(x,n)=d,x\in\Z_n\} \\
&\Z_n^*&&=&&\Z_{n,1}
\end{aligned}
\(p\) a large prime, \( n=2p, \ell=p-1\). \(p\) only covered by \(S_p\).
No
\(f^*(n,\ell)\) the smallest \(\ell\)-covering of \(\Z^*_n\) such that all segments has slope in \(\Z^*_n\).
Covering \(\Z_n^*\) with \(\Z_n^*\) slope segments
Every elment in \(\Z^*_n\) is covered by \(\varphi(n,\ell)\) segments with slope in \(\Z^*_n\).
Previous bound for \(\varphi(n,\ell)\)
\varphi(n,\ell) = \begin{cases}
\Omega(\frac{\ell}{n} \varphi(n))& \text{ if } \ell > n^{\Omega(\frac{1}{\log \log n})}\\
\Omega(\frac{\ell}{\log \ell}) & \text{ if } \ell > c \log n\\
\end{cases}
Constants \(c,u\geq 1\)
\varphi(n,\ell) = \begin{cases}
\Omega(\frac{\ell}{n} \varphi(n))& \text{ if } \ell > n^{\Omega(\frac{1}{\log \log n})}\\
\Omega(\frac{\ell}{\log \ell}) & \text{ if } \ell > c \log n\\
\end{cases}
c \log^u n
Previous bound for \(\varphi(n,\ell)\)
Case 1: \(\ell > c \log^u n\)
Sieve Theory
\(\pi(\ell)\geq c_1\frac{\ell}{\log \ell}\) (Prime Number Theorem)
\(\omega(n)\leq c_2\frac{\log n}{\log \log n}\).
\(\omega(n)\leq \frac{\pi(\ell)}{2} \) if \(\ell \geq c\log n\) for sufficiently large \(n\).
Case 2: \(\ell > c \log n\)
\varphi(n,\ell)\geq \pi(\ell)-\omega(n)
\(\pi(n)\): number of primes no larger than \(n\)
\(\omega(n)\): number of distinct prime factors of \(n\)
\varphi(n,\ell)=\Omega\left(\frac{\ell}{\log \ell}\right)
Bound for \(f^*(n,\ell)\)
f^*(n,\ell) = O\left(\frac{n}{\ell}\log n\right)
Using the fact about \(\varphi(n,\ell)\).
Case 1: \(\ell > c \log^u n\)
\begin{aligned}
f^*(n,\ell) &= O\left(\frac{\varphi(n)\log \ell}{\frac{\ell}{n}\varphi(n)}\right)\\
&= O\left(\frac{n}{\ell}\log n\right)
\end{aligned}
\text{each element covered by } b \text{ sets} \Rightarrow O(\frac{|S|\log \ell}{b})
\varphi(n,\ell) = \Omega(\frac{\ell}{n}\varphi(n))
\begin{aligned}
f^*(n,\ell) &= O\left(\frac{\varphi(n)\log \ell}{\frac{\ell}{\log \ell}}\right)\\
&= O\left(\frac{\varphi(n)}{\ell}(\log \ell)^2\right) \\
&= O\left(\frac{n}{\ell}(\log \log n)^2\right)\\
&= O\left(\frac{n}{\ell}\log n\right)
\end{aligned}
Case 2: \(c\log n<\ell < c \log^u n\)
O(\frac{|S|\log \ell}{b})
\varphi(n,\ell) = \Omega(\frac{\ell}{\log \ell})
\ell = O(\log^u n)
\begin{aligned}
f^*(n,\ell)&\leq \varphi(n) \\
&= O\left(\varphi(n)\frac{\log n}{\ell}\right) \\
&= O\left(\frac{n}{\ell}\log n\right)
\end{aligned}
Case 3: \(\ell < c \log n\)
We can cover \(\mathbb{Z}_{n,d}\) with \(f^*(\frac{n}{d},\ell) \) segments in \(\mathbb{Z}_{n,d}\).
We can cover \(\mathbb{Z}_n^*\) with \(f^*(n, \ell)\) segments in \(\mathbb{Z}_n^*\).
\(\mathbb{Z}_n\)
We can cover \(\mathbb{Z}_{n,d}\) with \(f^*(\frac{n}{d},\ell) \) segments in \(\mathbb{Z}_{n,d}\).
\(\mathbb{Z}_n\)
\(d \mid x\)
We can cover \(\mathbb{Z}_n^*\) with \(f^*(n, \ell)\) segments in \(\mathbb{Z}_n^*\).
We can cover \(\mathbb{Z}_{n,d}\) with \(f^*(\frac{n}{d},\ell) \) segments in \(\mathbb{Z}_{n,d}\).
\(\mathbb{Z}_n\)
\(d \mid x\)
\(\mathbb{Z}_{n,d}\)
We can cover \(\mathbb{Z}_n^*\) with \(f^*(n, \ell)\) segments in \(\mathbb{Z}_n^*\).
We can cover \(\mathbb{Z}_{n,d}\) with \(f^*(\frac{n}{d},\ell) \) segments in \(\mathbb{Z}_{n,d}\).
\(\mathbb{Z}_n\)
\(d \mid x\)
\(\mathbb{Z}_{n,d}\)
\(\mathbb{Z}^{*}_{\frac{n}{d}}\)
We can cover \(\mathbb{Z}_n^*\) with \(f^*(n, \ell)\) segments in \(\mathbb{Z}_n^*\).
Cover of \(\Z^*_\frac{n}{d}\) using \(\Z^*_{\frac{n}{d}}\) lifts to a cover of \(\Z_{n,d}\) using \(\Z_{n,d}\)
\mathbb{Z}^*_{12}
\mathbb{Z}_{12,2}
\mathbb{Z}_{12,3}
\mathbb{Z}_{12,4}
\mathbb{Z}_{12,6}
\mathbb{Z}_{12,12}
f^*(12,4)
\ell=4
\mathbb{Z}^*_{12}
\mathbb{Z}_{12,2}
\mathbb{Z}_{12,3}
\mathbb{Z}_{12,4}
\mathbb{Z}_{12,6}
\mathbb{Z}_{12,12}
f^*(12,4)
f^*(6,4)
\ell=4
\mathbb{Z}^*_{12}
\mathbb{Z}_{12,2}
\mathbb{Z}_{12,3}
\mathbb{Z}_{12,4}
\mathbb{Z}_{12,6}
\mathbb{Z}_{12,12}
f^*(12,4)
f^*(6,4)
f^*(3,4)
f^*(4,4)
f^*(2,4)
f^*(1,4)
f(12,4)
+
+
+
+
+
\leq
\ell=4
Bound for \(f(n,\ell)\)
\(d(n) = n^{O\left(\frac{1}{\log \log n}\right)}\)
\(\sigma(n) = O(n\log \log n)\)
f(n,\ell) \leq \sum_{d|n} f^*(\frac{n}{d},\ell)
\leq \sum_{d|n} \max(d \frac{\log n}{\ell},1)
\leq \sigma(n)\frac{\log n}{\ell} + d(n)
= \underbrace{ O\left(\frac{n}{\ell}\log n\log \log n\right)}_{\textbf{main term}} + {\color{red}\underbrace{n^{O\left(\frac{1}{\log \log n}\right)}}_{\textbf{error term}}}
Desired bound unless \(\ell\) is too large.
f^*(n,\ell) = O(\frac{n}{\ell}\log n)
\mathbb{Z}^*_{12}
\mathbb{Z}_{12,2}
\mathbb{Z}_{12,3}
\mathbb{Z}_{12,4}
\mathbb{Z}_{12,6}
\mathbb{Z}_{12,12}
Larger Cover
\(\ell = 4\)
Theorem:
If \(\ell\) is large, \(d\leq \ell'\), then each element in \(\Z_{n,d}\) is covered \(\Omega(\frac{\ell}{n}\varphi(n))\) times by segments with slope in \(\Z^*_n\).
\ell' := \frac{\ell}{c \log^u \frac{n}{\ell}}
Theorem:
If \(\ell\) is large, each element in \(\Z^*_n\) is covered \(\Omega(\frac{\ell}{n}\varphi(n))\) times by segments with slope in \(\Z^*_n\).
\(f^\dagger(n,\ell)\) the smallest \(\ell\)-covering of \(\bigcup_{d|n,d\leq \ell'} \Z_{n,d}\) such that all segments has slope in \(\Z^*_n\).
f^\dagger(n,\ell) = O(\frac{n}{\ell}\log n)
f^*(n,\ell) = O(\frac{n}{\ell}\log n)
What's the relation between \(f(n,\ell)\) and \(f^\dagger(n,\ell)\)?
\displaystyle f(n,\ell) \leq \sum_{d|m} f^\dagger(\frac{n}{d},\ell)
\displaystyle n = \prod_{i=1}^k p_i
\displaystyle m = \prod_{i=1}^j p_i \geq \frac{n}{\ell'}
If \(d|n\), then \(d=d_1d_2\), where \(d_1|m\), and \(d_2\leq \ell'\).
How large is \(d(m)\)?
\begin{aligned}
&\leq \sum_{d|m}O\left(\frac{n}{d}\frac{\log n}{\ell} \right)+1\\
&\leq O\left(\sigma(n)\frac{\log n}{\ell}\right) + d(m)
\end{aligned}
d(n) = n^{O(\frac{1}{\log \log n})}
d(m)
\leq 2d(m')
=m'^{O(\frac{1}{\log \log m'})}
= \left(\frac{n}{\ell}\log^u \frac{n}{\ell}\right)^{O\left(\frac{1}{\log \log \frac{n}{\ell}}\right)}
= \frac{n}{\ell} c\log^u \frac{n}{\ell}
< \frac{n}{\ell'}
m':=\frac{m}{p_j}
\begin{aligned}
f(n,\ell) &\leq \sum_{d|m} f^\dagger(\frac{n}{d},\ell)\\
&\leq \sum_{d|m}O\left(\frac{n}{d}\frac{\log n}{\ell} \right)+1\\
&\leq O\left(\sigma(n)\frac{\log n}{\ell}\right) + d(m)\\
&= \underbrace{ O\left(\frac{n}{\ell}\log n\log \log n\right)}_{\textbf{main term}} + {\color{red} \underbrace{\left(\frac{n}{\ell}\log^u \frac{n}{\ell}\right)^{O\left(\frac{1}{\log \log \frac{n}{\ell}}\right)}}_{\textbf{error term}}}\\
&= O\left(\frac{n}{\ell}\log n\log \log n\right)
\end{aligned}
\(f(n,\ell) = O(\frac{n}{\ell}\log n\log \log n)\)
\(f(n,\ell) = O(\frac{n}{\ell}\log n\log \log n)\)
Conclusion and Future Work
Better bound for \(f(n,\ell)\)?
What about \(\ell\)-covering of arbitrary set \(X\subseteq \Z_n\)?
Other interesting groups, e.g. \(\Z_n\times \Z_m\).
IP
\begin{aligned}
& \min_{x} & & \sum_{i} x_i & \\
& \text{s.t.} & & \sum_{i:j\in i[\ell]} x_i\geq 1 & j\in \Z_n \\
& & & x_i \in \{0,1 \} & i\in \Z_n \\
\end{aligned}
Integer programming form of set cover
LP (Relaxation)
\begin{aligned}
& \min_{x} & & \sum_{i} x_i & \\
& \text{s.t.} & & \sum_{i:j\in i[\ell]} x_i\geq 1 & j\in \Z_n \\
& & & 0 \leq x_i & i\in \Z_n \\
\end{aligned}
Symmetric LP
A linear program is symmetric if there exists a permutation of variables, so the linear program does not change it's value.
\begin{bmatrix} x_1 \\ x_2 \\.\\.\\.\\x_n \end{bmatrix}
\xRightarrow[\textbf{permutation group G}]{\forall \textbf{permutation } \pi \text{ in}}
\begin{bmatrix} x_{\pi(1)} \\ x_{\pi(2)} \\.\\.\\.\\x_{\pi(n)} \end{bmatrix}
\(\textbf{fesible}\) \(\xRightarrow[\textbf{ }]{}\) \({}\textbf{fesible}\)
\(\textbf{c}\cdot\textbf{x}\) \(\xRightarrow[\textbf{ }]{\textbf{same value}}\) \(\textbf{c}\cdot\pi\textbf{ x}\)
We call such a LP \(G\)-symmetric.
Reduce variables
Let \(G\) be a permutation group that consists of all permutations that swaps \(i\) and \(j\) if \(\gcd(i,n)=\gcd(j,n)\). LP2 is \(G\)-symmetric.
\begin{aligned}
\{x_i:i\in \Z_n\}&\longrightarrow &&\{y_d:d|n\}\\
\text{number:}\quad n&\longrightarrow &&d(n)
\end{aligned}
We let \(y_d\) to represent all variables in \(\{x_i|\gcd(i,n)=d\}\).
Reduce constraints
If \(\gcd(a,n) = \gcd(b,n)\), then the constraints that covers \(a\) and \(b\) have same format since all variables \(y_d\) have same coffcients in inequalities \(a\) and \(b\).
\begin{aligned}
\{\mathrm{constraint}_i:i\in \Z_n\}&\longrightarrow &&\{\mathrm{constraint}_d:d|n\}\\
n&\longrightarrow &&d(n)
\end{aligned}
LP (compact)
\begin{aligned}
& \min_{y} & & \sum_{d:d|n} \varphi(\frac{n}{d}) y_d & \\
& \text{s.t.} & & \sum_{d:d|c} \frac{\varphi(n/c, d \ell / c)\varphi(n/d)}{\varphi(n/c)} y_d\geq 1 & c|n \\
& & & 0\leq y_d & d|n \\
\end{aligned}
We have only \(d(n)\) variables and inequalities now!
\begin{aligned}
& \min_{y} & & \sum_{d:d|n} \color{red}{\varphi(\frac{n}{d}) y_d} & \\
& \text{s.t.} & & \sum_{d:d|c} \frac{\varphi(n/c, d \ell / c)\color{red}{\varphi(n/d)}}{\varphi(n/c)} {\color{red}y_d} \geq 1 & c|n \\
& & & 0\leq y_d & d|n \\
\end{aligned}
LP*
\begin{aligned}
& \min_{d} & & \sum_{d:d|n} u_d & \\
& \text{s.t.} & & \sum_{d:d|c} \frac{\varphi(n/c, d \ell / c)}{\varphi(n/c)} u_d\geq 1 & c|n \\
& & & 0\leq u_d & d|n
\end{aligned}
Let \(u_d=\varphi(n/d)y_d\), We have the following concise form.
Randomized algorithm
\(\underline{\textbf{Cover}(n,\ell)}\)
compute the prime factorization of \(n\)
compute \(\varphi(n/c,d\ell/c)\) and \(\varphi(n/c)\) for all \(d|c|n\)
compute the optimum for \(\mathrm{LP4}\)
\(C\leftarrow \empty\)
repeat \(4\log{n}\) times
for \(d|n\):
if \(u_d\geq1\):
\(C\leftarrow C\cup (S\sim {\Z_{n,d}\choose \lceil u_d \rceil})\)
else if \(u_d > (x\sim[0,1])\):
\(C\leftarrow C\cup (S\sim {\Z_{n,d}\choose 1})\)
return \(C\)
taking \(O(f(n,\ell)\mathrm{ploylog}(n)) \) time
taking \(n^{o(1)}\) time
taking \(n^{o(1)}\) time
taking \(n^{o(1)}\) time
Theorem:
Algorithm before takes \(O(f(n,\ell)\mathrm{polylog} n + n^{o(1)})\) time and computes a expected size \(O(f(n,\ell)\log n)\) \(\ell\)-covering set in expectation with constant probability.
Theorem: There is an algorithm which successfully returns \(X\sim {\Z^*_n \choose k}\) in \(O(k \mathrm{polylog}(n))\) time with probability at least \(1-\frac{1}{n}\).
Sampling in \(\Z_n^*\)
Running time analysis
Time of constructing the LP.
Step 1. Finding the prime factorization of \(n\) takes \(n^{o(1)}\) time through general number field sieve.
Step 2. Computing all \(\varphi(n/c),c|n\) through \(\varphi(n)=n\prod_{p|n}(1-1/p)\) takes \(n^{o(1)}\) time.
Sampling in \(\Z_n^*\)
Otherwise, \(k<\frac{\varphi(n)}{2}\). We will sample a random element \(x\sim \Z_n\), if \(x\in\Z_n^*\) and we hasn't get enough elements, pick it. Repeat it specific times.
We can show the runing time of the algorithm is \(O(k \mathrm{polylog}(n))\) and Chernoff bound can show the success probability at least \(1-\frac{1}{n}\).
Consider two cases. If \(k\geq \frac{\varphi(n)}{2}\), then find all elements in \(\Z_n^*\) by brute force and pick \(k\) from it.
Running time analysis
Step 3. Computing all \(\varphi(n/c,d\ell/c),d|c|n\) through the inclusion-exclusion principle takes \(n^{o(1)}\) time.
\varphi(n,\ell) = \sum_{i=0}^{\omega(n)} (-1)^{i} \sum_{S\subset P, |S|=i} \left\lfloor\frac{\ell}{\prod_{p\in S} p} \right\rfloor
The running time of computing \(\varphi(n,\ell)\) is \(\tilde{O}(2^{\omega(n)})=n^{O(\frac{1}{\log\log n})}=n^{o(1)}\). The number of all \((c,d)\) is also \(n^{o(1)}\).
Running time analysis
The total time of sampling is :
\(O(\sum_{d|n}u_d \mathrm{polylog}n)=O(f(n,\ell)\mathrm{ploylog}(n)) \).
Combining all above, the algorithm takes \((O(f(n,\ell)\mathrm{ploylog}(n)+n^{o(1)})\) time.
Thank you
Limited-magnitude errors:
Jiang A., Langberg M., Schwartz M., Bruck J.: Trajectory codes for flash memory. IEEE Trans. Inf. Theory 59(7), 4530–4541 (2013).
Sieve theory:
Alina Carmen Cojocaru, M Ram Murty, et al. An introduction to sieve methods and their applications, volume 66. Cambridge University Press, 2006.
References
l-covering
By Chao Xu
