Chao Xu
Professor at UESTC.
Almost optimal
\(\ell\)-covering of \(\mathbb{Z}_n\)
Ke Shi
Chao Xu
ke.shi@ustc.edu
\(\ell\)-covering
S_i^{\ell} = \{ 0i ,1i ,2i,\ldots,\ell i \pmod n \}
\(B\subseteq \mathbb{Z}_n\) is a \(\ell\)-covering of \(U\),
If \(U\subseteq \bigcup_{i\in B} S_i^{\ell}\)
We omit \(\ell\) and\(\pmod n\) when it is clear
A segment of length \(\ell\) with slope \(i\).
Find a small \(\ell\)-covering \(B\subseteq \mathbb{Z}_n\)
\(n=7, \ell=3\)
\(S_1\)
\(S_3\)
\(S_4\)
\(\mathbb{Z}_n\)
\(S_2\)
\(S_5\)
\(S_6\)
\(1\)
\(4\)
\(6\)
\(0\)
\(1\)
\(3\)
\(5\)
\(0\)
\(1\)
\(4\)
\(5\)
\(0\)
\(3\)
\(2\)
\(6\)
\(0\)
\(2\)
\(4\)
\(6\)
\(0\)
\(1\)
\(3\)
\(2\)
\(0\)
\(1\)
\(3\)
\(2\)
\(4\)
\(5\)
\(6\)
\(0\)
\(S_0\)
\(0\)
\(n=7, \ell=3\)
\(\mathbb{Z}_n\)
\(1\)
\(4\)
\(5\)
\(0\)
\(3\)
\(2\)
\(6\)
\(0\)
\(1\)
\(3\)
\(2\)
\(4\)
\(5\)
\(6\)
\(0\)
\(B=\{3,4\}\)
\(B\)
\(S_3\)
\(S_4\)
Find a small \(\ell\)-covering \(B\subseteq \mathbb{Z}_n\)
\(n=7, \ell=3\)
\(\mathbb{Z}_n\)
\(1\)
\(4\)
\(5\)
\(0\)
\(3\)
\(2\)
\(6\)
\(0\)
\(1\)
\(3\)
\(2\)
\(4\)
\(5\)
\(6\)
\(0\)
\(f(n,\ell)\) is the size of the smallest \(\ell\)-covering of \(\mathbb{Z}_n\).
\(f(7,3)\leq |B| = 2\)
\left\lceil \frac{n-1}{\ell} \right\rceil \leq f(n,\ell) \leq n-1
\( 2 = \left\lceil \frac{7-1}{3} \right\rceil \leq \)
\(B=\{3,4\}\)
Find a small \(\ell\)-covering \(B\subseteq \mathbb{Z}_n\)
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 \(\ell=2,3,4\), and when \(n\) and \(\ell\) are very special
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, construction time \(O(n\ell)\)
Chen et. al. 2013: \(f(n,\ell) \leq \frac{n^{1+o(1)}}{\sqrt{\ell}}\), similar construction time
Our results
\(f(n,\ell) = O(\frac{n}{\ell}\log n)\)
A \(f(n,\ell)\log n\) size \(\ell\)-covering can be found in time
\(\tilde{O}(\frac{n}{\ell}) + n^{o(1)}\)
There exist examples where
\(f(n,\ell)=\Omega(\frac{n}{\ell}\frac{\log n}{\log \log n})\)
Our results
\(f(n,\ell) = O(\frac{n}{\ell}\log n)\)
There exist examples where
\(f(n,\ell)=\Omega(\frac{n}{\ell}\frac{\log n}{\log \log n})\)
A \(f(n,\ell)\log n\) size \(\ell\)-covering can be found in time
\(\tilde{O}(\frac{n}{\ell}) + n^{o(1)}\)
Semigroup \((S,\diamond)\), \(A,B\subseteq S\). \(A\diamond B = \{a\diamond b | a\in A, b\in B\}\). \(B\) is an \(A\)-covering, if \(A\diamond B = S\).
Theorem Bollobas et. al. 2009
Let \(G\) be a group, for any \(A\subseteq G\), there is an \(A\)-covering of size \(\frac{|G|}{|A|}(\log |A|+1)\).
A more general covering
Our problem \(S=\Z_n\), \(\diamond = \cdot\), \(A = [\ell] \coloneqq \{0,1,2,\ldots, \ell\}\)
Idea: Reduce to group case
\begin{aligned}
&\Z_{n,d}&&=&&\{x|\gcd(x,n)=d,x\in\Z_n\} \\
& \displaystyle \Z_n &&=&& \cup_{d \mid n} \Z_{n,d}\\
&\Z_n^*&&=&&\Z_{n,1}
\end{aligned}
\((\Z^*_n,\cdot)\) forms a group. Find \(([\ell]\cap \Z^*_n)\)-covering of \(\Z^*_n\).
\(f^*(n,\ell)\) the smallest \(\ell\)-covering of \(\Z^*_n\) such that all segments has slope in \(\Z^*_n\).
\varphi(n,\ell) = |[\ell] \cap \Z^*_n|
\varphi(n,\ell) = \Omega(\frac{\ell}{n} \varphi(n)) \text{ if } \ell > n^{\Omega(\frac{1}{\log \log n})}
\varphi(n) = |\Z^*_n|
f^*(n,\ell) = O\left(\frac{n}{\ell}\log \ell\right)
When \(\ell = n^{\Omega(\frac{1}{\log \log n})}\)
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
\displaystyle f(n,\ell) \leq \sum_{d|n} f^*(\frac{n}{d},\ell)
\begin{aligned}
&\leq \sum_{d|n}O\left(\frac{n}{d}\frac{\log n}{\ell} \right)+1\\
&\leq O\left(\sigma(n)\frac{\log n}{\ell}\right) + d(n)\\
\end{aligned}
When \(\ell = n^{\Omega(\frac{1}{\log \log n})}\)
O\left(\frac{n}{\ell}\log n\log \log n\right) + n^{O\left(\frac{1}{\log \log n}\right)}
\leq
O\left(\frac{n}{\ell}\log n\log \log n\right)
=
When \(\ell = n^{1-\Omega(\frac{1}{\log \log n})}\)
Improvements
only using \(\Z^*_n\) to cover \(\Z_{n,d}\) for many \(d\)
A better estimate for \(\varphi(n,\ell)\) using sieve theory
\varphi(n,\ell) = \begin{cases}
\\
\\
\\
\end{cases}
\Omega(\frac{\ell}{n} \varphi(n))
\Omega(\frac{\ell}{\log \log n})
\begin{align*}
\text{ if } \ell & > c \log n\\
\text{ and } \ell & \leq c \log^5 n
\end{align*}
\text{ if } \ell > c \log^5 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
\(\ell = 4\)
\(g(n,\ell)\) the smallest \(\ell\)-covering of \(\bigcup_{d|n,d\leq \frac{\ell}{c\log^5 n}} \Z_{n,d}\) such that all segments has slope in \(\Z^*_n\).
g(n,\ell) = O(f^*(n,\ell))
\(\mathbb{D}_n\) be the set of divisors of \(n\).
\((\mathbb{D}_n,\odot)\), where \(a\odot b = \gcd(ab,n)\).
Let \(A = \{i \mid i\in \mathbb{D}_n, i\leq \frac{\ell}{c\log^5 n}\}\).
Find a \(A\)-covering \(B\) so the sum is small.
If \(B\) is an \(A\)-covering of \(\mathbb{D}_n\), then
\displaystyle f(n,\ell) \leq \sum_{d\in B} g\left(\frac{n}{d},\ell\right)
f(n,\ell) = O\left(\frac{n}{\ell}\log n\right)
Future Work
Conjecture:
\varphi(n,\ell) =
\Omega(\frac{\ell}{n} \varphi(n)) \text{ if } \ell > c \log n
Conjecture:
O(\frac{n}{\ell}\log n) \text{ tight? } f(n,\ell) = O(\frac{n}{\ell}\log \ell) ?
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
A \(O(f(n,\ell)\log n)\) size \(\ell\)-covering can be found in time
\(\tilde{O}(\frac{n}{\ell}) + n^{o(1)}\)
Symmetric LP
Randomized rounding
Sampling
Symmetric LP
Example:
\(S_n\)-symmetric LP
\begin{aligned}
& \min_{x} & & \sum_{i=1}^n x_i & \\
& \text{s.t.} & & \sum_{i=1}^n x_i\geq n
\end{aligned}
Symmetric LP
\(G\)-symmetric LP has a \(G\)-symmetric optimum.
c\cdot x^* = c\cdot \pi x^*
\begin{aligned}
& \min_{x} & & \sum_{i=1}^n x_i & \\
& \text{s.t.} & & \sum_{i=1}^n x_i\geq n
\end{aligned}
x = (n,0,0,\ldots,0)
Optimum
\(G\)-symmetric optimum
x = (1,1,1,\ldots,1)
x_i = x_j ~ \forall i,j
Let \(G\) be a permutation group that consists of all permutations that swaps \(i\) and \(j\) if \(\gcd(i,n)=\gcd(j,n)\). LP is \(G\)-symmetric.
Reduce variables
\begin{aligned}
& \min_{x} & & \sum_{i} x_i & \\
& \text{s.t.} & & \sum_{i:j\in S_i} x_i\geq 1 & j\in \Z_n \\
& & & 0 \leq x_i \leq 1 & i\in \Z_n \\
\end{aligned}
\begin{aligned}
& & & x_i = x_j & \text{ if }\gcd(i,n)=\gcd(j,n) \\
\end{aligned}
Replace each \(x_i\) with \(y_d\) where \(\gcd(i,n)=d\)
Reduce variables
\begin{aligned}
& \min_{y} & & \sum_{i} y_{\gcd(i,n)} & \\
& \text{s.t.} & & \sum_{i:j\in S_i} y_{\gcd(i,n)}\geq 1 & j\in \Z_n \\
& & & 0 \leq y_{\gcd(i,n)} \leq 1 & i\in \Z_n \\
\end{aligned}
Reduce variables
Reduce variables
\begin{aligned}
& \min_{y} & & \sum_{d:d|n} \varphi(\frac{n}{d}) y_d & \\
& \text{s.t.} & & \sum_{d:d|j} \frac{\varphi(n/c, d \ell / c)\varphi(n/d)}{\varphi(n/c)} y_d\geq 1 & j\in \Z_n \\
& & & 0\leq y_d\leq 1 & d|n \\
\end{aligned}
We have only \(d(n)\) variables.
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\leq 1 & d|n \\
\end{aligned}
We have only \(d(n)\) variables and inequalities.
\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\leq \varphi(n/d) & d|n
\end{aligned}
Let \(u_d=\varphi(n/d)y_d\), We have the following concise form.
Randomized rounding
\(\underline{\textbf{Cover}(n,\ell)}\)
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 \(\mathrm{LP}^*\)
\(C\leftarrow \empty\)
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 \(C\)
Sample ~\(\mathbb{E}[u_d]\)
distinct elements
from \(\Z_{n,d}\)
Construct and solve the LP
Sampling \(k\) elements in \(\Z_n^*\)
If \(k<\frac{\varphi(n)}{2}\). Sample a random element \(x\sim \Z_n\), if \(x\in\Z_n^*\) and we hasn't get enough elements, pick it. Repeat at most \(8k \log(n)\) times.
The runing time of the algorithm is \(O(k \mathrm{polylog}(n))\) and by Chernoff bound the success probability \(\geq 1-\frac{1}{n}\).
If \(k\geq \frac{\varphi(n)}{2}\), then find all elements in \(\Z_n^*\) by brute force and pick \(k\) from it.
Theorem: There is an algorithm which successfully returns \(X\sim {\Z^*_n \choose k}\) in \(\tilde{O}(k)\) time with probability at least \(1-\frac{1}{n}\).
Sampling \(k\) elements in \(\Z_n^*\)
Randomized rounding
\(\underline{\textbf{Cover}(n,\ell)}\)
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 \(\mathrm{LP}*\)
\(C\leftarrow \empty\)
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 \(C\)
\(\tilde{O}(u_d) \)
\(n^{o(1)}\)
\(\tilde{O}(\sum_{d} u_d) = \tilde{O}(f(n,\ell))\)
Theorem:
The algorithm takes \(\tilde{O}(f(n,\ell)) + n^{o(1)}\) time and computes an expected size \(f(n,\ell)\log n\) \(\ell\)-covering set with constant probability.
l-covering for cocoon
By Chao Xu
l-covering for cocoon
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