Chao Xu
Professor at UESTC.
Almost tight \(\ell\)-covering of \(\mathbb{Z}_n\)
Ke Shi
Chao Xu
cxu@uestc.edu.cn
- 博士生，硕士生
- 博士后
- 助理教授 （50w+年薪）
电子科技大学
理论计算机科学团队
https://tcsuestc.com/
逻辑，组合优化，图论，机制设计，计算代数，实验算法，精确算法，分配机制，算法博弈论
Flash Memory Storage
state \(\in [\ell]=\{0,\ldots,\ell\}\)
cell stores a state
block \(\sim 10^6\) cells
Operations
increase a state
= single cell update
set a state to 0
= clear entire block
write \(10^6\) cells
Cell stores a value in \(\N\)
- Read: read the number in the cell
- Increase: increase the cell by a number in \([\ell]\)
A simpler increment only model
\(\Z_n=\{0,1,2,\ldots,n-1\}\)
Store and update a number in \(\Z_n\)
1 update = 1 cell increase
Store a number in \(\Z_n\)
Store a number in \(\Z_n\)
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:
Cell storage
t=4
n=12
y=9
s=(1,7,2,9)
x
1 update = 1 cell increase
\(D(x) = (1,2,3,4) ·(1,7,2,9)\pmod{12} =9\)
Example:
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 increase per update is possible if and only if
\displaystyle \Z_n = \bigcup_{i\in s}\{ei\pmod{n} | e\in [\ell]\}
x_j\gets x_j + e \Leftrightarrow y \gets y+e s_j\pmod{n}
y={\color{red}1}
Cell storage
\(t\) should be as small as possible to save space
For a fixed \(\ell\) and \(n\), how large must \(t\) be?
1 increase per update is possible if and only if
\displaystyle \Z_n = \bigcup_{i\in s}\{ei\pmod{n} | e\in [\ell]\}
1 update = 1 cell increase
\(\ell\)-covering
S_i^{\ell} = \{0i,1i,2i,\ldots, \ell i\}
\(B\subseteq \mathbb{Z}_n\) is a \(\ell\)-covering of \(U\),
If \(U\subseteq \bigcup_{i\in B} S_i^{\ell}\)
We omit \(\ell\) 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 \(B\subseteq \mathbb{Z}_n\)
IP
\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 \\
& & & x_i \in \{0,1 \} & i\in \Z_n \\
\end{aligned}
f(n,\ell) = OPT(IP)
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
\(\tilde{O}(\frac{n}{\ell}) + n^{o(1)}\)
\(f(n,\ell) = O(\frac{n}{\ell}\log n)\)
* \(\ell\) 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)
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 \(\frac{|G|}{|A|}(\log |A|+1)\).
\((\Z_{2n},\cdot)\), \(A=\{2,4,6,\ldots,2n\}\). No \(A\)-covering exists.
A more general covering
Our problem \(S=\Z_n\), \(\diamond = \cdot\), \(A = [\ell]\)
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
Idea: Reduce to group case
\begin{aligned}
&\Z_{n,d}&&=&&\{x|\gcd(x,n)=d,x\in\Z_n\} \\
&\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 \(\ell = n^{\Omega(\frac{1}{\log \log n})}\)
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}
Constant \(c \geq 1\)
\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^5 n
Previous bound for \(\varphi(n,\ell)\)
Case 1: \(\ell > c \log^5 n\)
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 \(\varphi(n,\ell)\).
When \(\ell = n^{\Omega(\frac{1}{\log \log n})}\)
Case 1: \(\ell > c \log^5 n\)
\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: \(c\log n<\ell < c \log^5 n\)
O(\frac{|S|\log \ell}{b})
\varphi(n,\ell) = \Omega(\frac{\ell}{\log \ell})
\ell = O(\log^5 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
\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
\displaystyle f(n,\ell) \leq \sum_{d|n} f^*(\frac{n}{d},\ell)
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)}
\sigma(n) = O(n \log \log n)
d(n)
Dominates when \(n\) is large
Improvements
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}
\Omega(\frac{\ell}{n} \varphi(n))& \text{ if } \ell > c \log^5 n\\
\Omega(\frac{\ell}{\log \log n}) & \text{ if } \ell > c \log n\\
& \text{ and } \ell \leq c \log^5 n\\
\end{cases}
\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 \(\bigcup_{d|n,d\leq \ell'} \Z_{n,d}\) such that all segments has slope in \(\Z^*_n\).
\ell' := \frac{\ell}{c \log^5 n}
g(n,\ell) = O(f^*(n,\ell))
\displaystyle f(n,\ell) \leq \sum_{d\in B} g\left(\frac{n}{d},\ell\right)
\(\mathbb{D}_n\) be the set of divisors of \(n\).
\((\mathbb{D}_n,\odot)\), where \(a\odot b = \gcd(ab,n)\).
Let \(A = \{i \mid i\in \mathbb{D}_n, i\leq \ell'\}\).
Find a \(A\)-covering \(B\) so the sum is small.
If \(B\) is an \(A\)-covering of \(\mathbb{D}_n\), then
\(B = \{1\}\cup (\mathbb{D}_n\setminus A)\)
\displaystyle f(n,\ell) \leq \sum_{d\in B} g\left(\frac{n}{d},\ell\right)
\begin{aligned}
&\leq \sum_{d\in B}O\left(\frac{n}{d}\frac{\log n}{\ell} \right)+1\\
&\leq O(\frac{n}{\ell}\log n) + |B|\\
&\leq O(\frac{n}{\ell}\log n) + n^{O(\frac{1}{\log \log n})}\\
\end{aligned}
\begin{aligned}
\leq O(\frac{n}{\ell}\log n)
\end{aligned}
# of divisors
larger than \(\ell'\)
\(\ell\) is small: \(\ell=n^{1-\Omega(\frac{1}{\log \log n})}\)
\(B = \mathbb{D}_m\) for some \(m|n\) and \(m\geq \frac{n}{\ell'}\)
\(\sigma(m) = O(m)\)
\displaystyle f(n,\ell) \leq \sum_{d\in B} g\left(\frac{n}{d},\ell\right)
\begin{aligned}
&\leq \sum_{d|m}O\left(\frac{n}{d}\frac{\log n}{\ell} \right)+1\\
&\leq O\left(\frac{n}{m}\sigma(m)\frac{\log n}{\ell}\right) + d(m)\\
\end{aligned}
\begin{aligned}
&\leq O\left(\frac{n}{\ell}\log n\right) + \left(\frac{n}{\ell'}\right)^{\frac{c}{\log \log n}}\\
&\leq O\left(\frac{n}{\ell}\log n\right)\\
\end{aligned}
If
\(B = \mathbb{D}_m\) for some \(m|n\) and \(m\geq \frac{n}{\ell'}\)
Theorem
There exists a \(m|n\) such that \(m = n^{\Omega(\frac{1}{\log \log \log n})}\) and \(\sigma(m)=O(m)\).
Such \(B\) exists when \(\ell\) is large: \(\ell = n^{1-O(\frac{1}{\log \log \log n})}\)
f(n,\ell) = O\left(\frac{n}{\ell}\log n\right)
\(\sigma(m) = O(m)\)
and
Future Work
Conjecture: \(f(n,\ell) = O(\frac{n}{\ell}\log \ell)\) if \(\ell = \Omega(\log n)\)
What about \(\ell\)-covering of arbitrary set \(X\subseteq \Z_n\)?
Other interesting semigroups, e.g. \(\Z_n\times \Z_m\).
\varphi(n,\ell) =
\Omega(\frac{\ell}{n} \varphi(n)) \text{ if } \ell > c \log n
Conjecture:
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
\displaystyle f(n,\ell) \leq \sum_{d|m} f^*(\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^5 \frac{n}{\ell}
< \frac{n}{\ell'}
m':=\frac{m}{p_j}
\begin{aligned}
f(n,\ell) &\leq \sum_{d|m} f^*(\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}
\(f(n,\ell) = O(\frac{n}{\ell}\log n\log \log n)\)
\(f(n,\ell) = O(\frac{n}{\ell}\log n\log \log n)\)
\(f(n,\ell) = O(\frac{n}{\ell}\log n)\)
Better choice of \(m\)
Alternative choice of divisor cover
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
IP
\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 \\
& & & x_i \in \{0,1 \} & i\in \Z_n \\
\end{aligned}
LP Relaxation
\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}
Input size: \(O(n^2)\). \(n\) variables and \(n\) constraints.
Symmetric LP
Certain permutation of variables does not change the LP.
\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}
\(\textbf{fesible}\) \(\xRightarrow[\textbf{ }]{}\) \({}\textbf{fesible}\)
\(\textbf{c}\cdot\textbf{x}\) \(\xRightarrow[\textbf{ }]{\textbf{same value}}\) \(\textbf{c}\cdot\pi\textbf{ x}\)
\(G\)-symmetric LP
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 or
By Chao Xu
