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 $y\in \Z_n$ by storing $x\in \N^t$

$s\in \Z_n^t$ a fixed vector

D(x) = s\cdot x \pmod n = y

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

t=4

n=12

n=12

y=9

y=9

s=(1,7,2,9)

s=(1,7,2,9)

x

x

1 rewrite = 1 AMC update

$D(x) = (1,2,3,4) ·(1,7,2,9)\pmod{12} =9$

Example:

AMC storage

t=4

t=4

n=12

n=12

y=9

y=9

s=(1,7,2,9)

s=(1,7,2,9)

x

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\}

\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}

x_j\gets x_j + e \Leftrightarrow y \gets y+e s_j\pmod{n}

y={\color{red}1}

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\}

\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\}

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

\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})}

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

\displaystyle\varphi(n,\ell)=\sum_{\substack{ x\leq\ell \\ \gcd(x,n)=1}}1

\varphi(n)=\varphi(n,n)

\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}

\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 $\Z^*_n$ .

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)

\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)

\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)

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}

\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})

\text{each element covered by } b \text{ sets} \Rightarrow O(\frac{|S|\log \ell}{b})

\varphi(n,\ell) = \Omega(\frac{\ell}{n}\varphi(n))

\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}

\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})

O(\frac{|S|\log \ell}{b})

\varphi(n,\ell) = \Omega(\frac{\ell}{\log \ell})

\varphi(n,\ell) = \Omega(\frac{\ell}{\log \ell})

\ell = O(\log^u n)

\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}

\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: $\ell < c \log n$

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)

f(n,\ell) \leq \sum_{d|n} f^*(\frac{n}{d},\ell)

\leq \sum_{d|n} \max(d \frac{\log n}{\ell},1)

\leq \sum_{d|n} \max(d \frac{\log n}{\ell},1)

\leq \sigma(n)\frac{\log n}{\ell} + d(n)

\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}}}

= \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)

f^*(n,\ell) = O(\frac{n}{\ell}\log n)

\mathbb{Z}^*_{12}

\mathbb{Z}^*_{12}

\mathbb{Z}_{12,2}

\mathbb{Z}_{12,2}

\mathbb{Z}_{12,3}

\mathbb{Z}_{12,3}

\mathbb{Z}_{12,4}

\mathbb{Z}_{12,4}

\mathbb{Z}_{12,6}

\mathbb{Z}_{12,6}

\mathbb{Z}_{12,12}

\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}}

\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^\dagger(n,\ell) = O(\frac{n}{\ell}\log n)

f^*(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 f(n,\ell) \leq \sum_{d|m} f^\dagger(\frac{n}{d},\ell)

\displaystyle n = \prod_{i=1}^k p_i

\displaystyle n = \prod_{i=1}^k p_i

\displaystyle m = \prod_{i=1}^j p_i \geq \frac{n}{\ell'}

\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}

\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(n) = n^{O(\frac{1}{\log \log n})}

d(m)

d(m)

\leq 2d(m')

\leq 2d(m')

=m'^{O(\frac{1}{\log \log 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)}

= \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} c\log^u \frac{n}{\ell}

< \frac{n}{\ell'}

< \frac{n}{\ell'}

m':=\frac{m}{p_j}

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}

\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}

$f(n,\ell) = O(\frac{n}{\ell}\log n\log \log n)$

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. $\Z_n\times \Z_m$ .

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

Almost tight $\ell$ -covering of $\mathbb{Z}_n$

Store and rewrite a number in $\Z_n$

$\ell$ -covering

Main Questions

Previous Work

Our results

Set cover

Each elements covered by $b$ sets

$O\left(\frac{|\mathcal{S}|}{b}\log \ell\right)$ size set cover

No

Previous bound for $\varphi(n,\ell)$

Previous bound for $\varphi(n,\ell)$

Case 1: $\ell > c \log^u n$

Sieve Theory

Case 2: $\ell > c \log n$

Bound for $f^*(n,\ell)$

Case 1: $\ell > c \log^u n$

Case 2: $c\log n<\ell < c \log^u n$

Case 3: $\ell < c \log n$

Bound for $f(n,\ell)$

Conclusion and Future Work

Almost tight ℓ\ellℓ-covering of Zn\mathbb{Z}_nZn​

Almost tight $\ell$ -covering of $\mathbb{Z}_n$