Revision #1 Authors: Ravi Boppana, Johan Hastad, Chin Ho Lee, Emanuele Viola

Accepted on: 4th June 2018 20:03

Downloads: 67

Keywords:

For a test $T \subseteq \{0,1\}^n$ define $k^*$ to be the maximum $k$ such

that there exists a $k$-wise uniform distribution over $\{0,1\}^n$ whose

support is a subset of $T$.

For $T = \{x \in \{0,1\}^n : \abs{\sum_i x_i - n/2} \le t\}$ we prove $k^* =

\Theta(t^2/n + 1)$.

For $T = \{x \in \{0,1\}^n : \sum_i x_i \equiv c \pmod m\}$ we prove that $k^* =

\Theta(n/m^2 + 1)$. For some $k = O(n/m)$ we also show that any $k$-wise

uniform distribution puts probability mass at most $1/m + 1/100$ over $T$.

Finally, for any fixed odd $m$ we show that there is an integer $k =

(1-\Omega(1))n$ such that any $k$-wise uniform distribution lands in

$T$ with probability exponentially close to $|T|/2^n$; and this

result is false for any even $m$.

Tight bounds, extension to thresholds, and discussion.

TR16-102 Authors: Ravi Boppana, Johan HÃ¥stad, Chin Ho Lee, Emanuele Viola

Publication: 4th July 2016 12:55

Downloads: 672

Keywords:

Let $k=k(n)$ be the largest integer such that there

exists a $k$-wise uniform distribution over $\zo^n$ that

is supported on the set $S_m := \{x \in \zo^n : \sum_i

x_i \equiv 0 \bmod m\}$, where $m$ is any integer. We

show that $\Omega(n/m^2 \log m) \le k \le 2n/m + 2$. For

$k = O(n/m)$ we also show that any $k$-wise uniform

distribution puts probability mass at most $1/m + 1/100$

over $S_m$. Finally, for any fixed odd $m$ we show that

there is $k = (1-\Omega(1))n$ such that any $k$-wise

uniform distribution lands in $S_m$ with probability

exponentially close to $|S_m|/2^n$; and this result is

false for any even $m$.