Simple Affine Extractors Using Dimension Expansion

Abstract

Let $\F$ be the field of $q$ elements. An \emph{\afsext{n}{k}} is a mapping $D:\F^n\ar\B$ such that for any $k$-dimensional affine subspace $X\subseteq \F^n$, $D(x)$ is an almost unbiased bit when $x$ is chosen uniformly from $X$. Loosely speaking, the problem of explicitly constructing affine extractors gets harder as $q$ gets smaller and easier as $k$ gets larger. This is reflected in previous results: When $q$ is `large enough', specifically $q= \Omega(n^2)$, Gabizon and Raz \cite{GR05} construct affine extractors for any $k\geq 1$. In the `hardest case', i.e. when $q=2$, Bourgain \cite{Bour05} constructs affine extractors for $k\geq \delta n$ for any constant (and even slightly sub-constant) $\delta>0$. Our main result is the following: Fix any $k\geq 2$ and let $d = 5n/k$. Then whenever $q>2\cdot d^2$ and $p=char(\F)>d$, we give an explicit \afsext{n}{k}. For example, when $k=\delta n$ for constant $\delta>0$, we get an extractor for a field of constant size $\Omega(\left(\frac{1}{\delta}\right)^2)$. We also get weaker results for fields of arbitrary characteristic (but can still work with a constant field size when $k=\delta n $ for constant $\delta > 0$). Thus our result may be viewed as a `field-size/dimension' tradeoff for affine extractors. For a wide range of $k$ this gives a new result, but even for large $k$ where we do not improve (or even match) the previous result of \cite{Bour05}, we believe that our construction and proof have the advantage of being very simple: Assume $n$ is prime and $d$ is odd, and fix any non-trivial linear map $T:\F^n\mapsto \F$. Define $QR:\F\mapsto \B$ by $QR(x)=1$ if and only if $x$ is a quadratic residue. Then, the function $D:\F^n\mapsto \B$ defined by $D(x)\triangleq QR(T(x^d))$ is an \afsext{n}{k}. Our proof uses a result of Heur, Leung and Xiang \cite{HLX02} giving a lower bound on the dimension of products of subspaces.