Small-Bias is Not Enough to Hit Read-Once CNF

Abstract

Small-bias probability spaces have wide applications in pseudorandomness which naturally leads to the study of their limitations. Constructing a polynomial complexity hitting set for read-once CNF formulas is a basic open problem in pseudorandomness. We show in this paper that this goal is not achievable using small-bias spaces. Namely, we show that for each read-once CNF formula F with probability of acceptance p and with m clauses each of size c, there exists a δ-biased distribution μ on {0, 1}ⁿ such that δ = 2^{−Ω(logm log(1/p))} and no element in the support of μ satisfies F, where n = m c (assuming that $e^{-\sqrt {m}}\leq p \leq p_{0}$, where p ₀ > 0 is an absolute constant). In particular if p = n ^−Θ(1), the needed bias is 2^{−Ω(log 2 n)}, which requires a hitting set of size $2^{\Omega (\log ^{2}{n})}$. Our lower bound on the needed bias is asymptotically tight. The dual version of our result asserts that if $f_{low}:\{0, 1\}^{n}\rightarrow \mathbb {R}$ is such that and E[f _{l o w} ] > 0 and f _{l o w} (x) ≤ 0 for each x ∈ {0, 1}ⁿ such that F(x) = 0, then the L1-norm of the Fourier transform of f _{l o w} is at least E[f _{l o w} ]2^{Ω(logm log(1/p))}. Our result extends a result due to De, Etesami, Trevisan, and Tulsiani (APPROX-RANDOM 2010) who proved that the small-bias property is not enough to obtain a polynomial complexity PRG for a family of read-once formulas of Θ(1) probability of acceptance.

Appendix A

1.1 Proof of Lemma 1

Lemma 1

If μ is a probability distribution on {0, 1} ⁿ such that $\mu (F_{\mathfrak {p}}=1)=0$ , then there exists a $\mathfrak {p}$ -symmetric probability distribution μ ^∗ on {0, 1} ⁿ such that $\mu ^{*}(F_{\mathfrak {p}}=1)=0$ and bias(μ ^∗ )≤bias(μ).

Proof

Let $G\subset GL_{n}(\mathbb {F}_{2})$ be the group of n×n invertible $\mathfrak {p}$-block permutation matrices over $\mathbb {F}_{2}$, i.e., G consists of the invertible n×n matrices T over $\mathbb {F}_{2}$ such that: ∃ a permutation π : [m]→[m] and invertible c×c matrices $T^{(1)},\ldots , T^{(m)} \in GL_{c}(\mathbb {F}_{2})$ such that for each j ∈ [m], we have $(T x)|_{\mathfrak {p}(\pi (j))} = T^{(j)} (x|_{\mathfrak {p}(j)})$. Thus T is uniquely determined by the permutation π and the matrices T ⁽¹⁾,…,T ^(m).

For T ∈ G, define the probability distribution μ _T on {0, 1}ⁿ as μ _T (x): = μ(T x). Symmetrize μ by averaging: define the probability distribution μ ^∗ on {0, 1}ⁿ as μ ^∗(x): = E _{T ∈ G} μ _T (x). The key points are:

• i) $W_{\mathfrak {p}}(x) = W_{\mathfrak {p}}(Tx)$, $\forall x\in {\mathbb {F}_{2}^{n}}$ and ∀T ∈ G.

  This follows from the fact that the matrices T ⁽¹⁾,…,T ^(m) are invertible.

• ii) Conversely, $\forall x,y\in {\mathbb {F}_{2}^{n}}$ such that $W_{\mathfrak {p}}(x) = W_{\mathfrak {p}}(y)$, ∃T ∈ G such that y = T x.

  To construct T, choose the permutation π to arbitrarily map the clauses satisfied by x to those satisfied by y, i.e., $x|_{\mathfrak {p}(j)} \neq 0$ iff $y|_{\mathfrak {p}(\pi (j))} \neq 0$. Then for each j ∈ [m], choose T ^(j) so that $T^{(j)} (x|_{\mathfrak {p}(j)})=y|_{\mathfrak {p}(\pi (j))}$.

• iii) b i a s(μ _T ) = b i a s(μ) for each T ∈ G since the matrices in G are invertible.

  This follows from the fact that for any invertible matrix $T \in GL_{n}(\mathbb {F}_{2})$, we have $bias_{z}(\mu _{T})= bias_{{T^{-1}}^{*}z}(\mu )$, where ^∗ is the transpose operator. Namely,

  $$\begin{array}{@{}rcl@{}} bias_{z}(\mu_{T})&=& \sum\limits_{x}\mu(Tx)(-1)^{\langle x,z\rangle} \\ & =& \sum\limits_{x}\mu(x)(-1)^{\langle T^{-1}x,z\rangle} \\ &=& \sum\limits_{x}\mu(x)(-1)^{\langle x,{T^{-1}}^{*}z\rangle} \text{(since \(\langle T^{-1}x,z\rangle = \langle x,{T^{-1}}^{*}z\rangle\))}\\ &=& bias_{{T^{-1}}^{*}z}(\mu). \end{array} $$

Since $\mu (F_{\mathfrak {p}}=1)=0$, it follows from (i) that $\mu _{T}(F_{\mathfrak {p}}=1)=0$ for each T ∈ G. Hence $\mu ^{*}(F_{\mathfrak {p}}=1)=0$. The fact that μ ^∗ is $\mathfrak {p}$-symmetric follows from (ii).

Finally, for each nonzero z ∈ {0, 1}ⁿ, we have b i a s _z (μ ^∗) = E _{T ∈ G} b i a s _z (μ _T ), hence |b i a s _z (μ ^∗)|≤ maxT ∈ G|b i a s _z (μ _T )|≤ maxT ∈ G b i a s(μ _T ) = b i a s(μ), where the last equality follows from (iii). Therefore, b i a s(μ ^∗) ≤ b i a s(μ). □