(1 + eps)-Approximate Sparse Recovery

Eric Price, David P. Woodruff
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引用次数: 47

Abstract

The problem central to sparse recovery and compressive sensing is that of \emph{stable sparse recovery}: we want a distribution $\math cal{A}$ of matrices $A \in \R^{m \times n}$ such that, for any $x \in \R^n$ and with probability $1 - \delta >, 2/3$ over $A \in \math cal{A}$, there is an algorithm to recover $\hat{x}$ from $Ax$ with\begin{align} \norm{p}{\hat{x} - x} \leq C \min_{k\text{-sparse } x'} \norm{p}{x - x'}\end{align}for some constant $C >, 1$ and norm $p$. The measurement complexity of this problem is well understood for constant $C >, 1$. However, in a variety of applications it is important to obtain $C = 1+\eps$ for a small $\eps >, 0$, and this complexity is not well understood. We resolve the dependence on $\eps$ in the number of measurements required of a $k$-sparse recovery algorithm, up to polylogarithmic factors for the central cases of $p=1$ and $p=2$. Namely, we give new algorithms and lower bounds that show the number of measurements required is $k/\eps^{p/2} \textrm{polylog}(n)$. For $p=2$, our bound of $\frac{1}{\eps}k\log (n/k)$ is tight up to \emph{constant} factors. We also give matching bounds when the output is required to be $k$-sparse, in which case we achieve $k/\eps^p \textrm{polylog}(n)$. This shows the distinction between the complexity of sparse and non-sparse outputs is fundamental.
(1 + eps)-近似稀疏恢复
稀疏恢复和压缩感知的核心问题是\emph{稳定稀疏恢复}:我们想要一个矩阵$A \in \R^{m \times n}$的分布$\math cal{A}$,这样,对于任何$x \in \R^n$和概率$1 - \delta >, 2/3$超过$A \in \math cal{A}$,有一种算法可以从$Ax$和\begin{align} \norm{p}{\hat{x} - x} \leq C \min_{k\text{-sparse } x'} \norm{p}{x - x'}\end{align}对某些常数$C >, 1$和范数$p$恢复$\hat{x}$。对于常数$C >, 1$,这个问题的测量复杂性是很容易理解的。然而,在各种应用程序中,为一个小的$\eps >, 0$获取$C = 1+\eps$是很重要的,而且这种复杂性还没有得到很好的理解。我们解决了$k$ -稀疏恢复算法所需的测量数量对$\eps$的依赖,直至$p=1$和$p=2$中心情况的多对数因子。也就是说,我们给出了新的算法和下界,表明所需的测量次数是$k/\eps^{p/2} \textrm{polylog}(n)$。对于$p=2$, $\frac{1}{\eps}k\log (n/k)$的边界被\emph{常数}因子所限制。当输出要求为$k$ -sparse时,我们也给出了匹配边界,在这种情况下,我们实现了$k/\eps^p \textrm{polylog}(n)$。这表明,稀疏输出和非稀疏输出的复杂性之间的区别是根本的。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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