In this work, we present some new results for compressed sensing and phase retrieval. For compressed sensing, it is shown that if the unknown $n$-dimensional vector can be expressed as a linear combination of $s$ unknown Vandermonde vectors (with Fourier vectors as a special case) and the measurement matrix is a Vandermonde matrix, exact recovery of the vector with $2s$ measurements and $O(\mathrm{poly}(s))$ complexity is possible when $n \geq 2s$. From these results, a measurement matrix is constructed from which it is possible to recover $s$-sparse $n$-dimensional vectors for $n \geq 2s$ with as few as $2s$ measurements and with a recovery algorithm of $O(\mathrm{poly}(s))$ complexity. In the second part of the work, these results are extended to the challenging problem of phase retrieval. The most significant discovery in this direction is that if the unknown $n$-dimensional vector is composed of $s$ frequencies with at least one being non-harmonic, $n \geq 4s - 1$ and we take at least $8s-3$ Fourier measurements, there are, remarkably, only two possible vectors producing the observed measurement values and they are easily obtainable from each other. The two vectors can be found by an algorithm with only $O(\mathrm{poly}(s))$ complexity. An immediate application of the new result is construction of a measurement matrix from which it is possible to recover all $s$-sparse $n$-dimensional signals (up to a global phase) from $O(s)$ magnitude-only measurements and $O(\mathrm{poly}(s))$ recovery complexity when $n \geq 4s - 1$.

翻译：本文针对压缩感知和相位恢复问题提出了若干新结果。对于压缩感知而言，研究表明：若未知的$n$维向量可表示为$s$个未知范德蒙德向量（以傅里叶向量为特例）的线性组合，且测量矩阵为范德蒙德矩阵，则当$n \geq 2s$时，仅需$2s$次测量和$O(\mathrm{poly}(s))$复杂度的算法即可实现向量的精确恢复。基于此结果，本文构造了一种测量矩阵，使得当$n \geq 2s$时，能够以低至$2s$次测量和$O(\mathrm{poly}(s))$复杂度的恢复算法，实现$s$稀疏$n$维向量的恢复。在工作的第二部分，我们将这些成果推广至更具挑战性的相位恢复问题。该方向最显著的发现是：若未知$n$维向量由$s$个频率组成且至少包含一个非谐波频率，当$n \geq 4s - 1$且采集至少$8s-3$个傅里叶测量值时，观测值仅对应两个可能的向量（两者可轻松互推），且可通过$O(\mathrm{poly}(s))$复杂度的算法求解。该新成果的直接应用是构造一种测量矩阵，使得当$n \geq 4s - 1$时，能够从$O(s)$次模值测量中恢复所有$s$稀疏$n$维信号（至多相差一个全局相位），恢复复杂度为$O(\mathrm{poly}(s))$。