Extended OMP algorithm for sparse phase retrieval

Abstract

We consider the problem of phase retrieval, namely recovering a signal from the magnitude of its linear measurements. Due to the loss of phase information, additional structure information about the signal is necessary. In this work, we focus our attention on sparse signals, i.e., signals consist of a small number of nonzero elements in an appropriate basis. The main contribution of this paper is that a novel algorithm for sparse phase retrieval and its modified version which has high recovery rate are proposed. Moreover, the quartic coherence of measurement matrix is first put forward to analyze recovery condition. The numerical results show that the proposed algorithm is accurate than existing techniques.

Appendix

We need to prove that for any $S\in \{1,\ldots ,n\}$ with $|S|=s$, the right indices are selected at every iterations and $\mathscr {A}_{S}$ is injective.

• (a) For the first iteration, one has to prove

  $$\begin{aligned} \max \limits _{j\in S}(\hbox {diag}(\mathscr {A}^{*}(y)))_{j}>\max \limits _{l\in \overline{S}}(\hbox {diag}(\mathscr {A}^{*}(y)))_{l}. \end{aligned}$$

  (3)

Let $k\in S$ be chosen so that $x_{k}=\max \limits _{j\in S}|x_{j}|>0$. Then

$$\begin{aligned} \begin{aligned} \hbox {diag}(\mathscr {A}^{*}(y)))_{k}&=\langle y,b_{k}\odot b_{k}\rangle \\&=\langle \mathscr {A}(xx^{*}),b_{k}\odot b_{k}\rangle \\&=\left\langle \sum _{p\in S}\sum _{q\in S}x_{p}x_{q}b_{p}\odot b_{q},b_{k}\odot b_{k}\right\rangle \\ \ge&|x_{k}|^{2}\langle b_{k}\odot b_{k},b_{k}\odot b_{k}\rangle \\&\quad -\left| \left\langle \sum _{\begin{array}{c} p,q\in S\\ \text {except}~ p=q=k \end{array}}x_{p}x_{q}b_{p}\odot b_{q},b_{k}\odot b_{k}\right\rangle \right| \\&\ge |x_{k}|^{2}-|x_{k}|^{2}(s^{2}-1)\nu . \end{aligned} \end{aligned}$$

Meanwhile, for $l\in \overline{S}$, one has

$$\begin{aligned} \begin{aligned} \hbox {diag}(\mathscr {A}^{*}(y)))_{l}&=\langle y,b_{l}\odot b_{l}\rangle \\&=\left\langle \sum _{p\in S}\sum _{q\in S}x_{p}x_{q}b_{p}\odot b_{q},b_{l}\odot b_{l}\right\rangle \\&\le |x_{k}|^{2}s^{2}\nu . \end{aligned} \end{aligned}$$

Then, (3) is fulfilled provided $1-(s^{2}-1)\nu >s^{2}\nu$, i.e.,

$$\begin{aligned} \begin{aligned} \nu <\frac{1}{2s^{2}-1}. \end{aligned} \end{aligned}$$

(4)

• (b) For the t-th iteration ($t>1$), one need to prove inequality

  $$\begin{aligned} \begin{aligned}&\max \limits _{j\in S}|(\mathscr {A}^{*}(\mathscr {A}(zz^\mathrm{T})-y)\cdot z)_{j}|\\&\quad >\max \limits _{l\in \overline{S}}|(\mathscr {A}^{*}(\mathscr {A}(zz^\mathrm{T})-y)\cdot z)_{l}| \end{aligned} \end{aligned}$$

  (5)

holds for any vectors $z\in \mathbb {R}^{n}$ obtained by least-squares minimization in OMPPR algorithm. On the one hand, for $l\in \overline{S}$,

$$\begin{aligned} \begin{aligned}&|(\mathscr {A}^{*}(\mathscr {A}(zz^\mathrm{T})-y)\cdot z)_{l}|=|(\mathscr {A}^{*}(\mathscr {A}(zz^\mathrm{T})-\mathscr {A}(xx^\mathrm{T}))\cdot z)_{l}|\\&\quad =\left| \left( \sum _{p\in S}\sum _{q\in S}(z_{p}z_{q}b_{p}\odot b_{q}-x_{p}x_{q}b_{p}\odot b_{q})\right) \right) \\&\qquad \left. \left. \odot \left( \sum _{i\in S}z_{i}b_{i}\right) ,b_{l}\right\rangle \right| \\&\quad =\left. \left| \left( \sum _{p\in S}\sum _{q\in S}(z_{p}z_{q}b_{p}\odot b_{q}-x_{p}x_{q}b_{p}\odot b_{q})\right) \right) ,\right. \\&\qquad \left. \left. b_{l}\odot \left( \sum _{i\in S}z_{i}b_{i}\right) \right\rangle \right| \\&\quad =\left| \sum _{p\in S}\sum _{q\in S}\sum _{i\in S}(z_{p}z_{q}-x_{p}x_{q})z_{i}\langle b_{p}\odot b_{q},b_{l}\odot b_{i}\rangle \right| \\&\quad \le \max \limits _{p,q,i\in S}|(z_{p}z_{q}-x_{p}x_{q})z_{i}|s^{3}\nu . \end{aligned} \end{aligned}$$

On the other hand, let $k\in S$ be chosen so that $k=\text {arg}\max \limits _{p\in S}|(|z_{p}|^{2}-|x_{p}|^{2})z_{p}|$,

$$\begin{aligned}&|(\mathscr {A}^{*}(\mathscr {A}(zz^\mathrm{T})-y)\cdot z)_{k}|\\&\quad =\left| \sum _{p\in S}\sum _{q\in S}\sum _{i\in S}(z_{p}z_{q}-x_{p}x_{q})z_{i}\langle b_{p}\odot b_{q},b_{k}\odot b_{i}\rangle \right| \\&\quad =|(|z_{k}|^{2}-|x_{k}|^{2})z_{k}|\cdot |\langle b_{k}\odot b_{k},b_{k}\odot b_{k}\rangle |\\&\qquad -\left| \sum _{\begin{array}{c} p,q,i\in S\\ \text {except}~ p=q=i=k \end{array}}(z_{p}z_{q}-x_{p}x_{q})z_{i}\langle b_{p}\odot b_{q},b_{k}\odot b_{i}\rangle \right| \\&\quad \ge |(|z_{k}|^{2}-|x_{k}|^{2})z_{k}|-\max \limits _{p,q,i\in S}|(z_{p}z_{q}-x_{p}x_{q})z_{i}|(s^{3}-1)\nu . \end{aligned}$$

So, (5) holds true if

$$\begin{aligned} \begin{aligned}&\max \limits _{p,q,i\in S}|(z_{p}z_{q}-x_{p}x_{q})z_{i}|s^{3}\nu \\&\quad <|(|z_{k}|^{2}-|x_{k}|^{2})z_{k}|-\max \limits _{p,q,i\in S}|(z_{p}z_{q}-x_{p}x_{q})z_{i}|(s^{3}-1)\nu . \end{aligned} \end{aligned}$$

Equivalently,

$$\begin{aligned} \begin{aligned} \nu <\frac{\max \limits _{p\in S}|(|z_{p}|^{2}-|x_{p}|^{2})z_{p}|}{\max \limits _{p,q,i\in S}|(z_{p}z_{q}-x_{p}x_{q})z_{i}|}\cdot \frac{1}{2s^{3}-1}. \end{aligned} \end{aligned}$$

(6)

• (c) Now we prove $\mathscr {A}_{S}$ is injective.

Let $zz^\mathrm{T}\in \mathscr {H}^{n\times n}$ be an eigenvector of $\mathscr {A}_{S}^{*}\mathscr {A}_{S}$ associated with $\lambda$, and choose j such that $|z_{j}|$ is maximal, i.e., $|z_{j}|=||z||_{\propto }$. Then $\mathscr {A}_{S}^{*}(\mathscr {A}_{S}(zz^\mathrm{T}))_{jj}=\lambda |z_{j}|^{2}$, and a rearrangement gives

$$\begin{aligned} \begin{aligned}&(\lambda -\langle b_{j}\odot b_{j},b_{j}\odot b_{j}\rangle ) |z_{j}|^{2}\\&\quad =\left\langle \sum _{\begin{array}{c} p,q\in S\\ \text {except}~ p=q=j \end{array}}z_{p}z_{q}b_{p}\odot b_{q},b_{j}\odot b_{j}\right\rangle . \end{aligned} \end{aligned}$$

Hence,

$$\begin{aligned} \begin{aligned}&|\lambda -1| \cdot |z_{j}|^{2} \\&\quad \le \left| \left\langle \sum _{\begin{array}{c} p,q\in S\\ except~ p=q=j \end{array}}z_{p}z_{q}b_{p} \odot b_{q},b_{j}\odot b_{j}\right\rangle \right| \\&\quad \le (s^{2}-1)|z_{j}|^{2}\nu . \end{aligned} \end{aligned}$$

Equivalently,

$$\begin{aligned} \begin{aligned} |\lambda -1| \le (s^{2}-1)\nu . \end{aligned} \end{aligned}$$

One can obtain the minimum eigenvalue of $\mathscr {A}_{S}^{*}\mathscr {A}_{S}$ $\lambda _{min}$ satisfying $\lambda _{min}\ge 1-(s^{2}-1)\nu$. Therefore, if

$$\begin{aligned} \begin{aligned} \nu <\frac{1}{s^{2}-1}, \end{aligned} \end{aligned}$$

(7)

$\mathscr {A}_{S}^{*}\mathscr {A}_{S}$ is injective. To see that $\mathscr {A}_{S}$ is injective, simply observe that $\mathscr {A}_{S}(zz^\mathrm{T})=0$ yields $\mathscr {A}_{S}^{*}(\mathscr {A}_{S}(zz^\mathrm{T}))=0$, so that $zz^\mathrm{T}=0$.

Combining (4), (6) and (7), we require $\nu$ satisfies (6). This completes our proof.