Admissible Measurements and Robust Algorithms for Ptychography

Abstract

We study an approach to solving the phase retrieval problem as it arises in a phase-less imaging modality known as ptychography. In ptychography, small overlapping sections of an unknown sample (or signal, say $x_0\in \mathbb {C}^{d}$) are illuminated one at a time, often with a physical mask between the sample and light source. The corresponding measurements are the noisy magnitudes of the Fourier transform coefficients resulting from the pointwise product of the mask and the sample. The goal is to recover the original signal from such measurements. The algorithmic framework we study herein relies on first inverting a linear system of equations to recover a fraction of the entries in $x_0 x_0^*$ and then using non-linear techniques to recover the magnitudes and phases of the entries of $x_0$. Thus, this paper’s contributions are three-fold. First, focusing on the linear part, it expands the theory studying which measurement schemes (i.e., masks, shifts of the sample) yield invertible linear systems, including an analysis of the conditioning of the resulting systems. Second, it analyzes a class of improved magnitude recovery algorithms and, third, it proposes and analyzes algorithms for phase recovery in the ptychographic setting where large shifts—up to $50\%$ the size of the mask—are permitted.

Appendix

1.1 Interleaving Operators and Circulant Structure

To set the stage for the proof of Theorem 2, we introduce a certain collection of permutation operators and study their interactions with circulant and block-circulant matrices. The structure we identify here will be of much use to us in unraveling the linear systems we encounter in our model for phase retrieval with local correlation measurements. For $\ell , N_1, N_2 \in \mathbb {N}, v \in \mathbb {C}^{\ell N_1}, k \in [\ell ],$ and $H \in \mathbb {C}^{\ell N_1 \times N_2}$, we define the block circulant operator ${{\,\mathrm{circ}\,}}^{N_1}$ by

$$\begin{aligned} {{\,\mathrm{circ}\,}}_k^{N_1}(v)&= \begin{bmatrix} v&S^{N_1} v&\cdots&S^{(k - 1)N_1} v \end{bmatrix} \\ {{\,\mathrm{circ}\,}}_k^{N_1}(H)&= \begin{bmatrix} H&S^{N_1} H&\cdots&S^{(k - 1) N_1}H \end{bmatrix}, \end{aligned}$$

where, as with ${{\,\mathrm{circ}\,}}(\cdot )$, when we omit the subscript we define ${{\,\mathrm{circ}\,}}^{N_1}(H) = {{\,\mathrm{circ}\,}}_\ell ^{N_1}(H)$ and ${{\,\mathrm{circ}\,}}^{N_1}(v) = {{\,\mathrm{circ}\,}}_\ell ^{N_1}(v)$. We now proceed with the following lemmas; the first establishes the inverse of $P^{(d, N)}$.

Lemma 2

For $d, N \in \mathbb {N},$ we have

$$\begin{aligned} (P^{(d, N)})^{-1} = P^{(d, N) *} = P^{(N, d)}. \end{aligned}$$

Proof of Lemma 2

Simply take $v \in \mathbb {C}^{d N}$ and calculate, for $i \in [d], j \in [N]$,

$$\begin{aligned} (P^{(d, N)} P^{(N, d)} v)_{(i - 1) N + j}&= (P^{(d, N)} (P^{(N, d)} v))_{(i - 1) N + j}\\&= (P^{(N, d)} v)_{(j - 1) d + i} = v_{(i - 1) N + j}, \end{aligned}$$

with these equalities coming from the definition in (16). $\square$

We now observe some useful ways in which the interleaving operators commute with the construction of circulant matrices.

Lemma 3

Suppose $V_i \in \mathbb {C}^{k \times n}, v_{ij} \in \mathbb {C}^k, w_j \in \mathbb {C}^{k N_1}$ for $i \in [N_1], j \in [N_2]$ and

$$\begin{aligned} M_1= & {} \begin{bmatrix} {{\,\mathrm{circ}\,}}(V_1) \\ \vdots \\ {{\,\mathrm{circ}\,}}(V_{N_1}) \end{bmatrix},\quad M_2 = \begin{bmatrix} {{\,\mathrm{circ}\,}}^{N_1}(w_1)&\cdots&{{\,\mathrm{circ}\,}}^{N_1}(w_{N_2}) \end{bmatrix},\ \text {and} \\ M_3= & {} \begin{bmatrix} {{\,\mathrm{circ}\,}}(v_{11}) &{} \cdots &{} {{\,\mathrm{circ}\,}}(v_{1 N_2}) \\ \vdots &{} \ddots &{} \vdots \\ {{\,\mathrm{circ}\,}}(v_{N_1 1}) &{} \cdots &{} {{\,\mathrm{circ}\,}}(v_{N_1 N_2}) \end{bmatrix}. \end{aligned}$$

Then

$$\begin{aligned} P^{(k, N_1)} M_1&= {{\,\mathrm{circ}\,}}^{N_1}\left( P^{(k, N_1)} \begin{bmatrix} V_1 \\ \vdots \\ V_{N_1} \end{bmatrix}\right) \end{aligned}$$

(50)

$$\begin{aligned} M_2 P^{(k, N_2)*}&= {{\,\mathrm{circ}\,}}^{N_1}\left( \begin{bmatrix} w_1&\cdots&w_{N_2} \end{bmatrix}\right) \end{aligned}$$

(51)

$$\begin{aligned} P^{(k, N_1)}M_3P^{(k, N_2)*}&= {{\,\mathrm{circ}\,}}^{N_1}\left( P^{(k, N_1)} \begin{bmatrix} v_{11} &{} \cdots &{} v_{1 N_2} \\ \vdots &{} \ddots &{} \vdots \\ v_{N_1 1} &{} \cdots &{} v_{N_1 N_2} \end{bmatrix}\right) . \end{aligned}$$

(52)

Proof of Lemma 3

We index the matrices to check the equalities. For (50), we take $(a, b, \ell , j) \in [d] \times [N_1] \times [k] \times [n]$ and have

$$\begin{aligned} (P^{(k, N_1)} M_1)_{(a-1)N_1 + b, (\ell - 1) n + j}= & {} (M_1)_{(b - 1) k + a, (\ell - 1)n + j} = \begin{bmatrix} S^{\ell - 1} V_1 \\ \vdots \\ S^{\ell - 1} V_{N_1} \end{bmatrix}_{(b - 1)k + a, j}\\= & {} (S^{\ell - 1}V_b)_{a, j} = (V_b)_{a + \ell - 1, j} \end{aligned}$$

and

$$\begin{aligned} {{\,\mathrm{circ}\,}}^{N_1}\left( P^{(k, N_1)} \begin{bmatrix} V_1 \\ \vdots \\ V_{N_1} \end{bmatrix}\right) _{(a-1)N_1 + b, (\ell - 1) n + j}&= \left( P^{(k, N_1)} \begin{bmatrix} V_1 \\ \vdots \\ V_{N_1} \end{bmatrix}\right) _{(a - 1)N_1 + b + (\ell -1)N_1, j} \\&= (V_b)_{a + \ell - 1, j} \end{aligned}$$

For (51), we take $(a, b, j) \in [k] \times [N_2] \times [k N_1]$ and have

$$\begin{aligned} (P^{(k, N_2)} M_2^*)_{(a - 1)N_2 + b, j} = (M_2)_{j, (b - 1) k + a} = (w_b)_{j + (a - 1)N_1} \end{aligned}$$

and

$$\begin{aligned} \left( {{\,\mathrm{circ}\,}}^{N_1}\left( \begin{bmatrix} w_1&\cdots&w_{N_2} \end{bmatrix}\right) \right) _{j, (a - 1)N_2 + b} = (S^{N_1(a - 1)} w_b)_j = (w_b)_{j + N_1(a - 1)}, \end{aligned}$$

and (52) follows immediately by combining (50) and (51). $\square$

Lemma 4 introduces useful identities relating interleaving operators to kronecker products.

Lemma 4

For , and $B_i \in \mathbb {C}^{m \times k}, i \in [\ell ]$, we have

(53)

(54)

(55)

(56)

Proof of Lemma 4

For (53), we see that, for $i, j, k \in [d] \times [N] \times [m]$, we have

$$\begin{aligned} (P^{(d, N)} v \otimes A)_{(i - 1) N + j, k}&= (v \otimes A)_{(j - 1) d + i, k} = v_j A_{i k}, \ \text {while} \\ (A \otimes v)_{(i - 1) N + j, k}&= A_{i k} v_j, \end{aligned}$$

and (54) follows by considering that To get (55), we trace the positions of columns, considering that $(V \otimes A) e_{(i - 1) m + j} = V_j \otimes A_i$. From (54), we observe that $P^{(d, N)} (V \otimes A) e_{(i - 1) m + j} = A_j \otimes V_i,$ so

$$\begin{aligned} P^{(d, N)} (V \otimes A) P^{(m, \ell )} e_{(j - 1) \ell + i}&= P^{(d, N)} (V \otimes A) e_{(i - 1) m + j}\\&= A_j \otimes V_i = (A \otimes V) e_{(j - 1) \ell + i}. \end{aligned}$$

As for (56), we remark that

$\square$

The following lemma on the Kronecker product is standard (e.g., Theorem 13.26 in [39]).

Lemma 5

We have $\mathrm{vec}(A B C) = (C^T \otimes A) \mathrm{vec}(B)$ for any $A \in \mathbb {C}^{m \times n}, B \in \mathbb {C}^{n \times p}, C \in \mathbb {C}^{p \times k}.$ In particular, for $a, b \in \mathbb {C}^{d}, \mathrm{vec}(a b^*) = \overline{b} \otimes a$, and

$$\begin{aligned} \mathrm{vec}E_{jk} (\mathrm{vec}E_{j' k'})^* = E_{k k'} \otimes E_{j j'}. \end{aligned}$$

(57)

The next lemma covers the standard result concerning the diagonalization of circulant matrices, as well as a generalization to block-circulant matrices.

Lemma 6

For any $v \in \mathbb {C}^{d}$, we have

$$\begin{aligned} {{\,\mathrm{circ}\,}}(v) = F_d {{\,\mathrm{diag}\,}}(\sqrt{d} F_d^* v) F_d^* = \sqrt{d} \sum _{j = 1}^d (f_j^{d *} v) f_j^d f_j^{d *} \end{aligned}$$

(58)

Suppose $V \in C^{k N \times m}$, then ${{\,\mathrm{circ}\,}}^N(V)$ is block diagonalizable by

$$\begin{aligned} {{\,\mathrm{circ}\,}}^N(V) = \left( F_k \otimes I_N\right) \left( {{\,\mathrm{diag}\,}}(M_1, \ldots , M_k)\right) \left( F_k \otimes I_m\right) ^*, \end{aligned}$$

(59)

where

$$\begin{aligned} \sqrt{k}\left( F_k \otimes I_N\right) ^* V = \begin{bmatrix} M_1 \\ \vdots \\ M_k \end{bmatrix}, \quad \text {or} \quad M_j = \sqrt{k} (f_j^k \otimes I_N)^* V \end{aligned}$$

(60)

Proof of Lemma 6

The diagonalization in (58) is a standard result: see, e.g., Theorem 7 of [26].

To prove (59), we set $V_i$ to be the $k \times m$ blocks of V such that and begin by observing that, for $u \in \mathbb {C}^k$ and $W \in \mathbb {C}^{m \times p}$, the $\ell ^{\mathrm{th}}$ $k \times p$ block of ${{\,\mathrm{circ}\,}}^N(V)(u \otimes W)$ is

$$\begin{aligned} \left( {{\,\mathrm{circ}\,}}^N(V)(u \otimes W)\right) _{[\ell ]} = \sum _{i = 1}^k u_i (S^{N (i - 1)}V)_\ell W = \sum _{i = 1}^k u_i V_{\ell - i + 1} W. \end{aligned}$$

Taking $u = f_j^k$ and $W = I_m$, this gives

$$\begin{aligned} \left( {{\,\mathrm{circ}\,}}^N(V)(f_j^k \otimes I_m)\right) _{[\ell ]}&= \frac{1}{\sqrt{k}}\sum _{i = 1}^k \omega _k^{(j - 1) (i - 1)} V_{\ell - i + 1} I_m = \frac{1}{\sqrt{k}} \omega _k^{(j - 1) (\ell - 1)} \sum _{i = 1}^k \omega _k^{-(j - 1)(i - 1)} V_i \\&= (f_j^k)_\ell \left( \sqrt{k} (f_j^k \otimes I_N)^* V \right) = (f_j^k)_\ell M_j. \end{aligned}$$

This relation is equivalent to the statement of the lemma, i.e., having

$$\begin{aligned} {{\,\mathrm{circ}\,}}^N(V) (f_j^k \otimes I_m) = (f_j^k \otimes M_j) = (f_j^k \otimes I_N) M_j. \end{aligned}$$

$\square$

Lemma 6 immediately gives the following corollary regarding the conditioning of ${{\,\mathrm{circ}\,}}^N(V)$, with which we return to considering spanning families of masks.

Corollary 3

With notation as in Lemma 6, the condition number of ${{\,\mathrm{circ}\,}}^N(V)$ is

$$\begin{aligned} \dfrac{\max \limits _{i \in [k]} \sigma _{\max } (M_i)}{\min \limits _{i \in [k]} \sigma _{\min } (M_i)}. \end{aligned}$$

1.2 Lemmas on Block Circulant Structure

We begin with Lemma 7, which describes the transposes of block circulant matrices. For this lemma and the remainder of this section, the reader is advised to recall the definitions of $R_k$ and $P^{(d, N)}$ from Sect. 1.5 and (16), as well as $\mathcal {P}(P, \{k_i\})$ and $\mathcal {T}_N$ from Sect. 3.2.

Lemma 7

Given $k, N, m \in \mathbb {N}$ and $V \in \mathbb {C}^{kN \times m}$, we have

$$\begin{aligned} {{\,\mathrm{circ}\,}}^N(V)^* = {{\,\mathrm{circ}\,}}^m\left( (R_k \otimes I_m) \mathcal {T}_N(V) \right) . \end{aligned}$$

Proof of Lemma 7

Suppose $V_i$ are the $N \times m$ blocks of V, such that $V = \left[ V_1^T \cdots V_k^T\right] ^T$. Indexing blockwise, we have ${{\,\mathrm{circ}\,}}^N(V)_{[ij]} = V_{i - j + 1}$, so that ${{\,\mathrm{circ}\,}}^N(V)^*_{[ij]} = V_{j - i + 1}^*$. In other words,

$$\begin{aligned} {{\,\mathrm{circ}\,}}^N(V)^* = \begin{bmatrix} V_1^* &{} V_2^* &{} \cdots &{} V_N^* \\ V_N^* &{} V_1^* &{} \cdots &{} V_{N - 1}^* \\ \vdots &{} &{} \ddots &{} \vdots \\ V_2^* &{} V_3^* &{} \cdots &{} V_1^* \end{bmatrix} = {{\,\mathrm{circ}\,}}^m((R_k \otimes I_m) \mathcal {T}_N(V) ) . \end{aligned}$$

$\square$

Lemmas 8 and 9 provide identities for a few block matrix structures that will be of interest.

Lemma 8

Given $N_1, N_2, k, m \in \mathbb {N}$ and $V_i \in \mathbb {C}^{k N_1 \times m}$ for $i \in [N_2]$, we have

$$\begin{aligned} \begin{bmatrix} {{\,\mathrm{circ}\,}}^{N_1}(V_1)&\cdots&{{\,\mathrm{circ}\,}}^{N_1}(V_{N_2}) \end{bmatrix} (P^{(k, N_2)} \otimes I_m)^* = {{\,\mathrm{circ}\,}}^{N_1}\left( \begin{bmatrix} V_1&\cdots&V_{N_2}\end{bmatrix}\right) . \end{aligned}$$

Proof of Lemma 8

We quote (51) from Lemma 3 and consider that $P^{(k, N_2)} \otimes I_m$ is a permutation that changes the blockwise indices of $m \times p$ blocks (or, acting from the right, $p \times m$ blocks) exactly the way that $P^{(k, N_2)}$ changes the indices of a vector. $\square$

Lemma 9

Given $k, n \in \mathbb {N}$ and $V_j \in \mathbb {C}^{m_j \times n_j}$ for $j \in [n]$ and setting $M = \sum _{j = 1}^n m_j, N = \sum _{j = 1}^N n_j,$ and $D = {{\,\mathrm{diag}\,}}(I_k \otimes V_j)_{j = 1}^n \in \mathbb {C}^{k M \times k N}$, we have

where $P_1 = \mathcal {P}(P^{(n, k)}, (m_{j \mathrm{mod}_1 n})_{j = 1}^{kn})$ and $P_2 = \mathcal {P}(P^{(n, k)}, (n_{j \mathrm{mod}_1 n})_{j = 1}^{kn})$.

Proof of Lemma 9

We immediately reduce to the case $m_j = n_j = 1$ for all j by observing that $P_1$ and $P_2$ will act on blockwise indices precisely as $P^{(n, k)}$ acts on individual indices. Here, we replace $V_j$ with $v_j \in \mathbb {C}$, and note that ${{\,\mathrm{diag}\,}}(V_j)_{j = 1}^n = {{\,\mathrm{diag}\,}}(v)$. Hence, we need only remark that

while

$\square$