## Abstract

New methods are described for determining tighter upper bounds on the support of an object, given the support of its autocorrelation. These upper bounds are shown, in a digital experiment, to be useful as object-support constraints used with the iterative transform algorithm for solving the phase-retrieval problem.

© 1990 Optical Society of America

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### Equations (37)

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(1)
$$\begin{array}{lll}F(u)=|F(u)|exp[i\psi (u)]\hfill & =\hfill & \mathcal{F}|f|\hfill \\ \hfill & =\hfill & \mathit{\int}f(x)exp(-i2\pi ux)\text{d}x,\hfill \end{array}$$
(2)
$$\begin{array}{lll}r(x)\hfill & =\hfill & \mathit{\int}f({x}^{\prime})f*({x}^{\prime}-x)\text{d}{x}^{\prime}\hfill \\ \hfill & =\hfill & {\mathcal{F}}^{-1}[{|F|}^{2}],\hfill \end{array}$$
(3)
$$f(x)=\text{\u2211}_{1}^{N}{\alpha}_{i}\delta (x-{x}_{i})+{f}_{c}(x),$$
(4)
$${S}_{f}={\cup}_{1}^{N}\{{x}_{i}\}\cup supp({f}_{c}),$$
(5)
$${S}_{f\u2605f}\subset {S}_{f}-{S}_{f}=\{x-y:x,y\in {S}_{f}\}.$$
(6)
$$B\subset \cap \{{S}_{0}:{S}_{0}\in {\mathcal{\text{S}}}_{0}\}.$$
(7)
$$\begin{array}{cccc}B\subset S-x& \text{or}& B\subset x-S& \text{for some}\hspace{0.17em}x\in {\mathbf{\text{R}}}^{2}.\end{array}$$
(8)
$$E(A,u)=E(S,u)-E(S,-u)$$
(9)
$${E}_{l}(A,u)\subset {E}_{l}(S,u)-{E}_{l}(S,-u).$$
(10)
$$\begin{array}{cc}B\subset x-E(S,-u)& \text{for some}\hspace{0.17em}x\in E(S,u)\end{array}.$$
(11)
$$\begin{array}{cc}B\subset E(S,u)-x& \text{for some}\hspace{0.17em}x\in E(S,-u)\end{array}.$$
(12)
$$(x,y)=\left\{\begin{array}{cc}\{z:z=\mathit{\text{tx}}+(1-t)y,t\in (0,1)\}& ifx\ne y\\ \varnothing & ifx=y\end{array}\right\}$$
(13)
$$[x,y]=\{z:z=\mathit{\text{tx}}+(1-t)y,t\in [0,1]\},$$
(14)
$${a}_{1}+{a}_{2}\notin [E(A,u)\backslash \{{a}_{1},{a}_{2}\}]+[E(A,u)\backslash \{{a}_{1}+{a}_{2}\}],$$
(15)
$$d(B,u)=sup\{\u3008x-y,u\u3009:x,y\in B\},$$
(16)
$$\mathcal{\text{P}}(B)=\{({b}_{1},{b}_{2},{b}_{3},{b}_{4}:{b}_{i}\in B\hspace{0.17em}\text{for}\hspace{0.17em}i=1,2,3,4;{b}_{1}\ne {b}_{j}\hspace{0.17em}for\hspace{0.17em}j=2,3,4;\hspace{0.17em}\text{and}\hspace{0.17em}{b}_{1}+{b}_{3}={b}_{2}+{b}_{4}\}.$$
(17)
$${\mathcal{\text{P}}}^{\prime}=\{({a}_{1},{a}_{2},{a}_{3},{a}_{4})\in \mathcal{\text{P}}(I):{a}_{1}\in E(A,u),{a}_{3}={a}^{\prime}\}=\varnothing ,$$
(18)
$$sup\{\u3008x,u\u3009:x\in B\}=-inf\{\u3008x,u\u3009:x\in B\}=\frac{d(B,u)}{2}$$
(19)
$$sup\{\u3008x,\upsilon \u3009:x\in B\}=-inf\{\u3008x,\upsilon \u3009:x\in B\}=\frac{d(B,\upsilon )}{2},$$
(20)
$$S+x\subset {S}_{0}={S}_{0}-y+y\subset {S}_{0}-{S}_{0}+y=A+y.$$
(21)
$$S+x\subset \cap \{A+y:y\in {S}_{0}\}\subset \cap \{A+y:y\in B\}=L.$$
(22)
$$\begin{array}{lll}\u3008x-y,u\u3009\hfill & =\hfill & \u3008x,u\u3009+\u3008y,-y\u3009\hfill \\ \hfill & \ge \hfill & \u3008{x}^{\prime},u\u3009+\u3008{y}^{\prime},-y\u3009\hfill \\ \hfill & =\hfill & \u3008{x}^{\prime}-{y}^{\prime},u\u3009.\hfill \end{array}$$
(23)
$$({x}_{1}-{y}_{2})+({x}_{2}-{y}_{1})={a}_{1}+{a}_{2}.$$
(24)
$${C}_{u}(B)=\cap \{C:B\subset C;C\hspace{0.17em}\text{is}\hspace{0.17em}uconvex\}.$$
(25)
$${C}_{u}(B)=\cup \{[x,y]:x,y\in B,(x-y)\perp u\}.$$
(26)
$${x}_{1}-{y}_{2}=({x}_{1}-{b}_{1})+({b}_{1}-{b}_{2})+({b}_{2}-{y}_{2})\perp u.$$
(27)
$$[{x}_{1},{y}_{2}],[{x}_{1},{x}_{2}],[{y}_{1},{y}_{2}],[{x}_{2},{y}_{1}]\subset {C}_{0}.$$
(28)
$$\begin{array}{lll}m\hfill & =\hfill & \frac{{a}_{1}+{a}_{2}}{2}\hfill \\ \hfill & =\hfill & \frac{{x}_{1}-{y}_{1}+{x}_{2}-{y}_{2}}{2}\hfill \\ \hfill & =\hfill & \frac{{x}_{1}-{y}_{1}+{x}_{1}+\beta \upsilon -{y}_{1}+\gamma \upsilon}{2}\hfill \\ \hfill & =\hfill & {x}_{1}+\frac{\beta +\gamma}{2}\upsilon -{y}_{1}.\hfill \end{array}$$
(29)
$$\text{d}(A,u)=2d(S,u).$$
(30)
$$E[A\cap (A+a),u]=\{a\}.$$
(31)
$$E[A\cap (A+a),-u]=\{0\},$$
(32)
$$d[A\cap (A+a),u]=\frac{d(A,u)}{2}=d(S,u).$$
(33)
$$\u3008{x}^{\prime}-y-\frac{x-y}{2},\upsilon \u3009<-\frac{d}{4}$$
(34)
$$\u3008-{y}^{\prime}+x-\frac{x-y}{2},\upsilon \u3009<-\frac{d}{4}.$$
(35)
$$\u3008{y}^{\prime}-{x}^{\prime},\upsilon \u3009>\frac{d}{2},$$
(36)
$$sup\{\u3008x+{x}_{0},u\u3009:x\in {B}_{1}\}=\u3008{x}_{0},u\u3009+\frac{d({B}_{1},u)}{2}>\frac{d({B}_{2},u)}{2}$$
(37)
$$inf\{\u3008-x+{x}_{0},-u\u3009:x\in {B}_{1}\}=-\frac{d({B}_{1},u)}{2}-\u3008{x}_{0},u\u3009<-\frac{d({B}_{2},u)}{2}$$