$q_1 = \frac{2\pi}{\lambda} \frac{s_N}{NL},$
where $s_N$ - distance between zero maximum and maximum $N$; $L$ - distance between DG and screen.

From formula A1 one could note that intensity $I$ of maximum depends on its number $h$:

$$
I(h) = I_0 \left( \frac{\sin (\pi b/a \cdot h)}{ (\pi b/a \cdot h)} \right)^2.
$$

Measured intensities for two different maxima gives us a transcendental equation

$$
\pi b/a = \arcsin \left( \frac{1}{2} \sqrt{\frac{I(1)}{I(2)}} \sin (2 \pi b/a ) \right).
$$

Root of this equation ($\pi b/a$) could be found numerically.

For DG3 measured intensities are $I(1) = 790~mV, \quad I(2) = 630~mV,$ and solution of equation is $\pi b/a = 0.397.$ This leads to $a/b \approx 8.3,$ which is quite close to theoretical value $8$.

For DG4 and DG5 one could note that each 4th and each 2nd maxima (respectively) are dimmed. That's why $a/b = 4$ for DG4, and $a/b = 2$.

Structure factor for reflex $-h, -k$:

$$
F(-h,-k) = \int \rho(x,y) e^{-i(q_1 h x + q_2 k y)} dx dy = \left( \int \rho(x,y) e^{-i(q_1 h x + q_2 k y)} dx dy \right)^* = F^*(h, k).
$$

Here $F^*$ is a complex conjugate of $F$. This property of $-h, -k$ reflex is only possible because $\rho(x, y)$ is real function.

Intensities are equal

$$
I(-h,-k) = F(-h, -k) F^*(-h,-k) = F^*(h, k) F(h, k) = I(h, k).
$$

Let $\rho (x, y)$ be unit cell for initial crystal. Structure factor for it is:

$$
F(q_x, q_y) = \int_{-x_0}^{x_0} \int_{-y_0}^{y_0} \rho(x, y) e^{i(q_x x + q_y y)} dx dy,
$$

here $x_0 = y_0 = a/2$, $a$ – is a period of lattice. Below limits are presented only when necessary.

For symmetry $x=0$ unit cell is $\rho_2 (x, y) = \rho (-x, y)$, structure factor is

\begin{multline*}
f_2(q_x, q_y) = \int_{-x_0}^{x_0} \int_{-y_0}^{y_0} \rho_2(x, y) e^{i(q_x x + q_y y)} dx dy = \int_{-x_0}^{x_0} \int_{-y_0}^{y_0} \rho(-x, y) e^{i(q_x x + q_y y)} dx dy = \\ = \int_{x_0}^{-x_0} \int_{-y_0}^{y_0} \rho(x, y) e^{i(-q_x x + q_y y)} d(-x) dy = F(-q_x, q_y).
\end{multline*}

For symmetry $x=y$ unit cell is $\rho_3 (x, y) = \rho (y,x)$, structure factor is

\begin{multline*}
f_3(q_x, q_y) = \int \rho_3(x, y) e^{i(q_x x + q_y y)} dx dy = \int \rho(y, x) e^{i(q_x x + q_y y)} dx dy = \\ = \int \rho(x, y) e^{i(q_y x + q_x y)} dy dx = F(q_y, q_x).
\end{multline*}

Point $(x, y)$ moves to new position $(x', y') = (x+x_1, y+y_1)$. That's why $\rho_4(x', y') = \rho (x, y).$

\begin{multline*}
f_4(q_x, q_y) = \int \rho_4(x', y') e^{i(q_x x' + q_y y')} dx' dy' = \\ = \left( \int \rho(x, y) e^{i(q_x x + q_y y)} dx dy \right)e^{i(q_x x_1 + q_y y_1)} = F(q_x, q_y) e^{i(q_x x_1 + q_y y_1)}.
\end{multline*}

Let's consider two nearest atoms A and B (fig.), that means $\vec{AB}$ – lattice vector, $|\vec{AB}|=a$. If rotational symmetry of order $m$ is present, atom C could be obtained from B by rotation by angle $\theta = \frac{2\pi}{m}.$ The same way gives atom D from atom A. These two atoms C and D should also be connected by translation symmetry, it means $\vec{CD} = n \cdot \vec{AB}, \quad n\in\mathbb{Z}$.

It leads us to equation

$$
\cos\theta = \frac{1-n}{2}.
$$

All possible $n$, $\cos \theta$, $\theta$, $m$ are listed in table.

Symmetries of unit cells:

Diffraction patterns from PG4 and PG6 have $C_4$ symmetry (as unit cells R, T), and all other don't have. In this group symmetries are not enough to match unit cells and diffraction patterns. One needs to use sum rule and to understand that absences are presented for unit cells S and T (as for PG6 and PG9).

PG9. Elements are designated as on fig. Elements 2, 3, 4 were obtained by $C_2$ symmetry and move.

$$
f_1(h,k) = F(h,k).
$$

$$
f_2(h,k) = F(-h, -k) e^{i 5 \pi h/7}.
$$

$$
f_3(h,k) = F(-h, k) e^{- i \pi (2 h/7 + k)}.
$$

$$
f_4(h,k) = F(h, -k) e^{- i \pi (h + k)}.
$$

For reflexes $(0,k)$

$$
f_{PG9}(0,k) = \left( 1 + \cos (\pi k) \right) \left( F(0, k) + F(0, -k) \right),
$$

each odd $k = 2n+1, n \in \mathbb{Z}$ is absent. The same is true for another axis $(h,0)$.

\textbf{PG6.} Let $f_1(h,k) = F(h,k)$. Element 5 could be obtained by mirroring $x=y$ and by moved by $(-a/2, -a/2)$, that's why $f_5(h,k) = F(k, h) e^{-\pi (h+k)}.$ Each next pair of elements were obtained by rotating $90^ \circ$.

Structure factors
$$
f_1 + f_5 = F(h, k) + F(k, h) e^{-\pi (h+k)}
$$

$$
f_2 + f_6 = F(k, -h) + F(h, -k) e^{-\pi (h-k)}
$$

$$
f_3 + f_7 = F(-h, -k) + F(-k, -h) e^{-\pi (-h-k)}
$$

$$
f_4 + f_8 = F(-k, h) + F(-h, k) e^{-\pi (-h+k)}
$$

For $(0,k)$

$$
f_{PG6}(0,k) = \left( 1+ \cos (\pi k) \right) \cdot \left( F(0,k) + F(k,0) + F(0, -k) + F(-k,0) \right),
$$

each odd $k = 2n+1, n \in \mathbb{Z}$ is absent. The same is true for another axis $(h,0)$.

Electric field is $E_0 = \sqrt{I_0}$ in plane exactly before the diffraction grating. Electric field of the reflex $\vec{q} = 0$ is proportional to transparent area of an unit cell and should be $E_0$ if the whole unit cell is transparent (i.e. there is no any diffraction grating):

$$
E = \frac{N_{transparent}}{N_{all}} E_0.
$$

Intensity

$$
I = \left( \frac{N_{transparent}}{N_{all}} \right)^2 I_0 .
$$

One needs to measure intensities $I(h,k)$ of each reflexes $|h|\le2$, $|k|\le2$ of MR1.

Transparency of a square with position ($\chi$, $\gamma$):

\begin{equation*}
\rho(\chi, \gamma) = \sum_{h=-2}^{2} \sum_{k=-2}^{2} \sqrt{I(h, k)} \cdot e^{i \varphi(h, k)} \cdot \exp \left(- 2\pi i \left(\frac{\chi}{4} h + \frac{\gamma}{4} k \right) \right),
\end{equation*}

here $\chi \in {0,1,2,3}$, $\gamma \in {0,1,2,3}$.

Because $\rho(\chi, \gamma) \in \mathbb{R}$, its enough to calculate only real part:

\begin{equation}
\rho(\chi, \gamma) = \sum_{h=-2}^{2} \sum_{k=-2}^{2} \sqrt{I(h, k)} \cdot \cos \left( \varphi(h, k) - \frac{\pi}{2} (\chi h + \gamma k) \right).
\end{equation}

Calculated transparencies for MR1 are on figure. It is not necessary to calculate whole 16 squares. One needs to calculate transparency of squares (0,0), (3,3) to understand which values corresponds to transparent and non-transparent values. After that square (3, 2) should be evaluated. It has value close to transparent, that's why structure X is the answer.

The same formula should be used to find structure of MR2. Intensities of diffraction patterns and calculated transparencies for MR2 are shown on figures

Because it is known that two non-transparent squares become transparent one needs to calculate transparency only for 11 black squares. Two of them (0, 2) and (3,0) are transparent for MR2.