# Exercise 6
## Generating points according to Fraunhöfer diffraction
The diffraction of a plane wave through a round slit is to be simulated by
generating $N =$ 50'000 points according to the intensity distribution
$I(\theta)$ [@hecht02] on a screen at a great distance $L$ from the slit itself
(see @fig:slit):
$$
I(\theta) = \frac{E^2}{2} \left( \frac{2 \pi a^2 \cos{\theta}}{L}
\frac{J_1(x)}{x} \right)^2 \with x = k a \sin{\theta}
$$
where:
- $E$ is the electric field amplitude, default set $E = \SI{1e4}{V/m}$;
- $a$ is the radius of the slit aperture, default set $a = \SI{0.01}{m}$;
- $\theta$ is the angle specified in @fig:slit;
- $J_1$ is the Bessel function of first kind;
- $k$ is the wavenumber, default set $k = \SI{1e-4}{m^{-1}}$;
- $L$ default set $L = \SI{1}{m}$.
\begin{figure}
\hypertarget{fig:slit}{%
\centering
\begin{tikzpicture}
\definecolor{cyclamen}{RGB}{146, 24, 43}
% Walls
\draw [thick] (-1,3) -- (1,3) -- (1,0.3) -- (1.2,0.3) -- (1.2,3)
-- (9,3);
\draw [thick] (-1,-3) -- (1,-3) -- (1,-0.3) -- (1.2,-0.3) -- (1.2,-3)
-- (9,-3);
\draw [thick] (10,3) -- (9.8,3) -- (9.8,-3) -- (10,-3);
% Lines
\draw [thick, gray] (0.7,0.3) -- (0.5,0.3);
\draw [thick, gray] (0.7,-0.3) -- (0.5,-0.3);
\draw [thick, gray] (0.6,0.3) -- (0.6,-0.3);
\draw [thick, gray] (1.2,0) -- (9.8,0);
\draw [thick, gray] (1.2,-0.1) -- (1.2,0.1);
\draw [thick, gray] (9.8,-0.1) -- (9.8,0.1);
\draw [thick, cyclamen] (1.2,0) -- (9.8,-2);
\draw [thick, cyclamen] (7,0) to [out=-90, in=50] (6.6,-1.23);
% Nodes
\node at (0,0) {$2a$};
\node at (5.5,0.4) {$L$};
\node [cyclamen] at (5.5,-0.4) {$\theta$};
\node [rotate=-90] at (10.2,0) {screen};
\end{tikzpicture}
\caption{Fraunhöfer diffraction.}\label{fig:slit}
}
\end{figure}
Once again, the *try and catch* method described in @sec:3 was implemented and
the same procedure about the generation of $\theta$ was applied. This time,
though, $\theta$ must be evenly distributed on half sphere, hence:
\begin{align*}
\frac{d^2 P}{d\omega^2} = const = \frac{1}{2 \pi}
&\thus d^2 P = \frac{1}{2 \pi} d\omega^2 =
\frac{1}{2 \pi} d\phi \sin{\theta} d\theta \\
&\thus \frac{dP}{d\theta} = \int_0^{2 \pi} d\phi \frac{1}{2 \pi} \sin{\theta}
= \frac{1}{2 \pi} \sin{\theta} \, 2 \pi = \sin{\theta}
\end{align*}
\begin{align*}
\theta = \theta (x) &\thus
\frac{dP}{d\theta} = \frac{dP}{dx} \cdot \left| \frac{dx}{d\theta} \right|
= \left. \frac{dP}{dx} \middle/ \, \left| \frac{d\theta}{dx} \right| \right.
\\
&\thus \sin{\theta} = \left. 1 \middle/ \, \left|
\frac{d\theta}{dx} \right| \right.
\end{align*}
If $\theta$ is chosen to grew together with $x$, then the absolute value can be
omitted:
\begin{align*}
\frac{d\theta}{dx} = \frac{1}{\sin{\theta}}
&\thus d\theta \sin(\theta) = dx
\\
&\thus - \cos (\theta') |_{0}^{\theta} = x(\theta) - x(0) = x - 0 = x
\\
&\thus - \cos(\theta) + 1 =x
\\
&\thus \theta = \text{acos} (1 -x)
\end{align*}
The so obtained sample was binned and stored in a histogram with a customizable
number $n$ of bins (default set $n = 150$) ranging from $\theta = 0$ to $\theta
= \pi/2$ because of the system symmetry. In @fig:original an example is shown.
{#fig:original}
## Gaussian convolution {#sec:convolution}
In order to simulate the instrumentation response, the sample was then smeared
with a Gaussian function with the aim to recover the original sample afterwards,
implementing a deconvolution routine.
For this purpose, a 'kernel' histogram with an even number $m$ of bins and the
same bin width of the previous one, but a smaller number of them ($m \sim 6\%
\, n$), was generated according to a Gaussian distribution with mean $\mu = 0$
and variance $\sigma$. The reason why the kernel was set this way will be
discussed shortly.
Then, the original histogram was convolved with the kernel in order to obtain
the smeared signal. As an example, the result obtained for $\sigma = \Delta
\theta$, where $\Delta \theta$ is the bin width, is shown in @fig:convolved.
The smeared signal looks smoother with respect to the original one: the higher
$\sigma$, the greater the smoothness.
{#fig:convolved}
The convolution was implemented as follow. Consider the definition of
convolution of two functions $f(x)$ and $g(x)$:
$$
f * g (x) = \int \limits_{- \infty}^{+ \infty} dy f(y) g(x - y)
$$
Since a histogram is made of discrete values, a discrete convolution of the
signal ($s$) and the kernel ($k$) must be computed. Hence, the procedure boils
down to an element wise product between $s$ and the flipped histogram of $k$
(from the last bin to the first one) for each relative position of the two
histograms. Namely, if $c_i$ is the $i^{\text{th}}$ bin of the convolved
histogram:
$$
c_i = \sum_{j = 0}^{m - 1} k_j s_{i - j}
= \sum_{j' = m - 1}^{0} k_{m - 1 - j'} s_{i - m + 1 + j'}
\with j' = m - 1 - j
$$
For a better understanding, see @fig:dot_conv: the third histogram turns out
with $n + m - 1$ bins, a number greater than the original one.
\begin{figure}
\hypertarget{fig:dot_conv}{%
\centering
\begin{tikzpicture}
\definecolor{cyclamen}{RGB}{146, 24, 43}
% original histogram
\draw [thick, cyclamen, fill=cyclamen!05!white] (0.0,0) rectangle (0.5,2.5);
\draw [thick, cyclamen, fill=cyclamen!05!white] (0.5,0) rectangle (1.0,2.8);
\draw [thick, cyclamen, fill=cyclamen!25!white] (1.0,0) rectangle (1.5,2.3);
\draw [thick, cyclamen, fill=cyclamen!25!white] (1.5,0) rectangle (2.0,1.8);
\draw [thick, cyclamen, fill=cyclamen!25!white] (2.0,0) rectangle (2.5,1.4);
\draw [thick, cyclamen, fill=cyclamen!25!white] (2.5,0) rectangle (3.0,1.0);
\draw [thick, cyclamen, fill=cyclamen!25!white] (3.0,0) rectangle (3.5,1.0);
\draw [thick, cyclamen, fill=cyclamen!05!white] (3.5,0) rectangle (4.0,0.6);
\draw [thick, cyclamen, fill=cyclamen!05!white] (4.0,0) rectangle (4.5,0.4);
\draw [thick, cyclamen, fill=cyclamen!05!white] (4.5,0) rectangle (5.0,0.2);
\draw [thick, cyclamen, fill=cyclamen!05!white] (5.0,0) rectangle (5.5,0.2);
\draw [thick, cyclamen] (6.0,0) -- (6.0,0.2);
\draw [thick, cyclamen] (6.5,0) -- (6.5,0.2);
\draw [thick, <->] (0,3.3) -- (0,0) -- (7,0);
% kernel histogram
\draw [thick, cyclamen, fill=cyclamen!25!white] (1.0,-1) rectangle (1.5,-1.2);
\draw [thick, cyclamen, fill=cyclamen!25!white] (1.5,-1) rectangle (2.0,-1.6);
\draw [thick, cyclamen, fill=cyclamen!25!white] (2.0,-1) rectangle (2.5,-1.8);
\draw [thick, cyclamen, fill=cyclamen!25!white] (2.5,-1) rectangle (3.0,-1.6);
\draw [thick, cyclamen, fill=cyclamen!25!white] (3.0,-1) rectangle (3.5,-1.2);
\draw [thick, <->] (1,-2) -- (1,-1) -- (4,-1);
% arrows
\draw [thick, cyclamen, <->] (1.25,-0.2) -- (1.25,-0.8);
\draw [thick, cyclamen, <->] (1.75,-0.2) -- (1.75,-0.8);
\draw [thick, cyclamen, <->] (2.25,-0.2) -- (2.25,-0.8);
\draw [thick, cyclamen, <->] (2.75,-0.2) -- (2.75,-0.8);
\draw [thick, cyclamen, <->] (3.25,-0.2) -- (3.25,-0.8);
\draw [thick, cyclamen, ->] (2.25,-2.0) -- (2.25,-4.2);
% smeared histogram
\begin{scope}[shift={(0,-1)}]
\draw [thick, cyclamen, fill=cyclamen!05!white] (-1.0,-4.5) rectangle (-0.5,-4.3);
\draw [thick, cyclamen, fill=cyclamen!05!white] (-0.5,-4.5) rectangle ( 0.0,-4.2);
\draw [thick, cyclamen, fill=cyclamen!05!white] ( 0.0,-4.5) rectangle ( 0.5,-2.0);
\draw [thick, cyclamen, fill=cyclamen!05!white] ( 0.5,-4.5) rectangle ( 1.0,-1.6);
\draw [thick, cyclamen, fill=cyclamen!05!white] ( 1.0,-4.5) rectangle ( 1.5,-2.3);
\draw [thick, cyclamen, fill=cyclamen!05!white] ( 1.5,-4.5) rectangle ( 2.0,-2.9);
\draw [thick, cyclamen, fill=cyclamen!25!white] ( 2.0,-4.5) rectangle ( 2.5,-3.4);
\draw [thick, cyclamen] (3.0,-4.5) -- (3.0,-4.3);
\draw [thick, cyclamen] (3.5,-4.5) -- (3.5,-4.3);
\draw [thick, cyclamen] (4.0,-4.5) -- (4.0,-4.3);
\draw [thick, cyclamen] (4.5,-4.5) -- (4.5,-4.3);
\draw [thick, cyclamen] (5.0,-4.5) -- (5.0,-4.3);
\draw [thick, cyclamen] (5.5,-4.5) -- (5.5,-4.3);
\draw [thick, cyclamen] (6.0,-4.5) -- (6.0,-4.3);
\draw [thick, cyclamen] (6.5,-4.5) -- (6.5,-4.3);
\draw [thick, cyclamen] (7.0,-4.5) -- (7.0,-4.3);
\draw [thick, cyclamen] (7.5,-4.5) -- (7.5,-4.3);
\draw [thick, <->] (-1,-2.5) -- (-1,-4.5) -- (8,-4.5);
\end{scope}
% nodes
\node [above] at (2.25,-5.5) {$c_i$};
\node [above] at (3.25,0) {$s_i$};
\node [above] at (1.95,0) {$s_{i-j}$};
\node [below] at (1.75,-1) {$k_j$};
\end{tikzpicture}
\caption{Element wise product as a step of the convolution between the original signal
(above) and the kernel (center). The final result is the lower
fledging histogram.}\label{fig:dot_conv}
}
\end{figure}
## Unfolding with FFT
Two different unfolding routines were implemented, one of which exploiting the
Fast Fourier Transform (FFT).
This method is based on the convolution theorem, according to which, given two
functions $f(x)$ and $g(x)$:
$$
\hat{F}[f * g] = \hat{F}[f] \cdot \hat{F}[g]
$$
where $\hat{F}[\quad]$ stands for the Fourier transform of its argument.
Being the histogram a discrete set of data, the Discrete Fourier Transform (DFT)
was applied. When dealing with arrays of discrete values, the theorem still
holds if the two arrays have the same length and a cyclical convolution is
applied. For this reason the kernel was 0-padded in order to make it the same
length of the original signal. Besides, the 0-padding allows to avoid unpleasant
side effects due to the cyclical convolution.
In order to accomplish this procedure, both histograms were transformed into
vectors. The implementation lies in the computation of the Fourier transform of
the smeared signal and the kernel, the ratio between their transforms and the
anti-transformation of the result:
$$
\hat{F}[s * k] = \hat{F}[s] \cdot \hat{F}[k] \thus
\hat{F} [s] = \frac{\hat{F}[s * k]}{\hat{F}[k]}
$$
The FFT are efficient algorithms for calculating the DFT. Given a set of $n$
values {$z_j$}, each one is transformed into:
$$
x_j = \sum_{k=0}^{n-1} z_k \exp \left( - \frac{2 \pi i j k}{n} \right)
$$
where $i$ is the imaginary unit.
The evaluation of the DFT is a matrix-vector multiplication $W \vec{z}$. A
general matrix-vector multiplication takes $O(n^2)$ operations. FFT algorithms,
instead, use a *divide-and-conquer* strategy to factorize the matrix into
smaller sub-matrices. If $n$ can be factorized into a product of integers $n_1$,
$n_2 \ldots n_m$, then the DFT can be computed in $O(n \sum n_i) < O(n^2)$
operations, hence the name.
The inverse Fourier transform is thereby defined as:
$$
z_j = \frac{1}{n}
\sum_{k=0}^{n-1} x_k \exp \left( \frac{2 \pi i j k}{n} \right)
$$
In GSL, `gsl_fft_complex_forward()` and `gsl_fft_complex_inverse()` are
functions which allow to compute the forward and inverse transform,
respectively.
The inputs and outputs for the complex FFT routines are packed arrays of
floating point numbers. In a packed array, the real and imaginary parts of each
complex number are placed in alternate neighboring elements. In this special
case, the sequence of values to be transformed is made of real numbers, hence
Fourier transform is a complex sequence which satisfies:
$$
z_k = z^*_{n-k}
$$
where $z^*$ is the conjugate of $z$. A sequence with this symmetry is called
'half-complex'. This structure requires particular storage layouts for the
forward transform (from real to half-complex) and inverse transform (from
half-complex to real). As a consequence, the routines are divided into two sets:
`gsl_fft_real` and `gsl_fft_halfcomplex`. The symmetry of the half-complex
sequence requires only half of the complex numbers in the output to be stored.
This works for all lengths: when the length is odd, the middle value is real.
Thus, only $n$ real numbers are required to store the half-complex sequence
(half for the real part and half for the imaginary).
\begin{figure}
\hypertarget{fig:reorder}{%
\centering
\begin{tikzpicture}
\definecolor{cyclamen}{RGB}{146, 24, 43}
% standard histogram
\draw [thick, cyclamen] (0.5,0) -- (0.5,0.2);
\draw [thick, cyclamen, fill=cyclamen!25!white] (1.0,0) rectangle (1.5,0.6);
\draw [thick, cyclamen, fill=cyclamen!25!white] (1.5,0) rectangle (2.0,1.2);
\draw [thick, cyclamen, fill=cyclamen!25!white] (2.0,0) rectangle (2.5,1.4);
\draw [thick, cyclamen, fill=cyclamen!25!white] (2.5,0) rectangle (3.0,1.4);
\draw [thick, cyclamen, fill=cyclamen!25!white] (3.0,0) rectangle (3.5,1.2);
\draw [thick, cyclamen, fill=cyclamen!25!white] (3.5,0) rectangle (4.0,0.6);
\draw [thick, cyclamen] (4.5,0) -- (4.5,0.2);
\draw [thick, ->] (0,0) -- (5,0);
\draw [thick, ->] (2.5,0) -- (2.5,2);
% shifted histogram
\begin{scope}[shift={(7,0)}]
\draw [thick, cyclamen, fill=cyclamen!25!white] (0.5,0) rectangle (1.0,1.4);
\draw [thick, cyclamen, fill=cyclamen!25!white] (1.0,0) rectangle (1.5,1.2);
\draw [thick, cyclamen, fill=cyclamen!25!white] (1.5,0) rectangle (2.0,0.6);
\draw [thick, cyclamen, fill=cyclamen!25!white] (3.0,0) rectangle (3.5,0.6);
\draw [thick, cyclamen, fill=cyclamen!25!white] (3.5,0) rectangle (4.0,1.2);
\draw [thick, cyclamen, fill=cyclamen!25!white] (4.0,0) rectangle (4.5,1.4);
\draw [thick, ->] (0,0) -- (5,0);
\draw [thick, ->] (2.5,0) -- (2.5,2);
\end{scope}
\end{tikzpicture}
\caption{The histogram on the right shows how the real numbers histogram on the
left is handled by the dedicated GSL functions.}\label{fig:reorder}
}
\end{figure}
If the bin width is $\Delta \theta$, then the DFT domain ranges from $-1 / (2
\Delta \theta)$ to $+1 / (2 \Delta \theta$). As regards the real values, the
aforementioned GSL functions store the positive values from the beginning of
the array up to the middle and the negative backwards from the end of the array
(see @fig:reorder).
Whilst do not matters if the convolved histogram has positive or negative
values, the kernel must be centered in zero in order to compute a correct
convolution. This requires the kernel to be made of an ever number of bins
in order to be possible to cut it into two same-length halves.
When $\hat{F}[s * k]$ and $\hat{F}[k]$ are computed, they are given in the
half-complex GSL format and their normal format must be restored in order to
use them as standard complex numbers and compute the ratio between them. Then,
the result must return in the half-complex format for the inverse DFT
computation. GSL provides the function `gsl_fft_halfcomplex_unpack()` which
convert the vectors from half-complex format to standard complex format but the
inverse procedure is not provided by GSL and was hence implemented in the
code.
At the end, the external bins which exceed the original signal size are cut
away in order to restore the original number of bins $n$. Results will be
discussed in @sec:conv_results.
## Unfolding with Richardson-Lucy
The Richardson–Lucy (RL) deconvolution is an iterative procedure typically used
for recovering an image that has been blurred by a known 'point spread
function'.
Consider the problem of estimating the frequency distribution $f(\xi)$ of a
variable $\xi$ when the available measure is a sample {$x_i$} of points
drown not by $f(x)$ but by another function $\phi(x)$ such that:
$$
\phi(x) = \int d\xi \, f(\xi) P(x | \xi)
$$ {#eq:conv}
where $P(x | \xi) \, d\xi$ is the probability (presumed known) that $x$ falls
in the interval $(x, x + dx)$ when $\xi = \xi$. If the so-called point spread
function $P(x | \xi)$ follows a normal distribution with variance $\sigma$,
namely:
$$
P(x | \xi) = \frac{1}{\sqrt{2 \pi} \sigma}
\exp \left( - \frac{(x - \xi)^2}{2 \sigma^2} \right)
$$
then, @eq:conv becomes a convolution and finding $f(\xi)$ turns out to be a
deconvolution.
An example of this problem is precisely that of correcting an observed
distribution $\phi(x)$ for the effect of observational errors, which are
represented by the function $P (x | \xi)$.
Let $Q(\xi | x) d\xi$ be the probability that $\xi$ comes from the interval
$(\xi, \xi + d\xi)$ when the measured quantity is $x = x$. The probability that
both $x \in (x, x + dx)$ and $(\xi, \xi + d\xi)$ is therefore given by $\phi(x)
dx \cdot Q(\xi | x) d\xi$ which is identical to $f(\xi) d\xi \cdot P(x | \xi)
dx$, hence:
$$
\phi(x) dx \cdot Q(\xi | x) d\xi = f(\xi) d\xi \cdot P(x | \xi) dx
\thus Q(\xi | x) = \frac{f(\xi) \cdot P(x | \xi)}{\phi(x)}
$$
$$
\thus Q(\xi | x) = \frac{f(\xi) \cdot P(x | \xi)}
{\int d\xi \, f(\xi) P(x | \xi)}
$$ {#eq:first}
which is the Bayes theorem for conditional probability. From the normalization
of $P(x | \xi)$, it follows also that:
$$
f(\xi) = \int dx \, \phi(x) Q(\xi | x)
$$ {#eq:second}
Since $Q (\xi | x)$ depends on $f(\xi)$, @eq:second suggests an iterative
procedure for generating estimates of $f(\xi)$. With a guess for $f(\xi)$ and
a known $P(x | \xi)$, @eq:first can be used to calculate and estimate for
$Q (\xi | x)$. Then, taking the hint provided by @eq:second, an improved
estimate for $f (\xi)$ can be generated, using the observed sample {$x_i$} to
give an approximation for $\phi$.
Thus, if $f^t$ is the $t^{\text{th}}$ estimate, the $t^{\text{th + 1}}$ is:
$$
f^{t + 1}(\xi) = \int dx \, \phi(x) Q^t(\xi | x)
\with
Q^t(\xi | x) = \frac{f^t(\xi) \cdot P(x | \xi)}
{\int d\xi \, f^t(\xi) P(x | \xi)}
$$
from which:
$$
f^{t + 1}(\xi) = f^t(\xi)
\int dx \, \frac{\phi(x)}{\int d\xi \, f^t(\xi) P(x | \xi)}
P(x | \xi)
$$ {#eq:solution}
When the spread function $P(x | \xi)$ is Gaussian, @eq:solution can be
rewritten in terms of convolutions:
$$
f^{t + 1} = f^{t}\left( \frac{\phi}{{f^{t}} * P} * P^{\star} \right)
$$
where $P^{\star}$ is the flipped point spread function [@lucy74].
In this special case, the Gaussian kernel which was convolved with the original
histogram stands for the point spread function. Dealing with discrete values,
the division and multiplication are element wise and the convolution is to be
carried out as described in @sec:convolution.
When implemented, this method results in an easy step-wise routine:
- choose a zero-order estimate for {$f(\xi)$};
- create a flipped copy of the kernel;
- compute the convolutions, the product and the division at each step;
- proceed until a given number $r$ of iterations is reached.
In this case, the zero-order was set $f(\xi) = 0.5 \, \forall \, \xi$. Different
number of iterations where tested. Results are discussed in
@sec:conv_results.
## The earth mover's distance
With the aim of comparing the two deconvolution methods, the similarity of a
deconvolved outcome with the original signal was quantified using the earth
mover's distance.
In statistics, the earth mover's distance (EMD) is the measure of distance
between two distributions [@cock41]. Informally, if one imagines the two
distributions as two piles of different amount of dirt in their respective
regions, the EMD is the minimum cost of turning one pile into the other,
making the first one the most possible similar to the second one, where the
cost is the amount of dirt moved times the distance by which it is moved.
Computing the EMD is based on a solution to the transportation problem, which
can be formalized as follows.
Consider two vectors $P$ and $Q$ which represent the two distributions whose
EMD has to be measured:
$$
P = \{ (p_1, w_{p1}) \dots (p_m, w_{pm}) \} \et
Q = \{ (q_1, w_{q1}) \dots (q_n, w_{qn}) \}
$$
where $p_i$ and $q_i$ are the 'values' (that is, the location of the dirt) and
$w_{pi}$ and $w_{qi}$ are the 'weights' (that is, the quantity of dirt). A
ground distance matrix $D_{ij}$ is defined such as its entries $d_{ij}$ are the
distances between $p_i$ and $q_j$. The aim is to find the flow matrix $F_{ij}$,
where each entry $f_{ij}$ is the flow from $p_i$ to $q_j$ (which would be
the quantity of moved dirt), which minimizes the cost $W$:
$$
W (P, Q, F) = \sum_{i = 1}^m \sum_{j = 1}^n f_{ij} d_{ij}
$$
The $Q$ region is to be considered empty at the beginning: the 'dirt' present in
$P$ must be moved to $Q$ in order to reach the same distribution as close as
possible. Namely, the following constraints must be satisfied:
\begin{align*}
&\text{1.} \hspace{20pt} f_{ij} \ge 0 \hspace{15pt}
&1 \le i \le m \wedge 1 \le j \le n
\\
&\text{2.} \hspace{20pt} \sum_{j = 1}^n f_{ij} \le w_{pi}
&1 \le i \le m
\\
&\text{3.} \hspace{20pt} \sum_{i = 1}^m f_{ij} \le w_{qj}
&1 \le j \le n
\\
&\text{4.} \hspace{20pt} \sum_{j = 1}^n f_{ij} \sum_{j = 1}^m f_{ij} \le w_{qj}
= \text{min} \left( \sum_{i = 1}^m w_{pi}, \sum_{j = 1}^n w_{qj} \right)
\end{align*}
The first constraint allows moving dirt from $P$ to $Q$ and not vice versa; the
second limits the amount of dirt moved by each position in $P$ in order to not
exceed the available quantity; the third sets a limit to the dirt moved to each
position in $Q$ in order to not exceed the required quantity and the last one
forces to move the maximum amount of supplies possible: either all the dirt
present in $P$ has been moved or the $Q$ distribution has been obtained.
The total moved amount is the total flow. If the two distributions have the
same amount of dirt, hence all the dirt present in $P$ is necessarily moved to
$Q$ and the flow equals the total amount of available dirt.
Once the transportation problem is solved and the optimal flow is found, the
EMD is defined as the work normalized by the total flow:
$$
\text{EMD} (P, Q) = \frac{\sum_{i = 1}^m \sum_{j = 1}^n f_{ij} d_{ij}}
{\sum_{i = 1}^m \sum_{j=1}^n f_{ij}}
$$
In this case, where the EMD is to be measured between two same-length
histograms, the procedure simplifies a lot. By representing both histograms
with two vectors $u$ and $v$, the equation above boils down to [@ramdas17]:
$$
\text{EMD} (u, v) = \sum_i |U_i - V_i|
$$
where the sum runs over the entries of the vectors $U$ and $V$, which are the
cumulative vectors of the histograms. In the code, the following equivalent
iterative routine was implemented.
$$
\text{EMD} (u, v) = \sum_i |\text{EMD}_i| \with
\begin{cases}
\text{EMD}_i = v_i - u_i + \text{EMD}_{i-1} \\
\text{EMD}_0 = 0
\end{cases}
$$
In fact:
\begin{align*}
\text{EMD} (u, v) &= \sum_i |\text{EMD}_i| = |\text{EMD}_0| + |\text{EMD}_1|
+ |\text{EMD}_2| + |\text{EMD}_3| + \dots \\
&= 0 + |v_1 - u_1 + \text{EMD}_0| +
|v_2 - u_2 + \text{EMD}_1| +
|v_3 - u_3 + \text{EMD}_2| + \dots \\
&= |v_1 - u_1| +
|v_1 - u_1 + v_2 - u_2| +
|v_1 - u_1 + v_2 - u_2 + v_3 - u_3| + \dots \\
&= |v_1 - u_i| +
|v_1 + v_2 - (u_1 + u_2)| +
|v_1 + v_2 + v_3 - (u_1 + u_2 + u_3))| + \dots \\
&= |V_1 - U_1| + |V_2 - U_2| + |V_3 - U_3| + \dots \\
&= \sum_i |U_i - V_i|
\end{align*}
This simple formula enabled comparisons to be made between a great number of
results.
In order to make the code more flexible, the data were normalized before
computing the EMD: in doing so, it is possible to compare even samples with a
different number of points.
## Results comparison {#sec:conv_results}
### Noiseless results {#sec:noiseless}
Along with the analysis of the results obtained varying the convolved Gaussian
width $\sigma$, the possibility to add a Gaussian noise to the convolved
histogram was also implemented to check weather the deconvolution is affected
or not by this kind of interference. This approach is described in the next
subsection, while the noiseless results are given in this one.
The two methods were compared for three different values of $\sigma$:
$$
\sigma = 0.1 \, \Delta \theta \et
\sigma = 0.5 \, \Delta \theta \et
\sigma = \Delta \theta
$$
Since the RL method depends on the number $r$ of performed rounds, in order to
find out how many of them it was sufficient or necessary to compute, the earth
mover's distance between the deconvolved signal and the original one was
measured for different $r$s for each of the three tested values of $\sigma$.
To achieve this goal, a number of 1000 experiments (default and customizable
value) were simulated and, for each of them, the original signal was convolved
with the kernel, the appropriate $\sigma$ value set, and then deconvolved with
the RL algorithm with a given $r$ and the EMD was measured. Then, an average of
the so-obtained EMDs was computed together with the standard deviation. This
procedure was repeated for a few tens of different $r$s till a flattening or a
minimum of the curve became evident. Results in @fig:rounds-noiseless.
The plots in @fig:rless-0.1 show the average (red) and standard deviation (grey)
of the measured EMD for $\sigma = 0.1 \, \Delta \theta$. The number of
iterations does not affect the quality of the outcome (those fluctuations are
merely a fact of floating-points precision) and the best result is obtained
for $r = 2$, meaning that the convergence of the RL algorithm is really fast and
this is due to the fact that the histogram was modified pretty poorly. In
@fig:rless-0.5, the curve starts to flatten at about 10 rounds, whereas in
@fig:rless-1 a minimum occurs around \SI{5e3}{} rounds, meaning that, whit such a
large kernel, the convergence is very slow, even if the best results are close
to the one found for $\sigma = 0.5$.
The following $r$s were chosen as the most fitted:
\begin{align*}
\sigma = 0.1 \, \Delta \theta &\thus n^{\text{best}} = 2 \\
\sigma = 0.5 \, \Delta \theta &\thus n^{\text{best}} = 10 \\
\sigma = 1 \, \Delta \theta &\thus n^{\text{best}} = \SI{5e3}{}
\end{align*}
Note the difference between @fig:rless-0.1 and the plots resulting from $\sigma =
0.5 \, \Delta \theta$ and $\sigma = \, \Delta \theta$ as regards the order of
magnitude: the RL deconvolution is heavily influenced by the variance magnitude:
the greater $\sigma$, the worse the deconvolved result.
On the other hand, the FFT deconvolution procedure is not affected by $\sigma$
amplitude changes: it always gives the same outcome, which would be exactly the
original signal, if the floating point precision would not affect the result. In
fact, the FFT is the analytical result of the deconvolution.
{#fig:rless-0.1}
{#fig:rless-0.5}
{#fig:rless-1}
EMD as a function of RL rounds for different kernel $\sigma$ values. The
average is shown in red and the standard deviation in grey. Noiseless results.
{#fig:eless-0.1}
{#fig:eless-0.5}
{#fig:eless-1}
EMD distributions for different kernel $\sigma$ values. The plots on the left
show the results for the FFT deconvolution, the central column the results for
the RL deconvolution and the third one shows the EMD for the convolved signal.
Noiseless results.
For this reason, the EMD obtained with the FFT can be used as a reference point
against which to compare the EMDs measured with RL.
As described above, for a given $r$, a thousands of experiments were simulated:
for each of this simulations, an EMD was computed. Besides computing their
average and standard deviations, those values were used to build histograms
showing the EMD distribution.
Once the most fitted numbers of rounds $r^{\text{best}}$ were found, their
histograms were compared to the histograms of the FFT results, started from the
same convolved signals, and the EDM of the convolved signals themselves, in
order to check if an improvement was truly achieved. Results are shown in
@fig:emd-noiseless.
As expected, the FFT results are always of the same order of magnitude,
\SI{1e-15}{}, independently from the kernel width, whereas the RL deconvolution
results change a lot, ranging from \SI{1e-16}{} for $\sigma = 0.1 \, \Delta
\theta$ to \SI{1e-4}{} for $\sigma = \Delta \theta$.
The first result was quite unexpected: for very low values of $\sigma$, the RL
routine gives better results with respect to the FFT. This is because of the
math which lies beneath the two methods: apparently, the RL technique is less
subject to round-off errors.
Then, as regards the comparison with the convolved signal, which shows always
an EMD of \SI{1e-2}{}, both RL and FFT always return results closer to the
original signal, meaning that the deconvolution is working.
### Noisy results
In order to observe the effect of the Gaussian noise upon these two
deconvolution methods, a certain value of $\sigma$ ($\sigma = 0.8 \, \Delta
\theta$) was arbitrary chosen. The noise was then applied to the convolved
histogram as follow.
{#fig:noisy}
For each bin, once the convolved histogram was computed, a value $v_N$ was
randomly sampled from a Gaussian distribution with standard deviation
$\sigma_N$, and the value $v_n \cdot b$ was added to the bin itself, where $b$
is the content of the bin. An example with $\sigma_N = 0.05$ of the new
histogram is shown in @fig:noisy.
The following three values of $\sigma_N$ were tested:
$$
\sigma_N = 0.005 \et
\sigma_N = 0.01 \et
\sigma_N = 0.05
$$
The same procedure followed in @sec:noiseless was then repeated for noisy
signals. Hence, in @fig:rounds-noise the EMD as a function of the RL rounds is
shown, this time varying $\sigma_N$ and keeping $\sigma = 0.8 \, \Delta \theta$
constant.
In @fig:rnoise-0.005, the flattening is achieved around $r = 20$ and in
@fig:rnoise-0.01 it is sufficient $\sim r = 15$. When the noise becomes too
high, on the other hand, the more $r$ increases, the more the algorithm becomes
inefficient, requiring a very small number of iterations.
Hence, the most fitting values were chosen as:
\begin{align*}
\sigma_N = 0.005 &\thus r^{\text{best}} = 20 \\
\sigma_N = 0.01 &\thus r^{\text{best}} = 15 \\
\sigma_N = 0.05 &\thus r^{\text{best}} = 1
\end{align*}
As regards the order of magnitude, as expected, the more $\sigma_n$ increases,
the more the EMD rises, ranging from $\sim$ \SI{2e-4}{} in @fig:rnoise-0.005
to $\sim$ \SI{1.5e-3}{} in @fig:rnoise-0.05.
Since the FFT is no more able to return the original signal as close as before,
it is no more assumed to be a reference point. In fact, as shown in @fig:noisy,
the FFT algorithm, when dealing with noisy signals whose noise shape is
unknown, is as efficient as the RL, since they share the same order of magnitude
regarding the EMD.
An increasing noise entails the shift of the convolved histogram to slightly
greater values, still remaining in the same order of magnitude of the noiseless
signals. However, the deconvolution still works, being the EMD of the convolved
histogram an order of magnitude greater than the worst obtained with both
methods.
In conclusion, when dealing with smeared signals with a known point spread
function, the FFT proved to be the best deconvolution method, restoring
perfectly the original signal to the extent permitted by the floating point
precision. When the point spread function is particularly small, this precision
problem is partially solved by the RL deconvolution process, which employs a
more stable math.
However, in the real world signals are inevitably blurred by unknown-shpaed
noise. When the kernel is not known a-priori, either of them turns out to be as
good as the FFT in the aforementioned situation: only a poor approximation of
the original signal can be derived.
{#fig:rnoise-0.005}
{#fig:rnoise-0.01}
{#fig:rnoise-0.05}
EMD as a function of RL rounds for different noise $\sigma_N$ values with the
kernel $\sigma = 0.8 \Delta \theta$. The average is shown in red and the
standard deviation in grey. Noisy results.
{#fig:enoise-0.005}
{#fig:enoise-0.01}
{#fig:enoise-0.05}
EMD distributions for different noise $\sigma_N$ values. The plots on the left
show the results for the FFT deconvolution, the central column the results for
the RL deconvolution and the third one shows the EMD for the convolved signal.
Noisy results.