From 16c25112892f0175f8b69cec45df7c663e890759 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Gi=C3=B9=20Marcer?= <giuliamarcer@yahoo.it>
Date: Wed, 20 May 2020 16:28:19 +0200
Subject: [PATCH] ex-6: Written about the RL algorithm and generated some plots

---
 notes/docs/bibliography.bib |  18 +-
 notes/images/nome.pdf       | Bin 0 -> 12040 bytes
 notes/images/original.pdf   | Bin 0 -> 14756 bytes
 notes/images/prova.pdf      | Bin 0 -> 12040 bytes
 notes/images/smoothed.pdf   | Bin 0 -> 15612 bytes
 notes/sections/6.md         | 317 +++++++++++++++++-------------------
 notes/todo                  |   3 -
 7 files changed, 169 insertions(+), 169 deletions(-)
 create mode 100644 notes/images/nome.pdf
 create mode 100644 notes/images/original.pdf
 create mode 100644 notes/images/prova.pdf
 create mode 100644 notes/images/smoothed.pdf
 delete mode 100644 notes/todo

diff --git a/notes/docs/bibliography.bib b/notes/docs/bibliography.bib
index 80f9df4..50df816 100644
--- a/notes/docs/bibliography.bib
+++ b/notes/docs/bibliography.bib
@@ -118,7 +118,6 @@
   publisher={Elsevier}
 }
 
-
 @article{ridder17,
   title={Variance reduction},
   author={Ridder, AAN and Botev, ZI},
@@ -138,3 +137,20 @@
   year={1978},
   publisher={Elsevier}
 }
+
+@book{hecht02,
+  title={Optics},
+  year={2002},
+  publisher={Pearson},
+  author={Eugene Hecht}
+}
+
+@article{lucy74,
+  title={An iterative technique for the rectification of observed
+         distributions},
+  author={Lucy, Leon B},
+  journal={The astronomical journal},
+  volume={79},
+  pages={745},
+  year={1974}
+}
diff --git a/notes/images/nome.pdf b/notes/images/nome.pdf
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diff --git a/notes/sections/6.md b/notes/sections/6.md
index d707c67..b0a174f 100644
--- a/notes/sections/6.md
+++ b/notes/sections/6.md
@@ -2,10 +2,10 @@
 
 ## Generating points according to Fraunhöfer diffraction
 
-The diffraction of a plane wave through a round slit must be simulated by
+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
-$I(\theta)$ on a screen at a great distance $L$ from the slit itself:
-
+$I(\theta)$ [@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}
@@ -16,7 +16,7 @@ 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 a Bessel function of first kind;
+- $J_1$ is the Bessel function of first kind;
 - $k$ is the wavenumber, default set $k = \SI{1e-4}{m^{-1}}$;
 - $L$ default set $L = \SI{1}{m}$.
 
@@ -51,9 +51,8 @@ where:
 \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 employed. This time,
-though, $\theta$ must be evenly distributed on half sphere:
-
+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 =
@@ -73,7 +72,6 @@ though, $\theta$ must be evenly distributed on half sphere:
 
 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
@@ -85,44 +83,50 @@ omitted:
   &\thus \theta = \text{acos} (1 -x)
 \end{align*}
 
-The sample was binned and stored in a histogram with a customizable number $n$
-of bins default set $n = 150$. In @fig:original an example is shown.
+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 $\theta
+= \pi/2$ bacause of the system symmetry. In @fig:original an example is shown.
 
-![Example of an intensity histogram.](images/fraun-original.pdf){#fig:original}
+![Example of intensity histogram.](images/original.pdf){#fig:original}
 
 
 ## Gaussian convolution {#sec:convolution}
 
-The sample must then be 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 a odd number $m$ of bins and the
-same bin width of the previous one, but a smaller number of them ($m < n$), was
-filled with $m$ points according to a Gaussian distribution with mean $\mu$,
-corresponding to the central bin, and variance $\sigma$.  
+The sample has then to be 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 created according to a Gaussian distribution with mean $\mu$,
+and variance $\sigma$. The reason why the kernel was set this way will be
+descussed lately.  
 Then, the original histogram was convolved with the kernel in order to obtain
-the smeared signal. Some results in terms of various $\sigma$ are shown in
-[@fig:results1; @fig:results2; @fig:results3].  
+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.
+As expected, the smeared signal looks smoother with respect to the original
+one.
+
+![Convolved signal.](images/smoothed.pdf){#fig:convolved}
+
 The convolution was implemented as follow. Consider the definition of
 convolution of two functions $f(x)$ and $g(x)$:
-
 $$
-  f \otimes g (x) = \int \limits_{- \infty}^{+ \infty} dy f(y) g(x - y)
+  f * g (x) = \int \limits_{- \infty}^{+ \infty} dy f(y) g(x - y)
 $$
 
 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 reverse histogram of $k$
-for each relative position of the two histograms. Namely, if $c_i$ is the
-$i^{\text{th}}$ bin of the convoluted histogram:
-
+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:
 $$
-  c_i = \sum_j k_j s_{i - j}
+  c_i = \sum_j k_j s_{i - j} = \sum_{j'} k_{m - j'} s_{i - m + j'}
+  \with j' = m - j
 $$
 
 where $j$ runs over the bins of the kernel.  
-For a better understanding, see @fig:dot_conv. As can be seen, the third
-histogram was obtained with $n + m - 1$ bins, a number greater than the initial
-one.
+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}{%
@@ -195,27 +199,32 @@ one.
 ## Unfolding with FFT
 
 Two different unfolding routines were implemented, one of which exploiting the
-Fast Fourier Transform. This method is based on the property of the Fourier
-transform according to which, given two functions $f(x)$ and $g(x)$:
+Fast Fourier Transform (FFT).
 
+This method is based on the convolution theorem, according to which, given two
+functions $f(x)$ and $g(x)$:
 $$
-  \hat{F}[f \otimes g] = \hat{F}[f] \cdot \hat{F}[g]
+  \hat{F}[f * g] = \hat{F}[f] \cdot \hat{F}[g]
 $$
 
 where $\hat{F}[\quad]$ stands for the Fourier transform of its argument.  
-Thus, the implementation of this tecnique lies in the computation of the Fourier
-trasform of the smeared signal and the kernel, the ratio between their
-transforms and the anti-transformation of the result:
-
-$$
-  \hat{F}[s \otimes k] = \hat{F}[s] \cdot \hat{F}[k] \thus
-  \hat{F} [s] = \frac{\hat{F}[s \otimes k]}{\hat{F}[k]}
-$$
-
 Being the histogram a discrete set of data, the Discrete Fourier Transform (DFT)
-was emploied. In particular, the FFT are efficient algorithms for calculating
-the DFT. Given a set of $n$ values {$z_i$}, each one is transformed into:
+was appied. When dealing with arrays of discrete values, the theorems still
+holds if the two arrays have the same lenght and a cyclical convolution is
+applied. For this reason the kernel was 0-padded in order to make it the same
+lenght 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, every histogram was transformed into a
+vector. The implementation lies in the computation of the Fourier trasform 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 {$z_i$}, each one is transformed into:
 $$
   x_j = \sum_{k=0}^{n-1} z_k \exp \left( - \frac{2 \pi i j k}{n} \right)
 $$
@@ -227,7 +236,6 @@ 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:
-
 $$
   z_j = \frac{1}{n}
         \sum_{k=0}^{n-1} x_k \exp \left( \frac{2 \pi i j k}{n} \right)
@@ -235,20 +243,12 @@ $$
 
 In GSL, `gsl_fft_complex_forward()` and `gsl_fft_complex_inverse()` are 
 functions which allow to compute the foreward and inverse transform,
-respectively.
-
-In order to accomplish this procedure, every histogram was transformed into a
-vector. The kernel vector was 0-padded and centred in the middle to make its
-length the same as that of the signal, making it feasable to implement the
-division between the entries of the vectors one by one.
-
+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 which must be transformed is made
-of real numbers, but the Fourier transform is not real: it is a complex sequence
-wich satisfies:
-
+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,
+but the Fourier transform is not real: it is a complex sequence wich satisfies:
 $$
   z_k = z^*_{n-k}
 $$
@@ -258,14 +258,10 @@ where $z^*$ is the conjugate of $z$. A sequence with this symmetry is called
 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 implies that only half of the complex numbers in the output need to be
-stored. This works for all lengths: when the length is even, 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).  
-If the bin width is $\Delta \theta$, then the DFT domain ranges from $-1 / (2
-\Delta \theta)$ to $+1 / (2 \Delta \theta$). 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).  
+sequence requires only half of the complex numbers in the output to be stored.
+This works for all lengths: when the length is even, 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}{%
@@ -273,7 +269,6 @@ the negative backwards from the end of the array (see @fig:reorder).
 \begin{tikzpicture}
   \definecolor{cyclamen}{RGB}{146, 24, 43}
   % standard histogram
-  \begin{scope}[shift={(7,0)}]
   \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);
@@ -284,8 +279,8 @@ the negative backwards from the end of the array (see @fig:reorder).
   \draw [thick, cyclamen] (4.5,0) -- (4.5,0.2);
   \draw [thick, ->] (0,0) -- (5,0);
   \draw [thick, ->] (2.5,0) -- (2.5,2);
-  \end{scope}
   % 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);
@@ -294,94 +289,126 @@ the negative backwards from the end of the array (see @fig:reorder).
   \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{On the left, an example of the DFT as it is given by the gsl function
-         and the same dataset, on the right, with the rearranged "intuitive"
-         order of the sequence.}\label{fig:reorder}
+\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}
 
-When $\hat{F}[s \otimes k]$ and $\hat{F}[k]$ are computed, 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 application.  
-GSL provides the function `gsl_fft_halfcomplex_unpack()` which passes the
-vectors from half-complex format to standard complex format. The inverse
-procedure, required to compute the inverse transformation of $\hat{F}[s]$, which
-is not provided by GSL, was implemented in the code.  
-The fact that the gaussian kernel is centerd in the middle of the vector and
-not in the $\text{zero}^{th}$ bin causes the final result to be shifted of half
-the leght of the vector the same as it was produced by a DFT. This makes it
-necessary to rearrange the two halfs of the final result.
+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-lenght halves.
 
-At the end, the external bins which exceed with respect to the original signal
-are cut away in order to restore the original number of bins $n$. Results are
-shown in [@fig:results1; @fig:results2; @fig:results3].
+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 tipically used
-for recovering an image that has been blurred by a known point spread function.
-
-It is based on the fact that an ideal point source does not appear as a point
-but is spread out into the so-called point spread function, thus the observed
-image can be represented in terms of a transition matrix
-$P$ operating on an underlying image:
+for recovering an image that has been blurred by a known 'point spread
+function'.
 
+Consider the problem of estimating the frequeny distribution $f(\xi)$ of a
+variable $\xi$ when the available measure is a sample {$x_i$} of points
+dronwn not by $f(x)$ but by an other function $\phi(x)$ such that:
 $$
- d_i = \sum_{j} u_j \, P_{i, j} 
+  \phi(x) = \int d\xi \, f(\xi) P(x | \xi)
 $$
 
-where $u_j$ is the intensity of the underlying image at pixel $j$ and $d_i$ is
-the detected intensity at pixel $i$. Hence, the matrix describes the portion of
-signal from the source pixel $j$ that is detected in pixel $i$.  
-In one dimension, the transfer function can be expressed in terms of the
-distance between the source pixel $j$ and the observed $i$:
+where $P(x | \xi) \, d\xi$ is the probability (presumed known) that $x$ falls
+in the interval $(x, x + dx)$ when $\xi = \xi$. 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 $P (x | \xi)$,
+called point spread function.
+
+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:
 
 $$
-  P_{i, j} = \widetilde{P}(i-j) = P_{i - j}
+  \phi(x) dx \cdot Q(\xi | x) d\xi = f(\xi) d\xi \cdot P(x | \xi) dx
+  \thus Q(\xi | x) = \frac{f(\xi) \cdot P(x | \xi)}{\phi(x)}
 $$
 
-In order to estimate $u_j$ given {$d_i$} and $\widetilde{P}$, the following
-iterative procedure can be applied for the estimate $\hat{u}^t_j$ of $u_j$,
-where $t$ stands for the iteration number. The $t^{\text{th}}$ step is updated
-as follows:
-
 $$
-  \hat{u}^{t+1}_j = \hat{u}^t_j \sum_i \frac{d_i}{c_i} \, P_{i - j}
-  \with c_i = \sum_j \hat{u}^t_j \, P_{i - j}
+  \thus Q(\xi | x) = \frac{f(\xi) \cdot P(x | \xi)}
+                          {\int d\xi \, f(\xi) P(x | \xi)}
+$$ {#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 a reiterative
+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 $f (\xi)$ is 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:
+$$
+  f^{t + 1}(\xi) = \int dx \, \phi(x) Q^t(\xi | x)
+  \with
+  Q^t(\xi | x) = \frac{f^t(\xi) \cdot P(x | \xi)}
+                      {\int d\xi \, f^t(\xi) P(x | \xi)}
 $$
 
-where $c_i$ is thereby an estimation of the blurred signal obtained with the
-previous estimation of the clean signal.  
-It has been shown empirically that if this iteration converges, it converges to
-the maximum likelihood solution for $u_j$. Writing it in terms of convolution,
-it becomes:
-
+from which:
 $$
-   \hat{u}^{t+1} = \hat {u}^{t} \cdot \left( \frac{d}{{\hat{u}^{t}} \otimes P}
-   \otimes P^{\star} \right)
+  f^{t + 1}(\xi) = f^t(\xi)
+  \int dx \, \frac{\phi(x)}{\int d\xi \, f^t(\xi) P(x | \xi)}
+  P(x | \xi)
+$$ {#eq:solution}
+
+If the spread function $P(x | \xi)$ follows a normal distribution with variance
+$\sigma$, namely:
+$$
+  P(x | \xi) = \frac{1}{\sqrt{2 \pi} \sigma}
+  \exp \left( - \frac{(x - \xi)^2}{2 \sigma^2} \right)
 $$
 
-where the division and multiplication are element wise, and
-$P^{\star}$ is the flipped point spread function.
+then, @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 stands for the point spread function
+and, 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:
 
   - create a flipped copy of the kernel;
-  - choose a zero-order estimate for {$c_i$};
-  - compute the convolutions with the method described in @sec:convolution, the
-    product and the division at each step;
+  - choose a zero-order estimate for {$f(\xi)$};
+  - compute the convolutions, the product and the division at each step;
   - proceed until a given number of reiterations is achieved.
 
-In this case, the zero-order was set $c_i = 0.5 \, \forall i$ and it was
-empirically shown that the better result is given with a number of three steps,
-otherwise it starts returnig fanciful histograms. Results are shown in
-[@fig:results1; @fig:results2; @fig:results3].
+In this case, the zero-order was set $f(\xi) = 0.5 \, \forall \, \xi$.
 
 
-## Results comparison
+## Results comparison {#sec:conv_Results}
 
 In [@fig:results1; @fig:results2; @fig:results3] the results obtained for three
 different $\sigma$s are shown. The tested values are $\Delta \theta$, $0.5 \,
@@ -399,46 +426,14 @@ In the real world, it is unpratical, since signals are inevitably blurred by
 noise.  
 The same can't be said about the RL deconvolution, which, on the other hand,
 looks heavily influenced by the variance magnitude: the greater $\sigma$, the
-worse the deconvoluted result. In fact, given the same number of steps, the
+worse the deconvolved result. In fact, given the same number of steps, the
 deconvolved signal is always the same 'distance' far form the convolved one:
 if it very smooth, the deconvolved signal is very smooth too and if the
 convolved is less smooth, it is less smooth too.
 
 The original signal is shown below for convenience.
 
-![Example of an intensity histogram.](images/fraun-original.pdf)
 
-<div id="fig:results1">
-![Convolved signal.](images/fraun-conv-0.05.pdf){width=12cm}
-
-![Deconvolved signal with FFT.](images/fraun-fft-0.05.pdf){width=12cm}
-
-![Deconvolved signal with RL.](images/fraun-rl-0.05.pdf){width=12cm}
-
-Results for $\sigma = 0.05 \Delta \theta$, where $\Delta \theta$ is the bin
-width.
-</div>
-
-<div id="fig:results2">
-![Convolved signal.](images/fraun-conv-0.5.pdf){width=12cm}
-
-![Deconvolved signal with FFT.](images/fraun-fft-0.5.pdf){width=12cm}
-
-![Deconvolved signal with RL.](images/fraun-rl-0.5.pdf){width=12cm}
-
-Results for $\sigma = 0.5 \Delta \theta$, where $\Delta \theta$ is the bin
-width.
-</div>
-
-<div id="fig:results3">
-![Convolved signal.](images/fraun-conv-1.pdf){width=12cm}
-
-![Deconvolved signal with FFT.](images/fraun-fft-1.pdf){width=12cm}
-
-![Deconvolved signal with RL.](images/fraun-rl-1.pdf){width=12cm}
-
-Results for $\sigma = \Delta \theta$, where $\Delta \theta$ is the bin width.
-</div>
 
 It was also implemented the possibility to add a Poisson noise to the
 convolved histogram to check weather the deconvolution is affected or not by
@@ -451,14 +446,6 @@ everywhere on the curve and it is particularly evident on the edges, where the
 expected data are very small. On the other hand, the Richardson-Lucy routine is
 less affected by this further complication.
 
-<div id="fig:poisson">
-![Deconvolved signal with FFT.](images/fraun-noise-fft.pdf){width=12cm}
-
-![Deconvolved signal withh RL.](images/fraun-noise-rl.pdf){width=12cm}
-
-Results for $\sigma = \Delta \theta$, with Poisson noise.
-</div>
-
 In order to quantify the similarity of a deconvolution outcome with the original
 signal, a null hypotesis test was made up.  
 Likewise in @sec:Landau, the original sample was treated as a population from
@@ -519,7 +506,7 @@ $$
                   {\sum_{i = 1}^m \sum_{j=1}^n f_{ij}}
 $$
 
-In this case, where the EMD must be applied to two same-lenght histograms, the
+In this case, where the EMD has to be applied to two same-lenght histograms, the
 procedure simplifies a lot. By representing both histograms with two vectors $u$
 and $v$, the equation above boils down to [@ramdas17]:
 
diff --git a/notes/todo b/notes/todo
deleted file mode 100644
index 95a722e..0000000
--- a/notes/todo
+++ /dev/null
@@ -1,3 +0,0 @@
-- cambiare simbolo convoluzione
-- aggiungere citazioni e referenze
-- leggere l'articolo di Lucy