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Fix typos <noupdate>

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Tian Cheng (Riccardo) Xia
2024-06-16 19:46:58 +02:00
parent 27b1947218
commit 329a2a736c
5 changed files with 26 additions and 23 deletions
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6
src/year1/image-processing-and-computer-vision/module1/sections/_edge_detection.tex
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@ -72,8 +72,8 @@ In a 2D signal (e.g. an image), the gradient allows to determine the magnitude a
\end{remark}
\end{description}
\item[Discete magnitude approximation] \marginnote{Discete magnitude approximation}
The gradient magnitude can be approximated using the approximated partial derivatives:
\item[Discrete magnitude approximation] \marginnote{Discrete magnitude approximation}
The gradient magnitude can be approximated in different ways by using the approximated partial derivatives:
\[
\Vert \nabla I \Vert = \sqrt{(\partial_x I)^2 + (\partial_y I)^2} \hspace{1.5em}
\Vert \nabla I \Vert_+ = \vert \partial_x I \vert + \vert \partial_y I \vert \hspace{1.5em}
@ -124,7 +124,7 @@ In a 2D signal (e.g. an image), the gradient allows to determine the magnitude a
In practice, the signal of an image is not always smooth due to noise.
Derivatives amplify noise and are therefore unable to recognize edges.
Smoothing the signal before computing the derivative allows to reduce the noise but also blurs the edges making it more difficult to localize them.
Smoothing the signal before computing the derivative allows to reduce the noise but also blurs the edges, making it more difficult to localize them.
A solution is to smooth and differentiate in a single operation by approximating the gradient as a difference of averages.
\end{remark}

2
src/year1/image-processing-and-computer-vision/module1/sections/_image_acquisition.tex
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@ -253,7 +253,7 @@ Geometric model of a pinhole camera.\\
\subsection{Ratios and parallelism}
Given a 3D line of length $L$ lying in a plane parallel to the image plane at distance $z$,
then its length $l$ in the image plane is:
its length $l$ in the image plane is:
\[ l = L\frac{f}{z} \]
In all the other cases (i.e. when the line is not parallel to the image plane),

9
src/year1/image-processing-and-computer-vision/module1/sections/_instance_obj_detection.tex
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@ -64,7 +64,7 @@ Possible similarity/dissimilarity functions are:
before computing \texttt{NCC}:
\[ \mu(\tilde{I}_{i,j}) = \frac{1}{MN} \sum_{m=0}^{M-1} \sum_{n=0}^{N-1} I(i+m, j+n) \hspace{3em} \mu(T) = \frac{1}{MN} \sum_{m=0}^{M-1} \sum_{n=0}^{N-1} T(m, n) \]
\[
\texttt{NCC}(i, j) =
\texttt{ZNCC}(i, j) =
\frac{ \sum\limits_{m=0}\limits^{M-1} \sum\limits_{n=0}\limits^{N-1} \Big( \big(I(i+m, j+n) - \mu(\tilde{I}_{i,j})\big) \cdot \big(T(m, n) - \mu(T)\big) \Big) }
{ \sqrt{\sum\limits_{m=0}\limits^{M-1} \sum\limits_{n=0}\limits^{N-1} \big(I(i+m, j+n) - \mu(\tilde{I}_{i,j})\big)^2} \cdot \sqrt{\sum\limits_{m=0}\limits^{M-1} \sum\limits_{n=0}\limits^{N-1} \big(T(m, n) - \mu(T)\big)^2} }
\]
@ -97,7 +97,7 @@ Edge-based template matching that works as follows:
\nabla \tilde{I}_{i,j}(\tilde{P}_k) = \begin{pmatrix} \partial_x \tilde{I}_{i,j}(\tilde{P}_k) \\ \partial_y \tilde{I}_{i,j}(\tilde{P}_k) \end{pmatrix} \hspace{2em}
\tilde{\vec{u}}_k(\tilde{P}_k) = \frac{\nabla \tilde{I}_{i,j}(\tilde{P}_k)}{\Vert \nabla \tilde{I}_{i,j}(\tilde{P}_k) \Vert}
\]
\item Compute the similarity as the sum of the cosine similarities of each pair of gradients:
\item Compute the similarity as the mean of the cosine similarities of each pair of gradients:
\[ S(i, j) = \frac{1}{n} \sum_{k=1}^{n} \vec{u}_k(P_k) \cdot \tilde{\vec{u}}_k(\tilde{P}_k) = \frac{1}{n} \sum_{k=1}^{n} \cos \theta_k \in [-1, 1] \]
$S(i, j) = 1$ when the gradients perfectly match. A minimum threshold $S_\text{min}$ is used to determine if there is a match.
\end{enumerate}
@ -156,13 +156,14 @@ by means of a projection from the image space to a parameter space.
For instance, consider two points $p_1$, $p_2$ in the image space and
their projection in the parameter space.
If the two lines intersect at the point $(\tilde{m}, \tilde{c})$,
then the line parametrized on $\tilde{m}$ and $\tilde{c}$ passes through $p_1$ and $p_2$ in the image space.
then the line parametrized on $\tilde{m}$ and $\tilde{c}$ passes through both $p_1$ and $p_2$ in the image space.
\begin{figure}[H]
\centering
\includegraphics[width=0.4\linewidth]{./img/hough_line_parameter_space.png}
\end{figure}
\indenttbox
\begin{remark}
By projecting $n$ points of the image space, there are at most $\frac{n(n-1)}{2}$ intersections in the parameter space.
\end{remark}
@ -267,7 +268,7 @@ Hough transform extended to detect an arbitrary shape.
\item Compute its gradient direction $\varphi(\vec{x})$ discretized to match the step $\Delta \varphi$ of the R-table.
\item For each $\vec{r}_i$ in the corresponding row of the R-table:
\begin{enumerate}
\item Compute an estimate of the barycenter as $\vec{y} = \vec{x} - \vec{r}_i$.
\item Compute an estimate of the barycenter as $\vec{y} = \vec{x} + \vec{r}_i$.
\item Cast a vote in the accumulator array $A[\vec{y}] \texttt{+=} 1$
\end{enumerate}
\end{enumerate}

12
src/year1/image-processing-and-computer-vision/module1/sections/_local_features.tex
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@ -6,7 +6,7 @@
\begin{example}[Homography]
Align two images of the same scene to create a larger image.
Homography requires at least 4 correspondences.
An homography requires at least 4 correspondences.
To find them, it does the following:
\begin{itemize}
\item Independently find salient points in the two images.
@ -264,8 +264,8 @@ but this is not always able to capture the same features due to the details diff
\begin{enumerate}
\item Create a Gaussian scale-space by applying the scale-normalized Laplacian of Gaussian with different values of $\sigma$.
\item For each pixel, find the characteristic scale and its corresponding Laplacian response across the scale-space (automatic scale selection).
\item Filter out the pixels whose response is lower than a threshold and apply NMS.
\item The remaining pixels are the centers of the blobs.
\item Filter out the pixels whose response is lower than a threshold and find the peaks.
\item The found pixels are the centers of the blobs.
It can be shown that the radius is given by $r = \sigma\sqrt{2}$.
\end{enumerate}
\end{description}
@ -332,7 +332,7 @@ When detecting a peak, there are two cases:
\]
\begin{theorem}
It can be proven that the DoG kernel is a scaled version of the LoG kernel:
It can be proved that the DoG kernel is a scaled version of the LoG kernel:
\[ G(x, y, k\sigma) - G(x, y, \sigma) \approx (k-1)\sigma^2 \nabla^{(2)}G(x, y, \sigma) \]
\begin{remark}
@ -341,7 +341,7 @@ When detecting a peak, there are two cases:
\end{theorem}
\item[Extrema detection] \marginnote{DoG extrema}
Given three DoG images with scales $\sigma_i$, $\sigma_{i-1}$ and $\sigma_{i+1}$,
Given three DoG images with scales $\sigma_{i-1}$, $\sigma_i$ and $\sigma_{i+1}$,
a pixel $(x, y, \sigma_i)$ is an extrema (i.e. keypoint) iff:
\begin{itemize}
\item It is an extrema in a $3 \times 3$ patch centered on it (8 pixels as $(x, y, \sigma_i)$ is excluded).
@ -407,7 +407,7 @@ After finding the keypoints, a descriptor of a keypoint is computed from the pix
\[
\begin{split}
\vert \nabla L(x, y) \vert &= \sqrt{ \big( L(x+1, y) - L(x-1, y) \big)^2 + \big( L(x, y+1) - L(x, y-1) \big)^2 } \\
\theta_L(x, y) &= \tan^{-1}\left( \frac{L(x, y+1) - L(x, y-1)}{L(x+1, y) - L(x-1, y)} \right)
\theta_L(x, y) &= \arctan\left( \frac{L(x, y+1) - L(x, y-1)}{L(x+1, y) - L(x-1, y)} \right)
\end{split}
\]

20
src/year1/image-processing-and-computer-vision/module1/sections/_spatial_filtering.tex
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@ -5,7 +5,7 @@
The noise added to a pixel $p$ is defined by $n_k(p)$,
where $k$ indicates the time step (i.e. noise changes depending on the moment the image is taken).
It is assumed that $n_k(p)$ is i.i.d and $n_k(p) \sim \mathcal{N}(0, \sigma)$.
It is assumed that $n_k(p)$ is i.i.d. and $n_k(p) \sim \mathcal{N}(0, \sigma)$.
The information of a pixel $p$ is therefore defined as:
\[ I_k(p) = \tilde{I}(p) + n_k(p) \]
@ -33,7 +33,7 @@ where $\tilde{I}(p)$ is the real information.
Let $K_p$ be the pixels in a window around $p$ (included):
\[
\begin{split}
O(p) &= \frac{1}{\vert K_p \vert} \sum_{q \in K_p} I(p) \\
O(p) &= \frac{1}{\vert K_p \vert} \sum_{q \in K_p} I(q) \\
&= \frac{1}{\vert K_p \vert} \sum_{q \in K_p} \Big( \tilde{I}(q) + n(q) \Big) \\
&= \frac{1}{\vert K_p \vert} \sum_{q \in K_p} \tilde{I}(q) + \frac{1}{\vert K_p \vert} \sum_{q \in K_p} n(q) \\
&\approx \frac{1}{\vert K_p \vert} \sum_{q \in K_p} \tilde{I}(q)
@ -80,7 +80,7 @@ where $\tilde{I}(p)$ is the real information.
\end{descriptionlist}
\item[Dirac delta] \marginnote{Dirac delta}
The Dirac delta "function" $\delta$ is defined as follows \cite{wiki:dirac,book:sonka}:
The Dirac delta ``function" $\delta$ is defined as follows \cite{wiki:dirac,book:sonka}:
\[ \forall x \neq 0: \delta(x) = 0 \text{, constrained to } \int_{-\infty}^{+\infty} \delta(x) \,\text{d}x = 1 \]
Extended to the 2-dimensional case, the definition is the following:
@ -93,7 +93,7 @@ where $\tilde{I}(p)$ is the real information.
\begin{remark}
Exploiting the sifting property, the signal of an image can be expressed through an integral of Dirac deltas
(i.e. a linear combination) \cite{slides:filters,book:sonka}:
(i.e. a linear combination) \cite{slides:filters, book:sonka}:
\[ i(x, y) = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} i(\alpha, \beta) \delta(x-\alpha, y-\beta) \,\text{d}\alpha\,\text{d}\beta \]
\end{remark}
\end{description}
@ -173,7 +173,8 @@ where $\tilde{I}(p)$ is the real information.
& \text{linearity of $T\{ \cdot \}$} \\
%
&= \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} i(\alpha, \beta) h(x-\alpha, y-\beta) \,\text{d}\alpha\,\text{d}\beta
& \text{\small translation-equivariance of $T\{ \cdot \}$} \\
& \text{impulse response} \\
% & \text{\small translation-equivariance of $T\{ \cdot \}$} \\
%
&= i(x, y) * h(x, y)
& \text{definition of convolution} \\
@ -214,7 +215,8 @@ where $\tilde{I}(p)$ is the real information.
\begin{align*}
h(x, y) * i(x, y) &= \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} i(\alpha, \beta)h(x-\alpha, y-\beta) \,\text{d}\alpha\,\text{d}\beta \\
&= \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} i(\alpha, \beta)h(\alpha-x, \beta-y) \,\text{d}\alpha\,\text{d}\beta
& \parbox[b]{0.25\textwidth}{\raggedleft signs in $h$ swappable for Dirac delta} \\
% & \parbox[b]{0.25\textwidth}{\raggedleft signs in $h$ swappable for Dirac delta}
\\
&= h(x, y) \circ i(x, y)
\end{align*}
\end{remark}
@ -336,8 +338,8 @@ where $\tilde{I}(p)$ is the real information.
\begin{description}
\item[Sampling]
In practice, the kernel is created by sampling from the wanted Gaussian distribution.
One can notice that a higher $\sigma$ results in a more spread distribution and therefore a larger kernel is more suited,
on the other hand, a smaller $\sigma$ can be represented using a smaller kernel as it is more concentrated around the origin.
One can notice that a higher $\sigma$ results in a more spread distribution and therefore a larger kernel is more suited.
On the other hand, a smaller $\sigma$ can be represented using a smaller kernel as it is more concentrated around the origin.
As a rule-of-thumb, given $\sigma$, an ideal kernel is of size $(2\lceil 3\sigma \rceil + 1) \times (2\lceil 3\sigma \rceil + 1)$.
@ -448,7 +450,7 @@ where $\tilde{I}(p)$ is the real information.
\begin{split}
O(p) &= \sum_{q \in S_p} w(p, q) \cdot \texttt{intensity}(q) \\
\text{where }& w(p, q) = \frac{1}{Z(p)} e^{-\frac{\Vert \mathcal{N}_p - \mathcal{N}_q \Vert_2^2}{h^2}} \\
& Z(p) = \sum_{q \in I} e^{\frac{\Vert \mathcal{N}_p - \mathcal{N}_q \Vert_2^2}{h^2}}
& Z(p) = \sum_{v \in I} e^{\frac{\Vert \mathcal{N}_p - \mathcal{N}_v \Vert_2^2}{h^2}}
\end{split}
\]
\end{minipage}