stuff. We must also clarify what a perimeter is, and how it relates to a
contour later.
+\section{Anisotropic coarea formula}
+
+We define the thresholded image at level $s$ as
+\begin{equation}
+ u^s(x) = \begin{cases}
+ 1 & \text{if } u(x) > s, \\
+ 0 & \text{otherwise}.
+ \end{cases}
+\end{equation}
+This allows us to write a non-negative image $u \geq 0$ as an integral
+over all the layers
+\begin{equation}
+ u = \int_0^\infty u^s \, ds.
+\end{equation}
+
+For any image $u \in C^1_c(\Omega)$ define the anisotropic total
+variation functional
+\begin{equation}
+ J_A(u) = \int_\Omega \sqrt{\nabla u(x)^T A(x) \nabla u(x)} \, dx.
+\end{equation}
+\fixme{Why compact support?}
+Before introducing the anisotropic coarea formula, we extend the
+definition of this functional to the space $\BV(\Omega)$. Being
+symmetric positive definite, the matrix $A(x)$ can be factored into two
+symmetric matrices as $A(x) = Q(x) Q(x)$. We can then write
+\begin{align}
+ J_A(u) &= \int_\Omega \abs{Q \nabla u} \, dx \\
+ &= \sup_{\abs{\xi(x)}
+ \leq 1} \int_\Omega (Q \nabla u)^T \xi \, dx \\
+ &= \sup_{\abs{\xi(x)}\leq 1} \int_\Omega \nabla u \cdot Q\xi \, dx
+ \\
+ &= \sup_{\abs{\xi(x)}\leq 1} \int_\Omega -u \diver (Q\xi) \, dx \\
+ &= \sup_{\eta^T A^{-1} \eta \leq 1} \int_\Omega -u \diver \eta \,
+ dx.
+\end{align}
+If we let $\norm{\xi}_A = \sqrt{\xi^T A \xi}$ and $\norm{\eta}_A^* =
+\sqrt{\eta^T A^{-1} \eta}$, then our new extended definition reads
+\begin{equation}
+ J_A(u) = \sup_{\substack{\xi \in C_c^1(\Omega; \mathbb{R}^N) \\
+ \norm{\xi(x)}_A^* \leq 1}} \int_\Omega -u \diver \xi \, dx.
+\end{equation}
+
+\begin{theorem}[Anisotropic coarea formula]
+ Given an image $u \in \BV(\Omega)$, the anisotropic total variation
+ can be calculated written as an integral over all the levels
+ \begin{equation}
+ J_A(u) = \int_{-\infty}^\infty J_A(u^s) \, ds.
+ \end{equation}
+\end{theorem}
+\begin{proof}
+ Assume that $u \in C^1(\Omega) \cap \BV(\Omega)$. \fixme{Something
+ about the approximation arguments.}
+ \paragraph{First we prove that $J_A(u) \leq \int_{-\infty}^\infty
+ J_A(u^s) \, ds$.}
+ Assume that $u \geq 0$ such that the integral in \fixme{ref} holds,
+ then inserting \fixme{ref} into \fixme{ref} gives
+ \begin{align}
+ J_A(u) &= \sup_{\norm{\xi}_A^* \leq 1} \int_\Omega -\left(
+ \int_0^\infty u^s ds \right) \diver \xi \, dx
+ = \sup_{\norm{\xi}_A^* \leq 1} \int_\Omega \int_0^\infty -u^s
+ \diver \xi \, ds \, dx \\
+ &\leq \int_0^\infty \left( \sup_{\norm{\xi}_A^*} \int_\Omega
+ -u^s \diver \xi \, dx \right) \, ds
+ = \int_0^\infty J_A(u^s) \, ds.
+ \end{align}
+ For $u \leq 0$ we use that $J_A(-v) = J_A(v)$ and that $J_A(c + v) =
+ J_A(v)$ for any constant $c$. This allows us to show that
+ \fixme{also something about $u^{-r}$}
+ \begin{align}
+ J_A(u) &= J_A(-u) \leq \int_0^\infty J_A \big( (-u)^r \big) \,
+ dr \\
+ &= \int_0^\infty J_A(1 - u^{-r}) \, dr = \int_0^\infty
+ J_A(u^{-r}) \, dr = \int_{-\infty}^0 J_A(u^s) \, ds.
+ \end{align}
+ Next, we write a general $u$ as a difference between two positive
+ function $u = u_+ - u_-$ \fixme{They will not be differentiable
+ everywhere, but a.e.\ ?} and conclude that
+ \begin{align}
+ J_A(u) &\leq J_A(u_-) + J_A(u_+) \\
+ &\leq \int_{-\infty}^0 J_A\big((-u_-)^s\big) \, ds +
+ \int_0^\infty J_A(u_+^s) \, ds\\
+ &= \int_{-\infty}^0 J_A(u^s) \, ds + \int_0^\infty J_A(u^s) \,
+ ds = \int_{-\infty}^\infty J_A(u^s) \, ds
+ \end{align}
+ \paragraph{Then we prove that $J_A(u) \geq \int_{-\infty}^\infty
+ J_A(u^s) \, ds$.}
+ Define the function
+ \begin{equation}
+ m(t) = \int_{\{ x \in \Omega : u(x) \leq t\}} \norm{\nabla u}_A
+ \, dx,
+ \end{equation}
+ and note that $m(\infty) = J_A(u)$. Since $m(t)$ is non-decreasing
+ with $t$, we can apply the existence theorem of Lebesgue \fixme{ref}
+ to conclude that $m'(t)$ exists a.e.\ (w.r.t.\ Lebesgue measure) and
+ that Lebesgue's inequality holds:
+ \begin{equation}
+ \int_{-infty}^\infty m'(t)\, dt \leq m(\infty) - m(-\infty) =
+ J_A(u).
+ \end{equation}
+ Next, fix an $s \in \mathbb{R}$ and define the function
+ \begin{equation}
+ \eta_r(t) = \begin{cases}
+ 0 & \text{if } t < s \\
+ (t - s)/r & \text{if } s \leq t < s + r \\
+ 1 & \text{if } t > s + r
+ \end{cases}
+ \end{equation}
+ visualized in Figure \fixme{fig} such that its derivative takes the
+ form shown in Figure \fixme{fig2}. We then have
+ \begin{equation}
+ \int_\Omega - \eta_r(u) \diver \xi \, dx
+ = \int_\Omega \eta_r'(u) \nabla u\cdot \xi \, dx
+ = \frac{1}{r} \int_{\{ s \leq u \leq s + r \}} \nabla u\cdot \xi
+ \, dx.
+ \end{equation}
+ Assuming that $\norm{\xi}_A^* \leq 1$ we have
+ \begin{align}
+ \frac{m(s+r) - m(s)}{r}
+ &= \frac{1}{r} \int_{\{ s \leq u \leq s+r \}} \norm{\nabla u}_A
+ \, dx \\
+ &\geq \frac{1}{r} \int_{\{ s \leq u \leq s + r\}} \nabla u \cdot
+ \xi \, dx
+ = \int_\Omega -\eta_r(u) \diver \xi \, dx.
+ \end{align}
+ As the limit of the left-hand side as $r \to 0$ exists almost
+ everywhere, suppose it exists at $s \in \mathbb{R}$, then
+ \begin{equation}
+ m'(s) \geq - \int_\Omega u^s \diver \xi \, dx
+ \end{equation}
+ since $\eta_r(u) \to u^s$ when $r \to 0$. As this holds for any
+ $\norm{\xi}_A^* \leq 1$, we get from \fixme{ref def} that $m'(s)
+ \geq J_A(u_s)$ almost everywhere and conclude
+ \begin{equation}
+ J_A(u) \geq \int_{-\infty}^\infty m'(t) \, dt \geq
+ \int_{-\infty}^\infty J_A(u^s) \, ds.
+ \end{equation}
+ \fixme{something something set of measure zero, lebesgue integral
+ something something}
+\end{proof}
+
\section{Cauchy--Crofton formulas}
If we parametrize straight lines as shown in Figure \fixme{ref}, we can
The singular values are the roots of a polynomial which depend
continuously on the coefficients of the polynomial which depend
continuously on the entries of the tensor. Thusly, the singular
- values are also bounded.
+ values are also bounded and the eigenvalues of our metric tensor is
+ greater than $0$.
+
+ According to the Rayleigh principle \fixme{REF}
+ \begin{equation}
+ \sigma_2 = \min_{\norm{x} = 1} \sqrt{x^T A x},
+ \end{equation}
+ so the denominator of \fixme{ref} is greater than $0$.
\end{proof}
\chapter{Discrete formulation}
\subsection{Discrete Riemannian Cauchy--Crofton formula}
+\begin{figure}
+ \centering
+ \includegraphics[width=0.7\textwidth]{fig/neigh_lines.png}
+ \caption{
+ The set of lines $\mathcal{L}$ is discretized to $\mathcal{L}_D$
+ where each line belongs to a family given by the angle $\phi$
+ parameter. In the figure to the left we see a visualization of
+ \fixme{almost} all the families as a neighborhood. \fixme{Note
+ that the angle $\phi$ is normally taken as a parameter
+ giving the normal to the actual line, but since we only
+ consider the difference $\Delta\phi$, this does not matter?}
+ To the right we see all the lines of one family.
+ }
+ \label{fig:neigh_lines}
+\end{figure}
By approximating the integral in \fixme{REF} by a discrete sum we obtain
the approximation
\begin{align}
\Delta\rho \\
\end{align}
where $\mathcal{L}_D$ is a discretization of the set of lines in the
-plane $\mathcal{L}$ \fixme{figure}.
-\begin{align}
- \abs{C}_R &\approx \sum_{e} n_C(e) \, \frac{\det M(e)}{2\left(e^T
- \cdot M(e) \cdot e\right)^{\sfrac{3}{2}}} \, \Delta\phi \,
- \Delta\rho
-\end{align}
+plane $\mathcal{L}$ \fixme{figure}. Since the final goal is to work with
+digital images, it makes sense to discretize our domain $\Omega$ as a
+regular (?) lattice? $\mathcal{S}$. The set of lines $\mathcal{L}_D$ can
+then be made up of lines through points on this lattice as shown in
+Figure \fixme{some figure with lines}. Further working towards our graph
+cut representation of the problem later, each line is made up of
+\emph{edges} going between the points in the lattice. We will denote an
+edge by $e$. The curve length can then be approximated by
+\begin{equation}
+ \abs{C}_R \approx \sum_{e} n_C(e) \, \frac{\norm{e}^3 \det M(e)}{2
+ \left(e^T \cdot M(e) \cdot e\right)^{\sfrac{3}{2}}} \, \Delta\phi \,
+ \Delta\rho.
+\end{equation}
+\begin{figure}
+ \centering
+ \includegraphics[width=0.7\textwidth]{fig/area_argument.png}
+ \caption{
+ We see that for any reasonable family of lines (i.e.\ we do
+ not leave any unneccessary gaps), the area covered by a blue
+ square, $\delta^2$ is equal to the area covered by a red
+ rectangle $\Delta \rho \norm{e_k}$, as there will be the same
+ number of blue squares and red rectangles if we cover the whole
+ plane.
+ }
+ \label{fig:area_argument}
+\end{figure}
+Assuming some regularity on our curve, and that our discretization is
+somehow fine, we approximate $n_C(e)$ by a function which is $1$ if $C$
+crosses $e$ an odd number of times, and $0$ if not, meaning that we
+ignore multiple passes across $e$ and only care about if $C$ goes from
+one side of $e$ to the other. Further, as can be seen in Figure
+\fixme{some figure with lines etc.}, the distance between lines in line
+family $k$ is $\Delta \rho_k = \delta^2 / \norm{e_k}$. Thus the curve
+length is approximated by
+\begin{equation}
+ \abs{C}_R \approx \sum_{e} n_C(e) \, \frac{\norm{e}^2 \det M(e)
+ \, \delta^2 \, \Delta\phi}{2 \left(e^T \cdot M(e) \cdot
+ e\right)^{\sfrac{3}{2}}}.
+\end{equation}
+
+\fixme{We now have the problem that $C$ is a curve, but later we want it
+to be a cut...}
\section{Graph cut formulation}
projects / master / commitdiff
