\, \Delta \lambda.
\end{equation}
Note that we will later ignore the $\Delta \lambda$ difference, as we can
-just absorb this into the $\beta$ parameter of \fixme{ref}. The perimeter is then
+just absorb it into the $\beta$ parameter of
+\eqref{eq:first_anisotropic_functional}. The perimeter is then
calculated using a discretized version of the Cauchy--Crofton formula
introduced in Theorem \ref{thm:riemannian_cauchy_crofton}. Again, we
stop the sum at $L-2$ since the level set $\{ u > L - 1\}$ is empty and
\begin{figure}
\input{fig/line_disc}
\end{figure}
-By approximating the integral in \fixme{REF} by a discrete sum we obtain
+By approximating the integral Theorem
+\ref{thm:riemannian_cauchy_crofton} by a discrete sum we obtain
the approximation
\begin{equation}
\begin{aligned}
&= \sum_\nu \sum_\rho \sum_{x \in \ell_{\nu, \rho} \cap C} \,
\frac{\det M(x)}{2\left(\nu^T
\cdot M(x) \cdot \nu\right)^{\sfrac{3}{2}}} \, \Delta\rho \,
- \Delta\phi.
+ \Delta\nu.
\label{eq:cauchy_crofton_approx1}
\end{aligned}
\end{equation}
The choice of our discrete set of lines $\mathcal{L}_D$ is important, as
it will decide the accuracy of our approximation in
\eqref{eq:cauchy_crofton_approx1}. We will only consider lines going
-through points at least two in in our grid $\mathcal{G}$, and for now we will
+through at least two points in our grid $\mathcal{G}$, and for now we will
consider a discretization which is uniform throughout the domain,
meaning that $\Delta \rho$ is constant for each line family, and that in
each grid point, there is a line from each family. \fixme{moar
For each family of lines given by an angle parameter $\phi$ in the
regular grid of size $\delta$ we have the relation
\begin{equation}
- \delta^2 = \norm{e} \Delta \rho
+ \delta^2 = \norm{e} \Delta \rho.
\end{equation}
\label{lem:delta_rho}
\end{lemma}
\begin{proof}
- This is easy to verify for simple angles $\phi = \{0,
- \sfrac{\pi}{4}, \sfrac{\pi}{2}, \hdots\}$. For a general argument
- consider the two partitions of the plane $\mathbb{R}^2$ as shown in
- Figure \ref{fig:area_proof}. The blue squares have area $\delta^2$,
- while the red rectangles have length $\norm{e}$ and width $\Delta
- \rho$. The blue partition repeats with a period of $\delta$ in both
- directions.
-
- Each grid point is also the terminal of two edges going
- each in the directions $\phi$ and $-\phi$. And also from each grid
- point we draw a line of length $\Delta \rho$ to the next line. From
- this construction, the red partition is also periodic with period
- $\delta$ in both directions.
-
- Now consider a square of size $n\delta \cdot n\delta$, where we
- connect the right side to the left side, and the top to the bottom
- so that it \fixme{topologically} is shaped like a donut. Equipped
- with a grid structure as before, this surface will contain $n \cdot
- n$ distinct grid points. Since our blue and red partitions are
- $\delta$-periodic, we can use them to partition this surface,
- without trouble.
-
- We have one blue square for each grid point, and we also have one
- red rectangle for each grid point, thus we have $n^2$ of each, and
- their areas must be equal
+ Consider a line $\ell$ intersecting the point $(p, q)$ in the grid.
+ The distance $\Delta \rho$ from this line $\ell$ to the next line
+ can then be calculated as a minimum over the distance to all other
+ lines.
+
+ The given family of lines consists of edges in the grid, which we
+ write $e = (\delta s, \delta t)^T$ where $s$ and $t$ are coprime
+ such that $e$ does not intersect any other points than its two
+ endpoints.
+
+ Let $(p\prime, q\prime)$ be an arbitrary point not on the line
+ $\ell$. We can then calculate
\begin{equation}
- \delta^2 = \norm{e} \Delta \rho.
+ \begin{aligned}
+ \Delta \rho &= \min_{(p\prime, q\prime)} \left\{
+ \left\langle \delta [p - p\prime, q - q\prime],
+ \frac{e^\perp}{\norm{e^\perp}} \right\rangle \right\} \\
+ &= \min \left\{\delta^2 \cdot \frac{t(p-p\prime) - s(q -
+ q\prime)}{\norm{e}} \right\}.
+ \end{aligned}
+ \end{equation}
+ Since $s$ and $t$ are coprime, there exists $a, b \in \mathbb{Z}$
+ such that $at - bs = 1$, and since $(p\prime, q\prime)$ is any point
+ not on the same line as $(p, q)$ we obtain
+ \begin{equation}
+ \Delta \rho = \frac{\delta^2}{\norm{e}}.
\end{equation}
-
- %Our goal is to show that the areas are equal. Each $\rho$-rectangle
- %is partitioned by the blue grid into the subsets $A_i$ such that
- %\begin{equation}
- % A = \bigcup_{i=1}^{i \leq m} A_i,
- %\end{equation}
- %where $A$ is the set of points in a $\rho$-rectangle. All
- %$\rho$-rectangles start in a grid point, so the will decompose in the
- %same way. Since all $\delta$-squares look the same, we can find all
- %$A_i$ in a given $\delta$-square such that
- %\begin{equation}
- % \delta^2 \geq \abs{\bigcup_{i=1}^{i \leq m} A_i} = \Delta \rho \abs{e}.
- %\end{equation}
- %Further the $\rho$-rectangles partition the plane and thus each point in
- %a $\delta$-square is also in a $\rho$-rectangle, and each
- %$\rho$-rectangle consists of the same parts $A_i$, and thus
- %\begin{equation}
- % \delta^2 \leq \abs{\bigcup_{i=1}^{i \leq m} A_i} = \Delta \rho \abs{e}.
- %\end{equation}
- %\fixme{we are missing some argument that we don't have more than one
- %$A_i$ in the square}
- %We can then conclude that $\delta^2 = \Delta \rho \abs{e}$.
\end{proof}
Inserting
this \fixme{and the tensor approx} into the curve length approximation
projects / master / commitdiff