\centering
-\begin{tikzpicture}[scale=1.2]
- \foreach \x in {0,...,5} {
- \foreach \y in {0,...,5} {
+\begin{tikzpicture}[scale=2.0]
+ \foreach \x in {0,...,2} {
+ \foreach \y in {0,...,2} {
\node[tiny vertex] (\x\y) at (\x, \y) {};
}
}
- \clip (0, 0) rectangle (5, 5);
+ \clip (0, 0) rectangle (2, 2);
- \path[nedge=4cm] (0,5) -- (2,4);
- \path[nedge=4cm] (0,4) -- (2,3);
- \path[nedge=4cm] (0,3) -- (2,2);
- \path[nedge=4cm] (0,2) -- (2,1);
- \path[nedge=4cm] (0,1) -- (2,0);
+ \path[nedge=4cm] (0,0) -- (2,2);
+ \path[nedge=4cm] (0,1) -- (2,3);
+ \path[nedge=4cm] (1,0) -- (3,2);
- \path[nedge=4cm] (-1,1) -- (1,0);
+ \path (1,1) -- node[anchor=north west] {$b$} (2,2);
+ \path (0,0) -- node[anchor=south east] {$a$} (1,1);
- \path[nedge=4cm] (1,5) -- (3,4);
- \path[nedge=4cm] (1,4) -- (3,3);
- \path[nedge=4cm] (1,3) -- (3,2);
- \path[nedge=4cm] (1,2) -- (3,1);
- \path[nedge=4cm] (1,1) -- (3,0);
-
- \path[nedge=4cm] (2,5) -- (4,4);
- \path[nedge=4cm] (3,5) -- (5,4);
- \path[nedge=4cm] (4,5) -- (6,4);
-
- \draw [cyan, xshift=4cm] plot [smooth, tension=2] coordinates {
- (1,1) (2,1) (5,5) (3,3)};
+ \draw[cyan, very thick] plot [smooth, tension=1] coordinates {
+ (-1,1.0) (0.5,0.2) (0.8,1.4) (2,1.5)};
%\path[dash] (2, 2) -- (5,2);
%\path[dash] (2, 2) -- (5,5);
%\path[edge] (a) -- (h);
%\path[edge] (a) -- (i);
- \path (4,1) [edge, <->, anchor=center] -- node[anchor=south
- east,yshift=-3.5pt]
- {$\Delta \rho$}
- ($(5,0)!(4,1)!(3,1)$) ;
-
\end{tikzpicture}
-\caption{One family of lines having the same $\phi$ parameter.}
+\caption{
+ Here our intersection approximation would not be correct, as only
+ the intersection with edge $b$ is counted in \fixme{ref}, even
+ though the curve intersects edge $a$ twice.
+}
\label{fig:curve_edge}
``did $e$ cross $C$ or not?'' This amounts to checking whether the
terminals of $e$ lie on each side of the perimeter $C$, and the
approximation is exact for zero or one intersection points, but will, as
-we see in Figure \fixme{ref}, not be entirely correct when we have more.
+we see in Figure \ref{fig:curve_edge}, not be entirely correct when we
+have more.
\begin{figure}
\input{fig/curve_edge}
\end{figure}
-\fixme{curve or perimeter here, maybe perimeter because then we know it
-follows the boundaries of the pixels.}
-
The second difficulty is that in the discrete setting, we will only have
an approximation of the metric tensor $M(x)$ for each point $x \in
\mathcal{G}$, and it is thus not available for arbitrary intersection
\end{equation}
the component-wise average of the tensors in the two end points of the
edge. \fixme{really? componentwise? will that not mess up the
-eigenvalues?}
+eigenvalues? sure, a bit, but it won't change consistency..}
+
+\fixme{
+ we must define what we mean by a reasonable line family. meaning
+ each line goes through more than one grid point. and there are no
+ grid points withoug a line through it
+}
\begin{figure}
\input{fig/area_proof}
\end{figure}
+
+\begin{lemma}
For each family of lines given by an angle parameter $\phi$ we have the
relation
\begin{equation}
- \Delta \rho = \delta^2 / \norm{e},
+ \delta^2 = \norm{e} \Delta \rho
\end{equation}
-which is easy to verify for simple angles $\phi$. For a general argument
+\end{lemma}
+\begin{proof}
+ Consider two partitions of the plane $\mathbb{R}^2$ as shown in
+ Figure \ref{fig:area_proof}. The blue squares area $\delta^2$,
+ while the red rectangles have length $\norm{e}$ and width $\Delta
+ rho$. The blue grid 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.
+
+ Consider a \fixme{donut} tile of size $\delta \cdot \delta$ where
+ the left side is connected to the right, and the top to the bottom.
+
+ We can use a tile of size $\delta \cdot \delta$ of the original red
+ and blue partition to partition this \fixme{donut} tile as well.
+\end{proof}
+\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 Figure \ref{fig:area_proof}, and call the blue squares
$\delta$-squares, and the red rectangles $\rho$-rectangles. The
$\delta$-squares have area $\delta^2$, while the $\rho$-rectangles have
area $\Delta \rho \cdot \abs{e}$. Our goal is to show that these are equal.
First note that we have a periodicity in both directions with period
-$\delta$. Thus each $\delta$-square looks exactly the same. Further,
-each $\rho$-rectangle is partitioned by blue lines into the similar
-parts such that
+$\delta$, and each $\delta$-square looks exactly the same. Further,
+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
+ A = \bigcup_{i=1}^{i \leq m} A_i,
\end{equation}
-where $A$ is the set of points in a $\rho$-rectangle and $A_i$ is one of
-its parts. Since all $\delta$-squares look the same, we can find all
+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}
-Since the $\rho$-rectangles are a partition of the plane, each point in
+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}
-We can then conclude that $\delta^2 = \Delta \rho \abs{e}$. Inserting
+\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 into the curve length approximation we obtain
\begin{equation}
\abs{C}_M \approx \sum_{e \cap C} \frac{\det M(e) \norm{e}^2
projects / master / commitdiff