From: Bjørn Rustad <bjorn@rustad.me>
Date: Fri, 14 Nov 2014 10:48:34 +0000 (+0100)
Subject: Ugly figure-argument
X-Git-Url: http://git.rustad.me/?a=commitdiff_plain;h=88f40fbdf551f2c912da4b71041b66a1e1ce9107;p=master

Ugly figure-argument
---

diff --git a/results.tex b/results.tex
index 3a9bc06..ef05ed0 100644
--- a/results.tex
+++ b/results.tex
@@ -1,4 +1,10 @@
 \chapter{Results}
 
+We show some basic results for different parameters.
+
 \section{Comparison}
 
+Compare with the regular total variation method, and anisotropic
+diffusion. Question is, where do we get our anisotropic diffusion
+results from? Code our own? Or use some popular implementation?
+
diff --git a/theory.tex b/theory.tex
index adc4239..cfd4d1d 100644
--- a/theory.tex
+++ b/theory.tex
@@ -1142,9 +1142,37 @@ eigenvalues?}
 \begin{figure}
     \input{fig/area_proof}
 \end{figure}
-Further, as can be seen in Figure \ref{fig:area_proof}
-the distance between lines in a given line family is $\Delta \rho =
-\delta^2 / \norm{e}$. Thus the curve length is approximated by
+For each family of lines given by an angle parameter $\phi$ we have the
+relation
+\begin{equation}
+    \Delta \rho = \delta^2 / \norm{e},
+\end{equation}
+which is easy to verify for simple angles $\phi$. 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
+\begin{equation}
+    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
+$A_i$ 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
+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
+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
     \, \delta^2 \, \Delta\phi}{2 \left(e^T \cdot M(e) \cdot