From: Bjørn Rustad Date: Mon, 1 Dec 2014 10:15:04 +0000 (+0100) Subject: new line family thing proof X-Git-Url: http://git.rustad.me/?a=commitdiff_plain;h=590a50fed4e4b379e866e9bd9e0e016591205a56;p=master new line family thing proof --- diff --git a/theory.tex b/theory.tex index 5ce6755..a7355c2 100644 --- a/theory.tex +++ b/theory.tex @@ -1121,7 +1121,8 @@ the discrete levels to get \, \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 @@ -1132,7 +1133,8 @@ has zero perimeter. \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} @@ -1148,7 +1150,7 @@ the approximation &= \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} @@ -1166,7 +1168,7 @@ shown in Figure \fixme{ref}. 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 @@ -1230,61 +1232,38 @@ eigenvalues? sure, a bit, but it won't change consistency..} 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