\end{equation}
where $\#(l_{\phi, \rho} \cap C)$ is the number of times the line
$l_{\phi, \rho}$ intersects the curve $C$.
+ \label{thm:euclidean_cauchy_crofton}
\end{theorem}
This elegant formula is very useful when we later will discretize our
energy function. The set of lines $\mathcal{L}$ is then discretized
b\rangle_M = \langle a, M(p) b \rangle$. The length of a curve $\gamma$
parametrized by some parameter $t$ then becomes
\begin{equation}
- \abs{\gamma}_M = \int_\gamma \langle \dot{\gamma},
- M\big(\gamma(t)\big) \, \dot{\gamma} \rangle \, dt
+ \abs{\gamma}_M = \int_\gamma \sqrt{\langle \dot{\gamma},
+ M\big(\gamma(t)\big) \, \dot{\gamma} \rangle} \, dt
+ \label{eq:riemannian_length}
\end{equation}
We will now prove a Cauchy--Crofton formula in this case where we have a
metric tensor in each point.
\begin{theorem}[The Riemannian Cauchy--Crofton formula]
- In the case where the scalar product in each point depends on a
- metric tensor $M(x)$ varying continuously over our space, the
- Cauchy--Crofton formula becomes \fixme{assuming some stuff about
- stuff}
+ Assume that our space $\Omega$ is equipped with a continous metric
+ tensor $M(x)$, whose eigenvalues are bounded $0 < k \leq
+ \lambda_2 \leq \lambda_1 \leq K < \infty$ for all $x \in \Omega$.
+ The Cauchy--Crofton formula for a differentiable curve $C$ of finite
+ length then becomes
\begin{equation}
\abs{C}_M = \int_\mathcal{L} \sum_{x \in l_{\nu, \rho} \cap C}
\, \frac{\det M(x)}{2 \left( \nu^T \cdot M(x) \cdot \nu
\right)^{\sfrac{3}{2}}}
\, d\mathcal{L}(l_{\nu, \rho}).
+ \label{eq:riemannian_cauchy_crofton}
\end{equation}
+ \label{thm:riemannian_cauchy_crofton}
\end{theorem}
-\begin{proof}
+Before proving this we present an important result from measure theory
+that we will need.
+\begin{theorem}[The Dominated Convergence theorem]
+ Let $\{ f_n \}$ be a sequence of real-valued measurable functions on
+ a complete measure space $(S, \Sigma, \mu)$. Suppose that the
+ sequence converges pointwise to a function $f$ and is dominated by
+ some integrable function $g$ in the sense that
+ \begin{equation}
+ \abs{f_n(x)} \leq g(x)
+ \end{equation}
+ for all $n$ and almost all $x \in S$. Then $f$ is integrable and
+ \begin{equation}
+ \lim_{n \to \infty} \int_S \abs{f_n - f} \, d\mu = 0
+ \end{equation}
+ which further implies that
+ \begin{equation}
+ \lim_{n \to \infty} \int_S f_n \, d\mu = \int_S f \, d\mu.
+ \end{equation}
+\end{theorem}
+For a proof and further background on measure theory and Lebesgue
+integration theory see for example \fixme{ref}
+\begin{proof}[Proof of the Riemannian Cauchy--Crofton formula]
Assume first that our space is equipped with at constant metric
tensor $M$. The length of our curve using this tensor can be
calculated by transforming the curve and applying the Euclidean
= \frac{\sigma_1^2 \sigma_2^2
}{\left(\sigma_1^2 \nu_x^2 + \sigma_2^2
\nu_y^2\right)^{\sfrac{3}{2}}}
- = \frac{\det \Sigma^2}{(\nu^T \, \Sigma^2 \,
+ = \frac{\det \Sigma^2}{(\nu^T \cdot \Sigma^2 \cdot
\nu)^{\sfrac{3}{2}}}.
\end{equation}
We are interested in the Jacobian of the whole transformation, so
all that is left to do is insert $U^T l_{\nu, \rho}$ to obtain
\begin{equation}
\abs{J_M(l_{\nu, \rho})} = \abs{J_{\Sigma^2}(U^T l_{\nu, \rho})}
- = \frac{\det M}{\left(\nu^T U \Sigma^2 U^T \nu\right)^{\sfrac{3}{2}}}
- = \frac{\det M}{\left(\nu^T M \nu\right)^{\sfrac{3}{2}}}
+ = \frac{\det M}{\left(\nu^T \cdot U \Sigma^2 U^T \cdot
+ \nu\right)^{\sfrac{3}{2}}}
+ = \frac{\det M}{\left(\nu^T \cdot M \cdot \nu\right)^{\sfrac{3}{2}}}
\end{equation}
We have now proven that for a constant metric tensor $M$, the length
of the differentiable curve $C$ with regards to this tensor can be
calculated as
\begin{equation}
- \abs{C}_M = \int_C \langle \dot{C}, M \dot{C} \rangle \, dt
+ \abs{C}_M = \int_C \sqrt{\langle \dot{C}, M \dot{C} \rangle} \,
+ dt
= \int_\mathcal{L} \# (l_{\nu, \rho} \cap C)
- \frac{\det M}{\left(\nu^T M \nu \right)^{\sfrac{3}{2}}} \,
- d\mathcal{L} (l_{\nu, \rho})
+ \frac{\det M}{\left(\nu^T \cdot M \cdot \nu
+ \right)^{\sfrac{3}{2}}} \, d\mathcal{L} (l_{\nu, \rho})
+ \label{eq:riemannian_const_m}
\end{equation}
- We now argue that this also holds for a non-constant but continuous
- metric tensor $M(x)$. By partitioning the domain into disjoint sets
- $U_i$ such that $\Omega = \cup U_i$, we make a piecewise constant
- approximation $M_\pi(x)$ such that if $x \in U_i$ then $M_\pi(x) =
- M(x_i)$ for some fixed $x_i \in U_i$. We then approximate the right
- side of \fixme{ref} by
+ We now argue that the similar formula in
+ \eqref{eq:riemannian_cauchy_crofton} holds for a non-constant but
+ continuous metric tensor $M(x)$. By partitioning the domain into
+ disjoint sets $U_i$ such that $\Omega = \cup_i U_i$, we make a
+ piecewise constant approximation $M_\pi(x)$ such that if $x \in U_i$
+ then $M_\pi(x) = M(x_i)$ for some fixed $x_i \in U_i$. We then
+ approximate \eqref{eq:riemannian_const_m} by
\begin{equation}
\abs{C}_{M_\pi} = \sum_i \int_\mathcal{L} \#(l_{\nu, \rho} \cap
C \cap U_i) \, w_i(\nu) \, d\mathcal{L}(l_{\nu, \rho})
where $w_i$ is the weight-function used in the set $U_i$ which can
be written
\begin{equation}
- w_i(\nu) = \frac{\det M(x_i)}{\left(\nu^T M(x_i) \nu
+ w_i(\nu) = \frac{\det M(x_i)}{\left(\nu^T \cdot M(x_i) \cdot \nu
\right)^{\sfrac{3}{2}}}.
\end{equation}
We further simplify the approximation by introducing the global
weight-function $w_\pi(\nu, x)$ which is equal to
$w_i(\nu)$ when $x \in U_i$. It can be written
\begin{equation}
- w_\pi(\nu, x) = \frac{\det M_\pi(x)}{\left(\nu^T M_\pi(x) \nu
- \right)^{\sfrac{3}{2}}}.
+ w_\pi(\nu, x) = \frac{\det M_\pi(x)}{\left(\nu^T \cdot M_\pi(x)
+ \cdot \nu \right)^{\sfrac{3}{2}}}.
\label{eq:wpi_def}
\end{equation}
Using this weight in \eqref{eq:mpi_approx} we can get rid of the sum
U_i} w_\pi(\nu, x) \, d\mathcal{L}(l_{\nu, \rho}) \\
&= \int_\mathcal{L} \sum_{x \in l_{\nu, \rho} \cap C}
w_\pi(\nu, x) \, d\mathcal{L}(l_{\nu, \rho}).
+ \label{eq:riemannian_segments}
\end{align}
As our partition $\pi$ is refined the weight $w_\pi(x)$ converges
- pointwise to the continously varying weight $w(x)$. Now it only
- remains to show that the integral converges, and that the left side
- converges to the left side of \fixme{ref}.
+ pointwise to the continously varying weight
+ \begin{equation}
+ w(\nu, x) = \frac{\det M(x)}{\left( \nu^T \cdot M(x) \cdot \nu
+ \right)^{\sfrac{3}{2}}}
+ \end{equation}
+ found in \eqref{eq:riemannian_cauchy_crofton}. Now it only remains
+ to show that the left and right hand side of
+ \eqref{eq:riemannian_segments} converges to the left and right hand
+ side of \eqref{eq:riemannian_cauchy_crofton}.
- Recall from \fixme{ref} that the left hand side is calculated as
+ Recall from \eqref{eq:riemannian_length} that the left hand side is
+ calculated as
\begin{equation}
\abs{C}_{M_\pi}
= \int_C \abs{\dot{C}}_{M_\pi} \, dt
\, dt.
\end{equation}
We know that $M_\pi(x)$ converges pointwise to $M(x)$, and thus
- $\abs{\dot{C}}_{M_\pi}$ converges pointwise to $\abs{\dot{C}}_M$.
- The largest eigenvalue of $M$ is equal to 1 and we can parametrize
- $C$ by its arclength parameter such that $\abs{\dot{C}} = 1$. Thus
- the integrand is bounded and we can apply Lebesgue's dominated
- convergence theorem to see that $\abs{C}_{M_\pi} \to
- \abs{C}_{M}$.
+ $\abs{\dot{C}(x)}_{M_\pi}$ converges pointwise to
+ $\abs{\dot{C}(x)}_M$. We have assumed bounds on the eigenvalues
+ of $M(x)$, and we can parametrize $C$ by its arc length such that
+ $\abs{\dot{C}(x)} = 1$, and thus the integrand is bounded and we can
+ apply Lebesgue's dominated convergence theorem to see that
+ $\abs{C}_{M_\pi} \to \abs{C}_M$.
We apply the same theorem to show that the right hand side of
- \fixme{ref} converges, but showing that $\sum_x w_\pi(\nu, x)$ is
- bounded is a bit more involved. Recall the definition of $w_\pi$ in
- \eqref{eq:wpi_def}. The numerator is equal to $\sigma_1^2
- \sigma_2^2$ and thus from the construction in \fixme{ref} bounded
- from above by $1$.
+ \eqref{eq:riemannian_segments} converges. Recall the definition of
+ $w_\pi$ in \eqref{eq:wpi_def}. The numerator is equal to $\sigma_1^2
+ \sigma_2^2 = \lambda_1 \lambda_2$ and is by assumption bounded from
+ above by $K^2$.
- Next we need to bound $\nu^T M_\pi(x) \nu$ away from zero. According
- to the Rayleigh principle \fixme{sigma squared is a bit meh?}
- \begin{equation}
- \sigma_2^2 = \min_{\norm{\xi} = 1} \sqrt{\xi^T M_\pi(x) \xi}
- \end{equation}
- and thus $\nu^T M_\pi(x) \nu \geq \sigma_2^2$ where $\sigma_2^2$ is
- the smallest eigenvalue of $M_\pi(x)$. Recalling the construction in
- \fixme{ref} we know that
- \begin{align}
- \sigma_2^2 &= \left(1 + \frac{(s_1 -
- s_2)^2}{\gamma^2}\right)^{-1} \geq \left(1 +
- \frac{s_1^2}{\gamma^2}\right)^{-1},
- \end{align}
- where $s_1$ and $s_2$ are the largest and smallest eigenvalue of our
- structure tensor respectively. Bounding $s_1$ from above would then
- imply $\sigma_2^2 \geq K > 0$.
+ \fixme{sigma squared meh?}
- The structure tensor is constructed as follows:
+ Next we need to bound $\nu^T M_\pi(x) \nu$ away from zero. According
+ to the Rayleigh principle
\begin{equation}
- S(x) = \left(K_{\rho} * \left( \nabla \tilde{f}_{\sigma} \otimes \nabla
- \tilde{f}_{\sigma} \right)\right)(x).
+ \lambda_2 = \min_{\norm{\xi} = 1} \sqrt{\xi^T M_\pi(x) \xi}
\end{equation}
- This is a smooth continuous map from $\bar{\Omega}$ to
- $\mathbb{R}^{2\times 2}$. As we can see in \fixme{ref} the
- eigenvalues are the roots of a monic polynomial and thus depend
- continuously on the coefficients of the polynomial, which in turn
- are continuous functions of the elements in the structure tensor
- $S(x)$.
-
- This proves that our weight function $w_\pi$ is bounded from above,
- but the sum
+ and thus $\nu^T M_\pi(x) \nu \geq \lambda_2 \geq k$. The weight
+ function $w_\pi$ is then bounded, but not neccesarily the sum
\begin{equation}
- \sum_{x \in l_{\nu, \rho} \cap C} w_\pi(\nu, x)
+ \sum_{\mathclap{x \in l_{\nu, \rho} \cap C}} w_\pi(\nu, x).
\end{equation}
- might be infinite. However, if this happens for a set of lines with
- measure greater than zero, the Euclidean Cauchy--Crofton formula in
- \fixme{ref} implies that the curve is infinitely long.
-
- Assuming that our curve has finite length we can therefore bound the
- integrand of \fixme{ref} and conclude using Lebesgue's dominated
- convergence theorem \fixme{ref} that
+ However, if the $l_{\nu, \rho} \cap C$ is infinite for a set of
+ lines with measure greater than zero, the Euclidean Cauchy--Crofton
+ formula in Theorem \ref{thm:euclidean_cauchy_crofton} implies that
+ the curve is infinitely long \fixme{which we have assumed it is
+ not?} As the sum is also bounded we can apply Lebesgue's dominated
+ convergence theorem again and conclude that
\begin{equation}
\int_\mathcal{L} \sum_{x \in l_{\nu, \rho} \cap C}
w_\pi(\nu, x) \, d\mathcal{L}(l_{\nu, \rho})
\int_\mathcal{L} \sum_{x \in l_{\nu, \rho} \cap C}
w(\nu, x) \, d\mathcal{L}(l_{\nu, \rho})
\end{equation}
- \fixme{introduce $w$ somewhere}
which--as both sides of the equality has been shown to
converge--leaves us with what we wanted to prove
\begin{equation}
\end{equation}
\end{proof}
+\section{Yeeep}
+
+Here we finish up the consideration of the contiuous problem, woop.
+
+The structure tensor is constructed as described in \fixme{ref}
+\begin{equation}
+ S(x) = \left(K_{\rho} * \left( \nabla \tilde{f}_{\sigma} \otimes \nabla
+ \tilde{f}_{\sigma} \right)\right)(x),
+\end{equation}
+where $\tilde{f}$ is the symmetric extension of our input image $f$ to
+$\mathbb{R}^2$.
+Because of the convolutions with the Gaussian function, this is a smooth
+continuous map from $\bar{\Omega}$ to $\mathbb{R}^{2\times 2}$. As we
+can see in \fixme{ref} the eigenvalues are the roots of a monic
+polynomial and thus depend continuously on the coefficients of the
+polynomial, which in turn are continuous functions of the elements in
+the structure tensor $S(x)$. The extreme value theorem \fixme{ref}
+states that a continuous real-valued function on a nonempty compact
+space is bounded above. Thus the eigenvalues $s_1$ and $s_2$ of $S(x)$
+are bounded from above and by construction the smallest eigenvalue of
+our metric tensor $M(x)$ is bounded away from zero as
+\begin{equation}
+ \lambda_2 = \left(1
+ + \frac{(s_1 - s_2)^2}{\gamma^2}\right)^{-1} \geq \left(1 +
+ \frac{s_1^2}{\gamma^2}\right)^{-1} \geq Q > 0.
+\end{equation}
+Hence, our metric tensor $M(x)$ fulfills all the assumptions of Theorem
+\ref{thm:riemannian_cauchy_crofton}. \fixme{did we argue for
+continuous?}
+
+And here we write something about the perimeter not being the length of
+the boundary and all that stuff. We really tie the room together.
+
\chapter{Discrete formulation}
This is where we discretize! It will also be an important chapter, as
projects / master / commitdiff