From caecfb70ae3ce754a806fa7a75f339dba2ea4043 Mon Sep 17 00:00:00 2001 From: =?utf8?q?Bj=C3=B8rn=20Rustad?= Date: Mon, 27 Oct 2014 17:29:27 +0100 Subject: [PATCH] THRMS --- theory.tex | 115 ++++++++++++++++++++++++++++++++++------------------- 1 file changed, 74 insertions(+), 41 deletions(-) diff --git a/theory.tex b/theory.tex index 104068d..4019a30 100644 --- a/theory.tex +++ b/theory.tex @@ -372,10 +372,12 @@ 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 + Cauchy--Crofton formula becomes \fixme{assuming some stuff about + stuff} \begin{equation} - \abs{C}_M = \int_\mathcal{L} \sum_{p \in l_{\nu, \rho} \cap C} - \, \frac{\det M(x)}{2 \left( \nu^T \cdot M(x) \cdot \nu \right)} + \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}). \end{equation} \end{theorem} @@ -516,8 +518,8 @@ metric tensor in each point. C \cap U_i) \, w_i(\nu) \, d\mathcal{L}(l_{\nu, \rho}) \label{eq:mpi_approx} \end{equation} - where $w_i$ is the weight-function used in the set $U_i$ using the - constant tensor $M_\pi(x_i)$ and can be written + 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 \right)^{\sfrac{3}{2}}}. @@ -528,6 +530,7 @@ metric tensor in each point. \begin{equation} w_\pi(\nu, x) = \frac{\det M_\pi(x)}{\left(\nu^T M_\pi(x) \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 over the partition $i$ and form a sum of all intersection point of @@ -547,54 +550,84 @@ metric tensor in each point. Recall from \fixme{ref} that the left hand side is calculated as \begin{equation} - \abs{C}_{M_\pi} = \int_C \sqrt{ \dot{C}^T M_\pi \dot{C} } \, dt. + \abs{C}_{M_\pi} + = \int_C \abs{\dot{C}}_{M_\pi} \, dt + = \int_C \sqrt{ \dot{C}(t)^T M_\pi\big( C(t) \big) \dot{C}(t) } + \, dt. \end{equation} - We know that $M_\pi(x)$ converges pointwise to $M(x)$ and using the - fact that the largest eigenvalue of $M$ is equal to 1, we know that + 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 $\abs{C}_{M_\pi} \to - \abs{C}_{M(x)}$. + \abs{C}_{M}$. We apply the same theorem to show that the right hand side of - \fixme{ref} converges, but showing that $w_\pi(\nu, x)$ is bounded - is a bit more involved. - - If we manage to bound the singular values and keep them away from - $0$, that would be great. The singular values \fixme{or eigenvalues} - of our tensor is + \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 $1$. + + 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_1 &= 1 \\ - \sigma_2 &= \frac{1}{1 + \frac{(s_1 - - s_2)^2}{\gamma^2}} \geq \frac{1}{1 + \frac{s_1^2}{\gamma^2}} - \geq \Kappa > 0 + \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} - and thus we only have to prove that the largest eigenvalue of our - structure tensor $s_1$ is finite to show that the singular values of - our metric tensor is bounded. - - The structure tensor is made by taking the outer product of the - gradient of each point in a smoothed version of the noisy image $f$. - This tensor is then again smoothed component-wise with a Gaussian - kernel + 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 $\sigma_2^2 \geq K > 0$. + + The structure tensor is constructed as follows: \begin{equation} - \nabla f_{\rho_1} \otimes \nabla f_{\rho_1} \mapsto K_{\rho_2} * - \left( \nabla f_{\rho_1} \otimes \nabla f_{\rho_1} \right). + S(x) = \left(K_{\rho} * \left( \nabla \tilde{f}_{\sigma} \otimes \nabla + \tilde{f}_{\sigma} \right)\right)(x). \end{equation} - This is a smooth map from $\bar{\Omega}$ to $\mathbb{R}^{2\times - 2}$, and assuming that our original image $f$ is bounded (?), so is - this tensor. - - The singular values are the roots of a polynomial which depend - continuously on the coefficients of the polynomial which depend - continuously on the entries of the tensor. Thusly, the singular - values are also bounded and the eigenvalues of our metric tensor is - greater than $0$. + 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 + \begin{equation} + \sum_{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. - According to the Rayleigh principle \fixme{REF} + 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 + \begin{equation} + \int_\mathcal{L} \sum_{x \in l_{\nu, \rho} \cap C} + w_\pi(\nu, x) \, d\mathcal{L}(l_{\nu, \rho}) + \to + \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} - \sigma_2 = \min_{\norm{x} = 1} \sqrt{x^T A x}, + \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}) \end{equation} - so the denominator of \fixme{ref} is greater than $0$. \end{proof} \chapter{Discrete formulation} -- 2.47.3