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I am currently writing my masters thesis and my topic also touches on Stochastic Tomography for volume reconstruction presented in this paper. Now i understand most of the process described, but i just don't understand one aspect of calculating the objective function. On page 4 under "Evaluating the Objective Function" the author mentions:

This is essentially a density estimation problem in 2D, where new samples need to be added on the fly. A suitable algorithm is kernel density estimation [Parzen 1962], also known as splatting

How is the KDE used for calculating the new residual value here? Are the input values for the KDE calculation actually all sample emissitvities which were already incorporated into the residual image at the pixel position the sample gets projected to? Does someone know how it is used here? Sadly the author does not get any more specific about that.

Thank you!

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How is the KDE used for calculating the new residual value here?

Assume that you have already processed $n$ samples $x_1,x_2,\ldots,x_n \in \mathbb{R}^3$. Then the residual images $r_j : \mathbb{R}^2 \rightarrow \mathbb{R}$, $j=1,2,\ldots$, take the form $$r_j(x) = m_j(x) - \sum_{i=1}^n e_i K(x-proj_j(x_i))$$ where $m_j$ is the $j$-th projection image, where $e_i \in \{-e_d, +e_d\}$ is the emissivity for the $i$-the sample $x_i$, where $K$ is the kernel you have chosen, and where $proj_j$ is the coordinate of the projection of $x_i$ into the $j$-th projection image.

To incorporate a new sample $x_{n+1}$, define new residual images $$\widetilde{r}_j^{\pm}(x) := r_j(x) - (\pm e_d) K(x-proj_j(x_{n+1}))$$ for positive and negative emissivity.

Are the input values for the KDE calculation actually all sample emissitvities which were already incorporated into the residual image at the pixel position the sample gets projected to?

In the end, the KDE calculation depends on all sample emissivities, yes.

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  • $\begingroup$ Thank you for the detailed explanation. I already was able to successfully implement it, but it is nice to see that i undertstood correctly ;) $\endgroup$ – puelo Aug 17 '17 at 20:04

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