Kidney Detection and Segmentation in Contrast-Enhanced Ultrasound 3D Images

Abstract

Contrast-enhanced ultrasound (CEUS) imaging has lately benefited of an increasing interest for diagnosis and intervention planning, as it allows to visualize blood flow in real-time harmlessly for the patient. It complements thus the anatomical information provided by conventional ultrasound (US). This chapter is dedicated to kidney segmentation methods in 3D CEUS images. First we present a generic and fast two-step approach to locate (via a robust ellipsoid estimation algorithm) and segment (using a template deformation framework) the kidney automatically. Then we show how user interactions can be integrated within the algorithm to guide or correct the segmentation in real-time. Finally, we develop a co-segmentation framework that generalizes the aforementioned method and allows the simultaneous use of multiple images (here the CEUS and the US images) to improve the segmentation result. The different approaches are evaluated on a clinical database of 64 volumes.

Appendix: Choice of the Parameter μ for Ellipsoid Detection

The choice of μ in Eq. (3) is paramount as it controls the number of points that are taken into account for the ellipsoid matrix estimation. To find a suitable value, let us consider an ideal case of an image I ₀ in which there is one white ellipsoid (I ₀ = 1) on a black background (I ₀ = 0), whose implicit function is $\phi _{ {\bf c}_{0},{\bf M}_{0}}$. We also assume that the confidence weight is w ≡ 1 everywhere. Then the matrix estimated by our approach would be

$$\begin{array}{rl} {{\bf M}}^{{\ast}}& = \mathrm{ argmin}_{{\bf M}}\ E_{\mathrm{ det}}({\bf c}_{0},{\bf M},{\bf 1})\\ {} & ={ \left [ \frac{2} {\mu } \frac{1} {\int _{\varOmega }I_{0}}\int _{\varOmega }I_{0}({\bf x})\ \left ({\bf x} -{\bf c}_{0}\right ){\left ({\bf x} -{\bf c}_{0}\right )}^{T}d{\bf x}\right ]}^{-1}\\ \end{array}$$

(26)

Using the fact that $I_{0} = \boldsymbol{ 1}_{\{1-{({\bf x}-{\bf c}_{0})}^{T}{\bf M}_{0}({\bf x}-{\bf c}_{0})\geq 0\}}$ is the indicator of the ellipsoid yields

$$\displaystyle{{ {\bf M}}^{{\ast}} ={ \left [ \frac{2} {\mu } \frac{1} {\text{Vol}({\bf M}_{0})}\int _{\{1-{({\bf x}-{\bf c}_{0})}^{T}{\bf M}_{0}({\bf x}-{\bf c}_{0})\geq 0\}}\left ({\bf x} -{\bf c}_{0}\right ){\left ({\bf x} -{\bf c}_{0}\right )}^{T}d{\bf x}\right ]}^{-1} }$$

(27)

After a variable substitution ${\bf x} \leftarrow {\bf M}_{0}^{1/2}({\bf x} -{\bf c}_{0})$, this expression becomes

$$\displaystyle{{ {\bf M}}^{{\ast}} ={ \left [ \frac{2} {\mu } \frac{\det ({\bf M}_{0}^{-1/2})} {\text{Vol}({\bf M}_{0})} {\bf M}_{0}^{-1/2}\int _{ \{\|{\bf x}\|\leq 1\}}{\bf x}{{\bf x}}^{T}d{\bf x}\ \ {\bf M}_{ 0}^{-1/2}\right ]}^{-1} }$$

(28)

With $\text{Vol}({\bf M}_{0}) = \displaystyle\frac{4\pi } {3} \sqrt{\det ({\bf M} _{0 }^{-1 })} = \frac{4\pi } {3} \det ({\bf M}_{0}^{-1/2})$, we then obtain

$$\displaystyle{{ {\bf M}}^{{\ast}} ={ \left [ \frac{2} {\mu }\ \frac{3} {4\pi }\ {\bf M}_{0}^{-1/2}\int _{ \{\|{\bf x}\|\leq 1\}}{\bf x}{{\bf x}}^{T}d{\bf x}\ \ {\bf M}_{ 0}^{-1/2}\right ]}^{-1} }$$

(29)

Note that the integral $\int _{\{\|{\bf x}\|\leq 1\}}{\bf x}{{\bf x}}^{T}d{\bf x}$ denotes the covariance matrix of a 3D unit ball, which is actually a scalar matrix that can be easily computed

$$\int _{\{\|{\bf x}\|\leq 1\}}{\bf x}{{\bf x}}^{T}d{\bf x} = \left(\begin{array}{ccc} 2\pi\frac{2} {3}\frac{1} {5} & 0 & 0 \\ 0 & 2\pi \frac{2} {3}\frac{1} {5} & 0 \\ 0 & 0 & 2\pi \frac{2} {3}\ \frac{1} {5} \end{array} \right) = \frac{4\pi } {15} \left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right)$$

(30)

Combining Eqs. (29) and (30) leads to

$$\displaystyle{{ {\bf M}}^{{\ast}} ={ \left [ \frac{2} {\mu }\left (\frac{1} {5}\ {\bf M}_{0}^{-1}\right )\right ]}^{-1} }$$

(31)

which yields the following relationship between M ^∗ and M ₀ :

$$\displaystyle{{ {\bf M}}^{{\ast}} = \frac{5} {2}\mu {\bf M}_{0} }$$

(32)

This shows that the exact solution M ₀ is retrieved for $\mu = \frac{2} {5}$. This value actually depends on the dimension of Ω. Here we assumed $\varOmega \subset \mathbb{R}^{3}$ but for 2D images, the optimal value would rather be $\mu = \frac{1} {2}$.