Discrete Tomography with Unknown Intensity Levels Using Higher-Order Statistics

Jozsef Nemeth

doi:10.1007/s10851-015-0581-0

Journal of Mathematical Imaging and Vision

November 2015, Volume 53, Issue 3, pp 314–331 | Cite as

Discrete Tomography with Unknown Intensity Levels Using Higher-Order Statistics

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Jozsef NemethEmail author

Article

First Online: 17 April 2015

Received: 30 June 2014

Accepted: 28 March 2015

280 Downloads
2 Citations

Abstract

Discrete tomography focuses on the reconstruction of images containing only a limited number of different intensity values. Most of the methods assume that the intensities are a priori known. In practice, however, this information is usually not available. Therefore, the problem of the estimation of the intensity levels has been recently addressed by many researchers. In this paper, we present a novel approach for the tomographic reconstruction of binary images, when the two gray-levels are unknown. The problem is traced back to the minimization of an appropriate objective functional, in which a higher-order statistics-based discretization term enforces binary solutions. Instead of the gray-levels, the only parameter of this term is their mid-level value, which is iteratively approximated during the optimization process. Experiments on synthetic phantom images as well as on real data show that the proposed graduated optimization scheme can efficiently minimize the objective functional and the method provides accurate reconstructions. Compared to some of the state-of-the-art algorithms, the proposed method provides competitive results, while it requires less parameter settings, thus it can be considered as a valid alternative.

Keywords

Discrete tomography Binary tomography Reconstruction Higher-order statistics Kurtosis

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Notes

Acknowledgments

The author is grateful to Antal Nagy, László Varga, Márton Balaskó, Péter Balázs, and András London for providing data for the experiments and for their helpful comments and suggestions.

Supplementary material

10851_2015_581_MOESM1_ESM.pdf (214 kb)

Supplementary material 1 (pdf 215 KB)

Appendix

Here we provide the computational details on how to locally estimate the Lipschitz constant $K^*({\nabla D})$ of the gradient of the discretization term. Using this estimation, a gradient descent step-length is given by $t^*=1/K^*(\nabla E)=1/(K^*({\nabla C})+\mu K^*({\nabla D}))$, which does not increase the value of the objective functional.

The computations are independent of the value of the mid-level $\alpha $, thus for the sake of simplicity, we assume that $\alpha =0$. All the equations remain valid if we perform substitutions $x_i = z_i-\alpha $ and $\mathbf{x} = \mathbf{z}-\alpha \mathbf{1}_n$. The discretization term is now given by

$$\begin{aligned} D(\mathbf{x}) = n\frac{\Vert \mathbf{x}\Vert _4^4}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^2}-1, \end{aligned}$$

(19)

where we use the notation of the general vector norm $\Vert \mathbf{x}\Vert _p = \left( \sum _{i=1}^{n}x_i^p\right) ^{1/p}$. The first order derivatives can be computed as

$$\begin{aligned} \nabla _iD(\mathbf{x})=\frac{\partial D(\mathbf{x})}{\partial x_i} = 4n\left( \frac{x_i^3}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^2} - \frac{x_i\Vert \mathbf{x}\Vert _4^4}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^3}\right) . \end{aligned}$$

(20)

The discretization term $D$ and its first and second order derivatives are continuous everywhere except at $\mathbf{0}_n$. Thus on a subset $\varOmega \subseteq \mathbb {R}^n\setminus \mathbf{0}_n$, the Lipschitz constant of $\nabla D$ is given by $K(\nabla D)=\sup _{\mathbf{x}\in \varOmega }\Vert J(\nabla D)(\mathbf{x})\Vert $, where $J(\nabla D)$ is the Jacobian matrix of $\nabla D$ (we note that $J(\nabla D)$ is equal to the Hessian of $D$). The Jacobian $J(\nabla D)$ is a $n\times n$ symmetric matrix and its diagonal and non-diagonal entries are given by

$$\begin{aligned} J_{i,i}(\nabla D)(\mathbf{x})= & {} \frac{\partial ^2 D(\mathbf{x})}{\partial x_i^2}= 4n\left( \frac{6x_i^2\Vert \mathbf{x}\Vert _4^4}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^4} - \frac{8x_i^4}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^3}\right. \nonumber \\&\left. + \frac{3x_i^2}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^2} - \frac{\Vert \mathbf{x}\Vert _4^4}{(\Vert \mathbf{x}\Vert _2^2)^3} \right) \nonumber \\ \end{aligned}$$

and

$$\begin{aligned} J_{i,j}(\nabla D)(\mathbf{x})= & {} \frac{\partial ^2 D(\mathbf{x})}{\partial x_i\partial x_j}\nonumber \\= & {} 4n\left( \frac{6x_ix_j\Vert \mathbf{x}\Vert _4^4}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^4} -\frac{4x_i^3x_j+4x_ix_j^3}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^3} \right) ,\nonumber \\&\hbox {if }i\ne j. \end{aligned}$$

(21)

We will estimate an upper bound for the $\Vert \cdot \Vert _1$-norm of the Jacobian which is equal to its maximum absolute column sum (we note that $\Vert J(\nabla D)(\mathbf{x})\Vert _1=\Vert J(\nabla D)(\mathbf{x})\Vert _\infty $ since $J(\nabla D)$ is symmetric). For that purpose, consider the ith absolute column sum:

$$\begin{aligned} K_i= & {} |J_{i,i}(\nabla D)(\mathbf{x})| + \sum \limits _{\begin{array}{c} j=1 \\ j\ne i \end{array}}^{n}|J_{j,i}(\nabla D)(\mathbf{x})|\nonumber \\\le & {} 4n\left( \frac{6x_i^2\Vert \mathbf{x}\Vert _4^4}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^4} + \frac{8x_i^4}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^3} + \frac{3x_i^2}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^2} + \frac{\Vert \mathbf{x}\Vert _4^4}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^3} \right) \nonumber \\&+\, 4n\sum \limits _{\begin{array}{c} j=1 \\ j\ne i \end{array}}^{n}\left( \frac{6|x_ix_j|\Vert \mathbf{x}\Vert _4^4}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^4} + \frac{4|x_i^3x_j|+4|x_ix_j^3|}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^3} \right) \nonumber \\= & {} 4n\left( \frac{6x_i^2\Vert \mathbf{x}\Vert _4^4}{(\Vert \mathbf{x}\Vert _2^2)^4} + \frac{8x_i^4}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^3} + \frac{3x_i^2}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^2} + \frac{\Vert \mathbf{x}\Vert _4^4}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^3} \right) \nonumber \\&+\,4n\frac{6|x_i|\Vert \mathbf{x}\Vert _4^4}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^4} \left( \left( \sum \limits _{j=1}^{n}|x_j|\right) -|x_i|\right) \nonumber \\&+\,4n\frac{4|x_i|^3}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^3}\left( \left( \sum \limits _{j=1}^{n}|x_j|\right) -|x_i|\right) \nonumber \\&+\, 4n\frac{4|x_i|}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^3} \left( \left( \sum \limits _{j=1}^{n}|x_j|^3\right) -|x_i|^3\right) \nonumber \\= & {} 4n\left( \frac{3x_i^2}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^2} + \frac{\Vert \mathbf{x}\Vert _4^4}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^3} +\frac{6|x_i|\Vert \mathbf{x}\Vert _4^4}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^4} \left( \sum \limits _{j=1}^{n}|x_j|\right) \right. \nonumber \\&+\left. \frac{4|x_i|^3}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^3}\left( \sum \limits _{j=1}^{n}|x_j|\right) +\frac{4|x_i|}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^3}\left( \sum \limits _{j=1}^{n}|x_j|^3\right) \right) . \end{aligned}$$

(22)

Let $X = \max _i\{|x_i|\}$. Using the inequalities $|x_i|\le X$, $\Vert \mathbf{x}\Vert _4^4\le nX^4$, and $\sum _{j=1}^{n}|x_j|\le nX$, we obtain a simple upper bound for all of the absolute column sums, thus an upper bound for the $\Vert \cdot \Vert _1$-norm of the Jacobian:

$$\begin{aligned} \Vert J(\nabla D)(\mathbf{x})\Vert _1= & {} \max \limits _i\left\{ K_i\right\} \nonumber \\\le & {} 4n\left[ \frac{3X^2}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^2} + \frac{9nX^4}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^3} + \frac{6n^2X^6}{\left( \Vert \mathbf{x}\Vert _2^2\right) ^4} \right] \end{aligned}$$

(23)

Hence, knowing only the maximal absolute value $X$ and the norm $\Vert \mathbf{x}\Vert ^2_2$, we can compute an upper bound for $\Vert J(\nabla D)(\mathbf{x})\Vert $ using Eq. (23). It also follows that if on a set $\varOmega _{S,M}\subseteq \mathbb {R}^n$, the maximum absolute elements $\max _i\{|x_i|\}$ are upper bounded while $\Vert \mathbf{x}\Vert ^2_2$ has a lower bound, then $\nabla D$ is Lipschitz continuous on that set, i.e. if

$$\begin{aligned} \varOmega _{S,M} = \left\{ \mathbf{x} \in \mathbb {R}^n \;|\; \max \limits _i\{|x_i|\} \le M, S\le \Vert \mathbf{x}\Vert _2^2 \right\} , \end{aligned}$$

(24)

then

$$\begin{aligned} K(\nabla D)= & {} \sup _{\mathbf{x}\in \varOmega _{S,M}}\Vert J(\nabla D)(\mathbf{x})\Vert \nonumber \\\le & {} 4n\left[ \frac{3M^2}{S^2} + \frac{9nM^4}{S^3} + \frac{6n^2M^6}{S^4}\right] . \end{aligned}$$

(25)

Now we construct the set of the next possible solutions. Suppose that the current solution is $\hat{\mathbf{x}}$ and let $\hat{\mathbf{y}} = \hat{\mathbf{x}} - (1/K^*({\nabla C}))\nabla E_{\hat{\alpha }, \mu }( \hat{\mathbf{x}} )$. Consider the line segment that connects $\hat{\mathbf{x}}$ and $\hat{\mathbf{y}}$:

$$\begin{aligned} \varLambda =\{ \ell \hat{\mathbf{x}}+(1-\ell )\hat{\mathbf{y}} \;\;|\;\; 0\le \ell \le 1\}. \end{aligned}$$

(26)

Since we are looking for a step-length $t=1/(K^*(\nabla E))=1/(K^*({\nabla C})+\mu K^*({\nabla D}))$, it results $t\le 1/K^*({\nabla C})$, and after the gradient descent step, the next solution $\hat{\mathbf{x}}'$ will fall on the line segment $\varLambda $. Thus, we will compute an upper bound of $K({\nabla D})$ on $\varLambda $. For that purpose, consider the largest absolute value of the components of the vectors in $\varLambda $, i.e. let $M=\max _{\mathbf{x}\in \varLambda }\{ \max _i|x_i| \}\}$ and the square of the smallest vector norm in $\varLambda $, i.e. let $S=\min _{\mathbf{x}\in \varLambda }\{\Vert \mathbf{x}\Vert _2^2\}$. Since $\varLambda $ is a line segment, $M$ can be computed considering only its end points $\hat{\mathbf{x}}$ and $\hat{\mathbf{y}}$:

$$\begin{aligned} M=\max \left\{ \max _i\{|\hat{x}_i|\}, \max _i\{|\hat{y}_i|\} \right\} , \end{aligned}$$

(27)

while $S$ can be computed as the square of the minimal norm of the points on the line segment $\varLambda $:

$$\begin{aligned} S={\left\{ \begin{array}{ll} \Vert \hat{\mathbf{x}}\Vert _2^2 &{}\quad \hbox {if }\quad \theta \le 0 \\ \Vert \hat{\mathbf{y}}\Vert _2^2 &{}\quad \hbox {if } \quad \theta \ge 1 \\ \Vert \hat{\mathbf{x}}+k(\hat{\mathbf{y}}-\hat{\mathbf{x}})\Vert _2^2 &{} \quad \hbox {otherwise}, \end{array}\right. } \end{aligned}$$

(28)

where $\theta =\frac{\langle \hat{\mathbf{x}}, \hat{\mathbf{x}}-\hat{\mathbf{y}} \rangle }{ \Vert \hat{\mathbf{x}}-\hat{\mathbf{y}} \Vert _2^2}$, and $\langle \cdot |\cdot \rangle $ denotes the dot-product operation. It can be easily seen, that $\varLambda \subseteq \varOmega _{S,M}$, thus an upper bound of $K^*({\nabla D})$ on $\varLambda $ is given by Eq. (25).

In the first phase of the proposed method where the initial continuous reconstruction is obtained by the minimization of $E_{0,0}(\mathbf{x})\equiv C(\mathbf{x})$, setting $t=1/K^*({\nabla C})$ as the gradient descent step size will provide convergence to the global optimum. In the discretization phase where the objective function $E_{\hat{\alpha }, \mu }$ is minimized with $\mu >0$, before each gradient descent step, $M$ and $S$ can be computed for the current solution $\hat{\mathbf{x}}$ using Eqs. (27) and (28), and then $K^*(\nabla D)$ can be estimated using Eq. (25). Then using step size $t^*=1/K^*({\nabla E}) = 1/(K^*({\nabla C})+\mu K^*({\nabla D}))$ ensures that the gradient descent step will not increase the function value, thus the iterations will converge. In our experiments, the proposed method provided similar results using this step-length estimation as with the line search method, we found that the estimation of $S$ and $M$ before each descent step was computationally less efficient in our implementation. On the other hand, in a parallel implementation, the computation of $S$ and $M$ could be accelerated, and the step-length $t^*$ can be considered as an alternative to the line search.

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Jozsef Nemeth
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1.Department of Computer Algorithms and Artificial IntelligenceUniversity of SzegedSzegedHungary

Discrete Tomography with Unknown Intensity Levels Using Higher-Order Statistics

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