First order derivative regularization

A way of forcing a function to be piecewise constant is forcing its first order derivatives to be sparse. Thus, using a first order derivative regularization operator and forcing the model residuals to follow a Cauchy distribution should make the letters ``blocky'' and preserve the letter edges Youzwishen (2001). Obtaining model residuals following a Cauchy distribution can be achieve posing the inverse problem as Iterative Reweighed Least Squares (IRLS) Darche (1989).

The Cauchy norm first order derivative edge-preserving regularization fitting goal was set following the nonlinear iterations: starting with ${\bf Q_{x}}^{0}= {\bf Q_{y}}^{0} = \bf I$, at the k^th iteration the algorithm solves

$$\begin{eqnarray} {\bf Kf}^{k} - { \bf g \approx 0 } \nonumber \\ {\bf \epsilon ... ... \\ {\bf \epsilon Q_{y}}^{k-1} { \bf D_y f}^k { \bf \approx 0 }\end{eqnarray}$$ (1)

where

$$\begin{eqnarray} {\bf Q_{x}}^{k-1} = \frac{1}{\left[1+\left( \frac{{ \bf D_x f}^... ...frac{{ \bf D_y f}^{k-1}}{\alpha} \right)^2 \right]^{\frac{1}{2}}}.\end{eqnarray}$$ (2)

$\bf K$ is a non-stationary convolution matrix, ${\bf f}^{k}$ is the result of the k^th nonlinear iteration, ${\bf Q_{x}}^{k-1}$ and ${\bf Q_{y}}^{k-1}$ are the (k-1)^th diagonal weighting operators, and ${ \bf D_x}$ and ${ \bf D_y}$ are the first order derivative operator in the x and y directions, $\bf I$ is the identity matrix, the scalar $\alpha$ is the trade-off parameter controlling the discontinuities in the solution, and the scalar $\epsilon$ balances the relative importance of the data and model residuals.

 

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Figure 4 A) Original image, B) Deblurred image using LS with the first order derivative edge-preserving regularization

 

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Figure 5 Comparison between Figures A and B; A) Slice y=229 and B) Slice x=229.

We were successful in obtaining what we designed the algorithm to produce. The result is blocky in the x and y directions (Figures  and ). However, the derivatives in the x and y directions do not produce an isotropic result. We know that letters often have round shapes. Thus, the problem could benefit from using a more isotropic operator to calculate the diagonal weights.