Blind Image Deconvolution: When Patch-wise Minimal Pixels Prior Meets Fractional-Order Method

Abstract

Blind image deconvolution is a challenging issue in image processing. In blind image deconvolution, the typical approach involves iteratively estimating both the blur kernel and latent image until convergence to the blur kernel of the observed image is achieved. Recently, several approaches have been attempted to develop a sophisticated regularization to obtain the clean image. However, existing methods often struggle to effectively handle ringing artifacts and local blur. To overcome these limitations, we introduce a fractional-order variational model. This model alleviates the ringing artifacts through the selection of an optimal derivative. Subsequently, to refine the latent image further, we leverage the local prior, namely patch-wise minimal pixels (PMP) prior. Since the PMP prior of clean images blocks is much sparser than that of blurred ones, it is capable of discriminating between clean and blurred image blocks. We illustrate the effective integration of the fractional-order operations and the PMP prior within our proposed approach. Moreover, the convergence of our algorithm has been proved as the values of the objective function monotonically decrease. Extensive experiments on different datasets demonstrate the superiority of the proposed method compared with other methods in terms of reconstruction quality for blind deconvolution.

Appendix A Proof of (22)

Proof

For the problem (21a), we can split it into element-wise forms

$$\begin{aligned} \displaystyle \sum _{i}^{m}\displaystyle \sum _{j}^{n}{{\min } \, }( \nabla ^{\alpha } I{_{i,j}}^{t}-G{_{i,j}})^{2} +\frac{\mu }{{\lambda }_{1}}H( G{_{i,j}}), \end{aligned}$$

(A.1)

where $H( G{_{i,j}})$ is a binary function returning 1 if $G{_{i,j}}\ne 0$ and 0 otherwise. For a single element, we can establish the equation

$$\begin{aligned} \mathbb {S}{_{i,j}}=( \nabla ^{\alpha } I{_{i,j}}^{t}-G{_{i,j}})^{2} +\frac{\mu }{{\lambda }_{1}}H( G{_{i,j}}). \end{aligned}$$

(A.2)

When $(\nabla ^{\alpha }I{_{i,j}}^{t})^{2} <\mu /{\lambda }_{1}$ and $G{_{i,j}}=0$, we have $\mathbb {S}{_{i,j}}=(\nabla ^{\alpha }I{_{i,j}}^{t})^{2}$. Note that $G{_{i,j}}\ne 0$, then

$$\begin{aligned} \mathbb {S}{_{i,j}}=( \nabla ^{\alpha } I{_{i,j}}^{t}-G{_{i,j}})^{2} +\frac{\mu }{{\lambda }_{1}}\ge \frac{\mu }{{\lambda }_{1}}. \end{aligned}$$

(A.3)

Therefore, we have $\min \mathbb {S}{_{i,j}} = (\nabla ^{\alpha }{I}_{i,j}^{t})^{2}$ when $G{_{i,j}}=0$.

When $(\nabla ^{\alpha }I{_{i,j}}^{t})^{2} \ge \mu /{\lambda }_{1}$ and $G{_{i,j}}=0$, $\mathbb {S}{_{i,j}}=(\nabla ^{\alpha }I{_{i,j}}^{t})^{2}$. Note that $G{_{i,j}}\ne 0$, then

$$\begin{aligned} \mathbb {S}{_{i,j}}=( \nabla ^{\alpha } I{_{i,j}}^{t}-G{_{i,j}})^{2} +\frac{\mu }{{\lambda }_{1}}= \frac{\mu }{{\lambda }_{1}} \end{aligned}$$

(A.4)

as $G{_{i,j}}=\nabla ^{\alpha } I{_{i,j}}^{t}$. Thus $\min \mathbb {S}{_{i,j}} = \frac{\mu }{{\lambda }_{1}}$ when $G{_{i,j}}=\nabla ^{\alpha } I{_{i,j}}^{t}$.

As a consequence, when

$$\begin{aligned} G{_{i,j}}=\left\{ \begin{matrix}0,& (\nabla ^{\alpha }I_{i,j}^{t})^{2} <\mu /{\lambda }_{1}, \\ \nabla ^{\alpha }I_{i,j}^{t}, & otherwise, \end{matrix}\right. \end{aligned}$$

(A.5)

$\mathbb {S}{_{i,j}}$ has a minimum value, and then for the problem (21a), the solution can be explicitly obtained as follows:

$$\begin{aligned} G^{t+1}=\left\{ \begin{matrix}0,& (\nabla ^{\alpha }I^{t})^{2} <\mu /{\lambda }_{1}, \\ \nabla ^{\alpha }I^{t}, & otherwise. \end{matrix}\right. \end{aligned}$$

(A.6)

$\square$