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Sparse representation based image deblurring model under random-valued impulse noise

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Abstract

In this article, we introduce a new patch-based model for restoring images simultaneously corrupted by blur and random-valued impulse noise. The model involves a \(l_0\)-norm data-fidelity term, a sparse representation prior over learned dictionaries, and the total variation (TV) regularization. Unlike previous works Cai et al. (Inverse Probl Imaging 2(2):187–204, 2008), Ma et al. (SIAM J Imaging Sci 6(4):2258–2284, 2013), one-phase approach is utilized for random-valued impulse noise. As in Yuan and Ghanem (IEEE conference on computer vision and pattern recognition (CVPR), pp 5369–5377, 2015), the \(l_0\) data-fitting term plays an influential role for removing random-valued impulse noise. Moreover, the sparse representation prior enables to preserve textures and details efficiently, whereas TV regularization locally smoothes images while keeping sharp edges. To handle nonconvex and nondifferentiable terms, we adopt a variable splitting scheme, and then the penalty method and alternating minimization algorithm are employed. This results in an efficient iterative algorithm for solving our model. Numerical results are reported to show the effectiveness of the proposed model compared with the state-of-the-art methods.

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Acknowledgements

Myeongmin Kang was supported by the NRF (2016R1C1B1009808). Myungjoo Kang was supported by the NRF (2015R1A5A1009350, 2017R1A2A1A17069644), and IITP-MSIT (B0717-16-0107). Miyoun Jung was supported by Hankuk University of Foreign Studies Research Fund and the NRF (2017R1A2B1005363).

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Appendix

Appendix

Proof of Lemma 1

Define \(\varPhi _1\) and \(\varPhi _2\) as the objective functions in problems (10) and (11), respectively:

$$\begin{aligned} \varPhi _1(u,D,\{\alpha _s\})= & {} \lambda \Vert f-Au\Vert _0 + \sum _{s\in \mathcal {B}} \mu _s \Vert \alpha _s\Vert _0 + \sum _{s\in \mathcal {B}} \Vert D\alpha _s - R_s u\Vert _2^2+ \eta \Vert \nabla u\Vert _1,\\ \varPhi _2(u,v,D,\{\alpha _s\})= & {} \lambda \langle \mathbf {1}, \mathbf {1}-v \rangle + \sum _{s\in \mathcal {B}} \mu _s \Vert \alpha _s\Vert _0 + \sum _{s\in \mathcal {B}} \Vert D\alpha _s - R_s u\Vert _2^2 + \eta \Vert \nabla u\Vert _1. \end{aligned}$$

Assuming that \((u^*,D^*, \{\alpha _s^{*}\})\) is a global solution of problem (10), suppose that \((u^*, \mathbf {1}-sign(|Au^*-f|), D^*, \{\alpha _s^*\})\) is not a global solution of problem (11). Then, there exists \((u_1^*,v_1^*,D_1^*, \{\alpha _s^{1,*}\})\) satisfying that

$$\begin{aligned}&\varPhi _2(u_1^*,v_1^*,D_1^*, \{\alpha _s^{1,*}\}) < \varPhi _2(u^*, \mathbf {1}-sign(|Au^*-f|), D^*, \{\alpha _s^*\}),\\&v_1^* \odot |f-Au_1^*|=\mathbf { 0} \quad \text {and} \quad \mathbf {0}\le v_1^* \le \mathbf {1}. \nonumber \end{aligned}$$
(16)

Let \(v_2^* = \mathbf {1}-sign(|f-Au_1^*|)\). Then we have

$$\begin{aligned} \varPhi _1(u_1^*,D_1^*,\{\alpha _s^{1,*}\})= & {} \varPhi _2(u_1^*,v_2^*,D_1^*, \{\alpha _s^{1,*}\}) \\\le & {} \varPhi _2(u_1^*,v_1^*,D_1^*, \{\alpha _s^{1,*}\}) \\< & {} \varPhi _2(u^*,\mathbf {1}-sign(|Au^*-f|),D^*, \{\alpha _s^{*}\}) \\= & {} \varPhi _1(u^*,D^*, \{\alpha _s^{*}\}), \end{aligned}$$

where the first and last equalities are due to the definition of \(\varPhi _i\) (\(i=1,2\)) and \(l_0\) norm, and the second and third inequalities are obtained from Lemma 2 (given below this proof) and (16), respectively. This contradicts to the assumption that \((u^*,D^*, \{\alpha _s^{*}\})\) is a global solution of (10).

Conversely, assuming that \((u^*,v^*,D^*, \{\alpha _s^{*}\})\) is a global solution of problem (11), suppose that \((u^*,D^*, \{\alpha _s^{*}\})\) is not a global solution of problem (10). Then, we can have a tuple \((u_1^*,D_1^*,\{\alpha _s^{1,*}\})\) such that

$$\begin{aligned} \varPhi _1(u_1^*,D_1^*, \{\alpha _s^{1,*}\}) < \varPhi _1(u^*, D^*, \{\alpha _s^*\}). \end{aligned}$$
(17)

With setting \(v_1^* = \mathbf {1}-sign(|f-Au_1^*|)\) and \(v_2^* = \mathbf {1}-sign(|f-Au^*|)\), we have that \(\Vert Au^*_1-f\Vert _0 = \langle \mathbf {1},\mathbf {1}-v_1^* \rangle \) and \(\Vert Au^*-f\Vert _0 = \langle \mathbf {1},\mathbf {1}-v_2^* \rangle \). These yield that

$$\begin{aligned} \varPhi _2(u^*_1,v^*_1, D^*_1, \{\alpha _s^{1,*}\})= & {} \varPhi _1(u_1^*, D_1^*, \{\alpha _s^{1,*}\})\\< & {} \varPhi _1(u^*, D^*, \{\alpha _s^*\})\\= & {} \varPhi _2(u^*, v_2^*,D^*, \{\alpha _s^*\})\\\le & {} \varPhi _2(u^*, v^*,D^*, \{\alpha _s^*\}), \end{aligned}$$

where the second and last inequalities are obtained from (17) and Lemma 2, respectively. This is also a contradiction to the assumption that \((u^*,v^*,D^*, \{\alpha _s^{*}\})\) is a global solution of (11). \(\square \)

Lemma 2

For fixed u, assume that \(v_1\) satisfies the constraints:

$$\begin{aligned} v \odot |f-Au|=\mathbf { 0} \quad \text {and} \quad \mathbf {0}\le v \le \mathbf {1}, \end{aligned}$$
(18)

and let \(v_2 = \mathbf {1}-sign(|Au-f|)\). Then, it holds that

$$\begin{aligned} \Vert Au-f\Vert _0 = \langle \mathbf {1}, \mathbf {1}-v_2\rangle \le \langle \mathbf {1}, \mathbf {1}- v_1\rangle . \end{aligned}$$

Proof of Lemma 2

By the definition of \(v_2\), it trivially holds that \(\Vert Au-f\Vert _0 = \langle \mathbf {1}, \mathbf {1}- v_2\rangle \). For any v satisfying (18), \(v_s = 0\) with \(s\in \varTheta =\{ s\,:\, |(Au-f)_s| \ne 0\}\). Clearly, \(v_2\) satisfies (18), so \((v_1)_s=(v_2)_s =0 \) for all \(s\in \varTheta \). Thus we obtain

$$\begin{aligned} \langle \mathbf {1}, \mathbf {1}-v_2\rangle = |\varTheta | \le |\varTheta | + \sum _{s\in {\varTheta }^{c}} (1-(v_1)_s) = \langle \mathbf {1}, \mathbf {1}- v_1\rangle . \end{aligned}$$

This completes the proof of the Lemma 2. \(\square \)

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Kang, M., Kang, M. & Jung, M. Sparse representation based image deblurring model under random-valued impulse noise. Multidim Syst Sign Process 30, 1063–1092 (2019). https://doi.org/10.1007/s11045-018-0587-z

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