计算机系统应用  2020, Vol. 29 Issue (2): 228-232 PDF

Confocal Image Restoration Based on Hessian Matrix Norm Regularization Algorithm
CHEN Ji-Yi, HE Tao, HU Jie
School of Mechanical Engineering, Shanghai Jiao Tong University, Shanghai 200240, China
Foundation item: National Natural Science Foundation of China (51675329, 51675342, 51775332, 51975360); Special Fund for Innovative Methodology of Ministry of Science and Technology (2018IM020100); Major Project of National Social Science Foundation of China (17ZDA020); Collaborative Research Fund for Medicine and Engineering of Shanghai Jiao Tong University (IH2018QNB03, YG2017QN61); First Breakthrough Project of Shanghai (7Z119010004)
Abstract: For the problem of deblurred confocal image restoration under Poisson noise, a regularization method based on Hessian matrix norm was proposed to solve the stairs effect existing in traditional methods. Based on the Poisson probability model, the method introduced the Hessian matrix norm as a regular condition, and applied the alternating direction multiplier method and gradient projection method to solve the optimization model. The quality of the reconstructed image obtained in the confocal laser scanning microscopy experiment was better than that of the traditional method. This result proves that the method can effectively restore the deblurred confocal image under Poisson noise.
Key words: confocal     image restoration     regularization

1 基于Hessian矩阵范数正则化的图像复原模型

 ${{y}} = P\left( {{{Af}}} \right)$ (1)

 ${{Hf}} = \left[ {\begin{array}{*{20}{c}} {{{{f}}_{xx}}}&{{{{f}}_{xy}}}\\ {{{{f}}_{yx}}}&{{{{f}}_{yy}}} \end{array}} \right]$ (2)

 $\widehat {{f}} = \mathop {\arg \min }\limits_{{f}} \left\{ {\left[ {{{Af}} - {{y}}\ln \left( {{{Af}}} \right)} \right] + \mu {{\left\| {{{Hf}}} \right\|}_2}} \right\}$ (3)

2 图像复原模型的求解

 $\left( {\widehat {{f}},\widehat {{{{u}}_1}},\widehat {{{{u}}_2}}} \right) = \mathop {\arg \min }\limits_{{{f}},{{{u}}_1},{{{u}}_2}} \left\{ \begin{array}{l} \left[ {{{{u}}_1} - {{y}}ln\left( {{{{u}}_1}} \right)} \right] + \mu {\left\| {{{H}}{{{u}}_2}} \right\|_2}\\ + \dfrac{\alpha }{2}\left[ \begin{array}{l} \left\| {{{{u}}_1} - \left( {{{Af}} + {{{b}}_1}} \right)} \right\|_2^2\\ + \left\| {{{{u}}_2} - \left( {{{f}} + {{{b}}_2}} \right)} \right\|_2^2 \end{array} \right] \end{array} \right\}$ (4)

 图 1 解耦的图像复原模型

(1)在固定 ${{u}}_1^{(k - 1)}$ , ${{u}}_2^{(k - 1)}$ , ${{b}}_1^{(k - 1)}$ , ${{b}}_2^{(k - 1)}$ 的前提下, 求解 ${{{f}}^{(k)}}$ .

 ${{{f}}^{(k)}} = \mathop {\arg \min }\limits_{{f}} \frac{\alpha }{2}\left[ \begin{array}{l} \left\| {{{u}}_1^{(k - 1)} - \left( {{{Af}} + {{b}}_1^{(k - 1)}} \right)} \right\|_2^2\\ + \left\| {{{u}}_2^{(k - 1)} - \left( {{{f}} + {{b}}_2^{(k - 1)}} \right)} \right\|_2^2 \end{array} \right]$ (5)

(2)对于步骤(1)所得的 ${{{f}}^{(k)}}$ , 可用来求解 ${{u}}_1^{(k)}$ , ${{u}}_2^{(k)}$ .

 ${{u}}_1^{(k)} = \mathop {\arg \min }\limits_{{{{u}}_1}} \left\{ \begin{array}{l} \left[ {{{{u}}_1} - {{y}}\ln\left( {{{{u}}_1}} \right)} \right]\\ + \dfrac{\alpha }{2}\left\| {{{{u}}_1} - \left( {{{Af}^{(k)}} + {{b}}_1^{(k - 1)}} \right)} \right\|_2^2 \end{array} \right\}$ (6)
 ${{u}}_2^{(k)} = \mathop {\arg \min }\limits_{{{{u}}_2}} \left\{ {\mu {{\left\| {{{Hu}_2}} \right\|}_2} + \frac{\alpha }{2}\left\| {{{{u}}_2} - \left( {{{{f}}^{(k)}} + {{b}}_2^{(k - 1)}} \right)} \right\|_2^2} \right\}$ (7)

(3)更新 ${{b}}_1^{(k)}$ , ${{b}}_2^{(k)}$ .

 ${{b}}_1^{(k)} = {{b}}_1^{(k - 1)} + {{Af}^{(k)}} - {{u}}_1^{(k)}$ (8)
 ${{b}}_2^{(k)} = {{b}}_2^{(k - 1)} + {{{f}}^{(k)}} - {{u}}_2^{(k)}$ (9)
2.1 ${{{f}}^{(k)}}$ 的求解

 ${{{f}}^{(k)}} = {F^{ - 1}}\left\{ \begin{array}{l} \left[ \begin{array}{l} F{\left( {{A}} \right)^ * } \cdot \left( {F\left( {{{u}}_1^{\left( {k - 1} \right)}} \right) - F\left( {{{b}}_1^{\left( {k - 1} \right)}} \right)} \right)\\ + \left( {F\left( {{{u}}_2^{\left( {k - 1} \right)}} \right) - F\left( {{{b}}_2^{\left( {k - 1} \right)}} \right)} \right) \end{array} \right]\\ /\left[ {F{{\left( {{A}} \right)}^ * } \cdot F\left( {{A}} \right) + {{I}}} \right] \end{array} \right\}$ (10)

2.2 ${{u}}_1^{(k)}$ 的求解

 $C\left( x \right) = x - {{y}}\ln \left( x \right) + \frac{\alpha }{2}{\left[ {x - \left( {{{Af}^{\left( k \right)}} + {{b}}_1^{\left( {k - 1} \right)}} \right)} \right]^2}$ (11)

 $\frac{{{\rm{d}}C\left( x \right)}}{{{\rm{d}}x}} = 1 - \frac{{{y}}}{x} + \alpha x - \alpha \left( {{{Af}^{\left( k \right)}} + {{b}}_1^{\left( {k - 1} \right)}} \right)$ (12)

 $\frac{{{{\rm{d}}^2}C\left( x \right)}}{{{\rm{d}}{x^2}}} = \frac{{{y}}}{{{x^2}}} + \alpha$ (13)

 $\alpha {x^2} + \left[ {1 - \alpha \left( {{{Af}^{\left( k \right)}} + {{b}}_1^{\left( {k - 1} \right)}} \right)} \right]x - {{y}} = 0$ (14)

 ${{u}}_1^{(k)} \!= \!\frac{{\alpha \left( {{{Af}^{\left( k \right)}} \!+ \!{{b}}_1^{\left( {k - 1} \right)}} \right) \!-\! 1 \!+\! \sqrt {{{\left[ {1 \!-\! \alpha \left( {{{Af}^{\left( k \right)}} \!+\! {{b}}_1^{\left( {k - 1} \right)}} \right)} \right]}^2} \!+ \!4\alpha {{y}}} }}{{2\alpha }}$ (15)
2.3 ${{u}}_2^{(k)}$ 的求解

 ${\left\| {{{Hu}_2}} \right\|_2} \!=\! \mathop {\max }\limits_{{\omega } \in {B_{\infty ,2}}} {\left\langle {{{\omega }},{{Hu}_2}} \right\rangle _{{R^{N \times 2 \times 2}}}} \!=\! \mathop {\max }\limits_{{\omega } \in {B_{\infty ,2}}} {\left\langle {{{{H}}^ * }{{\omega }},{{{u}}_2}} \right\rangle _{{R^N}}}\!\!\!\!$ (16)

 ${{u}}_2^{(k)} = \mathop {\arg \min }\limits_{u \in T} \left\{ \begin{array}{l} \tau \mathop {\max }\limits_{\omega \in {B_{\infty ,2}}} {\left\langle {{H^ * }{\omega },{u_2}} \right\rangle _{{R^N}}}\\ + \dfrac{1}{2}\left\| {{u_2} - \left( {{f^{(k)}} + {{b}}_2^{(k - 1)}} \right)} \right\|_2^2 \end{array} \right\}$ (17)

 ${{u}}_2^{(k)} = {P_T}\left[ {\left( {{{{f}}^{(k)}} + {{b}}_2^{(k - 1)}} \right) - \tau {{{H}}^ * }{{\omega }}} \right]$ (18)

 {P_{{B_{\infty ,2}}}}\left( {{{{\omega }}_i}} \right) = \left\{ {\begin{aligned} &{{{{\omega }}_i},\;\;\left\| {{{{\omega }}_i}} \right\| \le 1}\\ &{\frac{{{{{\omega }}_i}}}{{\left\| {{{{\omega }}_i}} \right\|}},\;\;\left\| {{{{\omega }}_i}} \right\| > 1} \end{aligned}} \right. (19)

${{v}}^{(1)} = \omega ^{(1)} = 0$

while i < Maxiter(最大迭代次数) do

$\;\;\;\;\omega ^{(i)} \leftarrow {P_{{B_{\infty ,2}}}}{{v}}^{(i)} + \dfrac{1}{{64\tau }}{P_T}(({{f}}^{(k)} + {{b}}_2^{(k - 1)}) - \tau {{{H}}^*}{{v}}^{(i)})$

${t_{i + 1}} \leftarrow \dfrac{{1 + \sqrt {1 + 4t_i^2} }}{2}$

$\;\;\;\;{{v}}^{(i + 1)} \leftarrow \omega ^{(i)} + \dfrac{{{t_i} - 1}}{{{t_{i + 1}}}}(\omega ^{(i)} - \omega ^{(i - 1)})$

end do

${{u}}_2^{(k)} = {P_T}(({{f}}^{(k)} + {{b}}_2^{(k - 1)}) - \tau {{{H}}^*}{{\omega }}^{(i)})$

3 实验结果与分析

3.1 仿真验证

 图 2 “woman”仿真处理结果

 图 3 局部放大视图

 图 4 “cameraman”仿真处理结果

3.2 实验设置

 图 5 激光扫描共聚焦显微镜实验平台

3.3 图像复原结果

 图 6 共聚焦图像去模糊结果

4 结论

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