计算机系统应用  2020, Vol. 29 Issue (1): 119-129 PDF

1. 中国科学院 计算机网络信息中心, 北京 100190;
2. 中国科学院大学, 北京 100049;
3. 云南财经大学 云南省经济社会大数据研究院, 昆明 650221;
4. 四川卧龙国家级自然保护区管理局, 卧龙 623006

Image Edge Detection Model Based on Higher Order Regular and Nonsmooth Data Fitting Terms
LI Chun1,2, CHEN Jing-Si3, WANG Peng-Yan4, LI Jian1, LUO Ze1
1. Computer Network Information Center, Chinese Academy of Sciences, Beijing 100190, China;
2. University of Chinese Academy of Sciences, Beijing 100049, China;
3. Big Data Research Institute of Yunnan Economy and Society, Yunnan University of Finance and Economics, Kunming 650221, China;
4. Sichuan Wolong Natural Reserve Administration, Wolong 623006, China
Foundation item: CAS Science and Technology Service Network Initiative (STS Initiative)(Y82E01, KFJ-STS-QYZD-058); National Research and Development Infrastructure and Facility Development Program of China (DKA2018-12-02-XX); Strategy Priority Research Program of Chinese Academy of Sciences (XDA19060205); Special Project of Informatization of Chinese Academy of Sciences (XXH13505-03-205, XXH13506-305, XXH13506-303); Around Five Top Priorities of “One-Three-Five” Strategic Planning of Computer Network Information Center, Chinese Academy of Sciences (CNIC-PY-1408, CNIC_PY_1409); Giant Panda International Cooperation Fund
Abstract: In the modern science and technology society, with the rapid development of digital image processing technology, image segmentation, and object edge detection are widely used in the medical, military, public defense, computer vision, and agricultural meteorology field. In this study, based on the classical Chan-Vese (CV) model, a piecewise constant image edge detection model with L1 norm data fitting term and TV2 second-order regular term is introduced. The new model uses a high-order regular function to penalize the objective function as a constraint on the new objective function, so that the model enabling to segment and detect images with low contrast and containing additional noise. Theoretically, we give reasonable assumptions, and a partial convergence analysis of the model is carried out. In terms of computation load, we study the theoretical solvability of the new model. Compulationally, for the numerical implementation of the model, the model is numerically solved by ADMM algorithm, and a new solution method is designed. A large number of numerical experiments were carried out with grayscale images and real images, and compared with the original CV model. The experimental results show that many advantages of the model with wide applications.
Key words: image edge detection     image segmentation     Chan-Vese model     higher order regular     ADMM     variational level set method

1 图像分割简介

2 预备知识

 $\left\{ {\begin{array}{*{20}{c}} {\mathop {\min }\limits_{u \in {R^N}} {E_1}\left( u \right) + {E_2}\left( v \right)} \\ {{A_1}u + {A_1}v = f} \end{array}} \right.$ (1)

 \left\{ {\begin{aligned} & \left( {{u^{k + 1}}, {v^{k + 1}}} \right) \in \mathop {\min }\limits_{u, v} {E_1}\left( u \right) + {E_2}\left( v \right) + \frac{\gamma }{2}\left\| {{A_1}u + {A_1}v - f - {b^k}} \right\|_2^2 \\ & {{b^{k + 1}} = {b^k} - \left( {{A_1}{u^{k + 1}} + {A_1}{v^{k + 1}} - f - {b^k}} \right)} \end{aligned}} \right. (2)

for $\scriptstyle k = 1, 2, 3, \cdots , K$ do

(1)解问题

$\scriptstyle \mathop {\arg \min }\limits_u {E_1}\left( u \right) + \frac{\gamma }{2}\left\| {{A_1}u + {A_2}v - f - {b^k}} \right\|_2^2$ ;

(2)求解问题

$\scriptstyle\mathop {\arg \min }\limits_u {E_2}\left( v \right) + \frac{\gamma }{2}\left\| {{A_1}u + {A_2}v - f - {b^k}} \right\|_2^2$ ;

(3)拉格朗日乘子更新

$\scriptstyle{b^{k + 1}} = {b^k} - \left( {{A_1}{u^{k + 1}} + {A_2}{v^{k + 1}} - f} \right)$ ;

end

3 模型提出与数值求解

$f: {\left[ {0, 1} \right]^2} \to R \subset {L^1}\left( \Omega \right)$ 为观测到信号(图像), 且满足如下数学表达式, $f = u + n$ , 其中 $u$ 为未受噪声污染信号(图像), $n$ 为外加噪声(IN). 为了重建 $u$ , 可以考虑如下优化问题:

 $\mathop {\min }\limits_u \left\{ {\lambda {{\left\| {f - u} \right\|}_1} + {{\left\| {\nabla u} \right\|}_1}} \right\}$ (3)

 $\mathop {\min }\limits_u \left\{ {\lambda {{\left\| {f - u} \right\|}_1} + {H^1}\left( \Gamma \right)} \right\}$ (4)

 $\mathop {\min }\limits_{{c_1}, {c_2}} \left\{ {\lambda \int\limits_\Omega {\left| {f - {c_1}} \right|dx + \lambda \int\limits_{{\Omega ^c}} {\left| {f - {c_1}} \right|dx + \lambda \int\limits_{\partial \Omega } {ds} } } } \right\}$ (5)

 $\mathop {\min }\limits_{{c_1}, {c_2}, u} {\left\{ {\lambda {{\left\| {f - {c_1}} \right\|}_1}u + \lambda {{\left\| {f - {c_2}} \right\|}_1}\left( {1 - u} \right)} \right. } {\left. { + \alpha {{\left\| {\nabla u} \right\|}_1} + \beta {{\left\| {\Delta u} \right\|}_1}} \right\}}$ (6)

 $\begin{array}{*{20}{c}} {p = \left( {{p_1}, {p_1}} \right) \in {{\left( {{R^{M \times N}}} \right)}^2}} \\ {divp\left( {i, j} \right) = {D_x}{p_1}\left( {i, j} \right) + {D_y}{p_2}\left( {i, j} \right)} \end{array}$

 $\left\{ {\begin{array}{*{20}{c}} {\mathop {\min }\limits_{{c_1}, {c_2}, u, v, w} \left\{ {\begin{array}{*{20}{c}} {\lambda {{\left\| {f - {c_1}} \right\|}_1}u + \lambda {{\left\| {f - {c_2}} \right\|}_1}\left( {1 - u} \right)} \\ { + \alpha {{\left\| {\nabla u} \right\|}_1} + \beta {{\left\| {\Delta u} \right\|}_1}} \end{array}} \right\}} \\ {v = \nabla u, w = \Delta u} \end{array}} \right.$ (7)

 $\mathop {\min }\limits_{{c_1}, {c_2}, u, v, w} \left\{ {\begin{array}{*{20}{c}} {\lambda {{\left\| {f - {c_1}} \right\|}_1}u + \lambda {{\left\| {f - {c_2}} \right\|}_1}\left( {1 - u} \right)} \\ { + \alpha {{\left\| v \right\|}_1} + \beta {{\left\| w \right\|}_1}} \\ { + \dfrac{{{\gamma _1}}}{2}\left\| {v - \nabla u + b_1^k} \right\|_2^2 + \dfrac{{{\gamma _2}}}{2}\left\| {w - \Delta u + b_2^k} \right\|_2^2} \end{array}} \right\}$ (8)

Bregmen迭代算法总结如算法2.

for $\scriptstyle k = 1, 2, 3, \cdots , K$ do

计算(8)的最优解;

更新: $\scriptstyle b_1^{k + 1} = b_1^k + {v^{k + 1}} - \nabla {u^{k + 1}}$ ;

更新: $\scriptstyle b_2^{k + 1} = b_2^k + {w^{k + 1}} - \Delta {u^{k + 1}}$ ;

end

3.1 u-子问题

$u$ -子问题进行求解时首先固定 $r = \left| {f - {c_1}} \right| +$ $\left| {f - {c_2}} \right|$ . 得到如下目标泛函:

 $\mathop {\min }\limits_u \left\{ {\lambda \int {rudx + } \dfrac{{{\gamma _1}}}{2}\left\| {v\! - \!\nabla u \!+\! b_1^k} \right\|_2^2} {\! + \!\dfrac{{{\gamma _2}}}{2}\left\| {w \!-\! \Delta u \!+\! b_2^k} \right\|_2^2} \right\}$ (9)

 $\left( {{\gamma _1}\Delta + {\gamma _2}{\Delta ^2}} \right)u = \lambda r + {\gamma _1}div\left( {v + b_1^k} \right) + {\gamma _2}div\left( {w + b_2^k} \right)$ (10)

 ${u^{k + 1}} = {{\cal F}^{ - 1}}\left( {{\cal F}\left( {\frac{{ {\lambda r + {\gamma _1}div\left( {v + b_1^k} \right)} { + {\gamma _2}div\left( {w + b_2^k} \right)} }}{{{\gamma _1}\Delta + {\gamma _2}{\Delta ^2}}}} \right)} \right)$ (11)
3.2 v-子问题

$v$ -子问题等价于求解如下优化问题:

 ${v^{k + 1}} = \mathop {\min }\limits_v \alpha {\left\| v \right\|_1} + + \frac{{{\gamma _1}}}{2}\left\| {v - \nabla u + b_1^k} \right\|_2^2$ (12)

 ${v^{k + 1}} = shrink\left( {\nabla {u^{k + 1}} - b_1^k, \frac{\alpha }{{{\gamma _1}}}} \right)$ (13)

 $shrink\left( {{s_\alpha }, {t_\alpha }} \right) = \frac{{{s_\alpha }}}{{\left| {{s_\alpha }} \right|}}\max \left( {{s_\alpha } - {t_\alpha }, 0} \right)$ (14)
3.3 w-子问题

$w$ -子问题满足如下优化条件:

 ${w^{k + 1}} = \mathop {\min }\limits_w \beta {\left\| w \right\|_1} + \frac{{{\gamma _2}}}{2}\left\| {w - \Delta u + b_2^k} \right\|_2^2$ (15)

 ${w^{k + 1}} = shrink\left( {\Delta {u^{k + 1}} - b_2^k, \frac{\beta }{{{\gamma _2}}}} \right)$ (16)

 $\mathop {\min }\limits_x \left\{ {\frac{1}{{2\alpha }}{{\left( {x - h} \right)}^2} + \lambda \left| x \right|} \right\}$ (17)

 Q\left( x \right) = \left\{ {\begin{aligned} & {\frac{1}{{2\alpha }}\left( {x - 2hx + 2\lambda \alpha x + {h^2}} \right),{\text{若}}x > 0}\\ & {\frac{1}{{2\alpha }}\left( {x - 2hx - 2\lambda \alpha x + {h^2}} \right),{\text{若}}x < 0} \end{aligned}} \right. (18)

$h \ge \lambda \alpha$ 时, $Q\left( x \right)$ 的最小值为 $h - \lambda \alpha \ge 0$ ; 当 $h \le - \lambda \alpha$ 时, $Q\left( x \right)$ 的最小值为 $h + \lambda \alpha \le 0$ ; 当 $- \lambda \alpha < h < \lambda \alpha$ 时, $Q\left( x \right)$ 的最小值为0. 所以, 最小值问题(17)可以用如下收缩算子更新:

 {x^{k + 1}} = \left\{ {\begin{aligned} & {{h^k} - \lambda \alpha ,{\text{若}}{h^k} \ge \lambda \alpha } \\ & {0,{\text{若}} - \lambda \alpha < {h^k} < \lambda \alpha }\\ & {{h^k} - \lambda \alpha ,{\text{若}}{h^k} \le - \lambda \alpha } \end{aligned}} \right. (19)

3.4 ${c_1},{c_2}$ -子问题

 ${c_i} \in \mathop {\arg \min }\limits_{{c_i}} \left\{ {\lambda \int {\left| {f - {c_i}} \right|h\left( x \right){\rm d}x} } \right\}$ (20)

 ${c_i} \in \mathop {\arg \min }\limits_{{c_i}} {\left| {f - {c_i}} \right|_h}$ (21)

 ${\left| {f - {c_i}} \right|_h} = \int {\left| {f - {c_i}} \right|h\left( x \right){\rm d}x}$ (22)

 $\left\{ {\begin{array}{*{20}{c}} {\mathop {\min }\limits_{{c_i}, {e_i}} \left\{ {\lambda {{\left| {{e_i}} \right|}_u}} \right\}} \\ {{e_i} = f - \chi {c_i}} \end{array}} \right.$ (23)

 $\mathop {\arg \min }\limits_{{c_i}} \left\{ {\frac{{{\eta _i}}}{2}\int {{{\left( {\chi {c_i} + {e_i} - f - d_i^k} \right)}^2}{\rm d}x} } \right\}$ (24)
 $\mathop {\arg \min }\limits_{{e_i}} \left\{ { {\lambda {{\left| {{e_i}} \right|}_h}} { + \frac{{{\eta _i}}}{2}\int {{{\left( {\chi c_i^{k + 1} + {e_i} - f - d_i^k} \right)}^2}{\rm d}x} } } \right\}$ (25)
 $d_i^{k + 1} = d_i^k - \left( {\chi c_i^{k + 1} + e_i^{k + 1} - f} \right)$ (26)

 $c_i^{k + 1} = \frac{{\int {d_i^k - e_i^k + f{\rm d}x} }}{{\int {{\rm d}x} }}$ (27)
 $e_i^{k + 1} = shrink\left( {f + d_i^k - \chi c_i^{k + 1}, \frac{{\lambda u}}{{{\eta _i}}}} \right)$ (28)

for $\scriptstyle k = 1, 2, 3, \cdots , K$ do

$\scriptstyle {h^k} = {u^k}$ if $\scriptstyle i = 1$ , $\scriptstyle {h^k} = \left( {1 - {u^k}} \right), i = 2$ ,

计算 $\scriptstyle c_i^{k + 1}$ : 用式(27)计算 $\scriptstyle c_i^{k + 1}$ , $\scriptstyle i = 1, 2$ ;

计算 $\scriptstyle e_i^{k + 1}$ : 用式(28)计算 $\scriptstyle e_i^{k + 1}$ , $\scriptstyle i = 1, 2$ ;

更新 $\scriptstyle d_i^{k + 1}$ : 用式(26)计算 $\scriptstyle d_i^{k + 1}$ , $\scriptstyle i = 1, 2$ ;

计算 $\scriptstyle {r^{k + 1}}$ : 用 $\scriptstyle {r^{k + 1}} = \left| {f - c_1^{k + 1}} \right| - \left| {f - c_2^{k + 1}} \right|$ ;

计算 $\scriptstyle {v^{k + 1}}$ : 用式(13)计算 $\scriptstyle {v^{k + 1}}$ ;

计算 $\scriptstyle {w^{k + 1}}$ : 用式(16)计算 $\scriptstyle {w^{k + 1}}$ ;

计算 $\scriptstyle {u^{k + 1}}$ : 用式(11)计算 $\scriptstyle {u^{k + 1}}$ ;

更新 $\scriptstyle b_1^{k + 1}$ : $\scriptstyle b_1^{k + 1} = b_1^k + {v^{k + 1}} - \nabla {u^{k + 1}}$ ;

更新 $\scriptstyle b_2^{k + 1}$ : $\scriptstyle b_2^{k + 1} = b_2^k + {w^{k + 1}} - \Delta {u^{k + 1}}$ ;

end

4 算法部分收敛性分析

 $\left\{ {\begin{array}{*{20}{l}} {\mathop {\min }\limits_{{c_1}, {c_2}, u, {e_1}, {e_2}, v, w} \lambda \int {\left| {{e_1}} \right|} udx \!+ \! \lambda \int {\left| {{e_2}} \right|} \left( {1 \! - \! u} \right)dx \!+ \! \alpha {{\left\| v \right\|}_1} \!+ \beta {{\left\| w \right\|}_1}} \\ {{e_i} = f - \chi {c_i}, v = \nabla u, w = \Delta u, i = 1, 2} \end{array}} \right.$ (29)

 $\begin{array}{l} L\left( {{c_1}, {c_2}, u, {e_1}, {e_2}, v, w, {\mu _1}, {\mu _2}, {\mu _3}, {\mu _4}} \right) \\ = \lambda \displaystyle\int {\left| {{e_1}} \right|} udx + \lambda \displaystyle\int {\left| {{e_2}} \right|} \left( {1 - u} \right)dx + \alpha {\left\| v \right\|_1} + \beta {\left\| w \right\|_1} \\ - \displaystyle\int {{\mu _1}} \left( {v - \nabla u} \right)dx - \int {{\mu _2}} \left( {{e_1} + \chi {c_1} - f} \right)dx \\ - \displaystyle\int {{\mu _3}} \left( {{e_2} + \chi {c_2} - f} \right)dx - \int {{\mu _4}} \left( {w - \Delta u} \right)dx \\ + {\gamma _1}\left\| {v - \nabla u} \right\|_2^2 + {\gamma _2}\left\| {w - \Delta u} \right\|_2^2 \\ + {\gamma _3}\left\| {{e_1} + \chi {c_1} - f} \right\|_2^2 + {\gamma _4}\left\| {{e_2} + \chi {c_2} - f} \right\|_2^2 \end{array}$ (30)

${X^*} = \left( {c_1^*, c_2^*, {u^*}, e_1^*, e_2^*, {v^*}, {w^*}, \mu _1^*, \mu _2^*, \mu _3^*, \mu _4^*} \right)$ 为问题(29)的KKT点, 则点 ${X^*}$ 满足如下的KKT条件:

 $\left\{ {\begin{array}{*{20}{l}} {\int {\mu _2^*dx = 0} } \\ {\int {\mu _3^*dx = 0} } \\ {\lambda \left| {e_1^*} \right| - \lambda \left| {e_2^*} \right| - div\mu _1^* = 0} \\ {0 \in \lambda \partial \left| {e_1^*} \right| \odot {u^*} - \mu _2^*} \\ {0 \in \lambda \partial \left| {e_2^*} \right| \odot \left( {1 - {u^*}} \right) - \mu _3^*} \\ {0 \in \partial \left| {{v^*}} \right| - \mu _1^*} \\ {0 \in \partial \left| {{w^*}} \right| - \mu _4^*} \\ {{v^*} = \nabla {u^*}, {w^*} = \Delta {u^*}} \\ {e_i^* = f - \chi c_i^*, i = 1, 2} \end{array}} \right.$ (31)

 ${\tilde X^k} = \left( c_1^k, c_2^k, {u^k}, e_1^k, e_2^k, {v^k}, {w^k}, {\gamma _1}d_1^k, {\gamma _2}d_2^k, {\gamma _3}b_1^k, {\gamma _4}b_2^k \right)$

 $\mathop {\lim }\limits_{k \to \infty } \left( {{X^k} - {X^{k - 1}}} \right) = 0.$

 $\left( {{\gamma _1}\Delta \! +\! {\gamma _2}{\Delta ^2}} \right){u^{k + 1}} \! =\! \lambda r \!+\! {\gamma _1}div\left( {{v^k}\! +\! b_1^k} \right) \!+ \!{\gamma _2}div\left( {{w^k}\! +\! b_2^k} \right) \!\!\!$ (32)

 $\int {dx\left( {c_i^{k + 1} - c_i^k} \right)} = \int - e_i^k + f + d_i^k - \chi c_i^k{\rm d}x, i = 1, 2$ (33)
 \begin{aligned}[b] & \left( {{\gamma _1}\Delta + {\gamma _2}{\Delta ^2}} \right)\left( {{u^{k + 1}} - {u^k}} \right) = \lambda r + {\gamma _1}div\left( {{v^k} + b_1^k} \right) +\\ & {\gamma _2}div\left( {{w^k} + b_2^k} \right) - \left( {{\gamma _1}\Delta + {\gamma _2}{\Delta ^2}} \right){u^k} \end{aligned} (34)
 $e_1^{k + 1} - e_1^k = shrink\left( {f + d_1^k - \chi c_1^{k + 1}, \frac{{\lambda u}}{{{\gamma _3}}}} \right) - e_1^k$ (35)
 $e_2^{k + 1} - e_2^k = shrink\left( {f + d_2^k - \chi c_2^{k + 1}, \frac{{\lambda \left( {1 - u} \right)}}{{{\gamma _4}}}} \right) - e_2^k$ (36)
 ${v^{k + 1}} - {v^k} = shrink\left( {\nabla {u^{k + 1}} - b_1^k, \frac{\alpha }{{{\gamma _1}}}} \right) - {v^k}$ (37)
 ${w^{k + 1}} - {w^k} = shrink\left( {\Delta {u^{k + 1}} - b_2^k, \frac{\beta }{{{\gamma _2}}}} \right) - {w^k}$ (38)

 $e_1^* = shrink\left( {e_1^* + \frac{{\mu _2^*}}{{{\gamma _3}}}, \frac{{\lambda {u^*}}}{{{\gamma _3}}}} \right)$ (39)
 $e_2^* = shrink\left( {e_2^* + \frac{{\mu _3^*}}{{{\gamma _4}}}, \frac{{\lambda \left( {1 - {u^*}} \right)}}{{{\gamma _4}}}} \right)$ (40)
 ${v^*} = shrink\left( {{v^*} + \frac{{\mu _1^*}}{{{\gamma _1}}}, \frac{\alpha }{{{\gamma _1}}}} \right)$ (41)
 ${w^*} = shrink\left( {{w^*} + \frac{{\mu _4^*}}{{{\gamma _2}}}, \frac{\beta }{{{\gamma _2}}}} \right)$ (42)

5 实验结果

 $\frac{{{{\left\| {{u^{k + 1}} - {u^k}} \right\|}_2}}}{{{{\left\| {{u^{k + 1}}} \right\|}_2}}} < tol$

 图 1 CV模型和高阶正则模型对飞机图像的检测效果图

 图 2 CV模型对自然风景图像的分割效果图

 图 3 高阶正则模型对自然风景图像的分割效果图

 图 4 CV模型和高阶正则模型对自然风景图像分割的Loss函数曲线收敛图

 图 5 高阶正则模型对病人肺部CT图像的分割效果图

 图 6 CV模型和高阶正则模型对病人肺部CT图像分割的Loss函数曲线收敛图

6 总结与展望

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