Efficient Color Image Segmentation via Quaternion-based $L_1/L_2$ Regularization

Abstract

Color image segmentation is a key technology in image processing. In this paper, a two-stage image segmentation method is proposed that is based on the nonconvex $L_1/L_2$ approximation of the Mumford-Shah (MS) model. Wherein, the nonconvex regularization term $L_1/L_2$ on the gradient can approximate the Hausdorff measure and extract more boundary information. The first stage is to solve the nonconvex variant of the MS model and to obtain the smoothed image u. In our framework, we utilize the semi-proximal alternating direction method of multipliers (sPADMM) to sufficiently solve the proposed model. The second stage is segmenting the smoothed u into different phases with thresholds determined by the threshold clustering method. To better deal with the inherent color structures within different channels, we also apply the quaternion representation of the color image. Quantitative and qualitative results demonstrate clearly that our method is better than some state-of-the-art color image segmentation methods.

Appendix

To prepare for convergence analysis, we summarize some equivalent conditions for strong convexity and Lipschitz smooth functions in Lemma 4 and Lemma 5, respectively.

Lemma 4

A function f(x) is called strongly convex with parameter $\beta$ if and only if one of the following conditions holds

• (a) $g(x)=f(x)-\frac{\beta }{2}\left\| x\right\| _2^2$ is convex;

• (b) $\left\langle \nabla f(x)-\nabla f(y),x-y\right\rangle \ge \beta \left\| x-y\right\| _2^2$, $\forall x,y$;

• (c) $f(y)\ge f(x)+\left\langle \nabla f(x),y-x\right\rangle +\frac{\beta }{2}\left\| y-x\right\| _2^2$, $\forall x,y$.

Lemma 5

The gradient of f(x) is Lipschitz continuous with parameter $L>0$ if and only if one of the following conditions holds

• (a) $\left\| \nabla f(x)-\nabla f(y)\right\| _2 \le L\left\| x-y\right\| _2$, $\forall x,y$;

• (b) $g(x)=\frac{L}{2}\left\| x\right\| _2^2-f(x)$ is convex;

• (c) $f(y)\le f(x)+\left\langle \nabla f(x),y-x\right\rangle +\frac{L}{2}\left\| y-x\right\| _2^2$, $\forall x,y$.

1.1 Proof of Lemma 1

Proof

The h-subproblem in (13) follows the optimality condition that

$$\begin{aligned} \begin{aligned} -\frac{\left\| \nabla u^{k+1}\right\| _1}{\left\| h^{k+1}\right\| ^3_2 }h^{k+1}+\rho _2(h^{k+1}-\nabla u^{k+1}-r^k)+Z_2(h^{k+1}-h^k)=0. \end{aligned} \end{aligned}$$

(31)

By using the dual update $r^{k+1}=r^k+\nabla u^{k+1}-h^{k+1}$, we can obtain

$$\begin{aligned} \begin{aligned} r^{k+1}=-\frac{\left\| \nabla u^{k+1}\right\| _1}{\rho _2\left\| h^{k+1}\right\| _2^3}h^{k+1}+\frac{1}{\rho _2}Z_2(h^{k+1}-h^k). \end{aligned} \end{aligned}$$

(32)

We can estimate

$$\begin{aligned} \begin{aligned} \left\| r^{k+1}-r^{k}\right\| _2&\le \frac{1}{\rho _2}\left\| \frac{\left\| \nabla u^{k+1}\right\| _1}{\left\| h^{k+1}\right\| _2^3}h^{k+1}-\frac{\left\| \nabla u^k\right\| _1}{\left\| h^k\right\| ^3_2}h^k \right\| _2\\&\quad +\frac{\left\| Z_2\right\| _2}{\rho _2}\left( \left\| h^{k+1}-h^k\right\| _2+\left\| h^k-h^{k-1}\right\| _2\right) \\&\le \frac{1}{\rho _2} \left( \frac{\left| \left\| \nabla u^{k+1}\right\| _1-\left\| \nabla u^k\right\| _1\right| }{\left\| h^{k+1}\right\| _2^3}+\left\| \nabla u^k\right\| _1 \left\| \frac{h^{k+1}}{\left\| h^{k+1}\right\| _2^3}-\frac{h^k}{\left\| h^k\right\| _2^3}\right\| \right) \\&\quad +\frac{1}{\rho _2}\left( \left\| Z_2\right\| _2\left\| h^{k+1}-h^k\right\| _2+\left\| Z_2\right\| _2\left\| h^k-h^{k-1}\right\| _2 \right) . \end{aligned} \end{aligned}$$

(33)

For the first term in (33), $\left\| x\right\| _1 \le \sqrt{l}\left\| x\right\| _2$ for a vector x of the length of l and $\left\| \nabla \right\| _2\le 2\sqrt{2}$ are used, thus resulting to

$$\begin{aligned} \begin{aligned} \left| \left\| \nabla u^{k+1}\right\| _1 -\left\| \nabla u^k\right\| _1 \right|&\le \left\| \nabla (u^{k+1}-u^k)\right\| _1 \le \sqrt{2mn}\left\| \nabla (u^{k+1}-u^k)\right\| _2\\&\le \sqrt{2mn}\cdot \left\| \nabla \right\| _2\cdot \left\| u^{k+1}-u^k\right\| _2\\&\le 4\sqrt{mn}\left\| u^{k+1}-u^k\right\| _2. \end{aligned} \end{aligned}$$

(34)

Here are notes that $u\in {\mathbb {R}}^{m\times n}$ and $\nabla u\in {\mathbb {R}}^{m\times n}$ (thus of length 2mn). Then for the first term in (33), we have

$$\begin{aligned} \begin{aligned} \left\| \nabla u^k\right\| _1\left\| \frac{h^{k+1}}{\left\| h^{k+1}\right\| _2^3}-\frac{h^k}{\left\| h^k\right\| _2^3}\right\| _2\le \frac{2M}{\epsilon ^3}\left\| h^{k+1}-h^k\right\| _2, \end{aligned} \end{aligned}$$

(35)

for which we use the following inequality

$$\begin{aligned} \begin{aligned} \left\| \nabla \left( \frac{1}{\left\| x\right\| _2}\right) -\nabla \left( \frac{1}{\left\| y\right\| _2}\right) \right\| _2\le \frac{2}{\epsilon ^3}\left\| x-y\right\| _2. \end{aligned} \end{aligned}$$

(36)

For x and y, $\left\| x\right\| _2 \ge \epsilon$ and $\left\| y\right\| _2 \ge \epsilon$ are satisfied. Then by combining (33)-(35) and using the Cauchy-Schwarz inequality, we can obtain (23). $\square$

1.2 Proof of Lemma 2

Proof

With the same method in [53], we can easily verify that there is a constant ${\bar{c}}_1>0$, such that

$$\begin{aligned} \begin{aligned} {\mathcal {L}}_1 (u^{k+1},h^{k+1};r^{k+1})-{\mathcal {L}}_1 (u^k,h^k;r^k) \le - \frac{{\bar{c}}_1}{2}\left\| u^{k+1}-u^k\right\| _2^2. \end{aligned} \end{aligned}$$

(37)

Actually, the only difference between our augmented Lagrangian function ${\mathcal {L}}_1$ and that in [53] is the term $\frac{\mu }{2}\left\| \nabla u\right\| _2^2$. As this term is convex and combined with Lemma 4, then apparently (37) holds.

Denote $a=\left\| u^{k+1}\right\| _1$ and $L=\frac{2M}{\epsilon ^3}$. Lemma 5 (c) and (36) lead to

$$\begin{aligned} \frac{a}{\left\| h^{k+1}\right\| _2} \le \frac{a}{\left\| h^{k}\right\| _2}-\left\langle \frac{ah^k}{\left\| h^k\right\| _2^3},h^{k+1}-h^k\right\rangle +\frac{L}{2}\left\| h^{k+1}-h^k\right\| _2^2. \end{aligned}$$

(38)

Assigning $z=\nabla u^{k+1}+r^k$ and utilizing the optimality condition of $h^{k+1}$, we can obtain

$$\begin{aligned} \begin{aligned}&\frac{\rho _2}{2}\left( \left\| h^{k+1}-z\right\| _2^2-\left\| h^k-z\right\| _2^2\right) \\&\quad =\frac{\rho _2}{2}\left( \left\| h^{k+1}\right\| _2^2-\left\| h^k\right\| _2^2\right) -\left\langle -\frac{ah^{k+1}}{\left\| h^{k+1}\right\| _2^3}+\rho _2 h^{k+1}+Z_2(h^{k+1}-h^k), h^{k+1}-h^k\right\rangle \\&\quad =\left\langle -\frac{ah^{k+1}}{\left\| h^{k+1}\right\| _2^3}, h^{k+1}-h^k\right\rangle -\frac{\rho _2}{2}\left\| h^{k+1}-h^k\right\| _2^2-\left\| h^{k+1}-h^k\right\| _{Z_2}^2\\&\quad \le \left\langle -\frac{ah^{k+1}}{\left\| h^{k+1}\right\| _2^3}, h^{k+1}-h^k\right\rangle -\frac{\rho _2+2\alpha }{2}\left\| h^{k+1}-h^k\right\| _{Z_2}^2, \end{aligned} \end{aligned}$$

(39)

where $\alpha \ge 0$ is the smallest eigenvalue of $Z_2$. By putting together (38) and (39), we get

$$\begin{aligned} \begin{aligned}&{\mathcal {L}}_1 (u^{k+1},h^{k+1};r^{k+1})-{\mathcal {L}}_1 (u^k,h^k;r^k) \\&\quad \le \left\langle \frac{ah^{k+1}}{\left\| h^{k+1}\right\| _2^3}-\frac{ah^k}{\left\| h^k\right\| _2^3}, h^{k+1}-h^k \right\rangle -\frac{\rho _2+2\alpha -L}{2}\left\| h^{k+1}-h^k\right\| _2^2 \\&\quad \le \left\| \frac{ah^{k+1}}{\left\| h^{k+1}\right\| _2^3}-\frac{ah^k}{\left\| h^k\right\| _2^3}\right\| _2 \left\| h^{k+1}-h^k\right\| _2-\frac{\rho _2+2\alpha -L}{2}\left\| h^{k+1}-h^k\right\| _2^2\\&\quad \le -\frac{\rho _2+2\alpha -3L}{2}\left\| h^{k+1}-h^k\right\| _2^2. \end{aligned} \end{aligned}$$

(40)

Ultimately, from the update of $r^k$, we calculate

$$\begin{aligned} \begin{aligned}&{\mathcal {L}}_1 (u^{k+1},h^{k+1};r^{k+1})-{\mathcal {L}}_1 (u^{k+1},h^{k+1};r^k)\\&\quad =\frac{\rho _2+2\alpha }{2}\left( \left\| r^k\right\| _2^2-\left\| r^{k+1}-2r^k\right\| _2^2\right) \le \frac{\rho _2+2\alpha }{2}\left\| r^{k+1}-r^k\right\| _2^2. \end{aligned} \end{aligned}$$

(41)

When combining Lemma 1 with inequalities (37), (40), and (41), we get

$$\begin{aligned} \begin{aligned} {\mathcal {L}}_1 (u^{k+1},h^{k+1};r^{k+1})&\le {\mathcal {L}}_1 (u^k,h^k;r^k)-c_1 \left\| u^{k+1}-u^{k}\right\| _2^2 \\&\quad -c_2'\left\| h^{k+1}-h^{k}\right\| _2^2+\frac{3\left\| Z_2\right\| _2^3}{\rho ^2_2}\left\| h^k-h^{k-1}\right\| _2^2, \end{aligned} \end{aligned}$$

(42)

where $c_1=\frac{{\bar{c}}_1}{2}-\frac{24mn}{\rho _2\epsilon ^4}$ and $c_2'=\frac{\rho _2-3L+2\alpha }{2}-3\left( \frac{2M}{\rho _2\epsilon ^3}+\frac{\left\| Z_2\right\| _2}{\rho _2}\right) ^2$. Then it follows from (42) that

$$\begin{aligned} \begin{aligned}&{\mathcal {L}}^{k+1}_h (u^{k+1},h^{k+1};r^{k+1})\\&\quad \le {\mathcal {L}}^{k+1}_h (u^k,h^k;r^k)-c_1 \left\| u^{k+1}-u^{k}\right\| _2^2 -c_2\left\| h^{k+1}-h^{k}\right\| _2^2-c_3\left\| h^k-h^{k-1}\right\| _2^2, \end{aligned} \end{aligned}$$

(43)

where $c_2=c_2'-\frac{1}{2}$ and $c_3=\frac{1}{2}-\frac{3\left\| Z_2\right\| _2^2}{\rho ^2_2}$. For sufficiently large $\rho _2$, we have $c_1, c_2, c_3>0$. $\square$

1.3 Proof of Lemma 3

Proof

The optimality condition of sPADMM iteration are as follows

$$\begin{aligned}&\frac{p^{k+1}}{\left\| h^k\right\| _2}+\lambda A^T(Au^{k+1}-f)+\rho _2\nabla ^T(\nabla u^{k+1}-h^k+r^k) \nonumber \\&\qquad \qquad \quad +\mu \nabla ^T\nabla u^{k+1}+Z_1(u^{k+1}-u^k)=0, \end{aligned}$$

(44a)

$$\begin{aligned}&-\frac{\left\| \nabla u^{k+1}\right\| _1}{\left\| h^{k+1}\right\| _2^3}+\rho _2(h^{k+1}-\nabla u^{k+1}-r^k)+Z_2(h^{k+1}-h^k)=0, \end{aligned}$$

(44b)

$$\begin{aligned}&r^{k+1}=r^k+\nabla u^{k+1}-h^{k+1}, \end{aligned}$$

(44c)

where $p^{k+1}\in \partial \left\| \nabla u^{k+1}\right\| _1$. Let $\eta _1^{k+1}, \eta _2^{k+1}, \eta _3^{k+1}$ be

$$\begin{aligned}&\eta _1^{k+1}=\frac{p^{k+1}}{\left\| h^{k+1}\right\| _2}+\lambda A^T(Au^{k+1}-f)+\mu \nabla ^T\nabla u^{k+1}\nonumber \\&\qquad \qquad \quad +\rho _2\nabla ^T(\nabla u^{k+1}-h^k+r^k), \end{aligned}$$

(45a)

$$\begin{aligned}&\eta _2^{k+1}=-\frac{\left\| \nabla u^{k+1}\right\| _1}{\left\| h^{k+1}\right\| _2^3}+\rho _2(h^{k+1}-\nabla u^{k+1}-r^k), \end{aligned}$$

(45b)

$$\begin{aligned}&\eta _3^{k+1}=\rho _2(\nabla u^{k+1}-h^{k+1}). \end{aligned}$$

(45c)

Obviously, we have

$$\begin{aligned} \begin{aligned} {\eta }_1^{k+1} \in \partial _{u} {\mathcal {L}}_1(u^{k+1},h^{k+1};r^{k+1}), \\ {\eta }_2^{k+1} \in \partial _{h} {\mathcal {L}}_1(u^{k+1},h^{k+1};r^{k+1}),\\ {\eta }_3^{k+1} \in \partial _{r} {\mathcal {L}}_1(u^{k+1},h^{k+1};r^{k+1}). \end{aligned} \end{aligned}$$

By combining (42) and (43), we have

$$\begin{aligned}&\eta _1^{k+1}=\left( \frac{p^{k+1}}{\left\| h^{k+1}\right\| _2}-\frac{p^{k+1}}{\left\| h^k\right\| _2}\right) -\rho _2\nabla ^T(h^{k+1}-h^k)\nonumber \\&\quad +\rho _2\nabla ^T(r^{k+1}-r^k)-Z_1(u^{k+1}-u^k), \end{aligned}$$

(46a)

$$\begin{aligned}&\eta _2^{k+1}=-\rho _2(r^{k+1}-r^k)-Z_2(h^{k+1}-h^k), \end{aligned}$$

(46b)

$$\begin{aligned}&\eta _3^{k+1}=\rho _2(r^{k+1}-r^k). \end{aligned}$$

(46c)

The subgradient chain rule [54] implies that $\partial \left\| u\right\| _1=\nabla ^Tq$, wherein

$$\begin{aligned} q=\{q\vert \left\langle q, \nabla u\right\rangle _Y =\left\| \nabla u\right\| _1, \vert q_{ijk}\vert , \forall i,j,k\}. \end{aligned}$$

Consequently, we know what the upper bound is for $\left\| p^{k+1}\right\| _2\le \left\| \nabla ^T\right\| _2\left\| q^{k+1}\right\| _2\le 2\sqrt{2mn}$. Based on simple calculations, it appears

$$\begin{aligned} \begin{aligned} \left\| \frac{p^{k+1}}{\left\| h^k\right\| _2}-\frac{p^{k+1}}{\left\| h^{k+1}\right\| _2}\right\| _2&=\left| \frac{1}{\left\| h^k\right\| _2}-\frac{1}{\left\| h^{k+1}\right\| _2}\right| \left\| p^{k+1}\right\| _2 \\&\le \frac{1}{\epsilon ^2}\left\| h^{k+1}-h^k\right\| _2\left\| p^{k+1}\right\| _2 \\&\le \frac{2\sqrt{2mn}}{\epsilon ^2}\left\| h^{k+1}-h^k\right\| _2. \end{aligned} \end{aligned}$$

(47)

Lastly, by setting $\gamma =\max \{27\rho ^2_2,3\left\| Z_1\right\| _2^2,3(\frac{2\sqrt{2mn}}{\epsilon ^2}+2\sqrt{2}\rho _2)^2+2\left\| Z_2\right\| _2^2\}$ our claim follows instantly. $\square$