Image Segmentation Using a Local GMM in a Variational Framework

Abstract

In this paper, we propose a new variational framework to solve the Gaussian mixture model (GMM) based methods for image segmentation by employing the convex relaxation approach. After relaxing the indicator function in GMM, flexible spatial regularization can be adopted and efficient segmentation can be achieved. To demonstrate the superiority of the proposed framework, the global, local intensity information and the spatial smoothness are integrated into a new model, and it can work well on images with inhomogeneous intensity and noise. Compared to classical GMM, numerical experiments have demonstrated that our algorithm can achieve promising segmentation performance for images degraded by intensity inhomogeneity and noise.

Appendices

Appendix A: Proof of Lemma 1

Let $\mathcal{E}(\mathbf{u})=\sum_{k=1}^{K}(\mathcal{B}_{k}-\log\mathcal {A}_{k})u_{k}+\sum_{k=1}^{K}u_{k}\log u_{k}$. We use Lagrange multiplier method to calculate the minimal value of $\mathcal{E}$. Denote Lagrange functional $L(\mathbf{u})=\mathcal{E}(\mathbf {u})+\lambda(\sum_{k=1}^{K}u_{k}-1)$, then

$$ \frac{\delta L}{\delta u_k}=\mathcal{B}_k+\log\frac{u_k}{\mathcal {A}_k}+1+\lambda. $$

By setting $\frac{\delta L}{\delta u_{k}}=0$, and one can get a stationary point u ^∗ with component function

$$ u_k^{*}=\mathcal{A}_k \exp(-\mathcal{B}_k-\lambda-1). $$

(17)

Please note u ^∗∈Δ, we sum up k from 1 to K in the both sides of (17) and thus we obtain

$$ \lambda+1=\log\sum_{k=1}^{K} \mathcal{A}_k\exp(-\mathcal{B}_k). $$

(18)

Equations (17) and (18) produce

$$ u_k^{*}=\frac{\mathcal{A}_k\exp(-\mathcal{B}_k)}{\sum_{l=1}^{K}\mathcal {A}_l\exp(-\mathcal{B}_l)}. $$

(19)

It is easy to check that u ^∗ is a minimizer of $\mathcal{E}$. Substituting Eq. (19) into $\mathcal{E}$ and one can finally get

$$ \mathcal{E}\bigl(\mathbf{u}^{*}\bigr)=-\log\sum _{k=1}^{K}\mathcal{A}_k\exp (- \mathcal{B}_k), $$

which completes the proof.

Appendix B: Proof of Proposition 1

According to the first formulation of iteration scheme (4) and Lemma 1, we have

$$ \left \{ \begin{array}{l} -\mathcal{L}(\varTheta^{\nu})=\tilde{\mathcal{E}}(\varTheta^{\nu},\mathbf {u}^{\nu+1}),\\[5pt] -\mathcal{L}(\varTheta^{\nu+1})=\tilde{\mathcal{E}}(\varTheta^{\nu +1},\mathbf{u}^{\nu+2}). \end{array} \right . $$

On the other hand, the two equations in (4) can provide

$$ \tilde{\mathcal{E}}\bigl(\varTheta^{\nu},\mathbf{u}^{\nu+1}\bigr) \geq \tilde {\mathcal{E}}\bigl(\varTheta^{\nu+1},\mathbf{u}^{\nu+1} \bigr) \geq \tilde {\mathcal{E}}\bigl(\varTheta^{\nu+1}, \mathbf{u}^{\nu+2}\bigr), $$

and thus $-\mathcal{L}(\varTheta^{\nu+1})\leq -\mathcal{L}(\varTheta^{\nu})$.

Appendix C: Proof of Proposition 2

We use distribution function to show this.

Thus $p_{\mathfrak{Z}}(z)=\sum_{k=1}^{K}\frac{\gamma_{k}}{\sqrt{2\pi}\sigma_{k}\beta(x)}\exp \{-\frac{ [z-c_{k}\beta (x) ]^{2}}{2\sigma_{k}^{2}\beta^{2}(x)} \}$.

Appendix D: Proof of Proposition 3

The proof is motivated by [40]. We first give a schematic diagram in Fig. 9. Sine we suppose ω is relative small and Γ is smooth enough, then we can regard the curve AB as a line segment. As a result, one can get h=ωcosθ,S _shadow=θω ²−ω ²sinθcosθ. For a sufficient small ω, the intersections of more that two boundary curves can be ignored compared to the total length of Γ. According to all the suppositions, then we have

Fig. 9

A schematic diagram for the regularization term. Shadow area is the penalty area defined in (7)

Appendix E: The derivation of Eq. (12)

This derivation is very similar to the proof of Lemma 1. The Lagrangian function L for problem (10) is given by

$$ L(\mathbf{u},d)=\mathcal{J}(\mathbf{u})+\int_{\varOmega} d(x) \Biggl[\sum_{k=1}^K u_k(x)-1 \Biggr] \mathrm {d}x, $$

where d is the Lagrangian multiplier. Then

$$ \begin{aligned}[c] \frac{\delta L}{\delta u_k}&=\frac{1}{2(\sigma_k^2)^{\nu}}\int_{\varOmega }G_{\sigma}(y-x) \biggl[\frac{f(x)}{\beta^{\nu}(y)}-c_k^{\nu} \biggr]^2 \mathrm {d}y\\ &\quad {}-\log\frac{\gamma_k^{\nu}}{\sqrt{(\sigma_k^2)^{\nu}}}+\log u_k(x)+1+\lambda r_k^{\nu}(x)+d(x). \end{aligned} $$

By setting $\frac{\delta L}{\delta u_{k}}=0$ and with simplification, it becomes

(20)

exp{−1−d(x)} in (20) can be obtained by summing up k from 1 to K in the both sides of this equation and using the fact $\sum_{k=1}^{K}u_{k}(x)=1$. With the notations used in Sect. 3.3, we have

$$ \exp \bigl\{-1-d(x) \bigr\}=\frac{1}{\sum_{k=1}^K q_k^{\nu}(x)}. $$

Substituting it back to (20), this immediately leads to the updating Eq. (12).

Appendix F: The derivation of Eq. (13)

We only give the details of calculating the last equation in (13), others can be obtained in the same manner. Since

we get

By setting $\frac{\delta\mathcal{J}}{\delta\beta}=0$, then it becomes

$$ \beta^2(y)+s^{\nu+1}(y)\beta(y)-t^{\nu+1}(y)=0. $$

It is easy to get $\beta(y)=\frac{-s^{\nu +1}(y)+\sqrt{ [s^{\nu+1}(y) ]^{2}+4t^{\nu+1}(y)}}{2}$ is a positive root since we assume β(y)>0.