Jensen Shannon Divergence as Reduced Reference Measure for Image Denoising

Abstract

This paper focuses on the use the Jensen Shannon divergence for guiding denoising. In particular, it aims at detecting those image regions where noise is masked; denoising is then inhibited where it is useless from the visual point of view. To this aim a reduced reference version of the Jensen Shannon divergence is introduced and it is used for determining a denoising map. The latter separates those image pixels that require to be denoised from those that have to be leaved unaltered. Experimental results show that the proposed method allows to improve denoising performance of some simple and conventional denoisers, in terms of both peak signal to noise ratio (PSNR) and structural similarity index (SSIM). In addition, it can contribute to reduce the computational effort of some performing denoisers, while preserving the visual quality of denoised images.

Appendix

Let us consider Eqs. (8) and (9). Since $\frac{2x}{2 + x} \le \log (1 + x) \le x$, then $\frac{2 (a + b)}{(6\sigma + a + b)^2} \le \bar{t} \le \frac{1}{3 \sigma }$ and then

$$\begin{aligned} D_{JS} \le \frac{1}{2} \bigg ( e^{-\frac{2a (a + b)}{(6\sigma + a + b)^2}} + e^{-\frac{2b (a + b)}{6\sigma + a + b)^2}} \bigg ) \bigg ( 1 + \frac{\sigma _y^2}{2} \bigg ( \frac{1}{3\sigma } \bigg )^2 \bigg ), \end{aligned}$$

(14)

where $a = M_X - E [y]$ e $b = E [y] - m_X$.

In general, for $x \in [0, \sim 1.59]$, it holds $e^{-x} \le 1-\frac{x}{2}$. Hence, by assuming that

$$\begin{aligned} \frac{2a(a + b)}{(6\sigma + a + b)^2} \le \frac{3}{2}, \quad \frac{2b(a + b)}{(6\sigma + a + b)^2} \le \frac{3}{2}, \end{aligned}$$

(15)

$a + b\sim 6\sigma _I$ and $6\sigma + a + b \sim 6\sigma _y$, we have

$$\begin{aligned} D_{JS} \le \frac{1}{2} \bigg ( 1 - \frac{(a + b)^2}{2(6\sigma + a + b)^2} \bigg ) \bigg ( 1 + \frac{\sigma _y^2}{18\sigma ^2} \bigg ). \end{aligned}$$

Since $\sigma _Y^2 =\sigma _X^2 +\sigma ^2$, then

$$\begin{aligned} \bigg ( 1 - \frac{36 \sigma _X^2}{72 \sigma _Y^2}) \bigg ) \bigg ( 1 + \frac{\sigma _Y^2}{18\sigma ^2} \bigg ) \le 1 \Leftrightarrow - \frac{\sigma _X^2}{2\sigma _Y^2} + \frac{\sigma _Y^2}{18 \sigma ^2} - \frac{\sigma _X^2 \sigma _Y^2}{36\sigma _X^2 \sigma ^2} \le 0 \end{aligned}$$

$$\begin{aligned} \frac{17 - \sqrt{217}}{2}\sigma ^2 \le \sigma _Y^2 \le \frac{17 + \sqrt{217}}{2} \sigma ^2, \; \end{aligned}$$

i.e. $\quad 0.14 \le \frac{\sigma _X^2}{\sigma ^2} \le 14.87, \quad \text {with} \quad \sigma ^2 \ge 1.$

Observation

   If $\min (a,b) \le \max (a,b) \le 3 \min (a,b)$, then constraints in Eq. (15) are satisfied. More in general, by setting $\sigma = k(a + b), k \in \mathbf {R}$, then

$$\begin{aligned} \frac{2a(a + b)}{(6\sigma + a + b)^2} \le \frac{2(a + b)^2}{(a + b)(6k + 1)^2} \le \frac{3}{2} \Leftrightarrow k \ge \frac{2-\sqrt{3}}{6 \sqrt{3}} \Rightarrow \frac{\sigma _X}{\sigma } \le \frac{\sqrt{3}}{2 - \sqrt{3} } \sim 5.6. \end{aligned}$$