Algorithm 1: Proposed Model
1: Let $a_0$ be an ideal image. Define the noisy image corrupted with Gaussian noise ‘a’, noise factor n, and standard deviation σ using
$a=a_0+n$									(1)
2: Evaluate the conditional probability distribution using
$\mathrm{P}\left(a_0/a\right)={\left(2\pi {\sigma }^2\right)}^{-1}{e}^{-\left({\left(a-a_0\right)}^2/2{\sigma }^2\right)}$									(2)
3: Determine the observed noise version Pn given P0, the noiseless patch of a0 using
$\mathrm{P}\left({\mathrm{P}}_n/{\mathrm{P}}_0\right)=c{e}^{-\left({\left(P_n-P_0\right)}^2/2{\sigma }^2\right)}$where $c={\left(2\pi {\sigma }^2\right)}^{-1}$									(3)
4: Define the Euclidean norm of ${\mathrm{P}}_0$ as $\Vert {\mathrm{P}}_0\Vert$. Compute $\mathrm{P}\left({\mathrm{P}}_n/{\mathrm{P}}_0\right)$ using the Bayes rule.
$\mathrm{P}\left({\mathrm{P}}_n/{\mathrm{P}}_0\right)= \mathrm{P}\left({\mathrm{P}}_n/\Vert {\mathrm{P}}_0\Vert \right)\mathrm{P}\left(\Vert {\mathrm{P}}_0\Vert \right)/\mathrm{P}\left({\mathrm{P}}_n\right)$									(4)
5: For a normalization constant $\alpha$, the initial cluster ${\mathrm{Q}}_0$, iteration t, find the direct path, $\mathrm{P}\left(\mathrm{Q}\right)$ to construct cluster $\mathrm{Q}$ of Gaussian samples.
$\mathrm{P}\left(\mathrm{Q}\right)=\alpha c{\mathrm{P}}_0{e}^{\left({\mathrm{Q}}_0-{\mathrm{P}}_0\right)t-1}\left({\mathrm{Q}}_0-{\mathrm{P}}_0\right)/2$									(5)
6: Evaluate $\arg {\mathrm{P}}_0\max \mathrm{P}\left({\mathrm{P}}_n/{\mathrm{P}}_0\right)$.
$\arg {\mathrm{P}}_0\max \mathrm{P}\left({\mathrm{P}}_n/{\mathrm{P}}_0\right)=\arg {\mathrm{P}}_0\max e^{-{(\Vert {\mathrm{P}}_0-{\mathrm{P}}_n\Vert )}^2}c{\mathrm{P}}_0\Vert {\mathrm{P}}_0-{\mathrm{P}}_n\Vert {)}^2/4{\alpha }^2$									(6)
7: Find the posterior estimation $C_{p_n}.$
$C_{p_n}\cong C_{p_0}+{\alpha }^2$									(7)
8: Determine the maximum posterior estimation using
$\arg {\mathrm{P}}_0\max \mathrm{P}\left({\mathrm{P}}_n/{\mathrm{P}}_0\right)=\arg {\mathrm{P}}_0\max e^{-{(\Vert {\mathrm{P}}_0-{\mathrm{P}}_n\Vert )}^2}c{\mathrm{P}}_0\Vert {\mathrm{P}}_0-{\mathrm{P}}_n\Vert {)}^t/{\alpha }^2$									(8)
9: Evaluate $C_{p_n}.$
$C_{p_n}={\mathrm{P}}_n+\left(C_{p_{\left(n-1\right)}}-{\alpha }^2\right)c{\mathrm{P}}_n\left({\mathrm{P}}_n-{\mathrm{P}}_0\right)$									(9)
10: Update $C_{p_n}$ using
$C_{p_n}\cong \frac{1}{\mathrm{P}{\left({\widehat{p}}_n\right)}^{-1}}{\displaystyle \sum }_{\widehat{\mathrm{Q}}\in \mathrm{P}\left({\widehat{p}}_n\right)}\left({\widehat{\mathrm{Q}}}_{ }-\overline{{\widehat{\mathrm{P}}}_n}\right) {\left({\widehat{\mathrm{Q}}}_{ }-\overline{{\widehat{\mathrm{P}}}_n}\right)}^t\mathrm{Q}$									(10)
11: Estimate $C_{p_0}$using sampling estimates, neighbor patches ${\widehat{\mathrm{Q}}}_1$.
$C_{p_0}=\frac{1}{\mathrm{P}{\left({\widehat{p}}_1\right)}^{ }}{\displaystyle \sum }_{\widehat{\mathrm{Q}}\in \mathrm{P}\left({\widehat{p}}_1\right)}\left({\widehat{\mathrm{Q}}}_1-\overline{{\widehat{\mathrm{P}}}_1}\right) {\left({\widehat{\mathrm{Q}}}_1-\overline{{\widehat{\mathrm{P}}}_1}\right)}^t$									(11)
$\overline{{\widehat{\mathrm{P}}}_1}\cong \frac{1}{\mathrm{P}\left({\widehat{p}}_1\right)}{\displaystyle \sum }_{\widehat{\mathrm{Q}}\in \mathrm{P}\left({\widehat{p}}_1\right)}{\widehat{\mathrm{Q}}}_1$									(12)
12: Approximate $C_{p_n}$ using the covariance matrix of noisy patches
$c_{p_n}=c_{p_0}+{\nabla }^{2}I$									(13)
13: Apply the Bayes method.
${\widehat{\mathrm{P}}}_2=\overline{{\widehat{\mathrm{P}}}_1}+c_{{\widehat{p}}_1}{\left[c_{{\widehat{p}}_1}+{\nabla }^{2}I\right]}^{-1}({\mathrm{P}}_n-{\overline{\mathrm{P}}}_{n1})$									(14)
14: Perform grey level image—denoising process for different values of $\sigma$.
Type	$\mathit{\sigma }$= 2	$\mathit{\sigma }$= 5	$\mathit{\sigma }$= 10	$\mathit{\sigma }$= 20	$\mathit{\sigma }$= 30	$\mathit{\sigma }$= 40	$\mathit{\sigma }$= 60	$\mathit{\sigma }$= 80	$\mathit{\sigma }$= 100
Classic	46.14	40.89	35.57	32.46	30.31	27.63	26.03	26.03	25.41
Iteration	46.19	40.12	36.62	33.60	29.67	27.51	26.24	26.24	25.72
Patch size:
$e_1=\{\begin{array}{c}3, \sigma \le 20\\ 5, 20<\sigma \le 50\\ 7, \mathrm{otherwise}\end{array}$ $e_2=\{\begin{array}{c}3, \sigma \le 50\\ 5, 50<\sigma \le 70\\ 7, \mathrm{otherwise}\end{array}$									(15)
15: Find a similar pattern retained:
${\mathrm{N}}_1=3e_1^2$ ${\mathrm{N}}_2=3e_2^2$									(16)
16: Compute the size of the search area:
${\mathsf{\lambda }}_1=$ ${\mathrm{N}}_1/2$, ${\mathsf{\lambda }}_2={\mathrm{N}}_2/2$									(17)
17: Set the threshold as
${\mathsf{\tau }}_0=16e_2^2$									(18)
$C_{p_1}\cong \frac{1}{\mathrm{P}{\left({\widehat{p}}_n\right)}^{-1}}{\displaystyle \sum }_{\widehat{\mathrm{Q}}\in \mathrm{P}\left({\widehat{p}}_n\right)}\left({\widehat{\mathrm{Q}}}_1-\overline{{\widehat{\mathrm{P}}}_n}\right) {\left({\widehat{\mathrm{Q}}}_1-\overline{{\widehat{\mathrm{P}}}_n}\right)}^t$									(19)
18: Estimate ${\widehat{\mathrm{P}}}_2$ using
${\widehat{\mathrm{P}}}_1=\overline{{\widehat{\mathrm{P}}}_n}+c_{{\widehat{p}}_n}{\left[c_{{\widehat{p}}_n}+{\sigma }^{2}I\right]}^{-1}\left({\mathrm{P}}_n-{\overline{\mathrm{P}}}_n\right)$, ${\left({\widehat{\mathrm{P}}}_n\right)}^{-1}\cong \frac{1}{\mathrm{P}\left({\widehat{p}}_1\right)}{\displaystyle \sum }_{\widehat{\mathrm{Q}}\in \mathrm{P}\left({\widehat{p}}_1\right)}\tilde{\mathrm{Q}}$									(20)
$C_{{\widehat{P}}_1}\cong \frac{1}{\mathrm{P}{\left({\widehat{p}}_1\right)}^{-1}}{\displaystyle \sum }_{\widehat{\mathrm{Q}}\in \mathrm{P}\left({\widehat{p}}_1\right)}\left({\widehat{\mathrm{Q}}}_1-{\left({\widehat{\mathrm{P}}}_n\right)}^{-1}\right){\left({\widehat{\mathrm{Q}}}_1-{\left({\widehat{\mathrm{P}}}_n\right)}^{-1}\right)}^t$									(21)
${\widehat{\mathrm{P}}}_2={\left({\widehat{\mathrm{P}}}_n\right)}^{-1}+c_{{\widehat{p}}_1}{\left[c_{{\widehat{p}}_1}+{\sigma }^{2}I\right]}^{-1}\left({\mathrm{P}}_n-{\left({\mathrm{P}}_n\right)}^{-1}\right)$									(22)
19: Evaluate the huge set of original patches using minimum mean square estimation
$\mathrm{P}\left(\raisebox{1ex}{${\mathrm{P}}_n$}\!\left/ \!\raisebox{-1ex}{${\widehat{\mathrm{P}}}_n$}\right.\right)=\frac{1}{\left(2\pi {\sigma }^2\right)}{e}^{\frac{\Vert {\mathrm{P}}_0-{\mathrm{P}}_n{}^2\Vert }{2{\sigma }^2}}$									(23)
${\widehat{\mathrm{P}}}_0=E\left[\raisebox{1ex}{${\mathrm{P}}_n$}\!\left/ \!\raisebox{-1ex}{${\widehat{\mathrm{P}}}_n$}\right.\right]={\displaystyle \int }\mathrm{P}\left(\raisebox{1ex}{${\mathrm{P}}_n$}\!\left/ \!\raisebox{-1ex}{${\widehat{\mathrm{P}}}_n$}\right.\right){\mathrm{P}}_{0}d{\mathrm{P}}_0={\displaystyle \int }\frac{\mathrm{P}\left(\raisebox{1ex}{${\mathrm{P}}_0$}\!\left/ \!\raisebox{-1ex}{$p_n$}\right.\right)}{\mathrm{P}\left(p_n\right)}\mathrm{P}\left({\mathrm{P}}_0\right){\mathrm{P}}_{0}d{\mathrm{P}}_0$									(24)
${\widehat{\mathrm{P}}}_0\cong \frac{\frac{1}{m}{{\displaystyle \sum }}_i\mathrm{P}\left(\raisebox{1ex}{${\mathrm{P}}_n$}\!\left/ \!\raisebox{-1ex}{${\mathrm{P}}_0$}\right.\right){\mathrm{P}}_{0i}}{\frac{1}{m}{{\displaystyle \sum }}_i\mathrm{P}\left(\raisebox{1ex}{${\mathrm{P}}_n$}\!\left/ \!\raisebox{-1ex}{${\mathrm{P}}_i$}\right.\right)}$									(25)
20: Evaluate $E\left(\raisebox{1ex}{${\mathrm{P}}_0$}\!\left/ \!\raisebox{-1ex}{${\mathrm{P}}_n$}\right.\right)$ using wavelet neighborhood de-noising.
${\mathrm{P}}_0=\sqrt{zU}$									(26)
${\mathrm{P}}_n={\mathrm{P}}_0+\mathrm{N}=\sqrt{zU}+\mathrm{N}$									(27)
$E\left(\raisebox{1ex}{${\mathrm{P}}_0$}\!\left/ \!\raisebox{-1ex}{${\mathrm{P}}_n$}\right.\right)={{\displaystyle \int }}_0^{\infty }\mathrm{P}\left(\raisebox{1ex}{$z$}\!\left/ \!\raisebox{-1ex}{${\mathrm{P}}_n$}\right.\right)E\left(\raisebox{1ex}{${\mathrm{P}}_0$}\!\left/ \!\raisebox{-1ex}{${\mathrm{P}}_n$}\right.,z\right)dz$									(28)
$\arg \underset{\alpha }{\min }{\Vert \mathrm{P}\Vert }_0$ = $\Vert {\mathrm{P}}_0-D_{\alpha }\Vert {}_{ }^2\le e^{2{\left(c\sigma \right)}^2}$									(29)
21: Perform the restoration using
$\mathrm{P}\left({\mathrm{P}}_i\right)=\frac{1}{{\left(2\pi \right)}^{\frac{e^2}{2}}{\left|c_{e_i}\right|}^{\frac{1}{2}}} e^{-\frac{1}{2}{\left({\mathrm{P}}_i-\mu \right)}^T{c}_{e_i}\left({\mathrm{P}}_i-\mu \right)}$									(30)
22: Calculate ${\mathrm{P}}_i$ with index $k_i$ using
$\left({\mathrm{P}}_i,k_i\right)=\arg \underset{{\mathrm{p}}_k}{\max }\log \mathrm{P}\left(\raisebox{1ex}{${\mathrm{P}}_i$}\!\left/ \!\raisebox{-1ex}{${\mathrm{P}}_{ni}$}\right.\right)$									(31)
$\left({\mathrm{P}}_i,k_i\right)=\arg \underset{{\mathrm{p}}_k}{\max }\left(\mathrm{logP}\left(\raisebox{1ex}{${\mathrm{P}}_{ni}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{P}}_i$}\right.,c_k\right)+\mathrm{logP}\left(\raisebox{1ex}{${\mathrm{P}}_i$}\!\left/ \!\raisebox{-1ex}{$c_k$}\right.\right)\right)$									(32)