基本信息
源码名称:基于凸优化的快速单图像反射抑制
源码大小:2.25KB
文件格式:.m
开发语言:MATLAB
更新时间:2020-06-10
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   源码介绍

透过玻璃拍摄的照片往往有镜面反射,本程序可以实现消除拍摄图片不必要的镜面反射,简单高效

clc;
clear all;
close all;
%数据输入
Im = imread(''); %图片位置
T = reflection_suppress(Im, 0.05, 1e-8);%设置梯度阈值h
subplot(1,2,1); imshow(Im); subplot(1,2,2); imshow(T);

function T = reflection_suppress(Im, h, epsilon)
    
    Y = im2double(Im);     
    [m, n, r] = size(Y);
    T = zeros(m,n,r);
    Y_Laplacian_2 = zeros(m,n,r);
    
    for dim = 1:r
        GRAD = grad(Y(:,:,dim));
        GRAD_x = GRAD(:,:,1);
        GRAD_y = GRAD(:,:,2);
        GRAD_norm = sqrt(GRAD_x.^2 GRAD_y.^2);
        GRAD_norm_thresh = wthresh(GRAD_norm, 'h', h);                     % gradient thresholding
        ind = (GRAD_norm_thresh == 0);
        GRAD_x(ind) = 0;
        GRAD_y(ind) = 0;
        GRAD_thresh(:,:,1) = GRAD_x;
        GRAD_thresh(:,:,2) = GRAD_y;                                       
        Y_Laplacian_2(:,:,dim) = div(grad(div( GRAD_thresh)));             % compute L(div(delta_h(Y)))
    end
        
    rhs = Y_Laplacian_2 epsilon * Y;     
        
    for dim = 1:r
        T(:,:,dim) = PoissonDCT_variant(rhs(:,:,dim), 1, 0, epsilon);      % solve the PDE using DCT
    end

end



% solve the equation  (mu*L^2 - lambda*L epsilon)*u = rhs via DCT
% where L means Laplacian operator
function u = PoissonDCT_variant(rhs, mu, lambda, epsilon)   

[M,N]=size(rhs);

k=1:M;
l=1:N;
k=k';
eN=ones(1,N);
eM=ones(M,1);
k=cos(pi/M*(k-1));
l=cos(pi/N*(l-1));
k=kron(k,eN);
l=kron(eM,l);

kappa=2*(k l-2);
const = mu * kappa.^2 - lambda * kappa epsilon;
u=dct2(rhs);
u=u./const;
u=idct2(u);                       % refer to Theorem 1 in the paper

return
end



% compute the gradient of a 2D image array

function B=grad(A)

[m,n]=size(A);
B=zeros(m,n,2);

Ar=zeros(m,n);
Ar(:,1:n-1)=A(:,2:n);
Ar(:,n)=A(:,n);


Au=zeros(m,n);
Au(1:m-1,:)=A(2:m,:);
Au(m,:)=A(m,:);

B(:,:,1)=Ar-A;     
B(:,:,2)=Au-A;     

end


% compute the divergence of gradient
% Input A is a matrix of size m*n*2
% A(:,:,1) is the derivative along the x direction
% A(:,:,2) is the derivative along the y direction

function B=div(A)  

[m,n,~]=size(A);
B=zeros(m,n);

T=A(:,:,1);
T1=zeros(m,n);
T1(:,2:n)=T(:,1:n-1);

B=B T-T1;

T=A(:,:,2);
T1=zeros(m,n);
T1(2:m,:)=T(1:m-1,:);

B=B T-T1;
end