i need help to understand the steps of this code
2 vues (au cours des 30 derniers jours)
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hi everyone please can someone explain to me the steps of this program ?
its a coherence filter program
thanks for helping
%%%
function myCEDnew( )
myalpha = 0.001; sigma=0.7; T = 15; rho = 4; C= 1;
im = imread('../images/2.png');
[numrow, numcol] = size(im);
imorig = im;
stepT=0.15;
t = 0;
im=double(im);
while (t < (T-0.001))
t = t + stepT;
%% 1 gaussian K_sigma
limitX=-ceil(2*sigma):ceil(2*sigma);
kSigma = exp(-(limitX.^2/(2*sigma^2)));
kSigma = kSigma/sum(kSigma(:));
usigma=imfilter(imfilter(im,(kSigma'), 'same' ,'replicate'),kSigma, 'same' ,'replicate');
%% Gradient
[uy,ux]=gradient(usigma);
%% 3 gaussian K_rho
limitXJ=-ceil(3*rho):ceil(3*rho);
kSigmaJ = exp(-(limitXJ.^2/(2*rho^2)));
kSigmaJ = kSigmaJ/sum(kSigmaJ(:));
Jxx = imfilter(imfilter((ux.^2),(kSigmaJ'), 'same' ,'replicate'),kSigmaJ, 'same' ,'replicate');
Jxy = imfilter(imfilter((ux.*uy),(kSigmaJ'), 'same' ,'replicate'),kSigmaJ, 'same' ,'replicate');
Jyy = imfilter(imfilter((uy.^2),(kSigmaJ'), 'same' ,'replicate'),kSigmaJ, 'same' ,'replicate');
%% Principal axis transformation
% Eigenvectors of J, v1 and v2
v2x = zeros(numrow, numcol);
v2y = zeros(numrow, numcol);
lambda1 = zeros(numrow, numcol);
lambda2 = zeros(numrow, numcol);
for i=1:numrow
for j=1:numcol
pixel = [Jxx(i,j), Jxy(i,j); Jxy(i,j), Jyy(i,j)];
[pixelV, pixelD] = eig(pixel);
v2x(i,j) = pixelV(1,2);
v2y(i,j) = pixelV(2,2);
lambda1(i,j) = pixelD(1,1);
lambda2(i,j) = pixelD(2,2);
if((v2x(i,j)^2)+(v2y(i,j)^2)==0)
abcd=0;
else
v2x(i,j) = v2x(i,j)/(sqrt((v2x(i,j)^2)+(v2y(i,j)^2)));
v2y(i,j) = v2y(i,j)/(sqrt((v2x(i,j)^2)+(v2y(i,j)^2)));
end;
end;
end;
v1x = -v2y;
v1y = v2x;
%% Calculation of diffusion matrix
di=(lambda1-lambda2);
lambda1 = myalpha + (1 - myalpha)*exp(-C./(di).^(2));
lambda2 = myalpha;
Dxx = lambda1.*v1x.^2 + lambda2.*v2x.^2;
Dxy = lambda1.*v1x.*v1y + lambda2.*v2x.*v2y;
Dyy = lambda1.*v1y.^2 + lambda2.*v2y.^2;
%% Non negativity discretization scheme referred from http://citeseer.ist.psu.edu/viewdoc/download;jsessionid=16CCEAD1A72E6A1CC960DF99795D62B7?doi=10.1.1.21.632&rep=rep1&type=pdf
im=non_negativity_discretization(im,Dxx,Dxy,Dyy,stepT);
end;
%% output
figure(1);
subplot(1, 2, 1);
imagesc(imorig);
title('Original image');
colormap('Gray');
daspect ([1 1 1]);
subplot(1, 2, 2);
imagesc(im);
title('Coherence Enhancing Diffusion Filtering');
colormap('Gray');
daspect ([1 1 1]);
figure(2);
subplot(1, 2, 1);
imagesc(imorig);
title('Original image');
colormap('Gray');
daspect ([1 1 1]);
subplot(1, 2, 2);
imagesc(atan2(double(uy), double(ux)));
title('Orientation of smooth gradient');
colormap('Gray');
daspect ([1 1 1]);
imwrite((uint8(im)), '../images/2CED.png');
end
function im=non_negativity_discretization(im,Dxx,Dxy,Dyy,stepT)
% Make positive and negative indices
[numrow,numcol] = size(im);
px = [2:numrow,numrow]; nx = [1,1:numrow-1];
py = [2:numcol,numcol]; ny = [1,1:numcol-1];
% In literature a,b and c are used as variables
a=Dxx;b=Dxy;c=Dyy;
% Stencil Weights
wbR1 = (0.25)*((abs(b(nx, py))-b(nx,py)) + (abs(b)-b));
wtM2 = (0.5)*( (c(:, py)+c) -(abs(b(:,py))+abs(b)));
wbL3 = (0.25)*((abs(b(px, py))+b(px,py)) + (abs(b)+b));
wmR4 = (0.5)*( (a(nx,: )+a) -(abs(b(nx,:))+abs(b)));
wmL6 = (0.5)*( (a(px,: )+a) -(abs(b(px,:))+abs(b)));
wtR7 = (0.25)*((abs(b(nx, ny))+b(nx,ny)) + (abs(b)+b));
wmB8 = (0.5)*( (c(:, ny)+c) -(abs(b(:,ny))+abs(b)));
wtL9 = (0.25)*((abs(b(px, ny))-b(px,ny)) + (abs(b)-b));
im= im+stepT*(wbR1.*(im(nx,py) -im(:,:))+wtM2.*(im(:, py) -im(:,:))+wbL3.*(im(px,py) -im(:,:))+wmR4.*(im(nx,:) -im(:,:))+ ...
wmL6.*(im(px,:) -im(:,:))+ wtR7.*(im(nx,ny) -im(:,:))+ wmB8.*(im(:, ny) -im(:,:))+ wtL9.*(im(px,ny) -im(:,:)));
end
1 commentaire
Image Analyst
le 29 Mai 2019
I doubt any of us can/would, but the authors probably would. Have you asked them?
Réponses (0)
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