Effacer les filtres
Effacer les filtres

An Edge-Guided Image Interpolation Algorithm via Directional Filtering and Data Fusion

1 vue (au cours des 30 derniers jours)
Sony
Sony le 24 Avr 2012
Modifié(e) : qing zhao le 13 Nov 2017
function A=esintp(I)
%%%%%process A%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[n,m]=size(I);
n=2*n;m=2*m;
A=zeros(n,m);
A(1:2:n,1:2:m)=I;
Ax=A;Ay=A;
f=[-1/16 9/16 9/16 -1/16];
for i=4:2:n-4
for j=4:2:m-4
x=[A(i-3,j-3) A(i-1,j-1) A(i+1,j+1) A(i+3,j+3)];
y=[A(i+3,j-3) A(i+1,j-1) A(i-1,j+1) A(i-3,j+3)];
Ax(i,j)=sum(x.*f);
Ay(i,j)=sum(f.*y);
end
end
for i=4:2:n-4
for j=4:2:m-4
zx=[Ax(i-2,j-2) Ax(i-1,j-1) Ax(i,j) Ax(i+1,j+1) Ax(i+2,j+2)];
zy=[Ay(i+2,j-2) Ay(i+1,j-1) Ay(i,j) Ay(i-1,j+1) Ay(i-2,j+2)];
x=[A(i-1,j-1) A(i+1,j+1) A(i+1,j-1) A(i-1,j+1)];
mx=mean(x);
px=sum((x-mx).^2)/3+0.1;
pzx=sum((zx-mx).^2)/4+0.1;
pzy=sum((zy-mx).^2)/4+0.1;
pv1=pzx-px;pv1=max(0.01,pv1);
pv2=pzy-px;pv2=max(0.01,pv2);
R=[pv1 0;0 pv2];
y=[Ax(i,j) Ay(i,j)]';
H=[1;1];
A(i,j)=mx+px*H'*(H*px*H'+R)^(-1)*(y-H*mx);
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Ax=A;Ay=A;
f=[-1/16 0 9/16 1 9/16 0 -1/16];
Ax=conv2(Ax,f);
Ax=Ax(:,4:m+3);
Ay=conv2(Ay,f');
Ay=Ay(4:n+3,:);
c=1.5;
h=[-1 5 5 -1];h=h/sum(h);
for i=5:2:n-5
for j=4:2:m-4
y=[Ax(i-2,j) A(i-1,j) A(i+1,j) Ax(i+2,j)];
my=sum(h.*y);
x=[Ax(i,j-2) Ax(i,j-1) Ax(i,j) Ax(i,j+1) Ax(i,j+2)];
vx=cov(x)+0.1;
vy=c*cov([y my])+0.1;
A(i,j)=(Ax(i,j)*vy+my*vx)/(vx+vy);
end
end
for i=4:2:n-4
for j=5:2:m-5
x=[Ay(i,j-2) A(i,j-1) A(i,j+1) Ay(i,j+2)];
mx=sum(h.*x);
y=[Ay(i-2,j) Ay(i-1,j) Ay(i,j) Ay(i+1,j) Ay(i+2,j)];
vx=c*cov([x mx])+0.1;
vy=cov(y)+0.1;
A(i,j)=(mx*vy+Ay(i,j)*vx)/(vx+vy);
end
end
return;
can anyone explain this code???
  2 commentaires
Royi Avital
Royi Avital le 7 Juin 2014
Could you point exactly what do you don't understand?
qing zhao
qing zhao le 13 Nov 2017
this code have something wrong,haven't you found it?

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Réponses (2)

Image Analyst
Image Analyst le 7 Juin 2014
I bet the author could, but probably doesn't want to, or else he or she would have done so by adding comments. I'm not even going to attempt it.

qing zhao
qing zhao le 13 Nov 2017
Modifié(e) : qing zhao le 13 Nov 2017
Can someone provide some suggestions about above code.

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