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

조회 수: 1 (최근 30일)
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???

답변 (2개)

Image Analyst
Image Analyst 2014년 6월 7일
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 2017년 11월 13일
편집: qing zhao 2017년 11월 13일
Can someone provide some suggestions about above code.

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