Question

I understand that the loops should not be nested but I have done the same to further sub-divide the 5×5 window into four 3×3 sub-windows. Please help me with this, how to sub divide the 5×5 window into four 3×3 sub-windows without nested loop.

Code:

for i=1:m

for j=1:n

  if(i<=(m-4) && j<=(n-4))

      % Reading 5 x 5 window in an image

v=1; 
for p=i:i+4
    u=1;
    for q=j:j+4
        P(v,u)=L(p,q);
        u=u+1;
    end
    v=v+1;
end

    % Sub dividing the 5 x 5 window into four 3x3 sub windows 

k=1;
for r=1:3
    l=1;
    for s=1:3
        v2(k,l)=P(r,s);
        v1(k,l)=P(r,s+2);   
        v3(k,l)=P(r+2,s);
        v4(k,l)=P(r+2,s+2);
        l=l+1;    
    end
    k=k+1;
end

I need this four vector v1,v2,v3 and v4 for further processing. Link of my previous question about retaining the corner pixels:

https://stackoverflow.com/questions/13793103/retain-corner-pixels-in-an-image

Answer 1

Let me tell a bit more than just answering your question on sub-dividing your window.

1. You read in a 5×5 window from an image like this:

   P=L(i:(i+4),j:(j+4));

   so that (i,j) is the top-left corner of the window on the original image.

2. This “sub-division” looks for me like taking a 3×3 starting from the TL, TR, BL and BR(*) corner of the 5×5 window. That can be done simply like this:

   tl = P(1:3, 1:3); % this is v2 in your question
tr = P(1:3, 3:5); % this is v1 (didn't you mix up v2 and v1 by any chance?)
bl = P(3:5, 1:3); % this is v3
br = P(3:5, 3:5); % this is v4

   I suggest rather using this tl, tr, bl, br instead of v1, v2, v3, v4 because it is more expressive.

(*)(T=top, B=bottom, L=left, R=right)

---

UPDATE: (how to change the loops?)
You want to cover all the windows on L, so the first window will be P=L(1:5, 1:5), the last will be P=L((m-4):m,(n-4):n). All the others will be in between. So the first value of i is i=1, the last is i=m-4 and the first j is j=1, the last is j=n-4.

You can also verify this manually, typing a demo L into the command prompt of matlab, like this:

L=reshape([1:42], 6, 7); % a 6x6 test array
m=6; n=7;

i=1; j=1;
P_first=L(i:(i+4),j:(j+4)); % should contain 1

i=m-4; j=n-4;
P_last=L(i:(i+4),j:(j+4)); % should contain 42