not sure how to approach this

I have a 2D matrix. Let's say A=80x100; I need a way to visit each cell in this matix only once. Moreover my problem is a little more involved than that. Not only do I want to visit each cell only once but the user also defines a window of certain size. Let's say WINDOW=3x3. Then let's say I randomly choose a location on A that corresponds to indecies [2,3]. Then this is considered a central location and in my algorithm I will also consider points around this central location as specified by the window. Hence in this example I will also consider points at indecies [1,3],[3,3],[1,2],[2,2],[3,2],[1,4],[2,4],[3,4]. So in the next round when I am trying to figure out which cell in A to visit, I will not consider these 9 points in my random draw. Any suggestions? Thanks!

Answer 1

[Matt J]

        A=rand(80,100);

        [m,n]=size(A);
        winsize=3; 
         midwin=winsize/2+.5;

        [dx,dy]=ndgrid(1:winsize);

        jumps = sub2ind([m,n],dx,dy)-sub2ind([m,n],midwin,midwin);

        B=A; 
        B(:)=1:numel(A);
        B=B(midwin:end+1-midwin, midwin:end+1-midwin);
        points=B(:);
        numpoints=length(points);   

     while numpoints

         randlocation=points(randi(numpoints));

         window=randlocation+jumps;

         %%%%%%%%%%%%do stuff to A(window)  

         points=setdiff(points,window(:));
         numpoints=numel(points);    

     end

Answer 2

[Sean de Wolski]

idx = randperm(numel(A));

To give you a list of random linear indices to visit.

Then use a for-loop to loop over these. When you get to one index, use ind2sub() to convert it to row/column coordinates (or do this before hand for all of them and loop over that.)

doc ind2sub %for more info

Then make your window by adding and subtracting floor(window_size/2) to the indices and do whatever operation you have on this. You could also do this addition/subtraction before hand and loop over that instead.

Answer 3

[jenka]

This is sooo great! I am so thankful to you!

[Matt J]

I notice on your profile that you have asked 19 questions and accepted 0 of the answers. In case you didn't know about how the site works, you should click "Accept" on the Answer that served you best (and maybe vote for runner-ups).

[jenka]

Hi Matt,no, I did not know that. Will do this for sure. I have a quick question regarding border's effect of this problem. I am not sure how to quickly fix this? For example, assume

A=rand(4,5).

Then if randlocation is equal to 16.

The window is

window =
  11 15 19
  12 16 20
  13 17 21

Which is not entirely correct. As in this case the window should be

window =
  11 15 19
  12 16 20

Any suggestions? Thanks

[Matt J]

In the code I gave you, randlocation will never be on the boundary. I excluded all locations without a full winsize x winsize neighborhood in this line

 B=B(midwin:end+1-midwin, midwin:end+1-midwin);

If you're saying that you can't ignore boundary points, you could pre-embed your A matrix into a larger matrix, and modify as follows

        A=rand(80,100);
        winsize=3; 
         midwin=winsize/2+.5;

       A0=A; %save original A     
       A=nan(size(A0)+winsize-1);      
       A(midwin:end+1-midwin, midwin:end+1-midwin)=A0; %enlarge

        [m,n]=size(A);

        [dx,dy]=ndgrid(1:winsize);
        jumps = sub2ind([m,n],dx,dy)-sub2ind([m,n],midwin,midwin);
        B=A; 
        B(:)=1:numel(A);
        B=B(midwin:end+1-midwin, midwin:end+1-midwin);
        points=B(:);
        numpoints=length(points);  

     while numpoints

         randlocation=points(randi(numpoints));

         window=randlocation+jumps;

         window(isnan(A(window)))=[]; %THROW AWAY NANS!!!!!!!!!!!!!

         %%%%%%%%%%%%do stuff to A(window)  
         points=setdiff(points,window(:));
         numpoints=numel(points);    

     end