Question

I have a 6 * 6 matrix

A=
  3     8     8     8     8     8
  4     6     1     0     7    -1
  9     7     0     2     6    -1
  7     0     0     5     4     4
  4    -1     0     2     8     1
  1    -1     0     8     3     9

I am interested in finding row and column number of neighbors starting from A(4,4)=5. But They will be linked to A(4,4) as neighbor only if A(4,4) has element 4 on right, 6 on left, 2 on top, 8 on bottom 1 on top left diagonally, 3 on top right diagonally, 7 on bottom left diagonally and 9 on bottom right diagonally. TO be more clear A(4,4) will have neighbors if the neighbors are surrounding A(4,4) as follows:

 1     2     3;
 6     5     4;
 7     8     9;

And this will continue as each neighbor is found.
Also 0 and -1 will be ignored. In the end I want to have these cells’ row and column number as shown in figure below. Is there any way to visualize this network as well. This is sample only. I really have a huge matrix.

Answer 1

A = [3     8     8     8     8     8;
     4     6     1     0     7    -1;
     9     7     0     2     6    -1;
     7     0     0     5     4     4;
     4    -1     0     2     8     1;
     1    -1     0     8     3     9];

test = [1 2 3; 
        6 5 4; 
        7 8 9];

%//Pad A with zeros on each side so that comparing with test never overruns the boundries    
%//BTW if you have the image processing toolbox you can use the padarray() function to handle this
P = zeros(size(A) + 2);        
P(2:end-1, 2:end-1) = A;

current = zeros(size(A) + 2);
past = zeros(size(A) + 2);

%//Initial state (starting point)
current(5,5) = 1; %//This is A(4,4) but shifted up 1 because of the padding

condition = 1;

while sum(condition(:)) > 0;
      %//get the coordinates of any new values added to current
     [x, y] = find(current - past); 
     %//update past to last iterations current
     past = current; 

     %//loop through all the coordinates returned by find above
     for ii=1:size(x); 

         %//Make coord vectors that represent the current coordinate plus it 8 immediate neighbours.
         %//Note that this is why we padded the side in the beginning, so if we hit a coordinate on an edge, we can still get 8 neighbours for it!
         xcoords = x(ii)-1:x(ii)+1; 
         ycoords = y(ii)-1:y(ii)+1; 

         %//Update current based on comparing the coord and its neighbours against the test matrix, be sure to keep the past found points hence the OR
         current(xcoords, ycoords) = (P(xcoords, ycoords) == test) | current(xcoords, ycoords); 

     end 

     %//The stopping condition is when current == past
     condition = current - past;

 end 

%//Strip off the padded sides
FinalAnswer = current(2:end-1, 2:end-1)
[R, C] = find(FinalAnswer);
coords = [R C] %//This line is unnecessary, it just prints out the results at the end for you.

OK cool you got very close, so here is the final solution with the loops. It runs in about 0.002 seconds so it’s pretty quick I think. The output is

FinalAnswer =

     0     0     0     0     0     0
     0     1     1     0     0     0
     0     1     0     1     0     0
     1     0     0     1     1     1
     0     0     0     0     1     0
     0     0     0     0     0     1


coords =

     4     1
     2     2
     3     2
     2     3
     3     4
     4     4
     4     5
     5     5
     4     6
     6     6