binary mapping continuous ones or not

we have divided a 256*256 matrix into 4*4 matrix and we have got 4096 combination.now we want to map these 4*4 matrix to binay 1's and 0's in such a way that if there are minimum four continuous 1's in any direction or in any manner we want to map it to binary '1' else map it to binary '0'

ex: (1)
       [0 0 1 1;...
        0 0 1 0;...
        0 1 0 0;...
        1 0 0 0]      it must be mapped to binary '1'
(2)  [0 0 0 0;...
      0 0 1 1;...
      0 1 0 0;...
      1 0 0 0]      it must be mapped to binary '1'
(3)  [0 0 0 1;...
      0 1 0 0;...
      0 0 0 0;...
      1 0 0 1]      it must be mapped to binary '0'

[dpb]

The cases (1) and similar are pretty easy...it's

any(all(x).' | all(x,2) | all(diag(x)) | all(fliplr(diag(x))))

which simply checks all rows and columns plus the diagonals. The above has no attempt at simplification; just the logic. Of course, it's applied with arrayfun or similar construct over the full set of arrays, x, however they're stored.

The other cases are tougher; don't have time at the moment but look at

doc nearestNeighbor

for thoughts to get started, maybe. On thought is that you can reduce the dimensionality by reducing any arrays with zero columns/rows such as (2) to a 3x2 and (3) to two; one a row vector and the other 2x3 in these particular examples. Maybe the resulting strength reduction in search those reduced arrays outweighs the extra ones to keep track of, I don't know.

Answer 1

[Image Analyst]

If you have the Image Processing Toolbox, it's trivial. Just label and get the largest area.

% Label the 4-by-4 matrix.
labeledImage = bwlabel(matrix4x4, 8); 
% Measure areas
measurements = regionprops(labeledImage, 'Area');
% Get the areas of all the blobs in the image
allAreas = [measurement.Area];
% Find the largest area
maxArea = max(allAreas);
% See what it needs to be "mapped" to.
if maxArea >= 4
  % It must be mapped to binary '1'
else
  % It must be mapped to binary '0'
end

[dpb]

Good catch, IA...I've never had the Image Processing Toolbox; not sure even if had would've thought of the single line as an area altho suppose that with "time-in-grade" doing image processing it's a natural...