How to construct a binary image based on the most significant bit of the pixels?

The brightness intensity of the image fractal.jpg takes values between 0 and 255.Thus each pixel of the image can be represented with 8 bits.Since we have the image, how can i construct a binary image based on the most significant bit of the pixels.Specifically , if the MSB of a pixel is 1 , the pixel in the new image will take 255 , while if it is 0 ,the pixel in the new image will take value 0 .Repeat for the second most significant bit and the least significant bit (least significant bit, LSB). I have to implement this in Octave.

Answer 1

[Image Analyst]

For a gray scale image (not color), you can use bitget:

% Compute the bit plane image.
bitPlaneImage = bitget(grayImage, bp+1);

For a full demo, see my attached demo.

[Alexandros Ioannidis]

Thank you

Answer 2

[Alexandros Ioannidis]

 % Load image from specified file
    img1=imread('fractal.jpg');
 % Display initial image 
    figure (1);
    imshow(img1); 

    % Scan through all rows in the image
    for y = 1:size(img1, 1)
        % Scan col
        for x = 1:size(img1, 2)
            % img1 - MSB
            if img1(y,x) > 127
                % Turn one light pixel white
                img1(y,x) = 255;
            else
                % Turn one dark pixel black
                img1(y,x) = 0;
            end
        end
  end

    % Display image
    figure (2);   
    imshow(img1);

[Image Analyst]

Not sure what that code is for, but you can do the same thing as that double "for" loop, vectorized by doing:

img1 = uint8(255 * img1>127);

Of course thresholding gets far more complicated for the lower bits because you won't have just one range like here, you'll have 2, 4, ... 128 ranges depending on what bit you want to visualize. That's why I recommended bitget() instead of thresholding or dividing.