Decompose image into the sum of two images

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Jenalyn Palacios
Jenalyn Palacios el 26 de Mzo. de 2020
Comentada: Jenalyn Palacios el 26 de Mzo. de 2020
I have a black image with different sized white squares all over the image. If we let A denote this image, "decompose" A as the sum of two images A=A1+A2, such that A1 contains the larger sqares and A2 contains the smaller squares. Your code should display seperately A1 and A2.
This is my code so far. but i know its completely wrong. Can someone please help me with this!!
A=imread('small-squares.tif');
se = strel('square',7);
eroded = imerode(A,se);
figure
subplot (1,2,1), imshow(A), title('original')
subplot (1,2,2), imshow(eroded)

Respuesta aceptada

Image Analyst
Image Analyst el 26 de Mzo. de 2020
Hint: you need to segment the image (perhaps by thresholding),
allSquares = grayImage > 0;
or from imbinarize(). Then use regionprops to find out the size of all the squares and what threshold you want to use to distinguish small squares from big ones.
props = regionprops(.......
Then use bwareafilt() to create masks for smallSquareMask and bigSquareMask.
smallSquareMask = bwareafilt(............
bigSquareMask = bwareafilt(............
Then mask the gray image twice
% Mask the image using bsxfun() function to multiply the mask by each channel individually.
maskedRgbImage = bsxfun(@times, rgbImage, cast(mask, 'like', rgbImage));
to get two masked grayscale images or RGB images. When you add those two together you'll get the original image.

Más respuestas (1)

Guillaume
Guillaume el 26 de Mzo. de 2020
I don't see how your code even attempt to answer your assignment.
Which of the image processing function would you use to detect objects in an image?
Which other image processing function would you use to know the size of such objects once detected?
Once you know which pixel belong to which object (from the first function) and the size of such object, it's easy to copy them to the correct image.
I'm afraid I won't give you any more hint than that. It's not a particularly difficult task.

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