How do I change values in an array based upon its previous value?

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Michael Rowlinson
Michael Rowlinson 2019-5-15
评论： Michael Rowlinson 2019-5-16
Hi all! I'm new to MATLAB.
So basically I have a 2D array and a spreading pattern/fractal that I wish to execute, I can make the pattern happen when turning zeros to ones but then it get's messy after that. the image is a basic rundown of what I want to happen for each iteration of the pattern.
Thank you, Michael.Pattern MATLAB.PNG
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Guillaume
Guillaume 2019-5-15
Why isn't the pattern spread around the 1s in step 2? I.e. why isn't that last matrix:
0 0 1 0 0
0 1 2 1 0
1 2 3 2 1
0 1 2 1 0
0 0 1 0 0

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Guillaume
Guillaume 2019-5-15
编辑：Guillaume 2019-5-15
Assuming you've made a mistake in your second step (see comment), this is trivially achieved with imdilate (requires image processing toolbox
A = [0 0 0 0 0; 0 0 0 0 0;0 0 1 0 0; 0 0 0 0 0; 0 0 0 0 0]
nstep = 5;
for step = 1:nstep
A = imdilate(A > 0, [0 1 0; 1 1 1;0 1 0]) + A
end
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Guillaume
Guillaume 2019-5-16
编辑：Guillaume 2019-5-16
Ah, ok. You can use a simple 2d convolution to find the number of neighbours a value. Then it's a simple matter of a bit of arithmetic:
A = [0 0 0 0 0; 0 0 0 0 0;0 0 1 0 0; 0 0 0 0 0; 0 0 0 0 0]
nstep = 5;
for step = 1:nstep
A(A > 0) = A(A > 0) + 1; %increase existing values by 1
A = A + (conv2(A > 0, [0 1 0;1 0 1;0 1 0], 'same') == 1) .* ~A %find zeros with just one neighbour, set them to 1
end
Michael Rowlinson
Michael Rowlinson 2019-5-16
That looks about right, I can't check if it'll work for me right now but I'll get back to this soon, thank you!

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