z(1) = c
z(n+1) = z(n)^2 + c
For any complex c, we can continue this iteration until either abs(z(n+1)) > 2 or n == lim, then return the iteration count n.
- If c = 0 and lim = 3, then z = [0 0 0] and n = 3.
- If c = 1 and lim = 5, then z = [1 2], and n = length(z) or 2.
- If c = 0.5 and lim = 5, then z = [0.5000 0.7500 1.0625 1.6289] and n = 4.
For a matrix of complex numbers C, return a corresponding matrix N such that each element of N is the iteration count n for each complex number c in the matrix C, subject to the iteration count limit of lim.
If C = [0 0.5; 1 4] and lim = 5, then N = [5 4; 2 1]
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For c==4 and other numbers where abs(c)>2, I think the function should be defined to return 0 rather than 1.
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I am really stuck on this one. I know it is likley me not the question but I cannot figure out why in the final validation, -2i should give N=1, I get it to be N=2? any help is appriciated, this is the only situlation where code fails.
For people like me that hoped this challenge would end with a pretty picture:
`[X,Y]=meshgrid(-2:0.0025:2,-2:0.0025:2);C=X+i.*Y;N=mandelbrot(C,50);imagesc(N)`
@Ratchet_Hamster
for complex no, u need to take the absolute value to check if it is greater than 2
Broken ink to Cleve Moler's PDF
Thanks for noticing that, @Shlomo Geva. The link has been fixed.
Really nice problem, and great very simple solution by the community.