Is it part of the mandelbrot set?
+2
−0
Input is a number, you have to decide if it is part of the mandelbrot set or not, after at least 16 iterations.
This is done by applying this formula: $z_n = z_{n-1}^2 + c$ repeatedly. $c$ is the input number and $z_0 = 0$. Therefore:
- $z_1 = c$
- $z_2 = c^2 + c$
- $z_3 = ( c^2 + c )^2 +c$
- $z_4 = ( ( c^2 + c )^2 +c )^2 +c$
- ...
If $|z_{16}|>2$, then the input $c$ is not part of the mandelbrot set. If $|z_{16}| ≤ 2$, then we consider it part of the mandelbrot set for this challenge.
Rules
- The input is a complex number $|c|≤2$
- Output is a boolean value for inside (after 16 iterations) or not inside the mandelbrot set.
- The input format can be chosen how it is best for you, but input has to be representable with an error of $|E|≤ { 1 \over 1024} $ on each axis i.e. you need at least 11 bit for each axis.
- At least up to 16 iterations have to be checked. If more is easier to do, do more.
Optional
- If it is easier, you can check for the real part $Re(z_{n})$ to be $Re(z_{n})>2$ or $|Re(z_{n})|>2$ instead of $|z_{n}|>2$. But then it should be checked for each iteration. This includes some numbers that are not part of the mandelbrot set but we ignore that for this challenge.
- If it is easier, the limit can be anywhere between 2-6 and doesn't have to be 2, for example you could check for $|z_{16}|>4$ instead of $|z_{16}|>2$.
- If it is easier, an additional input for the number of iterations can be given, but then at least 3-64 iterations (chosen by the input) have to be supported.
- If it is easier, an additional input for $z_0$ or $z_1$ can be given.
- You can alternatively "output" a value by either returning very fast or running for a very long time (the factor between both possibilities should be at least 16).
Shortest code wins.
Non-golfed Python Example
def isPartOfMandelbrot(c):
z=c
for i in range(16):
z=z**2+c
if abs(z)>2:
return 0
return 1
print( isPartOfMandelbrot( float(input()) + float(input())*1j ) )
Testcases
-1.8620690 -0.3448276i: 0
-1.7241379 -0.0689655i: 0
-1.7241379 +0.0689655i: 0
-1.5862069 -0.0689655i: 0
-1.5862069 +0.0689655i: 0
-1.5862069 +1.0344828i: 0
-1.5862069 +1.1724138i: 0
-1.4482759 -1.1724138i: 0
-1.4482759 -0.0689655i: 0
-1.4482759 +0.0689655i: 0
-1.3103448 -0.8965517i: 0
-1.3103448 -0.3448276i: 0
-1.3103448 -0.2068966i: 0
-1.3103448 +0.2068966i: 0
-1.3103448 +0.3448276i: 0
-1.1724138 -1.0344828i: 0
-1.1724138 -0.0689655i: 1
-1.1724138 +0.0689655i: 1
-1.0344828 -0.2068966i: 1
-1.0344828 -0.0689655i: 1
-1.0344828 +0.0689655i: 1
-1.0344828 +0.2068966i: 1
-0.8965517 -1.1724138i: 0
-0.8965517 -0.4827586i: 0
-0.8965517 -0.3448276i: 0
-0.8965517 -0.2068966i: 1
-0.8965517 -0.0689655i: 1
-0.8965517 +0.0689655i: 1
-0.8965517 +0.2068966i: 1
-0.8965517 +0.3448276i: 0
-0.7586207 -0.6206897i: 0
-0.7586207 -0.4827586i: 0
-0.7586207 -0.3448276i: 0
-0.7586207 +0.3448276i: 0
-0.7586207 +0.4827586i: 0
-0.6206897 -0.6206897i: 0
-0.6206897 -0.4827586i: 0
-0.6206897 -0.3448276i: 1
-0.6206897 -0.2068966i: 1
-0.6206897 -0.0689655i: 1
-0.6206897 +0.0689655i: 1
-0.6206897 +0.2068966i: 1
-0.6206897 +0.3448276i: 1
-0.6206897 +0.4827586i: 0
-0.6206897 +0.6206897i: 0
-0.6206897 +0.8965517i: 0
-0.6206897 +1.3103448i: 0
-0.6206897 +1.8620690i: 0
-0.4827586 -1.7241379i: 0
-0.4827586 -0.4827586i: 1
-0.4827586 -0.3448276i: 1
-0.4827586 -0.2068966i: 1
-0.4827586 -0.0689655i: 1
-0.4827586 +0.0689655i: 1
-0.4827586 +0.2068966i: 1
-0.4827586 +0.3448276i: 1
-0.4827586 +0.4827586i: 1
-0.4827586 +1.5862069i: 0
-0.3448276 -1.3103448i: 0
-0.3448276 -0.7586207i: 0
-0.3448276 -0.4827586i: 1
-0.3448276 -0.3448276i: 1
-0.3448276 -0.2068966i: 1
-0.3448276 -0.0689655i: 1
-0.3448276 +0.0689655i: 1
-0.3448276 +0.2068966i: 1
-0.3448276 +0.3448276i: 1
-0.3448276 +0.4827586i: 1
-0.3448276 +0.7586207i: 0
-0.3448276 +1.3103448i: 0
-0.3448276 +1.8620690i: 0
-0.2068966 -1.0344828i: 0
-0.2068966 -0.8965517i: 0
-0.2068966 -0.7586207i: 1
-0.2068966 -0.6206897i: 1
-0.2068966 -0.4827586i: 1
-0.2068966 -0.3448276i: 1
-0.2068966 -0.2068966i: 1
-0.2068966 -0.0689655i: 1
-0.2068966 +0.0689655i: 1
-0.2068966 +0.2068966i: 1
-0.2068966 +0.3448276i: 1
-0.2068966 +0.4827586i: 1
-0.2068966 +0.6206897i: 1
-0.2068966 +0.7586207i: 1
-0.2068966 +0.8965517i: 0
-0.2068966 +1.0344828i: 0
-0.0689655 -1.0344828i: 0
-0.0689655 -0.7586207i: 1
-0.0689655 -0.6206897i: 1
-0.0689655 -0.4827586i: 1
-0.0689655 -0.3448276i: 1
-0.0689655 -0.2068966i: 1
-0.0689655 -0.0689655i: 1
-0.0689655 +0.0689655i: 1
-0.0689655 +0.2068966i: 1
-0.0689655 +0.3448276i: 1
-0.0689655 +0.4827586i: 1
-0.0689655 +0.6206897i: 1
-0.0689655 +0.7586207i: 1
-0.0689655 +1.0344828i: 0
-0.0689655 +1.1724138i: 0
-0.0689655 +1.3103448i: 0
+0.0689655 -0.7586207i: 0
+0.0689655 -0.4827586i: 1
+0.0689655 -0.3448276i: 1
+0.0689655 -0.2068966i: 1
+0.0689655 -0.0689655i: 1
+0.0689655 +0.0689655i: 1
+0.0689655 +0.2068966i: 1
+0.0689655 +0.3448276i: 1
+0.0689655 +0.4827586i: 1
+0.0689655 +0.7586207i: 0
+0.2068966 -1.8620690i: 0
+0.2068966 -0.6206897i: 0
+0.2068966 -0.4827586i: 1
+0.2068966 -0.3448276i: 1
+0.2068966 -0.2068966i: 1
+0.2068966 -0.0689655i: 1
+0.2068966 +0.0689655i: 1
+0.2068966 +0.2068966i: 1
+0.2068966 +0.3448276i: 1
+0.2068966 +0.4827586i: 1
+0.2068966 +0.6206897i: 0
+0.2068966 +0.7586207i: 0
+0.2068966 +1.0344828i: 0
+0.2068966 +1.4482759i: 0
+0.2068966 +1.5862069i: 0
+0.3448276 -0.4827586i: 0
+0.3448276 -0.3448276i: 1
+0.3448276 -0.2068966i: 1
+0.3448276 +0.2068966i: 1
+0.3448276 +0.3448276i: 1
+0.3448276 +0.4827586i: 0
+0.3448276 +1.7241379i: 0
+0.4827586 -0.3448276i: 0
+0.4827586 -0.2068966i: 0
+0.4827586 +0.0689655i: 0
+0.4827586 +0.2068966i: 0
+0.4827586 +0.3448276i: 0
+0.4827586 +1.7241379i: 0
+0.6206897 +0.4827586i: 0
+0.6206897 +1.7241379i: 0
+0.7586207 +0.4827586i: 0
+0.7586207 +1.4482759i: 0
+0.7586207 +1.7241379i: 0
+0.8965517 -1.7241379i: 0
+0.8965517 -0.4827586i: 0
+0.8965517 -0.2068966i: 0
+1.0344828 -1.5862069i: 0
+1.3103448 -1.0344828i: 0
+1.3103448 +1.1724138i: 0
+1.4482759 -0.4827586i: 0
+1.4482759 +0.3448276i: 0
+1.5862069 -0.6206897i: 0
+1.5862069 +0.2068966i: 0
+1.7241379 -0.4827586i: 0
+1.8620690 +0.2068966i: 0
I tried to eliminate testcases that are on a edge where different rounding methods, a higher number of iterations or different limits (other than $|z_{16}|>2$) yield different results. But it may still have some that are on a edge.
1 answer
+3
−0
Haskell, 66 bytes
-7 bytes thanks to Razetime
import Data.Complex
(\c->(<2).magnitude$(iterate(\z->z*z+c)c)!!16)
I use the fact, that if the result is bigger than 2, it will be forever then, so I don't need to check this relation in each iterations. It is enough to check it in the end.
iterate(\z->z*z+c)c gives us a list of the results, the original c is the first, then I take 16 elements from the list, so with the last function I get the last element which was iterated 15 times, that is $z_{16}$ (the first 0 times, the nth element n-1 times). Then I get the absolute value, and compare it with 2.
Testing:
import Data.Complex
f=(\c->(<2).magnitude $ last $ take 16$ iterate (\z -> z*z+c) c)
map f [(-1.862069) :+ (-0.3448276),0.3448276 :+ 0.3448276,0.2068966 :+ 0.7586207]
gives
[False,True,False]
Or if you like it better to print to the standard output:
main = do print $ map f [(-1.8620690) :+ (-0.3448276), 0.3448276 :+ 0.3448276, 0.2068966 :+ 0.7586207]
main
2 comment threads
Can I use a 3rd party package?
(3 comments)
using indexing to shorten the answer
(1 comment)
0 comment threads