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Challenges Is it part of the mandelbrot set?

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 ...

1 answer  ·  posted 1y ago by H_H‭  ·  last activity 1y ago by Arpad Horvath‭

Question code-golf
#4: Post edited by user avatar H_H‭ · 2023-09-22T07:16:14Z (about 1 year ago)
  • 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.
  • - 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.
  • - Optionally, if printing needs more characters, 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.
  • 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
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  • -0.0689655 +1.1724138i: 0
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  • +0.0689655 -0.7586207i: 0
  • +0.0689655 -0.4827586i: 1
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  • +0.0689655 +0.3448276i: 1
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  • +0.0689655 +0.7586207i: 0
  • +0.2068966 -1.8620690i: 0
  • +0.2068966 -0.6206897i: 0
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  • +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.
#3: Post edited by user avatar H_H‭ · 2023-09-22T07:14:51Z (about 1 year ago)
  • 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.
  • - 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 "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.
  • 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.
  • - 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.
  • - Optionally, if printing needs more characters, 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.
#2: Post edited by user avatar H_H‭ · 2023-09-20T16:04:32Z (about 1 year ago)
  • 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 $|c|≤2$
  • - At least up to 16 iterations have to be checked. If more is easier to do, do more.
  • - 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.
  • - 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.
  • - Output one value for inputs in the mandelbrot set (after 16 iterations) and a different value for inputs outside the mandelbrot set. If it is easier you can return the number of iterations till $|z_{n}|>2$ is reached.
  • Shortest code wins.
  • ## Non-golfed Examples
  • Python example using native complex number
  • ```
  • 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 ) )
  • ```
  • C example using fractions:
  • ```
  • #include <stdio.h>
  • #include <stdlib.h>
  • /*
  • Return 0 when the input is not in the mandelbrot set
  • Return 1 when the input is in the mandelbrot set
  • cr: real part of the input, as numerator of a fraction
  • ci: imaginary part of the input, as numerator of a fraction
  • denominator: the denominator of all fractions
  • iterations: Up to how many iterations we check.
  • */
  • int insideMandelbrot(int cr, int ci, int denominator, int iterations)
  • {
  • int zr=cr;
  • int zi=ci;
  • while( zr<2*denominator )
  • {
  • int t = zr*zr/denominator - zi*zi/denominator;
  • zi = zr*zi/denominator*2 + ci;
  • zr = t + cr;
  • if( !iterations-- )
  • { return 1; }
  • }
  • return 0;
  • }
  • int main(int argc, char **argv)
  • {
  • if( argc<3 )
  • { return 0; }
  • //We use 2 fraction to store the complex number
  • //One fraction for the real one fraction for the imaginary part
  • //The fraction have a fixed denominator of 512.
  • //Both are expected to be between -1024 (for -2) and 1024 (for 2)
  • //cr is the real part. ci the imaginary part
  • int cr = atoi( argv[1] );
  • int ci = atoi( argv[2] );
  • puts( insideMandelbrot(cr,ci,512,16) ? "Inside" : "Outside" );
  • }
  • ```
  • ## 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.
  • 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.
  • - 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 "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: Initial revision by user avatar H_H‭ · 2023-09-12T16:01:52Z (about 1 year ago)
Is it part of the mandelbrot set?
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 $|c|≤2$
- At least up to 16 iterations have to be checked. If more is easier to do, do more.
- 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.
- 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.
- Output one value for inputs in the mandelbrot set (after 16 iterations) and a different value for inputs outside the mandelbrot set. If it is easier you can return the number of iterations till  $|z_{n}|>2$ is reached.


Shortest code wins.

## Non-golfed Examples

Python example using native complex number
```
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 ) )

```

C example using fractions:
```
#include <stdio.h>
#include <stdlib.h>

/*
Return 0 when the input is not in the mandelbrot set
Return 1 when the input is in the mandelbrot set

cr: real part of the input, as numerator of a fraction
ci: imaginary part of the input, as numerator of a fraction
denominator: the denominator of all fractions
iterations: Up to how many iterations we check.
*/
int insideMandelbrot(int cr, int ci, int denominator, int iterations)
{
  int zr=cr;
  int zi=ci;
  while( zr<2*denominator )
    {
      int t = zr*zr/denominator - zi*zi/denominator;
      zi = zr*zi/denominator*2 + ci;
      zr = t + cr;
      if( !iterations-- )
        { return 1; }
    }
  return 0;
}

int main(int argc, char **argv)
{
  if( argc<3 )
    { return 0; }

  //We use 2 fraction to store the complex number
  //One fraction for the real one fraction for the imaginary part
  //The fraction have a fixed denominator of 512.
  //Both are expected to be between -1024 (for -2) and 1024 (for 2)
  //cr is the real part. ci the imaginary part
  int cr = atoi( argv[1] );
  int ci = atoi( argv[2] );

  puts( insideMandelbrot(cr,ci,512,16) ? "Inside" : "Outside" );
}

```

## 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.