Increase performance creating Mandelbrot set in python
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I created a program in python that generates an image of the mandelbrot set. The only problem I have is that the program is quite slow, it takes about a quarter of an hour to generate the following image of 2000 by 3000 pixels:
I first created a matrix of complex numbers using numpy according to amount of pixels. I also created an array for the image generation.
import numpy as np
from PIL import Image
z = 0
real_axis = np.linspace(-2,1,num=3000)
imaginary_axis = np.linspace(1,-1,num=2000)
complex_grid = [[complex(np.float64(a),np.float64(b)) for a in real_axis] for b in imaginary_axis]
pixel_grid = np.zeros((2000,3000,3),dtype=np.uint8)
Then I check whether each complex number is in the mandelbrot set or not and give it an RGB colour code accordingly.
for complex_list in complex_grid:
for complex_number in complex_list:
for iteration in range(255):
z = z**2 + complex_number
if (z.real**2+z.imag**2)**0.5 > 2:
pixel_grid[complex_grid.index(complex_list),complex_list.index(complex_number)]=[iteration,iteration,iteration]
break
else:
continue
z = 0
Finally I generate the image using the PIL library.
mandelbrot = Image.fromarray(pixel_grid)
mandelbrot.save("mandelbrot.png")
I am using jupyter notebook and python 3. Hopefully some of you can help me improve the performance of this program or other aspects of it.
performance python-3.x image fractals complex-numbers
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add a comment |
$begingroup$
I created a program in python that generates an image of the mandelbrot set. The only problem I have is that the program is quite slow, it takes about a quarter of an hour to generate the following image of 2000 by 3000 pixels:
I first created a matrix of complex numbers using numpy according to amount of pixels. I also created an array for the image generation.
import numpy as np
from PIL import Image
z = 0
real_axis = np.linspace(-2,1,num=3000)
imaginary_axis = np.linspace(1,-1,num=2000)
complex_grid = [[complex(np.float64(a),np.float64(b)) for a in real_axis] for b in imaginary_axis]
pixel_grid = np.zeros((2000,3000,3),dtype=np.uint8)
Then I check whether each complex number is in the mandelbrot set or not and give it an RGB colour code accordingly.
for complex_list in complex_grid:
for complex_number in complex_list:
for iteration in range(255):
z = z**2 + complex_number
if (z.real**2+z.imag**2)**0.5 > 2:
pixel_grid[complex_grid.index(complex_list),complex_list.index(complex_number)]=[iteration,iteration,iteration]
break
else:
continue
z = 0
Finally I generate the image using the PIL library.
mandelbrot = Image.fromarray(pixel_grid)
mandelbrot.save("mandelbrot.png")
I am using jupyter notebook and python 3. Hopefully some of you can help me improve the performance of this program or other aspects of it.
performance python-3.x image fractals complex-numbers
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If you want to skip points that are known to be inside the Mandelbrot set, the main cardioid and period bulbs can be skipped. If you skip processing points contained within the main cardioid and the main disk, you can significantly speed up the program. See also this resource for a further analysis of the main cardioid and disk. This optimization becomes less effective as you zoom in on the edges and becomes completely ineffective if these regions are off-screen.
$endgroup$
– Cornstalks
5 hours ago
add a comment |
$begingroup$
I created a program in python that generates an image of the mandelbrot set. The only problem I have is that the program is quite slow, it takes about a quarter of an hour to generate the following image of 2000 by 3000 pixels:
I first created a matrix of complex numbers using numpy according to amount of pixels. I also created an array for the image generation.
import numpy as np
from PIL import Image
z = 0
real_axis = np.linspace(-2,1,num=3000)
imaginary_axis = np.linspace(1,-1,num=2000)
complex_grid = [[complex(np.float64(a),np.float64(b)) for a in real_axis] for b in imaginary_axis]
pixel_grid = np.zeros((2000,3000,3),dtype=np.uint8)
Then I check whether each complex number is in the mandelbrot set or not and give it an RGB colour code accordingly.
for complex_list in complex_grid:
for complex_number in complex_list:
for iteration in range(255):
z = z**2 + complex_number
if (z.real**2+z.imag**2)**0.5 > 2:
pixel_grid[complex_grid.index(complex_list),complex_list.index(complex_number)]=[iteration,iteration,iteration]
break
else:
continue
z = 0
Finally I generate the image using the PIL library.
mandelbrot = Image.fromarray(pixel_grid)
mandelbrot.save("mandelbrot.png")
I am using jupyter notebook and python 3. Hopefully some of you can help me improve the performance of this program or other aspects of it.
performance python-3.x image fractals complex-numbers
$endgroup$
I created a program in python that generates an image of the mandelbrot set. The only problem I have is that the program is quite slow, it takes about a quarter of an hour to generate the following image of 2000 by 3000 pixels:
I first created a matrix of complex numbers using numpy according to amount of pixels. I also created an array for the image generation.
import numpy as np
from PIL import Image
z = 0
real_axis = np.linspace(-2,1,num=3000)
imaginary_axis = np.linspace(1,-1,num=2000)
complex_grid = [[complex(np.float64(a),np.float64(b)) for a in real_axis] for b in imaginary_axis]
pixel_grid = np.zeros((2000,3000,3),dtype=np.uint8)
Then I check whether each complex number is in the mandelbrot set or not and give it an RGB colour code accordingly.
for complex_list in complex_grid:
for complex_number in complex_list:
for iteration in range(255):
z = z**2 + complex_number
if (z.real**2+z.imag**2)**0.5 > 2:
pixel_grid[complex_grid.index(complex_list),complex_list.index(complex_number)]=[iteration,iteration,iteration]
break
else:
continue
z = 0
Finally I generate the image using the PIL library.
mandelbrot = Image.fromarray(pixel_grid)
mandelbrot.save("mandelbrot.png")
I am using jupyter notebook and python 3. Hopefully some of you can help me improve the performance of this program or other aspects of it.
performance python-3.x image fractals complex-numbers
performance python-3.x image fractals complex-numbers
asked 17 hours ago
IanIan
1415
1415
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If you want to skip points that are known to be inside the Mandelbrot set, the main cardioid and period bulbs can be skipped. If you skip processing points contained within the main cardioid and the main disk, you can significantly speed up the program. See also this resource for a further analysis of the main cardioid and disk. This optimization becomes less effective as you zoom in on the edges and becomes completely ineffective if these regions are off-screen.
$endgroup$
– Cornstalks
5 hours ago
add a comment |
$begingroup$
If you want to skip points that are known to be inside the Mandelbrot set, the main cardioid and period bulbs can be skipped. If you skip processing points contained within the main cardioid and the main disk, you can significantly speed up the program. See also this resource for a further analysis of the main cardioid and disk. This optimization becomes less effective as you zoom in on the edges and becomes completely ineffective if these regions are off-screen.
$endgroup$
– Cornstalks
5 hours ago
$begingroup$
If you want to skip points that are known to be inside the Mandelbrot set, the main cardioid and period bulbs can be skipped. If you skip processing points contained within the main cardioid and the main disk, you can significantly speed up the program. See also this resource for a further analysis of the main cardioid and disk. This optimization becomes less effective as you zoom in on the edges and becomes completely ineffective if these regions are off-screen.
$endgroup$
– Cornstalks
5 hours ago
$begingroup$
If you want to skip points that are known to be inside the Mandelbrot set, the main cardioid and period bulbs can be skipped. If you skip processing points contained within the main cardioid and the main disk, you can significantly speed up the program. See also this resource for a further analysis of the main cardioid and disk. This optimization becomes less effective as you zoom in on the edges and becomes completely ineffective if these regions are off-screen.
$endgroup$
– Cornstalks
5 hours ago
add a comment |
4 Answers
4
active
oldest
votes
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I'm going to reuse some parts of the answer I recently posted here on Code Review.
Losing your Loops
(Most) loops are damn slow in Python. Especially multiple nested loops.
NumPy can help to vectorize your code, i.e. in this case that more
of the looping is done in the C backend instead of in the Python
interpreter. I would highly recommend to have a listen to the talk
Losing your Loops: Fast Numerical Computing with NumPy by Jake
VanderPlas.
All those loops used to generate the complex grid followed by the nested loops used to iterate over the grid and the image are slow when left to the Python interpreter. Fortunately, NumPy can take quite a lot of this burden off of you.
For example
real_axis = np.linspace(-2, 1, num=3000)
imaginary_axis = np.linspace(1, -1, num=2000)
complex_grid = [[complex(np.float64(a),np.float64(b)) for a in real_axis] for b in imaginary_axis]
could become
n_rows, n_cols = 2000, 3000
complex_grid_np = np.zeros((n_rows, n_cols), dtype=np.complex)
real, imag = np.meshgrid(real_axis, imaginary_axis)
complex_grid_np.real = real
complex_grid_np.imag = imag
No loops, just plain simple NumPy.
Same goes for
for complex_list in complex_grid:
for complex_number in complex_list:
for iteration in range(255):
z = z**2 + complex_number
if (z.real**2+z.imag**2)**0.5 > 2:
pixel_grid[complex_grid.index(complex_list),complex_list.index(complex_number)]=[iteration,iteration,iteration]
break
else:
continue
z = 0
can be transformed to
z_grid_np = np.zeros_like(complex_grid_np)
elements_todo = np.ones((n_rows, n_cols), dtype=bool)
for iteration in range(255):
z_grid_np[elements_todo] =
z_grid_np[elements_todo]**2 + complex_grid_np[elements_todo]
mask = np.logical_and(np.absolute(z_grid_np) > 2, elements_todo)
pixel_grid_np[mask, :] = (iteration, iteration, iteration)
elements_todo = np.logical_and(elements_todo, np.logical_not(mask))
which is just a single loop instead of three nested ones. Here, a little more trickery was needed to treat the break
case the same way as you did. elements_todo
is used to only compute updates on the z
value if it has not been marked as done. There might also be a better solution without this.
I added the following lines
complex_grid_close = np.allclose(np.array(complex_grid), complex_grid_np)
pixel_grid_close = np.allclose(pixel_grid, pixel_grid_np)
print("Results were similar: {}".format(all((complex_grid_close, pixel_grid_close))))
to validate my results against your reference implementation.
The vectorized code is about 9-10x faster on my machine for several n_rows/n_cols
combinations I tested. E.g. for n_rows, n_cols = 1000, 1500
:
Looped generation took 61.989842s
Vectorized generation took 6.656926s
Results were similar: True
$endgroup$
add a comment |
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This will cover performance, as well as Python style.
Save constants in one place
You currently have the magic numbers 2000 and 3000, the resolution of your image. Save these to variables perhaps named X
, Y
or W
, H
.
Mention your requirements
You don't just rely on Python 3 and Jupyter - you rely on numpy and pillow. These should go in a requirements.txt if you don't already have one.
Don't save your complex grid
At all. complex_number
should be formed dynamically in the loop based on range
expressions.
Disclaimer: if you're vectorizing (which you should do), then the opposite applies - you would keep your complex grid, and lose some loops.
Don't use index
lookups
You're using index
to get your coordinates. Don't do this - form the coordinates in your loops, as well.
Mandelbrot is symmetrical
Notice that it's mirror-imaged. This means you can halve your computation time and save every pixel to the top and bottom half.
In a bit I'll show some example code accommodating all of the suggestions above. Just do (nearly) what @Alex says and I'd gotten halfway through implementing, with one difference: accommodate the symmetry optimization I described.
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add a comment |
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Mandelbrot-specific optimisations
These can be combined with the Python-specific optimisations from the other answers.
Avoid the redundant square root
if (z.real**2+z.imag**2)**0.5 > 2:
is equivalent to
if z.real ** 2 + z.imag ** 2 > 4:
(simply square both sides of the original comparison to get the optimised comparison)
Avoid squaring unless you are using it
Any points that get further than 2 from the origin will continue to escape towards infinity. So it isn't important whether you check that a point has gone outside a circle of radius 2, or that it has gone outside some other finite shape that fully contains that circle. For example, checking that the point is outside a square instead of a circle avoids having to square the real and imaginary parts. It also means you will need slightly more iterations, but very few and this should be outweighed by having each iteration faster.
For example:
if (z.real**2+z.imag**2)**0.5 > 2: # if z is outside the circle
could be replaced by
if not (-2 < z.real < 2 and -2 < z.imag < 2): # if z is outside the square
The exception to this suggestion is if the circle is important to your output. If you simply plot points inside the set as black, and points outside the set as white, then the image will be identical with either approach. However, if you count the number of iterations a point takes to escape and use this to determine the colour of points outside the set, then the shape of the stripes of colour will be different with a square boundary than with a circular boundary. The interior of the set will be identical, but the colours outside will be arranged in different shapes.
In your example image, not much is visible of the stripes of colour, with most of the exterior and interior being black. In this case I doubt there would be a significant difference in appearance using this optimisation. However, if you change to displaying wider stripes in future, this optimisation may need to be removed (depending on what appearance you want).
Hard-code as much of the interior as you can
The interior of the set takes far longer to calculate than the exterior. Each pixel in the interior takes a guaranteed 255 iterations (or more if you increase the maximum iterations for even higher quality images), whereas each pixel in the exterior takes less than this. The vast majority of the exterior pixels take only a few iterations.
If you want the code to be adaptable for zooming in to arbitrary positions, then you won't know in advance which parts of the image are going to be interior points. However, if you only want this code to generate this one image of the whole set, then you can get a significant improvement in speed by avoiding calculating pixels you know are interior. For example, if you check whether the pixel is in the main cardioid or one of the large circles, you can assign all those pixels an iteration count of 255 without actually doing any iteration. The more you increase the maximum iterations, the more circles it will be worthwhile excluding in advance, as the difference in calculation time between the average exterior pixel and the average interior pixel will continue to diverge dramatically.
I don't know the exact centres and radii of these circles, or an exact equation for the cardioid, but rough estimates that are chosen to not overlap the exterior will still make a big difference to the speed. Even excluding some squares chosen by eye that are entirely in the interior would help.
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add a comment |
$begingroup$
I'm not a python expert. I am pretty good with Mandlebrot generation (I've spent a lot of time on my custom Julia Set generator.)
So I'll say this: optimize the heck out of stuff that will be running many iterations. Forget about clean-code or nice OOP principles. For lots-of-iterations stuff like this, you want as nitty gritty as possible.
So let's take a look at your interior-most loop:
z = z**2 + complex_number
if (z.real**2+z.imag**2)**0.5 > 2:
pixel_grid[complex_grid.index(complex_list),complex_list.index(complex_number)]=[iteration,iteration,iteration]
break
else:
continue
Imagine what's happening behind the scenes in memory with just that very first line. You've got an instance of a complex number. You want to square it... so it has to create another instance of a complex object to hold the squared value. Then, you're adding another complex number to it - which means you're creating another instance of Complex to hold the result of the addition.
You're creating object instances left and right, and you're doing it on an order of 3000 x 2000 x 255 times. Creating several class instances doesn't sound like much, but when you're doing it a billion times, it kinda drags things down.
Compare that with pseudocode like:
px = num.real
py = num.imag
while
tmppx = px
px = px * px - py * py + num.real
py = 2 * tmppx * py + num.imag
if condition-for-hitting-escape
stuff
if condition-for-hitting-max-iter
moreStuff
No objects are getting created and destroyed. It's boiled down to be as efficient as possible. It's not as nice looking... but when you're doing something a billion times, shaving off even a millionth of a second results in saving 15 minutes.
And as someone else mentioned, you want to simplify the logic so that you don't have to do the square-root operation - and if you're okay with small variations in how the gradient is colored, changing the "magnitude" check with "are x or y within a bounding box".
Aka, the more things you can remove out of that runs-a-billion-times loop, the better off you'll be.
$endgroup$
add a comment |
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4 Answers
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4 Answers
4
active
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$begingroup$
I'm going to reuse some parts of the answer I recently posted here on Code Review.
Losing your Loops
(Most) loops are damn slow in Python. Especially multiple nested loops.
NumPy can help to vectorize your code, i.e. in this case that more
of the looping is done in the C backend instead of in the Python
interpreter. I would highly recommend to have a listen to the talk
Losing your Loops: Fast Numerical Computing with NumPy by Jake
VanderPlas.
All those loops used to generate the complex grid followed by the nested loops used to iterate over the grid and the image are slow when left to the Python interpreter. Fortunately, NumPy can take quite a lot of this burden off of you.
For example
real_axis = np.linspace(-2, 1, num=3000)
imaginary_axis = np.linspace(1, -1, num=2000)
complex_grid = [[complex(np.float64(a),np.float64(b)) for a in real_axis] for b in imaginary_axis]
could become
n_rows, n_cols = 2000, 3000
complex_grid_np = np.zeros((n_rows, n_cols), dtype=np.complex)
real, imag = np.meshgrid(real_axis, imaginary_axis)
complex_grid_np.real = real
complex_grid_np.imag = imag
No loops, just plain simple NumPy.
Same goes for
for complex_list in complex_grid:
for complex_number in complex_list:
for iteration in range(255):
z = z**2 + complex_number
if (z.real**2+z.imag**2)**0.5 > 2:
pixel_grid[complex_grid.index(complex_list),complex_list.index(complex_number)]=[iteration,iteration,iteration]
break
else:
continue
z = 0
can be transformed to
z_grid_np = np.zeros_like(complex_grid_np)
elements_todo = np.ones((n_rows, n_cols), dtype=bool)
for iteration in range(255):
z_grid_np[elements_todo] =
z_grid_np[elements_todo]**2 + complex_grid_np[elements_todo]
mask = np.logical_and(np.absolute(z_grid_np) > 2, elements_todo)
pixel_grid_np[mask, :] = (iteration, iteration, iteration)
elements_todo = np.logical_and(elements_todo, np.logical_not(mask))
which is just a single loop instead of three nested ones. Here, a little more trickery was needed to treat the break
case the same way as you did. elements_todo
is used to only compute updates on the z
value if it has not been marked as done. There might also be a better solution without this.
I added the following lines
complex_grid_close = np.allclose(np.array(complex_grid), complex_grid_np)
pixel_grid_close = np.allclose(pixel_grid, pixel_grid_np)
print("Results were similar: {}".format(all((complex_grid_close, pixel_grid_close))))
to validate my results against your reference implementation.
The vectorized code is about 9-10x faster on my machine for several n_rows/n_cols
combinations I tested. E.g. for n_rows, n_cols = 1000, 1500
:
Looped generation took 61.989842s
Vectorized generation took 6.656926s
Results were similar: True
$endgroup$
add a comment |
$begingroup$
I'm going to reuse some parts of the answer I recently posted here on Code Review.
Losing your Loops
(Most) loops are damn slow in Python. Especially multiple nested loops.
NumPy can help to vectorize your code, i.e. in this case that more
of the looping is done in the C backend instead of in the Python
interpreter. I would highly recommend to have a listen to the talk
Losing your Loops: Fast Numerical Computing with NumPy by Jake
VanderPlas.
All those loops used to generate the complex grid followed by the nested loops used to iterate over the grid and the image are slow when left to the Python interpreter. Fortunately, NumPy can take quite a lot of this burden off of you.
For example
real_axis = np.linspace(-2, 1, num=3000)
imaginary_axis = np.linspace(1, -1, num=2000)
complex_grid = [[complex(np.float64(a),np.float64(b)) for a in real_axis] for b in imaginary_axis]
could become
n_rows, n_cols = 2000, 3000
complex_grid_np = np.zeros((n_rows, n_cols), dtype=np.complex)
real, imag = np.meshgrid(real_axis, imaginary_axis)
complex_grid_np.real = real
complex_grid_np.imag = imag
No loops, just plain simple NumPy.
Same goes for
for complex_list in complex_grid:
for complex_number in complex_list:
for iteration in range(255):
z = z**2 + complex_number
if (z.real**2+z.imag**2)**0.5 > 2:
pixel_grid[complex_grid.index(complex_list),complex_list.index(complex_number)]=[iteration,iteration,iteration]
break
else:
continue
z = 0
can be transformed to
z_grid_np = np.zeros_like(complex_grid_np)
elements_todo = np.ones((n_rows, n_cols), dtype=bool)
for iteration in range(255):
z_grid_np[elements_todo] =
z_grid_np[elements_todo]**2 + complex_grid_np[elements_todo]
mask = np.logical_and(np.absolute(z_grid_np) > 2, elements_todo)
pixel_grid_np[mask, :] = (iteration, iteration, iteration)
elements_todo = np.logical_and(elements_todo, np.logical_not(mask))
which is just a single loop instead of three nested ones. Here, a little more trickery was needed to treat the break
case the same way as you did. elements_todo
is used to only compute updates on the z
value if it has not been marked as done. There might also be a better solution without this.
I added the following lines
complex_grid_close = np.allclose(np.array(complex_grid), complex_grid_np)
pixel_grid_close = np.allclose(pixel_grid, pixel_grid_np)
print("Results were similar: {}".format(all((complex_grid_close, pixel_grid_close))))
to validate my results against your reference implementation.
The vectorized code is about 9-10x faster on my machine for several n_rows/n_cols
combinations I tested. E.g. for n_rows, n_cols = 1000, 1500
:
Looped generation took 61.989842s
Vectorized generation took 6.656926s
Results were similar: True
$endgroup$
add a comment |
$begingroup$
I'm going to reuse some parts of the answer I recently posted here on Code Review.
Losing your Loops
(Most) loops are damn slow in Python. Especially multiple nested loops.
NumPy can help to vectorize your code, i.e. in this case that more
of the looping is done in the C backend instead of in the Python
interpreter. I would highly recommend to have a listen to the talk
Losing your Loops: Fast Numerical Computing with NumPy by Jake
VanderPlas.
All those loops used to generate the complex grid followed by the nested loops used to iterate over the grid and the image are slow when left to the Python interpreter. Fortunately, NumPy can take quite a lot of this burden off of you.
For example
real_axis = np.linspace(-2, 1, num=3000)
imaginary_axis = np.linspace(1, -1, num=2000)
complex_grid = [[complex(np.float64(a),np.float64(b)) for a in real_axis] for b in imaginary_axis]
could become
n_rows, n_cols = 2000, 3000
complex_grid_np = np.zeros((n_rows, n_cols), dtype=np.complex)
real, imag = np.meshgrid(real_axis, imaginary_axis)
complex_grid_np.real = real
complex_grid_np.imag = imag
No loops, just plain simple NumPy.
Same goes for
for complex_list in complex_grid:
for complex_number in complex_list:
for iteration in range(255):
z = z**2 + complex_number
if (z.real**2+z.imag**2)**0.5 > 2:
pixel_grid[complex_grid.index(complex_list),complex_list.index(complex_number)]=[iteration,iteration,iteration]
break
else:
continue
z = 0
can be transformed to
z_grid_np = np.zeros_like(complex_grid_np)
elements_todo = np.ones((n_rows, n_cols), dtype=bool)
for iteration in range(255):
z_grid_np[elements_todo] =
z_grid_np[elements_todo]**2 + complex_grid_np[elements_todo]
mask = np.logical_and(np.absolute(z_grid_np) > 2, elements_todo)
pixel_grid_np[mask, :] = (iteration, iteration, iteration)
elements_todo = np.logical_and(elements_todo, np.logical_not(mask))
which is just a single loop instead of three nested ones. Here, a little more trickery was needed to treat the break
case the same way as you did. elements_todo
is used to only compute updates on the z
value if it has not been marked as done. There might also be a better solution without this.
I added the following lines
complex_grid_close = np.allclose(np.array(complex_grid), complex_grid_np)
pixel_grid_close = np.allclose(pixel_grid, pixel_grid_np)
print("Results were similar: {}".format(all((complex_grid_close, pixel_grid_close))))
to validate my results against your reference implementation.
The vectorized code is about 9-10x faster on my machine for several n_rows/n_cols
combinations I tested. E.g. for n_rows, n_cols = 1000, 1500
:
Looped generation took 61.989842s
Vectorized generation took 6.656926s
Results were similar: True
$endgroup$
I'm going to reuse some parts of the answer I recently posted here on Code Review.
Losing your Loops
(Most) loops are damn slow in Python. Especially multiple nested loops.
NumPy can help to vectorize your code, i.e. in this case that more
of the looping is done in the C backend instead of in the Python
interpreter. I would highly recommend to have a listen to the talk
Losing your Loops: Fast Numerical Computing with NumPy by Jake
VanderPlas.
All those loops used to generate the complex grid followed by the nested loops used to iterate over the grid and the image are slow when left to the Python interpreter. Fortunately, NumPy can take quite a lot of this burden off of you.
For example
real_axis = np.linspace(-2, 1, num=3000)
imaginary_axis = np.linspace(1, -1, num=2000)
complex_grid = [[complex(np.float64(a),np.float64(b)) for a in real_axis] for b in imaginary_axis]
could become
n_rows, n_cols = 2000, 3000
complex_grid_np = np.zeros((n_rows, n_cols), dtype=np.complex)
real, imag = np.meshgrid(real_axis, imaginary_axis)
complex_grid_np.real = real
complex_grid_np.imag = imag
No loops, just plain simple NumPy.
Same goes for
for complex_list in complex_grid:
for complex_number in complex_list:
for iteration in range(255):
z = z**2 + complex_number
if (z.real**2+z.imag**2)**0.5 > 2:
pixel_grid[complex_grid.index(complex_list),complex_list.index(complex_number)]=[iteration,iteration,iteration]
break
else:
continue
z = 0
can be transformed to
z_grid_np = np.zeros_like(complex_grid_np)
elements_todo = np.ones((n_rows, n_cols), dtype=bool)
for iteration in range(255):
z_grid_np[elements_todo] =
z_grid_np[elements_todo]**2 + complex_grid_np[elements_todo]
mask = np.logical_and(np.absolute(z_grid_np) > 2, elements_todo)
pixel_grid_np[mask, :] = (iteration, iteration, iteration)
elements_todo = np.logical_and(elements_todo, np.logical_not(mask))
which is just a single loop instead of three nested ones. Here, a little more trickery was needed to treat the break
case the same way as you did. elements_todo
is used to only compute updates on the z
value if it has not been marked as done. There might also be a better solution without this.
I added the following lines
complex_grid_close = np.allclose(np.array(complex_grid), complex_grid_np)
pixel_grid_close = np.allclose(pixel_grid, pixel_grid_np)
print("Results were similar: {}".format(all((complex_grid_close, pixel_grid_close))))
to validate my results against your reference implementation.
The vectorized code is about 9-10x faster on my machine for several n_rows/n_cols
combinations I tested. E.g. for n_rows, n_cols = 1000, 1500
:
Looped generation took 61.989842s
Vectorized generation took 6.656926s
Results were similar: True
edited 15 hours ago
Reinderien
4,600823
4,600823
answered 15 hours ago
AlexAlex
866316
866316
add a comment |
add a comment |
$begingroup$
This will cover performance, as well as Python style.
Save constants in one place
You currently have the magic numbers 2000 and 3000, the resolution of your image. Save these to variables perhaps named X
, Y
or W
, H
.
Mention your requirements
You don't just rely on Python 3 and Jupyter - you rely on numpy and pillow. These should go in a requirements.txt if you don't already have one.
Don't save your complex grid
At all. complex_number
should be formed dynamically in the loop based on range
expressions.
Disclaimer: if you're vectorizing (which you should do), then the opposite applies - you would keep your complex grid, and lose some loops.
Don't use index
lookups
You're using index
to get your coordinates. Don't do this - form the coordinates in your loops, as well.
Mandelbrot is symmetrical
Notice that it's mirror-imaged. This means you can halve your computation time and save every pixel to the top and bottom half.
In a bit I'll show some example code accommodating all of the suggestions above. Just do (nearly) what @Alex says and I'd gotten halfway through implementing, with one difference: accommodate the symmetry optimization I described.
$endgroup$
add a comment |
$begingroup$
This will cover performance, as well as Python style.
Save constants in one place
You currently have the magic numbers 2000 and 3000, the resolution of your image. Save these to variables perhaps named X
, Y
or W
, H
.
Mention your requirements
You don't just rely on Python 3 and Jupyter - you rely on numpy and pillow. These should go in a requirements.txt if you don't already have one.
Don't save your complex grid
At all. complex_number
should be formed dynamically in the loop based on range
expressions.
Disclaimer: if you're vectorizing (which you should do), then the opposite applies - you would keep your complex grid, and lose some loops.
Don't use index
lookups
You're using index
to get your coordinates. Don't do this - form the coordinates in your loops, as well.
Mandelbrot is symmetrical
Notice that it's mirror-imaged. This means you can halve your computation time and save every pixel to the top and bottom half.
In a bit I'll show some example code accommodating all of the suggestions above. Just do (nearly) what @Alex says and I'd gotten halfway through implementing, with one difference: accommodate the symmetry optimization I described.
$endgroup$
add a comment |
$begingroup$
This will cover performance, as well as Python style.
Save constants in one place
You currently have the magic numbers 2000 and 3000, the resolution of your image. Save these to variables perhaps named X
, Y
or W
, H
.
Mention your requirements
You don't just rely on Python 3 and Jupyter - you rely on numpy and pillow. These should go in a requirements.txt if you don't already have one.
Don't save your complex grid
At all. complex_number
should be formed dynamically in the loop based on range
expressions.
Disclaimer: if you're vectorizing (which you should do), then the opposite applies - you would keep your complex grid, and lose some loops.
Don't use index
lookups
You're using index
to get your coordinates. Don't do this - form the coordinates in your loops, as well.
Mandelbrot is symmetrical
Notice that it's mirror-imaged. This means you can halve your computation time and save every pixel to the top and bottom half.
In a bit I'll show some example code accommodating all of the suggestions above. Just do (nearly) what @Alex says and I'd gotten halfway through implementing, with one difference: accommodate the symmetry optimization I described.
$endgroup$
This will cover performance, as well as Python style.
Save constants in one place
You currently have the magic numbers 2000 and 3000, the resolution of your image. Save these to variables perhaps named X
, Y
or W
, H
.
Mention your requirements
You don't just rely on Python 3 and Jupyter - you rely on numpy and pillow. These should go in a requirements.txt if you don't already have one.
Don't save your complex grid
At all. complex_number
should be formed dynamically in the loop based on range
expressions.
Disclaimer: if you're vectorizing (which you should do), then the opposite applies - you would keep your complex grid, and lose some loops.
Don't use index
lookups
You're using index
to get your coordinates. Don't do this - form the coordinates in your loops, as well.
Mandelbrot is symmetrical
Notice that it's mirror-imaged. This means you can halve your computation time and save every pixel to the top and bottom half.
In a bit I'll show some example code accommodating all of the suggestions above. Just do (nearly) what @Alex says and I'd gotten halfway through implementing, with one difference: accommodate the symmetry optimization I described.
edited 14 hours ago
answered 16 hours ago
ReinderienReinderien
4,600823
4,600823
add a comment |
add a comment |
$begingroup$
Mandelbrot-specific optimisations
These can be combined with the Python-specific optimisations from the other answers.
Avoid the redundant square root
if (z.real**2+z.imag**2)**0.5 > 2:
is equivalent to
if z.real ** 2 + z.imag ** 2 > 4:
(simply square both sides of the original comparison to get the optimised comparison)
Avoid squaring unless you are using it
Any points that get further than 2 from the origin will continue to escape towards infinity. So it isn't important whether you check that a point has gone outside a circle of radius 2, or that it has gone outside some other finite shape that fully contains that circle. For example, checking that the point is outside a square instead of a circle avoids having to square the real and imaginary parts. It also means you will need slightly more iterations, but very few and this should be outweighed by having each iteration faster.
For example:
if (z.real**2+z.imag**2)**0.5 > 2: # if z is outside the circle
could be replaced by
if not (-2 < z.real < 2 and -2 < z.imag < 2): # if z is outside the square
The exception to this suggestion is if the circle is important to your output. If you simply plot points inside the set as black, and points outside the set as white, then the image will be identical with either approach. However, if you count the number of iterations a point takes to escape and use this to determine the colour of points outside the set, then the shape of the stripes of colour will be different with a square boundary than with a circular boundary. The interior of the set will be identical, but the colours outside will be arranged in different shapes.
In your example image, not much is visible of the stripes of colour, with most of the exterior and interior being black. In this case I doubt there would be a significant difference in appearance using this optimisation. However, if you change to displaying wider stripes in future, this optimisation may need to be removed (depending on what appearance you want).
Hard-code as much of the interior as you can
The interior of the set takes far longer to calculate than the exterior. Each pixel in the interior takes a guaranteed 255 iterations (or more if you increase the maximum iterations for even higher quality images), whereas each pixel in the exterior takes less than this. The vast majority of the exterior pixels take only a few iterations.
If you want the code to be adaptable for zooming in to arbitrary positions, then you won't know in advance which parts of the image are going to be interior points. However, if you only want this code to generate this one image of the whole set, then you can get a significant improvement in speed by avoiding calculating pixels you know are interior. For example, if you check whether the pixel is in the main cardioid or one of the large circles, you can assign all those pixels an iteration count of 255 without actually doing any iteration. The more you increase the maximum iterations, the more circles it will be worthwhile excluding in advance, as the difference in calculation time between the average exterior pixel and the average interior pixel will continue to diverge dramatically.
I don't know the exact centres and radii of these circles, or an exact equation for the cardioid, but rough estimates that are chosen to not overlap the exterior will still make a big difference to the speed. Even excluding some squares chosen by eye that are entirely in the interior would help.
$endgroup$
add a comment |
$begingroup$
Mandelbrot-specific optimisations
These can be combined with the Python-specific optimisations from the other answers.
Avoid the redundant square root
if (z.real**2+z.imag**2)**0.5 > 2:
is equivalent to
if z.real ** 2 + z.imag ** 2 > 4:
(simply square both sides of the original comparison to get the optimised comparison)
Avoid squaring unless you are using it
Any points that get further than 2 from the origin will continue to escape towards infinity. So it isn't important whether you check that a point has gone outside a circle of radius 2, or that it has gone outside some other finite shape that fully contains that circle. For example, checking that the point is outside a square instead of a circle avoids having to square the real and imaginary parts. It also means you will need slightly more iterations, but very few and this should be outweighed by having each iteration faster.
For example:
if (z.real**2+z.imag**2)**0.5 > 2: # if z is outside the circle
could be replaced by
if not (-2 < z.real < 2 and -2 < z.imag < 2): # if z is outside the square
The exception to this suggestion is if the circle is important to your output. If you simply plot points inside the set as black, and points outside the set as white, then the image will be identical with either approach. However, if you count the number of iterations a point takes to escape and use this to determine the colour of points outside the set, then the shape of the stripes of colour will be different with a square boundary than with a circular boundary. The interior of the set will be identical, but the colours outside will be arranged in different shapes.
In your example image, not much is visible of the stripes of colour, with most of the exterior and interior being black. In this case I doubt there would be a significant difference in appearance using this optimisation. However, if you change to displaying wider stripes in future, this optimisation may need to be removed (depending on what appearance you want).
Hard-code as much of the interior as you can
The interior of the set takes far longer to calculate than the exterior. Each pixel in the interior takes a guaranteed 255 iterations (or more if you increase the maximum iterations for even higher quality images), whereas each pixel in the exterior takes less than this. The vast majority of the exterior pixels take only a few iterations.
If you want the code to be adaptable for zooming in to arbitrary positions, then you won't know in advance which parts of the image are going to be interior points. However, if you only want this code to generate this one image of the whole set, then you can get a significant improvement in speed by avoiding calculating pixels you know are interior. For example, if you check whether the pixel is in the main cardioid or one of the large circles, you can assign all those pixels an iteration count of 255 without actually doing any iteration. The more you increase the maximum iterations, the more circles it will be worthwhile excluding in advance, as the difference in calculation time between the average exterior pixel and the average interior pixel will continue to diverge dramatically.
I don't know the exact centres and radii of these circles, or an exact equation for the cardioid, but rough estimates that are chosen to not overlap the exterior will still make a big difference to the speed. Even excluding some squares chosen by eye that are entirely in the interior would help.
$endgroup$
add a comment |
$begingroup$
Mandelbrot-specific optimisations
These can be combined with the Python-specific optimisations from the other answers.
Avoid the redundant square root
if (z.real**2+z.imag**2)**0.5 > 2:
is equivalent to
if z.real ** 2 + z.imag ** 2 > 4:
(simply square both sides of the original comparison to get the optimised comparison)
Avoid squaring unless you are using it
Any points that get further than 2 from the origin will continue to escape towards infinity. So it isn't important whether you check that a point has gone outside a circle of radius 2, or that it has gone outside some other finite shape that fully contains that circle. For example, checking that the point is outside a square instead of a circle avoids having to square the real and imaginary parts. It also means you will need slightly more iterations, but very few and this should be outweighed by having each iteration faster.
For example:
if (z.real**2+z.imag**2)**0.5 > 2: # if z is outside the circle
could be replaced by
if not (-2 < z.real < 2 and -2 < z.imag < 2): # if z is outside the square
The exception to this suggestion is if the circle is important to your output. If you simply plot points inside the set as black, and points outside the set as white, then the image will be identical with either approach. However, if you count the number of iterations a point takes to escape and use this to determine the colour of points outside the set, then the shape of the stripes of colour will be different with a square boundary than with a circular boundary. The interior of the set will be identical, but the colours outside will be arranged in different shapes.
In your example image, not much is visible of the stripes of colour, with most of the exterior and interior being black. In this case I doubt there would be a significant difference in appearance using this optimisation. However, if you change to displaying wider stripes in future, this optimisation may need to be removed (depending on what appearance you want).
Hard-code as much of the interior as you can
The interior of the set takes far longer to calculate than the exterior. Each pixel in the interior takes a guaranteed 255 iterations (or more if you increase the maximum iterations for even higher quality images), whereas each pixel in the exterior takes less than this. The vast majority of the exterior pixels take only a few iterations.
If you want the code to be adaptable for zooming in to arbitrary positions, then you won't know in advance which parts of the image are going to be interior points. However, if you only want this code to generate this one image of the whole set, then you can get a significant improvement in speed by avoiding calculating pixels you know are interior. For example, if you check whether the pixel is in the main cardioid or one of the large circles, you can assign all those pixels an iteration count of 255 without actually doing any iteration. The more you increase the maximum iterations, the more circles it will be worthwhile excluding in advance, as the difference in calculation time between the average exterior pixel and the average interior pixel will continue to diverge dramatically.
I don't know the exact centres and radii of these circles, or an exact equation for the cardioid, but rough estimates that are chosen to not overlap the exterior will still make a big difference to the speed. Even excluding some squares chosen by eye that are entirely in the interior would help.
$endgroup$
Mandelbrot-specific optimisations
These can be combined with the Python-specific optimisations from the other answers.
Avoid the redundant square root
if (z.real**2+z.imag**2)**0.5 > 2:
is equivalent to
if z.real ** 2 + z.imag ** 2 > 4:
(simply square both sides of the original comparison to get the optimised comparison)
Avoid squaring unless you are using it
Any points that get further than 2 from the origin will continue to escape towards infinity. So it isn't important whether you check that a point has gone outside a circle of radius 2, or that it has gone outside some other finite shape that fully contains that circle. For example, checking that the point is outside a square instead of a circle avoids having to square the real and imaginary parts. It also means you will need slightly more iterations, but very few and this should be outweighed by having each iteration faster.
For example:
if (z.real**2+z.imag**2)**0.5 > 2: # if z is outside the circle
could be replaced by
if not (-2 < z.real < 2 and -2 < z.imag < 2): # if z is outside the square
The exception to this suggestion is if the circle is important to your output. If you simply plot points inside the set as black, and points outside the set as white, then the image will be identical with either approach. However, if you count the number of iterations a point takes to escape and use this to determine the colour of points outside the set, then the shape of the stripes of colour will be different with a square boundary than with a circular boundary. The interior of the set will be identical, but the colours outside will be arranged in different shapes.
In your example image, not much is visible of the stripes of colour, with most of the exterior and interior being black. In this case I doubt there would be a significant difference in appearance using this optimisation. However, if you change to displaying wider stripes in future, this optimisation may need to be removed (depending on what appearance you want).
Hard-code as much of the interior as you can
The interior of the set takes far longer to calculate than the exterior. Each pixel in the interior takes a guaranteed 255 iterations (or more if you increase the maximum iterations for even higher quality images), whereas each pixel in the exterior takes less than this. The vast majority of the exterior pixels take only a few iterations.
If you want the code to be adaptable for zooming in to arbitrary positions, then you won't know in advance which parts of the image are going to be interior points. However, if you only want this code to generate this one image of the whole set, then you can get a significant improvement in speed by avoiding calculating pixels you know are interior. For example, if you check whether the pixel is in the main cardioid or one of the large circles, you can assign all those pixels an iteration count of 255 without actually doing any iteration. The more you increase the maximum iterations, the more circles it will be worthwhile excluding in advance, as the difference in calculation time between the average exterior pixel and the average interior pixel will continue to diverge dramatically.
I don't know the exact centres and radii of these circles, or an exact equation for the cardioid, but rough estimates that are chosen to not overlap the exterior will still make a big difference to the speed. Even excluding some squares chosen by eye that are entirely in the interior would help.
edited 5 hours ago
answered 6 hours ago
trichoplaxtrichoplax
542316
542316
add a comment |
add a comment |
$begingroup$
I'm not a python expert. I am pretty good with Mandlebrot generation (I've spent a lot of time on my custom Julia Set generator.)
So I'll say this: optimize the heck out of stuff that will be running many iterations. Forget about clean-code or nice OOP principles. For lots-of-iterations stuff like this, you want as nitty gritty as possible.
So let's take a look at your interior-most loop:
z = z**2 + complex_number
if (z.real**2+z.imag**2)**0.5 > 2:
pixel_grid[complex_grid.index(complex_list),complex_list.index(complex_number)]=[iteration,iteration,iteration]
break
else:
continue
Imagine what's happening behind the scenes in memory with just that very first line. You've got an instance of a complex number. You want to square it... so it has to create another instance of a complex object to hold the squared value. Then, you're adding another complex number to it - which means you're creating another instance of Complex to hold the result of the addition.
You're creating object instances left and right, and you're doing it on an order of 3000 x 2000 x 255 times. Creating several class instances doesn't sound like much, but when you're doing it a billion times, it kinda drags things down.
Compare that with pseudocode like:
px = num.real
py = num.imag
while
tmppx = px
px = px * px - py * py + num.real
py = 2 * tmppx * py + num.imag
if condition-for-hitting-escape
stuff
if condition-for-hitting-max-iter
moreStuff
No objects are getting created and destroyed. It's boiled down to be as efficient as possible. It's not as nice looking... but when you're doing something a billion times, shaving off even a millionth of a second results in saving 15 minutes.
And as someone else mentioned, you want to simplify the logic so that you don't have to do the square-root operation - and if you're okay with small variations in how the gradient is colored, changing the "magnitude" check with "are x or y within a bounding box".
Aka, the more things you can remove out of that runs-a-billion-times loop, the better off you'll be.
$endgroup$
add a comment |
$begingroup$
I'm not a python expert. I am pretty good with Mandlebrot generation (I've spent a lot of time on my custom Julia Set generator.)
So I'll say this: optimize the heck out of stuff that will be running many iterations. Forget about clean-code or nice OOP principles. For lots-of-iterations stuff like this, you want as nitty gritty as possible.
So let's take a look at your interior-most loop:
z = z**2 + complex_number
if (z.real**2+z.imag**2)**0.5 > 2:
pixel_grid[complex_grid.index(complex_list),complex_list.index(complex_number)]=[iteration,iteration,iteration]
break
else:
continue
Imagine what's happening behind the scenes in memory with just that very first line. You've got an instance of a complex number. You want to square it... so it has to create another instance of a complex object to hold the squared value. Then, you're adding another complex number to it - which means you're creating another instance of Complex to hold the result of the addition.
You're creating object instances left and right, and you're doing it on an order of 3000 x 2000 x 255 times. Creating several class instances doesn't sound like much, but when you're doing it a billion times, it kinda drags things down.
Compare that with pseudocode like:
px = num.real
py = num.imag
while
tmppx = px
px = px * px - py * py + num.real
py = 2 * tmppx * py + num.imag
if condition-for-hitting-escape
stuff
if condition-for-hitting-max-iter
moreStuff
No objects are getting created and destroyed. It's boiled down to be as efficient as possible. It's not as nice looking... but when you're doing something a billion times, shaving off even a millionth of a second results in saving 15 minutes.
And as someone else mentioned, you want to simplify the logic so that you don't have to do the square-root operation - and if you're okay with small variations in how the gradient is colored, changing the "magnitude" check with "are x or y within a bounding box".
Aka, the more things you can remove out of that runs-a-billion-times loop, the better off you'll be.
$endgroup$
add a comment |
$begingroup$
I'm not a python expert. I am pretty good with Mandlebrot generation (I've spent a lot of time on my custom Julia Set generator.)
So I'll say this: optimize the heck out of stuff that will be running many iterations. Forget about clean-code or nice OOP principles. For lots-of-iterations stuff like this, you want as nitty gritty as possible.
So let's take a look at your interior-most loop:
z = z**2 + complex_number
if (z.real**2+z.imag**2)**0.5 > 2:
pixel_grid[complex_grid.index(complex_list),complex_list.index(complex_number)]=[iteration,iteration,iteration]
break
else:
continue
Imagine what's happening behind the scenes in memory with just that very first line. You've got an instance of a complex number. You want to square it... so it has to create another instance of a complex object to hold the squared value. Then, you're adding another complex number to it - which means you're creating another instance of Complex to hold the result of the addition.
You're creating object instances left and right, and you're doing it on an order of 3000 x 2000 x 255 times. Creating several class instances doesn't sound like much, but when you're doing it a billion times, it kinda drags things down.
Compare that with pseudocode like:
px = num.real
py = num.imag
while
tmppx = px
px = px * px - py * py + num.real
py = 2 * tmppx * py + num.imag
if condition-for-hitting-escape
stuff
if condition-for-hitting-max-iter
moreStuff
No objects are getting created and destroyed. It's boiled down to be as efficient as possible. It's not as nice looking... but when you're doing something a billion times, shaving off even a millionth of a second results in saving 15 minutes.
And as someone else mentioned, you want to simplify the logic so that you don't have to do the square-root operation - and if you're okay with small variations in how the gradient is colored, changing the "magnitude" check with "are x or y within a bounding box".
Aka, the more things you can remove out of that runs-a-billion-times loop, the better off you'll be.
$endgroup$
I'm not a python expert. I am pretty good with Mandlebrot generation (I've spent a lot of time on my custom Julia Set generator.)
So I'll say this: optimize the heck out of stuff that will be running many iterations. Forget about clean-code or nice OOP principles. For lots-of-iterations stuff like this, you want as nitty gritty as possible.
So let's take a look at your interior-most loop:
z = z**2 + complex_number
if (z.real**2+z.imag**2)**0.5 > 2:
pixel_grid[complex_grid.index(complex_list),complex_list.index(complex_number)]=[iteration,iteration,iteration]
break
else:
continue
Imagine what's happening behind the scenes in memory with just that very first line. You've got an instance of a complex number. You want to square it... so it has to create another instance of a complex object to hold the squared value. Then, you're adding another complex number to it - which means you're creating another instance of Complex to hold the result of the addition.
You're creating object instances left and right, and you're doing it on an order of 3000 x 2000 x 255 times. Creating several class instances doesn't sound like much, but when you're doing it a billion times, it kinda drags things down.
Compare that with pseudocode like:
px = num.real
py = num.imag
while
tmppx = px
px = px * px - py * py + num.real
py = 2 * tmppx * py + num.imag
if condition-for-hitting-escape
stuff
if condition-for-hitting-max-iter
moreStuff
No objects are getting created and destroyed. It's boiled down to be as efficient as possible. It's not as nice looking... but when you're doing something a billion times, shaving off even a millionth of a second results in saving 15 minutes.
And as someone else mentioned, you want to simplify the logic so that you don't have to do the square-root operation - and if you're okay with small variations in how the gradient is colored, changing the "magnitude" check with "are x or y within a bounding box".
Aka, the more things you can remove out of that runs-a-billion-times loop, the better off you'll be.
answered 4 hours ago
KevinKevin
1413
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$begingroup$
If you want to skip points that are known to be inside the Mandelbrot set, the main cardioid and period bulbs can be skipped. If you skip processing points contained within the main cardioid and the main disk, you can significantly speed up the program. See also this resource for a further analysis of the main cardioid and disk. This optimization becomes less effective as you zoom in on the edges and becomes completely ineffective if these regions are off-screen.
$endgroup$
– Cornstalks
5 hours ago