使用Python组合九次尖点多分支和九重图

恩尼亚布罗特

以下代码取自GitHub上的this Mandelbrot (dimension=2) project，我需要通过更改定义Mandelbrot集的递归方程中的指数来创建此分形项目的高维版本，如Triabrot（维=4）、Pentabrot（维=6）、Heptabrot（维=8）和Enneabrot（维=10）。换句话说，对于Mandelbrot分形，我们定义了一个维数变量d，而不是z_{n+1}=z_n*z_n+z_0，然后d维多重Brot的方程将是z_{n+1}=z_n^d+z_0，并且每个这样的方程将在其稳定区的图形中产生d-1尖点。要了解有关此主题的更多信息，请观看以下两个YouTube视频：Times Tables by Mathologer和Mandelbrot Set by Numberphile

每一块输出都需要改变轴；这是因为分形在复杂平面中移动，我们需要沿着镜头移动，通过Python观察它们。我们还可以使用哪些其他公式？也许axisFix=（（d^2-4）/d）/10？在这个建议的公式中，d^2-4是存在的，因为我希望Mandelbrot分形（d=2）在没有任何位移的情况下打印在中心。因此，轴平移的值为零。这很有趣，因为对于d=-2它也是零，这意味着我们也必须试着看类似z_{n+1}=z_n^{-2.0}+z_0的方程。目标是找到最平滑的函数f（d），使f（2）=0，并通过以较小的大小（0.01-0.1）增量增加d的值并打印此处定义为动画的mandelbrot（阈值、密度、尺寸）函数的输出来制作动画。函数越平滑，演示文稿中幻灯片动画中幻灯片的过渡就越平滑

import numpy as np import matplotlib.pyplot as plt # counts the number of iterations until the function diverges or # returns the iteration threshold that we check until def countIterationsUntilDivergent(c, threshold, d): z = complex(0, 0) for iteration in range(threshold): # here is the recurrence relation z_{n+1} = z_n^d + z_0, used for # drawing d-1-dimensional Mandelbrot creatures (growing fractals) z = z**d + c if abs(z) > 4: break pass pass return iteration # takes the iteration limit before declaring function as convergent and # takes the density of the atlas # create atlas, plot mandelbrot set, display set def mandelbrot(threshold, density, d): # it is necessary to change the axis for every patch of outputs; # this happens because the fractals move in the complex plane # and we need to move along our lens to watch them through Python ## what other formulas could we use? Maybe axisFix = ((d^2-4)/d)/10? ## d^2-4 is there because I want the Mandelbrot fractal (d=2) ## to be right there were it is printed without any replacement ## so the value of the axis translation would be zero. This is ## funny because it is also zero for d=-2 which means that ## we must also be trying to look at equations like ### z_{n+1}=z_n^{-2.0} + z_0 ### the goal is to find the smoothest function ### f(d) such that f(2)=0 and make an animation ### by increasing the value of d by increments of small size (0.01-0.1) ### and printing the output of the mandelbrot function defined here ### as an animation. The smoother the function, the smoother ### the transition of slides in the animation axisFix = d/10 # location and size of the atlas rectangle realAxis = np.linspace(-2.25+axisFix, 0.75+axisFix, density) imaginaryAxis = np.linspace(-1.5, 1.5, density) # realAxis = np.linspace(-0.22, -0.219, 1000) # imaginaryAxis = np.linspace(-0.70, -0.699, 1000) realAxisLen = len(realAxis) imaginaryAxisLen = len(imaginaryAxis) # 2-D array to represent mandelbrot atlas atlas = np.empty((realAxisLen, imaginaryAxisLen)) # color each point in the atlas depending on the iteration count for ix in range(realAxisLen): for iy in range(imaginaryAxisLen): cx = realAxis[ix] cy = imaginaryAxis[iy] c = complex(cx, cy) atlas[ix, iy] = countIterationsUntilDivergent(c, threshold, d) pass pass # plot and display mandelbrot set fig1 = plt.gcf() plt.axis('off') # plt.savefig('mandel.eps', format='eps') plt.imshow(atlas.T, interpolation="nearest") # plt.show() output_name = str(d)+'.pdf' fig1.savefig(output_name, format='pdf', bbox_inches='tight', dpi=2000) # time to party!! dimensions = np.arange(10, 100) / 10 # for d in dimensions: # mandelbrot(120, 1000, d) # Enneabrot mandelbrot(120, 1000, 10) # Heptabrot mandelbrot(120, 1000, 8) # Pentabrot mandelbrot(120, 1000, 6) # Triabrot mandelbrot(120, 1000, 4) # Mandelbrot mandelbrot(120, 1000, 2)

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1楼 · 发布于 2024-09-30 06:30:40

以下是结果，感谢Aryan Hemmati编辑了最后一张照片，将灵魂的九面形与十维Mandelbrot分形相结合

我需要一个Python和LaTeX（最好是矢量图形TikZ/渐近线/PGF/Metapost/GeoGebra，按顺序排列）通过在终端上运行简单代码生成此动画。我们可以很容易地改变参数来制作七叶草（d=8）、五叶草（d=6）甚至三叶草（d=4）。我附上了使用Python绘制Enneabrot（d=10）的代码，我修改了Danyaal Rangwala的this code，并在递归方程中定义了一个新的变量d（维度），用于计算解的精确平衡区，这最终显示Enneabrot是我最后一次尝试生成这种分形的数字（从d=1.0开始，增量为0.1，直到d=10.0）

在这里，我将发布Mandelbrot分形（d=2）以及我在上面定义的分形的输出：Enneabrot（d=10）、Heptabrot（d=8）、Pentabrot（d=6），甚至Triabrot（d=4）。其他图表将在本answer的下一版本中遵循

曼德布罗特（d=2）