在使用逃逸时间算法时,是否可以将mandelbrot集的公式(默认为f(z)=z^2+c)更改为不同的公式(f(z)=z^2+c*e^(-z)是我所需要的),如果可能的话,如何更改?
我现在用FB36的代码

# Multi-threaded Mandelbrot Fractal (Do not run using IDLE!)
# FB - 201104306
import threading
from PIL import Image
w = 512 # image width
h = 512 # image height
image = Image.new("RGB", (w, h))
wh = w * h
maxIt = 256 # max number of iterations allowed
# drawing region (xa < xb & ya < yb)
xa = -2.0
xb = 1.0
ya = -1.5
yb = 1.5
xd = xb - xa
yd = yb - ya
numThr = 5 # number of threads to run
# lock = threading.Lock()

class ManFrThread(threading.Thread):
    def __init__ (self, k):
          self.k = k
          threading.Thread.__init__(self)
    def run(self):
        # each thread only calculates its own share of pixels
        for i in range(k, wh, numThr):
            kx = i % w
            ky = int(i / w)
            a = xa + xd * kx / (w - 1.0)
            b = ya + yd * ky / (h - 1.0)
            x = a
            y = b
            for kc in range(maxIt):
                x0 = x * x - y * y + a
                y = 2.0 * x * y + b
                x = x0
                if x * x + y * y > 4:
                    # various color palettes can be created here
                    red = (kc % 8) * 32
                    green = (16 - kc % 16) * 16
                    blue = (kc % 16) * 16
                    # lock.acquire()
                    global image
                    image.putpixel((kx, ky), (red, green, blue))
                    # lock.release()
                    break

if __name__ == "__main__":
    tArr = []
    for k in range(numThr): # create all threads
        tArr.append(ManFrThread(k))
    for k in range(numThr): # start all threads
        tArr[k].start()
    for k in range(numThr): # wait until all threads finished
        tArr[k].join()
    image.save("MandelbrotFractal.png", "PNG")

最佳答案

从代码中我推断出z = x + y * ic = a + b * i。对应于f(z) - z ^2 + c。你想要f(z) = z ^2 + c * e^(-z)
回想一下e^(-z) = e^-(x + yi) = e^(-x) * e^i(-y) = e^(-x)(cos(y) - i*sin(y)) = e^(-x)cos(y) - i (e^(-x)sin(y))。因此,您应该更新您的行如下:

x0 = x * x - y * y + a * exp(-x) * cos(y) + b * exp(-x) * sin(y);
y = 2.0 * x * y + a * exp(-x) * sin(y) - b * exp(-x) * cos(y)
x = x0

如果你没有得到你想要的特征差分级别,你可能需要调整maxIt(现在可能需要更多或更少的迭代来转义,平均来说),但这应该是你想要的数学表达式。
正如评论中所指出的,你可能需要调整标准本身,而不仅仅是最大迭代,以获得所需的分化水平:改变最大值对那些永远无法逃脱的人来说是没有帮助的。
你可以尝试获得一个好的逃跑条件,或者只是尝试一些东西,看看你得到了什么。

关于python - 多线程mandelbrot集,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/44830302/

10-12 22:07