是否可以将曼德布洛特集合的公式(默认情况下为 f(z( = z^2 + c(更改为不同的公式(f(z( = z^2 + c * e^(-z( 是我需要的(在使用逃逸时间算法时，如果可能的话，如何？ 我目前正在通过FB36使用此代码
# 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")
从代码中我推断出z = x + y * i和c = a + b * i .这对应于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，您应该将行更新为以下内容：
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
如果你没有得到你所追求的特征区分水平，你可能需要调整maxIt(现在可能需要更多或更少的迭代才能转义，平均而言(，但这应该是你所追求的数学表达式。
正如评论中指出的那样，您可能需要调整标准本身，而不仅仅是最大迭代次数，以获得所需的差异化水平：更改最大值对永无止境的迭代无济于事。
你可以尝试推导出一个好的逃生条件，或者只是尝试一些东西，看看你得到了什么。