10

所以我正在尝试创建一个洪水填充算法,但我不断收到递归错误。该算法似乎具有无限递归,我无法确定原因。我浏览了整个互联网,但找不到解决方案,因为根据大多数消息来源,我的程序似乎是正确的。然而,似乎有什么不对劲。这是代码的编辑版本。错误消息仍然是最大递归。

我能得到一些帮助吗?

from PIL import Image, ImageTk
from random import *


w= 75
h= w

flood = Image.new("RGB", (w,h), (0,0,0))

x = 0
y = 0
count = 0

colorlist = []
i = 0

while x < w -1:
    y = 0
    while y < h-1:
        r = random()
        if r < .25:
            flood.putpixel((x,y), (0,0,0))
        else:
            flood.putpixel((x,y), (255,255,255))
        y += 1
    x += 1
x = 0
y = 0
while x < w-1:
    y = 0
    while y < h-1:
        r = random()
        if x == 0 or y == 0 or x == w-1 or y ==h-1:
            flood.putpixel((x,y), (0,0,0))
        y += 1
    x += 1


def floodfill(x,y, d,e,f, g,h,i, image, count):
        count+=1
        (a,b,c) = image.getpixel((x,y))
        if (a,b,c) == (255,255,255):
            (j,k,l) = image.getpixel((x-1,y))
            (m,n,o) = image.getpixel((x+1, y))
            (p,q,r) = image.getpixel((x,y-1))
            (s,t,u) = image.getpixel((x,y+1))
        if count > 990:
            return
        if (a,b,c) == (255,255,255):
            image.putpixel((x,y), (g,h,i))
            floodfill(x-1, y, d,e,f, g,h,i, image, count)
            floodfill(x+1, y, d,e,f, g,h,i, image, count)
            floodfill(x, y-1, d,e,f, g,h,i, image, count)
            floodfill(x, y+1, d,e,f, g,h,i, image,count)


floodfill(2,2, 0,0,0,255,0,0,flood, 0)

flood.save("flood.png")
print("done")
4

3 回答 3

12

Python 倾向于抛出maximum recursion depth exceeded错误,即使算法不会无限递归并且最终会自行停止。有两种解决方案:增加递归限制,或切换到迭代算法。

您可以使用 提高递归限制sys.setrecursionlimit。选择一个高于算法最坏情况递归深度的数字。在您的情况下,这将是图像中的像素数,length * height.

将您的算法更改为迭代算法相当简单,因为您绘制像素的顺序并不重要,只要您至少获得一次即可。Aset非常适合保存唯一的无序数据,所以让我们用它来存储我们需要绘制的像素。

def floodFill(x,y, d,e,f, g,h,i, image):
    toFill = set()
    toFill.add((x,y))
    while not toFill.empty():
        (x,y) = toFill.pop()
        (a,b,c) == image.getpixel((x,y))
        if not (a,b,c) == (255, 255, 255):
            continue
        image.putpixel((x,y), (g,h,i))
        toFill.add((x-1,y))
        toFill.add((x+1,y))
        toFill.add((x,y-1))
        toFill.add((x,y+1))
    image.save("flood.png")

如果您确实使用迭代方法,请务必在其中进行绑定检查。否则,它可能会永远运行!或者至少在您的硬盘驱动器被一toFill组巨大的驱动器填满之前。

于 2012-07-31T19:26:52.823 回答
3

为什么不以深度优先的方式进行泛洪填充,而不是递归呢?递归无论如何都使用隐式堆栈,所以你没有什么可失去的。

是的,正如评论中指出的那样,您应该检查 x 和 y 是否超出范围。

于 2012-07-31T19:21:13.173 回答
3

这尚未经过测试,但主要基于您提供的代码。floodfill它应该可以工作并提供实现算法的替代方法。该功能可能会更有效。

import PIL
import random
import collections

WHITE = 255, 255, 255
BLACK = 0, 0, 0
RED = 255, 0, 0

def main(width, height):
    flood = PIL.Image.new('RGB', (width, height), BLACK)
    # Create randomly generated walls
    for x in range(width):
        for y in range(height):
            flood.putpixel((x, y), BLACK if random.random() < 0.15 else WHITE)
    # Create borders
    for x in range(width):
        for y in range(height):
            if x in {0, width - 1} or y in {0, height - 1}:
                flood.putpixel((x, y), BLACK)
    floodfill(50, 25, RED, image)
    # Save image
    image.save('flood.png')

def floodfill(x, y, color, image):
    # if starting color is different from desired color
    #     create a queue of pixels that need to be changed
    #     while there are pixels that need their color changed
    #         change the color of the pixel to what is desired
    #         for each pixel surrounding the curren pixel
    #             if the new pixel has the same color as the starting pixel
    #                 record that its color needs to be changed
    source = image.getpixel((x, y))
    if source != color:
        pixels = collections.deque[(x, y)]
        while pixels:
            x, y = place = pixels.popleft()
            image.putpixel(place, color)
            for x_offset in -1, 1:
                x_offset += x
                for y_offset in -1, 1:
                    y_offset += y
                    new_place = x_offset, y_offset
                    if image.getpixel(new_place) == source:
                        pixels.append(new_place)

if __name__ == '__main__':
    main(100, 50)
于 2012-07-31T19:32:48.647 回答