function path = solve_maze(img_file)
 %% Init data
 img = imread(img_file);
 img = rgb2gray(img);
 maze = img > 0;
 start = [985 398];
 finish = [26 399];
 
 %% Init BFS
 n = numel(maze);
 Q = zeros(n, 2);
 M = zeros([size(maze) 2]);
 front = 0;
 back = 1;
 
 function push(p, d)
  q = p + d;
  if maze(q(1), q(2)) && M(q(1), q(2), 1) == 0
   front = front + 1;
   Q(front, <img src="/UploadFiles/2021-04-08/icon_smile.gif">

这是个简单的实现，应该很容易就能够改写为Python或其他语言，下面是程序的运行结果：

提问者更新：
我用Python实现了一下Mikhail的方法，其中用到了numpy库，感谢Thomas推荐。我感觉这个算法是正确的，但是效果不太如预期，以下是相关代码，使用了PyPNG库处理图片。
 译注：很遗憾，我用提问者提供的代码并没有跑通程序，并且似乎代码缩进有点问题，而下面其他参与者的代码能够执行通过，并且效果很好。


import png, numpy, Queue, operator, itertools
 
def is_white(coord, image):
 """ Returns whether (x, y) is approx. a white pixel."""
 a = True
 for i in xrange(3):
  if not a: break
  a = image[coord[1]][coord[0] * 3 + i] > 240
 return a
 
def bfs(s, e, i, visited):
 """ Perform a breadth-first search. """
 frontier = Queue.Queue()
 while s != e:
  for d in [(-1, 0), (0, -1), (1, 0), (0, 1)]:
   np = tuple(map(operator.add, s, d))
   if is_white(np, i) and np not in visited:
    frontier.put(np)
  visited.append(s)
  s = frontier.get()
 return visited
 
def main():
 r = png.Reader(filename = "thescope-134.png")
 rows, cols, pixels, meta = r.asDirect()
 assert meta['planes'] == 3 # ensure the file is RGB
 image2d = numpy.vstack(itertools.imap(numpy.uint8, pixels))
 start, end = (402, 985), (398, 27)
 print bfs(start, end, image2d, [])




回答by Joseph Kern：


#!/usr/bin/env python
 
import sys
 
from Queue import Queue
from PIL import Image
 
start = (400,984)
end = (398,25)
 
def iswhite(value):
  if value == (255,255,255):
  return True
 
def getadjacent(n):
  x,y = n
  return [(x-1,y),(x,y-1),(x+1,y),(x,y+1)]
 
def BFS(start, end, pixels):
 
  queue = Queue()
  queue.put([start]) # Wrapping the start tuple in a list
 
  while not queue.empty():
 
    path = queue.get()
    pixel = path[-1]
 
    if pixel == end:
      return path
 
    for adjacent in getadjacent(pixel):
      x,y = adjacent
      if iswhite(pixels[x,y]):
        pixels[x,y] = (127,127,127) # see note
        new_path = list(path)
        new_path.append(adjacent)
        queue.put(new_path)
 
  print "Queue has been exhausted. No answer was found."
 
if __name__ == '__main__':
 
  # invoke: python mazesolver.py [.jpg|.png|etc.]
  base_img = Image.open(sys.argv[1])
  base_pixels = base_img.load()
 
  path = BFS(start, end, base_pixels)
 
  path_img = Image.open(sys.argv[1])
  path_pixels = path_img.load()
 
  for position in path:
    x,y = position
    path_pixels[x,y] = (255,0,0) # red
 
  path_img.save(sys.argv[2])




动态执行效果：



回答by Jim
使用树搜索太繁杂了，迷宫本身就跟解路径是可分的。正因如此，你可以使用连通区域查找算法来标记迷宫中的连通区域，这将迭代搜索两次这些像素点。如果你想要更好地解决方法，你可以对结构单元使用二元运算(binary operations)来填充每个连通区域中的死路。
下面是相关的MATLAB代码及运行结果：

 




% read in and invert the image
im = 255 - imread('maze.jpg');
 
% sharpen it to address small fuzzy channels
% threshold to binary 15%
% run connected components
result = bwlabel(im2bw(imfilter(im,fspecial('unsharp')),0.15));
 
% purge small components (e.g. letters)
for i = 1:max(reshape(result,1,1002*800))
  [count,~] = size(find(result==i));
  if count < 500
    result(result==i) = 0;
  end
end
 
% close dead-end channels
closed = zeros(1002,800);
for i = 1:max(reshape(result,1,1002*800))
  k = zeros(1002,800);
  k(result==i) = 1; k = imclose(k,strel('square',8));
  closed(k==1) = i;
end
 
% do output
out = 255 - im;
for x = 1:1002
  for y = 1:800
    if closed(x,y) == 0
      out(x,y,:) = 0;
    end
  end
end
imshow(out);





回答by Stefano
stefano童鞋给出了生成搜索过程GIF及AVI文件的代码 maze-solver-python (GitHub)