如何在0和1的矩阵中找到两个随机点之间的路径

Question

I'm trying to create a method in which I get a square matrix of 0 and 1, where 1 can walk and 0 are the obstacles that must be circumvented, in addition to the random start and end points.

I needed this method to provide me with a sense of the path that should be followed, such as:

Where the return would be "right, up, right, right, down, right" and "down, right, down, right, right, up, right, up" respectively.

Thanks

Answer 1

One way could be to write a function which checks if the current field is the destination field and call this function recursively for all neighbors. In each recursion step you add the direction and return the full path when you are finished.

dest_x = 4
dest_y = 1
visited = []

def next (path, x, y):
    visited.append ([x, y])
    
    # obstacle
    if matrix[x][y] == 0:
        return None
    
    # found destination
    if x == dest_x and y == dest_y:
        return path
    
    for i in [{'x': x - 1, 'y': y, 'direction': 'left'},
              {'x': x + 1, 'y': y, 'direction': 'right'},
              {'x': x, 'y': y - 1, 'direction': 'up'},
              {'x': x, 'y': y + 1, 'direction': 'down'}]:
        if [i['x'], i['y']] not in visited and i['x'] >= 0 and i['x'] < len(matrix) and i['y'] >= 0 and i['y'] < len(matrix[x]):
            n = next (path + [i['direction']], i['x'], i['y'])
            if n != None:
                return n
    
    
    

matrix = [[1,1,1,1],
          [1,1,1,1],
          [1,0,0,1],
          [1,1,1,1],
          [1,1,1,1]]

print (next ([], 0, 1))

This will find one way if there is one. In case you need the shortest way you can store the path in an array when you reach the destination instead of returning it. When all paths are discovered you can print the path with minimum length