为什么A*算法不会卡在两个节点之间

Question

我是路径查找方面的初学者，虽然我理解 A* 的基本思想，但我仍然不明白为什么在回溯时，实现不会陷入循环在最近访问的两个节点之间。

为了更清楚，我一直在查看 here 中的代码（我将复制并粘贴它，以防 link 死掉）：

class Node():
    """A node class for A* Pathfinding"""

    def __init__(self, parent=None, position=None):
        self.parent = parent
        self.position = position

        self.g = 0
        self.h = 0
        self.f = 0

    def __eq__(self, other):
        return self.position == other.position


def astar(maze, start, end):
    """Returns a list of tuples as a path from the given start to the given end in the given maze"""

    # Create start and end node
    start_node = Node(None, start)
    start_node.g = start_node.h = start_node.f = 0
    end_node = Node(None, end)
    end_node.g = end_node.h = end_node.f = 0

    # Initialize both open and closed list
    open_list = []
    closed_list = []

    # Add the start node
    open_list.append(start_node)

    # Loop until you find the end
    while len(open_list) > 0:

        # Get the current node
        current_node = open_list[0]
        current_index = 0
        for index, item in enumerate(open_list):
            if item.f < current_node.f:
                current_node = item
                current_index = index

        # Pop current off open list, add to closed list
        open_list.pop(current_index)
        closed_list.append(current_node)

        # Found the goal
        if current_node == end_node:
            path = []
            current = current_node
            while current is not None:
                path.append(current.position)
                current = current.parent
            return path[::-1] # Return reversed path

        # Generate children
        children = []
        for new_position in [(0, -1), (0, 1), (-1, 0), (1, 0), (-1, -1), (-1, 1), (1, -1), (1, 1)]: # Adjacent squares

            # Get node position
            node_position = (current_node.position[0] + new_position[0], current_node.position[1] + new_position[1])

            # Make sure within range
            if node_position[0] > (len(maze) - 1) or node_position[0] < 0 or node_position[1] > (len(maze[len(maze)-1]) -1) or node_position[1] < 0:
                continue

            # Make sure walkable terrain
            if maze[node_position[0]][node_position[1]] != 0:
                continue

            # Create new node
            new_node = Node(current_node, node_position)

            # Append
            children.append(new_node)

        # Loop through children
        for child in children:

            # Child is on the closed list
            for closed_child in closed_list:
                if child == closed_child:
                    continue

            # Create the f, g, and h values
            child.g = current_node.g + 1
            child.h = ((child.position[0] - end_node.position[0]) ** 2) + ((child.position[1] - end_node.position[1]) ** 2)
            child.f = child.g + child.h

            # Child is already in the open list
            for open_node in open_list:
                if child == open_node and child.g > open_node.g:
                    continue

            # Add the child to the open list
            open_list.append(child)


def main():

    maze = [[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]

    start = (0, 0)
    end = (7, 6)

    path = astar(maze, start, end)
    print(path)


if __name__ == '__main__':
    main()

在这种特定情况下，这看起来很简单，但如果 "maze" 是这样的：

maze = 
   [[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 1, 1, 1, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 1, 0, 0, 0],
    [0, 0, 0, 0, 1, 1, 1, 0, 0, 0],
    [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]

当检查的节点是(5, 5)时，有一个死胡同，所以算法应该"backtrack"到节点(3, 5)，然后从那里往下走。

我的问题是我真的不明白这是怎么发生的。返回一个节点并检查邻居意味着再次检查所有内容，它只会返回到 (5, 5).

这不会发生，因为实现工作得很好，但我似乎真的无法掌握如何实现。有什么提示吗？

Answer 1

重点是运行Dijkstra或A*时每个节点最多访问一次。
这是因为每次访问一个节点时（在我们将其从队列中弹出之后），我们 'mark' 这个节点已经被访问过。在您提供的实现中，标记是通过将节点添加到 closed_list:

closed_list.append(current_node)

节点现在是 closed_list 的事实将允许在将节点推送到队列之前检查节点是否已经被访问过。这在代码中做得更进一步（笨拙）：

            # Child is on the closed list
            for closed_child in closed_list:
                if child == closed_child:
                    continue

此机制对于确保算法终止（对于 Dijkstra 或 A*）以及其复杂度在 O(n.log n).
中是必不可少的但是在大多数实现中，您不会看到 closed_list，而是与每个节点关联的 visited 布尔值，或者颜色（white 如果未访问，green 如果已经访问访问过）。它们在终止保证方面都是等效的（不一定在性能方面，因为列表查找可能会花费 O(n)）。

为什么A*算法不会卡在两个节点之间

Why does A* algorithm not get stuck between two nodes

python

algorithm

a-star