双向搜寻

Question

我正在尝试在python中实现双向搜索。

据我了解，我应该以某种方式合并两个广度优先搜索，一个从起始（或根）节点开始，另一个从目标（或结束）节点开始。当两个广度优先搜索都在同一顶点“相遇”时，双向搜索终止。我已经实现了BFS，下面给出了代码

def bfs(graph, start):
    path = []
    queue = [start]
    while queue:
        vertex = queue.pop(0)
        if vertex not in path:
            path.append(vertex)
            queue.extend(graph[vertex])
    return path

您能为我提供一个代码示例（在Python中）还是链接到用于双向图形搜索的代码？

Answer 1

这似乎是一个有趣的问题，所以我尝试了一下。这是我的尝试。您没有说图形的格式，所以我猜了一个像下面这样的字典：

example_graph = {0:[1,2], 1:[0,3,4], 3:[1], 4:[1], 2:[0,5,6], 5:[2], 6:[2]}

代码从指定的起始和目标顶点开始，遍历当前活动顶点的列表，并记住如何通过{vertices：lists}字典到达顶点的。如果一个活动顶点在具有从另一端开始的路径的另一个活动顶点旁边发现自己，则它将合并列表并返回结果，否则它将扩展所有路径并继续。

def bi_directional_search(graph, start, goal):
    # Check if start and goal are equal.
    if start == goal:
        return [start]
    # Get dictionary of currently active vertices with their corresponding paths.
    active_vertices_path_dict = {start: [start], goal: [goal]}
    # Vertices we have already examined.
    inactive_vertices = set()

    while len(active_vertices_path_dict) > 0:

        # Make a copy of active vertices so we can modify the original dictionary as we go.
        active_vertices = list(active_vertices_path_dict.keys())
        for vertex in active_vertices:
            # Get the path to where we are.
            current_path = active_vertices_path_dict[vertex]
            # Record whether we started at start or goal.
            origin = current_path[0]
            # Check for new neighbours.
            current_neighbours = set(graph[vertex]) - inactive_vertices
            # Check if our neighbours hit an active vertex
            if len(current_neighbours.intersection(active_vertices)) > 0:
                for meeting_vertex in current_neighbours.intersection(active_vertices):
                    # Check the two paths didn't start at same place. If not, then we've got a path from start to goal.
                    if origin != active_vertices_path_dict[meeting_vertex][0]:
                        # Reverse one of the paths.
                        active_vertices_path_dict[meeting_vertex].reverse()
                        # return the combined results
                        return active_vertices_path_dict[vertex] + active_vertices_path_dict[meeting_vertex]

            # No hits, so check for new neighbours to extend our paths.
            if len(set(current_neighbours) - inactive_vertices - set(active_vertices))  == 0:
                # If none, then remove the current path and record the endpoint as inactive.
                active_vertices_path_dict.pop(vertex, None)
                inactive_vertices.add(vertex)
            else:
                # Otherwise extend the paths, remove the previous one and update the inactive vertices.
                for neighbour_vertex in current_neighbours - inactive_vertices - set(active_vertices):
                    active_vertices_path_dict[neighbour_vertex] = current_path + [neighbour_vertex]
                    active_vertices.append(neighbour_vertex)
                active_vertices_path_dict.pop(vertex, None)
                inactive_vertices.add(vertex)

    return None

Answer 2

import queue

class Node:
  def __init__(self, val):
    self.val = val
    self.neighbors = None
    self.visited_right = False  # whether the node was reached by the BFS that started from source
    self.visited_left = False  # whether the node was reached by the BFS that started from destination
    self.parent_right = None  # used for retrieving the final path from start to the meeting point
    self.parent_left = None  # used for retrieving the final path from the meeting point to destination


def bidirectional_search(s, t):

  def extract_path(node):
    """return the path when both BFS's have met"""
    node_copy = node
    path = []

    while node:
      path.append(node.val)
      node = node.parent_right

    path.reverse()
    del path[-1]  # because the meeting node appears twice

    while node_copy:
      path.append(node_copy.val)
      node_copy = node_copy.parent_left
    return path


  q = queue.Queue()
  q.put(s)
  q.put(t)
  s.visited_right = True
  t.visited_left = True

  while not q.empty():
    n = q.get()

    if n.visited_left and n.visited_right:  # if the node visited by both BFS's
      return extract_path(n)

    for node in n.neighbors:
      if n.visited_left == True and not node.visited_left:
        node.parent_left = n
        node.visited_left = True
        q.put(node)
      if n.visited_right == True and not node.visited_right: 
        node.parent_right = n
        node.visited_right = True
        q.put(node)

  return False

n0 = Node(0)
n1 = Node(1)
n2 = Node(2)
n3 = Node(3)
n4 = Node(4)
n5 = Node(5)
n6 = Node(6)
n7 = Node(7)
n0.neighbors = [n1, n5]
n1.neighbors = [n0, n2, n6]
n2.neighbors = [n1]
n3.neighbors = [n4, n6]
n4.neighbors = [n3]
n5.neighbors = [n0, n6]
n6.neighbors = [n1, n3, n5, n7]
n7.neighbors = [n6]
print(bidirectional_search(n0, n4))