According to bfs, it is a search method to go through all the nodes layer by layer, until the gal has been found.

To make it simple, there is a visiting sequence of bfs:

Attention:

When we have goal test with node 0 we should create 1,2,3 node but nothing to do with goal test, as before, when we have goal test with node 1, we should create 4,5,6 immediately and so on.

Here comes an exercise:

A red bird wants to find the yellow bird in the shortest route and find this route in the shortest time.

we want to use the bfs, while due to the physical situation, we have to offer some methods to avoid heading back and visiting some position twice.

below is my code in python3:

import frontiers

def solve(problem) :
    state = problem.initial_state # the start location
    map = {} # a dic to store the whole map
    map[state] = set() # a set to store a node
    map[state].add(0) # add the parents information to a set
    map[state].add('root')  # initialize the root node
    queue_node_created = frontiers.Queue()  # store the position of nodes created
    queue_node_created.push(state)  # initialization

    path_stack = frontiers.Stack() # store the path in inverted order
    list_queue = []  # store the path in right order

    while True:
        node_value = queue_node_created.pop()  # pop the node in the queue for goal_test or expending new nodes
        if problem.goal_test(node_value):  # goal test
            while True:
                list_1 = list(map[node_value])
                if type(list_1[0]) == str:
                    parent_node_value = list_1[1]
                    last_action = list_1[0]# find action taken and parent node
                else:
                    parent_node_value = list_1[0]
                    last_action = list_1[1]  # find action taken and parent node
                if last_action == 'root':  # if find the root
                    break
                path_stack.push(last_action)  # store the action taken
                node_value = parent_node_value  # refresh the node_value
                print(last_action)
            while not path_stack.is_empty():
                list_queue.append(path_stack.pop())

            return list_queue  # return the path
        else:
            Next_steps = problem.get_successors(node_value)  # get successors
            for node in Next_steps:
                node_position = node[0]
                if node_position in map.keys():  # in case that the same node is visited more than twice
                    continue

                else:
                    map[node_position] = set()  # push the node info (parent's position and action) into map
                    map[node_position].add(node_value)
                    map[node_position].add(node[1])

                queue_node_created.push(node_position)  # push the new node into the queue

In the appendix, there are two pictures of the result and the shade of color means the frequency of visiting in our algorithm.

At the beginning, I used the list with dictionaries in it. The list represents the whole tree and a dictionary act as a layer. In this way, I can easily store the entire tree. But, after testing, the effect of it was quite low, like the 24th or 25th floor could be the deepest layer in 1 min's running.

I felt hopeless due to the perform of what I had programmed, which I spent 2days to finish. Thanks to my female friend who is much smarter than me, I found a better solution to deal with node storage and new node' test in sequence just as you can see in my code.

If I pay more attention to the scope of te variables, semantic problems and solutions to the problem, I am able to save lots of time.

Effect