Application of Breath-first search in AI(route search)

標籤：bfs   a   

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 the some position twice.

below is my code in python3:

import frontiersdef 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. I spent 2days to finish it. 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.

Pay more attention to the scope of the variables, semantic problems and solutions to the problem. we can save lots of time.

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Application of Breath-first search in AI(route search)