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bfs in python 3

Clumsy Crossbill answered on May 29, 2021 Popularity 1/10 Helpfulness 1/10

answer bfs in python 3

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bfs in python 3

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Popularity 1/10 Helpfulness 1/10 Language python

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Tags: python

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Contributed on May 29 2021

Clumsy Crossbill

0 Answers Avg Quality 2/10

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from collections import defaultdict

def bfs(graph, start):
    queue = []
    visited = set()
    queue.append(start) # Store root/starting node to visit
    visited.add(start) # Visit starting node 

    while queue: # As long as there is any node 
        node = queue.pop(0) # take that node from queue
        print(node, end=" ")

        for neighbor in graph[node]: # explore contents (children) of that node
            if neighbor not in visited: # check if that node is not visited
                queue.append(neighbor) # store that node for visit
                visited.add(neighbor) # visit that node


graph = defaultdict(list)
graph['A'] = ['B', 'C']
graph['B'] = ['A', 'D', 'E']
graph['C'] = ['A', 'F']
graph['D'] = ['B']
graph['E'] = ['B', 'F']
graph['F'] = ['C', 'E']

bfs(graph, 'A')  # Output: A B C D E F

### Using networkx : For network analysis problem
import networkx as nx
def path_exists(G, node1, node2): # G is the graph where multiple nodes are connected with edges
    """
    This function checks whether a path exists between two nodes (node1, node2) in graph G.
    """
    visited_nodes = set()
    queue = [node1]

    for node in queue:
        neighbors = list(G.neighbors(node))
        if node2 in neighbors:
            print('Path exists between nodes {0} and {1}'.format(node1, node2))
            return True
            break

        else:
            visited_nodes.add(node)
            queue.extend([n for n in neighbors if n not in visited_nodes])

        # Check to see if the final element of the queue has been reached
        if node == queue[-1]:
            print('Path does not exist between nodes {0} and {1}'.format(node1, node2))

            # Place the appropriate return statement
            return False

Popularity 9/10 Helpfulness 4/10 Language python

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Tags: python py

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Contributed on Oct 23 2023

Innocent Iguana

0 Answers Avg Quality 2/10

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