Explain the BSF (Breadth First Search) traversing method

Breadth-First Search (BFS) is a fundamental graph traversal algorithm used to explore nodes and edges of a graph in a systematic manner. It is particularly useful for finding the shortest path in unweighted graphs and for exploring all nodes at the present depth level before moving on to nodes at the next depth level.

How BFS Works

1. Initialization:

   • Start by selecting a source node (root node).
   • Create a queue data structure to keep track of nodes to be explored.
   • Mark the source node as visited and enqueue it.

2. Traversal:

   • While the queue is not empty:
     • Dequeue a node from the front of the queue.
     • Visit the node (e.g., print its value or process it).
     • Enqueue all its adjacent, unvisited nodes and mark them as visited.

Detailed Steps

1. Enqueue the Starting Node:

   • Begin by enqueuing the starting node and marking it as visited.
   • Example: If the starting node is $$A$$, enqueue $$A$$ and mark it as visited.

2. Dequeue and Explore:

   • Dequeue the front node from the queue.
   • Visit this node and process it as needed.
   • Enqueue all its adjacent nodes that have not been visited yet and mark them as visited.
   • Repeat this process until the queue is empty.

Example

Consider the following graph:

    A
   / \
  B   C
 / \   \
D   E   F

Starting BFS from node $$A$$:

1. Enqueue $$A$$ and mark it as visited.
2. Dequeue $$A$$, visit it, and enqueue its neighbors $$B$$ and $$C$$.
3. Dequeue $$B$$, visit it, and enqueue its neighbors $$D$$ and $$E$$.
4. Dequeue $$C$$, visit it, and enqueue its neighbor $$F$$.
5. Dequeue $$D$$, visit it (no new neighbors to enqueue).
6. Dequeue $$E$$, visit it (no new neighbors to enqueue).
7. Dequeue $$F$$, visit it (no new neighbors to enqueue).

The order of traversal will be: $$A, B, C, D, E, F$$.

Pseudocode

def bfs(graph, start):
    visited = set()
    queue = collections.deque([start])
    visited.add(start)

    while queue:
        vertex = queue.popleft()
        print(vertex, end=" ")

        for neighbor in graph[vertex]:
            if neighbor not in visited:
                visited.add(neighbor)
                queue.append(neighbor)

Complexity

• Time Complexity: $$O(V + E)$$, where $$V$$ is the number of vertice...