We saw in lecture that both full BFS and and DFS produce trees when run on connected graphs, and forests when run on disconnected graphs. Suppose a full BFS and a full DFS are run on an undirected graph with three connected components. True or False: the number of trees in the BFS forest will equal to number of trees in the DFS forest.

Choice 1 of 2:True

Choice 2 of 2:False

**True.**

When a full BFS and a full DFS are run on an undirected graph with three connected components, each traversal (whether BFS or DFS) will explore all vertices within each connected component before moving to another component. Since each connected component is separate and not connected to others, both BFS and DFS will result in a separate tree for each connected component. Therefore, the number of trees in the BFS forest will equal the number of trees in the DFS forest, which in this case is three.

Would you like more details or have any other questions?

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1. How do BFS and DFS differ in terms of traversal order?
2. What is the significance of connected components in a graph?
3. How does the structure of the resulting trees differ between BFS and DFS?
4. Can BFS or DFS result in a cycle? Why or why not?
5. What are some practical applications of BFS and DFS?

**Tip:** When analyzing graph algorithms, it’s important to understand the difference between trees and forests, especially in the context of connected and disconnected graphs.

Explore the relationship between Breadth-First Search (BFS) and Depth-First Search (DFS) in graph theory. Understand how these algorithms produce trees when run on connected graphs and forests on disconnected graphs. Learn why the number of trees in the BFS forest equals the number in the DFS forest when applied to an undirected graph with three connected components.

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