Question

Let G be an undirected graph. Consider a depth-first traversal of G, and let T be the resulting depth-first search tree. Let u be a vertex in G and let v be the first new (unvisited) vertex visited after visiting u in the traversal. Which of the following statements is always true?

A. {u, v} must be an edge in G, and v is a descendant of u in T
B. {u, v} must be an edge in G, and u is a descendant of v in T
C. If {u, v} is not an edge in G then u is a leaf in T
D. If {u} is not an edge in G then u and v must have the same parent in T

Answer 1

In a depth-first traversal of an undirected graph, if vertex v is the first new vertex visited after vertex u, {u, v} is an edge in the graph and v is a descendant of u in the resulting depth-first search tree.

Step-by-step explanation:

The student is asking about properties of a depth-first search (DFS) in an undirected graph, specifically in relation to the DFS tree that results from the traversal and the relationship between two vertices u and v. In DFS, when a new vertex v is visited immediately following the vertex u, it implies that there is a direct edge between u and v in the graph G. Therefore, {u, v} must indeed be an edge in G.

Additionally, since v is the next vertex visited after u, v becomes a child of u in the depth-first search tree T, which makes v a descendant of u.In a depth-first traversal of an undirected graph, the first new vertex visited after visiting a vertex u is a vertex v that is adjacent to u.

To determine which of the statements is always true, we need to consider the relationships between u and v in the depth-first search tree T.

The correct statement is Option C: If {u, v} is not an edge in G, then u is a leaf in T. This is because if {u, v} is not an edge in G, it means that vertex v is not adjacent to u in the graph. Therefore, u does not have any outgoing edges in the tree T and is therefore a leaf.

As a result, the correct statement is: A. {u, v} must be an edge in G, and v is a descendant of u in T.