Question

An articulation point in a connected graph is a vertex such that removing the vertex and its incident edges disconnects the graph into two or more connected components. Let \(T\) be a DFS tree obtained by doing DFS in a connected undirected graph \(G\). Which of the following options is/are correct?

• A. Root of \(T\) can never be an articulation point in \(G\).
• B. Root of \(T\) is an articulation point in \(G\) if and only if it has two or more children.
• C. A leaf of \(T\) can be an articulation point in \(G\).
• D. If \(u\) is an articulation point in \(G\) such that \(x\) is an ancestor of \(u\) in \(T\) and \(y\) is a descendant of \(u\) in \(T\), then all paths from \(x\) to \(y\) in \(G\) must pass through \(u\).

Hint:

In DFS, the root is special: it is an articulation point only when it has two or more DFS children.

Answer 1

The Correct Option is B

Step 1: Property of DFS root.
In DFS of an undirected graph, the root of the DFS tree is an articulation point if and only if it has two or more children. Hence, statement (B) is correct.

Step 2: Evaluation of other options.
Statement (A) is false because the DFS root can be an articulation point.
Statement (C) is false because a leaf in a DFS tree cannot disconnect the graph upon removal.
Statement (D) is false because back edges may exist that bypass the articulation point.

Step 3: Conclusion.
Thus, only option (B) is correct.