Let T be a depth first search tree in an undirected graph G. Vertices ...
Statement and Explanation:
The statement is asking which of the given options is true when considering a depth-first search tree T in an undirected graph G, where u and v are leaves with degrees of at least 2 in G.
Option A: There must exist a vertex w adjacent to both u and v in G.
Option B: There must exist a vertex w whose removal disconnects u and v in G.
Option C: There must exist a cycle in G containing u.
Option D: There must exist a cycle in G containing u and all its neighbors in G.
Analysis:
To determine which option is true, let's analyze each option:
Option A: There must exist a vertex w adjacent to both u and v in G.
- This option is not necessarily true because u and v could be leaves connected to different branches in T, making it possible for there not to be a vertex w adjacent to both u and v in G.
Option B: There must exist a vertex w whose removal disconnects u and v in G.
- This option is not necessarily true because u and v could be leaves connected to the same branch in T, and the removal of any vertex would not disconnect u and v in G.
Option C: There must exist a cycle in G containing u.
- This option is not necessarily true because u is a leaf and does not have any outgoing edges in T, so it cannot be part of a cycle in G.
Option D: There must exist a cycle in G containing u and all its neighbors in G.
- This option is true because u and v are leaves with degrees of at least 2 in G. This means that they have at least one neighbor each in G. Since u is a leaf, its only neighbor is v. Therefore, there exists a cycle in G containing u and all its neighbors (which is just v).
Conclusion:
Based on the analysis, the correct answer is option D.