Let T be a depth first search tree in an undirected graph G. Vertices ...
Below example shows that A and B are FALSE:
Below example shows C is false:
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Let T be a depth first search tree in an undirected graph G. Vertices ...
Understanding the Depth First Search Tree
In a depth first search (DFS) tree T, leaves represent vertices that have no children in the tree. In this case, both u and n are leaves, meaning that they were the last vertices visited in their respective branches of the DFS.
Vertex Degrees and Implications
- Both u and n have degrees of at least 2 in the original graph G.
- This implies that there are at least two edges connected to each vertex, allowing u and n to connect to other vertices apart from those in the DFS tree.
Analyzing the Statements
- Statement A: "There must exist a vertex w adjacent to both u and n in G."
- Not necessarily true as u and n could connect to completely different sets of vertices.
- Statement B: "There must exist a vertex w whose removal disconnects u and n in G."
- This is also not guaranteed. The removal of a single vertex does not ensure disconnection.
- Statement C: "There must exist a cycle in G containing u and n."
- While possible, it is not guaranteed as they may be connected differently.
- Statement D: "There must exist a cycle in G containing u and all its neighbors in G."
- Correct Statement: Since u is a leaf in the DFS tree but has a degree of at least 2, it must connect to other vertices. These neighbors can potentially connect to each other or n, forming a cycle.
Conclusion
Thus, the existence of a cycle involving u and all its neighbors is ensured by their connections, validating option D as the correct answer.
Let T be a depth first search tree in an undirected graph G. Vertices ...
Below example shows that A and B are FALSE:
Below example shows C is false: