Final answer:
To prove that one of the nodes x or y is an ancestor of the other in a depth-first search tree when an edge (x, y) in the graph G is not in T, a direct proof utilizing the characteristics of DFS and tree structures can be applied.
Step-by-step explanation:
The question pertains to a depth-first search tree (T), nodes (x and y), and an edge (x, y) that is present in graph G but not in T. To prove that one of x or y is an ancestor of the other in T, we must consider the nature of a depth-first search and the properties of a tree. A depth-first search (DFS) proceeds by exploring as far as possible along each branch before backtracking. In the context of a DFS tree, an edge that is not part of the tree but connects two nodes within it is known as a back edge. If there exists an edge (x, y) in the graph G that is not part of the tree T, it implies that during the depth-first search, either x was discovered and explored before reaching y, or vice versa. Since DFS explores nodes thoroughly before backtracking, this means that when (x, y) was encountered, one of the nodes had to be an ancestor of the other; otherwise, the edge would have been included in the tree during the initial search. Therefore, the proof for this question would be a direct proof, considering the definition and properties of a DFS tree and the relationship between nodes and edges in such trees.