Final answer:
Starting a depth-first search from vertex 0 and selecting the vertex with the smallest index first creates a DFS tree by exploring deeply each branch before backtracking.
Step-by-step explanation:
Conducting a depth-first search (DFS) from vertex 0 of a directed graph and choosing the vertex with the smallest index first, results in an underlying DFS tree that shows how the search progressed through the graph.
The process of DFS involves exploring as far as possible along each branch before backtracking, which means the search moves deep into the graph before visiting neighbors of the initial vertices. To start from vertex 0 and always choose the vertex with the smallest index ensures that the search is orderly and predictable, leading to a specific DFS tree. If the graph has multiple edges out of the vertices, the smallest indexed vertex is chosen first.
With the rules applied, the DFS tree would expand from vertex 0 to its lowest-indexed neighbor, then continue from that neighbor to its own lowest-indexed neighbor that has not yet been visited, and so on until all vertices reachable from vertex 0 have been included in the tree.
I apologize for any confusion, but it seems that the directed graph you mentioned isn't provided in your question. If you could share the details of the directed graph, such as the vertices, edges, and their connections, I'd be happy to help you determine the underlying depth-first search (DFS) tree when starting from vertex 0 and following the rule of always choosing the edge to the vertex with the smallest index first.