Question Problem 6 [15 pts] We have a connected undirected graph G = ( V , E ) G = (V, E) , and a specific vertex u ∈ V u \in V u ∈ V . Suppose we compute a depth-first search tree rooted at u u , and obtain a tree T T that includes all nodes of G G . Suppose we then compute a breadth-first search tree rooted at u u , and obtain the same tree T T . Prove that G = T G = T . (In other words, if T T is both a depth-first search tree and a breadth-first search tree rooted at u u , then G G cannot contain any edges that do not belong to T T . Hint: Consider a proof by contradiction. Answer To prove that G = T G = T , we will use a proof by contradiction. Proof by Contradiction Assume that G ≠ T G \neq T . This implies that there exists at least one edge e ∈ E e \in E e ∈ E (where G = ( V , E ) G = (V, E) that is not included in T T . Since G G is a connected, undirected graph, and T T is a tree that includes all nodes of G G , T T is a spanning tree of G G . By definition of a spanning tree, T T connects all vertices of G G with exactly ∣ V ∣ − 1 |V| - 1 edges (where ∣ V ∣ |V| is the number of vertices in G) G . Now, consider the additional edge e e . Adding e e to T T would create a cycle, because T T already spans all vertices and e e e provides an additional connection. Depth-First Search Tree (DFS Tree) In a DFS tree, each node is explored as deeply as possible before backtracking. If T T is a DFS tree rooted at u u u , then T T reflects the depth-first traversal from u u u . Breadth-First Search Tree (BFS Tree) In a BFS tree, each node is explored level by level. If T T is also a BFS tree rooted at u u , then T T reflects the breadth-first traversal from u u u . Contradiction Since both DFS and BFS start from the same root u u and T T is the same tree for both traversals: For DFS, the traversal would explore as deeply as possible. For BFS, the traversal would explore level by level. If T T is the same for both DFS and BFS, this implies that there cannot be any additional edge e e e that is not part of T T . This is because the presence of such an edge e e would mean that the tree structures from DFS and BFS would differ, as the traversal orders and tree shapes would differ. This contradicts the assumption that T T is both a DFS tree and a BFS tree rooted at u u . Thus, our assumption that G ≠ T G \neq T leads to a contradiction. Hence, it must be that G = T G = T . Conclusion Therefore, if T T is both a depth-first search tree and a breadth-first search tree rooted at u u , then G G cannot contain any edges that do not belong to T T , implying G = T G = T .

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