(a). For any connected graph G, all internal nodes of the BFS tree on G have the same number of children.
(b). For any connected graph G, the DFS tree on G and the BFS tree on G have the same number of edges.
(a) To prove that all internal nodes of the BFS tree on G have the same number of children, we can use contradiction.
Assume that there exists an internal node in the BFS tree on G that has a different number of children compared to other internal nodes. Let's call this node N.
Since N is an internal node, it must have a parent. Let's call the parent node of N as P.
If N has a different number of children compared to other internal nodes, it means that at least one of its children has a different number of children. This contradicts the property of the BFS tree, where all nodes at the same level in the tree have the same number of children.
Therefore, our assumption that there exists an internal node in the BFS tree on G with a different number of children is false. All internal nodes of the BFS tree on G indeed have the same number of children.
(b) To prove that the DFS tree on G and the BFS tree on G have the same number of edges, we can use induction.
Let's assume that G has n vertices and m edges.
For the base case, when n = 1 (a graph with only one vertex), both the DFS tree and the BFS tree will be the same single-node tree with 0 edges.
Now, let's assume that for all connected graphs with fewer than n vertices and m edges, the DFS tree and the BFS tree have the same number of edges.
Consider a connected graph G with n vertices and m edges. We can perform a DFS or BFS traversal on G to obtain the corresponding DFS tree or BFS tree.
Since both the DFS and BFS traversal visit every vertex exactly once, the DFS tree and BFS tree will have the same number of vertices.
Now, let's compare the number of edges in the DFS tree and the BFS tree. Both trees have the same number of vertices, so we need to show that they also have the same number of edges.
Assume that the DFS tree has k edges. We know that G has m edges, so removing one edge from G will split it into two or more connected components. This means that removing one edge from the DFS tree will also split it into two or more trees.
The same applies to the BFS tree. If we remove one edge from the BFS tree, it will be split into two or more trees.
By induction hypothesis, each of these trees (resulting from removing an edge) will have the same number of edges in the DFS tree and the BFS tree.
Therefore, adding back the removed edge will increase the number of edges in the DFS tree and the BFS tree by the same amount.
Since we can remove and add edges in the same way for both trees, this means that the DFS tree and the BFS tree will have the same number of edges.
By induction, we have shown that for any connected graph G, the DFS tree and the BFS tree have the same number of edges.
To understand and explain the given statements, we need to have a clear understanding of the Breadth-First Search (BFS) and Depth-First Search (DFS) algorithms.
BFS (Breadth-First Search) and DFS (Depth-First Search) are both graph traversal algorithms used to explore or search a graph or tree data structure.
(a) For any connected graph G, all internal nodes of the BFS tree on G have the same number of children:
To explain this statement, we first need to understand the structure of the BFS tree for a connected graph G.
In BFS, we start at a given source vertex and explore its neighboring vertices (children) before moving further to their children. The algorithm explores vertices level by level, moving outward from the source.
When the BFS algorithm is applied to a connected graph G, it generates a BFS tree. The BFS tree is a tree that represents the exploration of the graph, starting from a particular source vertex.
Now, let's assume that there is an internal node in the BFS tree of G that does not have the same number of children as the other internal nodes.
To get a contradicting scenario, we need to understand that in BFS, the algorithm traverses the graph level by level. It means that at each level, all the vertices that are adjacent to the vertices at the previous level are added to the BFS tree.
If there is a node in the BFS tree with a different number of children than the others, it implies that the algorithm has not explored all the vertices at that level, violating the BFS algorithm's level-based exploration.
Hence, we can conclude that in a connected graph G, all internal nodes of the BFS tree have the same number of children.
(b) For any connected graph G, the DFS tree on G and the BFS tree on G have the same number of edges:
To explain this statement, we need to understand the difference in the structure of the DFS tree and the BFS tree.
In DFS, we start at a given source vertex and explore as far as possible along each branch before backtracking. It goes deep into the graph and explores all vertices along a path before backtracking. The DFS tree represents the exploration as a tree.
In BFS, as mentioned earlier, we explore the graph level by level, moving outward from the source. The BFS tree represents the exploration as a tree.
Now, let's assume that the DFS tree and the BFS tree on a connected graph G have a different number of edges.
To contradict this, we need to understand that the DFS tree and the BFS tree are both generated by the exploration of the same graph G. They represent the same graph but with different traversal approaches.
If they have a different number of edges, it means that the exploration has missed some edges or explored extra edges, which is not possible because both DFS and BFS aim to explore all vertices and edges of the graph.
Therefore, we can conclude that for any connected graph G, the DFS tree and the BFS tree will have the same number of edges.
In summary:
(a) For any connected graph G, all internal nodes of the BFS tree on G have the same number of children.
(b) For any connected graph G, the DFS tree on G and the BFS tree on G have the same number of edges.