<｜begin▁of▁sentence｜>### Instruction:
Given a graph with an even number of vertices, where vertices with even identifiers only connect to vertices with odd identifiers and vice versa, how can you unfold this graph into a tree using a BFS algorithm, starting from a specified root vertex?
### Response:
from igraph import Graph

# Define a function that will determine the children of each node
# In this example, we assume a simplistic model where each vertex
# with an even identifier connects to vertices with odd identifiers only and vice versa.
def get_children(vertex):
    return [v for v in range(vertex_count) if v % 2 != vertex % 2]

vertex_count = 10  # The total number of vertices in the graph

# Create an empty graph
g = Graph()
# Add vertices
g.add_vertices(vertex_count)

# Use unfold_tree to create a tree based on the provided get_children function
# root specifies the root of the BFS tree
root = 0 # The root vertex where the BFS tree starts
unfolded_tree = g.unfold_tree(get_children, root=root)

# Now you can work with the unfolded_tree as usual, for example, print its summary
print(unfolded_tree.summary())
<|EOT|>
