"""Biconnected components and articulation points."""
from itertools import chain
from networkx.utils.decorators import not_implemented_for
__all__ = [
"biconnected_components",
"biconnected_component_edges",
"is_biconnected",
"articulation_points",
]
[docs]@not_implemented_for("directed")
def is_biconnected(G):
"""Returns True if the graph is biconnected, False otherwise.
A graph is biconnected if, and only if, it cannot be disconnected by
removing only one node (and all edges incident on that node). If
removing a node increases the number of disconnected components
in the graph, that node is called an articulation point, or cut
vertex. A biconnected graph has no articulation points.
Parameters
----------
G : NetworkX Graph
An undirected graph.
Returns
-------
biconnected : bool
True if the graph is biconnected, False otherwise.
Raises
------
NetworkXNotImplemented
If the input graph is not undirected.
Examples
--------
>>> G = nx.path_graph(4)
>>> print(nx.is_biconnected(G))
False
>>> G.add_edge(0, 3)
>>> print(nx.is_biconnected(G))
True
See Also
--------
biconnected_components
articulation_points
biconnected_component_edges
is_strongly_connected
is_weakly_connected
is_connected
is_semiconnected
Notes
-----
The algorithm to find articulation points and biconnected
components is implemented using a non-recursive depth-first-search
(DFS) that keeps track of the highest level that back edges reach
in the DFS tree. A node `n` is an articulation point if, and only
if, there exists a subtree rooted at `n` such that there is no
back edge from any successor of `n` that links to a predecessor of
`n` in the DFS tree. By keeping track of all the edges traversed
by the DFS we can obtain the biconnected components because all
edges of a bicomponent will be traversed consecutively between
articulation points.
References
----------
.. [1] Hopcroft, J.; Tarjan, R. (1973).
"Efficient algorithms for graph manipulation".
Communications of the ACM 16: 372–378. doi:10.1145/362248.362272
"""
bcc = list(biconnected_components(G))
if len(bcc) == 1:
return len(bcc[0]) == len(G)
return False # Multiple bicomponents or No bicomponents (empty graph?)
# if len(bcc) == 0: # No bicomponents (it could be an empty graph)
# return False
# return len(bcc[0]) == len(G)
[docs]@not_implemented_for("directed")
def biconnected_component_edges(G):
"""Returns a generator of lists of edges, one list for each biconnected
component of the input graph.
Biconnected components are maximal subgraphs such that the removal of a
node (and all edges incident on that node) will not disconnect the
subgraph. Note that nodes may be part of more than one biconnected
component. Those nodes are articulation points, or cut vertices.
However, each edge belongs to one, and only one, biconnected component.
Notice that by convention a dyad is considered a biconnected component.
Parameters
----------
G : NetworkX Graph
An undirected graph.
Returns
-------
edges : generator of lists
Generator of lists of edges, one list for each bicomponent.
Raises
------
NetworkXNotImplemented
If the input graph is not undirected.
Examples
--------
>>> G = nx.barbell_graph(4, 2)
>>> print(nx.is_biconnected(G))
False
>>> bicomponents_edges = list(nx.biconnected_component_edges(G))
>>> len(bicomponents_edges)
5
>>> G.add_edge(2, 8)
>>> print(nx.is_biconnected(G))
True
>>> bicomponents_edges = list(nx.biconnected_component_edges(G))
>>> len(bicomponents_edges)
1
See Also
--------
is_biconnected,
biconnected_components,
articulation_points,
Notes
-----
The algorithm to find articulation points and biconnected
components is implemented using a non-recursive depth-first-search
(DFS) that keeps track of the highest level that back edges reach
in the DFS tree. A node `n` is an articulation point if, and only
if, there exists a subtree rooted at `n` such that there is no
back edge from any successor of `n` that links to a predecessor of
`n` in the DFS tree. By keeping track of all the edges traversed
by the DFS we can obtain the biconnected components because all
edges of a bicomponent will be traversed consecutively between
articulation points.
References
----------
.. [1] Hopcroft, J.; Tarjan, R. (1973).
"Efficient algorithms for graph manipulation".
Communications of the ACM 16: 372–378. doi:10.1145/362248.362272
"""
yield from _biconnected_dfs(G, components=True)
[docs]@not_implemented_for("directed")
def biconnected_components(G):
"""Returns a generator of sets of nodes, one set for each biconnected
component of the graph
Biconnected components are maximal subgraphs such that the removal of a
node (and all edges incident on that node) will not disconnect the
subgraph. Note that nodes may be part of more than one biconnected
component. Those nodes are articulation points, or cut vertices. The
removal of articulation points will increase the number of connected
components of the graph.
Notice that by convention a dyad is considered a biconnected component.
Parameters
----------
G : NetworkX Graph
An undirected graph.
Returns
-------
nodes : generator
Generator of sets of nodes, one set for each biconnected component.
Raises
------
NetworkXNotImplemented
If the input graph is not undirected.
See Also
--------
k_components : this function is a special case where k=2
bridge_components : similar to this function, but is defined using
2-edge-connectivity instead of 2-node-connectivity.
Examples
--------
>>> G = nx.lollipop_graph(5, 1)
>>> print(nx.is_biconnected(G))
False
>>> bicomponents = list(nx.biconnected_components(G))
>>> len(bicomponents)
2
>>> G.add_edge(0, 5)
>>> print(nx.is_biconnected(G))
True
>>> bicomponents = list(nx.biconnected_components(G))
>>> len(bicomponents)
1
You can generate a sorted list of biconnected components, largest
first, using sort.
>>> G.remove_edge(0, 5)
>>> [len(c) for c in sorted(nx.biconnected_components(G), key=len, reverse=True)]
[5, 2]
If you only want the largest connected component, it's more
efficient to use max instead of sort.
>>> Gc = max(nx.biconnected_components(G), key=len)
To create the components as subgraphs use:
``(G.subgraph(c).copy() for c in biconnected_components(G))``
See Also
--------
is_biconnected
articulation_points
biconnected_component_edges
Notes
-----
The algorithm to find articulation points and biconnected
components is implemented using a non-recursive depth-first-search
(DFS) that keeps track of the highest level that back edges reach
in the DFS tree. A node `n` is an articulation point if, and only
if, there exists a subtree rooted at `n` such that there is no
back edge from any successor of `n` that links to a predecessor of
`n` in the DFS tree. By keeping track of all the edges traversed
by the DFS we can obtain the biconnected components because all
edges of a bicomponent will be traversed consecutively between
articulation points.
References
----------
.. [1] Hopcroft, J.; Tarjan, R. (1973).
"Efficient algorithms for graph manipulation".
Communications of the ACM 16: 372–378. doi:10.1145/362248.362272
"""
for comp in _biconnected_dfs(G, components=True):
yield set(chain.from_iterable(comp))
[docs]@not_implemented_for("directed")
def articulation_points(G):
"""Yield the articulation points, or cut vertices, of a graph.
An articulation point or cut vertex is any node whose removal (along with
all its incident edges) increases the number of connected components of
a graph. An undirected connected graph without articulation points is
biconnected. Articulation points belong to more than one biconnected
component of a graph.
Notice that by convention a dyad is considered a biconnected component.
Parameters
----------
G : NetworkX Graph
An undirected graph.
Yields
------
node
An articulation point in the graph.
Raises
------
NetworkXNotImplemented
If the input graph is not undirected.
Examples
--------
>>> G = nx.barbell_graph(4, 2)
>>> print(nx.is_biconnected(G))
False
>>> len(list(nx.articulation_points(G)))
4
>>> G.add_edge(2, 8)
>>> print(nx.is_biconnected(G))
True
>>> len(list(nx.articulation_points(G)))
0
See Also
--------
is_biconnected
biconnected_components
biconnected_component_edges
Notes
-----
The algorithm to find articulation points and biconnected
components is implemented using a non-recursive depth-first-search
(DFS) that keeps track of the highest level that back edges reach
in the DFS tree. A node `n` is an articulation point if, and only
if, there exists a subtree rooted at `n` such that there is no
back edge from any successor of `n` that links to a predecessor of
`n` in the DFS tree. By keeping track of all the edges traversed
by the DFS we can obtain the biconnected components because all
edges of a bicomponent will be traversed consecutively between
articulation points.
References
----------
.. [1] Hopcroft, J.; Tarjan, R. (1973).
"Efficient algorithms for graph manipulation".
Communications of the ACM 16: 372–378. doi:10.1145/362248.362272
"""
seen = set()
for articulation in _biconnected_dfs(G, components=False):
if articulation not in seen:
seen.add(articulation)
yield articulation
@not_implemented_for("directed")
def _biconnected_dfs(G, components=True):
# depth-first search algorithm to generate articulation points
# and biconnected components
visited = set()
for start in G:
if start in visited:
continue
discovery = {start: 0} # time of first discovery of node during search
low = {start: 0}
root_children = 0
visited.add(start)
edge_stack = []
stack = [(start, start, iter(G[start]))]
while stack:
grandparent, parent, children = stack[-1]
try:
child = next(children)
if grandparent == child:
continue
if child in visited:
if discovery[child] <= discovery[parent]: # back edge
low[parent] = min(low[parent], discovery[child])
if components:
edge_stack.append((parent, child))
else:
low[child] = discovery[child] = len(discovery)
visited.add(child)
stack.append((parent, child, iter(G[child])))
if components:
edge_stack.append((parent, child))
except StopIteration:
stack.pop()
if len(stack) > 1:
if low[parent] >= discovery[grandparent]:
if components:
ind = edge_stack.index((grandparent, parent))
yield edge_stack[ind:]
edge_stack = edge_stack[:ind]
else:
yield grandparent
low[grandparent] = min(low[parent], low[grandparent])
elif stack: # length 1 so grandparent is root
root_children += 1
if components:
ind = edge_stack.index((grandparent, parent))
yield edge_stack[ind:]
if not components:
# root node is articulation point if it has more than 1 child
if root_children > 1:
yield start