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Note

This documents the development version of NetworkX. Documentation for the current release can be found here.

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Source code for networkx.algorithms.components.biconnected

"""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