is_locally_k_edge_connected(G, s, t, k)
Is it impossible to disconnect s and t by removing fewer than k edges? If so, then s and t are locally k-edge-connected in G.
An undirected graph.
Source node
Target node
local edge connectivity for nodes s and t
True if s and t are locally k-edge-connected in G.
Tests to see if an edge in a graph is locally k-edge-connected.
is_k_edge_connected
func
>>> from networkx.algorithms.connectivity import is_locally_k_edge_connected
... G = nx.barbell_graph(10, 0)
... is_locally_k_edge_connected(G, 5, 15, k=1) True
>>> is_locally_k_edge_connected(G, 5, 15, k=2) False
>>> is_locally_k_edge_connected(G, 1, 5, k=2) TrueSee :
The following pages refer to to this document either explicitly or contain code examples using this.
networkx.algorithms.connectivity.edge_augmentation.is_k_edge_connected
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