def __satisfies_necessary_and_sufficient_conditions(g):
"""
Determines whether a digraph has an Eulerian cycle using necessary
and sufficient conditions (without computing the cycle itself):
- at least one edge
- indegree(v) = outdegree(v) for every vertex
- the graph is connected, when viewed as an undirected graph
(ignoring isolated vertices)
"""
# Condition 0: at least 1 Edge
if g.get_E() == 0:
return False
# Condition 1: indegree(v) == outdegree(v) for every vertex
for v in range(g.get_V()):
if g.outdegree() != g.indegree(v):
return False
# Condition 2: graph is connected, ignoring isolated vertices
h = Graph(g.get_V())
for v in range(g.get_V()):
for w in g.adj_vertices(v):
h.add_edge(v, w)
# check that all non-isolated vertices are connected
s = DirectedEulerianCycle.__non_isolated_vertex(g)
bfs = BreadthFirstPaths(h, s)
for v in range(g.get_V()):
if h.degree(v) > 0 and not bfs.has_path_to(v):
return False
return True
def __satisfies_necessary_and_sufficient_conditions(g):
"""
Determines whether a digraph has an Eulerian path using necessary
and sufficient conditions (without computing the path itself):
- indegree(v) = outdegree(v) for every vertex,
except one vertex v may have outdegree(v) = indegree(v) + 1
(and one vertex v may have indegree(v) = outdegree(v) + 1)
- the graph is connected, when viewed as an undirected graph
(ignoring isolated vertices)
"""
# Condition 0: at least 1 Edge
if g.get_E() == 0:
return True
# Condition 1: indegree(v) == outdegree(v) for every vertex,
# except one vertex may have outdegree(v) = indegree(v) + 1
deficit = 0
for v in range(g.get_V()):
if g.outdegree() > g.indegree(v):
deficit += (g.outdegree() - g.indegree(v))
if deficit > 1:
return False
# Condition 2: graph is connected, ignoring isolated vertices
h = Graph(g.get_V())
for v in range(g.get_V()):
for w in g.adj_vertices(v):
h.add_edge(v, w)
# check that all non-isolated vertices are connected
s = DirectedEulerianPath.__non_isolated_vertex(g)
bfs = BreadthFirstPaths(h, s)
for v in range(g.get_V()):
if h.degree(v) > 0 and not bfs.has_path_to(v):
return False
return True
