Warning

This documents an unmaintained version of NetworkX. Please upgrade to a maintained version and see the current NetworkX documentation.

# Source code for networkx.algorithms.flow.networksimplex

# -*- coding: utf-8 -*-
"""
Minimum cost flow algorithms on directed connected graphs.
"""

__author__ = """Loïc Séguin-C. <loicseguin@gmail.com>"""
# Copyright (C) 2010 Loïc Séguin-C. <loicseguin@gmail.com>

__all__ = ['network_simplex']

import networkx as nx
from networkx.utils import generate_unique_node

def _initial_tree_solution(G, demand = 'demand', capacity = 'capacity',
weight = 'weight'):
"""Find a initial tree solution rooted at r.

The initial tree solution is obtained by considering edges (r, v)
for all nodes v with non-negative demand and (v, r) for all nodes
with negative demand. If these edges do not exist, we add them to
the graph and call them artificial edges.
"""
H = nx.DiGraph((edge for edge in G.edges(data=True) if
edge.get(capacity, 1) > 0))
demand_nodes = (node for node in G.nodes_iter(data=True) if
node.get(demand, 0) != 0)
r = H.nodes()

T = nx.DiGraph()
y = {r: 0}
artificialEdges = []
flowCost = 0

n = H.number_of_nodes()
try:
maxWeight = max(abs(d[weight]) for u, v, d in H.edges(data = True)
if weight in d)
except ValueError:
maxWeight = 0
hugeWeight = 1 + n * maxWeight

for v, d in H.nodes(data = True)[1:]:
vDemand = d.get(demand, 0)
if vDemand >= 0:
if not (r, v) in H.edges():
H.add_edge(r, v, {weight: hugeWeight, 'flow': vDemand})
artificialEdges.append((r, v))
y[v] = H[r][v].get(weight, 0)
flowCost += vDemand * H[r][v].get(weight, 0)

else: # (r, v) in H.edges()
if (not capacity in H[r][v]
or vDemand <= H[r][v][capacity]):
H[r][v]['flow'] = vDemand
y[v] = H[r][v].get(weight, 0)
flowCost += vDemand * H[r][v].get(weight, 0)

else: # existing edge does not have enough capacity
newLabel = generate_unique_node()
H.add_edge(r, newLabel, {weight: hugeWeight, 'flow': vDemand})
H.add_edge(newLabel, v, {weight: hugeWeight, 'flow': vDemand})
artificialEdges.append((r, newLabel))
artificialEdges.append((newLabel, v))
y[v] = 2 * hugeWeight
y[newLabel] = hugeWeight
flowCost += 2 * vDemand * hugeWeight

else: # vDemand < 0
if not (v, r) in H.edges():
H.add_edge(v, r, {weight: hugeWeight, 'flow': -vDemand})
artificialEdges.append((v, r))
y[v] = -H[v][r].get(weight, 0)
flowCost += -vDemand * H[v][r].get(weight, 0)

else:
if (not capacity in H[v][r]
or -vDemand <= H[v][r][capacity]):
H[v][r]['flow'] = -vDemand
y[v] = -H[v][r].get(weight, 0)
flowCost += -vDemand * H[v][r].get(weight, 0)
else: # existing edge does not have enough capacity
newLabel = generate_unique_node()
{weight: hugeWeight, 'flow': -vDemand})
{weight: hugeWeight, 'flow': -vDemand})
artificialEdges.append((v, newLabel))
artificialEdges.append((newLabel, r))
y[v] = -2 * hugeWeight
y[newLabel] = -hugeWeight
flowCost += 2 * -vDemand * hugeWeight

return H, T, y, artificialEdges, flowCost, r

def _find_entering_edge(H, c, capacity = 'capacity'):
"""Find an edge which creates a negative cost cycle in the actual
tree solution.

The reduced cost of every edge gives the value of the cycle
obtained by adding that edge to the tree solution. If that value is
negative, we will augment the flow in the direction indicated by
the edge. Otherwise, we will augment the flow in the reverse
direction.

If no edge is found, return and empty tuple. This will cause the
main loop of the algorithm to terminate.
"""
newEdge = ()
for u, v, d in H.edges_iter(data = True):
if d.get('flow', 0) == 0:
if c[(u, v)] < 0:
newEdge = (u, v)
break
else:
if capacity in d:
if (d.get('flow', 0) == d[capacity]
and c[(u, v)] > 0):
newEdge = (u, v)
break
return newEdge

def _find_leaving_edge(H, T, cycle, newEdge, capacity = 'capacity',
reverse=False):
"""Find an edge that will leave the basis and the value by which we
can increase or decrease the flow on that edge.

The leaving arc rule is used to prevent cycling.

If cycle has no reverse edge and no forward edge of finite
capacity, it means that cycle is a negative cost infinite capacity
cycle. This implies that the cost of a flow satisfying all demands
is unbounded below. An exception is raised in this case.
"""
eps = False
leavingEdge = ()

# If cycle is a digon.
if len(cycle) == 3:
u, v = newEdge
if capacity not in H[u][v] and capacity not in H[v][u]:
raise nx.NetworkXUnbounded(
"Negative cost cycle of infinite capacity found. "
+ "Min cost flow unbounded below.")

if reverse:
if H[u][v].get('flow', 0) > H[v][u].get('flow', 0):
return (v, u), H[v][u].get('flow', 0)
else:
return (u, v), H[u][v].get('flow', 0)
else:
uv_residual = H[u][v].get(capacity, 0) - H[u][v].get('flow', 0)
vu_residual = H[v][u].get(capacity, 0) - H[v][u].get('flow', 0)
if (uv_residual > vu_residual):
return (v, u), vu_residual
else:
return (u, v), uv_residual

# Find the forward edge with the minimum value for capacity - 'flow'
# and the reverse edge with the minimum value for 'flow'.
for index, u in enumerate(cycle[:-1]):
edgeCapacity = False
edge = ()
v = cycle[index + 1]
if (u, v) in T.edges() + [newEdge]: #forward edge
if capacity in H[u][v]: # edge (u, v) has finite capacity
edgeCapacity = H[u][v][capacity] - H[u][v].get('flow', 0)
edge = (u, v)
else: #reverse edge
edgeCapacity = H[v][u].get('flow', 0)
edge = (v, u)

# Determine if edge might be the leaving edge.
if edge:
if leavingEdge:
if edgeCapacity < eps:
eps = edgeCapacity
leavingEdge = edge
else:
eps = edgeCapacity
leavingEdge = edge

if not leavingEdge:
raise nx.NetworkXUnbounded(
"Negative cost cycle of infinite capacity found. "
+ "Min cost flow unbounded below.")

return leavingEdge, eps

def _create_flow_dict(G, H):
"""Creates the flow dict of dicts of graph G with auxiliary graph H."""
flowDict = dict([(u, {}) for u in G])

for u in G.nodes_iter():
for v in G.neighbors(u):
if H.has_edge(u, v):
flowDict[u][v] = H[u][v].get('flow', 0)
else:
flowDict[u][v] = 0
return flowDict

[docs]def network_simplex(G, demand = 'demand', capacity = 'capacity',
weight = 'weight'):
"""Find a minimum cost flow satisfying all demands in digraph G.

This is a primal network simplex algorithm that uses the leaving
arc rule to prevent cycling.

G is a digraph with edge costs and capacities and in which nodes
have demand, i.e., they want to send or receive some amount of
flow. A negative demand means that the node wants to send flow, a
positive demand means that the node want to receive flow. A flow on
the digraph G satisfies all demand if the net flow into each node
is equal to the demand of that node.

Parameters
----------
G : NetworkX graph
DiGraph on which a minimum cost flow satisfying all demands is
to be found.

demand: string
Nodes of the graph G are expected to have an attribute demand
that indicates how much flow a node wants to send (negative
demand) or receive (positive demand). Note that the sum of the
demands should be 0 otherwise the problem in not feasible. If
this attribute is not present, a node is considered to have 0
demand. Default value: 'demand'.

capacity: string
Edges of the graph G are expected to have an attribute capacity
that indicates how much flow the edge can support. If this
attribute is not present, the edge is considered to have
infinite capacity. Default value: 'capacity'.

weight: string
Edges of the graph G are expected to have an attribute weight
that indicates the cost incurred by sending one unit of flow on
that edge. If not present, the weight is considered to be 0.
Default value: 'weight'.

Returns
-------
flowCost: integer, float
Cost of a minimum cost flow satisfying all demands.

flowDict: dictionary
Dictionary of dictionaries keyed by nodes such that
flowDict[u][v] is the flow edge (u, v).

Raises
------
NetworkXError
This exception is raised if the input graph is not directed,
not connected or is a multigraph.

NetworkXUnfeasible
This exception is raised in the following situations:
* The sum of the demands is not zero. Then, there is no
flow satisfying all demands.
* There is no flow satisfying all demand.

NetworkXUnbounded
This exception is raised if the digraph G has a cycle of
negative cost and infinite capacity. Then, the cost of a flow
satisfying all demands is unbounded below.

Notes
-----
This algorithm is not guaranteed to work if edge weights
are floating point numbers (overflows and roundoff errors can
cause problems).

--------
cost_of_flow, max_flow_min_cost, min_cost_flow, min_cost_flow_cost

Examples
--------
A simple example of a min cost flow problem.

>>> import networkx as nx
>>> G = nx.DiGraph()
>>> G.add_edge('a', 'b', weight = 3, capacity = 4)
>>> G.add_edge('a', 'c', weight = 6, capacity = 10)
>>> G.add_edge('b', 'd', weight = 1, capacity = 9)
>>> G.add_edge('c', 'd', weight = 2, capacity = 5)
>>> flowCost, flowDict = nx.network_simplex(G)
>>> flowCost
24
>>> flowDict # doctest: +SKIP
{'a': {'c': 1, 'b': 4}, 'c': {'d': 1}, 'b': {'d': 4}, 'd': {}}

The mincost flow algorithm can also be used to solve shortest path
problems. To find the shortest path between two nodes u and v,
give all edges an infinite capacity, give node u a demand of -1 and
node v a demand a 1. Then run the network simplex. The value of a
min cost flow will be the distance between u and v and edges
carrying positive flow will indicate the path.

>>> G=nx.DiGraph()
...                            ('u','v',1), ('u','x',2),
...                            ('v','y',1), ('x','u',3),
...                            ('x','v',5), ('x','y',2),
...                            ('y','s',7), ('y','v',6)])
>>> flowCost, flowDict = nx.network_simplex(G)
>>> flowCost == nx.shortest_path_length(G, 's', 'v', weight = 'weight')
True
>>> sorted([(u, v) for u in flowDict for v in flowDict[u] if flowDict[u][v] > 0])
[('s', 'x'), ('u', 'v'), ('x', 'u')]
>>> nx.shortest_path(G, 's', 'v', weight = 'weight')
['s', 'x', 'u', 'v']

It is possible to change the name of the attributes used for the
algorithm.

>>> G = nx.DiGraph()
>>> G.add_edge('p', 'q', cost = 7, vacancies = 5)
>>> G.add_edge('p', 'a', cost = 1, vacancies = 4)
>>> G.add_edge('q', 'd', cost = 2, vacancies = 3)
>>> G.add_edge('t', 'q', cost = 1, vacancies = 2)
>>> G.add_edge('a', 't', cost = 2, vacancies = 4)
>>> G.add_edge('d', 'w', cost = 3, vacancies = 4)
>>> G.add_edge('t', 'w', cost = 4, vacancies = 1)
>>> flowCost, flowDict = nx.network_simplex(G, demand = 'spam',
...                                         capacity = 'vacancies',
...                                         weight = 'cost')
>>> flowCost
37
>>> flowDict  # doctest: +SKIP
{'a': {'t': 4}, 'd': {'w': 2}, 'q': {'d': 1}, 'p': {'q': 2, 'a': 2}, 't': {'q': 1, 'w': 1}, 'w': {}}

References
----------
W. J. Cook, W. H. Cunningham, W. R. Pulleyblank and A. Schrijver.
Combinatorial Optimization. Wiley-Interscience, 1998.

"""

if not G.is_directed():
raise nx.NetworkXError("Undirected graph not supported.")
if not nx.is_connected(G.to_undirected()):
raise nx.NetworkXError("Not connected graph not supported.")
if G.is_multigraph():
raise nx.NetworkXError("MultiDiGraph not supported.")
if sum(d[demand] for v, d in G.nodes(data = True)
if demand in d) != 0:
raise nx.NetworkXUnfeasible("Sum of the demands should be 0.")

# Fix an arbitrarily chosen root node and find an initial tree solution.
H, T, y, artificialEdges, flowCost, r = \
_initial_tree_solution(G, demand = demand, capacity = capacity,
weight = weight)

# Initialize the reduced costs.
c = {}
for u, v, d in H.edges_iter(data = True):
c[(u, v)] = d.get(weight, 0) + y[u] - y[v]

# Print stuff for debugging.
# print('-' * 78)
# nbIter = 0
# print('Iteration %d' % nbIter)
# nbIter += 1
# print('Tree solution: %s' % T.edges())
# print(' Edge %11s%10s' % ('Flow', 'Red Cost'))
# for u, v, d in H.edges(data = True):
#     flag = ''
#     if (u, v) in artificialEdges:
#         flag = '*'
#     print('(%s, %s)%1s%10d%10d' % (u, v, flag, d.get('flow', 0),
#                                    c[(u, v)]))
# print('Distances: %s' % y)

# Main loop.
while True:
newEdge = _find_entering_edge(H, c, capacity = capacity)
if not newEdge:
break # Optimal basis found. Main loop is over.
cycleCost = abs(c[newEdge])

# Find the cycle created by adding newEdge to T.
path1 = nx.shortest_path(T.to_undirected(), r, newEdge)
path2 = nx.shortest_path(T.to_undirected(), r, newEdge)
join = r
for index, node in enumerate(path1[1:]):
if index + 1 < len(path2) and node == path2[index + 1]:
join = node
else:
break
path1 = path1[path1.index(join):]
path2 = path2[path2.index(join):]
cycle = []
if H[newEdge][newEdge].get('flow', 0) == 0:
reverse = False
path2.reverse()
cycle = path1 + path2
else: # newEdge is at capacity
reverse = True
path1.reverse()
cycle = path2 + path1

# Find the leaving edge. Will stop here if cycle is an infinite
# capacity negative cost cycle.
leavingEdge, eps = _find_leaving_edge(H, T, cycle, newEdge,
capacity=capacity,
reverse=reverse)

# Actual augmentation happens here. If eps = 0, don't bother.
if eps:
flowCost -= cycleCost * eps
if len(cycle) == 3:
if reverse:
eps = -eps
u, v = newEdge
H[u][v]['flow'] = H[u][v].get('flow', 0) + eps
H[v][u]['flow'] = H[v][u].get('flow', 0) + eps
else:
for index, u in enumerate(cycle[:-1]):
v = cycle[index + 1]
if (u, v) in T.edges() + [newEdge]:
H[u][v]['flow'] = H[u][v].get('flow', 0) + eps
else: # (v, u) in T.edges():
H[v][u]['flow'] -= eps

# Update tree solution.
T.remove_edge(*leavingEdge)

# Update distances and reduced costs.
if newEdge != leavingEdge:
forest = nx.DiGraph(T)
forest.remove_edge(*newEdge)
R, notR = nx.connected_component_subgraphs(forest.to_undirected())
if r in notR.nodes(): # make sure r is in R
R, notR = notR, R
if newEdge in R.nodes():
for v in notR.nodes():
y[v] += c[newEdge]
else:
for v in notR.nodes():
y[v] -= c[newEdge]
for u, v in H.edges():
if u in notR.nodes() or v in notR.nodes():
c[(u, v)] = H[u][v].get(weight, 0) + y[u] - y[v]

# Print stuff for debugging.
# print('-' * 78)
# print('Iteration %d' % nbIter)
# nbIter += 1
# print('Tree solution: %s' % T.edges())
# print('New edge:      (%s, %s)' % (newEdge, newEdge))
# print('Leaving edge:  (%s, %s)' % (leavingEdge, leavingEdge))
# print('Cycle:         %s' % cycle)
# print('eps:           %d' % eps)
# print(' Edge %11s%10s' % ('Flow', 'Red Cost'))
# for u, v, d in H.edges(data = True):
#     flag = ''
#     if (u, v) in artificialEdges:
#         flag = '*'
#     print('(%s, %s)%1s%10d%10d' % (u, v, flag, d.get('flow', 0),
#                                    c[(u, v)]))
# print('Distances: %s' % y)

# If an artificial edge has positive flow, the initial problem was
# not feasible.
for u, v in artificialEdges:
if H[u][v]['flow'] != 0:
raise nx.NetworkXUnfeasible("No flow satisfying all demands.")
H.remove_edge(u, v)

for u in H.nodes():
if not u in G:
H.remove_node(u)

flowDict = _create_flow_dict(G, H)

return flowCost, flowDict