# -*- coding: utf-8 -*- """ Minimum cost flow algorithms on directed connected graphs. """ __author__ = """Loïc Séguin-C. """ # Copyright (C) 2010 Loïc Séguin-C. # All rights reserved. # BSD license. __all__ = ['network_simplex', 'min_cost_flow_cost', 'min_cost_flow', 'cost_of_flow', 'max_flow_min_cost'] 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[2].get(capacity, 1) > 0)) demand_nodes = (node for node in G.nodes_iter(data=True) if node[1].get(demand, 0) != 0) H.add_nodes_from(demand_nodes) r = H.nodes()[0] 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) T.add_edge(r, v) 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) T.add_edge(r, v) 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 T.add_edge(r, newLabel) T.add_edge(newLabel, v) 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) T.add_edge(v, r) 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) T.add_edge(v, r) flowCost += -vDemand * H[v][r].get(weight, 0) else: # existing edge does not have enough capacity newLabel = generate_unique_node() H.add_edge(v, newLabel, {weight: hugeWeight, 'flow': -vDemand}) H.add_edge(newLabel, r, {weight: hugeWeight, 'flow': -vDemand}) artificialEdges.append((v, newLabel)) artificialEdges.append((newLabel, r)) y[v] = -2 * hugeWeight y[newLabel] = -hugeWeight T.add_edge(v, newLabel) T.add_edge(newLabel, r) 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'): """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, newEdge is always a reverse edge (otherwise, # there would be no leaving edge). if len(cycle) == 3: u, v = newEdge 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) # 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 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). See also -------- 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_node('a', demand = -5) >>> G.add_node('d', demand = 5) >>> 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() >>> G.add_weighted_edges_from([('s','u',10), ('s','x',5), ... ('u','v',1), ('u','x',2), ... ('v','y',1), ('x','u',3), ... ('x','v',5), ('x','y',2), ... ('y','s',7), ('y','v',6)]) >>> G.add_node('s', demand = -1) >>> G.add_node('v', demand = 1) >>> flowCost, flowDict = nx.network_simplex(G) >>> flowCost == nx.shortest_path_length(G, 's', 'v', weight = 'weight') True >>> [(u, v) for u in flowDict for v in flowDict[u] if flowDict[u][v] > 0] [('x', 'u'), ('s', 'x'), ('u', 'v')] >>> 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_node('p', spam = -4) >>> G.add_node('q', spam = 2) >>> G.add_node('a', spam = -2) >>> G.add_node('d', spam = -1) >>> G.add_node('t', spam = 2) >>> G.add_node('w', spam = 3) >>> 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[0]) path2 = nx.shortest_path(T.to_undirected(), r, newEdge[1]) 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[0]][newEdge[1]].get('flow', 0) == 0: path2.reverse() cycle = path1 + path2 else: # newEdge is at capacity 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) # Actual augmentation happens here. If eps = 0, don't bother. if eps: flowCost -= cycleCost * eps if len(cycle) == 3: u, v = newEdge H[u][v]['flow'] -= eps H[v][u]['flow'] -= 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.add_edge(*newEdge) 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[0] 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[0], newEdge[1])) # print('Leaving edge: (%s, %s)' % (leavingEdge[0], leavingEdge[1])) # 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 def min_cost_flow_cost(G, demand = 'demand', capacity = 'capacity', weight = 'weight'): """Find the cost of a minimum cost flow satisfying all demands in digraph G. 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. Raises ------ NetworkXError This exception is raised if the input graph is not directed or not connected. 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. See also -------- cost_of_flow, max_flow_min_cost, min_cost_flow, network_simplex Examples -------- A simple example of a min cost flow problem. >>> import networkx as nx >>> G = nx.DiGraph() >>> G.add_node('a', demand = -5) >>> G.add_node('d', demand = 5) >>> 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 = nx.min_cost_flow_cost(G) >>> flowCost 24 """ return network_simplex(G, demand = demand, capacity = capacity, weight = weight)[0] def min_cost_flow(G, demand = 'demand', capacity = 'capacity', weight = 'weight'): """Return a minimum cost flow satisfying all demands in digraph G. 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 ------- 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 or not connected. 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. See also -------- cost_of_flow, max_flow_min_cost, min_cost_flow_cost, network_simplex Examples -------- A simple example of a min cost flow problem. >>> import networkx as nx >>> G = nx.DiGraph() >>> G.add_node('a', demand = -5) >>> G.add_node('d', demand = 5) >>> 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) >>> flowDict = nx.min_cost_flow(G) >>> flowDict {'a': {'c': 1, 'b': 4}, 'c': {'d': 1}, 'b': {'d': 4}, 'd': {}} """ return network_simplex(G, demand = demand, capacity = capacity, weight = weight)[1] def cost_of_flow(G, flowDict, weight = 'weight'): """Compute the cost of the flow given by flowDict on graph G. Note that this function does not check for the validity of the flow flowDict. This function will fail if the graph G and the flow don't have the same edge set. Parameters ---------- G : NetworkX graph DiGraph on which a minimum cost flow satisfying all demands is to be found. 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'. flowDict: dictionary Dictionary of dictionaries keyed by nodes such that flowDict[u][v] is the flow edge (u, v). Returns ------- cost: Integer, float The total cost of the flow. This is given by the sum over all edges of the product of the edge's flow and the edge's weight. See also -------- max_flow_min_cost, min_cost_flow, min_cost_flow_cost, network_simplex """ return sum((flowDict[u][v] * d.get(weight, 0) for u, v, d in G.edges_iter(data = True))) def max_flow_min_cost(G, s, t, capacity = 'capacity', weight = 'weight'): """Return a maximum (s, t)-flow of minimum cost. G is a digraph with edge costs and capacities. There is a source node s and a sink node t. This function finds a maximum flow from s to t whose total cost is minimized. Parameters ---------- G : NetworkX graph DiGraph on which a minimum cost flow satisfying all demands is to be found. s: node label Source of the flow. t: node label Destination of the flow. 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 ------- 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 or not connected. NetworkXUnbounded This exception is raised if there is an infinite capacity path from s to t in G. In this case there is no maximum flow. This exception is also raised if the digraph G has a cycle of negative cost and infinite capacity. Then, the cost of a flow is unbounded below. See also -------- cost_of_flow, ford_fulkerson, min_cost_flow, min_cost_flow_cost, network_simplex Examples -------- >>> G = nx.DiGraph() >>> G.add_edges_from([(1, 2, {'capacity': 12, 'weight': 4}), ... (1, 3, {'capacity': 20, 'weight': 6}), ... (2, 3, {'capacity': 6, 'weight': -3}), ... (2, 6, {'capacity': 14, 'weight': 1}), ... (3, 4, {'weight': 9}), ... (3, 5, {'capacity': 10, 'weight': 5}), ... (4, 2, {'capacity': 19, 'weight': 13}), ... (4, 5, {'capacity': 4, 'weight': 0}), ... (5, 7, {'capacity': 28, 'weight': 2}), ... (6, 5, {'capacity': 11, 'weight': 1}), ... (6, 7, {'weight': 8}), ... (7, 4, {'capacity': 6, 'weight': 6})]) >>> mincostFlow = nx.max_flow_min_cost(G, 1, 7) >>> nx.cost_of_flow(G, mincostFlow) 373 >>> maxFlow = nx.ford_fulkerson_flow(G, 1, 7) >>> nx.cost_of_flow(G, maxFlow) 428 >>> mincostFlowValue = (sum((mincostFlow[u][7] for u in G.predecessors(7))) ... - sum((mincostFlow[7][v] for v in G.successors(7)))) >>> mincostFlowValue == nx.max_flow(G, 1, 7) True """ maxFlow = nx.max_flow(G, s, t, capacity = capacity) H = nx.DiGraph(G) H.add_node(s, demand = -maxFlow) H.add_node(t, demand = maxFlow) return min_cost_flow(H, capacity = capacity, weight = weight)