所有代码：

Question

为了更好地理解约束编程背后的工具路由，我创建了一个软件仓库的玩具示例和另外4个配置为允许两条路由的节点。

想法是车辆从仓库0前往1，然后选择2或3，继续4并返回仓库0;车辆要么采摘绿色或红色路径。我的实际问题更复杂，有多辆车，但也有类似的限制。

在这个例子中，我为成本创建了一个欧几里德距离函数：

class Distances:
    def __init__(self):
        self.locations = [
            [-1,  0], # source
            [ 0, -1], # waypoint 1
            [ 0,  1], # waypoint 2
            [ 1,  0], # destination
        ]

    def __len__(self):
        return len(self.locations) + 1

    def dist(self, x, y):
        return int(10000 * math.sqrt(
            (x[0] - y[0]) ** 2 +
            (x[1] - y[1]) ** 2))

    def __call__(self, i, j):
        if i == 0 and j == 0:
            return 0
        if j == 0 or i == 0:
            return 1 # very small distance between depot and non-depot, simulating 0
        return self.dist(self.locations[i - 1], self.locations[j - 1])


distance = Distances()

用于约束顺序的l0距离函数：

# l0-distance to add order constraints
class Order:
    def __call__(self, i, j):
        return 0 if i == j else 1


order = Order()

然后我创建模型并尝试解决这个问题：

search_parameters = pywrapcp.RoutingModel.DefaultSearchParameters()
search_parameters.first_solution_strategy = (
        routing_enums_pb2.FirstSolutionStrategy.ALL_UNPERFORMED)

search_parameters.local_search_metaheuristic = routing_enums_pb2.LocalSearchMetaheuristic.SIMULATED_ANNEALING
search_parameters.time_limit_ms = 3000

routing = pywrapcp.RoutingModel(len(distance), 1)

routing.SetArcCostEvaluatorOfAllVehicles(distance)
routing.SetDepot(0)
solver = routing.solver()

routing.AddDimension(order, int(1e18), int(1e18), True, "order")

# Since `ALL_UNPERFORMED` is used, each node must be allowed inactive
order_dimension = routing.GetDimensionOrDie("order")
routing.AddDisjunction([1], int(1e10))
routing.AddDisjunction([2, 3], int(1e10))
routing.AddDisjunction([4], int(1e10))

solver.AddConstraint(order_dimension.CumulVar(1) <= order_dimension.CumulVar(2))
solver.AddConstraint(order_dimension.CumulVar(1) <= order_dimension.CumulVar(3))

solver.AddConstraint(order_dimension.CumulVar(2) <= order_dimension.CumulVar(4))
solver.AddConstraint(order_dimension.CumulVar(3) <= order_dimension.CumulVar(4))

# routing.AddPickupAndDelivery(1, 2)
# routing.AddPickupAndDelivery(1, 3)
# routing.AddPickupAndDelivery(2, 4)
# routing.AddPickupAndDelivery(3, 4)

routing.CloseModelWithParameters(search_parameters)
assignment = routing.SolveWithParameters(search_parameters)

if assignment is not None:
    print('cost: ' + str(assignment.ObjectiveValue()))
    route = []
    index = routing.Start(0)
    while not routing.IsEnd(index):
        route.append(routing.IndexToNode(index))
        index = assignment.Value(routing.NextVar(index))
    for node in route:
        print(' - {:2d}'.format(node))
else:
    print('nothing found')

所以[1]和[4]是允许ALL_UNPERFORMED第一个解决方案工作的分离，而[2, 3]分离则表明应该选择绿色或红色路径

通过这些析取解算器找到解决方案，但是如果我添加2并且3应该在1之后和4之前访问，则解算器不会访问2或3。为什么会这样？为什么解算器无法找到更优的路由0 -> 1 -> 2/3 -> 4 -> 0，从而避免int(1e10) [2,3]的{{1}}分离惩罚？

修改

通过删除订单约束并添加到Distance.__call__：

来软约束订单约束

if (i == 2 or j == 1) or (i == 3 or j == 1) or (i == 4 or j == 2) or (i == 4 or j == 3):
    return int(1e10)

惩罚不允许的订单，导致路线0 -> 2 -> 1 -> 4 -> 0。所以我想知道为什么or-tools不会交换1和2，即使明确在use_swap_active中启用use_relocate_neighbors和search_parameters.local_search_operators。

注意：失败，因为它应该是：

if (i == 2 and j == 1) or (i == 3 and j == 1) or (i == 4 and j == 2) or (i == 4 and j == 3):
    return int(1e10)

总结：搜索空间很小，更好的解决方案是在返回的解决方案的use_relocate_neighbors附近，但是or-tools却找不到它。为什么呢？

所有代码：

import pandas
import os.path

import numpy
import math
from ortools.constraint_solver import pywrapcp
from ortools.constraint_solver import routing_enums_pb2


class Distances:
    def __init__(self):
        self.locations = [
            [-1,  0], # source
            [ 0, -1], # waypoint 1
            [ 0,  1], # waypoint 2
            [ 1,  0], # destination
        ]

    def __len__(self):
        return len(self.locations) + 1

    def dist(self, x, y):
        return int(10000 * math.sqrt(
            (x[0] - y[0]) ** 2 +
            (x[1] - y[1]) ** 2))

    def __call__(self, i, j):
        if i == 0 and j == 0:
            return 0
        if j == 0 or i == 0:
            return 1 # very small distance between depot and non-depot, simulating 0
        return self.dist(self.locations[i - 1], self.locations[j - 1])


distance = Distances()


# l0-distance to add order constraints
class Order:
    def __call__(self, i, j):
        return 0 if i == j else 1


order = Order()

search_parameters = pywrapcp.RoutingModel.DefaultSearchParameters()
search_parameters.first_solution_strategy = (
        routing_enums_pb2.FirstSolutionStrategy.ALL_UNPERFORMED)

search_parameters.local_search_metaheuristic = routing_enums_pb2.LocalSearchMetaheuristic.SIMULATED_ANNEALING
search_parameters.time_limit_ms = 3000

routing = pywrapcp.RoutingModel(len(distance), 1)

routing.SetArcCostEvaluatorOfAllVehicles(distance)
routing.SetDepot(0)
solver = routing.solver()

routing.AddDimension(order, int(1e18), int(1e18), True, "order")

# Since `ALL_UNPERFORMED` is used, each node must be allowed inactive
order_dimension = routing.GetDimensionOrDie("order")
routing.AddDisjunction([1], int(1e10))
routing.AddDisjunction([2, 3], int(1e10))
routing.AddDisjunction([4], int(1e10))

solver.AddConstraint(
    (routing.ActiveVar(2) == 0)
    or
    (order_dimension.CumulVar(1) <= order_dimension.CumulVar(2))
)
solver.AddConstraint(
    (routing.ActiveVar(3) == 0)
    or
    (order_dimension.CumulVar(1) <= order_dimension.CumulVar(3))
)


solver.AddConstraint(
    (routing.ActiveVar(2) == 0)
    or
    (order_dimension.CumulVar(2) <= order_dimension.CumulVar(4))
)
solver.AddConstraint(
    (routing.ActiveVar(3) == 0)
    or
    (order_dimension.CumulVar(3) <= order_dimension.CumulVar(4))
)

# routing.AddPickupAndDelivery(1, 2)
# routing.AddPickupAndDelivery(1, 3)
# routing.AddPickupAndDelivery(2, 4)
# routing.AddPickupAndDelivery(3, 4)

routing.CloseModelWithParameters(search_parameters)
assignment = routing.SolveWithParameters(search_parameters)

if assignment is not None:
    print('cost: ' + str(assignment.ObjectiveValue()))
    route = []
    index = routing.Start(0)
    while not routing.IsEnd(index):
        route.append(routing.IndexToNode(index))
        index = assignment.Value(routing.NextVar(index))
    for node in route:
        print(' - {:2d}'.format(node))
else:
    print('nothing found')

Answer 1

@furnon 在github上通过github-issues列表回答了我的问题：https://github.com/google/or-tools/issues/252#issuecomment-249646587

首先，约束编程在更严格的约束条件下表现更好，我想有些事情是在搜索范围内进行的。特别是，我不得不限制订单维度：

routing.AddDimension(order, int(1e18), int(1e18), True, "order")

通过

得到更好的约束

routing.AddDimension(order, len(distance) + 1 ,len(distance) + 1, True, "order")

随后，不需要检查2或3是否有效，因此我们可以简化这样的顺序约束：

solver.AddConstraint(order_dimension.CumulVar(1) <= order_dimension.CumulVar(2))
solver.AddConstraint(order_dimension.CumulVar(1) <= order_dimension.CumulVar(3))
solver.AddConstraint(order_dimension.CumulVar(2) <= order_dimension.CumulVar(4))
solver.AddConstraint(order_dimension.CumulVar(3) <= order_dimension.CumulVar(4))

正如我在内联版本中所做的那样，但不是全代码版本。现在返回一个可行的解决方案。

OR工具路由无法找到“琐碎的”工具。存在订单约束时的最优解决方案

所有代码：

1 个答案: