在存在顺序约束的情况下，OR 工具路由无法找到 'trivial' 最佳解决方案

Question

为了更好地理解 or-tools 路由背后的约束编程，我创建了一个 depot 的玩具示例和配置为允许两条路由的其他 4 个节点。

想法是车辆从站点 0 行驶到 1，然后选择 2 或 3，继续行驶到 4 并且returns 到仓库 0；车辆选择绿色或红色路径。我的实际问题比较复杂，有多辆车，但有类似的约束。

对于这个例子，我为成本创建了一个欧氏距离函数：

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()

和一个 l0 距离函数来约束顺序：

# 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 第一个解决方案起作用的析取，而析取 [2, 3] 说明绿色或红色路径应该被选中。

有了这些析取，求解器找到了一个解决方案，但是如果我添加 2 和 3 应该在 1 之后和 4 之前访问，求解器会根本不访问 2 或 3。为什么会这样？为什么求解器不能找到更优的路径 0 -> 1 -> 2/3 -> 4 -> 0 避免 [2,3] 的 int(1e10) 析取惩罚？

编辑：

通过删除顺序约束并添加到 Distance.__call__ 来软约束顺序约束：

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)

为了惩罚不允许的订单，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，即使在 search_parameters.local_search_operators.

中明确启用 use_swap_active 和 use_relocate_neighbors

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)

结论：search-space小，较好的解在返回解use_relocate_neighbors附近，但or-tools没有找到。为什么？

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-问题列表回答了我的问题：https://github.com/google/or-tools/issues/252#issuecomment-249646587

首先，约束规划在更严格的约束下表现更好，我想有些东西会被彻底搜索。特别是，我必须限制顺序维度：

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

通过

更好地约束

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

随后，不需要检查 2 或 3 是否处于活动状态，因此我们可以像这样简化顺序约束：

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，但在全代码版本中没有。现在返回一个可行解。

在存在顺序约束的情况下，OR 工具路由无法找到 'trivial' 最佳解决方案

OR-tools routing fails to find 'trivial' optimal solution in the presence of order-constraints

python

or-tools

所有代码：