Question

我最近从AMPL / CPLEX-CP迁移到Python / DOcplex.CP-MP。在AMPL中，可以使用“ expand _constraint_name”命令显示约束的扩展版本，这对于调试模型很有帮助。

DOcplex中是否有这种功能？例如：

在DOcplex中，可以通过某种方式对诸如all_diff之类的全局约束进行操作，如下所示：

import docplex.cp.model as cp

NB_QUEEN = 8
mdl = cp.CpoModel()
x = cp.integer_var_list(NB_QUEEN, 0, NB_QUEEN - 1, "X")

# add the constraints to the model
mdl.add(mdl.all_diff(x[i] + i for i in range(NB_QUEEN)))

# print the constraints
test_constraint = mdl.all_diff(x[i] - i for i in range(NB_QUEEN))
print(test_constraint)

将打印：

alldiff([X_0 - 0, X_1 - 1, X_2 - 2, X_3 - 3, X_4 - 4, X_5 - 5, X_6 - 6, X_7 - 7])

是否可以打印以下约束的扩展版本？

 mdl.add(mdl.sum( X[i]+j for j in R1) ==1 for i in R2)

仅在上述约束上调用print（）即可。

谢谢。

编辑

感谢Alex Fleischer，我制作了一个小型工作示例，展示了如何从一组约束中打印约束。

x = mdl.integer_var_list(5, name='X')
ct = ( [1 == sum( x[i]+j for i in range(0, 5) ) for j in range(0,5) ])
print(ct[2])

可打印

0 + (X_0 + 2) + (X_1 + 2) + (X_2 + 2) + (X_3 + 2) + (X_4 + 2) == 1

如果希望打印集合中的所有约束，则可以这样做

for j in range(0, 5)
    print(c[j])

Answer 1

是的，您可能会显示这样的约束，但是在您的情况下，您有一组约束。让我在CPLEX_Studio128 \ python \ examples \ cp \ basic

中的钢铁厂示例中为您提供一个小示例

from docplex.cp.model import CpoModel
from collections import namedtuple


#-----------------------------------------------------------------------------
# Initialize the problem data
#-----------------------------------------------------------------------------

# List of coils to produce (orders)
Order = namedtuple("Order", ['id', 'weight', 'color'])
ORDERS = (
            Order( 1, 22, 5),
            Order( 2,  9, 3),
            Order( 3,  9, 4),
            Order( 4,  8, 5),
            Order( 5,  8, 7),
            Order( 6,  6, 3),
            Order( 7,  5, 6),
            Order( 8,  3, 0),
            Order( 9,  3, 2),
            Order(10,  3, 3),
            Order(11,  2, 1),
            Order(12,  2, 5)
         )

# Max number of different colors of coils produced by a single slab
MAX_COLOR_PER_SLAB = 2

# List of available slab weights.
AVAILABLE_SLAB_WEIGHTS = [11, 13, 16, 17, 19, 20, 23, 24, 25,
                          26, 27, 28, 29, 30, 33, 34, 40, 43, 45]


#-----------------------------------------------------------------------------
# Prepare the data for modeling
#-----------------------------------------------------------------------------

# Upper bound for the number of slabs to use
MAX_SLABS = len(ORDERS)

# Build a set of all colors
allcolors = set(o.color for o in ORDERS)

# The heaviest slab
max_slab_weight = max(AVAILABLE_SLAB_WEIGHTS)

# Minimum loss incurred for a given slab usage.
# loss[v] = loss when smallest slab is used to produce a total weight of v
loss = [0] + [min([sw - use for sw in AVAILABLE_SLAB_WEIGHTS if sw >= use]) for use in range(1, max_slab_weight + 1)]


#-----------------------------------------------------------------------------
# Build the model
#-----------------------------------------------------------------------------

# Create model 
mdl = CpoModel()

total_loss=mdl.integer_var(0,10000000,"total loss")


# Index of the slab used to produce each coil order
production_slab = mdl.integer_var_list(len(ORDERS), 0, MAX_SLABS - 1, "production_slab")

# Usage of each slab
slab_use = mdl.integer_var_list(MAX_SLABS, 0, max_slab_weight, "slab_use")

# The orders are allocated to the slabs with capacity
mdl.add(mdl.pack(slab_use, production_slab, [o.weight for o in ORDERS]))

# Constrain max number of colors produced by each slab
for s in range(MAX_SLABS):
   su = 0
   for c in allcolors:
       lo = False
       for i, o in enumerate(ORDERS):
           if o.color == c:
               lo |= (production_slab[i] == s)
       su += lo
   mdl.add(su <= MAX_COLOR_PER_SLAB)

# Minimize the total loss
ct=(total_loss == sum([mdl.element(slab_use[s], loss) for s in range(MAX_SLABS)]))

mdl.add(ct)
print("ct=",ct)

mdl.add(mdl.minimize(total_loss))

# Set search strategy
mdl.set_search_phases([mdl.search_phase(production_slab)])


#-----------------------------------------------------------------------------
# Solve the model and display the result
#-----------------------------------------------------------------------------

# Solve model
print("Solving model....")
msol = mdl.solve(FailLimit=100000, TimeLimit=10)

# Print solution
if msol:
    print("Solution: ")
    for s in set(msol[ps] for ps in production_slab):
        # Determine orders using this slab
        lordrs = [o for i, o in enumerate(ORDERS) if msol[production_slab[i]] == s]
        # Compute display attributes
        used_weight = msol[slab_use[s]]          # Weight used in the slab
        loss_weight = loss[used_weight]          # Loss weight
        colors = set(o.color for o in lordrs)    # List of colors
        loids = [o.id for o in lordrs]           # List of order irs
        print("Slab weight={}, used={}, loss={}, colors={}, orders={}"
              .format(used_weight + loss_weight, used_weight, loss_weight, colors, loids))
else:
    print("No solution found")

给出

ct= "total loss" == 0 + element(slab_use0, [0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 0, 2, 1, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 5, 4, 3, 2, 1, 0, 2, 1, 0, 1, 0]) + element(slab_use1, [0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 0, 2, 1, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 5, 4, 3, 2, 1, 0, 2, 1, 0, 1, 0]) + element(slab_use2, [0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 0, 2, 1, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 5, 4, 3, 2, 1, 0, 2, 1, 0, 1, 0]) + element(slab_use3, [0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 0, 2, 1, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 5, 4, 3, 2, 1, 0, 2, 1, 0, 1, 0]) + element(slab_use4, [0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 0, 2, 1, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 5, 4, 3, 2, 1, 0, 2, 1, 0, 1, 0]) + element(slab_use5, [0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 0, 2, 1, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 5, 4, 3, 2, 1, 0, 2, 1, 0, 1, 0]) + element(slab_use6, [0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 0, 2, 1, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 5, 4, 3, 2, 1, 0, 2, 1, 0, 1, 0]) + element(slab_use7, [0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 0, 2, 1, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 5, 4, 3, 2, 1, 0, 2, 1, 0, 1, 0]) + element(slab_use8, [0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 0, 2, 1, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 5, 4, 3, 2, 1, 0, 2, 1, 0, 1, 0]) + element(slab_use9, [0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 0, 2, 1, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 5, 4, 3, 2, 1, 0, 2, 1, 0, 1, 0]) + element(slab_use10, [0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 0, 2, 1, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 5, 4, 3, 2, 1, 0, 2, 1, 0, 1, 0]) + element(slab_use11, [0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 0, 2, 1, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 5, 4, 3, 2, 1, 0, 2, 1, 0, 1, 0])

进行显示的行

# Minimize the total loss
ct=(total_loss == sum([mdl.element(slab_use[s], loss) for s in range(MAX_SLABS)]))

mdl.add(ct)
print("ct=",ct)

致谢

如何在DOcplex中扩展和打印非全局约束？

1 个答案: