【问题标题】:How to expand and print non-global constraints in DOcplex?如何在 DOcplex 中扩展和打印非全局约束?
【发布时间】:2023-03-14 15:30:01
【问题描述】:

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,这在调试模型时很有帮助。

DOcplex 中有这样的功能吗?例如:

在 DOcplex 中,可以通过某种方式对 all_diff 等全局约束执行此操作,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])

有没有办法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)

仅在上述约束下调用 print() 是不行的。

谢谢。

编辑

感谢 Alex Fleischer,我做了一个小的工作示例,展示了如何从一组约束中打印一个约束。

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])

打印出来的

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,他们可以这样做

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

【问题讨论】:

    标签: python docplex


    【解决方案1】:

    是的,您可以显示此类约束,但在您的情况下,您有一组约束。让我从 CPLEX_Studio128\python\examples\cp\basic 中的钢厂示例中举一个小例子

    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")
    

    给了

    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])
    

    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)
    

    【讨论】:

    • 太好了,所以问题是要打印的约束表达式必须分配给一个变量,并且应该在该变量上调用 print 函数。有关一个小的工作示例,请参阅 OP 对原始问题的编辑。
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