为什么ortools 严格设置约束？

如何解决为什么ortools 严格设置约束？

我经历了 google ortools scheduling tutorial 并创建了稍微不同的约束。

假设我们有一个 df，表示以下内容：

1. 每位护士只能在预先分配的班次上工作。根据表格，护士 0 必须在 0 或 1 班工作，护士 2 必须只在 2 班工作，依此类推。
2. 此外，我添加了以下约束：“如果nurse_0 需要shift_0，那么她必须需要shif_1”。它由“is_child”列显示 - shift_1 是 child_shift for shift_0 fornurse_0。

 df_dict = {
            "slot_number":[1,1,2,3],"asset_name":['0','1','0','2'],"is_child":["No","No","Yes",'No','No']
           }

  df = pd.DataFrame.from_dict(df_dict)

创建变量和模型：

num_nurses = 3
num_shifts = 3
all_nurses = range(num_nurses)
all_shifts = range(num_shifts)

model = cp_model.CpModel()

shifts = {}
for n in all_nurses:
    for s in all_shifts:
        shifts[(n,s)] = model.NewBoolVar('shift_n%is%i' % (n,s)) 

一名护士只能轮班：

for s in all_shifts:
    model.Add(sum(shifts[(n,s)] for n in all_nurses) == 1)

我的约束：

# whether the slot (key) have any children in other slots
children = {0: [1],1: [],2: []} 
#  what nurses are considered the children (values) in which slot (key)
nurse_child_sched = {0:[],1:[0],2:[]} 
# In this case if nurse 0 take slot 0,then she must take slot 1 too.
for s in all_shifts:
    for n in all_nurses:
        if (children[s]):
            for child in children[s]:
                if n in nurse_child_sched[child]:
                    model.Add((shifts[(n,s)] + shifts[(n,child)])==2)
                    print(f"{s}-parent_slot and {child}-child_slot were connected for nurse {n}")

用于创建时间表和显示解决方案的代码：

class NursesPartialSolutionPrinter(cp_model.cpsolverSolutionCallback):
"""Print intermediate solutions."""

def __init__(self,shifts,num_nurses,num_shifts,sols):
    cp_model.cpsolverSolutionCallback.__init__(self)
    self._shifts = shifts
    self._num_nurses = num_nurses
    self._num_shifts = num_shifts
    self._solutions = set(sols)
    self._solution_count = 0

def on_solution_callback(self):
    if self._solution_count in self._solutions:
        print('Solution %i' % self._solution_count)

        for n in range(self._num_nurses):
            is_working = False
            for s in range(self._num_shifts):
                if self.Value(self._shifts[(n,s)]):
                    is_working = True
                    print('  Nurse %i works shift %i' % (n,s))
            if not is_working:
                print('  Nurse {} does not work'.format(n))
        print()
    self._solution_count += 1

def solution_count(self):
    return self._solution_count

solver = cp_model.cpsolver()
solver.parameters.linearization_level = 0
a_few_solutions = range(5)
solution_printer = NursesPartialSolutionPrinter(shifts,a_few_solutions)

solver.SearchForAllSolutions(model,solution_printer)

最后，我得到以下结果：

Solution 0
  Nurse 0 works shift 0
  Nurse 0 works shift 1
  Nurse 1 works shift 2
  Nurse 2 does not work

Solution 1
  Nurse 0 works shift 0
  Nurse 0 works shift 1
  Nurse 1 does not work
  Nurse 2 works shift 2

Solution 2
  Nurse 0 works shift 0
  Nurse 0 works shift 1
  Nurse 0 works shift 2
  Nurse 1 does not work
  Nurse 2 does not work

不过，按照我的逻辑，至少也必须有下面的解决方案。我意识到，我的约束表明模型将nurse_0 放在shift_0 上，但我只需要放置两个插槽之间的关系，以防nurse_0 放在第一班，否则没有约束。提前致谢。

Solution x
  Nurse 1 works shift 0
  Nurse 0 works shift 1
  Nurse 0 works shift 2
  Nurse 1 does not work
  Nurse 2 does not work

解决方法

附注：

一名护士只能轮班：

  for s in all_shifts:
    model.Add(sum(shifts[(n,s)] for n in all_nurses) == 1)

这里你写了“每个班次只分配给一名护士”

如果你想表达“一个护士只能轮班”，你应该写：

for n in all_nurses:
    model.Add(sum(shifts[(n,s)] for s in all_shifts) == 1)

注意：但在这种情况下，由于护士只工作一个班次，因此不可能进行儿童班次，所以我猜您的代码没问题，而不是注释...