如何使用 Google 的 CP-SAT 求解器计算“AddAbsEquality”或“AddMultiplicationEqualit”以进行非线性优化？

Question

我的目标是根据预测序列恢复数据序列。 假设原始数据序列是x_org = [10, 20, 30, 40, 50]但我收到随机数据为x_ran = [50, 40, 20, 10, 30] 。 现在，我的目标是通过使它们最接近原始模式来恢复模式（最大限度地减少恢复损失）。

我使用了与 google OR-tool 网站 [https://developers.google.com/optimization/assignment/assignment_teams] 和 [https: //developers.google.com/optimization/cp/integer_opt_cp]。

我可以最小化损失（误差）的总和，但不能计算平方和/绝对和。

from ortools.sat.python import cp_model

x_org = [10, 20, 30, 40, 50]
x_ran = [50, 40, 20, 10, 30]
n = len(x_org)


model = cp_model.CpModel()

# Defidning recovered data
x_rec = [model.NewIntVar(0, 10000, 'x_rec_%i') for i in range(n)]

# Defidning recovery loss        
x_loss = [model.NewIntVar(0, 10000, 'x_loss_%i' % i) for i in range(n)]

# Defining a (recovery) mapping matrix 
M = {}
for i in range(n):
    for j in range(n):
        M[i, j] = model.NewBoolVar('M[%i,%i]' % (i, j)) 
    
# -----------------Constraints---------------%
# Each sensor is assigned one unique measurement.
for i in range(n):
    model.Add(sum([M[i, j] for j in range(n)]) == 1)

# Each measurement is assigned one unique sensor.
for j in range(n):
    model.Add(sum([M[i, j] for i in range(n)]) == 1)


# Recovering the remapped data x_rec=M*x_ran (like, Ax =b)
for i in range(n):   
    model.Add(x_rec[i] == sum([M[i,j]*x_ran[j] for j in range(n)]))

# Loss = orginal data - recovered data
for i in range(n):
    x_loss[i] = x_org[i] - x_rec[i]

    
# minimizing recovery loss
model.Minimize(sum(x_loss))

#--------------- Calling solver -------------%

# Solves and prints out the solution.
solver = cp_model.CpSolver()
status = solver.Solve(model)

print('Solve status: %s' % solver.StatusName(status))

if status == cp_model.OPTIMAL:
    print('Optimal objective value: %i' % solver.ObjectiveValue())
    for i in range(n):
        print('x_loss[%i] = %i' %(i,solver.Value(x_loss[i]))) 

那么 output 没有绝对误差之和为：

Solve status: OPTIMAL
Optimal objective value: 0
x_loss[0] = -10
x_loss[1] = -30
x_loss[2] = 0
x_loss[3] = 30
x_loss[4] = 10

这表明即使损失的总和为零，恢复也不正确。 但是，当我尝试添加另一个 int 变量来存储损失的绝对值时[如下所示]，编译器会报错。

# Defidning abs recovery loss        
x_loss_abs = [model.NewIntVar(0, 10000, 'x_loss_abs_%i' % i) for i in range(n)] 
# Loss = orginal data - recovered data
for i in range(n):
    model.AddAbsEquality(x_loss_abs[i], x_loss[i])
    #model.AddMultiplicationEquality(x_loss_abs[i], [x_loss[i], x_loss[i]])

错误是回溯是：

TypeError                                 Traceback (most recent call last)
<ipython-input-42-2a043a8fef8b> in <module>
      3 # Loss = orginal data - recovered data
      4 for i in range(n):
----> 5     model.AddAbsEquality(x_loss_abs[i], x_loss[i])

~/anaconda3/envs/tensorgpu/lib/python3.7/site-packages/ortools/sat/python/cp_model.py in AddAbsEquality(self, target, var)
   1217         ct = Constraint(self.__model.constraints)
   1218         model_ct = self.__model.constraints[ct.Index()]
-> 1219         index = self.GetOrMakeIndex(var)
   1220         model_ct.int_max.vars.extend([index, -index - 1])
   1221         model_ct.int_max.target = self.GetOrMakeIndex(target)

~/anaconda3/envs/tensorgpu/lib/python3.7/site-packages/ortools/sat/python/cp_model.py in GetOrMakeIndex(self, arg)
   1397         else:
   1398             raise TypeError('NotSupported: model.GetOrMakeIndex(' + str(arg) +
-> 1399                             ')')
   1400 
   1401     def GetOrMakeBooleanIndex(self, arg):

TypeError: NotSupported: model.GetOrMakeIndex((-x_rec_%i + 10))

您能否建议如何最小化恢复损失的绝对总和/平方和？ 谢谢。

Answer 1

AddAbsEquality要求 arguments 是变量（而不是诸如x_org[i] - x_rec[i]类的表达式。因此必须在使用它之前创建一个临时决策变量（此处为v ）。以下似乎有效：

# ...
x_loss_abs = [model.NewIntVar(0, 10000, 'x_loss_abs_%i' % i) for i in range(n)]

# ...
for i in range(n):
   # x_loss[i] = x_org[i] - x_rec[i] # Original
   v = model.NewIntVar(-1000,1000,"v") # Temporary variable
   model.Add(v == x_org[i] - x_rec[i] )
   model.AddAbsEquality(x_loss_abs[i],v)

# ....
model.Minimize(sum(x_loss_abs))

然后解决方案是（我更改了输出）：

Optimal objective value: 0
x_org: [[10, 20, 30, 40, 50]]
x_rec: [10, 20, 30, 40, 50]
x_loss: [0, 0, 0, 0, 0]