使用 Pulp 检索多个最优解

Question

我有兴趣使用 Pulp 获得多个最优解决方案（如果存在）。那里的许多文献让我相信编程包不可能做到这一点，但我确实找到了this promising example。

这里是代码

"""
The Looping Sudoku Problem Formulation for the PuLP Modeller

Authors: Antony Phillips, Dr Stuart Mitcehll
"""
# Import PuLP modeler functions
from pulp import *

# A list of strings from "1" to "9" is created
Sequence = ["1", "2", "3", "4", "5", "6", "7", "8", "9"]

# The Vals, Rows and Cols sequences all follow this form
Vals = Sequence
Rows = Sequence
Cols = Sequence

# The boxes list is created, with the row and column index of each square in 
# each box
Boxes =[]
for i in range(3):
    for j in range(3):
        Boxes += [[(Rows[3*i+k],Cols[3*j+l]) for k in range(3) for l in range(3)]]

# The prob variable is created to contain the problem data
prob = LpProblem("Sudoku Problem",LpMinimize)

# The problem variables are created
choices = LpVariable.dicts("Choice",(Vals,Rows,Cols),0,1,LpInteger)

# The arbitrary objective function is added
prob += 0, "Arbitrary Objective Function"

# A constraint ensuring that only one value can be in each square is created
for r in Rows:
    for c in Cols:
        prob += lpSum([choices[v][r][c] for v in Vals]) == 1, ""

# The row, column and box constraints are added for each value
for v in Vals:
    for r in Rows:
        prob += lpSum([choices[v][r][c] for c in Cols]) == 1,""

    for c in Cols:
        prob += lpSum([choices[v][r][c] for r in Rows]) == 1,""

    for b in Boxes:
        prob += lpSum([choices[v][r][c] for (r,c) in b]) == 1,""

# The starting numbers are entered as constraints
prob += choices["5"]["1"]["1"] == 1,""
prob += choices["6"]["2"]["1"] == 1,""
prob += choices["8"]["4"]["1"] == 1,""
prob += choices["4"]["5"]["1"] == 1,""
prob += choices["7"]["6"]["1"] == 1,""
prob += choices["3"]["1"]["2"] == 1,""
prob += choices["9"]["3"]["2"] == 1,""
prob += choices["6"]["7"]["2"] == 1,""
prob += choices["8"]["3"]["3"] == 1,""
prob += choices["1"]["2"]["4"] == 1,""
prob += choices["8"]["5"]["4"] == 1,""
prob += choices["4"]["8"]["4"] == 1,""
prob += choices["7"]["1"]["5"] == 1,""
prob += choices["9"]["2"]["5"] == 1,""
prob += choices["6"]["4"]["5"] == 1,""
prob += choices["2"]["6"]["5"] == 1,""
prob += choices["1"]["8"]["5"] == 1,""
prob += choices["8"]["9"]["5"] == 1,""
prob += choices["5"]["2"]["6"] == 1,""
prob += choices["3"]["5"]["6"] == 1,""
prob += choices["9"]["8"]["6"] == 1,""
prob += choices["2"]["7"]["7"] == 1,""
prob += choices["6"]["3"]["8"] == 1,""
prob += choices["8"]["7"]["8"] == 1,""
prob += choices["7"]["9"]["8"] == 1,""
prob += choices["3"]["4"]["9"] == 1,""
prob += choices["1"]["5"]["9"] == 1,""
prob += choices["6"]["6"]["9"] == 1,""
prob += choices["5"]["8"]["9"] == 1,""

# The problem data is written to an .lp file
prob.writeLP("Sudoku.lp")

# A file called sudokuout.txt is created/overwritten for writing to
sudokuout = open('sudokuout.txt','w')

while True:
    prob.solve()
    # The status of the solution is printed to the screen
    print("Status:", LpStatus[prob.status])
    # The solution is printed if it was deemed "optimal" i.e met the constraints
    if LpStatus[prob.status] == "Optimal":
        # The solution is written to the sudokuout.txt file
        for r in Rows:
            if r == "1" or r == "4" or r == "7":
                sudokuout.write("+-------+-------+-------+\n")
            for c in Cols:
                for v in Vals:
                    if value(choices[v][r][c])==1:
                        if c == "1" or c == "4" or c =="7":
                            sudokuout.write("| ")
                        sudokuout.write(v + " ")
                        if c == "9":
                            sudokuout.write("|\n")
        sudokuout.write("+-------+-------+-------+\n\n")
        # The constraint is added that the same solution cannot be returned again
        prob += lpSum([choices[v][r][c] for v in Vals
                                    for r in Rows
                                    for c in Cols
                                    if value(choices[v][r][c])==1]) <= 80
    # If a new optimal solution cannot be found, we end the program
    else:
        break
sudokuout.close()

# The location of the solutions is give to the user
print("Solutions Written to sudokuout.txt")

我最感兴趣的代码在这里

prob += lpSum([choices[v][r][c] for v in Vals
                                    for r in Rows
                                    for c in Cols
                                    if value(choices[v][r][c])==1]) <= 80

我的问题是，对于所有最佳解决方案，我认为带有 1 的所有平方和将是 81。那么为什么不使用<= 81 呢？另外，我不确定为什么在连续迭代求解后添加这个约束会产生不同的最优解，如果它们确实存在的话。谁能解释一下？

Answer 1

这种迭代只能用于纯二进制整数程序。

该约束将使当前的最优解不可行，因为其变量的总和将达到 81

那么约束<=80 将使该解决方案不可行，并且下一轮循环将不得不寻找另一个解决方案。