我有兴趣使用Pulp来获得多种最佳解决方案（如果存在的话）。那里的大部分文献都让我相信这对编程方案是不可能的，但我确实找到了this promising example。
这是代码
"""
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")
我最感兴趣的代码就在这里
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，我认为所有与1的正方形的总和将是所有最优解的81。那么为什么不使用<= 81呢？此外，我不确定为什么在连续迭代求解后添加此约束会产生不同的最优解，如果确实存在的话。有人可以解释一下吗？
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加起来为81
然后约束<=80将使该解决方案不可行，并且下一轮循环将不得不找到另一种解决方案。