Retrieving multiple optimal solutions with Pulp

Question

I am interested in using Pulp to get multiple optimal solutions if they exist. Much of the literature out there leads me to believe that this is not possible with a programming package, but I did find this promising example .

Here is the code

"""
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")

The code I'm most interested in is here

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

My question is, I would think the sum of all of the squares with 1 would be 81 for all optimal solutions. So why is <= 81 not used instead? Also, I am not sure why adding this constraint after solving at successive iterations produces different optimal solutions if they do indeed exist. Can someone please explain?

Answer 1

This kind of iteration can done only for pure binary Integer Programs.

That constraint will make the current optimal solution infeasible as the sum of its variables will add up to 81

Then the constraint <=80 will make that solution infeasible and the next round of the loop will have to find another solution.