Sudoku

Originally Contributed by: Iain Dunning

Sudoku is a popular number puzzle. The goal is to place the digits 1,...,9 on a nine-by-nine grid, with some of the digits already filled in. Your solution must satisfy the following rules:

• The numbers 1 to 9 must appear in each 3x3 square
• The numbers 1 to 9 must appear in each row
• The numbers 1 to 9 must appear in each column

Here is a partially solved Sudoku problem:

Solving a Sudoku isn't an optimization problem with an objective; its actually a feasibility problem: we wish to find a feasible solution that satisfies these rules. You can think of it as an optimization problem with an objective of 0.

Mixed-integer linear programming formulation

We can model this problem using 0-1 integer programming: a problem where all the decision variables are binary. We'll use JuMP to create the model, and then we can solve it with any integer programming solver.

using JuMP
using HiGHS

We will define a binary variable (a variable that is either 0 or 1) for each possible number in each possible cell. The meaning of each variable is as follows: x[i,j,k] = 1 if and only if cell (i,j) has number k, where i is the row and j is the column.

Create a model

sudoku = Model(HiGHS.Optimizer)

A JuMP Model
Feasibility problem with:
Variables: 0
Model mode: AUTOMATIC
CachingOptimizer state: EMPTY_OPTIMIZER
Solver name: HiGHS

Create our variables

@variable(sudoku, x[i = 1:9, j = 1:9, k = 1:9], Bin);

Now we can begin to add our constraints. We'll actually start with something obvious to us as humans, but what we need to enforce: that there can be only one number per cell.

for i in 1:9  ## For each row
    for j in 1:9  ## and each column
        # Sum across all the possible digits. One and only one of the digits
        # can be in this cell, so the sum must be equal to one.
        @constraint(sudoku, sum(x[i, j, k] for k in 1:9) == 1)
    end
end

Next we'll add the constraints for the rows and the columns. These constraints are all very similar, so much so that we can actually add them at the same time.

for ind in 1:9  ## Each row, OR each column
    for k in 1:9  ## Each digit
        # Sum across columns (j) - row constraint
        @constraint(sudoku, sum(x[ind, j, k] for j in 1:9) == 1)
        # Sum across rows (i) - column constraint
        @constraint(sudoku, sum(x[i, ind, k] for i in 1:9) == 1)
    end
end

Finally, we have the to enforce the constraint that each digit appears once in each of the nine 3x3 sub-grids. Our strategy will be to index over the top-left corners of each 3x3 square with for loops, then sum over the squares.

for i in 1:3:7
    for j in 1:3:7
        for k in 1:9
            # i is the top left row, j is the top left column.
            # We'll sum from i to i+2, e.g. i=4, r=4, 5, 6.
            @constraint(
                sudoku,
                sum(x[r, c, k] for r in i:(i+2), c in j:(j+2)) == 1
            )
        end
    end
end

The final step is to add the initial solution as a set of constraints. We'll solve the problem that is in the picture at the start of the tutorial. We'll put a 0 if there is no digit in that location.

The given digits

init_sol = [
    5 3 0 0 7 0 0 0 0
    6 0 0 1 9 5 0 0 0
    0 9 8 0 0 0 0 6 0
    8 0 0 0 6 0 0 0 3
    4 0 0 8 0 3 0 0 1
    7 0 0 0 2 0 0 0 6
    0 6 0 0 0 0 2 8 0
    0 0 0 4 1 9 0 0 5
    0 0 0 0 8 0 0 7 9
]
for i in 1:9
    for j in 1:9
        # If the space isn't empty
        if init_sol[i, j] != 0
            # Then the corresponding variable for that digit and location must
            # be 1.
            fix(x[i, j, init_sol[i, j]], 1; force = true)
        end
    end
end

solve problem

optimize!(sudoku)

Presolving model
151 rows, 115 cols, 466 nonzeros
0 rows, 0 cols, 0 nonzeros
Presolve: Optimal

Solving report
  Status            Optimal
  Primal bound      0
  Dual bound        0
  Gap               0% (tolerance: 0.01%)
  Solution status   feasible
                    0 (objective)
                    0 (bound viol.)
                    0 (int. viol.)
                    0 (row viol.)
  Timing            0.00 (total)
                    0.00 (presolve)
                    0.00 (postsolve)
  Nodes             0
  LP iterations     0 (total)
                    0 (strong br.)
                    0 (separation)
                    0 (heuristics)

Extract the values of x

x_val = value.(x);

Create a matrix to store the solution

sol = zeros(Int, 9, 9)  # 9x9 matrix of integers
for i in 1:9
    for j in 1:9
        for k in 1:9
            # Integer programs are solved as a series of linear programs so the
            # values might not be precisely 0 and 1. We can round them to
            # the nearest integer to make it easier.
            if round(Int, x_val[i, j, k]) == 1
                sol[i, j] = k
            end
        end
    end
end

Display the solution

sol

9×9 Matrix{Int64}:
 5  3  4  6  7  8  9  1  2
 6  7  2  1  9  5  3  4  8
 1  9  8  3  4  2  5  6  7
 8  5  9  7  6  1  4  2  3
 4  2  6  8  5  3  7  9  1
 7  1  3  9  2  4  8  5  6
 9  6  1  5  3  7  2  8  4
 2  8  7  4  1  9  6  3  5
 3  4  5  2  8  6  1  7  9

Which is the correct solution:

Constraint programming formulation

We can also model this problem using constraint programming and the all-different constraint, which says that no two elements of a vector can take the same value.

Because of the reformulation system in MathOptInterface, we can still solve this problem using HiGHS.

model = Model(HiGHS.Optimizer)
set_silent(model)
# HiGHS v1.2 has a bug in presolve which causes the problem to be classified as
# infeasible.
set_optimizer_attribute(model, "presolve", "off")

Instead of the binary variables, we directly define a 9x9 grid of integer values between 1 and 9:

@variable(model, 1 <= x[1:9, 1:9] <= 9, Int);

Then, we enforce that the values in each row must be all-different:

@constraint(model, [i = 1:9], x[i, :] in MOI.AllDifferent(9));

That the values in each column must be all-different:

@constraint(model, [j = 1:9], x[:, j] in MOI.AllDifferent(9));

And that the values in each 3x3 sub-grid must be all-different:

for i in (0, 3, 6), j in (0, 3, 6)
    @constraint(model, vec(x[i.+(1:3), j.+(1:3)]) in MOI.AllDifferent(9))
end

Finally, as before we set the initial solution and optimize:

for i in 1:9, j in 1:9
    if init_sol[i, j] != 0
        fix(x[i, j], init_sol[i, j]; force = true)
    end
end

optimize!(model)

Display the solution

csp_sol = round.(Int, value.(x))

9×9 Matrix{Int64}:
 5  3  4  6  7  8  9  1  2
 6  7  2  1  9  5  3  4  8
 1  9  8  3  4  2  5  6  7
 8  5  9  7  6  1  4  2  3
 4  2  6  8  5  3  7  9  1
 7  1  3  9  2  4  8  5  6
 9  6  1  5  3  7  2  8  4
 2  8  7  4  1  9  6  3  5
 3  4  5  2  8  6  1  7  9

Which is the same as we found before:

sol == csp_sol

true