Note: This page has been translated by MathWorks. Click here to see

To view all translated materials including this page, select Country from the country navigator on the bottom of this page.

To view all translated materials including this page, select Country from the country navigator on the bottom of this page.

This example shows how to solve a Sudoku puzzle using binary integer programming. For the solver-based approach, see Solve Sudoku Puzzles Via Integer Programming: Solver-Based.

You probably have seen Sudoku puzzles. A puzzle is to fill a 9-by-9 grid with integers from 1 through 9 so that each integer appears only once in each row, column, and major 3-by-3 square. The grid is partially populated with clues, and your task is to fill in the rest of the grid.

Here is a data matrix `B`

of clues. The first row, `B(1,2,2)`

, means row 1, column 2 has a clue 2. The second row, `B(1,5,3)`

, means row 1, column 5 has a clue 3. Here is the entire matrix `B`

.

```
B = [1,2,2;
1,5,3;
1,8,4;
2,1,6;
2,9,3;
3,3,4;
3,7,5;
4,4,8;
4,6,6;
5,1,8;
5,5,1;
5,9,6;
6,4,7;
6,6,5;
7,3,7;
7,7,6;
8,1,4;
8,9,8;
9,2,3;
9,5,4;
9,8,2];
drawSudoku(B) % For the listing of this program, see the end of this example.
```

This puzzle, and an alternative MATLAB® solution technique, was featured in Cleve's Corner in 2009.

There are many approaches to solving Sudoku puzzles manually, as well as many programmatic approaches. This example shows a straightforward approach using binary integer programming.

This approach is particularly simple because you do not give a solution algorithm. Just express the rules of Sudoku, express the clues as constraints on the solution, and then MATLAB produces the solution.

The key idea is to transform a puzzle from a square 9-by-9 grid to a cubic 9-by-9-by-9 array of binary values (0 or 1). Think of the cubic array as being 9 square grids stacked on top of each other, where each layer corresponds to an integer from 1 through 9. The top grid, a square layer of the array, has a 1 wherever the solution or clue has a 1. The second layer has a 1 wherever the solution or clue has a 2. The ninth layer has a 1 wherever the solution or clue has a 9.

This formulation is precisely suited for binary integer programming.

The objective function is not needed here, and might as well be a constant term 0. The problem is really just to find a feasible solution, meaning one that satisfies all the constraints. However, for tiebreaking in the internals of the integer programming solver, giving increased solution speed, use a nonconstant objective function.

Suppose a solution is represented in a 9-by-9-by-9 binary array. What properties does have? First, each square in the 2-D grid (i,j) has exactly one value, so there is exactly one nonzero element among the 3-D array entries . In other words, for every and ,

Similarly, in each row of the 2-D grid, there is exactly one value out of each of the digits from 1 to 9. In other words, for each and ,

And each column in the 2-D grid has the same property: for each and ,

The major 3-by-3 grids have a similar constraint. For the grid elements and , and for each ,

To represent all nine major grids, just add 3 or 6 to each and index:

where

Each initial value (clue) can be expressed as a constraint. Suppose that the clue is for some . Then . The constraint ensures that all other for .

Create an optimization variable `x`

that is binary and of size 9-by-9-by-9.

x = optimvar('x',9,9,9,'Type','integer','LowerBound',0,'UpperBound',1);

Create an optimization problem with a rather arbitrary objective function. The objective function can help the solver by destroying the inherent symmetry of the problem.

sudpuzzle = optimproblem; mul = ones(1,1,9); mul = cumsum(mul,3); sudpuzzle.Objective = sum(sum(sum(x,1),2).*mul);

Represent the constraints that the sums of `x`

in each coordinate direction are one.

sudpuzzle.Constraints.consx = sum(x,1) == 1; sudpuzzle.Constraints.consy = sum(x,2) == 1; sudpuzzle.Constraints.consz = sum(x,3) == 1;

Create the constraints that the sums of the major grids sum to one as well.

majorg = optimconstr(3,3,9); for u = 1:3 for v = 1:3 arr = x(3*(u-1)+1:3*(u-1)+3,3*(v-1)+1:3*(v-1)+3,:); majorg(u,v,:) = sum(sum(arr,1),2) == ones(1,1,9); end end sudpuzzle.Constraints.majorg = majorg;

Include the initial clues by setting lower bounds of 1 at the clue entries. This setting fixes the value of the corresponding entry to be 1, and so sets the solution at each clued value to be the clue entry.

for u = 1:size(B,1) x.LowerBound(B(u,1),B(u,2),B(u,3)) = 1; end

Solve the Sudoku puzzle.

sudsoln = solve(sudpuzzle)

LP: Optimal objective value is 405.000000. Cut Generation: Applied 3 strong CG cuts, and 2 zero-half cuts. Lower bound is 405.000000. Relative gap is 0.00%. Optimal solution found. Intlinprog stopped at the root node because the objective value is within a gap tolerance of the optimal value, options.AbsoluteGapTolerance = 0 (the default value). The intcon variables are integer within tolerance, options.IntegerTolerance = 1e-05 (the default value).

`sudsoln = `*struct with fields:*
x: [9x9x9 double]

Round the solution to ensure that all entries are integers, and display the solution.

sudsoln.x = round(sudsoln.x); y = ones(size(sudsoln.x)); for k = 2:9 y(:,:,k) = k; % multiplier for each depth k end S = sudsoln.x.*y; % multiply each entry by its depth S = sum(S,3); % S is 9-by-9 and holds the solved puzzle drawSudoku(S)

You can easily check that the solution is correct.

`type drawSudoku`

function drawSudoku(B) % Function for drawing the Sudoku board % Copyright 2014 The MathWorks, Inc. figure;hold on;axis off;axis equal % prepare to draw rectangle('Position',[0 0 9 9],'LineWidth',3,'Clipping','off') % outside border rectangle('Position',[3,0,3,9],'LineWidth',2) % heavy vertical lines rectangle('Position',[0,3,9,3],'LineWidth',2) % heavy horizontal lines rectangle('Position',[0,1,9,1],'LineWidth',1) % minor horizontal lines rectangle('Position',[0,4,9,1],'LineWidth',1) rectangle('Position',[0,7,9,1],'LineWidth',1) rectangle('Position',[1,0,1,9],'LineWidth',1) % minor vertical lines rectangle('Position',[4,0,1,9],'LineWidth',1) rectangle('Position',[7,0,1,9],'LineWidth',1) % Fill in the clues % % The rows of B are of the form (i,j,k) where i is the row counting from % the top, j is the column, and k is the clue. To place the entries in the % boxes, j is the horizontal distance, 10-i is the vertical distance, and % we subtract 0.5 to center the clue in the box. % % If B is a 9-by-9 matrix, convert it to 3 columns first if size(B,2) == 9 % 9 columns [SM,SN] = meshgrid(1:9); % make i,j entries B = [SN(:),SM(:),B(:)]; % i,j,k rows end for ii = 1:size(B,1) text(B(ii,2)-0.5,9.5-B(ii,1),num2str(B(ii,3))) end hold off end

Was this topic helpful?