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 = [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 tie breaking 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:
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. 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.
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