Документация

function [S,eflag] = sudokuEngine(B)
% This function sets up the rules for Sudoku. It reads in the puzzle
% expressed in matrix B, calls intlinprog to solve the puzzle, and returns
% the solution in matrix S.
%
% The matrix B should have 3 columns and at least 17 rows (because a Sudoku
% puzzle needs at least 17 entries to be uniquely solvable). The first two
% elements in each row are the i,j coordinates of a clue, and the third
% element is the value of the clue, an integer from 1 to 9. If B is a
% 9-by-9 matrix, the function first converts it to 3-column form.

%   Copyright 2014 The MathWorks, Inc. 

if isequal(size(B),[9,9]) % 9-by-9 clues
    % Convert to 81-by-3
    [SM,SN] = meshgrid(1:9); % make i,j entries
    B = [SN(:),SM(:),B(:)]; % i,j,k rows
    % Now delete zero rows
    [rrem,~] = find(B(:,3) == 0);
    B(rrem,:) = [];
end

if size(B,2) ~= 3 || length(size(B)) > 2
    error('The input matrix must be N-by-3 or 9-by-9')
end

if sum([any(B ~= round(B)),any(B < 1),any(B > 9)]) % enforces entries 1-9
    error('Entries must be integers from 1 to 9')
end

%% The rules of Sudoku:
N = 9^3; % number of independent variables in x, a 9-by-9-by-9 array
M = 4*9^2; % number of constraints, see the construction of Aeq
Aeq = zeros(M,N); % allocate equality constraint matrix Aeq*x = beq
beq = ones(M,1); % allocate constant vector beq
f = (1:N)'; % the objective can be anything, but having nonconstant f can speed the solver
lb = zeros(9,9,9); % an initial zero array
ub = lb+1; % upper bound array to give binary variables

counter = 1;
for j = 1:9 % one in each row
    for k = 1:9
        Astuff = lb; % clear Astuff
        Astuff(1:end,j,k) = 1; % one row in Aeq*x = beq
        Aeq(counter,:) = Astuff(:)'; % put Astuff in a row of Aeq
        counter = counter + 1;
    end
end

for i = 1:9 % one in each column
    for k = 1:9
        Astuff = lb;
        Astuff(i,1:end,k) = 1;
        Aeq(counter,:) = Astuff(:)';
        counter = counter + 1;
    end
end

for U = 0:3:6 % one in each square
    for V = 0:3:6
        for k = 1:9
            Astuff = lb;
            Astuff(U+(1:3),V+(1:3),k) = 1;
            Aeq(counter,:) = Astuff(:)';
            counter = counter + 1;
        end
    end
end

for i = 1:9 % one in each depth
    for j = 1:9
        Astuff = lb;
        Astuff(i,j,1:end) = 1;
        Aeq(counter,:) = Astuff(:)';
        counter = counter + 1;
    end
end

%% Put the particular puzzle in the constraints
% Include the initial clues in the |lb| array by setting corresponding
% entries to 1. This forces the solution to have |x(i,j,k) = 1|.

for i = 1:size(B,1)
    lb(B(i,1),B(i,2),B(i,3)) = 1;
end

%% Solve the Puzzle
% The Sudoku problem is complete: the rules are represented in the |Aeq|
% and |beq| matrices, and the clues are ones in the |lb| array. Solve the
% problem by calling |intlinprog|. Ensure that the integer program has all
% binary variables by setting the intcon argument to |1:N|, with lower and
% upper bounds of 0 and 1.

intcon = 1:N;

[x,~,eflag] = intlinprog(f,intcon,[],[],Aeq,beq,lb,ub);

%% Convert the Solution to a Usable Form
% To go from the solution x to a Sudoku grid, simply add up the numbers at
% each $(i,j)$ entry, multiplied by the depth at which the numbers appear:

if eflag > 0 % good solution
    x = reshape(x,9,9,9); % change back to a 9-by-9-by-9 array
    x = round(x); % clean up non-integer solutions
    y = ones(size(x));
    for k = 2:9
        y(:,:,k) = k; % multiplier for each depth k
    end

    S = x.*y; % multiply each entry by its depth
    S = sum(S,3); % S is 9-by-9 and holds the solved puzzle
else
    S = [];
end

LP:                Optimal objective value is 29565.000000.                                         

Cut Generation:    Applied 1 strong CG cut,                                                         
                   and 3 zero-half cuts.                                                            
                   Lower bound is 29565.000000.                                                     
                   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).

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