--- /dev/null
+## Copyright (C) 2009 Jaroslav Hajek <highegg@gmail.com>
+##
+## This program is free software; you can redistribute it and/or modify it under
+## the terms of the GNU General Public License as published by the Free Software
+## Foundation; either version 3 of the License, or (at your option) any later
+## version.
+##
+## This program is distributed in the hope that it will be useful, but WITHOUT
+## ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
+## FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more
+## details.
+##
+## You should have received a copy of the GNU General Public License along with
+## this program; if not, see <http://www.gnu.org/licenses/>.
+
+## -*- texinfo -*-
+## @deftypefn{Function File} [@var{x}, @var{ntrial}] = solvesudoku (@var{s})
+## Solves a classical 9x9 sudoku. @var{s} should be a 9x9 array with
+## numbers from 0:9. 0 indicates empty field.
+## Returns the filled table or empty matrix if no solution exists.
+## If requested, @var{ntrial} returns the number of trial-and-error steps needed.
+## @end deftypefn
+
+## This uses a recursive backtracking technique combined with revealing new singleton
+## fields by logic. The beauty of it is that it is completely vectorized.
+
+function [x, ntrial] = solvesudoku (s)
+
+ if (nargin != 1)
+ print_usage ();
+ endif
+
+ if (! (ismatrix (s) && ndims (s) == 2 && all (size (s) == [9, 9])))
+ error ("needs a 9x9 matrix");
+ endif
+
+ if (! ismember (unique (s(:)), 0:9))
+ error ("matrix must contain values from 0:9");
+ endif
+
+ if (! verifysudoku (s))
+ error ("matrix is not a valid sudoku grid");
+ endif
+
+ [x, ntrial] = solvesudoku_rec (s);
+
+endfunction
+
+function ok = verifysudoku (s)
+ [i, j, k] = find (s);
+ b = false (9, 9, 9);
+ b(sub2ind ([9, 9, 9], i, j, k)) = true;
+ okc = sum (b, 1) <= 1;
+ okr = sum (b, 2) <= 1;
+ b = reshape (b, [3, 3, 3, 3, 9]);
+ ok3 = sum (sum (b, 1), 3) <= 1;
+ ok = all (okc(:) & okr(:) & ok3(:));
+endfunction
+
+function [x, ntrial] = solvesudoku_rec (s)
+
+ x = s;
+ ntrial = 0;
+
+ ## Run until the logic is exhausted.
+ do
+ b = getoptions (x);
+ s = x;
+ x = getsingletons (b, x);
+ finished = isempty (x) || all (x(:));
+ until (finished || all ((x == s)(:)));
+
+ if (! finished)
+ x = [];
+ ## Find the field with minimum possibilities.
+ sb = sum (b, 3);
+ sb(s != 0) = 10;
+ [msb, i] = min (sb(:));
+ [i, j] = ind2sub ([9, 9], i);
+ ## Try all guesses.
+ for k = find (b(i,j,:))'
+ s(i,j) = k;
+ [x, ntrial1] = solvesudoku_rec (s);
+ ntrial += 1 + ntrial1;
+ if (! isempty (x))
+ ## Found solutions.
+ break;
+ endif
+ s(i,j) = 0;
+ endfor
+ endif
+
+endfunction
+
+## Given a 9x9x9 logical array of allowed values, get the logical singletons.
+function s = getsingletons (b, s)
+
+ n0 = sum (s(:) != 0);
+
+ ## Check for fields with only one option.
+ sb = sum (b, 3);
+ if (any (sb(:) == 0))
+ s = [];
+ return;
+ else
+ s1 = sb == 1;
+ ## We want to return as soon as some new singletons are found.
+ [s(s1), xx] = find (reshape (b, [], 9)(s1, :).');
+ if (sum (s(:) != 0) > n0)
+ return;
+ endif
+ endif
+
+ ## Check for columns where a number has only one field left.
+ sb = squeeze (sum (b, 1));
+ if (any (sb(:) == 0))
+ s = [];
+ return;
+ else
+ s1 = sb == 1;
+ [j, k] = find (s1);
+ [i, xx] = find (b(:, s1));
+ s(sub2ind ([9, 9], i, j)) = k;
+ if (sum (s(:) != 0) > n0)
+ return;
+ endif
+ endif
+
+ ## Ditto for rows.
+ sb = squeeze (sum (b, 2));
+ if (any (sb(:) == 0))
+ s = [];
+ return;
+ else
+ s1 = sb == 1;
+ [i, k] = find (s1);
+ [j, xx] = find (permute (b, [2, 1, 3])(:, s1));
+ s(sub2ind ([9, 9], i, j)) = k;
+ if (sum (s(:) != 0) > n0)
+ return;
+ endif
+ endif
+
+ ## 3x3 tiles.
+ bb = reshape (b, [3, 3, 3, 3, 9]);
+ sb = squeeze (sum (sum (bb, 1), 3));
+ if (any (sb(:) == 0))
+ s = [];
+ return;
+ else
+ s1 = reshape (sb == 1, 9, 9);
+ [j, k] = find (s1);
+ [i, xx] = find (reshape (permute (bb, [1, 3, 2, 4, 5]), 9, 9*9)(:, s1));
+ [i1, i2] = ind2sub ([3, 3], i);
+ [j1, j2] = ind2sub ([3, 3], j);
+ s(sub2ind ([3, 3, 3, 3], i1, j1, i2, j2)) = k;
+ if (sum (s(:) != 0) > n0)
+ return;
+ endif
+ endif
+
+endfunction
+
+## Given known values (singletons), calculate options.
+function b = getoptions (s)
+
+ ## Find true values.
+ [i, j, s] = find (s);
+ ## Columns.
+ bc = true (9, 9, 9);
+ bc(:, sub2ind ([9, 9], j, s)) = false;
+ ## Rows.
+ br = true (9, 9, 9);
+ br(:, sub2ind ([9, 9], i, s)) = false;
+ ## 3x3 tiles.
+ b3 = true (3, 3, 3, 3, 9);
+ b3(:, :, sub2ind ([3, 3, 9], ceil (i/3), ceil (j/3), s)) = false;
+ ## Permute elements to correct order.
+ br = permute (br, [2, 1, 3]);
+ b3 = reshape (permute (b3, [1, 3, 2, 4, 5]), [9, 9, 9]);
+ ## The singleton fields themselves.
+ bb = true (9*9, 9);
+ bb(sub2ind ([9, 9], i, j), :) = false;
+ bb = reshape (bb, [9, 9, 9]);
+ ## Form result.
+ b = bc & br & b3 & bb;
+ ## Correct singleton fields.
+ b = reshape (b, 9, 9, 9);
+ b(sub2ind ([9, 9, 9], i, j, s)) = true;
+
+endfunction
+
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