I believe that Andrew Stuart's solver uses a brute force algorithm to enumerate all the solutions.
The first puzzle in the 'Multiple Solutions?' thread is :
6.4.925.15.9.1.....2....7..4.2.3.......2.1.4.813..592..4.38.6.53..1..2.8..7.2.3..
Using Andrew's solver, we uncheck BUG, Unique Rectangles, Extended Unique Rectangles and Hidden Unique Rectangles, then advance the puzzle as far as his solver will go. This gives:
6.4.925.15.9.1.4.2128.5.7.9462938157795261843813..592624138.6.53561..2.8987526314
with C=3 solutions:
634792581579813462128654739462938157795261843813475926241387695356149278987526314
634792581579813462128654739462938157795261843813475926241389675356147298987526314
674892531539617482128453769462938157795261843813745926241389675356174298987526314
Andrew's solver shows all the incomplete cells with their candidates:
- Code: Select all
+--------------+--------------+--------------+
| 6 37 4 | 78 9 2 | 5 38 1 |
| 5 37 9 | 68 1 37 | 4 68 2 |
| 1 2 8 | 46 5 34 | 7 36 9 |
+--------------+--------------+--------------+
| 4 6 2 | 9 3 8 | 1 5 7 |
| 7 9 5 | 2 6 1 | 8 4 3 |
| 8 1 3 | 47 47 5 | 9 2 6 |
+--------------+--------------+--------------+
| 2 4 1 | 3 8 79 | 6 79 5 |
| 3 5 6 | 1 47 479 | 2 79 8 |
| 9 8 7 | 5 2 6 | 3 1 4 |
+--------------+--------------+--------------+
So r1c2 must be either 3 or 7, r1c4 must 7 or 8 and so on. The usual Sudoku rules determine the additional constraints. So, if r1c2=3, then r1c8=8, r1c4=7 and first band fills in. Now r7c8 must be either 7 or 9. r7c8=7 gives the first solution and r7c8=9 gives the second solution. Now, if r1c2=7, we get the third solution.
Now, if we had used UR with this puzzle, we would note r78c68 => r8c6=4 which would reveal the third solution and [more importantly] conceal the first two solutions.
I am now on the hunt for some other examples
Best Wishes
Gordon