Nice puzzle, Keith.
For this one-stepper game you need at least a solver which tells you, if eliminations leave singles to the end. On paper i would be satisfied with any solution and never repeat it to find a one-stepper.
I don't know, how familiar you are with these ALS-chains. Often solutions are easy to follow, because a link can be splitted into several easier ones, but for others you need to count many candidates.
Guess you know, that if you have n+1 candidates for n cells, there is always an either/or between any 2 of them, like (4=6)r8c27 or (4=2)r4c267.
- Code: Select all
*-------------------------------------------*
| 3 8 5 | 7 4 2 | 1 9 6 |
| 4 1 7 | 6 9 8 | 5 3 2 |
| 2 69 69 | 1 3 5 | 8 4 7 |
|---------------+-------------+-------------|
| 1 b469 2469 | 5 7 b69 | b269 8 3 |
| 5 369 2369 | 4 8 369 | 7 26 1 |
| 8 7 369 | 39 2 1 | 69 5 4 |
|---------------+-------------+-------------|
| 6 2 8 | 39 1 39 | 4 7 5 |
| 9 a34 134 | 2 5 7 | 6-3 16 8 |
| 7 5 1-3 | 8 6 4 | c23 12 9 |
*-------------------------------------------*
(3=4)r8c2-(4=2)r4c267-(2=3)r9c7 => r8c7,r9c3<>3, stte
btw i would prefer SteveC's solution, as you know.