Without assumption of uniqueness, here is a simple ALS-xz to solve it:
- Code: Select all
*--------------------------------------------------*
| 8 9 35 | 6 4 2 | 37 1 57 |
| 1 4 35 | 357 37 8 | 2 6 9 |
| 6 7 2 |-35 9 1 | 8 #35 4 |
|----------------+----------------+----------------|
| 9 5 6 | 1 2 47 | 37 34 8 |
| 23 1 4 | 8 37 6 | 5 9 27 |
| 7 23 8 |@39 5 @49 | 6 #24 1 |
|----------------+----------------+----------------|
| 235 23 1 | 79 8 79 | 4 #25 6 |
| 4 6 7 | 2 1 5 | 9 8 3 |
| 25 8 9 | 4 6 3 | 1 7 25 |
*--------------------------------------------------*
ALS A: r6c46={349}
ALS B: r367c8={2345}
x=4
z=3
Therefore r3c4 can't be 3.
Logic:
Either r6c68 must not be 4.
If r6c6<>4, r6c6=9 => r6c4=3.
If r6c8<>4, r6c8=2 => r7c8=5 => r3c8=3.
Therefore one of r6c4 & r3c8 must be 3.
Hence r3c4, seeing both these cells, can't be 3.
I'm looking forward to more elegant solutions.