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Some solvers may put a 6 at r8c1 but this can’t be true, so there are two 3-candidate cells.
The BUG-stopping candidates are 9 at r6c3 and also 9 at r9c1. What can be inferred from this?
That BOTH must be true, that AT LEAST ONE must be true, or that ONE ONLY must be true?
Is there some other technique which could be applied here?
Yogi

Code: Select all

+----------------+----------------+----------------+
| 7    3    5    | 2    6    8    | 4    9    1    |
| 1    8    4    | 9    7    3    | 5    6    2    |
| 2    69   69   | 5    4    1    | 3    8    7    |
+----------------+----------------+----------------+
| 6-9  1    3    | 68   2    4    | 89   7    5    |
| 4    7    89   | 3    89   5    | 2    1    6    |
| 5    2    68+9 | 7    1    69   | 89   4    3    |
+----------------+----------------+----------------+
| 3    4    2    | 1    89   7    | 6    5    89   |
| 89   5    1    | 68   3    69   | 7    2    4    |
| 68+9 69   7    | 4    5    2    | 1    3    89   |
+----------------+----------------+----------------+

Yes, you can solve it with a BUG+2. Either r6c3=9 or r9c1=9 (or both) => -9r4c1
Alternatively there is an xy-wing or a w-wing.

Great. Thanx.