Advanced technique · Uniqueness · 09
Three exceptional cells extend BUG reasoning into a broader candidate relationship.
The idea
A BUG+3 extends the same BUG idea to three exceptional cells. The rest of the unsolved grid is bivalue, while three cells contain the additional candidates needed to break the BUG state. There is no single universal placement rule like BUG+1; the useful deduction comes from proving which of those exceptional candidates must remain in order to preserve uniqueness.
The question: Which candidate relationship among the three exceptional cells prevents the BUG?
Walkthroughs
Start with the BUG structure, then verify there are three exceptional cells that can break the BUG before making the elimination.
Walkthrough
Three exceptional cells r4c5, r5c2 and r6c3 all contain number 6 plus extra two numbers each. One of them must contain number 6 to avoid BUG pattern, therefore, number 6 can be removed from any cell that sees all of them.
What to notice: BUG+3 is about the candidate network around the exceptions, not simply counting to three. Similar to BUG+2, three exceptional cells do not have to be in a same house.
Walkthrough
Here three exceptional cells are r7c5, r8c1 and r8c3 with number 5. Any cell that sees all of them should not have 5 as its candidate.
What to notice: Recognition and proof are separate skills. Seeing three exceptional cells is only the beginning.
When the three exceptional cells share a useful candidate, identify the cells that see every exception carrying that candidate. Those cells cannot contain it if at least one exception must take the number.
If no clean common relationship appears, stop. BUG+3 is a specialized pattern and should not be forced when ordinary subsets, chains, or other advanced relationships give a clearer proof.
How it works
A complete BUG would be non-unique. With three exceptional cells, the puzzle remains unique only because their additional candidates prevent that state. The deduction comes from identifying a candidate that must survive in at least one exception and then applying the resulting constraint to cells that see all of the relevant exceptions.
Confirm that the remaining unsolved cells are bivalue and that the three cells are the only breaks in the BUG-like pattern.
Check how those candidates are distributed through the affected rows, columns, and boxes.
Only remove a candidate when the three-cell relationship guarantees that at least one exception must contain it.
A useful reminder
The pattern is useful when it appears cleanly, but it is considerably more specialized than BUG+1. Keep the proof explicit and prefer a simpler established technique when one is available.
Watch & learn
Coming soon...
Practice
When almost every unsolved cell is bivalue, identify the three cells that break the pattern. Verify the BUG-like counts, then look for a candidate that must survive in at least one exception. If you cannot prove the shared constraint, keep searching.
