When every unsolved cell has exactly two candidates except one with three, that cell's extra candidate must be the answer.
Also known as: Bivalue Universal Grave + 1
A Bivalue Universal Grave (BUG) is a state where every remaining empty cell has exactly two candidates. In such a state, every candidate appears exactly twice in each unit (row, column, box) it occupies. This creates a deadly pattern, you could swap all instances of any two digits and get another valid solution, violating uniqueness.
BUG+1 occurs when every empty cell has two candidates except one cell that has three. The puzzle is one candidate away from a BUG state. Since a valid Sudoku has a unique solution, the trivalue cell must resolve to the candidate that breaks the BUG symmetry, the digit that appears an odd number of times in one of the cell's units.
This technique relies on the uniqueness assumption: a properly constructed Sudoku has exactly one solution. BUG+1 is a powerful shortcut that can instantly solve a cell that would otherwise require complex chain reasoning.
After extensive solving, check if all remaining empty cells have exactly two candidates. If one cell has three candidates and all others have two, you have a BUG+1.
To find which digit to place: examine the trivalue cell's row, column, and box. In a BUG state, every candidate appears exactly twice per unit. One of the three candidates in the trivalue cell will appear 3 times (odd count) in at least one of its units. That's the digit that must be placed to prevent the BUG.
Equivalently, the digit to place is the one whose removal would leave every candidate appearing exactly twice in every unit, completing the BUG pattern, which is forbidden.
Example 1: BUG+1
After basic solving, 12 cells remain empty. Notice something remarkable: 11 of these cells have exactly two candidates each. Only one cell (R9C9) has three candidates: {1, 3, 8}. This is a BUG+1 state.
In a valid Sudoku, a BUG state (all bivalue) would allow digit-swapping that creates multiple solutions, which is forbidden. The extra candidate in R9C9 is what prevents this.
Looking at row 9: digit 3 appears 3 times as a candidate (in R9C5, R9C7, and R9C9), while digits 1 and 8 each appear exactly twice. The digit breaking the symmetry is 3, it must be placed in R9C9.
Place 3 in R9C9. The remaining 11 cells each have exactly two candidates, and the puzzle resolves with basic techniques.
Explore all 28 solving techniques in our complete technique guide.