BUG+1 in Sudoku: The Bivalue Universal Grave Explained

BUG+1 (Bivalue Universal Grave plus one) is a shortcut for the moment when every unsolved cell has exactly two candidates except one cell with three. That cell must be the candidate that appears three times in its row, column and box, because any other choice would leave a grid with either no solution or several.

To spot it, scan the grid for cells with more than two candidates. If you find exactly one, and it has exactly three, count how often each of its candidates appears in its row. The digit that shows up three times is the answer. Place it and the puzzle usually falls apart in singles.

What a bivalue universal grave is

A bivalue universal grave, or BUG, is a grid state where every unsolved cell has exactly two candidates and every candidate appears exactly twice in each row, column and box where it appears at all. It is a larger version of the unique rectangle deadly pattern. In a BUG, each candidate can be paired off with the other in its cell, and the whole grid can flip between two fillings. Such a grid has either no solution or at least two.

A valid puzzle has exactly one solution, so it can never reach a BUG. When you are one candidate away from one, that extra candidate is what keeps the puzzle unique, and it must be true.

Worked example

The diagram above is from a real puzzle in our library, after the basic techniques have stalled. Seventeen cells are open. Sixteen of them have exactly two candidates, for example R1C1 {4, 9}, R3C3 {1, 2} and R8C7 {1, 4}. The only exception is R9C6, with {4, 6, 9}.

Now count each of R9C6's candidates in its three houses:

• Row 9: the open cells are R9C2 {1, 9}, R9C4 {4, 6}, R9C6 {4, 6, 9} and R9C7 {1, 4}. The 4 appears three times, the 6 twice, the 9 twice.
• Column 6: R1C6 {2, 4}, R4C6 {2, 6}, R8C6 {4, 9} and R9C6. Again the 4 appears three times, the 6 and 9 twice each.
• Box 8: R8C6 {4, 9}, R9C4 {4, 6} and R9C6. The 4 appears three times, 6 and 9 twice.

Suppose R9C6 were 6 or 9. Removing the 4 from it would leave every open cell with two candidates, and every candidate twice in every house: a BUG, which a unique puzzle cannot have. So R9C6 is 4. From there, R9C4 becomes 6, R9C7 becomes 1, and the rest follows with singles.

Puzzle 201191 from the Sudoku Royale library. Givens (row by row, dots are blanks): .......3.8..9..21..3..6..9...7.3......3.4562....79.5..2.618....75................. The candidates shown are what is left after singles, pointing and claiming, naked and hidden pairs, triples and quads, X-Wing and Swordfish, with nothing else applied.

How to check a BUG+1 safely

1. Make sure your candidates are complete. BUG+1 reads the whole grid, so one missing pencil mark gives a wrong answer. Only trust it with full, accurate pencil marks.
2. Find the odd cell. Every unsolved cell must have two candidates except one, which has three.
3. Count in one house. In the odd cell's row, find the candidate that appears three times. Check the column and box too; the same digit should appear three times there.
4. Place it. If the counts disagree between houses, something is off in your candidates, so recheck before placing anything.

If the grid has two or more cells with extra candidates, BUG+1 does not apply. Look for a unique rectangle in the two-candidate cells, or link them with an XY-Wing or an XY-Chain.

Where BUG+1 fits

BUG+1 is a uniqueness technique, like unique rectangles. It assumes the puzzle has exactly one solution, which every published puzzle does. It only shows up late in a solve, when the grid is full of two-candidate cells, and it is rare. When it does appear it is one of the fastest breaks in sudoku: one count, one placement.

For the full ladder, see our list of sudoku techniques in learning order and the overview of advanced sudoku strategies.