Bivalue Universal Grave Sudoku Technique

Updated 6 October 2026

A Bivalue Universal Grave, usually written BUG+1, is a uniqueness technique for the late game: when every empty cell holds exactly two candidates except one cell that holds three, the digit that appears three times in that cell’s row, column and box must be placed there.

When a BUG+1 applies

A bivalue universal grave is a whole-board deadly pattern. Imagine a position where every empty cell holds exactly two candidates, and every digit that is still open in a row, column or box appears there exactly twice. Such a position always has either no solution or at least two: every cell can flip to its other candidate together with the cells it pairs with, and no unit can tell the difference. A puzzle with exactly one solution can never reach it.

The “+1” is the single cell that keeps the board out of that grave. It has three candidates instead of two. In each of its three units, two of its digits appear the usual two times, while the third digit appears three times — the extra copy is the one in this cell. Strike that third digit from the cell and the board is exactly the grave, so the digit can never be ruled out there. The cell must take the digit that appears three times, and that is a placement, not just an elimination.

Like the Unique Rectangle, BUG+1 depends on the puzzle having one solution. Unlike it, the pattern is global: you have to confirm that every other empty cell on the board is bivalue, not just four corners.

Worked example

Count the candidates in row 8 first. r8c1 holds 1 and 8, r8c2 holds 1 and 9, and r8c3 holds 1, 8 and 9. Digit 8 appears twice, digit 9 appears twice, and digit 1 appears three times. Column 3 tells the same story: 1 sits in r2c3, r6c3 and r8c3, while 8 and 9 each appear twice. The bottom-left box agrees, with 1 in r8c1, r8c2 and r8c3.

Strike the 1 from r8c3 and every empty cell holds two candidates, with every digit appearing twice in every unit — the grave, which a single-solution puzzle can never reach. So the 1 cannot be ruled out of r8c3, and r8c3 must be 1. The highlighted cells are the ones in r8c3’s units that still carry a 1, which is exactly the count that proves the placement. On this particular board an XY-Chain would also make progress; the example isolates BUG+1 the same way every page in this guide isolates its own technique.

How to spot a BUG+1

1. Fill in every candidateBUG+1 is a whole-board pattern, so it needs complete candidates for every empty cell before you can trust it.
2. Confirm one cell has three candidatesEvery empty cell must be bivalue except one, which must have exactly three candidates.
3. Count that cell’s digits in its unitsIn the cell’s row, column and box, find the digit that appears three times while the other two appear twice.
4. Place the digit that appears three timesWrite that digit into the three-candidate cell. The rest of the board usually falls to singles right after.

Common mistakes

• Missing a second non-bivalue cell. BUG+1 needs exactly one cell with three candidates; two such cells, or any cell with four, means the pattern is not there.
• Working from incomplete pencil marks. A cell that looks bivalue because a candidate was never written in breaks the whole argument.
• Placing the wrong digit. It is the digit that appears three times in the cell’s units that goes in, not one of the digits that appear twice.
• Using it on a puzzle that might have more than one solution. Like every uniqueness technique, it assumes a single answer.