GridWits
Extreme tier Technique 39 of 46 Used by Smart Hint

BUG (Bivalue Universal Grave)

Assuming a unique solution: if every empty cell is bivalue except one trivalue cell, and every candidate has the required even house counts except one extra digit, that cell must hold the extra digit.

How it works

  1. 1Picture the whole board looking like this box: every empty cell is down to exactly two candidates except B2, which still carries three, and every row, column and box has the BUG parity — each live digit appears twice, except one extra digit in B2's three houses.
  2. 2That's BUG+1. Under the puzzle's unique-solution assumption, leaving the extra digit unused would reduce the grid to a Bivalue Universal Grave whose candidates can swap into two completions.
  3. 3Count each of B2's candidates inside the box: 3 appears twice and 7 appears twice, but 4 appears three times — at A1, B1 and B2 itself.
  4. 4The odd count is what breaks the tie, so B2 must be the 4.

Play a Sudoku that needs this technique →

A 1 B 2 C 3 D 4 E 5 F 6 G 7 H 8 I 9 1 4 2 4 5 3 9 3 4 7 6 2 9 1 7 8
  • Where the conclusion lands
Next in the ladder Unique Rectangle Types 2-6 →

The rest of the Unique Rectangle family: when only two corners are a bare pair, what the other two carry still gives the deadly pattern away.