Lesson 28

BUG+1

The rule

In a puzzle with a unique solution, suppose every unsolved cell has two candidates except one cell with three. If removing one candidate from that cell would leave every digit appearing as a candidate either zero or two times in each row, column and box, that candidate must be true: otherwise, any solution could be mirrored by swapping each unsolved cell to its other candidate.

Step by step

1. Every unsolved cell left on the board is bivalue: it holds exactly two candidates. The only exception is the highlighted cell, which holds three: 2, 4, 6.
2. Now look at the highlighted cell's row, column and box: the candidate 6 appears three times in each of them, while every other candidate appears exactly twice, if at all. If this cell were not 6, every candidate would appear exactly twice everywhere, a deadly pattern that could be completed in two different ways. The puzzle has only one solution, so 6 must go here, and 2 and 4 can be removed.