GridWits

BUG (Bivalue Universal Grave)

Extreme Used by Smart Hint

If every empty cell has exactly two candidates except one with three, that one cell must hold whichever digit appears an odd number of times.

A 1 B 2 C 3 D 4 E 5 F 6 G 7 H 8 I 9 1 4 2 4 1 2 3 4 3 4 7 1 7 2 3 2 7 3 7

How it works

  1. Imagine every empty cell on the board has exactly two candidates except B2, which has three: {3, 4, 7}. That's a Bivalue Universal Grave — left unresolved, the puzzle could be completed more than one way by cycling every bivalue cell between its two digits.
  2. Count how many times each of B2's candidates appears elsewhere on the board. Whichever one shows up an odd number of times — here, 4 — is the one B2 must actually be.

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