Uniqueness Deduction for Tough Sudoku of 08/07/08

The following illustrated proof for the Tough Sudoku of August 7, 2008 employs a very small group of Sudoku techniques, tips and tricks: Hidden Pairs, Locked Candidates, and a Uniqueness deduction.

Previous blog pages may be helpful. Links to these pages are found to the right, under Sudoku Techniques. This list is getting long, so specifically, one may want to refer to the following previous blog pages:

• Locked Candidates
• Pairs, Triples, Quads
• Definitions: Most terms used on this page have a defintion somewhere on the defintions page - although an update is long overdue.

Many steps not illustrated are possible. Only one combination step, besides hidden singles, is shown. This one step is sufficient to solve this puzzle, if one assumes that there exists only one solution to the puzzle.

Style Change

Previously, I had abandoned using the chessboard algebraic notation in favor of the more common rc notation. However, since many people have begun again to post proofs on the tough pages, I have returned to algebraic notation. For those unfamiliar with this notational style, grid coordinates are posted with the pictures.

The Puzzle

The Puzzle at UP 27 (27 cells filled either as givens, or solved)

Candidate 5

Candidate 2

Of the steps above, step 3 is not required. I view the other 3 steps as one, as the recognition of both (25) limited to def9, and the recognition of both (25) limited to df5, and the recognition of (5) limited to def1 is required for step 4, but also sufficient for steps 1 & 2.

Explanation of the Uniqueness Deducition

Suppose that candidates 2&5 are limited to only def1, def5, def9. Exactly three of those cells would be filled in, eventually, by some other candidate(s). There are some illegal ways to do this, let us presume that one considers only the legal manners of doing this. The remaining six cells in columns def and rows 159 can receive only candidate 2 or candidate 5. This group will form an independent sub-puzzle that cannot be resolved uniquely. One can arbitrarily insert 2 or 5 at any of the remaining cells and legally fill in all six remaining cells. More importantly, nothing that occurs in the rest of the puzzle can possibly determine which of (25) goes in any of those cells. To wit:

• (hidden pair 25) at 2 choose 3 of def9 => 25 is independent in row 9, box e8, and those 2 cells.
• (hp25) at 2 choose 3 of def5 (in this case exactly df5) => 25 is independent in row 5, box e5, and those 2 cells.
• (hp25) at 2 choose 3 of def1 =>25 is independent in row 1, box e2, and those 2 cells.
• Given those three occurrences, there must exist (hp25) in each of columns def using ony rows 159 => 25 is independent in columns def
• Since there are only four conditions in the rules of sudoku: rows, columns, boxes, cells => 25 is independent in those 6 cells

One step that unlocks this puzzle

After making the indicated eliminations, one can solve the puzzle using only hidden singles.

Solution