ronk wrote:RW wrote:You don't even need a solved cell, it's enough is one of the candidates in the potential deadly pattern is eliminated:

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`ab ab`

ab axy

Candidate 'b' is already eliminated from cell 'axy' => we may eliminate 'a' as well without assuming uniqueness.

I get it but IMO ...

not assuming uniqueness

while using a uniqueness technique will be perpetually confusing to many.

Like

RW said this is not an uniqueness technique, but a "pure logic" technique, albeit applied to a puzzle with possible multiple solutions.

The "pure logic" goes like this:

We have 2 possible scenarios. Scenario 1: the puzzle has one solution. Then the 4 cells can't have the pattern [ab]+[ba] because [ba]+[ab] would make an alternative solution. Therefore the 4th cell can't be

a.

Scenario 2: the puzzle has multiple solutions. However suppose the 4 cells are in the pattern [ab]+[ba]. And this pattern exists in one of the possible solutions, let's call it solution grid S1. Without changing the other 77 cells of S1, we change these 4 cells into [ba]+[ab]. With regard to the original given clues, this new solution grid, S2, must also be a legitimate solution to the original puzzle.

But remember earlier we have used other logical means to eliminate candidate

b from the 4th cell. Thus S2 must not be a legal solution to the original puzzle.

A contradiction has been reached. Hence the initial premise, i.e. the 4 cells being in the pattern [ab]+[ba], must not be allowed in any meaningful solution, and candidate

a can be eliminated from the 4th cell if our goal is to reach one of the (possibly many) valid solutions.

In practice, if you can eliminate candidate

b from the 4th cell, candidate

b must also be eliminated from either the 2nd or 3rd cell (or both), turning one or both of these 2 cells into

a, thus eliminating candidate

a from the 4th cell.

Note that this whole process is very easy to go wrong if you're careless about how to spot an "avoidable rectangle". For example, in a diagonal (Sudoku X) puzzle, if the

b in the 4th cell is eliminated from one of the diagonals, then these 4 cells no longer form an "avoidable rectangle", and you

can't eliminate

a from the 4th cell.

I'm recently doing a lot of work on emerald puzzles, which make heavy use of the uniqueness assumption, therefore I have a lot of interest on related theories like this.