tdillon wrote:If you have a puzzle that's been reduced to bi-value cells, and if there are no remaining hidden singles, then a simple pigeonhole argument shows that in each house each of the remaining candidates appear exactly twice. That's to say the Hodoku description of BUG+0 is implied by the simpler description "bi-value cells, no hidden singles".

You need to add that in each house, each missing cell value, has at at least one candidate.

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`+----------+----------+-----------+`

| 3 5 9 | 8 7 4 | 1 2 6 |

| 8 6 1 | 9 5 2 | 37 37 4 |

| 4 2 7 | 1 3 6 | 8 5 9 |

+----------+----------+-----------+

| 6 79 8 | 4 2 5 | 79 1 3 |

| 1 79 3 | 6 8 79 | 4 79 5 |

| 2 4 5 | 3 1 79 | 79 6 8 |

+----------+----------+-----------+

| 7 3 6 | 2 9 8 | 5 4 1 |

| 5 8 4 | 7 6 1 | 39 39 2 |

| 9 1 2 | 5 4 3 | 6 8 7 |

+----------+----------+-----------+

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rjamil wrote:Please also note that the zero solution puzzle is easy to check and not considered at the moment.

You say that it's easy to check for a zero solution puzzle.

I don't know what you mean by that exactly.

It's easy to check for cells with no candidates, and for a house where a missing cell value doesn't have a candidate.

Is that what you meant ?

The puzzle above, doesn't have a solution,

It's easy to see that it doesn't.

There's no candidate for '2' in box 6, for example.

The one below, has a BUG+0, but it doesn't have multiple solutions.

It doesn't have any solution, but that isn't immediately obvious.

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`+----------+---------+----------+`

| 6 9 14 | 5 3 2 | 7 8 14 |

| 45 8 3 | 9 7 1 | 6 45 2 |

| 7 2 15 | 8 6 4 | 13 35 9 |

+----------+---------+----------+

| 2 3 7 | 1 8 5 | 4 9 6 |

| 59 1 59 | 6 4 7 | 8 2 3 |

| 8 46 46 | 3 2 9 | 5 1 7 |

+----------+---------+----------+

| 39 7 69 | 4 1 8 | 2 36 5 |

| 1 46 2 | 7 5 3 | 9 46 8 |

| 34 5 8 | 2 9 6 | 13 7 14 |

+----------+---------+----------+

On the other hand, it has an easy elimination, that leads (via singles) to a state where it's clear that it doesn't have a solution.

There's a skycraper in 4's, for example.

There's a 2-string kite (in 4's).

Maybe that's what you were getting at ?

In any case, there's this:

1) Suppose a puzzle shows a BUG+0.

2) Suppose that your list of solving techniques, is powerful enough to produce an elimination for any candidate where guessing that the candidate is "true" (in some solution), and following it up with naked and hidden single placements, leads to a "zero solution" state.

3) In that case, you can be sure that if no eliminations are detected, then the puzzle has multiple solutions.

Sudoku Explainer's Dynamic Contradiction Forcing Chains, are powerful enough to cover (2).

Denis Berthier's (basic/singles) "braids" are powerful enough.

Added: Since it's all bi-value cells and bi-local candidates, 3D Medusa Coloring (is it?), is probably powerful enough too.

Added3:Actually you don't need anything as powerful as "braids" or "dynamic contradiction forcing chains".

Simple AICs are sufficient.

3D Medusa is suffiecient, see below, but any elimination coming from a 3D Medusa coloring, can done with a simple AIC too.

If 3D Medusa doesn't produce eliminations, then it can be used to produce solutions to the puzzle. On the other hand, if the puzzle doesn't have a solution, then 3D Medusa will produce an elimination, and the elimination could also come as an AIC elimination.

With that being true, if there are no (simple) AIC eliminations, then the puzzle has multiple solutions

Added2: 3D Medusa -- yeah, that's the one.

After choosing a starting cell, if the coloring stops with no conflicts or eliminations, then the every cell with a colored candidate, will contain a candidate with the opposite color, and for every colored candidate, and each house containing it, will the "other cell" in the house will have the candidate colored with the opposite color. The set of colored candidates (and containing cells) will be a "BUG-Lite + 0", and the colors will correspond to it's (only) two solutions.

If there are additional cells without uncolored candidates at that point, they will form another "BUG-Lite + 0", and choosing two more colors and repeating the process, will lead to either a conflict or elimination, or another fully-colored "BUG-Lite + 0".

If continuing on, leads to the entire grid being filled, then the number of solutions will be 2^N, where N is the number of color pairs.

At that point, solutions to the full puzzle, can be produced by combining one set of colored candidates from each color pair.

If, at any point, a conflict is seen (leading to either a candidate or a color being eliminated) then the entire puzzle will have zero solutions.

Note: That's so simple, it must have been discussed somewhere, long ago.

Cheers,

Blue.