kwdos wrote:... I just wonder whether, even in principle in a multi-solution Sudoku (i.e. an incorrectly set non-valid Sudoku), from among many complex forcing chains which could perhaps occur simultaneously, there might exist cases where one forcing chain leads to an apparently correct result for one of the many solutions whereas if one picked another forcing chain one might produce one of the other solutions.

No. All logical steps end at some point when there are dual solutions. If there are mutiple solutions, there will come a point there there simply is no logic reason to pick one path over another. You simple cannot have two valid forcing chains that give contraditory results -- which is what is required for them to lead to two different solutions. Take for example, this very simple puzzle, a 2x2 Latin Square:

- Code: Select all
`. .`

. .

Fill in the cells with the digits "1" or "2" so no digit appears twice in each column and row.

There are two solutions. There is no way around this -- there is no "better" or "right" solution.

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`1 2`

2 1

and

2 1

1 2

No "logical" step will favor one over the other. Logic will enable you to narrow your search down to two answers. There is no "there" there.

kwdos wrote:... If that were to be true then obviously one would need some other test to check that the Sudoku one thinks one has solved using forcing chains really has a unique solution and one could not necessarily trust such a solution without making an assumption about uniqueness. I see that I explained this badly originally but I hope this has made my niggling worry clearer.

If the puzzle has more than one solution, the solver will NEVER find a SINGLE solution using forcing chains, swordfish, etc, as demonstrated above.

kwdos wrote:I know you believe forcing chains can't hide the non-uniqueness of a multi-solution Sudoku - indeed I suspect, and hope, that they cannot - but I'd just like to see a rigourous proof of this!

The micro-mini example should stand instead of a proof.

Using the dual-solution puzzle posted further up in this thread:

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` `

1 . . | . . . | 5 8 .

8 . . | . . 1 | . . 6

. 6 . | . 2 . | . . .

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

. . . | 5 9 . | . . .

. . . | . 7 . | . . .

. 9 3 | . . 2 | . . .

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

. 1 . | . . . | 8 . 4

. . . | 9 . 7 | . 5 .

6 . . | 2 . . | 7 1 .

Using relatively simple logic, I come to:

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` 1 . . | . . . | 5 8 . `

8 . . | . . 1 | . . 6

. 6 . | . 2 . | . . .

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

. . . | 5 9 . | . . .

. . . | . 7 . | . . .

. 9 3 | . . 2 | . . .

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

. 1 . | . . . | 8 . 4

. . 8 | 9 1 7 | 6 5 .

6 . . | 2 . . | 7 1 .

{1} {237} {2479} {3467} {346} {3469} {5} {8} {279}

{8} {2357} {24579} {347} {345} {1} {2349} {23479} {6}

{3579} {6} {4579} {3478} {2} {34589} {1349} {3479} {179}

{247} {2478} {16} {5} {9} {3468} {1234} {23467} {1278}

{245} {2458} {16} {13468} {7} {3468} {12349} {23469} {12589}

{457} {9} {3} {1468} {468} {2} {14} {467} {1578}

{279} {1} {279} {36} {356} {356} {8} {29} {4}

{234} {234} {8} {9} {1} {7} {6} {5} {23}

{6} {35} {59} {2} {48} {48} {7} {1} {39}

For no logical reason, I can place a '3' in r9c2, leading easily to one of the two solutions:

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

8 5 2 | 7 4 1 | 3 9 6

3 6 9 | 8 2 5 | 1 4 7

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

7 4 6 | 5 9 8 | 2 3 1

2 8 1 | 4 7 3 | 9 6 5

5 9 3 | 1 6 2 | 4 7 8

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

9 1 7 | 3 5 6 | 8 2 4

4 2 8 | 9 1 7 | 6 5 3

6 3 5 | 2 8 4 | 7 1 9

If instead, I place a 5 in r9c2, with great difficulty, including forcing chains, I come up with the one and only other solution:

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

8 7 4 | 3 5 1 | 9 2 6

9 6 5 | 7 2 8 | 4 3 1

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

4 2 1 | 5 9 6 | 3 7 8

5 8 6 | 1 7 3 | 2 4 9

7 9 3 | 8 4 2 | 1 6 5

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

2 1 7 | 6 3 5 | 8 9 4

3 4 8 | 9 1 7 | 6 5 2

6 5 9 | 2 8 4 | 7 1 3

Forcing chains and all other logical tactics can only give the solver the value of the cells that are the same in both solutions, or eliminate possiblities that can be removed from both solutions, nothing further. Unless the solver makes an unsupported assumption, as I did by placing a digit in r9c2, the solving process comes to a dead end.

If whoever or whatever posed the puzzle had only one of the solutions in mind -- claiming for no good reason that this was the "correct" soluton and the other was not -- well, what exactly does that mean? Where reduced to guessing how many fingers they're holding up behind they're back.