I've found that manual searching for another unavoidable (permutation) of the given (strongly minimal) UA set is rather amusing logical game. You need not to have exact understanding - what is UA set to solve this puzzle, but you should follow simple logical rules.

For example, let's consider strongly minimal UA set U14, having 2 permutations (2-valent strongly minimal UA set) having such first permutation:

- Code: Select all
`4 5 7|. . 9`

. 1 9|. . 3

6 3 .|. . 4

-----+-----

5 6 .|. . .

. . .|. . .

. 7 1|. . .

We must find the second permutation of this UA set, i.e. we must rearrange given digits in such way that following conditions are met:

1. New configuration must contain digits in the same cells comparing with given configuration.

2. Sets of digits used for each row/column/box must be the same.

3. New configuration must not contain the same digits in the same cells comparing with given configuration.

Certainly everyone can simply find another permutation by means of computer, but finding it manually, in "pencilmark style" is rather interesting.

Here is one possible solution way.

1. We can see, that bottom row (r6) contain 2 digits only - "7" and "1". New permutation must contain these digits in the same cells of the same row, but in reversing order. The same consideration can be used for digits "5" and "6" in the r4 row. So, we know 4 digits of the target permutation:

- Code: Select all
`? ? ?|. . ?`

. ? ?|. . ?

? ? .|. . ?

-----+-----

6 5 .|. . .

. . .|. . .

. 1 7|. . .

2. Let's consider column c3. It must contain digits "1", "7", "9". We already have digit "7" in the r6c3 cell. So, digit "7" must go away from the cell r1c3 (from the column c3) to another cell of the row r1. Cell r1c2 of the row r1 is the only possible new place for the digit "7", because columns c1 and c6 don't contain this digit. So, we came to this configuration:

- Code: Select all
`? 7 ?|. . ?`

. ? ?|. . ?

? ? .|. . ?

-----+-----

6 5 .|. . .

. . .|. . .

. 1 7|. . .

3. When we placed digit "7" in the cell r6c3, we pushed out "old" digit "1" from this cell. This digit must go in another cell of the column c3, because this column must contain digit "1" somewhere. Cell r2c3 of the column c3 is the only possible new place for the digit "1", because row r1 don't contain this digit. So, we came to this configuration:

- Code: Select all
`? 7 ?|. . ?`

. ? 1|. . ?

? ? .|. . ?

-----+-----

6 5 .|. . .

. . .|. . .

. 1 7|. . .

4. Column c3 can contain digit "9" only in the cell r1c3, because positions of the rest digits ("1" and "7") are already known. We have

- Code: Select all
`? 7 9|. . ?`

. ? 1|. . ?

? ? .|. . ?

-----+-----

6 5 .|. . .

. . .|. . .

. 1 7|. . .

5. When we placed digit "1" in the cell r2c3, we pushed out "old" digit "9" from this cell. This digit must go in another cell of the row r2, because this row must contain digit "9" somewhere. Cell r2c6 of the row r2 is the only possible new place for the digit "9", because column c2 cannot contain this digit. So, we came to this configuration:

- Code: Select all
`? 7 9|. . ?`

. ? 1|. . 9

? ? .|. . ?

-----+-----

6 5 .|. . .

. . .|. . .

. 1 7|. . .

6. Row r2 can contain digit "3" only in the cell r2c2, because positions of the rest digits ("1" and "9") are already known. We have

- Code: Select all
`? 7 9|. . ?`

. 3 1|. . 9

? ? .|. . ?

-----+-----

6 5 .|. . .

. . .|. . .

. 1 7|. . .

7. Column c2 can contain digit "6" in the cell r3c2 only, because positions of the rest digits ("1", "3", "5", "7") are already known. We have

- Code: Select all
`? 7 9|. . ?`

. 3 1|. . 9

? 6 .|. . ?

-----+-----

6 5 .|. . .

. . .|. . .

. 1 7|. . .

7. If we check out row r3, we can see that cell r3c1 can contain digit "3" or digit "4", but digit "3" is absent in the column c1. So, cell r3c1 can contain digit "4" only. We have

- Code: Select all
`? 7 9|. . ?`

. 3 1|. . 9

4 6 .|. . 3

-----+-----

6 5 .|. . .

. . .|. . .

. 1 7|. . .

8. It is easy to find 2 missing digits in the cells r1c1 and r1c6 (we should check out columns c1 and c6). So, we have found puzzle solution:

- Code: Select all
`5 7 9|. . 4`

. 3 1|. . 9

4 6 .|. . 3

-----+-----

6 5 .|. . .

. . .|. . .

. 1 7|. . .

If you found this example being too simple, you can solve another puzzle (strongly minimal UA set U27):

- Code: Select all
`. . .|6 . .|4 . .`

. . 6|. 2 1|. 3 9

7 8 9|3 4 .|6 1 2

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

. 3 7|4 . 8|. . .

. 6 1|. . 3|. . .

8 . 4|1 6 7|. . .

Serg