Nontrivial automosphism of a given sudoku are not trivial to discover, unless the sudoku is given in such a way that it is easy to discover it; moreover, if one scrambles the sudoku it can be almost impossible for the human eye to see the automorphism.
When this technique can be applied, it can be very useful though. Furthermore, it is very useful when the automosphism has fixed cells, and so this technique excels with diagonal automorphisms.
If we assume uniqueness of the solution, we have the following proposition:
- If A is an automorphism of the puzzle , then A is too an automorphism of its solution.
To see how that proposition can be applied I give an example:
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#Example 1
. . .|. . 7|. 6 .
. 5 .|3 . .|8 . .
. . 9|. 4 .|. . 3
-----+-----+-----
. 7 .|1 . .|4 . .
. . 6|. 5 .|. 2 .
3 . .|. . .|. . 8
-----+-----+-----
. 2 .|6 . .|1 . .
4 . .|. 8 .|. . .
. . 7|. . 2|. . 5 ER 10.4
Note that after reflecting the sudoku along the main diagonal (r1c1-r9c9, r1 on top) and doing the permutation of digits 1->1,2->8,3->7,4->6,5->5,6->4,7->3,8->2,9->9, we obtain the same sudoku, so that must be an automorphism of the sudoku. If we assume uniqueness of the solution, then that automorphism is too an automorphism of the solution.
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.------------------.------------------.------------------.
| 128 1348 12348| 2589 129 7 | 259 6 1249 |
| 1267 5 124 | 3 1269 169 | 8 1479 12479|
| 12678 168 9 | 258 4 1568 | 257 157 3 |
:------------------+------------------+------------------:
| 2589 7 258 | 1 2369 3689 | 4 359 69 |
| 189 1489 6 | 4789 5 3489 | 379 2 179 |
| 3 149 1245 | 2479 2679 469 | 5679 1579 8 |
:------------------+------------------+------------------:
| 589 2 358 | 6 379 3459 | 1 34789 479 |
| 4 1369 135 | 579 8 1359 | 23679 379 2679 |
| 1689 13689 7 | 49 139 2 | 369 3489 5 |
'------------------'------------------'------------------'
Now we focus on r1c1, if we suppose the solution has r1c1=2, then applying the automorphism to the solution we obtain a solution in which r1c1=8, it will certainly be a different solution. In the same way, if r1c1=8, then we can obtain a solution in which r1c1=2. Therefore the only choice for r1c1 is 1. Analogously we can deduce that r6c6=9, r8c8=9. We have reached
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.------------------.------------------.------------------.
| 1 348 2348 | 2589 29 7 | 259 6 249 |
| 267 5 24 | 3 1269 16 | 8 147 12479|
| 2678 68 9 | 258 4 1568 | 257 157 3 |
:------------------+------------------+------------------:
| 2589 7 258 | 1 236 368 | 4 35 69 |
| 89 1489 6 | 478 5 348 | 379 2 179 |
| 3 14 1245 | 247 267 9 | 567 157 8 |
:------------------+------------------+------------------:
| 589 2 358 | 6 379 345 | 1 3478 47 |
| 4 136 135 | 57 8 135 | 2367 9 267 |
| 689 13689 7 | 49 139 2 | 36 348 5 |
'------------------'------------------'------------------'
Here we can apply a sworfish on 9's and the rest is singles. We did not use a single complicated chain that Sudoku Explainer sees.
In this example the automorphism had fixed cells, and we could take advantage of those fixed cells to fill them.
A more extreme example is:
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#Example 2
1 . .|. 6 .|. . 5
. 9 .|. . 2|. 7 .
. . .|8 . .|3 . .
-----+-----+-----
. . 2|1 . .|. 4 .
4 . .|. . .|. . 8
. 8 .|. . 9|6 . .
-----+-----+-----
. . 7|. . 4|. . .
. 3 .|6 . .|. 1 .
5 . .|. 2 .|. . 9 ER 10.5
We observe that this sudoku is automorphic along the main diagonal and along the antidiagonal. Applying the technique we obtain r3c3=r5c5=r7c7=5, and r6c4=r4c6={3,7}. The rest is singles.
When the automorphism has a lesser number of fixed cells, less information can be deduced:
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# Example 3
. . .|7 . .|3 . .
. 2 .|. . 6|. 5 .
. . 8|. 2 .|. . 9
-----+-----+-----
. 3 .|1 . .|. . 4
. . 9|. . .|1 . .
6 . .|. . 9|. 7 .
-----+-----+-----
1 . .|. 8 .|2 . .
. 5 .|4 . .|. 8 .
. . 7|. . 3|. . . ER 10.5
If we rotate this sudoku 180 degrees and do the permutation 1->9,2->8,3->7,4->6,5->5,6->4,7->3,8->2,9->1, the sudoku remains the same, the only fixed cell of that automorphism is r5c5, and the only fixed number is 5, so r5c5=5, and so
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. . .|7 . .|3 . .
. 2 .|. . 6|. 5 .
. . 8|. 2 .|. . 9
-----+-----+-----
. 3 .|1 . .|. . 4
. . 9|. 5 .|1 . .
6 . .|. . 9|. 7 .
-----+-----+-----
1 . .|. 8 .|2 . .
. 5 .|4 . .|. 8 .
. . 7|. . 3|. . . ER=9.3
In this example, the automorphism technique only allowed us to reduce the complexity of chains to use, but it is still a improvement.
So far I have only used the automorphism technique to fill in cells, it can be used too to fill cells that are the automorphic image of an already filled cell, but that information is not very useful (as a deduction to fill a cell can be morphed to fill another cell) and I have decided not to explain it.
To end this post, I'll leave you with a puzzle that has "sticks symmetry", a symmetry that ravel pointed out:
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#Exercise 1
1 . .|. . .|. . .
. . .|4 . .|. 6 5
. . .|. 9 8|. . .
-----+-----+-----
2 . 7|. 3 .|9 . .
. 5 .|2 . .|. . 3
. 1 .|. . 7|. 4 .
-----+-----+-----
3 7 .|. . 2|8 . .
. . 6|3 . .|. 2 .
. . 1|. 7 .|. . 4 ER 10.0
It has a nontrivial automorphism with 9 fixed cells, after we apply the technique, the puzzle can be solved with singles only.
Questions? Suggestions?
Mauricio.
Many thanks to sudocue, it provided the nice pencilmarks.
