The New Sudoku Players' Forum

[marek stefanik]

Also needed a complex net. Easy until this point (note that a bivalue cell where neither option is self-referencing allows us to eliminate the candidates from each other's reference cell):

Code: Select all

.--------------------.--------------------.---------------------.
|e1479   23467 f269  | 126   8      2346  | 127    5     e13    |
| 135   c28     158  | 9     135    7     | 6     d23     4     |
| 13457  23467  1256 | 1256  1345   2346  | 127    9      8     |
:--------------------+--------------------+---------------------:
| 1379   5      189  | 478   2      89    | 1489   34678  13679 |
| 6     b278    4    | 3    a79     1     | 258   b278    5–9   |
| 1379   2378   1289 | 478   6      5     | 12489  23478  1379  |
:--------------------+--------------------+---------------------:
| 2      1     g356  | 5678  34579  34689 | 4589   4678   5679  |
| 8     h46     7    |i25   i459   h269   | 3      1     j569   |
| 45     9     g356  | 1678  1357   3468  | 458    4678   2     |
'--------------------'--------------------'---------------------'

Note that each row needs a self-referencing digit, while each digit can only be self-referencing in one row, so each row has exactly one such digit.
(9=7)r5c5 – 7r5c28 = [8r5c2 & 2r5c8] – (8=2*)r2c2 – (2=3)r2c8 – 3r1c9 = [1r1c9 & 9r1c1] – (9*=*6)r1c3 – 6r79c3 = [6r8c2 & 2r8c6°] – 2r8c4 = [5r8c4 & 4r8c5] – 9r8c5 °=° 9r8c9 => –9r5c9
Basics and bivalues to the end.

[ryokousha]

Normal sudoku rules, uniqueness can be assumed.

Code: Select all

------------------------------
  .  .  5 | .  .  . | 8  .  7
  .  .  . | .  .  . | .  .  .
  .  8  2 | 3  .  . | 4  1  .
------------------------------
  .  9  . | 4  .  . | .  .  .
  .  4  . | .  6  . | 2  8  .
  6  .  . | .  1  . | .  .  .
------------------------------
  7  .  . | .  2  8 | 9  6  .
  .  .  . | .  .  . | .  .  .
  .  .  . | .  .  5 | .  2  3
------------------------------
..5...8.7..........823..41..9.4......4..6.28.6...1....7...2896...............5.23


[eleven]

Make some equivalence transformations to show the symmetry:
switch stacks 2<->3, cols 123->312, cols 56

Code: Select all

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

rotational symmetry (19)(28)(37)(46)(5) -> 5r5c5
So in the orignal puzzle r5c9=5, stte

[Added:]There was the question, how to find the symmetry.
If it's 2-fold (pairs of symmetrical digits), only boxes with the same number of digits can be symmetric. Here it is easy: b29,b39,b18(3 digits with 2 in a row),b45. So b6 must be in the center, etc.

[ryokousha]

Yes.
The reason I chose this particular puzzle is that there is another way to see the symmetry:
Transform to RN, which is symmetrical under 180° rotation and the permutation (14)(25)(36)(78)(9). The rotation and the permutation leave the RN puzzle - including the weird low/mid/high rule for vertical triples instead of boxes - invariant. So we can conclude 9r5c5 in RN and 5r5c9 in RC, stte.

The reason the deduction does not work for diagonal mirror symmetry is, I think, that an RN puzzle can never have a diagonal mirror symmetry due to the vertical triple rule. A 180° rotation together with a carefully chosen permutation that leaves the triple rule alone works, but as we have seen it results in a morphed symmetry in RC already (I think that must always be the case?).

For shye's OP the problem is that the RN is not a sudoku, unless we require the low/mid/high rule for vertical boxtriples already for the RC (which then solves it by singles). So again, GSP cannot be applied.