The New Sudoku Players' Forum

[mith]

Code: Select all

+-------+-------+-------+
| 9 . 8 | 7 . . | . . . |
| 6 . . | . . 5 | . . . |
| . . 4 | 8 . 3 | . . . |
+-------+-------+-------+
| . . . | . 2 . | 1 9 . |
| . . . | 1 . . | . 5 7 |
| . . 9 | . . . | . . 4 |
+-------+-------+-------+
| . . . | 3 . . | 7 . . |
| . . 2 | . . . | 4 6 . |
| . 1 . | . . . | . 3 2 |
+-------+-------+-------+
9.87.....6....5.....48.3.......2.19....1...57..9.....4...3..7....2...46..1.....32

[Cenoman]

My most reasonable solution, two steps (Almost ALS XZ rule and almost ALS Y-Wing) :

Code: Select all

 +------------------------+------------------------+----------------------+
 |  9      235     8      |  7      146     146    |  256-3  124   1356   |
 |  6      37      1      |  2      49      5      |  89-3   478   389    |
 |  257    257     4      |  8      169     3      |  2569   127   1569   |
 +------------------------+------------------------+----------------------+
 |  4578   45678   3567   |  456    2       4678   |  1      9     368    |
 |  248    2468    36     |  1      3468    9      |  2368   5     7      |
 |  1      25678   9      |  56     35678   678    |  2368   28    4      |
 +------------------------+------------------------+----------------------+
 |  458    45689   56     |  3      14568   2      |  7      18    1589   |
 |  3      5789    2      |  59     1578    178    |  4      6     1589   |
 |  4578   1       567    |  4569   45678   4678   |  589    3     2      |
 +------------------------+------------------------+----------------------+

1. [(3=164)r1c569 - (4=893)r2c589] = (5)r1c9 - r13c7 = (5-9)r9c7 = r9c4 - (9=5)r8c4 - r46c4 = (5-3)r6c5 = (3)r6c7 => -3 r1c7
2. [(3=7)r2c2 - (725=6)r3c127 - (6=283)b6p478] = (9)r3c7 - r9c7 = r9c4 - (9=5)r8c4 - r46c4 = (5-3)r6c5 = (3)r6c7 => -3 r2c7; lclste

Presented as kraken AALS's:

Hidden Text: Show

1. Almost-almost NT(164)r1c569
(164)r1c569 - (4=893)r2c579
(3)r1c9
(5)r1c9 - r13c7 = (5-9)r9c7 = r9c4 - (9=5)r8c4 - r46c4 = (5-3)r6c5 = (3)r6c7
------------------
=> -3 r1c7

2. Almost-almost NT(257)r3c7
(257)r3c127 - (7=3)r2c2
(6)r3c7 - (6=283)b6p478
(9)r3c7 - r9c7 = r9c4 - (9=5)r8c4 - r46c4 = (5-3)r6c5 = (3)r6c7
-------------------
=> -3 r2c7; lclste

[SpAce]

My most reasonable solution, two steps (Almost ALS XZ rule and almost ALS Y-Wing) :

Very nice!

[mith]

Agreed!

(There is a one step solution though. )

[mith]

Hint: This puzzle is a bit of a troll. It does not require any chains.

[eleven]

Ah, how could i miss that ?
So lets swap bands 1,2 and stacks 2,3 to get a 180° rotational symmetric puzzle in normal form,
with digit pairs 12,39,47,56 and 88 (you can always find the other digit in the cell "mirrored" by the center)

Code: Select all

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

We can immediately put an 8 into the center (the only digit symmetric to itself) -> stte
(in the original puzzle the 8 goes to r2c8)

[mith]

Yep, that's it.

It occurred to me that when Gurth's applies, it also applies to any morphs... it's just some are obviously easier to spot than others. For the particular case of being able to divide the grid into a 3x6, 3x3, 6x6, and 6x3 that are all rotational symmetric and all share the same digit pairings, it's not too hard to see that the symmetry applies. (And I happened to have a 9.0 which reduces to singles with Gurth's, so...)

I'm wondering if there is some simple generic process for determining whether a puzzle is a potential candidate for morphing in this way. In such a puzzle, there must necessarily be a pairing between rows, columns, and boxes, but maybe that's not a sufficient condition.

[eleven]

If i had studied it better, i should have had the idea, that it is symmetric, because the same move can always be done for 2 digits, starting with the singles for 1/2 and 3/9.
(Also in Cenoman's solution you can have the same, if you replace the digits/cells by their symmetric counterparts.)

My method to find the symmetry is basically, to look at the boxes. Only boxes with the same number of candidates can be symmetric to each other, and they must have patterns, which can be mapped to each other with row/column swaps.

[denis_berthier]

Great example. It's the first time I see symmetry being applied to find a solution.