Naked Sextuple

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Naked Sextuple

Postby hobiwan » Tue Jan 22, 2008 12:23 am

While looking for test cases for uniqueness tests I stumbled over the following Puzzle:

Code: Select all
.------------------.------------------.------------------.
| 2     57    8    | 3     49    19   | 17    45    6    |
| 46    35    9    | 7     24    1268 | 13    2458  28   |
| 46    37    1    | 468   5     268  | 37    248   9    |
:------------------+------------------+------------------:
| 8     9     26   | 45    1     3    | 45    26    7    |
| 3     46    5    | 2     48    7    | 9     68    1    |
| 7     1     24   | 9     6     58   | 45    3     28   |
:------------------+------------------+------------------:
| 5     26    67   | 16    3     269  | 8     179   4    |
| 19    68    3    | 168   789   4    | 2     179   5    |
| 19    248   47   | 158   2789  2589 | 6     179   3    |
'------------------'------------------'------------------'

unique rectangle: r89c1 r89c8
naked sextuple for digits 245678 in r1c8,r2c8,r3c8,r4c8,r5c8,r89c8 => r7c8<>7


Of course the same elimination can be made with a uniqueness test type 2, but how often do you see a sextuple (if it is even called that way)?
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Postby Ruud » Tue Jan 22, 2008 12:50 am

Sorry to disappoint you, but there is an additional candidate for digit 1 in r89c8, so that spoils the naked sextuple.

Also, the disjoint group {179} in r789c8 sortof jumps at you.

In a 9x9 Sudoku, the largest naked/hidden subset you need to look for is of size 4. In a unit of N unsolved cells, each naked subset of size M has a complementary hidden subset of size N-M. The smaller of these 2 sets is always 4 or less.

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Postby hobiwan » Tue Jan 22, 2008 7:58 am

You don't disappoint me. The Sextuple of cource exists only within the rules for a Uniqueness Test Type 3. From sudopedia:

Suppose both ceiling cells have extra candidates. By treating these two cells as one node,
find k - 1 other cells (as nodes) in the same house as these two cells so that the union
of the candidates for these k cells has exactly k unique digits. Then the Naked Subset
rule can be applied eliminate these k digits from the other cells in the house. A
rectangle that meets this test is also called a Type 3 Unique Rectangle.


By treating cells r89c8 as a single node and ignoring the candidates {19} in those cells since they belong to the unique rectangle you get 6 candidates in 6 nodes. The only thing I am not sure of is, wether the pseudo naked subset may include a digit from the unique rectangle itself (at least in this case the eliminiation is valid).

The disjoint group {179} does jump at you, but it doesn't allow for any eliminations.
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Postby ronk » Tue Jan 22, 2008 1:52 pm

hobiwan wrote:By treating cells r89c8 as a single node and ignoring the candidates {19} in those cells since they belong to the unique rectangle you get 6 candidates in 6 nodes.

Your deduction is valid, but nothing is gained by adding a naked quintet to the "naked single" of a Type 3 unique rectangle.
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Postby hobiwan » Tue Jan 22, 2008 7:26 pm

Your deduction is valid, but nothing is gained by adding a naked quintet to the "naked single" of a Type 3 unique rectangle.


You are right of course. I just thought it was funny. Never mind.
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Postby Smythe Dakota » Wed Jan 23, 2008 8:36 am

Ruud wrote:.... In a unit of N unsolved cells, each naked subset of size M has a complementary hidden subset of size N-M. ....

True, but some players (amateurs like me, at least) tend to look for naked things before we look for hidden things. On one occasion, in a newspaper puzzle of about average newspaper difficulty, I noticed a naked quint, in a 3x3 box with 7 unsolved cells. The existence of the quint surprised me, but then I realized this meant there was a hidden pair. Of course, the hidden pair proved more useful than the naked quint. Sometimes a backdoor approach can be quite effective!

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