The New Sudoku Players' Forum

[yzfwsf]

When I used the solver to solve this sukaku, I found the bug, but now I have no idea to fix it. How do the experts deal with the extended unique rectangles in sukaku?
Let's talk about the exact definition of extended unique rectangles

Code: Select all

1.3.....9.23..6789.23456789123456789.23456.89123.567....3...78912345..8....45..8912.45...9.23.5.7.9.23.5678912.456...12345..8...3..6..912.45678..2345.7..1.3..6.891234567..1234567891.3.5.7891..4567.91.345..8912345.7891.345.7.9..34567.91..4..7..12.4...8..2....78912.456....23.567..1234.6.8912....789123.5..8..2345678...3456...123456789.2345.7..12.4..7...2.4...89..34.....123456....23..6789123.56..9.234....9.2..567...23456.89.2.4.678.1.3.5..89.234.678.12.4.678...345.7...23..6.89123.56...12.4.6789..34.6..9123456789..3..67891.345678.1234.6.8.1..4...891.3456.8....45.7..12.45678.123.....91..456.89.23.....9..3456.89..3.56...1234567.9123456.8912....78912.4567891..4567.912.45678.1.345678..234..7891.3456..912345.7.9.2.45.7..1.345678.


bug pm

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.---------------------------------.---------------------------------.---------------------------------.
|  1          23678      23456789 |  2345678    234589     23567    |  3789       23458      4589     |
|  2459       2357       2356789  |  12456      123458     369      |  1245678    23457      13689    |
|  234567     2345678    35789    |  14567      134589     12345789 |  134579     34567      147      |
:---------------------------------+---------------------------------+---------------------------------:
|  248        9          2456     |  23567      12348      1278     |  12358      2345678    3456     |
|  2345678    2357       1        |  9          34         23456    |  23678      2356       234      |
|  256        234568     24678    |  1358       23478      124678   |  3457       9          12356    |
:---------------------------------+---------------------------------+---------------------------------:
|  24789      34         2345789  |  78         6          248      |  49         1          457      |
|  245678     1          45689    |  23         34589      35       |  234679     234568     2789     |
|  2456789    4567       245678   |  134578     234789     13459    |  234579     245        345678   |
'---------------------------------'---------------------------------'---------------------------------'

XR.png (41.92 KiB) Viewed 1226 times

[tarek]

The deadly pattern (DP) that is made up of 2 candidates forms a loop the smallest being 4 cells forming a rectangle.

The extended form appears to have more than 2 candidates in the DP (3 or more)

In a 9x9 vanilla I can't visualize a DP with 3 candidates that has more than 6 cells. So in short you will need 6 cells that have 3 different candidates in 2 lines x 3 lines x 3 boxes to have your DP. This DP can be prevented by looking at the extra candidates.

In the easier 2 candidate pattern you would have depending on the number of total cells (x the smallest being 4): (x/2) lines x (x/2) lines x (x/2) boxes with a potential maximum loop of 16 cells in a 9x9 sudoku (I've seen a 14 cell loop but nothing bigger in a real puzzle)

Any comments from others?

[eleven]

I have no idea, what "Extended Unique Rectangles" are.
There is no uniqueness pattern in the cells with green candidates and without a 3 in box1/column3.

This MUG has 3 digits and 7 cells:

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123 123 . | 123
123 .   . | 123
.   .   . | .
---------------
123 123 . | .

[tarek]

It appears that it refers to BUG-Lite technique … I had a quick look on the internet. The only person who is using this name "Extended UR" is Andrew Stewart!

[RW]

Edit: this does not apply in a pencilmark sudoku. See here for how to modify the technique for this variant.

This extended unique rectangle involves 7 digits. Maybe easiest to visualize like a variation of the Reverse Bug-lite. If all solved digits in two partially filled rows/columns within the same band form a deadly pattern, the unsolved digits form an unavoidable set. Or one could say that a 2xN area of unsolved cells within the same band must always contain at least N+1 different digits to yield a unique solution.

Imagine A and B are solved:

A.. B.. ...
B.. A.. ...

No matter how it is solved, we can always flip around the remaining 7 digits to find another solution:

Acd Bef ghi
Bgh Aic def

Agh Bic def
Bcd Aef ghi

In the example, one of the blue 3s must be true in order to avoid this, therefore we can eliminate 3 from r1c3.

Oh, and hi all! It's been a while! Nice to see so many familiar names are still around!

RW

Edit: fixed the example to make more sense.

[tarek]

RW ...

What!!!

A very nice surprise ...

Looking at Andrew Stuart Extended UR ... It appears that he is referring to the MUG ... doh

[RW]

Ah, seems this is actually exactly what I called a Reverse Bug-lite back in the days. Don't remember much of the old terminology anymore, nor do I know what new terminology you have come up with in the last 7 or so years.

RW

[eleven]

This is not RW, because that's just nonsense in this case.

[yzfwsf]

No matter how it is solved, we can always flip around the remaining 7 digits to find another solution:

Acd Bef ghi
Bgh Aic def

Agh Bic def
Bcd Aef ghi

In the example, one of the blue 3s must be true in order to avoid this, therefore we can eliminate 3 from r1c3.

Oh, and hi all! It's been a while! Nice to see so many familiar names are still around!

RW

Edit: fixed the example to make more sense.

In fact, it's wrong to eliminate 3r1c3, so I asked the experts.

[RW]

This is not RW, because that's just nonsense in this case.

Nah, just a rusty RW.

I see what you mean, didn't consider the fact that this is a pencilmark sudoku with eliminations already performed inside the pattern. Had to google 'sukaku', unfamiliar term to me. Seems I suffer from the same bug as the bugged solver mentioned by OP. I assume the solver implements some logic similar to what I described. A 2xN area in a band must include at least N+1 clues. In a pencilmark sudoku, an additional requirement to implement this technique would be that the two sets of N must have identical pencilmarks in all pairs of cells to be considered an unavoidable set.

To show what I mean, I altered the grid to show a situation where the elimination should be valid:

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.---------------------------------.
|  1          23678      2456789-3|
| #2459       2357      #2459+3   |
| #24567+3    2345678   #24567+3  |
:---------------------------------+
| #248        9         #248      |
|  2345678    2357       1        |
| #256        234568    #256      |
:---------------------------------+
| #24789      34        #24789+3  |
| #45678      1         #45678    |
| #245678     4567      #245678   |
'---------------------------------'

Comparing r1 to r3, apart from the 3s, they are identical and contain only 7 different digits in the 2x7 area. This should be a uniqueness pattern even in a pencilmark sudoku. The original version is obviously not a uniqueness pattern.

[eleven]

Please go away, fake RW.
Moderators, please throw him out.

[RW]

If you think it's not me because I made a mistake, you must have very selective memory. I can assure you I've been wrong around here many times before.

Anyway, my last example does show the correct way to apply the technique in a puzzle like this. If you were to put a three in r1c3 (forcing 3 in r5c1), you can no longer eliminate any candidate from any of the marked cells without also eliminating it from the other marked cell in the same row. In the end you'd have two candidates in all cells and no way to tell which goes left and which goes right - at least two solutions. But in a pencilmark sudoku this is an unavoidable set only if the sets of candidates in both cells in a row are identical, which they are not in the original puzzle. (In a regular sudoku puzzle the original elimination would be valid even with missing candidates. In fact, I believe I showed back in the days that with missing candidates we could make the elimination even without assuming uniqueness.)

[SpAce]

This is an interesting development. Without any hard evidence one way or the other, I'm actually ready to believe that RW is RW, or at least I can't rule that out That's mostly based on the similarly clear writing style that is a pleasure to read. The opposite conclusion would also require that someone actually hacked RW's account. Most fakes are new accounts with a user name very similar to an old legend, claiming they've lost access to the original account. Most would also give up and disappear when challenged.

That said, my skills in advanced uniqueness techniques (especially in a sukaku context) are so mediocre that I can't make any judgements based on those demonstrations. I would trust eleven's judgment in that regard, especially since he should be intimately familiar with RW's style and skills, having named him one of his top two favorite manual solvers. He seems to be convinced that RW is not RW, so we have a bit of a dilemma.

Oh well. I'm hoping for a happy ending, but we'll see how it turns out

[StrmCkr]

If it is RW heres a refrence to its application
http://forum.enjoysudoku.com/single-digit-deadly-patterns-t6106.html
However it would be easy to prove RW is RW by confirming how he became a supporter of this site in 2010.

The writting styles are similar albiet rusty from a 7+ year hiatious welcome back if it is you

[tarek]

It will be difficult to fill the shoes of a mathematician from Finland with skills like RW even if you were Arto Inkala