The New Sudoku Players' Forum

[Carcul]

Another Example of a Type 3 AUR

Here is an interesting example of a type 3 AUR. The initial grid:

Code: Select all

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


and after some steps we get:

Code: Select all

 6    2459  124 | 28   149 3   | 245  1458  7     
 245  2345  7   | 268  14  68  | 9    13458 1458
 249  8     1234| 7    5   19  | 234  6     14     
----------------+--------------+-----------------
 3    2479  24  | 59   6   5789| 1    2489  489   
 12   269   8   | 3    19  4   | 7    259   569   
 1479 4679  5   | 89   2   1789| 46   489   3     
----------------+--------------+-----------------
 45   1     346 | 4569 8   2   | 3456 7     4569   
 2457 23457 9   | 456  37  56  | 8    1345  1456   
 8    3457  46  | 1    37  569 | 456  3459  2     


There is a slightly different type 3 AUR in cells r8c25/r9c25, which can be used to eliminate a "2" from r2c2 (and r1c2):

[r2c2]-2-[r8c2|r9c2|r7c1](-4-[r7c3])-4-[r9c3]-6-[r7c3]-3-[r3c3]=3=[r3c7]=2=[r1c7]-2-[r1c4]=2=[r2c4]-2-[r2c2], => r2c2<>2.

As an exercise, I invite the reader to complete the solving of this grid (from the beginning).

Carcul

[Jeff]

Code: Select all

 6    2459  124 | 28   149 3   | 245  1458  7     
 245  2345  7   | 268  14  68  | 9    13458 1458
 249  8     1234| 7    5   19  | 234  6     14     
----------------+--------------+-----------------
 3    2479  24  | 59   6   5789| 1    2489  489   
 12   269   8   | 3    19  4   | 7    259   569   
 1479 4679  5   | 89   2   1789| 46   489   3     
----------------+--------------+-----------------
 45   1     346 | 4569 8   2   | 3456 7     4569   
 2457 23457 9   | 456  37  56  | 8    1345  1456   
 8    3457  46  | 1    37  569 | 456  3459  2     


[r2c2]-2-[r8c2|r9c2|r7c1](-4-[])-4-[]-6-[r7c3]-3-[r3c3]=3=[r3c7]=2=[r1c7]-2-[r1c4]=2=[r2c4]-2-[r2c2], => r2c2<>2.

Hi Carcul, Is [r8c2|r9c2|r7c1] a strong node to infer r7c3 and r9c3 <>4.

[Carcul]

Hi Jeff.

Yes: if r8c2 is not "2", then we have a unique rectangle that eliminates a "4" from r7c3 and r9c3.

Regards, Carcul

[Jeff]

Yes: if r8c2 is not "2", then we have a perfect type 3 AUR that eliminates a "4" from r7c3 and r9c3.

Hi Carcul, Could you explain how a perfect type 3 AUR works?

[ronk]

In your AUR type numbering, which type number is for the simplest of AURs ... the AUR+1?

This example has an AUR (45) + 1 at cells r1c26 and r3c26:

This is not an almost unique rectangle. The example in your grid is already an unique rectangle (of type 1), which imply immediatly r1c2=2.
Please check the definition of AUR that I have writen in the original post.

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Almost Unique Rectangle (AUR): a set of four cells populated in such a way that if one or two of them does not have a specific candidate 'x', then those four cells form a unique rectangle.

So apparently for four cells in a rectangular pattern ...

1. If the cells contain the identical bivalue candidates, we have non-uniqueness
2. If the cells contain one extra candidate (relative to non-uniqueness), we have a unique rectangle
3. If the cells contain two extra candidates (relative to non-uniqueness), we have an almost unique rectangle (AUR)

Therefore, if we refer to AURs with "AUR (xy) + n" notation, we need three extra candidates (relative to non-uniqueness) to have an AUR (xy) + 1. Is that correct

Ron

[ronk]

Is [r8c2|r9c2|r7c1] a strong node to infer r7c3 and r9c3 <>4?

Absolutely! From my more graphical-based POV, that appears to be a grouped 'Sue De Coq' or a grouped Two-Sector Disjoint Set.



The orange cells are a grouped almost-almost-locked-set and the pink and green cells are almost-locked-sets, abeit grouped in the green case. This permits elimination of 4s and 5s (if there were any) from box 7, excepting the cells that define the sets of course.

I don't think this graphical POV leads to the 2s eliminations that Carcul noted. However, r7c3<>4 and r9c3<>4 solve the puzzle.


Ron

[Carcul]

Hi Jeff.

What I wanted to write was "unique rectangle" (I have already edited the reply above). So, if r8c2 is not "2", then we have an unique rectangle in which the cells r7c1, r8c2, r9c2 function as a naked pair. Sorry for the error.

Regards, Carcul

[Carcul]

Hi Ronk.

3. If the cells contain two extra candidates (relative to non-uniqueness), we have an almost unique rectangle (AUR)

This depends on the identity of the candidates and how they are distributed among the four cells. See the first grid in the original post.

The orange cells are a grouped almost-almost-locked-set and the pink and green cells are almost-locked-sets, abeit grouped in the green case. This permits elimination of 4s and 5s (if there were any) from box 7, excepting the cells that define the sets of course. And r7c3<>4, r9c3<>4 solve the puzzle.

This is very nice. But, just by curiosity, what is the contradiction that results from, for example, r9c3=4? I think your reasoning with the Sue de Coq would be completely correct if "7" were not a candidate in r8c1, which is not the case.

Regards, Carcul

[ronk]

... just by curiosity, what is the contradiction that results from, for example, r9c3=4?

Only that 'if r9c3=4, there would be two 4s in box 7'.

I don't know how to do a formal proof, particularly with set notation, but my argument would go something like ...
Given sets:
A = {r8c1,r8c2,r9c2} = {2457,23457,3457} = {23457}
B = {r7c1} = {45}
C = {r8c5,r9c5} = {37,37} = {37} (actually a locked set as is, but read on)

... and noting that B and C do not share any values, then a locked set A must contain exactly one of B and exactly one of C. Even though C is grouped (two rows) and locked (two cells, two candidates), each different value {37} of C is "available" to all cells of A. Therefore, this 'grouped Sue De Coq' is equivalent to one that has all cells of A and C in the same row, with 3 cells for A and 1 cell for C, respectively, in the equivalent. C of the equivalent is then also an almost locked set. [edit: Note that this equivalency is entirely due to the naked pair in two corners of the AUR, i.e., set C. I don't mean to imply that the AUR is necessary, however, as r8c1 might contain the candidate 3 instead of r8c2, and then the grouped Sue De Coq would still apply but not the AUR.]

From this point on the standard 'Sue De Coq' proof would apply. Specific to your question: if locked B doesn't contain the 4, then locked A does. Therefore, any other candidate that 'sees' both 4s can be eliminated.

I think your reasoning with the Sue de Coq would be completely correct if "7" were not a candidate in r8c1, which is not the case.

Even with the candidate 7 in r8c1, the reasoning is completely correct, for the reason stated above, i.e., each of {r8c1,r8c2,r9c2} 'sees' both candidate values {37} of {r8c5,r9c5}.

Regards, Carcul[/quote]

[Jeff]

Is [r8c2|r9c2|r7c1] a strong node to infer r7c3 and r9c3 <>4?

Absolutely! From my more graphical-based POV, that appears to be a grouped 'Sue De Coq' or a grouped

Ronk, Do you understand my question? I don't mind you answered it for Carcul. But, you should at least get it right.

Thanks Carcul for the explanation.

[Carcul]

Hi Ronk.

Only that 'if r9c3=4, there would be two 4s in box 7'.

This is not correct. I have made r9c3=4 (ate least three or four times) and I don't get two "4s" in box 7. In fact, I get no contradiction.

I don't know how to do a formal proof,

The most important proof is simply to make the substitution: translating this to a given notation (loops, ALS, etc) is another problem. Consider the following:

If I have used a technique (which uses the set of cells "A") to elimininate candidate "x" from cell "B" (which can belong or not to A), then this would imply that making B=x leads to some sort of contradiction in/with the set A.

Do you (and Jeff, and everybody that reads this) agree with this? The thing is that I have been thinking for some time in the above statement and sometimes I get confused with it, and I would like to see this "question" solved. What do you think about this? I would like to see some opinions about this issue.

Regards, Carcul

[ronk]

Is [r8c2|r9c2|r7c1] a strong node to infer r7c3 and r9c3 <>4?

Absolutely! From my more graphical-based POV, that appears to be a grouped 'Sue De Coq' or a grouped

Ronk, Do you understand my question? I don't mind you answered it for Carcul. But, you should at least get it right.

I apologize for misreading one character in your post.

At the time, I thought your strong node was the same as the one I saw, i.e., [r8c2|r9c2|r8c1] ... and mine produced exactly the same eliminations ... so that statement for my strong node would have been exactly the same except for the one character in the entire line.

As to your strong node, I must admit I don't see it.

Ron

P.S. I can barely believe this apology was necessary and don't see a need to tolerate this kind of aggravation, so perhaps we should just mutually agree not to respond to each other's posts. Is it a deal?

[Jeff]

If I have used a technique (which uses the set of cells "A") to elimininate candidate "x" from cell "B" (which can belong or not to A), then this would imply that making B=x leads to some sort of contradiction in/with the set A.

Hi Carcul, I can't think of any reasons to suggest otherwise; may be our friend Ronk could.

[Jeff]

P.S. I can barely believe this apology was necessary and don't see a need to tolerate this kind of aggravation, so perhaps we should just mutually agree not to respond to each other's posts. Is it a deal?

Ronk, you really don't need to apologise or make any deal. Just feel free to ignore.

[ronk]

Only that 'if r9c3=4, there would be two 4s in box 7'.

This is not correct. I have made r9c3=4 (ate least three or four times) and I don't get two "4s" in box 7. In fact, I get no contradiction.

Sorry, I misunderstood your question. That there would be two 4s in box 7 is a deduction based on the Sue De Coq, which means it effectively depends upon the cells of the Sue De Coq being placed first which, err ... one can't deduce so guessing would be required, I guess. As I've never done it quite that way, I'll give it a try.

As to a contradiction resulting from first placing r9c4=4, I get two 9s in row 7 ...

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 6     59    1     | 2     4     3     | 5     58    7   
 2     25    7     | 6     1     8     | 9     35    4   
 24    8     3     | 7     5     9     | 2     6     1   
-------------------+-------------------+-----------------
 3     49    2     | 5     6     7     | 1     489   89 
 1     6     8     | 3     9     4     | 7     2     5   
 479   479   5     | 8     2     1     | 46    49    3   
-------------------+-------------------+-----------------
 5     1     4     | 9     8     2     | 3     7     9   
 27    237   9     | 4     37    5     | 8     1345  6   
 8     37    6     | 1     37    5     | 45    3459  2   


... but when presented with many choices of fills (based on singles, e.g.) the contradiction may vary depending upon the sequence of choices made.

I'm still trying to understand the balance of your post, and intend to respond to that portion later.

Regards, Ron