The New Sudoku Players' Forum

[ArkieTech]

Code: Select all

 *-----------*
 |...|.1.|...|
 |43.|.72|.1.|
 |2.9|6.5|...|
 |---+---+---|
 |8..|...|2..|
 |.26|.9.|54.|
 |..4|...|..6|
 |---+---+---|
 |...|1.6|8.7|
 |.8.|74.|.53|
 |...|.2.|...|
 *-----------*


Play/Print this puzzle online

[Cenoman]

Code: Select all

 +----------------------+---------------------+--------------------+
 |  57     6     57     |  348   1     348    |  39    389    2    |
 |  4      3     8      |  9     7     2      |  6     1      5    |
 |  2      1     9      |  6     3-8   5      |  347   378   a48*  |
 +----------------------+---------------------+--------------------+
 |  8      579   1357   | F345   6     1347   |  2     379    19   |
 |  137    2     6      | e38    9     1378   |  5     4     a18*  |
 |  1357   579   4      |  2   Ga358* b1378   |  379  a3789*  6    |
 +----------------------+---------------------+--------------------+
 |  9      4     35     |  1     35    6      |  8     2      7    |
 |  6      8     2      |  7     4     9      |  1     5      3    |
 |  1357   57    1357   |Ed358   2    c38     |  49    6      49   |
 +----------------------+---------------------+--------------------+
                                                        (8=3)r5c4   -   (3)r6c5
                                                      /     e               G  \\
[kite(8)r6c5==r6c8-r5c9=r3c9] = (8)r6c6 - r9c6 = r9c4                           (8)r6c5 => -8 r3c5; ste
             a                     b       c      d   \                        //   G   
                                                       (5)r9c4 = r4c4 - (5)r6c5
                                                           E      F         G

[SteveG48]

Code: Select all

 *------------------------------------------------------------*
 | 57    6     57    | b348   1     348   | 39    389   2     |
 | 4     3     8     |  9     7     2     | 6     1     5     |
 | 2     1     9     |  6    c38    5     | 347   378  b48    |
 *-------------------+--------------------+-------------------|
 | 8     579   1357  |  345   6     1347  | 2     379   19    |
 | 137   2     6     |be38    9     1378  | 5     4    b18    |
 | 1357  579   4     |  2   de358   1378  | 379   3789  6     |
 *-------------------+--------------------+-------------------|
 | 9     4     35    |  1   de35    6     | 8     2     7     |
 | 6     8     2     |  7     4     9     | 1     5     3     |
 | 1357  57    1357  | a8-35  2     38    | 49    6     49    |
 *------------------------------------------------------------*


8r9c4 = [Skyscraper 8c49] - (8=3)r3c5 - 3r67c5 = (5r7c5)&(8r6c5)&(3r5c4) => -35 r9c4 ; stte

[SpAce]

Code: Select all

.------------------.----------------------.--------------------.
| 57     6    57   |  348    1       348  | 39    389     2    |
| 4      3    8    |  9      7       2    | 6     1       5    |
| 2      1    9    |  6      3-8     5    | 347   378   dB4(8) |
:------------------+----------------------+--------------------:
| 8      579  1357 |  345    6       1347 | 2     379     19   |
| 137    2    6    | A38     9       1378 | 5     4     dB18   |
| 1357   579  4    |  2   caA35(8)  c1378 | 379  c3789    6    |
:------------------+----------------------+--------------------:
| 9      4    35   |  1     a35      6    | 8     2       7    |
| 6      8    2    |  7      4       9    | 1     5       3    |
| 1357   57   1357 |  358    2      b38   | 49    6       49   |
'------------------'----------------------'--------------------'

Kraken Cell (358)r6c5


(38)b5p84 - 8r(5=3)c9                                [A-B]
||
(5*3)r67c5 - (3=8)r9c6 - 8r6c(*56=8) - 8r(5=3)c9     [a-d]
||
(8)r6c5

=> -8 r3c5; stte

The same decrypted: Show

(3)r6c5 - (3=8)r5c4 - r5c9 = (8)r3c9
||
(5*)r6c5 - (5=3)r7c5 - (3=8)r9c6 - r6c*56 = r6c8 - r5c9 = (8)r3c9
||
(8)r6c5

Edit: corrected the PM thanks to blue and the discussion here

[blue]

Code: Select all

+-----------------+-------------------+------------------+
| 57    6    57   | a348  1     a348  | 39   389     2   |
| 4     3    8    | 9     7     2     | 6    1       5   |
| 2     1    9    | 6     a3-8  5     | 347  378     e48 |
+-----------------+-------------------+------------------+
| 8     579  1357 | 345   6     1347  | 2    379     19  |
| 137   2    6    | d38   9     1378  | 5    4       e18 |
| 1357  579  4    | 2     c358  c1378 | 379  c3789   6   |
+-----------------+-------------------+------------------+
| 9     4    35   | 1     35    6     | 8    2       7   |
| 6     8    2    | 7     4     9     | 1    5       3   |
| 1357  57   1357 | 358   2     b38   | 49   6       49  |
+-----------------+-------------------+------------------+

Code: Select all

Kraken Duo  => -8r3c5; stte

  a       b           c
3r1c6 - (3=8)r9c6 - 8r6c6
  ||                  ||
  ||                8r6c5 ---------------
  ||                  ||          e       \
  ||                8r6c8 - 8r5c9 = 8r3c9 - 8r3c5
  ||      d               /               /
3r1c4 - (3=8)r5c4 -------                /
  ||                                    /
3r3c5 ---------------------------------

(machine generated)

[eleven]

Code: Select all

 *-----------------------------------------------------------------*
 |  57     6     57     | c3+48  1     348    |  39    389    2    |
 |  4      3     8      |  9     7     2      |  6     1      5    |
 |  2      1     9      |  6    a38    5      |  347   378   #48   |
 |----------------------+---------------------+--------------------|
 |  8      579   1357   | c3+45  6     1347   |  2     379    19   |
 |  137    2     6      | b38    9     1378   |  5     4     #18   |
 |  1357   579   4      |  2    c3+58  1378   |  379   3789   6    |
 |----------------------+---------------------+--------------------|
 |  9      4     35     |  1     35    6      |  8     2      7    |
 |  6      8     2      |  7     4     9      |  1     5      3    |
 |  1357   57    1357   |  358   2     38     |  49    6      49   |
 *-----------------------------------------------------------------*

w-wing 38r3c5,r5c4, SL 8r3c59 -> 3r3c5=r5c4
xy-wing 84-45-58 r14c4,r6c5 -> -8r3c5
In one step: 3r3c5 =w-wing= 3r5c4 - 3r14c4,r6c5 = xy-wing => -8r3c5, stte

[Sudtyro2]

Code: Select all

+----------------+-------------------+-----------------+
| 57   6   57    | 348*  1     348#  | 39   389*   2   |
| 4    3   8     | 9     7     2     | 6    1      5   |
| 2    1   9     | 6     3-8*  5     | 347  378#   48  |
+----------------+-------------------+-----------------+
| 8    579 1357  | 345   6     1347  | 2    379    19  |
| 137  2   6     | 38    9     1378  | 5    4      18  |
| 1357 579 4     | 2     358*  1378# | 379  3789*  6   |
+----------------+-------------------+-----------------+
| 9    4   35    | 1     35    6     | 8    2      7   |
| 6    8   2     | 7     4     9     | 1    5      3   |
| 1357 57  1357  | 358   2     3-8   | 49   6      49  |
+----------------+-------------------+-----------------+

In 8s, a simple 5-link oddagon(*) having three guardians(#).
Guardians 8r1c6,r3c8 directly see 8r3c5, while guardian 8r6c6 needs a network to remotely see 8r3c5. But, guardian 8r6c6 does directly see 8r9c6. Since at least one guardian must be true, let's write the following kraken column:

Code: Select all

8r1c6,r3c8 => -8r3c5; stte
  ||
8r6c6      => -8r9c6; stte


We've avoided the long network solution, but is the logic valid? My old brain wants to say yes, but SpaCe pointed out for a similar situation here that this is probably not the case. There does exist a derived weak inference between the two stte digits, so at least one must be false, but we have not shown a "common outcome" that would definitively eliminate one of the two digits. Do the Forum rules for stte eliminations specifically require that common outcome? Comments would be most welcome!

SteveC

[Cenoman]

My old brain wants to say yes, but SpaCe pointed out for a similar situation here that this is probably not the case. There does exist a derived weak inference between the two stte digits, so at least one must be false, but we have not shown a "common outcome" that would definitely eliminate one of the two digits. Do the Forum rules for stte eliminations specifically require that common outcome? Comments would be most welcome!

Hi Steve,
I noticed SpAce's doubt at the time and thought I had to dig it. Your question is a good opportunity. The situation is different between your link and the above.

First a side remark: what you need between the two ste digits isn't a weak link "digit1 - digit2" but an inference link "digit1 => digit2", such that if digit1 is True, set1 of guardians is false, digit2 is true, set2 of guardians is false; all guardians are false (contradiction). On September 18 puzzle, such an inference link was available trough a kite. But in the above puzzle, it is not immediate (there is one through a kraken, as you mention). My suggestion of short-writing on September 18 was made with the hypothesis of implicit existence of the link.

So SpAce was right, the logic needs to explicit some inference link between "the two stte digits". The common outcome must be: "anyhow, all guardians are false".

For example, the above puzzle would be solved this way:

Code: Select all

In 8s, 5-link oddagon(*) having three guardians(#):

Kraken box (3)b8p279
(3)r9c6 - (8)r9c6
(3-5)r7c5 = r6c5  - (8)r6c5 = (8)r3c5     demonstrating " 8r9c6 => 8r3c5 " or equivalent " -8r3c5 => -8r9c6 "
(3)r9c4 - (3=8)r5c4 /
Therefore:
8r1c6,r3c8 => -8r3c5 => -8r9c6 }
  ||                           } -8 r9c6; ste
8r6c6      => -8r9c6           }


If you find me a disappointing guy, I will not protest.

[SpAce]

In 8s, a simple 5-link oddagon(*) having three guardians(#).
Guardians 8r1c6,r3c8 directly see 8r3c5, while guardian 8r6c6 needs a network to remotely see 8r3c5. But, guardian 8r6c6 does directly see 8r9c6. Since at least one guardian must be true, let's write the following kraken column:

Code: Select all

8r1c6,r3c8 => -8r3c5; stte
  ||
8r6c6      => -8r9c6; stte


We've avoided the long network solution, but is the logic valid? My old brain wants to say yes, but SpaCe pointed out for a similar situation here that this is probably not the case. There does exist a derived weak inference between the two stte digits, so at least one must be false, but we have not shown a "common outcome" that would definitely eliminate one of the two digits. Do the Forum rules for stte eliminations specifically require that common outcome? Comments would be most welcome!

I would welcome others' comments too, because I already said my piece earlier. I haven't changed my mind, either. In my mind there are two valid ways to achieve a definite conclusion with any kraken-like logic (by "kraken" I mean any SIS, such as the set of oddagon guardians): 1) either all members of the SIS reach a consensus about some conclusion, or 2) some do and the rest self-destruct with a contradiction (but that variant doesn't seem as elegant). In both of those cases we're left with a single conclusion that must be true and we can confidently take that. If it also happens to be an stte-solution, then great, we're done.

However, if we reach two or more different conclusions (none of which is a contradiction), as in your case, it doesn't prove anything directly useful. All you know is that at least one of those conclusions must be true but that's not any different from our starting point (at least one of the SIS members must be true, but we don't know which one(s) -- that's why we try to find a point where they all agree). As I said before, it shouldn't play any part if you happen to know that all of those conclusions are valid stte-eliminations, because that's not something that the proof tells you or what a purely manual solver would be able to know. You should actually run (and notate) your chains all the way to the end to prove that they all do lead to the solution, but that wouldn't be a very satisfactory proof. On the other hand, if you use that prior knowledge without proving it, then it's a case of circular reasoning, as I said before.

Therefore, I still think that we can't accept as a solution two or more different stte-conclusions that have an OR relationship, no matter how elegant it might look, because it depends on knowing beforehand that they in fact are all stte-conclusions. That's not something a manual solver would know without running trials for all paths to the end.

(Oh, I didn't notice Cenoman had already replied.)

[Cenoman]

Hi eleven, blue, hi all,

eleven's solution is the nicest for a single line writing.
The almost-almost-almost xy-wing is a great, great finding !

To be sure I caught it, I have translated it into a matrix:

Hidden Text: Show

Code: Select all

3r3c5 8r3c5
      8r3c9 8r5c9
            8r5c4 3r5c4
8r1c4             3r1c4 4r1c4
                  3r4c4 4r4c4 5r4c4
8r6c5             3r6c5       5r6c5
=> -8 r3c5

I also put blue's solution into a matrix:

Hidden Text: Show

Code: Select all

8r3c9 8r5c9
8r6c5 8r6c8 8r6c6
            8r9c6 3r9c6
      8r5c4             3r5c4
3r3c5             3r1c6 3r1c4
=> -8 r3c5

blue's solution has the smallest matrix, i.e. uses the least native strong links. A challenge that JC Van Hay was fond of. So, kudos for blue !

For comparison, here are matrices of SpAce's, Steve's and my own solutions:

Hidden Text: Show

Code: Select all

 SpAce's
8r3c9 8r5c9
8r6c5 8r6c8 8r6c6
            8r9c6 3r9c6
                  3r7c5 5r7c5
      8r5c4                   3r5c4
8r6c5                   5r6c5 3r6c5
=> -8 r3c5

Code: Select all

 Steve's
3r5c4 8r5c4           not a triangular matrix but juxtaposition of two smaller triangular matrices
5r7c5        3r7c5                |                                |      3r5c4 8r5c4
      8r6c5 35r67c5               |    5r7c5 3r7c5                 |            8r6c5 35r67c5
             3r3c5   8r3c5        |          3r3c5   8r3c5         |                   3r3c5   8r3c5
                     8r3c9 8r5c9  |                  8r3c9 8r5c9   |                           8r3c9 8r5c9
8r9c4                8r1c4 8r5c4  |    8r9c4         8r1c4 8r5c4   |      8r9c4                8r1c4 8r5c4
=> -35 r9c4                       |    => -5 r9c4                  |      => -3 r9c4

Code: Select all

 Cenoman's
8r3c9 8r5c9
8r6c5 8r6c8 8r6c6
            8r9c6 8r9c4
                  8r5c4 3r5c4
                  5r9c4       5r4c4
8r6c5                   3r6c5 5r6c5
=> -8 r3c5

[Sudtyro2]

Special thanks to Cenoman and SpaCe for their generous and positive feedback. As always, it's very much appreciated!

SteveC

[SpAce]

Code: Select all

 *------------------------------------------------------------*
 | 57    6     57    | b348   1     348   | 39    389   2     |
 | 4     3     8     |  9     7     2     | 6     1     5     |
 | 2     1     9     |  6    c38    5     | 347   378  b48    |
 *-------------------+--------------------+-------------------|
 | 8     579   1357  |  345   6     1347  | 2     379   19    |
 | 137   2     6     |be38    9     1378  | 5     4    b18    |
 | 1357  579   4     |  2   de358   1378  | 379   3789  6     |
 *-------------------+--------------------+-------------------|
 | 9     4     35    |  1   de35    6     | 8     2     7     |
 | 6     8     2     |  7     4     9     | 1     5     3     |
 | 1357  57    1357  | a8-35  2     38    | 49    6     49    |
 *------------------------------------------------------------*


8r9c4 = [Skyscraper 8c49] - (8=3)r3c5 - 3r67c5 = (5r7c5)&(8r6c5)&(3r5c4) => -35 r9c4 ; stte

Hi Steve! Related to the recent matrix discussion here I did some practicing with this interesting puzzle and the various solutions (partly because Cenoman had so nicely provided matrices to compare with). I like yours very much, especially since it's the most difficult one to turn into a matrix, but I can't quite agree with the last combo node of the chain. I understand that it's just taking a shortcut, but being a nitpicker, I wanted to try some options:

1. AB:(8)r9c4 = [Skyscraper] - (8=3)r3c5 - A:(3=58)r76c5 - B:(8=3)r5c4 => -35 r9c4
2. (8)r9c4 = [Skyscraper] - (8=35)r37c5 - (3|5=8)r6c5 - (8)r5c4|(5)r6c5 = (35)r5c4,r7c5 => -35 r9c4
3. (8)r9c4 = [Skyscraper] - r3c5 = r6c5 - (8)r5c4|(5)r6c5 = (35)r5c4,r7c5 => -35 r9c4
4. (8)r9c4 = [Skyscraper] - r3c5 = (83)b5p84&(85)r67c5 => -35 r9c4

Needless to say I like the last one best (and the first multi-header the least). What do you think? (It can also be combined nicely with eleven's solution, but that's behind the above link.)

Cenoman, I ran into the same problems as you when trying to convert that into a matrix. I guess your double-matrix (is it a case of BTM?) is the cleanest way to do it. The only alternative I can think of is something like this:

Code: Select all

4x4 TM:

       |  3c4,5b8,9n4  8b5,6n5      8r3,8b2  8r5
-------+------------------------------------------
5N4,5C5|  35r5c4,r7c5  8r5c4|5r6c5
8C5    |               8r6c5        8r3c5
8C9    |                            8r3c9    8r5c9
8C4    |  8r9c4                     8r1c4    8r5c4
-------+------------------------------------------
       | -35r9c4

Alien 5x8-Fish (Mixed Rank 3/1): {8C459 5C5 5N4} \ {8r35 3c4 8b25 5b8 6n5 9n4} => -35 r9c4 (Rank 1)

(Don't ask me to explain the rank regions -- it has so many triplets that I get confused.)


Would that work? The difficulty seems to arise from the fact that we have two link triplets (8r3c5 and 8r6c5) working in opposite directions (pointing at each other). My solution is to represent one of them as an OR-node. It's more obvious if we look at it as a net:

Code: Select all

(8)r1c4 -------------                     (5)r6c5 = r7c5 - (5)r9c4
||                    \                 /                   ||
(8)r9c4                 - r3c5 = r6c5 -                    (8)r9c4
||                    /                 \                   ||
(8)r5c4 - r5c9 = r3c9                     (8=3)r5c4 ------ (3)r9c4

=> +8r9c4 (because I closed the (discontinuous) loop for symmetry; obviously not necessary)


(Interestingly, 8r5c4 is both a set and a link triplet and also contradicts itself.)

[Cenoman]

Only minor comments:
First, an opinion on a point I am not asked to comment: from the four options tried by SpAce, the one I like best is #3. (my reason: it is the closest of the net and the matrix)
Second,

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4x4 TM:

       |  3c4,5b8,9n4  8b5,6n5      8r3,8b2  8r5
-------+------------------------------------------
5N4,5C5|  35r5c4,r7c5  8r5c4|5r6c5
8C5    |               8r6c5        8r3c5
8C9    |                            8r3c9    8r5c9
8C4    |  8r9c4                     8r1c4    8r5c4
-------+------------------------------------------
       | -35r9c4

Would that work?

Yes. To me, this TM is OK.
Not surprising, I'd rather write the chain (8)r9c4 = [Skyscraper] - r3c5 = r6c5 - (8)r5c4|(5)r6c5 = (3)r5c4&(5)r7c5 => -35 r9c4
and the TM

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       |  3c4,5b8,9n4    8b5,6n5    8r3,8b2  8r5
-------+------------------------------------------
5N4,5C5|  3r5c4&5r7c5  8r5c4|5r6c5
8C5    |                  8r6c5     8r3c5
8C9    |                            8r3c9    8r5c9
8C4    |     8r9c4                  8r1c4    8r5c4
-------+------------------------------------------
       |   -35r9c4


but this comment is nothing but my taste...

Third,

I guess your double-matrix (is it a case of BTM?) is the cleanest way to do it

The combined matrix is not a TM, nor a BTM because it is 6x5, not a square matrix (which is the first requirement for all types in Steve K's paper)

[SpAce]

Only minor comments:

Once again, thank you for them! I only disagree with... well, nothing!

First, an opinion on a point I am not asked to comment: from the four options tried by SpAce, the one I like best is #3. (my reason: it is the closest of the net and the matrix)

Yes, but you know my obsession with shortening things ad absurdum Here's one more variant that I think might be pretty close to Steve's original intention:

(8)r9c4 = [Skyscraper] - (8=3)r3c5 - (35=8)r76c5 - (8)r5c4|(83)r37c5 = (35)r5c4,r7c5 => -35 r9c4

...which I guess turns into a matrix using the same trick as before (since you approved it):

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Code: Select all

5x5 TM:

        | 3c4,5b8,9n4    8b5,38c5     35c5    8r3,8b2   8r5   
--------+----------------------------------------------------
5N4,37N5| 3r5c4&5r7c5  8r5c4|38r37c5
67N5    |                 8r6c5       35r67c5
3N5     |                             3r3c5    8r3c5
8C9     |                                      8r3c9    8r5c9
8C4     |    8r9c4                             8r1c4    8r5c4
--------+----------------------------------------------------
        |  -35r9c4

Alien 6x10-Fish (Mixed Rank 4/1) {8C49 367N5 5N4} \ {8r35 3c4 358c5 8b25 5b8 9n4} => -35 r9c4 (Rank 1)

If I counted right it has an impressive rank of 4, which makes it a pretty interesting specimen. Maybe one of these days I'll figure out (or someone tells me) how those triplet rules work here (I'm just assuming the eliminations must be Rank 1). Interestingly, 8r3c5 is being linked in three different ways (r3,c5,b2).

Not surprising, I'd rather write the chain (8)r9c4 = [Skyscraper] - r3c5 = r6c5 - (8)r5c4|(5)r6c5 = (3)r5c4&(5)r7c5 => -35 r9c4

I have no problem with that. It does make it more obvious that the two cells at the end aren't connected and where the digits lie. The same information can be inferred from my style but it's not as explicit.

and the TM

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       |  3c4,5b8,9n4    8b5,6n5    8r3,8b2  8r5
-------+------------------------------------------
5N4,5C5|  3r5c4&5r7c5  8r5c4|5r6c5
8C5    |                  8r6c5     8r3c5
8C9    |                            8r3c9    8r5c9
8C4    |     8r9c4                  8r1c4    8r5c4
-------+------------------------------------------
       |   -35r9c4


I'm glad about that! In fact, I wrote it exactly the same way at first. I don't know why I changed it. I definitely prefer this formatting.

Third,

I guess your double-matrix (is it a case of BTM?) is the cleanest way to do it

The combined matrix is not a TM, nor a BTM because it is 6x5, not a square matrix (which is the first requirement for all types in Steve K's paper)

Yes, of course. I think I counted the first two lines as one : 3r5c4&5r7c5 8r5c4 3r7c5. I guess that still wouldn't make it a BTM, though its exact definition still eludes me.

[Cenoman]

I counted the first two lines as one : 3r5c4&5r7c5 8r5c4 3r7c5.
I guess that still wouldn't make it a BTM, though its exact definition still eludes me.

Oups ! In my focusing on native strong links, I forgot to explore that one !
Edit: after reading SpAce further comments, the matrix cannot be identified as a BTM
Applying the process that you have explored in the other thread (here), I guess this matrix can be identified as a BTM I first thought this matrix could be identified as a BTM:

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3r5c4&5r7c5 8r5c4  3r7c5               
            8r6c5 35r67c5               
                   3r3c5   8r3c5       
                           8r3c9 8r5c9 
   8r9c4                   8r1c4 8r5c4 
=> -35 r9c4

TM a
   3r5c4    8r5c4                 
            8r6c5 35r67c5               
                   3r3c5   8r3c5       
                           8r3c9 8r5c9 
   8r9c4                   8r1c4 8r5c4
=> -3 r9c4

TM b
   5r7c5     3r7c5               
             3r3c5   8r3c5       
                     8r3c9 8r5c9 
   8r9c4             8r1c4 8r5c4
=> -5r9c4

Of course, I am not sure whether the first term in the first row can be split that way. It seems to make sense with the subsequent OR of the two matrices, which maintains an "OR" operator between 8r5c4 and 3r7c5 and thus requires the AND operator to link 3r5c4 and 5r7c5.
I now think the first term in the first row cannot be split that way. It does not make sense with a subsequent OR of the two matrices, as there is no native (or derived) "OR" operator between 8r5c4 and 3r7c5. The AND operator to link 3r5c4 and 5r7c5 is required to demonstrate both eliminations -3r9c4 & -5r9c4 and would result of such an OR between 8r5c4 and 3r7c5, which unfortunately is not present.

Note another way to escape:
Following Steve's chain closely:

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   8r9c4    SS 8c49
             8r3c5   3r3c5
                    35r67c5 8r6c5
3r5c4&5r7c5          3r7c5  8r6c4
=> -35r9c4

I guess that including a pattern such as a wing, a fish or other, is valid. This is similar to what can be done with ALS's