The New Sudoku Players' Forum

[daj95376]

_

I see blue's solution as a 2x Kraken Cell on r7c83. He just partitioned the combinations instead of treating the candidates serially.

What caught my attention was his "Skyscraper". Viewed another way, it's an oddagon with guardian cells r1c1 and r6c4.

Here's where it gets interesting ... and uses blue's partitioning.

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 +--------------------------------------------------------------+
 | g189   3     189   |  5     69    4     |  26    7     268   |
 |  6     5     4     |  7     8     2     |  3     19    19    |
 | *89    2     7     |  3     1    *69    |  4     68    5     |
 |--------------------+--------------------+--------------------|
 |  7     8     3     |  1     4    *69    |  269   5     269   |
 | *29    1     6     |  8    *29    5     |  7     3     4     |
 |  5     4     29    | g269   3     7     |  8     169   169   |
 |--------------------+--------------------+--------------------|
 |  1248  679   128   |  2469  269   3     |  5     2689  6789  |
 |  3     679   28    |  269   5     18    |  169   4     6789  |
 |  248   69    5     |  2469  7     18    |  169   2689  3     |
 +--------------------------------------------------------------+
 # 68 eliminations remain

  9   r6c4 - 9r5c5
       ||
 (9-1)r1c1 = r1c3 - 1r7c3 = 28r7c38 - 2r7c5 = (2-9)r5c5
                            ||
                          = 169r267c8 - 6r3c8 = r3c6 - (6=9)r4c6 - 9r5c5


_

[DonM]

This is correct, technically, but it leaves it to the reader wondering about the "missing 8's", to divine that if none of the items in the '||' column were true, there would be a conflict with an 8 in each cell.

It is shown that 4 digits (1269) of the 5 total cells in 2 cells will result in r5c5=9. The 8s don't really matter. One of the two cells will have to contain 1,2,6, or 9 so we know that r5c5=9 no matter what. Classic AAALS.

[JC Van Hay]

What would this be called ?
AAALS or AAANS ?
Kraken something ?

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28r7c38 - 2r7c5 = 2r5c5 ------------------------
  ||                                             \
1r7c3 - r1c3 = (1-9)r1c1 = [Skyscraper: 9c16\r3] - 9r5c5    => r5c5<>9; stte
  ||                                             /
69r7c8 -(NT 169r267c8)- 6r3c8 = r1c79 - (6=9)r1c5

Replacing 1N5 + 6B3 by 6R3, one has the choice !

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+-------------------+-------------------+-------------------+
| 8(19)  3    89(1) | 5     69     4    | 26   7       268  |
| 6      5    4     | 7     8      2    | 3    (19)    19   |
| 8(9)   2    7     | 3     1      (69) | 4    8(6)    5    |
+-------------------+-------------------+-------------------+
| 7      8    3     | 1     4      6(9) | 269  5       269  |
| 2(9)   1    6     | 8     -9(2)  5    | 7    3       4    |
| 5      4    29    | 269   3      7    | 8    (169)   169  |
+-------------------+-------------------+-------------------+
| 1248   679  (128) | 2469  69(2)  3    | 5    (2689)  6789 |
| 3      679  28    | 269   5      18   | 169  4       6789 |
| 248    69   5     | 2469  7      18   | 169  2689    3    |
+-------------------+-------------------+-------------------+


#1. Kraken 2r57c5 + (128)r7c3 :

2r5c5
||
12r7c35-1r1c3=(1-9)r1c1=Skyscraper(9C16)
||
82r7c35-(82=169)r267c8-6r3c8=(6-9)r3c6=9r4c6

or

1r7c3=*[2r5c5=*NP(28)r7c35-(28=169)r267c8-6r3c8=(6-9)r3c6=9r4c6]
|
1r1c3=(1-9)r1c1=Skyscraper(9C16)

:=> [2r5c5==9r5c1==9r4c6]-(9=2)r5c5; that is : whatever the solution of 2r57c5 + (128)r7c3, r5c5<>9.

#2. Kraken AANT(169 28)r2678 + (128)r7c3 :

NT(169)r267c8-6r3c8=(6-9)r3c6=9r4c6
||
2r7c8-2r7c5=2r5c5
||
8r7c8-8r7c3=*[2r5c5=2r7c5-(2=*1)r7c3-1r1c3=(1-9)r1c1=Skyscraper(9C16)]

or

8r7c3=*[2r5c5=2r7c5-(2=*1)r7c3-1r1c3=(1-9)r1c1=Skyscraper(9C16)]
|
8r7c8=*[2r5c5=2r7c5-2r7c8=*NT(169)r267c8-6r3c8=(6-9)r3c6=9r4c6]


:=> [2r5c5==9r5c1==9r4c6]-(9=2)r5c5; that is : whatever the solution of AANT(169 28)r2678 + (128)r7c3, r5c5<>9.

Note : these representations are easily derived from the following exclusion matrix :

Hidden Text: Show

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9r5c5
9r4c6=9r3c6        *
9r5c1=9r3c1=9r1c1  * ASS(9C16)
            1r1c1=1r1c3
      6r3c6=============6r3c8
                        6r6c8=1r6c8=9r6c8              *
                              1r2c8=9r2c8              *
                        6r7c8=======9r7c8=8r7c8=2r7c8  * AANT(619 82)r267c8
                  1r7c3===================8r7c3=2r7c3
2r5c5===========================================2r7c5

[DonM]

What I like about Blue's AAALS version is that it avoids using a Kraken-anything for a change.

Not having to do with the above: I still continue to really dislike the term 'Kraken' which describes nothing. These 'Krakens' are in-the-end, almost-AICs (AAIC). Steve K was the first one to use the term AAIC during the period when he was the first one to graphically present the pattern in manually-derived solutions. Being the nice guy he was, he subsequently changed to using 'Kraken' in deference to Mike Barker who first used the term in his computer-derived versions of the same pattern. I wish Steve had stuck with AAIC.

I've always used AAIC and for awhile several years ago, I had some company, but I appear to be the only standout now...

[denis_berthier]

It is interesting to note that (as remarked by JC Van Hay) this puzzle can be solved by a very easy sequence of the most elementary rules (not involving uniqueness), so that the extremely complicated pseudo one-step solutions proposed above (and in many threads of this section of the forum) may be totally misleading to a newcomer unaware that the goal here is precisely not to find a natural solution but to find a pseudo one-step one.

I say "pseudo one-step" because a real one-step solution would involve only natural, indecomposable patterns (instead of complicated nets, AHAHAH-ICs, OHOHOH-ICs, embedded URs and the like). Of course, only very few puzzles allow a real one-step solution.

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Starting from the original puzzle: .3.5.4.7.6.4782....2..1...578.1......16..5..454..3.8.......35..3......4...5.7...3
(For simplicity, I don't write the details for singles and whips[1])

singles
whips[1] (= claiming-pointing / row-block and column-block interactions )
naked-pairs-in-a-column: c6{r3 r4}{n6 n9} ==> r9c6 ≠ 9, r9c6 ≠ 6, r8c6 ≠ 9, r8c6 ≠ 6
x-wing-in-rows: n2{r6 r8}{c3 c4} ==> r9c4 ≠ 2, r7c4 ≠ 2, r7c3 ≠ 2
biv-chain[2]: b2n9{r3c6 r1c5} - r5n9{c5 c1} ==> r3c1 ≠ 9
singles
biv-chain[3]: r9n2{c1 c8} - r9n8{c8 c6} - r8n8{c6 c3} ==> r8c3 ≠ 2
singles to the end


However, the point I'd like to stress here is different.
It is about how using combinations of simple patterns in order to make the resolution path look shorter (here, one-step) quickly leads to overly complicated and artificial patterns.
Some time ago, there were discussions about defining a rating based on the complete path instead of the standard approaches (based on the hardest step). The above example illustrates how this may be hard, even for the simplest puzzles.