The New Sudoku Players' Forum

[Cenoman]

Weekly "extreme" puzzle (7.2), proposed by the site sudoku.org.uk

Code: Select all

 +---+---+---+
 |..3|...|.8.|
 |6..|..7|...|
 |2..|1.4|...|
 +---+---+---+
 |...|46.|.7.|
 |.2.|...|.9.|
 |.1.|.73|...|
 +---+---+---+
 |8.1|2.6|..3|
 |...|7..|..2|
 |.4.|.5.|8..|
 +---+---+---+
..3....8.6....7...2..1.4......46..7..2.....9..1..73...8.12.6..3...7....2.4..5.8..

[pjb]

I needed 3 steps, must be something better out there.

1. Kraken cell
chain1: (4)r1c9 - r5c9
chain2: (5)r1c9 - (5=9)r1c6 - (59=7)r1c2 - r3c3 = r5c3 - (7=4)r5c1 - r5c9
chain3: (7)r1c9 - r1c7 = r7c7 - r7c2 = r3c2 - r3c3 = (7-6)r5c3 = (6-4)r5c9
chain4: (9)r1c9 - (9=5)r1c6 - (59=7)r1c2 - r3c3 = r5c3 - (7=4)r5c1 - r5c9 => -4 r5c9

2. (5=9)r3c3 - (9)r23c2 = (9)r7c2 - (9=4)r7c5 - (4=5)r7c8 => -5 r3c8

3. (8=9)r2c2 - (9=5)r7c2 - (5=4)r7c8 - (4=9)r7c5 - (9=8)r3c5 => -8 r2c4, r3c2 => stte

Phil

[jco]

Hello,

I found a solution with 4 steps. I tried to find a UR(18) move
instead of steps 3 and 4, but did not manage to make it work.

After basics

Code: Select all

.------------------------------------------------------.
|  1  B579   3   | 6   29    59  | c24579  8     b4579 |
|  6   589   4   | 58  2389  7   |  1259   1235   159  |
|  2  B5789 A579 | 1   389   4   |  569-7  356    5679 |
|----------------+---------------+---------------------|
|  59  3     89  | 4   6     2   |  15     7      158  |
| g47  2     67  | 58  18    158 |  3      9     a6-4  |
|  45  1     68  | 9   7     3   |  2456   2456   4568 |
|----------------+---------------+---------------------|
|  8 eC579   1   | 2   49    6   |dD4579   45     3    |
|  3   6     59  | 7   1489  189 |  1459   145    2    |
| f79  4     2   | 3   5     19  |  8      16     1679 |
'------------------------------------------------------'

1. ER (7)r3c3=r13c2-r7c2=r7c7 => -7 r3c7
2. (6=4)r5c9-r1c9=(4-7)r1c7=r7c7-r7c2=r9c1-(7=4)r5c1 => -4 r5c9 (& 15 placements)

Code: Select all

.-----------------------------------------------------.
| 1   7    3  | 6    29    c59  | d259     8      4   |
| 6   589  4  | 58   2389   7   |  1259    123-5  159 |
| 2   589  59 | 1    389    4   |  6       3-5    7   |
|-------------+-----------------+---------------------|
| 9   3    8  | 4    6      2   |  15      7      15  |
| 4   2    7  | 58   18     158 |  3       9      6   |
| 5   1    6  | 9    7      3   |  24      24     8   |
|-------------+-----------------+---------------------|
| 8   59   1  | 2   a49     6   |  7      a45     3   |
| 3   6    59 | 7    1489  b189 |  149-5   145    2   |
| 7   4    2  | 3    5     b19  |  8       6      19  |
'-----------------------------------------------------'

3. (5=49)r7c58-(9)r89c6=(9-5)r1c6=(5)r1c7 => -5 r23c8 (3 placements & -5r2c2 [LC(5)r3c23])

Code: Select all

.--------------------------------------------------.
| 1   7     3  | 6    2     f59   | 59    8    4   |
| 6   89    4  | 58   3      7    | 1259  12   159 |
| 2   589  d59 | 1   e8-9    4    | 6     3    7   |
|--------------+------------------+----------------|
| 9   3     8  | 4    6      2    | 15    7    15  |
| 4   2     7  | 58   18     158  | 3     9    6   |
| 5   1     6  | 9    7      3    | 24    24   8   |
|--------------+------------------+----------------|
| 8  b59    1  | 2   a49     6    | 7     45   3   |
| 3   6    c59 | 7    1489   18-9 | 149   145  2   |
| 7   4     2  | 3    5      1-9  | 8     6    19  |
'--------------------------------------------------'

4. (9)r7c5=r7c2-r8c3=r3c3-r3c5=(9)r1c6 => -9 r3c5, -9 r89c6; ste

Regards,
jco

[DEFISE]

2 steps after basics:

11 Singles
Intersec row/bloc : 1r4b6 => -1r5c9
Intersec col/bloc : 5c1b4 => -5r4c3 -5r5c3 -5r6c3
Intersec col/bloc : 8c2b1 => -8r3c3
Intersec row/bloc: 5b5r5 => -5r5c1 -5r5c9
Intersec row/bloc : 8b5r5 => -8r5c3 -8r5c9

whip[5]: r9n7{c1 c9}- r7n7{c7 c2}- r1n7{c2 c7}- r1n4{c7 c9}- r5n4{c9 .} => -7r5c1
=> 15 singles
whip[6]: r7n9{c2 c5}- b2n9{r1c5 r1c6}- r1n5{c6 c7}- r4c7{n5 n1}- r8c7{n1 n4}- r7n4{c8 .} => -9r8c3
=> STTE

[Cenoman]

I had also a two step solution:

Code: Select all

 +--------------------+--------------------+------------------------+
 |  1   c579    3     |  6    29     59    | d24579   8     d4579   |
 |  6    589    4     |  58   2389   7     |  1259    1235   159    |
 |  2    5789   579   |  1    389    4     |  5679    356    5679   |
 +--------------------+--------------------+------------------------+
 |  59   3      89    |  4    6      2     |  15      7      158    |
 | f4-7  2      67    |  58   18     158   |  3       9     e46     |
 |  45   1      68    |  9    7      3     |  2456    2456   4568   |
 +--------------------+--------------------+------------------------+
 |  8   b579    1     |  2    49     6     |  4579    45     3      |
 |  3    6      59    |  7    1489   189   |  1459    145    2      |
 |fa79   4      2     |  3    5      19    |  8       16    e1679   |
 +--------------------+--------------------+------------------------+

1. (7)r9c1 = r7c2 - r1c2 = (74)r1c79 - (7r9c9 & 4r5c9) = (7r9c1 | 4r5c1) => -7 r5c1; 15 placements & lcls

Code: Select all

 +------------------+--------------------+----------------------+
 |  1    7     3    |  6    29     59    |  259    8      4     |
 |  6    589   4    |  58   2389   7     |  1259   1235   159   |
 |  2    589   59*  |  1    38-9   4     |  6      3-5    7     |
 +------------------+--------------------+----------------------+
 |  9    3     8    |  4    6      2     |  15     7      15    |
 |  4    2     7    |  58   18     158   |  3      9      6     |
 |  5    1     6    |  9    7      3     |  24     24     8     |
 +------------------+--------------------+----------------------+
 |  8    59*   1    |  2    49*    6     |  7      45*    3     |
 |  3    6     59*  |  7    1489   189   |  1459   145    2     |
 |  7    4     2    |  3    5      19    |  8      6      19    |
 +------------------+--------------------+----------------------+

2. Remote pair (59)r3c3\r7c58 (through r8c3, r7c2) => -5r3c8, -9r3c5; ste

[denis_berthier]

.
Solution using only Subsets and the most elementary bivalue-chains (length ≤ 4).

Resolution state after 11 singles and 5 whips[1]:

Code: Select all

   +-------------------+-------------------+-------------------+
   ! 1     579   3     ! 6     29    59    ! 24579 8     4579  !
   ! 6     589   4     ! 58    2389  7     ! 1259  1235  159   !
   ! 2     5789  579   ! 1     389   4     ! 5679  356   5679  !
   +-------------------+-------------------+-------------------+
   ! 59    3     89    ! 4     6     2     ! 15    7     158   !
   ! 47    2     67    ! 58    18    158   ! 3     9     46    !
   ! 45    1     68    ! 9     7     3     ! 2456  2456  4568  !
   +-------------------+-------------------+-------------------+
   ! 8     579   1     ! 2     49    6     ! 4579  45    3     !
   ! 3     6     59    ! 7     1489  189   ! 1459  145   2     !
   ! 79    4     2     ! 3     5     19    ! 8     16    1679  !
   +-------------------+-------------------+-------------------+

finned-x-wing-in-rows: n7{r7 r1}{c2 c7} ==> r3c7 ≠ 7
biv-chain[3]: r5c9{n4 n6} - b4n6{r5c3 r6c3} - r6n8{c3 c9} ==> r6c9 ≠ 4
biv-chain[3]: c9n4{r1 r5} - r5c1{n4 n7} - r9n7{c1 c9} ==> r1c9 ≠ 7
x-wing-in-rows: n7{r1 r7}{c2 c7} ==> r3c2 ≠ 7 ;;; now that the fin has been eliminated, we have a simple x-wing where we had previously a finned one
biv-chain[3]: c9n7{r9 r3} - c3n7{r3 r5} - r5n6{c3 c9} ==> r9c9 ≠ 6
hidden-single-in-a-block ==> r9c8 = 6
biv-chain[3]: b3n7{r3c9 r1c7} - r1n4{c7 c9} - r5c9{n4 n6} ==> r3c9 ≠ 6
hidden-single-in-a-block ==> r3c7 = 6
hidden-pairs-in-a-row: r6{n6 n8}{c3 c9} ==> r6c9 ≠ 5
biv-chain[4]: r5c1{n4 n7} - b7n7{r9c1 r7c2} - c7n7{r7 r1} - b3n4{r1c7 r1c9} ==> r5c9 ≠ 4
singles ==> r5c9 = 6, r5c3 = 7, r5c1 = 4, r6c1 = 5, r4c1 = 9, r4c3 = 8, r6c3 = 6, r9c1 = 7, r6c9 = 8, r1c2 = 7, r3c9 = 7, r7c7 = 7, r1c9 = 4
finned-x-wing-in-rows: n5{r7 r3}{c8 c2} ==> r2c2 ≠ 5
whip[1]: b1n5{r3c3 .} ==> r3c8 ≠ 5
singles ==> r3c8 = 3, r2c5 = 3, r1c5 = 2
finned-x-wing-in-columns: n9{c3 c5}{r3 r8} ==> r8c6 ≠ 9
finned-x-wing-in-columns: n9{c6 c9}{r9 r1} ==> r1c7 ≠ 9
stte

[jco]

Hi,

I had also a two step solution:

Code: Select all

 +--------------------+--------------------+------------------------+
 |  1   c579    3     |  6    29     59    | d24579   8     d4579   |
 |  6    589    4     |  58   2389   7     |  1259    1235   159    |
 |  2    5789   579   |  1    389    4     |  5679    356    5679   |
 +--------------------+--------------------+------------------------+
 |  59   3      89    |  4    6      2     |  15      7      158    |
 | f4-7  2      67    |  58   18     158   |  3       9     e46     |
 |  45   1      68    |  9    7      3     |  2456    2456   4568   |
 +--------------------+--------------------+------------------------+
 |  8   b579    1     |  2    49     6     |  4579    45     3      |
 |  3    6      59    |  7    1489   189   |  1459    145    2      |
 |fa79   4      2     |  3    5      19    |  8       16    e1679   |
 +--------------------+--------------------+------------------------+

1. (7)r9c1 = r7c2 - r1c2 = (74)r1c79 - (7r9c9 & 4r5c9) = (7r9c1 | 4r5c1) => -7 r5c1; 15 placements & lcls

Code: Select all

 +------------------+--------------------+----------------------+
 |  1    7     3    |  6    29     59    |  259    8      4     |
 |  6    589   4    |  58   2389   7     |  1259   1235   159   |
 |  2    589   59*  |  1    38-9   4     |  6      3-5    7     |
 +------------------+--------------------+----------------------+
 |  9    3     8    |  4    6      2     |  15     7      15    |
 |  4    2     7    |  58   18     158   |  3      9      6     |
 |  5    1     6    |  9    7      3     |  24     24     8     |
 +------------------+--------------------+----------------------+
 |  8    59*   1    |  2    49*    6     |  7      45*    3     |
 |  3    6     59*  |  7    1489   189   |  1459   145    2     |
 |  7    4     2    |  3    5      19    |  8      6      19    |
 +------------------+--------------------+----------------------+

2. Remote pair (59)r3c3\r7c58 (through r8c3, r7c2) => -5r3c8, -9r3c5; ste

Beautiful solution to a very interesting puzzle!
I had difficulties with the large number of possible chain eliminations and the fact that many of them were ineffective!

Regards,
jco

[AnotherLife]

Code: Select all

 +--------------------+--------------------+------------------------+
 |  1   c579    3     |  6    29     59    | d24579   8     d4579   |
 |  6    589    4     |  58   2389   7     |  1259    1235   159    |
 |  2    5789   579   |  1    389    4     |  5679    356    5679   |
 +--------------------+--------------------+------------------------+
 |  59   3      89    |  4    6      2     |  15      7      158    |
 | f4-7  2      67    |  58   18     158   |  3       9     e46     |
 |  45   1      68    |  9    7      3     |  2456    2456   4568   |
 +--------------------+--------------------+------------------------+
 |  8   b579    1     |  2    49     6     |  4579    45     3      |
 |  3    6      59    |  7    1489   189   |  1459    145    2      |
 |fa79   4      2     |  3    5      19    |  8       16    e1679   |
 +--------------------+--------------------+------------------------+

1. (7)r9c1 = r7c2 - r1c2 = (74)r1c79 - (7r9c9 & 4r5c9) = (7r9c1 | 4r5c1) => -7 r5c1; 15 placements & lcls

I think this is a disguised forcing chain.

Code: Select all

.-----------------.---------------.---------------------.
| 1     a579  3   | 6   29    59  | A24579  8    Ba4579 |
| 6     589   4   | 58  2389  7   | 1259   1235  159    |
| 2     5789  579 | 1   389   4   | 5679   356   5679   |
:-----------------+---------------+---------------------:
| 59    3     89  | 4   6     2   | 15     7     158    |
| dD4-7 2     67  | 58  18    158 | 3      9     C46    |
| 45    1     68  | 9   7     3   | 2456   2456  4568   |
:-----------------+---------------+---------------------:
| 8     b579  1   | 2   49    6   | 4579   45    3      |
| 3     6     59  | 7   1489  189 | 1459   145   2      |
| c79   4     2   | 3   5     19  | 8      16    b1679  |
'-----------------'---------------'---------------------'

Kraken Row r1c279
(7)r1c2 - r7c2 = r9c1 - r5c1
(7-4)r1c7 = r1c9 - r5c9 = (4-7)r5c1
(7)r1c9 - r9c9 = r9c1 - r5c1
______________
=> -7 r5c1

EDIT
Sorry, my second chain was incorrect.

[jco]

Hello AnotherLife,

I think this is a disguised forcing chain.

The way I see that move is as described in its writing, that is, if 7 is not at r9c1, then it must be at r7c2, then it cannot be at r1c2, which implies that (47) must be at r1c79 (the key being the strong link of 4s at row 1). That being the case, we cannot have both 7 and 4 at the other two available positions in column 9 (since 7 or 4 will be at r1c9). However, this in turn implies (due to the other two key strong links on 4s at row 5, and on 7s at row 9) that either r5c1=4 or r9c1=7. So, the three strong links mentioned allow this beautiful inference, specially the one on 4s at row 1.
For me, it is only natural that it can be written as a Kraken (forcing chain *verity*), but I have no doubt that Cenoman's first move is more informative than a Kraken description.
BTW, I have no prejudice against Krakens, on the contrary! They are a very powerful weapon for a manual solver. I see them as generalized AICs (any AIC can be written as a Kraken). It is also nice that many moves cannot be written as AICs. Often to open the way in some very difficult puzzles, one uses Krakens (tons of examples in this puzzle section and elsewhere) and no nice AIC is available to replace them!
I do not like as much the forcing chain contradictions, but of course they are just as valid as the verity type.

Regards,
jco

Edit: small corrections in the text.

[AnotherLife]

The way I see that move is as described in its writing, that is, if 7 is not at r9c1, then it must be at r7c2, then it cannot be at r1c2, which implies that (47) must be at r1c79 (the key being the strong link of 4s at row 1). That being the case, we cannot have both 7 and 4 at the other two available positions in column 9 (since 7 or 4 will be at r1c9).

Hello, jco
You come to this conclusion because if r1c2<>7 then either r1c7=7 (=> r1c7<>4 => r1c9=4) or r1c9=7. On the contrary, I start from the premise that one of the possibilities r1c2=7, r1c7=7, or r1c9=7 is true. Maybe our interpretations are slightly different but we make use of the same strong and weak links.

I have no prejudice against Krakens, on the contrary! They are a very powerful weapon for a manual solver. I see them as generalized AICs (any AIC can be written as a Kraken).

I would rather consider a Kraken (a forcing chain) as a bunch of non-branching AICs that are independent of each other.

It is also nice that many moves cannot be written as AICs.

I agree with you. In the example discussed above we cannot eliminate the 7 from r5c1 using one AIC even with groups and ALS's.

Regards,
Bogdan

[jco]

Hello AnotherLife,

(...) In the example discussed above we cannot eliminate the 7 from r5c1 using one AIC even with groups and ALS's.

I do not understand this comment. The AIC written by Cenoman proves that 7 is false at r5c1. It is a valid AIC: each part of it implies the next (in the way I have explained). So, in this case the move can be described by an AIC, while patterns like jellyfish eliminations, for instance, can't.

Regards,
jco

[Cenoman]

1. (7)r9c1 = r7c2 - r1c2 = (74)r1c79 - (7r9c9 & 4r5c9) = (7r9c1 | 4r5c1) => -7 r5c1; 15 placements & lcls

I think this is a disguised forcing chain.

Kraken Row r1c279
(7)r1c2 - r7c2 = r9c1 - r5c1
(7-4)r1c7 = r1c9 - r5c9 = (4-7)r5c1
(7)r1c9 - r9c9 = r9c1 - r5c1
______________
=> -7 r5c1

Hi AnotherLife,

If I write the Triangular Matrix (TM) of your kraken, I get:

Code: Select all

4r5c1 4r5c9
      4r1c9 4r1c7
7r9c1             7r7c2
            7r1c7 7r1c2 7r1c9
7r9c1                   7r9c9
-----------
=>-7r5c1

If I write the TM of my chain, I get:

Code: Select all

7r9c1 7r7c2
4r5c1        4r5c9
7r9c1                    7r9c9
      7r1c2 (7,4)r1c79 (4,7)r1c79
-----------
=>-7r5c1

where (7,4)r1c79 and (4,7)r1c79 represent each of the two configurations of the AHP (47)r1c79

Unfolding this condensed writing, I get:

Code: Select all

7r9c1 7r7c2
4r5c1        4r5c9
7r9c1              7r9c9
             4r1c9       4r1c7
      7r1c2        7r1c9 7r1c7
-----------
=>-7r5c1

which is exactly the same as the kraken TM (except the order of rows/columns) .

You wrote yourself:

we make use of the same strong and weak links.

Every piece of logic is explicited, so, what is disguised in this way of writing ?
I wonder which qualification you would have used, had I written a memory chain !

[jco]

Hello AnotherLife,

Just one more comment.

I would rather consider a Kraken (a forcing chain) as a bunch of non-branching AICs that are independent of each other.

Two short quotes on krakens:

"The elimination is what I've called a Kraken Fish. "Kraken" comes from the extra "tentacles" on the fish creating the elimination. (...) "
(Source: Mike Barker » Mon Feb 04, 2008. This Forum)

"I guess the term originated with finned fishes whose fins were used as starting points for forcing chains, but since then it's been used (at least on this forum) for any kind of verity producing forcing chains. In addition to fish fins (+the fish itself), the starting hub can be any strong-inference-set (SIS), such as all instances of a digit in a house ("Unit Forcing Chains") or all candidates in a cell ("Cell Forcing Chain") or all extra candidates of a deadly pattern (Oddagons, BUGs, URs, etc) -- what's common to all of them is that (at least) one starting point must be true, so if you can prove the same elimination (or placement) with all of them, it must be true. All of those different kinds are called Krakens here, even though most of them have nothing to do with fishes."
(Source: Re: need help with this puzzle, and technique in general. Postby SpAce » Wed Aug 29, 2018 (this Forum))

and a reference on forcing chains

http://forum.enjoysudoku.com/forcing-chains-terminology-and-definition-t2859.html#p18340


Regards,
jco

Edit: posted almost at the same time as Cenoman's post! Replaced my last comment by link to thread on forcing chains by Jeff.

[AnotherLife]

1. (7)r9c1 = r7c2 - r1c2 = (74)r1c79 - (7r9c9 & 4r5c9) = (7r9c1 | 4r5c1) => -7 r5c1; 15 placements & lcls

Hi, Cenoman,
I did not mean to criticize your way of writing. My point was that we cannot take this one as a conventional AIC, but we must either allow branches in the logical chain, as in your interpretation, or consider several independent AICs, as in my interpretation.

Two short quotes on krakens...

Thanks for your more extensive coverage of the krakens. I did not try to give a precise definition but to emphasize that we need several AICs to construct a kraken.