The New Sudoku Players' Forum

[ArkieTech]

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 *-----------*
 |...|1..|47.|
 |.3.|.6.|..9|
 |1..|.73|6..|
 |---+---+---|
 |...|.1.|.8.|
 |..5|...|3..|
 |.4.|.8.|...|
 |---+---+---|
 |..4|65.|..7|
 |2..|.4.|.9.|
 |.61|..9|...|
 *-----------*


Play/Print this puzzle online

[Leren]

Code: Select all

*---------------------------------------------------------------------------------*
| 56    Dd25      268      | 1       9      C258      |  4       7       3        |
| 4       3       7        | 28      6      C258      |Da258b    1       9        |
| 1     Fc259d   c289      | 4       7       3        |  6     Eb25c   Eb258c     |
|--------------------------+--------------------------+---------------------------|
| 367     279     269      | 35      1       4        |  79      8       25       |
| 8       1       5        | 9       2       7        |  3       6       4        |
| 37      4       29       | 35      8       6        |  79      25      1        |
|--------------------------+--------------------------+---------------------------|
| 9       8       4        | 6       5      B12       | A12      3       7        |
| 2    EGe7-5ec   3        | 78      4       18       |  158b    9       6        |
| 57      6       1        | 278     3       9        |  258a    4       258      |
*---------------------------------------------------------------------------------*

Kraken Column 7 Digit 2:

2 r2c7 - r3c89 = r3c23 - (2=5) r1c2                  - 5 r8c2;

2 r7c7 - r7c6 = r2c6 - 5 r2c6 = r2c7 - r3c89 = r3c2  - 5 r8c2;
              | r1c6 - (2=5) r1c2                    - 5 r8c2;

2 r9c7 - 5 r9c7 = 5 r2c7 - r3c89 = r3c2 - 5 r8c2;
                | 5 r8c7                             - 5 r8c2; => - 5 r8c2; stte

Leren

[SteveG48]

Code: Select all

 *--------------------------------------------------*
 | 56  a25   268  | 1    9    258  | 4    7    3    |
 | 4    3    7    |e28   6    258  |d258  1    9    |
 | 1   a259 b289  | 4    7    3    | 6    25  c258  |
 *----------------+----------------+----------------|
 | 367  279  269  | 35   1    4    | 79   8    25   |
 | 8    1    5    | 9    2    7    | 3    6    4    |
 | 37   4    29   | 35   8    6    | 79   25   1    |
 *----------------+----------------+----------------|
 | 9    8    4    | 6    5    12   | 12   3    7    |
 | 2   a57   3    | 8-7  4    18   | 158  9    6    |
 | 57   6    1    |f278  3    9    | 258  4   d258  |
 *--------------------------------------------------*

(7=259)r138c2 - (29=8)r3c3 - r3c9 = (8)r2c7 - (8=2)r2c4 - (28=7)r9c4 => -7 r8c4 ; stte
                                  = (8)r9c9 -------------/



Or it can be written as Kraken cell [278]r9c4 => -7 r8c4 using the same paths.

[gurth]

[gurth]

After looking at the solutions posted before mine, I decided to edit my solution by adding the 'details' paragraph, but I did not change anything in the first version.

[eleven]

To stay with the broken wings.

Code: Select all

+----------------+----------------+----------------+
| 56   25   268  | 1    9    258  | 4    7    3    |
| 4    3    7    |#28   6   *258  |#258  1    9    |
| 1    259  289  | 4    7    3    | 6    25  #258  |
+----------------+----------------+----------------+
| 367  279  269  | 35   1    4    | 79   8    25   |
| 8    1    5    | 9    2    7    | 3    6    4    |
| 37   4    29   | 35   8    6    | 79   25   1    |
+----------------+----------------+----------------+
| 9    8    4    | 6    5    12   | 12   3    7    |
| 2    57   3    |*78   4    18   | 158  9    6    |
| 57   6    1    |#278  3    9    |*258  4   #258  |
+----------------+----------------+----------------+

BW 8r2c47,r3c9,r9c49, guardians r2c6,r8c4,r9c7
(8-5)r2c6=5r1c6
(8-7)r8c4=(7-5)r8c2=5r9c1
8r9c7-r9c9=r3c9-r3c3=(8-6)r1c3=6r1c1
=> r1c1<>5
[Edit:] made it a bit shorter.

[blue]

Code: Select all

+--------------------+----------------+----------------+
| (56)  25    2-6(8) | 1     9  25(8) | 4      7   3   |
| 4     3     7      | 2(8)  6  25(8) | 25(8)  1   9   |
| 1     259   289    | 4     7  3     | 6      25  258 |
+--------------------+----------------+----------------+
| 367   279   269    | 35    1  4     | 79     8   25  |
| 8     1     5      | 9     2  7     | 3      6   4   |
| 37    4     29     | 35    8  6     | 79     25  1   |
+--------------------+----------------+----------------+
| 9     8     4      | 6     5  12    | 12     3   7   |
| 2     5(7)  3      | (78)  4  1(8)  | 15(8)  9   6   |
| (57)  6     1      | 278   3  9     | 258    4   258 |
+--------------------+----------------+----------------+

Four presentations of the "same logic": ((any one) => [6r1c1=8r1c3] => -6r1c3; stte)

(6=5)r1c1 - (5=7)r9c1 - r8c2 = (7-8)r8c4 = (Grouped Skyscraper: r28\c7) - r1c6 = 8r1c3 [my favorite]

8r1c3 = r1c6 -* r2c46 = r2c7 - r8c6 *= (8-7)r8c4 = r8c2 - (7=5)r9c1 - (5=6)r1c1

Code: Select all

8r1c3 = r1c6 - r2c46 = r2c7 - r8c7 = (8-7)r8c4 = r8c2 - (7=5)r9c1 - (5=6)r1c1
             \                 ||
               -------------- r8c6

Code: Select all

8r8c4 - 7r8c4 = r8c2 - (7=5)r9c1 - (5=6)r1c1 - 6r1c3
  ||                                         /
8r8c7 - r2c7 = r2c46 - r1c6 = 8r1c3 --------
  ||                 /
8r8c6 --------------


Edit: fixed typo: r25\c7 -> r28\c7 [ Thank you, eleven and Steve ]

[eleven]

Nice one blue,

i also prefer the almost skyscraper (in rows 28).


Gurth,

did you know, that the method you are generally using here, was called "extended Medusa coloring" in the dailysudoku forum ?. See e.g. here.

[gurth]

Eleven, I didn't know about Medusa: when I tried to look it up in the collection of techniques lists, I found the links no longer valid.
I do know that the method of Conjugate Worlds which I use here, is logically identical to David P Bird's GEM (graduated equivalence markings). We developed these methods simultaneously and independently of each other. I believe that my representation by colours is vastly preferable to David's markings, which are difficult to spot.
When David got stuck, he used to start a fresh grid. I wouldn't do that, I'd simply introduce another set of conjugate colours, not immediately connectible to the original colours. In fact I just did this yesterday, while solving a SER 7.2 puzzle I had composed. The situation was that I had placed several candidates and eliminated Red, but that still wasn't enough. Introducing another pair of colours soon fixed that.

[daj95376]

Using the guardian cells and value of an oddagon pattern to define the start of a forcing chain elimination. Interesting!

Grouped/almost Skyscraper. Hmmm!

Anyway ...

Code: Select all

 +-----------------------------------------------------+
 | e56   25  d268  |  1    9    258  |  4    7    3    |
 |  4    3    7    |  2-8  6    258  | a258  1    9    |
 |  1    259 c289  |  4    7    3    |  6    25  b258  |
 |-----------------+-----------------+-----------------|
 |  367  279  269  |  35   1    4    |  79   8    25   |
 |  8    1    5    |  9    2    7    |  3    6    4    |
 |  37   4    29   |  35   8    6    |  79   25   1    |
 |-----------------+-----------------+-----------------|
 |  9    8    4    |  6    5    12   |  12   3    7    |
 |  2   g57   3    | h78   4    18   |  15-8 9    6    |
 | f57   6    1    |  278  3    9    |  258  4    258  |
 +-----------------------------------------------------+
 # 46 eliminations remain

 8r2c7 = r3c9 - r3c3 = (8-6)r1c3 = (6-5)r1c1 = r9c1 - (5=7)r8c2 - (7=8)r8c4  =>  -8 r2c4,r8c7


[David P Bird]

Eleven, I didn't know about Medusa: when I tried to look it up in the collection of techniques lists, I found the links no longer valid.
I do know that the method of Conjugate Worlds which I use here, is logically identical to David P Bird's GEM (graduated equivalence markings). We developed these methods simultaneously and independently of each other.....

Thanks for the honourable mention Gurth!
I use a Sudoku Drudge spreadsheet as a helper using conditional formatting to colour the cells holding the current focus digit according to its Graded Equivalence Mark. Unfortunately this restricts me to two colours. I discovered that if I allowed myself to take advantage of the forcing chains and branched AICs this can find, it completely took the challenge out of the simpler puzzles so I became a bit of puritan and self-imposed a strict discipline on how I use it.

To tackle the harder problems I need to recognise the inferences I can use from patterns – finned fish, deadly patterns etc – that can be considered Boolean nodes, but this still puts some extreme puzzles out of my reach, but that's where the (masochistic) challenge remains for me. As I remember it, your Conjugate Worlds approach is more liberal.

<Medusa> as originally specified only colours conjugate links (but does allow group nodes). Presumably it could be extended to allow pattern inferences to be coloured too, but I don't know how well it can follow AICs containing a number of weak-only links.

David

[gurth]

Thanks for the honourable mention Gurth!
I use a Sudoku Drudge spreadsheet as a helper using conditional formatting to colour the cells holding the current focus digit according to its Graded Equivalence Mark. Unfortunately this restricts me to two colours. I discovered that if I allowed myself to take advantage of the forcing chains and branched AICs this can find, it completely took the challenge out of the simpler puzzles so I became a bit of puritan and self-imposed a strict discipline on how I use it.

To tackle the harder problems I need to recognise the inferences I can use from patterns – finned fish, deadly patterns etc – that can be considered Boolean nodes, but this still puts some extreme puzzles out of my reach, but that's where the (masochistic) challenge remains for me. As I remember it, your Conjugate Worlds approach is more liberal.

David

Interesting to compare our different approaches! I am more liberal to simple techniques, say with naked trips as the hardest: I expect colours to work with those only! So I took a look at an SER 8.3 puzzle. It really didn't look like I'd get anywhere with two colours, so I picked on 3 conjugate pairs from the start, using 6 colours! That's my "Colt" 6-shooter. Working on one page, with no erasures, I placed several numbers: by that time all the colours had disappeared, but a whole lot of markings, from the false colours, and meaning therefore nothing, remained. SE now rated the position at 7.2 ! So I had reduced the puzzle to a rather easier one. Next I suppose I should deign to start afresh (I thought of saving the coloured pic in black and white, to get rid of the false and distracting colours, but I thought why put up with such a B&W mess, rather continue on a fresh sheet).

I might as well append my worksheet, for what it's worth:

[David P Bird]

Gurth, thanks for your reply. I've got the general idea of your approach if not the detail as I only managed to decrypt some of your notation. BTW I have to shrink the display to 75% to stop the notation on the right of your grid being cropped which then makes the details in diagram harder discern.

As you provided a taster of your approach let me reciprocate. I started to solve the puzzle you posted but stopped when I realised it would be somewhat of a slog. My opening chains were
(1)r2c3 = (1)r1c1 - (1)r1c7 = (1-9)r2c7 = (9)r8c7 - (9)r8c2 = (9-3)r4c2 = (3)r6c3 => r2c3 <> 3, r6c3 <> 1
(7)r89c3 = (7-9)r8c2 = (9-3)r4c2 = (3)r6c3 => r6c3 <> 7
(1)r1c7 = (1)r1c1 - (1=9)r6c1 - (9)r9c1 = (9)r8c12 - (9)r8c7 = (9)r2c7 => r2c7 <> 1
(I've omitted the trivial follow-on eliminations after each step.)

I then marked up this grid which serves to show why I exercise restraint in how I use the information available:

Code: Select all

 *----------------------------*-----------------------------------*--------------------------------*
 | 456.       3.5.6=  <9>     | 457        457.       <8>         | 1         <2>        3=4.5.7.  |
 | <7>        8       1       | 24569!     2456       2.3=4.5.9.  | 3.5.9=    3.45.9.    3.45.     |
 | 2.45       3=5.    2=4.    | 457.9:     <1>        3.457.9.    | <8>       3.4.5.7=9. <6>       |
 *----------------------------*-----------------------------------*--------------------------------*
 | 259.       579.    <6>     | <8>        2457.      2.4.5.7.9=  | 3=5.7.    <1>        3.45.7    |
 | 1"2.5.8.   57      2.7.8=  | 1'2:4:5:7! <3>        2457.       | 5.67      45.67.     <9>       |
 | 1'9"       <4>     3       | 1"9'       5.7=       <6>         | <2>       5=7.       <8>       |
 *----------------------------*-----------------------------------*--------------------------------*
 | <3>        <1>     <5>     | 467        <9>        47          | 67        <8>        <2>       |
 | 4.689.     6.7.9=  4=7.8.  | 2567       2567.8     257         | 3.5.679.  3=5.6.7.9. <1>       |
 | 689.       <2>     7=8.    | <3>        5.67.8     <1>         | <4>       5.6.7.9=   5=7.      |
 *----------------------------*-----------------------------------*--------------------------------*
Symbols  0'  0"  Candidate true at own parity false at the other
         0-  0=  Must be true at own parity, may be true at the other
         0.  0:  May be true at own parity, must be false at the other
         0!      False at both parities (AIC to prove this available)

This grid was marked starting from the (19)pair r6c14 from which (9)r2c4 & (7)r5c4 can be eliminated. Marking from (3')r1c2, (3")r3c2 would be better which gives 5 available eliminations but it doesn't demonstrate the point I want to make. At parity II (5)r6c8, (5)r9c9, & (9)r2c7 will be true which together will eliminate every (5) candidate in box 3,therefore parity I must be true. It takes three chains to accumulate these killer candidates which (with a little practice) can be tracked backwards in the GEM marked grid:
(1=9)r6c1 - (9)r89c1 = (9-6)r8c2 = (6-3)r1c2 = (3-7)r1c9 = (7)r3c8 - (7=5)r6c8
(1)r6c1 = (1-8)r5c1 = (8)r5c3 - (8=7)r9c3 - (7=5)r9c9
(1=9)r6c1 - (9)r89c1 = (9)r8c2 - (9)r9c7 = (9)r2c7
Now for me not only is that a net deduction it's also assumptive because all it shows is that if (1)r6c1 is false it produces a contradiction.