The New Sudoku Players' Forum

[Cenoman]

eleven's (very nice !) eliminations in AIC's:

Note the derived strong link (9)r4c2 == (6)r9c4 from the XY-Chain (9=5)r4c2 - (5=9)r5c3 - (9=7)r2c3 - (7=5)r9c3 - (5=6)r9c4

(#65=9)r49c4 - (*9)r4c2 == (6#)r9c4 - (6=59*)r4c24 => -9 r4c57*, -6r3c4#
-9 r4c7 is the effective elimination for ste solving.

TM 6x6 for the 9s

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9r4c2 5r4c2
      5r5c3 9r5c3
            9r2c3 7r2c3
                  7r9c3 5r9c3
                        5r9c4 6r9c4
9r4c4 5r4c4                   6r4c4
------------
=> -9 r4c57; ste

This matrix is a TM, and not a PM, because of its column 2

A similar TM exists for the 6s

Hidden Text: Show

Code: Select all

6r9c4 5r9c4
      5r9c3 7r9c3
            7r2c3 9r2c3
                  9r5c3 5r5c3
                        5r4c2 9r4c2
6r4c4 5r4c4                   9r4c4
------------
=> -6 r3c4

All this, AICs and matrices, is rank 1 logic. I can't see how to writeit as MSLS.

[marek stefanik]

I'm wondering what would be the easiest way to write this as a msls?

All this, AICs and matrices, is rank 1 logic. I can't see how to write it as MSLS.

While the pattern is rank1, big part of it (everything but the 5s) is rank0. There is a simple way to prove this hidden rank0 logic.

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   +---------+---------+---------+
   | .  .  . | .  .  . | .  .  . |
   | .  .  . | .  .  . | .  .  . |
   | .  .  . | .  .  . | .  .  . |
   +---------+---------+---------+
   | .  5  . | 5  .  . | .  .  . |
   | .  .  5 | .  .  . | .  .  . |
   | .  .  . | .  .  . | .  .  . |
   +---------+---------+---------+
   | .  .  . | .  .  . | .  .  . |
   | .  .  . | .  .  . | .  .  . |
   | .  .  5 | 5  .  . | .  .  . |
   +---------+---------+---------+

Each of the 5s we're trying to cover is part of two houses from this set: r49c34b4. If three of them were true, we'd essentially have six 5s in 5 houses.
We can therefore derive a 2-link (at most 2 true candidates) that covers them:
A = 2L(5r4c1234, 5r56c3, 5r9c34) = 5r49c34b4 / 2
Note that the 2-link can also be proven using a finned X-Wing (5r4c1234, 5r9c34 => -5r56c3).
With this link the pattern reduces to rank0.

We can do the same with other patterns, too. The following (theoretical) Almost MSLS contains an XY-Wing:

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   +---------------+---------------+---------------+
   | 12   .    .   | .    234  .   | .    .    .   |
   | .    .    .   | .    .   .    | .    .    .   |
   | .    .    .   | .    .    .   | .    .    .   |
   +---------------+---------------+---------------+
   | .    .    .   | .    .    .   | .    .    .   |
   | 135  .    .   | .    345  .   | .    .    56  |
   | .    .    .   | .    .    .   | .    .    .   |
   +---------------+---------------+---------------+
   | .    .    .   | .    .    .   | .    .    .   |
   | .    .    .   | .    .    .   | .    .    .   |
   | .    .    .   | .    47   .   | .    .    67  |
   +---------------+---------------+---------------+

Only three out of the candidates 12r1c1, 23r1c5, 13r5c1, 3r5c5 can be true.
This can be easily seen as an XY-Wing, the more rigorous proof derives the 3-link as (r1c1, r1c5, r5c1, 1c1, 2r1, 3r5, 3c5) / 2.
With that link we uncover the hidden rank0 logic, allowing the eliminations on 4567 to be made.

Marek

[jco]

Regarding the previous question of 999_Springs, considering the comments by marek stefanik, it seems to me that one can reason as follows:

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*---------------------------------------------------------------*
|  5  6     1    |  3     24      7      |  8     24     9      |
|  2  79   #79   |  1     456     8      |  46    45     3      |
|  8  3     4    |  59-6  2569    2569   |  16    1257   2567   |
|----------------+-----------------------+----------------------|
|  7 #59    2    | #(5)69 1356-9  4      |  16-9  1358   568    |
|  3  4    #59   |  7     8       1569   |  2     15     56     |
|  6  1     8    |  2     359     59     |  49    3457   57     |
|----------------+-----------------------+----------------------|
|  4  2     6    |  8     7       3      |  5     9      1      |
|  1  58    3    |  4     259     259    |  7     6      28     |
|  9  578  #57   | #56    1256    1256   |  3     28     4      |
*---------------------------------------------------------------*


. if r4c4=5, then immediately follows that -9 r4c57, -6 r3c4,
. otherwise we have
Truths: r2c3,r4c2,r5c3,r9c3,r9c4,r4c4
Links: 7c3,5b4,9r4,9c3,5r9,6c4
and so, rank=0, producing the same eliminations
(Cenoman's matrices can be replaced by one PM with two result columns).

(in my previous post I forgot to mention that I was looking at different ways
to see eleven's move getting both eliminations, though knowing that removing
the 9s is enough for ste, as Cenoman pointed out)

Edit: corrected text regarding statement on PM.

[denis_berthier]

.

Code: Select all

Resolution state after Singles and whips[1]:
   +-------------------+-------------------+-------------------+
   ! 5     6     1     ! 3     24    7     ! 8     24    9     !
   ! 2     79    79    ! 1     456   8     ! 46    45    3     !
   ! 8     3     4     ! 569   2569  2569  ! 16    1257  2567  !
   +-------------------+-------------------+-------------------+
   ! 7     59    2     ! 569   13569 4     ! 169   1358  568   !
   ! 3     4     59    ! 7     8     1569  ! 2     15    56    !
   ! 6     1     8     ! 2     359   59    ! 49    3457  57    !
   +-------------------+-------------------+-------------------+
   ! 4     2     6     ! 8     7     3     ! 5     9     1     !
   ! 1     58    3     ! 4     259   259   ! 7     6     28    !
   ! 9     578   57    ! 56    1256  1256  ! 3     28    4     !
   +-------------------+-------------------+-------------------+

The simplest-first solution has only 3 steps:

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finned-x-wing-in-columns: n2{c9 c6}{r3 r8} ==> r8c5 ≠ 2
biv-chain[3]: r4c2{n5 n9} - b6n9{r4c7 r6c7} - r6c6{n9 n5} ==> r4c4 ≠ 5, r4c5 ≠ 5
finned-swordfish-in-columns: n5{c4 c3 c9}{r3 r9 r5} ==> r5c8 ≠ 5
stte

As for the 1-step solutions:
There are 18 W1-anti-backdoors: n7r2c2 n9r2c3 n5r2c5 n6r2c7 n4r2c8 n9r3c4 n5r3c9 n9r4c2 n9r4c7 n5r5c3 n1r5c6 n9r5c6 n5r5c8 n6r5c9 n4r6c7 n9r8c5 n7r9c3 n1r9c5
15 of which give 1-step solution with a whip[≤6]. The simplest two of them need only a typed-whip[4] (in rc-space or rn-space):

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whip-rc[4]: r5c3{n9 n5} - r5c9{n5 n6} - r4c7{n6 n1} - r5c8{n1 .} ==> r4c2 ≠ 9
stte

OR:

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whip-rn[4]: r2n5{c8 c5} - r6n5{c5 c6} - r4n5{c5 c2} - r8n5{c2 .} ==> r5c8 ≠ 5
stte

The latter whip can better be seen in the rn-view of the resolution state:
rn-view:

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Physical rows are rows, physical columns are digits. Data are columns.
   3         58        4         58        1         2         6         7         9         
   4         1         9         578       58        57        23        6         23       
   78        5689      2         3         45689     45679     89        1         456       
   578       3         58        6         24589     4579      1         89        2457     
   68        7         1         2         3689      69        4         5         36       
   2         4         58        78        5689      1         89        3         567       
   9         2         6         1         7         3         5         4         8         
   1         569       3         4         256       8         7         29        56       
   56        568       7         9         23456     456       23        28        1

[jco]

Hello,

Regarding the previous question of 999_Springs, considering the comments by marek stefanik, it seems to me that one can reason as follows:

Code: Select all

*---------------------------------------------------------------*
|  5  6     1    |  3     24      7      |  8     24     9      |
|  2  79   #79   |  1     456     8      |  46    45     3      |
|  8  3     4    |  59-6  2569    2569   |  16    1257   2567   |
|----------------+-----------------------+----------------------|
|  7 #59    2    | #(5)69 1356-9  4      |  16-9  1358   568    |
|  3  4    #59   |  7     8       1569   |  2     15     56     |
|  6  1     8    |  2     359     59     |  49    3457   57     |
|----------------+-----------------------+----------------------|
|  4  2     6    |  8     7       3      |  5     9      1      |
|  1  58    3    |  4     259     259    |  7     6      28     |
|  9  578  #57   | #56    1256    1256   |  3     28     4      |
*---------------------------------------------------------------*


. if r4c4=5, then immediately follows that -9 r4c57, -6 r3c4,
. otherwise we have
Truths: r2c3,r4c2,r5c3,r9c3,r9c4,r4c4
Links: 7c3,5b4,9r4,9c3,5r9,6c4
and so, rank=0, producing the same eliminations
(Cenoman's matrices can be replaced by one PM with two result columns).

(in my previous post I forgot to mention that I was looking at different ways
to see eleven's move getting both eliminations, though knowing that removing
the 9s is enough for ste, as Cenoman pointed out)

Edit: corrected text regarding statement on PM.

Feedback on this from experienced players would be very helpful,
since I am still learning Allan Barker theory (being exactly on the link triplet's part).
I am using the analysis of eleven's move as a learning tool, since it is related to the subject just mentioned.

Many thanks in advance!

[marek stefanik]

I am probably not the right person to reply, since I try to avoid both base and link triplets in my solutions – I usually double the link in base triplets and the whole point of my multilinks is to get rid of link triplets – but what you wrote seems consistent with Allan Barker's description on his website.
I don't know how to use these principles with more advanced patterns though, and I probably couldn't use this to prove junior exocets with a link triplet on every digit, let alone many senior exocets which, I think, give us a base triplet per digit as a bonus.
Nonetheless I think they can serve as a good indication of whether a pattern can create an area of lower rank and what the best rank could be.

[jco]

@marek stefanik

I recall right after reading your comments, I wrote a PM to thank you.
Since then I wanted to register also here my thanks for the above explanation.
After I finished studying Allan Baker approach (had that comments in mind),
and learned the details on the difficulties with "triplets", that improved my understanding about
your proposed multilinks approach (that I studied before Allan's material - wrong order!).
My thanks also for explaining further my few questions (by PM) on multilinks.