The tridagon rule

Advanced methods and approaches for solving Sudoku puzzles

Re: The tridagon rule

Postby mith » Tue Mar 15, 2022 8:00 pm

xzccs, I didn't totally follow your argument, since your images are not loading for me. But playing around with this in the 4 box case, I've come up with the following (which I assume is just a specific case of your proof):

1. Considering only row- and column- links, there is exactly one rectangle in the pattern (4-loop), and the other cells form a bigger loop (8-loop).
2. Because we are limited to 3 digits, the 4-loop must have at least one case of a digit appearing twice, in diagonally opposite boxes. Let's call this digit A.
3. The other cells in those boxes are from the same two digits, say B and C.
4. There are four more cells in the other two boxes which are part of the 8-loop. Let's call these cells BB, BC, CB, and CC, based on how they are linked to the B and C cells in the original two boxes. (We know there is a unique cell for each case due to the 8-loop; if "BB" were two different cells, we would have a 4-loop.)
5. BB and CC must be in the same box; if they were in different boxes, the other 8-loop cell in the CC box must also see BB.
6. Likewise BC and CB must be in the same box. This a contradiction, since they would both need to contain A.
Last edited by mith on Tue Mar 15, 2022 8:06 pm, edited 1 time in total.
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Re: The tridagon rule

Postby mith » Tue Mar 15, 2022 8:04 pm

Code: Select all
.  .  B  | BC .  .
.  A  .  | .  r  .
C  .  .  | .  .  CB
---------+---------
CC .  .  | C  .  .
.  r  .  | .  A  .
.  .  BB | .  .  B


An illustration of one case. A in b1 and b5, r in b2 and b4 as the other cells in the 4-loop. b2 contains both BC and CB, #
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Re: The tridagon rule

Postby xzccs » Wed Mar 16, 2022 3:31 am

Sorry, I forgot to change the permission of my google drive. I think the image is available now :D .
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Re: The tridagon rule

Postby xzccs » Wed Mar 16, 2022 7:05 am

mith wrote:xzccs, I didn't totally follow your argument, since your images are not loading for me. But playing around with this in the 4 box case, I've come up with the following (which I assume is just a specific case of your proof):

1. Considering only row- and column- links, there is exactly one rectangle in the pattern (4-loop), and the other cells form a bigger loop (8-loop).
2. Because we are limited to 3 digits, the 4-loop must have at least one case of a digit appearing twice, in diagonally opposite boxes. Let's call this digit A.
3. The other cells in those boxes are from the same two digits, say B and C.
4. There are four more cells in the other two boxes which are part of the 8-loop. Let's call these cells BB, BC, CB, and CC, based on how they are linked to the B and C cells in the original two boxes. (We know there is a unique cell for each case due to the 8-loop; if "BB" were two different cells, we would have a 4-loop.)
5. BB and CC must be in the same box; if they were in different boxes, the other 8-loop cell in the CC box must also see BB.
6. Likewise BC and CB must be in the same box. This a contradiction, since they would both need to contain A.

That's right. It is a specific case. I've tried to prove it by that way before, but it's still a kind of enumeration. When fucusing more on the relationship between loops and within loops, I found this more general proof.
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Re: The tridagon rule

Postby denis_berthier » Wed Mar 16, 2022 7:53 am

Hi xzccs
Welcome to this forum

I'm not sure to understand all the details, but the main problem seems to be to find a real example. There may not be enough room on the 9x9 board for allowing such patterns.
Long ago, I tried something similar with the sk-loop: in theory, there's no obvious reason why it couldn't have a version in 6 blocks. But I couldn't properly place givens to create such a pattern.
An example would be very welcome.
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Re: The tridagon rule

Postby xzccs » Wed Mar 16, 2022 8:47 am

denis_berthier wrote:Hi xzccs
Welcome to this forum

I'm not sure to understand all the details, but the main problem seems to be to find a real example. There may not be enough room on the 9x9 board for allowing such patterns.
Long ago, I tried something similar with the sk-loop: in theory, there's no obvious reason why it couldn't have a version in 6 blocks. But I couldn't properly place givens to create such a pattern.
An example would be very welcome.

Thanks. I just took 6 boxes for image. I think it's quite hard to find an useful example since there should be too many clues in 6 boxes case.
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Re: The tridagon rule

Postby mith » Wed Mar 16, 2022 5:06 pm

It's pretty easy to demonstrate that a 6-box pattern in a 9x9 will always simplify to a 4-box pattern if there is only one guardian cell. Just as an example, let's consider the following parities in the boxes:

Code: Select all
//.
/./
./\


This is a broken pattern. Let's say there is a guardian cell in b9. Then there is no guardian cell in b124, and b5 is forced to have the same parity:

Code: Select all
//.
///
./\


But now we have a 4-box pattern in b5689.

If there are multiple guardian cells, in "opposing" boxes (boxes which don't see each other and don't see a third box that is part of the 6-box pattern; b19, b26, b48 in this example), then it's perhaps possible the 6-box pattern isn't reducible to a 4-box pattern, but all the useful examples of the pattern seen so far have at most two guardian cells in either the same box or adjacent boxes. Maybe the weak link between the guardian candidates could still give something, I dunno.
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Re: The tridagon rule

Postby eleven » Thu Mar 17, 2022 12:43 am

In valid sudokus such 3-digit patterns in 6 boxes cannot exist, because 2 of the 3 digits must be givens and therefore would be missing in the pattern.
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Re: The tridagon rule

Postby mith » Mon Mar 21, 2022 7:24 pm

eleven wrote:In valid sudokus such 3-digit patterns in 6 boxes cannot exist, because 2 of the 3 digits must be givens and therefore would be missing in the pattern.


Is it necessarily the case that having a single digit ruled out from a few cells of the pattern collapses it? (I know when looking at 4-box patterns it has been my experience that a given from the triple seeing into those boxes tends to collapse it to bivalue oddagons, but I haven't really thought about whether that will always happen.)
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Re: The tridagon rule

Postby denis_berthier » Fri Mar 25, 2022 10:48 am

.
Here is one more example (#2 in mith 246-list http://forum.enjoysudoku.com/the-hardest-sudokus-new-thread-t6539-1190.html)
Code: Select all
     +-------+-------+-------+
     ! . . . ! . . . ! . . 1 !
     ! . . . ! . . 2 ! . 3 . !
     ! . . . ! . 4 . ! 5 6 . !
     +-------+-------+-------+
     ! . . . ! . . 7 ! . . . !
     ! . . 4 ! 8 1 . ! 2 . . !
     ! 1 9 . ! . 2 4 ! 8 . . !
     +-------+-------+-------+
     ! . 8 9 ! . . 1 ! . . 7 !
     ! 4 2 . ! . 7 8 ! . . 9 !
     ! 7 . 1 ! . 9 . ! . . . !
     +-------+-------+-------+
........1.....2.3.....4.56......7.....481.2..19..248...89..1..742..78..97.1.9....;28


Code: Select all
Resolution state after Singles and whips[1]:
   +----------------------+----------------------+----------------------+
   ! 235689 34567  235678 ! 35679  3568   3569   ! 479    2489   1      !
   ! 5689   14567  5678   ! 15679  568    2      ! 479    3      48     !
   ! 2389   137    2378   ! 1379   4      39     ! 5      6      28     !
   +----------------------+----------------------+----------------------+
   ! 23568  356    23568  ! 3569   356    7      ! 13469  1459   3456   !
   ! 356    3567   4      ! 8      1      3569   ! 2      579    356    !
   ! 1      9      3567   ! 356    2      4      ! 8      57     356    !
   +----------------------+----------------------+----------------------+
   ! 356    8      9      ! 23456  356    1      ! 346    245    7      !
   ! 4      2      356    ! 356    7      8      ! 136    15     9      !
   ! 7      356    1      ! 23456  9      356    ! 346    2458   234568 !
   +----------------------+----------------------+----------------------+
196 candidates.


When whips are enabled, a full solution is obtained after an application of the tridagon elimination rule:
Code: Select all
hidden-pairs-in-a-column: c4{n2 n4}{r7 r9} ==> r9c4≠6, r9c4≠5, r9c4≠3, r7c4≠6, r7c4≠5, r7c4≠3
hidden-pairs-in-a-row: r4{n2 n8}{c1 c3} ==> r4c3≠6, r4c3≠5, r4c3≠3, r4c1≠6, r4c1≠5, r4c1≠3
t-whip[4]: c7n1{r4 r8} - r8c8{n1 n5} - r6c8{n5 n7} - r5c8{n7 .} ==> r4c7≠9
whip[1]: b6n9{r5c8 .} ==> r1c8≠9
hidden-pairs-in-a-block: b3{n7 n9}{r1c7 r2c7} ==> r2c7≠4, r1c7≠4
t-whip[5]: r5n7{c2 c8} - c8n9{r5 r4} - r4n1{c8 c7} - r4n4{c7 c9} - r2n4{c9 .} ==> r2c2≠7
t-whip[5]: r4n9{c4 c8} - r4n1{c8 c7} - r4n4{c7 c9} - r2n4{c9 c2} - r2n1{c2 .} ==> r2c4≠9
t-whip[6]: r5n7{c2 c8} - c8n9{r5 r4} - r4n1{c8 c7} - r4n4{c7 c9} - r2n4{c9 c2} - c2n1{r2 .} ==> r3c2≠7
t-whip[6]: r6n7{c8 c3} - r3n7{c3 c4} - r2n7{c4 c7} - r2n9{c7 c1} - r3n9{c1 c6} - r5n9{c6 .} ==> r5c8≠7
hidden-single-in-a-block ==> r6c8=7
hidden-single-in-a-block ==> r5c2=7
whip[4]: b5n9{r4c4 r5c6} - r5c8{n9 n5} - r8n5{c8 c3} - r6n5{c3 .} ==> r4c4≠5
whip[11]: b5n9{r4c4 r5c6} - r5c8{n9 n5} - b9n5{r7c8 r9c9} - c6n5{r9 r1} - c6n6{r1 r9} - b8n3{r9c6 r7c5} - c5n5{r7 r4} - c2n5{r4 r2} - r2n4{c2 c9} - c9n8{r2 r3} - c9n2{r3 .} ==> r4c4≠3
tridagon type diag for digits 3, 5 and 6 in blocks:
        b5, with cells: r5c6 (target cell), r4c5, r6c4
        b4, with cells: r5c1, r4c2, r6c3
        b8, with cells: r9c6, r7c5, r8c4
        b7, with cells: r9c2, r7c1, r8c3
 ==> r5c6≠3,5,6
stte


If only Subsets and tridagons are activated, no tridagon appears, but there is a tridagon-link:
Code: Select all
hidden-pairs-in-a-column: c4{n2 n4}{r7 r9} ==> r9c4≠6, r9c4≠5, r9c4≠3, r7c4≠6, r7c4≠5, r7c4≠3
hidden-pairs-in-a-row: r4{n2 n8}{c1 c3} ==> r4c3≠6, r4c3≠5, r4c3≠3, r4c1≠6, r4c1≠5, r4c1≠3
extended tridagon for digits 3, 5 and 6 in blocks:
        b4, with cells: r6c3 (link cell), r5c1, r4c2
        b5, with cells: r6c4, r5c6 (link cell), r4c5
        b7, with cells: r8c3, r7c1, r9c2
        b8, with cells: r8c4, r7c5, r9c6
 ==> tridagon-link(n7r6c3, n9r5c6)

With no other rule, nothing more can be deduced.
However, as mentioned by others in the "hardest" thread the tridagon link can be combined with chains in order to produce some eliminations.
I'll deal later with this situation.
What I wanted to show here is only an interesting case (IMO) where previous eliminations allow to reach a situation where the mere elimination rule applies.
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Re: The tridagon rule

Postby denis_berthier » Sat Mar 26, 2022 5:48 am

.
TRIDAGON LINKS

The first post of this thread was about
- making explicit a full resolution rule (i.e. a rule that allows immediate eliminations) based on the impossible "trivalue-oddagon" pattern
- and formulating it in a way that can easily be transcribed into a logical formula or any implementation.
Notice that this modest approach has led to consider the rule as an elimination one (rather than an assertion one) and to raise a previous restriction on the target cell: it may contain more than one additional candidate.

Other applications of the impossibility pattern are known. Again, the goal of this post is not to review all of them, but only the simplest case, where two and only two different cells of the impossible pattern have one and only one additional candidate. In this case, the conclusion is that one of the additional candidates must be True - which doesn't define a resolution rule in and of itself, but can be used to build new types of Forcing Chains.
Based on the analysis of the first post, it is easy to see that there are the following cases (and only them).

By the same isomorphisms as before, we can always suppose one of the two additional candidates is in r1c1 and that the pattern of cells is as in the first post:
Code: Select all
# . . | + . .
. + . | . + .
. . + | . . +
-------------
+ . . | x x x
. + . | x x x
. . + | x x x

with the same conditions on the pattern in the third block:
Code: Select all
. . +          . + .          + . .
. + .          + . .          . . +
+ . .          . . +          . + .

(where, at this point, the last two patterns are equivalent under r2/r3, c2/c3, and r5/r6 permutations).


Now, the question is: where can the second candidate be?

First case: in the same block. Only one possible place modulo isomorphisms:
Code: Select all
# . . | + . .
. # . | . + .
. . + | . . +
-------------
+ . . | x x x
. + . | x x x
. . + | x x x


Second case: in a different block but the same band. Only two possible places modulo isomorphisms:
same row:
Code: Select all
# . . | # . .
. + . | . + .
. . + | . . +
-------------
+ . . | x x x
. + . | x x x
. . + | x x x


different row:
Code: Select all
# . . | + . .
. + . | . # .
. . + | . . +
-------------
+ . . | x x x
. + . | x x x
. . + | x x x



Third case: in a different block but the same stack. Only two possible places modulo isomorphisms (deduced from the previous two by row/column symmetry)


Fourth case: in opposite blocks. It can be almost anywhere in the 4th block (which must still satisfy one of the three patterns found in the first post).
Code: Select all
# . . | + . .
. + . | . + .
. . + | . . +
-------------
+ . . | x x x
. + . | x x x
. . + | x x x


Notice that the above four cases are possible and have real examples.
.
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Re: The tridagon rule

Postby denis_berthier » Mon Mar 28, 2022 6:46 am

.
As an appetiser for more formal definitions of Tridagon-Forcing-Whips, here is a solution of puzzle #39 in mith's list of 246 puzzles not in T&E(2) (http://forum.enjoysudoku.com/the-hardest-sudokus-new-thread-t6539-1190.html), for which I've already given some analyses in the next posts of the same thread.
Code: Select all
     +-------+-------+-------+
     ! . . . ! . . . ! . . 1 !
     ! . . . ! . . 2 ! 3 4 . !
     ! . . 5 ! . 1 3 ! 6 . 2 !
     +-------+-------+-------+
     ! . . . ! . 7 . ! . 3 6 !
     ! . . . ! . 8 9 ! 4 . . !
     ! . . 4 ! 6 . . ! . . . !
     +-------+-------+-------+
     ! . 1 2 ! 5 . . ! . 6 . !
     ! 4 . 3 ! 1 . . ! 5 2 . !
     ! 5 6 . ! . . . ! 1 . . !
     +-------+-------+-------+
........1.....234...5.136.2....7..36....894....46......125...6.4.31..52.56....1..;29

Code: Select all
Resolution state after Singles and whips[1]:
   +----------------------+----------------------+----------------------+
   ! 236789 234789 6789   ! 4789   4569   45678  ! 789    5789   1      !
   ! 16789  789    16789  ! 789    569    2      ! 3      4      5789   !
   ! 789    4789   5      ! 4789   1      3      ! 6      789    2      !
   +----------------------+----------------------+----------------------+
   ! 1289   2589   189    ! 24     7      145    ! 289    3      6      !
   ! 12367  2357   167    ! 23     8      9      ! 4      157    57     !
   ! 123789 235789 4      ! 6      235    15     ! 2789   15789  5789   !
   +----------------------+----------------------+----------------------+
   ! 789    1      2      ! 5      349    478    ! 789    6      34789  !
   ! 4      789    3      ! 1      69     678    ! 5      2      789    !
   ! 5      6      789    ! 234789 2349   478    ! 1      789    34789  !
   +----------------------+----------------------+----------------------+
193 candidates.

Here, only Subsets, whips[≤12] and Tridagon-Forcing-Whips are activated.
Code: Select all
hidden-pairs-in-a-column: c9{n3 n4}{r7 r9} ==> r9c9≠9, r9c9≠8, r9c9≠7, r7c9≠9, r7c9≠8, r7c9≠7
hidden-pairs-in-a-row: r1{n2 n3}{c1 c2} ==> r1c2≠9, r1c2≠8, r1c2≠7, r1c2≠4, r1c1≠9, r1c1≠8, r1c1≠7, r1c1≠6
hidden-single-in-a-block ==> r3c2=4
z-chain[4]: r6c6{n5 n1} - c8n1{r6 r5} - c8n5{r5 r6} - c5n5{r6 .} ==> r1c6≠5
whip[1]: c6n5{r6 .} ==> r6c5≠5
naked-pairs-in-a-block: b5{r5c4 r6c5}{n2 n3} ==> r4c4≠2
naked-single ==> r4c4=4
hidden-pairs-in-a-column: c4{n2 n3}{r5 r9} ==> r9c4≠9, r9c4≠8, r9c4≠7
whip[1]: b8n7{r9c6 .} ==> r1c6≠7
whip[1]: b8n8{r9c6 .} ==> r1c6≠8
whip[1]: b8n9{r9c5 .} ==> r1c5≠9, r2c5≠9
whip[6]: r2n1{c1 c3} - b1n6{r2c3 r1c3} - r5c3{n6 n7} - r5c9{n7 n5} - r2n5{c9 c5} - r2n6{c5 .} ==> r2c1≠7
whip[6]: r2n1{c1 c3} - b1n6{r2c3 r1c3} - r5c3{n6 n7} - r5c9{n7 n5} - r2n5{c9 c5} - r2n6{c5 .} ==> r2c1≠8
whip[6]: r2n1{c1 c3} - b1n6{r2c3 r1c3} - r5c3{n6 n7} - r5c9{n7 n5} - r2n5{c9 c5} - r2n6{c5 .} ==> r2c1≠9
whip[6]: r5c9{n5 n7} - r5c8{n7 n1} - r5c3{n1 n6} - c1n6{r5 r2} - r2c5{n6 n5} - c9n5{r2 .} ==> r6c8≠5
whip[8]: r2n1{c3 c1} - b1n6{r2c1 r1c3} - c6n6{r1 r8} - c5n6{r8 r2} - r2n5{c5 c9} - r5c9{n5 n7} - c3n7{r5 r9} - r8n7{c2 .} ==> r2c3≠8
whip[8]: r2n1{c3 c1} - b1n6{r2c1 r1c3} - c6n6{r1 r8} - c5n6{r8 r2} - r2n5{c5 c9} - r5c9{n5 n7} - c3n7{r5 r9} - r8n7{c2 .} ==> r2c3≠9
whip[11]: b3n5{r1c8 r2c9} - r2c5{n5 n6} - r8c5{n6 n9} - c9n9{r8 r6} - c9n8{r6 r8} - b9n7{r8c9 r7c7} - c7n9{r7 r1} - r1c4{n9 n8} - r2n8{c4 c2} - b1n9{r2c2 r3c1} - r7n9{c1 .} ==> r1c8≠7

extended tridagon for digits 7, 8 and 9 in blocks:
        b3, with cells: r2c9 (link cell), r1c7, r3c8
        b1, with cells: r2c2, r1c3 (link cell), r3c1
        b9, with cells: r8c9, r7c7, r9c8
        b7, with cells: r8c2, r7c1, r9c3
 ==> tridagon-link[12](n5r2c9, n6r1c3)

tridagon-forcing-whip-elim[13] based on tridagon-link(n6r1c3, n5r2c9)
....for n6r1c3: -
....for n5r2c9: partial-whip[1]: c5n5{r2 r1} -
 ==> r1c5≠6, r2c3≠6

biv-chain[3]: r2c3{n7 n1} - r2c1{n1 n6} - b4n6{r5c1 r5c3} ==> r5c3≠7
biv-chain[3]: r5c3{n1 n6} - c1n6{r5 r2} - b1n1{r2c1 r2c3} ==> r4c3≠1

tridagon-forcing-whip-elim[13] based on tridagon-link(n5r2c9, n6r1c3)
....for n5r2c9: -
....for n6r1c3: partial-whip[1]: r2n6{c1 c5} -
 ==> r2c5≠5
stte


Two very short chains (partial-whips[1]) are enough to allow eliminations based on the tridagon link. (Notice that the Tridagon-link is available at the start and part of the whips appearing before it in this path may not be necessary to solve the puzzle.)
I planned to post a second example showing the use of longer partial-whips, but I have no time right now.
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Re: The tridagon rule

Postby pjb » Tue Mar 29, 2022 2:00 am

denis_berthier wrote:
Here is one more example (#2 in mith 246-list the-hardest-sudokus-new-thread-t6539-1190.html)
Code: Select all
     +-------+-------+-------+
     ! . . . ! . . . ! . . 1 !
     ! . . . ! . . 2 ! . 3 . !
     ! . . . ! . 4 . ! 5 6 . !
     +-------+-------+-------+
     ! . . . ! . . 7 ! . . . !
     ! . . 4 ! 8 1 . ! 2 . . !
     ! 1 9 . ! . 2 4 ! 8 . . !
     +-------+-------+-------+
     ! . 8 9 ! . . 1 ! . . 7 !
     ! 4 2 . ! . 7 8 ! . . 9 !
     ! 7 . 1 ! . 9 . ! . . . !
     +-------+-------+-------+
........1.....2.3.....4.56......7.....481.2..19..248...89..1..742..78..97.1.9....;28



This actually solves in a single step after basics:
Code: Select all
 235689  34567   235678 | 35679  3568   3569   | 479    2489   1     
 5689    14567   5678   | 15679  568    2      | 479    3      48     
 2389    137     2378   | 1379   4      39     | 5      6      28     
------------------------+----------------------+---------------------
 28     *356     28     | 3569  *356    7      | 13469  1459   3456   
*356     3567C   4      | 8      1     *356+9A | 2      57-9D  356   
 1       9      *356+7B |*356    2      4      | 8      57     356   
------------------------+----------------------+---------------------
*356     8       9      | 24    *356    1      | 346    245    7     
 4       2      *356    |*356    7      8      | 136    15     9     
 7      *356     1      | 24     9     *356    | 346    2458   234568


(9)r5c6 = (7)r6c3 - (7)r5c2 = (7-9)r5c8 => -9 r5c8; btte

Phil
pjb
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Re: The tridagon rule

Postby denis_berthier » Tue Mar 29, 2022 4:10 am

.
Hi pjb
It depends on which order you apply the rules.
denis_berthier
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Re: The tridagon rule

Postby denis_berthier » Tue Mar 29, 2022 4:24 am

.
Here is now the promised example using longer partial-whips.
It's #217 in the same list as above:
Code: Select all
+-------+-------+-------+
! . . . ! . . . ! . . 1 !
! . . . ! . . 1 ! 2 3 . !
! . . 1 ! . 2 4 ! 5 . 6 !
+-------+-------+-------+
! . . 3 ! . . . ! 6 . . !
! . 6 . ! . 5 2 ! . . . !
! 7 . 8 ! . . 6 ! . . . !
+-------+-------+-------+
! . 1 . ! . 6 5 ! 4 . 3 !
! . 3 . ! 2 4 . ! 1 5 . !
! . 4 . ! 1 . 3 ! . 6 2 !
+-------+-------+-------+
........1.....123...1.245.6..3...6...6..52...7.8..6....1..654.3.3.24.15..4.1.3.62;32
SER = 11.6

Code: Select all
Resolution state after Singles and whips[1]:
   +----------------------+----------------------+----------------------+
   ! 234569 25789  24569  ! 356789 3789   789    ! 789    4789   1      !
   ! 4569   5789   4569   ! 56789  789    1      ! 2      3      4789   !
   ! 39     789    1      ! 3789   2      4      ! 5      789    6      !
   +----------------------+----------------------+----------------------+
   ! 12459  259    3      ! 4789   1789   789    ! 6      124789 45789  !
   ! 149    6      49     ! 34789  5      2      ! 3789   14789  4789   !
   ! 7      259    8      ! 349    139    6      ! 39     1249   459    !
   +----------------------+----------------------+----------------------+
   ! 289    1      279    ! 789    6      5      ! 4      789    3      !
   ! 689    3      679    ! 2      4      789    ! 1      5      789    !
   ! 589    4      579    ! 1      789    3      ! 789    6      2      !
   +----------------------+----------------------+----------------------+
179 candidates.

Here also, a trigon-link is available at the start, but let's so as previously and apply it only at its place in the complexity hierarchy.
Code: Select all
hidden-pairs-in-a-column: c4{n5 n6}{r1 r2} ==> r2c4≠9, r2c4≠8, r2c4≠7, r1c4≠9, r1c4≠8, r1c4≠7, r1c4≠3
whip[7]: b4n4{r5c3 r4c1} - r1n4{c1 c3} - c3n2{r1 r7} - c1n2{r7 r1} - r1n6{c1 c4} - r1n5{c4 c2} - b4n5{r4c2 .} ==> r5c8≠4
whip[11]: c5n3{r1 r6} - r6c7{n3 n9} - r6c4{n9 n4} - c4n3{r6 r3} - r3c1{n3 n9} - r9n9{c1 c3} - b4n9{r5c3 r4c2} - r2n9{c2 c9} - r6c9{n9 n5} - r4n5{c9 c1} - r9n5{c1 .} ==> r1c5≠9

extended tridagon for digits 7, 8 and 9 in blocks:
        b2, with cells: r3c4 (link cell), r2c5, r1c6
        b3, with cells: r3c8, r2c9 (link cell), r1c7
        b8, with cells: r7c4, r9c5, r8c6
        b9, with cells: r7c8, r9c7, r8c9
 ==> tridagon-link[12](n3r3c4, n4r2c9)

whip[12]: c5n3{r1 r6} - r6n1{c5 c8} - r6n2{c8 c2} - r6n5{c2 c9} - r6n4{c9 c4} - c4n3{r6 r3} - r3c1{n3 n9} - c2n9{r3 r4} - r4n5{c2 c1} - r9c1{n5 n8} - c7n8{r9 r5} - r5n3{c7 .} ==> r1c5≠8

tridagon-forcing-whip-elim[15] based on tridagon-link(n4r2c9, n3r3c4)
....for n4r2c9: -
....for n3r3c4: partial-whip[3]: r5n3{c4 c7} - r6c7{n3 n9} - r6c4{n9 n4} -
 ==> r6c9≠4

t-whip[5]: r6n4{c4 c8} - c8n2{r6 r4} - c8n1{r4 r5} - c1n1{r5 r4} - r4n4{c1 .} ==> r5c4≠4
t-whip[6]: r5n7{c9 c4} - r5n3{c4 c7} - r6c7{n3 n9} - r6c9{n9 n5} - r6c2{n5 n2} - c8n2{r6 .} ==> r4c8≠7
t-whip[6]: r5n8{c9 c4} - r5n3{c4 c7} - r6c7{n3 n9} - r6c9{n9 n5} - r6c2{n5 n2} - c8n2{r6 .} ==> r4c8≠8
whip[6]: c8n2{r4 r6} - r6n1{c8 c5} - r4n1{c5 c1} - r4n2{c1 c2} - r4n5{c2 c9} - r6c9{n5 .} ==> r4c8≠9

tridagon-forcing-whip-elim[17] based on tridagon-link(n4r2c9, n3r3c4)
....for n4r2c9: -
....for n3r3c4: partial-whip[5]: r5n3{c4 c7} - r6n3{c7 c5} - r1c5{n3 n7} - c7n7{r1 r9} - c7n8{r9 r1} -
 ==> r2c9≠8

tridagon-forcing-whip-elim[19] based on tridagon-link(n4r2c9, n3r3c4)
....for n4r2c9: -
....for n3r3c4: partial-whip[7]: r5n3{c4 c7} - r6n3{c7 c5} - r1c5{n3 n7} - c7n7{r1 r9} - c7n8{r9 r1} - r3n8{c8 c2} - c2n7{r3 r2} -
 ==> r2c9≠7

The end of the path is in W9 and has nothing noticeable: Show
hidden-pairs-in-a-row: r2{n7 n8}{c2 c5} ==> r2c5≠9, r2c2≠9, r2c2≠5
whip[5]: r2n4{c3 c9} - r5n4{c9 c1} - r5c3{n4 n9} - r2n9{c3 c1} - b7n9{r7c1 .} ==> r1c3≠4
whip[9]: c8n2{r6 r4} - c8n1{r4 r5} - c1n1{r5 r4} - r4n4{c1 c4} - c9n4{r4 r2} - c3n4{r2 r5} - r5c1{n4 n9} - r2n9{c1 c3} - b7n9{r7c3 .} ==> r6c8≠4
hidden-single-in-a-row ==> r6c4=4
t-whip[6]: r2c9{n9 n4} - c3n4{r2 r5} - r4n4{c1 c8} - c8n2{r4 r6} - c8n1{r6 r5} - r5c1{n1 .} ==> r5c9≠9, r2c1≠9
whip[5]: r2n9{c9 c3} - r3c1{n9 n3} - c4n3{r3 r5} - b5n9{r5c4 r6c5} - c2n9{r6 .} ==> r4c9≠9
whip[5]: r2n9{c9 c3} - r3c1{n9 n3} - c4n3{r3 r5} - r5n9{c4 c1} - b7n9{r7c1 .} ==> r6c9≠9
naked-single ==> r6c9=5
t-whip[5]: c8n2{r4 r6} - r6c2{n2 n9} - r5c3{n9 n4} - r5c1{n4 n1} - c8n1{r5 .} ==> r4c8≠4
singles ==> r1c8=4, r2c9=9
t-whip[6]: r5n3{c4 c7} - r6n3{c7 c5} - r1c5{n3 n7} - c7n7{r1 r9} - r8c9{n7 n8} - b6n8{r4c9 .} ==> r5c4≠8
whip[1]: r5n8{c9 .} ==> r4c9≠8
t-whip[7]: r4c9{n4 n7} - r5n7{c9 c4} - c4n3{r5 r3} - r3c1{n3 n9} - b2n9{r3c4 r1c6} - r8n9{c6 c3} - r5c3{n9 .} ==> r5c9≠4, r4c1≠4
hidden-single-in-a-row ==> r4c9=4
whip[1]: b6n7{r5c9 .} ==> r5c4≠7
whip[8]: r5c4{n9 n3} - c7n3{r5 r6} - c5n3{r6 r1} - c1n3{r1 r3} - r3n9{c1 c2} - r6n9{c2 c8} - b9n9{r7c8 r9c7} - c5n9{r9 .} ==> r4c4≠9
whip[7]: r4c4{n8 n7} - r7c4{n7 n9} - b2n9{r3c4 r1c6} - c6n7{r1 r8} - r8c9{n7 n8} - r7c8{n8 n7} - r3c8{n7 .} ==> r3c4≠8
biv-chain[3]: r2c2{n7 n8} - r3n8{c2 c8} - b3n7{r3c8 r1c7} ==> r1c2≠7
whip[7]: r5c4{n9 n3} - r3c4{n3 n7} - r3c8{n7 n8} - r7c8{n8 n7} - r8c9{n7 n8} - b8n8{r8c6 r9c5} - r2c5{n8 .} ==> r7c4≠9
naked-pairs-in-a-column: c4{r4 r7}{n7 n8} ==> r3c4≠7
naked-pairs-in-a-row: r3{c1 c4}{n3 n9} ==> r3c2≠9
naked-pairs-in-a-block: b1{r2c2 r3c2}{n7 n8} ==> r1c2≠8
z-chain[4]: b8n9{r8c6 r9c5} - b9n9{r9c7 r7c8} - b9n8{r7c8 r9c7} - r1n8{c7 .} ==> r8c6≠8
biv-chain[3]: b3n8{r3c8 r1c7} - c6n8{r1 r4} - c4n8{r4 r7} ==> r7c8≠8
biv-chain[3]: r5c9{n7 n8} - b9n8{r8c9 r9c7} - r1c7{n8 n7} ==> r5c7≠7
z-chain[2]: c7n7{r9 r1} - b2n7{r1c6 .} ==> r9c5≠7
biv-chain[3]: r7n8{c1 c4} - r9c5{n8 n9} - b9n9{r9c7 r7c8} ==> r7c1≠9
biv-chain[3]: b9n8{r8c9 r9c7} - r9c5{n8 n9} - r8c6{n9 n7} ==> r8c9≠7
singles ==> r8c9=8, r5c9=7
biv-chain[3]: r9c7{n7 n9} - b8n9{r9c5 r8c6} - r8n7{c6 c3} ==> r9c3≠7
stte


In this example, we have 3 different eliminations based on the same tridagon link, but using partial-whips of lengths 3, 5 and 7.
Same remark as before: some of the whips before these may be unnecessary.
I'm working on a functionality that provides the choice of giving some priority to exotic patterns such as tridagons.
denis_berthier
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