The New Sudoku Players' Forum

[denis_berthier]

.
This is a puzzle published yesterday by mith in the "hardest" thread (p. 92). It has a non-degenerated anti-tridagon pattern at the start, but contrary to all the examples before, it's in T&E(2).
It is indeed in B2B (low part of T&E(2)).
There's a trivial elimination at the start (which doesn't change the B2B classification).
I'm curious to see if you can find more tridagon-related eliminations.

Code: Select all

    +-------+-------+-------+
     ! . . . ! . . . ! . . . !
     ! . . 1 ! . . 2 ! . 3 4 !
     ! . . 2 ! . 1 3 ! 5 6 . !
     +-------+-------+-------+
     ! . 1 7 ! . 8 6 ! . . 3 !
     ! . 6 . ! 2 . 1 ! . 8 . !
     ! 2 . 8 ! 7 3 . ! 6 . . !
     +-------+-------+-------+
     ! . 7 . ! . 2 8 ! . . . !
     ! 1 2 . ! 3 6 . ! . 7 . !
     ! 8 . 6 ! 1 . 7 ! 3 . . !
     +-------+-------+-------+
...........1..2.34..2.1356..17.86..3.6.2.1.8.2.873.6...7..28...12.36..7.8.61.73..  ER=10.6/1.2/1.2
SER = 10.6

Code: Select all

Resolution state after Singles and whips[1]:
   +-------------------+-------------------+-------------------+
   ! 45679 3     459   ! 45689 4579  459   ! 12789 129   12789 !
   ! 5679  589   1     ! 5689  579   2     ! 789   3     4     !
   ! 479   489   2     ! 489   1     3     ! 5     6     789   !
   +-------------------+-------------------+-------------------+
   ! 459   1     7     ! 459   8     6     ! 249   2459  3     !
   ! 3459  6     3459  ! 2     459   1     ! 479   8     579   !
   ! 2     459   8     ! 7     3     459   ! 6     1459  159   !
   +-------------------+-------------------+-------------------+
   ! 3459  7     3459  ! 459   2     8     ! 149   1459  6     !
   ! 1     2     459   ! 3     6     459   ! 489   7     589   !
   ! 8     459   6     ! 1     459   7     ! 3     2459  259   !
   +-------------------+-------------------+-------------------+

Code: Select all

OR2-anti-tridagon[12] for digits 4, 5 and 9 in blocks:
        b4, with cells: r4c1, r5c3, r6c2
        b5, with cells: r4c4, r5c5, r6c6
        b7, with cells: r7c1, r8c3, r9c2
        b8, with cells: r7c4, r8c6, r9c5
with 2 guardians: n3r5c3 n3r7c1

Trid-OR2-whip[1]: OR2{{n3r7c1 n3r5c3 | .}} ==> r7c3≠3

[Edit, after Leren's post]: after the previous elimination:

Code: Select all

hidden-single-in-a-block ==> r7c1=3
hidden-single-in-a-block ==> r5c3=3
   +-------------------+-------------------+-------------------+
   ! 45679 3     459   ! 45689 4579  459   ! 12789 129   12789 !
   ! 5679  589   1     ! 5689  579   2     ! 789   3     4     !
   ! 479   489   2     ! 489   1     3     ! 5     6     789   !
   +-------------------+-------------------+-------------------+
   ! 459   1     7     ! 459   8     6     ! 249   2459  3     !
   ! 459   6     3     ! 2     459   1     ! 479   8     579   !
   ! 2     459   8     ! 7     3     459   ! 6     1459  159   !
   +-------------------+-------------------+-------------------+
   ! 3     7     459   ! 459   2     8     ! 149   1459  6     !
   ! 1     2     459   ! 3     6     459   ! 489   7     589   !
   ! 8     459   6     ! 1     459   7     ! 3     2459  259   !
   +-------------------+-------------------+-------------------+

The question now is: can you find any other elimination due to the OR2-anti-tridagon or, at least, too what remains of it?
.

[Leren]

Well, I'm no Trigadon expert but just by inspection r5c3 or r7c1 = 3, so r5c1 and r7c3 <> 3, so in fact both r5c3 and r7c1 are both = 3. Leren

[denis_berthier]

Well, I'm no Trigadon expert but just by inspection r5c3 or r7c1 = 3, so r5c1 and r7c3 <> 3, so in fact both r5c3 and r7c1 are both = 3. Leren

OK. I should have written them also. Immediately after the OR2 elimination, there are 2 Singles:

Code: Select all

hidden-single-in-a-block ==> r7c1=3
hidden-single-in-a-block ==> r5c3=3

There are also long whips and g-whips.
But what I'm looking for is any elimination due to the OR2-anti-tridagon - if any.

I'll complete my first post.

[Leren]

OK I'll bite. Since the anti Trigadon is in Box 3, the Trigadon digits must actually conform to the Trigadon box pattern, so r1c8 = 9. That's just pure speculation. It's True in the solution but it might be a fluke.

Did I escape destruction or did Thor bash my head in with his hammer ? Leren

[denis_berthier]

OK I'll bite. Since the anti Trigadon is in Box 3, the Trigadon digits must actually conform to the Trigadon box pattern, so r1c8 = 9. That's just pure speculation. It's True in the solution but it might be a fluke.
Did I escape destruction or did Thor bash my head in with his hammer ? Leren

Box 3 is "opposite" the 4 blocks of the tridagon pattern. Mith's criterion says that this box is the only one that can contain any of the 3 digits as a decided value - when the non-degenerated pattern is still present.
Here, the pattern is highly degenerated. But even if it was still there, I can't follow your argument about r1c8 = 9.

My idea is, at this point, nothing more can be deduced from the remnants of the tridagon. As a result, the tridagon is almost useless to simplify the puzzle. Let's see if anyone has better insights.

[marek stefanik]

I managed to solve the puzzle using TH, however it wasn't much simpler than an optimised solution in B2B could be.
Note that the puzzle is also susceptible to relabeling, for example placing 4r8c3, 5r8c6 we get down to 7.7 skfr.
To better demonstrate my findings, I will show how one can accomplish a few placements even without the 45 in b3.
...........1..2.3...2.13.6..17.86..3.6.2.1.8.2.873.6...7..28...12.36..7.8.61.73..;108 solutions

After TH and singles:

Code: Select all

.------------------.------------------.-------------------------.
| 45679  3     459 | 45689  4579  459 | 1245789  12459  1245789 |
| 45679  4589  1   | 45689  4579  2   | 45789    3      45789   |
| 4579   4589  2   | 4589   1     3   | 45789    6      45789   |
:------------------+------------------+-------------------------:
|#459    1     7   |#459    8     6   | 2459     2459   3       |
|#459    6     3   | 2     #459   1   | 4579     8      4579    |
| 2     #459   8   | 7      3    #459 | 6        1459   1459    |
:------------------+------------------+-------------------------:
| 3      7    #459 |#459    2     8   | 1459     1459   6       |
| 1      2    #459 | 3      6    #459 | 4589     7      4589    |
| 8     #459   6   | 1     #459   7   | 3        2459   2459    |
'------------------'------------------'-------------------------'

Notice that the 459 in b47 form vertical triples of opposite parities.
Exactly one digit is therefore in the same minirow of both boxes.
Obviously, this cannot the last row because of c2.

Suppose b4p4=b7p6, and WLOG let them be 4s.

Code: Select all

.----------------.------------------.-------------------------.
| 5679  3    *59 | 45689  4579 *459 | 1245789 *12459  1245789 |
| 5679  4589  1  | 45689  4579  2   | 45789    3      45789   |
| 579   4589  2  | 4589   1     3   | 45789    6      45789   |
:----------------+------------------+-------------------------:
| 59    1     7  | 459    8     6   | 2459    *2459   3       |
| 4     6     3  | 2    #*59    1   |*579      8     *579     |
| 2    *59    8  | 7      3    *4–59| 6       *1459   1459    |
:----------------+------------------+-------------------------:
| 3     7    *59 | 459    2     8   | 1459    *1459   6       |
| 1     2     4  | 3      6  #**59  |*589      7     *589     |
| 8    *59    6  | 1      4–59  7   | 3       *2459   2459    |
'----------------'------------------'-------------------------'

59r58c2368 \ r16b679
Each of 59 must appear at least once in #-marked cells => they are a RP => -59r6c6, r9c5
(Note that r8c6 has to be covered twice, ie. the pattern is not fully rank0.)

Code: Select all

.----------------.----------------.-------------------------.
| 5679  3    #59 | 45689  579 #59 | 1245789  12459  1245789 |
| 5679  4589  1  | 45689  579  2  | 45789    3      45789   |
| 579   4589  2  | 4589   1    3  | 45789    6      45789   |
:----------------+----------------+-------------------------:
| 59    1     7  | 59     8    6  | 2459     2459   3       |
| 4     6     3  | 2      59   1  | 579      8      579     |
| 2     59    8  | 7      3    4  | 6        159    159     |
:----------------+----------------+-------------------------:
| 3     7    #59 |#59     2    8  | 1459     1459   6       |
| 1     2     4  | 3      6   #59 | 589      7      589     |
| 8     59    6  | 1      4    7  | 3        259    259     |
'----------------'----------------'-------------------------'

bivalue oddagon, ie. contra.
Therefore b4p1=b7p3.

Let us now relabel in b4 (459 vertically) and accordingly in b7 (495 vertically).
Note that the relabeled puzzle has 108/(3!)=18 solutions, as the permutation of 459 is now fixed.
After basics (just taking the 67 in r1 and putting them into the original puzzle simplifies it to 9.1 skfr):

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.-----------.------------.-----------------.
| 6   3   5 | 48  7   9  |G1248  124  128–4|
| 79  48  1 | 6   5   2  |G4789  3    789–4|
| 79  48  2 | 48  1   3  | 579   6    579  |
:-----------+------------+-----------------:
| 4   1   7 | 59  8   6  | 259   259  3    |
| 5   6   3 | 2  #49  1  |#479   8   G479  |
| 2   9   8 | 7   3  #45 | 6     145  145  |
:-----------+------------+-----------------:
| 3   7   4 | 59  2   8  | 159   159  6    |
| 1   2   9 | 3   6  #45 |#458   7   G458  |
| 8   5   6 | 1   49  7  | 3     249  249  |
'-----------'------------'-----------------'

broken wing 4r58c67b5 => -4r12c9
(Or either option for 4b58 results in claiming candidates in b1 from one of c78, as discussed by StrmCkr here.)

If we now restrict b3b67 to 459 as in the original puzzle, we reach the (relabeled) solution with singles.
Alternatively, the 4 in the original puzzle can only correspond to our 9.
After placing 4s in the original puzzle where our 9s are, we only need a skyscraper to finish.

Marek

[denis_berthier]

.
Thanks Leren & Marek for trying.

Yes, we can use relabelling, but as this is in T&E(2), this doesn't show any real simplification that would be due to Tridagons.
As for your second approach:
- "Suppose b4p4=b7p6" puts it in the T&E-ish category.
- "Let us now relabel in b4" puts it in the T&E(2)-ish solutions as before.

I proposed this puzzle in order to check if one could find a pattern-based Tridagon-related elimination that would allow to reduce the puzzle complexity.
It seems that the tridagon pattern is useless in this example.

This is not specific to Tridagons. For any pattern, there are cases with huge impact and cases with no impact. Indeed, more generally, any elimination of a single candidate can have a huge impact (e.g. making it reducible to Singles) or no impact at all. I think it's nice to have an example for Tridagons (thanks mith).
.

[eleven]

Notice that the 459 in b47 form vertical triples of opposite parities.
Exactly one digit is therefore in the same minirow of both boxes.
Obviously, this cannot the last row because of c2.

Nice observation.

Suppose b4p4=b7p6, and WLOG let them be 4s.

Code: Select all

.----------------.------------------.-------------------------.
| 5679  3    *59 | 45689  4579 *459 | 1245789 *12459  1245789 |
| 5679  4589  1  | 45689  4579  2   | 45789    3      45789   |
| 579   4589  2  | 4589   1     3   | 45789    6      45789   |
:----------------+------------------+-------------------------:
| 59    1     7  | 459    8     6   | 2459    *2459   3       |
| 4     6     3  | 2    #*59    1   |*579      8     *579     |
| 2    *59    8  | 7      3    *4–59| 6       *1459   1459    |
:----------------+------------------+-------------------------:
| 3     7    *59 | 459    2     8   | 1459    *1459   6       |
| 1     2     4  | 3      6  #**59  |*589      7     *589     |
| 8    *59    6  | 1      4–59  7   | 3       *2459   2459    |
'----------------'------------------'-------------------------'

59r58c2368 \ r16b679
Each of 59 must appear at least once in #-marked cells => they are a RP => -59r6c6, r9c5
(Note that r8c6 has to be covered twice, ie. the pattern is not fully rank0.)

That's hard to verify for me, so i tried to find another way for the contradiction.

Code: Select all

+------------------+-------------------+--------------------------+
|  5679  3    x59  | 45689  4579 y459  | 1245789 *12459  1245789  |
|  5679  4589  1   | 45689  4579  2    | 45789    3      45789    |
|  579   4589  2   | 4589   1     3    | 45789    6      45789    |
+------------------+-------------------+--------------------------+
| x59    1     7   |y459    8     6    | 2459    *2459   3        |
|  4     6     3   | 2     x59    1    | 579      8      579      |
|  2    y59    8   | 7      3     4-59 | 6       #1459   1459     |
+------------------+-------------------+--------------------------+
|  3     7    y59  | 459    2     8    | 1459    #1459   6        |
|  1     2     4   | 3      6    x59   | 589      7      589      |
|  8    x59    6   | 1     y459   7    | 3       *2459   2459     |
+------------------+-------------------+--------------------------+

The digit x from 59 in r4c1 goes to r8c2 and r1c3 too, the other (y) to r6c2 and r7c3
So in c8 x is left in r67. Both imply x in r5c5, r8c6 (and 4r6c6), and y in r1c6,r4c4,r9c5 killing all y's in c8.