The New Sudoku Players' Forum

[AnotherLife]

This puzzle seems much harder than regular puzzles of Unfair 6.* level. Who can find a good solution to it? As usual, human-friendly methods are appreciated.

Code: Select all

|.2.|.7.|.1.|
|13.|6.9|.57|
|...|...|...|
|---+---+---|
|.1.|7.2|.9.|
|6..|...|..1|
|.5.|3.8|.7.|
|---+---+---|
|...|...|...|
|76.|8.5|.43|
|.8.|.3.|.2.|

.2..7..1.13.6.9.57..........1.7.2.9.6.......1.5.3.8.7..........76.8.5.43.8..3..2.


[denis_berthier]

.

Code: Select all

Resolution state after Singles and whips[1]:
   +----------------------+----------------------+----------------------+
   ! 4589   2      6      ! 45     7      3      ! 489    1      489    !
   ! 1      3      48     ! 6      248    9      ! 248    5      7      !
   ! 4589   49     7      ! 245    2458   1      ! 234689 368    24689  !
   +----------------------+----------------------+----------------------+
   ! 348    1      348    ! 7      6      2      ! 3458   9      458    !
   ! 6      7      238    ! 59     59     4      ! 238    38     1      !
   ! 249    5      249    ! 3      1      8      ! 246    7      246    !
   +----------------------+----------------------+----------------------+
   ! 2349   49     123459 ! 1249   249    67     ! 156789 68     5689   !
   ! 7      6      129    ! 8      29     5      ! 19     4      3      !
   ! 49     8      1459   ! 149    3      67     ! 15679  2      569    !
   +----------------------+----------------------+----------------------+
143 candidates.

The natural solution uses only Subsets and bivalue-chains of size ≤ 3 and corresponds perfectly to the SER.

Code: Select all

naked-pairs-in-a-block: b7{r7c2 r9c1}{n4 n9} ==> r9c3≠9, r9c3≠4, r8c3≠9, r7c3≠9, r7c3≠4, r7c1≠9, r7c1≠4
hidden-single-in-a-column ==> r6c3=9
finned-x-wing-in-columns: n4{c2 c5}{r7 r3} ==> r3c4≠4
swordfish-in-columns: n2{c1 c4 c9}{r6 r7 r3} ==> r7c5≠2, r7c3≠2, r6c7≠2, r3c7≠2, r3c5≠2
naked-pairs-in-a-row: r7{c2 c5}{n4 n9} ==> r7c9≠9, r7c7≠9, r7c4≠9, r7c4≠4
biv-chain[3]: r8c7{n1 n9} - r8c5{n9 n2} - r7c4{n2 n1} ==> r7c7≠1
hidden-triplets-in-a-row: r7{n1 n2 n3}{c3 c4 c1} ==> r7c3≠5
hidden-single-in-a-block ==> r9c3=5
biv-chain[3]: c1n5{r3 r1} - r1c4{n5 n4} - r9n4{c4 c1} ==> r3c1≠4
biv-chain[3]: r9c1{n9 n4} - c4n4{r9 r1} - r1n5{c4 c1} ==> r1c1≠9
whip[1]: r1n9{c9 .} ==> r3c7≠9, r3c9≠9
biv-chain[3]: c9n9{r1 r9} - r9c1{n9 n4} - c4n4{r9 r1} ==> r1c9≠4
finned-swordfish-in-rows: n4{r9 r1 r6}{c1 c4 c7} ==> r4c7≠4
biv-chain[3]: r1c9{n8 n9} - r9c9{n9 n6} - c8n6{r7 r3} ==> r3c8≠8
biv-chain[3]: r1c9{n8 n9} - r9c9{n9 n6} - r7c8{n6 n8} ==> r7c9≠8
biv-chain[3]: c7n5{r4 r7} - r7n8{c7 c8} - r5c8{n8 n3} ==> r4c7≠3
whip[1]: b6n3{r5c8 .} ==> r5c3≠3
biv-chain[3]: c7n2{r2 r5} - r5c3{n2 n8} - r2c3{n8 n4} ==> r2c7≠4
finned-x-wing-in-rows: n4{r7 r2}{c5 c2} ==> r3c2≠4
stte

Restrictions on the number of steps require longer chains. That doesn't imply anything on the SER, which isn't designed to take the number of steps into account.

Here are the simplest two 2-step solutions I found:

Code: Select all

whip[4]: r2c3{n8 n4} - c2n4{r3 r7} - c5n4{r7 r3} - c5n8{r3 .} ==> r2c7≠8
whip[7]: c9n2{r6 r3} - r2c7{n2 n4} - r2c3{n4 n8} - r5c3{n8 n3} - c8n3{r5 r3} - r3n6{c8 c7} - r6c7{n6 .} ==> r5c7≠2
stte

OR:

Code: Select all

whip[4]: r2c3{n8 n4} - c2n4{r3 r7} - c5n4{r7 r3} - c5n8{r3 .} ==> r2c7≠8
whip[7]: c9n2{r6 r3} - r2c7{n2 n4} - r2c3{n4 n8} - r5c3{n8 n3} - c8n3{r5 r3} - r3n6{c8 c7} - r6c7{n6 .} ==> r6c1≠2
stte

The following is 1-step for those who consider Pairs as no-step:

Code: Select all

naked-pairs-in-a-block: b7{r7c2 r9c1}{n4 n9} ==> r7c1≠4, r9c3≠9, r9c3≠4, r8c3≠9, r7c3≠9, r7c3≠4, r7c1≠9
hidden-single-in-a-column ==> r6c3=9
whip[8]: c3n4{r2 r4} - r6c1{n4 n2} - c9n2{r6 r3} - r2c7{n2 n4} - r6c7{n4 n6} - r3n6{c7 c8} - c8n3{r3 r5} - r5c3{n3 .} ==> r2c3≠8
stte

Also a 1-step solution using nukes:

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FORCING-T&E applied to bivalue candidates n4r2c3 and n8r2c3 :
===> 11 values decided in both cases: n5r1c4 n9r5c4 n5r5c5 n7r5c2 n5r3c1 n6r1c3 n7r3c3 n2r3c4 n9r6c3 n6r6c7 n5r9c3
===> 81 candidates eliminated in both cases: n4r1c1 n5r1c1 n4r1c3 n5r1c3 n8r1c3 n9r1c3 n4r1c4 n6r1c7 n9r1c7 n4r1c9 n6r1c9 n8r1c9 n2r2c5 n4r2c5 n8r2c7 n4r3c1 n8r3c1 n9r3c1 n7r3c2 n4r3c3 n5r3c3 n6r3c3 n8r3c3 n9r3c3 n4r3c4 n5r3c4 n2r3c5 n5r3c5 n2r3c7 n4r3c7 n6r3c7 n8r3c7 n8r3c8 n2r3c9 n4r3c9 n9r3c9 n4r4c1 n3r4c3 n4r4c7 n8r4c7 n8r4c9 n9r5c2 n7r5c3 n8r5c3 n9r5c3 n5r5c4 n9r5c5 n3r5c7 n5r5c7 n9r6c1 n2r6c3 n4r6c3 n2r6c7 n4r6c7 n6r6c9 n4r7c1 n5r7c1 n9r7c1 n2r7c3 n4r7c3 n5r7c3 n9r7c3 n2r7c4 n9r7c4 n4r7c5 n1r7c7 n6r7c7 n8r7c7 n9r7c7 n6r7c9 n9r7c9 n9r8c3 n5r9c1 n1r9c3 n4r9c3 n9r9c3 n9r9c4 n5r9c7 n6r9c7 n9r9c7 n5r9c9
stte

[Cenoman]

In five "simple" steps (one basic 3-fish, two Wings, two AIC's with ALS/AHS):

Code: Select all

 +---------------------+---------------------+-------------------------+
 |  4589   2    6      |  45     7      3    |  489      1     489     |
 |  1      3   a48     |  6     a248*   9    | A24-8*    5     7       |
 |  4589  d49   7      |  245    458-2  1    |  34689-2  368  B24689   |
 +---------------------+---------------------+-------------------------+
 |  348    1    348    |  7      6      2    |  3458     9     458     |
 |  6      7    238*   |  59     59     4    |  238*     38    1       |
 | D24     5    9      |  3      1      8    |  6-24     7    C246     |
 +---------------------+---------------------+-------------------------+
 |  23    c49   135-2  |  1249  b49-2   67   |  156789   68    5689    |
 |  7      6    12*    |  8     b29*    5    |  19       4     3       |
 |  49     8    15     |  149    3      67   |  15679    2     569     |
 +---------------------+---------------------+-------------------------+

1. Swordfish: (2)r258\c357 => -2 r3c57, r6c7, r7c35
2. (8=42)r2c35 - (2=94)r78c5 - r7c2 = r3c2 - (4=8)r2c3 => -8 r2c7
3. W-Wing: (4=2)r2c7 - r3c9 = r6c9 - (2=4)r6c1 =>-4r6c7; one placement

Code: Select all

 +--------------------+-------------------+-----------------------+
 |  4589   2    6     |  45    7     3    |  489    1     489     |
 |  1      3   C48    |  6     248   9    | C24     5     7       |
 |  4589   49   7     |  245   458   1    |  3489  d368  d24689   |
 +--------------------+-------------------+-----------------------+
 |  348    1    348   |  7     6     2    |  3458   9     458     |
 |  6      7  Aa2-38  |  59    59    4    |Bb238   e38    1       |
 |  24     5    9     |  3     1     8    |  6      7    c24      |
 +--------------------+-------------------+-----------------------+
 |  23     49   135   |  12    49    67   |  1578   68    568     |
 |  7      6    12    |  8     29    5    |  19     4     3       |
 |  49     8    15    |  149   3     67   |  1579   2     569     |
 +--------------------+-------------------+-----------------------+

4. (2)r5c3 = r5c7 - r6c9 = (26-3)r3c89 = (3)r5c8 =>-3r5c3
5. H-Wing: (2)r5c3 = r5c7 - (2=48)r2c37 =>-8r5c3; ste

[RSW]

After basics:

Code: Select all

 +-------------+------------+-----------------------+
 | 4589 2  6   | 45  7   3  | 489      1      489   |
 | 1    3  48  | 6   248 9  | 248      5      7     |
 | 4589 49 7   | 25  458 1  |*3489(6?) 368    24689 |
 +-------------+------------+-----------------------+
 | 348  1  348 | 7   6   2  | 3458     9      458   |
 | 6    7  238 | 59  59  4  | 238      38     1     |
 | 24   5  9   | 3   1   8  | 46       7      246   |
 +-------------+------------+-----------------------+
 | 23   49 135 | 12  49  67 | 15678    68     568   |
 | 7    6  12  | 8   29  5  | 19       4      3     |
 | 9+4? 8  15  | 149 3   67 | 15679    2      569   |
 +-------------+------------+-----------------------+


Unfortunately, my human solving skills are much less elegant than my computer programming skills. So, my human solutions tend to be much uglier than the output of my computer solver.

However, honouring the intent of the OP, I applied a pencil and paper technique that I used in my early days. Essentially: pick a likely candidate based on an heuristic, and follow the inferences, crossing out the candidates in the PM grid, until it leads to a contradiction. My heuristic chose the following, most likely possibilities, in this order:
+5r3c4, +4r6c9, +6r6c9, +4r7c5, +2r7c4, +4r9c1... and more.

The 6th one, +4r9c1, leads, with minimum effort, to a contradiction in the state of candidate 6 in cell r3c7, and thus => -4r9c1; stte

Not pretty, but no computer.

------
Edit: Thanks for this puzzle. I need more of these in order to expand my skills.

[denis_berthier]

I applied a pencil and paper technique that I used in my early days. Essentially: pick a likely candidate based on an heuristic, and follow the inferences, crossing out the candidates in the PM grid, until it leads to a contradiction. My heuristic chose the following, most likely possibilities, in this order:
+5r3c4, +4r6c9, +6r6c9, +4r7c5, +2r7c4, +4r9c1... and more.
The 6th one, +4r9c1, leads, with minimum effort, to a contradiction in the state of candidate 6 in cell r3c7, and thus => -4r9c1; stte.

As each try sounds like T&E and a T&E elimination can be done by some braid (and most of the time also by some whip) of unrestricted length, I tried your 6th case, starting from the same PM (where a Pair has already been applied):

Code: Select all

   +-------------------+-------------------+-------------------+
   ! 4589  2     6     ! 45    7     3     ! 489   1     489   !
   ! 1     3     48    ! 6     248   9     ! 248   5     7     !
   ! 4589  49    7     ! 25    458   1     ! 34689 368   24689 !
   +-------------------+-------------------+-------------------+
   ! 348   1     348   ! 7     6     2     ! 3458  9     458   !
   ! 6     7     238   ! 59    59    4     ! 238   38    1     !
   ! 24    5     9     ! 3     1     8     ! 46    7     246   !
   +-------------------+-------------------+-------------------+
   ! 23    49    135   ! 12    49    67    ! 15678 68    568   !
   ! 7     6     12    ! 8     29    5     ! 19    4     3     !
   ! 49    8     15    ! 149   3     67    ! 15679 2     569   !
   +-------------------+-------------------+-------------------+

Using CSP-Rule fonction "try-to-eliminate-candidates", I got:
whip[10]: b4n4{r6c1 r4c3} - r2c3{n4 n8} - c1n8{r3 r4} - b4n3{r4c1 r5c3} - r5n2{c3 c7} - c9n2{r6 r3} - r2c7{n2 n4} - r6c7{n4 n6} - r3n6{c7 c8} - c8n3{r3 .} ==> r9c1≠4
stte


Could you say more about your heuristics? It's a side of Sudoku that's rarely mentioned.

[RSW]

Using CSP-Rule fonction "try-to-eliminate-candidates", I got:
whip[10]: b4n4{r6c1 r4c3} - r2c3{n4 n8} - c1n8{r3 r4} - b4n3{r4c1 r5c3} - r5n2{c3 c7} - c9n2{r6 r3} - r2c7{n2 n4} - r6c7{n4 n6} - r3n6{c7 c8} - c8n3{r3 .} ==> r9c1≠4
stte

Without looking at it in deep detail, that looks roughly like the elimination that I got.

As each try sounds like T&E

Yes, it's certainly T&E. Though we may not want to admit it, that's the way humans solve problems.

Could you say more about your heuristics? It's a side of Sudoku that's rarely mentioned.

Embarrassingly, it's not very sophisticated at all. I just compile a list of the strong links: the bilocal digits, and the bivalue cells. Then, for each of these, I look at the number of candidates that would be eliminated (at the first step) for either of the the two choices in each case. Then, I sort the list by the mean number of eliminations. The step that is most likely to give the most eliminations goes to the top of the list. Although it's a rather crude method, I find that the shortest solution is nearly always close to the top of the list.

[denis_berthier]

As each try sounds like T&E

Yes, it's certainly T&E. Though we may not want to admit it, that's the way humans solve problems.

I have no problem with admitting this. I have many testimonies to confirm it (friends, students...)
Braids or whips can be seen as a way of replacing "abominable T&E" by pure logic chain patterns that do the same eliminations ("lovely braids" - the name of an old thread of mine - and most of the time by structurally much simpler whips).

Could you say more about your heuristics? It's a side of Sudoku that's rarely mentioned.

Embarrassingly, it's not very sophisticated at all. I just compile a list of the strong links: the bilocal digits, and the bivalue cells. Then, for each of these, I look at the number of candidates that would be eliminated (at the first step) for either of the the two choices in each case. Then, I sort the list by the mean number of eliminations. The step that is most likely to give the most eliminations goes to the top of the list. Although it's a rather crude method, I find that the shortest solution is nearly always close to the top of the list.

Apart from the bivalue restriction and a few more details, that's what Defise and I also do in the fewer-steps method.
As any heuristics, there's no guarantee it will work every time, but it does quite often.

[RSW]

Apart from the bivalue restriction...

Now, I'm very curious. What is the difference between a bivalue restriction and a bilocal restriction?

[denis_berthier]

Apart from the bivalue restriction...

Now, I'm very curious. What is the difference between a bivalue restriction and a bilocal restriction?

None. What I call bivalue in my super-symmetric view of Sudoku, is what you call bivalue (rc-bivalue for me) or bilocal (rn, cn or bn bivalue for me).

[eleven]

Similar to Cenoman's, 5 or 6 steps. I could not see a nice, effective move to shorten it.
So for me it's an easy, but long puzzle.