The New Sudoku Players' Forum

[vidarino]

There are 3 candidates for digit 6 in row 6. The first one is at R6C5. The other 2 would both force R3C6 to 6. R1C5, R2C5 and R4C6 are the shared peers for R6C5 and R3C6. Those can be eliminated.

I don't see how you get the "other 2 would both force R3C6 to 6" without involving the ER in box 9. Would you please clarify?

Either of the two would eliminate the one at R4C7, which would lock the candidates in box 3 to col 7, leaving a single in row 3 in R3C6, zapping the one in R4C6.

[ronk]

Either of the two would eliminate the one at R4C7, which would lock the candidates in box 3 to col 7, leaving a single in row 3 in R3C6, zapping the one in R4C6.

That's a lot less obscure than what I saw at first. Due to the ER in box 9, my reasoning was that grouped candidates r3c89 and r6c89 behaved like a conjugate pair. Then r3c6 and r6c5, both seen by r12c5 and r4c6, would also bear opposite conjugate colors.

Code: Select all

 .  6  . | .  6  6 | 6  6  . 
 .  .  6 | .  6  . | 6  .  . 
 .  .  . | .  .  A | .  a  a 
---------+---------+---------
 .  .  . | .  6  6 | 6  6  6 
 .  .  . | .  .  . | .  .  . 
 .  .  . | .  a  . | .  A  A 
---------+---------+---------
 .  6  . | .  .  . | .  B  . 
 .  .  . | .  .  . | .  .  . 
 .  .  6 | .  .  . | .  B  b 

Valid but obscure, I'm afraid. So obscure, in fact, that I lost track of it myself in my prior (now deleted) post.

Ron

[ravel]

Hmm, probably not quite what you're asking ...

Yes, not exactly what i wanted, but it gives another way for a nice loop, thanks:
[R4C6]-6-[R4C7]=6=[R1C7|R2C7]-6-[R3C89]=6=[R3C6]-6-[R4C6]

This is a puzzle (menneske 168386) that can be solved with grouped coloring of number 2:

Code: Select all

-------------
|.9.|27.|.43|
|.6.|..9|...|
|..4|...|9.7|
-------------
|...|.8.|...|
|..3|4.5|7..|
|..9|...|1.8|
-------------
|..5|82.|...|
|...|..3|...|
|27.|...|.5.|
-------------



[Myth Jellies]

Code: Select all

 .  6  . | .  6  6 | 6  6  . 
 .  .  6 | .  6  . | 6  .  . 
 .  .  . | .  .  6b| .  6B 6B
---------+---------+---------
 .  .  . | .  6  6 | 6  6  6 
 .  .  . | .  .  . | .  .  . 
 .  .  . | .  6a . | .  6A 6A
---------+---------+---------
 .  6  . | .  .  . | *  6  . 
 .  .  . | .  .  . | *  .  . 
 .  .  6 | .  .  . | *  6  6 

I see it more like this. You have two conjugate color pairings, (A, a) and (B, b). The starred empty cells in box 9 mean that the A-group in r6c89 and the B-group in r3c89 cannot both contain a true candidate. (They could both be false though.) Thus, via multi-coloring with groups, either a or b (or both) must be true. Therefore we can eliminate candidate 6 from r12c5 and r4c6, which results in a lot of other reductions as well. Lets try this new rule for grouped multi-coloring on an old friend from the filet-o-fish thread

Code: Select all

+-----------+-----------+-----------+
| .   2X  2B| .   2   2A| .   2   2 |
| .   2X  2B| .   2   2A| .   2   2 |
| .   2   2B| .   2   . | *   *   * |
+-----------+-----------+-----------+
| .   2   . | .   2   . | .   2   2 |
| .   2   2b| .   2   2a| .   2   2 |
| .   2   . | .   2   . | .   2   2 |
+-----------+-----------+-----------+
| .   .   . | .   .   . | 2   .   . |
| .   .   . | 2   .   . | .   .   . |
| 2   .   . | .   .   . | .   .   . |
+-----------+-----------+-----------+


Here the new grouping rule works in a slightly different fashion. In this case we have colors 'a' and 'b' sharing a group, thus they both can't be true. Therefore either A or B or both are true. If B is true, then obviously the cells marked with an X can't be true. And since the starred cells do not contain candidates, we also know by the new rule that if A is true then X can't be true. So we can remove the candidates in the cells marked with X.

[Neunmalneun]

This is the moment it requires the forcing chain:

Code: Select all

.------------.------------.------------.
| 57  9   457| 2   6   3  | 8   1   47 |
| 6   12  14 | 7   5   8  | 9   24  3  |
| 27  8   3  | 9   1   4  | 27  6   5  |
:------------+------------+------------:
| 3   4   2  | 6   9   5  | 1   78  78 |
| 178 5   17 | 3   278 127| 4   9   6  |
| 178 6   9  | 4   78  17 | 5   3   2  |
:------------+------------+------------:
| 4   12  15 | 18  3   27 | 6   578 9  |
| 9   3   8  | 5   47  6  | 27  247 1  |
| 125 7   6  | 18  24  9  | 3   458 48 |
'------------'------------'------------'

What simple alternative is available at this stage?

I think the most simple step would be a rather short forcing chain between R9C1 and R1C3.

If R1C9 = 1, R9C9 = 4, R1C9 = 7, so R1C3 cannot be 7.

If R9C1 isn't 1, we have a naked triple in C1 (257) which kills the 7s in R56C1 and forces R5c3 to be 7. In that case R1C3 cannot be 7 either.

[ronk]

Code: Select all

 .  6  . | .  6  6 | 6  6  . 
 .  .  6 | .  6  . | 6  .  . 
 .  .  . | .  .  6b| .  6B 6B
---------+---------+---------
 .  .  . | .  6  6 | 6  6  6 
 .  .  . | .  .  . | .  .  . 
 .  .  . | .  6a . | .  6A 6A
---------+---------+---------
 .  6  . | .  .  . | *  6  . 
 .  .  . | .  .  . | *  .  . 
 .  .  6 | .  .  . | *  6  6 

I see it more like this. You have two conjugate color pairings, (A, a) and (B, b). The starred empty cells in box 9 mean that the A-group in r6c89 and the B-group in r3c89 cannot both contain a true candidate. (They could both be false though.) Thus, via multi-coloring with groups, either a or b (or both) must be true. Therefore we can eliminate candidate 6 from r12c5 and r4c6 ...

"The starred empty cells in box 9 mean that the A-group in r6c89 and the B-group in r3c89 cannot both contain a true candidate" is another way of saying the 6s in box 9 are also a grouped conjugate link. Illustrating that chain with colors C-c ...

Code: Select all

 .  6  . | .  6- 6 | 6  6  . 
 .  .  6 | .  6- . | 6  .  . 
 .  .  . | .  .  6b| .  6B 6B
---------+---------+---------
 .  .  . | .  6  6-| 6  6  6 
 .  .  . | .  .  . | .  .  . 
 .  .  . | .  6a . | .  6A 6A
---------+---------+---------
 .  6  . | .  .  . | .  6C . 
 .  .  . | .  .  . | .  .  . 
 .  .  6 | .  .  . | .  6C 6c 

... how would you express the eliminations using implication chains?

Ron

[ravel]

As said above it can be done with this chain and box elimination then:
[R4C6]-6-[R4C7]=6=[R1C7|R2C7]-6-[R3C89]=6=[R3C6]-6-[R4C6]
Since i am no expert in chain expressions, i do not know, how to formulate Myth Jellies/ronks method. Another alternative is the following:
[r12c5|r4c6]-6-[r3c6]=6=[r3c8|r3c9]-6-[r12c7]=6=[r4c7] (-6-[r4c6]) -6-[r6c89]=6=[r4c5]-6-[r12c5]

[ronk]

As said above it can be done with this chain and box elimination then:
[R4C6]-6-[R4C7]=6=[R1C7|R2C7]-6-[R3C89]=6=[R3C6]-6-[R4C6]

Yes, that grouped turbot fish is definitely much simpler, so I'm rapidly losing even academic interest in the other POV.

Ron

[Myth Jellies]

"The starred empty cells in box 9 mean that the A-group in r6c89 and the B-group in r3c89 cannot both contain a true candidate" is another way of saying the 6s in box 9 are also a grouped conjugate link. Illustrating that chain with colors C-c ...

Code: Select all

 .  6  . | .  6- 6 | 6  6  . 
 .  .  6 | .  6- . | 6  .  . 
 .  .  . | .  .  6b| .  6B 6B
---------+---------+---------
 .  .  . | .  6  6-| 6  6  6 
 .  .  . | .  .  . | .  .  . 
 .  .  . | .  6a . | .  6A 6A
---------+---------+---------
 .  6  . | .  .  . | .  6C . 
 .  .  . | .  .  . | .  .  . 
 .  .  6 | .  .  . | .  6C 6c 

... how would you express the eliminations using implication chains?

I find it hard to work with your C's. Instead I use the empty box 9 column 7 cells to note that D and d must have a conjugate relationship

Code: Select all

 .  6  . | .  6  6 | 6D 6d . 
 .  .  6 | .  6  . | 6D .  . 
 .  .  . | .  .  6 | .  6d 6d
---------+---------+---------
 .  .  . | .  6  6 | 6d 6D 6D
 .  .  . | .  .  . | .  .  . 
 .  .  . | .  6  . | .  6D 6D
---------+---------+---------
 .  6  . | .  .  . | *  6  . 
 .  .  . | .  .  . | *  .  . 
 .  .  6 | .  .  . | *  6  6 

Therefore any subgroups formed from the d's in box 3 and the D's in box 6 must have at least a weak link relationship. You can't guarantee that either of your subgroups contains a true cell, but if one does then that excludes the other subgroup from containing a true cell.

[Jeff]

Code: Select all

 .  6  . | .  6  6 | 6  6  . 
 .  .  6 | .  6  . | 6  .  . 
 .  .  . | .  .  6b| .  6B 6B
---------+---------+---------
 .  .  . | .  6  6 | 6  6  6 
 .  .  . | .  .  . | .  .  . 
 .  .  . | .  6a . | .  6A 6A
---------+---------+---------
 .  6  . | .  .  . | *  6  . 
 .  .  . | .  .  . | *  .  . 
 .  .  6 | .  .  . | *  6  6 

Here is a possible nice loop for MJ's starred empty cell deduction. Since the starred empty cells cannot be used in a nice loop, the conjugate relationship [r4c7]=6=[r12c7] resulted in column 7 are to be included.

[r12c5|r4c6]-6-[r6c5]=6=[r6c89]-6-[r4c7]=6=[r12c7]-6-[r3c89]=6=[r3c6]-6-[r12c5|r4c6]
implies r1c5<>6, r2c5<>6, r4c6<>6

[Myth Jellies]

Basically a (new?) rule for groups is if two groups have a conjugate (strong) link, then if you form two subgroups from those conjugate groups, those subgroups are weakly linked.

[ravel]

I got the impression here that with grouped coloring each possible one-digit elimination can be done, as detected with templates or nishio. There are notation problems with a 3x3 (ev. finned) swordfish or a jellyfish, which could be solved by definition, e.g. allowing a notation like
[X]-n-([FIN]=n=)[SWORDFISH]-n-[X].
As alternative i would assume that a combination of grouped coloring and filet-o-fish could do that all.
Is there a proof or counter example ?
It should not be very hard to write a program that finds all grouped colorings and filet-o-fishes and thus gives a human reproducible output for each possible elimination (which is not the case for sussers nishio reductions), where the chain or pattern with the least needed cells could be taken.

[Havard]

I was wondering if anyone could see a pattern here:

Code: Select all

. . . | . 6 . | . 6 6
. . . | . . 6 | . . 6
. . 6 | . . . | . . .
------+-------+------
. 6 . | 6 . 6 | . .-6
. 6 . | . 6 . | 6 6 .
. . . | 6 . . | 6 . .
------+-------+------
6 . . | . 6 . | 6 . 6
6 . . | 6 . . | . . .
6 . . | . . 6 | . 6 .


Now I know using nishio that R4C9 can be eliminated, and I can also show this by trying both possibilities in the strong link in R45C2. If R4C2 is "true", the elimination is obvious, and by setting R5C2 as "true" I get:

Code: Select all

. . . | . 6 . | . 6 6
. . . | . | 6-------6
. . 6 | . | . | . . .
------+---|---+------
. . . | 6 | 6 | . .-6
. 6 . | . | . | . . .
. . . | 6 | . | 6 . .
------+---|---+-|----
6 . . | . 6 . | 6 . 6
6 . . | 6 . . | . . .
6 . . | . . 6 | . 6 .


which I guess proves that it can be eliminated, but I am not happy with this "two-step" solution... So can anyone see a direct pattern that will allow elimination of that candidate?

I really like this thread! Finding patterns where Nishio tells us elimination can be done is like my favorite thing about sudoku!

here is the puzzle by the way:

Code: Select all

. . 2 | . . 1 | 8 . .
. 8 4 | 2 . . | 5 9 .
. 9 . | 5 . . | . 2 .
------+-------+------
9 . . | . 5 . | 2 3 .
. . . | 1 . 3 | . . .
. 7 8 | . 9 . | . . 1
------+-------+------
. 2 . | . . 5 | . 1 .
. 1 7 | . . 9 | 3 4 .
. . 3 | 7 . . | 9 . .


basic eliminations pluss a Turbot Fish and xyz-wing takes us to the spot described above. Once that 6 is killed it allows for a lot of ALS reductions.

havard

[ronk]

So can anyone see a direct pattern that will allow elimination of that candidate?

There may be one using candidate grouping, but I don't see it. Though I realize this thread is about alternatives to POM, your puzzle yields to POM's vulnerable pairs. Partially coloring the 4s and 8s we have:

Code: Select all

  5    3    2    | 9    467  1    | 8    67   467 
  1    8    4    | 2    367  67   | 5    9    367 
  7    9    6    | 5    348A 4c8a | 1    2    34   
 ----------------+----------------+----------------
  9    46   1    | 468  5    678  | 2    3    678a
  2    46   5    | 1    678a 3    | 467  678A 9   
  3    7    8    | 46   9    2    | 46   5    1   
 ----------------+----------------+----------------
 4C68a 2    9    | 3    4c6  5    | 67   1    678A
  68   1    7    | 68   2    9    | 3    4    5   
  468  5    3    | 7    1    4C68 | 9    68a  2   

In cell r3c6, (color) 4c (true) excludes 8a, and in cell r7c1, 4C excludes 8a. Since either 4c or 4C must be true, 8a must be false, which leads to cascading singles to solve the puzzle.

The deduction may also be expressed with nice loop:
r3c6-8-r3c5=8=r5c5-8-r5c8=8=r4c9-8-r7c9=8=r7c1=4=r7c5-4-r9c6=4=r3c6, implying r3c6<>8

Ron

[ravel]

I was wondering if anyone could see a pattern here:

Code: Select all

. . . | . 6 . | . 6 6
. . . | . . 6 | . . 6
. . 6 | . . . | . . .
------+-------+------
. 6 . | 6 . 6 | . .-6
. 6 . | . 6 . | 6 6 .
. . . | 6 . . | 6 . .
------+-------+------
6 . . | . 6 . | 6 . 6
6 . . | 6 . . | . . .
6 . . | . . 6 | . 6 .


I see no pattern and I cannot find a nice cycle, but with grouped coloring both 6's in row 6 can be eliminated:
[r4c9](-6-[r6c7])-6-[r7c9,r12c9]=6=[r1c8](-6-[r1c5]=6=[r2c6])-6-[r9c8]=6=[r7c7]-6-[r7c5]=6=[r8c4]-6-[r6c4]
Text version: From 6 in r4c9 you get a 6 in r1c8 and r2c6, then in r6c7 and finally in r8c4.