The New Sudoku Players' Forum

[Neunmalneun]

Code: Select all

 *-----------*
 |58.|.3.|...|
 |...|...|.21|
 |9..|...|...|
 |---+---+---|
 |76.|2..|...|
 |.4.|...|8..|
 |...|.9.|5..|
 |---+---+---|
 |...|4.1|.7.|
 |3..|...|...|
 |...|6..|...|
 *-----------*

That's where I was stuck.
 *-----------------------------------------------------------*
 | 5     8     1     | 9     3     2     | 7     4     6     |
 | 46    37    3467  | 5     468   678   | 9     2     1     |
 | 9     27    2467  | 1     46    67    | 3     58    58    |
 |-------------------+-------------------+-------------------|
 | 7     6     59    | 2     18    58    | 14    3     49    |
 | 2     4     59    | 3     16    56    | 8     19    7     |
 | 18    13    38    | 7     9     4     | 5     6     2     |
 |-------------------+-------------------+-------------------|
 | 68    9     268   | 4     5     1     | 26    7     3     |
 | 3     127   46    | 8     27    9     | 1246  15    45    |
 | 14    5     247   | 6     27    3     | 124   89    89    |
 *-----------------------------------------------------------*


I thought a "46" in R2C3 would lead to a UR pattern with "72" in the six cells R3C23, R8C25 and R9C35. So eliminating both candidates 46 in R2C3 solved the puzzle but I am not sure if that was just a lucky guess. Are there alternatives for solving it at that point?

Thanks in advance and kind regards

[vidarino]

Good thinking, but there's an elimination too many there, I think. The Unique Swordfish also has an extra 1 in R8C2.

As far as I can see, you can only safely eliminate 6 from R3C2, since that would make R2C1=4, which in turn would make R9C1=1, which eventually would rid the Unique Swordfish of all extras, leaving a deadly pattern.

I can't seem to get rid of the 1 in R8C2 just by setting R3C2=4, so I don't think it can be validly eliminated by looking at the unique pattern alone.

[Neunmalneun]

Thanks for your answer, Vidar. That's part of my question: can you eliminate a pair (not necessarily each single candidate of that pair)?

"46" in R2C3 would force R8C2<>1. (R3C3=7 => R7C3=8 => R7C1=6 => R2C1=4 => R9C1=1).

[ronk]

I can't seem to get rid of the 1 in R8C2 just by setting R3C2=4, so I don't think it can be validly eliminated by looking at the unique pattern alone.

Assuming you mean R2C3, there is (besides the obvious)...

r2c3=4 -> r2c3<>3 -> r2c2=3 -> r6c2<>3 -> r6c2=1 -> r8c2<>1

... but both r2c3<>4 and r2c3<>6 together don't go far.

Ron

[Carcul]

Hi Neunmalneun.

You can use your pattern (a not almost unique loop) in cells r3c2/r3c3/r8c2/r9c3/r8c5/r9c5 in different ways:

[r3c6](-6-[r3c3])-6-[r3c5]-4-[r3c3]=(Almost Unique Loop:r3c2|r3c3|r8c2|r9c3|r8c5|r9c5)=4|1,4=[r8c2|r9c1|r9c3]-4-[r8c3]-6-[r7c1]-8-[r6c1]-1-[r6c2]-3-[r2c2]-7-[r2c6]=7=[r3c6], => r3c6<>6

and that does not solve the puzzle. However, we can use the same pattern in a different way:

[r9c7]-1-[r9c1](-4-[r9c3])-4-[r8c3]-6-[r3c3]={Nice Loop: [r3c3]=(Almost Unique Loop:r3c2|r3c3|r8c2|r9c3|r8c5|r9c5)=4|1=[r8c2]-1-[r6c2]-3-[r2c2](-7-[r3c3])-7-[r3c2]-2-[r3c3]}=4=[r3c3]-4-[r3c5]=4=[r2c5]=8=[r4c5]=1=[r4c7]-1-[r9c7],

which implies r9c7<>1 and that solve the puzzle.

Regards, Carcul

[ronk]

That's part of my question: can you eliminate a pair (not necessarily each single candidate of that pair)?

According to the uniqueness principle, one or more of r2c3=4, r2c3=6, r9c3=4 and r8c2=1 must be true. If you can find a placement or elimination common to all of those, then that placement or elimination is valid.

Here, because r2c1=46, you can validly say ... if r3c3=46, then r2c3<>46. But you still would need both r9c3=4 and r8c2=1, and likely individually, to also cause r2c3<>4 and r2c3<>6.

Ron

[ronk]

Code: Select all

 | 3     127   46    | 8     27    9     | 1246  15    45    |


My row 8 looks like ...

Code: Select all

 | 3     127   2467  | 8     27    9     | 1246  15    45    |


How did you exclude the 2 and 7 from r8c3?

TIA, Ron

[Neunmalneun]

How did you exclude the 2 and 7 from r8c3?

TIA, Ron

Basically by the same reasoning. If R8C3 was 27 R9C3 would have to be 4 (uniqueness rule). That forces two 1s in box 7 (in R8C2 due to the naked pair 27 in R8 and in R9C1 due to the 4 in R9C3)

So R8C3<>27 => R8C3=46