The New Sudoku Players' Forum

[newuser]

I need help determining if the following puzzle can be solved without trial-and error. It appeared last friday (4/21) in the Washington Post (#86 Hard)

| 7 3 4 | 158 15 9 | 16 2 68 |
| 58 1 58 | 3 2 6 | 4 7 9 |
| 6 2 9 | 418 7 14 | 13 5 38 |
|---------------------+-------------------+---------------------|
| 139 6 17 | 15 8 2 | 379 4 35 |
| 278 5 278 | 9 4 3 | 67 1 26 |
| 12349 49 12 | 7 6 15 | 39 8 235 |
|---------------------+-------------------+----------------------|
| 24 8 6 | 24 3 7 | 5 9 1 |
| 1245 7 3 | 1245 9 145 | 8 6 4 |
| 149 49 15 | 6 15 8 | 2 3 7 |
*--------------------------------------------------------------------*

I can't get any further on logic. Am I overlooking something? Thanks for your help.










[/u]

[RW]

Am I overlooking something?

Probably the two 6s in row 8. I'd recommend you to start over.

RW

[newuser]

My typo. I edited - second 6 should be a 4. Now what do you think?

[RW]

You have a naked pair in r9c2 and r9c5. This lets you remove candidate 1 from r9c1. Now you have candidates 49 in r9c1, r9c2 and r6c2 => if 4 or 9 went in r6c1 the puzzle would have two solutions => you can remove those candidates from r6c1 => only cell to place 4 in the middle left box=r6c2.

[edit: typo: naked pair in r9c3 and r9c5]

Did that help? Should be other possibilities also, but I just tend to like uniqueness reductions!

RW

[Sped]

Code: Select all

 *--------------------------------------------------------------------*
 | 7      3      4      | 158    15     9      | 16     2      68     |
 | 58     1      58     | 3      2      6      | 4      7      9      |
 | 6      2      9      | 148    7      14     | 13     5      38     |
 |----------------------+----------------------+----------------------|
 | 139    6      17     | 15     8      2      | 379    4      35     |
 | 28     5      278    | 9      4      3      | 67     1      26     |
 | 12349  49     12     | 7      6      15     | 39     8      235    |
 |----------------------+----------------------+----------------------|
 | 24     8      6      | 24     3      7      | 5      9      1      |
 | 125    7      3      | 125    9      15     | 8      6      4      |
 | 149    49     15     | 6      15     8      | 2      3      7      |
 *--------------------------------------------------------------------*


It's all singles... There's just one place for a 4 in box 8, for instance. But a few exclusions based on naked pairs makes things easier. There's 1,5 in box 8, row 9 , and column 6.

[newuser]

Why do you disqualify 1 from r9c1? That's the logic that eludes me...there are plenty of opportunities in the puzzle where there are naked pairs that match up with a three-some that includes the pair).

r9c3, r9c5, and r1c5 are all 15. Since r1c4 is 158, I can eliminate the 15, leaving 8 in r1c4, and the puzzle solves itself.

[newuser]

Thanks to all. I'm just glad it didn't involve trial and error.

[Sped]

Why do you disqualify 1 from r9c1? That's the logic that eludes me...there are plenty of opportunities in the puzzle where there are naked pairs that match up with a three-some that includes the pair).

r9c3, r9c5, and r1c5 are all 15. Since r1c4 is 158, I can eliminate the 15, leaving 8 in r1c4, and the puzzle solves itself.

As you said, r9c3 and r9c5 are both 1,5 and they're both in row 9. That means that 1s and 5s can be excluded from all other cells in row 9. That's how naked pairs work.

If naked pairs haven't sunk in yet.. imagine a 1 in r9c1. What would that mean to r9c3 ? Why, it'd have to be a 5. And r9c5? it'd have to be a 5 also.

We can't have two 5s in row 9. Therefore r9c1 cannot be a 1. In fact, no cell in row 9 can be a 1 except r9c3 and r9c5. No cell but those can be a 5, either.

That's how naked pairs work.

[newuser]

I'm with you; I didn't catch your typo and thought you were working with the 49 pair and not the 15 pair. I thought you were going around corners with the elimination.

Thanks for your patience.

Have you encountered any puzzles that required trial and error?

[TKiel]

Have you encountered any puzzles that required trial and error?

Puzzles that 'require' trial and error are generally regard as invalid because that means they have more than one solution.

I'd be interested in the starting grid for this puzzle because when I entered only the bolded numbers in your edited first post, the software I use said it was invalid. However, when I just cut and pasted the whole grid, it did not indicate it was invalid, which leads me to believe that your bolded numbers don't match up exactly with the starting clues.

Tracy

[newuser]

you're right. r4c6 should have been a bolded 2.

My interest in the game remains as long as each has only one solution - obtained without trial and error.

[QBasicMac]

Have you encountered any puzzles that required trial and error?

I suppose I should answer. I am the only regular at this site that loves Trial and Error.

I, personally, think any puzzle that cannot be solved with reasonably simple techniques, including Swordfish, X-Tree, Hidden triples, and simple coloring (One variable) is a good candidate for trial and error.

Instead of one puzzle which requires you assume it has only one solution to use BUG, etc. or incredibly complex coloring or logic strings that are the equivalent of T&E anyway, you can divide the puzzle into several fun-to-solve reasonable puzzles. Wheee!

But remember the rule: you must fully follow every path to prove the puzzle had one and only one solution.

I see nothing wrong with Trial and Error other than excess work if the puzzle can be solved with logic. But the key idea is "less work". If it is more work to find some boring logic solution than to do the Trial and Error, then by all means I recommend the latter.

Why not use Trial and Error immediately and not mess around with naked triples, X-Wings, etc.? Simple: T&E would lead to a zillion puzzles and thus be more work. Reduce work - increase pleasure. IMHO.

Mac

[Cec]

"..r9c3, r9c5, and r1c5 are all 15. Since r1c4 is 158, I can eliminate the 15, leaving 8 in r1c4, and the puzzle solves itself."

Hi newuser,
If I'm wrong I hope someone will correct me but the elimination of the [15] at this stage is incorrect. The [15]'s in column 5 do form a "naked" pair but the [15]'s in row1 do not form either a "naked" or "hidden" pair which requires the two candidates to occupy only two cells in a group (row, column or box). Note in row1 that cell r1c7 also includes a candidate 1 which means the [15]'s are in three cells.
Also, in a hypothetical situation where the candidate 1 did not exist in r1c7 then you would have a "hidden" pair [15] in row1. This would enable candidate 8 to be excluded (not 15) from r1c4 thus leading to the placement of 8 in r1c9 and the 6 in r1c7 and a "naked" pair [15] still left in row1.

To complement Sped's above nice explanation of "naked" pairs you might wish to look Here for further explanations of pairs, triples, etc.
Cec

[newuser]

Cec is right and I realized it after I detected the oversight of a redundant 4 in Box 8 and the naked pair of 15s in row 9. My earlier solution relying on row 1's 15s was a lucky guess, I suppose.

Thanks for your attention.

newuser

[Cec]

"... and I realized it after I detected the oversight of a redundant 4 in Box 8 and ..."

As Sped pointed out above "there's just one place for a 4 in box 8". It's certainly not "redundant " - it's the only 4 (a "hidden" single) in this box.
Cec