The New Sudoku Players' Forum

[turnthepage]

Could someone please help me with the following puzzle? I need to know what the next step would be (what, why, how, etc.). I hope that I'm not asking too much. Thanks in advance for any help that you have to offer.

Code: Select all

 *-----------*
 |.9.|6..|2.5|
 |...|4..|.8.|
 |7..|.8.|6..|
 |---+---+---|
 |.7.|1.8|4..|
 |.3.|5.4|.1.|
 |1.4|3.2|.6.|
 |---+---+---|
 |..7|.1.|..8|
 |.1.|..6|...|
 |3.2|...|196|
 *-----------*


[Ruud]

Hi,

Here is the candidate grid:

Code: Select all

.---------------------.---------------------.---------------------.
| 48     9      138   | 6      37     137   | 2      347    5     |
| 256    256    1356  | 4      23579  13579 | 379    8      1379  |
| 7      245    135   | 29     8      1359  | 6      34     1349  |
:---------------------+---------------------+---------------------:
| 2569   7      569   | 1      69     8     | 4      235    239   |
| 2689   3      689   | 5      679    4     | 789    1      279   |
| 1      58     4     | 3      79     2     | 5789   6      79    |
:---------------------+---------------------+---------------------:
| 4569   456    7     | 29     1      359   | 35     2345   8     |
| 4589   1      589   | 2789   234579 6     | 357    23457  2347  |
| 3      458    2     | 78     457    57    | 1      9      6     |
'---------------------'---------------------'---------------------'

Digit 7 is locked in the intersection of row 9 and box 8. Remove digit 7 from the other cells in box 8 and check the consequences.

cheers, Ruud.

[QBasicMac]

Could someone please help me with the following puzzle

You might not be using pencilmarks or you would notice that r3c4=29 and r7c4=29. Now since r8c4=289 (the 7 having been eliminated by Ruud) then r8c4=8 (since 2 and 9 are not possible)

I hate pencilmarks myself. It changes a fun puzzle into a rote step-by-step process where I wear my zombie hat.

But it would be pretty hard to get the 8 in r8c4 without them. That's why I choose books with easy puzzles

You could theoretically find the 8 without pencilmarks as follows.

Checking column 4 for singles, you find r3c4 is not a single because both a 2 and a 9 can go there. Rats! So you go to the next open cell, r7c4 and note the same case. Hey, you think to yourself, 2 and 9 are accounted for. So when you get to r8c4 you eliminate everything but 7 and 8. Still no single, so you go on to r9c4 and note that it also reduces to 7 and 8.

Hmmm, you think "One of those two cells is 7 and the other is 8." But the 7 would have to go in r9 because there is no place on row 9 for a 7 in box 9. So you can place 7 and 8 and the rest of the puzzle is easy.

Mac

[turnthepage]

Ruud and Mac,

Thanks to both of your for your help! With your help I was able to solve the puzzle. Now I'm moving on to the next one.

Have a great day

Shonda

[ravel]

You could theoretically find the 8 without pencilmarks as follows..

Code: Select all

 *-----------*
 |.9.|6..|2.5|
 |...|4..|.8.|
 |7..|.8.|6..|
 |---+---+---|
 |.7.|1.8|4..|
 |.3.|5.4|.1.|
 |1.4|3.2|.6.|
 |---+---+---|
 |..7|.1.|..8|
 |.1.|..6|...|
 |3.2|...|196|
 *-----------*

Maybe this way is a bit easier:
The (hidden) 78-pair in column 4 (r89) is rather easy to see, because you have both 7 and 8 in rows 3 and 7.
Looking at row 9, the 7 cannot go r9c2, therefore must be in box 8. From the pair we have only one 7 in this row there.

[Pat]

Code: Select all

 *-----------*
 |.9.|6..|2.5|
 |...|4..|.8.|
 |7..|.8.|6..|
 |---+---+---|
 |.7.|1.8|4..|
 |.3.|5.4|.1.|
 |1.4|3.2|.6.|
 |---+---+---|
 |..7|.1.|..8|
 |.1.|..6|...|
 |3.2|...|196|
 *-----------*


I hate pencilmarks -
But it would be pretty hard to get the 8 in r8c4 without them.

as Ruud had pointed out,
the 7 for r9 must be in box8,
so the 7 for c4 can only be at r9 (not r8).

at this point,
there's only one place remaining for the 8 for c4
- should be quite easy to see if you avoid pencilmarks.

~ Pat

[QBasicMac]

should be quite easy to see

True! Right as rain, Pat.

As you can see from previous posts, I pointed out a way to work without pencilmarks, but it was unnecessarily complex because I got distracted by Ruuds pencilmark display and lost sight of the puzzle

Obviously a 7 must go in column 5 of box 5. And then, because r3c1=7, a 7 must go in column 6 of box 2. Thus 7 must go in column 4 of box 8.

But because of r7c3=7 and full row 9 in box 9, 7 must go in row 9 of box 8.

Hence r8c4=7. I like to think that if I were doing this puzzle originally, I would have seen that immediately. It is so clear.

After r8c4=7, the rest is simple singles. Never had to mess with "8" at all. This was, in fact, an easy puzzle. LOL.

Mac