The New Sudoku Players' Forum

[ido117]

I'm new here and this is my first post..

I'm trying to solve this puzzle for more then 2 hours and can't succeed.
I found just 5 numbers and I want to know what to do next
this is the puzzle:

. . . | 1 . . | . . .
. . . | 4 . 3 | . 8 5
7 . . | . . . | 1 9 .
------+-------+------
. 5 . | 3 . . | . 2 9
. . . | . 2 . | . . .
1 8 . | . . 4 | . 3 .
------+-------+------
. 4 1 | . . . | . . 2
6 2 . | 5 . 7 | . . .
. . . | . . 8 | . . .


For better view you can see it here:
http://img221.imageshack.us/my.php?image=sudoku1mm5.jpg
It contains the numbers I found.

Thanks for your help..

[submacrolize]

Just go here.

Click on the gray squares that you want to see. Right click to view the whole solution

[wintder]

Code: Select all

.------------------------.------------------------.------------------------.
| 234589  369     345689 | 1       56789   2569   | 23467   467     3467   |
| 29      1       69     | 4       679     3      | 267     8       5      |
| 7       36      34568  | 268     568     256    | 1       9       346    |
:------------------------+------------------------+------------------------:
| 4       5       467    | 3       1678    16     | 4678    2       9      |
| 349     3679    34679  | 6789    2       1569   | 45678   14567   14678  |
| 1       8       2      | 679     5679    4      | 567     3       67     |
:------------------------+------------------------+------------------------:
| 3589    4       1      |*69      369    *69     | 356789  567     2      |
| 6       2       389    | 5       1349    7      | 3489    14      1348   |
| 359     379     3579   | 269     13469   8      | 345679  14567   13467  |
'------------------------'------------------------'------------------------'

See the 69 pair in row 7? They are in your markup.
What eliminations do they allow you to make?

[ido117]

Code: Select all

.------------------------.------------------------.------------------------.
| 234589  369     345689 | 1       56789   2569   | 23467   467     3467   |
| 29      1       69     | 4       679     3      | 267     8       5      |
| 7       36      34568  | 268     568     256    | 1       9       346    |
:------------------------+------------------------+------------------------:
| 4       5       467    | 3       1678    16     | 4678    2       9      |
| 349     3679    34679  | 6789    2       1569   | 45678   14567   14678  |
| 1       8       2      | 679     5679    4      | 567     3       67     |
:------------------------+------------------------+------------------------:
| 3589    4       1      |*69      369    *69     | 356789  567     2      |
| 6       2       389    | 5       1349    7      | 3489    14      1348   |
| 359     379     3579   | 269     13469   8      | 345679  14567   13467  |
'------------------------'------------------------'------------------------'

See the 69 pair in row 7? They are in your markup.
What eliminations do they allow you to make?

Sorry but i didn't understand what you mean..
It is just simple "Very hard puzzle" from the internet

[wintder]

Code: Select all

.------------------------.------------------------.-------------------------.
| 234589  369     345689 | 1       56789   2569   | 23467   467      3467   |
| 29      1       69     | 4       679     3      | 267     8        5      |
| 7       36      34568  | 268     568     256    | 1       9        346    |
:------------------------+------------------------+-------------------------:
| 4       5       467    | 3       1678    16     | 4678    2        9      |
| 349     3679    34679  | 6789    2       1569   | 45678   14567    14678  |
| 1       8       2      | 679     5679    4      | 567     3        67     |
:------------------------+------------------------+-------------------------:
| 358-9   4       1      |*69      3-6-9  *69     | 3578-6-9 57-6    2      |
| 6       2       389    | 5       134-9   7      | 3489     14      1348   |
| 359     379     3579   | 2-6-9   134-6-9 8      | 345679   14567   13467  |
'------------------------'------------------------'-------------------------'

The pair of cells in row 7, column 4 and 6 use up all the 69s that can see the two. That solves two cells instantly leaving you here.

Code: Select all

.------------------------.------------------------.-------------------------.
| 23589   369     345689 | 1       56789   2569   | 23467   467      3467   |
| 29      1       69     | 4       679     3      | 267     8        5      |
| 7       36      34568  | 68      568     256    | 1       9        346    |
:------------------------+------------------------+-------------------------:
| 4       5        67    | 3       678     16     |  678    2        9      |
| 39      3679    3679   | 6789    2       1569   | 45678   14567    14678  |
| 1       8       2      | 679     5679    4      | 567     3        67     |
:------------------------+------------------------+-------------------------:
| 358     4       1      | 69      3       69     | 3578     57      2      |
| 6       2       389    | 5       14      7      | 3489     14      1348   |
| 359     379     3579   | 2       14      8      | 345679   14567   13467  |
'------------------------'------------------------'-------------------------'

[Bigtone53]

(Whoops, posted without seeing Wintner's further explanation but left here in case it helps)

ido117

Wintner is a very experienced member of this site and I know that he will not be offended if I simplify his explanation.

1. Refer back to his first grid of penciled possibilities. In the seventh row down and in the middle 'box' there are two 6-9 possibilities and those are the only possibilities for these two cells

2. We don't yet know which is which, but for sure those two cells in row 7 column 4 and row 7 column 6 (r7c4, r7c6) have between them the 6 and the 9 in both row 7 and the bottom middle box of nine cells

3. This means that you can eliminate 6s and 9s elsewhere in row 7 and the bottom middle box od 9. This is what Wintner is doing with his minus signs in his second pencilmark grid. This shows for instance that cell r7c5, which could be 3, 6 or 9, can only in fact be 3.

... and so on. I hope that this helps.

[RW]

Assuming you're not used to staring at a full pencilmark grid, here's how to do it without them:

Code: Select all

 *-----------*
 |...|1..|...|
 |.1.|4.3|.85|
 |7..|..2|19.|
 |---+---+---|
 |45.|3.1|.29|
 |...|.2.|...|
 |182|..4|.3.|
 |---+---+---|
 |.41|...|..2|
 |62.|5.7|...|
 |...|..8|...|
 *-----------*

1) Look at row 9. Where can you place digit 2? There's only one possibility, r9c4, which is called hidden single. That brings you here:

Code: Select all

 *-----------*
 |...|1..|...|
 |.1.|4.3|.85|
 |7..|..2|19.|
 |---+---+---|
 |45.|3..|.29|
 |...|.2.|...|
 |182|..4|.3.|
 |---+---+---|
 |.41|...|..2|
 |62.|5.7|...|
 |...|2.8|...|
 *-----------*

2) Look at box 8 and row 7. In row 7, digits 1 and 4 are in columns 2 and 3. Which 2 cells do these digits have to occupy in box 8? That's right, r89c5. These two cells have to hold these digits, no other digits may go in those two cells (hidden pair). What's the remaining possibility for digit 3 in box 8?

Once you have figured that out, the next step is easiest described if I only show the essential information:

Code: Select all

 *-----------*
 |.|.|...|...|
 |-+-|4--|-85|
 |7|.|...|...|
 |---+---+---|
 |.5.|...|...|
 |.|.|...|...|
 |.8.|...|...|
 |---+---+---|
 |.4.|...|...|
 |...|...|...|
 |...|...|...|
 *-----------*

3) You already have digits 458 in both row 2 and column 2. In box 1, none of the cells in row 2 or column 2 may hold any of those three digits. This leaves only three possible cells for three digits, those three cells have to hold those three digits and no other digits may go in those three cells (hidden triplet). Knowing that r1c1, r1c3 and r3c3=458 and nothing else, what is the remaining possibility for digit 2 in box 1? When you've found out that, you can also solve the last digit 2 in box 3, which brings you here:

Code: Select all

 *-----------*
 |...|1..|2..|
 |21.|4.3|.85|
 |7..|..2|19.|
 |---+---+---|
 |45.|3..|.29|
 |...|.2.|...|
 |182|..4|.3.|
 |---+---+---|
 |.41|.3.|..2|
 |62.|5.7|...|
 |...|2.8|...|
 *-----------*

4) Remember step 1), r8c5 must be either 1 or 4. Then look at r8c8, what are the possibilities there? As both r8c5 and r8r8 must be either 1 or 4, it means that those two digits must be in those two cells, and may be in no other cells of the same row (naked pair). No solved digit yet, but remember this in the next step.

5) Look at box 3. Where can you place digit 3? As the 3 in box 3 has to be in either r1c9 or r3c9, digit 3 may be in no other cells of column 9 (locked candidates).

6) Now, what digit could be in r8c9? It can already see digits 2,5,6,7,9 and cannot be any of these. It cannot be 1 or 4 (step 4), it cannnot be 3 (step 5). Only remaining possibility is 8 (naked single). That also gives you the location of the 8 in box 7, leading here:

Code: Select all

 *-----------*
 |...|1..|2..|
 |21.|4.3|.85|
 |7..|..2|19.|
 |---+---+---|
 |45.|3..|.29|
 |...|.2.|...|
 |182|..4|.3.|
 |---+---+---|
 |841|.3.|..2|
 |62.|5.7|..8|
 |...|2.8|...|
 *-----------*

7) Remember step 3). r1c1 must be either 4, 5 or 8, but now you already have 4 and 8 in the column, leaving only one possibility.

The rest of the puzzle is easy, even without pencilmarks.

RW

[Cec]

Kudos all round for the above excellent help offered to ido 117. I'd like to add that newcomers have sometimes become confused when "boxes" are mentioned without reference to these being the nine 3X3 boxes within the grid and not the cells . I would suggest ido 117 that you click on Here to read up on the Forum's recommended Terminology.

On another matter, as to "overlapping" of the above "Pencilmark" grid which I suspect would make it more confusing for newcomers to understand, I notice that when I copy and paste this puzzle between the "Code" tags the grid is clear from viewing the "Preview" shot but becomes messy by overlapping as seen above when submitting to the Forum page. Is there any way to overcome this problem.?

Cec

[Bigtone53]

What I wait for more and more with these initial requests is some sort of acknowledgement or thank you from the asker that people have tried to help. This are rare these days.

[Cec]

What I wait for more and more with these initial requests is some sort of acknowledgement or thank you from the asker that people have tried to help. This are rare these days.

Yes Bigtone, I can well relate to your above feelings. For what it's worth I believe that most "askers" do appreciate the help given but unfortunately there are always some who don't even think about replying with a simple "thank you". There is also the possibility that unforeseen circumstances could arise which prevents the "asker" replying.

Cec

[Pat]

What I wait for more and more with these initial requests is some sort of acknowledgement or thank you from the asker that people have tried to help.

we had that discussion a couple of years ago --

QBasicMac calls such a person "ephemeralus"

[Cec]

"...we had that discussion a couple of years ago .."

Well done Pat. Your "filing" system is much more organised than mine. I did actually remember this discussion being initiated by QBasicMac as you pointed out and it was interesting to read the responses. I suspect Bigtone is not alone in voicing his feelings.

Cec