The New Sudoku Players' Forum

[Lama]

Logic no longer works on this puzzle, as far as I have figured. If you are able to solve this, please tell me how, cause I keep getting stuck on these types of sudokus. Thank you for your time.

| 1478 | 18 | 6 || 149 | 2 | 147 || 3 | 89 | 5 |
| 147 | 3 | 5 || 1469 | 679 | 8 || 2 | 69 | 46 |
| 9 | 2 | 48 || 346 | 5 | 346 || 7 | 1 | 468 |
| 3 | 56 | 9 || 156 | 8 | 156 || 4 | 2 | 7 |
| 2 | 68 | 1 || 7 | 4 | 9 || 5 | 3 | 68 |
| 568 | 4 | 7 || 256 | 3 | 256 || 9 | 68 | 1 |
| 56 | 7 | 3 || 256 | 1 | 256 || 8 | 4 | 9 |
| 46 | 9 | 2 || 8 | 67 | 467 || 1 | 5 | 3 |
| 1458 | 158 | 48 || 345 | 9 | 345 || 6 | 7 | 2 |

[tarek]

hi there Lama,

1. How many 9s do you need in column 5 ????

2. Then look at 6s in Column 2 Then in Box 4, Because Column 2 ultimately MUST have a 6..... Can you deduce the next step ????

3. After that, you need some advanced techniques (advanced fishes or colouring)

Tarek

[Lama]

So here is the changed version, though it is a small change. Tarek, thanks for your help, but I don't see where to go after step 1. And I don't know advanced techniques.

Code: Select all

| 1478 | 18  | 6  || 149  | 2  | 147 || 3 | 89 |  5  |
| 147  |  3  | 5  || 1469 | 67 |  8  || 2 | 69 | 46  |
|  9   |  2  | 48 || 346  | 5  | 346 || 7 | 1  | 468 |
|  3   | 56  | 9  || 156  | 8  | 156 || 4 | 2  |  7  |
|  2   | 68  | 1  ||  7   | 4  |  9  || 5 | 3  | 68  |
|  58  |  4  | 7  || 256  | 3  | 256 || 9 | 68 |  1  |
|  56  |  7  | 3  || 256  | 1  | 256 || 8 | 4  |  9  |
|  46  |  9  | 2  ||  8   | 67 | 467 || 1 | 5  |  3  |
| 1458 | 158 | 48 || 345  | 9  | 345 || 6 | 7  |  2  |

[/code]

[tso]

Code: Select all

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

Tarek's step 2:

Code: Select all

 *-----------------------------------------------------------*
 | 1478  18    6     | 149   2     147   | 3     89    5     |
 | 147   3     5     | 1469  67    8     | 2     69    46    |
 | 9     2     48    | 346   5     346   | 7     1     468   |
 |-------------------+-------------------+-------------------|
 | 3     5+6   9     | 156   8     156   | 4     2     7     |
 | 2    +68    1     | 7     4     9     | 5     3     68    |
 | 5-68  4     7     | 256   3     256   | 9     68    1     |
 |-------------------+-------------------+-------------------|
 | 56    7     3     | 256   1     256   | 8     4     9     |
 | 46    9     2     | 8     67    467   | 1     5     3     |
 | 1458  158   48    | 345   9     345   | 6     7     2     |
 *-----------------------------------------------------------*

One of the two cells marked with + must be 6 because these are the only two cells in column 2 that can hold a 6. Therefore, no other cell in box 4 can hold a 6. (A similar deduction can be made using r8c1 and r9c1 -- one of these must be a 6 since these are the only two cells in box 7 that can hold a 6 -- therefore, no other 6 can be in column 1.)

One of the simplest of the so-called advanced techniques is coloring -- and coloring 8's will solve the puzzle.

Code: Select all

 *-----------------------------------------------------------*
 | 1478  18    6     | 149   2     147   | 3     89    5     |
 | 147   3     5     | 1469  67    8     | 2     69    46    |
 | 9     2     48    | 346   5     346   | 7     1     468   |
 |-------------------+-------------------+-------------------|
 | 3     56    9     | 156   8     156   | 4     2     7     |
 | 2     6+8   1     | 7     4     9     | 5     3     6-8   |
 | 58    4     7     | 256   3     256   | 9     68    1     |
 |-------------------+-------------------+-------------------|
 | 56    7     3     | 256   1     256   | 8     4     9     |
 | 46    9     2     | 8     67    467   | 1     5     3     |
 | 1458  158   48    | 345   9     345   | 6     7     2     |
 *-----------------------------------------------------------*

There are only 2 cells in row 5 that can hold an 8 -- color one PLUS and one MINUS. Keep coloring in the same way -- any time there are only two 8's in a row, column or box, color them opposite. Eventually, you'll find a cell that is in the same group (row, column or box) with both a PLUS and a MINUS. That cell cannot hold an 8.

Don't look at the next diagram if you want to try to find this exclusion yourself.

Code: Select all

 *-----------------------------------------------------------*
 | 1478  1-+8  6     | 149   2     147   | 3    -89    5     |
 | 147   3     5     | 1469  67    8     | 2     69    46    |
 | 9     2     4-8   | 346   5     346   | 7     1     46+8  |
 |-------------------+-------------------+-------------------|
 | 3     56    9     | 156   8     156   | 4     2     7     |
 | 2     6+8   1     | 7     4     9     | 5     3     6-8   |
 | 5-8   4     7     | 256   3     256   | 9     6+8   1     |
 |-------------------+-------------------+-------------------|
 | 56    7     3     | 256   1     256   | 8     4     9     |
 | 46    9     2     | 8     67    467   | 1     5     3     |
 | 1458  158   4+8   | 345   9     345   | 6     7     2     |
 *-----------------------------------------------------------*

[tarek]

Hi there Lama,

The method I tried to simplify in step 2 is what is referred to as "Box-Line reduction or elimination" aka locked candidates.

Let us say in you puzzle, looking at Column 2, the 6s all fall in Box 4 this gives us an interesting situation & hence the Column 2 & Box 4 interaction.....

From the basic rules of Sudoku.. Column 2 must have a 6, Wherever its final resting place would be it will be still in Box4......Look now at Box 4, in addition to the 6s in in Column 2, there is a 6 in r6c1 (you can safely eliminate that one, knowing from what we mentioned that the 6 is in intersection of Column 2 & box 4)......

As for the advanced techniques....
There is a plethora of advanced techniques which you will fined over the net & in the advanced techniques forum.

Tarek

[Lama]

Thanks for your help guys.

I have a question about the colouring method you used, tso.

How did you colour the 8's in boxes 1 and 7? I thought you could only colour if there were two 8's in each group, but there are three 8's for the groups in 1 and 7.

[tso]

Thanks for your help guys.

I have a question about the colouring method you used, tso.

How did you colour the 8's in boxes 1 and 7? I thought you could only colour if there were two 8's in each group, but there are three 8's for the groups in 1 and 7.

There ARE only two 8's in each group I used.

Column 3 has only two cells that can be 8 -- r3c3 and r9c3. It's irrelevant that there are other 8s in box 7 -- they weren't (and cannot be) colored.

Row 3 has only two cells that can be 8 -- r3c3 and r3c9. It's irrelevant that there are other 8s in box 1 -- they weren't colored.

R1c2 is in the same box and row with a MINUS and the same column as a PLUS.

[Lama]

Thanks again tso.

I still don't understand though. I understand how the 8's in r3 and c3 are coloured, what I don't see is how you coloured only two of the 8's in c2 and two of the 8's in r1. Maybe I am just misunderstanding completely. If you could clear this up for me I would be very grateful, its driving me nuts.

[QBasicMac]

I don't get it either, TSO

OK, I color r5c2+ which means r1c2, r6c1, r9c2, and r5c9 are -
That leads to r6c8+ and r1c8-
That leads to r1c1+ and r3c3 and r9c1 -
Finally, I get r3c9+ and r9c3+

All the 8's are colored and there is no problem.

Where is the conflict that would eliminate something?

Mac

[QBasicMac]

On the other hand, maybe you meant this:

r1c2+ means r1c1, r5c2, r9c2, and r1c8 are -
So r3c9+ and thus r3c3 and r5c9 are -

So either r6c1+ and r6c8- leaving no eights in box 6,
or r6c8+ and r6c1- leaving no eights in box 3

Thus r1c2 is not 8 and the puzzle solves

Yeah, that's probably it. Sorry.

Mac

[GreenLantern]

Alternating colors in 8's, follow this path: r9c3,r3c3,r3c9,r5c9,r5c2

Now, the 8 in r1c2 can be eliminated because r3c3 and r5c2 are
alternate colors.

[Lama]

How about the 8 in r1c1 and r9c2?

[GreenLantern]

Extending my path by 1 more cell: r9c3,r3c3,r3c9,r5c9,r5c2,r6c1
would allow the 8 in r9c1 to be eliminated based on colors.

Not sure if the 8 in r1c1 can also be eliminated by colors, but in any case,
the puzzle can be solved in a straightforward fashion after eliminating the
8 in r1c2.

[tso]

How about the 8 in r1c1 and r9c2?

The reason you can eliminate the 8 from r1c2 is that it is in the same column as a PLUS (r5c2) and the same row as a MINUS (r1c8). (It's also in the same box as a MINUS (r3c3); that's over kill. If all the cells marked with MINUS were 8, then r1c2 could not be 8. If all the cells marked PLUS were 8, then r1c2 could not be 8. Therefore, r1c2<>8. (I should have marked r1c2 with an 'x' instead of a '-+' to avoid confusion.

You cannot say this about r1c1 or r9c2 -- neither shares a row, column or box with one of each sign.

Greenlantern is correct -- 8 can also be excluded from r9c1 from the same coloring pattern:

Code: Select all

 *-----------------------------------------------------------*
 | 1478  1x8   6     | 149   2     147   | 3    -89    5     |
 | 147   3     5     | 1469  67    8     | 2     69    46    |
 | 9     2     4-8   | 346   5     346   | 7     1     46+8  |
 |-------------------+-------------------+-------------------|
 | 3     56    9     | 156   8     156   | 4     2     7     |
 | 2     6+8   1     | 7     4     9     | 5     3     6-8   |
 | 5-8   4     7     | 256   3     256   | 9     6+8   1     |
 |-------------------+-------------------+-------------------|
 | 56    7     3     | 256   1     256   | 8     4     9     |
 | 46    9     2     | 8     67    467   | 1     5     3     |
 | 145x8 158   4+8   | 345   9     345   | 6     7     2     |
 *-----------------------------------------------------------*

However, contrary to what Greenlantern says, coloring does not require following a path. Start in ANY cell with an 8. Label it PLUS. If the cell's row contains exactly one other cell that can hold an 8, label it MINUS. Repeat for the column and the box. Then pick on of the cells you labeled MINUS and repeat the procedure until you can no longer continue. Then:

If there is a cell that is at the intersection of a PLUS and a MINUS, exclude it.

OR

If there are two cells in one group that are the same sign -- exclude ALL those cells with that sign.

After these exclusions, you may be able to color additional cells. I this particular case, after excluding the 8 from r1c2 and r9c1, no other exclusions from coloring.