The New Sudoku Players' Forum

[kickindaspeaker]

Here's my current grid. I have found the solution using Trial and Error, but coming back to that grid where I was stuck, I can't find the "logical" way to get any further, so I would be interested in any explanation on how to move forward from there?

Code: Select all

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

[Hud]

I found a naked pair (1,9) in row 6. There may have been an x wing before that but I can't recall. Once you find the naked pair it's toast.

[kickindaspeaker]

Thanks!! Just before you posted your answer I found the same naked pair (I can't believe I couldn't see it before).

I'm stuck again though:

Code: Select all

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

[kickindaspeaker]

Excellent. Thanks a lot!!!

[Hud]

I'm just curious, but do you use pcncilmarks? If not, you'll need them on more difficult puzzles.
9 is the only number that can go in r5c7. That's my final answer.
Regis

[QBasicMac]

I'm just curious, but do you use pcncilmarks?

Good question, Regis.

kickindaspeaker: Here is your puzzle using pencilmarks


Here is your original puzzle with pencilmarks

Code: Select all

+--------------+---------------+-----------------+
| 4   1     5  | 7     3   89  | 269   689   269 |
| 7   3     9  | 48    6   2   | 1     48    5   |
| 2   8     6  | 49    1   5   | 3479  3479  79  |
+--------------+---------------+-----------------+
| 6   249   12 | 3489  47  189 | 5     379   279 |
| 3   59    8  | 2     57  6   | 79    1     4   |
| 19  2459  7  | 349   45  19  | 2369  369   8   |
+--------------+---------------+-----------------+
| 5   67    4  | 1     9   3   | 8     2     67  |
| 8   69    3  | 5     2   7   | 469   469   1   |
| 19  279   12 | 6     8   4   | 79    5     3   |
+--------------+---------------+-----------------+


Locked candidate 4 in box 2
Translation: In box 2, all 4's are in column 4; hence the 4 of column 4 must be one of those cells. Therefore all other 4's can be erased from your pencilmarks. Erase 4 from r4c4 and r6c4.

Hidden pair 24 in row 6
Translation: In row 6 there are two cells that contain candidates 2 and 4. No other cell has them. Therefore one of those two cells must be a 2 and the other a 4. Thus r6c2 cannot be 5 or 9, erase those two.

Naked pair 19 in row 6
Translation: There are two cells in row 6 that contain the pencilmarks 1 and 9. There are no other pencilmarks in those two cells. Therefore, 1 and 9 can be removed from all other cells in row 6. It turns out there are only some 9's to erase. Erase them.

Ah! r6c4 has only one candidate! Write 3 in that cell and erase all candidate 3's from any other cells in the box, row and column.

Here are your pencilmarks now.

Code: Select all

+-------------+-------------+-----------------+
| 4   1    5  | 7   3   89  | 269   689   269 |
| 7   3    9  | 48  6   2   | 1     48    5   |
| 2   8    6  | 49  1   5   | 3479  3479  79  |
+-------------+-------------+-----------------+
| 6   249  12 | 89  47  189 | 5     379   279 |
| 3   59   8  | 2   57  6   | 79    1     4   |
| 19  45   7  | 3   45  19  | 26    6     8   |
+-------------+-------------+-----------------+
| 5   67   4  | 1   9   3   | 8     2     67  |
| 8   69   3  | 5   2   7   | 469   469   1   |
| 19  279  12 | 6   8   4   | 79    5     3   |
+-------------+-------------+-----------------+


(Just to show what it looks like)

Mac

[kickindaspeaker]

Thanks a lot for all your help. I do use pencil marks but with paper and pencil and sometimes I must have missed some things that appear obvious.

I don't underrstand this part:

Hidden pair 24 in row 6
Translation: In row 6 there are two cells that contain candidates 2 and 4. No other cell has them. Therefore one of those two cells must be a 2 and the other a 4. Thus r6c2 cannot be 5 or 9, erase those two.

In row 6, many cells contain 2 and 4, and only one contains both...??

[Animator]

The explanation (IMHO) is wrong.

Based on the pencilmarks you posted I say that there is no hidden 24 pair. (That is, after applying the locked candidate 4.)

Based on your explanation I say that it is wrong.

Thus r6c2 cannot be 5 or 9, erase those two.

Yet your later pencilmarks do show a 5 in r6c2.

I would take this approach:

• Locked candidate 4 in box 2;
• Naked 19 pair in row 6;
• Naked/Hidden pair 45 in row 6;
• Single 3 in r6c4 + Hidden single 2 in r6c7 + ...
  

[QBasicMac]

The explanation (IMHO) is wrong.

It sure is!! I wonder what I was smoking last night.

Sorry for the confusion.

Mac

[QBasicMac]

In comparing what I said:

Hidden pair 24 in row 6
Translation: In row 6 there are two cells that contain candidates 2 and 4. No other cell has them. Therefore one of those two cells must be a 2 and the other a 4. Thus r6c2 cannot be 5 or 9, erase those two.

with what I meant (and actually did on my pencilmark sheet)

Hidden pair 45 in row 6
Translation: In row 6 there are two cells that contain candidates 4 and 5. No other cell has them. Therefore one of those two cells must be a 4 and the other a 5. Thus r6c4 cannot be 2 or 9, erase those two.

I am amazed at how different those statements are. I guess I remembered cell r6c2 had 2459 and mixed up which two remained and which two were erased when I wrote up the after-the-fact explanation.

Again, kickindaspeaker (and Animator), I'm sorry you wasted a lot of time trying to understand my garbled step.

Mac

[MCC]

I'm stuck again though:

Code: Select all

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

kickindaspeaker, from this position look at where the 4 can go in c7, once you've placed this it's singles all the way.

MCC