I'm stuck with this "evil" puzzle from websudoku.c

Post the puzzle or solving technique that's causing you trouble and someone will help

I'm stuck with this "evil" puzzle from websudoku.c

Postby kickindaspeaker » Tue Jan 03, 2006 1:58 am

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
kickindaspeaker
 
Posts: 7
Joined: 02 January 2006

Postby Hud » Tue Jan 03, 2006 2:29 am

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.
Hud
 
Posts: 570
Joined: 29 October 2005

Postby kickindaspeaker » Tue Jan 03, 2006 2:45 am

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
 
Posts: 7
Joined: 02 January 2006

Postby kickindaspeaker » Tue Jan 03, 2006 2:58 am

Excellent. Thanks a lot!!!:D
kickindaspeaker
 
Posts: 7
Joined: 02 January 2006

Postby Hud » Tue Jan 03, 2006 4:29 am

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
Hud
 
Posts: 570
Joined: 29 October 2005

Postby QBasicMac » Tue Jan 03, 2006 5:12 am

Hud wrote: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
QBasicMac
 
Posts: 441
Joined: 13 July 2005

Postby kickindaspeaker » Tue Jan 03, 2006 7:49 am

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...??
kickindaspeaker
 
Posts: 7
Joined: 02 January 2006

Postby Animator » Tue Jan 03, 2006 9:03 am

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.
QBasicMac wrote: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 + ...
Animator
 
Posts: 469
Joined: 08 April 2005

Postby QBasicMac » Tue Jan 03, 2006 2:01 pm

Animator wrote:The explanation (IMHO) is wrong.


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

Sorry for the confusion.

Mac
QBasicMac
 
Posts: 441
Joined: 13 July 2005

Postby QBasicMac » Tue Jan 03, 2006 2:57 pm

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
QBasicMac
 
Posts: 441
Joined: 13 July 2005

Postby MCC » Tue Jan 03, 2006 6:11 pm

kickindaspeaker wrote: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
MCC
 
Posts: 1275
Joined: 08 June 2005


Return to Help with puzzles and solving techniques