I was totally stuck...

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

I was totally stuck...

Postby NoClue » Fri Apr 01, 2011 9:16 pm

Hi, members of this forum,

recently I tried to solve the following Sudoku (Sudoku # 642400 very hard):

Code: Select all
+-----+-----+-----+
|6 . 9|. 2 7|. . .|
|7 . 3|. 8 .|. . .|
|. . .|. . .|4 . 9|
+-----+-----+-----+
|. . 8|. 4 5|3 . .|
|. 9 .|. 1 .|. 4 .|
|. . 7|8 6 .|2 . .|
+-----+-----+-----+
|9 . 4|. . .|. . .|
|. . .|. 3 .|9 . 4|
|. . .|2 9 .|7 . 5|
+-----+-----+-----+


now the Sudoku, until I got stuck:

Code: Select all
+----------------+-----------+----------------+
|6... 1458.. 9...|134. 2 7...|158 1358.. 138..|
|7... 12458. 3...|1469 8 1469|156 1256.. 126..|
|128. 128... 12..|136. 5 136.|4.. 123678 9....|
+----------------+-----------+----------------+
|12.. 126... 8...|79.. 4 5...|3.. 1679.. 167..|
|356. 9..... 56..|37.. 1 2...|568 4..... 678..|
|1345 1345.. 7...|8... 6 39..|2.. 1359.. 13...|
+----------------+-----------+----------------+
|9... 123568 4...|156. 7 168.|168 12368. 12368|
|1258 125678 1256|156. 3 168.|9.. 1268.. 4....|
|138. 1368.. 16..|2... 9 4...|7.. 1368.. 5....|
+----------------+-----------+----------------+


As far you can see, I could only find 4 hidden singles, the rest of the Sudoku had to be filled with pencilmarked.

Any help with hints is appreciated.

Thanks in advance
NoClue
 
Posts: 48
Joined: 01 April 2011

Re: I was totally stuck...

Postby daj95376 » Fri Apr 01, 2011 9:40 pm

You made a mistake and the grid is not valid. Start over!
daj95376
2014 Supporter
 
Posts: 2624
Joined: 15 May 2006

Re: I was totally stuck...

Postby aran » Fri Apr 01, 2011 10:45 pm

NoClue wrote:Hi, members of this forum,

recently I tried to solve the following Sudoku (Sudoku # 642400 very hard):

Code: Select all
+-----+-----+-----+
|6 . 9|. 2 7|. . .|
|7 . 3|. 8 .|. . .|
|. . .|. . .|4 . 9|
+-----+-----+-----+
|. . 8|. 4 5|3 . .|
|. 9 .|. 1 .|. 4 .|
|. . 7|8 6 .|2 . .|
+-----+-----+-----+
|9 . 4|. . .|. . .|
|. . .|. 3 .|9 . 4|
|. . .|2 9 .|7 . 5|
+-----+-----+-----+


now the Sudoku, until I got stuck:

Code: Select all
+----------------+-----------+----------------+
|6... 1458.. 9...|134. 2 7...|158 1358.. 138..|
|7... 12458. 3...|1469 8 1469|156 1256.. 126..|
|128. 128... 12..|136. 5 136.|4.. 123678 9....|
+----------------+-----------+----------------+
|12.. 126... 8...|79.. 4 5...|3.. 1679.. 167..|
|356. 9..... 56..|37.. 1 2...|568 4..... 678..|
|1345 1345.. 7...|8... 6 39..|2.. 1359.. 13...|
+----------------+-----------+----------------+
|9... 123568 4...|156. 7 168.|168 12368. 12368|
|1258 125678 1256|156. 3 168.|9.. 1268.. 4....|
|138. 1368.. 16..|2... 9 4...|7.. 1368.. 5....|
+----------------+-----------+----------------+


As far you can see, I could only find 4 hidden singles, the rest of the Sudoku had to be filled with pencilmarked.

Any help with hints is appreciated.

Thanks in advance

Despite daj's remark, the grid is valid, and is solvable with singles.
Your pencil marks haven't been whittled down enough though, and this may be complicating the picture eg
r6c9=1
r3c8=7 (hidden single b3)
r8c2=7 (hidden single b7)
r6c1=4 (forced by 4r12c2)
Starting so to speak from there :
r4c12=hidden pair 12, hence r5c3=6, (nb no candidate 6 at r5c1 because of 6r1c1) hence r8c3=5
and it all unravels.
PS but of course daj was posting on 1 April
aran
 
Posts: 334
Joined: 02 March 2007

Re: I was totally stuck...

Postby daj95376 » Sat Apr 02, 2011 1:31 am

aran wrote:PS but of course daj was posting on 1 April

Not the problem. I ran the grid through my solver and HoDoKu and both reported multiple solutions because r4c7=3 was treated as a solved cell instead of a given. Essentially, I fouled up and forgot to include the initial puzzle.
daj95376
2014 Supporter
 
Posts: 2624
Joined: 15 May 2006

Re: I was totally stuck...

Postby NoClue » Sat Apr 02, 2011 6:46 am

@ aran

at first I want to thank you very much for your assistance. But - at the moment - I don´t understand why r6c9 has to be "1". Either block 6 or row 6 have many candidates "1" - so I couln´t follow, why they can be excluded.

Thanks in advance to clarify this "gap of knowledge" of mine...:-)
NoClue
 
Posts: 48
Joined: 01 April 2011

Re: I was totally stuck...

Postby aran » Sat Apr 02, 2011 8:17 am

NoClue wrote:@ aran
at the moment - I don´t understand why r6c9 has to be "1". Either block 6 or row 6 have many candidates "1" - so I couln´t follow, why they can be excluded.

r4c7 being 3 eliminates any other 3s it can see...such as 3r6c9.
amazing what can escape one on a heavily pencil-marked grid !
aran
 
Posts: 334
Joined: 02 March 2007

Re: I was totally stuck...

Postby NoClue » Sat Apr 02, 2011 3:15 pm

Hi, aran,

yes, you are right - I have overlooked some candidates, which could have been eliminated. There were so many candidates in some cells, so that I couldn´t see it clearly. But with your help - I can! thanks again.

In r12c2 was there a pointing pair? In r5c2 I had the candidate 345 and r12c2 was 1458 / 12458 - so I suppose that there was a "pointing pair" with the candidate 4 - I hope I saw it right.

after eliminating some candidates the new "intermediate" Sudoku is:

Code: Select all
+--------+-------+----------+
|6. 45. 9|14 2 7.|18. 158 3.|
|7. 45. 3|14 8 9.|16. 125 26|
|18 18. 2|3. 5 6.|4.. 7.. 9.|
+--------+-------+----------+
|12 12. 8|9. 4 5.|3.. 6.. 7.|
|5. 9.. 6|7. 1 2.|5.. 4.. 8.|
|4. 3.. 7|8. 6 3.|2.. 9.. 1.|
+--------+-------+----------+
|9. 268 4|5. 7 18|168 3.. 26|
|28 7.. 5|6. 3 18|9.. 128 4.|
|3. 68. 1|2. 9 4.|7.. 68. 5.|
+--------+-------+----------+
NoClue
 
Posts: 48
Joined: 01 April 2011

Re: I was totally stuck...

Postby aran » Sat Apr 02, 2011 9:49 pm

NoClue wrote:In r12c2 was there a pointing pair? In r5c2 I had the candidate 345 and r12c2 was 1458 / 12458 - so I suppose that there was a "pointing pair" with the candidate 4 - I hope I saw it right.

In box 1 (b1), the pair 45 in r12c2 can be seen in two ways :
- r3c123 consists of the naked triple 128, therefore all of those candidates can be removed from r12c1, leaving the naked pair 45
- since the only cells available for 4 and 5 in b1 are r12c2, these must necessarily form a pair - known as a hidden pair - in those cells.
As this little example shows, "hidden" logic is usually one step ahead of "naked" logic

NoClue wrote:after eliminating some candidates the new "intermediate" Sudoku is:

Code: Select all
+--------+-------+----------+
|6. 45. 9|14 2 7.|18. 158 3.|
|7. 45. 3|14 8 9.|16. 125 26|
|18 18. 2|3. 5 6.|4.. 7.. 9.|
+--------+-------+----------+
|12 12. 8|9. 4 5.|3.. 6.. 7.|
|5. 9.. 6|7. 1 2.|5.. 4.. 8.|
|4. 3.. 7|8. 6 3.|2.. 9.. 1.|
+--------+-------+----------+
|9. 268 4|5. 7 18|168 3.. 26|
|28 7.. 5|6. 3 18|9.. 128 4.|
|3. 68. 1|2. 9 4.|7.. 68. 5.|
+--------+-------+----------+

You have gone wrong at this point as the presence in row 5 of 5 at r5c1 and r5c7 shows
aran
 
Posts: 334
Joined: 02 March 2007

Re: I was totally stuck...

Postby ronk » Sat Apr 02, 2011 11:17 pm

aran wrote:In box 1 (b1), the pair 45 in r12c2 can be seen in two ways :
- r3c123 consists of the naked triple 128, therefore all of those candidates can be removed from r12c1, leaving the naked pair 45
- since the only cells available for 4 and 5 in b1 are r12c2, these must necessarily form a pair - known as a hidden pair - in those cells.
As this little example shows, "hidden" logic is usually one step ahead of "naked" logic

NoClue, for every naked locked set, there is a complementary hidden locked set. Sometimes the naked set is smaller and sometimes the hidden set is smaller. Neither is "usually one step ahead" of the other.
ronk
2012 Supporter
 
Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Re: I was totally stuck...

Postby aran » Sun Apr 03, 2011 8:57 am

ronk wrote:
aran wrote:"hidden" logic is usually one step ahead of "naked" logic

for every naked locked set, there is a complementary hidden locked set. Sometimes the naked set is smaller and sometimes the hidden set is smaller. Neither is "usually one step ahead" of the other.

I don't think you've taken my point.
Take bilocation as the most simple example of hidden logic : x bilocal in r1 at r1c1+r1c7
then xr1c1=xr1c7
but suppose you miss that (yes it can happen) and take the route : (x=y)r1c1-(y=NTabc r459c1)-(ab=x)r1c7.
aran
 
Posts: 334
Joined: 02 March 2007

Postby Pat » Sun Apr 03, 2011 11:54 am

NoClue wrote:recently I tried to solve the following Sudoku
(Sudoku # 642400 very hard):

Code: Select all
 6 . 9 | . 2 7 | . . .
 7 . 3 | . 8 . | . . .
 . . . | . . . | 4 . 9
-------+-------+------
 . . 8 | . 4 5 | 3 . .
 . 9 . | . 1 . | . 4 .
 . . 7 | 8 6 . | 2 . .
-------+-------+------
 9 . 4 | . . . | . . .
 . . . | . 3 . | 9 . 4
 . . . | 2 9 . | 7 . 5

I could only find 4 hidden singles,
the rest of the Sudoku had to be filled with pencilmarked.

Any help with hints is appreciated.


hi NoClue,

there are 7 "hidden singles" (not 4)
bringing us to
Code: Select all
 6 . 9 | . 2 7 | . . .
 7 . 3 | . 8 . | . . .
 . . . | . 5 . | 4 7 9
-------+-------+------
 . . 8 | . 4 5 | 3 . .
 . 9 . | . 1 2 | . 4 .
 4 . 7 | 8 6 . | 2 . .
-------+-------+------
 9 . 4 | . 7 . | . . .
 . 7 . | . 3 . | 9 . 4
 . . . | 2 9 4 | 7 . 5

    now in box 1,
    the 5 must be in column 2
    -- exclude 5 elsewhere in that column

      this solves the 5 for row 6

        and "hidden singles" finishes it --
        certainly no need for "pencilmarks"
User avatar
Pat
 
Posts: 4056
Joined: 18 July 2005

Re: I was totally stuck...

Postby NoClue » Sun Apr 03, 2011 5:08 pm

Hi, Pat,

many thanks for giving me a lot opf tips. That helped me further.
NoClue
 
Posts: 48
Joined: 01 April 2011


Return to Help with puzzles and solving techniques