dailysudoku.com Forum Index dailysudoku.com
Discussion of Daily Sudoku puzzles
 
 FAQFAQ   SearchSearch   MemberlistMemberlist   UsergroupsUsergroups   RegisterRegister 
 ProfileProfile   Log in to check your private messagesLog in to check your private messages   Log inLog in 

april 13th any logical solution?

 
Post new topic   Reply to topic    dailysudoku.com Forum Index -> Daily Sudoku puzzles
View previous topic :: View next topic  
Author Message
George Woods



Joined: 28 Mar 2006
Posts: 304
Location: Dorset UK

PostPosted: Thu Apr 13, 2006 11:07 pm    Post subject: april 13th any logical solution? Reply with quote

At the "crunch point" a 6 in row 5 cols 1 2 or 3 leads to a conflict that row 4 col 7 must hold both 6 or 9. Henace row 5 col1 must be 5 BUT does anyone out there have a logical line of analysis?
Back to top
View user's profile Send private message
Steve R



Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England

PostPosted: Thu Apr 13, 2006 11:15 pm    Post subject: Reply with quote

If we have the same crunch point, all candidates but 5 are excluded from r5c1 ... as long as you spot that column 3 splits (35)(469) and that the 3 or 5 in r2c3 forces the 6 of box1 into the first column.

Steve
Back to top
View user's profile Send private message
George Woods



Joined: 28 Mar 2006
Posts: 304
Location: Dorset UK

PostPosted: Thu Apr 13, 2006 11:49 pm    Post subject: Reply with quote

Thanks - My fault - most puzzles can be done on doublets so I ignored the triplet 469 - that lead to the 35 etc etc... At least I realised that there must be a more direct route!!!!
Back to top
View user's profile Send private message
Guest






PostPosted: Fri Apr 14, 2006 4:27 am    Post subject: Reply with quote

Code:
as long as you spot that column 3 splits (35)(469)


Keith, would you be so kind as to explain how you came to that conclusion? I don't get it...

Cheers,
Sarah
Back to top
George Woods



Joined: 28 Mar 2006
Posts: 304
Location: Dorset UK

PostPosted: Fri Apr 14, 2006 8:23 am    Post subject: Reply with quote

The "crunch point is
008 200 397
070 000 182
201 008 465
700 005 008
000 987 014
180 420 750
017 000 006
002 000 040
360 802 570

The first observation is that the 3 in row 5 must lie in box 4
so in col 3 the cells in r4,r6,and r9 are all "attacked" by 35 so the only places left for 35 are r2 and r5! etc.......


Last edited by George Woods on Sat Apr 15, 2006 9:45 pm; edited 1 time in total
Back to top
View user's profile Send private message
dotdot



Joined: 07 Dec 2005
Posts: 29
Location: oberseen

PostPosted: Fri Apr 14, 2006 10:12 am    Post subject: Reply with quote

George Woods wrote:
so I ignored the triplet 469 - that lead to the 35

I was going to remark that it is not necessary to recognise the triplet, but you have now demonstrated that yourself.

On the other hand this was a pretty beastly constellation for direct recognition of {3,5}.
It is nicer when the constraints come directly from existing outliers, as in row9 here.
Row4 and row6 already have the 5s but no 3s yet which makes normal scanning pretty ineffective here.
Back to top
View user's profile Send private message
Sarah



Joined: 04 Apr 2006
Posts: 6
Location: Brooklyn

PostPosted: Fri Apr 14, 2006 1:54 pm    Post subject: Reply with quote

OK, I'm feeling like a dummy, but I have to continue this until I get it.

I understand that you can isolate the 5 in box 4 to lie in r5. But I still find the options 3,5,6 for r5c3, and 3,5,6, 9 for r2c3. I am just not getting how the 3,5 pair in c3...

Help!
Thanks
Back to top
View user's profile Send private message
dotdot



Joined: 07 Dec 2005
Posts: 29
Location: oberseen

PostPosted: Fri Apr 14, 2006 2:45 pm    Post subject: double negative? Reply with quote

Sarah,
Seeing what can be in r2c3 and r5c3 isn't conclusive here.
The argument goes in two steps:
- see what can't go in certain cells
- what can't go there must be accomodated in the other cells.

In col3 we already know 4 cells so the other five are free.
- 3 and 5 are seen to be excluded from r4c3, r6c3 and r9c3 (cf George above)
- so the 3 and 5 must occur in the remaining free cells (of col3).
Back to top
View user's profile Send private message
Sarah



Joined: 04 Apr 2006
Posts: 6
Location: Brooklyn

PostPosted: Fri Apr 14, 2006 7:27 pm    Post subject: Reply with quote

Thanks dotdot. Sometimes it seems the logic comes almost naturally, and other times I have an unexplained block that all of sudden gets unclogged ... as happened here.
Thanks for the patient explanation.
Back to top
View user's profile Send private message
keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA

PostPosted: Fri Apr 14, 2006 8:10 pm    Post subject: A simpler way? Reply with quote

Here is the situation posted by George, with the possibilities:

Code:

+----------------------+----------------------+----------------------+
| 456    45     8      | 2      1456   146    | 3      9      7      |
| 4569   7      34569  | 356    34569  3469   | 1      8      2      |
| 2      39     1      | 37     379    8      | 4      6      5      |
+----------------------+----------------------+----------------------+
| 7      2349   3469   | 136    136    5      | 269    23     8      |
| 56     235    356    | 9      8      7      | 26     1      4      |
| 1      8      369    | 4      2      36     | 7      5      39     |
+----------------------+----------------------+----------------------+
| 4589   1      7      | 35     3459   349    | 289    23     6      |
| 589    59     2      | 13567  135679 1369   | 89     4      39     |
| 3      6      49     | 8      49     2      | 5      7      1      |
+----------------------+----------------------+----------------------+



There is a triple <589> in R8, which means R8C9 is <3>.

The rest is, I think, pretty straightforward.

Keith
Back to top
View user's profile Send private message
George Woods



Joined: 28 Mar 2006
Posts: 304
Location: Dorset UK

PostPosted: Fri Apr 14, 2006 10:27 pm    Post subject: Reply with quote

"Quelle Domage" Keith--- I had filled in my crunch point in ink, finishing the puzzle in pencil--- erased the pencil to create the "crunch point" that I published above. The 1 in the bottom right hand corner was erroneously entered (incompletely erased) so your beautifully simple solution would not have worked on a puzzle started from the beginning!
Back to top
View user's profile Send private message
keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA

PostPosted: Sat Apr 15, 2006 12:15 am    Post subject: Reply with quote

Dang!

Keith
Back to top
View user's profile Send private message
TKiel



Joined: 22 Feb 2006
Posts: 292
Location: Kalamazoo, MI

PostPosted: Sat Apr 15, 2006 11:27 am    Post subject: Reply with quote

(George Woods--You can edit your post above to make the posted grid correct.)

With the 1 changed from a placement to a candidate in r9c9, the grid looks like this:
Code:

 *-----------*
 |..8|2..|397|
 |.7.|...|182|
 |2.1|..8|465|
 |---+---+---|
 |7..|..5|..8|
 |...|987|.14|
 |18.|42.|75.|
 |---+---+---|
 |.17|...|..6|
 |..2|...|.4.|
 |36.|8.2|57.|
 *-----------*
 
 *-----------------------------------------------------------------------------*
 | 456     45      8       | 2       1456    146     | 3       9       7       |
 | 4569    7       34569   | 356     34569   3469    | 1       8       2       |
 | 2       39      1       | 37      379     8       | 4       6       5       |
 |-------------------------+-------------------------+-------------------------|
 | 7       2349    3469    | 136     136     5       | 269     23      8       |
 | 56      235     356     | 9       8       7       | 26      1       4       |
 | 1       8       369     | 4       2       36      | 7       5       39      |
 |-------------------------+-------------------------+-------------------------|
 | 4589    1       7       | 35      3459    349     | 289     23      6       |
 | 589     59      2       | 13567   135679  1369    | 89      4       139     |
 | 3       6       49      | 8       149     2       | 5       7       19      |
 *-----------------------------------------------------------------------------*

After making the exclusions from the locked candidates in box 4 and from the naked triple/hidden pair in column 3, the grid looks like this:
Code:

  *-------------------------------------------------------------------------*
 | 456     45      8       | 2       1456    146     | 3       9       7    |
 | 4569    7       35      | 356     34569   3469    | 1       8       2    |
 | 2       39      1       | 37      379     8       | 4       6       5    |
 |----------------------------+-------------------------+-------------------|
 | 7       249     469     | 136     136     5       | 269     23      8    |
 | 56      235     35      | 9       8       7       | 26      1       4    |
 | 1       8       69      | 4       2       36      | 7       5       39   |
 |-------------------------+-------------------------+----------------------|
 | 4589    1       7       | 35      3459    349     | 289     23      6    |
 | 589     59      2       | 13567   135679  1369    | 89      4       139  |
 | 3       6       49      | 8       149     2       | 5       7       19   |
 *--------------------------------------------------------------------------*

The locked candidate 6's in box 1 and/or 4 allow the exclusion of 6 in r5c1 and that solves the puzzle.
Back to top
View user's profile Send private message
PGordon



Joined: 13 Apr 2006
Posts: 2
Location: New York City

PostPosted: Thu Apr 20, 2006 6:48 pm    Post subject: Apr 13 Reply with quote

Having reached the same crunch point as George, I managed to miss the obvious point that Box 1 "owns" the 6 in column 1 , so r5c1 can only be 5. So in a total fury I chose to apply the old rule "don't force it, get a bigger hammer" and decided to "see what happens if" r8c9 is a 3 rather than the 1 which was its only other possibility. This with over 40 cells still empty. Oddly enough, this actually led to the correct solution, using only logic and no further guesswork. How embarrassing!
Back to top
View user's profile Send private message
keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA

PostPosted: Thu Apr 20, 2006 9:18 pm    Post subject: Reply with quote

Would you rather be good, or lucky?

Keith
Back to top
View user's profile Send private message
Display posts from previous:   
Post new topic   Reply to topic    dailysudoku.com Forum Index -> Daily Sudoku puzzles All times are GMT
Page 1 of 1

 
Jump to:  
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum


Powered by phpBB © 2001, 2005 phpBB Group