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 

Puzzle 11 August - don't understand the hint

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





PostPosted: Thu Aug 11, 2005 12:18 pm    Post subject: Puzzle 11 August - don't understand the hint Reply with quote

I normally manage to solve these puzzles comfortably in my lunchbreak, and even when time pressure means I resort to 'cheating' I understand the logic behinf a hint once I've seen it.

Today I'm obviously having an off day. I have got the puzzle this far

010 346 089
803 275 000
054 981 200

940 607 800
300 109 007
075 438 029

037 064 090
000 012 304
400 093 060

The hint says that the bottom row, cell3 is a 1.
Anyone enlighten me why the 1 cannot be between the 4 and 6 on row 4, or between the 3 and 6 on the bottom row?

As I said, I'm obviously having a thick day.
Back to top
Kim
Guest





PostPosted: Thu Aug 11, 2005 1:10 pm    Post subject: Reply with quote

<The hint says that the bottom row, cell3 is a 1.
Anyone enlighten me why the 1 cannot be between the 4 and 6 on row 4, or between the 3 and 6 on the bottom row? >

Try this...
in r7c7 it can be a (1 5) r7c9 (1 2 Cool
in r8c8 is can be a (5 7)
in r9c7 it can be a (1 5 7) and r9c9 (1 2 8)

see the pattern the (1 5 7) in r7c7, r8c8 and r9c7???

That will elimatine the 1 in r7c9 and r9c9 giving you fixed pairs of (2 8)

In doing so, it gives you a fixed pair of (2 Cool in r9c2 and r9c9 making the only number left in r9c3 a 1....

Understand??? I have a tough time explaining this stuff like the pros do. Smile

Kim
Back to top
Guest






PostPosted: Thu Aug 11, 2005 1:13 pm    Post subject: Reply with quote

<<Try this...
in r7c7 it can be a (1 5) r7c9 (1 2
in r8c8 is can be a (5 7)
in r9c7 it can be a (1 5 7) and r9c9 (1 2

see the pattern the (1 5 7) in r7c7, r8c8 and r9c7??? >>

Darn it...It left out the 8's.

It should look like ....
In r7c7, it can be a (1 5) and in r7c9 (1 2 Cool
In r8c8, it can be a (5 7)
In r9c7 it can be a (1 5 7) and in r9c9 (1 2 8)

See the pattern?

Hope that clarifies!!! Smile
Kim
Back to top
Kim
Guest





PostPosted: Thu Aug 11, 2005 1:16 pm    Post subject: Reply with quote

Darn smileys are everywhere....

Wherever you see the stupid smiley with dark glasses, replace that stupid piece of crap with an 8 or eight!!!

Note to self...Never put an 8 right next to a ) or you will get a smiley with glasses!!!

Nothing like 15 posts to answer your question....Good luck
Kim
Back to top
Alan
Guest





PostPosted: Thu Aug 11, 2005 1:27 pm    Post subject: Reply with quote

Doh!!!

I'd done most of the hard work, then missed the easy bit.
I had the 157 triplet and the matched 28's in c9.
I'd then been too blind to see the 5 in r3c2, leaving 258 as possibilities for r9c2 so didn't get the matched pair of 28's in the bottom row.

Again, I say, Doh!!!

and thanks.
Back to top
Susie22
Guest





PostPosted: Thu Aug 11, 2005 2:18 pm    Post subject: Reply with quote

Can someone please explain to me about the triplet (1 5 7) in r7c7, r8c8 and r9c7??? Why will this elimatine the 1 in r7c9 and r9c9 giving you fixed pairs of 28's.
Back to top
David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Thu Aug 11, 2005 2:33 pm    Post subject: Note to Kim Reply with quote

Kim wrote:
Darn smileys are everywhere....

Wherever you see the stupid smiley with dark glasses, replace that stupid piece of crap with an 8 or eight!!!

Note to self...Never put an 8 right next to a ) or you will get a smiley with glasses!!!

Nothing like 15 posts to answer your question....Good luck
Kim


Say, Kim -- next time you post, please notice the checkbox just above the "submit" button labeled "Disable Smilies in this post." dcb Shocked
Back to top
View user's profile Send private message Send e-mail Visit poster's website
Guest






PostPosted: Thu Aug 11, 2005 3:58 pm    Post subject: Reply with quote

Susie22 wrote:
Can someone please explain to me about the triplet (1 5 7) in r7c7, r8c8 and r9c7??? Why will this elimatine the 1 in r7c9 and r9c9 giving you fixed pairs of 28's.


Because the cells with a 15, 57 and 157 cancel themselves out. Therefore, no other cells in that block can contain a 157. Make sense? I am sure someone can explain it better.

Kim
Back to top
someone_somewhere



Joined: 07 Aug 2005
Posts: 275
Location: Munich

PostPosted: Thu Aug 11, 2005 4:06 pm    Post subject: Reply with quote

Susie22 wrote:
Can someone please explain to me about the triplet (1 5 7) in r7c7, r8c8 and r9c7??? Why will this elimatine the 1 in r7c9 and r9c9 giving you fixed pairs of 28's.


If you have in a 3x3 (or on a horizontal row or on a vertical column), in 3 cells exact 3 numbers as possible (in our case 1, 5 and 7) than this 3 numbers are not possible in some other cell of this 3x3 (or on horizontal, or vertical respective).
Is it clear? If not, I can try to explain it in more detail.

have fun,
Back to top
View user's profile Send private message
someone_somewhere



Joined: 07 Aug 2005
Posts: 275
Location: Munich

PostPosted: Thu Aug 11, 2005 6:25 pm    Post subject: Reply with quote

Or we could have the situation for a 3x3>


157 9 12

3 57 4

157 6 12

Now, if the numer 1 would be in r1c3 than we would habe the pair of numbers 1 and 5 left to be in 3 cells: r1c1 r2c2 and r3c1 and this would be impossible !
So, we can eliminate / exclude number 1 from cells r1c3 and r3c3.

have fun,
Back to top
View user's profile Send private message
David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Fri Aug 12, 2005 4:47 pm    Post subject: Reply with quote

someone_somewhere wrote:
Or we could have the situation for a 3x3>


157 9 12

3 57 4

157 6 12

Now, if the numer 1 would be in r1c3 than we would habe the pair of numbers 1 and 5 left to be in 3 cells: r1c1 r2c2 and r3c1 and this would be impossible !
So, we can eliminate / exclude number 1 from cells r1c3 and r3c3.

have fun,


Shouldn't that be

157 9 18

3 57 4

157 6 12

?? The example would still be impossible, otherwise! dcb Shocked
Back to top
View user's profile Send private message Send e-mail Visit poster's website
Guest






PostPosted: Fri Aug 12, 2005 8:44 pm    Post subject: Reply with quote

column 3 should be 128, so after eliminating 1 should make it 28
Back to top
someone_somewhere



Joined: 07 Aug 2005
Posts: 275
Location: Munich

PostPosted: Sat Aug 13, 2005 5:15 am    Post subject: Reply with quote

Ok, Ok, if you want to correct things ...
Let's do it proper:

The initial position has a 9 in r1c9 and r6c9. At least if you take a look at the first message posted here.

. . .

see u,
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