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Alan Guest

Posted: Thu Aug 11, 2005 12:18 pm Post subject: Puzzle 11 August  don't understand the hint 


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. 

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Kim Guest

Posted: Thu Aug 11, 2005 1:10 pm Post subject: 


<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
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 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.
Kim 

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Guest

Posted: Thu Aug 11, 2005 1:13 pm Post subject: 


<<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
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!!!
Kim 

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Kim Guest

Posted: Thu Aug 11, 2005 1:16 pm Post subject: 


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 

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Alan Guest

Posted: Thu Aug 11, 2005 1:27 pm Post subject: 


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. 

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Susie22 Guest

Posted: Thu Aug 11, 2005 2:18 pm Post subject: 


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. 

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

Posted: Thu Aug 11, 2005 2:33 pm Post subject: Note to Kim 


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 

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Guest

Posted: Thu Aug 11, 2005 3:58 pm Post subject: 


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 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Thu Aug 11, 2005 4:06 pm Post subject: 


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, 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Thu Aug 11, 2005 6:25 pm Post subject: 


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, 

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

Posted: Fri Aug 12, 2005 4:47 pm Post subject: 


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 

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Guest

Posted: Fri Aug 12, 2005 8:44 pm Post subject: 


column 3 should be 128, so after eliminating 1 should make it 28 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Sat Aug 13, 2005 5:15 am Post subject: 


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, 

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