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PoppaPoppa
Joined: 06 Nov 2005 Posts: 21 Location: Arkansas USA

Posted: Thu Dec 29, 2005 4:42 am Post subject: Reply to Someone_Somewhere 


sorry, Munich, I don't understand. I don't see a naked pair of 4 5 atr3c7  there is also a possible 9 there, nicht wahr? 

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

Posted: Thu Dec 29, 2005 7:51 am Post subject: 


Hi Popa,
Same response as I have posted on the other thread:
Ok.
Let's take a deep look at the 6th 3x3 block, with the corners: r4c7r4c9r6c9r6c7.
Digit 7 is in r2c8. So it can't be in r7c8 and can't be in r9c8. So, it must be r7c7, r8c7 or r9c7 (it has to be one of the boxes of the 9th 3x3 block).
This leaves us for digit 7 in the 6th block only: r5c9 or r6c9.
Now let's concetrate on digit 4.
From 1st 3x3 box we know that it has to be in r1c1 or r1c4. So it can't be in r1c9.
From 2dt 3x3 box we know that it has to be in r2c4 or r2c5. So it can't be in r2c9.
4 is in r4c6.
So we have to have a 4 in column 9 and the only posibilities left are r5c9 or r6c9.
So 4 can't be in any other place of the 6th 3x3 block.
Now we have our Hidden Pair 4, 7 in r5c9 and r6c9.
Hope this could help, if not I will try again.
see u, 

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PoppaPoppa
Joined: 06 Nov 2005 Posts: 21 Location: Arkansas USA

Posted: Fri Dec 30, 2005 1:12 am Post subject: Again, don't understand 


Someone_Somewhere: I'm determined to follow your logic but I cannot see why, for example where you say
"From 1st 3x3 box we know that it has to be in r1c1 or r1c4. So it can't be in r1c9. "
First, I think you mean 'From 1st 3x3 box we know that it has to be in r1c1 or r1c**3**'
Secondly, your logic escapes me  there can also be 4s in row 2, c1 and 3, and r 3, c3... therefore I see no reason to exclude the 4 from row 1 c9!
Do you use Skype? I'd love to talk to you about this . My Skype name is poppapinebluff 

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

Posted: Fri Dec 30, 2005 9:17 am Post subject: 


Hi,
Sorry, I have Skype not installed.
I will let you know, if and when I install it.
Hope, you solved menwhile the Sudoku puzle.
I can only recommand you:
 the usage of the "candidate table", till you can solve puzles without it.
 the reading and training of the "solving techniques":
Row on 3x3 Block interaction
Column on 3x3 Block interaction
3x3 Block on Row/Column
Hidden Pair in Row, Column or 3x3 block
Naked Pair in Row, Column or 3x3 block
Hidden Triple in Row, Column or 3x3 block
Naked Triple in Row, Column or 3x3 block
Hidden Quadruples in Row, Column or 3x3 block
Naked Quadruples in Row, Column or 3x3 block
XWing on Row or Column
XYWing
Turbotfish
Swordfish on Row or Column
Jellyfish on Row or Column and
Squirmbag on Row  are not so important, in my opinion
and if you are mastering also the "double implication" technique (or a similar/equivalent one), there will be no Sudoku's left that you can not crack if you work long enough on them ...
see u, 

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

Posted: Fri Dec 30, 2005 9:49 pm Post subject: Someone_Somewhere had it right ... 


PoppaPoppa wrote:  Code:  070 520 630
050 000 070
900 008 002
017 004 000
090 000 060
000 800 310
109 600 005
040 000 096
086 095 003 
sorry, Munich, I don't understand. I don't see a naked pair of 4 5 atr3c7  there is also a possible 9 there, nicht wahr? ...
I'm determined to follow your logic but I cannot see why, 
Hi, PoppaPoppa! Happy New Year!
Someone_Somewhere had it nailed  let me take a stab at it.
First, there cannot be a "9" at r3c7 because of the "9" at r3c1. Let's take a look at the initial possibilities in the top right 3x3 box:
Code:  6 3 1489
489 7 1489
145 45 2 
Now there can't be a "1" in r3c7, because if there were it would be impossible to place a "1" in the bottom right 3x3 box. So there's a pair {4, 5} in r3c7 & r3c8. Eliminating the "4" from the other 3 unresolved cells in the top right 3x3 box gives us this revised matrix:
Code:  6 3 189
89 7 189
45 45 2 
The only possibilities at r4c9 are {8, 9}  you have {1, 4, 7} in row 4, and {2, 3, 6, 5} in column 9. So you can see a triplet {1, 8, 9} in r1c9, r2c9, & r4c9, leaving the hidden pair {4, 7} in r5c9 & r6c9. This means that r6c6 is the ONLY spot in which the "9" in row 6 can go. dcb 

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PoppaPoppa
Joined: 06 Nov 2005 Posts: 21 Location: Arkansas USA

Posted: Sat Dec 31, 2005 6:44 pm Post subject: Happy New Year 


Thanks David I finally got it with the help of several folks who I thanked in another thread  Happy New year to you too! 

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alanr555
Joined: 01 Aug 2005 Posts: 198 Location: Bideford Devon EX39

Posted: Wed Mar 01, 2006 12:48 am Post subject: 


We appear not to have heard from our Munchen friend for a while.
He will be missed. Meanwhile it is good to see some newcomers
engaging with the leading edge  in addtion to our Colorado friend
who is a real mainstay to this forum.
Alan Rayner BS23 2QT 

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