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BrendaS
Joined: 08 Jan 2006 Posts: 3 Location: Canada

Posted: Sun Jan 08, 2006 5:57 pm Post subject: January 7 Very Hard Classic 


I have the following done:
629 xxx xxx
135 9xx xx2
847 1xx x39
x98 5x3 x14
xxx x1x 39x
31x 4x9 8x5
x8x xx7 9xx
96x xx1 753
xxx x9x x48
The hint at this point is r1c5 = 3
I can't understand the logic of this  can someone help me?
Brenda 

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dotdot
Joined: 07 Dec 2005 Posts: 29 Location: oberseen

Posted: Sun Jan 08, 2006 8:10 pm Post subject: 


Hallo Brenda
Looking at col5, there are 3s scattered to the left and right of it, each doing its own bit of exclusion. In fact they exclude 3 from every free cell in col5 except the hint's row1 cell
 and unfortunately the row7 cell too.
This is where the going in this puzzle gets tough (I got pretty fed up here) because you have to see one pair exclusion and then another pair exclusion.
The first involves 1 and 3 in box7: col1 and col2 and row8 each exclude 1 and 3 from the parts they share with box7. This leaves just two cells, r7c3 and r9c3, unaffected; so they are {1,3}.
The second involves 4 and 5 in row7: col4 and col8 and col9 each exclude 4 and 5 from where they cross row7. This isn't quite good enough because it leaves not two but three cells unaffected.
But one of these cells is r7c3 which we found to be {1,3} so it can't be 4 or 5 either. This now leaves just the two cells r7c1 and r7c5; so they are {4,5}.
And the upshot of all that is that r7c5, being {4,5}, cannot be 3. 

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

Posted: Sun Jan 08, 2006 8:58 pm Post subject: Another way to look at it 


Here's another way to look at it, Brenda.
There is, as dotdot says, a hidden pair {1, 3} in r7c3 & r9c3. And this leads directly to the identification of another hidden pair {4, 5} in row 7  at r7c1 & r7c5.
But now the value "3" in column 4 must lie in the bottom center 3x3 box  there's no other way to fit a "3" in that box. So the candidate list at r1c4 is just {7, 8}  this pairs up with the {7,8} possibility at r1c8, from which we infer that r1c9 = 1. Now we have the {4, 5} pair in r1c6 & r1c7, and the only place left for a "3" in row 1 is at r1c5. dcb 

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new2sudoku
Joined: 10 Jan 2006 Posts: 1

Posted: Tue Jan 10, 2006 1:14 am Post subject: Jan 7th puzzle... 


I'm new to sudoku and was looking for some help. First of all, where could I start reading about this hidden pair stuff? I am great with numbers but don't understand how you can come up with numbers when they have multiple solutions? Do you have to guess and go on from there? Where can I learn how to do the harder puzzles? 

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

Posted: Tue Jan 10, 2006 2:40 am Post subject: Re: Jan 7th puzzle... 


new2sudoku wrote:  I'm new to sudoku and was looking for some help. First of all, where could I start reading about this hidden pair stuff? 
Hi, new2sudoku! Welcome to the forum.
There's a solid introduction to basic Sudoku techniques on the Sadman Software site.
With particular reference to "hidden pairs," here's a great example  it's the October 24, 2005 Daily Sudoku classic puzzle.
Code:  470 203 900
060 810 020
021 079 500
500 001 000
200 000 005
000 700 006
702 900 450
090 547 032
054 102 079 
If you look at this closely you'll notice a pair of numbers, {6, 7}, lying in column 2 (in r1c2 & r2c2). The same pair appears in row 6  at r6c4 & r6c9.
Now observe how row 2 and column 6 intersect in the middle left 3x3 box. Clearly neither of the values "6" or "7" can lie in r4c2, r5c2, r6c1, r6c2, or r6c3. Therefore the pair {6, 7} must lie in r4c3 & r5c3, in some order. This leaves only 4 unknowns  3, 5, 8, 9  in the rest of column 3.
Now if you look at r8c3 you'll notice that 3, 5, & 9 are not possible in that spot, so it must be "8".
There's an example of a hidden pair  I hope it's helpful. dcb 

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