5
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I currently have a sudoku puzzle that looks like this:

[7][ ][4] | [6][ ][ ] | [1][9][3]
[ ][2][ ] | [ ][7][4] | [5][8][6]
[6][ ][ ] | [ ][ ][9] | [4][7][2]
--------- | --------- | ---------
[ ][ ][ ] | [ ][ ][ ] | [6][4][7]
[2][ ][ ] | [4][5][7] | [9][3][1]
[ ][4][7] | [ ][6][ ] | [2][5][8]
--------- | --------- | ---------
[1][ ][ ] | [7][ ][ ] | [3][2][ ]
[ ][9][2] | [ ][3][ ] | [7][6][ ]
[ ][7][ ] | [ ][ ][6] | [8][1][ ]

I've been looking at it for awhile but it seems like my next move is to guess where the next number goes, as there are many places where it's a 50/50 between 2 numbers. From my understand though is that sudoku is a logic puzzle and you shouldn't need to resort to guessing to figure out the next move. I was wondering if somebody could find and explain my next logical move ( or 2 ).

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  • 1
    $\begingroup$ Do you have your real puzzle marked up? $\endgroup$ – Josh Caswell Mar 5 '15 at 21:21
  • $\begingroup$ @JoshCaswell I do but I wasn't sure how to convey that in a nice format here. $\endgroup$ – Howdy_McGee Mar 5 '15 at 21:29
  • 1
    $\begingroup$ Gotcha. Looked for pairs and multiples? $\endgroup$ – Josh Caswell Mar 5 '15 at 21:32
5
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Well, you can figure out a 1 (top center group):

 7     4  |  6        |  1  9  3 
    2     |     7  4  |  5  8  6 
 6        |        9  |  4  7  2 
--------- | --------- | ---------
 -  1111  |           |  6  4  7 
 2  -  -  |  4  5  7  |  9  3  1 
 -  4  7  |     6     |  2  5  8 
--------- | --------- | ---------
 1        |  7        |  3  2    
    9  2  |     3     |  7  6    
    7     |        6  |  8  1    
Dashes are blocked off, the consecutive 1's are where a 1 has to be. 

So since all space by 1 are blocked off from having a 1 there (as below)

 7     4  |  6 - - - -|- 1  9  3 
    2     |     7  4  |  5  8  6 
 6        |        9  |  4  7  2 
--------- | --------- | ---------
 -  1111 -|- - - -    |  6  4  7 
 2  -  -  |  4  5  7  |  9  3  1 
 -  4  7  |     6     |  2  5  8 
--------- | --------- | ---------
 1 - - - -|- 7 - -    |  3  2    
    9  2  |     3     |  7  6    
    7     |    - - 6 -|- 8  1    

you can see that there is only 1 space left in the center column for a 1

 7     4  |  6        |  1  9  3 
    2     |     7  4  |  5  8  6 
 6        |    (1) 9  |  4  7  2 
--------- | --------- | ---------
          |           |  6  4  7 
 2        |  4  5  7  |  9  3  1 
    4  7  |     6     |  2  5  8 
--------- | --------- | ---------
 1        |  7        |  3  2    
    9  2  |     3     |  7  6    
    7     |        6  |  8  1    

to

 7     4  |  6        |  1  9  3 
    2 (1) |     7  4  |  5  8  6 
 6        |     1  9  |  4  7  2 
--------- | --------- | ---------
          |           |  6  4  7 
 2        |  4  5  7  |  9  3  1 
    4  7  |     6     |  2  5  8 
--------- | --------- | ---------
 1        |  7        |  3  2    
    9  2  |     3     |  7  6    
    7     |        6  |  8  1    

to

 7     4  |  6        |  1  9  3 
    2  1  |     7  4  |  5  8  6 
 6        |     1  9  |  4  7  2 
--------- | --------- | ---------
   (1)    |           |  6  4  7 
 2        |  4  5  7  |  9  3  1 
    4  7  |     6     |  2  5  8 
--------- | --------- | ---------
 1        |  7        |  3  2    
    9  2  |     3     |  7  6    
    7     |        6  |  8  1    
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  • $\begingroup$ Now that the boxes are gone I see / understand it now. Thanks! $\endgroup$ – Howdy_McGee Mar 5 '15 at 21:32
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    $\begingroup$ I was thinking the same thing, that with all the brackets the parenthesis get lost. $\endgroup$ – JonTheMon Mar 5 '15 at 21:36
  • $\begingroup$ @Howdy_McGee (also directed at Jon, he's auto-notified) I extended this answer to get a 9, a 3, and another 3. Just check my answer :D $\endgroup$ – warspyking Mar 5 '15 at 21:52
  • $\begingroup$ I must be missing something. I don't see how your first step leads to the second step $\endgroup$ – Ivo Beckers Mar 6 '15 at 12:12
  • $\begingroup$ wait I do see it now, but personally I would add a step in between showing that the center block and the bottomcenter block needs to have the 1 in the left and right column so the top one has to be in the center column $\endgroup$ – Ivo Beckers Mar 6 '15 at 12:15
1
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Continuing further you can find the 9 in the top left corner, and then finish off that line.

[7][_][4] | [6][ ][ ] | [1][9][3]
[9][2][1] | [3][7][4] | [5][8][6]
[6][_][_] | [ ][1][9] | [4][7][2]
--------- | --------- | ---------
[ ][1][ ] | [ ][ ][ ] | [6][4][7]
[2][ ][ ] | [4][5][7] | [9][3][1]
[ ][4][7] | [ ][6][ ] | [2][5][8]
--------- | --------- | ---------
[1][ ][ ] | [7][ ][ ] | [3][2][ ]
[ ][9][2] | [ ][3][ ] | [7][6][ ]
[ ][7][ ] | [ ][ ][6] | [8][1][ ]

After the above is done you can find a 3 by the process of elimination:

[7][_][4] | [6][ ][ ] | [1][9][3]
[9][2][1] | [3][7][4] | [5][8][6]
[6][3][ ] | [ ][1][9] | [4][7][2]
--------- | --------- | ---------
[ ][1][ ] | [ ][ ][ ] | [6][4][7]
[2][_][ ] | [4][5][7] | [9][3][1]
[ ][4][7] | [ ][6][ ] | [2][5][8]
--------- | --------- | ---------
[1][_][ ] | [7][ ][ ] | [3][2][ ]
[ ][9][2] | [ ][3][ ] | [7][6][ ]
[ ][7][ ] | [ ][ ][6] | [8][1][ ]
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0
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You can find the value for the cell in the fourth row, first column. The only possible values are (3, 5, 8, 9).
The fifth row has only two values left, (6, 8). Since they're both in the same block, we can eliminate the 8 from our target.
Since the first column already has two cells with (3, 9) as their only possible values, we can eliminate those as well, leaving only the 5.

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