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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3
  • 1
    $\begingroup$ Do you have your real puzzle marked up? $\endgroup$
    – jscs
    Commented Mar 5, 2015 at 21:21
  • $\begingroup$ @JoshCaswell I do but I wasn't sure how to convey that in a nice format here. $\endgroup$
    – Howdy_McGee
    Commented Mar 5, 2015 at 21:29
  • 1
    $\begingroup$ Gotcha. Looked for pairs and multiples? $\endgroup$
    – jscs
    Commented Mar 5, 2015 at 21:32

3 Answers 3

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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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5
  • $\begingroup$ Now that the boxes are gone I see / understand it now. Thanks! $\endgroup$
    – Howdy_McGee
    Commented Mar 5, 2015 at 21:32
  • 1
    $\begingroup$ I was thinking the same thing, that with all the brackets the parenthesis get lost. $\endgroup$
    – JonTheMon
    Commented Mar 5, 2015 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
    Commented Mar 5, 2015 at 21:52
  • $\begingroup$ I must be missing something. I don't see how your first step leads to the second step $\endgroup$
    – Ivo
    Commented Mar 6, 2015 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
    Commented Mar 6, 2015 at 12:15
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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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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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