As you can see in the image, I've been struggling with this sudoku for quite some time and I'm completely stuck. I'm sure I'm missing one standard trick which would make it easy to solve, but I can't find it! I only mark when there are only two possibilities, so AFAIK that's the case in this sudoku.

What I'm looking for is exactly how to find the next number.

PS: I would like to make the image smaller, but I don't know how.

  • $\begingroup$ If anybody wants to try to finish this sudoku, they can do it here: asudoku.com/… $\endgroup$ – VLAS Feb 8 '16 at 13:43

You are missing what are known as pointing pairs. If you're not sure where to look, here's a hint:

you need two of these in the same box

  • 1
    $\begingroup$ I've checked with a solver, and this Sudoku can be solved using pointing pairs and box-line reduction alone (along with other basic methods). I agree that these are the next two places to look. $\endgroup$ – user20 Jul 2 '15 at 6:15
  • $\begingroup$ Could you point out your solution? After completing all the cells trying to find it (thing that I should start doing!) with all the possibilities, I realised there was a so-called 'hidden pair' with (1,2) in the sixth row 2nd and 7th columns. $\endgroup$ – hjhjhj57 Jul 2 '15 at 6:58
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    $\begingroup$ That's right! You use pointing pairs when eliminating candidates, you just didn't know the name. The only cells in the second column that can contain a 1 are in the middle box, just as the only cells in the second column that can contain a 2 are in the middle box. These are called pointing pairs. This means that the only cells in the sixth row that can contain a 1 or 2 are the second and seventh, your hidden pair. $\endgroup$ – Ben Jul 2 '15 at 7:31

Look in the bottom left 3 x 3 block, and what I will tell you is based on the fact the puzzle MUST have a unique solution. For reference, I will number the squares across as 1-9 and the squares down A-I.

You can see that the 1 can only be place in either H3 or I3. Now since the squares H4 and I4 form a naked double (the numbers being 1 and 2), it will be obvious that the other number in H3 and I3 cannot be the 2. This is because it would be impossible for the puzzle to have a unique solution. Now, where can the 2 go?

The 2 therefor has to be in either H1 or I1, and that eliminates the 2 in the possibilities you have for A1 meaning you can immediately write 4 there. How is that then?

  • $\begingroup$ Why did you choose to answer an older question that already had an accepted answer? $\endgroup$ – Patrick N Aug 20 '15 at 18:39
  • $\begingroup$ @PatrickN an important part of the SE system consists in having more than one answer to each question when it's possible. In this case, this answer actually provides insight into a different and interesting way to solve the puzzle, namely using the uniqueness of its solution. Even though I still need to see it's correctness, I like this answer a lot. +1 $\endgroup$ – hjhjhj57 Aug 20 '15 at 19:56
  • $\begingroup$ I certainly understand that, but the underlying principle there is that the alternate answers may be informative and helpful, both to the poster and other readers. Considering that this is an answer tailored to a unique puzzle that (I assume) has already been solved, its relevance is lower than in other cases. $\endgroup$ – Patrick N Aug 20 '15 at 20:06

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