# Sudoku NYT medium difficulty

I keep getting stuck on the NY Times Sudoku medium difficulty puzzles. I go through one number at a time and narrow down large box where that number can only be in two spaces. When I finish the nines I start back at the ones using the information from what numbers were determined in the first step. Then I go to the rows and columns and find ones that have around three or less boxes left, then I fill those candidates in. Then I'm stuck. Any thoughts on what strategy I'm not understanding or what things I could look for?

Stiv's answer is right (+1), but let me share another technique which might help you discover this in the future.

You've rightfully concluded that the last three columns in the second row are 1-5-8 (in no particular order). That means you can put a mark like this freehand circle around them:

and conclude the other three cells in the top right 3x3 must be filled with the other remaining numbers, 6-7-9 (again in no particular order).
6 and 7 are already out for the cell in the third row, so it must be a 9.
This trick works with pairs as well, perhaps even more often than with triples.

• Agreed - another useful way to think about it! +1
– Stiv
Commented May 12, 2020 at 11:59
• As a consequence of which, the other two boxes (6,7,9) cannot be 9 leaving only one place for 9 in the top row. Finding one often leads to a chain of dominoes falling. Commented May 12, 2020 at 16:45
• Yes, that's often the case with sudokus :) Commented May 12, 2020 at 16:49
• For the number between 6 and 2, I would have marked that with a 5 and 9. But you're saying that the 1,5,8 are forced in the second row? Is that because they all share those three numbers but the 5,9 has the 9 as the odd man out; therefore, it has to be a 9? Commented May 12, 2020 at 17:48
• @tazboy if you look at the second row only, you know that the last three columns have to be some combination of 1-5-8 (doesn't matter which order). That means the positions of 1, 5 and 8 in the top right 3x3 are already known; they're in the freehand circle. Since 2, 3 and 4 in that 3x3 are also known, the three remaining cells in that 3x3 (including the one between 6 and 2) can only be 6-7-9. You don't need any information from the first row or the last three columns to deduce that. Commented May 12, 2020 at 17:54

Sometimes you have to look at rows and columns which have more than 3 empty spaces left! Consider the space on the third row between the 6 and 2...

The row is missing 5, 7, 8 and 9 (four numbers). Rule out 7 and 8 immediately because they exist in the same column already. Then notice it cannot be a 5 because there would be no legal spaces left to place a 5 in the top-left 3x3 box. Thus this space must be a 9.

Your strategy should work for most sudokus if you don't limit yourself in this arbitrary way. Consider ALL rows and ALL columns, even if they have 4 or more empty squares, and you should find places you can make progress, like in my example above. Good luck!

• Dang that's tough. I'm not used to thinking like that. Thanks for that strategy. Commented May 12, 2020 at 17:43
• What Stiv's answer really leads towards is the idea that, sometimes in logic puzzles like Sudoku, a way forward is to take a guess somewhere and see if that leads to contradictions several moves down the line. This is a simple example of that: the R3C7 cell does not itself disallow either 5 or 9, but trying to continue the puzzle with a 5 leads to the contradiction in the top left box, allowing you to deduce the 5 was incorrect. Sometimes, that may require several moves worth of forward looking, similar to how placing 1 number can cause a chain of forced moves elsewhere with that same number. Commented May 13, 2020 at 3:07

Sometimes you need to look not only to separate numbers, rows and columns, but also the whole rows and columns of 3x3 blocks. In your example, you can notice that in top right 3x3 square, number 7 can be only in the top row. This cancels the possibility of number 7 being in the first row in the left block, leaving the number 9 the only option in the top left corner.

Spoiling all the candidates in the attached image, but useful tool that can help you with hints, and the strategy behind the hint: https://www.sudoku-solutions.com/index.php?page=sudoku9by9

I think it usually gives you the easiest hints first, and then more and more complex strategies as needed.

When you find rows and columns that have only three boxes left you fill those numbers in (green in the image). Switch contexts and realize you now have only three empty boxes in the top-right square.

2,3,4 are given and you know the middle row is 1,5,8 (green), so the other three boxes have to be 6,7,9. The lower-left box (red) has a 6 in the row already and a 7 in the column so it has to be 9. Using that you can write in the options for the the lower row of the top-left box and what the value in the top-left square of that box must be.

• This is essentially the technique I'm describing in my answer, isn't it? Commented May 12, 2020 at 13:57
• Why couldn't the bottom left be a 5? Commented May 12, 2020 at 17:51
• It can't be 5 because you know a 5 has to be in the row above it for this square, since there is one per row and that row doesn't have a 5 yet, with all cells outside this square, for that row, already filled Commented May 12, 2020 at 22:59

Starting with the givens in the original puzzle (not taking your eliminations into account) and after propagating the obvious constraints from the givens, the puzzle is solved with:

• a hidden-single-in-a-column (==> r3c1 = 8),

• followed by two interactions between rows and blocks (the most elementary rules after singles):

5 in row r9 ==> r7c2 ≠ 5

6 in row r5 ==> r4c8 ≠ 6, r4c7 ≠ 6

• followed by singles
• An explanation of what these techniques are (link?) and why they apply here would be useful. Commented Sep 26, 2020 at 14:53
• "interactions" are also called "claiming" and "pointing". You can see their definitions on any Sudoku website. Commented Sep 27, 2020 at 7:01
• Please don't assume that we know the terminology that you're using. Answers should be self-contained, and I can't tell what this answer is saying without extra research. Commented Sep 27, 2020 at 14:44

My answer might appeal to beginners. I use letters for rows. C1=8. Since C7 can see both 5 & 7, C7=9. Thus, A1=9. Either B7=1 or E7=1. I assume B7=1 and that leads quite easily to a complete solution. My only technique is to consider cells with exactly 2 possible entries. For example, C2, C3, E1 and I1 are all (57).