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Please explain the logic. I am not trying to solve it using trial and error or using any software. I am just looking for the next step. This sudoku is from apple 'Sudoku King' app.

  • $\begingroup$ It looks like there are two solutions, so this is as far as you can get using pure logic. $\endgroup$
    – venus
    Apr 13, 2021 at 23:20
  • $\begingroup$ r7c6 and r9c3 are unambiguous, though. $\endgroup$
    – loopy walt
    Apr 14, 2021 at 0:18
  • 3
    $\begingroup$ Delete the app. Any sudoku app with a broken sudoku in it is worse than no sudoku app at all. $\endgroup$
    – Bass
    Apr 14, 2021 at 8:14

1 Answer 1


Let's take one more step, with logic.

spots where 4s could go
The green squares are where a 4 could go, based on your pencil marks. Let's establish a series of facts.
1. If a 4 goes in R7C6, then a 4 must go in R9C3 (by hidden single in the bottom-left 3x3 box)
2. By #2, if a 4 goes in R7C6, then there will be 4s in C3 and in C6
3. No more than one 4 may be used in a column (by column-restrictions)
4. Exactly 4 must be used in R6 (by row-restrictions)
5. The possible spots for a 4 in R6 are in C3 and C6
6. By #2, #3 and #5, if a 4 goes in R7C6 then a 4 cannot go in R6. This contradicts #4, therefore a 4 does not go in R7C6
7. By #4 and #6, if a 4 goes in R7C6 then a contradiction arises.

Or, look at this picture:

spots where 4s could go if R7C6 is a 4
If R7C6 is a 4, then R9C3 must be a 4 (#1). I have colored in those spots blue. There is now no way to place a 4 in R6's green squares without it being in the same column as an already-placed 4, and two 4s in one column is not allowed (#3).


R7C6 and R9C3 are 9s (R7C6 we just found couldn't be a 4, so it must be a 9, and simple hidden single for the bottom-left 3x3 box determines where its 9 goes)

But from here, there's no way to progress by logic, because there are two solutions. (Thanks to @venus in the comments for alerting me to this!)

enter image description here enter image description here
I got the first solution by assuming R2C4 was a 7 (one of its possibilities) and the second by assuming it was a 6 (its other possibility). Note how no numbers (beyond those determined in the last logic-step) are shared.

  • $\begingroup$ could you explain this part - Well, we can't block off both of the ones in R6. The ones in R6 are in C3 and C6, so the 4s in the two bottom-row 3x3 boxes can't be in both those columns. But... if a 4 goes in R7C6, then a 4 must go in R9C3. Having a hard time $\endgroup$ Apr 14, 2021 at 13:22
  • 1
    $\begingroup$ @user2543622 I have reorganized that section and added a picture. Does it make sense now? $\endgroup$
    – bobble
    Apr 14, 2021 at 14:25
  • $\begingroup$ yes...is there a good sudoku iphone app which doesnt have more than 1 solution for the same puzzle $\endgroup$ Apr 14, 2021 at 15:44
  • $\begingroup$ I don't know anything about iPhone apps (I use Android). Try asking in chat. $\endgroup$
    – bobble
    Apr 14, 2021 at 15:47

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