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Stuck in this sudoku. If anyone can solve this then please share the method.


As Glorfindel has found, you can

eliminate the 9 in R1C6 and the 2's in R3C6 and R7C6

Then, there's a chain:

If R5C4 = 9, then R5C9 = 8, R7C9 = 9, R7C8 = 8, R1C8 = 9, R1C5 = 7, R6C5 = 2, R4C5 = 9, R5C4 is not 9.
This is a contradiction, so you can eliminate the 9 from R5C4.

Then another chain:

If R6C4 = 7, then R6C5 = 2, R4C5 = 9, R1C5 = 7, R1C6 = 8, R3C6 = 5, R3C4 = 2, R7C4 = 7, R6C4 is not 7.
This is a contradiction, so you can eliminate the 7 from R6C4.

The rest is trivial.
Not an easy puzzle. I could not find easier techniques to use.

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  • $\begingroup$ This is not too long, but it is still trial and error to me. "If ... (trial) then .... then contradiction (error)". $\endgroup$ – Florian F Jun 7 at 13:58
  • $\begingroup$ @FlorianF If it can be done without trial and error, I'd like to know. I couldn't find any easier ways to solve this one. $\endgroup$ – Dennis_E Jun 7 at 20:55
  • $\begingroup$ @Dennis_E When it comes to identifying chains with contradictions, is there a pattern or technique involved in doing so? For example, how did you know to start with R5C4 and R6C4. I get stuck on puzzles like this all the time. $\endgroup$ – Anita Taylor Jun 25 at 19:14
  • $\begingroup$ @AnitaTaylor I'm not an expert, so I don't know what the best way is. But I look at chains of cells that contain pairs, like the 89, 89, 89, 89, 79, 72, 29. R5C4 is connected to both the start and the end of that chain, so I try one of those numbers. (Yes, I suppose this is trial and error but all the basic techniques I know were exhausted) $\endgroup$ – Dennis_E Jun 26 at 8:40

A simple hint:

In column 6, you've correctly identified that rows 2 and 8 can only be 2 and 9. That means row 1 can't be a 9; row 3 can't be a 2; and row 7 can't be a 2 either.

This technique is called a naked pair; it seems you've used it already in row 7 to eliminate the 9's in row 4 and 6.

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