Hello,

I am just a pastime sudoku solver. I sovle mostly classic puzzles from the newspapers or from some websites. I want to know if all the great/champion sudoku solvers resort to guessing or backtracking. Or is there some way/technique that it can be reduced.

I am attaching a screenshot to this post. I was solving a puzzle and got stuck at a point. Usually I would assume one value and then keep on solving unless I finish the puzzle or get my assumption wrong.

Please have a look at the image and let me know if I can proceed in the same puzzle without guessing at this stage. Or if it seems the only option.

Thanking you in advance





Attachments
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duvidha.PNG (60KB - 13 downloads)

P.S ... it should be 7/9 in the top left grid.

Not sure about "great" solvers, but I think most of the time it isn't necessary to guess. At least, not if the Sudoku is from a trusted source. On this site for instance, I've competed in many monthly tests and haven't used guesswork much at all. As for techniques, check sudokuwiki.org



Out of curiosity, did you guess to get a 1 in box 5 or was that just a marking?

I'm not sure about the entire Sudoku, because I haven't solved it entirely, but I did get a few numbers without any need for guessing, for instance the 3 in R9C4 (I didn't use your marking for this, the 1 in the 1st box is enough to see that 1 can only be in two places on row 7, and so can't be in the rest of the 8th box), 3 in R1C6. There's also an easy way to get the one remaining 8 in the grid, since all others are given. In general though, it does seem difficult. But if you're missing the basic techniques too, then its better to start with easier ones and become good with using those techniques, in my opinion.

Thanks a lot Prasanna for replying :)

The 1 in box 5 was wrongly written. It was for the 2 places I circled in row 7. I want to know how did you derive that 3 would be in R9c4. Coz 3 can also go in R9C6.

Once you've determined the 1 in R7 has to be in C4 or C5, 3 is the only number left that could possibly be in R9C4. 1, 4, 5, 6, and 8 are already present in the box, 2 and 9 are already in the column, and 7 is already in the row. Even though the 3 isn't eliminated from R9C6, everything but the 3 is eliminated from R9C4.

Got it now! Thanks for explaining in so detail. That really really helps. I somehow missed that logic. I now recollect having solved puzzles using this technique. Maybe I am out of practice or just becoming non-intelligent. :D