I keep staring at this sudoku puzzle and can't seem to find the next move.
Is the next move a guessing move? I thought sudoku always has clear solutions, you just need to figure them out.
2 Answers
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No you shouldn't guess. The puzzle has a unique solution. However, you can't continue unless you calculate all possibilities. Once you do that, solving the puzzle will be easy. I have added the possibilities below. If you want to do that on your own, don't look down.
Once you do that, I think the next step is clear:
Note:
others pointed out in the comments that it is unclear why that particular cell cannot have a 5. To explain, in the red column there are 2 cells indicated in blue. They have two possibilities: (3,5). No matter how the possibilities are distributed, any of the cells will definitely have a 3 and the other a 5. Which makes it impossible for any other cell in that column to have a 3 or a 5.
Additional explanation (requested by the OP):
Why can't there be a 7 in the last row? Because it can only be in row 7.Explained below:
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2$\begingroup$ How is it clear that in that place it's a 7? Why is it not a 5? $\endgroup$– AdelinJan 16, 2018 at 10:53
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$\begingroup$ @Adelin if you look at thd possible numbers at that cell, you can see there is only a 7. What makes you think there should be a 5 $\endgroup$– MohammadJan 16, 2018 at 11:00
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1$\begingroup$ Ah, that clarifies it. Thanks for the explanation. $\endgroup$– votbearJan 17, 2018 at 2:05
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1$\begingroup$ The 7s in row 7 don't look right. 7 can't be the only possibility in two places in the same row/box. Your chart must be incomplete, or I'm misunderstanding something. $\endgroup$– ChipsterApr 13, 2020 at 19:47
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For a start:
2,4 and 7 cannot be in the middle square bottom row, which means that there has to be 1 and 6.
This, in turn, means that:
the 8 must be in the top middle and that there must be a 5 to the middle left in the middle square, and so on.
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$\begingroup$ Welcome to Puzzling! This question already had an answer accepted 11 months ago - be sure to read other answers to make sure you aren't duplicating content. $\endgroup$– Deusovi ♦Dec 5, 2018 at 22:51
5is for sure in R6C1, dicatated by top middle 3x3 cell. There, you can have8and3exclusively in the R1 C4 respectively C5, which tells us that the remaining1and5are on R3, therefore telling us that the5in top left 3x3 cell is on C3 which, working togheter with the5from R8C6, cuts the bottom left 3x3 in such a way that5is only possible in R6C1. The remaining two (7and2) are easy $\endgroup$