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I am fairly new to Sudoku and couldn't find how to progress with this one without guessing.enter image description here

HyperX
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2 Answers2

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For those that may have seen a manual or website on Sudoku solving, it is worth trying to spot a Y-Wing on this board. This is instructive because guessing in general is an art, while attempting to spot X and Y wings(for example) is something that solvers can be trained specifically to look out for.

The other answer is sound as well: given that the bottom right corner has many cells with only two candidates, forcing chains are a natural strategy.

Here's the Y-Wing:

[56] on R5C1, [67] on R8C1 and [57] on R7C2 form a Y-Wing, which shows that 5 cannot be a candidate for R4C2. Thus, R4C2 = 1.

Note that after one uses the Y-Wing to resolve a cell, the rest of the puzzle can easily be solved without any more complicated techniques.

bobble
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Sarvesh Ravichandran Iyer
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Not sure if this counts as guessing, but R4C2 can be solved:

Look at R5C1.
If it's a 5, then R4C2 is a 1.
If it's a 6, then R8C1 is a 7, R7C2 is a 5 and R4C2 is a 1.

Dennis_E
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  • What number would be R6C1 ? – IISkullsII Sep 28 '22 at 14:24
  • @IISkullsII It will come out to be a $4$, once you begin simplification from R4C2 (I'm assuming that you're asking what the final value would be, and not some way to get it before you get R4C2). – Sarvesh Ravichandran Iyer Sep 28 '22 at 16:16
  • More or less. I wasnt precise enough. Assuming, the OP has entered all possible Values into each Cell. I'm still not convinced, that if R5C1 is either 5 or 6, that R4C2 must be 1. If i assume R5C1 is 6, then R5C2 has to be 2, which results in R6C3 to be 4 and R6C1 has to be 1, which would render R4C2 into nothing else than 5. – IISkullsII Sep 29 '22 at 06:11
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    @IISkullsII That's actually really good deduction, and you're absolutely right so far. However, you're actually on your way to showing that R5C1 is not a $6$, which you can probably post as an answer, it's a very good observation. Indeed, once you assume that R5C1 is a $6$, the you've correctly shown in your comment that R4C2 is a $5$, but then R7C2 is a $7$ and then R8C1 must be a $6$. That leads to two $6$s in the same column (R5C1 and R8C1), hence giving a contradiction. Therefore, you've actually shown that R5C1 must be a $5$. – Sarvesh Ravichandran Iyer Sep 29 '22 at 12:09