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Related to this Q&A. I am looking for a pattern that could be applied in order to solve this Sudoku.

The current patterns that I apply:

  • Ensure that every digit appears once in every block
  • Ensure that there is a unique sequence from 1 to 9.

Now I am blocked as the block top right could contain a 8 and a 5, but I cannot rationalize where the 8 and where the 5 should be placed. What pattern am I missing?

As encouraged by @legorhin I tried to come up with possible solutions for the Sudoku cells.

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  • $\begingroup$ solve other parts of the puzzle then come back to it later $\endgroup$ – Legorhin Oct 14 at 18:01
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    $\begingroup$ Why have you got a 7 in r7c2? What about r9c3? $\endgroup$ – JMP Oct 14 at 18:27
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The first thing I saw was:

In the middle column, the 6 has to be on the top row, because the other two rows already contain a 6.

The second thing I saw:

The cell at R7C2 can only be a 3 because all other numbers are in a shared row/column/sector.
This makes R8C6 a 3 also, which makes R2C5 a 3. I think after that it gets easier.

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Look at the bottom middle block. The 6's in row 6 column 4 and row 7 column 8 mean that the $6$ in this block is on the right hand side. Using this with the 6 at r2c9 and the 6 at r6c4 mean that r1c5=6.

Also:

The 6's at r4c3 and r7c8 collude with the 9's at r5c3 and r7c7 to means that r8c1 and r1c9 are both 6 and 9. This means the remainder of the lower left block is {1,3,8}, and therefore r7c2=3, r7c1=1 and r8c3=8.

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  • $\begingroup$ You and I are seeing the same thing, at the same time, lol. $\endgroup$ – JS1 Oct 14 at 18:15
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My late contribution:

The $6$ supplied by other answers could have been solved without filling in any possibles. There are three numbers $3,6,7$ missing from the centre column, and the $6$ can only go in the top cell.

Similarly there are three numbers $3,7,9$ missing from the second column, and the $3$ can only go in the cell at the top of the bottom block (yet you marked the possibilities $3,7$).

So this is one of my techniques: for any row, column or block already well populated, see where the missing ones can fit.

Allied to this, is that when you have a duplicate pair possible in any row, column or block, as you have with the $3,7$ in the centre column, you can remove all other $3$s and $7$s in that element - here leaving that lone $6$.

The same goes in the rarer case when you spot a set of three numbers appearing as the only possibles in three cells.

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