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I found an X-Wing pattern using 1's on r29c48 and eliminated the 1's on r1c8, r3c8, r7c8, and r3c4. Lots of elimination but that was not helpful for me to crack it.

I think I am very close to solving the puzzle but there is one step which stops me at the moment. Would be helpful if you could solve this puzzle using sudoku logic.

  • 1
    $\begingroup$ Is there a uniqueness argument to be made here in C3: if C3R3 is a 2, either combination of the 89 pair is possible if the rest of the grid checks out? $\endgroup$
    – Braegh
    Nov 11, 2021 at 10:10
  • $\begingroup$ No I don't think so. I check with that we can't add 2 at r3c3. $\endgroup$ Nov 11, 2021 at 10:41

2 Answers 2


There is a connection between R3C7 (top right box, options $1,8$) and R9C8 (bottom right box, options $1,2$):

they are either both $1$ or neither $1$, via their joint link with R7C7 (bottom right box, options $1,8$).

In the first case,

that they are both $1$, we get a contradiction:

enter image description here

From this fact, you can fully solve the bottom three rows immediately, and then continue on from there.


Here's a possible way forward

Notice that if r3c5 is a 3 then r3c9 is a 9 and r1c6 is a 1.
This means that r2c4 is a 5 and so r2c8 is a 1, r3c7 is an 8 and r3c4 is a 2.
But now row 3 already has a 2,8,9 and we haven't filled in column 3.
Hence, r3c5 is not a 3 and r3c9 must be.

After this

Notice that if r3c2 is 5 then r3c4 is 2 and r3c5 is 6.
This means that r3c8 is 8 and r3c7 is 1.
But this would leave us with putting 5 in r2c4 (since there is nowhere else to put the 5 in this box) and in r2c8.
Hence r3c2 is 1.


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