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Oct 22, 2021 at 15:53 vote accept Nusi
Oct 22, 2021 at 15:53 comment added Nusi It should be good right now. Great explanation! :D
Oct 22, 2021 at 11:30 comment added Stiv Thanks for flagging that up @Retudin - yes, that was an error in my explanation; I did say it was a difficult puzzle! :) I've replaced the problematic section with a better proof by contradiction that should definitely work.
Oct 22, 2021 at 11:29 history edited Stiv CC BY-SA 4.0
Contradictory contradiction contraindicated (i.e. all fixed now, I hope...)
Oct 22, 2021 at 10:07 comment added Retudin Step 2: "those that will be present in the two 2x2 squares in rows 3-4". Maybe I am missing something, but this seems incorrect reasoning: The squares can use column 1+6 of the same row.
Oct 22, 2021 at 8:38 history edited Stiv CC BY-SA 4.0
Minor typo corrected and rewording
Oct 22, 2021 at 8:03 comment added Stiv @Nusi I think that should now all be fixed! (Unless your eagle eye spots anything else I have foolishly done!) Thanks for pointing that out :)
Oct 22, 2021 at 8:01 history edited Stiv CC BY-SA 4.0
Fixed logic
Oct 22, 2021 at 7:25 comment added Nusi I think I'm missing something. In the third picture of step 2, why r3c1 cannot be 4? Can you please elaborate it? Thanks! :D
Oct 21, 2021 at 23:04 comment added Avi Very tricky puzzle indeed - I wasn't even able to make an initial breakthrough
Oct 21, 2021 at 22:19 comment added Stevo You quite literally beat me to it; was just starting writing up my answer.
Oct 21, 2021 at 22:13 history answered Stiv CC BY-SA 4.0