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#1: Kurt Kowalczyk (bahabro) on Jun 19, 2022 [HINT]

this does actually solve without deep look-ahead....only adv logic, but you have to do it like 6 times. el the 5's on either end.... then iel inside of that on both sides. followed by 2 instances of smile logic to finish the middle.... no guessing, I swear!

Hey, he's upside down!

Thanks Karen

thanks, Kurt

a lot of smile logic and edge logic

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Found to be solvable with moderate lookahead by gator.

EL on C1 makes R1-2,9C1 dots. LL.

EL on C20 makes R1-2,9C20 dots. LL.

EL on R10 makes R10C7,14 dots. LL.

Looking at R3, if you place the two 1 clues at R3C7 and R3C9, then this would cause R1 to be invalid on the right side. The same thing happens if placed at R3C12 and R3C14, but with the left side being invalid. So we know that a 1 clue must go in columns 7-9 and the other 1 clue goes in columns 12-14. Now look at R1 and how it affects R3. If the 1 clue is placed at R1C8, this would force R3C7 and R3C9 to be black. But we already know this causes a problem. So therefore, R1C9 is black. The same kind of logic makes R1C12 black. LL.

Smile logic on the bottom 2 rows. Then smile logic again to finish.

Nice puzzle!

I enjoyed this a lot too! I'm sort of restating what Gator, Gary and Kurt said with this solution, but I think it provides a different enough perspective using counting.

Line logic gets you to 63%. Then in R1-R3 C7-C14 you will need 6 black dots, one from each of the first 2's in C7-C9 and C12-C14.

Two of those will be from R3 so 4 must be from R1. You can't have all 3 in C7-9 or C12-14 so there must be 2 black dots in each side in row 1 The only way to get 1 is then to place the 1 in R7/9 and R12/14.

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