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#1: Bill Eisenmann (Bullet) on Jan 17, 2023 [SPOILER]

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No guessing needed here, but there was a lot of lookahead. What cracked it for me was the 4 in R15.

From earlier LL placement and some moderate lookahead that ruled out R12/C8-9, I was able to place R15/C8 as black. Using that, and with some deeper lookahead, I was able to determine that C10 and C11 cannot BOTH be white, which meant the 4 had to cross C11.

Once I had that, I knew the R15/C8 square had to be the 1 or the 4. If the 1, R15/C7 is white. If the 4, then R15/C8-11 are the 4, and R15/C7 is white. Either way, I can rule it white, and the 3 on C7 is R11-13.


Proof on the lookahead for R15:

LL already ruled C15-20 white, so the 2-1-4 are spread across C1-C14. I also already had R16/C10-12 as part of the 5, and the black R15/C8 I stated above.

Case 1: R15/C10-11 are both white
The 2-1-4 all have to fit within C1-9. This violates clues in C1-5, because either R14 or R16 needs to be black on C2, and same row for C3, and either way R15 needs to be white on C3. This either makes R17/C3 black (i) or C13/C3 black (ii).
(i) The black 3 on R17 needs to be C10-12 in this scenario due to R15 white in those columns.
(ii) The 4 on R13 would be C3-6, violating the 2 in C4, which now needs to cover both R13 and R15.

Case 2: R15/C11 is white, C10 black
The 2-1-4 is across C1-10. The 3 on R17 has to be C10-12. Following LL, C4 and up are all filled, with the 2-1 on R15 having to fill C1-3. Not enough room, thus violates this clue.

Based on these cases, R15/C11, at the very least, has to be black. Since C15 is white, the 4 cannot go farther than C11-C14, making R15/C11 part of the 4, thus allowing all the logic at the top of this post.

very nice

I saw a kicking horse.

Nice image.

Seems like deep lookahead to me:
LL
EL on C17 (look at C15 and C16): C17R3,11-17 is white
LL
EL on 5 in R16 (look at R15): R16C15 is white
LL
EL on R17: R17C4-7 is white
EL on R12 (look at R13 and R14): R12C9 is white
EL on R13 (look at R14): R13C9 is white
EL on R12 (look at R13 and R14): R12C8 is white
LL
The 1 in R15 can be C4-6,8:
If it is C5 then R15C2 is also black. These make R13C2,5 white, contradicting R13.
If it is C4 then R13C4 is white, R15C3 is white, and R15C2 is black. Then C2R13,17 are white, so C3R13,17 are white, contradicting C3 (deep lookahead).
So the 1 in R15 is C6 or C8. Either way R15C7 is white.
LL to finish.

You had a different way to get to it from how I did it, but for both of us, it came down to confirming R15C7 is white to crack the rest of the puzzle. :)

I wasn't sure what the difference was between moderate or deep lookahead, so I didn't specify deep in my synopsis, but if your solution was deep lookahead, then mine definitely counts as deep as well.

thank you all for taking the time i really appreciate it

Found to be solvable with deep lookahead by wombatilim.

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