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#1: besmirched tea (Besmirched Tea) on Sep 20, 2021 [SPOILER]

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LOL Bill.

Zelgerath, the solve was great, and the picture is intriguing. But we have no idea what it is. Would you care to explain what you had in mind?

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I see a person running holding a torch? This was difficult because I had to guess to get it started. Otherwise, quite fun.

Nice drawing - enigmatic title.

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Yonah, those are certainly plausible :)

I've done extensive EL with some pretty deep look ahead but I'm stuck atm. Trying to make something work with columns 4 5 6 but the lookahead becomes deep enough it feels like guessing/trial and error.

I'm at what feels like the same point, Emimonster. After all line logic and moderate lookahead edge logic, I used deep lookahead edge logic on the bottom row. In my puzzle, the rightmost of the two sixes in R20 could go anywhere between these two zones: C10-15 to C19-25. I placed it in all 10 spots, and no matter where I placed it, the leftmost 6 had to fill in R20C4 through R20C8. Then regular edge logic makes R20C19 through R20C21 black.

After that, even deep lookahead fails for me. I can't find anything more to latch onto except guessing and seeing how far it gets.

Initial LL only completes 3%.

C25: EL marks R1-2 and R13-17 white.
R20: EL marks C1-2 and C16 white. LL.

R20: We have 3 rough positions for the 2 6s: C3-9, C10-15, and C17-25. If we try to use positions 2 and 3, it makes R19 invalid. Therefore position 1 must be used, and we can put blacks in C4-C8. LL.

C1: EL marks R11-13 white. LL.

R20: Extending the 2nd 6 into C17 makes R18 invalid, so R20C23 is black. Whichever way that 6 does go, R19C24 is white.

R1: EL marks C7 white.

R18: The black in R17C9 can be any of the first 3 clues in R17 but it can't be the 3 or the 5. If the black in R18C9 is part of a 3, C10-11 will both need to go into R17 because the white at R18C12 would block those columns in R19. Since this move is invalid, R18C10 must be white. LL.

Deep lookahead C6: The lower 2 must be in either R12-13 or in R17-18. If it is in R12-13, then R13 C9-12 are black and R14 C9-11 are black. This nudges the 7 in R12 into C2, which makes R13C4 white. This would leave no room for the 2 in C4 since R14 has been blocked, so the lower 2 in C6 must be in R17-18 instead.

LL to finish.

Found to be solvable with moderate lookahead by wombatilim.

Found to be solvable with deep lookahead by wombatilim.

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