Puzzle Description:

(This puzzle was recovered from the 2016 database crash. Puzzle creators, please edit this puzzle and put the real description here!)

#1: Jota (jota) on Jun 11, 2019

Thanks!

Found to have a unique solution by wombatilim.

Found to be solvable with deep lookahead by wombatilim.

Initial line & color logic completes 42% with only black remaining.

C1: EL marks R4 and R15-20 white.
C16: EL marks R10-11 white.
R20: EL marks C2-4 white.

Deep lookahead C1: If the 5 is placed in R5-R9, the 7 in R5 will get placed in C1-7 and the 5 in R4 gets placed in C3-7. This will leave a black 2 in R6 that will need to be placed somewhere in C8-11 and a black 1 in R7. There is nowhere to put that 2 in R6 that does not make R7 invalid. Therefore, R5C1 must be white, and EL on C1 now marks R6-8 white as well. LL.

R4: EL v.R6 marks C2 white. LL.
R5: EL marks C2 white. LL.

R10: If C12 or C13 is black, it pushes the 8 in C15 down into R19, which makes whichever of those columns the black was placed invalid. R10 C12-13 must both be white. The same logic can be repeated for R11. LL.

R4: If the 5 is placed in C3, it doesn't leave room for both 2s in R6. LL.

R19: If the 7 extends to C16, it will make C14 invalid in 2 steps, so R19C16 is white. LL.

C16: If the 4 extends to R12, C14 will be invalid in 2 steps, so R12C16 is white. LL.

Deep lookahead R13-14:
If the 2nd 2 in R14 goes to C14, the 2nd 2 in R13 goes with it. The 4 in C16 goes to R18, bringing the 8 in C15 with it. This blocks R10 C14-15, which in turn blocks R11 C14-15, pushing the 8 down to at least R19. The 7 in R19 is placed in C9-15.
If the 2nd 2 in R13 goes into C16, the 2nd 2 in R14 goes with it. The first 2 in C14 will place black in R11, and when joining with the 8 in C15, black will be in R10 C14-15 as well. In 3 more moves the 7 in R19 is placed in C7-13.
The overlap between these two options is R19 C9-13: these can be marked black. LL.

C2: EL v.C4 marks R16-18 white. LL.

R18: EL marks C3-4 white.

C3: If the 3 is placed in R9-11, it places the 3 in C4 into R10-12, completing both columns. the 3 in R17 would place black into C6-7, which makes C15 invalid. Therefore R9-11 C3 are white.

LL to finish.

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