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#1: Wombat (wombatilim) on Mar 7, 2023 [HINT]

Initial line & color logic completes 38% with only black remaining.

C6: EL marks R1-6 white.
C35: EL marks R1-6 white.
C34: EL marks R1-6 white.
C33: EL marks R1-2 white.
C32: EL marks R1-2 white.

R24: If C10 is black, R24C20 must be black, but R23C20 must be white, and we need a 2 in C20. Therefore R24C10 and R24C20 are both white.

Stuck here for now, currently at 40% complete.

C7: EL marks R4-5 white.

R23: If C20 is black, it makes R22 invalid. Minimal LL.
R24: If C21 is black, it makes R23 invalid.
R23: If C21 is black, it makes R22 invalid.
R22: If C20 is black, it makes R21 invalid. LL.

R23: If the 5 goes farther left than C25, R24 would be invalid. R23 C22-24 are white.

R23: If the 5 is in C25-C29, the 2 in C25 would need to go to R24, starting the 4 and completing both rows. R25-26 would then be made invalid. R23C25 must therefore be white.

R23: Any remaining position for the 5 will place at least 2 blacks into R22. R22 C21-23 can be marked white.

LL to finish.

Found to have a unique solution by wombatilim.

Found to be solvable by line and color logic alone by wombatilim.

Found to be solvable with moderate lookahead by wombatilim.

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