Puzzle Description:

#1: Raymond Fuller (rfuller4) on Aug 11, 2020

Such a flirt!

Indeed!

Any hints after line logic runs out? Even with moderate lookahead I've only filled in a few more squares.

Cute. Did have to guess a couple of times..

After line logic runs out, the key is the 6 on row 16: using the vertical 1s and that you have to fit a four or a five next to it, you can get a black spot on line 16. This forces a spot from the four in row 15, which gives you much of the five in row 17. From there it is fairly straightforward.

R28: Wherever you put the 5, the 7 in R27 spans C18-21.
R28: If the 5 is farther right than C24, it creates a contradiction in R26, so R28 C19-20 are black. LL.

R15: EL eliminates C18-23.
R16: EL eliminates C14.

R16: If the 6 is placed in C22-27, the 2s in C24-27 *could* go into R17, but doing so would create a contradiction in R15. If it is in C23-28, the 2s in those columns would need to go into R15. Any other valid placement for the 6 starts the 5 in R17, so the farthest right the 5 can be is C19-23. This means we can mark R17 C24-28 white.

LL to finish.

Found to be solvable with moderate lookahead by wombatilim.

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