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#1: Joe (infrapinklizzard) on Jun 2, 2019 [HINT]

This solve is wrong -- see comment 10 for a correct one (I hope :P )

After Line Logic does only the bottom,

Edge Logic 6 r1 = c1-3, c14-17 white
LL
EL 7 r18 = c1-12 w
LL
EL 3 r17 = c1-10 w
LL
EL 3 c15 = r2-4 w
LL
EL 3 r15 = c13 w
LL
EL 2 r14 = c12-13 w
LL
Internal EL 3 c11 = r12-13 w
LL
Extended EL 2 c14 = no matter where it goes, it triggers the 2 in c13, so c13r1 can never be black.
LL

Deep Lookahead: If the 3 in r2 were in c12-14, it would cause a conflict in r2c11. So r2c14 must be white
LL

Extended EL 2 c14 = no matter where it goes, it triggers the 2 in c13, so c13r2 can never be black.
LL to finish

Found to be solvable with deep lookahead by infrapinklizzard.

Whoever voted this as trivial is trolling.

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Cool solve! (HUGS)

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I can't see how you made R17C10 a dot with edge logic.

Fun puzzle and great image with minimal pixels. Excellent

You're right, Gator. And it definitely changes the solve.

After Line Logic does only the bottom,

Edge Logic 6 r1 = c1-3, c14-17 white
LL
EL 7 r18 = c1-12 w
LL
EL 3 r17 = c1-9 w
LL
EL 2 c17 = r2-5 w

Distant EL 3 r2 (vs r4) = c16 w

Extended EL 3 r2: if it were in c13-15, it would force the 2 in c13 up into r1, fixing the 6 which would then conflict with r2. So r2c15 must be white.

EEL 3 r2 (again): basically the same. the 2 in either c12 or 13 will be forced up creating the same conflict as before, so r2 c14 must be white
LL
Two-way logic on the 2 in r5: It can go in either c13-14 or c15-16. Either way r6c14 must be black.
LL
(Deep Lookahead) If the 2 in r5 were in c15-16,
> The 2 in c14 would be forced down into r7,
> forcing the 1 in r7 into c17,
> forcing the 2 in c17 into r8,
> forcing the 3 in r8 into c15, creating a conflict.
= So r5c16 must be white
LL (Don't miss that r10c16 must be white)
Two-way logic on the 2 in c17: It can go in either r10-11 or r12-13. Either way c16r13-14 must be white.
LL to finish

Joe - What is the LL that you did after making R2C14 white? I could only make R2C11 black from what I can see.

Sigh. I have no idea. Based on my next hint I must have assumed that r2's 3 must be in c9-11, but there is no basis for that.

Try # three:

After Line Logic does only the bottom,

Edge Logic 6 r1 = c1-3, c14-17 white
LL
EL 7 r18 = c1-12 w
LL
EL 3 r17 = c1-9 w
LL
EL 2 c17 = r2-5 w

Distant EL 3 r2 (vs r4) = c16 w

Extended EL 3 r2: if it were in c13-15, it would force the 2 in c13 up into r1, fixing the 6 which would then conflict with r2. So r2c15 must be white.

EEL 3 r2 (again): basically the same. the 2 in either c12 or 13 will be forced up creating the same conflict as before, so r2 c14 must be white
LL

NEW STUFF:
No matter where the 2 in c17 goes, it will trigger the 1 in c16. Therefore c16r3-5, r7, r10, and r13-14 must be white
DEL 2 r3 (vs r5) = c15 white
(Deep)The 3 in c15 cannot go in r4-6. It would force c16r6 to be black, leaving nowhere for the 2 in c17 to fit. So c15r4 must be white

And there I'm stuck.

Feel free to continue on if possible, or to change the determination to guessing if you can't find a way

Let's see. No matter where the 6 clue is placed in R1 (it can't go in C5-10 without making R2 invalid), R2C1-4 and R2C6-7 will be dots.

I looked for a while but couldn't find anything past that point. I'll change to guessing for now.

Found to require some guessing by gator.

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