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#1: Andrew Schultz (blurglecruncheon) on Jan 23, 2020 [HINT]

EL on the bottom 10 and top 6 place all the black dots there. A lot of alternating 2/1's provide some quick contradictions. Blue logic is trivial and reds use smile logic.

But R7-R15 still need filling in.

I used EL on column 5. R10 must be black otherwise r8 is black, C6r* is white, c6R11-12 are black, c7R12 is black, and c7r11 is black.

Do the same for c36. We can fill in rows 14 and 15 now.

And now we can use edge logic on R12 to place the 6, much as we did for the top and bottom rows. The rest falls with line logic.

Nice solve!

Here's my solve:

Initial line & color logic solves 32% with black and red remaining.

Smile logic finishes the red. LL places 8 of the 10 blacks in R18 and EL can finish the 10 in that row. LL.

R1: EL places the 6 in C18-23. LL.

So far that's the same as Andrew's solve but I'm not tracking his EL on C5. R10C5 is the only square EL can mark in that column, and it has to be white or it creates a contradiction in C6. I'm guessing there's a typo. However, any remaining placement for the 3 puts black in R10-11 C6. After LL the same can be repeated for C35-36. LL to 82%.

R11: EL places the 6 in C18-23. (Back to being the same as Andrew's solve.)

LL to finish.

Found to be solvable with moderate lookahead by wombatilim.

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