Puzzle Description Suppressed:Click below to view spoilers

#1: besmirched tea (Besmirched Tea) on Sep 20, 2020 [HINT]

I rated it moderate lookahead.
I used a technique that I don't know the name of for diagonals, where you see how many ones are between the twos to know if the two is going up or down.

Like if you have one square filled of a 2 and don't know whether it goes up or down. Say one way connects it to a 6, and the other to a 4. If from that column with a 2 the numbers up top are 2-1-1-2, it's probably the 4.
I use this method for solving circles where its a bunch of segments of 2 or 3, I look for 2-2-2 which is a 45 degree diagonal, or 2-1-2 which means a 3. Not sure how to explain it clearer?

I call that 2-way logic, because you're looking at the clues from two directions. :)

Not sure I'll be able to get to the end, but I'm going to start typing out what I find.

After the initial color/line logic, look at the 2 clue in R6. It has to go in either C28 or C30. In either case, R4C30 will end up as a dot.

If you place the 2 clue in R5 at either C5-6 or C31-32, it will make R4 invalid. So R5C5 and R5C32 are dots. LL.

Imagine the 2 clue in C30 at R7-8. This would force both R6C28 and R8C28 to be black making C28 invalid. So R8C30 is a dot. LL.

Look at the 2 clue in C31. It has to either be in R4-5 or R7-8. This causes either R7C28 or R8C28 to be black respectively. So we can mark R4-5C28 as dots.

If the 5 clue in R23 extends to C14, this will make R24 invalid a couple of steps later. So R23C9 must be black. LL.

Look at the 3 clue in R24 at right side. Use edge logic to make R24C24-27 dots.

Now look at the 4 clue in R23 at the right side. Use edge logic to make R23C25-27 dots.

And the same logic in R22 to make R22C25-27 dots.

Also the right 3 clue in R22 cannot be at R22C24 as it would either make R21 or R23 invalid. So that is a dot. LL.

Back to R21, and use edge logic to make R21C25-27 dots. You can also make R20C27 a dot with edge logic.

I'm going to stop there for now and come back to this later.

Nice image, but I think the square wheels will make the ride a bit bumpy.

Nice puzzle.

Picking up after Gator's hint:

R23: EL can mark C14 white.
C13: If the top 2 is in R4-5, it makes C14 invalid, so these squares are white.

C28: If the 1 is in R7, it makes R8C29 the 1 in that row, and R7C27 must be black as well. If the 1 is in R8, then R8C27 is also black. Either way it accounts for the top 1 in C27, so the rest of the column can be marked white (R4-5, R12, and R16-19 C27).

C26: If the top 2 is in R4-5, it makes C25 invalid, so these squares are white.

C26: If we put a black in R12, it puts blacks in R9, R12, and R15 in C25. This is invalid, so R12C26 must be white.

C26: EL marks R19-20 white.
R20: EL marks C25 white.
R21: If C24 is black, R20 is invalid, so this square is white.
C26: EL now also marks R18 white.

Deep lookahead C31: If the top 2 is in R4-5, it marks R4-5 C14-25 white, and R7 C26-28 black. This triggers the top 2 in C26, and R8 C24-26 are black. This makes the black in R10C25 part of the 2 in C25. On the other hand, if the top 2 in C31 is in R7-8, it makes R8 C27-28 black, which makes the black in R10C26 the top 2 in that column. This triggers either R9 or R11 in C26, either of which still makes the black in R10C25 part of the 2 in C25. Knowing that this 2 is accounted for, we can now use EL in C26 to mark R16-17 white. LL.

C26: If the bottom black 2 extends to R13, it makes R14 invalid, so it goes to R15C26 instead. LL.

C24: Whether the black 2 goes to R14 or R16, it accounts for the remaining 1 in C23. R7, R9, R13, and R17-24 in C23 are white.

R7: EL places blacks in C9-11. LL.

R7: No matter how the 7 is finished, R8 is black in C7-8 and in C12. LL. (Make sure you mark R10C6 white, it's an easy LL spot to miss.)

C6: If the 2 is placed in R12-13 or in R17-18, it makes C5 invalid, so these squares are white.

LL to finish.

Picking up after Gator's hint:

R23: EL can mark C14 white.
C13: If the top 2 is in R4-5, it makes C14 invalid, so these squares are white.

C28: If the 1 is in R7, it makes R8C29 the 1 in that row, and R7C27 must be black as well. If the 1 is in R8, then R8C27 is also black. Either way it accounts for the top 1 in C27, so the rest of the column can be marked white (R4-5, R12, and R16-19 C27).

C26: If the top 2 is in R4-5, it makes C25 invalid, so these squares are white.

C26: If we put a black in R12, it puts blacks in R9, R12, and R15 in C25. This is invalid, so R12C26 must be white.

C26: EL marks R19-20 white.
R20: EL marks C25 white.
R21: If C24 is black, R20 is invalid, so this square is white.
C26: EL now also marks R18 white.

Deep lookahead C31: If the top 2 is in R4-5, it marks R4-5 C14-25 white, and R7 C26-28 black. This triggers the top 2 in C26, and R8 C24-26 are black. This makes the black in R10C25 part of the 2 in C25. On the other hand, if the top 2 in C31 is in R7-8, it makes R8 C27-28 black, which makes the black in R10C26 the top 2 in that column. This triggers either R9 or R11 in C26, either of which still makes the black in R10C25 part of the 2 in C25. Knowing that this 2 is accounted for, we can now use EL in C26 to mark R16-17 white. LL.

C26: If the bottom black 2 extends to R13, it makes R14 invalid, so it goes to R15C26 instead. LL.

C24: Whether the black 2 goes to R14 or R16, it accounts for the remaining 1 in C23. R7, R9, R13, and R17-24 in C23 are white.

R7: EL places blacks in C9-11. LL.

R7: No matter how the 7 is finished, R8 is black in C7-8 and in C12. LL. (Make sure you mark R10C6 white, it's an easy LL spot to miss.)

C6: If the 2 is placed in R12-13 or in R17-18, it makes C5 invalid, so these squares are white.

LL to finish.

Found to be solvable with deep lookahead by wombatilim.

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