Puzzle Description:

(This puzzle was recovered from the 2016 database crash. Puzzle creators, please edit this puzzle and put the real description here!)

#1: Shawn Sheppard (morn) on Sep 9, 2021 [SPOILER]

Carrefour

Found to have a unique solution by valerie.

Found to be solvable with deep lookahead by valerie.

I tried solving it with the helper, but it got stuck. So I figured the problem was just color logic. That got me farther, but then I got stuck again. Finally I tried the awful "fill in a square and run the helper until it gets an error, then revert and mark the square with a dot" method. That worked, but I never feel good about using that method to solve a puzzle. Anyway, the puzzle *does* have a unique solution, but as far as I could tell it needs a lot of lookahead. Or assuming symmetry to solve it -- I didn't try that.

Anyway, if someone has a logical way to solve it, I would like to learn it!

I think this is how I solved it, lots of deep lookahead:
LL/CL on blue
if C12 R2 (and R5) is blue, that conflicts with the 3 blue in C18, so C12 R2 is white
LL
if C12 R4 (and R3) is white, that conflicts with the 1 blue in C19, so C12 R4 is blue
LL
if the 2 blue in R3 is in C21-22, that conflicts with the 2 blue in R1, so R3 C21 is white
LL
EL on 2 blue in C22 (looking at C20) makes C22 R1 white
LL
Blue is finished!

In C3 R9-12, 3 consecutive whites would conflict with the 2s in C8, so C3 R1-4,17-20 are white
EL on the 1s in C11 and the 2s in C10: C10 R9-12 are white
LL
EL on the 4 in C2 (looking at the 2 in C1): C2 R1-2,19-20 are white
LL
If R7 C5-6 are both white, that makes R17 C5 and R19 C6 red, making R17 C7 white and R19 C7 red, conflicting with the 4s in C7: so R7 C1,10,11 are white
If R19 C5-6 are both white, that makes R4 C5 and R2 C6 red, making R4 C7 white and R2 C7 red, conflicting with the 4s in C7: so R19 C1,10,11 are white
LL
C8 cannot have 3 consecutive red, which would happen if C2 R10-11 are white and C3 R9 or R12 are white. So R6 C2-3 cannot both be red, nor can R20 C2-3. So C2 R6,20 are white.
LL
If R5 C6-7 are both white, that makes C6 R3-4 white and C7 R1-4 red, conflicting with the 2 in C8. So R5 C4,9-11 are white.
If R16 C6-7 are both white, that makes C6 R17-18 white and C7 R17-20 red, conflicting with the 2 in C8. So R16 C4,9-11 are white.
LL
EL on the 1s in C11: the 2s in C9 must be somewhere in R1-4 and R17-20, so C9 R6,7,14,15 are white.
LL
EL on 4s in C7: C7 R1,20 are white.
LL
If C10 R3-4 is red, then C10 R1-2 is white, then C9 R1-2 is red, then C9 R3-4 is white, conflicting with the 1 in C11. So C10 R4 is white.
LL
If C9 R1-2 is white that conflicts with the 1 in C11, so C9 R2 is red.
LL
Done!

Oh wow! Thanks Philip!

This is easily solvable using symmetry. I know a lot of people think that's guessing, but if you know it's a unique solution symmetry is perfectly acceptable.

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