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#1: Jota (jota) on Aug 17, 2016

Couldn't solve w/o guessing!

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I'm sorry you are having so much trouble try starting the right side with R1 C20 skip C21 and then C22. That will give you the first row on the right side. Only the black and blue (except one red) are on the right.

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Found to be solvable with deep lookahead by infrapinklizzard.

ll/cl solves left side
el 6 r13=c30 w
ll
cl:
>c29r9 must be blue;
>c21r7 must be white;
>the 4 in c23 can't be in r7 so c23r3-4 is black;
>note the blue 4 in c18. Note that all the remaining rows start with a black clue. Therefore the blue 4 in c18 must have a black pixel next to it in c17 in each row. So the black 4 in c17 must be exactly next to it. That means neither can be in r10. So c17r1, r10-11 are white

Then deep lookahead:
look at the blue clues in c18-19. With all 1s in the crossing rows except r9, the top 1 in c19 must be above the 4 in c18. So the lowest it can go is r6. That limits the 3 in c19 to [r1-5] making c19r3 black.
That also means that the 4 in c18 must be in the range [r5-10] making c18r7-8 blue.
careful ll/cl finishes it

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