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#1: annalivia (annalivia) on Sep 11, 2013

It fell to earth, I knew not where

nice!

I shot the sheriff (but I did not shoot no deputy).

It better not fall on me! Fun.

thus apple on head he made his dare,
though worm inside showed much despair.

I thought of the same thing as Kristen. Now that song will be in my head the rest of the day!

Good one, Brian!

Couldn't solve with logic, didn't mark it, I'll wait for the experts.

Comment Suppressed:Click below to view spoilers

after all line and color logic,
edge logic on the red 4 in r9 makes c6-7 white. Then color logic finishes the red.


Deep lookahead can finish the black by trying the three positions of the black 1 in r10 for zigzags.*



However it can be finished with moderate lookahead:

Some summing:
>in columns 7-11 there is a total of nine black pixels to be placed. (Note that they *all* must be within the rows of 16-20)
>in rows 16-20 there are five black 2 clues - totaling ten black pixels to be placed.
>that means that one of the 2s in r16-20 must extend into c6. Thus as the 2 in that column couldn't reach that far, c6r12-13 must be white and c6c14 /cannot be black/(though it still could be blue).

However, color logic says that it can't even be blue (due to the column clues), so c6r14 is also white.

Now sum again:
>c1-5 have five black 2s for a total of ten pixels to be placed
>r10-15 has a total of eleven black pixels to be placed.
>therefore c6r15 must be black as it is the only unknown space within r10-15 that is not within c1-5.

A little line logic and then more summing:
>c1-4 has four black 2s for a total of eight pixels to be placed
>r10-14 has a total of nine black pixels to be placed
>thus c5r14 is black as it's the only unknown pixel in r10-14 that is not in c1-4.

And then the puzzle finishes with line logic.



====
* (Remember that since you're starting by assumption you can only mark dots for when you run into a conflict in both zigzag directions - you cannot ever mark a black.) In this way, you can rule out r10c3 when it conflicts in r13 on the left zig and r14 on the right zag. Then you can rule out r10c2 when in conflicts in r12 zigging and in r15 when zagging. Thus r10c1 must be black. Then LL finishes the puzzle.

Found to be solvable with moderate lookahead by infrapinklizzard.

It's funny that this is actually easier to solve with deep lookahead than moderate lookahead.

lol Joe

Thanks!

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