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#1: Kristen Vognild (Kristen) on Feb 11, 2014 [SPOILER]

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Follow this way, Bálint! :)

Yes, I haven't seen yet that it's a simple puzzle if you can see... Thanks, great solve!

bear?

I don't think so, but I don't know.

Either I'm doing it wrong or this is surprisingly tough.

You can do alternating smile logic in c1-2, but since the number of clues are the same in each column, there are two possible arrangements. So no good starting there.

Here's what I did:

Sum rows 1-3. That's six.
Sum the minimum clues that can fit in there. (The top clues in c1-4 must be within the top three rows.) So that's five.

That means that a maximum of one pixel of the 2 in r5 can be within the first three rows and c5r1-2 must be white. (Then line logic makes r4 black.)


Summing the first two columns gives five. This means that all the left 1s in rows 1-5 must be within the first two columns. Since the rightmost pixel in r4 is set, r4c3 must be white.

Now look at c2. The top 1 can go in only two places. If it were to go in c2r3, it that would force the lower 1 into r5.
This would force the pixels next to them in c3 to be white, not leaving enough space for c3's clues. Therefore the top 1 in c2 must not be in r3 and is r1.

And LL to finish.

Found to be solvable with moderate lookahead by infrapinklizzard.

Joe...these puzzles have been tough on you, I wish you luck, I feel as I have to guess somewhat ...who knows

it's kind of funny, joe, we we took basically the same steps but were looking at them differently. here's what i typed :)

after LL:

if C3R5 is black, C3R4 is a dot. if C5R5 is black, C5R4 is black...either way, C5R4 is black.

there are only two rows for the top 1 in C2 to go and the second pixel in the 2 in C1 will go in whichever row that one doesn't go in. both clues have a 1 under and both bottom two rows start with 1, so either C1R4 or C2R4 have to be black...dotting C3R4.

C3R1 and C3R3 can't both be black or it violates C2, so C3R5 is black.

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