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#1: Joe (infrapinklizzard) on Apr 22, 2013 [SPOILER]

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I believe this is moderate look-ahead (barely), but I'm sure there'll be arguments. Perfect for this guy.

After initial line logic, the only obvious advanced logic is edge logic on the right 2 in r14, ruling out c18-19. A little line logic.

Now some unnamed logic:
In r15 there is only one 1 yet to be placed (we're just not sure which one). Now notice that in r16 there is also only one 1 left.
There are three places the 1 in r15 can be placed (c11, 15 & 19), and all of them have clues that will force the black down into r16.
Since any place we put the 1 in r15 will fix the place of the 1 in r16, the rest of r16 other than c11, 15 & 19 must be white, so dot r16c12, 14, 16 & 18.

There are a few things we can learn about the 3 in c15:
- It can be in one of three positions, r15-17, r16-18 or r17-19.
- If it were in either r15-17 or 16-18, then r16 would be fulfilled and edge logic could be worked on the 5 in r17. So let's see if we can falsify the r17-19 position:

Since there are remaining only two 1s in c14 and one 1 in c16, if the 3 were in r17-19 there'd be only one way the crossing clues could fit, and they would use up all the vertical clues in c14&16.

That would mean that r12's 1 would have to be in c12 and that r14's 2 would have to be in c12 as well. This makes two blocks of one in c12 -- contrary to the clues.

Therefore, the 3 in c15 cannot be in r19. Dot that and do some line logic.

Now simple edge logic on the 5 rules out c11 and c19. Line logic to finish.

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Yes, great image.

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Why am I not surprised?

See? He's not a happy person. How sad.

good one joe :)

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Great puzzle Joe

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Great solve! I had to find the 5 on R17 to finish.

I solved it a little differently.

Continuing from the "three-way" logic, look at columns 12-13 and 17-18. In either case both 2 clues cannot go into row 20, so either R18C12 or R18C13 must be black and either R18C17 or R18C18 must be black. This will let us dot R18C15. Line logic, then the edge logic with the 5 clue.

Great puzzle!

Found to be solvable with moderate lookahead by gator.

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