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#1: Carol Brand (KarylAnn) on May 31, 2013 [SPOILER]

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Very nice. I solved it with logic only, no guessing, etc.

I was curious because I really thought I had solved it with line and color logic only, so solved it again after seeing Joe's comments. I did end up as he said with some greens without a straightforward solution. Apparently I placed some green squares in the right place with educated guessing.

After line and color logic, there are just 5 green pixels left to be placed in an L-shaped area consisting of three 2x2 squares.

There are a few ways this can be solved.

>With lookahead:

The two 1s in c9 & c10 cannot be next to each other because the crossing clues are also 1s. So one of them must be in r9.
Therefore r9c7-8 must be white. Then LL to finish.


>With contradictions:

Look at the 1|1 in c8. The top 1 must be within r6-7 and the bottom within r8-9.
If the bottom 1 is in c8r9, then the rest of that row must be white and there is then not enough room for both the 1s in c9 & c10.
Therefore it can't be in that position and c8r9 is white. Then LL to finish.

>or my favorite, with summing:

The clues in c9-10 mean that there are two green pixels in the 'top' square c9r8-c10r9.
The clues in r6-7 mean that there are two green pixels in the 'right' square c7r6-c8r7.
That leaves only one green pixel in the 'joint' square c7r8-c8r9.

Since both c8 and r8 need an "extra" pixel, the green pixel in the 'joint' square must be in c8r8. Then LL to finish.

Found to be solvable with moderate lookahead by infrapinklizzard.

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learned something new.... I like how diff towns try to steal it and a ransom of beer, etc must be paid :) reminds me of college frat shenanigans. we'd steal stuff from the other frats and ransom it back like that(and vise versa)

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