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#1: Kadou (Kadou) on Feb 25, 2012 [SPOILER]

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Can't 'ear you! CONGRATULATIONS, Kadou!! You WIN ABSOLUTELY NOTHING!!!

logic only...line, edge and two-way.

after line logic:
look at the 7s in c3 and c7 (you have 3 of each one already filled in)
they cant both extend DOWN into r11, r10, or r9 and they both cant extend UP into r4, r3, r2 or r1 so this gives you quite a bit of information
1) they already share r5-7 but they can at most share only more row, (r8) so exactly one of those 7 clues must be in c2-7 (with the 7th pixel in either c1 or c8) and the other 7 clue must be in c5-11
2) the 1 clue in r9 must go into c3 or c7 so you can dot the rest of that row (this gives you a black in r2c13)
3) whichever of those 7 clues that extends up into r4 must be the 1 clue in that row, so you can dot c1-2, 4-6, 8-9, 16-17, and 20
4) the 7 clue that extends into r3 is part of the first 3 clue, so you can dot c11-12, and 14-15
more line logic
5) the 7 clue that extends into r2 is the 1 clue, so you can dot c1-2, 4-6, 8-9, 16-17, and 19-20
more line logic
the two 3 clues in r19-20 cant both go into c4 so they both go into c1
look at the first 3 clues in r6-7: they cant both go into c4 so at least one of them goes into c2 so r7c2 is black
more line logic
edge logic on the 3 clue in r11 to dot c3
line logic to finish
great solve
no guessing

there is probably an easier way to solve this

Thanks David and Bugaboo. I did use lots of two-way when proving it. The ear was originally white, but I couldn't solve. I didn't write down the steps, but bugaboo's solution is familiar.

Ripped off yet again,i was all over this one,but of course got beat out yet again,enjoy your award Kadou.:)

Found to be solvable with moderate lookahead by jan.

I really loved how the edge logic needed jumped around. Lots of similar stuff but nothing identical. Placing the 2 in columm 1 or the 3's scattered across the board did the trick, eventually.

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