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#1: Paula (fuzziebear3) on Mar 21, 2008 [SPOILER]

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I used some edge logic. When you are done line solving, look at column 3. The 3 clue can either go in rows 18-20 or 22-24. In order for the 3 clue to go in rows 18-20, then R8C3, R12C3, and R14C3 would all have to be filled in to satisfy column 3. But the 2 clues in rows 12 and 14 will cause column 4 to be invalid since column 4 only needs one more cell, not two more. Therefore the 3 clue must go in rows 22-24.

I also looked at both columns 21 and 22. Notice how there are three 5 clues that need to be filled in between those two columns. If you look at the rows they could go in (rows 6-10, 16-20, and 26-30), you can see that no more than one 5 clue can go in each spot (also the column 21 5 clue cannot go in rows 6-10). For example, if you tried to place the column 21 5 clue in rows 26-30 AND you tried to place the column 22 5 clue in rows 26-30, then this would make rows 27 and 29 invalid. The same thing would happen in rows 16-20. Looking at this all together, we can see that the top 5 clue in column 22 has to go in rows 6-10.

There is also a look two move ahead contradiction when you look at the row 37. You have a 1 clue left to place, and it can either go in R37C26 or R37C30. If you try to fill in R37C30, then R36C30, R38C30, and R37C26 will all be dots. Next you would have to fill in R36C26 AND R38C26, but this makes column 26 invalid. So R37C30 must be a dot and R37C26 will be filled in.

Now use similar logic for row 40, and you will find R40C30 has to be a dot.

And with that I'm done trying to find a logical way to solve this one. Ugh.

I had to use patterns to solve it. It's like the PBN version of an ugly Christmas sweater. :)

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