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#1: Adam Nielson (monkeyboy) on Nov 2, 2010 [SPOILER]

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Ron: not sure about this ? ... i didn't select that option, but it was kinda rough to get this one.

I could not find a "normal" advanced technique to use to finish this one, but I did find a logic constraint. Let me explain it and I'd like to hear feedback on whether you all think this is solvable or not.

When all of the blue, red, and green are done, you should have 8 black cells filled in. Now look at columns 7 and 8. They both have 3 clues in them. Now look at rows 2 and 5. They both only have 1 clues in them.

If you look at all the possible places that 3 clue in column 7 can go, then you will notice two things. If R2C7 is filled in, then R5C8 is filled in. If R5C7 is filled in, then R2C8 is filled in. This happens in every case.

So from this we know that either R2C7 or R2C8 will always be filled in. You can now apply the exact same logic on columns 14 and 15 to see that either R2C14 or R2C15 will always be filled in.

Knowing this enables us to dot most of row 2. It is also enough to let us finish the puzzle.

Thoughts on this?

i don't believe that your solution Gator is thinking ahead too much

I think that's a sort of internal edge/two-way logic, without the jumping-off point of the two-way being visible--or the edge. I remember using that technique for a puzzle at least once (I don't think I ultimately solved the puzzle, but it helped get me a little further along).

I agree that this one is definitely logically solvable.

i admit i got stuck and couldnt find a way forward logically but after reading what you wrote gator i agree that this is logically solvable
nice pick up

I saw that with the 3s, but my brain could not hold all the info long enough to solve it without using trial and error.

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Gator, an easier first step might be to look at row 5 vs columns 7&8.

Since there are only ones in r5, there are only three possibilities for those two pixels. -- The could be both white, or black & white, or white & black.

If we assume they are both white, then both 3s would go up and conflict in r2. So, one must be black and the other white. This would bring one of the 3s up into r2.

And the same in c14-15.

The rest per you.


I'm glad I don't wield the power of adjudication as I am also ambivalent. This is definitely on the edge of wpbnlogical. I feel longer chains can be considered logical if the next step is forced by the previous step (like zig-zag logic). Even though this is about 3 steps, my biggest problem with this is that the last step is not really forced by the previous logic. While this sounds simple in retrospective explanation, the previous steps are a bit much to keep in mind as you look for their ramifications.

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