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#1: bugaboo (bugaboo) on Jan 5, 2011 [HINT]

cute dogs
line/color logic for over half
then edge logic on the blue 3 in column 7 (placing 2 of them)
then diagonal logic (guessing) to finish

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This would fall under "image logic", i.e. guessing. :)

I'm still working on a way to find a way to finish the blue with guessing.

Maybe I guessed and didn't realize it but the blue3 in column 7 can only go one place, and then it's straightforward after that. Unless I missed something?

Not sure how much of all of this is necessary. I'll let someone else analyze that. OK, here goes:

Once the color/line solving is done, you will be left with only blue to place. I used edge logic on the 3 clue in column 7 to make R8C7-R11C7 dots. More line logic.

Next R9C17 and R12C17 can both be marked as dots. If either one of these are filled in, then it would be impossible to complete the 2 clue in column 16.

Next I looked at the 2 clue in row 13 to see how this would affect row 12. Any place that you place the 2 clue, R12C9-R12C10 will always be dots.

Next I looked at the 2 clue in column 16 to see how column 15 would be affected. If the 2 clue is in R8C16-R9C16, R9C16-R10C16, R11C16-R12C16, or R12C16-R13C16, then either R9C15 or R12C15 will be blue. The only place left to look at is R10C16-R11C16. With that filled in, either R10C15 or R11C15 will be blue. So far every possible case of the 2 clue in column 16, R4C15, R8C15, and R13C15 will always be dots as a result. Now look again at the 2 clue in column 16, but this time focus on how column 17 is affected. No matter where you place the 2 clue, R4C17 will always be a dot.

Next I looked at the 2 clue in row 8. If I place the 2 clue at R8C13-R8C14, this will make row 12 invalid. So R8C14 is a dot.

Next I found a logical constraint with the 3 clues in rows 9 and 12. As there are only 1 and 2 clues left in the columns where the two 3 clues can go AND there is only one clue in each column, then each 3 clue cannot occupy the same column. For example, if part of the 3 clue in row 9 was at R9C14 and part of the 3 clue in row 12 was as R12C14, then column 14 would be invalid as it would take a 4 clue in column 14 to make that possible. But there is only a 2 clue.

So look at the possible places that the 3 clue in row 12 can go, and look how this affects the 3 clue in row 9. What you find in common is that either R9C11 or R12C11 are blue in every case. Since we know one or the other has to be blue, then R4C11, R8C11, R10C11, and R11C11 have to be dots.

I believe this is no more than looking two moves ahead, but it requires A LOT of "looking at every possibility" kind of logic.

So is this one to be considered logically solvable? Opinions?

I saw the question mark, so I was very careful to only use logic and not rely on visual logic (really just guessing, because it doesn't rule out other possibilities).

I was able to finish everything but the blue on the right. But looking at the clues told me it had to be a diagonal. Which way? It was very easy to follow the diagonal in my head to figure out which way it would have to go in order to fit without contracitions. Yes, it is more than two moves ahead, but its so easy to see mentally without having to mark any squares, that I think it should count as logical.

If this puzzle is so difficult that most people have to resort to guessing, and it needs lots of logic to solve, then why are people rating it as easy (2)?

Oh - I just checked the information on this puzzle, and there is only one person who has rated it so far. Out of 20 who say they completed it, only one person rated it? Weird.

Teresa - the world may never know... :)

gator i went back and followed your logic and it counts for me as being logical
when solving this the first time i had initially missed what you described in paragraph 8 or else i would have solved this myself without guessing
fantastic find there on the logic

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I really enjoyed and think is super cute!

Very cute

i agree with jota - very fun solve.

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Had to guess, but it wasn't complicated guessing or maybe I was lucky.

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Color logic clears out all but the blues in C7-18 R4-13. If there is a blue in C6R8 or below then there is a 3 in C7. Helper gets you to 86%.

The 1 in row 4 is a nuisance. It can go a lot of places potentially. That makes this puzzle harder, but it's interesting.

The 3 in R12 cannot go in C9 or the 2 in R13 is trapped. It can't go in C16 due to 2-way logic.

Note wherever the 2 goes in C13 the 3 must be attached to it and can't go to C10. (If the 2 is in C12-13 then C12R13 is blue.)

If the 3 is at C11-13 then the 2 in C14 will cause a contradiction in C15.

That places a blue dot at C14R12.

Now though C16R7 as a blue dot gives a contradiction. Same for C16R8. We can place R9C10.

Now if R8C12/13 is blue, one-way logic shows the 3 in R132 can't make a 2 in R11 or R13.

Finally a blue in R12C12 means there will be a blue in R11 C12/13/14 or R13 C12/13/14. (C12-14 has 2's but they can't all face the same way or R11/13 would have a 3.) Then C16's 2 can't fit.

Line logic.

I *think* this is moderate lookahead but it's really pushing it so I won't mark it yet.

To me, it looks like straightforward 2-way logic. The blue 3 in column 7 corresponds to the blue 1 1 2 in Rows 4-6, and proceeds logically from there, down to R17 C13.

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