Puzzle Description:

ho!

#1: bugaboo (bugaboo) on Dec 9, 2010 [HINT]

nice image and tough solve
when you get stuck look at row 4 where you should have 2 black pixels already filled in
if the 6 extends to the left in either column 1 or 2 then row 10 can never be filled in
some extended edge logic after that but i think it took looking at least 3 moves ahead

Placing the 6 did it for me.

fun solve and image, but I too had to do some extended edge logic.

I found this one really challenging. Agree on the extended edge logic. I'm interested to see whether Gator marks this as solvable, because I really had to look ahead more than a couple of steps at at least two points.

Cool puzzle. A little beyond my capability. I couldn't do the extended edge logic without guessing, as it was too many steps ahead to keep straight in my brain. But I did like it.

I haven't given up on this one yet, but I'm inclined to say that it requires guessing. I have not been able to find a way that does not require looking 3 or more moves ahead yet.

Definitely have to look several pixels in, but since it's a small puzzle, that was a good exercise.

Don't know what qualifies as guessing, but "extended" edge logic down the left side definitely leads pretty quickly to contradictions that "force" blank spaces and eventually results in some "forced" black spaces which lead to a solution.

tough ten/ten puzzle

I took another look at this, but everything I find is looking 3 or more moves ahead. If/when Jan implements to revised Solvability system, I'll come back and mark this one as "much lookahead required".

thanks for supporting what i initially found

here was my method...the last move being my "look-ahead" move, not sure if it is 2 moves or 3 moves ahead though:

initial LL = 4 blacks and 4 dots

EL 2 in C10 = 3 more dots

No matter where 2 in C10 goes = dots in C9R8 and C9R9

No matter where 2 in R1 goes = dots in C6,9,10R2

LL = 3 blacks and 8 dots

same move as bugaboo with 6 = 2 dots and then 2 blacks

If 3 in C2 is high, two of the 2's in R1-3 have to go into C1, puts enough dots in C3 that 3&1 clues would compact at bottom...OR if the 3 in C2 goes low, it will give you the same black in C3R10.

LL to end.

Yep, definitely 3 moves ahead.

How about this move: Looking at the right 3 clue in row 10, imagine if it was all the way to the right (columns 8-10). Looking one move ahead, we would have to dot R4C9, R6C8-R6C9, and R7C8-R7C9. Now looking 2 moves ahead, R4C4, R6C4, and R7C4 would have to be black which makes column 4 invalid. So R10C10 is a dot.

Thoughts?

That does make sense, but I noticed something while following your description that I think is even more straightforward:

Looking at the right 3 clue in row 10, imagine if it was all the way to the right (columns 8-10). Looking one move ahead, we would have to dot (among others) C4R6 and C8R6 which makes R6 invalid. So R10C10 is a dot.

Or R7C4 and R7C8 would be dots making R7 invalid. :)

OK, I think we have enough hints on this one to mark it as solvable.

Found to be logically solvable by gator.

easy enough ;)

Love it!

Nice image for a ten by ten.

Column 1 really opened it up for me. Go through the possibilities of the 2 being in R3 and R8, after dotting the middle of Columns 1 & 2 (follow the 6 in R4). After that it will take some massive lookahead at the two remaining possibilities of where the 2 in C1 can go - only one is possible.

This is a nice challenge for the brain. I had to keep telling myself it is a small puzzle, I should be able to solve it. When I was finished, it made me think of the "Push, push it real good" song.

https://www.youtube.com/watch?v=Eiz_0bOt__I

Neat little puzzle.

Here is my way of solution that I found by solving this puzzle several times.

After initial LL, we only get 4 blacks and 4 dots.

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Now suppose that R10C1 is a dot.

First step: In C10, this makes R4C10 and R8C10 black.
Second step: In C4, (among other pixels) C4R4 and C4R7 are dots.
Third step: C8R4 is black in R4, and C8R7 is black in R7.

This causes a contradiction in C8, because the 2 and the 1 cannot cover the three black pixels in R4C8, R7C8 and R10C8.

Therefore R10C1 is black.
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Line logic solves the puzzle easily from this point.

(This method uses a 3-move lookahead that is hard to see for the first time, and maybe it is considered deep lookahead here.)

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