Puzzle Description:

Someone else made a dumbell puzzle but used a background. I wanted to see if it was solvable without it. I can't seem to get it started. Any suggestions?

#1: Tom O'Connell (sensei69) on Aug 3, 2013 [HINT]

Brian... I solved it first try, but I guess it was guessing. I figured all the nines had to take up a certain space. Also the 1-8's. Can't wait to see what Joe & Gator have to say.

For me it is not solvable without guessing.

I solved it without guessing. I took the 9 in c5 and used edge logic to eliminate most of the squares in that column. After that, the rest was fairly simple. Thanks.

Internal edge logic on the 9 in c5 shows that it cannot cross the 5s in r10 and r17 at the same time. However, it still can be placed in r11-19 as well as in the range of r2-12 with no immediate conflicts. Thus nothing can be placed with moderate lookahead.

If we could disprove the r11-19 position, it would be line logic to finish.

There is a four-step deep lookahead to disprove the r11-19 position, but it is rather daunting to keep in the head.
>Place c5's 9 in r11-19 and make c2-10 dots
>that forces the 9 in r5 to or past the 5 in c14
>the 5 in c14 then goes down to a maximum of r9 putting a dot in r10
>That dot then invalidates r10's clues, keeping in mind the dot in r5 place in the first step.
Whew.

Thus the 9 in c5 is within the range r2-12 (making r4-10 black). Then line logic to finish.

So it definitely solveable with deep lookahead, but it's (at least this method) very deep. I'll keep looking for a simpler method.

Found to be solvable with deep lookahead by infrapinklizzard.

Here's a shorter one using r10 again. It's simpler, but really, I'm not sure I would have found it without the other method first to point out where I should be looking.

>Place r5's 9 in r11-19 and dot above
>the 5 in r10 must be in the range of c6-11 making c7-10 black.
>the 9 in r16 must also extend through c7 and creates a conflict in that column (it's too far away.)

So this is now really only on the edge of deep-lookahead as this is just a touch past "extended edge logic". The major problem being finding it as there's no standard pointers to it.

So I'll leave the deep lookahead label. What do you think, Gator?

I got to this point (http://www.flickr.com/photos/99790840@N05/9430711541/) by, I think, using internal edge logic with any of the 9's starting on both sides of the line. You can easily and quickly apply the results to the other rows/columns of 9 because the set of clues is the same, just in different directions. From this point, it solves with simple line logic.

What someone should check on my work is whether the point pictured above is truly reachable logically and whether that counts as moderate or deep lookahead. I don't know the nuances between the two like others on the site do.

Noticing the symmetry lets you place the 1-8's making this easy to solve.

@Joshua - in order to do that you must have ruled out the 9 in c5 from going in r11-19 somehow. (Or one of the other 9s with an equivalent set of rows/ columns.) Ruling out that position and getting the dot at c5r19 (or c19r5, or c16r2, or c2r16) is the point of my hints.

@turous - Using symmetry is not PuristApprovedâ„¢. There can still be more than one solution in a puzzle with symmetric clues. The point for a logical solver is not to get A solution, but to get THE solution. Or to prove that you can't get THE solution.

I tried starting the 9 in R2 C5. Eventually I came to a conflict, so I was able to mark that square white, which made the rest of the puzzle solvable.

Getting a conflict from r2 is beyond my keep-it-in-the-head.

Yeah, I had to save and revert, but I can see it as deep lookahead; perhaps *someone* can keep all of those steps in their head.

@Joe Do you have an example of a puzzle with multiple soltuions that has rotational symmetry but not mirror image symmetry orthogonal to the puzzle axes? Obviously something with both mirror image and rotational symmetry (like 11,11) can have multiple solutions.

No, and there may not be one. However, nonexistent proof is not proof of nonexistence. The burden on a logical solver is to prove it, and we don't have proof for that assertion.

I would be very interested in a proof if you have one.

my get-started, lookahead move:

one of the 9's has to extend into R11, either one results in C15R11 being black. (same logic blacks C6R10, C10R6 and C11R15)

LL to finish

Sheesh, moderate lookahead it is! -- but good luck finding it!

This is one where the moderate lookahead label is a teaser. (Of course, I've been guilty of a few of those, too.)

Found to be solvable with moderate lookahead by infrapinklizzard.

"Sheesh"?
I wasn't trying to convince you one way or the other Joe ;)

Maybe "Yikes!" is a better phrasing?

This came out so smoothly after a little guess ("9s don't extend to the edge") that I forgot I had guessed at all.

Decided to accept the author's opinion and did not solve.

Decided to accept the author's opinion and did not solve.

Look at the 9s in column 5 and column 16. One of them must cross the 5 in row 7, and the other must cross the 5 in row 14. Either way, R14C10-11 (also R7C10-11) are dots. LL from there.

Gary, of all the methods here, yours is the only one I can make work without some form of trial and error. Your method also works for columns 7 and 14, making R10-11C7 and R10-11C14 dots/white.

Interesting puzzle!

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