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#1: Byrdie (byrdie) on Jan 11, 2009

Frustrating.

Hmmm. The red and blue solve easily. The green is hard.

You can do either-or logic on the first green in row six, which lets you mark one white cell in row seven, which gives you a green in that row and a white on the bottom.

Then you don't have to look ahead very far to see that a green at row 8, column 4 would give a contradiction, so that cell must be white. Same is true for row 8, column 6 for the same reason.

Either or logic again, this time in column 4. Either place you put the green cell makes the cell in row 10, column 5 green. Line solving from here gets you several more whites and one more green on the right side. Row 7 and column 5 are complete.

We can say row 6,column 6 is white because if green column 7 doesn't work, though it's the third case of a contradiction that requires a fair amount of looking ahead to see.

And then I run out of steam.

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I basically employed the same method as Jan. The red & blue solved quickly. Between a big hint from the title and looking at plausible cominations (either-or logic), I got it on the first try.

It's not a terrible image but one would think that some effort would've been applied (with the help of the checker) to make it more easily solvable.

I'm pretty sure Shay doesn't test solve his(?) puzzles. Some turn out to be fun to solve, but that's pretty much by accident. I wish he would take the extra step ti refine his puzzles a bit more.

This one is kind of interesting though. It feels like there ought to be a way to logically solve it. The solution is unique. But I can't quite find a way.

I got it on the first try by just guessing and I am usually a bad guesser. Jan's logic seems to be just a step away from a solution.

I followed Jan's logic and it seems very good. I think putting a green in the 10th row, 3rd column leads to a relatively quick contradiction after a few steps, so it must be white, and then you can finish the bottom row. Then a green in 9th row, 8th column also leads to a quick contradiction [it would force position (8th,8th) to be green and then positions (7,8) and (7,9) are white, but then there is not enough room for the greens in the 7th column], so it must be white.

Then with a green in 6th row, 8th column, you can finish the 6th row, and regular simple logic should bring it home.

It weird, I could have sworn I solved this without guessing, but now I'm not sure. Maybe I did use the "look ahead" sort of logic discussed above.

Or maybe I was guessing and only thought I was being logical.

I guessed but isn't it double smile logic? or perhaps triple.

I went as far as I could with regular logic, then just used "Shay logic" and solved it rather quickly.

Found to be logically solvable by jan.

This puzzle has been revised and is now black and white.

Curiously, it became logically solvable when that change was made.

i found that i was still needing to guess in the black and white version. a lot of trial and error.

Yes guessing unless you count multiple smile logic.

lol #9
i only see b/w
still had to do smile logic

I do not agree that this is logically solvable. There is too much look-ahead needed.

If the 1s in c3&7 and r9 could be confined to one on either side of the white space in its column or row, then smile logic would work.

However, the vertical ones can both be placed on the bottom side of their columns' r7. Also, the horizontal ones can be placed on either side of r9c5.

Using the criteria that seeing three or more moves ahead is guessing, then guessing is needed here.

Comment Suppressed:Click below to view spoilers

jan found it logically solvable in comment #10, but now it says "requires some guessing"? also, see a lot of people saying guessing is needed. three-way smile logic works, but for people who prefer more traditional methods, here is a walk-through:

- LL
- whether C2R5 or C3R5 is black, C3R6 is a dot.
- whether C7R5 or C8R5 is black, C7R6 is a dot.
- whether C3R9 or C4R9 is black, C6R10 is black.
- LL to finish

Found to be solvable with moderate lookahead by jan.

Note that there have been multiple versions of this puzzle, so some of the earlier comments may have referred to the older version. But I'm convinced by David.

I agree with David and Jan now. It's amazing how our "common solving intelligence" has gone up over the intervening years. There are many comments from around 2008 or before that proclaimed fairly straightforward edge logic puzzles as being unsolvable without guessing.

Looking back at my puzzles made before my comment date I see that only a few of them had two-way logic in their solution. (Khepri and 'Head in the Sky' stick out in my memory.)

Two-way logic is easy to overlook in a puzzle, but this one has all the obvious markers for trying it. Not only do you have several places with clues that can go in one of two places (a prerequisite, but not a guarantee), but there are lots of 1s which makes the likelihood that much greater.

(Note that the "smile logic" method still suffers from the faults I mentioned before.)

I temproarily made the color version of this puzzle to try it out. Tricky.

This thread is an interesting read! While you can guess the solution is symmetric, I actually found smile logic at the end the ay I did it.

You can sum the total dots on the left once line logic breaks down. There need to be 5.

So there must be 2 dots in rows 9 and 10 total on each side.

If there are 0 on the left in row 10 and this 2 in row 9, we get a quick contradiction, since column 4 needs to place 2 isolated dots in rows 5 and 6. Similarly, there can't be 0 on the right in row 10.

Then you can use smile logic on columns 2 and 9.

David's way is shorter and nicer, I think. But I wanted to share. I was surprised how much I learned from this puzzle.

i rarely think to try summing, so thank you for my reminder :)

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