Puzzle Description:

my drawing of a :P smilie face

#1: Liz P (lizteach) on Jan 22, 2011 [HINT] [SPOILER]

I found it solvable with just smile logic on the eyebrows to finish.

And a :P to you, too, Jayne. :D

Ack! The emoticons are coming to life. Thanks for the puzzle.

is it really straight smile logic?
you still have unaccounted for cells in the nose area
i am not sure about this one

After the "eyebrow" login the rest solves including the nose.

Agreed with bugaboo that it is not straight smile logic. Let's look at the nose areas first.

As both R9C9 and R14C9 cannot be black, then either R9C11 or R14C12 must be black. This tells us that R2C11 and R4C11 are dots or R2C12 and R4C12 are dots. In both cases, R2C11 and R4C11 have to be dots. This then enables us to mark R9C11 as black.

The next part is sort of like a Venn diagram. Individually, there are many possibilities, but there are very few places that overlap where it works in all cases. Each possibility is looking 2 or less moves ahead, but then the possibilities are analyzed collectively.

Ignoring row 14 for a moment, let's do some summing on rows 2 and 4. On the left, there are 7 columns open with 1 cell left to fill in each column. On the right, there are 8 columns open with 1 cell left to fill in each column.

Now look at row 4 and consider which of the 2 clues will be on the left or right side. Just from a "can it fit" point of view, there are two possibilities: one 2 clue on the left and three on the right, or two 2 clues on the left and two on the right.

Now look at row 2 and consider the possibilities. The two 3 clues can be on the left, one on each side, or on the right. If they are both on the left, then no 2 clues can be entered on row 4. From the row 4 analysis though, we know this is not possible. If they are both on the right, then only one 2 clue in row 4 can be placed on the right. Again, from the row 4 analysis, this is not possible. So we can conclude that one 3 clue will be on the left and one will be on the right.

Furthermore, since placing three 2 clues in row 4 on the right makes it so that a 3 clue cannot be placed in row 2 on the right, we can conclude that two 2 clues must be on the left and two 2 clues must be on the right.

Now that we know that one 3 clue in row 2 must be on the left and that two 2 clues in row 4 must be on the left, we can use smile logic. Next the nose can be finished and, finally, smile logic can be used on the right.

This does seem like a bit much for a person to keep in their head when solving. If you were taking notes as you did it, then it is not that bad. So should I mark this one as solvable or not?

I mis-wrote. What I used was not smile logic. It was two-way logic; at least, I think that's what I mean.

So using the "purple logic" term, when you run out of purple logic, and have only the nose and eyebrows left, you should be able to use two-way logic on the 1 in C6. There's a choice between R2 and R4. It has to be in R2 because placing it in R4 means that you can't put a row of 3 in R2. Perhaps that is a bit of a look-ahead, but I thought it was a fairly obvious one (certainly no more look-ahead than most edge logic).

From there, purple logic solves the nose and then all is left is to do the same two-way trick on the other eyebrow.

If anyone can follow the above, I'd like to know whether it makes sense, or whether my logic is somehow faulty.

(I find the logic discussions these puzzles engender really interesting.)

(And I should add, I guess I said it was smile logic because it *felt* a bit like smile logic while I was doing it, plus, it looked at bit like an upside-down smile!) :D

I looked at your logic Liz, but I think it requires looking more than 2 moves ahead to get to a contradiction.

How about this? Looking in column 3, we have a 1 clue. It can go in either R2C3 or R4C3. Look how this affects column 5. In both cases, R2C5 will be black and R4C5 will be a dot. You can do the same logic with the 1 clue in column 19 to make R2C17 black and R4C17 a dot. The rest solves with line logic. Much easier this way.

Found to be logically solvable by Gator.

Goto next topic

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