Puzzle Description:

You can't travel through the country of Turkey without seeing this amulet. It is called the evil eye and is used to ward off the effects of someone giving you the evil eye.

#1: Norma Dee (norm0908) on Nov 22, 2012

I think I need one of those.:) Nice puzzle.

thought it was a wreath at first
not sure how i solved it, but i got it the first try..guess i do solves afterall.

Thanks for the puzzle, Brian. Happy Turkey Day to you, too. I couldn't logically place the individual pixels in the center logically. I had to guess. I didn't rate solvabiity.

Re. the center pixels - by using... um... I don't know what it's called. Addition logic? Anyway, I could see that there were only two possible ways rows 13, 14, and 15 could be arranged. I wound up guessing from there, but only one of those two ways actually worked as I placed them (the wrong one created an error in, I think, C13), so I imagine there's a decent chance that a little more lookahead could fit them in place.

I'm sure there's a way....but after gorging, didn't feel like the trouble. when you look at the c's and r's with (3)1's, you know it has to go one of 2 ways. guessed one way and saw a contradiction immediately, so it sorted itself out.

if someone takes the time to find the logic, pls post. I'd like to check it out(or re-work myself if no one does) after a nice nap.

Nice puzzle, Brian

I guessed correctly the first time, too. You may need to add an extra pixel, so there's no guessing involved. :)

Well, with considerable look-ahead:

Notice that the space left is a 6x6 pixel square from r10c9 to r15c14. This can profitably be thought of as a tic-tac-toe grid of nine 2x2 squares, the center one of which is black. (Let's refer to them with directions - N,S,E,W, NE,SE,SW,NW)

Summing to make some basic observations:
-The squares North, South, East, and West of the black square must each have two blue pixels. NOTE this takes up eight of the thirteen pixels left to be placed, leaving only five to be distributed among the four corner squares.

Now look at the rectangles that take up the entirety of each side. (eg. the North rectangle would be the NW, N, & NE squares)
-The rectangles West & South of the black square (r10c9-r15c10 & r14c9-r15c14 respectively) each need four blue pixels.
-The rectangles North & East of the black square each need five blue pixels.

Subtracting the two blue pixels from its center square, each of those rectangles would have either two or three leftover pixels for its two "outside squares" (which correspond to the corners of the big free-space square). S & W rectangles = two pixels each; N & E rectangles = three pixels each.

In a manner not unlike smile logic's process, it is obvious then that the only way to distribute the extra five blue pixels is with one pixel in the NW, SW, & SE squares, and two in the NE square.


Now that we know that the place that is out of symmetry, and thus most vulnerable to attack, is the NE corner, let's review some things established:

-There are two blue pixels in the NE corner square. Since all the clues are 1s, those two pixels must both be in different rows and columns (i.e. diagonal to each other) and so one pixel *must be in c14*.
-There are two pixels in the E side square. As per the NE square, this means one pixel *must be in c14*.

This now means that c14 is satisfied and none of c14 in the SE corner square can have a pixel. Thus c14r14&15 are dotted. Then line logic to finish.

(This is more a brain-work look-ahead than a placement look-ahead. One could do this in pen without fear of "having to erase", but it takes a string of logic to place that first dot.)

When solving it myself, I intentionally placed c12 below the black in the wrong spot and got a contradiction almost immediately. I couldn't do it without look ahead but not much.
http://relaxdreamlive.blogspot.com/2011/06/evil-eye-turkish-superstitions.html

If you make c12r14 blue it forces blue at c11r15 and not enough space left to put the three remaining blues in c13. Therefore, c12r15 is the blue and the rest solves easily.

Joe making my mind go BOOM!!!

Joe remember the scene in the princess bride where the battle wits? Reminds of that:

http://www.youtube.com/watch?v=U_eZmEiyTo0

"Wait 'til I get going! Now, where was I?"

lmao

It was easy enough to guess. There were only two ways the 3 1s in column could go. I think some guessing or deep lookahead. Intuitively, it was easy, but that doesn't count.

Yeah, there are a few ways to do it with a several-step placement look-ahead. Both Tom's and Brian's are sound. You can also misplace any of the pixels in either row 11 or 12 above and below the black square.

I like mine because it was a fun divide-and-conquer holistic solution, rather than a tedious poke-and-prod trial-and-error pixel hunt.

Joe: I like your logic as well, however it requires more than a two move lookahead. I can go 5 or 6 steps ahead to prove (and have done so with a very few puzzles), but it qualifies as deep look ahead after 2.

come on people check this out:

after initial ll/cl all you should have left are the blue 1 clues in the center area, right?

look at the 2 black pixels in r13 - each of them needs a blue pixel below it but they cant both go in r14 so at least one of them goes into r15; because of this you can dot the rest of r15 except for c11-12
ll to finish
no look ahead at all
no guessing

bug: nice.

Tom: I didn't claim less than deep look-ahead. See my first comment.

Found to be solvable with moderate lookahead by infrapinklizzard.

As per bugaboo in comment 17

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