Puzzle Description:

no hint

#1: Kristen Vognild (Kristen) on Jul 1, 2013

I had to rely on symmetry to solve this one.

Not very much imagination went into making this. And who or what does "maya" refer to?

get 'em synthia, sink your teeth into that crap puzzle!

I found a deep lookahead method. Here's how I did it.

After initial line logic,

Edge logic on the 3 in c13 makes r1-4 white.
Edge logic on the 3 in c14 makes r1-3 white.
Edge logic on the 4 in c15 makes r1-3 white.
Then some line logic.
Edge logic on the 3 in c20 makes r1-3 white.

Now the same, mirrored:

Edge logic on the 3 in c37 makes r1-4 white.
Edge logic on the 3 in c36 makes r1-3 white.
Edge logic on the 4 in c35 makes r1-3 white.
Then some line logic.
Edge logic on the 3 in c30 makes r1-3 white.

*There are quite a few bits of tiny edge logic that can be done here, but they aren't necessary for this, and I couldn't find a moderate lookahead including them.

Now the deep lookahead:
|There are only two places for the right 3 in r2: c27-29 and c32-34.
|If it were to be in c27-29 then
|=the right 2 in r2 must be in c25 making the 6 come up that far
|=the 3s in c27&29 must go down into r4.
|=making a conflict with the crossing 2s in r4
|
|So it can't be in c27-29, and r2c32-34 must be black.

Then lotsa line logic.

Edge logic on the 2 in c22 makes r4 white.

Then line logic to finish.

~~~~~~~~~~~~~~~~~
* Here are some more bits of edge logic that can be done before the deep lookahead, but aren't needed to do it:

el 3 in c14 makes r4 white
el 3 in c16 makes r1 white
similarly el 3 in c18 makes r1 white
el 3 in c21 makes r1 white

and mirrored:
el 3 in c36 makes r4 white
el 3 in c34 makes r1 white
similarly el 3 in c32 makes r1 white
el 3 in c29 makes r1 white

and some extended edge logic:
If the 3 in c14 is in r5 then it will force the clues in both c13 and c15 down into r8.
That conflicts with the 2s in r8, so c14r5 must be white. (Then LL makes c14r8 black.)

If the 3 in c14 is in r10, then it will force the clues in both c13 and c15 up into r6.
That conflicts with the 2s in r6, so c14r10 must be white. (Then LL makes c14r7 black.)

and mirrored:
If the 3 in c36 is in r5 then it will force the clues in both c35 and c37 down into r8.
That conflicts with the 2s in r8, so c36r5 must be white. (Then LL makes c36r8 black.)

If the 3 in c36 is in r10, then it will force the clues in both c35 and c37 up into r6.
That conflicts with the 2s in r6, so c36r10 must be white. (Then LL makes c36r7 black.)

But at this point I'm stuck for moderate lookahead, so I had to do the deep stuff above.

Found to be solvable with deep lookahead by infrapinklizzard.

Well I quite enjoyed this solve--fun finding the little bits of edge logic to unlock it.

Weak. No idea what this is supposed to refer to.

Oh, it's fine. Some are easy, some are hard; some are obvious, some are huh? If you don't like this one, there'll be others. "Maya" is a common enough sound that it shows up in various languages. What comes to mind for me is Maya Angelou and the Maya culture of Central America.

MAYA naise

I also used symmetry, and the thing solved in no time flat.

Goto next topic

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