Puzzle Description:

I believe her name is Kim Kardashian

#1: Nancy (nbarsi) on Mar 15, 2012 [SPOILER]

The guessing part is if K.K. is the leader or the ET!

Cute! And I agree with Nancy :)

after line logic (with line logic after each step):

at least one of r12c2/r12c3 (or both) must be green, AND at least one of r14c2/r14c3 (or both) must be green but they cont both be in c3 so at least one of them (or both) must be in c2 so c2r12-14 are green (left half is now done)

at least one of r9c13/r9c14 (or both) must be green, AND
at least one of r10c13/r10c14 (or both) must be green, AND
at least one of r12c13/r12c14 (or both) must be green, AND
at least one of r13c13/r13c14 (or both) must be green but they cant all go in c12 so we know that the green pixel in r11c14 is part of the 5 clue

no matter where you place the two green 2 clues in r12 and r13, a green will always be in either r12c13 or r13c13 which means that the green pixel in r11c13 is part of the 3 clue and you can dot r3-8c13

then i felt stuck and resorted to using a summing technique in the upper right corner looking at the box from r3-8 with c15-20
there are 6 clues in those 6 rows which correspond to 1 clue in each of those 6 columns but the 2 dots in r8c19-20 only allow c3-7 to be used (5 columns); the sum of the 6 clues in the columns is 12 and the sum of the 6 clues in the rows is 11 so the 2 in c20 must start in r3 since there are fewer available rows to fit the greater number of clues in that region
line logic to finish

there might be an easier way to solve this (hopefully)

I cannot solve this without a lot of look-ahead. I agree with bugaboo up to the summing step. However, I do not think the summing step works.

There is an open pixel at r3c12 that thwarts it. the 1 in r3 could be put there, taking it out of the summing region given. Also, the 8 in c7 could extend up into that space. Either way can be ruled out, but only after the placement of about 5 more clues.

This is very frustrating as I know what it should look like. The 1 in r3 with the 2s below it, and with 2s crossing them are a clear indicator of zig-zaggyness. If one allows the chain of events of following a zig-zag as acceptable logic, then just looking at the 1 in r3 and following it to the possible ends rules out all positions but c20. then line logic to finish. But that's an iffy sell on webpbn.

Here's what I got before I got stuck:

"bugaboo, thanks for a needed clue. Here's what I did after the first half.

Your clue about the two green 2s in r12&13 led me to this:
"three-way" logic (all three places you can put that 2 makes one of the following conditions true) on the 2 in r12 makes either c13r12 or c13r13 green. Even though there's two possible outcomes, both of those outcomes force the (current) green in r11c13 to be part of the 3 clue in that column. Thus you can dot c13r3-8 (as bugaboo did).

Now I differed. I did edge logic on the 5 in c14 (along those dots we just placed). There can be a block of two maximum in c15, so this forces dots into c14r5-6. This also makes the (current) green at c14r11 part of the 5.

Further edge logic on that 5 now shows that if it were to go all the way up to r7, the 3 in c13 would be forced into r9-11. This puts two dots in r12c13-14, leaving no place for the 2 in that row. So c14r7 must be dotted. (LL)

Placing the 2 in r10 all the way left to c12 creates a conflict in c14, which then must be both green and white. So r10c12 must be dotted.(LL)

-back to upper corner: no matter where the 2 in c20 is, c19r3 is dotted.

but from here i'm stuck"

http://postimage.org/image/4h7snr85h/

(note: the blue is not filled in as i was doing it KRGB - but the blue does not affect things one way or another.)

i did overlook the possibility of the 8 clue extending up into the summing region but if that were not a factor it could still work; if the green 1 clue goes in r3c12 then the summing is even more obvious and easier to see

anyway the step where you dotted r7c14 is easily more than a 2 step look ahead but assuming we dont even use that step
how about this: if the 8 goes up to r6 the first "move" after that is placing the 2 clue above it in r3-4 and the second "move" is placing dots in r4c19 and r6c19 (which violates c19) so the 8 clue goes down at least one more row into r14 (more line logic)

then the green pixels in r9c14 and r10c14 are part of "2" clues in their rows but they cant both go into c15 so at least one goes into c13 (make r10c13 green)
more line logic

see if you can do more with that but i am stuck again (maybe just too tired)

After Joe's and bugaboo's hints:

R13: If the 2 is in C12-13, it makes R20 invalid, so R13C14 must be green instead. Minimal LL.

R14: If C18 is green, R11C12 and R13C13 would both be green, which again makes R20 invalid, so R14C18 is white. Minimal LL.

R13: Whichever way the 2 goes, R15C17 is green. Minimal LL.

C15: If R14 is a 1, it makes R13C13 green and completes C13. R16 and R18 in C15 would be white. This marks R20C12 and R19C14 green, and now R18 doesn't have room for its remaining green. Therefore R13C15 must be green.

LL to finish.

Found to be solvable with moderate lookahead by wombatilim.

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