Comments on Puzzle #29167: Not Happy at all
By Sylvain "WCPman" (qwerty)
peek at solution solve puzzle
version: 2 quality: 
difficulty: 
solvability: moderate lookahead
peek at solution solve puzzle
version: 2 quality: difficulty: solvability: moderate lookahead
Puzzle Description:
Kind of a close-up look at Earthworm Jim But not in a Groovy mood
#1: Donna McFarland (jade8114) on Dec 4, 2016
Never heard of him, so I looked him up good likeness. Would not want to meet him after dark or daylight or anytime.#2: Web Paint-By-Number Robot (webpbn) on Dec 6, 2016
New version published by qwerty.#3: Kristen Vognild (kristen) on Dec 7, 2016
Is Earthworm Jim a vampire? :)#4: Sylvain "WCPman" (qwerty) on Dec 7, 2016 [SPOILER]
No just a earthworm in a space suit shooting at aliens and cow#5: David Bouldin (dbouldin) on Dec 9, 2016 [HINT]
love the advanced logic in this one!#6: Web Paint-By-Number Robot (webpbn) on Dec 9, 2016
near the end, if you are stuck and still have most of the upper right quadrant to go, the IEL move that took me the longest to see was that the 1st 2 in C28 can't start in R2 because C27 would be violated in R2 and R6. this dots C28R2.
some LL later, you'll notice that whether the black at C21R8 is part of 2 or 1, C23R8 is a dot.
...i know i skipped a LOT of advanced logic in the middle part of the solve, but i hope this helps someone.
Found to be solvable with moderate lookahead by infrapinklizzard.#7: Joe (infrapinklizzard) on Dec 9, 2016 [HINT]
el 4 c1 = only one position possible.#8: Ailsa Hebert (bazette3) on Dec 22, 2022
ll
two-way logic 2 r14 = whether it goes into c11 or c13, it will trigger the 3 in r13 and so r13c4-6 must be white.
ll
el 7 c30 = r1-2, r11-15 w
ll
two way logic 3 r12 = whether it goes in c16 or c19 it will trigger the 2 in r11 and so r11c11-14 must be white.
(brief, useless) ll
el 1 c3 = if it were in r8 the 2 in c3 would cause an immediate conflict in c4. So c3r8 is white.
el 2 c1 = only one position possible.
ll
el 5 r10 = c15-17 w
ll
el 4 c14 = r8 w
ll
el 4 c17 = r3 w
ll
EL 3 r1 = c27-28 w
Two-clue edge logic c38 - if the top 2 were to be in r2-3, then both it and the second 2 (in r5-6) would force a pixel in c37 where there's only one clue left. So c38r2-3 must be white.
EL 3 r1 = c36 w
Weird edge logic 3 r2 = if it were to be anywhere in c24+ there would be nowhere for the 3 in r1 to go without causing a conflict in r2. So r2c24-27 must be white.
Two-way logic 2 c28 - whether it goes up or down, it will trigger the 1 in c27. Therefore all unknown rows in c27 except r4 and r6 must be white.
ll
Two-way logic 2 r28 - whether it goes up or down, c26r6 will be white.
EL 2 c26 = r5 w
ll to finish
Nice puzzle...really enjoyed the solve...thanks!!
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