Comments on Puzzle #12857: adage two
By annalivia (annalivia)
peek at solution solve puzzle
quality: 
difficulty: 
solvability: some guessing
peek at solution solve puzzle
quality: difficulty: solvability: some guessing
Puzzle Description Suppressed:Click below to view spoilers
#1: bugaboo (bugaboo) on Mar 22, 2011
technically there is a pixels worth of distance between where the 2 different bolts of lightning are striking so it could actually strike there#2: Kristen Vognild (kristen) on Mar 23, 2011 [HINT]
haha
the right side solved easily with color logic
the left side required extended edge logic/guessing
I made a guess, that the left blue squiggle would end at the black square, and it seems I guessed correctly.#3: Liz P (lizteach) on Mar 23, 2011 [HINT] [SPOILER]
Comment Suppressed:Click below to view spoilers#4: Kristen Vognild (kristen) on Mar 23, 2011 [SPOILER]
Comment Suppressed:Click below to view spoilers#5: Gator (Gator) on Mar 28, 2011 [HINT]
The left side requires looking too many moves ahead to be considered logically solvable.#6: Web Paint-By-Number Robot (webpbn) on Mar 28, 2011
Found to require some guessing by Gator.#7: Avgvstvs (avgvstvs) on Apr 21, 2011 [HINT] #8: Avgvstvs (avgvstvs) on Apr 21, 2011 [HINT]
I need to disagree. I believe this can be logically solved, and I explain my method:#9: Gator (Gator) on Apr 21, 2011 [HINT]
C10-20 can all be done with line logic and color logic (something that I particularly consider as a sub-species of line logic). The problem begins when we try to solve the left blue portion.
The key is to understand how the 4 on R16 and the 3 on R17 interact with each other. We need to see few moves ahead.
Let me follow this in detail. The 3 on R17 can go on seven different places: C1-3, C2-4, C3-5, C4-6, C5-7, C6-8 and C7-9. From the start, we can eliminate C-4-6 and C5-7 because both clearly generate an invalid R16. The other cases go as follow:
- C1-3: this would force the 4 on R16 to occupy C4-7, C5-8 or C6-9, but all of them would be invalid by the 1 on C6. R17 can't go on C1-3.
- C2-4: The 4 on R16 would be forced to occupy C4-7 and would be invalidated by the 1 on C6.
- C3-5: Same as previous case.
- C6-8: This time the 1 on C6 belongs to the 3 on R17, invalidating the 4 on R16 that should occupy (by the 2 on C7) C4-7.
This way we've eliminated all options with exception of C7-9. The rest can be done with simple line logic.
To be considered logically solvable, you can look up to 2 moves ahead (this is the constraint I work under when determining if a puzzle is solvable or not). I believe some of your steps are looking 3 or more moves ahead.
Also I think you may have looked at this wrong. Look again at columns 4-6. You have to get many moves ahead before you run into any contradiction. So this one cannot be ruled out. This still leaves us with 6 open cells in R17.
Show: Spoilers
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