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Comments on Puzzle #5113: Feelin' Flowery?
By Deanna W (Silkbear)

peek at solution       solve puzzle
  quality:   difficulty:   solvability: moderate lookahead  

Puzzle Description Suppressed:Click below to view spoilers

#1: Adam Nielson (monkeyboy) on Feb 23, 2009

For you, I hope. :-)
#2: Deanna W (Silkbear) on Apr 3, 2009
Yup! xD
#3: Gator (Gator) on May 1, 2009 [HINT]
I believe that one does take some guessing. I can get about 90 percent done, before having to guess.
#4: Web Paint-By-Number Robot (webpbn) on Sep 8, 2009
Found to require some guessing by jan.
#5: Jota (jota) on Nov 14, 2010
No option for me to guess. Weird?
#6: David Bouldin (dbouldin) on Sep 16, 2011
Very challenging, fun solve! Had to clear the board a couple of times, but made it through...and I wouldn't consider anything I did to be "guessing". Too many complicated steps to go back through and type out my logic, though. If someone posts a screen shot of their "logic has ended" point, I'll comment.
#7: Gator (gator) on Sep 18, 2011 [HINT]
Here you go:
http://i220.photobucket.com/albums/dd289/Gator621/WebPBN-HG/WEBPbn_FeelinFlowery.jpg

This is where I can get with line logic. I also can get R19C2 to be black by seeing that both R19C7 and R19C9 cannot both be black without making row 18 invalid.
#8: David Bouldin (dbouldin) on Sep 21, 2011 [HINT]
Is this two-move look-ahead?

Since C4R7 and C4R8 can't both be black, one or both of C14R7 and C14R8 must be black.
...which means C14R5 is a "1"?
#9: Gator (gator) on Sep 21, 2011 [HINT]
That works. So after some more line logic, you will reach other point where there is no more line logic. At this point, I looked at row 19. If both R19C7 and R19C9 are black, then R18C7 and R18C9 would have to be dots. This makes row 18 invalid as I would not be able to place the 4 remaining 1 clues. This makes R19C2 black. A little line logic.

Next I did some two-way logic with the 2 clue in row 19. If R19C1 or R19C3 is black (completing the 2 clue), this will cause us to fill in R18C7 AND R18C9. So those two cells have to be black. Some more line logic.

I then looked at column 2. If both R10C2 and R12C2 are black, then both R10C3 and R12C3 would have to be dots. This would make column 3 invalid though. This makes it so that R17C2 has to be black. The rest solves with line logic.

Thanks for getting this going again David!
#10: Web Paint-By-Number Robot (webpbn) on Sep 21, 2011
Found to be logically solvable by gator.
#11: David Bouldin (dbouldin) on Sep 21, 2011
my pleasure...love these puzzles!

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