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:#8: David Bouldin (dbouldin) on Sep 21, 2011 [HINT]
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.
Is this two-move look-ahead?#9: Gator (gator) on Sep 21, 2011 [HINT]
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"?
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.#10: Web Paint-By-Number Robot (webpbn) on Sep 21, 2011
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!
Found to be logically solvable by gator.#11: David Bouldin (dbouldin) on Sep 21, 2011
my pleasure...love these puzzles!
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