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Comments on Puzzle #24182: A Tentative Friendship
By Tonia Bergh (tonia)

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

Puzzle Description Suppressed:Click below to view spoilers

#1: Norma Dee (norm0908) on Apr 5, 2014

How very cute!

I gave a small kitten, who loved to curl up in your lap, to a lady who wanted a lap kitty, but it was not to be. They had a huge dog and were concerned about how the kitten would react to it. But, not to worry. The kitten adopted the dog at first sniff and they became inseparable. But to keep peace in the family the kitten would crawl up in the lady's lap every now and then.
#2: Aldege Cholette (aldege) on Apr 5, 2014
So,so cute Tonia,nice image.:)
#3: valerie o..travis (bigblue) on Apr 5, 2014
ditto aldege :)
#4: Heather M (AuntieH) on Apr 6, 2014
ditto Travis :)
#5: Kurt Kowalczyk (bahabro) on Apr 6, 2014
nice puzz and fun solve!

absolutely no guessing required, people....
#6: Susan Duncan (medic25733) on Apr 6, 2014
Very cute and how true!
#7: Heather (heatherkewl) on Apr 6, 2014
sweet
#8: Tonia Bergh (tonia) on Apr 6, 2014 [SPOILER]
Comment Suppressed:Click below to view spoilers
#9: teresa dickens (trdickens2) on Apr 6, 2014
What a charming picture!
#10: Jota (jota) on Apr 7, 2014
Loved the solve!
#11: Joe (infrapinklizzard) on Apr 8, 2014 [HINT]
LL/CL leaves only black.

Some non-obvious line logic: in r7, the 3,1 at the right side makes r7c29 white. (It would be white whether the black pixel in c26 is part of the 3 or the 1.) Then LL

More non-obvious line logic: the unknown at c24r7 cannot be part of either the 3 or the 1 in r7, so it must be white. Then LL

Edge logic on the 5 in r1 makes c4-14, c16, c22, and c30 white.

Then LL finishes the right side.

Edge logic on the 3 in r11 makes c13-15 white.

Edge logic on the 2 in c12 makes r8 white.

Some summing:
-The left 3s in r10 & 11 must be within c1-5 due to the black pixels in c3. That means a total of six pixels must be within the rectangle of c1-5,r9-10.

-Now, summing all the black clues in c1-5 gives us six pixels for all of c1-5.

-That means that all of c1-5 outside of r9-10 must be white.


Then immediately edge logic on the 4 in c6 makes r2-4 white. Then LL


Edge logic on the 2 in r9 makes c8-9 white. Then LL to finish.
#12: Web Paint-By-Number Robot (webpbn) on Apr 8, 2014
Found to be solvable with moderate lookahead by infrapinklizzard.
#13: Joe (infrapinklizzard) on Apr 8, 2014 [SPOILER]
Comment Suppressed:Click below to view spoilers
#14: Wombat (wombatilim) on Apr 15, 2014 [HINT]
Alternate method:

I solved this the same way as Joe did up to (and including) this line:
"Edge logic on the 3 in r11 makes c13-15 white."

Next I looked at the 3 clue in R15. No matter which way I went, the 1 clues in C1-2 and C4-5 were going to be in either R14 or R15, eliminating R2-13 in these columns.

I then went immediately to edge logic in C5, which eliminates R2-4 in that column. More line logic.

Next I took a look at C14, and saw that if I placed the 2 clue in R14-15, it would cause a contradiction in C13, and eliminated those two squares.

I then repeated the exercise with the 2 clue in C13 (placing it in R14-15 causes a contradiction in C12).

Line logic to finish.
#15: BlackCat (blackCat) on May 6, 2017
Cute but needed to guess at least 3 times.
#16: Bill Eisenmann (Bullet) on Nov 18, 2021
Adorable image, but not a smooth solve. Gets much better toward the end
#17: Steven Paradise (gossamerica) on Dec 20, 2021
Such a cute image, but getting there is a challenge!

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