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Comments on Puzzle #22429: When I was 7.
By Aldege Cholette (Aldege)

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

Puzzle Description:

I use to love walking downtown to look at the new for 1959 Cadillacs,Buicks,and Oldsmobiles thru the showroom window at Hamilton Motor Products.

#1: Jota (jota) on Jul 24, 2013

That's an awesome memory! I loved the solve!
#2: Norma Dee (norm0908) on Jul 24, 2013
I forgot to click on post. I loved the puzzle. It brought back fond memories. My son loved to go see all the new cars as soon as they his the show rooms. But he also loved auctions.
#3: Aldege Cholette (aldege) on Jul 25, 2013 [SPOILER]
Thx Jota and Norma. I don't know if you could make out the image,but I was trying to portray what a little boy could see from the outside looking in,which is just partial images of 2 cars. The front of one at top left,and the back of another with one big fin and taillights.:)
#4: Joe (infrapinklizzard) on Jul 25, 2013 [HINT]
After line and color logic solves all but the upper left red and center green,

Smile logic finishes the red.

Two-way logic on the 1 in c20, which can be in either r12 or 13 makes c19r13 green.
Line logic.

Then I cannot find any moderate lookahead with our "normal" advanced logic. There are several ways to do it with deep lookahead.

However, there is an out-of-the-box way with only moderate lookahead. It is a variant of summing.

There are now only eight more (green) pixels to be placed. They must all be in the rectangle from c14r10 to c21r13. Lets look carefully at the row clues.

In r10, both 2s are fixed by one pixel each. In r11 the single 2 can be in c16-19, a total of three positions. In r12, both 1s can only be in a two-pixel spot each. And in r13, both 2s are again fixed by one pixel each.

If you look closely at r10, 12, and 13, you can see that each of them have two clues, and each of those clues must be on opposite sides of the overall rectangle.

Let's sum the halves of the rectangle: The left half (c14-17) must have four more green pixels, and the right half (c18-21) must also have another four.

Each row 10, 12, and 13 will add one more pixel to each side of the rectangle. That's 3 more pixels per half, but each half needs four more.

Therefore the 2 in r11 must have one pixel in each half (and thus be in c17-18).

Then line logic to finish.
#5: Web Paint-By-Number Robot (webpbn) on Jul 25, 2013
Found to be solvable with moderate lookahead by infrapinklizzard.
#6: Aldege Cholette (aldege) on Jul 25, 2013
Thx Joe.:)
#7: Joel Lynn (furface1) on Jul 29, 2013 [SPOILER]
Yaaay! My 10,000th puzzle solved!

I had to squint to get the final image, but the fin is great!
#8: Aldege Cholette (aldege) on Jul 29, 2013
Thx Joel Lynn and congratulations.:)
#9: Koreen (mom24plus) on May 12, 2019
I thought I was solving some kind of spaceship(s) until then end. ;)

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