All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Algorithms
Painter's Dilemma (Posted on 2011-03-08) Difficulty: 3 of 5
A painter desires to paint the 4 walls of a rectangular shaped room. Each wall has 100 sq. ft. of paintable surface, and a gallon of paint will cover 250 sq. ft. He has two full 1 gallon cans of paint. As he is ready to start, he realizes that one of the gallons of paint has 6% more pigment than the other. Since the paint store is closed and he wants to finish right away, he decides to mix the paint together. Alas, he has no additional containers to assist with the mixing, and no measuring device. However he is a good judge of surface area, and can accurately estimate the percentage of a given wall he has painted to any proportion.

Being an expert, he knows that the paint on any one wall must have exactly the same amount of pigment throughout, but adjacent walls can differ by up to 2% and not be noticed.

What is the step by step plan that the painter can use that satisfies the conditions and also minimizes the volume of paint transferred during mixing?

BONUS: What if the initial difference in pigment is 10%?

No Solution Yet Submitted by Kenny M    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution Comment 2 of 2 |
(In reply to Hint by Kenny M)

I loved this puzzle!

Solution:

Let P be the percentage of pigment in can #1.
Begin with can #1 containing 1 gallon with P% pigment,
  and can #2 containing 1 gallon with P+6% pigment.
Paint the north wall from can #1, leaving 3/5 gallon with P% pigment.
Pour from can #2 into can #1 until can #1 is full,
  leaving can #1 with 1 gallon with P+2.4% (3/5*P%+2/5*(P+6%)) pigment,
  and can #2 with 3/5 gallon with P+6% pigment.
Paint the north wall from can #1 having P+2.4% pigment,
  leaving can #1 with 3/5 gallon with P+2.4% pigment.
Pour from can #2 into can #1 until can #1 is full,
  leaving can #1 with 1 gallon with P+3.84% (3/5*(P+2.4%)+2/5*(P+6%)) pigment,
  and can #2 with 1/5 gallon with P+6% pigment.
Paint the west wall from can #1 having P+3.84% pigment,
  leaving can #1 with 3/5 gallon with P+3.84% pigment,
  and can #2 with 1/5 gallon with P+6% pigment.
Pour the remaining 1/5 gallon of can #2 into can #1,
  leaving can #1 with 4/5 gallon with P+4.38% (3/5*(P+3.84%)+1/5*(P+6%)) pigment.
Paint the east and south walls from can #1.

The north wall is now painted with P+2.4% pigment;
the west wall with P+3.84% pigment;
and the east and south walls with P+4.38% pigment.

Bonus answer: replace the first 2 "paint" and "pour" steps above with:
Paint the north wall from can #2.
Pour from can #1 into can #2 until can #2 is full. Mix can #2 completely.
Pour from can #2 into can #1 until can #1 is full.
  Posted by Grant Stevens on 2011-04-02 19:03:17

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information