davidrferreira/balancedcentrifugeproblem

The Balanced Centrifuge Problem implemented in the C language.

balancedCentrifugeProblem_C

Description

Implementation of The Balanced Centrifuge Problem using the the C language.

This is one of the, probably many, ways to simulate this mathematical problem.

The Problem:

A centrifuge is a laboratory device which spins liquid samples at high speed in order to separate them based on density. It’s very important to balance a centrifuge before operating it - not doing it by using, for example, just a tube on only one side, the rotating mass will not be balanced causing the centrifuge to shake when rotating and may even damage the centrifuge or the tube. A tube on the opposite side is used to balance the mass for smooth operation.

So, a centrifuge is balanced if the center of mass of the collection of test tubes coincides with the center of mass of the centrifuge itself.

The problem is - given the number of holes in the centrifuge - for which different numbers of test tubes can you obtain a balanced-configuration?

The Solution:

(1) Number of Test Tubes.
(2) Number of Empty Spots in the Centrifuge.
(3) Number of Holes in the Centrifuge.

You can balance the centrifuge if and only if both (1) and (2) can be written as sums of prime factors of (3).

Example 1:

For the case of a centrifuge with 12-holes.

The prime factors of 12 are 2 and 3.

You can have a balanced-configuration for 2, 3, 4, 5, 6, 7, 8, 9, 10, 12 test tubes in a 12-hole centrifuge.

*Notice that, for example, for 5 test tubes, you have
(1) Number of Test Tubes = 5;
(2) Number of Empty Holes = 12 - 5 = 7.

Being that the prime factors of 12 are 2 and 3, you can easily see that you can write (1) and (2) as sums of those prime factors.
(1) 5 = 2 + 3;
(2) 7 = 2 + 2 + 3;

So, you can have a 5 tubes balanced in a 12-hole centrifuge.

Example 2:

In the other hand, it's easy to understand that if you have 11 test tubes you will never be able to have a balanced-configuration.

(1) Number of Test Tubes = 11.
(2) Number of Empty Holes = 12 - 11 = 1.

You can't write (1) or (2) as sums of the prime factors of 12 (which are, again, 2 and 3).