Formulating the Paired Heavy and Light Balls Balance Puzzle

Summary

The document presents a generalized balance-scale puzzle framed as a quantitative interview question. There are pairs of balls for each color, with one heavy and one light ball in every pair; all heavy balls share one weight, and all light balls share another. The task is to determine the minimum number of weighings needed to identify every heavy ball. The questioner gives a possible arrangement for the three-color case and observes that the outcomes appear to allow identification within a small number of additional weighings.

The text asks how to extend the reasoning to an arbitrary number of colors, but it includes no answer, proof, strategy, or lower bound. As a result, it serves as a problem statement rather than a worked solution. It does not specify further constraints such as whether weighings may be adaptive or whether every ball must be used at most once per weighing, so a complete treatment would need to settle those details before establishing an optimal count.

Key ideas

• Each color has a pair containing one heavy ball and one light ball.
• All heavy balls have equal weight, and all light balls have equal weight.
• A beam balance is used to identify the heavy member of every pair.
• The document asks for a general solution but does not provide a weighing strategy or proof.

Tags

# Solution of extension of six ball puzzle


# Solution of extension of six ball puzzle












A quant interview problem:

We have $2n$ identical size balls containing $n$ colors. For each color there are two balls, one ball is heavy and the other is light. All heavy balls weigh the same. All light balls weigh the same. How many weighings on a beam balance are necessary to identify all of the heavy balls? I know how to calculate the result for $n=3$, like we start with colors = $[\text{white}, \text{red}, \text{blue}]$. Then the first time we weigh $\text{white}_1, \text{red}_2$ comparing to $\text{white}_2$ and $\text{blue}_1$. So depending on the first outcome, we only need 2 weighs at most. But how about $2n$?

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.