Cubeception

Recently, I was playing with my two year old brother. There was an old $$3\times 3$$ Rubik's cube lying around and my dad posed a simple problem:

How many selections of $$2\times 2$$ cubes can be made from a $$3\times 3$$ Rubik's cube?

The answer to the above question is $$8$$ and it is so because each corner cube is part of a unique $$2\times 2$$ selection.

Now, how many selections of $$1 \times 1$$ cubes can be made from a $$3\times 3$$? The answer here is too simple and is $$27$$.

And the number of selections of a $$3\times 3$$ cube from a $$3\times 3$$ is $$1$$.

$$1, 8, 27$$.... Reminded me of something. This note is an attempt to prove or disprove the following hypothesis.

In an $$n \times n$$ Rubik's cube, the number of selections of solid cubes of edge length $$m(\le n)$$ is $$(n-m+1)^3$$

I am convinced that there is an elegant solution to this problem and that the hypothesis is true.

Note by Raghav Vaidyanathan
3 years ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Consider one dimension first. Given an array of length $$n$$, we can place a smaller array of length $$m$$ in $$n-m+1$$ positions. This is the same for all three dimensions, bringing the total number of placings to be $$(n+m-1)^3$$.

- 3 years ago

Great explanation! Thanks!

- 3 years ago