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# 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
1 year, 3 months ago

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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 $$(n+m-1)^3$$. · 1 year, 3 months ago

Great explanation! Thanks! · 1 year, 3 months ago

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