2
$\begingroup$

The unit cube in $\mathbb{R}^n$ is the set of points $(x_1,...,x_n)$ such that $0 \leq x_i \leq 1$ for all $1 \leq i \leq n$. The surface of this cube is the set of points of the cube such that at least one of the $x_i$ equals either $0$ or $1$. What is the shortest path to travel from $(0,0,...,0)$ to $(1,1,...,1)$ only along points on the surface of the cube?

In $\mathbb{R}^3$ the shortest path is $\sqrt{5}$, obtained by flattening the cube and drawing a diagonal.

However, what would the shortest path in $\mathbb{R}^n$ be?

$\endgroup$
3
$\begingroup$

You can still flatten the hypercube and draw a diagonal that goes straight through two cubes (two hyperfaces).

The shortest path is then the straight path from $(0,0,\ldots,0,0)$ to $(0,\frac 12,\ldots, \frac 12,1)$ and from there the straight path to $(1,1,\ldots,1,1)$.

Both segments have length $\sqrt {1+\frac{n-2}4} = \frac 12\sqrt{n+2}$, so the length of the shortest path is $\sqrt{n+2}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.