# Fewest inequalities to define a hollow cube

These four inequalities define a hollow (zero volume) cube. Is it possible to describe a hollow cube using a smaller system of (real scalar generalized-polynomial) inequalities?

\begin{eqnarray} x^2 & \leq & 1 \\ y^2 & \leq & 1 \\ z^2 & \leq & 1 \\ (x^2 - 1)^2(y^2 - 1)^2(z^2 - 1)^2 & \leq & 0 \end{eqnarray}

EDIT: After Rahul made me think in terms of norms, I came up with the following very simple form based on the 1-norm:

$$\left ( \sqrt{x^2} + \sqrt{y^2} + \sqrt{z^2} - 1 \right ) ^2 \leq 0$$

Rahul is right, the 1-isonorm and $\infty$-isonorm are similar in $\Re^2$ but not in $\Re^3$

$$\left ( \left ( \sqrt{x^{2p}} + \sqrt{y^{2p}} + \sqrt{z^{2p}} \right )^{1/p} - 1 \right ) ^2 \leq 0$$

approaches a cube in the limit as $p \rightarrow \infty$

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The equation in your edit is not correct; it gives you a hollow octahedron instead. – Rahul Feb 6 '11 at 17:51
@Rahul: Thanks. I was thinking it would be a rotated cube, but you're right that it would have 8 faces and 6 vertices. – Ben Voigt Feb 6 '11 at 19:36

Note: When I first posted this answer, the question did not contain the condition that the equations should involve polynomials.

Certainly there exist infinitely many possibilities of a single scalar equation that will do the job, using any function that is zero on the boundary of the unit cube and nonzero elsewhere. Here is one explicit example:

$$\max(|x|,|y|,|z|) - 1 = 0$$

If you prefer inequalities, you can just square the left-hand side and set it $\le 0$.

But is this just a mathematical version of code golf, or is there some other reason for this question?

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No particular reason, I guess it could be described as math golf... but then many mathematical problems are. And I'm remembering that the $\max$ function can be implemented as a limit of polynomials (definition of $\infty$-norm), so that's really cool. – Ben Voigt Feb 6 '11 at 16:50

What sort of functions do you allow? If you allow square roots then you can encode multiple inequalities in a single one:

$A \geq 0 \wedge B \geq 0 \Leftrightarrow (\sqrt{A^2}-A)^2 + (\sqrt{B^2}-B)^2 \leq 0$

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