Note, a polyhedron is the intersection of finitely many half spaces in $\mathbb{R}^n$ and a polytope is a bounded polyhedron.

Let $M$ be an $m \times n$ matrix of integers. Let $P$ be the (possibly unbounded) polyhedron in $\mathbb{R}^n$ given by $$M \cdot v \geq 0 \quad \textrm{and} \quad v \geq 0.$$ Clearly $0 \in P$ and $P$ is bounded iff $P = \{0\}$. By solving a linear programming problem we can determine if $0$ is the only point in $P$, and so if $P$ is a polytope. Doing this takes $O(m^n)$. Is there a faster way to determine if $P$ is actually a polytope?

  • $\begingroup$ "Doing this takes $O(m^n)$." Are you sure about this? Linear programs can be solved in time that is polynomial in $m$ and $n$ (see, e.g., the ellipsoid algorithm). $\endgroup$ – Gabor Retvari Nov 25 '12 at 23:18

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.