The Volume of a Crosspolytope Truncated by a Halfspace

Abstract

In this paper, we consider the computation of the volume of an n-dimensional crosspolytope truncated by a halfspace. Since a crosspolytope has exponentially many facets, we cannot efficiently compute the volume by dividing the truncated crosspolytope into simplices. We show an $O(n^6)$ time algorithm for the computation of the volume. This makes a contrast to the 0−1 knapsack polytope, whose volume is $\#P$-hard to compute. The paper is interested in the computation of the volume of the truncated crosspolytope because we conjecture the following question may have an affirmative answer: Does the existence of a polynomial time algorithm for the computation of the volume of a polytope K imply the same for K’s geometric dual? We give one example where the answer is yes.

This research was supported by research grant of Information Sciences Institute of Senshu University.

Appendix Supplemental Proof

To make this paper self-contained, we prove the following proposition which is used in the proof of Proposition 3.

Proposition 4

Let $P\subseteq \mathbb {R}^n$ be a convex n-dimensional polytope. Then there exists a set of m simplices $S_1,\dots ,S_m$ satisfying the following three conditions:

• 1. $P=\bigcup _{i=1,\dots ,m} S_i$;

• 2. any vertex of the simplices $S_1,\dots ,S_m$ is a vertex of P;

• 3. $\mathrm {Vol}(S_i\cap S_j)=0$ for any $1\le i < j \le m$.

Proof

The proof is the induction on n. As for the base case, we consider the case $n=1$. In this case, P is always a bounded interval, which is a simplex. Therefore, the proposition holds for the base case.

We proceed to the induction step. We assume that we have the claims of the proposition in case $n=k$. Then, in case $n=k+1$, we have that any facet of P can be divided into a set of k-dimensional simplices, satisfying the three conditions of the claim. Let $S'_1,\dots ,S'_M$ be the k-dimensional simplices obtained by dividing the P’s facets satisfying the three conditions for each facet. Let $\varvec{v}$ be one vertex of P. Then, we obtain the $(k+1)$-dimensional simplices as the convex hulls $S_i=\mathrm {conv}(S'_i\cup \{\varvec{v}\})$ for $i=1,\dots , M$.

As for the first condition of the proposition, we show that for any internal point $\varvec{p}\in P$, there exists a point $\varvec{q}\in S'_i$ for some $1\le i \le M$ such that $\varvec{p}\in \mathrm {conv}(\{\varvec{v},\varvec{q}\})$. We consider a point given by $\varvec{r}(t)=t(\varvec{p}-\varvec{v})+\varvec{v}$, where $t>0$. Since P is bounded, we have that $\varvec{r}(t)$ is on a facet F of P for some $t>1$. We have $t>1$ since $\varvec{p}$ is an internal point of P. Since each facet F can be divided into simplices, $\varvec{q}=\varvec{r}(t)$ is in one of these simplices.

Since the second condition of the claim clearly holds for $S_1,\dots ,S_M$ by definition, we proceed to the proof of the third condition. That is, $\mathrm {Vol}(S_i\cap S_j)=0$ for any $1\le i < j \le M$. Let $\varvec{p}\in S_i\cap S_j$. We consider the point $\varvec{r}(t)=t(\varvec{p}-\varvec{v})+\varvec{v}$ as in the above for $t>1$. Let $t_0$ be the value of t such that $\varvec{r}(t_0)$ is on the surface of P. Since $S_i=\mathrm {conv}(S'_i\cup \{\varvec{v}\})$ and $S_j=\mathrm {conv}(S'_j\cup \{\varvec{v}\})$, we have that $\varvec{r}(t_0)\in S'_i\cap S'_j$. Since the k-dimensional volume of $S'_i\cap S'_j$ is 0 by the assumption, we have that $\mathrm {Vol}(\mathrm {conv}(\{\varvec{v}\}\cup (S'_i\cap S'_j)))=0$, which shows the claim.   $\square$