A Fast and Practical Method to Estimate Volumes of Convex Polytopes

Abstract

The volume is an important attribute of a convex body. In general, it is quite difficult to calculate the exact volume. But in many cases, it suffices to have an approximate value. Volume estimation methods for convex bodies have been extensively studied in theory, however, there is still a lack of practical implementations of such methods. In this paper, we present an efficient method which is based on the Multiphase Monte-Carlo algorithm to estimate volumes of convex polytopes. It uses the coordinate directions hit-and-run method, and employs a technique of reutilizing sample points. We also introduce a new result checking method for performance evaluation. The experiments show that our method can efficiently handle instances with dozens of dimensions with high accuracy.

Appendices

Appendix

A About the Number of Sample Points

From Formula (1),

$$\begin{aligned} \frac{vol(P)}{vol(B(0,1))} = \prod _{i=0}^{l-1} \alpha _i = \prod _{i=0}^{l-1} \frac{step\_size}{c_i} = \frac{step\_size^l}{\prod _{i=0}^{l-1} c_i}, \end{aligned}$$

which shows that to obtain confidence interval of $vol(P)$, we only have to focus on $\prod _{i=0}^{l-1} c_i$. For a fixed $P$, $\{\alpha _i\}$ are fixed numbers. Let $c = \prod _{i=1}^{l}c_i$ and $\mathbb {D}(l, P)$ denote the distribution of $c$. With statistical results of substantial expriments on concentric balls, we observe that, when $step\_size$ is sufficiently large, the distribution of $c_i$ is unbiased and its standard deviation is smaller than twice of the standard deviation of binomial distribution in dimensions below 80. Though such observation sometimes not holds when we sample on convex bodies other than balls, we still use this to approximate the distribution of $c_i$. Consider random variables $X_i$ following binomial distribution $\mathbb {B}(step\_size, 1 / \alpha _i)$, we have

$$\begin{aligned} E(c) = E(c_1) \dots E(c_l) = E(X_1) \dots E(X_l) = step\_size^l \prod _{i=1}^{l}\frac{1}{\alpha _i}, \end{aligned}$$

$$\begin{aligned} D(c)= & {} E((c_1 \dots c_l)^2) - E(c)^2 = \prod _{i=1}^{l}(D(c_i)+E(c_i)^2) - E(c)^2\\= & {} \prod _{i=1}^{l}(4D(X_i)+E(X_i)^2) - E(c)^2\\= & {} \prod _{i=1}^{l}\frac{step\_size^2}{\alpha _i^2}(1+\frac{4\alpha _i}{step\_size}(1-\frac{1}{\alpha _i})) - E(c)^2\\= & {} E(c)^2(\beta - 1), \end{aligned}$$

where $\beta = \prod _{i=1}^{l}(1 + \frac{4\alpha _i}{step\_size} - \frac{4}{step\_size})$.

Suppose $\{\xi _1, \dots , \xi _{t}\}$ is a sequence of i.i.d. random variables following $\mathbb {D}(l, P)$. Notice $D(c)$, the variance of $\mathbb {D}(l, P)$, is finite because $\beta - 1 \rightarrow 0$ as $t\rightarrow \infty$. According to central limit theorem, we have

$$\begin{aligned} \frac{\sum _{i=1}^{t}\xi _i - tE(c)}{\sqrt{t} D(c)} \mathop {\rightarrow }\limits ^{d} N(0, 1). \end{aligned}$$

So we obtain the approximation of $95\,\%$ confidence interval of $c$, $[E(c) - \sigma \sqrt{D(c)}, E(c) + \sigma \sqrt{D(c)}]$, where $\sigma = 1.96$. And

$$\begin{aligned} Pr(\frac{vol(B(0, 1))step\_size^l}{E(c) + \sigma \sqrt{D(c)}} \le \overline{vol(P)} \le \frac{vol(B(0, 1))step\_size^l}{E(c) - \sigma \sqrt{D(c)}}) \approx 0.95. \end{aligned}$$

Let $\epsilon \in [0, 1]$ denote the ratio of confidence interval’s range to exact value of $vol(P)$, that is

$$\begin{aligned}&\frac{vol(B(0, 1))step\_size^l}{E(c) + \sigma \sqrt{D(c)}} - \frac{vol(B(0, 1))step\_size^l}{E(c) - \sigma \sqrt{D(c)}} \le vol(P) \cdot \epsilon \end{aligned}$$

(5)

$$\begin{aligned}\Longleftrightarrow & {} \frac{1}{E(c) - \sigma \sqrt{D(c)}} - \frac{1}{E(c) + \sigma \sqrt{D(c)}} \le \frac{\epsilon }{E(c)}\end{aligned}$$

(6)

$$\begin{aligned}\Longleftrightarrow & {} \frac{1}{1 - \sigma \sqrt{\beta - 1}} - \frac{1}{1 + \sigma \sqrt{\beta - 1}} \le \epsilon \end{aligned}$$

(7)

$$\begin{aligned}\Longleftrightarrow & {} 4\sigma ^2(\beta - 1) \le \epsilon ^2 (1 + \sigma ^2 - \sigma ^2\beta )^2\end{aligned}$$

(8)

$$\begin{aligned}\Longleftrightarrow & {} \epsilon ^2\sigma ^2\beta ^2 - 2\epsilon ^2(1+\sigma ^2)\beta - 4\beta + (\frac{1}{\sigma }+\sigma )^2 + 4 \ge 0. \end{aligned}$$

(9)

Solve inequality (9), we get $\beta _1(\epsilon , \sigma )$, $\beta _2(\epsilon , \sigma )$ that $\beta \le \beta _1$ and $\beta \ge \beta _2$ (ignore $\beta \ge \beta _2$ because $1 - \sigma \sqrt{\beta _2-1} < 0$). $\beta \le (1 + \frac{4}{step\_size})^l$, since $1 \le \alpha _i \le 2$.

$$\begin{aligned} (1 + \frac{4}{step\_size})^l \le \beta _1 \Longleftrightarrow step\_size \ge \frac{4}{\beta _1^{1 / l} - 1}, \end{aligned}$$

(10)

(10) is a sufficient condition of $\beta \le \beta _1$. Furthermore, $4 / (l\beta _1^{1 / l} - l)$ is nearly a constant as $\epsilon$ and $\sigma$ are fixed. For example, $4 / (l\beta _1^{1 / l} - l) \approx 1569.2 \le 1600$ when $\epsilon = 0.2$, $\sigma = 1.96$. So $step\_size = 1600l$ keeps the range of $95\,\%$ confidence interval of $vol(P)$ less than $20\,\%$ of the exact value of $vol(P)$.