4.3.2 Explicitly Modeling $ {\cal C}_{obs}$: The Translational Case

It is important to understand how to construct a representation of $ {\cal C}_{obs}$. In some algorithms, especially the combinatorial methods of Chapter 6, this represents an important first step to solving the problem. In other algorithms, especially the sampling-based planning algorithms of Chapter 5, it helps to understand why such constructions are avoided due to their complexity.

The simplest case for characterizing $ {\cal C}_{obs}$ is when $ {\cal C}= {\mathbb{R}}^n$ for $ n = 1$, $ 2$, and $ 3$, and the robot is a rigid body that is restricted to translation only. Under these conditions, $ {\cal C}_{obs}$ can be expressed as a type of convolution. For any two sets $ X, Y \subset
{\mathbb{R}}^n$, let their Minkowski difference4.10 be defined as

$\displaystyle X \ominus Y = \{x-y \in {\mathbb{R}}^n \;\vert\; x \in X$    and $\displaystyle y \in Y \},$ (4.37)

in which $ x-y$ is just vector subtraction on $ {\mathbb{R}}^n$. The Minkowski difference between $ X$ and $ Y$ can also be considered as the Minkowski sum of $ X$ and $ -Y$. The Minkowski sum $ \oplus$ is obtained by simply adding elements of $ X$ and $ Y$ in (4.37), as opposed to subtracting them. The set $ -Y$ is obtained by replacing each $ y \in Y$ by $ -y$.

In terms of the Minkowski difference, $ {\cal C}_{obs}= {\cal O}\ominus {\cal A}(0)$. To see this, it is helpful to consider a one-dimensional example.

Example 4..13 (One-Dimensional C-Space Obstacle)   In Figure 4.12, both the robot $ {\cal A}= [-1,2]$ and obstacle region $ {\cal O}= [0,4]$ are intervals in a one-dimensional world, $ {\cal W}=
{\mathbb{R}}$. The negation, $ -{\cal A}$, of the robot is shown as the interval $ [-2,1]$. Finally, by applying the Minkowski sum to $ {\cal O}$ and $ -{\cal A}$, the C-space obstacle, $ {\cal C}_{obs}= [-2,5]$, is obtained. $ \blacksquare$

Figure 4.12: A one-dimensional C-space obstacle.
\begin{figure}\centerline{\psfig{file=figs/1dobst.eps,width=4.25in}}\end{figure}
The Minkowski difference is often considered as a convolution. It can even be defined to appear the same as studied in differential equations and system theory. For a one-dimensional example, let $ f
: {\mathbb{R}}\rightarrow \{0,1\}$ be a function such that $ f(x) = 1$ if and only if $ x \in {\cal O}$. Similarly, let $ g : {\mathbb{R}}\rightarrow \{0,1\}$ be a function such that $ g(x) = 1$ if and only if $ x \in {\cal A}$. The convolution

$\displaystyle h(x) = \int_{-\infty}^{\infty} f(\tau) g(x - \tau) d\tau ,$ (4.38)

yields a function $ h$, for which $ h(x)>0$ if $ x \in \operatorname{int}({\cal C}_{obs})$, and $ h(x)
= 0$ otherwise.



Subsections
Steven M LaValle 2020-08-14