Information-Seeking Control Under Visibility-Based Uncertainty

Abstract

We address the problem of exploring an unknown environment using a range sensor. In the presence of opaque objects of arbitrary shape and topology, exploration is driven by minimization of residual uncertainty, that is primarily due to lack of visibility. The resulting optimal control is approximated with a computationally efficient heuristic, that is proven to complete exploration of a compact environment in a finite number of steps.

Appendix: Proofs

Proof

Fix some x,y∈D. If y=x then ϕ(y,x)=ψ(x). Assume x≠y. The visibility ϕ(y,x) is given by:

$$ \phi(y,x) = \min_{t \in [0,1]} \psi(x+t(y-x)) $$

(7.1)

For t∈[0,1] define f(t)=ψ(x+t(y−x)). Since A is invertible, ψ is strictly convex on D and f is strictly convex on [0,1]. Therefore f reaches its minimum in t ₀∈(0,1) if and only if we have f′(t ₀)=0 (first order optimality condition). Suppose this is the case, then since f′(t)=2〈x−x ⁰+tA(y−x),A(y−x)〉, t ₀ is necessarily given by

$$ t_0 = \frac{\langle A(x^0 - x), A(y - x) \rangle }{\| A(y-x)\|^2} $$

(7.2)

• If $y \notin K_{x}^{1} \cup K_{x}^{2}$, then 〈A(x ⁰−x),A(y−x)〉>0 and 〈A(x ⁰−x),A(y−x)〉<〈A(y−x),A(y−x)〉 so t ₀∈(0,1) is a minimum of f. We then have $\phi(y,x) = \psi ( x+\frac{ \langle A(x^{0} - x),A(y - x) \rangle}{\| A(y-x)\|^{2}} (y - x) )$. If the condition $y \notin K_{x}^{1} \cup K_{x}^{2}$ is not satisfied, then ϕ(y,x)=min{ψ(x),ψ(y)}.

• if $x \in K_{x}^{1}$ then ϕ(y,x)=min{f(0),f(1)}=f(0)=ψ(x).

• if $y \in K_{x}^{2}$ then ϕ(y,x)=min{f(0),f(1)}=f(1)=ψ(y).

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Proof

It is clear that if $y \in \mathring{K_{x}^{i}}$ for any i=1,2,3, then when x moves within some small ball y still belongs to $\mathring{K_{x}^{i}}$. As a result, using that

$$ \phi(y,x) = \left \{ \begin{array}{l@{\quad}l} \psi(x) &\mbox{ if } y \in \mathring{K_x^1} \\ \psi(y) &\mbox{ if } y \in \mathring{K_x^2} \end{array} \right. $$

(7.3)

we obtain immediately that

$$ \nabla_x \phi(y,x) = \left \{ \begin{array}{l@{\quad}l} 2A^T A(x - x^0) &\mbox{ if } y \in \mathring{K_x^1} \\ 0 &\mbox{ if } y \in \mathring{K_x^2} \end{array} \right. $$

(7.4)

For the rest of the proof assume that $y \in \mathring{K_{x}^{3}}$, so that

$$\phi(y,x) = \psi(x+t_0 (y-x))\quad t_0 = \frac{\langle A( x - x^0) ,A( x - y) \rangle}{\| A(x-y) \|^2}$$

The gradient of x→t ₀(x,y) is

(7.5)

(7.6)

Now, define f(x,y)=x+t ₀(y−x). The derivative of f in x in the h direction is:

$$ d f_x (h) = h + \nabla_x t_0 ^T h (y-x) + t_0 (-h) $$

(7.7)

which can be written as:

$$ df_x = 1-t_0 + (y-x) \nabla_x t_0 ^T $$

(7.8)

Using next the fact that ϕ(y,x)=ψ(f(x,y)), we obtain that:

(7.9)

(7.10)

(7.11)

(7.12)

So

(7.13)

that we rewrite

(7.14)

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Proof

Consider two paths $\gamma_{1}, \gamma_{2} \in \mathcal{P}$. For any t∈[0,T], we have:

(7.15)

(7.16)

(7.17)

(7.18)

So by maximizing both sides over t:

(7.19)

(7.20)

Similary,

$$ \phi(y,\gamma_1) \leq \phi(y,\gamma_2) +K \| \gamma_2 - \gamma_1 \|_{\infty} $$

(7.21)

So |ϕ(y,γ ₂)−ϕ(y,γ ₁)|≤K∥γ ₂−γ ₁∥_∞ for any γ ₁,γ ₂∈D. □