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.
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Acknowledgements
We acknowledge discussions with Yanina Landa, Ryo Takei. We thank Stanley Osher, Lieven Vandenberghe for feedback and comments.
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Research supported by DARPA under the MSEE Program FA8650-11-1-7156 and ARO MURI W911NF-11-1-0391.
Appendix: Proofs
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:
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
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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)}.
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if \(x \in K_{x}^{1}\) then ϕ(y,x)=min{f(0),f(1)}=f(0)=ψ(x).
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if \(y \in K_{x}^{2}\) then ϕ(y,x)=min{f(0),f(1)}=f(1)=ψ(y).
□
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
we obtain immediately that
For the rest of the proof assume that \(y \in \mathring{K_{x}^{3}}\), so that
The gradient of x→t 0(x,y) is
Now, define f(x,y)=x+t 0(y−x). The derivative of f in x in the h direction is:
which can be written as:
Using next the fact that ϕ(y,x)=ψ(f(x,y)), we obtain that:
So
that we rewrite
□
Proof
Consider two paths \(\gamma_{1}, \gamma_{2} \in \mathcal{P}\). For any t∈[0,T], we have:
So by maximizing both sides over t:
Similary,
So |ϕ(y,γ 2)−ϕ(y,γ 1)|≤K∥γ 2−γ 1∥∞ for any γ 1,γ 2∈D. □
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Valente, L., Tsai, YH.R. & Soatto, S. Information-Seeking Control Under Visibility-Based Uncertainty. J Math Imaging Vis 48, 339–358 (2014). https://doi.org/10.1007/s10851-013-0423-x
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DOI: https://doi.org/10.1007/s10851-013-0423-x