Optimal solution of the Generalized Dubins Interval Problem: finding the shortest curvature-constrained path through a set of regions

Abstract

The Generalized Dubins Interval Problem (GDIP) stands to determine the minimal length path connecting two disk-shaped regions where the departure and terminal headings of Dubins vehicle are within the specified angle intervals. The GDIP is a generalization of the existing point-to-point planning problem for Dubins vehicle with a single heading angle per particular location that can be solved optimally using closed-form expression. For the GDIP, both the heading angles and locations need to be chosen from continuous sets which makes the problem challenging because of infinite possibilities how to connect the regions by Dubins path. We provide the optimal solution of the introduced GDIP based on detailed problem analysis. Moreover, we propose to employ the GDIP to provide the first tight lower bound for the Dubins Touring Regions Problem which stands to find the shortest curvature-constrained path through a set of regions in the prescribed order.

Appendix

The Dubins Interval Problem (DIP) was initially proposed and solved by Manyam et al. (2017), and the authors provided a list of all possible optimal solutions that are summarized in Table 1. The authors considered originally $\text {R}_{\psi }\text {L}_{\psi }$ and $\text {L}_{\psi }\text {R}_{\psi }$ maneuver types to be the candidates to the optimal solution of the DIP, but claim here that these types are not local minima. Although it does not affect the enumeration of possible cases, we consider it important because we can exclude these two types in the proposed optimal solution of the GDIP. Therefore, a formal proof of the following lemma is provided.

Lemma 11

A maneuver of the ${\text {C}_\psi \overline{\text {C}}_\psi }$ type with two equally long turns cannot be an optimal solution of the DIP if heading angles $\theta _1, \theta _2$ remains unbounded, i.e., $\theta _1 \in \varTheta _1 {\setminus } \{ \theta _1^{\text {min}}, \theta _1^{\text {max}} \}$, $\theta _2 \in \varTheta _2 {\setminus } \{ \theta _2^{\text {min}}, \theta _2^{\text {max}} \}$ .

Proof

Let us consider an $\text {R}_{\psi }\text {L}_{\psi }$ maneuver with two equally long turns with origins $o_1$ and $o_2$ and corresponding turn angles $\alpha ,\beta \in (\pi , 2\pi )$, where w.l.o.g., we assume the minimum turning radius $\rho =1$ for simplicity and better readability. The distance between the maneuver endpoints is denoted $l = \Vert p_1 - p_2 \Vert$ and both $\theta _1$ and $\theta _2$ are not bounded by $\varTheta _1$ and $\varTheta _2$, respectively, see Fig. 16.

Fig. 16

$\text {R}_\psi \overline{\text {L}}_\psi$ maneuver of the $\text {C}_\psi \overline{\text {C}}_\psi$ type as a candidate solution for the DIP

To prove the ${\text {R}_\psi \overline{\text {L}}_\psi }$ maneuver is not a candidate solution, the problem is considered as a constrained optimization of the trajectory length

$$\begin{aligned} f(\alpha , \beta ) = {\mathcal {L}}_{\text {C}_\psi \overline{\text {C}}_\psi } = \alpha + \beta . \end{aligned}$$

(36)

The geometrical constraint is constructed for the distance between the endpoints (see Fig. 16) such that

$$\begin{aligned} l^2 = (2-\cos \alpha -\cos \beta )^2 + (\sin \alpha + \sin \beta )^2. \end{aligned}$$

(37)

The constraint is encoded into a function g(a, b) that equals to zero, i.e., $g(\alpha ,\beta )=0$, that can be expressed as

$$\begin{aligned} \begin{aligned} g(\alpha ,\beta )&= 3 - \frac{l^2}{2} - 2(\cos \alpha +\cos \beta ) + \cos \alpha \cos \beta \\&\quad + \sin \alpha \sin \beta . \end{aligned} \end{aligned}$$

(38)

Local extremes can be identified using Lagrangian defined by the functions f, g, and the Lagrange multiplier $\lambda$ (Bertsekas 2014)

$$\begin{aligned} L(\alpha ,\beta ,\lambda ) = f(\alpha ,\beta ) + \lambda g(\alpha ,\beta ). \end{aligned}$$

(39)

The necessary condition for a critical point $\nabla _{\alpha , \beta ,\lambda } L = 0$ holds for the case $\alpha = \beta$, but the point can be a local minimum, local maximum, or a saddle point. The second partial derivative test is utilized to distinguish these three cases. First, Lagrange multiplier $\lambda$ is determined based on

$$\begin{aligned} \lambda = -\frac{\partial f}{\partial \alpha } \left( \dfrac{\partial g}{\partial \alpha }\right) ^{-1} = -\frac{\partial f}{\partial \beta } \left( \dfrac{\partial g}{\partial \beta }\right) ^{-1}, \end{aligned}$$

(40)

and its value for the specific case when $\alpha =\beta$ is

$$\begin{aligned} \left. \lambda \right| _{\alpha =\beta } = \frac{-1}{2\sin \alpha }. \end{aligned}$$

(41)

The second partial derivative test is based on the Hessian ${\tilde{H}}$ of the Lagrangian function, also called bordered Hessian in the literature

$$\begin{aligned} {\tilde{H}} = \nabla ^2_{\alpha , \beta ,\lambda } L = \left[ \begin{array}{ccc} \frac{\partial ^2 L}{\partial a^2} &{}\quad \frac{\partial ^2 L}{\partial a \partial b} &{}\quad \frac{\partial g}{\partial a} \\ \frac{\partial ^2 L}{\partial b \partial a} &{}\quad \frac{\partial ^2 L}{\partial b^2} &{}\quad \frac{\partial g}{\partial b} \\ \frac{\partial g}{\partial a} &{}\quad \frac{\partial g}{\partial b}&{}\quad 0 \\ \end{array} \right] . \end{aligned}$$

(42)

The elements of bordered Hessians for the case $\alpha = \beta$ are

$$\begin{aligned}&\left. \frac{\partial ^2 L}{\partial \alpha ^2} \right| _{\alpha =\beta } = \left. \frac{\partial ^2 L}{\partial \beta ^2} \right| _{\alpha =\beta } = \lambda (2\cos \alpha - 1), \end{aligned}$$

(43)

$$\begin{aligned}&\left. \frac{\partial ^2 L}{\partial \alpha \partial \beta } \right| _{\alpha =\beta } = \lambda , \end{aligned}$$

(44)

$$\begin{aligned}&\left. \frac{\partial g}{\partial \alpha } \right| _{\alpha =\beta } = \left. \frac{\partial g}{\partial \beta } \right| _{\alpha =\beta } = 2\sin \alpha . \end{aligned}$$

(45)

The second partial derivative test states that a point is a local maximum of function $f(\alpha , \beta )$ alongside $g(\alpha , \beta ) = 0$ if $(-1)^k\det ({\tilde{H}}) < 0$, where $k=1$ stands for the number of constraints. The determinant for $\alpha = \beta$

$$\begin{aligned} \det \left( \left. {\tilde{H}} \right| _{\alpha =\beta } \right) = 16 \lambda \overbrace{(1-\cos \alpha )}^{\ge 0} \overbrace{(\sin \alpha )^2}^{\ge 0} \end{aligned}$$

(46)

is positive for $\alpha \in (\pi , 2\pi )$ based on (41). Therefore, the case $\alpha = \beta$ is a local maximum and the trajectory length can be shorten by a changing $\theta _1 \in \varTheta _1$ and $\theta _2 \in \varTheta _2$ angles if the angles are not bounded by $\varTheta _1$ and $\varTheta _2$. The proof for $\text {L}_{\psi }\text {R}_{\psi }$ is analogous. $\square$

Since ${\text {C}_\psi \overline{\text {C}}_\psi }$ maneuver type can be optimal only if at least one of the angles need to be bounded, i.e., $\theta _1 \in \{ \theta _1^{\text {min}}, \theta _1^{\text {max}} \}$, $\theta _2 \in \{ \theta _2^{\text {min}}, \theta _2^{\text {max}} \}$. Therefore, this type can be seen as a particular case of $\text {C}\overline{\text {C}}_{\varvec{\psi }}$ for which both turns are larger than $\pi$, and at least one angle is constrained, see Table 1.