Boundary Sketching with Asymptotically Optimal Distance and Rotation

Abstract

We address the problem of designing a distributed algorithm for two robots that sketches the boundary of an unknown shape. Critically, we assume a certain amount of delay in how quickly our robots can react to external feedback. In particular, when a robot moves, it commits to move along path of length at least $\lambda$, or turn an amount of radians at least $\lambda$ for some positive $\lambda \le 1/2^6$, that is normalized based on a unit diameter shape. Then, our algorithm outputs a polygon that is an $\epsilon$-sketch, for $\epsilon = 8\sqrt{\lambda }$, in the sense that every point on the shape boundary is within distance $\epsilon$ of the output polygon. Moreover, our costs are asymptotically optimal in two key criteria for the robots: total distance travelled and total amount of rotation.

Additionally, we implement our algorithm, and illustrate its output on some specific shapes.

This work is supported by the National Science Foundation grant IIS 2024520.

Appendices

A Simulation Figures

1.1 A.1 Intersecting Gaussians

Gradient Estimation. To estimate the gradient at a point of intersection, we formulate a least squares error (LSQ) problem and solve it using the pseudo-inverse, as described below.

Least Squares Error Formulation. Assume we have three points a, b, c that are close together, and that we have the values f(a), f(b), f(c). For the sake of the $\textsc {BOUNDARY}\text {-}\textsc {SKETCH}$, these three points can be said to be successive points of a robot’s path with the middle one being where it crosses the shape boundary. Then we estimate the gradient $\nabla f(b) = (x', y')$ as a vector that minimizes the following sum:

$$(f(a) - f(b) - \nabla f(b) \cdot (a-b))^2 + (f(c) - f(b) - \nabla f(b) \cdot (c-b))^2$$

Fig. 5.

$\textsc {BOUNDARY}\text {-}\textsc {SKETCH}$ for $\lambda = 0.000025$, the green border marks the shape boundary sandwiched by the inner and outer robots. (Color figure online)

Setting the rows of a $2\times 2$ matrix A to be $a-b$, $c-b$ respectively and the entries of vector $\beta$ to be $f(a)-f(b), f(c)-f(b)$ respectively, this problem is formally solving for the vector v such that, $|Av - \beta |^2$ is smallest.

Finally, we utilize the pseudo-inverse method for solving LSQ as discussed in numerous sources e.g. in Sect. 4.5 of [25].

Fig. 6.

$\textsc {BOUNDARY}\text {-}\textsc {SKETCH}$ for $\lambda = 0.000001$, the green border marks the shape boundary sandwiched by the inner and outer robots. (Color figure online)

1.2 A.2 Real World Plume Shapes

Fig. 7.

Volcanic plume in La Palma observed on September 20, 2021.

Fig. 8.

$\textsc {BOUNDARY}\text {-}\textsc {SKETCH}$ for $\lambda = 0.000025$, the green border marks the shape boundary sandwiched by the inner and outer robots (red and blue border). (Color figure online)

Fig. 9.

$\textsc {BOUNDARY}\text {-}\textsc {SKETCH}$ for $\lambda = 0.000001$, the green border marks the shape boundary sandwiched by the inner and outer robots (red and blue border). (Color figure online)

B Helper Lemmas

Lemma 17

$\ell$ is $\varOmega (1)$

Proof

Since the diameter of the shape is scaled to be 1, $\ell \ge 1$ and the lemma follows.

Lemma 18

The number of vertices in $\gamma$ is at most $\ell /\sqrt{\lambda }$.

Proof

This is immediate from the assumption that each of the vertices are at least $\sqrt{\lambda }$ distance apart.

Lemma 19

If a curve $\gamma$ is a polygon and $\phi$ is the total rotation of this polygon, then $\ell (\gamma ) \le \phi$.

Proof

Suppose the vertices of the polygon P are given by a list of n points $v_1, v_2, ..., v_n$. Then $\ell = \sum _{i=1}^{n} |v_i - v_{i+1}|$ where we set $v_{n+1} = v_1, v_{n+2} = v_2$.

Now fix three successive vertices, $v_i, v_{i+1}, v_{i+2}$ for $1 \le i \le n$ and denote them A, B, C respectively. In addition, let $a = |AB|, b = |BC|, c = |CA|, \alpha = \angle {BAC}, \beta = \angle {ABC}, \omega = \angle {BCA}$.

By the law of sines,

$$\frac{a}{\sin \alpha } = \frac{b}{\sin \beta } = \frac{c}{\sin \omega } = 2 \rho $$

where $\rho$ is the radius of the circumcircle of the triangle ABC.

Since the polygon is bounded by a unit square, $\rho \le 1$. By the inequality $\sin x \le x$ for all $x \in \mathbb R$, we have,

$$a \le 2\alpha $$

$$b \le 2\beta $$

Hence,

$$a+b \le 2(\alpha + \beta ) = 2(\pi - \omega ) = 2\phi _i$$

where $\phi _i$ is the i-th exterior angle of P.

Summing over all $i \in [1,n]$ we get,

$${2\ell (\gamma ) = \sum _{i=1}^{n} |v_{i} - v_{i+1}}| + |v_{i+1} - v_{i+2}| \le 2\sum _{i=1}^{n} \phi _i = 2\phi $$

This implies,

$$\ell (\gamma ) \le \phi $$

Lemma 20

Let $\gamma : [0,L] \rightarrow \mathbb R^2$ be a regular curve parametrized by its arc length L such that $\gamma (0) \ne \gamma (L)$. Then there exists $c \in (0,L)$ such that $\gamma '(c)$ is parallel to the line segment joining $\gamma (0)$ and $\gamma (L)$.

Proof

Let $\gamma (t) = (x(t), y(t))$ for all $t \in [0,L]$ where x, y are differentiable single valued real functions defined over [0, L].

Let u be the vector from $\gamma (0)$ to $\gamma (L)$. That is,

$$u = \gamma (0) - \gamma (0) = (x(L)-x(0), y(L)-y(0))$$

Let v be a vector perpendicular to $\gamma (L) - \gamma (0)$. Since, $\langle u, v \rangle = 0$, we can write,

$$v = (y(0) - y(L), -x(0) + x(L))$$

Now consider the function f defined as follows over [0, L],

$$f(t) = \gamma (t) \cdot v = x(t) (y(0) - y(L)) + y(t) (x(L) - x(0))$$

Observe that, $f(0) = f(L)$. Hence by Rolle’s theorem there exists $c \in (0,L)$ such that, $f'(c) = 0$. Since, for all $t \in (0,L)$, $f'(t) = \langle \gamma '(t), v \rangle + \langle \gamma (t), v' \rangle = \langle \gamma '(t), v \rangle$, we conclude, $\langle \gamma '(c), v \rangle = 0$.

This means $\gamma '(c)$ is perpendicular to v. Since u is perpendicular to v as well, we conclude that, $\gamma '(c)$ is parallel to u.

C Asymptotic Analysis

Lemma 21

Let $j>1$ be an index such that, after crossing the boundary at point D (Fig. 10), $D_1, D_2$ are both outside or $D_1, D_2$ are both inside. Then the number of times the While loop in Step 7 of Algorithm 2 executes before the robots cross the line DB is at most $\phi _{j-1}/\sqrt{\lambda }+ 1$.

Proof

Define $\psi _j$ to be the change of gradient in radians between $\gamma (t_{j-1})$ and $\gamma (t_j)$. Clearly, $\phi _{j-1} \ge \psi _j$.

The vertical distance from the robot at the beginning of the execution of Algorithm 2 to the line DB is at most $\lambda \sin (\psi _{j} + \sqrt{\lambda })$.

Thus after $\psi _{j} / \sqrt{\lambda }+ 1 \le \phi _{j-1}/\sqrt{\lambda }+ 1$ steps, the robots will cross DB, which concludes the proof.

Lemma 22

Suppose $\gamma$ defines a polygon. Given an instance of crossing the shape at a point D, let $\beta$ be the first exterior angle of the shape continuing from D. If $\beta$ is the only exterior angle of $\gamma$ from D to the next point of crossing and $\sqrt{\lambda }\le \beta \le \pi /8$, then the While loop in Step 7 of Algorithm 2 executes at most $8\beta /\sqrt{\lambda }$ times to cover the distance from B to C.

Fig. 10.

Figure for Lemmas 21 and  22, the red color indicates the shape boundary, blue color indicates path of the nearest robot, black colored segment is a construction for the proof.

Proof

Figure 10 illustrates this lemma where $\angle BAC = \beta$. Note that the radius of the circumcircle of the triangle $\triangle ABC$ is at most the radius $\rho$ of the circumcircle of the regular polygon in the diagram indicated by dotted lines. Given the exterior angle $\sqrt{\lambda }$ and side length $\lambda$ of the regular polygon, the radius of the circumcircle is $\rho = \lambda /2\sin (\sqrt{\lambda })$. By Lemma 6, $\rho \le \sqrt{\lambda }$.

Now by the law of sines, $|BC|/\sin \beta = 2\rho$ and this implies $|BC| \le 2\beta \sqrt{\lambda }$.

Next, the angle $\beta '$ formed by the chord BC with the center of the regular polygon is at most,

$$ 2 \arcsin \left( \frac{2\beta \sin (\sqrt{\lambda })}{ \sqrt{\lambda }} \right) \le 2 \arcsin (2\beta ) \le \frac{4\beta }{\sqrt{1-(2\beta )^2}} \le \frac{4\beta }{\sqrt{1-\pi ^2/16}} \le 8\beta $$

Next we multiply this angle with the radius to get the arc length between B and C,

$$s = \rho \beta ' \le 8\beta \sqrt{\lambda }$$

Finally, noting that the arc length covered by every step of the robots is at least $\lambda$, we get that after $8\beta /\sqrt{\lambda }$ steps from B the robots will cross AC.

Lemma 23

If $\gamma$ is a polygon and if for some $i \le m$, $\sqrt{\lambda }\le \phi _i \le \pi /8$, then $f(i) \le 8\phi _i/\sqrt{\lambda }+ \phi _{i-1}/\sqrt{\lambda }+ 1$.

Proof

The trivial case is where Algorithm 2 is not executed at all i.e. $f(i) = 0$.

Observe that the motion of the robots during the execution of Algorithm 2 forms part of the perimeter of a convex polygon with side lengths and exterior angles being $\lambda$ and $\sqrt{\lambda }$ respectively (except for the first and last sides).

Now suppose there are j vertices of $\gamma$ defined over $[t_i, t_{i+1}]$. Let these vertices be indexed $\gamma (a_k)$ where $a_k \in [t_i, t_{i+1}]$ for all $k \in [1,j]$. Finally, let $\beta _k$ be the exterior angle at $\gamma (a_k)$. By Lemma 21, there will be at most $\phi _{i-1}/\sqrt{\lambda }+ 1$ before the robot crosses the nearest side of the exterior angle at $\gamma (a_1)$.

Next, observe that by Lemma 22 the nearest robot to the shape will cross one of the sides of the exterior angle at $\gamma (a_k)$ by at most $8\beta _k/\sqrt{\lambda }$ iterations of the While loop in Algorithm 2.

In addition, the shape boundary turns either in convex or concave manner. If the turn at an index changes from convex to concave or concave to convex, it may actually move the sides of $\gamma$ closer for the robot and hence the amount the angle $\beta _{k}$ contributes to the overall iterations run inside the While loop of Algorithm 2 is at most $8\beta _k/\sqrt{\lambda }$. Therefore,

$$f(i) \le \sum _{k=1}^{j} 8\beta _{k}/\sqrt{\lambda }+ \phi _{i-1}/\sqrt{\lambda }+ 1= 8 \phi _i / \sqrt{\lambda }+ \phi _{i-1}/\sqrt{\lambda }+ 1$$

D Synchronization Details

Our final two lemmas show that the other robot do not rotate or traverse asymptotically more than the robot nearest to the boundary. In addition, we discuss briefly the synchronization steps in Algorithm 1.

Fig. 11.

Figure for synchronization steps in CROSS-BOUNDARY, green and blue curves indicate robot path, red curve indicates the shape boundary. (Color figure online)

Observe that in Fig. 11, after reorientating itself to the curve gradient direction, the green and blue curves are “closer” to each other. But this can easily be handled by letting the blue or green curve based robot traverse a little longer, in particular greater than $\sqrt{\lambda }- \lambda > \lambda$ distance and then turn. This synchronization that maintains the distance of $\sqrt{\lambda }$ between the two robots guarantees that between successive executions of Algorithm 1, the robots will cover $\varOmega (\sqrt{\lambda })$ distance of the shape boundary, in turn we are able to prove Lemma 12.

Lemma 24

For each execution of Algorithm 2, $D_2$ traverses a distance of $O(\sqrt{\lambda })$.

Fig. 12.

Figure for Lemmas 24 and  25

Proof

In Fig. 12, the line segment AC describes the path of $D_1$ and $D_2$ can move either directly from B to D or B to D via E. Now, by Algorithm 2 description, $|AC| = |DE| = \lambda$. Note that, $\angle BAE = \sqrt{\lambda }$ and $|BA| = |AE| = \sqrt{\lambda }$. Thus by the law of sine, $|EB|/\sin \sqrt{\lambda }= \sqrt{\lambda }/\sin ((\pi -\sqrt{\lambda })/2)$ Hence,

$$|EB| \le \lambda /\cos (\sqrt{\lambda }/2) \le 2\lambda .$$

Thus, the path length of $D_2$ at each step is at most $3\lambda$.

In the last step independent of which robot encountered the shape boundary and whether they are both inside or outside or one inside and one outside, $D_2$ will traverse at most a constant factor of $\sqrt{\lambda }$ by the above discussion based on Fig. 11. Finally, noting that $D_1$ traverses $O(\sqrt{\lambda })$ distance according to Lemma 7, $D_2$ traverses $O(\sqrt{\lambda })$ during each execution of Algorithm 2.

Lemma 25

For each execution of Algorithm 2, $D_2$ does not rotate asymptotically more than $D_1$.

Proof

We only need to check that the rotation of the green path towards BE and then ED is asymptotically bounded by $D_1$’s rotation. For brevity, we simply note this follows by elementary euclidean geometric comparison of various angles, beginning from the rotation of the red path to AC.

Finally, in the last step, both robots turn towards a common direction i.e. gradient at a point of crossing the boundary.