Continuous Time Gathering of Agents with Limited Visibility and Bearing-only Sensing

Abstract

A group of mobile agents, identical, anonymous, and oblivious (memoryless), having the capacity to sense only the relative direction (bearing) to neighboring agents within a finite visibility range, are shown to gather to a meeting point in finite time by applying a very simple rule of motion. The agents’ rule of motion is: set your velocity vector to be the sum of the two unit vectors in $\mathbb {R}^2$ pointing to your “extremal” neighbours determining the smallest visibility disc sector in which all your visible neighbors reside, provided it spans an angle smaller than $\pi$, otherwise, since you are “surrounded” by visible neighbors, simply stay put. Of course, the initial constellation of agents must have a visibility graph that is connected, and provided this we prove that the agents gather to a common meeting point in finite time, while the distances between agents that initially see each other monotonically decreases.

Appendix 1: Proof of Lemma 1

We shall first prove the following facts:

Fact a

Let $0 \le a \le b \le \pi$ and $0 \le a + b \le \pi$. Then we have

$$\begin{aligned} \begin{array}{rl} \sqrt{2(1+\cos (a+b))} =&{} 2\cos \left( \frac{a+b}{2}\right) \ge \cos (a) + \cos (b) \ge \\ &{}2 \cos ^2\left( \frac{a+b}{2}\right) = 1 + \cos (a+b) \end{array} \end{aligned}$$

Proof

The function cosine is decreasing in $[0,\pi ]$, and given that $\frac{a+b}{2} \ge \frac{b-a}{2}$:

$$\begin{aligned} 1 \ge \cos \left( \frac{b-a}{2}\right) \ge \cos \left( \frac{a+b}{2}\right) \end{aligned}$$

multiplying by $2\cos \left( \frac{a+b}{2}\right) \ge 0$:

$$\begin{aligned} \begin{array}{lllll} 2\cos \left( \frac{a+b}{2}\right) &{} \ge &{} 2\cos \left( \frac{a+b}{2}\right) \cos \left( \frac{b-a}{2}\right) &{} \ge &{} 2 \cos ^2\left( \frac{a+b}{2}\right) \\ 2\cos \left( \frac{a+b}{2}\right) &{} \ge &{} \cos (a) + \cos (b) &{} \ge &{} 1 + \cos (a+b) \end{array} \end{aligned}$$

A direct consequence is the following fact.

Fact b

Let $0 \le a,b \le \pi$. Then

$$\begin{aligned} \cos (a) + \cos (b) \ge \left\{ \begin{array}{lr} 1 + \cos (a+b) &{}: a+b \le \pi \\ 2\cos \left( \frac{a+b}{2}\right) &{}: a+b \ge \pi \end{array} \right. \end{aligned}$$

Proof

The first line is already part of Fact a. The second line can be proven by using the left inequality of Fact a with $\pi - a$ and $\pi -b$, noticing that $0 \le \pi - a \le \pi$, $0 \le \pi - b \le \pi$, and $\pi - a + \pi - b \le \pi$ for $a+b \ge \pi$.

Now we can prove Lemma 1. Suppose any given initial configuration of the polygon with interior angles $0 \le x_1,\ldots ,x_n \le \pi$. We then have $x_1 + \ldots + x_n = (n-2)\pi$.

Now repeat the following step: Go through all the pairs of non-zero values ($x_i,x_j$). As long as there is still a pair verifying $x_i + x_j \le \pi$, transform it from $(x_i,x_j)$ to $(0,x_i+x_j)$. When there are no such pairs, then among all the non-zero values, take the minimum value and the maximum value, say $x_i$ and $x_j$ (they must verify $x_i + x_j \ge \pi$ due to the previously applied process), and transform the pair from $(x_i,x_j)$ to $\left( \frac{x_i+x_j}{2},\frac{x_i+x_j}{2}\right)$.

Repeat the above process until convergence. We prove that the process converges and that we can get as close as desired to a configuration where all non-zero values are equal. Note that after each step, the sum of the values equals $(n-2)\pi$, and that the values of all $x_i$’s remain between 0 and $\pi$.

The number of values that the above process set to zero must be less or equal to 2 in order to have the sum of the n positive values equal to $(n-2)\pi$. Therefore it is guaranteed that after a finite number of iterations, there will be no pairs of nonzero values whose sum is less than $\pi$ (otherwise this would allow us to add a zero value without changing the sum).

Once in this situation, all we do is replacing pairs of “farthest” non-zero values $(x_i,x_j)$ with the pair $\left( \frac{x_i+x_j}{2},\frac{x_i+x_j}{2}\right)$. Let us show that all the nonzero values converge to the same value, specifically to their average.

Let k be the number of remaining non-zero values after the iteration $t_0$ which sets the “last value” to zero. Denote these values at the i-th iteration by $({x_1^{(i)}, \ldots , x_k^{(i)}})$. Define:

$$\begin{aligned} m = \frac{x_1^{(i)} + \ldots + x_k^{(i)}}{k} = \frac{(n-2)\pi }{k} \end{aligned}$$

$$\begin{aligned} E_i = (x_1^{(i)} - m)^2 + \ldots + (x_k^{(i)} - m)^2 \end{aligned}$$

Without loss of generality, suppose that at the i-th iteration the extreme values were $x_1$ and $x_2$ and so we transformed $(x_1^{(i)},x_2^{(i)})$ into $\left( x_1^{(i+1)} = \frac{x_1^{(i)}+x_2^{(i)}}{2},\right.$ $\left. x_2^{(i+1)} = \frac{x_1^{(i)}+x_2^{(i)}}{2}\right)$. So we have:

$$\begin{aligned} \begin{array}{ccc} E_{i+1} - E_i &{} = &{} 2(\frac{x_1^{(i)} + x_2^{(i)}}{2} - m)^2 - (x_1^{(i)} - m)^2 - (x_2^{(i)} - m)^2) \\ &{} = &{} -\frac{1}{2}(x_1^{(i)} - x_2^{(i)})^2 \end{array} \end{aligned}$$

But $x_1^{(i)}$ and $x_2^{(i)}$ being the extreme values, we have for any $1 \le l \le k$:

$$\begin{aligned} (x_1^{(i)} - x_2^{(i)})^2 \ge (x_l^{(i)} - m)^2 \end{aligned}$$

and by summing over l we get that:

$$\begin{aligned} k (x_1^{(i)} - x_2^{(i)})^2 \ge E_i \end{aligned}$$

Hence

$$\begin{aligned} \begin{array}{ccccc} &{} &{} E_{i+1} - E_i &{} = &{} -\frac{1}{2}(x_1^{(i)} - x_2^{(i)})^2 \le -\frac{E_i}{2k} \\ &{} &{} E_{i+1} &{} \le &{} \left( 1 - \frac{1}{2k}\right) E_i \\ 0 &{} \le &{} E_i &{} \le &{} \left( 1 - \frac{1}{2k}\right) ^{i-t_0} E_{t_0} \end{array} \end{aligned}$$

proving that $E_i$ converges to zero, i.e. all the non-zero values converge to m.

At each step of the above described process, according to Fact b, the sum of cosines can only decrease. Therefore from any given configuration we can get as close as possible to a configuration in which all non-zero values are equal, without increasing the sum of the cosines. Hence, the minimum value must be reached in a configuration in which all non-zero values are equal.

Since there can be at most only two zero values, the minimum value of the sum of the cosines is the minimum of the following:

• $2 + (n-2)\cos \left( \frac{(n-2)\pi }{n-2}\right) = -(n-4)$ (case with 2 zeros)

• $1 + (n-1)\cos \left( \frac{(n-2)\pi }{n-1}\right)$ (case with 1 zero)

• $n\cos \left( \frac{(n-2)\pi }{n}\right)$ (case with no zero)

An analytical comparison of these values depending on n leads to the result stated in Lemma 1.