On the right combination of altruism and randomness in the motion of homogeneous distributed autonomous agents

Abstract

We demonstrate the role of randomness and altruism in the motion of artificial agents in a deterministic environment. A swarm of distributed autonomous agents with no possibility of coordination tracks a unique target. The goal is to reach the target as efficiently as possible, i.e. with as few agents moving from their current position as possible. We show in two models how, by adopting features of randomness and altruism in the agent’s motion (which, in our case, translates to yielding to other agents), this objective can be reached. In the first, simplistic representation, agents are dimensionless. The system is formulated as a unidimensional Markov chain, and we show how correctly setting the level of randomness in agents’ movement enables optimization of the swarm total energy expenditure. In the second representation, the agent embodiment raises the question of interference in movement. Again, we show how with no possibility of coordination, and based solely on a partial knowledge of the current system state, it is possible to optimize the swarm movements by dynamically adapting the agent level of randomness in movement. These are but two possible representations of a swarm of simple non-cooperating agents sharing a common target. Yet, they demonstrate how moderating the individual agent attraction to the target by introducing the precise level of randomness in the decision to move or not, helps in breaking ties between agents in their race to the target, and to optimize the overall swarm energy expenditure.

Appendix: Proof of Eq. 4

Let $k$ be the number of agents that decide to make a step. Then, under an assumption of independent decisions, the probability ${q}_{m,i}^{k}$ that $k$ agents decide to move when the system state is $(m,i)$ is

$$q_{m,i}^{k} = \left( {\begin{array}{*{20}c} i \\ k \\ \end{array} } \right)\left( {p_{m,i} } \right)^{k} \left( {1 - p_{m,i} } \right)^{i - k}$$

Assuming $0<k$, let $j$, $0\le j\le k$ be the number of agents that survived after the step to sector of $m-1$, and denote by ${r}_{k,j}^{m-1}$ the probability that among $k$ agents that decided to move from sector $m$ to $m-1$, exactly $j$ agents "survived" (i.e., did not collide).

In order to calculate probability ${r}_{k,j}^{m-1}$, let us associate the segments of the current sector with an alphabet of size $m-1$ and the group of agents with a word of $k$ letters from this alphabet. Each position in the word (agent) receives a letter (segment assignment). In this formulation, ${r}_{k,j}^{m-1}$ is the probability for a word of $k$ letters from an alphabet of size $m-1$ to include exactly $j$ unique letters (repetitions mean that two agents are assigned to the same segment and, therefore, collide).

Denote by $f\left(a,b\right)$ the number of possibilities of combining words of length $a$ from an alphabet of $b$ letters such that the letters in the words appear either zero or strictly more than one time each. Let ${E}_{i}$ be the event that the $i$ th letter among $b$ available ones appears exactly once in the word of length $a$ and consequently ${\overline{E} }_{i}$ be the event that the $i$ th letter either does not appear in the word or appears strictly more than one time. We have:

$$\begin{aligned} & f\left( {a,b} \right) = \left| { \cap_{i} \overline{E}_{i} } \right| = b^{a} - \left| { \cup_{i} E_{i} } \right| \\ & \quad = b^{a} - ab\left( {b - 1} \right)^{a - 1} + \left( {\begin{array}{*{20}c} b \\ 2 \\ \end{array} } \right)a\left( {a - 1} \right)\left( {b - 2} \right)^{a - 2} - \left( {\begin{array}{*{20}c} b \\ 3 \\ \end{array} } \right)a\left( {a - 1} \right)\left( {a - 2} \right)\left( {b - 3} \right)^{a - 3} \\ & \quad = \mathop \sum \limits_{i = 0}^{a} \left( { - 1} \right)^{i} \left( {\begin{array}{*{20}c} b \\ i \\ \end{array} } \right)\frac{a!}{{\left( {a - i} \right)!}}\left( {b - i} \right)^{a - i} \\ \end{aligned}$$

Now consider the number of words with the required property. There are $\left(\begin{array}{c}b\\ j\end{array}\right)$ possibilities of choosing $k$ letters that appear only one time, $\left(\begin{array}{c}a\\ j\end{array}\right)$ possibilities of choosing the spaces in word that are occupied by these letters, and $j!$ possible orders of the letters’ appearance. The remaining $a-j$ spaces are filled by the remaining $b-j$ unused letters, which either do not appear in the word or appear strictly more than one time. Then, the number of required words is $\left(\begin{array}{c}b\\ j\end{array}\right)\left(\begin{array}{c}a\\ j\end{array}\right)j!f(a-j, b-j)$.

Note that the number of all possible words of the length $a$ over the alphabet of the size $b$ is ${b}^{a}$. Then, the probability of obtaining the word with the required property is $\left(\begin{array}{c}b\\ j\end{array}\right)\left(\begin{array}{c}a\\ j\end{array}\right)j!f(a-j, b-j)/{b}^{a}$.

Recall that according to the proposed formulation, the ${r}_{k,j}^{m-1}$ is equal to the probability of obtaining the word of length $k$ with exactly $j$ non-repeating letters from a $(m-1)$-alphabet.

$$r_{k,j}^{m - 1} = \frac{{\left( {\begin{array}{*{20}c} {m - 1} \\ j \\ \end{array} } \right)\left( {\begin{array}{*{20}c} k \\ j \\ \end{array} } \right)j!f\left( {k - j,m - 1 - j} \right)}}{{\left( {m - 1} \right)^{k} }}$$

Q.E.D.