Optimal Gathering by Asynchronous Oblivious Robots in Hypercubes

Abstract

We consider the problem of gathering a set of autonomous, identical, oblivious, asynchronous, mobile robots at a vertex of an anonymous hypercube. The robots operate in Look-Compute-Move cycles. In each cycle, a robot takes a snapshot of the current configuration (Look), then based on the perceived configuration, decides whether to stay idle or to move to an adjacent vertex (Compute), and in the later case makes an instantaneous move accordingly (Move). We have shown that the problem is unsolvable if the robots do not have multiplicity detection capability. With weak multiplicity detection capability, the problem is solvable in an oriented hypercube for any initial configuration of $2k+1 (k > 0)$ number of robots. For $4k (k > 0)$ number of robots, the problem is solvable under the same assumptions if and only if the group of automorphism of the configuration is trivial. Our proposed algorithms are optimal with respect to the total number of moves executed by the robots.

Appendix A Proof of Theorem 7

Consider a configuration $(Q_d, f)$ that has no partitive automorphism. Since we have assumed that the robots are positioned at distinct vertices, there is no distinction between the configuration and the perceived configuration. In other words, we have $\tilde{f} = f$. Assume that the configuration has no fixed Weber point. Let us assume that there is an algorithm $\mathcal {A}$ that deterministically solves LeadingWeberPoint. Let $w_1 \in \mathcal {W}$ be the leading Weber point elected by the robots. Since $w_1$ is not a fixed vertex, there is a $w_2 \ne w_1$ such that $\varphi (w_1) = w_2$, for some $\varphi \in Aut(G, f)$.

Each robot observes the positions of other robots in its local coordinate system. A local coordinate system of a robot is just an assignment $\varPsi : V(Q_d) \longrightarrow \{0,1\}^d$, respecting the rule that $u, v \in V(Q_d)$ are adjacent if and only if $\varPsi (u), \varPsi (v)$ differ in precisely one bit. Since there is no global agreement, the local coordinate system of each robot is arbitrary, and is chosen by the adversary. Let us formally define the view of a robot. The view of a robot is given by the triplet $\mathcal {V}_{\varPsi } = (\varPsi , \tilde{f}, me)$, where $\varPsi : V(Q_d) \longrightarrow \{0,1\}^d$ is the local coordinate system, $\tilde{f}: \{0,1\}^d \longrightarrow \{0,1\}$ is the multiplicity function defined on the set of vertices expressed in local coordinates, and $me \in \{0,1\}^d$ is the coordinates of the vertex on which the robot resides. The view $\mathcal {V}_{\varPsi }$ is the input for algorithm $\mathcal {A}$. The output $\mathcal {A}(\mathcal {V}_{\varPsi }) \in \{0,1\}^d$ is the coordinates of the required leading Weber point, i.e., the returned leading Weber point is the vertex $\varPsi ^{-1}(\mathcal {A}(\mathcal {V}_{\varPsi }))$.

Consider a robot $r_1$ in the configuration residing at vertex $v_1$. The robot $r_1$, using a local coordinate system $\varPsi _1: V(Q_d) \longrightarrow \{0,1\}^d$, elects $w_1$ as the leading Weber point. That is, given the input in the coordinate system $\varPsi _1$, the output of $\mathcal {A}$ is $\varPsi _1(w_1)$. Now consider the following cases.

Case 1: Suppose that $\varphi (v_1) = v_1$. Consider the local coordinate system $\varPsi _2 = \varPsi _1 \,\circ \, \varphi ^{-1}$. Note that the view of $r_1$ in coordinate systems $\varPsi _1$ and $\varPsi _2$ are exactly the same, i.e., $\mathcal {V}_{\varPsi _1} = \mathcal {V}_{\varPsi _2}$. Since $\mathcal {A}$ is a deterministic algorithm, $\mathcal {A}(\mathcal {V}_{\varPsi _1}) = \mathcal {A}(\mathcal {V}_{\varPsi _2})$. Since the elected leading Weber point in local coordinate system $\varPsi _1$ is $w_1$, we have $\mathcal {A}(\mathcal {V}_{\varPsi _1}) = \varPsi _1(w_1)$. So we have,

$$\begin{aligned}&\mathcal {A}(\mathcal {V}_{\varPsi _2}) = \mathcal {A}(\mathcal {V}_{\varPsi _1}) = \varPsi _1(w_1)\\ \Rightarrow \quad&\varPsi _2^{-1}(\mathcal {A}(\mathcal {V}_{\varPsi _2})) = \varPsi _2^{-1}(\varPsi _1(w_1)) = \varphi \circ \varPsi _1^{-1} \circ \varPsi _1(w_1) = \varphi (w_1) = w_2 \end{aligned}$$

Hence, we see that in local coordinate system $\varPsi _1$ the robot $r_1$ elects $w_1$ as the leading Weber point, while in $\varPsi _2$ it elects $w_2$. This is a contradiction.

Case 2: Now assume that $\varphi (v_1) = v_2 \ne v_1$. Then there must be a robot $r_2$ in $v_2$. Suppose that the adversary sets the local coordinate system of $r_2$ as $\varPsi _2 = \varPsi _1 \circ \varphi ^{-1}$. Then the view of $r_1$ and $r_2$ will be identical, i.e., $\mathcal {V}_{\varPsi _1} = \mathcal {V}_{\varPsi _2}$. Again we have, $\mathcal {A}(\mathcal {V}_{\varPsi _2}) = \mathcal {A}(\mathcal {V}_{\varPsi _1}) = \varPsi _1(w_1)$ and hence, $\varPsi _2^{-1}(\mathcal {A}(\mathcal {V}_{\varPsi _2})) = w_2$. Therefore, $r_2$ will elect $w_2$, while $r_1$ elects $w_1$ as the leading Weber point. This is again a contradiction.    $\square$