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Sniffing Helps to Meet: Deterministic Rendezvous of Anonymous Agents in the Grid
Distributed, Parallel, and Cluster Computing · Submitted 20 Jul 2024
abstract compiled · 2026-09-23 · no language-model review
The authors' main claim The above negative result shows a separation between the optimal complexity in the two easier scenarios and the optimal complexity in the most difficult scenario.
The compiled record
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The abstract, sentence by sentence
Two identical anonymous mobile agents have to meet at a node of the infinite oriented grid whose nodes are unlabeled. This problem is known as rendezvous. The agents execute the same deterministic algorithm. Time is divided into rounds, and in each round each agent can either stay idle at the current node or move to an adjacent node. An adversary places the agents at two nodes of the grid at a distance at most $D$, and wakes them up in possibly different rounds. Each agent starts executing the algorithm in its wakeup round. If agents cannot leave any marks on visited nodes then they can never meet, even if they start simultaneously at adjacent nodes and know it. Hence, we assume that each agent marks any unmarked node it visits, and that an agent can distinguish if a node it visits has been previously marked or not. The time of a rendezvous algorithm is the number of rounds between the wakeup of the later agent and rendezvous. We ask the question whether the capability of marking nodes enables the agents to meet, and if so, what is the fastest rendezvous algorithm. We consider this problem under three scenarios. First, agents know $D$ but may start with arbitrary delay. Second, they start simultaneously but do not have any {\em a priori} knowledge. Third, most difficult scenario, we do not make any of the above facilitating assumptions. Agents start with arbitrary delay and they do not have any a priori knowledge. We prove that in the first two scenarios rendezvous can be accomplished in time $O(D)$. This is clearly optimal. For the third scenario, we prove that there does not exist any rendezvous algorithm working in time $o(D^{\sqrt{2}})$, and we show an algorithm working in time $O(D^2)$. The above negative result shows a separation between the optimal complexity in the two easier scenarios and the optimal complexity in the most difficult scenario.
background contribution method result interpretation main claim
| # | sentence | role | the authors' own work | states the main conclusion | strength |
|---|---|---|---|---|---|
| s0 | Two identical anonymous mobile agents have to meet at a node of the infinite oriented grid whose nodes are unlabeled. | background 1.00 | 0.08 | 0.02 | asserted 0.95 |
| s1 | This problem is known as rendezvous. | background 1.00 | 0.04 | 0.01 | asserted 1.00 |
| s2 | The agents execute the same deterministic algorithm. | background 0.90 | 0.09 | 0.01 | asserted 1.00 |
| s3 | Time is divided into rounds, and in each round each agent can either stay idle at the current node or move to an adjacent node. | background 0.82 | 0.09 | 0.02 | asserted 0.97 |
| s4 | An adversary places the agents at two nodes of the grid at a distance at most $D$, and wakes them up in possibly different rounds. | background 0.87 | 0.13 | 0.02 | asserted 0.93 |
| s5 | Each agent starts executing the algorithm in its wakeup round. | background 0.77 | 0.09 | 0.02 | asserted 0.99 |
| s6 | If agents cannot leave any marks on visited nodes then they can never meet, even if they start simultaneously at adjacent nodes and know it. | background 0.81 | 0.22 | 0.04 | asserted 0.98 |
| s7 | Hence, we assume that each agent marks any unmarked node it visits, and that an agent can distinguish if a node it visits has been previously marked or not. | background 0.52 | 0.18 | 0.03 | asserted 0.95 |
| s8 | The time of a rendezvous algorithm is the number of rounds between the wakeup of the later agent and rendezvous. | background 0.75 | 0.22 | 0.02 | asserted 0.99 |
| s9 | We ask the question whether the capability of marking nodes enables the agents to meet, and if so, what is the fastest rendezvous algorithm. | gap 0.50 | 0.16 | 0.04 | unclear 0.85 |
| s10 | We consider this problem under three scenarios. | method 0.56 | 0.53 | 0.02 | asserted 0.70 |
| s11 | First, agents know $D$ but may start with arbitrary delay. | method 0.38 | 0.25 | 0.03 | asserted 0.70 |
| s12 | Second, they start simultaneously but do not have any {\em a priori} knowledge. | background 0.54 | 0.15 | 0.02 | asserted 0.85 |
| s13 | Third, most difficult scenario, we do not make any of the above facilitating assumptions. | method 0.41 | 0.33 | 0.03 | asserted 0.92 |
| s14 | Agents start with arbitrary delay and they do not have any a priori knowledge. | background 0.78 | 0.15 | 0.03 | asserted 0.88 |
| s15 | We prove that in the first two scenarios rendezvous can be accomplished in time $O(D)$. This is clearly optimal. | result 0.99 | 0.93 | 0.28 | asserted 0.99 |
| s16 | For the third scenario, we prove that there does not exist any rendezvous algorithm working in time $o(D^{\sqrt{2}})$, and we show an algorithm working in time $O(D^2)$. | result 0.98 | 0.96 | 0.67 | asserted 1.00 |
| s17 | The above negative result shows a separation between the optimal complexity in the two easier scenarios and the optimal complexity in the most difficult scenario. | interpretation 1.00 | 0.89 | 0.27 | asserted 1.00 |
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The paper
Original abstract
Two identical anonymous mobile agents have to meet at a node of the infinite oriented grid whose nodes are unlabeled. This problem is known as rendezvous. The agents execute the same deterministic algorithm. Time is divided into rounds, and in each round each agent can either stay idle at the current node or move to an adjacent node. An adversary places the agents at two nodes of the grid at a distance at most $D$, and wakes them up in possibly different rounds. Each agent starts executing the algorithm in its wakeup round. If agents cannot leave any marks on visited nodes then they can never meet, even if they start simultaneously at adjacent nodes and know it. Hence, we assume that each agent marks any unmarked node it visits, and that an agent can distinguish if a node it visits has been previously marked or not. The time of a rendezvous algorithm is the number of rounds between the wakeup of the later agent and rendezvous. We ask the question whether the capability of marking nodes enables the agents to meet, and if so, what is the fastest rendezvous algorithm. We consider this problem under three scenarios. First, agents know $D$ but may start with arbitrary delay. Second, they start simultaneously but do not have any {\em a priori} knowledge. Third, most difficult scenario, we do not make any of the above facilitating assumptions. Agents start with arbitrary delay and they do not have any a priori knowledge. We prove that in the first two scenarios rendezvous can be accomplished in time $O(D)$. This is clearly optimal. For the third scenario, we prove that there does not exist any rendezvous algorithm working in time $o(D^{\sqrt{2}})$, and we show an algorithm working in time $O(D^2)$. The above negative result shows a separation between the optimal complexity in the two easier scenarios and the optimal complexity in the most difficult scenario.
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