Collision-free exploration
Agents start at distinct nodes of a graph and must explore it so that no two agents are ever at the same node; how long does it take, and what must the agents sense of each other?
Problem statement and measuresexpand
\(k \ge 2\) agents start at distinct nodes of a connected undirected graph \(G\) with \(n\) nodes and move in synchronous rounds: in each round an agent stays or crosses one edge. A collision is two agents at the same node in the same round; two agents that cross one edge in opposite directions do not collide and do not notice it. Exploration is collision-free if no collision ever happens, also with an agent that an adversary has not yet woken up. Unless a setting says otherwise, every agent must visit every node and then stop. The agents run the same deterministic algorithm, cannot mark nodes and send no messages; what they perceive of each other (their vision) and what they know of \(G\) are fixed by the setting, and in some settings an oracle places coloured pebbles on nodes before the start. Time is the number of rounds from the wake-up of the later agent until every agent has stopped. \(X(n)\) is the time of a single-agent exploration procedure for graphs with at most \(n\) nodes, used as a building block (polynomial in \(n\), for instance with a universal exploration sequence).
Measures: Time, rounds from the wake-up of the later agent until all agents have stopped; Colours, number of pebble colours the oracle needs, when pebbles are used.
Two agents sensing presence, no pebbles
verifiedTwo identical agents that only sense each other's presence cannot explore even a path of three nodes without help.
Assumed (added on the last edge are outlined)
Not assumed
Formal definitions of this setting
- Graph dynamics: static graph
- \(G_t=G\) for all \(t\): the edge set never changes.
- Edge direction: undirected
- \(G=(V,E)\) with \(E\subseteq\binom V2\); every edge can be traversed in both directions.
- Node identifiers: anonymous nodes
- Nodes have no identifiers; an agent at \(v\) sees only \(\deg(v)\) and the port numbers \(1,\dots,\deg(v)\), which are assigned locally and independently at each endpoint of an edge.
- Timing: synchronous
- Time proceeds in rounds \(t=1,2,\dots\); in every round every agent communicates, computes and moves (at most one edge) simultaneously.
- Number of agents: two agents
- Exactly two agents.
- What must be visited: the team
- Every node must be visited by at least one agent, and all agents stop.
- Knowledge of the graph: bound on \(n\)
- The agents know an upper bound on \(n\) (Das et al. 2025a give them \(n\) itself).
- Vision of the other agents: presence
- An agent notices that the other agent is on a neighbouring node, but not on which one Das et al. 2025a.
- Help from an oracle: no help
- Nothing is placed on the nodes.
- Start: adversarial wake-up
- Each agent starts in a round chosen by the adversary; until then it stays at its start node, where no other agent may enter.
- Topology: any topology
- \(G\) is any connected graph in the class allowed by the other axes.
What is known
- a path of three nodes with the agents on its ends, started together: they act alike, so they either collide in the middle or never visit it (Section 1.2)
Results, in order of publication
No result recorded yet.
Open questions
None recorded for this variant.
Notes
Checked against the full text of Das et al. 2025a (arXiv v1).
What carries over from neighbouring settings
Read off the order of assumptions, not checked case by case: results may carry extra assumptions (shown on their pages), and asynchronous time is measured in epochs.
Algorithms from weaker settings work here
No catalogued weaker setting.
Lower bounds from stronger settings hold here
Branches from here
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