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, anonymous graphs, vision of radius 2
standard settingdraftTwo identical agents that see two edges around them and know a bound on \(n\) explore any graph with more than two nodes without collisions, at most a factor \(\log n\) slower than one agent.
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: every agent
- Every agent must visit every node and stop (in Czyzowicz et al. 2017a also return to its start node).
- 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: radius 2
- An agent sees the subgraph induced by the nodes at distance at most 2, with its port numbers and the agents on these nodes Bhagat & Pelc 2025.
- 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.
Statement for this variant
Nodes are anonymous, the adversary wakes the agents, and the graph has at least three nodes: on two nodes the adversary's wake-up alone forces a collision Bhagat & Pelc 2025.
What is known
Results, in order of publication
Theorem 3.1: each agent follows a single-agent exploration walk; when the agents come within distance 2 they synchronise and then take turns, each step of the walk costing \(O(\log n)\) rounds of signalling by moves. Any single-agent procedure can be used.
Theorem 9.1: at most \(6X(n) + O(1)\) rounds after the later wake-up, by a priority rule that lets both agents move in the same round. The paper also reports that the stopping procedure of Bhagat & Pelc 2025 can lead to a collision when the agents finish their walks in different rounds (Section 3). Not checked here.
Open questions
Can two agents explore without collisions as fast as one agent explores alone? Asked by Bhagat & Pelc 2025; Papin 2026 claims a constant factor.
For which number of agents below \(n\) is collision-free exploration of anonymous graphs feasible, and with which vision radius Bhagat & Pelc 2025?
Notes
Draft because the two papers disagree: Papin reports a collision in the stopping procedure of Bhagat and Pelc and gives a corrected algorithm, but his model section describes vision of the nodes at distance less than 2, under which two agents cannot explore a star Bhagat & Pelc 2025; his algorithm reuses Bhagat and Pelc's synchronisation, which needs radius 2. Neither the reported collision nor the corrected algorithm was checked here. Bhagat & Pelc 2025 read in arXiv v1 (2024); Papin 2026 is a preprint.
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
Lower bounds from stronger settings hold here
No catalogued stronger setting.
Know a better result for this setting, or found a mistake? Suggest a result · Report an error. Accepted contributions are acknowledged.