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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.

Labelled trees, no knowledge, any number of agents

verified
Time: tightColours: tight

Without a map or a bound on \(n\), agents on a tree with labelled nodes explore without collisions in \(O(n^2)\) rounds, which is optimal in the worst case.

Assumed (added on the last edge are outlined)

+Graph dynamicsstatic graph+Edge directionundirected+Node identifierslabelled nodes (an agent can recognise a node it has visited)+Timingsynchronous rounds+Startsimultaneous start+Topologytree

Not assumed

Number of agentsany number of agentsWhat must be visitedevery agent visits every nodeKnowledge of the graphnothing knownVision of the other agentsnone: agents never perceive each otherHelp from an oracleno help
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: labelled nodes
Each node \(v\) carries a distinct identifier that agents at \(v\) can read; hence an agent can recognise a node it has visited.
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: any \(k\)
Any number \(k\) of agents, \(2 \le k \le n\), which they do not know.
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: nothing known
The agents know nothing about \(G\), not even \(n\).
Vision of the other agents: no vision
An agent never perceives another agent.
Help from an oracle: no help
Nothing is placed on the nodes.
Start: simultaneous start
All agents start in the same round.
Topology: tree
\(G\) is connected and acyclic: \(|E|=n-1\).

Statement for this variant

Nodes carry the distinct identifiers \(1, \dots, n\), and an agent sees those of its node and of the neighbours; the agents are not told \(n\) or anything else about the tree. Every agent must return to its start node.

What is known

Time · rounds from the wake-up of the later agent until all agents have stopped
tight
\(\Theta(n^2)\) in the worst case over trees with \(n\) nodes.
lower bound
\(\Omega(n^2)\) Czyzowicz et al. 2017a
bottleneckcongestion at a node
holds when
  • a star: even with a map, \(n\Delta = n(n-1)\) rounds are needed (Theorem 3.2)
introductory—
current best
\(O(n^2)\)
Colours · number of pebble colours the oracle needs, when pebbles are used
tight
No pebbles.
lower boundnone known
introductory—
current best—

Results, in order of publication

Time \(O(n^2)\)

Introductory and current best (Theorem 4.1): an edge is coloured by the sum of the identifiers of its ends, which every agent sees; in phases that double a bound on the identifiers, the agents cross the edges of one colour at a time, and at the end every agent retraces its moves.

Open questions

None recorded for this variant.

Notes

The lower bound specialises Theorem 3.2 to a star. Checked against the full text (the 2012 report on HAL).

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

Know a better result for this setting, or found a mistake? Suggest a result · Report an error. Accepted contributions are acknowledged.