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Graph exploration by a single agent

One agent starts at a node of an unknown anonymous graph and must visit every node; how much memory does it need, which pebbles or knowledge make it possible, and how fast can it be?

Problem statement and measuresexpand

A single deterministic agent is placed at a node \(s\) of a connected graph \(G\) with \(n\) nodes, \(m\) edges and maximum degree \(\Delta\). Nodes are anonymous (they carry no names the agent can read), and the edges at each node are numbered locally; these numbers are the node's ports. The agent sees the degree of its node and the port by which it entered. It must visit every node of \(G\). Three termination requirements are standard:

  • perpetual exploration: every node is eventually visited; the agent never has to stop;
  • exploration with stop: every node is visited and the agent then stops, anywhere;
  • exploration with return: every node is visited and the agent stops at \(s\).

Return implies stop, and stop implies perpetual exploration. Memory is the number of bits of the agent's state; when the agent carries pebbles (objects it can drop on a node and pick up again) their number is reported with the memory. Time is the number of moves (edge traversals) until exploration is complete; for periodic exploration it is the period. Unless a variant says otherwise the agent knows nothing about \(G\), not even \(n\).

Exploration is the oldest problem of the mobile-agent literature: it started as maze search by finite automata, and the question of how little memory suffices runs from Budach's impossibility (1978) to Reingold's log-space algorithm (2005). Several agents exploring together form the sibling problem, collaborative exploration.

Measures: Time, edge traversals until every node is visited (for periodic exploration: the period); Memory, bits of agent memory; pebbles, if any, are counted separately.

Dynamic rings

verified
Time: tightMemory: unknown

On a \(T\)-interval-connected ring whose edges may be missing for up to \(\delta\) steps, the worst-case exploration time is known exactly up to additive terms.

Assumed (added on the last edge are outlined)

+Graph dynamicsdynamic graph, connected in every round (1-interval connected)+Edge directionundirected+Termination requirementvisit every node; never required to stop (perpetual)+Topologyring or path

Not assumed

Node identifiersanonymous nodesKnowledgenothing known about the graphRandomnessdeterministicMarkersno markersPreprocessingports and nodes as given by an adversary
Formal definitions of this setting
Graph dynamics: connected every round
\(\mathcal G=(G_t)_{t\ge0}\) is 1-interval connected if every snapshot \(G_t=(V,E_t)\) is connected. More generally it is \(T\)-interval connected if for every \(t\) the graph \(\bigl(V,\bigcap_{i=t}^{t+T-1}E_i\bigr)\) is connected.
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.
Knowledge: nothing known
The agent knows nothing about \(G\): not \(n\), \(m\), \(\Delta\) or \(D\).
Randomness: deterministic
The next state and move of an agent are a function of its current state and what it observes.
Markers: no markers
The agent cannot mark nodes.
Preprocessing: adversarial ports
Port numbers (and any node labels) are chosen by an adversary.
Termination requirement: perpetual
Every node is visited at least once (and, for periodic exploration, infinitely often); the agent never needs to halt.
Topology: ring
\(G\) is a cycle \(C_n\) (every node of degree 2), or a path when stated.

What is known

Time · edge traversals until every node is visited (for periodic exploration: the period)
tight
Matching up to \(\pm\Theta(\delta)\).
lower bound
\(n+\tfrac{n}{\max\{1,T-1\}}(\delta-1)-O(\delta)\) Ilcinkas & Wade 2018
bottleneckadversarial edge changes
holds when
  • underlying graph a ring
  • the agent does not know the dynamics
  • \(T\)-interval connectivity; \(\delta\) = longest absence of an edge
introductory—
current best
\(n+\tfrac{n}{\max\{1,T-1\}}(\delta-1)+O(\delta)\)
Memory · bits of agent memory; pebbles, if any, are counted separately
unknown
Memory not studied.
lower boundnone known
introductory—
current best—

Results, in order of publication

Time \(n+\tfrac{n}{\max\{1,T-1\}}(\delta-1)+O(\delta)\)

Introductory and current best. If the agent knows the dynamics, the worst case is \(2n-T-\Theta(1)\).

Open questions

None recorded for this variant.

Notes

Multi-agent exploration of dynamic rings and graphs, where the measure is how many agents are needed (Di Luna, Dobrev, Flocchini and Santoro; Gotoh, Flocchini, Masuzawa and Santoro), is not catalogued yet.

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