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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.

Temporal graphs, schedule known

verified
Time: tightMemory: unknown

If the whole schedule of an always-connected temporal graph is known, \(\Theta(n^2)\) steps are necessary and sufficient in the worst case.

Assumed (added on the last edge are outlined)

+Graph dynamicsdynamic graph, connected in every round (1-interval connected)+Edge directionundirected+Knowledgea complete map is known (for a dynamic graph: the whole schedule)+Termination requirementvisit every node; never required to stop (perpetual)

Not assumed

Node identifiersanonymous nodesRandomnessdeterministicMarkersno markersPreprocessingports and nodes as given by an adversaryTopologyarbitrary connected graph
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: knows the map
The agent is given a graph isomorphic to \(G\), with port numbers, and (where stated) its own position on it; for a dynamic graph, the whole sequence \((G_t)\).
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: any topology
\(G\) is any connected graph in the class allowed by the other axes.

What is known

Time · edge traversals until every node is visited (for periodic exploration: the period)
tight
\(\Theta(n^2)\) in the worst case.
lower bound
\(\Omega(n^2)\) Erlebach et al. 2021
bottleneckadversarial edge changes
holds when
  • an explicit family of always-connected temporal graphs
  • holds for centralised, offline exploration
introductory—
current best
\(O(n^2)\)
Memory · bits of agent memory; pebbles, if any, are counted separately
unknown
Not a memory question.
lower boundnone known
introductory—
current best—

Results, in order of publication

Time \(O(n^2)\)

The \(n^2\) upper bound is attributed by the authors to Michail and Spirakis. They also give \(O(n^{1.5}\mathrm{tw}^{1.5}\log n)\) for treewidth \(\mathrm{tw}\), \(O(n^{1.8}\log n)\) for planar and \(2n-2\) for cycles, and show inapproximability within \(O(n^{1-\varepsilon})\).

Open questions

  1. Tight bounds when the underlying graph is a grid, has bounded degree or is planar Erlebach et al. 2021.

Notes

This is a centralised planning problem rather than a distributed one, but it gives the baseline for what any agent could do in a temporal graph.

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