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Gathering by oblivious robots on graphs

Oblivious Look–Compute–Move robots must all reach one node of a graph and stay there; which starts can be gathered depends on the symmetries of the graph, on multiplicity detection and on the scheduler.

Problem statement and measuresexpand

\(k \ge 2\) anonymous, oblivious robots stand on the nodes of a connected anonymous graph \(G\) with \(n\) nodes, which every robot sees whole. A robot runs Look–Compute–Move cycles: in a Look it takes a snapshot of the graph with its occupied nodes, it computes from that snapshot only, and it stays or moves to a neighbour; moves are instantaneous, so robots are always seen on nodes. A node with two or more robots is a tower. With multiplicity detection a snapshot shows which nodes are towers (weak) or how many robots each node holds (strong); without it a tower looks like one robot. The robots must reach a configuration in which all of them are on one node and none moves again. A setting fixes the scheduler, the multiplicity detection, whether towers may exist at the start, and the graph class; an algorithm must gather from every start of the setting except the starts listed as unsolvable. Under round robin the scheduler activates one robot at a time in a fixed cyclic order. Time is counted in epochs: an epoch is an interval in which every robot completes at least one cycle (under round robin, \(k\) consecutive activations).

Measures: Time, epochs until all robots are on one node and no robot moves again.

Infinite grid, round robin

verified
Time: tight

On the infinite grid, robots that cannot see towers gather under round robin in a number of epochs linear in their largest distance, from every start except two adjacent occupied nodes and one kind of three-node path.

Assumed (added on the last edge are outlined)

+Graph dynamicsstatic graph+Edge directionundirected+Faultsno faults+Networkinfinite square grid+Schedulersequential: exactly one robot is activated per round+Activation orderround robin: one robot per round in a fixed cyclic order

Not assumed

Multiplicity detectionnone: a tower looks like one robotInitial configurationarbitrary (agents on any \(\ell \le k\) nodes)
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.
Faults: no faults
All agents follow the algorithm forever.
Network: infinite grid
The infinite square grid: nodes are the points of \(\mathbb Z^2\), neighbours at Manhattan distance one; finitely many robots.
Multiplicity detection: no multiplicity detection
A snapshot shows which nodes are occupied, not how many robots each holds.
Initial configuration: any start
Agents start at arbitrary nodes; \(\ell\le k\) nodes are initially occupied, possibly with several agents each.
Scheduler: sequential
Sequential: an SSYNC scheduler that activates exactly one robot per round, fairly; which robot is the adversary's choice unless the activation order is fixed (see the order axis).
Activation order: round robin
With the sequential scheduler: the robots are activated one per round in a cyclic order that the adversary fixes at the start, that the robots do not know, and that repeats forever; an epoch is \(k\) consecutive rounds, one activation per robot.

Statement for this variant

\(d\) is the largest distance between two initially occupied nodes.

What is known

Time · epochs until all robots are on one node and no robot moves again
tight
\(\Theta(d)\) epochs.
lower bounds
\(\text{impossible}\) Cicerone et al. 2025b preprint
bottleneckindistinguishable multiplicities
holds when
  • two adjacent occupied nodes (Theorem 1)
  • one of the two kinds of three-node paths must also be excluded (Corollary 1): the algorithm of Theorem 5 excludes three occupied nodes of a \(2 \times 2\) square; a variant that is only sketched excludes three consecutive nodes of a row instead (Theorem 6)
\(\Omega(d)\) Cicerone et al. 2025b preprint
bottleneckdistance
holds when
  • the two robots at distance \(d\) must meet, and each moves at most one edge per epoch (proof of Lemma 1)
introductory—
current best
\(O(d)\)
the start is neither two adjacent occupied nodes nor three occupied nodes of a \(2 \times 2\) square

Results, in order of publication

Cicerone et al. 2025b preprint current best
Time \(O(d)\)
the start is neither two adjacent occupied nodes nor three occupied nodes of a \(2 \times 2\) square

Introductory and current best (Theorem 5): shrink the smallest rectangle enclosing the robots, keeping a corner empty, down to two opposite nodes of a \(2 \times 2\) square, then gather on one of its two empty nodes within one epoch. Stated as \(O(n+m)\) epochs for the \(m \times n\) enclosing rectangle (their \(m\), \(n\) count rows and columns), which is \(O(d)\). Preprint. Read in the arXiv version (v1).

Open questions

None recorded for this variant.

Notes

The characterisation of Theorem 6 (exactly one of the two kinds of three-node paths must be excluded) depends on the sketched variant of the algorithm. Checked against the full text of Cicerone et al. 2025b (arXiv v1).

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