Rendezvous
Two agents start at different nodes of an unknown graph and must meet; how long does it take, and how much memory or symmetry breaking does it need?
Problem statement and measuresexpand
Two mobile agents start at different nodes of an unknown connected port-labelled graph. Nodes are anonymous unless a setting gives them node labels: distinct positive integers that an agent reads when it is at the node. In the synchronous model they must be at the same node in the same round; in the asynchronous model an adversary controls their speeds and they may also meet inside an edge. Agents are either labelled (distinct labels from \(\{1,\dots,L\}\); \(\lambda\) is the smaller label) or anonymous. They may start at different times; \(\tau\) is the delay. Time is the number of rounds from the start of the later agent (synchronous) or the total number of moves, that is edge traversals (asynchronous cost). Memory is bits per agent.
Measures: Time, rounds after the later agent starts (synchronous); edge traversals (asynchronous); Memory, bits per agent.
Labelled agents in infinite trees, simultaneous start
verifiedWith simultaneous start, orientation helps: \(O(d\log L)\) in oriented trees when a bound on \(L\) or \(d\) is known, \(O(d^2+\log^2 L)\) without; unoriented regular trees still need \(\Omega(\mathrm{ball}(d))\).
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.
- Graph size: locally finite
- \(G\) is finite or countably infinite with finite degrees.
- 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.
- Identifiers: distinct IDs (eponymous)
- Distinct identifiers (eponymous agents): the label function \(\mathrm{id}:\{1,\dots,k\}\to\{1,\dots,L\}\) is injective; usually \(L=k^{O(1)}\), so a label takes \(O(\log k)\) bits.
- Timing: synchronous
- Time proceeds in rounds \(t=1,2,\dots\); in every round every agent communicates, computes and moves (at most one edge) simultaneously.
- Start: simultaneous start
- Both agents start in the same round (\(\tau=0\)).
- Topology: tree
- \(G\) is connected and acyclic: \(|E|=n-1\).
What is known
- regular trees of degree \(\Delta \ge 2\) with an adversarial port numbering, even with simultaneous start and known \(d\) and \(L\) (Theorem 2.2 and Corollary 2.3)
- \(\mathrm{ball}(d)\) is the number of nodes within distance \(d\) of a node, exponential in \(d\) when \(\Delta > 2\)
- in oriented trees only \(\Omega(d\log L)\) is known, from Dessmark et al. 2006 (Section 3)
Results, in order of publication
Theorems 3.1 and 3.2; optimal.
Theorem 3.3; optimal when \(d\) and \(\log L\) are of the same order.
Open questions
In oriented trees with simultaneous start, can agents meet in time \(O(d\log L)\) without knowing anything about \(d\) or \(L\)? Posed in Bhagat & Pelc 2024.
Notes
Orientation has no axis here; it is recorded on the results. With arbitrary delay, orientation does not help (Proposition 3.1). Checked against the full text of Bhagat & Pelc 2024 (arXiv v1).
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.