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. 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.
Anonymous agents in trees, simultaneous start
verifiedWith simultaneous start, memory drops to \(\Theta(\log \nu+\log\log n)\) for trees with \(\nu\) leaves; with delay it is \(\Theta(\log n)\), an exponential gap.
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: finite graph
- \(G\) has finitely many nodes.
- Identifiers: anonymous
- All agents run the same deterministic algorithm on the same input; no agent has a label. Formally the label function is constant: \(\mathrm{id}(a)=\mathrm{id}(b)\) for all agents \(a,b\).
- 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
- trees of maximum degree 3
- \(\nu\) = number of leaves
Results, in order of publication
Introductory and current best. With arbitrary delay, \(\Omega(\log n)\) even on lines.
Open questions
None recorded for this variant.
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.
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