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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 measurescollapse
agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 01agent 1: 11 01agent 3: 11 11 011133Start: agents 1 and 3 at the ends of a tree (n = 4)Block 0: both explore; they learn n (block = 6 rounds)Block 1: bits 1 and 1; both exploreBlock 2: bits 1 and 1; both exploreBlock 3: bits 0 and 1; agent 1 waits, agent 3 exploresFinal: agent 3 walks onto the waiting agent 1 in round 21
Rendezvous of two labelled agents in a tree, simultaneous start, by the explore-or-wait idea of Dessmark et al. 2006 (simplified): after one depth-first tour to learn \(n\), each agent reads its label in binary with every bit doubled and \(01\) appended, and spends one block of \(2(n-1)\) rounds per bit, exploring and returning on a 1, waiting at its start node on a 0. Labels 1 and 3 give \(11\,01\) and \(11\,11\,01\); in block 3 agent 3 explores while agent 1 waits, and they meet. One frame per round.

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.

A first look

the basic problem, a simple algorithm and why some things cannot be done, for a newcomer

Two friends lose each other in a city with no phones and want to meet at the same spot. The city is a graph: street corners are nodes and the streets between them are edges; the friends are two agents that start at different nodes of this unknown connected graph. At each node the edges at that node carry local numbers, its ports. Time passes in synchronous rounds; in each round an agent stays or makes a move along one edge, and the agents meet when they are on the same node in the same round (crossing inside an edge does not count). In the basic version each agent knows its own label, a distinct integer from \(\{1,\dots,L\}\), but not the other's label and nothing about the graph; \(\lambda\) is the smaller of the two labels, \(n\) the number of nodes, and the later agent may start \(\tau\) rounds after the earlier one (the delay). The task is to meet, and the cost is the number of rounds from the start of the later agent.

1AlgorithmExplore or wait, bit by bit (trees, simultaneous start) Labelled agents in trees → Dessmark et al. 2006

If one agent knew it should wait, the other could simply search the graph ("wait for mommy" in Dessmark et al. 2006). Without knowing the other label, use the bits of your own label to take turns. This is a simplified version of their idea, for a tree (no cycles), both agents starting in the same round.

  1. Explore the tree by depth-first search and return to your start node (the node where you began); you learn \(n\), and this takes \(2(n-1)\) rounds. Call \(2(n-1)\) rounds a block.
  2. Write your label in binary, write every bit twice, and append \(0\,1\). For example \(5 = 101\) becomes \(11\,00\,11\,01\).
  3. Go through this string one bit per block: on a bit 1, explore the tree and return to your start node; on a bit 0, wait at your start node for the whole block.
Why it works and what it costs. Both agents compute the same block length, so their blocks start and end in the same rounds. Two different labels give strings that differ at some position before the shorter one ends (the final \(0\,1\) marks the end, and no doubled bit looks like it). In that block one agent explores while the other waits at its start node, so the exploring agent walks onto it. The cost is \(O(\log\lambda)\) blocks of \(O(n)\) rounds, \(O(n\log\lambda)\) in all. Dessmark et al. 2006 reach \(O(n+\log\lambda)\) on trees even with a delay \(\tau\), which is optimal; on general graphs the best known is \(\tilde O(n^5\log\lambda)\) Ta-Shma & Zwick 2014, against a lower bound of \(\Omega(n^2)\).

Why it matters. Turning distinct labels into different move/wait patterns is the basic symmetry-breaking device of deterministic rendezvous.

2ImpossibilityIdentical agents in symmetric positions Anonymous agents: memory → Czyzowicz et al. 2012b

Take a ring of 6 nodes where at every node port 0 leads clockwise and port 1 counterclockwise. Put two agents without labels (anonymous: same algorithm, same initial memory) on opposite nodes, 3 edges apart, starting in the same round. Each agent sees exactly what the other sees: a node of degree 2 (two edges at it), the same entry ports, nobody else around. So in every round they make the same choice: both stay, both go clockwise, or both go counterclockwise.

Why it holds. After any number of rounds they are still exactly 3 edges apart, so they never meet, whatever the algorithm and however much memory (bits kept between rounds) they have. Identical agents can meet only when their starting positions are not symmetric in this sense, and for those positions \(\Theta(\log n)\) bits of memory are exactly what is needed Czyzowicz et al. 2012b.

Why it matters. It explains why most rendezvous work gives the agents labels, and why for identical agents the question becomes which starting positions can be told apart.

3Lower boundTwo nodes and the length of a label Labelled agents, arbitrary finite graphs → Dessmark et al. 2006

Take the smallest graph: two nodes joined by one edge, one agent at each end, both starting together. Until they meet, an agent sees nothing new (always degree 1, always port 0), so its behaviour in the first \(t\) rounds is a string of \(t\) choices, stay or move, fixed by its label alone. They meet in a round where exactly one of them moves; if both move they just swap ends.

Why it holds. There are only \(2^t\) strings of \(t\) choices. Among \(2^t + 1\) labels, two give the same string; give these labels to the two agents and they do the same thing in each of the first \(t\) rounds, so they do not meet within \(t\) rounds. Hence some pairs of labels need about \(\log\) (number of labels) rounds, which gives the \(\Omega(\log\lambda)\) bound of Dessmark et al. 2006. Together with the \(\Omega(n)\) needed to cover the graph it shows their \(O(n+\log\lambda)\) algorithm for trees is optimal.

Why it matters. It shows that the cost of breaking symmetry is the length of a label, a counting argument that recurs in every model with identifiers.

What is known

The best upper bound against the best lower bound in each catalogued setting. Open a setting for its results, proofs and notes, or see the progress map with the branch points.

tight near-optimal polynomial gap unknown impossible· one row per catalogued setting, in tree order; each cell: best upper bound vs lower bound, and what forces the lower bound
Setting
Time
Memory
Asynchronous anonymous agents
Which pairs of starting nodes allow identical asynchronous agents to meet is characterised.
unknown \(\text{finite}\) vs —
unknown — vs —
Asynchronous labelled agents
An adversary controls speeds; agents may meet inside an edge. Feasible everywhere, and at polynomial cost, but with very large exponents.
unknown \(\text{poly}(n,\log \lambda)\) vs —
unknown — vs —
Labelled agents, arbitrary finite graphs standard
Polynomial in \(n\) and \(\log \lambda\) and independent of the delay, but far from the known lower bounds.
polynomial gap \(\tilde O(n^5\log \lambda)\) vs \(\Omega(n^2)\) exploration
unknown — vs —
Labelled agents in trees
In trees, \(O(n+\log \lambda)\) even with arbitrary delay, and this cannot be improved.
tight \(O(n+\log \lambda)\) vs \(\Omega(n+\log \lambda)\) distance
unknown — vs —
Labelled agents in rings, simultaneous start
With simultaneous start on a ring, rendezvous takes \(\Theta(d\log \lambda)\) for initial distance \(d\).
tight \(O(d\log \lambda)\) vs \(\Omega(d\log \lambda)\) symmetry breaking
unknown — vs —
Anonymous agents: memory
Identical agents can meet whenever their positions are not symmetric, or are symmetric but start with a long enough delay; \(\Theta(\log n)\) bits suffice when the positions are not symmetric.
polynomial gap \((n+\tau)^{O(n+\tau)}\) vs \(2^{\Omega(d)}\) symmetry breaking
tight \(O(\log n)\) vs \(\Omega(\log n)\) symmetry breaking
Anonymous agents in trees, simultaneous start
With 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.
unknown — vs —
tight \(O(\log \nu+\log\log n)\) vs \(\Omega(\log \nu+\log\log n)\) symmetry breaking
Labelled agents in infinite trees
In infinite regular trees the time grows with the number of nodes within distance \(d\): \(O(\mathrm{ball}(d)\log L)\) against \(\Omega(\mathrm{ball}(d)+d\log L)\).
near-optimal \(O(\mathrm{ball}(d)\log L)\) vs \(\Omega(\mathrm{ball}(d)+d\log L)\) port addressing
unknown — vs —
Labelled agents in infinite trees, simultaneous start
With 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 \(\Omega(\mathrm{ball}(d))\).
near-optimal \(O(d\log L)\) vs \(\Omega(\mathrm{ball}(d)+d\log L)\) port addressing
unknown — vs —
Anonymous agents on paths and rings with labelled nodes
On paths with labelled nodes, two anonymous agents meet in \(O(d\log^*\sigma_{\min})\) rounds, \(\sigma_{\min}\) the smallest node label near their start nodes, and no algorithm is faster in the worst case.
tight \(O(d\log^*\sigma_{\min})\) vs \(\Omega(d\log^*\sigma_{\max})\) symmetry breaking
unknown — vs —
Labelled agents, graphs with infinite degrees
A universal algorithm works even when nodes have infinite degree; it is within a cubic factor of the lower bound.
polynomial gap \(O((W\log L)^3)\) vs \(\Omega(\max(W,\log L))\) port addressing
unknown — vs —

Open questions 7

Asynchronous labelled agents

  1. Find an optimal-cost asynchronous rendezvous algorithm in arbitrary graphs (open problem of Pelc 2019).

Labelled agents, arbitrary finite graphs

  1. Find an optimal-time synchronous rendezvous algorithm for labelled agents in arbitrary graphs (open problem of Pelc 2019).

Anonymous agents: memory

  1. Is there a universal algorithm that meets for every feasible start in time polynomial in \(n\) and \(\tau\)? Posed in Pelc & Yadav 2019b.

Labelled agents in infinite trees

  1. Close the gap between \(O(\mathrm{ball}(d)\log L)\) and \(\Omega(\mathrm{ball}(d)+d\log L)\) in unoriented regular trees. Posed in Bhagat & Pelc 2024.
  2. Find efficient rendezvous algorithms for arbitrary infinite graphs without extra knowledge such as the agents' positions. Posed in Bhagat & Pelc 2024.

Labelled agents in infinite trees, simultaneous start

  1. 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.

Labelled agents, graphs with infinite degrees

  1. Find an optimal rendezvous algorithm, "even for finite graphs" Pattanayak & Pelc 2024b.

What the community started looking at

Deterministic rendezvous in graphs started with the question of feasibility and cost polynomial in the graph size and the label length, for labelled agents that may start at different times Dessmark et al. 2006. The community then removed the dependence on the delay Kowalski & Malinowski 2008, made the algorithms constructive Ta-Shma & Zwick 2014, asked how little memory anonymous agents need Czyzowicz et al. 2012b Fraigniaud & Pelc 2013, and moved to asynchronous agents, where even feasibility in every graph took years De Marco et al. 2006 Czyzowicz et al. 2012a Dieudonné et al. 2015. Pelc's survey records this evolution Pelc 2019. Node labels are another source of symmetry breaking: on paths with labelled nodes, anonymous agents meet in time proportional to \(d\) times the iterated logarithm of a nearby node label Bourreau et al. 2026.

Subproblem of Graph exploration by a single agent

If the agents can break symmetry (for example by their labels), one may wait while the other explores, so exploration solves rendezvous; the work is in doing better than that, or in meeting when neither knows who should wait Pelc 2019.

Derived problems

Gathering in graphs: Rendezvous is gathering of two agents; with more agents the questions become which teams can gather, and how many faulty agents a team tolerates.

Perspectives

  • Why compute by moving The atlas collects four models that grew out of four different wishes: software that travels through a network, swarms of robots that are as weak as possible, searchers that must find something at an unknown place, and matter that reshapes itself. They share an adversary, locality and symmetry; they differ in what they count.
  • Symmetry is the adversary Identical agents running the same deterministic program can be kept in lockstep forever. Across graphs, the plane and programmable matter, the papers break this symmetry with identifiers, with randomness, with asymmetric starts, or with the scheduler, and leader election keeps reappearing as the common core.
  • Finding your way A single agent in an anonymous graph: how much memory it needs, what a pebble, an advice string or knowledge of n buys, why stopping is harder than exploring, and how treasure hunt asks for exploration only as far as the target. Each line of papers, in its own words, on why the question matters.
  • Meeting at one place Rendezvous and gathering on graphs, gathering and convergence in the plane, and evacuation through an unknown exit all ask entities to end at one point. The papers give the same reasons for meeting, but each model makes a different part of it hard: symmetry and exploration on graphs, capability in the plane, information in continuous search.
  • What forces the bounds Every lower bound in the atlas names the resource the adversary exploits. Read side by side, a few bottlenecks (distance, symmetry, missing information, an adversary that acts) explain most tight results, and the open entries cluster where no bottleneck beyond distance is known.
  • The most consequential open problems Ten open questions ranked by how far a solution would propagate through the atlas, with the chain of problems each one feeds.

Further reading

Papers on this problem that are not yet placed in a setting.

  • Kammer et al. 2026 Temporal Exploration and Rendezvous via Weisfeiler-Leman Colorings. Temporal rendezvous: two synchronous agents know the whole always-connected temporal graph but share no labelling of its nodes and cannot communicate; the adversary picks the starting nodes and the labellings. They meet within \(O(n\log^3 n)\) steps using polynomial computation, for a long enough lifetime (Theorem 1.4), against an \(\Omega(n\log n)\) lower bound from the paper that posed the problem (Dogeas et al., ICALP 2024, not in the catalogue), as reported. Preprint; read.
  • Das 2026 Universal Rendezvous of Anonymous Agents with Footprints. Footprints: every node an agent visits keeps a permanent mark, the same for both agents, which an agent sees when it is at the node. For two anonymous agents with simultaneous start, Das gives a universal algorithm, which meets in every feasible instance (a graph and two start nodes), in every connected graph, finite or countably infinite with finite degrees (Theorem 2.1, Corollary 2.1); this answers an open question of Das and Pelc (SPAA 2026, not in the bibliography). By their Proposition 2.1, recalled here, an instance is feasible only if the views from the two start nodes differ or the footprints seen along some path differ. No bound on time is given. The problem has no axis for marking nodes yet. Preprint; read.
  • Das & Pelc 2026 Rendezvous of Mobile Deterministic Automata in Graphs. Two anonymous agents modelled as copies of one deterministic finite automaton, synchronous, with arbitrary delay, in finite graphs whose degrees are bounded by a constant known to the designer; the automaton must meet in every feasible instance of a class. Without marking no such automaton exists even for paths (Proposition 2.1); with any finite number of pebbles that can be picked up again, none exists for all graphs (Proposition 2.2, through the graphs of Rollik 1980); with finitely many pebbles that cannot be picked up, none exists even for paths (Theorem 3.1, Corollary 3.1); with one pebble per agent that can be picked up, one automaton meets in every feasible instance on trees, within \(O(n^2)\) rounds from the start of the earlier agent (Theorem 4.1, Corollary 4.1). Without pebbles, \(\Theta(\log n)\) bits are needed in trees with delay Fraigniaud & Pelc 2013, so one pebble replaces them by constant memory. The problem has no axis for marking nodes yet. Preprint; read.
  • Baaziz & Pelc 2026a Asynchronous Rendezvous of Anonymous Deterministic Mobile Automata in the Plane. Rendezvous in the plane rather than in a graph. Two anonymous agents, copies of one deterministic finite automaton with a compass, see the disc of radius 1 around them before each move, move asynchronously (the adversary sets the speed during each move), and carry pebbles that they can pick up again and that the other agent can tell apart from its own. With at most 8 pebbles each they always meet, at cost (total length of both trajectories) \(O(d^2)\) when the start points are at distance at most \(d\) (their \(D\)) (Theorem 2); the authors argue in Section 5 that \(\Omega(d^2)\) is necessary, since the adversary can hold one agent back until the other finds it. The same algorithm elects a leader between the two agents. Read (LIPIcs version).
  • Gao & Pelc 2026 Gathering teams of bounded memory agents on a line. Two labelled agents that are finite automata (states depending only on \(L\)), started in the same round on distinct nodes of the infinite oriented path, can be placed so that they never meet (Theorem 3.1, for any set of such agents); teams of two or more agents per starting node gather (Theorems 4.1–6.2). Read (arXiv v1).
  • Yamada et al. 2026 Rendezvous and merging for two metamorphic robotic systems without global compass. A different model: two metamorphic robotic systems (connected oblivious modules on the square grid) without a common compass meet until every module sees all the others, then merge into one connected system; five modules per system suffice (Theorems 1 and 2) and are necessary for rendezvous (Theorem 3). Read (arXiv v1).
  • Bampas et al. 2019a On asynchronous rendezvous in general graphs. Characterises, in polynomial time, the pairs of walks that force rendezvous against an asynchronous adversary within a given number of edge traversals, with a certificate of why. Abstract read.
  • Bouchard et al. 2018 On deterministic rendezvous at a node of agents with arbitrary velocities. Rendezvous at a node (not inside an edge) of labelled agents whose speed the adversary fixes per agent and per edge: polynomial in \(n\), the smaller label and the largest traversal time. Abstract read.
  • Chalopin et al. 2016a Rendezvous in networks in spite of delay faults. Labelled agents with delay faults, measured by cost: polynomial cost with random faults; impossible with unbounded adversarial faults even on rings (cost \(O(n\lambda)\) on trees, \(\Omega(\lambda)\) needed); polynomial with faults bounded by an unknown \(c\). Abstract read.
  • Dereniowski et al. 2015b Rendezvous of heterogeneous mobile agents in edge-weighted networks. Agents that know the map but traverse edges at different, private speeds: \(O(n\,T_{\mathrm{OPT}})\) in general, \(\Theta(T_{\mathrm{OPT}})\) if they exchange \(\Theta(n)\) bits at the start, with \(T_{\mathrm{OPT}}\) the offline meeting time. Abstract read.
  • Miller & Pelc 2016 Time versus cost tradeoffs for deterministic rendezvous in networks. Trade-off between time and cost (edge traversals) for labelled agents, as functions of an exploration bound \(E\): cost \(O(E)\) with time \(O(EL)\), or both \(O(E\log L)\), with lower bounds on the trade-off. Abstract and introduction read.
  • Miller & Pelc 2015a Fast rendezvous with advice. How much advice gives rendezvous in time \(O(d)\): \(\Theta(d\log(n/d)+\log\log L)\) bits. Abstract and introduction read.
  • Eguchi et al. 2022 Fast Neighborhood Rendezvous. Neighbourhood rendezvous (agents start at adjacent nodes) in \(o(\Delta)\) rounds, with randomisation, node identifiers, neighbourhood identifiers and bounded minimum degree. Abstract read.
  • Cosson 2024a Ariadne and Theseus: Exploration and Rendezvous with Two Mobile Agents in an Unknown Graph. Two agents with whiteboards: exploration in \(m\) steps from a common start and rendezvous in at most \(\lceil 3m/2\rceil\) steps, also against an asynchronous adversary, with matching lower bounds there. A preprint; abstract and main results read.
  • Gao & Pelc 2025 Sniffing helps to meet: Deterministic rendezvous of anonymous agents in the grid. Identical agents in the infinite oriented grid that mark the nodes they visit: time \(O(D)\) if they know the distance bound \(D\) or start together, and \(O(D^2)\) with neither, where a superlinear lower bound separates the cases. Abstract read.
  • Das et al. 2015a Mobile Agents Rendezvous in Spite of a Malicious Agent. Rendezvous of several asynchronous anonymous finite-state agents despite a fast malicious agent that blocks nodes: necessary conditions, solved on rings except for even numbers of agents on unoriented rings, and on oriented meshes iff agents see two hops. Abstract read.
  • Dogeas et al. 2024 Exploiting Automorphisms of Temporal Graphs for Fast Exploration and Rendezvous. Temporal graphs with \(r\) automorphism orbits are explored in \(O(rn^{1+\varepsilon})\) steps, with lower bounds; the paper also poses rendezvous in temporal graphs. Abstract read.
  • Mironowicz 2023 Entangled rendezvous: a possible application of Bell non-locality for mobile agents on networks. Location-aware agents on small networks that share entangled quantum states: optimal strategies with a limited number of steps, with an advantage over classical agents on cubic graphs and cycles. Introduction read.
  • Pelc & Yadav 2020 Latecomers Help to Meet: Deterministic Anonymous Gathering in the Plane. Gathering of anonymous agents in the plane with compasses and clocks, where delays can break symmetry: characterisation of gatherable configurations, no common algorithm for the 'bad' ones, and the knowledge of the team size needed for the others. Abstract read.
  • Dieudonné et al. 2023 Almost Universal Anonymous Rendezvous in the Plane. Rendezvous of two anonymous agents in the plane whose attributes (coordinates, clock rate, speed, wake-up) may differ: exact characterisation of feasible instances, no single algorithm for two small sets of them, and one algorithm for all others. Abstract read.
  • Elouasbi & Pelc 2018 Deterministic Meeting of Sniffing Agents in the Plane. Labelled disc agents in the plane that sense their distance but not the direction to each other: deterministic rendezvous under two models of distance estimation. Abstract read.
  • Czyzowicz et al. 2018b Linear Rendezvous with Asymmetric Clocks. Two anonymous robots on the line whose speeds or clock units may differ: a universal algorithm meets whenever at least one parameter differs. Abstract read.
  • Georgiou et al. 2018 Symmetric Rendezvous with Advice: How to Rendezvous in a Disk. Randomised symmetric rendezvous of two robots at distance 2 with a common reference point (in a disk): expected time below the 4.25 of the line. Abstract read.
  • Chuangpishit et al. 2017 Rendezvous on a Line by Location-Aware Robots Despite the Presence of Byzantine Faults. Location-aware robots on a line, some of them Byzantine: a central authority plans trajectories with bounded competitive ratio, better when the number of faulty robots is bounded. Abstract read.
  • Ozsoyeller et al. 2021 Symmetric Rendezvous in Planar Environments With and Without Obstacles. Symmetric rendezvous of two robots in planar environments with and without obstacles. Abstract read.
The assumption tree (11 settings, 4 branch points)

The root assumes nothing. Going down, each edge adds the assumptions on its chips (+blue); a grey chip names the branch that leaves that axis unassumed while a sibling assumes something. Solid nodes are catalogued variants with their status per measure (Time, Memory); dashed nodes are branch points with no catalogued variant, named after the axis on which their branches differ. Axis order: Graph dynamics › Edge direction › Graph size › Node identifiers › Identifiers › Timing › Start › Topology.

tight near-optimal polynomial gap unknown impossible no variant

Assumptions specific to this problem

Graph size

0finite or countably infinite, degrees may be infinite1finite or countably infinite, every degree finite2finite

Deterministic rendezvous in graphs was posed for finite graphs, where an agent can afford to explore the whole graph and the cost is a function of \(n\) Dessmark et al. 2006 Pelc 2019. Bhagat and Pelc argued that this approach "is inefficient for very large graphs, and cannot be used for infinite graphs", and gave the first synchronous algorithm for connected graphs with finite degrees and finitely or countably many nodes, with time polynomial in \(d\), \(\log L\) and the number of paths of length \(d\) from the starting nodes, a quantity that also bounds rendezvous from below in trees Bhagat & Pelc 2022. For asynchronous agents, Czyzowicz, Labourel and Pelc had already shown that rendezvous is feasible in every connected countable graph Czyzowicz et al. 2012a (as summarised in Pelc 2019); like all earlier work it assumed finite degrees Pattanayak & Pelc 2024b. Pattanayak and Pelc then allowed countably infinite degrees, where the path count is infinite, and obtained a universal synchronous algorithm with time polynomial in a weight of the path types between the starting nodes and in \(\log L\), matched up to a polynomial by a lower bound in the tree whose degrees are all infinite Pattanayak & Pelc 2024b. The axis thus separates three cost measures: \(n\), the local path count, and path-type weights.

Formal definitions
0infinite degrees allowed
\(G\) is finite or countably infinite; degrees may be infinite.
1locally finite
\(G\) is finite or countably infinite with finite degrees.
2finite graph
\(G\) has finitely many nodes.

Start

0arbitrary delay between the two starts1simultaneous start

Dessmark, Fraigniaud, Kowalski and Pelc treated two startup scenarios for deterministic rendezvous: simultaneous startup, and an arbitrary delay \(\tau\) chosen by the adversary, in which each agent counts rounds from its own wake-up without knowing \(\tau\) Dessmark et al. 2006. The delay is the stronger adversary: it can delay one agent long enough that the other must find it while it sleeps, which is treasure hunt Pelc 2019 Ta-Shma & Zwick 2014. It changes cost: on a ring simultaneous start gives \(\Theta(d\log \lambda)\), while with arbitrary delay \(\Omega(n+d\log \lambda)\) is a lower bound even when \(n\) is known; and the first polynomial algorithm for arbitrary graphs had cost growing with \(\sqrt\tau\), leaving open whether the dependence could be removed Dessmark et al. 2006, which Kowalski and Malinowski and then Ta-Shma and Zwick did Kowalski & Malinowski 2008 Ta-Shma & Zwick 2014. It also changes memory: Fraigniaud and Pelc showed that identical agents need \(\Theta(\log n)\) bits in trees with arbitrary delay, even on a line, but only \(O(\log\nu+\log\log n)\) bits with simultaneous start in trees with \(\nu\) leaves, an exponential gap when \(\nu\) is small Fraigniaud & Pelc 2013.

Formal definitions
0any delay
An adversary wakes the agents at times \(t_1\) and \(t_2\); the delay \(\tau=|t_1-t_2|\) is unknown.
1simultaneous start
Both agents start in the same round (\(\tau=0\)).

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