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Exploration of dynamic graphs by teams

An adversary changes the graph over time; how many agents are needed to explore it?

Problem statement and measurescollapse
round 0missinground 1missinground 2missinground 3missinground 4missinground 5missinground 6missinground 7abStart: a and b on one node; a goes clockwise, b the other wayThe adversary removes a's edge: a waits, b movesNow b is blocked and a moves: someone always advancesAfter N - 1 = 7 rounds every node is visited; both stop
Live exploration of a dynamic ring Di Luna et al. 2020b: in every round an adversary removes at most one edge of the ring (1-interval connectivity; the missing edge is dashed red). Two agents start on one node, know \(N = n = 8\) and walk in opposite directions, waiting while their next edge is missing; the adversary can block only one of them, so every round visits a new node, and after \(N - 1\) rounds both stop. One agent alone can be blocked forever.

Agents must visit every node of a dynamic graph: a sequence of graphs \(G_t = (V, E_t)\) on one node set, whose edges are removed round by round by an adversary subject to a connectivity constraint (1-interval connectivity: every \(G_t\) is connected; temporal connectivity: every node can reach every other through edges present at non-decreasing times). The primary measure is the number of agents; the secondary is time in rounds.

Measures: Agents, number of agents needed; Time, rounds.

A first look

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

Imagine guards who must check every station on a circular railway, while the operator closes some piece of track at any moment without warning. We model the railway as a graph: stations are nodes, the pieces of track between neighbouring stations are edges, and each guard is an agent. In the basic version the graph is a ring of \(n\) nodes (the nodes form one cycle, each joined by an edge to its two neighbours). Time runs in rounds: in each round every agent may cross one edge or wait. An adversary (an imagined opponent, the operator of the picture) chooses, in every round, which edges are missing (removed for that round), with one rule: the graph must stay connected in every round (1-interval connectivity); in a ring this means that at most one edge is missing per round. A graph that changes like this is a dynamic graph. The agents must together visit every node; in the version recorded here they also know an upper bound \(N \ge n\) and must stop. We count the agents needed and the rounds.

1AlgorithmTwo agents walking in opposite directions 1-interval-connected rings → Di Luna et al. 2020b

The simplest case: two agents start on the same node and know \(N \ge n\).

  1. The two agents set off in opposite directions around the ring (the paper shows how two identical agents can agree on this).
  2. In every round each agent tries to cross the next edge in its direction; if the edge is missing, it waits (it is blocked) and tries again in the next round.
  3. After \(N - 1\) rounds both agents stop.
Why it works and what it costs. At most one edge is missing per round, and until the two agents have between them covered the whole ring they are trying to cross two different edges, so at least one of them moves forward, onto a node that nobody has visited. Each round therefore visits at least one new node, and the \(n - 1\) other nodes are all visited within \(n - 1 \le N - 1\) rounds. Di Luna et al. 2020b handle the harder cases (different starting nodes, no common sense of left and right) with one extra rule, turning back after being blocked for long, and both agents stop after \(3N - 6\) rounds.

Why it matters. It shows that a second agent turns an impossible task into an easy one, because the adversary can only block one of them at a time.

2ImpossibilityOne agent can always be blocked 1-interval-connected rings → Di Luna et al. 2020b

A single agent stands on a node of the ring and, in some round, decides to cross one of the two edges at its node. The adversary removes exactly that edge in that round. Removing one edge from a ring leaves a path, which is still connected, so this is allowed. The adversary can do this in every round, so the agent never leaves its starting node.

Why it holds. So at least two agents are needed Di Luna et al. 2020b. The same paper also shows that the adversary can prevent two agents that start on different nodes from ever meeting, by removing an edge only at the moment the two would otherwise meet: the agents cannot rely on meeting to coordinate.

Why it matters. The number of agents becomes the main measure on dynamic graphs, because a few agents may be unable to explore at all, whatever time they are given.

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
Agents
Time
Connected at every step, any graph
Against an adaptive adversary that keeps every snapshot connected, \(2m-n+1\) silent agents explore any graph, and \(m-n+1\) are needed even with a central offline scheduler.
polynomial gap \(2m-n+1\) vs \(m-n+1\) adversarial edge changes
unknown \(O(n^9)\) vs —
1-interval-connected rings standard
One agent cannot explore a dynamic ring; two agents with a known bound on \(n\) explore in \(3N-6\) rounds.
tight \(3\) vs \(2\) adversarial edge changes
unknown \(3N-6\) vs —
Temporally connected graphs, arbitrary topology
Perpetual exploration of a temporally connected graph with \(\eta\) transient edges needs exactly \(2\eta+1\) agents.
tight \(2\eta+1\) vs \(2\eta+1\) adversarial edge changes
unknown — vs —

Open questions 2

Connected at every step, any graph

  1. Does the algorithm with \(2m-n+1\) agents explore in polynomial time? Posed in Dobrev et al. 2020a.

1-interval-connected rings

  1. With a landmark but no chirality, is \(O(n\log n)\) time necessary? Di Luna et al. 2020b

What the community started looking at

In dynamic graphs the community first asked how many agents make exploration possible at all, under each connectivity assumption Di Luna et al. 2020b Gotoh et al. 2021 Saxena & Mondal 2026; time came second.

Variant of Collaborative exploration

When an adversary removes edges, a single agent can be blocked forever; the question shifts from speed to how many agents are needed to visit every node.

Related problems

Broadcast by agents in dynamic graphs: The same adversary: in every round it may remove any edges as long as the snapshot stays connected (1-interval connectivity), and the primary measure is the number of agents needed in both problems.

Further reading

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

  • Adamson & Spirakis 2026 Brief Announcement: Exploration of Always S-Connected Temporal Graphs. Exploration by teams when the whole sequence of snapshots is given in advance (offline), each agent crossing at most one present edge per round. In an always \(S\)-connected temporal graph every snapshot joins every node by a path to some node of a fixed set \(S\); \(|S|\) agents, one starting on each node of \(S\), explore every such graph whose lifetime is polynomial in \(n\), \(|S|\) and the average degree of the underlying graph (Theorem 1, Corollary 1). Through a division of the underlying graph into parts with few boundary nodes, a single agent explores an always connected temporal graph whose underlying graph has treewidth \(\mathrm{tw}\) within \(O(n^{4/3}\,\mathrm{tw}^{9/2}\log^{5/2} n)\) rounds (Theorem 2). When every edge is present in exactly one snapshot, deciding whether one agent can explore is NP-hard (Theorem 5), and the minimum number of agents that explore a path is computed in \(O(n)\) time, that for a cycle in polynomial time (Theorems 6 and 7). Full version (arXiv v2) read; Carnevale 2026 cites \(\mathrm{tw}^{11/2}\) for the brief announcement, which was not read.
  • Bournat et al. 2017 Computability of perpetual exploration in highly dynamic rings. Perpetual exploration of connected-over-time rings by anonymous FSYNC robots with persistent memory and local weak multiplicity detection: \(k \ge 3\) robots for \(n > k\) (Theorem III.1); two robots fail for \(n > 3\) and one for \(n > 2\) (Theorems IV.1, V.1), and suffice for \(n = 3\) and \(n = 2\) (Theorems IV.2, V.2). Read.
  • Bournat et al. 2019 Self-stabilizing robots in highly dynamic environments. Self-stabilizing perpetual exploration of connected-over-time rings by FSYNC robots with distinct identifiers: three robots for rings of at least four nodes (Theorem 4.1); two robots are impossible for \(n \ge 4\) (Theorem 3.1). Read.
  • Feletti et al. 2026d Safe Exploration of Arbitrary Dynamic Dangerous Networks. Extends the temporally connected setting of Gotoh et al. 2021 to graphs with any number of black holes, unknown to the agents, which must visit every safe node while at least one agent survives. Semi-synchronous oblivious anonymous agents starting on arbitrary safe nodes explore perpetually with \(2\eta' + \sum_{x \in B}\deg(x) + 1\) agents, where \(B\) is the set of black holes and \(\eta'\) the number of transient edges between safe nodes, using whiteboards of \(O(\Delta)\) bits (Theorem 11); without black holes this is the \(2\eta + 1\) of Gotoh et al. 2021. Agents with identifiers that know the number of safe nodes also terminate (Theorem 16). Preprint; read.
  • Saxena et al. 2026b Dynamic Graph Exploration: Semi-Synchrony and Dynamic Port Labeling. Exploration of 1-interval connected graphs of arbitrary topology, with port numbers reassigned in every round, under a semi-synchronous scheduler: in each round an adversary deactivates up to \(\alpha\) agents (their \(p\)), and every agent is activated infinitely often. Agents have distinct identifiers, start on arbitrary nodes and know neither \(n\) nor \(k\); nodes have no storage. With \(k \le (n-2)(\alpha+1)\) agents exploration is impossible, even with full visibility, global communication, node identifiers and storage, unbounded memory and knowledge of every parameter (Theorem 2.1, \(n \ge 3\), \(\alpha \ge 1\)). With \((n-2)(\alpha+1)+1\) agents it stays impossible for agents that see their neighbouring nodes but communicate only face to face (Theorem 2.2, \(n \ge 9\)), and for agents that see only their own node but communicate globally (Theorem 2.3). Agents that see their neighbouring nodes and communicate globally explore whenever \(k \ge (n-2)(\alpha+1)+1\), that is \(\alpha \le \lfloor (k-1)/(n-2) \rfloor - 1\), without knowing \(n\), \(k\) or \(\alpha\); every agent eventually knows that exploration is complete, the total number of moves is \(O(k)\) times the largest diameter of a snapshot, and each agent uses \(O(\max\{\log n, \log\alpha\})\) bits (Theorem 3.1, \(n \ge 4\)). The number of agents is therefore tight for every \(\alpha \ge 1\); whether global communication is necessary is left open. For \(\alpha = 0\) the threshold \(n-1\) is the bound for synchronous agents of Saxena & Mondal 2025, as reported by Saxena et al. 2026b. The timing axis has no semi-synchronous value, so the result is not placed in a setting. Read in the arXiv version (v1); theorem numbers refer to it.
  • Sudo et al. 2026c Online Graph Exploration under T-Interval Connectivity. A single deterministic agent on \(T\)-interval-connected graphs of arbitrary topology whose nodes carry identifiers: the window must be \(\Omega(m)\) (Theorem 3), and with a window of \(O(\zeta m + n\log^2 n)\), where \(\zeta = \ln n/(1+\ln(m/n))\), the agent explores and stops within \(O((m-n+1)\,n + n\log^2 n)\) moves, optimal when \(m-n+1 = \Omega(\log^2 n)\) (Theorems 1 and 4). Placed in the setting \(T\)-interval-connected graphs, labelled nodes of single-agent exploration. Preprint; read.
  • Ayoubi et al. 2026 Online Treasure Hunt in Vertex-Permuted Dynamic Rings. In vertex-permuted rings, where every snapshot is a cycle through all \(n\) nodes in an order the adversary chooses at each step, up to \(n-3\) agents starting at one node cannot even find a treasure, so they cannot explore, even with memory, chirality, a view of the whole snapshot and knowledge of the node labels (Theorem 1). If every two nodes are adjacent within every \(\xi\) steps and agents flag the nodes they visit, \(k\) agents visit all nodes within \(\xi\lceil 2n/k\rceil\) steps (proof of Theorem 8). Preprint; read.
  • Gotoh et al. 2020 Dynamic Ring Exploration with (H,S) View. A single agent on a 1-interval connected ring with partial knowledge of the future: it sees which edges within \(H\) hops will be present in each of the next \(S\) rounds. It can explore exactly when \(H + S \ge n\) and \(S \ge \lceil n/2\rceil\) (Theorem 2), in \(O(n^2)\) rounds, \(O(n^2/H + nH)\) when \(S\) is large enough, and \(O(n^2/H + n\log H)\) when \(S \ge n-1\) (Theorems 1, 3), against \(\Omega(n^2/H)\) for every \(S\) (Theorem 4). Read in full.
The assumption tree (3 settings, 1 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 (Agents, Time); dashed nodes are branch points with no catalogued variant, named after the axis on which their branches differ. Axis order: Graph dynamics › Edge direction › Timing › Topology.

tight near-optimal polynomial gap unknown impossible no variant

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