Improved exploration of temporal graphs
Fuente:
arXiv
Salvato in:
| Autori principali: | , , , |
|---|---|
| Natura: | Preprint |
| Pubblicazione: |
2025
|
| Soggetti: | |
| Accesso online: | |
| Tags: |
Aggiungi Tag
Nessun Tag, puoi essere il primo ad aggiungerne!!
|
| _version_ | 1866914173540106240 |
|---|---|
| author | Bastide, Paul Groenland, Carla Michel, Lukas Rambaud, Clément |
| author_facet | Bastide, Paul Groenland, Carla Michel, Lukas Rambaud, Clément |
| contents | A temporal graph $G$ is a sequence $(G_t)_{t \in I}$ of graphs on the same vertex set of size $n$. The \emph{temporal exploration problem} asks for the length of the shortest sequence of vertices that starts at a given vertex, visits every vertex, and at each time step $t$ either stays at the current vertex or moves to an adjacent vertex in $G_t$. Bounds on the length of a shortest temporal exploration have been investigated extensively. Perhaps the most fundamental case is when each graph $G_t$ is connected and has bounded maximum degree. In this setting, Erlebach, Kammer, Luo, Sajenko, and Spooner [ICALP 2019] showed that there exists an exploration of $G$ in $\mathcal{O}(n^{7/4})$ time steps. We significantly improve this bound by showing that $\mathcal{O}(n^{3/2} \sqrt{\log n})$ time steps suffice.
In fact, we deduce this result from a much more general statement. Let the \emph{average temporal maximum degree} $D$ of $G$ be the average of $\max_{t \in I} d_{G_t}(v)$ over all vertices $v \in V(G)$, where $d_{G_t}(v)$ denotes the degree of $v$ in $G_t$. If each graph $G_t$ is connected, we show that there exists an exploration of $G$ in $\mathcal{O}(n^{3/2} \sqrt{D \log n})$ time steps. In particular, this gives the first subquadratic upper bound when the underlying graph has bounded average degree. As a special case, this also improves the previous best bounds when the underlying graph is planar or has bounded treewidth and provides a unified approach for all of these settings. Our bound is subquadratic already when $D=o(n/\log n)$. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2511_22604 |
| institution | arXiv |
| publishDate | 2025 |
| record_format | arxiv |
| spellingShingle | Improved exploration of temporal graphs Bastide, Paul Groenland, Carla Michel, Lukas Rambaud, Clément Data Structures and Algorithms Combinatorics A temporal graph $G$ is a sequence $(G_t)_{t \in I}$ of graphs on the same vertex set of size $n$. The \emph{temporal exploration problem} asks for the length of the shortest sequence of vertices that starts at a given vertex, visits every vertex, and at each time step $t$ either stays at the current vertex or moves to an adjacent vertex in $G_t$. Bounds on the length of a shortest temporal exploration have been investigated extensively. Perhaps the most fundamental case is when each graph $G_t$ is connected and has bounded maximum degree. In this setting, Erlebach, Kammer, Luo, Sajenko, and Spooner [ICALP 2019] showed that there exists an exploration of $G$ in $\mathcal{O}(n^{7/4})$ time steps. We significantly improve this bound by showing that $\mathcal{O}(n^{3/2} \sqrt{\log n})$ time steps suffice. In fact, we deduce this result from a much more general statement. Let the \emph{average temporal maximum degree} $D$ of $G$ be the average of $\max_{t \in I} d_{G_t}(v)$ over all vertices $v \in V(G)$, where $d_{G_t}(v)$ denotes the degree of $v$ in $G_t$. If each graph $G_t$ is connected, we show that there exists an exploration of $G$ in $\mathcal{O}(n^{3/2} \sqrt{D \log n})$ time steps. In particular, this gives the first subquadratic upper bound when the underlying graph has bounded average degree. As a special case, this also improves the previous best bounds when the underlying graph is planar or has bounded treewidth and provides a unified approach for all of these settings. Our bound is subquadratic already when $D=o(n/\log n)$. |
| title | Improved exploration of temporal graphs |
| topic | Data Structures and Algorithms Combinatorics |
| url | https://arxiv.org/abs/2511.22604 |