The Canadian Traveller Problem on outerplanar graphs

March 04, 2024 Β· Declared Dead Β· πŸ› International Symposium on Mathematical Foundations of Computer Science

πŸ‘» CAUSE OF DEATH: Ghosted
No code link whatsoever

"No code URL or promise found in abstract"

Evidence collected by the PWNC Scanner

Authors Laurent Beaudou, Pierre BergΓ©, Vsevolod Chernyshev, Antoine Dailly, Yan Gerard, AurΓ©lie Lagoutte, Vincent Limouzy, Lucas Pastor arXiv ID 2403.01872 Category cs.DS: Data Structures & Algorithms Citations 0 Venue International Symposium on Mathematical Foundations of Computer Science Last Checked 5 months ago
Abstract
We study the $k$-Canadian Traveller Problem, where a weighted graph $G=(V,E,Ο‰)$ with a source $s\in V$ and a target $t\in V$ are given. This problem also has a hidden input $E_* \subsetneq E$ of cardinality at most $k$ representing blocked edges. The objective is to travel from $s$ to $t$ with the minimum distance. At the beginning of the walk, the blockages $E_*$ are unknown: the traveller discovers that an edge is blocked when visiting one of its endpoints. Online algorithms, also called strategies, have been proposed for this problem and assessed with the competitive ratio, {\em i.e.}, the ratio between the distance actually traversed by the traveller divided by the distance he would have traversed knowing the blockages in advance. Even though the optimal competitive ratio is $2k+1$ even on unit-weighted planar graphs of treewidth 2, we design a polynomial-time strategy achieving competitive ratio 9 on unit-weighted outerplanar graphs. This value 9 also stands as a lower bound for this family of graphs as we prove that, for any $\varepsilon > 0$, no strategy can achieve a competitive ratio $9-\varepsilon$ on it. This comes actually from a strong connexion with another well-known online problem called the cow-path problem. Finally, we show that it is not possible to achieve a competitive ratio $e^{W(\frac{\ln k}{2})} - 1$ on arbitrarily weighted outerplanar graphs, where $W$ is the Lambert W function. This lower bound is asymptotically greater than $\frac{\ln k}{\ln \ln k}$.
Community shame:
Not yet rated
Community Contributions

Found the code? Know the venue? Think something is wrong? Let us know!

πŸ“œ Similar Papers

In the same crypt β€” Data Structures & Algorithms

Died the same way β€” πŸ‘» Ghosted