Spanning eulerian subdigraphs avoiding k prescribed arcs in tournaments
A digraph is {\bf eulerian} if it is connected and every vertex has its in-degree equal to its out-degree. Having a spanning eulerian subdigraph is thus a weakening of having a hamiltonian cycle. A digraph is {\bf semicomplete} if it has no pair of non-adjacent vertices. A {\bf tournament} is a semi...
Gespeichert in:
Hauptverfasser: | , , |
---|---|
Format: | Artikel |
Sprache: | eng |
Schlagworte: | |
Online-Zugang: | Volltext bestellen |
Tags: |
Tag hinzufügen
Keine Tags, Fügen Sie den ersten Tag hinzu!
|
Zusammenfassung: | A digraph is {\bf eulerian} if it is connected and every vertex has its
in-degree equal to its out-degree. Having a spanning eulerian subdigraph is
thus a weakening of having a hamiltonian cycle. A digraph is {\bf semicomplete}
if it has no pair of non-adjacent vertices. A {\bf tournament} is a
semicomplete digraph without directed cycles of length 2. Fraise and Thomassen
\cite{fraisseGC3} proved that every $(k+1)$-strong tournament has a hamiltonian
cycle which avoids any prescribed set of $k$ arcs. In \cite{bangsupereuler} the
authors demonstrated that a number of results concerning vertex-connectivity
and hamiltonian cycles in tournaments and have analogues when we replace vertex
connectivity by arc-connectivity and hamiltonian cycles by spanning eulerian
subdigraphs. They showed the existence of a smallest function $f(k)$ such that
every $f(k)$-arc-strong semicomplete digraph has a spanning eulerian subdigraph
which avoids any prescribed set of $k$ arcs. They proved that $f(k)\leq
\frac{(k+1)^2}{4}+1$ and also proved that $f(k)=k+1$ when $k=2,3$. Based on
this they conjectured that $f(k)=k+1$ for all $k\geq 0$. In this paper we prove
that $f(k)\leq (\lceil\frac{6k+1}{5}\rceil)$. |
---|---|
DOI: | 10.48550/arxiv.1907.00853 |