# Spanning eulerian subdigraphs avoiding k prescribed arcs in tournaments

## TL;DR

This paper investigates the existence of spanning eulerian subdigraphs in tournaments that avoid a set number of prescribed arcs, providing new bounds on the arc-strongness needed for their existence.

## Contribution

The authors establish a new upper bound on the arc-strongness function f(k), improving previous bounds and supporting the conjecture that f(k)=k+1 for all k.

## Key findings

- Proved that f(k) ≤ ⌈(6k+1)/5⌉ for all k.
- Improved the upper bound on the minimum arc-strongness for spanning eulerian subdigraphs.
- Supported the conjecture that f(k)=k+1 for all k.

## Abstract

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 $f(k)\leq (\lceil\frac{6k+1}{5}\rceil)$.