Abstract

Let $G$ be a bridgeless multigraph with $m$ edges and $n_2$ vertices of degree two and let $cc(G)$ be the length of its shortest cycle cover. It is known that if $cc(G) < 1.4m$ in bridgeless graphs with $n_2 \le m/10$, then the Cycle Double Cover Conjecture holds. Fan (2017) proved that if $n_2 = 0$, then $cc(G) < 1.6258m$ and $cc(G) < 1.6148m$ provided that $G$ is loopless; morever, if $n_2 \le m/30$, then $cc(G) < 1.6467m$. We show that for a bridgeless multigraph with $m$ edges and $n_2$ vertices of degree two, $cc(G) < 1.6148m + 0.0741n_2$. Therefore, if $n_2=0$, then $cc(G) < 1.6148m$ even if $G$ has loops; if $n_2 \le m/30$, then $cc(G) < 1.6173m$; and if $n_2 \le m/10$, then $cc(G) < 1.6223|E(G)|$. Our improvement is obtained by randomizing Fan's construction.

Highlights

  • We show that for a bridgeless multigraph with m edges and n2 vertices of degree two, cc(G) < 1.6148m + 0.0741n2

  • A cycle is a graph with all vertices of even degree and a circuit is an inclusion-wise minimal nonempty cycle

  • Let us contract all trees which we have introduced in the beginning the proof

Read more

Summary

Introduction

A cycle is a graph with all vertices of even degree and a circuit is an inclusion-wise minimal nonempty cycle. Let G be a bridgeless graph with m edges and n2 vertices of degree 2. Let G be a bridgeless graph (loops allowed) with m edges and n2 vertices of degree 2. Let G be a bridgeless graph of minimum degree 3 which contains m non-loop edges and s loops. There exists a probability distribution over proper spanning cycles F of G such that for every edge e we have P (e ∈ F ) = 2/3 if it is a non-loop and P (e ∈ F ) = 1/3 if it is a loop. If we contract an edge of the new 2-circuit in Ge into a single vertex v, we get Gs. Otherwise, we replace v by a tree in such a way that the new vertices have degree 3 and Ge is bridgeless (Figure 1b).

To bound the number of components of
Thus the length of the cover is at most
Full Text
Published version (Free)

Talk to us

Join us for a 30 min session where you can share your feedback and ask us any queries you have

Schedule a call