Skip to main content
6 events
when toggle format what by license comment
Dec 4, 2021 at 19:13 comment added Fedor Petrov Finally, if $k$ is even but $f$ is odd, take any graph with odd number of vertices
Dec 4, 2021 at 14:33 comment added Manfred Weis @FedorPetrov very nice; that construction may yield "irreducible instances" for vertex-cover heuristics for which regular graphs are hard, but in some cases the regular graph may be reduced by deleting the edges of an $f$-factor.
Dec 4, 2021 at 12:10 comment added Fedor Petrov If $k$ is odd, take a graph $G$ on $k+2$ vertices whose complement has degree sequence $2,1,...,1$. Take $k$ copies of $G$ and another vertex $u$ and join $u$ with vertices of degree $k-1$ in $G$. This graph does not have a non-trivial factor.
Dec 4, 2021 at 12:04 comment added Fedor Petrov For the record: If $k$ and $f$ are even then it does not exist
Dec 4, 2021 at 11:18 history edited Manfred Weis CC BY-SA 4.0
fixed an inequality direction
Dec 3, 2021 at 11:09 history asked Manfred Weis CC BY-SA 4.0