Skip to main content
edited body
Source Link
hbm
  • 1k
  • 7
  • 14

Let $c(n,k)$ be the least integer such that if $G$ is a simple graph on $n$ vertices with $n + c(n,k) - 1$ edges then $G$ has $k$ edge-disjoint cycles.

Clearly, $c(n, 1) = 1$ and it not very hard to prove that $c(n,k) \le 5$$c(n,2) \le 5$ (an exercise in Bondy's book.)

How does $c(n,k)$ depends on $n$ and $k$?

Let $c(n,k)$ be the least integer such that if $G$ is a simple graph on $n$ vertices with $n + c(n,k) - 1$ edges then $G$ has $k$ edge-disjoint cycles.

Clearly, $c(n, 1) = 1$ and it not very hard to prove that $c(n,k) \le 5$ (an exercise in Bondy's book.)

How does $c(n,k)$ depends on $n$ and $k$?

Let $c(n,k)$ be the least integer such that if $G$ is a simple graph on $n$ vertices with $n + c(n,k) - 1$ edges then $G$ has $k$ edge-disjoint cycles.

Clearly, $c(n, 1) = 1$ and it not very hard to prove that $c(n,2) \le 5$ (an exercise in Bondy's book.)

How does $c(n,k)$ depends on $n$ and $k$?

Source Link
hbm
  • 1k
  • 7
  • 14

The least number of edges to add to a tree that would force a certain number of edge-disjoint cycles

Let $c(n,k)$ be the least integer such that if $G$ is a simple graph on $n$ vertices with $n + c(n,k) - 1$ edges then $G$ has $k$ edge-disjoint cycles.

Clearly, $c(n, 1) = 1$ and it not very hard to prove that $c(n,k) \le 5$ (an exercise in Bondy's book.)

How does $c(n,k)$ depends on $n$ and $k$?