0
$\begingroup$

Hamilton path: goes through every node/vertex exactly once.

Hamilton circuit: goes through every vertex exactly one and ends at the starting node/vertex.

So i am wondering is possible to use degree to detect if a graph have hamilton paths or circuits. Like we do with eulers circuit , if every vertex in the graph have a even degree then we know that eulers circuit exists.

$\endgroup$
1
$\begingroup$

Check the theorems of Ore and Dirac.

According to the theorem of Ore:

Let $G$ be a (finite and simple) graph with $n ≥ 3$ vertices. We denote by $deg v$ the degree of a vertex $v$ in $G$, i.e. the number of incident edges in $G$ to $v$. Then, Ore's theorem states that if

$deg v + deg w ≥$ $n$ for every pair of non-adjacent vertices $v$ and $w$ of $G$

then $G$ is Hamiltonian.

According to the theorem of Dirac:

A simple graph with $n$ vertices $(n ≥ 3)$ is Hamiltonian if every vertex has degree $n / 2$ or greater.

$\endgroup$
  • $\begingroup$ What do you mean by Hamiltonian? path or circuit $\endgroup$ – humble24 Oct 29 '15 at 19:00
  • $\begingroup$ Dirac = circuit. Ore = path $\endgroup$ – kjanko Oct 29 '15 at 19:14
  • $\begingroup$ alright, for Ore do i just pick a random pair of non-adjacent vertices ? do i have to check for every possible pair of vertices , or isit enough with one random pair of non-adjacent vertices ? $\endgroup$ – humble24 Oct 29 '15 at 19:36
  • $\begingroup$ Every pair of non-adjacent vertices. $\endgroup$ – kjanko Oct 29 '15 at 19:39
  • $\begingroup$ alright, thanks $\endgroup$ – humble24 Oct 29 '15 at 19:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.