# Hamiltonian circuit

Let us consider a disk with one labelled point on the boundary and $n$ labelled points in the interior. Let T be a triangulation of the whole disk with vertices on the labelled points such that T contains no self-loops except the boundary of the disk, no multiple edges between two points in the interior of T, arcs of T can be curved.

1) Does T have an Hamiltonian circuit ?

2) If R is another triangulation of the same time, how many flips (as a function of $n$ should one perform at most in order to get from R to T ?

• I don't get it. The face inside and adjacent to the boundary can only be a triangle if its other two edges join the same pair of vertices. But you ruled out multiple edges.. Oct 15, 2013 at 4:06
• You're right. No multiple edges between two points in the interior of the disk. I'll edit. Oct 15, 2013 at 12:38