  We study an old geometric optimization problem in the plane. Given a perfect
matching $M$ on a set of $n$ points in the plane, we can transform it to a
non-crossing perfect matching by a finite sequence of flip operations. The flip
operation removes two crossing edges from $M$ and adds two non-crossing edges.
Let $f(M)$ and $F(M)$ denote the minimum and maximum lengths of a flip sequence
on $M$, respectively. It has been proved by Bonnet and Miltzow (2016) that
$f(M)=O(n^2)$ and by van Leeuwen and Schoone (1980) that $F(M)=O(n^3)$. We
prove that $f(M)=O(n\Delta)$ where $\Delta$ is the spread of the point set,
which is defined as the ratio between the longest and the shortest pairwise
distances. This improves the previous bound if the point set has sublinear
spread. For a matching $M$ on $n$ points in convex position we prove that
$f(M)=n/2-1$ and $F(M)={{n/2} \choose 2}$; these bounds are tight.
  Any bound on $F(\cdot)$ carries over to the bichromatic setting, while this
is not necessarily true for $f(\cdot)$. Let $M'$ be a bichromatic matching. The
best known upper bound for $f(M')$ is the same as for $F(M')$, which is
essentially $O(n^3)$. We prove that $f(M')\le n-2$ for points in convex
position, and $f(M')= O(n^2)$ for semi-collinear points.
  The flip operation can also be defined on spanning trees. For a spanning tree
$T$ on a convex point set we show that $f(T)=O(n\log n)$.
