Importance: Medium ✭✭
Subject: Geometry
Recomm. for undergrads: no
Posted by: David Wood
on: October 19th, 2009

\begin{conjecture} Every complete geometric graph with an even number of vertices has a partition of its edge set into plane (i.e. non-crossing) spanning trees. \end{conjecture}

For a set $P$ of $n$ points in the plane with no three collinear, the \emph{complete geometric graph} $K_P$ has vertex set $P$ and edge set consisting of the $\binom{n}{2}$ straight line-segments between each pair of points in $P$.

Since each subtree of $K_P$ has at most $n-1$ edges, every partition of $E(K_P)$ into subtrees has at least $\frac{n}{2}$ parts. The conjecture asks for such a partition into exactly $\frac{n}{2}$ subtrees, such that in addition, no two edges in each subtree cross.

It is folklore that the conjecture is true if $P$ is in convex partition. In fact, the edge set of the complete convex graph can be partitioned into plane Hamiltonian paths. Bose et al. [BHRW] characterised all possible partitions of the complete convex graph into plane spanning trees. Bose et al. [BHRW] also proved that every complete geometric graph on $n$ vertices can be partitioned into at most $n-\sqrt{\frac{n}{12}}$ plane subtrees.

I heard about this conjecture from Ferran Hurtado in 2003, but the problem is much older than that.


[BHRW] Prosenjit Bose, Ferran Hurtado, Eduardo Rivera-Campo, David R. Wood. Partitions of complete geometric graphs into plane trees, \emph{Computational Geometry: Theory & Applications} 34(2):116-125, 2006. \MRhref{MR2222887}

* indicates original appearance(s) of problem.


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