TITLE: Drawing trees on small number of lines in 2D and 3D
QUESTION [7 upvotes]: Problem. Given a tree do we need fewer lines in 3D than in 2D in order to draw it straightline and crossing-free?

(Asked 01.10.2016 by Alexander Wolff on page 20 of Volume 1 of the Lviv Scottish Book). 
Clarification. For a given finite simple graph $G$ by $\rho^1_2(G)$ (resp. $\rho^1_3(G)$) we denote the minimal number of straight lines needed to cover all edges in a straight-line crossing-free drawing of the graph $G$ in 2D (resp. 3D). We are asking for a tree $T$ such that $\rho^1_3(T)<\rho^1_2(T)$. 
The problem was originally asked 11.06.2016 by Oleg Verbitsky.

REPLY [2 votes]: I'm not 100% sure how to prove this, but here is my candidate for an embedding in 2D that minimizes the number of lines used:
For each node in the tree if it has an even number of edges, $2e$ then embed pairs of edges on the same line (i.e. put the edge $\frac{\pi}{e}$ radians apart) so that the node uses $e$ lines in total; if the node has an odd number of edges, $2e + 1$ then do the same, but have one edge be on its own line (put it at some angle not used by the other edges) to use $e+1$ lines in total. This process is possible by making the embedding spread out enough that there is no crossing.
I believe that you cannot do better than this because an edge can be on the same line of at most one incident edge. Furthermore I believe you cannot do better in 3D as well for the same reason.