Let $T$ be a planar triangulation. It is known that one can guard the faces of $T$ using at most $\lfloor n/3 \rfloor$ edge-guards (Worst-case-optimal algorithms for guarding planar graphs and polyhedral surfaces). I am trying to obtain a similar upper bound for an extension of this problem, as follows.
Now, let $T$ be a three-dimensional triangulation (a tetrahedralization), and let $S$ be a subset of its edges. We say that $S$ strongly guards $T$ if, for every tetrahedron in $T$, one of the six edges of that tetrahedron lies in $S$. Is there a known nontrivial upper bound for the number of edges required to strongly guard all tetrahedra of a tetrahedralization?
Obviously, this problem can be solved via edge-coloring with no monochromatic tetrahedra. Is there any upper bound on the edge chromatic number for three-dimensional triangulations better than $\Delta + 1$? Maybe the assumption that $T$ is a Delaunay triangulation can conduct to a probabilistic bound.