Throughout, by finite triangulation I mean a triangulation consisting of a finite number of triangles.
Suppose $T$ and $T'$ are finite triangulations of a 3-manifold $M$. We will say that $T'$ is simpler than $T$ iff $T'$ consists of the same number or fewer triangles than $T$ and that $T'$ is a simplest triangulation of $M$ iff $\forall$ triangulation $T$ of $M$, $T'$ is simpler than $T$.
Note: If a 3-manifold $M$ has a finite triangulations, then clearly it has a simplest triangulation.
By a theorem of Pachner (Theorem A.1.1. in 'The geometry of dynamical triangulations') any two triangulations of a manifold can be transformed from one to another by a finite number of stellar subdivisions. As we are only dealing with 3-manifolds, there are only 4 stellar subdivisions; known as the $1 \to 4$, $2 \to 3$, $3 \to 2$ and $4 \to 1$ moves as described in http://at.yorku.ca/t/a/i/c/45.pdf and hereafter called the Pancher moves. So clearly, there exists a finite sequence of Pancher moves from any finite triangulaiton $T$ of $M$ to $T'$, a simplest triangulation of $M$.
If $T$ is a finite triangulation of $M$, does the greedy algorithm of just applying as many $4 \to 1$ and $3 \to 2$ Pancher moves to $T$ as possible always result in a simplest triangulation of $M$?
Is there a finite triangulation $T$ of a 3-manifold $M$ such that repeatedly applying only the $4 \to 1$ and $3 \to 2$ Pancher moves does not eventually result in a simplest triangulation of $M$?